THE THEORY OF FUNCTIONS OF
REAL VARIABLES
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VOLUME II
. JAMIE: S Pif;RPONT
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LECTURES
ON
THE THEORY OF FUNCTIONS OF
REAL VARIABLES
c
VOLUME II
BY
JAMES PIERPONT, LL.D.
PROFESSOR OF MATHEMATICS ix YALE UNIVERSITY
GINN AND COMPANY
BOSTON • NEW YORK • CHICAGO • LONDON
Qfi
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COPYRIGHT, 1912, BY
JAMES PIERPONT
ALL RIGHTS RESERVED
923.4
fltftenaum
GINN AND COMPANY • PRO
PRIETORS • BOSTON • U.S.A.
TO
ANDREW W. PHILLIPS
THESE LECTURES
ARE INSCRIBED
WITH AFFECTION AND ESTEEM
773668
PREFACE
THE present volume has been written in the same spirit that
animated the first. The author has not intended to write a
treatise or a manual ; he has aimed rather to reproduce his uni
versity lectures with necessary modifications, hoping that the
freedom in the choice of subjects and in the manner of presenta
tion allowable in a lecture room may prove helpful and stimulating
to a larger audience.
A distinctive feature of these Lectures is an attempt to develop
the theory of functions with reference to a general domain of
definition. The first functions to be considered were simple
combinations of the elementary functions. Riemann in his great
paper of 1854, " Ueber die Darstellbarkeit einer Function (lurch
eine trigonometrische Reihe," was the first to consider seriously
functions whose singularities ceased to be intuitional. The re
searches of later mathematicians have brought to light a collection
of such functions, whose existence so long unsuspected has revolu
tionized the older notion of a function and made imperative the
creation of finer tools of research. But while minute attention
was paid to the singular character of these functions, practically
none was accorded to the domain over which a function may be
defined. After the epoch-making discoveries inaugurated in 1874
by G. Cantor in the theory of point sets, it was no longer neces
sary to consider a function of one variable as defined in an in
terval, a function of two variables as defined over a field bounded
by one or more simple curves, etc. The first to make use of this
new freedom was C. Jordan in his classic paper of 1892. He
has had, however, but few imitators. In the present Lectures the
author has endeavored to develop this broader view of Jordan,
persuaded that in so doing he is merely carrying a step farther
the ideas of Dirichlet and Riemann.
Often such an endeavor leads to nothing new, a mere statement
for any n of what is true for n = 1, or 2. A similar condition
vi PREFACE
prevails in the theory of determinants. One may prefer to treat
only two and three rowed determinants, but he surely has no
ground of complaint if another prefers to state his theorems and
demonstrations for general n. On the other hand, the general
case may present unexpected and serious problems. For example,
Jordan has introduced the notion of functions of a single variable
having limited variation. How is this notion to be extended to
two or more variables ? An answer is far from simple. One was
given by the author in Volume I ; its serviceableness has since
been shown by B. Camp. Another has been essayed by Lebesgue.
The reader must be warned, however, against expecting to find
the development always extended to the general case. This,
in the first place, would be quite impracticable without greatly
increasing the size of the present work. Secondly, it would often
be quite beyond the author's ability.
Another feature of the present work to which the author would
call attention is the novel theory of integration developed in
Chapter XVI of Volume I and Chapters I and II of Volume II.
It rests on the notion of a cell and the division of space, or in fact
any set, into unmixed partial sets. The definition of improper
multiple integrals leads to results more general in some respects
than yet obtained with Riemann integrals.
Still another feature is a new presentation of the theory of
measure. The demonstrations which the author has seen leave
much to be desired in the way of completeness, not to say rigor.
In attempting to find a general and rigorous treatment, he was
at last led to adopt the form given in Chapter XI.
The author also claims as original the theory of Lebesgue
integrals developed in Chapter XII. Lebesgue himself considers
functions such that the points e at which a <f(x) <6, for all a, b
form a measurable set. His integral he defines as
lim 2lmerm
W=« 1
where lm<f(x)<lm+l in em whose measure is e'm, and each
lm+1 — lm= 0, as n = oo. The author has chosen a definition which
occurred to him many years ago, and which to him seems far
more natural. In Volume I it is shown that if the metric field ^
PREFACE vii
be divided into a finite number of metric sets Sv S2 ••• of norm d,
then
= Min
where m^ M, are the minimum and maximum of/ in St. What
then is more natural than to ask what will happen if the cells
Bv S2>" are infinite instead of finite in number? From this
apparently trivial question results a theory of ^-integrals which
contains the Lebesgue integrals as a special case, and which,
furthermore, has the great advantage that riot only is the relation
of the new integrals to the ordinary or Riemannian integrals
perfectly obvious, but also the form of reasoning employed in
Riemann's theory may be taken over to develop the properties
of the new integrals.
Finally the author would call attention to the treatment of
the area of a curved surface given at the end of this volume.
Though the above are the main features of novelty, it is hoped
that the experienced reader will discover some minor points, not
lacking in originality, but not of sufficient importance to em
phasize here.
It is now the author's pleasant duty to acknowledge the in
valuable assistance derived from his colleague and former pupil,
Dr. W. A. Wilson. He has read the entire manuscript and
proof with great care, corrected many errors and oversights in
the demonstrations, besides contributing the substance of §§ 372,
373, 401-406, 414-424.
Unstinted praise is also due to the house of Ginn and Com
pany, who have met the author's wishes with unvarying liberality,
and have given the utmost care to the press work.
JAMES PIERPONT
NEW HAVEN, December, 1911
CONTENTS
CHAPTER I
POINT SETS AND PROPER INTEGRALS
ARTICLES
1-10. Miscellaneous Theorems . . . . .
11-15. Iterable Fields . . . . . ...
16-25. Union and Divisor of Point Sets . . .
PAGE
1
14
22
26-28.
29.
30.
31-61.
62-69.
70-78.
CHAPTER II
IMPROPER MULTIPLE INTEGRALS
Classical Definition . . . . . ...
Definition of de la Valle"e-Poussin ....
Author's Definition . . ...
General Theory . .' . ' . .
Relation between Three Types . . . .
Iterated Integrals . . .. . .
30
31
32
32
59
63
CHAPTER III
SERIES
79-80. Preliminary Definitions and Theorems . . . . r. . .77
81. Geometric, General Harmonic, Alternating, and Telescopic Series . 81
82. Dini's Series . . . , . '...... .86
83. Abel's Series . . . . . ... . .. . ' . 87
84. Trigonometric Series ......... 88
85. Power Series . . . 89
86. Cauchy's Theorem on the Interval of Convergence .... 90
87-91. Tests of Convergence. Examples . . . . . . .91
92. Standard Series of Comparison . . . . . . . 101
93-98. Further Tests of Convergence . . 104
99. The Binomial Series . .110
100. The Hypergeometric Series 112
101-108. Pringsheim's Theory . . .113
109-113 Arithmetic Operations on Series . ...... 125
114-115. Two-way Series . .133
ix
CONTEXTS
CHAPTER IV
MULTIPLE SERIES
ARTICLES
PAGE
116-125.
General Theory
. 137
126-133.
148
CHAPTER V
SERIES OF FUNCTIONS
134-145.
156
146.
The Moore-Osgood Theorem .......
. 170
147-149.
Continuity of a Series
. 173
150-152.
Termwise Integration ........
. 177
153-156.
Termwise Differentiation . ...
. 181
CHAPTER VI
POWER SERIES
157-158.
Termwise Differentiation and Integration
. 187
159.
Development of log (1 + E), arcsin x, arctan », e1, sin x, cosse
. 188
160.
Equality of two Power Series .......
. 191
161-162.
Development of a Power Series whose Terms are Power Series
. 192
163.
Multiplication and Division of Power Series .
. 196
164-165.
Undetermined Coefficients .......
. 197
166-167.
Development of a Series whose Terms are Power Series .
. 200
168.
Inversion of a Power Series
. 203
169-171.
Taylor's Development . . .
. 206
172.
Forms of the Remainder . . .
. 208
173.
Development of (l+sc)'* ........
. 210
174.
Development of log (1 -f x), etc. . . . .
. 212
175-181.
Criticism of Current Errors . . ...
. 214
182.
Pringsheim's Necessary and Sufficient Condition
. 220
183.
Circular Functions .........
. 222
184.
Hyperbolic Functions ........
. 228
185-192.
Hypergeometric Function .......
. 229
193.
Bessel Functions .........
. 238
CHAPTER VII
INFINITE PRODUCTS
195-202.
242
203-206.
260
207-212.
Uniform Convergence ........
. 254
213-218.
Circular Functions .........
. 257
CONTENTS
XI
ARTICLES PAGE
219. Bernouillian Numbers 266
220-228. B and T Functions 267
CHAPTER VIII
AGGREGATES
229-230. Equivalence " • '• .v • -• • • 276
231. Cardinal Numbers . . . ... . ... . . . . 278
232-241. Enumerable Sets . *; • • • • . • • • -280
242. Some Space Transformations . . . . . • • 286
243-250. The Cardinal c v • • • 287
251-261. Arithmetic Operations with Cardinals . . • . ... .292
262-264. Numbers of Liouville . . * . . . . • -299
CHAPTER IX
ORDINAL NUMBERS
265-267. Ordered Sets . . • • • • • : ' • ' .302
268-270. Eutactic Sets . . ... . . . . . . .304
271-279. Sections . . • • • 307
280-284. Ordinal Numbers . . . . ..' . . . . .310
285-288. Limitary Numbers . * ." . . . ... • • • 314
289-300. Classes of Ordinals . . . . . . • * . . 318
CHAPTER X
POINT SETS
301-312. Pantaxis . . . • 824
313-320. Transfinite Derivatives . . . . • • • • • 330
321-333. Complete Sets . . . . . . . . . .337
CHAPTER XI
MEASURE
334-343. Upper Measure . . . . . . • • • • .343
344-368. Lower Measure . . . . ... . . - 348
369-370. Associate Sets . . ... ... . . • • 365
371-376. Separated Sets . . . . . . .. • • • .366
CHAPTER XII
LEBESGUE INTEGRALS
377-402. General Theory _• • .371
403-406. Integrand Sets • .... 385
Xll
CONTENTS
ARTICLES
407-409. Measurable Functions
410. Quasi and Semi Divisors
411-413. Limit Functions
414-424. Iterated Integrals
IMPROPER L-!NTEGRALS
425-428. Upper and Lower Integrals
429-431. Z-Integrals
432-435. Iterated Integrals .
PAGE
388
390
392
394
402
405
409
CHAPTER XIII
FOURIER'S SERIES
436-437. Preliminary Remarks ....
438. Summation of Fourier's Series .
439-442. Validity of Fourier's Development .
443-446. Limited Variation
447-448. Other Criteria
449-456. Uniqueness of Fourier's Development
415
420
424
429
437
438
CHAPTER XIV
DISCONTINUOUS FUNCTIONS
457-462. Properties of Continuous Functions .
463-464. Pointwise and Total Discontinuity . . .
465-473. Examples of Discontinuous Functions
474-489. Functions of Class 1
490-497. Semicontinuous Functions ....
452
457
459
468
485
CHAPTER XV
DERIVATES, EXTREMES, VARIATION
498-518. Derivates .
519-525. Maxima and Minima
526-534. Variation .
535-537. Non-intuitional Curves
538-539. Pompeiu Curves
540-542. Faber Curves
493
521
531
537
542
546
CHAPTER XVI
SUB- AND INFRA-UNIFORM CONVERGENCE
543-550. Continuity
551-556. Integrability
557-561. Differentiability
555
562
570
CONTENTS
Xlll
CHAPTER XVII
GEOMETRIC NOTIONS
ARTICLES PAGE
562-563. Properties of Intuitional Plane Carves 578
564. Motion 579
565. Curve as Intersection of Two Surfaces 579
566. Continuity of a Curve 580
567. Tangents 580
568-572. Length '. . . .581
573. Space-filling Curves . . ~ . . 588
.574. Hilbert's Curve . . . . . . . . . .590
575. Equations of a Curve • '• . 593
576-580. Closed Curves . . . . .594
581. Area 599
582. Osgood's Curve . . . . . . . . •' .600
583. Re-sums' . .... 603
584-585. Detached and Connected Sets 603
586-591. Images ' . . . . ' . .605
592-597. Side Lights on Jordan Curves . . . . . . '• . • • • 610
598-600. Brouwer's Proof of Jordan's Theorem .-'•-,. . • . . . 614
601. Dimensional Invariance . .. . . ... . . 619
602. Schonfliess' Theorem . . . . . . ' . . . 621
603-608. Area of Curved Surfaces 623
Index . . . . . . . .639
List of Symbols 644
FUNCTION THEORY OF REAL
VARIABLES : v^ U;
CHAPTER I
POINT SETS AND PROPER INTEGRALS
1. In this short chapter we wish to complete our treatment of
proper multiple integrals and give a few theorems on point sets
which we shall either need now or in the next chapter where we
take up the important subject of improper multiple integrals.
In Volume I, 702, we have said that a limited point set whose
upper and lower contents are the same is measurable. It seems
best to reserve this term for another notion which has come into
great prominence of late. We shall therefore in the future call
sets whose upper and lower contents are equal, metric sets. When
a set 51 is metric, either symbol
I or 51
expresses its content. In the following it will be often con
venient to denote the content of 51 by
a.
This notation will serve to keep in mind that 51 is metric, when
we are reasoning with sets some of which are metric, and some
are not.
The frontier of a set as 51, may be denoted by
Front 51.
2. 1. In I, 713 we have introduced the very general notion of
cell, division of space into cells, etc. The definition as there
1
2 POINT SETS AND PROPER INTEGRALS
given requires each cell to be metric. For many purposes this
is not necessary ; it suffices that the cells form an unmixed divi
sion of the given set 51. Such divisions we shall call unmixed di
visions of norm 8. [I, 711.] Under these circumstances we have
now theorems analogous to I, 714, 722, 723, viz :
2. Let 3? contain the limited point set 51. Let A denote an un
mixed division of $8 of norm 8. Let 515 denote those cells of 33 con
taining points of 5L Then
8=0
The proof is entirely analogous to I, 714.
3. Let $8 contain the limited point set 51. Let f(x^ • • • xm) be
limited in 51. Let A be an unmixed division of ^Q of norm 8 into
cells Sr S2, • • • . Let 9ftt, mt be respectively the maximum and mini
mum off in St. Then
lim £A = lira 29HA = f /«, (1
6=0 8=0 */SI
lira flA = lim 2mA = f/* (2
5=0 S=0 ^H
Let us prove 1) ; the relation 2) may be demonstrated in a similar
manner. In the first place we show in a manner entirely analo
gous to I, 722, that
. (3
The only modifications necessary are to replace 8t, 8[, Sl)C, by their
upper contents, and to make use of the fact that A is unmixed, to
establish 5).
To prove the other relation
, (4
we shall modify the proof as follows. Let E be a cubical division
of space of norm e < eQ. We may take eQ so small that
PROPER INTEGRALS 3
The cells of E containing points of 21 fall into two classes.
1° the cells eltt containing points of the cell St but of no other cell
of A ; 2° the cells e( containing points of two or more cells of A.
Thus we have
where MM M(, are the maxima of / in et(C, e[. Then as above we
have
if eQ is taken sufficiently small.
On the other hand, we have
Now we may suppose S0, eQ are taken so small that
differ from 51 by as little as we choose. We have therefore for
properly chosen S0, e0,
This with 6) gives
which with 5) proves 4).
4. Let f(xl • • - xm) be limited in the limited field 21. Let A be
an unmixed division oftyiof norm 8, into cells Sv S2 • • • . Let
where as usual mt, Jf[ are the minimum and maximum of f in 8t.
Then
= Max £A, yrfa = Min S*.
The proof is entirely similar to I, 723, replacing the theorem
there used by 2, 3.
5. In connection with 4 and the theorem I, 696, 723 it may be
well to caution the reader against an error which students are apt
to make. The theorems I, 696, 1, 2 are not necessarily true if /
4 POINT SETS AND PROPER INTEGRALS
has both signs in 51. For example, consider a unit square S
whose center call O. Let us effect a division E of S into 100
equal squares and let 51 be formed of the lower left-hand square s
and of (7. Let us define / as follows :
/ = 1 within s
= - 100 at 0.
For the division E,
Hence, Min ^ < _
On the other hand, ^m g __ _\
d=0
The theorems I, 723, and its analogue- 4 are not necessarily true
for unmixed divisions of space. The division A employed must
be unmixed divisions of the field of integration 51. That this is
so, is shown by the example just given.
6. In certain cases the field 51 may contain no points at all.
In such a case we define „
f/=o.
J%
7. From 4 we have at once :
Let A be an unmixed division of 51 into cells Sv £2, ••• Then
% = Min 2£t,
with respect to the class of all divisions A.
8. We also have the following :
Let D be an unmixed division of space. Let d±, d^-- denote those
cells containing points of 51. • Then
^ = Min 24,
with respect to the class of the divisions D.
For if we denote by S, the points of 51 in d, we have obviously
5<2,.
Also by I, 696, S = Min 2et
PROPER INTEGRALS 5
with respect to the class of rectangular division of space E = \e,\.
But the class E is a subclass of the class D.
Thus
Min 28t < Min 2^ <Min 2«c.
A D E
Here the two end terms have the value 51.
3. Let/^j ••• zm), g(x^ ••• xm) be limited in the limited field 51.
We have then the following theorems :
1. Letf = g in 51 except possibly at the points of a discrete set £).
Then, _ _
//=//•
For let | / 1, \g\<M. Let D be a cubical division of norm d.
Let MU N, denote the maximum of/, g in the cell d,. Let A de
note the cells containing points of £), while A may denote the
other cells of 5l/>.
Then,
Hence,
^ _ ^-^ < 2 -flf, - # dt < 2
and the term on the right = 0 as d = 0.
2. Let f > g in 51 except possibly at the points of a discrete set
Then
Forlet
But in J., f>g, hence
The theorem now follows at once.
POINT SETS AND PROPER INTEGRALS
,_.,,
For in any cell dt
•,- Max • cf=c Max/; Min • cf = c Min/
when c > 0 ; while
Max • </ = <?Min/; Min • <?/=<? Max/
when c < 0.
4. If g is integrable in 51,
For from
Max / + Min g < Max (/ + g) < Max / + Max g,
we have
(2
But g being integrable,
Hence 2) gives
which is the first half of 1). The other half follows from the
relation
Min / + Min g<M'm (/ + g) < Min / + Max g.
5. The integrands /, g being limited,
For in any cell d,
PROPER INTEGRALS
6. Let f = g + h, \h\<H a constant, in 51 Then,
Then by 2 and 4
or
fff< /< <7 + ftf,
^/2i ^l ~2l J%
r
g< \ f<
4. Letf(xl ••• a:TO) be limited in limited 51. Then,
|/ 1 < Jf, wg have also,
fl/l<.JrtC
(2
(3
(4
(5
Let us effect a cubical division of space of norm S.
To prove 1) let JVt = Max|/| in the cell dL. Then using the
customary notation,
Hence
Letting 8=0, this gives
\vhich is 1).
<
8 POINT SETS AND PROPER INTEGRALS
To prove 3), we use the relation
-I/I </<!/!•
Hence
from which 3) follows on using 3, 3.
The demonstration of 4) is similar.
To prove 5), we observe that
5. 1. Let f> 0 be limited in the limited fields 33, (L Let 91 be
the aggregate formed of the points in either 33 or (L Then
This is obvious since the sums
may have terms in common. Such terms are therefore counted
twice on the right of 1) and only once on the left, before passing
to the limit.
Remark. The relation 1) may not hold when / is not > 0.
Example. Let $ = (0, 1), 55 = rational points, and (£ = irra
tional points in $. Let/= 1 in 55, and - 1 in g. Then
and 1) does not now hold.
2. Let 51 be an unmixed partial aggregate of the limited field
Let (5 = 55 - 51. If
g =/ in 51
= 0 in (5,
PROPER INTEGRALS
For
* * *
J 0=J 7+JL*
«/23 */2i »£(£
But
and obviously
3. The reader should note that the above theorem need not be
true if 51 is not an unmixed part of 33.
Example. Let 51 denote the rational points in the unit square
8.
Let
/ = 9 = ~ *-
Then
^M
=-L
4. Let Vibe a part of the limited field $. Letf^O be limited in
. Let g=f in 51 and = 0 in (£ = 33 - 51.
//> r^. (2
aar «^s
For let Jft, JVt be the maxima of/, g in the cell c?t. Then
21
Passing to the limit we get 1).
To prove 2) we note that in an}^ cell containing a point of 51
Min/> Min </.
6. 1. Letf(x^ "- xm) be limited in the limited field 51. Let
be an unmixed part of 51 such that §M == 51 as u = 0. Then
f/=Hmf/.
»/2l "^J^Su
10 POINT SETS AND PROPER INTEGRALS
For let /< Jfin^l. Let £ = 51 - $. Then
But
by 4, 1), 5).
Hence passing to the limit u = 0 in 2) we get 1).
2. We note that 1 may be incorrect if the $&u are not unmixed.
For let 51 be the unit square. Let $QU be the rational points in a
concentric square whose side is 1 — u. Let/= 1 for the rational
points of 51 and = 2 for the other points. Then
ff-2 limff-l
L/--» „=*! /-i.
7. In I, 716 we have given a uniform convergence theorem
when each 33M < 5(. A similar theorem exists when each 33 M >.
viz. :
33M < 33M, ^7 M < t/. .L^ 51 be a part of each $u. Let <gM =
as u == 0. Then for each e> 0, there exists a pair %0, dQ such that
For Stto < 1 + ^, % sufficiently small.
2
Also for any division D of norm ^ < some d0.
»^ z><#«0+ |-
But
Hence
8. 1. Let 51 be a point set in m = r + s way space. Let us set
certain coordinates as xr+l .» xm= 0 in each point of 51. The
resulting points 53 we call a projection of 51- The points of 51
PROPER INTEGRALS 11
belonging to a given point b of 33, we denote by (£6 or more shortly
by (5- We write
* = «•€,
and call 33, & components of 51.
We note that the fundamental relations of I, 733
hold not only for the components £, $, etc., as there given, but
also for the general components 21, 33.
In what follows we shall often give a proof for two dimensions
for the sake of clearness, but in such cases the form of proof will
admit an easy generalization. In such cases 33 will be taken as
the ^-projection or component of 21-
2. 7f 21 = 33 • (£ is limited and 33 is discrete, 21 is also discrete.
For let 21 lie within a cube of edge J (7 > 1 in m — r + s way
space. Then for any d < some d^
Then 1^ < 0s® D < e.
3. That the converse of 2 is not necessarily true is shown by
the two following examples, which we shall use later :
Example 1. Let 2( denote the points re, y in the unit square
determined thus :
For
0»~, n=l, 2, 3, ..-, m odd and < 2n,
let
'
Here 21 is discrete, while 33 = 1, where 33 denotes the projection
of 21 on the z-axis.
4. Example 2. Let 21 denote the points x< y in the unit square
determined thus :
12 POINT SETS AND PROPER INTEGRALS
For
/yy*
x = —, m, n relatively prime,
ft
let 1
o<,<l.
Then, 33 denoting the projection of 5l on the #-axis, we have
f=0, # = 1.
9. 1. Let 91 = $ • (£ be a limited point set. Then
For let/=l in 5L Let g = l at each point of 51 and at the
other points of a cube A = B - 0 containing 51, let g = 0. Then
By I, 733, '
But by 5, 4,
Thus
which gives 1), since
2. 7/i case 51 ^s metric we have
S=J8S, (2
is an integrable function over 3$.
This follows at once from 1).
PROPER INTEGRALS 13
3. In this connection we should note, however, that the converse
of 2 is not always true, i.e. if (£ is integrable, then 51 has content
and 2, 2) holds. This is shown by the following :
Example. In the unit square we define the points x, y of 51 thus :
For rational x,
For irrational x, 1 < < 1
Then (£ = J for every x in 23. Hence
But 51 = 0, 51 = 1.
10. 1. Letf(xl ••• x^ be limited in the limited field 51 = 23
Let us first prove 1). Let 51, 23, £ lie in the spaces 9?m, 9^ 9?a,
r + s = m. Then any cubical division D divides these spaces into
cubical cells c?t, d[, d[' of volumes d, cZ', d" respectively. Ob
viously d = d'd". D also divides 23 and each Q. into unmixed cells
8', 8". Let ^t = Max/ in one of the cells d» while JtfJ' = Max/
in the corresponding cell &['. Then by 2, 4,
since M« M' > 0. Hence
p'jT/<|
Letting the norm of D converge to zero, we get 1). We get
2) by similar reasoning or by using 3, 3 and 1).
14 POINT SETS AND PROPER INTEGRALS
2. To illustrate the necessity of making/ >0 in 1), let us take
51 to be the Pringsheim set of I, 740, 2, while / shall = - 1 in 51.
Then
On the other hand
Hence
and the relation 1) does not hold here.
Iterable Fields
11. 1. There is a large class of limited point sets which do not
have content and yet —
«=(*£. (1
•£»
Any limited point set satisfying the relation 1) we call iterable,
or more specifically iterable with respect to 33-
Example 1. Let 51 consist of the rational points in the unit
square. Obviously — — _
8= f<£= fg»l,
ji/33 ji/Q£
so that 5( is iterable both with respect to 33 and (£.
Example 2. Let 51 consist of the points x, y in the unit square
defined thus :
For rational x let 0 < y < 1.
For irrational x let 0 < y < 1.
Here 1 = 1.
Thus 51 is iterable with respect to (£ but not with respect to 33-
ITERABLE FIELDS 15
Example 3. Let 51 consist of the points in the unit square de
fined thus: ,, . i -I . A ^ ^ o
b or rational x let 0 < y < |.
For irrational a; let -J < ?/ < 1.
Here 51 = 1, while -
Hence 51 is iterable with respect to d but not with respect to 33.
Example 4- Let 51 consist of the sides of the unit square and
the rational points within the square.
Here 51 = 1, while
and similar relations for (L Thus 51 is not iterable with respect
to either 33 or (£.
Example 5. Let 51 be the Pringsheim set of I, 740, 2.
Here 51 = 1, while
Hence 51 is not iterable with respect to either 33 or Q.
2. Every limited metric point set is iterable with respect to any of
its projections.
This follows at once from the definition and 9, 2.
12. 1. Although 51 is not iterable it may become so on remov
ing a properly chosen discrete set £).
Example. In Example 4 of 11, the points on the sides of the
unit square form a discrete set £) ; on removing these, the deleted
set 51* is iterable with respect to either 33 or (£.
2. The reader is cautioned not to fall into the error of suppos
ing that if 5lx and 5^ are unmixed iterable sets, then 51 = 5lj + 512
is also iterable. That this is not so is shown by the Example in 1.
For let 5ti = 51*, 512= 1) in that example. Then £5 being dis
crete has content and is thus iterable. But 5( = 5^ -h 512 is not
iterable with respect to either 33 or (L
16 POINT SETS AND PROPER INTEGRALS
13. 1. Let 51 be a limited point set lying in the m dimensional
space ffim. Let 33, (£ be components of 51 in 9?r, 9^, r + s = m.
A cubical division D of norm 8 divides 3?OT into cells of volume
d and 9?r and 9?s into cells of volume dr, d# where d = drds. Let
b be any point of 33, lying in a cell dr. Let ^ds denote the sum
6
of all the cells ds, containing points of 51 whose projection is b.
Let 2c?5 denote the sum of all the cells containing points of 51
dr
whose projection falls in dr, not counting two da cells twice.
We have now the following theorem :
If 51 is iterable with respect to 33>
F°r
Hence
Let now 8=0. The first and third members = 51, using I, 699,
since 51 is iterable. Thus, the second and third members have
the same limit, and this gives 1).
2. If 51 is iterable with respect to 33»
lim 2dr2d. = g.
fi=o 33 6
This follows at once from 1).
3. Let 51 = 33 • (5 be a limited point set, iterable with respect to 33-
Then any unmixed part & of 51 is also iterable with respect to the
^-component o
For let b = a point of 33 ; (5' points of 51 not in (5 ; Cb= points
of @j in (5, (7^ = points of (E6 in (£'. Then for each @>Q there
exist a. pair of points, bv 52, distinct or coincident in any cell dr
such that as b ranges over this cell,
+ /3', C = Max Ob + 13",
ITERABLE FIELDS 17
Let S denote, as in 13, l, the cells of 2c£, which contain points of G?',
and F the cells containing points of both (5, & whose projections
fall in dr. Then from
Cb^C'^<Cb^S<Cb
we have
5*, < Obi + Cbi < Min Cb + /3' + #< Max
Multiplying by c?r and summing over 33 we have,
i < 2 rfr Min ^ + 2 £;dr 4- (!i < 2 d r Max <76
Passing to the limit, we have
+ 7;" + S<i, (2
the limit of the last term vanishing since (5, ($' are unmixed parts
of 21. Here r)', rjrf are as small as we please on taking 0 sufficiently
small. From 2) we now have
f (7=S-(!' =
J
4. Let 51 = 33 • £ £e iterable with respect to 33. Ze£ £ be a part
of 33 a/ic?^l aZ£ those points of 21 wA0S£ projection falls on B. Then
A is iterable with respect to B.
For let D be a cubical division of space of norm d. Then
where the sum on the right extends over those cells containing
no point of A. Also
where the second sum on the right extends over those cells dr
containing no point of B.
Subtracting 1), 2) gives
0 = lim { AD - ZdM] + lim { ^drds -
d=U I B » d=& ^r.s
18 POINT SETS AND PROPER INTEGRALS
As each of the braces is > 0 we have
14. We can now generalize the fundamental inequalities of I,
733 as follows :
Let /(#! ... xm) be limited in the limited field $ = 23 • (5, iterable
with respect to 33- Then
For let us choose the positive constants A, B such that
, n 31.
Let us effect a cubical division of the space of $ftm of norm 8 into
cells d. As in 13, this divides 9^., $ts into cells which we denote,
as well as their contents, by c?r, ds. Let b denote any point of 33.
As usual let w, M denote the minimum and maximum of / in the
cell d containing a point of 31. Let ra', M' be the corresponding
extremes of / when we consider only those points of 51 in d whose
projection is b. Let |/| < F in $.
Then for any b, we have by I, 696,
or
- B(^ds - g) + ^mds < )J, (2
since m^m1.
In a similar manner
-<£)• (3
Thus for any b in 33, 2), 3) give
6
Let /3 > 0 be small at pleasure. There exist two points bv b% dis
tinct or coincident in the cell dr, for which
ITERABLE FIELDS 19
where | ^ , /32 1 < £ and 6X, and Ct2 stand for (£v @6j, and finally
where
, J-=MaxJ/
for all points b in <fr.
Let c = Min £ in dr, then 4) gives
, - c) + 2widf < y + /?! < <7 + /32 < ^Mds + 4(2d, - c)
11 22
where the indices 1, 2 indicate that in 2 we have replaced b by
Multiplying by dr and summing over all the cells dr containing
points of 33, the last relation gives
- 0. (5
20 50 33 2 33 2
NOW aS 8 = 0, V,7V7 • O? V 7 V,7 • W 1,-IQo
Zdrz,as = 51, 2,ar2,as = VI, by lo, 2.
33 1 33 2
= 1 S = 5t, since 51 is iterable.
Thus the first and last sums in 5) are evanescent with 8. On
the other hand
2(dr2d,wi - 2<*,w) I < F^dr(^ds - 2ds)
% df I 33 ds 1
= 0 as 8=0, by 13, 1.
Thus
^
lim 2rfr2d,w =| /. (6
(7
Hence passing to the limit 3=0 in 5) we get 1), since 2£1dfr,
2/32t?r have limits numerically <yQ(5 which may be taken as small
as we please as /3 is arbitrarily small.
20 POINT SETS AND PROPER INTEGRALS
2. If 51 is not iterable with respect to 33, let it be so on remov
ing the discrete set T). Let the resulting field A have the com
ponents B, 0. Then 1 gives
since - -
(/=/>•
c/51 JA
3. The reader should guard against supposing 1) is correct if
only 51 is iterable on removing a discrete set £). For consider
the following :
Example. Let the points of 51 = 5lx + T) lie in the unit square.
Let 5lj consist of all the points lying on the irrational ordinates.
Let £) lie on the rational ordinates such that, wrhen
m, n relatively prime,
Let us define/ over 51 thus :
•<„<!
/= in 5Ir
/=0 in £).
The relation 1) is false in this case. For
//=!,
^21
while —
f f /=°-
*/s */e
15. 1. Let /(ajj • • • #TO) be limited in the limited point set 51.
Let D denote the rectangular division of norm d. All the points
of 5t/> except possibly those on its surface are inner points of 51.
[1,702.]
The limits Hm Cf , nm f / (1
rf=o «/9i rf=o «/2f
-D -D
exist and will be denoted by
/*/ , JV , (2
-^ ^21
and are called the inner, lower and upper integrals respectively.
ITERABLE FIELDS 21
To show that 1) exist we need only to show that for each 6 > 0
there exists a dQ such that for any rectangular divisions D1, D'f of
norms < d 0 ( - -
<€.
To this end, we denote by E the division formed by superimpos
ing D" on Df. Then E is a rectangular division of norm < d0.
T pf
«,-!,, = 4', *,-«,, = 4".
If d0 is sufficiently small, ., .„
-A , A. <« ^
an arbitrarily small positive number. Then
A = |ff -/ V-(f -f
|V-;^ Jjyy Wgy j5
if 97 is taken small enough.
2. The integrals
Cf, f /,
%/2I Jw
heretofore considered may be called the outer, lower and upper in
tegrals, in contradistinction.
3. Let f be limited in the limited metric field 51. Then the inner
and outer lower (upper) integrals are equal.
For W.D is an unmixed part of 51 such that
Then by 6, l, r r
limj /-//
rf=o J%D J%
But the limit on the left is by definition
4. When 51 has no inner points,
JV=o.
t/SM
22. POINT SETS AND PROPER INTEGRALS
For each $/> = 0, and hence each
Point Sets
16. -Le£ 51 = 33 + £ £>e metric. Then
g = 33 + g. (1
For let .Z) be a cubical division of space of norm d. The cells
of 51^ fall into three classes : 1°, cells containing only points of 33;
these form 33/>. 2°, cells containing points of (£ ; these form &D.
3°, cells containing frontier points of 33, not already included in 1°
or 2°. Call these \D. Then
^ = %D + &D + \D. (2
Let now d = 0. As 51 is metric, f^ = 0, since \D is a part of
Front 51 and this is discrete. Thus 2) gives 1).
17. 1. Let 51, 33, G - (1
be point sets, limited or not, and finite or infinite in number.
The aggregate formed of the points present in at least one of the
sets 1) is called their union, and may be denoted by
U(K, 33, (5 •••),
or more shortly by ,<* ™ ^ x
If 51 is a general symbol for the sets 1), the union of these sets
may also be denoted by 77 5 91 [
or even more briefly by r^>
If no two of the sets 1) have a point in common, their union
may be called their sum, and this may be denoted by
51 + 33 + S+ -
The set formed of the points common to all the sets 1) we call
their divisor and denote by
POINT SETS 23
if 51 is a general symbol as before.
2. Examples.
Let 51 be the interval (0, 2); S3 the interval (1, oo). Then
Z7(2l, «) = (0, oo), Dv($(, 33) = (1, 2).
Let ^ = (0, 1), 5I2 = (1, 2).-.
Then
Let ^=(1, i), 5i2=(i j), 5
Then ^(512,512-0 = (0*, 1),
!)»(«!, «a -) = 0.
Let *i =
Then
3' Let
Let
Let 21 = ^ + ^, 2li = ^2 + ^-
Then 21 = -D + (E! + (£2 H- -
Let us first exclude the = sign in 1). Then every element of
51 which is not in £) is in some 5ln but not in 5ln+r It is therefore
in (EB+1 but not in £n+2> £n+3^ "• The rest now follows easily.
4. Some writers call the union of two sets 51, 53 their sum,
whether 51, 33 have a point in common or not. We have not done
this because the associative property of sums, viz. :
does not hold in general for unions.
24 POINT SETS AND PROPER INTEGRALS
Example. Let 51 = rectangle (1234),
<8 = (5678),
(£ = (5 8 « /3) =
Then 0W»
and (OX* »)-<£), (2
are different.
Thus if we write -f- for £7", 1), 2) give
18. 1. -Le£ 511>512>513 "* ^e a 8e^ °/ limited complete point
aggregates. Then
33 = ^(51!, 512 -'OX)'
Moreover 33 as complete.
Let an be a point of 3ln, w = 1, 2, ... and 51 = «p «2' ^3 '**
Any limiting point a of 51 is in every 5ln. For it is a limit
ing point of
But all these points lie in 5lm, which is complete. Hence a lies in
5lm, and therefore in every y(v $[2, ••• Hence a lies in 23, and
» > 0.
33 i* complete. For let /3 be one of its limiting points. Let
V 2' 3' -
As each bm is in each 2ln, and 5(n is complete, y8 is in 5ln- Hence ft
is in 33-
2. Let W. be a limited point set of the second species. Then
s complete.
For 5((n) is complete and > 0. Also 5l(n) .>
19. i«« 5lr 212 ••• &« *^ 33; let 51 = ^7S5lni. i«< An be the com
plement of 5ln with respect to 33> so that An + 2In = 33- ^
^1 = Dv\An\ . Then A and 51 are complementary, so that A 4- 51 = 33
POINT SETS 25
For each point b of 33 lies in some 5ln, or it lies in no 5In, and
hence in every An. In the first case b lies in 51, in the second in
A. Moreover it cannot lie in both A and 51-
20. 1. Let 2lj <512< 513 ••• (1
be an infinite sequence of point sets whose union call 51. This
fact may be more briefly indicated by the notation
Obviously when 51 is limited,
1 > lim ln. (2
That the inequality may hold as well as the equality in 2) is
shown by the following examples.
Example 1. Let 5ln = the segment (-, 1 ] •
\H I
Then 5l = Z7j5U = (0*, 1).
Example 2. Let an denote the points in the unit interval whose
abscissae are given by
x = — , m < n = 1, 2, 3, ••• m, n relatively prime.
n
Let 5ln = a + - + an.
Here 2i
is the totality of rational numbers in (0*, 1*).
A c
21 = 1 and 5ln = 0, we see
51 > lim ln.
2. Let «i>_«2>-
Let 33 be their divisor. This we may denote briefly by
Obviously when 33X is limited,
8 < lim Sn.
26 POINT SETS AND PROPER INTEGRALS
Example 1. Let 33n = the segment ( 0, -V
Then ^ = ^533^ = (0), the origin.
Here • • • 1
n
and S = lim gn.
Example 2. Let 2ln be as in 1, Example 2. Let bn = 21 — 2ln.
$A — f~\ 9^\ _l_ fi
Here ^ _ the gegment (^ 2) and «„ = 2.
Hence g < lim »..
3. ie^ 33j < 332 < ••• ^^ unmixed parts of 21. -£e£ Sn = 21.
LJ. Then (£ = 21 — 33 i* discrete.
For let 31 = $n + (£n ; then (5n is an unmixed part of 21. Hence
s=sw+c.
Passing to the limit n = oo, this gives
limgn=0.
Hence (5 is discrete by 2.
4. We may obviously apply the terms monotone increasing,
monotone decreasing sequences, etc. [Cf. 1, 108, 211] to sequences
of the type 1), 3).
21. Let& = 2( + 33. If 51, 33 are complete,
£ = S+S. (1
For S = Dist (21, 33) > 0,
since 21, 33 are complete and have no point in common. Let D be
a cubical division of space of norm d. If d is taken sufficiently
small 2l/), 33/> have no cells in common. Hence
Letting d= 0 we get 1).
POINT SETS 27
22. 1. If 51, 33 fl/*e complete, so are also
£ =(21,33), £)
Let us first show that £ is complete. Let c be a limiting point
of £. Let Cj, £2, ••« be points of (5 which = c. Let us separate
the cn into two classes, according as they belong to 51, or do not.
One of these classes must embrace an infinite number of points
which = c. As both 51 and 33 are complete, c lies in either 51 or
33- Hence it lies in g.
To show that £) is complete. Let dv d2, ••• be points of £) which
= d. As each dn is in both 5( and 33, their limiting point d is in
51 and 33, since these are complete. Hence d is in £).
2. If 51, 23 are metric so are
For the points of Front £ lie either in Front 51 or in Front 33,
while the points of Front £) < Front 51 and also < Front 33. But
Front 5( and Front 33 are discrete since 51, 33 are metric.
23. Let the complete set 51 have a complete part 33. Then hoiv-
ever small e > 0 is taken, there exists a complete set & in 51, having no
point in common with 33 such that
£>5(-5-e. (1
Moreover there exists no complete set d, having no point in common
with 33 such that
The second part of the theorem follows from 21. To prove 1)
let D be a cubical division such that
5^=51 + 6', ^ = 2) + e", 0<e', €"<e. (2
Since 33 is complete, no point of 33 lies on the frontier of 33z>-
Let £ denote the points of 51 lying in cells containing no point of
33. Since 51 is complete so is (S, and 33, (£ have no point in common.
Thus
. (3
28 POINT SETS AND PROPER INTEGRALS
But the cells of &D may be subdivided, forming a new division A,
which does not change the cells of $QD, so that $8D = $3A, but so that
1A = ! + €'", 0<e'"<€. (4
. Thus 2), 3), 4) give
24. Let 51, 33 be complete. Let
Forlet II = 51 + A.
Then A contains complete sets (7, such that
^>U-S-e, (2
but no complete set such that
c>u-a, (3
by 23. On the other hand,
»*A + $.
Hence ^4 contains complete sets (7, such that
a>®-s>-€, (4
but no complete set such that
<7>S-£. (5
From 2), 3), and 4), 5) we have 1), since 6 is arbitrarily small.
25. Let
each $n being complete and such that 5ln > some constant k.
Then
POINT SETS 29
For suppose l = k-®>0.
Let Z = e+7,; e, 7,>0.
Then by 23 there exists in 5lj a complete set Q^, having no point
in common with £) such that
Sj > Sj - T) - e ;
or as 5tj > A:, such that ^ >
Let (5,), U = (2I2, eO-
by 24,
Thus
Thus $2 contains the non-vanishing complete set @2 having no
point in common with £). In this way we may continue. Thus
5Ir $2, ••• contain a non-vanishing complete component not in £),
which is absurd.
Corollary. Let 31 = (5Ij< 512< •••) Je complete. Then Sn = 3.
This follows easily from 23, 25.
CHAPTER II
IMPROPER MULTIPLE INTEGRALS
26. Up to the present we have considered only proper multiple
integrals. We take up now the case when the integrand f(x1 - • • xm)
is not limited. Such integrals are called improper. When m = 1,
we get the integrals treated in Vol. I, Chapter 14. An important
application of the theory we are now to develop is the inversion
of the order of integration in iterated improper integrals. The
treatment of this question given in Vol. I may be simplified and
generalized by making use of the properties of improper multiple
integrals.
27. Let 51 be a limited point set in w-way space $lm. At each
point of 51 let f(x1 •••#m) have a definite value assigned to it.
The points of infinite discontinuity of / which lie in 51 we shall
denote by $. In general $ is discrete, and this case is by far the
most important. But it is not necessary. We shall call $ the
singular points.
Example. Let 51 be the unit square. At the point # = —,
y = -, these fractions being irreducible, let f=ns. At the other
s
points of 51 let /= 1. Here every point of 51 is a point of infinite
discontinuity and hence Q = 51-
Several types of definition of improper integrals have been
proposed. We shall mention only three.
28. Type I. Let us effect a division A of norm 8 of 9?m into
cells, such that each cell is complete. Such divisions may be
called complete. Let 515 denote the cells containing points of 51,
but no point of Q , while 5l§ may denote the cells containing a
point of $• Since A is complete, / is limited in 5ls- Hence /
admits an upper and a lower proper integral in 5ls« The limits,
when they exist, -^
lim f /, lim ) /, (1
6=0 c/2l 5=0 ^21
GENERAL THEORY 31
for all possible complete divisions A of norm S, are called the
lower and upper integrals of / in 51, and are denoted by
CfdK,
«
(2
«l «*
or more shortly by
/•
When the limits 1) are finite, the corresponding integrals 2)
are convergent. We also say/ admits a lower or an upper improper
integral in 51- When the two integrals 2) are equal, we say that
/ is integrable in 51 and denote their common value by
or by J/. (3
We call 3) the improper integral of f in 51 ; we also say that
/ admits an improper integral in 51 and that the integral 3) is
convergent.
The definition of an improper integral just given is an extension
of that given in Vol. I, Chapter 14. It is the natural develop
ment of the idea of an improper integral which goes back to the
beginnings of the calculus.
It is convenient to speak of the symbols 2) as upper and lower
integrals, even when the limits 1) do not exist. A similar remark
applies to the symbol 3).
Let us replace /by l/| in one of the symbols 2), 3). The
resulting symbol is called the adjoint of the integral in question.
We write
(4
When the adjoint of one of the integrals 2), 3) is convergent,
the first integral is said to be absolutely convergent. Thus if 4) is
convergent, the second integral in 2) is absolutely convergent, etc.
29. Type II. Let X, /u,>0. We introduce a truncated func
tion /Afx defined as follows :
/AM =/Oi •••#») when - X </< /*
= — X when / < — X
= p when / > fjL.
32 IMPROPER MULTIPLE INTEGRALS
We define now the lower integral as
c c
J f = Iim J /AM.
A similar definition holds for the upper integral. The other
terms introduced in 28 apply here without change.
This definition of an improper integral is due to de la Vallee
Poussin. It has been employed by him and R. G-. D. Richardson
with great success.
30. Type III. Let a, /3 > 0. Let 5la/3 denote the points of 51
We define now
f / = Iim f f ; f / = Km f /. (1
tl2l a,/3 = °o^la/3 ^21 •,0—>*/*«0
The other terms introduced in 28 apply here without change.
This type of definition originated with the author and has been
developed in his lectures.
31. When the points of infinite discontinuity -3 are discrete
and the upper integrals are absolutely convergent, all three defini
tions lead to the same result, as we shall show.
When this condition is not satisfied, the results may be quite
different.
Example. Let 51 be the unit square. Let 5lr 512 denote respec
tively the upper and lower halves. At the rational* points 53,
x = —, y = ~, in 5L, let/= ns. At the other points (£ of 5lr let
n ' s
/=-2. In 51
1° Definition. Here
Hence
2° Definition. Here
T .
- + oo.
* Here as in all following examples of this sort, fractions are supposed to be
irreducible.
GENERAL THEORY 33
3° Definition. Here 3la/3 embraces all the points of 2J2, £ and a
finite number of points of 33 for a > 2, /3 arbitrarily large. Hence
//=-!, //=-!,
*£i ^n
and thus
32. In the following we shall adopt the third type of definition,
as it seems to lead to more general results when treating the im
portant subject of inversion of the order of integration in iterated
integrals.
We note that if /is limited in 21,
lim I / = the proper integral 1 /.
a, /3=»_2l ^l2l
For a, ft being sufficiently large, 2la/J = 21-
Also, if 31 is discrete,
//=//= °-
For 2la8 is discrete, and hence
r
Hence the limit of these integrals is 0.
33. Let m=|Min/| , JJf=|Max/| in 51.
Then - -
lim I / = lim I /, m finite.
a,/3=oc^2laj3 ft-»*?Mm,p
lim f /=lim f /, M finite.
+fi-<*!mmj a=^2ia)J/
For these limits depend only on large values of a, ft, and when
is finite.
$OT ^ = $a ^ , for all a > 7/1.
Similarly, when Mis finite
^.6 = 3L* f or all /3 > M.
34 IMPROPER MULTIPLE INTEGRALS
Thus in these cases we may simplify our notation by replacing
CUT (\f
2U, Jit » 2im/3
bj 3L. , %ft ,
respectively.
2. Thus we have:
Xf = lim I /, when Min/ is finite.
j *~A£
> J / = lim J / , when Max/ i« finite.
3. Sometimes we have to deal with several functions /,#, • ••
In this case the notation 5la/3 is ambiguous. To make it clear we
let 21/, a> p denote the points of 51 where
Similarly, 51^, a> ^ denotes the points where
34. I f is a monotone decreasing function of a for each ft.
*f*«i
I f is a monotone increasing function of ft for each a.
J**fi
If Max / is finite
| / are monotone decreasing functions of a.
•/«-«
If Uin fit finite
I / are monotone increasing functions of ft.
J*e
Let us prove the first statement. Let a' > a.
Let D be a cubical division of space of norm d.
Then ft being fixed,
( /= lirn 2 wtc?t, (1
•f«M <*=« 2la/3
f /=lim2w;^;, (2
•/tU'/l rf=° «a'^
using the notation so often employed before.
GENERAL THEORY 35
But each cell dt of 5la/3 lies among the cells dj of 2L /s. Thus we
can break up the sum 2), getting
Here the second term on the right is summed over those cells
not containing points of 5la/3. It is thus < 0. In the first term
on the right ml <mL. It is thus less than the sum in 1). Hence
Thus
L^L:
«'>«.
In a similar manner we may prove the second statement ; let
us turn to the third.
We need only to show that
| / is monotone decreasing.
J2l_'a
Let «'>«. Then
= lim SJIf^. (3
As before 'S.M'd' = ^M'd' + ^Mndn f 5
*-• «-/
But in the cells d,, MJ = ML. Hence the first term of 5) is
the same as 2 in 3). The second term of 5) is <0. The proof
follows now as before.
35. If Max / is finite and I fare limited, I f is convergent and
_5tt ^_2l
f/< f /.
•/• j/2T_a
If Min f is finite and I are limited, I tf t« convergent and
36 IMPROPER MULTIPLE INTEGRALS
For by 34 -
f / , f /
J«L. J**
are limited monotone functions. Their limits exist by I, 277, 8.
36. If M= Max/ is finite, and Jf is convergent, the correspond
ing upper integral is convergent and
where f >_ — a in 5Ltt.
Similarly, if m = Min / is finite and J f is convergent, the corre
sponding lower integral is convergent and
Let us prove the first half of the theorem.
We have /* 7*
I /= lim I
Jjf a=»J5l_a
Now
We have now only to pass to the limit.
37. If if is convergent, and $8 < 51,
"
does not need to converge. Similarly
_
does not need to converge, although I / does.
Example. Let 21 be the unit square ; let 8 denote the points
for which x is rational.
GENERAL THEORY
when x is irrational
when x is rational
37
Then
/= 1
= -
y
r r
I /=! ; hence I /=!.
J«/ Jtf
On the other hand,
Hence
is divergent.
X= lim I = lim log & = + oo
i s=* ^»s
38. 1. In the future it will be convenient to let ty denote the
points of 51 where /> 0, and 9? the points where/ <. 0. We may
call them the positive and negative components
2. If J /converges, so do \ f.
If I / converges, so do I /.
*/H *AR
For let us effect a cubical division of space of norm d. Let
fi' > /3. Let e denote those cells containing a point of ^ ; e'
those cells containing a point of ^ but no point of ^ ; 5 those
cells containing a point of 9la/3 but none of ^,.
Then
= limJS^f.
*/2la/3 d=0
SJ.
Obviously
Hence
, M,,<0.
~ ~
\ -\ »li
^2l' ^
38 IMPROPER MULTIPLE INTEGRALS
We find similarly
f - f =lim{2(^-Jf.> + 2JfXi-
*Ap/3/ ^?B d=°
Now
I -
ML*
<
for a sufficiently large a, and for any & £' > j3Q.
Hence the same is true of the left side of 1).
As corollaries we have :
3. If the upper integral off is convergent in 51, then
If the lower integral off is convergent in
For
<
etc.
4. -?// >: 0 ancZ I / is convergent, so is
»/5l
(1
Moreover the second integral is < the first.
This follows at once from 3, as 51 = ty.
39. If J / and J f converge, so do J /.
We show that I / converges ; a similar proof holds for I .
*/2t ?±2l
this end we have only to show that
e>0;
L -L
<e; «<«'<«",
To
(1
GENERAL THEORY 39
Let D be a cubical division of space of norm d. Let typ , typ-
denote cells containing at least one point of 2la'|8' , 2la"/j" at which
/>0. Let tta- i tla" denote cells containing only points of 2la-/3' ,
2la-0" at which/< 0. We have
Subtracting,
(2
Let jjf [ = Max / for points of ^ in c?t. Then since / has one
sign in 9fc,
| . (3
Letting d = 0, 2) and 3) give
"
Now if 13 is taken sufficiently large, the first term on the right is
< 6/2. On the other hand, sincej / is convergent, so is J / by
36. Hence for a sufficiently large, the last term on the right is
<e/2. Thus 4) gives 1).
40. Iff is integrable in 21, it is in any $3 < 21-
Let us first show it is integrable in any 2laj3.
Let D be a cubical division of space of norm d.
Then Aap = lira 2o>A , <*>, = DSC/ in dt.
<*=o 2la/3
Let a' > a, £' > 0. Then
A'> - A = lim 2a>W: - 2«A.
40 IMPROPER MULTIPLE INTEGRALS
Now any cell dt of 5laj8 is a cell of 5la-j3', and in d^ co( > co,.
Hence Aa>p > Aap. Thus Aa$ is a monotone increasing function
of a, /3. On the other hand
lim Aap = 0,
by hypothesis. Hence Aaft = 0 and thus /is integrable in 5la/3-
Next let / be limited in 33, then |/|<some 7 in 33. Then
33 < 5lv, r But / being integrable in 5ly, Y, it is in 33 by I, 700, 3.
Let us now consider the general case. Since / is integrable in 51
f/ , f /,
J J.
both converge by 38. Let now P, N be the points of ^P, %l lying
in 33. Then
both converge. Hence by 39,
both converge. But if 33a,6 denote the points of 33 at which
-«</<&,
r/= iim r /,
JjQ a,b=*J®ab
by definition.
But as just seen, C _
£*#
//-
and /is integrable in 33.
41. As a corollary of 40 we have :
1 . Iff is integrable in 51, it admits a proper integral in any part
of 51 in which f is limited.
2. Iff is integrable in any part of 51 in which f is limited, and if
either the lower or upper integral off in 51 is convergent, f is integra
ble in 51.
GENERAL THEORY 41
For let T: ^
J /=lim j / (1
JK .tf—JiM
exist. Since
necessarily
exists and 1), 2) are equal.
42. 1. In studying the function/ it is sometimes convenient to
introduce two auxiliary functions defined as follows :
g=f where/>0,
= 0 where /<0.
h=-f where/<0,
= 0 where />0.
Thus #, h are both > 0 and
We call them the associated non-negative functions.
2. As usual let 5L/s denote the points of 51 at which — a <f<@.
Let 510 denote the points where g<&, and 5la the points where h<a.
Then - -
I g = lim I g, (1
•'SI a,^=oo»/$laj3
f A = lim f h. (2
C/QT /> »y 91
For - -
I 9= I ^ by 5, 4.
^% ^2la^
Letting a, /Q = oo, this last gives 1).
A similar demonstration establishes 2).
3. We cannot say always
C C C 7 C
J # = lim J g ; t A — lim I A,
as the following example shows.
42 IMPROPER MULTIPLE INTEGRALS
Let / = 1 at the irrational points in 21 = (0, 1),
= — n, for x = — in 21.
n
Then ~ ~
^ ^2ta/3
Again let / = — 1 for the irrational points in 21,
= n for the rational points x = - -
n
Then
43. 1.
f A=0. f A=l.
JM* J.Kap
1) f</=f/; /</<//; (2
^2t *7^ ^ ^
3) f A = - f/; f^<- f/; (4
c/21 Jsjj17 c/^ ^/g(j
provided the integral on either side of the equations converges, or
provided the integrals on the right side of the inequalities converge.
Let us prove 1); the others are similarly established. Effecting
a cubical division of space of norm d, we have for a fixed ft,
= lim 2 Jft* = f /. (5
d=Q ^ ^0
Thus if either integral in 1) is convergent, the passage to the
limit ft = QO in 5), gives 1).
2. If I f is convergent, I g converge.
J<£ v/21
If I f is convergent, \ h converge.
*/2l c/21
This follows from 1 and from 38.
GENERAL THEORY 43
3. If I / is convergent, we cannot say that I / is always con-
•/g *Syi
vergent. A similar remark holds for the lower integral.
For let jf=:l at the rational points of 51 = (0, 1)
= at the irrational points.
x
Then
4. That the inequality sign in 2) or 4) may be necessary is
shown thus :
Let -I
/= -L for rational x in 51 = (0, 1)
= for irrational x.
Then r r
J>=0 , J/-i.
44. 1. ff=fg- lim f A, (1
*^2I *^2I a fi = xc'2I
f/ = lim f 0- (\ (2
«^2t a i3 = x*^2t **
provided, 1° ^e integral on the left exists, or 2° fAe integral and the
limit on the right exist.
For let us effect a cubical division of norm d. The cells con
taining points of 51 fall into two classes :
a) those in which /is always < 0,
5) those in which /is >0 for at least one point.
In the cells a), since /= g — A,
Max/ = Max (g-h)= Max g - Min h, (3
as Max# = 0. In the cells b) this relation also holds as Min h = 0.
Thus 3) gives
f /= f £- J A. (4
•^a? "X0 -^
44 IMPROPER MULTIPLE INTEGRALS
Let now a, /3 = oo. If the integral on the left of 1) is conver
gent, the integral on the right of 1^ is convergent by 43, 2. Hence
the limit on the right of 1) exists. Using now 42, 2, we get 1).
Let us now look at the 2° hypothesis. By 42, 2,
lim
g— 1 g.
* *V
Thus passing to the limit in 4), we get 1).
2. A relation of the type
£-£-/.*
does not always hold as the following shows.
Example. Let/ = n at the points x — —
A n
f m
= — n for x =
22W-4-1
= — 1 at the other points of 51 = (0, 1).
Then f/=-l f# = 0 fA = 0.
Ja JsT J»
45. If ( f is convergent, it is in any unmixed part 33 0/21.
c/2t
Let us consider the upper integral first. By 43, 2,
exists. Hence a fortiori,
Jt'
exists. Since 51 = 48 -f (£ is an unmixed division,
C h= C h+ C h.
J*+ J®afi JSafi
Hence C h< C h.
GENERAL THEORY 45
As the limit of the right side exists, that of the left exists also.
From this fact, and because 1) exists,
exists by 44, 1.
A similar demonstration holds for the lower integral over $3.
46. If&v 212 ••• ^imform an unmixed division 0/51, then
f/= f/+ -+ f /,
JKm
provided the integral on the left exists or all the integrals on the
right exist.
For if 2lm a/3 denote the points of 2la/3 in 5lm, we have
JL-JU--+.L- <2
Now if the integral on the left of 1) is convergent, the integrals
on the right of 1) all converge by 45. Passing to the limit in 2)
gives 1). On the other hypothesis, the integrals on the right of
1) existing, a passage to the limit in 2) shows that 1) holds in
this case also.
47. If i f and ( f converge, so does C I/I, and
Jty Jyi c/21
fl/l< f/- f/ (i
si 9s yi
<£g + Jh. (2
For let AB denote the points of 51 where
Then since
X i'i =1
*/AB «/A
< C g+ Ch
JAB *^AB
<fff+ f* (3
c/21 c/5(
<f/-f/ by 43,1. (4
Jy JM
Passing to the limit in 3), 4), we get 1), 2).
46 IMPROPER MULTIPLE INTEGRALS
48. 1. If | I/ 1 converges, loth f '/ converge.
*% JK
For as usual let ^3 denote the points of 5( where />0.
Then '
is convergent by 38, 3, since f \f\ is convergent.
Similarly,
f (-/)=- f/
•% */3?
is convergent. The theorem follows now by 39.
2. If \ | /| converges, so do
fff , f*. (i
Ji/2t j^2t
For by 1,
J%
both converge. The theorem now follows by 43, 2.
3. For
f/ (2
V*
converge it is necessary and sufficient that
is convergent.
For if 3) converges, the integrals 2) both converge by 1.
On the other hand if both the integrals 2) converge,
converge by 38, 2. Hence 3) converges by 47.
4. If/is integrable in 21, so is \f\.
For let Aft denote the points of 5( where 0 < |/| < f}. Then
and the limit on the right exists by 3.
GENERAL THEORY 47
But by 41, I,/ is integrable in A^. Hence |/| is integrable in
ft by I, 720. Thus
49. From the above it follows that if both integrals
. b
converge, they converge absolutely. Thus, in particular, if
converges, it is absolutely convergent.
We must, however, guard the reader against the error of sup
posing that only absolutely convergent upper and lower integrals
exist.
Example. At the rational points of 31 = (0, 1) let
At the irrational points let
Here
A,)-!
//
Thus, / admits an upper, but not a lower integral. On the
other hand the upper integral of / does not converge absolutely.
For obviously
50. We have just noted that if
J|/| = + 00.
is convergent, it is absolutely convergent. For m = 1, this result
apparently stands in contradiction with the theory developed in
Vol. I, where we often dealt with convergent integrals which do
not converge absolutely.
48 IMPROPER MULTIPLE INTEGRALS
Let us consider, for example,
sin-
If we set x = -, we get
sinu -,
u
which converges by I, 667, but is not absolutely convergent by
I, 646.
This apparent discrepancy at once disappears when we observe
that according to the definition laid down in Vol. I,
J = R Km I fdx,
a=0 *^a
while in the present chapter
«7= lim J fdx.
Now it is easy to see that, taking a large at pleasure but fixed,
J fdx = 00 as y3 = oo,
so that <7does not converge according to our present definition.
In the theory of integration as ordinarily developed in works
on the calculus a similar phenomenon occurs, viz. only absolutely
convergent integrals exist when m > 1.
51. 1. If j |/| is convergent,
I// £Xl/!- : (1
For 5la/3 denoting as usual the points of 51 where — «</</3
we have - - -
f / < J |/|< f I/I-
'.5/21 ^21 21
Passing to the limit, we get 1).
GENERAL THEORY 49
2. If § l/l is convergent, C f are convergent for any 33 <2(.
Ja J%
For I (/I is convergent by 38, 4.
*/$
Hence
converge by 48, 3.
3. If, 1°, f" | / 1 is convergent and Min / is finite, or if, 2°, f / is
convergent and Max / is finite, then
is convergent.
This follows by 36 and 48, 3.
52. Letf>Q inW. Let the integral
converge. If
then for any unmixed part 33 < 51,
(2
0 <«'<«. (3
For let 21 = 55 4- G>. Then 51^ = ^ + ^/3 is an unmixed division.
Also
*/9I */« c/ff
= f + « by 1)
s!*fi
=/
V»
+ +«.
50
IMPROPER MULTIPLE INTEGRALS
Hence
From 2)
C + C = C + C + a.
.5/33 ±:& — %$p — £/3
- -
a'= f - f
« J
(4
by4)
which establishes 3).
53, If the integral
converges, then
< 31
J" I/I
that
><r.
C1
(2
(3
Let us suppose first that/>0. If the theorem is not true,
there exists, however small o->0 is taken, a 53 satisfying 3) such
that
Then there exists a cubical division of space such that those
points of 51, call them (£, which lie in cells containing a point of
53, are such that (5<<r also. Moreover (£ is an unmixed part of 51.
Then from 4) follows, as/_>0, that
also.
Let us now take /3 so that
Then
and
by 52. But
1=1+ «''
GENERAL THEORY 51
Let now /3<r < e, then
/ <••
*/CFo
which contradicts 5).
Let us now make no restrictions on the sign of/. We have
/
But since 1) converges, the present case is reduced to the pre
ceding.
54. 1. Let I |/ 1 converge.
%/g
Let as usual 2la/3 denote the points of^at which —«</<#. Let
Aab be such that each 5la/3 lies in some Aab in which latter f is limited.
Let £)a/3 = Aab — 2la/3 and let a, b = ac with a, /3. Then
lim £)a/3 = 0.
a, 0=30
For if not, let
a, /3=a>
Then for any 0 < X < Z, there exists a monotone sequence \an, ftn\
such that
£W« > X for n > some w.
Let /in=Min(an, y8n), then \f > fj,n in X)an/3n^ an(l Mn^00*
Hence ^
J |/|>ASX^oo. (1
^D^fc
X)an/3n being a part of 31
by 38, 3. This contradicts 1).
2. Definition. We say .Afli 6 is conjugate to 5la/3 with respect to/.
55. 1. J.s wswaZ let - a </ < /3 m 3la^. Let 0<f</3 in 51^.
i^^ ^4.^ 5e conjugate to 2(a0 ^^A reference to f ; and Ab conjugate
to 51^ with respect to |/|.
52 IMPROPER MULTIPLE INTEGRALS
V, 1°,
or (f, 2°,
lim f /= f/.
a, 6=00 *d^a, 6 ^«l
For, if 2° holds, 1° holds also, since
Thus case 2° is reduced to 1°. Let then the 1° limit exist.
We have
f /» f tf- f *• (2
J2W J*./ s4g
as 4) in 44, 1 shows. Let now
£>,*.<- 4* -**•
Then,
But ^^ = 0, as «, ft = oo, by 54. Let us now pass to the limit
a, yS = QO in 3). Since the limit of the last term is 0 by 53, 54, we
- -
lim L #=lim ) 9- (4
«, 0=00 ^^ a, 6=ao ^Aa, b
Similarly, ~ ~
lim I h = lim I h. (5
o, 0=c» ^a/3 a, 5=0= ?Z^a6
Passing to the limit in 2), we get, using 4), 5),
f /= lim | f a- f h I
Ax/ «,6=oo (^«/ ll^ J
= lim f /
a, ft = ao ^^aft
In a similar manner we may establish 1) for the lower integrals.
GENERAL THEORY 53
2. The following example is instructive as showing that when
the conditions imposed in 1 are not fulfilled, the relation 1) may
not hold.
Example. Since ^ ,
I —=+oo,
«/0 X
there exists, for any bn > 0, a 0 < bn+1 < 5n, such that if we set
then n ^ n
Cr1<Cr2< *•• =00,
as bn = 0. Let now
/ = 1 for the rational points in 51 = (0, 1),
= - for the irrational.
x
Then
Let
Let An denote the points of 51 in (£>n, 1) and the irrational points
Then C r ^
JA» > ^n+l
But obviously the set An is conjugate to 510. On the other hand,
while /•
lim 1 /= + ».
56. Jf £fo integral
. //
converges, then
e > 0, a- > 0,
/or aw/ unmixed part $8 of 51
/!<€ (2
v
54 IMPROPER MULTIPLE INTEGRALS
Let us establish the theorem for the upper integral ; similar
reasoning may be used for the lower. Since 1) is convergent,
// (3
and X= lim C h (4
a(0==o«/5la/3
exist by 44, l. Since 3) exists, we have by 53,
for any 33 < 5t such that 33 < some cr'.
Since 4) exists, there exists a pair of values a, b such that
X= f h + r) , 0<77<^-, (6
— 2ta6 '
since the integral on the right side of 4) is a monotone increasing
function of a, b.
Since 51 = 53 + (£ is an unmixed division of 51,
C h= C h-}- C h.
^5lal8 ^a/3 ^<£aj3
Since h > 0, and the limit 4) exists, the above shows that
p = lim j h , v = lim I h
a, /3=ao »^SJ3a|3 a, /3=co ^_(Sa(3
exist and that
x=M+". a
Then a, ft being the same as in 6),
M = J A + V, • (8
and we show that
n ^ ' ^ „ (Q
\j <^ 77 <», 77 ^«-'
as in 52. Let now c > a, ft ; then
if we take ™ < _e
4
GENERAL THEORY 55
Thus, M<1 : fe B (n
But ry = f^_^
by 44, 1. Thus 2) follows on using 5), 11) and taking a- <</, cr".
57. If the integral ( f converges and $&u is an unmixed part of
21 such that $$u = 21 as u = 0, then
limf/=f/. (1
u = 0 *Aw *^9I
For if we set 21 = 93U 4- £u, the last set is an unmixed part of 21
and (£„ = 0. Now
J%u J.S*
Passing to the limit, we get 1) on using 56.
58. 1. Let ®afi
If, 1°, the upper contents of
f «P ^ ^/a^ - £>a0 ? 9a^ = ^ a,3 ~ ^a^ > ^a^ = ^/+* a/3 ~ ^a
= 0 as a, /3 =00 ,
(f, 2°, fAe upper integrals off, g,f+g are convergent, then
7/^1° Ao?c?s, awe? {f, 3°, ^Ae Zower integrals off, g,f + g are conver
gent, then
Let us prove 2) ; the relation 3) is similarly established. Let
Z)a, ft be a cubical division of space. Let (5a/3 denote the points of
£)a|8 lying in cells of Z>a/3, containing no point of the sets 1). Let
56 IMPROPER MULTIPLE INTEGRALS
Then Daft may be chosen so that gaj3 = 0.
Now
since the fields are unmixed. By 56, the second integral on the
right = 0 as a, & = oo . Hence
lim JT /= lim / /.
i,H-«D«au«a «,^-«»/«««
Similar reasoning applies to
Again,
Thus, letting a, @ =00 we get 2).
2. TTAe/i fAe singular points off, g are discrete, the condition 1°
holds.
3. 7/# is integrable and the conditions 1°, 2°, 3° are satisfied^
4. If f, g are integrable and condition 1° e's satisfied, f + g is in
tegrate and
5.
provided the integral off in question converges or is definitely infinite.
LV+C^LS+L«
Also
lim £)a/J = lim 2la/J
where 2la/3 refers to/.
6. When condition 1° is not satisfied, the relations 2) or 3)
may not hold.
GENERAL THEORY , 57
Example. Let 51 consist of the rational points in (0, 1).
Let /. ., . ..
/= \+n , g = \ —n
at the point x = ™. Then
n
f + g=2 in 21.
Now qr w
*fi «/i » *> «/5
embrace only a finite number of points for a given a, /9. On the
other hand,
Thus the upper content of the last set in 1) does not = 0 as
«, ft = oo and condition 1° is not fulfilled. Also relation 2) does
not hold in this case. For
, jT/=o ,
59. #c>0, Om-fjf; (1
(2
provided the integral on either side is convergent.
For
/ ifo<0. (4
Let c> 0. Since
therefore ~
--</<- in this set.
Hence any point of Hc/, ap, is a point of SI/, -. ^ and conversely.
Thus 2<cAa,= 2l,,^ whenoO.
58 IMPROPER MULTIPLE INTEGRALS
Similarly ^=SI,,?,| when c < 0.
Thus 3), 4) give
f e/=ef / c>0
• J
?,,/ c<0-
c' c
We now need only to pass to the limit a, 0 = <x> .
60. Let one of the integrals
converge. Iff = #, arc?^ a« a discrete set £) m 51, 60^ integrals
converge and are equal. A similar theorem holds for the lower
integrals.
For let us suppose the first integral in 1) converges. Let
*-.
then
Now
= lim = m g
f, a/3
Thus the second integral in 1) converges, and 2), 3) show that
the integrals in 1) are equal.
61. 1. Let f /, f g (1
£8 ~2l
converge. Let f '> g except possibly at a discrete set. Let
««,,,*) ; .f = 2l/,aa-3X0 ; Qa0 = 21^ -3V
V - -
fojs = 0, ga^ =0, as a, yQ = oo,
then — _
RELATION BETWEEN THE INTEGRALS OF TYPES I, II, III 59
For let (Sa/3 be defined as in 58, 1. Then
Let a, ft = <x>, we get 2) by the same style of reasoning as in
58.
2. If the integrals 1) converge, and their singular points are dis
crete, the relation 2) holds.
This follows by 58, 2.
3. If the conditions of 1 do not hold, the relation 2) may not
be true.
Example. Let 51 denote the rational points in (0*, 1*). Let
/• m • ca
f=n at x = — in 51.
n
g=l in 51.
Then f>g in 51.
But
Relation betioeen the Integrals of Types I, //, ///
62. Let us denote these integrals over the limited field 51 by
0« , V* , At
respectively. The upper and lower integrals may be denoted by
putting a dash above and below them. When no ambiguity arises,
we may omit the subscript 51. The singular points of the inte
grand/, we denote as usual by $•
63. If one of the integrals P is convergent, and $ is discrete, the
corresponding 0 integral converges, and both are equal.
T7^ *
1 P% = p%s + paj, using the notation of 28,
= e* + p«j.
Now o as a = 0 by 56.
60 IMPROPER MULTIPLE INTEGRALS
Hence p _ j. Q
5=0 5
== C%, by definition.
64. If O is convergent, we cannot say that P converges. A
similar remark holds for tjie lower integrals.
Example. For the rational points in 51 = (0, 1) let
for the irrational points let
/oo = - -
x
Then
a=0
On the other hand, _
PH = lim f /
does not exist. For however large /3 is taken and then fixed,
I /= — oo as a = oo.
65. If O is absolutely convergent and $ is discrete, then both P
converge and are equal to the corresponding 0 integrals.
For let D be any complete division of 51 of norm 8. Then
/~Jt>j[L (1
using the notation of 28. Now since
£k |/| converges, C%'& |/| = 0 as 8 = 0.
But
Again, D being fixed, if «0/30 are sufficiently large,
f f=C^f «>«0, /S>/30.
*/9I «
l«M
RELATION BETWEEN THE INTEGRALS OF TYPES I, II, III 61
Hence 1), 2) give
f /= #a, + e' le'j < - for any 8 < some S0.
J2lap 2
On the other hand, if S0 is sufficiently small,
ft= <£, + €" |e"|<| for 8<S0.
Hence f /=<7a + €'" |e'"|<e.
^ajS
Passing to the limit a, /3 = GO, we get
66. 7/ Fa/ ^s absolutely convergent, the singular points $ are
discrete.
For suppose 3 > 0. Let 53 denote the points of 51 where
\f\>0. Then g > ^ for any ^. Hence
as 13 = QO unless 3 = 0.
67. If V%f is absolutely convergent, so is C.
For let D be a cubical division of space of norm d.
Then
|/ 1 < some /3in 2ld.
Hence
Hence (7 is absolutely convergent.
68. Letf>0. If VyJ is convergent, there exists for each e>0,
a a- > 0 such that
for any 53 such that
. (2
62 IMPROPER MULTIPLE INTEGRALS
'
for X sufficiently large. Let X be so taken, then
P»/=jQ/* + «" , 0<«»<|. (3
AIso>
if o- is taken sufficiently small in 2).
From 3), 4) follows 1).
69. If V^f is absolutely convergent, both 0 converge and are
equal to the corresponding V integrals.
For by 67, 0 is absolutely convergent. Hence (/converge by 65.
Thus
(!<»f= I f 4- « . \n.' <?
Also
= 1 /+« » |«'<- for some d.
®d 3
L 4- /3 , I /3 1 < | for some X, ,
Hence
Now
But
and 7 < | if d is sufficiently small, and for any X, /A, by 68.
o
Taking a division of space having this norm, we then take X,
so large that
/AM=/ mSIc,.
Then
T; = « _ p - 7,
and hence
|77|<e.
From this and 1) the theorem now follows at once.
ITERATED INTEGRALS 63
Iterated Integrals
70. 1. We consider now the relations which exist between the
integrals
and - ?
If/, (2
~.<B«Ze
where 51 = 53 • (5 lies in a space 9?m, m = p -f ^, and 53 is a projection
of 51 in the space 9?p.
It is sometimes convenient to denote the last q coordinates of a
point x = (o^ ••• xp xp+l ••• xp+q) by yl--> yq. Thus the coordinates
x1 ••• xp refer to 53 and y^ • •• yg to (5. The section of 51 correspond
ing to the point x in 53 may be denoted by (£x when it is desirable
to indicate which of the sections (£ is meant.
2. Let us set
*(*!
then the integral 2) is
It is important to note at once that although the integrand / is
defined for each point in 51, the integrand <f> in 4) may not be.
Example. Let 51 consist of the points (z, y) in the unit square :
771 A ^ ^ I
x = - , 0<y<-.
?i n
Then 51 is discrete. At the point (x, y)in 51, let
Then i * A by 32
On the other hand
for each point of 53- Thus the integrals 2) are not defined.
64 IMPROPER MULTIPLE INTEGRALS
To provide for the case that <f> may not be defined for certain
points of 53 we give the symbol 2) the following definition.
f ( /= lim f I/, (5
J»J<r a,0=ooJ|8a0Jr
where F = (£ when the integral 3) is convergent, or in the con
trary case F is such a part of (£ that
-«</*/< A (6
^r
and such that the integral in 6) .is numerically as large as 6) will
permit.
Sometimes it is convenient to denote F more specifically by Fa/3.
The points 53a0 are the points of 53 at which 6) holds. It will
be noticed that each 53a/3 in 5) contains all the points of 53 where
the integral 3) is not convergent. Thus
Hence when 53 is complete or metric,
lim Sais=S. (7
a, /3=ao
Before going farther it will aid the reader to consider a few
examples.
71. Example 1. Let 51 be as in the example in 70, 2, while/ = n2
at x = -. We see that
n
On the other hand 53a/s contains but a finite number of points
for any a, @. Thus
Thus the two integrals 1), 2) exist and are equal.
Example 2. The fact that the integrals in Ex. 1 vanish may
lead the reader to depreciate the value of an example of this kind.
This would be unfortunate, as it is easy to modify the function so
that thee.e integrals do not vanish.
ITERATED INTEGRALS 65
Let 21 denote all the points of the unit square. Let us denote
the discrete point set used in Ex. 1 by £). We define / now as
follows : / shall have in £) the values assigned to it at these points
in Ex. 1. At the other points A = 21 — £),/ shall have the value 1.
Then C C C C
On the other hand 33a/3 consists of the irrational points in $3 and
a finite number of other points. Thus
• = 1. _ (4
Hence again the two 3), 4) exist and are equal.
Let us look at the results we get if we use integrals of types I
and II. We will denote them by C and F as in 62.
We see at once that
Let us now calculate the iterated integrals
$B Od, (5
and Fs F<£. (6
We observe that
<7e = 1 for x irrational
= + oo for x rational.
Thus the integral 5) either is not defined at all since the field
33s does not exist, or if we interpret the definition as liberally as
possible, its value is 0. In neither case is
Let us now look at the integral 6). We see at once that
-
does not exist, as Fe = 1 for rational #, and = +00 for irrational
x. On the other hand
Hence in this case
66 IMPROPER MULTIPLE INTEGRALS
Example 3. Let 51 be the unit square.
Let
/f m
= n tor x = — n even
n
= - n for x = ~ n odd.
At the other points of 51 let/= 1.
Then „ „ r
//=/Je/=1-
Here every point of 51 is a point of infinite discontinuity and
thus $ = 51.
Here (7^ is not defined, as 515 does not exist ; or giving the
definition its most liberal interpretation,
<7st=0.
The same remarks hold for O^Og.
On the other hand f^
FSI = 4- 00,
while jr y
does not exist, since Jr f m
Vg = ± n for x = —
n
= 1 for irrational x.
Moreover 77 TA V V
Example 4. Let 51 denote the unit square. Let
~ 9 £ m A^^l
f = n* lor a? = — , 7i even, 0<y< —
n n
= — n2 for # = — , n odd, 0 < y < - .
n n
At the other points of 5Ilet/= 1.
Then
ITERATED INTEGRALS 67
Let us look at the corresponding C and V integrals.
We see at once that
n -- v — i
ca- v*~ *•
Again the integral C^Cg does not exist, or on a liberal interpre
tation it has the value 0. Also in this example
(7^ and C9Cz
do not exist or on a liberal interpretation, they = 0.
Turning to the F' integrals we see that
while V<g Fjg does not exist finite or infinite.
Example 5. Let our field of integration 51 consist of the unit
square considered in Ex. 4, let us call it Gr, and another similar
square g, lying to its right. Let / be defined over (S as it was
defined in Ex. 4, and let/= 1 in g.
Then
Also . <t-*k-* •
Then „ „ 1
^^B^S ~ *'
while FsFg does not exist,
and
72. 1. In the following sections we shall restrict ourselves as
follows:
1° 51 shall be limited and iterable with respect to 53.
2° 53 shall be complete or metric.
3° The singular points $ of the integrand /shall be discrete.
2. Let us effect a sequence of superposed cubical divisions of
space
A» A* •••
whose norms dn = 0.
68 IMPROPER MULTIPLE INTEGRALS
Let 5ln = 33n • (£n denote the points of 31 lying in cells of Dn
which contain no point of $. We observe that we may always
take without loss of generality
»„=».
For let us adjoin to 51 a discrete set £) lying at some distance
from 51 such that the projection of £) on $ftp is precisely 33.
Let ^ = 5l4-£) = 33-<7 , (?=£+c , c = 0.
We now set ^ =/ in ^
•=0 in £).
JL hen 7* 7*
J 4,=j 4
•£1 ^-^
=//
Similarly
Hence
~
3. The set ^n being as in 2, we shall write
73. Ze£ BtT< n denote the points of 33 at which cn > cr. Then if 51
is iterable, with respect to 33,
lim^.^O. (1
n=ao
For since 51 is iterable,
8 = r S by definition.
Hence (5 considered as a function of x is an integrable function
in 33.
Similarly
and (£„ is an integrable function in 33.
ITERATED INTEGRALS 69
We have now 7? 7? - - ^ n
£ = @n + cn , cn>0
as (£n, cn are unmixed. Hence cn is an integrable function in 53.
But
As the left side = 0 as n = oo ,
limj.= 0. (2
But
As the left side = 0, we have for a given a
lim S^ = 0,
which is 1).
74. Let % = <$>& be iterable. Let the integral
' /^° a
be convergent and limited in complete 53. Let (gn denote the points
of 53 at which
lim (gn = S. (3
For let . „
••• =0.
Since jB^B = 0 as n = oo by 48, we may take v1 so large, and
then a cubical division of $RP of norm so small that those cells con
taining points of SViVi have a content <rj/'2. Let the points of
53 lying in these cells be called Bv and let ^ = SB - Bv Then
Bv 53X form an unmixed division of 53 and
is complete since 53 is.
70 IMPROPER MULTIPLE INTEGRALS
, We may now reason on ^ as we did on $, replacing 7?/2 by ??/22.
We get a complete set 332 <$$i such that
Continuing we get ™ ^
Thus
Let now b
Then b>23-77 (4
by 25.
Let bn denote those points of b for which 2) does hold. Then
b = Jbn(. For let b be any point of b. Since 1) is convergent,
there exists a o~L such that
at
, f/<€,
^c
for any c such that c <crt. Thus b is a point of b^ and hence of
{bj. Thus B~n=Fas b is complete. But (Sn>bn.
Hence lim"gn>"b,
which with 4), gives 3).
75. LetW = 53 • (£ 5e iterable. Let the integral
convergent and limited in complete 53.
Then /* -^
lim I ( /=0. (1
For let D be a cubical division of $RP of norm d.
Then C C C
Let d( denote those cells of D containing a point of (5n where
is defined as in 74. • '
ITERATED INTEGRALS 71
Let d[' denote the other cells containing points of 53. Then
St<Zd*€ + 2d?M,
where —
0< \f<M.
•35
Hence
Letting d = 0, we get
JJ>
c/33 Jc»
Letting now n == oo and using 3) of 74, we get 1), since e is
small at pleasure.
76. Let 51 = 33 • (E be iterable with respect to $3, which last is com
plete or metric. Let the singular points $ of f be discrete. Then
; :: .if, f<
Here any one of the members in 1) may be infinite. Then all
that follow are also infinite. A similar remark applies to 2).
Let us first suppose :
f> 0 , 33 is complete , j j f is convergent.
We have by 14,
Passing to the limit gives
f f < lim f f / (3
Jr7 .J^^s/
and also r T r"
lim I I / < I / , finite or infinite. (4
—
Now e>0 being small at pleasure, there exists a GrQ such that
^ff- (5
72 IMPROPER MULTIPLE INTEGRALS
But for a fixed n ^
I is limited in 53.
*/(£
Hence for 6r0 sufficiently large,
f f<jf , at each point of 53, G-Q<G-. (6
Then
where Fn, yn are points of F in (£», cn.
Hence
J»<?J>»'
Now 330 may not be complete ; if not let B0 be completed 53^.
As 53 is complete,
We may therefore write 8), using 5)
- . + f / </ / + f / <//+//-
c/5B^ J&oJS* JLBG^n ^»^Sn J.BGJ.yn
By 75, the last term on the right = 0 as n = oo. Thus passing
to the limit,
n/<lim f f/, (9
_._y -n=ao»/33^S/
since e > 0 is small at pleasure.
On the other hand, passing to the limit G- = oo in 7), and then
n = QO, we get
lim f f < f f . (10
n^JjQj&n— Jj&J&
Thus 3), 10), 9), and 4) give 1).
Let us now suppose that the middle term of 1) is divergent.
We have as before
f fciim f <(/
Jj8GJT n=*>J%J£n JK
Hence the integral on the right of 1) is divergent.
ITERATED INTEGRALS 73
Let us now suppose 33 is metric. We effect a cubical division
of ftp of norm rf, and denote by Bd those cells containing only
points of 23. Then Bd is complete and
Let Ad denote those points of 31 whose projections fall on Bd.
Then Ad is iterable with respect to Ed by 13, 3, and we have as
in the preceding case
/<//</• en
^Ad ^Bd^£ ^Ad
If the middle integral in 11) is divergent, I is divergent and 1)
*/2f
holds, also if the last integral in 11) is divergent, 1) holds. Sup
pose then that the two last integrals in 11) are convergent.
Then by 57
r=r f.
J J
Iimf = f.
„ d=t)JAd •**
Thus passing to the limit e?= 0 in 11) we get 1).
Let us now suppose f > — (r, G > 0.
Then
and we can apply 1) to the new function g.
Now
(13
by 58, 5, since 3 is discrete. Also by the same theorem,
Cg=, Cf+G- lim fT = f/ + GT, (14
J& J& y = V J_(l
denoting by (Ey the points of (£ where
and setting
T = lim gy.
y = x
74 IMPROPER MULTIPLE INTEGRALS
Now for any n
Hence &% = lim f f #=#li
n = oo c/58 *ASn «=
or a = lim fSn. (15
^ = 00 »/33
Now for a fixed w, 7 may be taken so large that for all points
of 33,
Hence
S > lim
Hence
Hence
31
and thus F is integrable in 58.
This result in 14) gives, on using 58, 3,
= c r
*/58 Jg
From 12), 13), and 17) follows 1).
77. As corollaries of the last theorem we have, supposing H to
be as in 76,
1. Iffis integrable in 51 andf> — Gr, then
If f< ^ then f/= f f/.
*/2l -/5I3 -^(S
2. J?// > — ^ #nd I is divergent, then
are divergent,
ITERATED INTECxRALS
75
3. If f~> — G- and one of the integrals I I f is convergent, then
JJ
is convergent.
78. Let 51 = 33 • (E he iterable with respect to 33, which last is com
plete or metric. Let the singular points Q be discrete. If
f/, (i
J»a
f f* <2
•/*/(
both converge, they are equal.
For let Dr D2 -•• be a sequence of superimposed cubical divisions
as in 72, 2. We may suppose as before that each $3re = 33-
Since 1) is convergent
e>0, »„, f/_f/<l „<«„. (3
(4
(5
Since /is limited in 5ln, which latter is iterable,
This shows that
*
<2 »>«r
is an integrable function in 58, and hence in any part of
From 3), 4) we have
f-ff
*/2l *^23 Jdr
We wish now to show that
JJ-IX
*^33*^C£ S3^Sj
When this is done, 6) and 7) prove the theorem.
To establish 7) we begin by observing that
(6
n > nn.
ff=lim C f.
J%J<Z ..s^oo^as^r
76
IMPROPER MULTIPLE INTEGRALS
Now for a fixed n, a, 0 may be taken so that T shall embrace all
the points of (gw for every point of $. Let us set
Then
JVM
(8
As
lira f (
a,^ = oo^J
On the other hand,
I 724
'-'
Thus 7) is established when we show that
,£jj/l<! ">"„• (9
To this end we note that |/| is integrable in 31 by 48, 4. Hence
by 77, l,
Also by I, 734,
-C |/| =-CX>|/!' (i1
From 10), 11) we have for n > n0,
since the left side == 0.
But as in 8)
. LI i/i =4J> +jro.j[i/i
Passing to the limit (7= oo gives
This in 12) gives 9).
CHAPTER III
SERIES
Preliminary Definitions and Theorems
79. Let av a2, ag • • • be an infinite sequence of numbers.
The symbol An ^1
J JL = al + «2 + a3 H- • • • (1
is called an infinite series. Let
An=a1 + a2 + -' + an. (2
If lim An (3
n=w
is finite, we say the series 1) is convergent. If the limit 3) is infi
nite or does not exist, we say 1) is divergent. When 1) is conver
gent, the limit 3) is called the sum of the series. It is customary
to represent a series and its sum by the same letter, when no con
fusion will arise. Whenever practicable we shall adopt the fol
lowing uniform notation. The terms of a series will be designated
by small Roman letters, the series and its sum will be denoted by
the corresponding capital letter. The sum of the first n terms of a
series as A will be denoted by An. The infinite series formed by
removing the first n terms, as for example,
4- a,,+2 + #»+3+ "• 4
will be denoted by An, and will be called the remainder after n
terms.
The series formed by replacing each term of a series by its nu
merical value is called the adjoint series. We shall designate it
by replacing the Roman letters by the corresponding Greek or
German letters. Thus the adjoint of 1) would be denoted by
A = «! -f- «2 -f- «3 + • . . = Adj A (5
where
«» =
77
78 SERIES
If all the terms of of a series are 5 0» it is identical with its
adjoint.
A sum of p consecutive terms as
an+l + an+2+ f-n+p
we denote by An,p.
Let ^
B= atl + #4 + at3 + ••• , t1<*2<*"
£g £Ae 8irie* obtained from A by omitting all its terms that vanish.
Then A and B converge or diverge simultaneously, and when conver
gent they have the same sum.
For B --A
<&* — -*«»»
Thus if the limit on either side exists, the limit of the other side
exists and both are equal.
This shows that in an infinite series we may omit its zero terms
without affecting its character or value. We shall suppose this
done unless the contrary is stated.
A series whose terms are all > 0 we shall call a positive term
series; similarly if its terms are all < 0, we call it a negative term
series. If an > 0, n > m we shall say the series is essentially a pos
itive term series. Similarly if an < 0, n>m we call it an essen
tially negative term series.
If A is an essentially positive term series and divergent,
lim An = + QO ; if it is an essentially negative term series and di
vergent, lim An= -co.
When lim An— ±00, we sometimes say A is ±00.
80. 1. For A to converge, it is necessary and sufficient that
e>0, m, \An<p\<e, n>m, ^ = 1,2,— (1
For the necessary and sufficient condition that
lim An
72 = 30
exists is ~ . , ,0
€>0, m, \AV- An\ <e, v, n> m. (2
But if v = n+ p
Av- An= An<p=an+1+ a-n+2+ ••• +an+p.
Thus 2) is identical with 1).
PRELIMINARY DEFINITIONS AND THEOREMS 79
2. The two series A, A, converge and diverge simultaneously.
When convergent,
A = A, + A.. (3
For obviously if either series satisfies theorem 1, the other
must, since the first terms of a series do not enter the relation 1).
On the other hand, . A A
-"•«+p — •"•« T •***, p"
Letting p = oo we get 3).
3. If A is convergent, An = Q.
For lim An = lim (A - An)
— A — lim An = A — A
= 0.
For A to converge it is necessary that an == 0.
For in 1) take p = 1 ; it becomes
e n > m
We cannot infer conversely because an = 0, therefore A is con
vergent. For as we shall see in 81, 2,
1 + J + 1+ -
is divergent, yet lim an = 0.
4. The positive term series A is convergent if An is limited.
For then lim An exists by I, 109.
5. A series whose adjoint converges is convergent.
For the adjoint A of A being convergent,
e>0, m, |AB|P <€, n>m, p =1, 2, 3 •••
But
Thus ' l^d<«
and ^1 is convergent.
Definition. A series whose adjoint is convergent is called
absolutely convergent.
80 SERIES
Series which do not converge absolutely may be called, when
necessary to emphasize this fact, simply convergent.
6. Let A = al + a^+ •••
be absolutely convergent.
Let B = a^ + alt+ — ; i1<i2< •••
be any series whose terms are taken from A, preserving their relative
order. Then B is absolutely convergent and
choosing n so large that An contains every term in Bm. Moreover
for m > some m' , An — Bm > some term of A. Thus passing to the
limit in 1), the theorem is proved.
7. Let A = a1 + az + ••• The series B = ka1 + ka2 + ••-
converges or diverges simultaneously with A. When convergent,
B = kA.
We have now only to pass to the limit.
From this we see that a negative or an essentially negative term
series can be converted into a positive or an essentially positive
term series by multiplying its terms by k = — 1.
8. If A is simply convergent, the series B formed of its positive
terms taken in the order they occur in A, and the series C formed of the
negative terms, also taken in the order they occur in A, are both
divergent.
If B and 0 are convergent, so are B, F. Now
An = Bnv + Fna, n = ^ + n2.
Hence A would be convergent, which is contrary to hypothesis.
If only one of the series B, 0 is convergent, the relation
shows that A would be divergent, which is contrary to hypothesis.
PRELIMINARY DEFINITIONS AND THEOREMS 81
9. The following theorem often affords a convenient means of
estimating the remainder of an absolutely convergent series.
Let A = al -f- #2 ~f~ '" ^e an o^^olutely convergent series. Let
B = bl + b2 + ••- be a positive term convergent series whose sum is
known either exactly or approximately. Then if \ an \ < bn, n > m
<Bn<B.
For
<Bn<B.
Letting p == oo gives the theorem.
EXAMPLES
81. 1. The geometric series is defined by
The geometric series is absolutely convergent when |#|<1 and di
vergent when |$r|>l. When convergent,
a=~ (2
Hence ., „
ft- — &
— — ~~
When |#|<1, lim gn = 0, and then
When |0|>1, lim#ra is not 0, and hence by 80, 3, 6r is not conver
gent.
2. The series ff= 1 1. J_ 1_ ^. (3
82 SERIES
is called the general harmonic series of exponent /x. When /-i = 1,
it becomes J=l + i + . + { + ... (4
the harmonic series. We show now that
The general harmonic series is convergent ivhen IJL > 1 and is di
vergent for fJL < 1.
Let /*>!. Then
Let n < 2". Then
Thus lim Hn exists, by I, 109, and
'J<iI5' C5
Let /*<!. Then 1 -,
Thus 3) is divergent for fi < 1, if it is for /z, = 1.
But we saw, I, 141, that
lim J
n = oo,
hence J is divergent.
It is sometimes useful to know that
hm-^-=l. (6
log n
In fact, by I, 180, 1
log n log n - log O - 1) loef—
*\n-
n-l,
= 1.
PRELIMINARY DEFINITIONS AND THEOREMS 83
Since * n > log n > lzn • • - we have
lim —5=0 ; lim — - = oo , r > 1. (7
n lrn
Another useful relation is
11 1
ffn= 1 -h - + - -h h - >log(7i -h 1). (8
2 6 n
For log(l + TW) — Iog7w = logf 1 H — )<— .
\ mj m
Let 7/1 = 1, 2---71. If we add the resulting inequalities we
get 8).
3. Alternating Series. This important class of series is defined
as follows. Let a1 > a2 > a3 > ••• =0.
Then ^ = a1-a2 + a3-a4 + ... (9
whose signs are alternately positive and negative, is such a series.
The alternating series 9) is convergent and
\An\<an+1. (10
For let p > 3. We have
If p is even,
P = On-fl - «n+2) + •" + On+P-l ~ ^
If p is odd,
Thus in both cases,
P
Again, if p is even,
P = «n+1 - (an+2 - «n+a) — — — («n+p_2 - #n+P-i) ~ «n+P-
* In I, 461, the symbol " lim " in the first relation should be replaced by Urn.
84 SERIES
If p is odd,
P = an+1 — («n+2 — aB+3) — ... — («n+p_! — an+p).
Thus in both cases,
From 11), 12) we have
0 < an+1 - an+2
Hence passing to the limit p= oo,
0 < an+1 - an+2 < | An, p | < an+l -
moreover, . n
«n+l = 0-
Example 1. The series
i-i+t-l+.v-
being alternating, is convergent. The adjoint series is
which being the harmonic series is divergent. Thus 13) is an
example of a convergent series which is not absolutely convergent.
Example 2. The series
_ _1_
' V2 - 1 V2 + 1 V3 - 1 V3 + 1
is divergent, although its terms are alternately positive and nega
tive, and an = 0.
m —
If now ^4. were convergent,
lim A = lim
by I, 103, 2.
n
PRELIMINARY DEFINITIONS AND THEOREMS 85
4. Telescopic Series. Such series are
A=(al- aa) + (aa - «3) + («8 - a4) + -
We note that
An=(a1-az) + - + (a»-an41)
= «i -<W (14
Thus the terms of any An cancelling out in pairs, An reduces to
only two terms and so shuts up like a telescope.
The relation 14) gives us the theorem :
A telescopic series is convergent when and only ivhen lim an exists.
Let , ,
A = al + «2 + ... denote any series.
Then
an = An - An_^ , A0 = 0.
Hence
This shows us that
-Awy se?*^s can be written as a telescopic series.
This fact, as we shall see, is of great value in studying the
general theory of series.
Example!. A = — — I — - — u -i ____
^^^
— in
Thus A is a telescopic series and
n
Example 2. Let av av a8, •-• > 0. Then
(1 + a,) ...
is telescopic. Thus
and J. is convergent and < 1.
86 SERIES
Examples. A = ^] #=^0, —1, —2, •••
-*• *^ /" „. i i \ f ™ i „„ \
•4- n — 1 a: H- n
is telescopic.
^ -1 1 -1.
- i , — • ^— — *
x x 4- n x
82. Dini's Series. Let A = al-\-a2-\- ••• 6e a divergent positive
term series. Then
is divergent.
For
_ m+l ,
— .
> , ^r =1
^ p Am+p
Letting m remain fixed and _p = oo, we have Dm>l, since
+p == °°- Hence D is divergent.
Lefc
Hence 2) = i + i + i+...is divergent.
Let A-1 + J.4-J+.-:
Then - - -<
is divergent, and hence, a fortiori,
v_l
But -^-n-l > log W'
Hence -. -, -.
V _J__ = _JL_ 4. 1 _t_
^7 n log ^ 2 log 2 3 log 3
is divergent, as J.5e/ first showed.
PRELIMINARY DEFINITIONS AND THEOREMS 87
83. 1. Abels Series.
An important class of series have the form
As Abel first showed how the convergence of certain types of
these series could be established, they may be appropriately called
in his honor. The reasoning depends on the simple identity
(Abel's identity),
p~l\fn+p-l~ ^n+p) H~ ^n+p-^n, pi (^
where as usual An^m is the sum of the first m terms of the re
mainder series An. From this identity we have at once the fol
lowing cases in which the series 1) converges.
2. Let the series A = a^ + a2 + ••• and the series 2 tn+l — tn\
converge. Let the tn be limited. Then B = afa + #2£2 4- ••• converges.
For since A is convergent, there exists an m such that
An,p\<e\ n>m, j0 = 1, 2, 3 .-
Hence
3. Z/e^ the series A = a^ + a% 4- ••• converge. Let tr tz, t3 -• be a
limited monotone sequence. Then B is convergent.
This is a corollary of 2.
4. Let A = al + a2 + • • • 5e swcA iAa* | An < (7, w = 1, 2, • • •
2 1 ^n+1 — ^w | converge and tn = 0. Then B is convergent.
For by hypothesis there exists an m such that
I tn+i — tn+z | 4- | tn+2 — tn+z | H- • • • + tn+Ji < e
for any n>m.
5. Let \An\<& and ^ > £2 > ts > >- =0. Then B is convergent.
This is a special case of 4.
88 SERIES
6. As an application of 5 we see the alternating series
is convergent. For as the A series we may take A = l — 1+1 —
1 + ••• as \An\<l.
84. Trigonometric Series.
Series of this type are
(7= «0 + a1 cos x + a% cos 2 x + aB cos 3 x+ ••• (1
S = al sin x + a2 sin 2 x + az sin 3 x + ••« (2
As we see, they are special cases of Abel's series. Special cases
of the series 1), 2) are
r = J + cos x + cos 2 x + cos 3 2; -f- . • .
2 = sin x + sin 2 z + sin 3 x + •••
It is easy to find the sums Tn, Sn as follows. We have
2 sin mx sin j -a? = cos m ~ x - cos
2
Letting w = 1, 2, ... w and adding, we get
2 sin 1 x • 2n = cos | 2: — cos ^ n + a:. (5
«
Keeping x fixed and letting n = oo, we see Sn oscillates between
fixed limits when x ^= 0, ± 2 TT,
Thus 2 is divergent except when x = 0, ± TT, ...
Similarly we find when x =£ 2 WTT,
r _sinO- j)a? /«
1 71 — — ;: — : - r^^ — • ( D
2 sin I #
Hence for such values Tn oscillates between fixed limits. For
the values x = 2 rmr the equation 3) shows that I\ = + oo.
From the theorems 4, 5 we have at once now
V 2 1 an+l — an \ converges and an = 0, and hence in particular if
ai>a2~ •" == 0, the series 1) converges for every x, and 2) converges
for x^2 mir.
If in 3) we replace x by x 4- TT, it goes over into
A = \ — cos # -J- cos 2 a; — cos 3 x + ... (7
PRELIMINARY DEFINITIONS AND THEOREMS 89
Thus An oscillates between fixed limits if x =£ ± (2 m — 1) TT,
when n = GO . Thus
7f 2 | 0B+1 + an converges and an = 0, arcd Aewce in particular if
a1>a2> • • « =0, the series a0 — ax cos x + «2 cos 2 z — a3 cos 3 x -f
converges for x j= (2 m — 1) TT.
85. Power Series.
An extremely important class of series are those of the type
called power series. Since P reduces to aQ if we set x = a, we see
that every power series converges for at least one point. On the
other hand, there are power series which converge at but one
point, e.g.
a0 + l!(*-tf) + 2!(:r-a)2 + 3!(*-a)3+ ... (2
For if x =£ a, lim n\ x — a \ n — ao, and thus 2) is divergent.
1. If the power series P converges for x = b, it converges absolutely
within -^ , , , ,
VK(a) , \=\a-b\.
If P diverges for x=b, it diverges without DA(a).
Let us suppose first that P converges at b. Let # be a point in
Z>A, and set | x — a \ — f . Then the adjoint of P becomes for this
point
But T A
lim «nXn = 0,
since series P is convergent for x = b.
Hence . ,,.-
and II is convergent. X
If P diverges at x = 5, it must diverge for all 5' such that
| b' — a | > X. For if not, P would converge at b by what we have
just proved, and this contradicts the hypothesis.
90 SERIES
2. Thus we conclude that the set of points for which P con
verges form an interval (a — p, a + p) about the point #, called
the interval of convergence ; p is called its norm. We say P is
developed about the point a. When a = 0, the series 1) takes on
the simpler form
a
which for many purposes is just as general as 1). We shall
therefore employ it to simplify our equations.
We note that the geometric series is a simple case of a power
series.
86. Cauchy's Theorem on the Interval of Convergence.
The norm p of the interval of convergence of the power series,
is given by i
- = hmVan an =
We show H diverges if f > p. For let
Then by I, 338, 1, there exist an infinity of indices iv i2 ••• for
which
Hence
and thus
since 3>l. Hence
v fcln
^ain£
71
is divergent and therefore II.
We show now that II converges if f < p. For let
Then there exist only a finite number of indices for which
Let m be the greatest of these indices. Then
V«n </3 n>m.
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 91
Hence
d «.?
Thus
and H is convergent.
Example 1. 2 *
Here -V^ = _L = 0 by I, 185, 4.
-faTf
Hence p = ao and the series converges absolutely for every x.
Example 2. * *
f-f+f--
1 O c>
Here V^"n = -^L = 1 by I, 185, 3.
Hence /o= 1, and the series converges absolutely for | x |< 1.
Tests of Convergence for Positive Term Series
87. To determine whether a given positive term series
is convergent or not, we may compare it with certain standard
series whose convergence or divergence is known. Such com
parisons enable us also to establish criteria of convergence of
great usefulness.
We begin by noting the following theorem which sometimes
proves useful.
1. Let A, B be two series which differ only by a finite number of
terms. Then they converge or diverge simultaneously.
This follows at once from 80, 2. Hence if a series A whose
convergence is under investigation has a certain property only
92 SERIES
after the mill term, we may replace A by Am, which has this
property from the start. •
2. The fundamental theorem of comparison is the following :
Let A = al + #2 + ••-, B = b1 + 62 -f ••;• be two positive term series.
Let r>0 denote a constant. If an< rbn, A converges if B does and
A < rB. If an > rbn, A diverges if B does.
For on the first hypothesis
An<rBn.
On the second hypothesis
An>rBn.
The theorem follows on passing to the limit.
3. From 2 we have at once :
Let A = a1 -f «2 -f- •••, B = bl + bz + ••• be two positive term series.
Let r, s be positive constants. If
a/n ., "i o
lim -^
exists and is 3=- 0, A and B converge or diverge simultaneously. If
B converges and — = 0, A also converges. If B diverges and -r=^
A also diverges.
4. Let A = a1 -f- a2 -f •••, B = b^ + b2 + ••• be positive term series.
If B is convergent and
an+l ^ °n+l _ 1 O Q .,
<— — n= i, ^, o •••
A converges. If B is divergent and
an ~~ °n
A diverges.
For on the first hypothesis
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 93
We may, therefore, apply 3. On the second hypothesis, we
have
and we may again apply 3.
Exanplel. A = ^2
is convergent. For
< "
n • n -f
and V — is convergent. The series A was considered in 81, 4, Ex. 1.
^ n2
Example 2. A = e~x cos x + e~2x cos 2 x -f •••
is absolutely convergent for x > 0.
For
which is thus < the nth term in the convergent geometric series
= V 1 log
is convergent.
For
0< «„ = - <.
n2 \ n y w2
Thus ^4. is comparable with the convergent series 2~V
~~l TL
88. We proceed now to deduce various tests for convergence
and divergence. One of the simplest is the following, obtained
by comparison with the hyperharmonic series.
Let A = al + a2+ ••• be a positive term series. It is convergent if
lim ann^ < oo , JJL > 1,
and divergent if
lim nan > 0.
94 SERIES
For on the first hypothesis there exists, by I, 338, a constant
Gr > 0 such that
Thus each term of A is less than the corresponding term of the
convergent series GrZ^ — •
On the second hypothesis there exists a constant c such that
an>- ft = l, 2, ...
n
and each term of A is greater than the corresponding term of the
divergent series e V -.
^ n
Example 1. A = V - w>0.
^ log™ n
Here ^=-^ = + 00, by I, 463.
log™ ^
Hence A is divergent.
Example 2. A = V — - .
^* n log ft
Here nan=^- = 0.
log ft
Thus the theorem does not apply. The series is divergent
by 82.
Example 8.
where /A is a constant and | 6n < Gr.
From I, 413, we have, setting r = 1 + *,
Hence
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 95
and L is divergent. If ^ > 0, L is an essentially positive term
series. Hence L = + 00. If /* < 0, L = — oo.
Let = 0. Then
which is comparable with the convergent series
Thus Z is convergent in this case.
Example 4- The harmonic series
1 + J + J+ •••
is divergent. For limnan = l.
Example 5. -,
A = 2 *i a ^ arbitrary.
^^ Wa lOgp 71
Here wl_a
nan = - — -— = 00 , a < 1
by I, 463, 1. Hence A is divergent for a< 1.
Example 6. -,
A =v
Here -<
wan = -fb = 1 by I, 185, Ex. 3.
vn
7.
Here, if /i > 0,
log(l + i
\ /€-
log n log
Hence A is divergent.
l\n
since n* > loe^ n and ( 1 H — ] == e
nj
96 SERIES
89. D'Alemberfs Test. The positive term series A = a± + #2 + • • '
converges if there exists a constant r < 1 for which
^±l<r, 71 = 1, 2, .-
an
/£ diverges if
«a±l>l.
«»
This follows from 87, 4, taking for B the geometric series
Corollary. Let 2a±la^.f. J/* / < 1, ^4. converges. If
««
diverge*.
Example 1. The Exponential Series.
Let us find for what values of x the series
is convergent. Applying D'Alembert's test to its adjoint, we find
xn n-ll
Thus ^converges absolutely for every x.
Let us employ 80, 9 to estimate the remainder En. Let x > 0.
The terms of E are all > 0. Since
^4-j[? 9^ fft
we have
xn f x V
-yl T~i~J '
(2
However large x may be, we may take n so large that x<n+ 1.
Then the series on the right of 2) is a convergent geometric series.
Let x < 0. Then however large | z | is, En is alternating for
some m. Hence by 81, 3 for n>^m,
\W\^\X" C\
-#nl<J— r- 0*
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 97
Example 2. The Logarithmic Series.
Let us find for what values of x the series
is convergent. The adjoint gives
"tn *+i '
Thus L converges absolutely for any |z|<l, and diverges for
When x = 1, L becomes
which is simply convergent by 81, 4.
When x = — 1, L becomes
which is the divergent harmonic series.
Example 3. A = — + — + — + —
-
an \n + ij
As A is convergent when /x>l and divergent if p<1, we see
that D'Alembert's test gives us no information when 1 = 1. It is,
however, convergent for this case by 81, 2.
Example 4-
Here
an4.i » -h
an n + 1 + x
and D'Alembert's test does not apply.
Example, 5.
A = 2
Here
*±!Y|-
98 SERIES
Thus A converges for |#|<1 and diverges for \x\ > 1. For
\x\ =1 the test does not apply. For x = 1 we know by 81, 2
that A is convergent for /-i < — 1, and is divergent for /t > — 1.
For x — — 1, A is divergent for /i > 0, since an does not = 0. A
is an alternating series for p <0, and is then convergent.
90. Cauchy's Radical Test. Let A = al + «2 -f ••• be a positive
term series. If there exists a constant r < 1 such that
A is convergent. If, on the other hand,
A is divergent.
For on the first hypothesis,
an<rn
so that each term of A is <, the corresponding term in
r _|_ r2 _j_ rs _j_ . . . a convergent geometric series. On the second
hypothesis, this geometric series is divergent and an>^rn.
Corollary. If lim \/an = I, and I < 1, A is convergent. Ifl>\,
A is divergent.
Example 1. The series
oo 1
mr— \ -*- \ sv j
Alogn7l 2
is convergent. For
1 =0.
log 71
Example 2.
A.
is convergent. For
n- nn 1 ^1
e
Example 3. In the elliptic functions we have to consider series
of the type
6(v) = 1 + 2 i0n' cos ZTrnv 0<q<l.
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 99
This series converges absolutely if
q + q* + q*+ "-
does. But here
Va* = Vq* = qn = Q.
Thus 6(v) converges absolutely for every v.
Example 4- Let 0 < a <6 < 1. The series
n = 2 m
is convergent. For if
If w = 2m-f-l, n_ 2m+i —
Thus for all n
Va~n<b<l.
Let us apply D'Alernbert's test. Here
Thus the test gives us no information.
91. Cauchy's Integral Test.
Let </>(#) be a positive monotone decreasing function in the interval
(«, QO ). The series
is convergent or divergent according as
is convergent or divergent.
For in the interval (n, n + 1), n>m>a,
100 SERIES
Hence
rn+l
<#>O + 1)< I ^dx<4)(n).
\s n
Letting n = m, m + 1, ••• m + j?, and adding, we have
Passing to the limit p = GO, we get
*«< fV*^*-!. (1
*^m
which proves the theorem.
Corollary. When <£ is convergent
Example 1. We can establish at once the results of 81, 2. For,
taking *(
is convergent or divergent according as /i > 1, or /-i < 1, by I,
635, 636.
We also note that if
A- JL J_ . J_ .
~
— (*™ 1 11
then ^< I JL.i-.JL,
*^n a;1+'x /x w**
Example 2. The logarithmic series
1
_ 1 9
wZf w "I *V ^ fl\Y Vf\
JN
are convergent if /A > 1 ; divergent if /4<1. t ^f1^ ^43
We take here
and apply I, 637, 638.
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 101
92. 1. One way, as already remarked, to determine whether
a given positive term series A = al + a2+ ••• is convergent or
divergent is to compare it with some series whose convergence or
divergence is known. We have found up to the present the
following standard series S:
The geometric series
The general harmonic series
1 + 1 + 1+- • (2
1 n £)(*. QJU, ^
The logarithmic series
We notice that none of these series could be used to determine
by comparison the convergence or divergence of the series follow
ing it.
In fact, let an, bn denote respectively the nth terms in 1), 2).
Then for g < 1, p > 0,
A_ = J_ = ^I±!!=00 by I, 464,
«n+i n»g» n»
or using the injinitary notation of I, 461,
bn > an.
Thus the terms of 2) converge to 0 infinitely slower than the
terms of 1), so that it is useless to compare 2) with 1) for conver
gence. Let#>l. Then
«.>»..
This shows we cannot compare 2) with 1) for divergence.
102 SERIES
Again, if «n, bn denote the nth terms of 2), 3) respectively, we
have, if /i > 1,
^ = -^- = 00 by I, 463,
an log'* n
or 7 ,
^5 = log n = QO,
Thus the convergence or divergence of 3) cannot be found
from 2) by comparison. In the same way we may proceed with
the others.
2. These considerations lead us to introduce the following
notions. Let A = al + a2+ •••, B = bl + 62 + ••• be positive term
series. Instead of considering the behavior of an/bn, let us gen
eralize and consider the ratios An : Bn for divergent and An : Bn
for convergent series. These ratios obviously afford us a measure
of the rate at which An and Bn approach their limit. If now A,
B are divergent and . ^
An ~ J5n->
we. say A, B diverge equally fast ; if
An<Bn,
A diverges slower than B, and B diverges faster than A. From
I, 180, we have :
Let A, B be divergent and
According as I is 0, =£0, oo, A diverges slower, equally fast, or
faster than B.
If A, B are convergent and
we say A, B converge equally fast ; if A converges and
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 103
B converges faster than A, and A converges slower than B. From
I, 184, we have :
Let A, B be convergent and
lira ^=1
on
According as I is 0, =£ 0, oc, A converges faster, equally fast, or slower
than B.
Returning now to the set of standard series #, we see that each
converges (diverges) slower than any preceding series of the set.
Such a set may therefore appropriately be called a scale of con
vergent (divergent) series. Thus if we have a decreasing positive
term series, whose convergence or divergence is to be ascertained,
we may compare it successively with the scale S, until we arrive
at one which converges or diverges equally fast. In practice such
series may always be found. It is easy, however, to show that there
exist series which converge or diverge slower than any series
in the scale S or indeed any other scale.
Forlet A, S, 0,... ' (2
be any scale of positive term convergent or divergent series.
Then, if convergent,
if divergent, An> Bn> Cn> ...
Thus in both cases we are led to a sequence of functions of the
Thus to show the existence of a series ft which converges (di
verges) slower than any series in 2, we have only to prove the
theorem :
3. (Du Bois Reymond.^) In the interval (a, oo) let
denote a set of positive increasing functions which =00 as x = oo .
Moreover, let /.
J\
104 SERIES
Then there exist positive increasing functions which = oc slower than
anyfn.
Foras/1>/2 there exists an a1>a such that /1>/2 + l for
x> ar Since /2 >/3, there exists an a2 > a1 such that /2 >/3 + 2
for x>a%. And in general there exists an an>an_1 such that
fn >fn+l + n f°r X > an' Let HOW
g(x) = fn(x)+n-l in (an_x, <*„)•
Then g is an increasing unlimited function in (a, oo) which
finally remains below any fm(x) + m — 1, w arbitrary but fixed.
Thus Q<ijm yvv =ijm y\^i <lim jm+i-r"« = 0>
Hence g<fm-
93. From the logarithmic series we can derive a number of
tests, for example, the following :
1. (Bertram's Tests.) Let A = al + a2 + ••• be a positive term
series.
Let log- -1-y—
ft(M)= «n^- l.-in s==l, 2, ... I0n=l.
If for some s and m,
Q8(.n) ^ A1 > 1 71 > w, (1
^4. ^s convergent. If, however,
&O)<1, (2
^1 ^s divergent.
For multiplying 1) by Z8+1n, we get
i
Hence
or
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 105
Thus A is convergent.
The rest of the theorem follows similarly.
2. For the positive term series A=al + a% + --to converge it is
necessary that, for n = oo,
lim an = 0, lim nan = 0, lira na^n = 0, lim^naj^nl^n = 0, •••
We have already noted the first two. Suppose now that
lim nanl^n ••• l,n > 0.
Then by I, 338 there exists an m and a c > 0, such that
naj-^n ••• l,n > c , n > m,
or
Hence A diverges.
Example 1.
««>^-r
nl^n ••• l,n
-
na log*3 n
We saw, 88, Ex. 5, that A is divergent for a < 1. For «= 1,
A is convergent for fi > 1 and divergent if yS ^ 1, according to
91, Ex. 2.
If a >1, let a=a' + a" a">l
Then if /3 > 0, -, -,
<— , n > 2,
and A is convergent since ^ — is. If ft < 0, let
Then 1 „, n 1
a"~ n0-' ' n?''
But log^'n<n«' by I, 463, 1 ;
and A is convergent since 2 — ~ is-
106 SERIES
Example 2. -, .,
A \^ 1 ^ 1
^ = x — — : jT = 2, — F'
I 1 -f- -4 ...-!__ 1 ^^ ->i ^/> tl
^?M/? 2 ^
Here .
^ aMft - log n + A* log 7i+ IT,
Vl ; ~ ^
62ft t2/l
/x-14--^}^^^ by 81, 6).
log ft J
Hence J. is convergent for ft > 0 and divergent for /A < 0. No
test for fi = 0.
But for /i = 0, 1
log _
/Tf ^77 >7, /•/ __r / 'M / ,vi
2 Lw Z0^
since I2n > I3n. Thus A is divergent for /a = 0.
94, A very general criterion is due to Kummer, viz. :
Let A = al + a2 4- •>• be a positive term series. Let kv &2, ••• be a
set of positive numbers chosen at pleasure. A is convergent, if for
some constant k > 0.
<w . = 1 9 .
«*+!
^4. ^s divergent if
is divergent and
For on the first hypothesis
1,
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 107
Hence adding,
and A is convergent by 80, 4.
On the second hypothesis,
^±1>«1
«» "^n1
Hence JL diverges since ^ is divergent.
95. 1. From Rummer's test we may deduce D'Alembert's test
at once. For take
k1 = kt= ..- =1.
Then A = a^ + «2 -f- ••• converges if
t.«. if
?2
«
Similarly A diverges if ^»±i >^1.
«•
2. To derive Raabe's test we take
~kn — n.
Then A converges if
j. if
Similarly A diverges if
108 SERIES
96. 1. Let A = a1 + a2 + ••• be a positive term series. Let
X2O) = I2n
Then A converges if there exists an s such that
\(n) > S > 1 for some n>m\
it diverges if ^ / N . ^ /.
We have already proved the theorem for X0(w). Let us show
how to prove it for Xj(^). The other cases follow similarly.
For the Kummer numbers kn we take
kn = n log n.
Then A converges if
As
- log(
l 4- -
1V+1
n)
Thus ^1 converges if Xj(^) > 8 > 1 for n>m.
In this way we see that .A diverges if \(n) < 1, n>m.
2. Oahens Test. For the positive term series to converge it is
necessary that
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 109
For if this upper limit is not + oo,
for all n. Hence
But the right side =0. Hence X1(^)<1 for w>some w, and
A is divergent by 1.
Example. We note that Raabe's test does apply to the harmonic
series i 4. i _i_ i 4. (\
Here
Hence Pn = 0, and
lim Pn = 0.
Hence the series 1) is divergent.
97. Grauss' Test. Let A = «j + «2 -f ••• be a positive term series
such that ,
an =ns + aji*-i+ ••• + a,
where s, a^ ••• b1 ••• do not depend on n. Then A is convergent if
al — bl > 1, and divergent if al — b1 < 1.
Using the identity I, 91, 2), we have
T _L ( 1
Thus lim X0(V) = a1 — br Hence, if a1 — b1 > 1, A is Conver
gent ; if a1 — #!<!, it is divergent. If al — b1 = 1, Raabe's test
does not always apply. To dispose of this case we may apply
the test corresponding to ^(V). Or more simply we may use
Cahen's test which depends on X^n). We find at once
lim Pn — #2 — b2 — bl < QO ;
and A is divergent.
110 SERIES
98. Let A = a± -f #2 -f- ••• be a positive term series such that
.A «* convergent if a > 1 cwd divergent if a< 1.
For
_^L__1 = « + J^ =
an^ J n^'1
and A converges if a > 1 and diverges if a < 1. If a = 1,
and A is divergent.
EXAMPLES
99. The Binomial Series. Let us find for what values of x and
Ai the series
converges. If ft is a positive integer, B is a polynomial of degree ft,
For ft = 0, 5=1. We now exclude these exceptional values of ft.
Applying D'Alembert's test to its adjoint we rind
— n+ 1
* = \x .
Thus 5 converges absolutely for x \ < 1 and diverges for \x\ > 1
Letx = l. Then
Here D'Alembert's test applied to its adjoint gives
As this gives us no information unless p< — 1, let us apply
Raabe's test. Here
, for sufficiently large n
n
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 111
Thus B converges absolutely if /* > 0, and its adjoint diverges
if fi< 0. Thus B does not converge absolutely for ft < 0.
But in this case we note that the terms of B are alternately
positive and negative. Also
1-
n
so that the |an| form a decreasing sequence from a certain term.
We investigate now when an = 0. Now
In I, 143, let a = — /*, /3 = 1. We thus find that lim an = 0 only
when /*> — 1. Thus -B converges when /*> — 1 and diverges
when /!< — !.
= -l. Then
1-2 1-2.3
If /A > 0, the terms of B finally have one sign, and
"(fc-'>-
Hence B converges absolutely.
If n < 0, let fj, = - X. Then B becomes
.* , X-X + 1 , X-X
~r A, -| 1
1-2
1.2-3
Here
Hence ^ diverges in this case. Summing up :
The binomial series converges absolutely for \x\<1 and diverges
for \x\ > 1. When x = \ it converges for p > — 1 and diverges for
fji < — 1 ; it converges absolutely only for /JL > 0. When x = — 1, &
converges absolutely for p > 0 awd diverges for p < 0.
112 SERIES
100. The Hypergeometric Series
A 7, *)=l + s + «
o
Let us find for what values of x this series converges. Passing
to the adjoint series, we find
X = X .
Thus .F converges absolutely for x \ < 1 and diverges for | x \ > 1.
Let x=1. The terms finally have one sign, and
qn+1 _ n2 + n(l + 7) + 7
Applying Gauss', test we find ^ converges when and only when
a; = — 1. The terms finally alternate in sign. Let us find
when an = 0. We have
Now
m
— ) ,
mj
Thus
K+2| = n^
But by I, 91, 1),
1 1
mj\ m
m
1 +
m
1+^
m<
where crm = 1, rm = 72 as w = oo.
PRIXGSHELM'S THEORY 113
Hence
m
Hence
and thus
Z = lira log
i
Now for an to = 0 it is necessary that Ln = — so. In 88, Ex. 3,
we saw this takes place when and only when « + /3 — 7— 1<0.
Let us find now when an+l \ < | an \. Now 1) gives
Thus when « + /3 — 7 — 1<0, | «n+2 1 < | an+l . Hence in this
case F is an alternating series. We have thus the important
theorem :
The hypergeometric series converges absolutely when x\<\ and
diverges when \x\>\. When x = \, F converges only when a + /3
— 7<0 and then absolutely. When x = — 1, F converges only
when ct + /3 — 7 — 1<0, and absolutely if a. -f- @ — 7 < 0.
Pnngsheims Theory
101. 1. In the 35th volume of the Mathematische Annalen
(1890) Pringsheim has developed a simple and uniform theory oi
convergence which embraces as special cases all earlier criteria,
and makes clear their interrelations. We wish to give a brief
sketch of this theory here, referring the reader to his papers for
more details.
Let Mn denote a positive increasing function of n whose limit
is 4- oo for n = DO . Such functions are, for example, p > 0,
n» , log'1 n , l$n , l-^nl^n ••• Ig^nl^n
114 SERIES
A^t where A is any positive term divergent series.
Bn~* where B is any positive term convergent series.
It will be convenient to denote in general a convergent positive
term series by the symbol
0=^+%+ ...
and a divergent positive term series by
D = dl + di + ...
2. The series
is convergent, and conversely every positive term convergent series
may be brought into this form.
For
and C is convergent.
Let now conversely C— cl -f- c2 4- ..-be a given convergent
positive term series. Let
tf»-l = — •
Then
3.
rn M.+1
1-^n) - (2
i* divergent, and conversely every positive term divergent series may
be brought into this form.
For
PRINGSHEIM'S THEORY 115
Let now conversely D = dl + d^-{- ••• be a given positive term
divergent series. Let M - T)
Then d =M -M
102. Having now obtained a general form of all convergent
and divergent series, we now obtain another general form of a
convergent or divergent series, but which converges slower than
1) or diverges slower than 101, 2). Let us consider first con
vergence. Let M'n < Mn, then
-< \
(i
is convergent, and if M'n is properly chosen, not only is each
term of 1) greater than the corresponding term of 101, 1), but 1)
will converge slower than 101, 1). For example, for M'n let us
take M*, 0 < /* < 1. Then denoting the resulting series by
C' = 2c', we have
n urn nr* w M
On M^M*+l lYLn+l — Mn
= ^ ^n"M » r = < I* (2
1 - r Mn+1
Thus C' converges slower than 0. But the preceding also
shows that 0' and ,.. ,-
^ ' ~M*H (B
converge equally fast. In fact 2) states that
Since Mn is any positive increasing function of n whose limit
is oo, we may replace Mn in 3) by lrMn so that
is convergent and a fortiori
^lrMn+]-lrMn r = 1 2 .. (4
A Q+"3f.+1
is convergent.
116 SERIES
Now by I, 413, for sufficiently large n,
log Mu+l - log Mn = - log(l -
V -"
Replacing here Mn by log 7ffn, we get
Z M
fc2^"»+l ~
and in general
Thus the series
V
f f\
V
converges as is seen by comparing with 4). We are thus led to
the theorem :
MnMa+l
.
= ' ' •••'
form, an infinite set of convergent series; each series converging
slower than any preceding it.
The last statement follows from I, 463, l, 2.
Corollary 1 {Abel). Let D = d-^ + d2 -f- ••• denote a positive term
divergent series. Then
is convergent.
Follows from 3), setting Mn+1 = Dn.
Corollary 2. If we take Mn = n we get the series 91, Ex. 2.
Corollary 3. Being given a convergent positive term series
C = cl + c2 4- ••• we can construct a series which converges slower
than C.
PRINGSHEIM'S THEORY 117
For by 101, 2 we may bring Q to the form
Then any of the series 7) converges slower than 0.
103. 1. Let us consider now divergent series. Here our
problem is simpler and we have at once the theorem :
The series ™ TUT ^
J>-|^^=?4, a
diverges sloiver than
(2
That 1) is divergent is seen thus : Consider the product
fA
i M
which obviously = oc.
Now
Hence Dn = oo and Z> is divergent.
As ^ = JL=0
di Mn
we see that 1) converges slower than 2).
2. Any given positive term series D = d^ + d2 + .-• caw
the form -Z).
For taking J^ > 0 at pleasure, we determine Jf2, J[f3 ... by the
relations T
118 SERIES
Then Mn+1 > Mn and
, _ Mn+l — Mn
— n/r
•L'J-n
Moreover Mn = oo. For
>! + !>„ by I, 90, 1.
But Dn = oo.
3. The series
t^m '='0«1'2'"
form an infinite set of divergent series, each series divergent slower
than any preceding it. l^Mn = Mn.
For log Mn+l - log Mn = log (l + Mn^~Mn^
Mn
This proves the theorem for r = 0. Hence as in 102 we find,
replacing repeatedly Mn by log Mm
Corollary 1. If we take Mn = n, we get the series 91, Ex. 2.
Corollary 2 {Abel). Let D = d1 4- d2 + ••« be a divergent positive
term series. Then ,
is divergent.
We take here Jlfn = Dw.
Corollary 3. Being given a positive term divergent series D, we
can construct a series which diverges slower than D.
For by 101, 3 we may bring D to the form
Then 1) diverges slower than D.
PRINGSHEIM'S THEORY 119
104. In Ex. 3 of I, 454, we have seen that Mn+i is not always ~
Mn. In case it is we have
1. The series
^s convergent.
Follows from 102, 3).
2. The series
is convergent if \i > 0 ; # ?'s divergent if /x< 0.
For ^« > I /iWn2 - Ml p> 0.
Thus
If/i<0
3. jy J^+1 ~ Mn, we have
I M -I M
For by 102, 5), 103, 3),
7 W 7 AT ~
~
7 TI^- 7 AT
r+l n+l~ r+l n'
Now since Mn+l~ Mn, we have also obviously
lmMn~lmMn+l m=l,2,-r.
105. Having obtained an unlimited set of series which converge
or diverge more and more slowly, we show now how they may be
employed to furnish tests of ever increasing strength. To ob
tain them we go back to the fundamental theorems of comparison
of 87. In the first place, if A= a^ + a2 -f ••• is a given positive
term series, it converges if
120 SERIES
It diverges if
(2
**»
In the second place, A converges if
^±i_£n±a<0, (3
«n Cn
and diverges if ,
an dn
The tests 1), 2) involve only a single term of the given series
and the comparison series, while the tests 3), 4) involve two
terms. With Du Bois Reymond such tests we may call respec
tively tests of the first and second kinds. And in general any
relation between p terms
&ni ^«+l» *"* an+p-l
of the given series and p terms of a comparison series,
•' Cn+p-\'> Or n> <+i "' <+p-l
which serves as a criterion of convergence or divergence may be
called a test of the pth kind.
Let us return now to the tests 1), 2), 3), 4), and suppose we
are testing A for convergence. If for a certain comparison
series 0
— not always <_6r , n > m
cn
it might be due to the fact that cn = 0 too fast. We would then
take another comparison series (7'= ^cfn which converges slower
than (7. As there always exist series which converge slower than
any given positive term series, the test 1) must decide the con
vergence of A if a proper comparison series is found. To find
such series we employ series which converge slower and slower.
Similar remarks apply to the other tests. We show now how
these considerations lead us most naturally to a set of tests which
contain as special cases those already given.
106. 1. G-eneral Criterion of the First Kind. The positive term
series A = a^ -f «2 4- ••• converges 'if
PRINGSHEIM'S THEORY 121
It diverges if }[m Mn > Q< 2
-Mn+1-Mn
This follows at once from 105, 1), 2); and 101, 2; 103, 1.
2. To get tests of greater power we have only to replace the
^ nr M '
series
just employed in 1), 2) by the series of 102 and 103, 3 which con
verge (diverge) slower. We thus get from 1 :
The positive term series A converges if
It diverges if ^ Mnl,Mn .- lrMn ^ > ^
Mn^ - Mn
Bonnets Test. The positive term series A converges if
lira n^n ••• lr-^nl}.+ttn • an < oo , IJL > 0.
It diverges if ihp ^ ... ^ . fln>0.
Follows from the preceding setting Mn = n.
3. The positive term series A converges or diverges according as
^""V < 1 , M>0, (3
Mn+l-Mn Mn+1~Mn.
<1 , /*<0.
For in the first case
and in the second case
The theorem follows now by 104, 2.
4. The positive term series A converges if
log #.+!-•%. loff - _
- g
122 SERIES
It diverges if
,
t
p- n <() orl rnn
hm - TUT - <() or inn - - — -j^ - <o.
2Un (,r+llUn
Here r = 0, 1, 2, ••• awe? as 6e/0re /0M"n = 7ffn.
For taking the logarithm of both sides of 3) we have for con
vergence Mn+l - Mn
As p is an arbitrarily small but fixed positive number, A con
verges if lim q > 0. Making use of 104, 3 we get the first part
of the theorem. The rest follows similarly.
Remark. If we take Mn — n we get Cauchy's radical test 90
and Bertram's tests 93.
Forif log! .
it is necessary that
Also if
annlln~-lrn_
0 7 7 1 1U6 7 "
annl^n ••• lr^n lrn
lr+1n
lr+ln
loo*1
1 4. n 1 ' 'r-l>
r+l
it is necessary that -^
lr+ln
107. In 94 we have given Rummer's criterion for the conver
gence of a positive term series. The most remarkable feature
about it is the fact that the constants kv k2-~ which enter it are
subject to no conditions whatever except that they shall be positive.
On this account this test, which is of the second kind, has stood
entirely apart from all other tests, until Pringsheim discovered its
counterpart as a test of the first kind, viz. :
PRINGSHEIM'S THEORY 123
Pringsheims Criterion. Let pr p2--- be a set of positive numbers
chosen at pleasure, and let Pn—p^ + ••• + pn. The positive term
series A converges if
For A converges if
lim _ ^L>0. (1
*
lim — ^ > 0 , by 106, 4. (2
But Mn+1 — Mn = dn is the general term of the divergent series
Thus 2) may be written
~#T
Moreover A converges if
an~
that is, if v cn ^
lim — >Q,
where as usual 0= cl-\- c2 + ••• is a convergent series.
Hence A converges if Cn
lim^>0. (4
But now the set of numbers pv p2 ••• gives rise to a series
P = p1 -f p2 + ••• which must be either convergent or divergent.
Thus 3), 4) show that in either case 1) holds.
108. 1. Let us consider now still more briefly criteria of the
second kind. Here the fundamental relations are 3), 4) of 105,
which may be written :
cn+l— cn > 0 for convergence; (1
dn+l— — dn <^0 for divergence. (2
124 SERIES
Or in less general form :
The positive term series A converges if
___. (3
«»+i /
It diverges if
(4
Here as usual 0=^+%+ ••• is a convergent, and D=d1 + d%-\ ----
a divergent series.
2. Although we have already given one demonstration of
Rummer's theorem we wish to show here its place in Pringsheim's
general theory, and also to exhibit it under a more general form.
Let us replace <?n, cn+l in 1) by their values given in 101, 2.
We get
Jfn - JEf a M-M
_
an+1 Mn
or snce
or by 103, 2
Mn+l an+1 Mn
where Z> = a1-fc?2+ ••• is any divergent positive term series.
Since any set of positive numbers &x, &2, ••• gives rise to a series
&! + &2 -f ••• which must be either convergent or divergent, we see
from 1) that 5) holds when we replace the cTs by the &'s. We
have therefore:
The positive term series A converges if there exists a set of positive
numbers &x, &2 •••such that
kn+lf*—k*>0. (<3
It diverges if
where as usual 6?j + d% -f ••• denotes a divergent series.
ARITHMETIC OPERATIOXS ON SERIES 125
Since the Fs are entirely arbitrary positive numbers, the rela
tion 6) also gives
A converges if
as is seen by writing
*» = F
K n
reducing, and then dropping the accent.
3. From Rummer's theorem we may at once deduce a set of
tests of increasing power, viz. :
The positive term series A is convergent or divergent according as
M-M a M - M
L an+1 Mnl^Mn ... Zr^n
is >Q or is <^ 0.
For kl, &2 ... we have used here the terms of the divergent
series of 103, 3.
Arithmetic Operations on Series
109. 1. Since an infinite series
is not a true sum but the limit of a sum
A= lim J.n,
we now inquire in how far the properties of polynomials hold for
the infinite polynomial 1). The associative property is expressed
in the theorem :
Let A = al 4- a2 + • • • be convergent. Let b^ = a1 -j- • • • -f am ,
^)2 — am1+i+ •" +«/»,!-•• Then the series £ = bl + b2-}- ••> is con
vergent and A = B. Moreover the number of terms which bn em
braces may increase indefinitely with n.
For B -A
and lim A =A by I, 103, 2.
126 SERIES
This theorem relates to grouping the terms of A in parentheses.
The following relate to removing them.
2. Let B = b1 + b2 + •» be convergent and let b1 = al + • • • + am ,
62 = anh+l + ••• + am^ ••- If 1° A = a1 + a2 + ••• is convergent,
A = B. 2° Jf *Ae terms an>0, J. i« convergent. 3° 7f ea^A
mn — mn_i <_p a constant, and an = 0, A is convergent.
On the first hypothesis we have only to apply 1, to show
A — B. On the second hypothesis
e > 0, m, Bn < e, n>m.
Then 5-^.<e, *>»*„.
On the third hypothesis we may set
A, = Br+Vr+l
where b'r+1 denotes a part of the a-terms in br+l. Since br+l con
tains at most p terms of A, b'r+l = 0.
Hence lim ^1. = lira J9r , or A = B.
Example 1. The series
JB=(l-l) + (l-l) + (l-l)+-
is convergent. The series obtained by removing the parentheses
4 = 1-1 + 1-1+ ...
is divergent.
Example 2.
As ^ is comparable with *?. — , it is convergent. Hence A is
""* nz
convergent by 3°.
110. 1. Let us consider now the commutative property.
Here Riemann has established the following remarkable
theorem :
ARITHMETIC OPERATIONS OX SERIES 127
The terms of a simply convergent series A = al + a2 -f- ••• can be
arranged to form a series S, for which lim Sn is any prescribed
number, or ±00.
For let 7? i , i
B = ol + o2 + •••
(7= <?! + <?,,+ •«•
be the series formed respectively of the positive and negative
terms of A, the relative order of the terms in A being preserved.
To fix the ideas let I be a positive number ; the demonstration
of the other cases is similar. Since Bn = +00, there exists an m1
such that
Bmi > I. (1
Let ml be the least index for which 1) is true. Since Cn= — oo,
there exists an m^ such that
Bmi+C^<l. (2
Let m2 be the least index for which 2) is true. Continuing,
we take just enough terms, say m3 terms of B, so that
Then just enough terms, say m± terms of (7, so that
J^+0«+^* +<&».»< 4
etc. In this way we form the series
whose sum is I. For
| aa | < e s > <r ;
rn = ml + wa + ••• +™
A = al + a2 + • * be absolutely convergent. Let the terms
of A be arranged in a different order, giving the series B. Then B
is absolutely convergent and A = B.
For we may take m so large that
Am < e.
128 SERIES
We may now take n so large that An — Bn contains no term
whose index is <_ m. Thus the terms of An — Bn taken with
positive sign are a part of ATO and hence
An — Bn
n > m.
Thus B is convergent and B = A.
The same reasoning shows that B is convergent, hence B is
absolutely convergent.
3. If A = tfj -h a2 -f ••> enjoys the commutative property, it is
absolutely convergent.
For if only simply convergent we could arrange its terms so as
to have any desired sum. But this contradicts the hypothesis.
Addition and Subtraction
111. Let A = a1 + a2 + -- , B = bl + b2 4- ••• be convergent.
The series
are convergent and O=A±B.
For obviously Cn = An ± Bn. We have now only to pass to the
limit.
Example. We saw, 81, 3, Ex. 1, that
^=1-1 + 1-1 + ...
is a simply convergent series. Grouping its terms by twos and
by fours [109, l] we get
^\2n — 1 2nJ —?\-±n — 3 4n — 2 4 n — 1 4fi/
Let us now rearrange A, taking two positive terms to one nega
tive. We get
ADDITION AND SUBTRACTION 129
We note now that
n_a 4n-2 4n-l 4n/ --U - 2
n
n - 1 4 n - 3 2
= B by 109, 2.
Thus B = I A.
This example, due to Dirichlet, illustrates the non-commutative
property of simply convergent series. We have shown the con
vergence of B by actually determining its sum. As an exercise let
us proceed directly as follows :
The series 1) may be written :
Sn- 3 ^1 ~n
n
Comparing this with
we see that it is convergent by 87, 3. Since 1) is convergent, B
is also by 109, 2.
112. 1. Multiplication. We have already seen, 80, 7, that we
may multiply a convergent series by any constant. Let us now
consider the multiplication of two series. As customary let
2at^ i, * = 1, 2, 3, - (1
IK
denote the infinite series whose terms are all possible products
a, • bK without repetition. Let us take two rectangular axes as in
analytic geometry ; the points whose coordinates are x = t, y = K
are called lattice points. Thus to each term aJbK of 1), cor-
130 SERIES
responds a lattice point t, K and conversely. The reader will find
it a great help here and later to keep this correspondence in mind.
Let A = «! -f #2 H ---- •> •& — ^i + ^2 + '" be absolutely convergent.
Then O = ^LaJ)K is absolutely convergent and A • B = C.
Let m be taken large at pleasure ; we may take n so large that
Tn — Am • Bm contains no term both of whose indices are <. m.
Then — — —
Tn - ATOBm < «1Bm + «2BOT + .. . + «mBro
AmBm
< e for m sufficiently large.
Hence limr
and C is absolutely convergent.
To show that 0= A - B, we note that
\Cn-AmBm <Tn-AmBm<€
2. We owe the following theorem to Mertens.
If A converges absolutely and B converges (not necessarily abso
lutely), then
0 = a1b1
is convergent and 0 = A • B.
We set C— c1-\- <?2 +
where cl = a1b1
4-
Adding these equations gives
ADDITION AND SUBTRACTION 131
But
Bm = B-Bm w = l, 2, -.
Hence _
where
The theorem is proved when we show dn = 0. To this end let
us consider the two sets of remainders
sl ,
= n.
Let * each one in the first set be | < Mv and each in the second
set < M. Then since
Now for each e > 0 there exists an n^ such that
also a i/, such that
Thus 1) shows that , , ,
I da | < e.
3. When neither A nor B converges absolutely, the series C
may not even converge. The following example due to Cauchy
illustrates this.
VI V2 V3 V4
5 = J=-^-_ + A__JL+...=A
VI V2 V3 V4
*The symbols | < |, | < | mean numerically <, numerically <.
132 SERIES
The series A being alternating is convergent by 81, 3. Its
adjoint is divergent by 81, 2, since here /LI = 1. Now
VI VI VV1V2 V2V1
^J_J
VVlVS V2V2 VSV
= c2 + c8 + £4 + -
and
Vw-1 VI
By I, 95,
Hence
2
1 -^2 , , 2O-1)
Hence (7 is divergent since cn does not = 0, as it must if 0
were convergent, by 80, 3.
4. In order to have the theorems on multiplication together,
we state here one which we shall prove later.
If all three series A, B, 0 are convergent, then C = A • B.
113. We have seen, 109, 1, that we may group the terms of a
convergent series A = a^ + «2 4- — into a series B = bl + 52 + •••
each term bn containing but a finite number of terms of A. It is
easy to arrange the terms of A into a finite or even an infinite
number of infinite series, B1 ', B", B1" ••• For example, let
B' = «j + ap+1
B" =
Then every term of A lies in one of these p series B. To decom
pose A into an infinite number of series we may proceed thus :
In B' put all terms an whose index n is a prime number ; in B"
put all terms whose index n is the product of two primes ; in
TWO-WAY SERIES 133
B(m) all terms whose index is the product of m primes. We ask
now what is the relation between the original series A and the
series B' , B" >~
If A = a1 + #2 -h ••• is absolutely convergent, we may break it up
into a finite or infinite number of series Bf, B", Bfrf, • •• Each of
these series converges absolutely and
'"
That each B(m} converges absolutely was shown in 80, 6. Let
us suppose first that there is only a finite number of these series,
say p of them. Then
An = £'nt + JB£ + - 4- B$ n = n, + ... 4- »,.
As n=oo, each nr 7i2-..=oo. Hence passing to the limit
n = oo , the above relation gives
= + + ...
Suppose now there are an infinite number of series B(m).
Set B = B' + £" + £'" + ...
We take v so large that A — Bn, n>v, contains no term an of
index < m, and m so large that
Am<e.
Then
I A — Bn | < Am < €. n > v.
Two-way Series
114. 1. Up to the present the terms of our infinite series have
extended to infinity only one way. It is, however, convenient
sometimes to consider series which extend both ways. They are
of the type
••• a_3 + a_2 + a_! H- a0 + a1 + a2 + «3 -f- ...
which may be written
or
134 SERIES
Such series we called two-way series. The series is convergent
if
lim Zan (2
is finite. If the limit 2) does not exist, A is divergent. The ex
tension of the other terms employed in one-way series to the
present case are too obvious to need any comment. Sometimes
n = 0 is excluded in 1) ; the fact may be indicated by a dash,
thus 2'an.
— 00
2. Let m be an integer ; then while n ranges over
o o 1 0 1 ^ Q
— O, — 4, — JL) v, 1, *j, O
v = n + m will range over the same set with the difference that v
will be m units ahead or behind n according as m ^ 0. This
shows that
Similarly, So. = fa...
fl=-GC 7J=-00
3. Examine 1. @ = |^x+an2
— ao
4-
This series is fundamental in the elliptic functions,
Example 2.
. ^ » / -^ -^x
a; ^ Va; + w n/
The sum of this series as we shall see is TT cot TTX.
TWO-WAY SERIES 135
115. For a two way series A to converge, it is necessary and
sufficient that the series B formed with the terms with negative indices
and the series C formed with the terms with non-negative indices be
convergent. If A is convergent, A = B 4- C.
It is necessary. For A being convergent,
\A-Br-Cs\<e/'2 , \A-Br- C8, <e/2
if s, s' > some a- and r > some p. Hence adding,
\C,-C, <e,
which shows C is convergent. Similarly we may show that B is
convergent.
It is sufficient. For B, C being convergent,
B-BT <e/2 , \0-0.\<t/-2
for r, s > some p. Hence
r\ T»
Thus lim2a.= B+C.
Example 1. The series
rr> ^"^ \ >• i_ vi /vi
x ^ \x-\- n n>
is absolutely convergent if x ^ 0, ±1, ± 2,
For
I a. I =
+ n n
Hence * , ~J
2an and Zan
o -»
are comparable with V — •
7*
Example 2. The series
®(a;) = igna;+a"2 rr arbitrary (2
is convergent absolutely if a < 0. It diverges if a > 0.
136 SERIES
Here
n > 0, Van = e*ean = 0 if a < 0
== QO if a > 0 ;
n = — n',nr>0 n^/~an = e~xean' = 0 ifa<0
= 00 if a > 0.
The case a = 0 is obvious.
Thus the series defines a one-valued function of x when a < 0.
As an exercise in manipulation let us prove two of its properties.
1° <H)(V) is an even function.
For
®(-20=2«-«*+"*. (3
- 00
If we compare this series with 2) we see that the terms corre
sponding to n = m and n = — m have simply changed places, as the
reader will see if he actually writes out a few terms of 2), 3).
Of. 114, 2.
2° <8>O + 2ma-)= e-^x+ma^x). m=±l, ±2, ...
For we can write 2) in the form
_x* ^ (.r+2na)2
e(aO=e 4a 2e 4a • (4
n =— oo
(#+2/na)2 ,. (r+2(m+n)o)8
g 4a ^^ 4a
f=— 00
which with 4) gives 3).
• CHAPTER IV
MULTIPLE SERIES
116. Let x = xr •" xm be a point in ra-way space 9?m. If the
coordinates of x are all integers or zero, x is called a lattice point,
and any set of lattice points a lattice system. If no coordinate of
any point in a lattice system is negative, we call it a non-negative
lattice system, etc. Let f(x^ ••• z-m) be denned over a lattice
system 1 = ^,...^. The set {/(vOJ is called an
sequence. It is customary to set
Then the sequence is represented by
The terms Um A , IfS A , lim
as tj • • • im converges to an ideal point have therefore been denned
and some of their elementary properties given in the discussion
of I, 314-328 ; 336-338.
Let x = x1 — xn y = yi-~ym be two points in Wm. If
y\ > x\ "• Vm > xm we shall write more shortly y > x. If x
ranges over a set of points x' > x" > x'" ••• we shall say that x is
monotone decreasing. Similar terms apply as in I, 211.
Ifnow /(*••• y.)>/(*,~<0
when y _> a?, we say f is a monotone increasing function. If
we say f is a monotone decreasing function.
Similar terms apply as in I, 211.
137
138 MULTIPLE SERIES
117. A very important class of multiple sequences is connected
with multiple series as we now show. Let «tl...tm be defined over
a non-negative lattice system. The symbol
or 2atl...lm , or AVl...;m
denotes the sum of all the a's whose lattice points lie in the rec
tangular cell Q<XI<VI -0<xm<vm.
Let us denote this cell by RVl...vn or by Rv. The sum 1) may be
effected in a variety of ways. To fix the ideas let m = 3. Then
A, „ „ =
etc. In the first sum, we sum up the terms in each plane and
then add these results. In the second sum, we sum the terms on
parallel lines and then add the results. In the last sum, we sum
the terms on the parallel lines lying in a given plane and add the
results; we then sum over the different planes.
Returning now to the general case, the symbol
A = 2all...tn iv ••• im= 0, 1, ••• oo,
00
or 4 = 2^...^
o
is called an w-tuple infinite series. For m = 2 we can write it
out more fully thus
^12+
In general, we may suppose the terms of any m-tuple series dis
played in a similar array, the term atl...lw occupying the lattice
point t = (i1-«-tm). This affords a geometric image of great
service. The terms in the cell Rv may be denoted by Av.
If lim AVl...Vm = lim A, (2
GENERAL THEORY 139
is finite, A is convergent and the limit 2) is called the sum of the
series^.. When no confusion will arise, we may denote the series
and its sum by the same letter. If the limit 2) is infinite or does
not exist, we say A is divergent.
Thus every w-tuple series gives rise to an ra-tuple sequence
{AVl...Vm\. Obviously if all the terms of A are > 0 and A is diver
gent, the limit 2) is -f oo. In this case we say A is infinite.
Let us replace certain terms of A by zeros, the resulting series
may be called the deleted series. If we delete A by replacing all
the terms of the cell RVl...VM by zero, the resulting series is called
the remainder and is denoted by AVl...vm or by Av. Similarly if
the cell Rv contains the cell R^ the terms lying in Rv and not in
Rp may be denoted by A^ „.
The series obtained from A by replacing each term of A by its
numerical value is called the adjoint series. In a similar manner
most of the terms employed for simple series may be carried over
to m-tuple series. In the series 2atl...llB the indices t all began
with 0. There is no necessity for this; they may each begin with
any integer at pleasure.
118. The Geometric Series. We have seen that
= 1 -f a + a2 + ••• a < 1,
1- b
Hence
1
for all points a, b within the unit square.
In general we see that
1 "2 *•
is absolutely convergent for any point x within the unit cube
and
140 MULTIPLE SERIES
119. 1. It is important to show how any term of A = 2atl...lm can
be expressed by means of the AVr..Vn.
Let Dl/il/2...%_i = AVlVt...Vm — Av^...Vm_^ (1
Then D^2. ..I/m_1_1 = AVlVt... ,m ^Vm - AVlVs...VM_^lt ,m_r (2
Let &w.'i'm-2 = A^-^-i - ^w^-i-r (3
Similarly
",-,-n
(5
Finally ^ = DViVf - !>„,„,_!, (6
and «IV,...I/M = A't — DVl-i* (7
If now we replace ,the D's by their values in terms of the As,
the relation 7) shows that «„,...„„ may be expressed linearly in
terms of a number of AIJLl...IJim where each pr = vr or vr — 1. • .
For m = 2 we find
aVlVa = AVtVf + ^-1,^-1 — AVlt V2_x — AVl_^Vf. (8
2. From 1 it follows that we may take any sequence \Alt...ltlt\
to form a multiple series
-^i — ^^ij ••• im*
This fact has theoretic importance in studying the peculiarities
that multiple series present.
120. We have now the following theorems analogous to 80.
1. For A to be convergent it is necessary and sufficient that
2. If A is convergent, so is A^ and
A -A \ — 1 1 tv» 1
-/Xj^ — ^1. ^~ *^-*-/x ~~" ^•**1I -*jLix j/»
Conversely if A^ is convergent, so is A.
GENERAL THEORY 141
3. For A to converge it is necessary and sufficient that
lim Av = 0.
4. A series ivhose adjoint converges is convergent.
5. Let A be absolutely convergent. Any deleted series B of A is
absolutely convergent and \ B \ < A.
6. If A = 2ati...lm is convergent, so is B — 2&«ti...,,n and
B = kA, k a constant.
121. 1. For A to converge it is necessary that
-A^...,.*,-! , -Z>ri,2...,m_2 , —Dvi , avil.2. ..,, = (), as v = cc .
For by 120, l A ,
I ylA, ••• Am ^fM-'Mm | V c
if X1..-Aro, /*! .../*m >jo.
Thus by 119, 1)
Hence passing to the limit p = 00 ,
IrniA,..., „_,<.*.
As e is small at pleasure, this shows that DVl...Vm_^ = 0. In this
way we may continue.
2. Although ,. n
lim av ... ,,m = 0
»i ••• >'m=»
when A converges, we must guard against the error of supposing
that av= 0 when v = (yl ••• vm) converges to an ideal point, all of
whose coordinates are not cc as they are in the limits employed
in 1.
This is made clear by the following example due to Pringsheim.
Then by 119, 8) I ^,-1 + 1
ar as
142 MULTIPLE SERIES
AS lim A-,. = 0
r, 5=»
A is convergent. But
lim aftS\ =— , lim | ars \ =-.
r=x> a s=™ ar
That is when the point (r, s) converges to the ideal point
(oo, s), or to the ideal point (r, GO ), ars does not = 0.
3. However, we do have the theorem :
Let . ~ ^ ~
A = 2at ... l at > 0
converge. Then for each e > 0 ^m? e:m£s a X swcA that «tl...t < e
/or «?i?/ t outside the rectangular cell R^.
This follows at once from 120, 1, since
*£•*»*•
122. 1. Letf(x^ ••• xm) be monotone. Then
^ ••• xm) =1 xl< av ••• xm < am, a may be ideal. (1
exists, finite or infinite. Iff is limited, I is finite. If f is unlim
ited, I = 4- QO when f is monotone increasing, and I = — GO whenf is
monotone decreasing.
For, let/ be limited. Let A = «: < <*2 < ••• = a.
Then T />x N 7
l]m/(«n) = I
n=cc
is finite by I,. 109.
Let now B = ftv /32, ••• = a be any other sequence.
T pf
lim/G8n) = I lim f (&) = f.
~«~ ^ "
Then there exists by I, 338 a partial sequence of B, say
O=j1, 72 ••• such that
also a partial sequence D— S1, S2 ••• such that
lim/(S,,) = I
GENERAL THEORY 143
But for each an there exists a yln _> an ;
/(7 J >/(«»)
and therefore J > I- (2
Similarly, for each dn there exists an a,a > Sn ;
hence /(8.) </(«,.)
and therefore 7 < Z (3
Thus 2), 3) give lim/(~)'=/.
.8
Hence by I, 316, 2 the relation 1) holds.
The rest of the theorem follows along the same lines.
2. As a corollary we have
The positive term series A = ^,ali,..ln is convergent if Av^tfl,m is
limited.
123. 1. Let A=2all...ls = 2at , B = 26tl...t- = 2J4 be tivo non-
negative term series. If they differ only by a finite number of
terms, they converge or diverge simultaneously.
This follows at once from 120, 2.
2. Let A, B be two non-negative term series. Let r > 0 denote
a constant. If aL < rbL , A converges if B is convergent and A < rB.
If a, > rbL, A diverges if B is divergent.
For on the first hypothesis
AK
and on the second A
A*
3. Let A, B be two positive term series. Let r, s be positive
constants. If
r fl|
hm — l
1=00 6t
exists and is =£ 0, A and B converge or diverge simultaneously. If
B converges and — = 0, A is convergent. If B diverges and -* = <»,
bn bn
A is divergent.
144
MULTIPLE SERIES
4. The infinite non-negative term series
2#t,...lg and 2 log (1
converge or diverge simultaneously.
This follows from 2.
5. Let the potver series
P — S/> rw7irm, .
; — ^^-'--
converge at the point a = («j, ••• «a), then it converges absolutely for
all points x within the rectangular cell R whose center is the origin,
and one of whose vertices is a; that is for \ x, \ < | a, , i=l, 2, ••• a.
For since P converges at a,
lim cmim2...a™i •••<'• = 0.
Thus there exists an J^such that each term
I c a7!*1 "• am* I "^ ]\T
| TO! ••• "~l "»j | — •<•'-'
Hence
, nm\ nm*
1
mi
xa
m\ ... ^J **s
«T
"
OH
x.\
""...
X.
ma
ai
aa
Thus each term of P is numerically < than M times the cor
responding term in the convergent geometric series
We apply now 2.
We shall call R a rectangular cell of convergence.
124. 1. Associated with any w-tuple series A = ^aL ...,_ are
an infinite number of simple series called associate simple series,
as we now show.
Let 7? 7? 7?
be an infinite sequence of rectangular cells each lying in the
following. Let
be the terms of ^1 arranged in any order lying in
Let
GENERAL THEORY 145
be the terms of A arranged in order lying in R^— R\^ and so on
indefinitely.
Then SI = ^ 4- «a + - + a.v + a.l+1 + -
is an associate simple series of A.
2. Conversely associated with any simple series 51 = 2an are an
infinity of associate m-tuple series. In fact we have only to arrange
the terms of 51 over the non-negative lattice points, and call now
the term an which lies at the lattice point ^ ••• im the term atl...lw.
3. Let 51 be an associate series of A = 2ah ... l|B. If 51 is convergent,
so is A and J. = 51
For AVl...Vm = %n.
Let now v = oo, then n = <x>. But 5l7i = 51, hence ^ ... Vm = 51.
4. If the associate series 51 is absolutely convergent, so is A.
Follows from 3.
5 If A = 2ari ... Vfn is a non-negative term convergent series, all its
associate series 51 converge.
For, any 51^^ lies among the terms of some A^v. But for X
sufficiently large A*9<* \<n<v.
Hence ar
*«•.*<« m>mQ.
6. Absolutely convergent series are commutative.
For let B be the series resulting from rearranging the given
series A.
Then any associate 53 of B is simply a rearrangement of an
associate series 51 of A. But 51 = SB, hence A = B.
7. A simply convergent m-tuple series A can be rearranged,
producing a divergent series.
For let 51 be an associate of A. 51 is not absolutely convergent,
since A is not. We can therefore rearrange 51, producing a series
53 which is divergent. Thus for some 53
lim $„
does not exist. Let 53' be the series formed of the positive, and
53" the series formed of the negative, terms of 53 taken in order.
146 MULTIPLE SERIES
Then either Wn = + oo or ^ = -00, or both. To fix the ideas
suppose the former. Then we can arrange the terms of 33 to
form a series (£ such that &n = -h oo. Let now (£ be an associate
series of O. Then
t; = cv,... ,„ = <£„
and thus
Km Cv = lim (£„ = + oo.
Hence 0 is divergent.
8. If the, multiple series A is commutative, it is absolutely con
vergent.
For if simply convergent, we can rearrange A so as to make the
resulting series divergent, which contradicts the hypothesis.
9. In 121, 2 we exhibited a convergent series to show that
«tl...lw does not need to converge to 0 if ^ ••• im converges to an ideal
point some of whose coordinates are finite. As a counterpart we
have the following :
Let A be absolutely convergent. Then for each e > 0 there exists
a \, such that any finite set of terms B lying without E^ satisfy the
relation
and conversely.
For let 51 be an associate simple series of Adj A. Since 51 is
convergent there exists an n, such that
But if X is taken sufficiently large, each term of B lies in 5ln,
which proves 1).
Suppose now A were simply convergent. Then, as shown in 7,
there exists an associate series £) which is infinite.
Hence, however large n is taken, there exists a p such that
Hence, however large X is taken, there exist terms J?= £)Wip which
do not satisfy 1).
10. We have seen that associated with any m-tuple series
GENERAL THEORY 147
extended over a lattice system 2ft in 9f?m is a simple series in 9?r
We can generalize as follows. Let 2ft = \i\ be associated with a
lattice system 2ft = \j\ in ^Rn such that to each i corresponds a,j and
conversely.
If i ~y we set «4i ... lm = a^ ... jn .
Then A gives rise to an infinity of w-tuple series as
We say B is a conjugate n-tuple series.
We have now the following :
Let A be absolutely convergent. Then the series B is absolutely
convergent and A = B.
For let A\ B' be associate simple series of A, B. Then A', B1
are absolutely convergent and hence A'=B'. But A = A\ B = B'.
Hence A = B, and B is absolutely convergent.
11. Let A = 2atl... tm be absolutely convergent. Let B= Sa^...^
be any p-tuple series formed of a part or all the terms of A. Then
B is absolutely convergent and
Ad A.
For let A', B' be associate simple series of A and B. Then B'
converges absolutely and | B' \ < Adj A.
125. 1. Let 4 = 2«ll...... (1
Set
in the cell
Then
(2
Let 72 denote that part of 9?m whose points have non-negative
coordinates. Let ^
J=\ fdX,:-dXm. (3
•X ^2
If «7is convergent, A = J. We cannot in general state the con
verse, for A is obtained from Av by a special passage to the limit, viz.
148 MULTIPLE SERIES
by employing a sequence of rectangular cells. If, however,
av >_ 0 we may, and we have
For the non-negative term series 1) to converge it is necessary and
sufficient that the integral 3) converges.
2. Let /(#! ••• XM) > 0 be a monotone decreasing function of
x in R, the aggregate of points all of whose coordinates are non-
negative. Let
The series A = ^
is convergent or divergent with
J»
fdxl "- dxm.
R
For let R}, R2, ••• be a sequence of rectangular cubes each Rn
contained in Rn+i.
Let RntS=Rt-Ra s>n.
Then X, ft being taken at pleasure but > some v, there exist an
Z, m such that
But the integral on the right can be made small at pleasure if J
is convergent on taking I > m > some n. Hence A is convergent
if J"is. Similarly the other half of the theorem follows.
Iterated Summation of Multiple Series
126. Consider the finite sum
One way to effect the summation is to keep all the indices but
one fixed, say all but tx, obtaining the sum
Then taking the sum of these sums when only *2 is allowed to
vary obtaining the sum
2 2ah...t m
ia=0 «j=0
ITERATED SUMMATION OF MULTIPLE SERIES 149
and so on arriving finally at
ml
...^ (2
whose value is that of 1). We call this process iterated summa
tion. We could have taken the indices ^ ••• im in any order
instead of the one just employed ; in each case we would have
arrived at the same result, due to the commutative property of
finite sums.
Let us see how this applies to the infinite series,
^i = 2ati...,m, ij — tm=0, 1, ...oo. (3
The corresponding process of iterated summation would lead us
to a series 31= 2 2 ••• fa (4
_.. ll * * " lm ' ^
which is an m-tuple iterated series. Now by definition
"m "m • 1 "i
31 = lim 2 lim 2 ••• lim 2ati...(m (5
= lim lim ••• lim AVi...Vm^ (6
while A ,. /rj
A= lim AVi...Vm. (J
Thus A is defined by a general limit while 31 is defined by an
iterated limit. These two limits may be quite different. Again
in 6) we have passed to the limit in a certain order. Changing
this order in 6) would give us another iterated series of the type
4) with a sum which may be quite different. However in a large
class of series the summation may be effected by iteration and this
is one of the most important ways to evaluate 3).
The relation between iterated summation and iterated integra
tion will at once occur to the reader.
127. 1. Before going farther let us note some peculiarities of
iterated summation. For simplicity let us restrict ourselves to
double series. Obviously similar anomalies will occur in m-tuple
series.
150 MULTIPLE SERIES
^ = «00 + a01 + S2+ -+«10+all +
be a double series. The mih row forms a series
and the wth column, the series
»i=0
7M=0 MJ=O »=0
are the series formed by summing by rows and columns, respec
tively.
2. A double series may converge although every row and every
column is divergent.
This is illustrated by the series considered in 121, 2. For A
00 00
is convergent while 2ar,, 2ar, are divergent, since their terms are
r=0 s=Q
not evanescent.
3. A double series A may be divergent although the series R ob
tained by summing A by rows or the series C obtained by summing
by columns is convergent.
For let Ars=0 if r or s = 0
r .»
= if r, s>0.
r -f s
Obviously by I, 318, lim Ars does not exist and A = 2#ra is di
vergent.
On the other hand,
R = lim lim Ars — 0,
r=oo 5=3
(7= lim lim Ars = 1.
S=00 r=3
Thus both R and (7 are convergent.
ITERATED SUMMATION OF MULTIPLE SERIES 151
4. In the last example R and C converged but their sums were
different. We now show :
A double series may diverge although both R and C converge and
have the same sum.
For let Art, = 0 if r or s = 0
rs
if r, « > 0.
Then by I, 319, lim A,, does not exist and A is divergent. On
the other hand, R = \{m \{m Ar, = 0,
(7= lim lim Ara = 0.
Then R and S both converge and have the same sum.
128. We consider now some of the cases in which iterated sum
mation is permissible.
Let A = 2ati ... ,w be convergent. Let i\, i2,-~ i'm be any permutation
of the indices ir iv ••• im. If all the m - l-tuple series
.'=0 t' =0
are convergent, A =
This follows at once from I, 324. For simplicity the theorem
is there stated only for two variables ; but obviously the demon
stration applies to any number of variables.
129. 1. Let f(x1-"Xm) be a limited monotone function. Let the
point a= (04 •••#,») be finite or infinite. When f is limited, all the
s-tuple iterated limits
exist. When s = m, these limits equal
..-^). (2
In these limits we suppose x<a.
152 MULTIPLE SERIES
For if /is limited, Hm/ > ^^
exists by 122, l. Moreover 3) is a monotone function of the re
maining m — 1 variables.
Hence similarly ]im Hm f
JTia_l=ais_l xis=ctis
exists and is a monotone function of the remaining m — 2 vari
ables, etc. The rest of the theorem follows as in I, 321.
2. As a corollary we have
Let Abe a non-negative term m-tuple series. If A or any one of
its m-tuple iterated series is convergent, A and all the ml iterated
m-tuple series are convergent and have the same sum. If one of these
series is divergent, they all are.
3. Let a be a non-negative term m-tuple series. Let s<m. All
the s-tuple iterated series of A are convergent if A is, and if one of
these iterated series is divergent, so is A.
130. 1. Let A = 2ati...tm be absolutely convergent. Then all its
s-tuple iterated series s = l, 2 ••• m, converge absolutely and its
m-tuple iterated series all = A.
For as usual let «ti...lm= atl...im\. Since A = Adj A is con
vergent, all the s-tuple iterated series of A are convergent.
OD ao
Thus sl = 2 fltl...im is convergent since 2 atl. . (TO = tr1. Moreover
II=Q l 4=0
00 00
I sl I < crr Similarly 2 2«lf...lm = 2st is convergent since
i,=0 4=0 i?
OO QO
2 2«M... ,w =20-^8 convergent; etc. Thus every s-tuple iter-
'2=0 4=0 i2
ated series of ^4. is absolutely convergent. The rest follows now
by 128.
2. Let A = 2alt...,m. /f one of i^e m-tuple iterated series B
formed from the adjoint A of A is convergent, A is absolutely con
vergent.
Follows from 129, 2.
3. The following example may serve to guard the reader against
a possible error.
ITERATED SUMMATION OF MULTIPLE SERIES 153
Consider the series
and R = ea + e*a + e3a -h
This is a geometric series and converges absolutely for a < 0.
Thus one of the double iterated series of A is absolutely conver
gent. We cannot, however, infer from this that A is convergent,
for the theorem of 2 requires that one of the iterated series formed
from the adjoint of A should converge. Now both those series
are divergent. The series A is divergent, for | arg \ = <x> , as
r, s = oo .
131. 1. Up to the present the series
2*.,..... (1
have been extended only over non-negative lattice points. This
restriction was imposed only for convenience ; we show now how
it may be removed. Consider the signs of the coordinates of a
point x= (xv ••• xm). Since each coordinate can have two signs,
there are 2m combinations of signs. The set of points x whose
coordinates belong to a given one of these combinations form a
quadrant for m = 2, an octant for m = 3, and a 2m-tant or polyant
in 9?m. The polyant consisting of the points all of whose coordi
nates are > 0 may be called the first or principal polyant.
Let us suppose now that the indices i in 1) run over one or more
polyants. Let R^ be a rectangular cell, the coordinates of each of
its vertices being each numerically < \. Let A^ denote the terms
of A lying in 72A. Then I is the limit of A^ for X = oo, if for each
e > 0 there exists a \0 such that
\A,-A,n\<e X>\0. (2
154 MULTIPLE SERIES
If lim At (3
A=ao
exists, we say A is convergent, otherwise A is divergent. In a
similar manner the other terms employed in multiple series may
be extended to the present case. The rectangular cell 72Ao which
figures in the above definition may without loss of generality be
replaced by the cube
Moreover the condition necessary and sufficient for the exist
ence of the limit 3) is that
| A - 4* I < e X, p > X0.
132. The properties of series lying in the principal polyant
may be readily extended to series lying in several polyants. For
the convenience of the reader we bring the following together,
omitting the proof when it follows along the same lines as before.
1. For A to converge it is necessary and sufficient that
lim AI = 0.
A=oo
2. A series whose adjoint converges is convergent.
3. Any deleted series B of an absolutely convergent series A is
absolutely convergent and
\B\< Adj A.
4. If A = Sdtij ... ,n is convergent, so is B = 2&#(J ...,M and A = kB.
5. The non-negative term series A is convergent if A^ is limited,
X = oo.
6. If the associate simple series 51 of an m-tuple series A converges,
A is convergent. Moreover if 21 is absolutely convergent, so is A.
Finally if A converges absolutely, so does 21.
7. Absolutely convergent series are commutative and conversely.
8. Let f(x1 •••a;m)>.0 be a monotone decreasing function of the
distance of x from the origin.
Let ~
ITERATED SUMMATION OF MULTIPLE SERIES 155
Then A = Zat.
1 m
converges or diverges ivith
the integration extended over all space containing terms of A.
133. 1. Let B, (7, D ••• denote the series formed of the terms of A
lying in the different poly ants. For A to converge it is sufficient
although not necessary that B, C, ••• converge. When they do,
For if i?A, <7A ••• denote the terms of B, C ••• which lie in a
rectangular cell 72A,
AI = BL+CI + -
Passing to the limit we get 1).
That A may converge when B, (?, ••• do not is shown by the
following example. Let all the terms of A = 2atl...tm vanish ex
cept those lying next to the coordinate axes. Let these have the
value +1 if ir L2-~ im>0 and let two #'s lying on opposite sides
of the coordinate planes have the same numerical value but opposite
signs. Obviously, Ax = 0, hence A is convergent. On the other
hand, every B, C ••• is divergent.
2. Thus when B, C ••• converge, the study of the given series
A may be referred to series whose terms lie in a single polyant.
But obviously the theory of such series is identical with that of
the series lying in the first polyant.
3. The preceding property enables us at once to extend the
theorems of 129, 130 to series lying in more than one polyant.
The iterated series will now be made up, in general of two-way
simple series.
CHAPTER V
SERIES OF FUNCTIONS
134. 1. Let i — (tr i2 ••• ip) run over an infinite lattice system &.
Let the one-valued functions
/.,... .POi •••*») = /<»=/.
be denned over a domain 51, finite or infinite. If the jt?-tuple series
F=F(x)=F(xi »• aw) =2/., ...„(*! •»*»,) (1
extended over the lattice system % is convergent, it defines a one-
valued function F(x1 • • • #m) over 51. We propose to study the
properties of this function with reference to continuity, differen
tiation and integration.
2. Here, as in so many parts of the theory of functions depend
ing on changing the order of an iterated limit, uniform convergence
is fundamental.
We shall therefore take this opportunity to develop some of its
properties in an entirely general mariner so that they will apply
not only to infinite series, but to infinite products, multiple inte
grals, etc.
3. In accordance with the definition of I, 325 we say the series
1) is uniformly convergent in 51 when F^ converges uniformly to its
Kmit F. Or in other words when for each e>0 there exists a X
such that |J--j;|<« M>X,
for any x in 51. Here, as in 117, F^ denotes the terms of 1) lying
in the rectangular cell 72M, etc.
As an immediate consequence of this definition we have :
Let 1) converge in 51. For it to converge uniformly in 51 it is
necessary and sufficient that \ F^ is uniformly evanescent in 51, or in
other words that for each e > 0, there exists a \ such that F^ > e for
x in 51, and /JL>\.
166
GENERAL THEORY 157
135. 1. Let
in 51. Here 51, T may be finite or infinite. If there exists an
77 >0 such that f = (f> uniformly in ^(V), a finite or infinite, we
shall say f converges uniformly at a ; if there exists no 77 < 0. \ve
say / does not converge uniformly at a.
2. Let now a range over 51. Let 53 denote the points of 51 at
which no ?? exists or those points, they may lie in 51 or not, in
whose vicinity the minimum of 77 is 0. Let D denote a cubical
division of space of norm d. Let $8D denote as usual the cells of
D containing points of 53. Let &D denote the points of 51 not in
53^. Then/=<£ uniformly in &D however small d is taken, but
then fixed. The converse is obviously true.
3. Iff converges uniformly in 51, and if moreover it converges at a
finite number of other points -03, it converges uniformly in 51 + 53.
For if / = (f> uniformly in 51,
f-<f>\<e x in 51, t in F5o*(r).
Then also at each point b, of 53,
\f -4>\< e x=bt <inFif*(T).
If now 8 < S0, 8V S2 ••• these relations hold for any x in 51 + 53
and any t in Vs*(r).
4. Let /(#! ••• xm, fj ••• tn) = $ (xl ••• xm^ uniformly in 51. Let
f be limited in 51 for each t in F6*(r). Then <f> is limited in 51-
For </>=/(*, 0 +€' |«'|<« C1
for any x in 51 and t in F5*(r). Let us therefore fix t. The
relation 1) shows that (f> is limited in 51.
5. If 2 \f^ ... i (tfj ••• #m) | converges uniformly in 51, so c?oes 2/ti ...t .
For any remainder of a series is numerically < than the corre
sponding remainder of the adjoint series.
6. Let the s-tuple series
158 SERIES OF FUNCTIONS
converge uniformly in 51. Then for each e > 0 there exists a X
such that , _.
\^\<e (1
for any Rv > R^ > R^. When s = 1, these rectangular cells re
duce to intervals, and thus we have in particular
I /«Oi — aw) | < e for any n > n'.
When s > 1 we cannot infer from 1) that
\f^t.^i-xm)\<€ , in 51, (2
for any i lying outside the above mentioned cell Rk.
A similar difference between simple and multiple series was
mentioned in 121, 2.
However if fL > 0 in 51, the relation does hold. Cf. 121, 3.
136. 1. Let f(xv~xm^ t1"- tn) be defined for each x in 51, and t
T finite or infinite. The convergence is uniform if for any x in 51
I/ - <t> I < V<*i • • • O tin Vt* (r), B fixed
while Km -fy = 0.
t=T
For taking e>0 at pleasure there exists an ?;>0 such that
|VI<* » ^in ^,*(T).
But then if S < T/,
l/-*l<«
for any ^ in V&*(r) and any 2: in 51.
Example.
sin a; sin n . Qr .„ N
° = *' m 51 = (0, oo).
Is the convergence uniform ?
Let
y
then u = 0, as ^ = -
GENERAL THEORY 159
Then f_ A | _ | sin x cos u \ _ \ sin x cos u sin2 u \
I + x cot2 u I sin2 u + x cos2 w |
^ sin x sin2 w I ^ , 0 . A
< < tan2 u = 0.
Hence the convergence is uniform in 51-
2. As a corollary we have
Weierstrass"1 Test. For each point in 51, let l/^... tp| ^Mtl...lp
The series ^/^...^(^ •••#„,) is uniformly convergent in 51 if ^Mtl...ip
is convergent.
Example 1.
Here
and _F is uniformly convergent in 51 since
2-
— on
is convergent.
Example 2.
F(x) = ^an sin An
is uniformly convergent for ( — 00, QO) if
Z.|o.|
is convergent.
137. 1. 7%e power series P = 2amt...mpa;jli ••• x™v converges
uniformly in any rectangle R lying within its rectangle of con
vergence.
For let b = (6r ••• 6P) be that vertex of R lying in the principal
polyant. Then P is absolutely convergent at 6, i.e.
is convergent. Let now x be any point of R. Then each term in
2<V...,nP £?-&"
is < than the corresponding term in 1).
160 SERIES OF FUNCTIONS
2. If the power series P — aQ + a^x + a2x2 + ••• converges at an
end point of its interval of convergence, it converges uniformly at
this point.
Suppose P converges at the end point x = R > 0. Then
however large n is taken. But for 0 < x < R
< e by Abel's identity, 83, i.
Thus the convergence is uniform at x — R. In a similar
manner we may treat x = — R.
3. Let/n^j ••• xm), n = 1, 2 ••• be denned over a set 5L If each
\fn < some constant cn in 21, fn is limited in 51. If moreover the
cn are all < some constant 0, we say the fn(x) are uniformly
limited in 51. In general if each function in a set of functions
\f\ denned over at point set 51 satisfy the relation
\f \ < a fixed constant 0, x in 51,
we say the jf's are uniformly limited in 51.
The series F= ^gnhn is uniformly convergent in 51, if G =ffi -h </2 -f- • • •
is uniformly convergent in 51, while 2 1 ha+1 — hn \ and hn \ are
uniformly limited in 51.
This follows at once from Abel's identity as in 83, 2.
4. The series F=2gnhn is uniformly convergent in 51, if in 51,
2 | h,l+l — hn | is uniformly convergent, hn is uniformly evanescent,
and the Grn uniformly limited.
Follows from Abel's identity, 83, l.
5. The series F= ^gnhn is uniformly convergent in 51 if
Gr = gl -f- g% + •• • is uniformly convergent in 51 while hv h2 ••• are
uniformly limited in 51 and \hn\ is a monotone sequence for each
point of 51.
For by 83, l, , ^ , , ^ , n .
** <J** &n<
(JEXERAL THEORY 101
6. The series F = ^gnhn is uniformly convergent in 51 if Gl = gv
6r2 = gl + <72, ~- are uniformly limited in 51 and if hr A2, ••• not only
form a monotone decreasing sequence for x in 51 but also are uni
formly evanescent.
For by 83,1, | J^, | < | ^ G. .
Example. Let A = al + a2 -f- ••• be convergent. Let 5r 62 •-• ^ 0
be a limited monotone sequence. Then
converges uniformly in any interval 51 which does not contain a
point of | — [ •
For obviously the numbers
form a monotone sequence at each point of 51. We now apply 5.
7. As an application of these theorems we have, using the re
sults of 84,
The series 0
fl0 -f «! cos x + a2 cos 2 x +
converges uniformly in any complete interval not containing one of
the points ± 2 mir provided 2 an+l — an is convergent and an = 0,
and hence in particular if a^^ #2 _> ••• = 0.
8. The series ^
«0 — a1 cos x + a 2 cos 2 # —
converges uniformly in any complete interval not containing one of
the points ± (2 m — I)TT provided 2 | an+1 + an is convergent and
an = 0, and hence in particular if al > a% > • • • = 0.
9. The series . . 0 . 0
al sin x + «2 sin is 2T 4- ^3 sin 00:+ •••
converges uniformly in any complete interval not containing one of
the points ± 2 mir provided 2 | an+l — an \ is convergent and an = 0,
and hence in particular if a± >^ «2 21 ••• = 0.
162 SERIES OF FUNCTIONS
10. The series ^ sill ^ _ ^ 2 * + ^sin 3* - ...
converges uniformly in any complete interval not containing one of
the points ±(2w — I)TT provided £ | an+l + an is convergent and
an = 0, and hence in particular if a1^a2 >_ ••• =0.
138. 1. Let F=^...^Xl...Xm)
be uniformly convergent in 51. Let A, B be two constants and
4«*)<^<X><^/.<X> m «•
Then (? = 2^..,.(^...^)
is uniformly convergent in 51.
For then j T-T ^ ^ ^
AJb A, M < CrA? M <
But ^ being uniformly convergent,
2. =l,...l8^
converge uniformly in 51. Then
is uniformly convergent in 31. Moreover if F is limited in 51, so
is L.
For /t > 0 in 51, hence
l/i|<«
for any i outside some rectangular cell M\.
Thus for such i
Af< < log (1 +/) < £/ in «.
139. 1. Preserving the notation of 136, let g1, #2, ••• ^m fo chosen
such that if we set
formly in 51,
lim A = lim ... * M. - •- ^nS = 0.
GENERAL THEORY 163
For if /= <f> uniformly in 31,
e>0, S>0 |/_<£|<€
for any x in 51 and any t in FS*(T), 8 independent of x.
But then | A | < e tin Fi*(r).
2. As a corollary we have :
Let «j, #2, ... = a. Ze£ ^= l/a fo uniformly convergent at a.
Then . JU<o=o.
140. Example 1.
lim/=lim_8muBin2tL ,( [2 for*=0,
«=o «=o sin2 -M + # cos2 M [ 0 for x ^= 0.
The convergence is not uniform at 2; = 0. For
/._ 2cosw
1 + ^ cot2 w
Hence if we set x = u2
lim/= 1, since u2 cot2 u= 1.
w = 0
Thus on this assumption
Example 2. F= l-x + x(l-x^ + X2(l _ x)+ ^(1 - a:) + •••
Here _ »
^=1(1-^) -2:".
o
Hence F is uniformly convergent in any (— r, r), 0 < r < 1, by
136, 2.
We can see this directly. For
Hence F is convergent for — 1<#<1, and then .F(V)=1,
except at x = 1 where F = 0.
Thus | Fn (x) | = | x |n, except at x = 1.
But we can choose m so large that rm<e.
Then | ^,(2:) | < e for any x in (— r, r).
164 SERIES OF FUNCTIONS
We show now that F does not converge uniformly at x=\
For let
n
Then
and F does not converge uniformly at x — 1, by 139, 2.
Example 3. « 2
Here -^ -^
^B = l + na;2~l+0i +
and F is telescopic. Hence
x,
= 0 , a; = 0.
Thus -,
Let us take
Then
and JP is not uniformly convergent at x = 0. It is, however, in
(—00, co ) except at this point. For let us take x at pleasure
such, however, that | x \ > &. Then
We now apply 136, 1.
Example 4.
GENERAL THEORY 165
Here
. -, T
=x __ *_ n + 1
11 + 7iV 1 +(7i+ I)2*2 J
and F is telescopic. Hence
F _ x O
1 +(W + 1)2^2
in 51 = (-R,
if
1 +
The convergence is not uniform at x = 0.
For set an = — — -. Then
71+1
I ^»(O I = I, does not = 0.
It is, however, uniformly convergent in 51 except at 0. For
(* + l
(n
=-1 + (W + 1)2£2
< e for 7i > some m.
141. Let us suppose that the series F converges absolutely and
uniformly in 51. Let us rearrange F, obtaining the series 6r.
Since F is absolutely convergent, so is 6r and F = (7. We can
not, however, state that 6r is uniformly convergent in 51, as Bocher
has shown.
Example.
jl- 1 +x- x
Here
Hence F is uniformly convergent in 31 = (0, 1).
Let .
Then
166 SERIES OF FUNCTIONS
Let 1
n
Then
as n ==i oo.
-Yi-1^
A ~e)
Hence Gr does not converge uniformly at x = 1.
142. 1. Let /==<#> uniformly in a finite set of aggregates 5lx,
512' '" ^P- Then f converges uniformly in their union (5Ij, ••• 5lp).
For by definition
€ > 0, Bs > 0, |/ - <f> I < 6 a: in 5I8, « in F5a*(r). (1
Since there are only p aggregates, the minimum B of S^ ••• 8p
is > 0. Then 1) holds if we replace 88 by S.
2. The preceding theorem may not be true when the number
of aggregates 5fx, 512 ••• is infinite. For consider as an example
F= 2(1- sX,
which converges uniformly in 51 = (0, 1) except at x = 1. Let
-19.
s s +
Then .F is uniformly convergent in each 5la» but is not in their
union, which is 5l-
3. Letf= </>, g = >/r uniformly in 51.
TAew /± # == <#> ± ^ uniformly.
If ^>, ^ remain limited in 51,
fg = (^^ uniformly. (1
Jf moreover \ ^r \ > some positive number in 51,
•£ = 2. uniformly. (2
The demonstration follows along the lines of I, 49, 50, 51.
GENERAL THEORY 167
4. To show that 1), 2) may be false if <£, ty are not limited.
Let -
/»jap*+*1 2l = (0*, 1), r = 0.
QC>
Then <£ = -^ = - and the convergence is uniform.
x
But 9 .
Let x = t. Then A = 2 as t = 0, and fg does not
uniformly.
Again, let -^
the rest being as before.
Then
But setting x = £,
Al =
= — oo as t == 0
9
and •- does not converge uniformly to -£• •
^ t
143. 1. As an extension of I, 317, 2 we have :
Let
uniformly in 51.
lim^C^ ••• tn) = ij
t=T
Let y ^ rj in F*(T). Then
t=T
= ^(^ ••• «m), uniformly.
The demonstration is entirely analogous to that of I, 292.
uniformly in 51. ie^ ^Ae points
v= (vv vv — vp)
168 SERIES OF FUNCTIONS
form a limited set 33. Let F^ ««• up) be continuous in a complete
set containing 33. Then
Mj ... up) = F(v^^ Vp)
uniformly in 51.
For F, being continuous in the complete set containing 33, is
uniformly continuous. Hence for a given e > 0 there exists a
fixed a- > 0, such that
| F(u) - F(v) <e u in V9(v) , v in 2$.
But as wt = t\ uniformly there exists a fixed 8 > 0 such that
K - vt | < €' , x in 51 , t in
Thus if e' is sufficiently small, w=(wj, — MP) lies in
when x is in 51 and t in Fi*(r).
144. 1. Let
t-r
uniformly in 51. 5T. v ,
lim e7 =
t = T
uniformly in 51, if (j> is limited.
This is a corollary of 143, 2.
-. Let
uniformly in 51. X^^ </> 6g greater than some positive constant in 51-
TOert .. | - ,
lim log/ = log <p,
/=T
uniformly in 51, if c/> remains limited in 51.
Also a corollary of 143, 2.
3. Let f = c£ awe? g = ^ uniformly, as t == T.
ig^ </>, i/r 5g limited in 5t, an(i (/> > some positive number. Then
fa = ^ uniformly in 51. (1
For
GENERAL THEORY 169
But by 2), log/=log$ uniformly in 21; and by 142, 3
g log/ = T/T log <£, uniformly in 21. Hence 2) gives 1) by 1.
145. 1. The definition of uniform convergence may be given a
slightly different form which is sometimes useful. The function
is a function of two sets of variables x and t, one ranging in an 9?m
the other in an 9?n.
Let us set now w = (x^ — xm, ^ — £n) and consider w as a point in
m -f p way space.
As x ranges over 21 and t over FS*(T)> let w range over $&&.
Then
t=T
uniformly in 21 when and only when
e>0, S>0 |/-<£ <e winSBa, 8 fixed.
By means of this second definition we obtain at once the follow
ing theorem:
2. Instead of the variables xl — xm, fj — tn let us introduce the
variables y^ ••• ym, u± ••• un so that as w ranges over 58s,
z = (y1~-ym,u1...un)
ranges over (Es, the correspondence between 33s, (5s being uniform.
Thenf = <f> uniformly in 21 when and only when
e>0, ^>0 \f — <f>\<e , zmgfi, 8 fixed.
3. Example. Let /(#, TI) =
where a /
Then ^(a;) = lim/(2J, ri)=Q , in 21 = (0, oo).
Let us investigate whether the convergence is uniform at the
point x in 21-
First let x > 0. If 0 < a < x < b, we have
170 SERIES OF FUNCTIONS
As the term on the right = 0 as n = oo , we see/=</> uniformly
in (#, b).
When, however, a = 0, or b = oo , this reasoning does not hold.
In this case we set _ nfixp
which gives ^ logl/|3 t
n*&
As the point (x, ri) ranges over £ denned by
z^>0 , n>\,
the point (£, ?i) ranges over a field £ defined by
and the correspondence between £ and X is uniform. Here
The relation 2) shows that when # > 0, t = en asw-^oo; also
when x = 0, £ = 1 for any n. Thus the convergence at x = 0 is
uniform when
H-
The convergence is not uniform at x = 0 when 3) is not satisfied.
For take -,
• -•JL , n=l,2,...
n\/a.
For these values of a: ^
which does not = 0 as n = oo .
146. 1. (Moore, Osgood.) Let
uniformly in 3(. Let a be a limiting point of 21 and
lim/
a-=a
/or gac^ f iw Fi*(r).
c^lim
ar=a
exist and are equal. Here a, T are finite or infinite.
GENERAL THEORY 171
We first show <I> exists. To this end we show that
€>0 , S>0 , | <£<V) - <£<V') | < € a/, x" in F6*(» (1
Now since /(a:, £) converges uniformly, there exists an 77 >0
such that for any x', x" in 51
) (2
On the other hand, since/ = -\/r there exists a & >0 such that
/(*', 0 = ^(0+*'"' (4
for any a;', xrl in V&*(a) ; t fixed.
From 2), 3), 4), 5) we have at once 1). Having established
the existence of <l>, we show now that 3> = \f. For since / con
verges uniformly to 0, we have
/O, 0 - £O) I < | * in « •• , t in r,*(r). (6
Since /= i/r, we have
| /(z, 0 - ^ (0 I < I * in TV (a) , t fixed in F,* (T). (7
Since <f> = <£,
I*W~* <| *»^»(a). (8
Thus 7), 8) hold simultaneously for 8 < 8', 8".
Hence , , , ,s _ - ,
or lim
2. Thus under the conditions of 1)
lim lim / = lim lim / ;
x=a t=r t=r x—a
in other words, we may interchange the order of passing to the
limit.
172 SERIES OF FUNCTIONS
3. The theorem in 1 obviously holds when we replace the un
restricted limits, by limits which are subjected to some condition ;
e.g. the variables are to approach their limits along some curve.
4. As a corollary we have :
Let F = 2/80*i ••• %m) t>e uniformly convergent in 21, of which x = a
is a limiting point. Let Iimf8 = 18, and set L = ^L18. Then
X=(l
Urn F = L ; a finite or infinite,
x=a
or in other words
lim 2jf, = 2 limft.
Example 1.
** %nenx
converges uniformly in 51 = (0, oo) as we saw 136, 2, Ex. 1. Here
iimA=i=?n) I ; '
i -r "? 7 X^ 1 ^
and L = 2L = >, — =1.
Hence lim F(x) = 1.
a?=oo
Also 1
hence R\\mF(x)=Q.
Example 2.
in x
converges uniformly in any interval finite or infinite, excluding
x — 0, where F is not defined. For
and -j y\ 1 _
^ n !
Hence lim JYaf) = e.
GENERAL THEORY 173
Example 3.
= 0 forz=0.
Here limJ(a
*=0
while
Slim /.(*)= 20 = 0.
Thus here Hm
«=0 35 = 0
But F does not converge uniformly at x = 0. On the other
hand, it does converge uniformly at x— ± oo.
Now
and lim 2/n(V> = 2 lim /„(*),
a?=±x at=j-x>
as the theorem requires.
Example 4- Wf \ _ V f nx2 (
--
which converges about x = 0 but not uniformly.
However, iim 2/j|(!B) = 2lim/.(*) = 0.
Thus the uniform convergence is not a necessary condition.
147. 1. Let lim /(^•••^ffl, ^ ••• O = <£0i •••#,») uniformly at
t-T
x = a. Let /(#, f) 6^ continuous at x= a for each t in Vf>*(r).
Then <f> is continuous at a.
This is a corollary of the Moore-Osgood theorem.
For by 146, 1
lim lim /(a + h, f) = lim lim /(a + h,t).
h=o t=r t=r h=o
Hence ]im ^^ + A) = lim /(a, f) = <£(a). Q.E.D.
174 SERIES OF FUNCTIONS
A direct proof may be given as follows :
/(«, 0 = j>(x) + e' | €' | < c, x in
<K^-<K^')=/<X, 0 -/(*", 0+6
But |/(*",0-/<y>0 <* , if I *' -
2. Ze£ F = ^fSl...Sp(^x1 "• xm) be uniformly convergent at x=a.
Let each fSl...Sp be continuous at a. Then F(xl ••• xm*) is continuous
at x = a.
Follows at once from 1).
3. In Ex. 3 of 140 we saw that
= V _ ?L _
is discontinuous at x = 0 and does not converge uniformly there.
In Ex. 4 of 140 we saw that
does not converge uniformly at x = 0 and yet is continuous there.
We have thus the result : The condition of uniform convergence in
1, is sufficient but not necessary.
Finally, let us note that
xa
is a series which is not uniformly convergent at x = 0, although
F(x) is continuous at this point.
4. Let each term, of F= ^fSi...Sp(^x^ ••• xm) be continuous at x = a
while F itself is discontinuous at a. Then F is not uniformly
convergent.
For if it were, F would be continuous at a, by 2.
Remark. This theorem sometimes enables us to see at once
that a given series is not uniformly convergent. Thus 140,
Exs. 2, 3.
GENERAL THEORY 175
5. The power series P= 2aSl...Sm afr ••• x£ is continuous at any
inner point of its rectangular cell of convergence.
For we saw P converges uniformly at this point.
6. The power series P = aQ + a^x + a2x2 + ••• is a continuous
function of x in its interval of convergence.
For we saw P converges uniformly in this interval. In par
ticular we note that if P converges at an end point x = e of its
interval of convergence, P is continuous at e.
This fact enables us to prove the theorem on multiplication of
two series which we stated 112, 4, viz. :
148. Let A = ^ + fli + a^ + _ ^ 5 = &0 + &! + fta + -
0= aQbQ + (fl^j + a^) + (00£2 + afa + a260) + ...
converge. Then AB = C.
For consider the auxiliary series
F(x) = a0 + «!•£ + atf? + —
x) = aQb0 -f (a/1 + aj^x + •"
Since A, E, 0 converge, F, G-, H converge for x = 1, and hence
absolutely for | x \ < 1. But for all | x \ < 1,
Thus L lim H(x) = L lira F(x) • L lira G- (x),
x=\ x=l x = i
or
149. 1. We have seen that we cannot say that/= $ uniformly
although / and <£ are continuous. There is, however, an impor
tant case noted by Dini.
Let f(x± — xm, ^ — tn) be a function of two sets of variables
such that x ranges over 51, and t over a set having r as limiting
point, T finite or ideal. Let
lim/O, f)= </>O) in 51.
t=r
Then we can set ... ,. ,, ^ , , f ,,
f(x,t)=4>(x)++(x, 0-
176 SERIES OF FUNCTIONS
Suppose now | ty(x, t')\<\ ^(x, t) \ for any t' in the rectangu
lar cell one of whose vertices is t and whose center is r. We say
then that the convergence of / to <£ is steady or monotone at x.
If for each x in 51, there exists a rectangular cell such that the
above inequality holds, we say the convergence is monotone or
steady in 51.
The modification in this definition for the case that T is an ideal
point is obvious. See I, 314, 315.
2. We may now state Dini's theorem.
Let /(#! ••• zmi *i •••£»)== 0(^i '•• xm) steadily in the limited com
plete field 51 as t = T; T finite or ideal. Let f and <j> be continuous
functions of x in 51. Then f converges uniformly to <£ in 51.
For let x be a given point in 51, and
We may take t' so near r that | *fy(x, t') \ < -•
o
Let x' be a point in V^x), Then
f(x<, o = <K*') + tO', O-
As /is continuous in #,
Similarly,
Thus
Hence | ./r<V> 0 I < € for any *' in F.O)
and for any £ in the rectangular cell determined by t' .
As corollaries we have :
3. Let Gr = 2 |/tl...ia(^i ••• Xm) | converge in the limited complete
domain 51. Let G- and each ft be continuous in 51- Then G- and
a fortiori F= 2/tl...t- converge uniformly in 51, furthermore ftl...ls= 0
uniformly in 51.
4. Let Gr = 2 |/tl... ^(^ ••• rrm) | converge in the limited complete
domain 51, having a as limiting point. Let Gr and each f, be con-
GENERAL THEORY 177
tinuous at a. Then G and a fortiori F= 2/tl...la converge uniformly
at a.
5. Let 6r = 2 |/tl...ta(^i ••• 2"m) | converge in the limited complete
domain 21, having a as limiting point. Let lim Cr and each lim/"t
x=a x=a
exist. Moreover, let
Then Gr is uniformly convergent at a.
For if in 4 the function had values assigned them at x = a dif
ferent from their limits, we could redefine them so that they are
continuous at a.
150. 1 . Let lim f(xt • • • a?m, ^ • • • tn~) = <£ (x^ • • • xm) uniformly in
t=r
the limited field 21. Xe£ 0 be limited in 21.
lim f/= f<£ = film/.
t=TJ% ju% «/g # = T
For let /=<^ + ^r.
Since /= <#> uniformly |^| < €
for any t in some V*(r) and for any z in 21.
Thus
Remark. Instead of supposing <£ to be limited we may suppose
that/(#, t) is limited in 21 for each t near r.
2. As corollary we have
Let lim/(^1 ••• xm, tl"-t^) = <t>(%i •- xni) uniformly in the limited
t=T
field 2L Letf be limited and integrable in 2f for each t in Tr5*(r).
Then <f> is integrable in 21 and
lim f/= f*= flim/.
«=T ^$1 ^21 ^21 t=r
3. From 1, 2, we have at once:
Let F =^flv..Ls(xl"-xm) be uniformly convergent in the limited
field 21- Let each /tl...ls be limited and integrable in 2(. Then F is
integrable and
r» /»
I F— ^ ( f
I •* — ~ I /»!•-«•
Jgt ^
178 SERIES OF FUNCTIONS
If the f^...^ are not integrable, we have
Example. J&7— V
~
does not converge uniformly at x = 0. Cf . 140, Ex. 3.
Here _ J_
and jYrN fl for
~ 10 for x = 0.
Hence
'*
'*.*= 1- /'
*/o 1
+ nx*
— 1 _ arctg Vn ^_ .j
Thus we can integrate F termwise although F does not converge
uniformly in (0, 1).
151. That uniform convergence of the series
with integrable terms, in the interval 51 = (a < 5) is a sufficient
condition for the validity of the relation
Xb fb pb
Fdx = \ f^dx + I fjdx + • • •
is well illustrated graphically, as Osgood has shown.*
Since 1) converges uniformly in H by hypothesis, we have
and
| Fn (x) | < e n > m (3
for any x in 51.
* Bulletin Amer. Math. Soc. (2), vol. 3, p. 59.
GENERAL THEORY
179
In the figure, the graph of F(x) is drawn heavy. On either
side of it are drawn the curves F— e, F -\- e giving the shaded
band which we call the e-band.
From 2), 3) we see that the graph of
each Fn< n>m lies in the e-band. The
figure thus shows at once that
and
X
_
can differ at most by the area of the a
e-band, i.e. by at most
Cb'2edx = 2e(b-
•Jo,
152. 1. Let us consider a case where the convergence is not
uniform, as
F(x^ -^ [nx _ (n-l)x\ _
•* - -
Here
nx
If we plot the curves y = Fn(x), we observe that they flatten
out more and more as n == oo, and approach the #-axis except
near the origin, where
they have peaks which
increase indefinitely in
height. The curves
Fn(x), n>m, and m suf
ficiently large, lie within
an e-band about their
limit F(x) in any inter
val which does not in
clude the origin.
If the area of the
region under the peaks
could be made small at
pleasure for m sufficiently large, we could obviously integrate
termwise. But this area is here
180 SERIES OF FUNCTIONS
= - as n = QO .
Thus we cannot integrate the F series termwise.
2. As another example in which the convergence is not uniform
let us consider
Here
-r, nx
The convergence of F is uniform in 51 = (0, 1) except at x = 0.
The peaks of the curves Fn(x) all have the height e~l.
Obviously the area of the
region under the peaks can be
made small at pleasure if m is
taken sufficiently large. Thus
in this case we can obviously
integrate termwise, although
the convergence is not uniform
in 21.
We may verify this analytically. For
CXFndx= CX—dx = --1 + nx=Q asn = oo.
Jo Jo enx n nenx
3. Finally let us consider
Here , n**
1 + *M
The convergence is not uniform at x = 0.
The peaks of Fn(x) are at the points x = »"*, at which points
Fn = l -Vn.
GENERAL THEORY 181
Their height thus increases indefinitely with n. But at the
same time they become so slender that the area under them = 0.
In fact
We can therefore integrate termwise in (0 < a).
153. 1. Let Urn G-(x, ^ ... fn) =g(x) in 51 = (a, a + $), T /m'te
<=T
or infinite. Let each G-'x(x, t) be continuous in 51 ; also let Gr'x(x, £)
converge to a limit uniformly in 51 as t = r. Then
Urn a'x(x,t)=g'(x) m5l, (1
t=T
and g\x} is continuous.
For by 150, 2,
By I, 538, ~,
I (jrxdx = CrCx, t^) — GrCa* £).
*/ a
Also by hypothesis, ^ { & ^ f} _ &^ ()l=ffi
Hence
g(x)-g(a) = film ^(a;, «)<£«. (2
*^a /=T
But by 147, l, the integrand is continuous in 51.
Hence by I, 537, the derivative of the right side of 2) is this in
tegrand. Differentiating 2), we get 1).
2. Let F(x) = 2/tl ... lt(x) converge in 51 = (a, a + S). Let each
f'(x) be continuous, also let
2/; (*)
be uniformly convergent in 51. Then
l?'(ar) = 2//(aj), in %.
This is a corollary of 1.
182 SERIES OF FUNCTIONS
3. The more general case that the terms /4...ls are functions of
several variables xv ••• xm follows readily from 2.
154. Example.
Here
a function whose uniform convergence was studied, 145, 3. We saw
F(x) = 0 for any x > 0.
Hence F'(x) = 0 x>0.
Let (?O) = ?//,(*)• (1
hence J"(^)=2/^), (2
and we may differentiate the series termwise.
If a; = 0, and «=1, X>0; (7n(0) = - wx = - oo as n = oo.
In this case 2) does not hold, and we cannot differentiate the
series termwise.
For x=Q, and a>l, 6rn(0)=0, and now 2) holds; we may
therefore differentiate the series termwise. But if we look at the
uniform convergence of the series 1), we see this takes place only
when «-l X
/S //
155. 1. (Porter.) Let F^=^ ^
converge in 51 = (a, J). For every x in 51 let \fl(x)\ < g^ a constant.
Let G- = ^g, converge. Then F(x) has a derivative in 51 and
F'W^-Sf „.....(*) i C1
or we may differentiate the given series termwise.
GENERAL THEORY 183
For simplicity let us take 8 = 1. Let the series on the right of
1) be denoted by <#>(V). ^or eac^ x m ^ we snow
e>0, .,., _ ^
< e, I Az I < S.
For
where fn lies in V?>(x
Thus D=.
i
But G- being convergent, 6rm < e/3 if m is taken sufficiently large.
Hence
|J>»I< 2 i/;a)l+ 2 |/;c*)|< 2 <?.<}«.
ff»+l m-fl
On the other hand, since =« ==/»(^) and since there are only ?w
Az
terms in Z)m, we may take 5 so small that
|l>J<e/8.
Thus |^>
2. Example 1. Let
TO---WtfO ->l 0
This series converges uniformly in 51 = (0 < 6), since
I /» s \ I
Also x -^^ ^n
-.•
n ! (1 + «r
Hence
As l^rn converges, we may differentiate 1) termwise. In
general we have
valid in 51.
184 SERIES OF FUNCTIONS
3. Example 2. The fl functions.
These are defined by
0j (x) = 2 i ( - 1) Yn+i)2 sin (2 n + 1) TO
o
= 2 jl sin THE — 2 2* sin 3 THE -f
,V2 (a;) == 2 l^n+=)2 cos (2/1 + 1) TTX
0
= 2 <?* cos vrx + 2q* cos 3 TT.E +
,98 (a:) = 1 4- 2 2?n2 cos 2
= 14-2^ cos 2 ?ra; + 2 ^* cos 4 TTX -f-
»V0 (a;) = 1 H- 2 ( - l)n£n2 cos 2
i
= 1 — 2 <7 cos 2 THE + 2 g-4 cos 4 THE — • • •
Let us take , . -^
Then these series converge uniformly at every point x. For
let us consider as an example »vr The series
is convergent since the ratio of two successive terms is
9*
and this = 0. Now each term in »vx is numerically
<\q (n+^< q\n\
and hence < the corresponding term in T.
Thus &i(x) is a continuous function of x for every x by 147, 2.
The same is true of the other .(/'s. These functions were discovered
by Abel, and were used by him to express the elliptic functions.
Let us consider now their derivatives.
Making use of 155, 1 it is easy to show that we may differentiate
these series termwise. Then
d ' O) = 2 -IT! ( - l)n(2 n + 1) 0<»+*>* cos (2 w + 1) TTX
o
= 2?r(g* cos TTX — 3<2* cos 3 TTX + •••).
GENERAL THEORY 185
^ (z) = _ 2 TT (2 n + 1) 2<"+*)2 sin (2 71 + 1) TTZ
o
= — 2^(9! sin TT# •+- 3 q$ sin 3 THE + •••)•
00
#g' (a;) = — 4 7r2 w^n2 sin 2 71772;
i
= — 4 TT (5- sin 2 TTZ + 2 ^* sin 4 TTX + • • •) .
,^ (a;) = - 4 Tr ( - l)nnq^ sin 2
i
= + 4 TT (^ sin 2 TTX — 2q* sin 4 TTX + •••).
To show the uniform convergence of these series, let us con
sider the first and compare it with
The ratio of two successive terms of this series is
,2n+1
which = 0. Thus S is convergent. The rest follows now as
before.
156. 1. Let
t=r h
uniformly for 0 < | A | < 17, T finite or infinite.
Let
'x(a, 0 ftw«*
/or each t near r. Then g' (a) exists and
This is a corollary of 146, 1. Here
&O + A, t)-GKa, 0
A
takes the place of f(x, £).
2. From 1 we have as corollary :
186 . SERIES OF FUNCTIONS
converge for each x in 51 which has x = a as a proper limiting point.
Letf((a) exist for each i = (iv ••• tn). Let
converge uniformly with respect to h. Then
CHAPTER VI
POWER SERIES
157. On account of their importance in analysis we shall
devote a separate chapter to power series.
We have seen that without loss of generality we may employ
the series a0 + alX + a^ + ... (1
instead of the formally more general one
aQ + al(x — a)+ a^x-a)2-\- -
We have seen that if 1) converges for x = c it converges abso
lutely and uniformly in (—7, 7) where 0 < 7 < | c |. Finally,
we saw that if c is an end point of its interval of convergence, it
is unilaterally continuous at this point. The series 1) is, of course,
a continuous function of x at any point within its interval of
convergence.
158. 1. Let P(x) = a0 + a^x + a2x2 + ••• converge in the interval
$1 = (— «, «) which may not be complete. The series
P.-l .' 2 • ... M. + 2 • ft • — (f»+ l)an+lx + -
obtained by differentiating each term of P n times is absolutely and
uniformly convergent in $8 = (— A /3), ft< a, and
P.C?), ins.
For since P converges absolutely for x = /3,
an{3n<M, n=l,2,...
Let now x lie within 33. Then the adjoint series of P^(x} is
Now its mih term
j _ mctmQm f g\m~l . mM
& \£) • ft\ft
187
188 POWER SERIES
But the series whose general term is the last term of the pre
ceding inequality is convergent.
2. Let P=a0 + a1x + a%xz + •••
converge in the interval 21. Then
where «, x lie in 21. Moreover Q considered as a function of x con
verges uniformly in 21.
For by 137, P is uniformly convergent in (a, x). We may
therefore integrate term wise by 150, 3. To show that Q is uni
formly convergent in 21 we observe that P being uniformly con
vergent in 21 we may set
where i D" i n i
I Pm I < <r, m>m0, o- small at pleasure.
Then
on taking <r sufficiently small.
159. 1. Let us show how the theorems in 2 may be used to
obtain the developments of some of the elementary functions in
power series.
The Logarithmic Series. We have
1 — x
for any x in 21 = (- 1*, 1*). Thus
l-x
Hence
This gives also
a; = z- + - - ; a; in
GENERAL THEORY 189
The series 1) is also valid for x = 1. For the series is conver
gent for x = 1, and log (1 + x) is continuous at x = 1. We now
apply 147, 6.
For x = 1, we get
2. The Development of arcsin x. We have by the Binomial
for x in 51 = (-!*, 1*). Thus
It is also valid for x = 1. For the series on the right is conver
gent for x — 1. We can thus reason as in 1.
For x = 1 we get
Zr_ 1 1.3 1.3.5
2~ ^2-3 2.4.5 2.4.6-7
3. The Arctan Series. We have
_J — =l_^ +
1+ x2
for x in 51 = (- 1*, 1*). Thus
X . X xo
= *~3-+5-
valid in 31. The series 3) is valid for x = 1 for the same reason as
in 2.
For x = 1 we get TT _ -, _ 1 , 1 _ 1 ,
4" 3 + 5 7~^
4. The Development of ex. We have seen that
converges for any #. Differentiating, we get
190 POWER SERIES
Hence E'(x)=E(x) (a)
for any x. Let us consider now the function
by (a). Thus by I, 400,/(V) is a constant. For x = 0,/(V) = 1.
Hence r ^ ~3
?x — 1 4- — -4- — 4- -|- .
hl! + 2! + 3!^
valid for any #.
5. Development of cos #, sin x.
The series 9 d r
(7=1-^ + ^-^+...
2 ! ^ 4 ! 6 !
converges for every x. Hence, differentiating,
Hence adding, (7+ (7" = 0. (b)
Let us consider now the function
f(x) = (7 sin x + C" cos #.
Then
sn ^ _ sn ^ + cos a;
= ((7+ <7")cosa;
= 0 by (b).
Thus /(V) is a constant. But (7=1, C" = 0, for a; = 0, hence
/(*)-o,
or C sin a: + C' cos 2;= 0. (c)
In a similar manner we may show that
or #0*0= Ccosx— 0' sin x= 1. (d)
GENERAL THEORY 191
If we multiply (c) by sin x and (d) by cos x and add, we get
0= cos x. Similarly we get C' — — sin x. Thus finally
..
valid for any #.
160. 1. Let P = amxm + am+1xm+1 + ••• , am =£ 0, converge in
some interval 51 about the origin. Then there exists an interval
53 < 51 in which P does not vanish except at x = 0.
= xmQ.
Obviously Q converges in 51. It is thus continuous at x = 0.
Since Q =£ 0 at x = 0 it does not vanish in some interval 33 about
3=0 by I, 351.
In analogy to polynomials, we say P has a zero or root of order
m at the origin.
2. Let P — aQ 4- a^x + a2z2 4- ••• vanish at the points bv J2, ... = 0.
Then all the coefficients an = 0. The points bn are supposed to be
different from each other and from 0.
For by hypothesis P(bn) = 0. But P being continuous at x = 0,
Hence P(0)=0,
and thus
Hence
p =
Thus P1 vanishes also at the points bn. We can therefore
reason on Pl as on P and thus a1 = 0. In this way we may
continue.
3. If P = a0 + aiz+-
192 POWER SERIES
be equal for the points of an infinite sequence B whose limit is x = 0,
then an= bn, n = 0, 1, 2 «••
For P — Q vanishes at the points B.
Hence i A A -i o
«n- bn = v , w = 0, 1, 2-..
4. Obviously if the two series P, § are equal for all x in a
little interval about the origin, the coefficients of like powers are
equal; that is ^ = ^ ? n = 0,l,2...
161. 1. Let y = ao + <v; + v3+... (1
converge in an interval 51. As # ranges over 51, let i/ range over
an interval $&. Let
'" (2
converge in 35. Then z may be considered as a function of x de
nned in 51. We seek to develop z in a power series in x.
To this end let us raise 1) to the 2°, 3°, 4° ... powers ; we get
series „ o
y = ^20 + a2lx + ^22^ + "•
wx* +'~ (3
which converge absolutely within 51.
We note that amn is a polynomial.
in «0, ••- #n with coefficients which are positive integers.
If we put 3) in 2), we get a double series
D= (b + b) + b^x -f b^x2 +
a21'7; + *2a222'2 +
si^ + V82232 +
+ ..... . V •
V20
If we sum by rows, we get a series whose sum is evidently 0,
since each row of D is a term of z. Summing by columns we get
a series we denote by
C = t'0 + cx + czx* + • • • (5
GENERAL THEORY 193
cl = blal + 62«21 4- Vsi -! ---- (6
We may now state the following theorem, which is a solution of
our problem.
. Let the adjoint y-series,
converge for £ = £0 to the value T) = 7/0. Let the adjoint z series
- (8
converge for 77 = ?;0. T^e/i the z series 2) caw be developed into a
power series in x, viz. the series 5), which is valid for \ x \ < £0.
For in the first place, the series 8) converges for 77 <.TJO. We
show now that the positive term series
converges for £ < f0. We observe that ^) differs from Adj D,
at most by its first term. To show the convergence of X) we
have, raising 7) to successive powers,
7,2 =
We note that Amn is the same function Fm%n of «0, «j, ••• «n as
amn is of a0, ••• aB, i.e.
As the coefficients of Fm^ n are positive integers,
194 POWER SERIES
Putting these values of ?;, ??2, rf ••• in 8), we get
A = 8 +
Summing by rows we get a convergent series whose sum is ?
or 8). But this series converges for £ < f0 since then rj<rj^
and 8) converges by hypothesis for rj = IJQ. Now by 9) each
term of £) is < than the corresponding term in A. Hence £)
converges for f < £0.
2. As a corollary of 1 we have :
Let
y = c
converge in H, and
converge for all — oo < y < -f °o« 2%ew 2 can 5e developed in a
poiver series in x,
z— cQ + 0^+ c%x2 + ... = O
for all x within 51.
3. Let the series
y = amxm -f am+1xm+l H ---- , w > 1
converge for some x > 0. T/ fAg series
z = bQ+bly + btf* + •••
converges for some y > 0, i'£ caw ft^ developed in a power series
Z= CQ+ CjX + C2X2-\ ----
convergent for some x > 0.
For we may take f = | x \ > 0 so small that
V = «mfm + «»+if ™+1 H- -•
has a value which falls within the interval of convergence of
4. Another corollary of 1 is the following :
y= a9 + a]x + a^x2 + ...
GENERAL THEORY 195
converge in 21 = (— A, A). Then y can be developed in a power
series about any point c of 21,
which is valid in an interval 53 whose center is c and lying within 21.
162. 1. As an application of the theorem 161, l let us take
s^ + jL + ^ + r +
i ! 2 r 3 r
_ z__z^ , ^__
y 1 ! 3 ! 5 !
As the reader already knows,
z = ev , y — sin a?,
hence z considered as a function of x is
z =
We have
z = 1 + z + 0 • z2 - i r3 + 0.*4 + T|o ^+ 0 -a6 4-
+ -i.^+ 0 - \x* + 0 +^+..
H-j^-h 0 -iV^5 + 0 + ..
Summing by columns, we get
Z=^n*=1 + ^+j^_i^__l_^5__l_^...
2. As a second application let us consider the power series
z =
convergent in the interval 21 = (- R, R). Let z be a point in 21.
Let us take 77 > 0 so small that y = x + h lies within 21 for all
|*|<L*
Then ,.
z = a0 + flj (x + h)
+ a2 (z2 + 2 zA + A2)
196 POWER SERIES
This may be regarded as a double series. By 161, l it may be
summed by columns. Hence
P(x + h) = «0 + ap + a2x2 + a&* + •••
+ A(at + 2 «2z + 3 OgZ2 + •••)
+ |l(2«a + 2.3a^ + 3.4fl42Ja + •••) (2
+ ............
= P(x)+hP>(x)+^P"(x) + j;P'»W+.~ (3
on using 158, 1.
This, as the reader will recognize, is Taylor's development of
the series 1) about the point x. We thus have the theorem :
A power series 1) may be developed in Taylor s series 3) about
any point x within its interval of convergence. It is valid for all h
such that x+ h lies within the interval of convergence 0/1).
163. 1. The addition, subtraction, and multiplication of power
series may be effected at once by the principles of 111, 11:2. We
have if P = a + a
converge in a common interval 31 :
P+ Q = (aQ + b()) + (a1 + b1)x
P - Q = («o - Jo) + Oi -
P ' Q = «0^o + (aibo + aob
These are valid within SI, and the first two in 31.
2. Let us now consider the division of P by R. Since
£-P.I
R R
the problem of dividing P by R is reduced to that of finding the
reciprocal of a power series.
P = a ax a ••• « ^ 0
converge absolutely in R =(— -#,
§ = ^2; +
fo numerically < \ a0 1 m 33 = ( — r, r) r < R.
GENERAL THEORY 197
Then 1/P can be developed in a power series
1
P
valid in 53. The first coefficient CQ = — .
ao
For 1111
p
for all x in 53. We have now only to apply 161, 1.
3. Suppose
To reduce this case to the former, we remark that
Then ±^\_ I
P~ xm ' Q'
But \/Q has been treated in 2.
164. 1. Although the reasoning in 161 affords us a method of
determining the coefficients in the development of the quotient of
two power series, there is a more expeditious method applicable
also to many other problems, called the method of undetermined
coefficients. It rests on the hypothesis that /(#) can be developed
in a power series in a certain interval about some point, let us say
the origin. Having assured ourselves on this head, we set
f(x) = aQ + a^x + atf? + •••
where the a's are undetermined coefficients. We seek enough
relations between the a's to determine as many of them as we
need. The spirit of the method will be readily grasped by the
aid of the following examples.
Let us first prove the following theorem, which will sometimes
shorten our labor.
198 POWER SERIES
2' ^ f(x~) = a0 + alx + a<ix2 + •••; -E<x<R, (1
is an even function, the right-hand t<ide can contain only even powers
of x; iff(x) is odd, only odd powers occur on the right.
For if/ is even, /(») -/<-•). (2
But f(-x) = a0-a1z+<V*- ... (3
Subtracting 3) from 1), we have by 2)
0 = 2(a1a; + a#? + a#? + ..-)
for all x near the origin. Hence by 160, 2
«i = «8=«6= ••' =°-
The second part of the theorem is similarly proved.
165. Example 1. /(a-) = tan a.
Since gin 3.
tan x = — — ,
cos x
and ^
we have ^3 ^
Since cos x > 0 in any interval 33 = f — - + S, — — S j , S > 0, it
follows that | § | < 1 in 33.
Thus by 163, 2, tana; can be developed in a power series about
the origin valid in $&. We thus set
tan x = a^c. + fl^r3 + a^ -f ••• (2
GENERAL THEORY 199
since tana: is an odd function. From 1), 2) we have, clearing
fractions,
Comparing coefficients on each side of this equation gives
- ' a -
3 2!~ FT 3~3'
a a.
r Q ,
a- -- ^H — a -- 1= -- . /.«_
7 2! 4! 6! 7! 7 315
1 17
7!
__ __. __
9 2!4! 6! 8! 9!' 9 2835
Thus 21
(3
•)
Example 2.
sin x
\
Since
we see that I O I < 1
when 2; is in ^=(-7r + S, TT-^), 3>0. Thus xf(x) =
can be developed in a power series in 58. As f(x) is an odd
function, xf(x} is even, hence its development contains only even
powers of x. Thus we have
xf(x) = rt0 4- a^ -}- a±x* H ----
200 POWER SERIES
Hence
a - + -t _ s + ,
3!5! 7I> 8 3! 5! 7!+9!
Comparing like coefficients gives
q-|
8 ! 5 ! 7 ! 3 • 7
Thus 111 7 QI
sina; a; 6 360 3-7!
valid in (— TT*, TT*).
166. Let J<>) =/!(*) +/.CO+ -
/.(*) = «n0 + «.i* + «n2^2 + - » = 1, 2
Let the adjoint series
(4
converge for % = R and have </>n as sums for this value of f .
converge. Then ^ converges uniformly in H = (— ^2, ^2) and F
may be developed as a power series, valid in 21, by summing by
columns the double series
GENERAL THEORY 201
F converges uniformly in 21. For as \x\ < f,
4 «nl + «n2 + — = <V
We now apply 136, 2 as 2$n is convergent for % = R.
To prove the latter part of the theorem we observe that
« 4 « R 4 «
4
is convergent, since summing it by rows it has <E> as sum. Thus
the double series 1) converges absolutely for |z|<f, by 123, 2.
Thus the series 1) may be summed by columns by 130, l and has
F(x) as sum, since 1) has F as sum on summing by rows.
167. Example.
This series we have seen converges in 51 = (0, 5), b positive and
arbitrarily large.
Since it is impossible to develop the fn(x) in a power series about
the origin which will have a common interval of convergence, let
us develop F in a power series about XQ > 0.
We have
1 1 1
= F~W 1 1 ~ ^l^a^ + (14 o»O2 ~ '" J
where A (—!)"«""
202 POWER SERIES
Thus F give rise to the double series
D = A' - *
where ,, _(— l)n A
An* — - A.n K .
The adjoint series to/n(a:) is, setting f = [a — a;0|,
, ^ ^2-P
T n \.9 / i I -j
nl \1 + c
This is convergent if
^ a n < 1 or if
that is, if 02
For any x such that #0<#<2#0 , £ = 2
Then for such an x
*"=;ni+a-(L -
and the corresponding series
is evidently convergent, since <£n <
7Z 1
We may thus sum D by columns ; we get
K=0
where 0 ^r_l)n+K an«
The relation 1) is valid for 0 < x < 2 :r0.
GENERAL THEORY 203
168. Inversion of a Power Series.
Let the series » . 7 . . T ,2 . si
v = o0 + b1t + If 4- • •- (1
have 6X =£ 0, and let it converge for t = tQ. If we set
t = xt0, U=V7+ °*
°ico
it goes over into a series of the form
u = x — a^x2 — a33? — ••• (2
which converges for x = 1. Without loss of generality we may
suppose that the original series 1) has the form 2) and converges
for x— 1. We shall therefore take the given series to be 2). By
I, 437, 2 the equation 2) defines uniquely a function x of u which
is continuous about the point u = 0, and takes on the value x= 0,
for u = 0.
We show that this function x may be developed in a power
series in w, valid in some interval about u = 0.
To this end let us set
x = u 4- c2u? 4- c3ifi 4- ••• (3
and try to determine the coefficient <?, so that 3) satisfies 2)
formally. Raising 3) to successive powers, we get
x2 = u? 4- 2 c&* 4- (*22 +' 2 <?3>4 + (2 c± + 2 <y3X + -
a* = M3 + 3 ^4 + (3 ^2 + 3 <gw5 + ... (4
x^ = u* -f 4 £2w5 -f- • • •
Putting these in 2) it becomes
u = u + (ea - <z2)w2 -f (<?8 - 2 a2c2 - a3)w3
+ (*4 - «2(^22 4- 2 c8) - 3 «3^2 - a4)w*
+ (<?6 - 2 a2(c4 + ^2^3) - 3 a3(6-22 4- <?8) - 4 «4^2 - a6) w5 (5
4-
Equating coefficients of like powers of w on both sides of this
equation gives
c2 — #2
c3 = 2 a2e2 4- «8
^4 = «2(^22 + 2 c8) 4- 3 a^ 4- «4 (6
* = 2 «^ + CC + 3 ^^2 + * + 4 ^^ + « •
204 POWER SERIES
This method enables us thus to determine the coefficient c in
3) such that this series when put in 2) formally satisfies this
relation. We shall call the series 3) where the coefficients c have
the values given in 6), the inverse series belonging to 2).
Suppose now the inverse series 3) converges for some wo9bO;
can we say it satisfies 2) for values of u near the origin ? The
answer is, Yes. For by 161, 3, we may sum by columns the
double series which results by replacing in the right side of 2)
r rr2 <r&
Ox, ^/ , ^/ ,
by their values in 3), 4). But when we do this, the right side of
2) goes over into the right side of 5), all of whose coefficients
= 0 by 6) except the first.
We have therefore only to show that the inverse series con
verges for some u =£ 0. To show this we make use of the fact that
2) converges for x = 1. Then an = 0, and hence
|fln|<some« ft = 2, 3,... (7
On the other hand, the relations 6) show that
*n=/»(«2, «8»— *») (8
is a polynomial with integral positive coefficients. In 8) let us
replace #2, az>" by a, getting
7»=/n(a» «> — «)• (9
Obviously | cn \ < yn. (10
Let us now replace all the a's in 2) by a ; we get the geometric
series 2^4 x-n
u = x — ax? — ax3 — ax* — ... (11
The inverse series belonging to 11) is
x = u + y2u2 + 73^3 + 74^4 H — (13
where obviously the y's are the functions 9).
We show now that 11) is convergent about u = 0. For let us
solve 12) ; we get
_ 1 + u + VI - 2(2 q +
2(1
GENERAL THEORY 205
Let us set 1 — 2(2 a -f l)w •+- u2 = 1 — v. For u near u = 0,
< 1. Then by the Binomial Theorem
VI - v = 1 + d^v + d2v2 H ----
Replacing v by its value in w, this becomes a power series in u
which holds for u near the origin, by 161, 3. Thus 14) shows that
x can be developed in a power series about the origin. Thus 13)
converges about u = 0. But then by 10) the inverse series 3)
converges in some interval about u = 0.
We may, therefore, state the theorem :
Let U==b + b1x + b2x* + b3x* + ... , fl^O, (15
converge about the point x = 0. TJien this relation defines x as a
function of u which admits the development
x=(u- 5)(i +al(u-b) + a2(u-b)*+ ••• }
I o\ j
about the point u = b. The coefficients a may be obtained from 15)
by the method of undetermined coefficients.
Example. We saw that
T2 r3 r4 ^5
M = log(l + ^) = ^-|+|-| + |-- (1
If we set
u = x + tf2z2 + asx3 + a4z4 + ••• (2
we have l l i _ i
«2 = -l , rt3 = l , «4=-t , «5=i"
If we invert 2), we get
x = u -f- ^2'w2 + c>3 ^3 + • • •
where c's are given by 6) in 168. Thus
- ^= - Ki + 2 • D+3 •*•*-* = - A- ••• ^ = A-
-^=2(-i)(-2L + |.i)+3
= -120- •*• Co =120'
206 POWER SERIES
Thus we get
*=M+S+S+£+£+
But from 1) we have
i+*='"=i + n+f^+-
which agrees with 3).
Taylor s Development
169. 1. We have seen, I, 409, that if f (x) together with its
first n derivatives are continuous in 21 = (a < 6), then
+ ^/""O + OK) (1
Where a<a + A<6 , 0<0<1.
Consider the infinite power series in h.
^=/(«) + /j/'(*) + f*/>)+ - (2
We call it the Taylor s series belonging to f(x). The first n
terms of 1) and 2) are the same. Let us set
JZ.= ^/*0»
IV •
We observe that Rn is a function of n, 7i, a and an unknown
variable 0 lying between 0 and 1.
We have ... , ,, m , r>
/(a -f A) = rn + ^u.
F"rom this we conclude at once':
•
//" 1°, f(x) and its derivatives of every order are continuous in
51 = (a, 6), and 2°
limJRn = lim -^/(B)(a+ ^A) = 0 , n = ac, (4
TAYLOR'S DEVELOPMENT 207
Then 7 TO
/(« + A) =/(") + -/(«)+ |j/"O) + - (5
The above theorem is called Taylor's theorem; and the equa
tion 5) is the development of f(x) in the interval 51 by Taylor's
series.
Another form of 5) ?s
/ O) =/(«) + ZLo) + ^-/"(a) + ... (6
When the point a is the origin, that is, when a = 0, 5) or 6)
gives 2
/(*) =/(°) + */'(<>) + fr/"(0) + - (7
This is called Maclauriris development and the right side of 7)
Maclaurirfs series. It is of course only a special case of Taylor's
development.
2. Let us note the content of Taylor's Theorem. It says :
If 1° f(x) can be developed in this form in the interval
2l = (a<&);
2° if f(x) and all its derivatives are known at the point
x = a ;
then the value of / and all its derivatives are known at every
point x within 51.
The remarkable feature about this result is that the 2° condi
tion requires a knowledge of the values of f(x) in an interval
(a, a -h S) as small as we please. Since the values that a func
tion of a real variable takes on in a part of its interval as (a < e),
have no effect 011 the values thatjf(#) may have in the rest of the
interval (c < 6), the condition 1° must impose a condition on f(x)
which obtains throughout the whole interval 51-
v
170. Let f(x) be developable in a power series about the point a,
viz. let
Then ,(n)
"n ~n^ ?i = 0, 1, ... (2
i.e. the above series is Taylor's series.
208 POWER SERIES
For differentiating 1) n times, we get
/»>(*) = »!«„ + "±11 a^(x - a) + ...
Setting here x= a, we get 2).
The above theorem says that if f(x) can be developed in a
power series about x = a, this series can be no other than Taylor's
series.
171. 1. Let f(n\x) exist and be numerically less than some con
stant M for all a < x < 6, and for every n. Then f(x) can be
developed in Taylor s series for all x in (a, 5).
For then I 7? I M ''"
n !
But obviously ^. hn _ ^
n=x> n !
2. The application of the preceding theorem gives at once :
(3
which are valid for every x.
SinCe
we have
«*=l+*l°£« + *21^+... (4
valid for all x and a > 0.
172. 1. To develop (1 + #)'x awe? log (1 + #) we need another
expression of the remainder 72n due to Cauchy. We shall con
duct our work so as to lead to a very general form for Rn.
From 169, 1 we have
TAYLOR'S DEVELOPMENT 209
We introduce the auxiliary function defined over (a, £>).
KO =AO +/' (00* -0 +
Then </(*)=/(*)
and
Hence . «.=K*)-K«0. • (2
(f we differentiate 1), we find the terms cancel in pairs, leaving
-/no- (3
We apply now Cauchy's theorem, I, 448, introducing another
arbitrary auxiliary function &(x) which satisfies the conditions
of that theorem.
Then 9(x)-g(a) = /(g), a < c < x.
Using 2) and 3), we get, since x = a + A,
where 0 < ^ < 1.
2. If we set (?<» = (6-*)",
we have a function which satisfies our conditions. Then 4) becomes
a formula due to Schlomilch and Roche.
For /A = 1, this becomes
A-O-^
r> — 1 .
which is Cauchy's formula.
210 POWER SERIES
For IJL = n, we get from 5)
n l
which is Lagrange's formula already obtained.
k
173. 1. We consider now the development of
(1 + x)* x ^ — 1 , p arbitrary.
The corresponding Taylor's series is
We considered this series in 99, where we saw that :
T converges for | x \ < 1 and diverges for | x \ > 1.
When x = 1, T converges only when /*>—!; when x = — 1^
^converges only when ft^O.
We wish to know when
The cases when T diverges are to be thrown out at once. Con
sider in succession the cases that T converges. We have to
investigate when lim Rn = 0.
Case 1°. 0< | x \ < 1. It is convenient to use here Cauchy's
form of the remainder. This gives
1 . 2 • •••
setting .,
_^.M-1 . ...M-M + 1
*"
hence lim TFn = 0.
TAYLOR'S DEVELOPMENT 211
x,
which is finite. Hence Un is < some constant M.
To show that Km Sn = 0, we make use of the fact that the series
T converges for the values of x under consideration. Thus for
every /*
lim J"M-l--M-n + 2 ^ =
1 • 2 - ... n— 1
since the limit of the nih term of a convergent series is 0. In
this formula replace /-t by ^ — 1, then
1-2 n-l fMx
Hence ,. 0 A
hm Sn = 0.
Thus n . D
hm Rn = 0.
Hence 1) is valid for | x j < 1.
Case 2. x= 1, IJL>— 1. We employ here Lagrange's form of
the remainder, which gives
±1(1 + *)*-»
1.2- ...
setting
TT _
1.2. ...
Consider Wn. Since /i increases without limit, fi — n becomes
and remains negative. As 6 > 0
lim Wn = 0.
For Z7B, we use I, 143. This shows at once that
lim Un = 0.
Hence ,. ,,
hm Rn = 0
and 1) is valid in this case, i.e. for x— 1, /x >— 1.
212 POWER SERIES
Case 3. x = — 1, ft ^> 0. We use here for p > 0 the Schlomilch-
Roche form of the remainder 172, 5) . We set a = 0, h = — 1 and get
( _ ly/*-1 • /*-2 • .../x-rc+1
> l,2...-n-l
Applying I, 143, we see that lim -Rn = 0.
Hence 1) is valid here if p > 0.
When /i = 0 equation 1) is evidently true, since both sides
reduce to 1.
Summing up, we have the theorem :
The development of (1 + x)^ in Taylor's series is valid when
| x | < 1 for all /x. When x = -h 1 it is necessary that /a > — 1 ;
when x= — 1 it is necessary that fJi^>0.
2. We note the following formulas obtained from 1), setting
x = 1 and — 1.
174. 1. We develop now log (1 + #). The corresponding
Taylor's series is
* + 1 ~~ 2" + ¥ "
We saw, 89, Ex. 2, that T7 converges when and only when
\x\<\ or x = 1.
Let 0 < x < 1. We use Lagrange's remainder, which gives here
=
n
Thus
Hence lim Rn = 0.
TAYLOR'S DEVELOPMENT 213
Let _ l < x < 0. We use here Cauchy's remainder, which
gives, setting x = — £, 0 < f < 1,
if
Evidently
Also
Finally 1 — 0
lim TFn = 0 since < 1.
1 — 6%
We can thus sum up in the theorem :
Taylor s development of log (1 -f x) is valid when and only when
x\ < 1 or x = 1. That is, for such values of x
2. We note the following special case :
i - * + \ - \ + - = log 2.
The series on the left we have already met with.
175. We add for completeness the development of the follow
ing functions for which it can be shown that lim Rn = 0.
which is valid for (—1, 1).
arctan x == x \- — — —+... (2
which is valid for (— 1*, 1).
1 a* 1 . 3 *• 1.3.52?
which is valid for (— 1*, 1*).
214 POWER SERIES
176. We wish now to call attention to various false notions
which are prevalent regarding the development of a function in
Taylor's series.
Criticism 1. It is commonly supposed, if the Taylor's series T
belonging to a function /(#) is convergent, that then
/(*) = T.
That this is not always true we proceed to illustrate by various
examples.
Example 1. For f(x) take Cauchy's function, I, 335,
-i
*•+!,
(7 00= lime' n
w=oo
_JL
Forz=£0 O(x)=e x* ; for x = 0 (7(a;)=0.
1° derivative. For x * 0, C'(x) = - C(x).
For, = 0,
A=O
2° derivative, x * 0, £"'(*) = (7 (a;) ( 4 ~ 4 I '
[x*> x* )
x = 0, <7"0) = lim ^Wll_^i2) = lim 1 «"*' = 0.
3° derivative, x * 0, C"' (x) = 0 (x) \ -^ - ^ + ^
Id;9 a:' z5
* = 0, C""(0)=lim-^jp=0.
/TI general we have :
# ^= 0, C(n)(x') = C (x) | h terms of lower degree i
I x? J
Thus the corresponding Taylor's series is
<r r*
T — (7(0) + — <7'(0) + — (7"(0) + •••
1 ! 2 !
= 0 + 0 -x+ 0 -x*+ 0 • a^+ —
TAYLOR'S DEVELOPMENT 215
That is, T is convergent for every #, but vanishes identically.
It is thus obvious that C (x) cannot be developed about the origin
in Taylor's series.
Example 2. Because the Taylor's series about the origin be
longing to <7(#) vanishes identically, the reader may be inclined
to regard this example with suspicion, yet without reason.
Let us consider therefore the following function,
/<» = C(x) + > . G(x) + g(x).
Then f(x) and its derivatives of every order are continuous.
Smce /(n)(a;) = c<n)^ + g<n)^
71 = 1, 2 ...
and =0
we have „
Hence Taylor's development for f (x) about the origin is
This series is convergent, but it does not converge to the right
value since rp_ x
-L — & •
177. 1. Example 3. The two preceding examples leave noth
ing to be desired from the standpoint of rigor and simplicity.
They involve, however, a function, namely, (7(#), which is not
defined in the usual way; it is therefore interesting to have ex
amples of functions defined in one of the ordinary everyday
ways, e.g. as infinite series. Such examples have been given by
Pringsheim.
The infinite series
defines, as we saw, 155, 2, a function in the interval 21 = (0, 5),
b >0 but otherwise arbitrary, which has derivatives in 51 of every
order, viz. :
° /• -f ~\n nn\
;p « (2
216 POWER SERIES
The Taylor's series about the origin for F(x) is
T(x) = ^—F^^) ; Xl = lforX=0,
A=0 X •
and by 2)
Hence
(3
As £*>0 and lim ^=0, £A+I<£A, this series is an alternate series
for any x in 51. Hence T converges in 51.
2. Readers familiar with the elements of the theory of func
tions of a complex variable will know without any further reason
ing that our Taylor's series T given in 3) cannot equal the given
function F in any interval 51, however small b is taken. In fact,
F(x) is an analytic function for which the origin is an essentially
singular point, since F has the poles -- - n = 1, 2, 3 •••, whose
limiting point is 0.
3. To show by elementary means that F(x) cannot be devel
oped about the origin in a Taylor's series is not so simple. We
prove now, however, with Pringsheim :
If we take a^fi±lY«4.68 •••, T(x) does not equal F(x)
\e — I/
throughout any interval 51 = (0, 5), however small b > 0 is taken.
We show 1° that if F(x) = T(x) throughout 51, this relation is
true in 53 = (0, 25*).
In fact let 0<xQ<b.
By 161, 4 we can develop I7 about #0, getting a relation
^0*0 = i<7JC<>-3b)'' (1
0
valid for all x sufficiently near x9. On the other hand, we saw in
167 that
F(x) = ^BK(x-x«Y (2
o
is also valid for 0<z<2#0. But by hypothesis, the two power
series 1) and 2) are equal for points near x9. Hence they are
TAYLOR'S DEVELOPMENT 217
equal for 0<x<2x0. As we can take XQ as near b as we choose,
F=Tin 33.
By repeating the operation often enough, we can show that F =
T in any interval (0, B) where B > 0 is arbitrarily large.
To prove our theorem we have now only to show F=£ T for
some one x >0.
Since
1 1 V /I 1 11
we have -. -,
,FO)>— L — L.
1 + x l + ax
On the other hand
Hence T(x)<--
e
To find a value of x for which Gr^- take x— a~*. For this
value of x
Observe that Cr considered as a function of a is an increasing
function. For /e _j_ -|\ 2 ^
"~ \e - lj ~ e '
Hence F> T tor x >«-*.
178. Criticism 2. It is commonly thought if f(x) and its
derivatives of every order are continuous in an interval H, that
then the corresponding Taylor's series is convergent in 31.
That this is not always so is shown by the following example,
due to Pringsheim.
It is easy to see that
converges for every x _>0, and has derivatives of every order for
these values of #, viz. :
218 POWER SERIES
Taylor's series about the origin is
The series T is divergent for x > 0, as is easily seen.
179. Criticism 3. It is commonly thought if f(x) and all its
derivatives vanish for a certain value of x, say for x = a, that
then/(V) vanishes identically. One reasons thus:
The development of/(V) about x— a is
/oo^/oo+^fiVoo + (a;r,a)2/"(a)+ -
As /and all its derivatives vanish at a, this gives
f(x) = 0 + 0 - O - a) + 0 . (x - a)2 + • - -
= 0 whatever x is.
There are two tacit assumptions which invalidate this conclusion.
First, one assumes because / and all its derivatives exist and
are finite at x = a, that therefore f(x) can be developed in
Taylor's series. An example to the contrary is Cauchy's function
C(x). We have seen that C(x) and all its derivatives are 0 at
x = 0, yet 0(x) is not identically 0 ; in fact (7 vanishes only once,
viz. at x = 0.
Secondly, suppose f(x) were developable in Taylor's series in a
certain interval 51 = (a— A, a + h). Then / is indeed 0 through
out H, but we cannot infer that it is therefore 0 outside H. In
fact, from Dirichlet's definition of a function, the values that /has
in 51 nowise interferes with our giving / any other values we
please outside of 51.
180. 1. Criticism 4- Suppose f(x) can be developed in Taylor's
series at a, so that
for H =(«<*).
TAYLOR'S DEVELOPMENT 219
Since Taylor's series I7 is a power series, it converges not only
in H, but also within 53 = (2 a — b, a). It is commonly supposed
that f(x) = T also in 53. A moment's reflection shows such an
assumption is unjustified without further conditions on f(x).
2. Example. We construct a function by the method considered
in I, 333, viz.
Then f(x) = cos z, in 5i = (0, 1)
= 1 + sin x, within 53 = (0, - 1).
We have therefore as a development in Taylor's series valid
in 91 ~2 ~4 ~6
/(*)-i-f>+fi-f> +...-*
It is obviously not valid within 53, although T converges in 53.
3. We have given in 1) an arithmetical expression for/ (x).
Our example would have been just as conclusive if we had said :
Let /(#) = cos x in H,
and = 1 + sin # within 53-
181. 1. Criticism 5. The following error is sometimes made.
Suppose Taylor's development
valid in 51 = (a <
It may happen that jT is convergent in a larger interval
One must not therefore suppose that 1) is also valid in 53.
2. Example.
Let f(x)=e* in 51 = (a, 5),
and = e* + sin (x - b) in SB = (6, 5).
Then Taylor's development
is valid for 51. The series T converging for every x converges in
53 but 1) is not valid for 53.
220 POWER SERIES
182. Let f(x) have finite derivatives of every order in
$[ = (#< 5). In order that/(V) can be developed in the Taylor's
series 7 2
/(*)=/(« + *)=/(«) + ¥'00 + ^ /"(«) + - (1
valid in the interval 51 we saw that it is necessary and sufficient
that
lim Rn = 0.
But Rn is not only a function of the independent variable A, but
of the unknown variable 6 which lies within the interval (0, 1)
and is a function of n and h.
Pringsheim has shown how the above condition may be replaced
by the following one in which 6 is an independent variable.
For the relation 1) to be valid for all h such that §<Ji < H, it is
necessary and sufficient that Cauchy'sform of the remainder
the h and 0 being independent variables, converge uniformly to zero
for the rectangle D whose points (h, 6) satisfy
1° It is sufficient. For then there exists for each e > 0 an m
such that
| Rn(h, 0) | < e n 5 m
for every point (A, 6) of D.
Let us fix h ; then Rn \ < e no matter how 6 varies with n.
2° It is necessary. Let hQ be ah arbitrary but fixed number in
H = (0,JJ*>
We have only to show that, from the existence of 1), for h<_hQ,
it follows that
uniformly in the rectangle Z>, defined by
TAYLOR'S DEVELOPMENT 221
The demonstration depends upon the fact that Rn(h, 0) is h
times the wth term /„(«, &) of the development of f'(x) about the
point a + «. In fact let h = a + k. Then by 158
/'(a + A) =/'(<, + « + *) = /'(* + «) + .••+i^/^a + «)+---
whose nth term is
Let « = #A, then
«.(A,«)=A/m(o,*)
as stated.
The image A0, of _Z>0 is the half of a square of side A0, below the
diagonal.
To show that Rn converges uniformly to 0 in DQ we have only
to show that ^^ ^ ^ Q uniformly in A0 . (2
To this end we have from 1) for all t in H
/' (a + 0 =/'(«) + (TOO + £ /"'(«)+- (3
Its adjoint
/'(a) | +«;/»(«) | + •••
also converges in 31.
By 161, 4 we can develop 4) about t = «, which gives
But obviously (r(«, Ar) is continuous in A0, and evidently all its
terms are also continuous there. Therefore by 149, 3,
("~])(«) = 0 uniformly in A0 . (5
n — 1 !
But if we show that
|/<»>(a + «) <_&*-» (a) (6
it follows from 5) that 2) is true. Our theorem is then
established.
222 POWER SERIES
To prove 6) we have from 1)
/(n) f f, I ~\ f(n)(n\ _l_ r/f (w-f-l)/ \ _i
\u i **/ J V^y T" *~/ v^v T
and from 4)
The comparison of 7), 8) proves 6).
Circular and Hyperbolic Functions
183. 1. We have defined the circular functions as the length
of certain lines ; from this definition their elementary properties
may be deduced as is shown in trigonometry.
From this geometric definition we have obtained an arithmeti
cal expression for these functions. In particular
= F!~37 + 5]~f! + Q
cosrr=l — — +- — +... /.?
2 ! 4 ! 6 ! l
valid for every x.
As an interesting and instructive exercise in the use of series
we propose now to develop some of the properties of these func
tions purely from their definition as infinite series. Let us call
these series respectively 8 and O.
Let us also define tan x = sm x , sec x = — - , etc.
cos x cos x
2. To begin, we observe that both #and C converge absolutely
for every x, as we have seen. They therefore define continuous
one-valued functions for every x. Let us designate them by the
usual symbols .
We could just as well denote them by any other symbols, as
4>O) , *(*).
3. Since
n ^ £
, (7=1 forz = 0,
we have . ft A
smO = 0 , cos 0 = 1.
CIRCULAR AND HYPERBOLIC FUNCTIONS 223
4. Since S involves only odd powers of x, and C only even
powers,
sin x is an odd, cos x is an even function.
5. Since S and C are power series which converge for every x,
they have derivatives of every order. In particular
dC =_z,??__3?_,&__ _ c f
dx" 18! 5! 7!
Hence dsinx dcoxx
— = cos x , — = — sin #. (3
6. To get the addition theorem, let an index as x, y attached to
S, C indicate the variable which occurs in the series. Then
__ _
7!5!2!8!4! 6! )
7! o!2! 314!
Adding,
1
= x + y _ (a; + y )3 (^ + 5)5
" 1! 3! 5!
Thus for every x, y
sin (a; -f- y) = sin # cos y -f- cos # sin y .
In the same way we find the addition formula for cos x.
224 POWER SERIES
7. We can get now the important relation
sin2 x -h cos2 x = 1 (4
directly from the addition theorem. Let us, however, find it by
aid of the series. We have
11 11
+ + +
11 11 11
l + 6!2!+4!4!+6!2!+8!
Hence
Now by I, 96,
Thus £2 O2 = ^2 x cofc2 x = 1
8. In 2 we saw sin #, cos 2; were continuous for x\ 4) shows
that they are limited and indeed that they lie between ± 1.
For the left side of 4) is the sum of two positive numbers and
thus neither can be greater than the right side. •
9. Let us study the graph of sin #, cos #, which we shall call 2
and F, respectively.
Since sin x = 0, — s'n x = cos x = 1, for x = 0, 2 cuts the #-axis at
ax
0 under an angle of 45 degrees.
CIRCULAR AND HYPERBOLIC FUNCTIONS 225
Similarly we see y — 1 for x = 0. F crosses the #-axis there
and is parallel to the #-axis.
and each parenthesis is positive for 0 < x2< 6,
sin.r>0 for 0<:r<Vti= 2.449 ...
Since
,_ l __ l
21^4!^
T 1 7 p O O Ck
cosa:>0 for 0<*< V2 = 1.414 «•
Since
cosa;<0 for x = 2.
Since Dx cos a: = — sin # and sin x > 0 for 0 < x < V6, we see
cos x is a decreasing function for these values of x. As it is con
tinuous and > 0 for x = V2, but < 0 for x = 2, cos x vanishes once
and only once in (V2, 2).
This root, uniquely determined, of cos x we denote by - • As a
first approximation, we have
From 4) we have sin2 ?= !• As we saw sin#>0 for x< V6,
we have
Thus sin x increases constantly from 0 to 1 while cos x decreases
from 1 to 0 in the interval ( 0, ^ )= Ir We thus know how sin #,
cos x behave in Ir
From the addition theorem
sin ( ~ + x ) = sin ^ cos x + cos ^ sin x = cos z.
\<A / 2 J
-|- a;= cos cos a: — sin sin a; = — sin a:.
226 POWER SERIES
Knowing how sin#, cosx march in 7j, these formulae tell us
how they march in J2 = (-, 77-).
From the addition theorem,
sin (TT + x) = — sin x, cos (TT + x) = — cos x.
Knowing how sin x, cos x march in (0, TT), these formulae inform
us about their march in (0, 2 TT).
The addition theorem now gives
sin (x + 2 TT) = sin x, cos (x + 2 TT) = cos x.
Thus the functions sin ar, cos # are periodic and have 2 TT as period.
The graph of sin x cos x for negative x is obtained now by
recalling that sin x is odd and cos x is even.
10. As a first approximation of TT we found
V2 < | < 2.
By the aid of the development given 159, 3
= *-! + i-+- 5)
we can compute TT as accurately as we please.
In fact, from the addition theorem we deduce readily
n4==vl ' >S4=vl'
Hence TT
tan — = 1.
This in 5) gives Leibnitz s formula,
4 = ' ~ 3 + 5 ~ 7 + '
The convergence of this series is extremely slow. In fact by
81, 3 we see that the error committed in stopping the summation
at the nth term is not greater than — ' — . How much less the
error is, is not stated. Thus to be sure of making an error less
than — — it would be necessary to take | (10W + 2) terms.
CIRCULAR AND HYPERBOLIC FUNCTIONS 227
11. To get a more rapid means of computation, we make use
of the addition theorem.
To start with, let ,
a = arctg £.
Then 5) gives 1^11^11 11
~5 35S + 555 757 ^
a rapidl}7 converging series.
The error Ea committed in breaking off the summation at the
wth term is ., _,
By virtue of the formula for duplicating the argument
0 2 tan «
tan 2 a = — - — —,
1 —
wehave
Similarly . tan 4 « = if-o.
Let
^ = 4«-|. (7
The addition theorem gives
tan 4 a - 1 1
tan 13
1 + tan 4 « 239 '
Then 5) gives 111 11
r) | -I xo
~~ 289 3239* 5289*
also a very rapidly converging series.
We find for the error
"2w-12392"-i'
The formula 7) in connection with 6) and 8) gives - . The
error on breaking off the summation with the wth term is
228 POWER SERIES
184. The Hyperbolic Functions. Closely related with the cir
cular functions are the hyperbolic functions. These are defined
by the equations
£>X I a - X
cosh x = - ' - - (2
tanh x =
2
sinh x ex — e~x
cosh a; ex + e~x
sech x = — : — , cosech x-
cosh x sinh x
Since „
we have , ,
= _+|j+|I+... . . (3
=l + g + i£ + ... -'. (4
valid for every x. From these equations we see at once :
sinh ( — x) = — sinh x ; cosh ( — x) = cosh x.
sinh 0 = 0. coshO= 1.
-— sinh a; =1+ — + ^7+ — = cosh x. (5
dx 214!
Let us now look at the graph of these functions. Since sinh #,
cosh x are continuous functions, their graph is a continuous curve.
For x > 0, sinh x > 0 since each term in 3) is > 0. The relation
4) shows that cosh x is positive for every x.
If x' > x > 0, sinh x' > sinh #, since each term in 3) is greater
for x' than for x. The same may be seen from 5).
THE HYPERGEOMETRIC FUNCTION 229
Evidently from 3), 4)
lim sinh x = + oo , lim cosh x = + oo .
X=+30 X=+X>
At x = 0, cosh 2: has a minimum, and sinh x cuts the 2>axis
at 45°.
For x > 0, cosh x > sinh # since
The two curves approach each other asymptotically as x = -hoc .
For the difference of their ordinates is e~x which = 0 as x = + oo .
The addition theorem is easily obtained from that of ex. In fact
. , , e* — e~x
sinh x cosh y = - - --
-)imi ar y
= ^y— ex~v + e~x+v — e~x~v).
Hence
sinh x cosh y + cosh 2: sinh y = \(ex^v — e~(x+y)) = sinh (x +
Similarly we find
cosh (2: + ?/) = cosh a: cosh ?/ + sinh x sinh y.
In the same way we may show that
cosh2 x — sinh2 x = 1.
Hyper geometric Function
185. This function, although known to "Wallis, Euler, and the
earlier mathematicians, was first studied in detail by Gauss. It
may be defined by the following power series in x:
-
The numbers a, /3, 7 are called parameters. We observe that
a, (B enter symmetrically, also when a. = 1, ft = 7 it reduces to
the geometric series. Finally let us note that 7 cannot be zero or
a negative integer, for then all the denominators after a certain
term = 0.
230 POWER SERIES
The convergence of the series F was discussed in 100. The
main result obtained there is that F converges absolutely for all
| x | < 1, whatever values the parameters have, excepting of course
7 a negative integer or zero.
186. For special values of the parameters, F reduces to ele
mentary functions in the following cases :
1. If a or /? is a negative integer — n, F is a polynomial of
degree n.
2. ^(1,1, 2; -x)= log (1 + rc). (1
Also
The relation 1) is now obvious.
Similarly we have
,l, 2; *) = — log (I-*)-
4. s-FC}, -|-, |, a:2) = arcsin x.
5. 2^(1> 1» f » — ^2) = arc tan ^.
6. Km jpf «, 1, 1, -} = e*. (2
a=+oo \ Ct/
For
• 2 • 1 • 2
t t* - t* j •»- i* i * • - /I I
~1 . 2 . 3 " 1 .2-8W "
THE HYPERGEOMETRIC FUNCTION 231
Let 0 < Gr < 0. Then
is convergent since its argument is numerically < 1. Comparing
3), 4) we see each term of 3) is numerically < the corresponding
term of 4) for any \x < G- and any a > 0. Thus the series 3)
considered as a function of a is uniformly convergent in the
interval (/3 + co ) by 136, 2 ; and hereby x may have any value
in (— G, G-). Applying now 146, 4 to 3) and letting a= + 00,
we see 3) goes over into 2).
7. lim xF( «,«,-; - — - ] = sin x. (5
a=+» \ A 4 arj
For
Let x = G- > 0 and a = a. Then
is convergent by 185. We may now reason as in 6.
8. Similarly we may show :
/ "I 2 \
lim F( «, a, - ; - -—5] = cos x.
a=+oo \ 2 4 «2/
lim El a, «, 1 , — - ) = sinh x.
a=+oo \ 2 4 a2/
lim F(a, «, -, T— i) = cosh ^.
a=+30 V 2 4«V
187. Contiguous Functions. Consider two F functions
-F(a,&7; x) , JT(«f, /3', 7' ; *).
If a differs from a!, by unity, these two functions are said to be
contiguous. The same holds for 0, and also for 7. Thus to
F (afiyx) correspond 6 contiguous functions,
-F(a±l,£±l, 7±1; x).
232 POWER SERIES
Between F and two of its contiguous functions exists a linear
relation. As the number of such pairs of contiguous functions is
6-5,,
172- L5'
there are 15 such linear relations. Let us find one of them.
We set 4- 1 • a -I- 2 • a 4- — 1 S R 4-1 fl 4- — 9
1 • 2 • • • • ft • 7 • 7 4- 1 • •••74-71— 1
Then the coefficient of xn in F(a.@vx} is
in F(a + 1, ft 7, x) it is
in ^(«, ft 7 — 1, x) it is
I n
,
Thus the coefficient of #n in
(7 _ « _ 1)^(«, ft 7, x) + a.F(a + 1, ft 7, z)
+ (l-7)JF(a, ft 7-1,
is 0. This being true for each n, we have
(7 _ « _ 1) Jf (a, ft 7, z) + «JP (a + 1, ft 7, x)
a, ft 7-1, aO=
Again, the coefficient of a;n in JP(a, /3 — 1, 7, a;) is «(/3 —
in a?^(a -f- 1, ft 7, a;) it is 71(7 + w — 1) §n.
Hence using the above coefficients, we get
«, ft 7, ») + «(1 - a;)F(a + 1, ft 7, »)
From these two we get others by elimination or by permuting
the first two parameters, which last does not alter the value of
the function
Thus permuting «, ft in 1) gives
(7 - £ - 1)^(«, ft % s) +ft*T(«1 £ + 1, 7, s)
ft 7 ~ 1, *0 = 0. (3
THE HYPERGEOMETRIC FUNCTION 233
Eliminating F '(<*, /3, 7— 1, x) from 1), 3) gives
(£-a)jF(«, & 7, a-) + «.F(a + l, & 7, z)
- AF(a, £ + 1, 7, a:) = 0. (4
Permuting a, /9 in 2) gives
OX- *- /3)^(«, £, 7, *) + £(1 - x)F(a, 0 + 1, 7, *0
a-l, £, 7,2r) = 0. (5
From 3), 5) let us eliminate jP(«, /S + 1, 7, a;), getting
(« _ 1 _ (7 _ ff _ 1»^(>, A 7, a;) + (7 - «)JP(« - 1, A 7, a;)
+ (l_7)(l_aOJFT(«, /3, 7-l,af)=0. (6
In 1) let us replace a by a — 1 and 7 by 7 + 1 ; we get
(7 - a + 1)^(« - 1, & 7 + 1* *) + O - 1)^(«' A 7 + 1, a)
_7^(«-l, A 7, x)=Q. (a)
In 6) let us replace 7 by 7 + 1 ; we get
/3,7, aO = 0. (b)
Subtracting (b) from (a), eliminates F(^a — 1, /8, 7 + 1, a;) and
gives
7(1 - aO^O/^z) - 7^(« - 1, A 7, a;)
+ (7 - /8)^(«, A 7 + 1, *) = 0. (7
From 6), 7) we can eliminate F(a— 1, /?, 7, z), getting
7!7-l+C«H-/8+l-27>SJF(«, A 7,3;)
+ (7 - «)(7 - P)zF(a, /3, 7 + 1. *0
+ 7(1 - 7)(1 - aO^Ca, A 7 - 1,«) = 0. (8
In this manner we may proceed, getting the remaining seven.
188. Conjugate Functions. From the relations between con
tiguous functions we see that a linear relation exists between any
three functions
whose corresponding parameters differ only by integers. Such
functions are called conjugate.
234 POWER SERIES
For let jo, q, r be any three integers. Consider the functions
JF(a#y30, JF(« + 1, ft 7, x) ••• F(a+p, ft 7, x),
+ 2> 7, *) •'• ^
We have jt? + q + r + 1 functions, and any 3 consecutive ones
are contiguous. There are thus p + q -f r — 1 linear relations
between them. We can thus by elimination get a linear relation
between any three of these functions.
189. Derivatives. We have
1.2. ...
~~r~ ^^ /„ _i 1 \ ' i ~,n
== X V "l 7 — -i ?» — ~~ -i -i —
= ^JF(a + l, /3+1, 7 + 1, x~).
7
Hence
7^// /^.. Q , _N a ' P Til S~ i 1 /Oil . . i 1 ^,\
^
7 • 7
and so on for the higher derivatives. We see they are conjugate
functions.
190. Differential Equation for F. Since F, F1 , F" are conju
gate functions, a linear relation exists between them. It is found
to be
+ {(« + 13 4- l)z - 7} ^' + «/3^= 0. (1
To prove the relation let us find the coefficient of xn on the left
side of 1). We set
p _«•« + !• — « + n-l-£-ff + l-— ff + ro-l_
~~
THE HYPERGEOMETRIC FUNCTION 235
The coefficient of xn in x2F" is
n(n-l)PB,
in — xF" it is
7+n
in (a + £ + l)a;jF' it is
w« + +
in — 7.F' it is
in a/3P it is
Adding all these gives the coefficient of xn in the left side of 1).
We find it is 0.
191. Expression of F(a/3yx') as an Integral.
We show that for | x \ < 1,
B(fri-P)-F(ap<ix)= fV-^l -uy-P-id-xu^-'du (1
«^o
where ^(jt?, ^) is the Beta function of I, 692,
B(p, q) = f nr-i(l - u)«-ldu.
For by the Binomial Theorem
N -, .a ,«•«+! •«.«»«4-l««
(1 - zw)-* = 1 + - xu + -y- 1- A2 + - 1.2.3
for | xu \ < 1. Hence
,7
7 - /?) + ^5(/3 + 1, 7 ~ ^)
2,7-/3)+... (2
236 POWER SERIES
Now from I, 692, 10)
Hence
7+1 ry . ry _|_
etc. Putting these values in 2) we get 1).
192. Value of F (a, £, 7, x) for x = I.
We saw that the F series converges absolutely for x = 1 if
a + P — 7 < 0. The value of F when x = 1 is particularly in
teresting. As it is now a function of a, & 7 only, we may denote
it by .F(«, /3, 7). The relation between this function and the T
function may be established, as Gauss showed, by means of 187, 8)
viz *
7!7-l+O + /3 + l-27)zJ F(afax)
+ (7 - «) (7 - &)xF(a, #7 + 1, x)
7(1 ~ 7) (1 - *0^(«, /3, 7 - 1, *) = 0. (1
Assuming that n A
« + £ - 7 < 0, (2
we see that the first and second terms are convergent for x = 1 ;
but we cannot say this in general for the third, as it is necessary
for this that a + /3 — (7 — 1) < 0. We can, however, show that
L lim (1 - x) JF(«, /?, 7 - 1, x) = 0, (3
supposing 2) to hold. For if x\ < 1,
^(«, & 7 - 1» *) = «0 + V + a2z2 + ... (4
Now by 100, this series also converges for x = — 1. Thus
lim an = 0. (5
n= oo
From 4) we have
£, 7 - 1, «) = «0 + O - «)^ + (« - ^)^ + •"
Let the series on the right be denoted by 6r(». As
6?n+1(l) = an, we see 6r(l) is a convergent series, by 5), whose
sum is 0. But then by 147, 6, Cr(x) is continuous at x = 1.
Hence ilirn 6^jc
THE HYPERGEOMETRIC FUNCTION 237
and this establishes 3). Thus passing to the limit x = 1 in 1)
gives
7 O + /3 - 7) F(«, & 7) + (7 - «) (7 - £)-F(«, /3, 7 + 1) = 0,
*(«, A 7) =
7(7 - a-
Replacing 7 by 7 + 1, this gives
etc. Thus in general
7) =
-.?(«,& 7 + »)•
Gauss sets now
71 ! 71^
; >5 I} = («+!)(* + 2) ...(* + »)'
Hence the above relation becomes
jfro ) = n(>i.y-l-)n(n.7-«-ff-l) J( p } (g
n(w,7-«-i)n(w,7-/3-i)
N°W
lira J(«, A 7 + ») = 1. (7
7l = OC
For the series
)=l+ + IL±±l+.. (8
converges absolutely when 2) holds. Hence
'!*"'+1|
In n o n n \ 1
• Cr 1 • 1 - Or • Or + 1
is convergent. Now each term in 8) is numerically < the corre
sponding term in 9) for any 7 > 6r. Hence 8) converges uni
formly about the point 7 = -f oc. We may therefore apply 146, 4.
As each term of 8) has the limit 0 as 7 = + oc, the relation 7)
is established.
238 POWER SERIES
We shall show in the next chapter that
lim II (n, x)
n=oo
exists for all x different from a negative integer. Gauss denotes
it by II (x) ; as we shall see,
rO) = n<>-l) , fora>0.
Letting n = GO, 6) gives
*(«, A 7) = n c-y - i)n(7 -«- 0-1)
n(y-«-l)n(7-/3-l)
We must of course suppose that
7, 7 - «, 7 - & 7 - a - £,
are not negative integers or zero, as otherwise the corresponding
II or F function are not defined.
Bessel Functions
193. 1. The infinite series
converges for every x. For the ratio of two successive terms of
the adjoint series is \x 2
which = 0 as s = GO for any given x.
The series 1) thus define functions of x which are everywhere
continuous. They are called Bessel functions of order
n=0, 1, 2 ...
In particular we have
^oW = * - 7772 + 2271^ ~ 22 . 42 • 62 + '" ^
j (x\ = %_ x8 x5 tf_ sn
2 22 • 4 22 • 42 • 6 22 • 42 • 62 • 8
Since 1) is a power series, we may differentiate it termwise and
' _ V (-lV(
BESSEL FUNCTIONS 239
2. The following linear relation exists between three consecutive
Bessel functions :
For ,. xn~l .1. sr-iv afr+n~1 rr
71-1 -'- ; »+*-1- '
Hence
— 1 « ,~2a+n— 1 (
4-2C— IV- _ | _
-l! iv y 2«+2*-M8:-
3. We show next that
2^2(8!) = «/_!(*) -J-.+1(*) «>0. (8
For subtracting 7) from 6) gives
From 8) we get, on replacing Jn+1 by its value as given by 5) :
Ji(x) = - Vn(» + ^(2:), TI > 0. (9
X
From 5) we also get
«£(*)=• -Jr,,co-«W*) w>°- (10
X
4. The Bessel function Jn satisfies the following linear homo
geneous differential equation of the 2° order :
n = o. (ii
240 POWER SERIES
This may be shown by direct differentiation of 1) or more sim
ply thus : Differentiating 9) gives
,. (12
X X
Equation 10) gives
Replacing here Jn_^ by its value as given by 9), we get
Putting this in 12) gives 11).
5. e*-2~ = ^unJn(x) (13
— 00
for any #, and for u =£ 0.
For
e 2 = e e
(1 _i_ xu j_ ^u*
1 "T + ^TTi
Now for any x and for any u =£ 0, the series in the braces are
absolutely convergent. Their product may therefore be written
in the form ' ^ (x* 1 \
22 \2y 2! 2! 'V
3 ! 2 ! V2
Jx _ j^
V2 r\
(
BESSEL FUNCTIONS 241
194. 1. Expression of Jn(x} as an Integral.
' (X COS
Hence -r, , _ -, x g
cos (a; cos <£) = ]T ^- — ^ a^* cos2* </>
o (-s)-
and thus
CC X- -J N 8
cos (a: cos </>) sin2n ^ = V ^ ~ ^ 2^« cos2* (^ sin2n </>.
o (^ s) •
As this series converges uniformly in (0, TT) for any value of re,
we may integrate termwise, getting
j[cos (a? cos </>) sin2" <j>d<t> = ^ ^~ 1)f* <& j'cos2' (/> sin2"
o (2«)I
We shall show in 225, 6, that
r(l£±l).l.8.5....8.-lv5:
Thus the last series above
9
0
Thus
™n
2 )
CHAPTER VII
INFINITE PRODUCTS
195. 1. Let {a^... tj be an infinite sequence of numbers, the
indices i = (t1-..t8) ranging over a lattice system £ in s-way
space. The symbol p —T\ - TT (1
= gatl ...,,_ ^
is called an infinite product. The numbers at are its factors. Let
PM denote the product of all the factors in the rectangular cell
** " ' lirnP, (2
is finite or definitely infinite, we call it the value of P. It is
customary to represent a product and its value by the same letter
when no ambiguity will arise.
When the limit 2) is finite and =£ 0 or when one of the factors
= 0, we say P is convergent, otherwise P is divergent.
We shall denote by P^ the product obtained by setting all the
factors a, = 1, whose indices t lie in the cell R^. We call this the
co-product of PM.
The products most often occurring in practice are of the type
P = a1-a2.aB. ••• = TLan. (3
The factor P^ is here replaced by
and the co-product PM by
•* m == &m+l ' ^m+2 " ^wi+3 "
Another type is +00
P=Han. . (4
The products 3), 4) are simple, the product 1) is s-tuple. The
products 3), 4) may be called one-way and tivo-way simple products
when necessary to distinguish them.
242
GENERAL THEORY 243
p — i . 1 . 2 . 3 . 4. . . .
* *1 2 3 ¥ 6
Obviously the product P = 0, as
p. = I = o.
/i
Hence P = 0, although no factor is zero. Such products are
called zero products. Now we saw in I, 77 that the product of a
finite number of factors cannot vanish unless one of its factors
vanishes. For this reason zero products hold an exceptional posi
tion and will not be considered in this work. We therefore have
classed them among the divergent products. In the following
theorems relative to convergence, we shall suppose, for simplicity,
that there are no zero factors.
196. 1. For P = Tlatl...lt to converge it is necessary that each PM
is convergent. If one of these PM converges, P is convergent and
P — P • P
M — -*• fj. -*• fj. •
The proof is obvious.
2. If the simple product P = al • «2 • a3 >•• is convergent, its fac
tors finally remain positive.
For, when P is convergent, Pn j > some positive number, for
n > some m. If now the factors after am were not all positive, Pn
and Pv could have opposite signs v>n, however large n is taken.
Thus Pn has no limit.
197. 1. To investigate the convergence or divergence of an
infinite product P = Ila^...,, when at > 0, it is often convenient to
consider the series
called the associate logarithmic series. Its importance in this con
nection is due to the following theorem :
The infinite product P with positive factors and the infinite series
L converge or diverge simultaneously. When convergent, P = eL,
L = log P.
For logPM = £M, (1
P, = *V (2
244 INFINITE PRODUCTS
If P is convergent, P^ converges to a finite limit =£ 0. Hence
L^ is convergent by 1). If L^ is convergent, P^ converges to a
finite limit =?t 0 by 2).
2. Example 1.
p = n ( i + - V » = nan 7i = i, 2, ...
\ »/
is convergent for every x.
For, however large | x \ is taken and then fixed, we can take m
so large that
n>m.
n
Instead of P we may therefore consider Pm.
Then
But by I, 413
Hence Lm = 2^ Mnz? • —
m+\ %«
which is convergent.
The product P occurs in the expression of sin x as an infinite
product.
Let us now consider the product
Q = nf 1 + -}e n n = ± 1, ± 2,
V n)
The associate logarithmic series L is a two-way simple series.
We may break it into two parts U, L", the first extended over
positive ft, the second over negative n. We may now reason on
these as we did on the series 3), and conclude that Q converges
for every x.
3. Example 2.
n
is convergent for any x different from
0, - 1, - 2, - 3,
GENERAL THEORY 245
For let p be taken so large that \x\ < p. We show that the
co-product
converges for this 2;. The corresponding logarithmic series is
QO
L = Z, I x logf 1 + 1) - logf 1 + ^ 1
j>+i I V n/ V »/ J
As each of the series on the right converges, so does L. Hence
Gr converges for this value of x.
198. 1. When the associate logarithmic series
Zr=21ogatl...lf , aL>0
is convergent, Um log ^ _ _ Oj b 121> ^
|t|=QO
and therefore -,.
limati...ls = l.
|l|=00
For this reason it is often convenient to write the factors
ati...l4 of an infinite product P in the form 1 + 6tl...tj. When P is
written in the form
p = n(i + ^...ts),
we shall say it is written in its normal form. The series
2J,,..,, = 2J.
we shall call the associate normal series of P.
2. The infinite product
z'^s associate normal series
converge or diverge simultaneously.
246 INFINITE PRODUCTS
For P and £ = 2 log (1 + «.)
converge or diverge simultaneously by 197. But A and L con
verge or diverge simultaneously by 123, 4.
3. If the simple product P — al - a2 • a3 ••• is convergent, an=l.
For by 196, 2 the factors an finally become > 0, say for n > m.
Hence by 197, l the series
oo
2 log an an > 0
n=m
is convergent. -Hence log an = 0. .-. an = 1.
199. Let R^ < R^ < • •• X | = oo be a sequence of rectangular
cells. Then if P is convergent,
For P is a telescopic series and
200. 1. Let P
We call (?=n(l + all...l8 , «t = «4
the adjoint of P, and write
¥ = Adj P.
2. P converges, if its adjoint is convergent. We show that
€ > 0, X, \Pp-Pv\<€ p,V>\.
Since $ is convergent,
is also convergent by 199. Hence
0<^,-(ipM<€ \<n<v.
But Pv — PM is an integral rational function of the a's with
positive coefficients. Hence
IP, -.p. i <*,-$,• ci
GENERAL THEORY 247
3. When the adjoint of P converges, we say P is absolutely
convergent.
The reader will note that absolute convergence of infinite
products is defined quite differently from that of infinite
series. At first sight one would incline to define the adjoint of
P = Hatl... 18
to be «i __ jj I a I
With this definition the fundamental theorem 2 would be false.
For let P=n(-l)»;
its adjoint would be, by this definition,
Now $n = 1. .*. $ is convergent. On the other hand,
_Pn=(— l)n and this has no limit, as n=cc. Hence P is
divergent.
4. In order that P= TI(1 + atl...ls) converge absolutely, it is
necessary and sufficient that ^
converges absolutely.
Follows at once from 198, 2.
Example. / ^
nd--.
converges absolutely for every x.
For y*2. avl
^ n2 *~ n2
is convergent.
201. 1. Making use of the reasoning similar to that employed
in 124, we see that with each multiple product
are associated an infinite number of simple products
and conversely.
248 INFINITE PRODUCTS
We have now the following theorems :
2. If an associate simple product Q is convergent, so is P, and
P=Q.
For since Q is convergent, we may assume that all the a's are
> 0 by 196, 2. Then
_ g2 log CV i. by 124, 3,
= P by 197, 1.
3. If the associate simple product Q is absolutely convergent, so
is P.
Forlet
Since Q is absolutely convergent,
is convergent. Hence 11(1 + atl...t ) is convergent by 2.
4. Let P= H(l H- «it...i ) fo absolutely convergent. Then each
associate simple product Q= 11(1 + #n) ^s absolutely convergent and
p=<?.
For since P is absolutely convergent,
2«^...^
converges by 200, 4. But then by 124, 5
2«n
is convergent. Hence Q is absolutely convergent.
5. If P= natl...lg is absolutely convergent, the factors flj.-i, >0
^y ^Aei/ ^'g outside of some rectangular cell R^.
For since P converges absolutely, any one of its simple associ
ate products Q=llan converges. But then an>Q for n>m, by
198, 3. Thus a(i...t > 0 if i lies outside of some R^.
6. From 5 it follows that in demonstrations regarding abso
lutely convergent products, we may take all the factors > 0,
without loss of generality.
GENERAL THEORY 249
For
and all the factors of PM are > 0, if JJL is sufficiently large. This
we shall feel at liberty to do, without further remark.
T. 4 =11(1 + *,,...„) «t>0
converge or diverge simultaneously.
For if A is convergent,
2a,,.....
is convergent by 200, 4. But then L is convergent by 123, 4.
The converse follows similarly.
202. 1. As in 124, 10 we may form from a given m- tuple
product A=Ua
l ' m
as infinite number of conjugate w-tuple products
*-™*~t.
where a, = bj if i andy are corresponding lattice points in the two
systems.
We have now :
2. If A is absolutely convergent, so is B, and A — B.
For by 201, 6, without loss of generality, we may take all the
factors > 0.
Then sioga...
= B.
3. Let A-\\a
A - llah...lwi
be an absolutely convergent m-tuple product.
be any p-tuple product formed of a part of or all the factors of A.
Then B is absolutely convergent.
250 INFINITE PRODUCTS
For 2 log a, is convergent.
Hence 2 log fy is.
Arithmetical Operations
203. Absolutely convergent products are commutative, and con
versely.
For let , , ,
^. = IK...lm
be absolutely convergent. Then its associate simple product
2l=Han
is absolutely convergent and A = 51, by 201, 4. Let us now re
arrange the factors of A, getting the product B. To it corre
sponds a simple associate series 23 and B = $8. But 21 = 23 since
51 is absolutely convergent. Hence A = B.
Conversely, let A be commutative. Then all the factors atl...,
finally become > 0. For if not, let
E1<E2< ... =00 (1
be a sequence of rectangular cells such that any point of 9?m lies
in some cell. We may arrange the factors a, such that the partial
products corresponding to 1),
1 ' ^2 ' 3 ""
have opposite signs alternately. Then A is not convergent, which
is a contradiction. We may therefore assume all the a's > 0.
Then A 2iogat...t
*a. — e m
remains unaltered however the factors on the left are rearranged.
Hence v n
21og<v..lwt
is commutative and therefore absolutely convergent by 124, 8.
Hence the associate simple series
« = 2 log a, = 2 log (1 + 6.)
is absolutely convergent by 124, 5. Hence
is convergent and therefore A is absolutely convergent.
ARITHMETICAL OPERATIONS 251
204. 1. Let A = u
ll lt
be absolutely convergent. Then the s-tuple iterated product
is absolutely convergent and A = B where i\ •- ij is a permutation of
ir ia ••• i,.
For by 202, 3 all the products of the type
Ha UaL
Vl« li"-'«
i«-ii« i«
are absolutely convergent, and by I, 324
n = nn.
»«-i»« i«-i i«
Similarly the products of the type
n
'«-ll«-2t«-S
are absolutely convergent and hence
n= n n n.
In this way we continue till we reach A and B.
2. We may obviously generalize 1 as follows :
Let A-IK--
5e absolutely convergent. Let us establish a 1 to 1 correspondence
between the lattice system £ oi'er which i = (ix — O ranges, and the
lattice system 9)1 over which
ranges. Then the p-tuple iterated product
B = tt • n - ... Ila. ,
1 2 r ;n>n"
2*s absolutely convergent, and
252 INFINITE PRODUCTS
3. An important special case of 2 is the following:
Let A = Tlan , w=l,2,...
converge absolutely. Let us throw the an into the rectangular array
an , ala...
^21 ' ^22 *'*
converge absolutely, and
4. 7%e convergent infinite product
associative.
Forlet '""«!< OT2<-..= oo.
•L^et -i i z. /"i i — \ /^i i
We have to show that
<?=(i+
is convergent and P = Q.
This, however, is obvious. For
= P,, v = ml+ ... +wn.
But when n = oo so does v.
Hence ; lim(?n = limPn.
Remark. We note that mm+1 — mm may = oo with n.
ARITHMETICAL OPERATIONS 253
205. Let A = IIati...t. , B = Tlb^...^
be convergent. Then
C=UaL-bt , D=U^
ft.
are convergent and
C=A.B , D = 4
B
Moreover if A, B are absolutely convergent, so are <?, D.
Let us prove the theorem regarding (7; the rest follows simi
larly. We have n D
Cjt = ^cL^ • .o^.
Now by hypothesis A^ = A, B^ = B as p = oo.
Hence C, = A.B.
To show that O is absolutely convergent when A, B are, let us
write a, = 1 + at , 5t = 1 + bt and set | at | = «t , bt | = ft.
Since A, B converge absolutely,
+ O , 2 log (1 + &)
are convergent. Hence
2 {log (1 + <*) + log (1 + #) i = 2 log (1 + «t) (1 + &)
is absolutely convergent. Hence C is absolutely convergent
by 201, 7.
206. Example. The following infinite products occur in the
theory of elliptic functions :
<?2 = n (i + j2"-1)
§3 =n (i-j2-1).
They are absolutely convergent for all | q\ < 1.
For the series ^^ ^ ^\qZn~l\
are convergent. We apply now 200, 4.
As an exercise let us prove the important relation
P = <?,<?2<?3 = 1-
254 INFINITE PRODUCTS
For by 205, p = n (1 + ?2n)(l 4- fn~l)(l - q2n~l
Now all integers of the type 2 w, are of the type 4^ — 2 or 4
Hence by 204, 3,
n (i - 0a») = n (i - ?4n) n (i - ^-2),
Thus p^n
= 1
Uniform Convergence
207. Jn ^Ae limited or unlimited domain 51, let
uniformly convergent and limited. Then
^8 uniformly convergent in H.
For L
J^A = eLx-
Now LK = L uniformly. Hence by 144, 1, F is uniformly con
vergent.
208. If the adjoint of
is uniformly convergent in 51 (finite or infinite), F is uniformly
convergent.
For if the adjoint product,
is uniformly convergent, we have
|^-^|<e
for any x in 51-
UNIFORM CONVERGENCE 255
But as already noticed in 200, 2, 1)
I P — P I < $ — % I
I •* M -* v I — rV Fr |«
Hence .F is uniformly convergent.
209. The product
uniformly convergent in the limited or unlimited domain 51, if
* = 2*., .....Ov..<0 , *.= !/,!
limited and uniformly convergent in 51.
For by 138, 2 the series
is uniformly convergent and limited in 51- Then by 207, the
adjoint of F is uniformly convergent, and hence by 208, F is.
210. Let -p, A TT r
**(*, — O = n/H ...t.(*i ... rrm)
fo uniformly convergent at x = a. // eaeA /t t* continuous at a, F
is also continuous at a.
This is a corollary of 147, 1.
211. 1. Let G = S | /^...^(X ••• zw) | converge in the limited
complete domain 51 having a as a limiting point. Let Gr and each
/t be continuous at a. Then
is continuous at a.
For by 149, 4, Gr is uniformly convergent. Then by 209, F is
uniformly convergent, and therefore by 210, F is continuous.
2. Let Gr =2 l/i^-i.C*?! ••• #m) | converge in the limited complete
domain 51, having x = a as limiting point. Let
lim/c=at , lira G = 2at.
lira
256 INFINITE PRODUCTS
For by 149, 5, 6r is uniformly convergent at x = a. It is also
limited near x— a. Thus by 209,
is uniformly convergent at a. To establish 1) we need now only
to apply 146, l.
212. 1. Let F=TlfLi...,a(x) , />0 (1
converge in 51= (a, a + 8). Then
logJ=i = 21og/.. (2
If we can differentiate this series termwise in 51 we have
Thus to each infinite product 1) of this kind corresponds an infi
nite series 3). Conditions for termwise differentiation of the series
2) are given in 153, 155, 156. Other conditions will be given in
Chapter XVI.
2. Example. Let us consider the infinite product
0(x) =2q*Q sin 7nzfl(l - 2 q2n cos 2 TTX + q*n) (1
which occurs in the elliptic functions.
Let us set
1 - un = 1 - 2 <fn cos 2 TTX + (fn.
Then \un\<2\q\*n+\q\*n.
Thus if | q \ < 1, the product 1) is absolutely convergent for any x.
It is uniformly convergent for any x and for | q \ < r< 1.
If it is permissible to differentiate termwise the series obtained
by taking the logarithm of both sides of 1), we get
0(x) ^1 — 2 (fn COS 2 7TX + <?4"
If we denote the terms under the 2 sign in 2) by vn we have
THE CIRCULAR FUNCTIONS
257
Now the series 2an converges if q \ < 1. For setting bn= \ <?2n|,
the series 25ra is convergent in this case. Moreover,
lira £» = 1.
7i-=oo 0 ^
H
Thus we may differentiate termwise.
The Circular Functions
213. 1. Sin z and cos x as Infinite Products.
From the addition theorem
sin (nix -f x) = sin (m -f 1) a: = sin raz cos 2: 4- cos mx sin 2:
m = 1, 2, 3 ••• we see that for an odd n
sin rcz = a0 sinn z + a1 sin""1 a? + ••• 4- «n-i sin x
where the coefficients a are integers. If we set t = sin x, we get
sin nx = Fn(t) = aQtn + a^ + . . . + «n_^. (1
Now J"n being a polynomial of degree n, it has n roots. They are
0, ±sin^ ±sin^, ... isini^,
n
n
corresponding to the values of x which make sin nx = 0. Thus
t* - sin2 - • • . i* - sin* -
» \ n
Dividing through by
sin - sin
n n
. n
sin2
n — TT
n
and denoting the new constant factor by a, 1), 2) give
sin nx = a sin x
I _
sin2 x
77
sm* —
n
!1 Il2
..l*i ss!* 1
. 97l — 1 7T
S1"^-2j
258
INFINITE PRODUCTS
To find a we observe that this equation gives
sin nx
sin x
I"., sin2 x
a 1
. 27T
snr —
n_
Letting x = 0 we now get a = n. Thus putting this value of a.
y
in 3), and replacing x by -, we have finally
sin x = n sin - P (x, n)
where
1
sin2-"
n
r 1 ° . ,. U
• 2T7r
n -
• ~ 2
We note now that as n = oo,
Similarly
x
sin -
. # n .
n sin - = x = x.
n x
n
sm " ^
n . x*
It seems likely therefore that if we pass to the limit n = oo in
4), we shall get gin x _ xp^ (5
where P( ^ — \
The correctness of 5) is easily shown.
Let us set
L (x, n) = log P (x, ri) = 2 log
1-
n
,'r.2
sn
= log P(x) = 2 log l -
THE CIRCULAR FUNCTIONS
We observe that
lim P (x, ri) = lim e**> n) = e™ = P (x)
259
provided
lim L(x, ti) = L(x).
We have thus only to prove 7). Let us denote the sum of the
first_w terms in 6) by Ln(x, n) and the sum of the remaining
byZm(z, rc). Then
|ZO,M)-£O)|<
Since for
we have
(8
- < sin x < x,
sin2-
n 4 n2 x2
n n
and hence for an m1 so large that - - < 1, we have,
-log
But the series
1-
sin2-
n
I'TT
sin2
n _
r > m.
is convergent. Hence for a sufficiently large m
mx,n<- , m*<-
Now giving m this fixed value, obviously for all n > some v the
first term on the right of 8) is < e/3, and thus 7) holds.
260 INFINITE PRODUCTS
2. In algebra we learn that every polynomial
«0 + a^x + a2z2 + ... + anxn
can be written as a product
««O - «i)0* - «2) ••• O - «„),
where a « "" are ^s r°°ts. Now
is the limit of a polynomial, viz. the first n terms of 9). It is
natural to ask, Can we not express sin x as the limit of a product
which vanishes at the zeros of sin x ? That this can be done we
have just shown in 1.
3. If we set x = TT/% in 5), it gives,
2"V 4r*J 2 2r.2r
Hence TT ,, 2r.2r 2- 2. 4- 4- 6- 6-
2/9 ,». 1 \/9 « _|_ 1 \ 1 Q Q e c 7 *
c ^ / — j. ) i ^ / "f" x) A*O*O*O*O* i ...
a formula due to Wallis.
4. From 5) we can get another expression for sin z, viz. :
sin x = #11(1 — —}ern r=±l, ±2, ••• (11
V rirj
For the right side is convergent by 197, 2. If now we group
the factors in pairs, we have
rvr,
This shows that the products in 5) and 11) are equal.
5. From 5) or 11) we have
sin x = lim Pn(x) = lim x IT' x+87r (12
n=oo s=-n 87T
where the dash indicates that s = 0 is excluded.
\
THE CIRCULAR FUNCTIONS 261
214. We now show that
coss = nfi- 4a; V (i
To this end we use the relation
sin 2 x == 2 sin x cos x.
Hence
n?r
2
(2«-i)
from which 1) is immediate.
From 1) we have, as in 213, 4,
cos x = ul - n,1)g' n = 0, ±1, ±2, ... (2
215. From the expression of sin #, cos x as infinite products,
their periodicity is readily shown. Thus from 213, 12)
sina: = lim Pn(x).
Hence
or • > N •
sin (a: -f TT) = — sin a:.
Hence sin(x + 2 TT) = sin x
and thus sin a; admits the period 2 TT.
216. 1. Infinite Series for tana:, cosec #, e^c.
If 0<a;<7r, all the factors in the product 213, 5) are positive.
TllUS log sin x = log x + log l- , 0<a:<7r. (1
262 INFINITE PRODUCTS
Similarly 214, 1) gives
log cos. = log l-2 , 0<,<. (2
To get formulae having a wider range we have only to square
the products 213, 5) and 214, 1). We then get
log sin2 z = log z2 + 2 log (l - -f^-Y, (3
i \
valid for any x such that sin x =£ 0 ; and
A ^,2 \ 2
l-^-M^), (4
valid for any x such that cosx=£ 0.
If we differentiate 3), 4) we get
. 9-
(5
(6
,
j a;2 — « V
*
'c? «s iw 3), 4).
Remark. The relations 5), 6) exhibit cot x, tan x as a series of
rational functions whose poles are precisely the poles of the given
functions. They are analogous to the representation in algebra
of a fraction as the sum of partial fractions.
2. To get developments of sec #, cosec a?, we observe that
cosec x = tan \ x + cot x.
Hence
1 ^ 4
= ~~*~ *j fv 0 _ 1
s-l)27T2-
a^
valid for x^± STT.
THE CIRCULAR FUNCTIONS
3. To get sec x, we observe that
cosec f ^ — x J = sec x.
263
Now
Hence
cosec
x i (STT — x STT + X]
H.cfe-sV— +±(-
cosec
=s.
^ J |_, ,
7T
Let us regroup the terms of S, forming the series
As
= o,
:TT — a;
we see that T7 is convergent and = S. Thus
wr—'Zf
sec a; — 2,^ —
valid for all a; such that cos a; ^= 0.
217. As an exercise let us show the periodicity of cot x from
216, 5). We have
Now
TO ^
cot a; = lim Fn(x) = lim V a: =£ STT,
"=» ^nx H- *T
1
a; 4- (n+ l)?r x — nir
Letting TI = oo we see that
lim ^(2: + TT) = lim Fn(x)
cot (x 4- ?r) = cot a;.
and hence
264 INFINITE PRODUCTS
218. Development of log sin #, tan x, etc., in power series.
From 216, 1)
sn x
If we give to -^-^its limiting value 1 as x = 0, the relation 1)
x
holds for | x \ < TT.
Now for I x I < TT
Thus
sin x x2 . 1 x* . 1 XQ
227T2 2247T4 3267T6
^2 i 1 ^ , 1 a*
327T2 2347T4 3367T6
provided we sum this double series by rows. But since the series
is a positive term series, we may sum by columns, by 129, 2.
Doing this we get
+... , . (2
where 1111
^=i+i+i^+- ' ,;•";;
The relation 2) is valid for \ x | < TT.
In a similar manner we find
96 ^6
£(?6+... (3
valid for | »| <• Here
VT^ 7T
THE CIRCULAR FUNCTIONS 265
The terms of G-n are a part of Hn. Obviously
These coefficients put in 3) give
valid for | x \ < - • If we differentiate 4) and 2), we get
Zt
tan x= 2(22 _ 1>F ^ + 2(2< _ 1)^ + 2(2«- 1)^ + ... (5
7T* 7T* 7T"
valid for |a?| <— ;
2i
eot*=i-2^-2J?i^-2J.g-... (6
£ 7T 7T 7T
valid for 0 < \x\ < TT.
Comparing 5) with the development of tan x given 165, 3)
gives
a,
i
"12
1
1
1 ,
1
32
1
= 6
7T4
1 2?r2-»
6*2!"
1 237T4
. 2?r2.
L t^ y
1* '
1 ,
2*
1 ,
„ . i
i ,
90
7T6
30 4!
1 257T6
3 4!
i? 257T6
'I'1
1 ,
1
36 +
!+...
945
7T8
42 6!
1 27 7T8
5 6!
07 ,,.8
z?
8
1' '
28 '
9450
30 8!
7 8!
Let u
.s set
jy —
2*-V-
1 J&2
Then
n T. —
5) gi\
res
:24-i)
-1)
(8
valid for | a; | < ^ • The coefficients B^ B3 - • - are called Ber-
nouillian numbers. From 7) we see
A =6 ' ^3 = A ' ^5 = A ' A = A'
266 INFINITE PRODUCTS
From 6), 8) we get
cota»H=-|(f^»-^-' do
valid for 0 < x \ < ?r.
219. Recursion formula for the Bernouillian Numbers.
If we set f(x) = tan x,
we have by Taylor's development
o ! 51
where /2n-i,(0) 2(2'- - I)g2n 22"(22" - 1) R
(2w-l)! ,T2» (2re)! 2-1
Now by I, 408,
/o»-i>(0)- (2 Y
From 1), 2) we get
2
^Y1)^^^^,A..(-1) (3
We have already found ^, ^3, .B., ^7 ; it [s now easy to find
successively :
Thus to calculate J59, we have from 3)
2io - 1) B _ ^8 27(28 - 1) J_ 9.8.7.6 25(26-1) ^
5 " 1 • 2 " 4 ' 30 1 . 2 . 3 . 4 3 42
Q . Q . 7 93/^94 1\ 1 -t
_ V O j £ ^ - JJ J_ 1
1-2.3 2 30 ^ C } ' 6
Thus
512
5 . 7936 _5^
512 . 1023 ~ 66'
~ 2°16
THE B AND T FUNCTIONS 267
The B and F Functions
220. In Volume I we defined the B and F functions by means
of integrals: r* u_1 -,
B (u, v) = \ — (1
r<V)= Ce-xxu~ldx (2
(u)=
^0
which converge only when u, v > 0. Under this condition we saw
We propose to show that T(u) can be developed in the infinite
product / j
(4
n «
This product converges, as we saw, 197, 3, for any u =£ 0, — 1,
— 2, ••• From 201, 7 and 207 it is obvious that G- converges abso
lutely and uniformly at any point u different from these singular
points. Thus the expression 4) has a wider domain of definition
than that of 2). Since G- = F^ as we said, for w>0, we shall ex
tend the definition of the F function in accordance with 4), for
negative u.
It frequently happens that a function f(x) can be represented
by different analytic expressions whose domains of convergence
are different. For example, we saw 218, 9), that tan x can be de
veloped in a power series
92/92
^ ^~
92
valid for a?<?« On the other hand,
tan x =
X X* X°
II 81 51* '" sin a;
r2
cosz
_
214!
268 INFINITE PRODUCTS
and w x
tanz = 227oT — 1x2 ^ 216' 6)
are analytic expressions valid for every x for which the function
tan x is denned.
221. 1. Before showing that Gr and F have the same values for
u > 0, let us develop some of the properties of the product Gr given
in 220, 4). In the first place, we have, by 210:
The function Cr(u) is continuous, except at the points u = 0, — 1,
-2, ...
Since the factors of 4) are all positive for u > 0, we see that
Gr(u) is positive for u > 0.
2. In the vicinity of the point x = — m, m = 0, 1, •••
x + m
where H(u) is continuous near this point, and does not vanish at
this point.
For
'! + -
m
m
where If is the infinite product G- with one factor left out. As we
may reason on .ZTas we did on G-, we see .ZT converges at the point
x = — m. Hence ff=£ 0 at this point. But ^Talso converges uni
formly about this point; hence H is continuous about it.
222> „ r 1 1.2.-O-1)
G- = lim ^ ^ nu. (1
n=* u (u + 1) (u + 2) • . - (u + n — 1)
To prove this relation, let us denote the product under the limit
sign by Pn. We have
2 3 4 n
THE B AND T FUNCTIONS 269
Also
Thus Pn = Grn. But Grn = G-, hence Pn, is convergent and Gr =
limPn.
223. Uuler's Constant. This is defined by the convergent series
It is easy to see at once that
\ n.
where C is the Eulerian constant.
For when a > 0, au = eu loga.
Hence
1 *'U<
u
Now
and
by 218, 7). By calculation it is found that
C^ .577215 —
224. Another expression of G- is
"c . , ^ = 1,2,... (i
270 INFINITE PRODUCTS
are convergent. Hence
/
from which 1 ) follows at once, using 223.
225. Further Properties of G-.
Let us use the product
P.00-1 __ fr"1" __ »•
u (« + !)••• ,(«+•» -?!)
employed in 222. Then
Pn(tt + l) = ^P^i». (2
w 4- n
As
= u as w = QO
M 4- n
we get 1) from 2) at once on passing to the limit.
2. #O + rc)=tt(i* + l) ... O + ra-l)#O). (3
This follows from 1) by repeated applications.
3. #00=1.2. ...w-l=(w-l)I (4
where n is a positive integer.
sin TTU
For G(l-u)=-uG(-u) b 1
= — , by 224, 1).
Hence
n
THE B AND T FUNCTIONS 271
We now use 213, 5).
Let us note that by virtue of 1, 2 the value of Gr is known for
all u > 0, when it is known in the interval (0, 1). By virtue of
5) G- is known for u < 0 when its value is known for u > 0.
Moreover the relation 5) shows the value of 6r is known in (J, 1)
when its value is known in (0, J).
As a result of this we see G- is known when its values in the
interval (0, J) are known ; or indeed in any interval of length J.
Gauss has given a table of log Gr(u) for l<w<1.5 calculated
to 20 decimal places. A four-place table is given in " A Short
Table of Integrals " by B. 0. Peirce, for 1 < u < 2.
5. &($) = V^. (6
For in 5) set u = ^. Then
Hence &(£) =
We must take the plus sign here, since G- > 0 when u > 0, by 221
1-3. 5- ..27i-
where n is a positive integer.
For
226. Expressions for log 6r(w), a?ic? ^8 Derivatives
From 224, 1) we have for w > 0,
ZO) = log (? O) = - CW
Differentiating, we get
U l VH M +
That this step is permissible follows from 155, 1.
272 INFINITE PRODUCTS
We may write 2)
(3
That the relations 2), 3) hold for any u=£ 0, - 1, - 2 ••• follows
by reasoning similar to that employed in 216. In general we have
^-(-lyO-')'^,1.!), . r>L (4
In particular,
- (7. (5
-iyo-i)iSi.(-i)'oi-i)!Jt.
227. Development of log #(%) m a Power Series. If Taylor's
development is valid about the point u = 1, we have
log 0(«) = £
or using 226, 5), and setting u = 1 + #,
log
n
We show now this relation is valid for — J < x < 1, by proving
that
converges to 0, as s = oo .
For, if 0<z<l, then
Also if -i<o;<0,
x
Ox
The relation 1) is really valid for — 1 < a;< 1, but for our pur
pose it suffices to know that it holds in 51 = (— J, 1). Legendre
THE B AND T FUNCTIONS 273
has shown how the series 1) may be made to converge more
rapidly. We have for any x in 51
2 n
This on adding and subtracting from 1) gives
log 0(1 + x) = - log (1 + x) + (1 - C)x + I( - !
2
Changing here x into — a; gives
log <?(! - T) = - log (1 - *) - (1 - C)x + 2(J5T.- I)?"
71
Subtracting this from the foregoing gives
log 0(1 + a) - log (7(1 - *)
From 225, 4
log (7(1 + x)
sin TTX
This with the preceding relation gives
log (7(1 + z)
valid in 51.
This series converges rapidly for 0<#<i, and enables us to
compute G-(u) in the interval l<w<|. The other values of (7
may be readily obtained as already observed.
228. 1. We show now with Pringsheim* that Gr(u) =r(w),/0r
u>0.
We have for 0<><ll,
r(w + ?i) = f Vzzu+n-1<fa
0
. Annalen, vol. 31, p. 455.
274 INFINITE PRODUCTS
Now for any x in the interval (0, ri),
xu<nu , xu>xnu~1
since u > 0 and u — 1 < ; 0.
Also for any x in the interval (n, oo )
xu<xnu~l , xu>^nu.
Hence
/•n
nu-ij e
r*n s+<*>
<nu\ e~xxn-ldx + nu-l\ e~xxndx.
»/0 Jn
Thus
/*ra 1 /^a°
I e-^71-1^^- I
^0 ^^/O
— ± I e-*x*dx.
Let us call these integrals ^4, ^, (7 respectively.
We see at once that
= n\ =
n n
Also, integrating by parts,
Thus
Similarly
Hence
where
e*(n-l)\ enn\
Now
THE B AND T FUNCTIONS 275
But
svi mm
vn > 1 H T + — + 7 — ^ , for any m
n+l O + l) ... (n + m)
mnm m m
(n + 1) -. O + m) A + ^ ~ /t ~ wA fl + m\n
\ nj \ nj \ n)
Let us take
n>m* or *<
?i 7W
Then
m m
Since m may be taken large at pleasure,
lim vn = oo
and hence ,.
lim qn = 0.
Thus
But from T(u + 1) = uT(u) we have
T(u + 1 +ri) _ u + n Y(u + n) __ -,
also, as n = oo . Thus the relation 1) holds for 1 < u<_ 2, and in
fact for any u >0.
A«a
-iXO Tl /^/i* ) .,_ "\ /\ f I "1 \ /^ I
we have
Hence using 1), 1N . ^.
(M.-l)!n« F(M + n)
u(u + 1) ..« (w + n- 1) (w-1)!^
Letting w= oo , we get F(w)= (r(w) for any w>0, making use
of 1) and 222, 1).
2. Having extended the definition of F(w) to negative values
of w, we may now take the relation
as a definition of the B function. This definition will be in
accordance with 220, 1) for w, v > 0, and will define B for negative
w, v when the right side of 2) has a value.
CHAPTER VIII
AGGREGATES
Equivalence
229. 1. Up to the present the aggregates we have dealt with
have been point aggregates. We now consider aggregates in
general. Any collection of well-determined objects, distinguish
able one from another, and thought of as a whole, may be called
an aggregate or set.
Thus the class of prime numbers, the class of integrable func
tions, the inhabitants of the United States, are aggregates.
Some of the definitions given for point aggregates apply obvi
ously to aggregates in general, and we shall therefore not repeat
them here, as it is only necessary to replace the term point by
object or element.
As in point sets, 51 = 0 shall mean that 51 embraces no elements.
Let 51, 33 be two aggregates sucli that each element a of 51 is
associated with some one element b of 33, and conversely. We say
that 51 is equivalent to 33 and write
51-33.
We also say 51 and 33 are in one to one correspondence or are in
uniform correspondence. To indicate that a is associated with b
in this correspondence we write
a~b.
2. If 51 ~ 33 and 33 ~ 6, then 51 ~ {£.
For let a ~ b, b ~ c. Then we can set 51, (£ in uniform corre
spondence by setting a ~ c.
3. Let 5( = 33 + £ + £)+ -
A = B + C+D + •••
// 33 ~ B, 6 ~ C\ •••, then 51 ~ A.
276
EQUIVALENCE 277
For we can associate the elements of 51 with those of A by
keeping precisely the correspondence which exists between the
elements of 33 and B, of £ and (7, etc.
Example 1. 51 = 1, 2, 3, ...
33 = aj, &2, a3, ...
If we set an~ n, 51 and 33 will stand in 1, 1 correspondence.
Example 2. 51 = 1, 2, 3, 4, •••
33 = 2,4,6,8, -
If we set n of 51 in correspondence with 2 n of 33, 51 and 33 will
be in uniform correspondence.
We note that 33 is a part of 5t ; we have thus this result : An
infinite aggregate may be put in uniform correspondence with a
partial aggregate of itself.
This is obviously impossible if 51 is finite.
Example 3. 51 = 1, 2, 3, 4, ...
33 = 101, 102, 103, 104, ...
If we set n ~ 10n, we establish a uniform correspondence be
tween 51 and 33. We note again that 31 ~ 33 although 51 > 33.
Example 4- Let (E = \% j, where, using the triadic system,
f=-«A- f.= 0,2
denote the Cantor set of I, 272. Let us associate with % the point
where xn = 0 when fn =0, and =1 when f n = 2 and read 1) in
the dyadic system.
Then \x\ is the interval (0, 1). Thus we have established a
uniform correspondence between (E and the points of a unit interval.
In passing let us note that if f < f ' and a;, a/ are the correspond
ing points in \x\, then x < x' .
This example also shows that we can set in uniform correspond
ence a discrete aggregate with the unit interval.
We have only to prove that (E is discrete. To this end consider
the set of intervals C marked heavy in the figure of I, 272. Ob-
278 AGGREGATES
viously we can select enough of these deleted intervals so that
their lower content is as near 1 as we choose. Thus
Cont 0=1.
As Cont C < 1, O is metric and its content is 1. Hence (£ is
discrete.
230. 1. Let tyi=a + A, 33 = /3 -f .#, where a, b are elements
of SI, 33 respectively. 7/1 SI ~- 33, then A~B avid conversely.
For, since SI ~ 33, each element a of 51 is associated with some
one element b of 33, and the same holds for 33.' If it so happens
that a ~ /3, the uniform correspondence of A, B is obvious. If
on the contrary « ~ b' and /3 ~ a', the uniform correspondence be
tween A, B can be established by setting a' ~ b' and having the
other elements in A, B correspond as in 51 ~ 33.
2. We state as obvious the theorems:
No part 33 of a finite set SI can be ~ SI.
No finite part 33 of an infinite set 51 can be ~ SI.
Cardinal Numbers
231. 1. We attach now to each aggregate SI an attribute
called its cardinal number, which is defined as follows :
1° Equivalent aggregates have the same cardinal number.
2° If SI is — to a part of 33, but 33 is not ~ SI or to any part
of SI, the cardinal number of SI is less than that of 33, or the
cardinal number of 33 is greater than that of SI. The cardinal
number of SI may be denoted by the corresponding small letter
a or by Card SI.
The cardinal number of an aggregate is sometimes called its
power or potency.
If SI is a finite set, let it consist of n objects or elements.
Then its cardinal number shall be n. The cardinal number of
a finite set is said to be finite, otherwise transfinite. It follows
from the preceding definition that all transfinite cardinal num
bers are greater than any finite cardinal number.
CARDINAL NUMBERS 279
2. It is a property of any two finite cardinal numbers a, b that
either
a = b , or a > b , or a < b. (1
This property has not yet been established for transfinite car
dinal numbers. There is in fact a fourth alternative relative to
21, 33, besides the three involved in 1). For until the contrary
has been shown, there is the possibility that :
No part of 21 is ~ 23, and no part of 23 is ~ 21.
The reader should thus guard against expressly or tacitly
assuming that one of the three relations 1) must hold for any
two cardinal numbers.
3. We note here another difference. If 21, 23 are finite with
out common element,
Card (21 + 23) > Card 21. (2
Let now 21 denote the positive even and 23 the positive odd
numbers. Obviously
Card (21 + 23) = Card 21 = Card 23
and the relation 2) does not hold for these transfinite numbers.
4. We have, however, the following :
Let 21 > 23, then
Card 21 > Card 23.
For obviously 23 is ~ to a part of 21, viz. 23 itself.
5. This may be generalized as follows :
Let
If Card 23 < Card B , Card <£ < Card (7, etc.,
then Card 21 < Card A.
For from Card 23 <. Card B follows that we can associate in 1,
1 correspondence the elements of 23 with a part or whole of B.
The same is true for (£, (7; $), D ; •••
Thus we can associate the elements of 21 with a part or the
whole of A.
280 AGGREGATES
Enumerable Sets
232. 1. An aggregate which is equivalent to the system of
positive integers $ or to a part of 3 ig enumerable.
Thus all finite aggregates are enumerable. The cardinal num
ber attached to an infinite enumerable set is K0, aleph zero.
At times we shall also denote this cardinal by e, so that
e = V
2. Every infinite aggregate 51 contains an infinite enumerable set $&.
For let a1 be an element of 51 and
Then 5tj is infinite ; let az be one of its elements and
Then 512 is infinite, etc.
Then $ = a
is a part of 21 and forms an infinite enumerable set.
3. From this follows that
K0 is the least transfinite cardinal number.
233. The rational numbers are enumerable.
For any rational number may be written
r = ™ (1
n
where, as usual, m is relatively prime to n.
The equation
admits but a finite number of solutions for each value of
^ = 2,3,4, .-
Each solution m, n of 2), these numbers being relatively prime,
gives a rational number 1). Thus we get, e.g.
p=2 , ±1.
O i O i 1
P = o •> ±2, ± -J.
P = 4: , ±3, ±J.
J> = 5 , ±4, ±J , ±| ±J.
ENUMERABLE SETS 281
Let us now arrange these solutions in a sequence, putting those
corresponding to p = q before those corresponding to p = q + 1.
We get
r\ > r2 » r3 .- (3
which is obviously enumerable.
234. Let the indices tj, t2, '•• *p range over enumerable sets. Then
«VK-^J
z* enumerable.
For the equation
"i + "2 + "• + "* = w,
where the i/'s are positive integers, admits but a finite number
of solutions for each n = p, jo + 1, p + 2, jt? + 3 «•« Thus the
elements of $»_$/, ?
may be arranged in a sequence
by giving to n successively the values p, p + 1, ••• and putting the
elements bVl...Vp corresponding to n = q+ 1 after those correspond
ing to n = q.
Thus the set 35 is enumerable. Consider now 51. Since each
index im ranges over an enumerable set, each value of im as i'm is
associated with some positive integer as m' and conversely. We
may now establish a 1, 1 correspondence between 31 and 33 by
setting
£/»x-™;~aiX-v
Hence 21 is enumerable.
235. 1. An enumerable set of enumerable aggregates form an
enumerable aggregate.
For let 5t, 33, (£ ••• be the original aggregates. Since they form
an enumerable set, they can be arranged in the order
*!,«,**,, - (1
But each $Im is enumerable ; therefore its elements can be
arranged in the order
am\ > «m2 » ^m3 » ^m* » '"
282 AGGREGATES
Thus the a-elements in 1) form a set
\amn\ m, n,= 1, 2, •••
which is enumerable by 234.
2. The real algebraic numbers form an enumerable set.
For each algebraic number is a root of a uniquely determined
irreducible equation of the form
xn+aiXn-l+ ... +an=0,
the a's being rational numbers. Thus the totality of real algebraic
numbers may be represented by
\Pn, aidf- an\
where the index n runs over the positive integers and a1 ••• an range
over the rational numbers.
3. Let 51, 33 ^ two enumerable sets. Then
Card 51=
And in general if^, 512 — are an enumerable set of enumerable
aggregates, Card (5^ , 512 ,•••) = K0.
This follows from 1.
236. Every isolated aggregate 91, limited or not, forms an enumer
able set.
For let us divide 9?m into cubes of side 1. Obviously these form
an enumerable set C\, <72'». About each point a of 91 in any <7n
as center we describe a cube of side er, so small that it contains no
other point of 91. This is possible since 91 is isolated. There are but
a finite number of these cubes in On of side a- = -, v = 1, 2, 3, •••
v
for each v. Hence, by 235, l, 91 is enumerable.
237. 1. Every aggregate of the. first species 91, limited or not, is
enumerable.
For let 91 be of order n. Then
91 = 91 + 9i;
ENUMERABLE SETS 283
where 5lt denotes the isolated points of 51 and 51^ the proper limit
ing points of 51-
Similarly,
Thus,
5i = 5t; + 5i;,t + 5r;,t + - + a?).
But 5l(n) is finite and 51<;1) < 5I(n).
Thus 51 being the sum of n 4- 1 enumerable sets, is enumerable.
2. .Tjf 51' is enumerable, so is 51.
For as in 1,
* » X + 9£
and 5i;<5{'.
238. 1. Every infinite aggregate 51 contains a part 53 swc?A £^a£
53-51.
For let & = («j, a2, «3 •••) be an infinite enumerable set in 51,
so that
51 = (5 + g.
Let (g = d + .#.
To establish a uniform correspondence between .#, (5 let us
ociate an in
We now set
associate an in @ with an+1 in J^. Thus
Obviously 51 ~ 53 since .Z7~ (5, and the elements of 5 are common
to 51 and 53. -
2. -Zjf 5l~53 are infinite, each contains a part 5lj, 53i such that
For by 1, 51 contains a part 5^ such that 51 — 51J. Similarly,
53 contains a part 53: such that 53— 53r As 51 — 33, we have the
theorem.
284 AGGREGATES
239. 1. A theorem of great importance in determining
whether two aggregates are equivalent is the following. It is
the converse of 238, 2.
then ^^^
In the correspondence 3^ ~ 33, let 512 be the elements of 5Ij
associated with 33X . Then
5I2 ~ $! ~ 31
and hence or <ar /1
& «v eig* (.1
But as Slj > $12 , we would infer from 1) that also
21 ~V (2
As ^ ~ 33 by hypothesis, the truth of the theorem follows at
once from 2).
To establish 2) we proceed thus. In the correspondence 1), let
5(3 be that part of $12 which ~ ^ in 51. In the correspondence
^ ~ 513, let 514 be that part of 5t3 which ~ 512 in 5lx .
Continuing in this way, we get the indefinite sequence
51 > 5Ij > 512 > 313 > ••.
SUCh that rw rvf af
S4 ~ <!12 ~ ^14 ~ •••
3^-213-21,-...
Let now
Then 3l = D + G1H-G2 + 68 + C4+ - (3
and similarly «, = ® + G, + G, + G4 + S5 + •"
We note that we can also write
2l1 = £)+G3 + l£2 + e6 + G4+ ». (4
Now from the manner in which the sets 513, 514 ••• were obtained,
it follows that
£i~£3 » e8-@6- (5
Thus the sets in 4) correspond uniformly to the sets directly
above them in 3), and this establishes 1).
ENUMERABLE SETS 285
2. In connection with the foregoing proof, which is due to
Bernstein, the reader must guard against the following error. It
does not in general follow from
that £ ~ (5
which is the first relation in 5).
Example. Let 51 = (1, 2, 3, 4, •••).
511=(2, 3, 4, 5".) , 512=(3, 4, 5, 6 ...)
513=(5, 6, 7, 8 •••).
Then £ = 1 6 = (3, 4).
Now 5(, 51 r 51 2 , 513 are all enumerable sets ; hence
9T <~ 91 91 ~- 31
But obviously C^ is not equivalent to (E3, since a set containing
only one element cannot be put in 1 to 1 correspondence with a
set consisting of two elements.
240. 1. If%>$>
For by hypothesis a part of 23, viz. (£, is ~2L But a part of
is ~33, viz. 33 itself. We apply now 239.
2. Let a be any cardinal number. If
then a = Card $.
For let Card 51 = «. Then from
a < Card 33
it follows that 51 ~ a part or the whole of 33 ; while from
Card 33 < «
it follows that 33 is ~ a part or the whole of 51.
3. Any part 33 of an enumerable set 51 is enumerable.
For if 33 is finite, it is enumerable. If infinite,
Card33>K0-
On the other hand
Card 33 < Card 51 = K0.
286 AGGREGATES
4. Two infinite enumerable sets are equivalent.
For both are equivalent to $, the set of positive integers.
241. 1. Let (£ be any enumerable set in 51 ; set 51 = (£ + 23. If
23 ^s infinite, 51 — 53.
For 23 being infinite, contains an infinite enumerable set g.
Let 23 = g + ®. Then
51 = (5 + g + ®,
8 -. 8 +'ft
But <g + g ~ g. Hence 51-23.
2. We may state 1 thus :
Card (31 -(S)= Card 51
provided 51 — (5 i« infinite.
3. From 1 follows at once the theorem :
.Let 51 be any infinite set and (5 an enumerable set. Then
Card (51 + <g) = Card 51.
Some Space Transformations
242. 1. Let T be a transformation of space such that to each
point x corresponds a single point XT, and conversely.
Moreover, let #, y be awy two points of space. After the trans
formation they go over into XT, yT. If
Dist(>, y)= DistOy, yT)
we call jTa displacement.
If the displacement is defined by
x'1 = xl + al , ... x'm = xm + am
it is called a translation.
If the displacement is such that all the points of a line in space
remain unchanged by T, it is called a rotation whose axis is the
fixed line.
THE CARDINAL c 287
If 9? denotes the original space, and 9?r the transformed space
after displacement, we have, obviously,
2. Let =tx = tx t Q ^
Then when x ranges over the w-way space 3E, y ranges over an
w-way space 9). If we set x ~ y as defined by 1),
Alcn
Dist (0, y) =t Dist (0, x).
We call 1) a transformation of similitude. If t > 1, a figure in
space is dilated ; if t < 1, it is contracted.
3. Let Q be any point in space. About it as center, let us de
scribe a sphere S of radius R. Let P be any other point. On the
join of P, Q let us take a point P' such that
Dist (P', Q) =
Dist (P, Q)
Then P' is called the inverse of P with respect to S. This trans
formation of space is called inversion. Q is the center of inversion.
Obviously points without S go over into points within, and con
versely. As P = x , P' = Q.
The correspondence between the old and new spaces is uniform,
except there is no point corresponding to Q.
TJie Cardinal c
243. 1. All or any part of space & may be put in uniform cor
respondence with a point set lying in a given cube C.
For let @t denote the points within and on a unit sphere S about
the origin, while &e denotes the other points of space. By an in
version we can transform @e into a figure @/ lying in S. By a
transformation of similitude we can contract @t, &f as much as we
choose, getting ^', (gy- We may now displace these figures so
as to bring them within C in such a way as to have no points in
common, the contraction being made sufficiently great. The
288 AGGREGATES
correspondence between <S and the resulting aggregate is obviously
uniform since all the transformations employed are.
As a result of this and 240, l we see that the aggregate of all
real numbers is ~ to those lying in the interval (0, 1); for example,
the aggregate of all points of SRm is ~ to the points in a unit cube,
or a unit sphere, etc.
244. 1. The points lying in the unit interval 5t = (0*, 1*) are
not enumerable.
For if they were, they could be arranged in a sequence
«i, «2' a3 "• 0-
Let us express the as as decimals in the normal form. Then
an = ' anian2ana '"
Consider the decimal
b = • &J&2&3 ...
also written in the normal form, where
^1^^1,1 ' ^2^^2,2 ' ^3^^3,3 i "•
Then b lies in 5( and is yet different from any number in 1).
2. We have (0*, 1*) ~ (0, 1) , by 241, 3,
-(a, 6) , by 243,
where a, b are finite or infinite.
Thus the cardinal number of any interval, finite or infinite,
with or without its end points is the same.
We denote it by c and call it the cardinal number of the recti
linear continuum, or of the real number system $1.
Since 9^ contains the rational number system R, we have
OKo-
3. The cardinal number of the irrational or of the transcendental
numbers in any interval 51 is also c.
For the non-irrational numbers in 51 are the rational which are
enumerable ; and the non-transcendental numbers in 51 are the
algebraic which are also enumerable.
THE CARDINAL C 289
4. The cardinal number of the Cantor set (£ of I, 272 is c.
For each point a of (E has the representation ' in the triadic
system n 0
J a = • a-,a9ao ••• , a = U, ^.
But if we read these numbers in the dyadic system, replacing
each an = 2 by the value 1, we get all the points in the interval
(0, 1). As there is a uniform correspondence between these two
sets of points, the theorem is established.
245. An enumerable set 51 is not perfect, and conversely a perfect
set is not enumerable.
For suppose the enumerable set
51 = ^, «a — (1
were perfect. In D^^a^) lies an infinite partial set 51 x of 51,
since by hypothesis 51 is perfect. Let a^ be the point of lowest
index in 5lj . Let us take r2 < rl such that D^(a^) lies in
D* (ai)- In -A-fC^mi) ^es an infinite partial set 512 of 5lr Let
am3 be the point of lowest index in 512, etc.
Consider now the sequence
It converges to a point a by I, 127, 2. But a lies in 51, since this
is perfect. Thus a is some point of 1), say a = a,. But this
leads to a contradiction. For a, lies in every .Z)r*n(#™n); on the
other hand, no point in this domain has an index as low as mn
which = oo, as n = GO. Thus 51 cannot be perfect.
Conversely, suppose the perfect set 51 were enumerable. This
is impossible, for we have just seen that when 51 is enumerable it
cannot be perfect.
246. Let 51 be the union of an enumerable set of aggregates 5ln
each having the cardinal number c. Then Card 51 = c.
For let 33n denote the elements of 5ln not in 5lj,5l2 ••• 5ln_j.
Let (£B denote the interval (w — 1, w*). Then the cardinal
number of Q -f- £ + ... is c.
290 AGGREGATES
But Card33n< Card£n.
Hence Card 21 < c , by 231, 6. (1
On the other hand,
Card SI > Card 21! = c. (2
From 1), 2) we have the theorem, by 240, 2.
247. 1. As already stated, the complex x= (2^, #2, ••• #n) de
notes a point in n- way space. Let x±, #2, ••• denote an infinite
enumerable set. We may also say that the complex
x— (2^, #2, • •• in inf.)
denotes a point in oo -way space 9^.
2. Let 21 denote a point set in $Rn, n finite or infinite. Then
Card 21 < c. (1
For let us first consider the unit cube (£ whose coordinates xm
range over 33 = (0*, 1*). Let £) denote the diagonal of (L Then
c = Card £) < Card <£. (2
On the other hand we show Card (£ < c.
For let us express each coordinate xm as a decimal in normal
form. Then
Let us now form the number
obtained by reading the above table diagonally. Let 9) denote the
set of ?/'s so obtained as the #'s range over their values. Then
For the point y, for example, in which aln= 0, n = 1, 2, ••• lies
in 33 but not in 9) as otherwise xl = 0. Let us now set x ~ y.
Then |£ ~ 9 and hence Car(J g < £ (3
From 2), 3) we have Card G> = c.
THE CARDINAL c 291
Let us now complete (E by adding its faces, obtaining the set C.
By a transformation of similitude T WQ can bring CT within (£.
Hence Card S> Card C.
On the other hand, (£ is a part of (7, hence
Card £ < Card C.
Thus Card C = c. The rest of the theorem follows now easily.
248. Let g = j/J denote the aggregate of one-valued continuous
functions over a unit cube & in 9?n.
Then
Let C denote the rational points of (5, i.e. the points all of
whose coordinates are rational. Then any / is known when its
values over 0 are known. For if a is an irrational point of (5,
we can approach it over a sequence of rational points alf #2 ••• = a.
But / being continuous, /(a) = lim/(an), and / is known at «.
On the other hand, 0 being enumerable, we can arrange its points
in a sequence n
O=c^ c^ ...
Let now 9^ be a space of an infinite enumerable number of
dimensions, and let y = (^, y^ • ••) denote any one of its points.
Let / have the value vl at c1, the value 7/2 at cz and so on for
the points of C. Then the complex T/J, ?;2, ... completely deter
mines / in (£. But this complex also determines the point
V = C7?!^ ^2 '") i'n ^x- W° now associate / with 77. Thus
Card g <. Card ft = c.
But obviously Card g > c, for among the elements of g there
is an/ which takes on any given value in the interval (0, 1), at
a given point of (L
249. There exist aggregates whose cardinal number is greater
than any given cardinal number.
Let 33= \b\ be an aggregate whose cardinal number b is given.
Let a be a symbol so related to 33 that it has arbitrarily either
the value 1 or 2 corresponding to each b of 33. Let 51 denote the
292 AGGREGATES
aggregate formed of all possible a's of tins kind, and let a, be its
cardinal number.
Let ft be an arbitrary element of 33. Let us associate with ft
that a which has the value 1 for b = ft and the value 2 for all
other 5's. This establishes a correspondence between 33 and a
part of 51. Hence
a^b.
Suppose a = b. Then there exists a correspondence which
associates with each b some one a and conversely. This is
impossible.
For call ab that element of 51 which is associated with b. Then
ab has the value 1 or 2 for each ft of 33. There exists, however,
in 51 an element a' which for each ft of 33 has just the other
determination than the one ab has. But a' is by hypothesis
associated with some element of 33, say that
Then for b = br, a1 must have that one of the two values 1, 2
which ab' has. But it has not, hence the contradiction.
250. The aggregate of limited integrable functions $ defined over
5( = (0, 1) has a cardinal number f > c.
For let f(x) = 0 in 51 except at the points (5 of the discrete
Cantor set of I, 272, and 229, Ex. 4. At each point of {£ let /
have the value 1 or 2 at pleasure. The aggregate ® formed of
all possible such functions has a cardinal number > c, as the
reasoning of 249 shows. But each / is continuous except in (5,
which is discrete. Hence / is integrable. But gf > ®. Hence
.': ."••".. .' ;. f><- '•-' " .'"'.'•• ,
Arithmetic Operations with Cardinals
251. Addition of Cardinals. Let 51, 33 be two aggregates with
out common element, whose cardinal numbers are a, b. We define
the sum of a and b to be
Card (51, 33)= a + b.
ARITHMETIC OPERATIONS WITH CARDINALS 293
We have now the following obvious relations:
K0 + n = K0 , n a positive integer. (1
+ ••• + K0 = K0 , w terw*. (2
+ K0 + •" = KO , aw. infinite enumerable set of terms. (3
cardinal numbers of 51, 23, (? are a, b, c,
The first relation states that addition is associative, the second
that it is commutative.
252. Multiplication.
1. LetH = {a{, ^3 = J5J have the cardinal numbers a, b. The
union of all the pairs (a, b) forms a set called the product of 51 and
23. It is denoted by 51 • 23. We agree that (a, 6) shall be the
same as (5, a). Then
51-23 = 23-51.
We define the product of a and b to be
Card 51 - 23 = Card 23 • 5T = a . b = b - a.
2. We have obviously the following formal relations as in finite
cardinal numbers :
a(b -c) = (a -b)c,
a • b = b • a,
a(b + c) = ab + ac,
which express respectively the associative, commutative, and dis-
tripulative properties of cardinal numbers.
Example 1. Let 5l=jaj, %> = \l\ denote the points on two
indefinite right lines. Then
If we take a, b to be the coordinates of a point in a plane 9?2,
* The reader should note that here, as in the immediately following articles, c is
simply the cardinal number of (£ which is any set, like 31, 53 •••
294 AGGREGATES
Example 2. Let 21 = \a\ denote the family of circles
Let 33 = 5^! denote a set of segments of length b. We can
interpret (#, 6) to be the points on a cylinder whose base is 1)
and whose height is I. Then 91-33 is the aggregate of these
cylinders.
253. 1. K0 = ra.K0 , or nt = t. (1
For let ^ , ^ _v
(§ = (<?!, £2 ••• in inf.)
Then K--c~<;«t,'o , («i,«a) , («i» «•)•••
(/7 ^ ^ ^/7 /> ^ r/7 ^ "\ . .
2' lx ' v^' 2^ ' VM/21 3x
The cardinal number of the set on the left is n£0, while the
cardinal number of the set on the right is K0 .
2. ec = c. (2
For let (£ = {c} denote the points on a right line, and (5 = (1, 2,
Then (S£
may be regarded as the points on a right line ln. Obviously,
Card JU=c.
Hence
ec = Card (g(£ = c.
254. Exponents. Before denning this notion let us recall a
problem in the theory of combinations, treated in elementary
algebra.
Suppose that there are 7 compartments
n n n
^ii v25 *" ^y>
and that we have k classes of objects
ARITHMETIC OPERATIONS WITH CARDINALS 295
Let us place an object from any one of these classes in (7r an
object from any one of these classes in <72--- and so on, for each
compartment. The result is a certain distribution of the objects
from these k classes K, among the y compartments O.
The number of distributions of objects from k classes among y
compartments is &.
For in O1 we may put an object from any one of the k classes.
Thus (7j may be tilled in k ways. Similarly (72 may be filled in
k ways. Thus the compartments (Tp <72 may be filled in k2 ways.
Similarly (7-p (72, <73 may be filled in A? ways, etc.
255. 1. The totality of distributions of objects from k classes
K among the 7 compartments 0 form an aggregate which may be
denoted by TTC
We call it the distribution of K over O. The number of distri
bution of this kind may be called the cardinal number of the set,
and we have then Card K° = &
2. What we have here set forth for finite 0 and .STmay be ex
tended to any aggregates, 51 = |a|, 33 = \b\ whose cardinal num
bers we call a, b. Thus the totality of distributions of the a's
among the 5's, or the distribution of 51 over 93, is denoted by
21®,
and its cardinal number is taken to be the definition of the symbol
ab- Thus'
256. Example 1. Let
X* + djZ*-1* >•' + «n= 0 (1
have rational number coefficients. Each coefficient a8 can range
over the enumerable set of elements in the rational number
system R = jr{, whose cardinal number is K0. The n coefficients
form a set 21 = (at, ••• an) = \a\. To the totality of equations 1)
corresponds a distribution of the r's among the a's, or the set
R*
whose cardinal number is
K5 = c».
296 AGGREGATES
As Card R* = *0 = e
we have the relation :
K£ = K0 , or en = e
for any integer n.
On the other hand, the equations 1) may be associated with
the complex
(flj, ... an),
and the totality of equations 1) is associated with
<£ = fOi> •'• «n)j-
But K«i.4)1-f«ii»f*j«
JOi, «2' as)i = IOi» °2){ • \az\ » etc.
N C^f^^KJvJa.].
Card (£ = e • e « ••« e , n times as factor.
But Card (^ = Card J2*
since each of these sets is associated uniformly with the equations
1). Thus .. -
en = e-e--"e , n times as factor.
257. Example 2. Any point x in m-way space $Rm depends on
m coordinates xl^ x%, ••• xm, each of which may range over the set
of real numbers 9t, whose cardinal number is c. The m coordi
nates xl ... xm form a finite set
I =(>!, -.. a;,,).
Thus to $lm = \x\ corresponds the distribution of the numbers in
$R, among the m elements of £, or the set
K*
whose cardinal number is
cm.
As Card M* = c
we have
cm = c for any integer m. (1
As in Example 1 we show
cm = c • c * • « • c , m times as factor.
ARITHMETIC OPERATIONS WITH CARDINALS 297
258. ab+c = ab.ac. (1
To prove this we have only to show that
and 31s8 • 21s
can be put in 1-1 correspondence. But this is obvious. For
the set on the left is the totality of all the distributions of the
elements of 51 among the sets formed of 33 and (L On the other
hand, the set on the right is formed of a combination of a distri
bution of the elements of 51 among the 93, and among the (£. But
such a distribution may be regarded as the distribution first con
sidered.
259. (a*)' = aK (1
We have only to show that we can put in 1-1 correspondence
the elements of
(3158/ and 2l*-e. (2
Let 51= \a\, 33 = {6j, 6= \c\. We note that 51s is a union of
distributions of the as among the J's, and that the left side of 2)
is formed of the distributions of these sets among the c's. These
are obviously associated uniformly with the distributions of the
a's among the elements of 33 • (L
260. 1. cn = (mc)n = mn* = m* = c (1
where m, n are positive integers.
For each number in the interval & = (0, 1*) can be represented
in normal form once and once only by
• a^a^ •••in the ra-adic system, (2
where the 0 < as < m. [I, 145] .
Now the set of numbers 2) is the distribution of 3ft =(0, 1, 2,
• •• m — 1) over (5 = (ax, a2, «3 •••), or
whose cardinal number is
m<
On the other hand, the cardinal number (£ is C.
298 AGGREGATES
Hence, mc = c.
As n* = e, we have 1), using 1) in 257.
2. The result obtained in 247, 2 may be stated:
V = c. (3
3. t* = c. (4
For obviously ne < ec < ce.
But by 3), cc = c and by 1) n* = c.
261. 1. The cardinal number t of all functions f (xl ••• xm) which
take on but two values in the domain of definition 31, of cardinal num
ber a, is 2 2*.
Moreover, 2 21 > a.
This follows at once from the reasoning of 249.
2. Let f be the cardinal number of the class of all functions de
fined over a domain 31 whose cardinal number is c. Then
f = cc=2c>c. (1
For the class of functions which have but two values in 21 is by
1, 2c.
On the other hand, obviously
f = cc.
But
cc = (2e)c, by 260, 1)
= 2", by 259, 1)
= 2S by 253, 2).
Thus, cc = 2c.
That f > c
follows from 250, since the class of functions there considered lies
in the class here considered. '
3. The cardinal number t of the class of limited integrable func
tions in the interval 31 is = f, the cardinal number of all limited
functions defined over 31.
NUMBERS OF LIOUVILLE 299
For let & be a Cantor set in 51 [I, 272]. Being discrete, any
limited function defined over £) is integrable. But by 229, Ex. 4,
the points of 51 may be set in uniform correspondence with the
points of £).
4. The set of all functions
/O) =/,(*) +/,(*)+... (2
which are the sum of convergent series, and whose terms are continu
ous in 5(, has the cardinal number c.
For the set g of continuous functions in 51 has the cardinal
number c by 248. These functions are to be distributed among
the enumerable set & of terms in 2). Hence the set of these
functions is ~
u >
whose cardinal number is c
Remark. Not every integrable function can be represented by
the series 2).
For the class of integrable functions has a cardinal number > c,
by 250.
5. TJie cardinal number of all enumerable sets in an m-way space
Wm is c.
For it is obviously the cardinal number of the distribution of
9?m over an enumerable set (5, or
Card ft® = ce = c.
Numbers of Liouville
262. In I, 200 we have defined algebraic numbers as roots of
equations of the type
where the coefficients a are integers. All other numbers in $ft we
said were transcendental. We did not take up the question
whether there are any transcendental numbers, whether in fact,
not all numbers in 9? are roots of equations of the type 1).
o'X' AGGREGATES
The first to actually show the existence of transcendental num
bers was Liouville. He showed how to form an infinity of such
numbers. At present we have practical means of deciding
whether a given number is algebraic or not. It was one of the
signal achievements of Hermite to have shown that e = 2.71818 •••
is transcendental.
Shortly after Lindemann. adapting Hermite's methods, proved
that TT = 3.14159 ••• is also transcendental. Thereby that famous
problem the Quadrature of the Circle was answered in the negative.
The researches of Hermite and Lindemann enable us also to form
an infinity of transcendental numbers. It is, however, not our pur
pose to give an account of these famous results. We shall limit
our considerations to certain numbers which we call the numbers
of Liouville.
In passing let us note that the existence of transcendental num
bers follows at once from 285, 2 and 244, 2. fe -X ^ ?
For the cardinal number of the set of real algebraic number is
e, and that of the set of all real numbers is c, and c > e.
263. In algebra it is shown that any algebraic number a is a
root of an irreducible equation,
f- + <i. = o a
whose coefficients are integers without common divisor. We say
the order of a is ni.
We prove now the theorem
r« = ^ , Pn*qm relatively prime,
= a, an algebraic number of order m, as n = x . Then
]«-r.]>-i_ , „>„. (2
For let a be a root of 1). We may take S>0 so small that
/(x)=#=0 in Dj*(«), and * so large that rn lies in ^(a), for n>9.
, (3
NUMBERS OF LIOUVILLE 301
for n > i, since the numerator of the middle member is an integer,
and hence > 1 .
On the other hand, by the Law of the Mean [I,
where £ lies in !>«<» Now /(«)=0 and /'(£)< some M.
Hence -, x -
--"^4- (4
on using 3). Bat however large Jf in, there exists a v, such that
qn>M, for any n > v. This in 4) gives 2).
264. 1. The numbert
where an< 10», and not aR of them vanish after a certain index, are
transcendental.
For if L is algebraic, let its order be m. Then if Ln denotes
the sum of the first n terms of 1), there exists a v such that
^.li-AOjjjjl- , forn>,. (2
But ^ l
^"lO-'-V: + C 10(»+l>»! ' »»>v;,
v^ being taken sufficiently large. But 3) contradicts 2).
The numbers 1) we call the numbers of Lioumtte.
2. The set of Lioumlle numbers AM the cardinal number c.
For all real numbers in the interval (0*, 1) can be represented
/i-w+fe+^+- • °^9'
where not all the 6Ts vanish after a certain index. The numbers
*-
can obviously be put in uniform correspondence with the set
Thus Card \\\ =c. But \L\ > }X(, hence Card \L\ >c. On the
other hand, the numbers \L\ form only a part of the numbers in
(0*, 1). Hence Card {L\ < c.
CHAPTER IX
ORDINAL NUMBERS
Ordered Sets
265. An aggregate 51 is ordered, when a, b being any two of
its elements, either a precedes 5, or a succeeds 5, according to some
law ; such that if a precedes 5, and b precedes <?, then a shall pre
cede c. The fact that a precedes b may be indicated by
a<b.
Then a>b
states that a succeeds b.
Example 1. The aggregates
1, 2, 3, ...
2,4,6,...
-3, -2, -1,0,1,2,3,
are ordered.
Example 2. The rational number system R can be ordered in
an infinite variety of ways. For, being enumerable, they can be
arranged in a sequence - «. *
ri •> ri ' ra » * " rn • •
Now interchange r^ with rn. In this way we obtain an infinity
of sequences.
Example 3. The points of the circumference of a circle may be
ordered in an infinite variety of ways.
For example, let two of its points Pl, P2 make the angles a+6r
a 4- 02 with a given radius, the angle 6 varying from 0 to 2 ?r.
Let Pl precede P2 when 61 < 02.
302
ORDERED SETS 303
Example 4> The positive integers $ may be ordered in an infi
nite variety of ways besides their natural order. For instance, we
may write them in the order
1, 3, 5, ... 2, 4, 6, -.
so that the odd numbers precede the even. Or in the order
1, 4, 7, 10, ... 2, 5, 8, 11, ... 3, 6, 9, 12, ...
and so on. We may go farther and arrange them in an infinity
of sets. Thus in the first set put all primes ; in the second set
the products of two primes ; in the third set the products of
three primes; etc., allowing repetitions of the factors. Let any
number in set m precede all the numbers in set n>m. The num
bers in each set may be arranged in order of magnitude.
Example 5. The points of the plane 9?2 may be ordered in an
infinite variety of ways. Let Ly denote the right line parallel to
the z-axis at a distance y from this axis, taking account of the sign
of y. We order now the points of 9?2 by stipulating that any
point on Lv, precedes the points on any L^, when y1 <y" , while
points on any Ly shall have the order they already possess on that
line due to their position.
266. Similar Sets. Let 21, 33 be ordered and equivalent. Let
a ~ b, a ~ y3. If when a < a in 21, b < /9 in 33, we say 21 is similar
to 33, and write or m
VI — 1O'
Thus the two ordered and equivalent aggregates are similar
when corresponding elements in the two sets occur in the same
relative order.
Example 1. Let ar 100
g = 1, J, d, ...
33 = a1, a2, az, •••
In the correspondence 21 ~ 33, let n be associated with an. Then
31^33-
Example 2. Let or _ 1 9 Q
VI — A, ^j, o, «•
33= «x a2 ... am, b1, 52, b3 «•«
304 ORDINAL NUMBERS
In the correspondence 21 ~ 33, let ar~r for r = 1, 2, ••• m; also
let bn~m + n,n = 1, 2 ••• Then 51 ^ 53.
Example 3. Let 91 — 1 9 3
53 = ^, 62 ••• a1? «2 ... am.
Let the correspondence between 21 and 53 be the same as in
Ex. 2. Then 21 is not similar to 53. For 1 is the first element in
21 while its associated element al is not first in 53.
Example 4. Let w — 1 9 q
<\ — i, .<£, o, •««
53= a1? «2 ••• 6j, 52 ...
Let an ~ 2 w, bn ~ 2 n - 1. Then 21 - 53 but 51 is not a* 53.
267. Let 21 ^53, 53 ^(L 7%^ 21 ^ £.
For let a ~ 5, a'~6' in 21-53. Let b ~ c, V ~ c' in 53 ~ &. Let
us establish a correspondence 21 ~ S by setting a ~ c, a! ~c' . Then
if a <a' in 21, £< c' in S. Hence 21 ~ (5.
Eutactic Sets
268. Let 21 be any ordered aggregate, and 53 a part of 21, the
elements of 53 being kept in the same relative order as in 21. If 21
and each 53 both have a first element, we say that 21 is well ordered,
or eutactic.
Example 1. 21 = 2, 3, ••• 500 is well ordered. For it has a first
element 2. Moreover any part of 21 as 6, 15, 25, 496 also has a
first element.
Example 2. 21 = 12, 13, 14, ..-in inf. is well ordered. For it
has a first element 12, and any part 53 of 21 whose elements pre
serve the same relative order as in 21, has a first element, viz.
the least number in 53.
The condition that the elements of 53 should keep the same rel
ative order as in 21 is necessary. For B = ••• 28, 26, 24, 22, 20,
21, 23, 25, 27, ... is a partial aggregate having no first element.
But the elements of B do not have the order they have in 21.
EUTACTIC SETS 305
Example 3. Let 51 = rational numbers in the interval (0, 1)
arranged in their order of magnitude. Then 51 is ordered. It
also has a first element, viz. 0. It is not well ordered however.
For the partial set 53 consisting of the positive rational numbers of
51 has no first element.
Example 4- An ordered set which is not well ordered may some
times be made so by ordering its elements according to another
law.
Thus in Ex. 3, let us arrange 51 in a manner similar to 233.
Obviously 51 is now well ordered.
Example 5. 51 = ar a2 • - b^ b% ••• is well ordered. For al is the
first element of 51 ; and any part of 51 as
has a first element.
269. 1. Every partial set 53 of a well-ordered aggregate 51 is well
ordered.
For 53 has a first element, since it is a part of 51 which is well
ordered. If £ is a part of 53, it is also a part of 51, and hence has
a first element.
2. If a is not the last element of a well-ordered aggregate 51, there
is an element of 51 immediately following a.
For let 53 be the part of 51 formed of the elements after a. It
has a first element b since 51 is well ordered. Suppose now
a < c < b.
Then b is not the first element of 53, as c < b is in 53.
3. When convenient the element immediately succeeding a may
be denoted by
a + 1.
Similarly we may denote the element immediately preceding a,
when it exists, by
a-1.
306 ORDINAL NUMBERS
For example, let
5( = a^a2 •'• 5j£>2 ••'
Then an+l = an+1 , bm + 1 = bm+1
an—l = an-l » bm— l = bm_v
There is, however, no b1 — 1.
270. 1. If 51 is well ordered, it is impossible to pick out an in
finite sequence of the type
a1 > a2 > a3 > ••• (1
For ™
20 = ••• a8, 0^, Oj
is a part of 51 whose elements occur in the same relative order as
in 51, and 33 has no first element.
2. A sequence as 1) may be called a decreasing sequence, while
a1 < a2 < as ...
may be called increasing.
In every infinite well ordered aggregate there exist increasing
sequences.
3. Let 51, 33, (£, ••• be a well ordered set. Let 5l = .JaJ be well
ordered in the a's, -53 = \b\ be well ordered in the b's, etc. The set
U = Sl, «, 6 -
is well ordered with regard to the little letters a, b
For U has a first element in the little letters, viz. the first ele
ment of 51. Moreover, any part of U, as S3, has a first element in
the little letters. For if it has not, there exists in 33 an infinite
decreasing sequence
t > s > r > •••
This, however, is impossible, as such a sequence would deter
mine a similar sequence in U as
£><S>ft > •••
which is impossible as U is well ordered with regard to 21, 33 •••
4. Let 9l<33<£< -.. (1
Let each element of 51 precede each element of 33, etc.
SECTIONS 307
Let each 51, 33, ••• be well ordered.
Let
& = % + £+ C+ .-
is a well ordered set, 3 preserving the relative order of elements
intact.
For @ has a first element, viz. the first element of 51. Any
part S of & has a first element. For, if not, there exists in @
an infinite decreasing sequence
r > q> p > ••• (2
Now r lies in some set of 1) as 9?. Hence <?, jt?, ••• also lie in
Sft. But in $ft there is no sequence as 2).
5. Let 5(, 23, S, ••• be an ordered set of well ordered aggre
gates, no two of which have an element in common. The reader
must guard against assuming that 51 + 33 + Q[ + •••» keeping the
relative order intact, is necessarily well ordered.
For let us modify Ex. 5 in 265 by taking instead of all the
points on each Lv only a well ordered set which we denote by 5lj,-
Then the sum 0, Var
21 = <fcflg
has a definite meaning. The elements of 51 we supposed arranged
as in Ex. 5 of 265.
Obviously 51 is not well ordered.
Sections
271. We now introduce a notion which in the theory of well-
ordered sets plays a part analogous to Dedekind's partitions in
the theory of the real number system $ft. Cf. I, 128.
Let 51 be a well ordered set. The elements preceding a given
element a of 51 form a partial set called the section of 51 generated
by a. We may denote it by
So,
or by the corresponding small letter a.
308 ORDINAL NUMBERS
Example 1. Let 51 = 1 2 3
Then
£100=1, 2, -99
is the section of 51 generated by the element 100.
Example 2. Let
51 = 1^, a2--. 6j, b2->
Then
Sb5 = ala2'"blb2b3b4i
is the section generated by b5.
tSb-^ = a^a^ • • •
that generated by b1, etc.
272. 1. Every section of a well ordered aggregate is well ordered.
For each section of 51 is a partial aggregate of 51, and hence
well ordered by 269, 1.
2. In the well ordered set 51, let a<b. Then Sa is a section
ofSb.
3. Let @ denote the aggregate of sections of an infinite well
ordered set 51. If we order © such that Sa < Sb in © wAew a < b in
51, @ is well ordered.
For the correspondence between 51 and @ is uniform and similar.
273. ie£ 51, 93 fo we# ordered and 51^33. If a^b, then
Sa ~ £6.
For in 51 let a"<a'>a. Let b' ~ a1 and b"^an. Since
51 ^ 53, we have
&"<£'<&;
hence the theorem.
274. If 51 is well ordered, 51 is not similar to any one of its
sections.
For if 51 ^ Sa, to a in 51 corresponds an element a1 < a in Sa.
To al in 51 corresponds an element a2 in Sa, etc. In this way we
obtain an infinite decreasing sequence
a> a1>a2> •••,
which is impossible by 270, 1.
SECTIOXS 309
275- Let 51, 53 be well ordered and 51 =* 53. 2%ew to #a w 31 can
not correspond two sections Sb, S/3 each ^ $a.
For let b < & and £a ^ £6, £a ^ /SJS. Then
Sb a* $8, by 267. (1
But 1) contradicts 274.
276. Let 51, 53 be two well ordered aggregates. It is impossible
to establish a uniform and similar correspondence betiveen 51 and 53
in more than one way.
For say Sa ^ Sb in one correspondence, and Sa ^ S/3 in an
other, b, /3 being different elements of 53. Then
Sb ^ Sfr by 267.
This contradicts 275.
277. 1. We can now prove the following theorem, which is
the converse of 273.
Let 51, 53 be well ordered. If to each section of 81 corresponds one
similar section of 53, and conversely, then 53 — 51.
Let us first show that 51 ~ 53. Since to any Sa of 51 corre
sponds a similar section Sb in 53, let us set a ~ b. No other
a1 ~ 5, and no other b' ~ a, as then Sa' ^ Sb or Sb' ^ Sa, which
contradicts 274. Let the first element of 51 correspond to the
first of 53. Thus the correspondence we have set up between 21
and 53 is uniform and 51 ~ 53.
We show now that this correspondence is similar. For let
a ~ b and a' ~~ b', a' < a.
Then b' < b. For a' lies in Sa ^ Sb and br — a' lies in Sb.
2. From 1 and 273 we have now the fundamental theorem :
In order that two well-ordered sets 51, 53 be similar, it is necessary
and sufficient that to each section of 51 corresponds a similar section
of 53, and conversely.
278. Let 51, 53 be well ordered. If to each, section of 51 corre
sponds a similar section of 53i but not conversely, then 5( is similar to
a section of 53.
310 ORDINAL NUMBERS
Let us begin by ordering the sections of 51 and 53 as in 272, 3.
Let B denote the aggregate of sections of 53 to which similar sec
tions of 51 do not correspond. Then B is well ordered and has a
first section, say Sb. Let /9 < b. Then to <S/3 in 53 corresponds
by hypothesis a similar section Sa in 5L On the other hand, to
any section Sa' of 51 corresponds a similar section Sb' of 53. Ob
viously b' < b. Thus to any section of 51 corresponds a similar
section of Sb and conversely. Hence 51^>S7> by 277, 1.
279. Let 51, 53 be well ordered. Either 5t is similar to 53 or one
is similar to a section of the other.
For either :
1° To each section of 51 corresponds a similar section of 53
and conversely ;
or 2° To each section of one corresponds a similar section of
the other but not conversely ;
or 3° There is at least one section in both 51 and 53 to which no
similar section corresponds in the other.
If 1° holds, 51 =* 53 by 277, 1. If 2° holds, either 51 or 53 is similar
to a section of the other.
We conclude by showing 3° is impossible.
For let A be the set of sections of 51 to which no similar section
in 53 corresponds. Let B have the same meaning for 53. If we
suppose 51, 53 ordered as in 272, 3, A will have a first section say
Sa, and B a first section Sft.
Let a < a. Then to Sa in 51 corresponds by hypothesis a sec
tion Sb of Sfl as in 278. Similarly if b' < #, to Sb' of 53 corre
sponds a section Sa' of Sa. But then Sa^-Sft by 277, l, and this
contradicts the hypothesis.
Ordinal Numbers
280. 1. With each well ordered aggregate 51 we associate an
attribute called its ordinal number, which we define as follows :
1° If 5(^53, they have the same ordinal number.
2° If 5t 2-a section of 53, the ordinal number of 5( is less than
that of 53.
ORDINAL NUMBERS 311
3° If a section of 51 is ^ 23, the ordinal number of 51 is greater
than that of S3.
The ordinal number of 5( may be denoted by
Orel 51,
or when no ambiguity can arise, by the corresponding small letter a.
As any two well ordered aggregates 51, 53 fall under one and only
one of the three preceding cases, any two ordinal numbers a, b
satisfy one of the three following relations, and only one, viz. :
a = b , a<b , a>b,
and if a < b, it follows that b > a-
Obviously they enjoy also the following properties.
2 If
J a = b , b = c , then a = c.
For if c = Ord £, the first two relations state that
•
31^33 , SB^S.
But then ^g t by267_
Hence
n = c.
3 If
J a > b , b > c , then a > c.
281. 1. Let 51 be a finite aggregate, embracing say n elements.
Then we set ~ -, w
Ord 51 = n.
Thus the ordinal number of a finite aggregate has exactly similar
properties to those of finite cardinal numbers. The ordinal num
ber of a finite aggregate is called finite, otherwise transfinite.
The ordinal number belonging to the well ordered set formed
of the positive integers • c\ _ 1 9 Q
x) — ^ ^-> **l "*
we call co.
2. The least transfinite ordinal number is co.
For suppose a = Ord 51 < co, is transfinite. Then 51 is ^ a
section of Q. But every section of $ is finite, hence the
contradiction.
312 ORDINAL NUMBERS
3. The cardinal number of a set 51 is independent of the order
in which the elements of 51 occur. This is not so in general for
ordinal numbers.
For example, let Qr -, 0 Q
&SS 1, Z, $, «••
33=1,3,5,... 2,4,6, ...
Here Card 1 = Card % = K0.
But Ord 21 < Ord 33,
since 51 is similar to a section of 33, viz. the set of odd numbers,
1, 3, 5, ...
282. 1. Addition of Ordinals. Let 51, 33 be well ordered sets
without common elements. Let (£ be the aggregate formed by
placing the elements of 33 after those of 51, leaving the order in 33
otherwise unchanged. Then the ordinal number of (£ is called the
sum of the ordinal numbers of 51 and 33, or
•
Ord <£ = Ord 51 + Ord 33,
or c = a + b.
The extension of this definition to any set of well-ordered aggre
gates such that the result is well ordered is obvious.
2. We note that a + j > „, a + f, > b. . . .
For 51 is similar to a section of (£, and 33 is equivalent to a part
of ft
3. The addition of ordinal numbers is associative.
This is an immediate consequence of the definition of addition.
4. The addition of ordinal numbers is not always commutative.
Thus if a= (^2... in inf.), Ord 5i = *>,
33 = (^2 ... 5n), Ord 33 = rc;
let
$)=:(&!... bnafy ..-), Ord£)
Then c = a +n , b = ™ + &>.
ORDINAL NUMBERS 313
But 51 — a section of (5, viz. : ^ ^S^, while £) ^ 51. Hence
a < c , « = b,
<w + n > &> , n + a) = a).
5. Tjf a > b, fAett c -f a > c -f b, and a + c > b 4- c.
For let a = Ord 5(, b = Ord 33, c = Ord (L
Since a > b, we can take for 53 a section Sb of 51. Then c -f a is
the ordinal number of AT , or /^i
^ + zl? V-L
and c + b is the ordinal number of
6 + Sb, (2
preserving the relative order of the elements.
But 2) is a section of 1), and hence c + a > c -h b.
The proof of the rest of the theorem is obvious.
283. 1. The ordinal number immediately following a is a + 1.
For let a = Ord 51. Let 53 be a set formed by adding after all
the elements of 51 another element b. Then
a + 1 = Ord 53 = b.
Suppose now
a<c<b , c=Ord£. (1
Then (5 is similar to a section of 53. But the greatest section
of 53 is Sb = 51. Hence
c < a,
which contradicts 1).
2. Let a > b. Then there is one and only one ordinal number b
such that , ,
a = D + o.
Forlet a = Ord 51 , b = Ord53.
We may take 53 to be a section Sb of 51. Let X) denote the set
of elements of 51, coming after Sb. It is well ordered and has an
ordinal number b- Then
3l = 53 + £),
preserving the relative order, and hence
a = b + b.
There is no other number, as 282, 5 shows.
314 ORDINAL NUMBERS
284. 1. Multiplication of Ordinals. Let 51, 53 be well-ordered
aggregates having a, b as ordinal numbers. Let us replace each
element of 51 by an aggregate ^ 53. The resulting aggregate (£
we denote by «* w
53 • 21.
As (5 is a well-ordered set by 270, 3 it has an ordinal number c.
We define now the product b • a to be c, and write
b • a = c.
We say c is the result of multiplying a by b, and call a, b factors.
We write
a « a = a2 , a • a • a = a3 , etc.
2. Multiplication is associative.
This is an immediate consequence of the definition.
3. Multiplication is not always commutative.
For example, let
53 = (1, 2, 3 ... in inf.).
-L llGH fv\ CYf /• 7 7 7
Hence Ord (53 • 51) = a, . 2 > o>,
Ord(H«53)= 2 &> = &).
4. If a < b, ^ew ca < cb.
For (5 • 51 is a section of £ • 53.
Limitary Numbers
285. 1. £««
al< «2<«3< ... (
&e «n infinite increasing enumerable sequence of ordinal numbers
There exists a first ordinal number a greater than every an.
J-^v3u /~\ i rw"
an= Ord5ln.
LIMITARY NUMBERS 315
Since an_i < «n, 2ln_! is similar to a section of 2ln. For simplicity
we may take 2ln_! to be a section of 2ln. Let, therefore,
Consider now w qr ™ ™
21 = Slj + *B2 + £}g + ...
keeping the relative order of the elements intact. Then 21 is well
ordered and has an ordinal number «.
As any 5ln is a section of 21,
«„<«.
Moreover any number y8<« is also < some am. For if 33 has
the ordinal number & 33 must be similar to a section of 21. But
there is no last section of 21.
2. The number a we have just determined is called the limit of
the sequence 1). We write
a = Km an , or an = a.
We also say that a corresponds to the sequence 1).
All numbers corresponding to infinite enumerable increasing
sequences of ordinal numbers are called limitary.
3. If every an in 1) is < & then a < ft.
For if £< a, a is not the least ordinal number greater than
every «n.
4. If /3<«, fi is < some an.
286. In order that n
'
define the same number \ it is necessary and sufficient that each
number in either sequence is surpassed by a number in the other.
For let . o • Q
an = a , pn— p-
If no Pn is greater than «TO, /3 < «m < a, by 285, 3, and « =£ /3.
On the other hand, if each «m< some /3n, a</3. Similarly
316 ORDINAL NUMBERS
287. Cantor's Principles of Generating Ordinals. We have now
two methods of generating ordinal numbers. First, by adding 1
to any ordinal number a. In this way we get
a, a + 1, a +2, ...
Secondly, by taking the limit of an infinite enumerable increas
ing sequence of ordinal numbers, as
«! < «2 < «3 < "•
Cantor calls these two methods the first and second principles
of generating ordinal numbers.
Starting with the ordinal number 1, we get by successive appli
cations of the first principle the numbers
1,2,3,4,...
The limit of this sequence is co by 285, l. Using the first prin
ciple alone, this number would not be attained ; to get it requires
the application of the second principle. Making use of the first
principle again, we obtain
<w + l, a> + 2, w + 3, ...
The second principle gives now the limitary number CD -f o> = a>2
by 285, 1. From this we get, using the first principle, as before,
tt>2 + l, 6)2 + 2, 0)2 + 3,...
whose limit is ft)3. In this way we may obtain the numbers
com + n , m, n finite.
The limit of any increasing sequence of these numbers as
CO , ft) 2 , ft)3 , 0)4:, «••
is a) « ft) = a)2, by 285, 1.
From a)2 we can get numbers of the type
caPl + com 4- n Z, m, n finite.
Obviously we may proceed in this way indefinitely and obtain
all numbers of the type
where a0, al •>• an are finite ordinals.
LIMITARY NUMBERS 317
But here the process does not end. For the sequence
O) < ft)2 < O)3 < •••
has a limit which we denote by &)w.
Continuing we obtain
(i)"™, ft)"" , etc.
288. It is interesting to see how we may obtain well ordered
sets of points whose ordinal numbers are the numbers just con
sidered.
In the unit interval 51 = (0, 1), let us take the points
• :•*.'.. t » * '» if-- .a
These form a well ordered set whose ordinal number is to.
The points 1) divided 51 into a set of intervals,
*! , «a , 213- (2
In m of these intervals, let us take a set similar to 1). This
gives us a set whose ordinal number is com.
In each interval 2), let us take a set similar to 1). This gives
us a set whose ordinal number is &>2. The points of this set
divide 51 into a set of o>2 intervals. In each of these intervals,
let us take a set of points similar to 1). This gives a set of
points whose ordinal number is &>3, etc.
Let us now put in 5Ij a set of points 33X whose ordinal number
is co. In 512 let us put a set 332 whose ordinal number is o>2, and
so on, for the other intervals of 2).
We thus get in 51 the well ordered set
# = 2^ + ®a + 88 + -
whose ordinal number is the limit of
CO , ft) 4- CO2 , CO + CO2 + CO3 , ...
This by 286 has the same limit as
a) , co2 , w3 , ••• or co".
With this set we may now form a set whose ordinal number is
o)wft>, etc.
318 ORDINAL NUMBERS
Classes of Ordinals
289. Cantor has divided the ordinal numbers into classes.
Class 1, denoted by Z^ embraces all finite ordinal numbers.
Class 2, denoted by Z2, embraces all transfinite ordinal numbers
corresponding to well ordered enumerable sets ; that is, to sets
whose cardinal number is K0 .- For this reason we also write
It will be shown in 293, 1 that Z^ is not enumerable. Moreover
if we set ~ , „
K! = Card Z2 ,
there is no cardinal number between K0 and Kj as will be shown in
294. We are thus justified in saying that Class 3, denoted by
Z% or ^(Kj), embraces all ordinal numbers corresponding to well
ordered sets whose cardinal number is Kj, etc.
Let j3 = Ord 33 be any ordinal number. Then all the numbers
a < 0 correspond to sections of 33. These sections form a well
ordered set by 272, 3. Therefore if we arrange the numbers
« < /3 in an order such that a1 precedes a when Sa' < Sa, they are
well ordered. We shall call this the natural order. Then the
first number in Z-^ is 1, the first number of Z% is at. The first
number in Z3 is denoted by H.
290. As the numbers in Class 1 are the positive integers, they
need no comment here. Let us therefore turn to Class 2.
If a is in Z2 , so is a -f 1 •
For let a = Ord 51. Let 53 be the well ordered set obtained
by placing an element b after all the elements of 21. Then
a + 1 = Ord 53.
But 53 is enumerable since 31 is.
Hence a -f 1 lies in Z% .
291. Let •^«1<«2<«3<...
be an enumerable infinite set of numbers in Z2. Then a = lim an lies
in Z2.
CLASSES OF ORDIXALS 319
For using the notation employed in the proof of 285, 1, a is the
ordinal number of
«=«! + », + »,+ •••
But 2lr SBj, 332"* are eacn enumerable.
Hence 21 is enumerable by 235, 1, and a lies in Zv
292. We prove now the converse of 290 and 291.
Each number a in Z^, except &>, is obtained by adding 1 to some
number in Z%; or it is the limit of an infinite enumerable increasing
set of numbers in Z^
For, let a = Ord 21. Suppose first, that 21 has a last element,
say a. Since 21 is enumerable, so is Sa. Hence
yS=Ord - Sa
is in Z%. Then a= 8 + 1
Suppose secondly, that 21 has no last element. All the numbers
ft < a in Z% belong to sections of 21. Since 21 is enumerable, the
numbers ft are enumerable. Let them be arranged in a sequence
/?!, /32, /V" C1
Since they have no greatest, let $[ be the tirst number in it
>/?!, let ft% be the first number in it > ft{, etc. We get thus the
sequence ^ ^<^< ... (2
whose limit is X, say.
Then X = «. For X is > any number in 1), which embraces all
the numbers of Z^ < a. Moreover it is the least number which
enjoys this property.
293. 1. The numbers of Z2 are not enumerable.
For suppose they were. Let us arrange them in the sequence
«i» «a» as •" C1
Then, as in 292, there exists in this sequence the infinite enu-
merable sequence ^ < «; < «^ < . . . (2
such that there are numbers in 2) greater than any given number
in 1).
320 ORDINAL NUMBERS
Let < = «'. Then a' lies in Z2 by 291. On the other hand, by
285, a' is > any number in 2), and therefore > any number in
1). But 1) embraces all the numbers of Z2, by hypothesis. We
are thus led to a contradiction.
2. We set
294. There is no cardinal number between K0 and $r
For let «=Card 5( be such a number. Then 5Hs ~ an infinite
partial aggregate of Z%, which without loss of generality may be
taken to be a section of Z^. But every such section is enumer
able. Hence 51 is enumerable and «=X0, which is a contradiction.
295. We have just seen that the numbers in Z2 are not enumer
able. Let us order them so that each number is less than any
succeeding number. We shall call this the natural order.
1. The numbers of Z% when arranged in their natural order form
a well ordered set.
For Z% has a first element co. Moreover any partial set Z, the
relative order being preserved, has a first element. For if it has
not, there exists an infinite enumerable decreasing sequence
This, however, is not possible. For /3, 7, ••• form a part of Sa
which is well ordered.
There is thus one well ordered set having Xj as cardinal num-
ber- Let
Let now 51 be an enumerable well ordered set whose ordinal
number is a. The set ^ w
^2 4- «i
the elements of 51 coming after Z^, has the cardinal number Kj by
241, 3. It is well ordered by 270, 3. It lias therefore an ordinal
number which lies in Z3, viz. n -f « by 282, l. Thus Z3 embraces
an infinity of numbers.
2. The least number in Z% is H.
For to any number a< H corresponds a section 51 of Z^. Hence
a lies in Z
CLASSES OF ORDINALS
321
296. 1. An aggregate formed of an Kj set of ^1 sets is an Kj set.
Consider the set
A =
la
«<
Here each row is an ^ set. As there are an Kx set of rows, A
is an Kj set of Xj sets. To show that A is an ^ set, we associate
each allt with some number in the first two number classes.
In the first place the elements alK where i tc < o> may be associ
ated with the numbers 1, 2, 3, ••• < <w. The elements al<0, aM
lying just inside the &>th square and which are characterized
by the condition that i = 1, 2, ••• o>; /c=l, 2 ••• < o> form an
enumerable set and may therefore be associated with the ordinals
co, o> + 1, ••• < G)2. For the same reason the elements just inside
the a) + 1st square may be associated with the ordinals o>2, o>2 + 1,
... < 6)3. In this way we may continue. For when we have
arrived at the «th row and column (edge of the ath square) we
have only used up an enumerable set of numbers in the sequence
i, 2, ... a,... <n (i
in our process of association. There are thus still an Kx set left
in 1) to continue the process of association.
2. As a corollary of 1 we have :
The ordinal numbers
n2, n3, o4, ...
lie in Z~ .
297. I. Let a < /3 < 7 < •••
be an increasing sequence of numbers in Z3 having
(1
as cardinal
number and such that any section of 1) has K0 as its cardinal.
There exists a first ordinal number \ in Z3 greater than any number
in 1).
For let
= Ord
7=Ord6 —
322 ORDINAL NUMBERS
Since a < /3 we may take 51 to be a section of 53. Similarly
we may suppose 53 is a section of (£, etc.
Let now % = % + B, <5 = »+(7,...
Consider now 8 = « + .B+ (7+ -
keeping the relative order intact. Then 8 is well ordered by
270, 4. Let
X = Ord 8.
Since Card 8 = Kx, by 296, 1, X lies in Z3.
As any 51, 53, ••• is a section of 8,
«< /3< ••• < X.
Moreover, any number n < X is also < some a, /3, 7 ••• For if
9ft has ordinal number /A, 9ft must be similar to a section of 8.
But there is no last section in 8.
2. We shall call sequences of the type 1), an Kj sequence.
The number X whose existence we have just established, we shall
call the limit ofV). We shall also w^rite
a < @< 7 ••• =X
to indicate that a, /3, ••• is an Kj sequence whose limit is X.
298. 1. The preceding theorem gives us a third method of
generating ordinal numbers. We call it the third principle.
We have seen that the first and second principles suffice to gen
erate the numbers of the first two classes of ordinal numbers but
do not suffice to generate even the first number, viz. O in Z3. We
prove now the following fundamental theorem :
2. The three principles already described are necessary and suffi
cient to generate the numbers in Z3.
For let a = Ord be any number of Z3. If 31 has a last element,
reasoning similar to 292, i shows that
« = £ + 1.
If 51 has no last element, all the numbers of Z3<a form an X0
or K set. In the former case
CLASSES OF ORDINALS 323
where ft lies in Z2. In the latter case, reasoning similar to 292, l
shows that we can pick out an Kj increasing sequence
299. 1. The numbers of Z3 form a set whose cardinal number a
is >Kr
The proof is entirely similar to 293, 1. Suppose, in fact, that
a = Kj . Let us arrange the elements of Z3 in the Kx sequence
«i , «2 •••««••• «n ••• (1
As in 292, there exists in this sequence an ^ increasing sequence
a[<ai< .-. -«'. (2
Then «' lies in Z% by 297, l. On the other hand ex! is greater than
any number in 2) and hence greater than any number in 1).
But 1) embraces all the numbers in Z3 by hypothesis. We are
thus led to a contradiction.
2. We set K2
3. There is no cardinal number between Kj and K2 .
For let a = Card 51 be such a number. Then 31 is equivalent to
a section of Z3. But every such section has the cardinal num
ber Xr
300. The reasoning of the preceding paragraphs may be at
once generalized. The ordinal numbers of Zn corresponding to
well ordered sets of cardinal number Kn_2 form a well ordered set
having a greater cardinal number a than Kn_2 • Moreover there is
no cardinal lying between Xn_2 and «. We may therefore ap
propriately denote a by KB_r The KM_2 sequence of ordinal
numbers
lying in Zn has a limit lying in Zn, and this fact embodies the
nth principle for generating ordinal numbers. The first n prin
ciples are just adequate to generate the numbers of Zn. They do
not suffice to generate even the first number in Zn+l .
Finally we note that an Kn set of Kn sets forms an KM set.
n
CHAPTER X
POINT SETS
Pantaxis
301. 1. (Borel.) Let each point of the limited or unlimited set
51 lie at the center of a cube (L Then there exists an enumerable set
of non- overlapping cubes \c\ such that each c lies within some (5, and
each point of 51 lies in some c. If 51 is limited and complete, there
is a finite set jc| having this property.
For let D-p D2 ••• be a sequence of superposed cubical divisions
of norms = 0. Any cell of Dl which lies within some (5 and
which contains a point of 21 we call a black cell ; the other cells
of D we call white. The black cells are not further subdivided.
The division D2 divides each white cell. Any of these subdivided
cells which lies within some (£ and contains a point of 51 we call a
black cell, the others are white. Continuing we get an enumer
able set of non-overlapping cubical cells Jc[.
Each point a of 51 lies within some c. For a is the center of
some (5. But when n is taken sufficiently large, a lies in a cell of
Dn, which cell lies within ($.
Let now 51 be limited and complete. Each a lies within a cube c,
or on the faces of a finite number of these c. With a we associ
ate the diagonal 8 of the smallest of these cubes. Suppose
MinS = 0 in 51. As 51 is complete, there is a point a in 51 such
that Min S = 0, in any F^(cc). This is not possible, since if ?? is
taken sufficiently small, all the points of Vj, lie in a finite number
of the cubes c.
Thus Min S > 0. As the c's do not overlap, there are but a
finite number.
2. In the foregoing theorem the points of 51 are not necessarily
inner points of the cubes c. Let a be a point of 51 on the face of
one of these c. Since a lies within some (5, it is obvious that the
324
PANTAXIS 325
cells of some Dn, n sufficiently large, which surround a form a
cube (?, lying within d. Thus the points of 51 lie within an
enumerable set of cells J<?j, each c lying within some (£. The
cells c of course will in general overlap. Obviously also, if 51 is
complete, the points of 51 will lie within a finite number of
these <?'s.
302. If 51 is dense, 51' is perfect.
For, in the first place, 51' is dense. In fact, let a be a point of
51'. Then in any D*(«) there are points of 51. Let a be such a
point. Since 51 is dense, it is a limiting point of 51 and hence is a
point of 51'. Thus in any D*(a) there are points of 51'.
Secondly, 51' is complete, by I, 266.
303. Let $8 be a complete partial set of the perfect aggregate 51.
Then 6 = 51 - 53 is dense.
For if (£ contains the isolated point c, all the points of 51 in Dr*(c)
lie in 53, if r is taken sufficiently small. But 53 being com
plete, c must then lie in 53.
Remark. We take this occasion to note that a finite set is to be
regarded as complete.
304. 1. J/51 does not embrace all 9?n, it has at least one frontier
point in $ln.
For let a be a point of 51, and b a point of $Rn not in 5(. The
points on the join of a, b have coordinates
2^ = ^ + 0(^-0 = ^(0), 0<0<1, i = l, 2,. ..n.
Let 6' be the maximum of those 0's such that #(0) belongs to
5t if 9 < 0'. Then x(6') is a frontier point of 51.
2. Let 51, 33 have no point in common. If Dist (51, 53) >0, we
say 51, 53 are exterior to each other.
305. 1. Let 51 = \a] be a limited or unlimited point set in $lm.
We say 53 < 51 is pantactic in 51, when in each Dg(a) there is a
point 53.
We say 53 is apantactic in 51 when in each D^a) there is a point
a of 51 such that some D^t) contains no point of 53.
326 POINT SETS
Example 1. Let 21 be the unit interval (0, 1), and $8 the ra
tional points in 21. Then $8 is pantactic in 21.
Example 2. Let 21 be the interval (0, 1), and 33 the Cantor set
of I, 272. Then 53 is apantactic in 21.
2. If 53 < 21 is pantactic in 21, 21 contains no isolated points not
in®.
For let a be a point of 21 not in 53. Then by definition, in any
Z)5(#) there is a point of 53. Hence there are an infinity of points
of 53 in this domain. Hence a is a limiting point of 21.
306. Let 21 be complete. We say 53 < 21 is of the 1° category
in 21, if 53 is the union of an enumerable set of apantactic sets
in 21.
If 53 is not of the 1° category, we say it is of the 2° category.
Sets of the 1° category may be called Baire sets.
Example. Let 21 be the unit interval, and 53 the rational
points in it. Then 53 is of the 1° category.
For 53 being enumerable, let 53 = \on\. But each bn is a single f k
point and is thus apantactic in 21.
The same reasoning shows that if 53 is any enumerable set in
21, then 53 is of the 1° category.
307. 1. If 53 is of the 1° category in 21, 21 - 53 = B is > 0.
For since 53 is of the 1° category in 21, it is the union of an
enumerable set of apantactic sets {53nj. Then by definition there
exist points a^ a2, ••• in 21 such that
where D(a^) contains no point of 53X, -#(«2) no point of 532? etc.
Let b be the point determined by 1). Since 21 is complete by
definition, b is a point of 21. As it is not in any 53n, it is not
in 53. Hence B contains at least one point.
2. Let 21 be the union of an enumerable set of sets J2lnj, each 2ln
being of the 1° category in 53- Then 21 is of the 1° category in 53.
This is obvious, since the union of an enumerable set of enu
merable sets is enumerable.
PANTAXIS 327
3. Let 33 be of the 1° category in 21. Then B = 21 -33 is of the
2° category in 21.
For otherwise $8 + -5 would be of the 1° category in 21- But
51 - (53 + B) = 0,
and this violates 1.
4. It is now easy to give examples of sets of the 2° category.
For instance, the irrational points in the interval (0, 1) form a
set of the 2° category.
308. Let 21 be a set of the 1° category in the cube Q. Then
A = Q — 21 Aas the cardinal number c.
If A has an inner point, Da(a), for sufficiently small 8, lies in .A.
As Card D& = c, the theorem is proved.
Suppose that A has no inner point. Let 51 be the union of the
apantactic sets 2lj < 2(2 < ••• in Q. Let An = Q — 2ln. Let qn be
the maximum of the sides of the cubes lying wholly in An. Ob
viously qn = 0, since by hypothesis A has no inner points. Let Q
be a cube lying in Al. As qn = 0, there exists an n^ such that Q
has at least two cubes lying in Ani ; call them $0, Qlt There ex
ists an n2 > Wj such that $0, Ql each have two cubes in A^\ call
them Q Q • O Q
Vo, o » Vo, i > Vi, o i Vi, i »
or more shortly (>tli tj.
Each of these gives rise similarly to two cubes in some An&,
which may be denoted by ftit tji ls, where the indices as before have
the values 0, 1. In this way we may continue getting the cubes
Qn » ftji, » &**•"•
Let a be a point lying in a sequence of these cubes. It obvi
ously does not lie in 21, if the indices are not, after a certain stage,
all 0 or all 1. This point a is characterized by the sequence
which may be read as a number in the dyadic system. But these
numbers have the cardinal number c.
309. Let 21 be a complete apantactic set in a cube Q. Then there
exists an enumerable set of cubical cells jq} such that each point of
21 lies on a face of one of these q, or is a limit point of their faces.
328 POINT SETS
For let D^ > D2 > ••• be a sequence of superimposed divisions
of O, whose norms 5n = 0. Let
be the cells of Dl containing no point of 51 within them. Let
dzl, d22, c723 ... (2
denote those cells of D2 containing no point of 51 within them and
not lying in a cell of 1). In this way we may get an infinite se
quence of cells £) = \dmn\, where for each m, the corresponding n
is finite, and m = cc. Eacli point a of A lies in some dm^ n . For 51
being complete, Dist (a, 51) > 0. As the norms Bn = 0, a must lie
in some cell of Z>n, for a sufficiently large n. The truth of the
theorem is now obvious.
310. Let 33 be pantactic in 51. Then there exists an enumerable
set (£<_ 33 which is pantactic in 51.
For let D1 > Z>2> «•• be a set of superimposed cubical divisions
of norms dn = 0. In any cell of Dl containing within it a point
of 51, there is at least one point of 53. If the point of 51 lies on
the face of two or more cells, the foregoing statement will hold
for at least one of the cells. Let us now take one of these points
in each of these cells; this gives an enumerable set (S^. The
same holds for the cells of D2. Let us take a point in each of
these cells, taking when possible points of @j. Let (52 denote the
points of this set not in (S^. Continuing in this way, let
(S = (£1 + (*2+ ...
Then (5 is pantactic in 51, and is enumerable, since each (5n is.
Corollary. In any set 51, finite or infinite, there exists an enumer
able set & which is pantactic in 51.
For we have only to set 53 = 51 in the above theorem.
311. 1. The points (£ where the continuous function f(xl • • • xm)
takes on a given value g in the complete set tyi,form a complete set.
For let cv c2 ... be points of (5 which = c. We show c is a
point of (5. For -- -
PANTAXIS 329
As /is continuous, »f \^ff\
J\Cn) —J\C)'
Hence /<*)**
and c lies in (L
2. Letf(xl "• xm) be continuous in the limited or unlimited set 51.
If the value of f is known in an enumerable pantactic set & in 51,
which contains all the isolated points of 51, in case there be such, the
value off is known at every point of 5(.
For let a be a limiting point of 51 not in (£. Since (5 is pantactic
in 51, there exists a sequence of points e^ e2-- in (S which = a.
Since / is continuous, /(#„) = /(a). As / is known at each en,
it is known at a.
3. Let g = J/j be the class of one-valued continuous functions
defined over a limited point set 51. Then
f = Card g = c.
For let SH^ be a space of an infinite enumerable number of
dimensions, and let A
r— (TII #2» •••)
denote one of its points. Let/ have the value rjl at e^ the value
?72 at £2 ••• for the points of (5 defined in 2. Then the complex
0?i» ^2 ••')
completely determines /. But this complex determines also a
point 77 in ^ whose coordinates are rjn. We now associate / with
• Hence
On the other hand, f>c, since in g there is the function
/(#! ••• xm) = g in 51, where g is any real number.
312. Let 53 denote the class of complete or perfect subsets lying in
the infinite set 51, which latter contains at least one complete set.
Then « n ^ cu
b = Card 53 = c.
For let alt a2, • •• = a, all these points lying in 51. Then
But for tj we may take any number in ^ = (1, 2, 3, •••) ; for i%
we may take any number in $2 = (^ + 1. ^ + 2, •••), etc.
330 POINT SETS
Obviously the cardinal number of the class of these sequences
l)ise< = c. But («,«^;.«;,..)
is a complete set in 51. Hence b.>c. On the other hand, b<c.
Forlet A >-»•>••• (2
be a sequence of superimposed cubical division of norms = 0.
Each Dn embraces an enumerable set of cells. Thus the set of
divisions gives an enumerable set of cells. Each cell shall have
assigned to it, for a given set in 53, the sign + or — according as
33 is exterior to this cell or not. This determines a distribution
of two things over an enumerable set of compartments.
The cardinal number of the class of these distributions is 2c = c.
But each 53 determines a distribution. Hence b<c.
Transfinite Derivatives
313. 1. We have seen, I, 266, that
%'>%">%'"> .-.
Thus 2l(n) = Dv(2l', a" •.. 2irn)> (1
Let now 51 be a limited point aggregate of the second species.
It has then derivatives of every finite order. Therefore by 18,
D<21', 21", 51'", •••) (2
contains at least one point, and in analogy with 1), we call the
set 2) the derivative of order co of 21, and denote it by
2l(w),
or more shortly by
2K
Now we may reason on 2lw as on any point set. If it is infinite,
it must have at least one limiting point, and may of course have
more. In any case its derivative is denoted by
(«+i) or 5lw+1.
The derivative of 5lw+1 is denoted by
or
Making use of o> we can now state the theorem :
TRANSFINITE DERIVATIVES 331
In order that the point set 51 is of the first species it is necessary
and sufficient that 5lu; = 0.
2. We have seen in 18 that 51"' is complete. The reasoning of
I, 266 shows that 5lw+1, 5lw+2, •••, when they exist, are also complete.
Then 18 shows that, if 5lw+71 n = 1, 2, ... exist,
Dw(aw>aw+1>aw+2> •••) (3
exists and is complete. The set 3) is called the derivative of order
ft> • 2 and is denoted by
§j(«2) or 5^
Obviously we may continue in this way indefinitely until we
reach a derivative of order a containing only a finite number of
points. Then ^a+1 = Q
That this process of derivation may never stop is illustrated by
taking for 51 any limited perfect set, for then
3. We may generalize as follows : Let a denote a limitary ordi
nal number. If each 51^ > 0, /3 < a, we set
when it exists.
4. If 5la > 0, while 51 + 1 = 0, we say 5t is of order a.
314. 1. Let a be a limiting point of 51. Let
Obviously as is monotone decreasing with 8. Suppose that
there exists an a and a BQ > 0, such that for all 0 < 8 < S0
«= Card F<».
We shall say that a is a limiting point of rank a.
If every «5 > a, we shall say that
Rank a > «.
If every as > a, we shall say that
Rank a > a.
332 POINT SETS
2. Let 51 be a limited aggregate of cardinal number a. Then there
is at least one limiting point of 5(, of rank a.
The demonstration is entirely similar to I, 264. Let S1 >
£2 > ... =0. Let us effect a cubical division of 51 of norm 8^. In
at least one cell lies an aggregate 5lx having the cardinal num
ber a. Let us effect a cubical division of 5^ of norm £2 . In at
least one cell lies an aggregate 512 having the cardinal number a,
etc. These cells converge to a point #, such that
Card F5O) = a,
hoAvever small S is taken.
3. If Card 51 > e, there exists a limiting point oftyt of rank > e.
The demonstration is similar to that of 2.
4. If there is no limiting point of 51 of rank > e, 51 is enumerable.
This follows from 3.
5. Let Card 51 be > e. Let 33 denote the limiting points of 51
whose ranks are > e. Then 33 is perfect.
For obviously 33 is complete. We need therefore only to show
that it is dense. To this end let b be a point of 33. About b let
us describe a sequence of concentric spheres of radii rn = 0. These
spheres determine a sequence of spherical shells S$n5, no two of
which have a point in common. If 5ln denote the points of 51 in $n,
we have y = y^ .= ^ + ^ + ^ + ...
Thus if each 5(m were enumerable, V is enumerable and hence
Rank b is not > e. Thus there is one set 5lm which is not enu
merable, and hence by 3 there exists a point of 33 in Sm. But then
there are points of 33 in any F^*(5), and b is not isolated.
6. A set 51 which contains no dense component is enumerable.
For suppose 51 were not enumerable. Let ty denote the proper
limiting points of 51. Then ^ contains a point whose rank is > e.
But the set of these points is dense. This contradicts the hy
pothesis of the theorem.
315. Let a lie in Zn. -Zf5la > 0, it is complete.
For if a is non-limitary, reasoning similar to I, 266 shows that
5la is complete. Suppose then that a is limitary, and 5la is not
TRANSFINITE DERIVATIVES 333
complete. The derivatives of 51 of order < a which are not com
plete, form a well ordered set and have therefore a first element
91^, where & is necessarily a limitary number. Then
But every point of 51^ lies in each 5IY. Hence every limiting
point of W is a limiting point of each 51* and hence lies in 2F3.
Hence 2C3 is complete, which is a contradiction.
316. Let a be a limitary number in Zn. If ^ > 0 for each
/3 < «, 5la exists.
For there exists an Km, ra j< n — 2, sequence
7< & <e< 77 < •••=«. (1
Let c be a point of 21?, d a point of 21s, e a point of 51% etc.
Then the set ^ ^ ,,/,...)
has at least one limiting point I of rank XTO. Let e be any number
in 1). Then Hs a limiting point of rank Xm of the set
(«./»•••)-
Thus I is a limiting point of every 21^, /3 < «, and hence of 5la.
317. Let us show how we may form point sets whose order a
is any number in Z1 or Z2.
We take the unit interval 51 = (0. 1) as the base of our con
siderations.
In 51, take the points
»i-i . I . f • tf ••• C1
Obviously $[ = 1, ${' = 0. Hence ^ is of order 1. The set
SBj divides 51 into a set of intervals
5lx , 5I2 , 513 -. (2
In 5lj = (0, l) take a set of points similar to 1) which has as
single limiting point, the point J. In 512 = Q-, |) take a set of
points similar to 1) which has as single limiting point, the point
|, etc. Let us call the resulting set of points 332.
334 POINT SETS
Obviously ™, l 3 *
^2 = i » f ' t ' "*
Hence ^ = ^ = j and $ w = 0
Thus <B2 is of order 2.
In each of the intervals 2) we may place a set of points similar
to 332, such that the right-hand end point of each interval 5{n is a
limiting point of the set. The resulting set 233 is of order 3, etc.
This shows that we may form sets of every finite order.
Let us now place a set of order 1 in 5lx, a set of order 2 in 512,
etc. The resulting set 5L is of order o>. For 33^ has no points
in Slj, 212 ... $„_!, while the point 1 lies in every ^?l).
Hence $(.+i> = o,
and ^w is of order o>.
Let us now place in each 5ln a set similar to $BW, having the
right-hand end point of $ln as limiting point. The resulting set
33W+1 is of order o> -f 1. In this way we may proceed to form sets
of order o> + 2, o> + 3, ••• just as we did for orders 2, 3, ••• We
may also form now a set of order o>2, as we before formed a set
of order co.
Thus we may form sets of order
w , a) • 2 , a) • 3 , a) • 4
and hence of order &)2, etc.
318. 1. Let 51 be limited or not, and let 2ltw denote the isolated
points of 2F. Then
ft
For t i 'i
.L flUS nfl c\fi . nv; / . CW/'YI—-!'*
that is, 51' is the sum of the points of 51' not in 51", of the points
of 51" not in 2T", etc. If now there are points common to every
H<->wehave <», M ..
TRANSFINITE DERIVATIVES 335
On 5lw we can reason as on 51', and in general for any a < fl we
have a,
/3<a
which gives 1).
2. #*5P = 0, 51 tmd 51' are enumerable.
For not every ^ > Q .<ftb
Hence there is a first a, call it 7, such that 51Y = 0. Then 1)
reducesto
But the summation extends over an enumerable set of terms,
each of which is enumerable by 289. Hence 51' is enumerable.
But then 51 is also enumerable by 237, 2.
3. Conversely, if 51' is enumerable, 5ln = 0.
For if 51" > 0, there is a non-enumerable set of terms in 1), if
no 5l(^) is perfect ; and as each term contains at least one point,
51' is not enumerable. If some 5l(/3) is perfect, 51' contains a per
fect partial set and is therefore not enumerable by 245.
4. From 2, 3, we have :
For 51' to be enumerable, it is necessary and sufficient that there
exists a number a in Z1 or Z2 such that 5la = 0.
5. If 51 is complete, it is necessary and sufficient in order that 51
be enumerable, that there exists an a in Z^ or Z2 such that 5la = 0.
F°r 51 = 5lt + 51',
and the first term is enumerable.
6. If 51^ = 0 for some ft < H, we say 51 is reducible, otherwise it
is irreducible.
319. If 5In > 0, it is perfect.
By 315 it is complete. We therefore have only to show that
its isolated points 51" = 0. Suppose the contrary ; let a be an
isolated point of 5ln.
Let us describe a sphere S of radius r about a, containing no
other point of 5ln. Let 53 denote the points of 51' in S. Let
r >r >r> ••• = 0.
POINT SETS
Let Sn denote a sphere about a of radius rn. Let 53n denote the
points of 53 lying between /S^, $n, including those points which
may lie on Sn_lf Then
-i- a.
Each 53m is enumerable. For any point of 53," is a point of
53" = a. Hence 53" = 0 and 53TO is enumerable by 318, 2.
Thus 53 is enumerable. This, however, is impossible since
53" = a, and is thus > 0.
320. 1. In the relation
5T = S^ + 51" £=1,2,-.. <fl,
ft
the first term on the right is enumerable.
For let us set
«* _ ?9fo) .
r1>rt>- =0. • ' ' V' .
Let 53n denote the points of 53 whose distance B from 5P satis
fies the relation ^ ?> ^
n+1
Then the distance of any point of 53J> from 51" is > rn+1 . If 530
includes all points of 53 whose distance from $P is > r1, we have
Each 53n is enumerable. For if not, 53" > 0. Any point of
53£ as b lies in 51". Hence
Dist (5, 51") = 0.
On the other hand, as b lies in 53^, its distance from 51° is
> rn+1, which is a contradiction.
2. If 51' is not enumerable, there exists a first number a in Zl or
Z% such that 5la is perfect.
This is a corollary of 1.
3. If 51 is complete and not enumerable, there exists a first number
a in Zl + Z% such that 5T is perfect.
is complete, .
where (5 is enumerable, and ^ is perfect. If 51 is enumerable, ^ = 0.
COMPLETE SETS 337
Complete Sets
321. Let us study now some of the properties of complete point
sets. We begin by considering limited perfect rectilinear sets.
Let 51 be such a set. It has a first point a and a last point b. It
therefore lies in the interval /=(«, £). If 51 is pantactic in any
partial interval J= (a, /3) of J, 51 embraces all the points of «7,
since 51 is perfect. Let us therefore suppose that 51 is apantactic
in /. .An example of such sets is the Cantor set of I, 272.
Let D = \ S I be a set of intervals no two of which have a point
in common. We say D is pantactic in an interval /, when I con
tains no interval which does not contain some interval 8, or at
least a part of some 8.
It is separated when no two of its intervals have a point in ^£/
common.
322. 1. Every limited rectilinear apantactic perfect set 51 deter
mines an enumerable pantactic set of separated intervals D— \S\,
whose end points alone lie in 51.
For let 51 lie in /=(«, $), where «, ft are the first and last
points of 51. Let 93 = /— 51. Each point b of 33 falls in some in
terval S whose end points lie in 51. For otherwise we could
approach b as near as we chose, ranging over a set of points of 51.
But then b is a point of 51, as this is perfect. Let us therefore
take these intervals as large as possible and call them B.
The intervals 8 are pantactic in J, for otherwise 51 could not be
apantactic. They are enumerable, for but a finite set can have
lengths > I/n + 1 and < I/n, n = 1, 2 •••
It is separated, since 51 contains no isolated points.
2. The set of intervals D = \S\ just considered are said to be I v
adjoint to 51, or determined by 51, or belonging to 51. I ^1
323. Let 51 be an apantactic limited rectilinear perfect point set, to
ichich belongs the set of intervals D= |8j. Then 51 is formed of the
end points E= \e\ of these intervals, and their limiting points E' .
For we have just seen that the end points e belong to 51. More
over, 51 being perfect, E' must be a part of 51.
338 POINT SETS
51 contains no other points. For let a be a point of 51 not in E,
E' . Let a be another point of 31. In the interval (a, a) lies an
end point e of some interval of D. In the interval (a, e) lies an
other end point er In the interval (a, gj) lies another end point
02, etc. The set of points e, e^ e^-- = a. Hence a lies in E' ,
which is a contradiction.
324. Conversely, the end points E = \e\ and the limiting points of
the end points of a pantactic enumerable set of separated intervals
D = j 8 \ form a perfect apantactic set 51.
For in the first place, 51 is complete, since 51 = (E, E'). 51 can
contain no isolated points, since the intervals S are separated.
Hence 51 is perfect. It is apantactic, since otherwise 51 would em
brace all the points of some interval, which is impossible, as D is
pantactic.
325. Since the adjoint set of intervals D = \S\ is enumerable, it
can be arranged in a 1, 2, 3, ••• order according to size as follows.
Let 8 be the largest interval, or if several are equally large, one
of them. The interval 8 causes Jto fall into two other intervals.
The interval to the left of 8, call IQ, that to the right of £, call Ir
The largest interval in /0, call S0, that in Iv call Sv In this way
we may continue without end, getting a sequence of intervals
S, 80, Sj, S00, S01, S10, Sn-.. (1
and a similar series of intervals
•*i ^0' 1» 00' 01"*
The lengths of the intervals in 1) form a monotone decreasing
sequence which = 0.
If v denote a complex of indices ij/c -••
326. 1. The cardinal number of every perfect limited rectilinear
point set 51 is c.
For if 51 is not apantactic, it embraces all the points of some in
terval, and hence Card 51 = c. Let it be therefore apantactic.
COMPLETE SETS 339
Let D= JS,; be its adjoint set of intervals, arranged as in 325.
Let (£ be the Cantor set of I, 272. Let its adjoint set of intervals
be H= \r]v\, arranged also as in 325. If we set 8V^> 77,, we have
D^H. Hence Card 51 = Card £.
But Card 6 = c by 244, 4.
2. The cardinal number of every limited rectilinear complete set 51
i %o. is either e or c.
For we have seen, 320, 4, that
where (£ is enumerable and ^3 is perfect,
If $ = 0, Card 51 = e.
If $>0, Card 51 = c.
For Card 51 = Card (g + Card $ = e + c = c.
327. The cardinal number of every limited complete set 51 in ^Rn is
either e or c. It is c, ^/5^ Aas a perfect component.
The proof may be made by induction.
For simplicity take m = 2. By a transformation of space [242],
we may bring 51 into a unit square S. Let us therefore suppose
51 were in S originally. Then Card 51 < c by 247, 2.
Let (£ be the projection of 51 on one of the sides of #, and $3 the
points of 51 lying on a parallel to the other side passing through a
point of £. If 33 has a perfect component, Card 33 = c, and hence
Card 51 = c. If 33 does not have a perfect component, the cardinal
number of each 33 is e. Now (£ is complete by I, 717, 4. Hence
if (E contains a perfect component, Card (S = c, otherwise Card
(5 = e. In the first case Card 51 = c, in the second it is e.
328. 1. Let 51 be a complete set lying within the cube Q. Let
DI > D% > ... denote a set of superimposed cubical divisions of Q
of norms = 0. Let d1 be the set of those cubes of Dl containing
no point of 51. Let d2 be the set of those cubes of D2 not in dl ,
which contain no point of 51. In this way we may continue. Let
& = \dn\. Then every point of A = O - 51 lies in 33. For 51 being
340 POINT SETS
complete, any point a of A is an inner point of A. Hence 2>p(a)
lies in A, for some p sufficiently small. Hence a lies in some dm.
We have thus the result :
Any limited complete set is uniquely determined by an enumerable
set of cubes \dn\, each of which is exterior to it.
We may call 33 = \dn\ the border of 51, and the cells dn, border
cells.
2. The totality of all limited perfect or complete sets has the car
dinal number c.
For any limited complete set (E is completely determined by its
border \dn\. The totality of such sets has a cardinal number
< ce = c. Hence Card j(S J < c. Since among the sets (5 is a c-set
of segments, Card (£ > c.
329. If 5lt denote the isolated points of 51, and 5(A its proper
limiting points, we may write
51 = 51 + 5(A.
Similarly we have
5lA=5lAl
51A, = 5U + HA3, etc.
We thus have
51 = 5It + 5lAt + 5U + - - - + 5IAn-it + ?1A« .
At the end of each step, certain points of 51 are sifted out. They
may be considered as adhering loosely to 51, while the part which
remains may be regarded as cohering more closely to the set. Thus
we may call 2lA«-it, the nih adherent, and 21AW the nth coherent.
If the nth coherent is 0, 51 is enumerable.
If the above process does not stop after a finite number of steps,
let 5
If 5lw > 0, we call it the coherent of order &>.
Then obviously 9f — "S9f n 4- 9f
We may now sift 5L as we did 51.
COMPLETE SETS 341
If a is a limitary number, defined by
we set 5la
and call it, when it exists, the coherent of order a. Thus we can
write + 2^ «=1, 2, ...</3 (1
where /3 is a number in Z2.
330. 1. When 51 is enumerable,
, 2, .-.
ivhere $ i§ the sum °f an enumerable set of isolated sets, and £), when
it exists, is dense.
For the adherences of different orders have no point in common
with those of any other order. They are thus distinct. Thus the
sum $ can contain but an enumerable set of adherents, for other
wise 51 could not be enumerable. Thus there is a first ordinal
number j3 for which
5lA = 0.
As now in general
8t^= 2U + 8x0+1,
we have ^ = ^/+1 = ^,+2 = ...
As 21A0 thus contains no isolated points, it is dense, when not 0,
by I, 270.
2. Wlien 51 is not enumerable, X) > 0. For if not, 21 = 3' and 3
is enumerable.
331. 8 = S'. (1
For let D be a cubical division of space. As usual let
or Of'
«•/) 9 <^Z>
denote those cells of D containing a point of 51, 5T respectively.
The cells of 5(i not in 51^ will be adjacent to those of 51^, and
342 POINT SETS
these may be consolidated with the cells of D, forming a new di
vision A of norm 8 which in general will not be cubical. Then
21A = 211 + 21
The last term is formed of cells that contain only a finite number
of points of 21. These cells may be subdivided, forming a new
division U such that in
%E = 2fi + %E* (2
the last term is < e/3- Now if 8 is sufficiently small,
aA-S<! , H -«'<!• (3
o o
Hence from 2), 3) we have 1).
332. -7f2l>0, Card 21 = c.
For let 93 denote the sifted set of 21 [I, 712]. Then $ is per
fect. Plence Card 53 = c, hence Card 21 = c.
333. Let 21 = JaJ, where each a is metric and not discrete. If no
two of the as have more than their frontiers in common, 21 is an
enumerable set in the d's. 21 may be unlimited.
Let us first suppose that 21 lies in a cube O. Let a denote a on
removing its proper frontier points. Then no two of the a's have
a point in common. Let
where the first term ql = Q. There can be but a finite number of
sets a, such that their contents lie between two successive ^'s.
For if ^ ^
«4 » "*~'>q,
we have ^+^+ .~ +«tn>nq..
But the sum on the left is < O, for any n.
As n may = oo, this makes O = oo, which is absurd.
If 21 is not limited, we may effect a cubical division of 9?m.
This in general will split some of the a's into smaller sets b. In
each cube of this division there is but an enumerable set of the b's
by what has just been proved.
CHAPTER XI
MEASURE
Upper Measure
334. 1. Let 51 be a limited point set. An enumerable set of
metric sets D= {e?tj, such that each point of 51 lies in some d^ is
called an enclosure of 51. If each point of 51 lies within some c?0 D
is called an outer enclosure. The sets dt are called cells. To each
enclosure corresponds the finite or infinite series
23, (1
which may or may not converge. In any case the minimum of all
the numbers 1) is finite and <_0. For let A be a cubical division
of space, 51A is obviously an enclosure and the corresponding sum
1) is also 51A, since we have agreed to read this last symbol either
as a point set or as its content.
We call ,». v ,
Mm 2at,
with respect to the class of all possible enclosures D, the upper
measure of 51, and write
I = Meas 51 = Min 2rft.
D
2. The minimum of the sums 1) is the same when we restrict our
selves to the class of all outer enclosures.
For let D= \dt\ be any enclosure. For each c?t, there exists a
cubical division of space such that those of its cells, call them d^
*~* £
containing points of d^ have a content differing from dt by < «;•
Obviously the cells \diK\ form an outer enclosure of 51, and
343
344 MEASURE
As e is small at pleasure, Min 2c?t over the class of outer en
closures = Min 2c?t over the class of all enclosures.
3. Two metric sets whose common points lie on their frontiers
are called non-overlapping. The enclosure D = 2dt is called non-
overlapping, when any two of its cells are non-overlapping.
Any enclosure D may be replaced by a non-overlapping enclosure.
For let U(d1^ c?2) = d1 4- «2,
U(d^ dz< d3) = d1 + e2 + 08,
U(dl d2 d3 d±) = d1 -h ez + e3 + «4, etc.
Obviously each en is metric. For uniformity let us set dj = er
Then E = \en\ is a non-overlapping enclosure of 31. As
£%.<*&
we see that £^# minimum of the sums 1) ^s the same, when we restrict
ourselves to the class of non- overlapping enclosures.
Obviously we may adjoin to any cell en, any or all of its
improper limiting points.
4. In the enclosure E= \en\ found in 3, no two of its cells
have a point in common. Such enclosures may be called distinct.
335. 1. Let D = \dt\, E = \eK\ be tivo non- overlapping enclosures
of%. Let
^ *!>»(<, O-
Then
A= 5^1, *,* = !, 2, ...
is a non- over lapping enclosure of 51.
For SiK is metric by 22, 2. Two of the S's are obviously non-
overlapping. Each point of 51 lies in some dL and in some eK,
hence a lies in 8t(C.
2. We say A is the divisor of the enclosures D, E.
336.
For let E= \e,\ be an enclosure of 53. Those of its cells dt con
taining a point of H form an enclosure D = \dt\ of H. Now the
class of all enclosures A = \B^ of 51 contains the class D as a sub
class.
UPPER MEASURE 345
As 2<2?t,
we have
Min 2St< Min Zdt< Min
A /> £•
from which 1) follows at once.
337. If 21 w
21 = 51. (l
For let D be a cubical division of space such that
a/)_a<€ , a -«/><€. (2
Let us set 33 = 1^. Let -#=JetJ be an outer enclosure of 33.
Since 33 is complete, there exists a finite set of cells in E which
contain all the points of 33 by 301. The volume of this set is
obviously > 33 ; hence a fortiori
2?t > 8.
Hence = -
33>33.
BUt I>i,by336,
>» = !i,
>g-€,by2). (3
On the other hand, =
2l<2U<2l + e, by 2). (4
From 3), 4) we have 1), since e is arbitrarily small.
338. If% is complete, =
31 = 21.
For by definition =
21 = Min 2aM
with respect to all outer enclosures D = \dt\. But 21 being com
plete, we can replace D by a finite set of cells F= \f,\ lying in D,
such that F is an enclosure of 21. Finally the enclosure F can be
replaced by a non-overlapping enclosure G- = \g,} by 334, 3.
with respect to the class of enclosures 6r. But this minimum
value is also 21 by 2, 8.
346 MEASURE
339. Let the limited set 21 = \Un\ be the union of a finite or infinite
enumerable set of sets 2In. Then
For to each ?Jfn corresponds an enclosure Dn = \ dni \ such that
Sc?m < 2In + — •> e > 0, arbitrarily small.
But the cells of all the enclosures Z>n, also form an enclosure.
Hence
This gives 1), as e is small at pleasure.
340. Let n lie in the metric set 2ft. Let A = 2ft - 21, be the
complementary set. Then
For from m = K + A,
follows = _ =
m<K + A, by 339.
But . _
9W = 2ft, by 337.
341. If 51 = 53 + 6, awe? 53, (£ are exterior to each other,
1 = S + f . (1
For, if any enclosure D — \d^ of 51 embraces a cell containing
a point of 53 and (£, it may be split up into two metric cells c?[,
c?[', each containing points of 53 only, or of (£ only. Then
Thus we may suppose the cells of D embrace only cells
D' = \d(\ containing no point of (£, and cells Df' = \d['l con
taining no point of 53. Then
s5t = 25; + 25;'. (2
UPPER MEASURE 347
By properly choosing D, we may crowd the sum on the left
down toward its minimum. Now the class of enclosures D' is
included in the class of all enclosures of 33, and a similar remark
holds for D".
Thus from 2) follows that
This with 339 gives 1).
342. If 51 = 23 + 9ft, 2ft being metric,
For let D be a cubical division of norm d. Let n denote points
of 9ft in the cells containing points of Front 9ft. Let tn denote
the other points of Oft- Then tn and 33 are exterior to each other,
and by 337 and 341,
Meas(33 + m) = i + m.
As 2l = 33 + tn + tt,
Meas(33 + m)<I by 336.
Als° l<i + m + n by 339.
Thus S + m<I<i + m + n. (2
Now if d is sufficiently small,
9ft-e<m ; tt<e.
Thus 2) gives, as rn<9ft,
which gives 1), as e>0 is arbitrarily small.
343. 1. Let 51 lie in the metric set 53, and also in the metric set
£• Let £ = %-% , <7=<S-a.
Then §-J=§-5.
For let
3) =
348 MEASURE
Thus
§ - H = £ + 81 - (§1 + 5) = S - 5
g - <7 = S + Sj - (ii 4- 5) = S - S.
2. If 5l<53, the complement of 51 with respect to 53 will
frequently be denoted by the corresponding English letter. Thus
Mod 53
Lower Measure
344. 1. We are now in position to define the notion of lower
measure. Let 51 lie in a metric set 2ft. The complementary set
A = 2ft — 51 has an upper measure A. We say now that 2ft — 3
is the lower measure of 51, and write
By 343 this definition is independent of the set 2ft chosen.
When | _ ^
we say 51 is measurable, and write
A set whose measure is 0 is called a null set.
2. Let U= [e,] be an enclosure of A.
Then 51 = Max (§* - 25,) »
m'^ respect to the class of all enclosures E.
3. If (g = |et} is an enclosure of 51, the enclosures ^ and <g may
obviously, without loss of generality, be restricted to metric cells
which contain no points not in 2ft. If this is the case, and if @,
^are each non-overlapping, we shall say they are normal enclosures.
If (§:, 8 are two normal enclosures of a set 51, obviously their
divisor is also normal.
LOWER MEASURE 349
345. 1. 51 >0.
For let 51 lie in the metric set $[)?.
Then / 2 = ®-Z "'.-
But by 336, 2<§,
hence 2K-1>0.
2. 5l<i.
For let 51 lie in the metric set 9ft.
Then
by 340.
Hence 51 = §?-
346. 1. For any limited set 51,
51 < 5j[*< I < 5l (1
For let Z>= §dtj be an enclosure of 51. Then
51 = Min 2<?t,
,0
when Z> ranges over the class F of all ^wzVe enclosures. On the
other hand,
I = Min 2 d,
D
when D ranges over the class E of all enumerable enclosures.
But the class E includes the class F. Hence I < S.
To show that ™ ^ or f 9
21 < <H, V^
we observe that as just shown
A>I.
Hence, A ^^
sw-A<sw»^ = a. (3
A + 5t = % by 16.
This with 3) gives 2).
350 MEASURE
2. If 51 is metric, it is measurable, and
This follows at once from 1).
347. Let 51 be measurable and lie in the metric set 9ft. Then A
is measurable, and «, «> ^
For
A = 9ft - 5t. (2
since 51 is measurable. This last gives
This with 2) shows that J = A ; hence ^4. is measurable. From
2) now follows 1).
348. If %<%,then 5T<<8. (1
For as usual let A, B be the complements of 51, 23 with respect
to a metric set 9ft. Since 51 < 53, A > B.
Hence, by 336,
A>B.
Thus, ^ = ^
9ft - 4 < 9ft - B,
which gives 1).
349. For 51 to be measurable, it is necessary and sufficient that
where 9ft is any metric set > 51, and A = 9ft — 51.
It is sufficient, for then 1) shows that
! = §£-!.
But the right side is by definition 51 ; hence 3 = 5(.
It is necessary as 347 shows.
350. Let 51 = {aj be the union of an enumerable set of non
overlapping metric sets. Then 51 is measurable, and
S = 2an. 1
LOWER MEASURE 351
Let S denote the infinite series on the right of 1). As usual
let Sn denote the sum of the first n terms. Let 5ln = (c^, ••• an).
Then 5ln < 51 and by 336,
§n = #n<! , for any n. (2
Thus S is convergent and
£<I. (3
On the other hand, by 339,
1 < S. (4
From 3), 4) follows that
S = I = lim Sn = lim Sn. (5
We show now that 51 is measurable. To this end, let 9ft be a
metric set > 51, and 2ln + An = 9ft as usual.
Then I in + 2n = St . (6
But A < An , hence A < An.
Thus 6) gives = ^ ^
A + 2ln < 3ft,
for any n. Hence
Hence by 339, J+8aafi|.
Thus by 349, 51 is measurable.
351. Let
For let 9ft be a metric set > 51. Let A, B, 0 be the comple
ments of 51, 53, S, with reference to 9ft.
Let E= \em\ , F= \fn\
be normal enclosures of B, C. Let
and D = j^mn| the divisor of E, F.
352 MEASURE
As all the points of A are in B, and also in (7, they are in both
E and F, and hence in the cells of D, which thus forms a normal
enclosure of A. Let
Let us set _ /. _ . j
m '~ /in I t/ wi ' */ TI in I ?i *
Then by 350, - _ - -
By 347, _ ! ^ I ^
?» = 7m + 9m , /n = ??n + AB .
Hence ^ ^
,
Hence adding,
(^-2/n)
rfmn)]. (2
Thus by 339, the term in [ ] is < 0. Thus 2) gives
Now ^ = U\gm, hn, dmn\ m,n=l, 2,
2n) < - 2wn < a. (3
But ^
33 = Max (2ft - 2?m)
2 = Max (-
Thus 3) gives 1) at once.
Measurable Sets
352. 1. Let H =$ + (£. 7f 53, S are measurable, then 51 i
measurable, and
f = 5 + i. (
For <£ + (£< 51 , by 351
< I < i + S , by 339.
*D -i.
LOWER MEASURE 353
2. Let 51 = 33 + (£. If 51, 53 are measurable, so is & and
i = i-5. (2
For let 51 lie in the metric set 9}?. Then
2fl - a = 2K - OB + S) = (2» - G) - 53-
Thus A = C-S;
Hence <7=S + A
Thus (7 is measurable by 1. Hence £ is measurable by 347,
and
5l = 33 + G.
From this follows 2) at once.
353. 1. Let 51 = 25ln be the sum of an enumerable set of measur
able sets. Then 51 is measurable and
If 51 is the sum of a finite number of sets, the theorem is obvi
ously true by 352, 1. In case 51 embraces an infinite number of
sets, the reasoning of 350 may be employed.
2. Let $1 = \yin] be the union of an enumerable set of null sets.
Then $1 is a null set.
Follows at once from 1.
3. Let 51= J5InJ be the union of an enumerable set of measurable
sets whose common points two and two, form null sets. Then 51 is
measurable and
§ = 25ln.
4. Let G = \tn\ be a non- overlapping enclosure 0/51. Then (§: is
measurable, and
S = 2en.
5. Let 53 < 51. Those cells of (£ containing a point of 33 may
be denoted by 33e, and their measure will then be of course
If $ = 51, this will be §. This notation is analogous to that
used in volume I when treating content.
354 MEASURE
6. If g= \\n\ is another non-overlapping enclosure of some set
then
T) = Dw((g, g)
is measurable.
For the cells of £) are
&.
Thus Stlt is metric, and
354. 1. Harnack Sets. Let 21 be an interval of length I. Let
X = Zx + Z2 + ...
be a positive term series whose sum X > 0 is <_ I. As in defining
Cantor's set, I, 272, let us place a black interval of length ^ in the
middle of 21. In a similar manner let us place in each of the re
maining or white intervals, a black interval, whose total lengths
= Z2. Let us continue in this way; we get an enumerable set of
black intervals 53, and obviously
§ =X.
If we omit the end points from each of the black intervals we get
a set 53*, and obviously
The set £ = 51 - 53*
we call a Harnack set. This is complete by 324 ; and by 338, 347,
When X = Z, ^p is discrete, and the set reduces to a set similar
to Cantor's set. When \<l, we get an apantactic perfect set
whose upper content is I — X > 0, and whose lower content is 0.
2. Within each of the black intervals let us put a set of points
having the end points for its first derivative. The totality of
these points form an isolated set $ anc^ 3' = €>• But b}r 331,
3 = $• If now ^j is not discrete, <J is n°t- We have thus the
theorem :
There exist isolated point sets which are not discrete.
LOWER MEASURE 355
3. It is easy to extend Harnack sets to $„. For example, in 9?2,
let S be the unit square. On two of its adjacent sides let us place
congruent Harnack sets $>. We now draw lines through the end
points of the black intervals parallel to the sides. There results
an enumerable set of black squares @ = \<Sn\. The sides of the
squares @ and their limiting points form obviously an apantactic
perfect set $.
Let af + a|+... = w
be a series whose sum 0 < m < 1.
We can choose § such that the square corresponding to its larg
est black interval has the area a\ ; the four squares corresponding
to the next two largest black intervals have the total area &\, etc.
Then =
Henoe ...
355. 1. If (g = \tm\ is an enclosure of % such that
it is called an e-enclosure. Let A be the complement of SI with
respect to the metric set 9D?. Let E = \en\ be an e-enclosure of A.
We call (g, E complementarg e-enclosures belonging to SI.
2. //^ SI is measurable, then each pair of complementary e/2
normal enclosures @, ^, wAosg divisor $) =
!5) < e, e small at pleasure. (1
For let (£, J^ be any pair of complementary e/2 normal enclo
sures. Then
t-f<{ , !-.!.< f
Adding, we get 0 < S + 2- (84-2)< c;
0<S-h^-^<e. (2
But the points of 9Q? fall into one of three classes : 1° the points
of S) ; 2° those of (g not in £) ; 3° those of j^ not in £). Thus
i + .f = ^ + S.
This in 2) gives 1).
356 MEASURE
356. 1. Up to the present we have used only metric enclosures
of a set 21. If the cells enclosing 21 are measurable, we call the
enclosure measurable.
Let (£ = \tn\ be a measurable enclosure. If the points common
to any two of its cells form a null set, we say (£ is non-
overlapping. The terms distinct, normal, go over without
change.
2. We prove now that |= Min ^ (1
with respect to the class of non-overlapping measurable enclosures.
For, as in 339, there exists a metric enclosure mn = \dnK\ of
each en such that 2 dnll differs from ?n by < e/2n. But the set
•
\mn\ forms a metric enclosure of 21. Tims
< 2 < K < 2 en + = 2n 4- e,
ft , K
which establishes 1).
357. Z0£ @ be a distinct measurable enclosure of 21. Let \ denote
those cells containing points of the complement A. If for each e > 0
there exists an (£ such that f <• e, £ Aew 21 ^s measurable.
For let (g = e + f. Then e < 21. Hence e < 21 by 348. But
Hence l
and thus S = 21
358. 1. . The divisor £) of two measurable sets 2(, 53 ^
For let ($, ^ be a pair of complementary e/4 normal enclosures
belonging to 21 ; let g, F be similar enclosures of 53. Let
e = Dt;(<g, J£) , f =
Then
?<e/2 , ?<€/:, by 355, 2,
LOWER MEASURE 357
Now © = Dv(&, g) is a normal metric enclosure of £). More
over its cells g which contain points of 3) and (7(3) ) lie among
the cells of e, f. Hence
Thus b}r 357, 2) is measurable.
2. Let 51, $8 be measurable.
Let ) , II =
For
Hence
U = + Meas (53 -
359. Xe/ SI = U \ 5Im j be the union of an enumerable set of
measurable cells ; moreover let 51 be limited. Then 51 is measurable.
If we set
For 2) = Dy^, 512) is measurable by 358.
T »f
51, = ^ + ^ , 5l2 = £ + a2.
Then ax, a2 are measurable by 352, 2.
AQ
U = (5^,512) = 2)4-^ + 02,
U is measurable. As U and ^31 are measurable, so is 332. In a
similar manner we show that $3, 534 ••• are measurable. As
SI is measurable by 353, 1, and the relation 1) holds by the same
theorem.
360. Let 5Tj < 5I2 < ••• be a set of measurable aggregates whose
union 51 is limited. Then 51 is measurable, and
5=limin.
358 MEASURE
For let Qf «r w or
a2 = ^2-^i > ^3 = ^3-^2 •••
For uniformity let us set a: = 21. Then
a = 2o_.
As each an is measurable
t = 2aw
= lim (aj+ ••• + an)
n=oo
= lim Sn .
361. .Le£ Stj, 5I2 "* ^ measurable and their union H limited. If
£) = Dv 52IBj > 0, & is measurable.
For let 21 lie in the metric set $ft;
let 2) + D = 2ft , %n+An = m
as usual.
Now $) denoting the points common to all the 2ln, no point of
D can lie in all of the $(n, hence it lies in some one or more of the
An. Thus D<\An\. (1
On the other hand, a point of \An\ lies in some Am, hence it
does not lie in 2lm. Hence it does not lie in £). Thus it lies in
D. Hence \An\<D. (2
From 1), 2) we have D = $A \
As each An is measurable, so is D. Hence 3) is.
362. If Hj^E^ ••• is an enumerable set of measurable aggre
gates, their divisor $) is measurable, and
For as usual let D, An be the complements of 2), 2ln with respect
to some metric set 3ft.
Then D=\An\ , An<A»+l.
Hence by 360,
LOWER MEASURE 359
As $D = m - D,
wehave 5 = ^-5
= lim §?-
363. 1. The points x = (x± ••> xm) such that
form a standard rectangular cell, whose edges have the lengths
e1 = bl — a1 , ••• , ^m = ^ — am.
When e^ = ez = ••• = em, the cell is a standard cube. A normal
enclosure of the limited set SI, whose cells S = Jenj are standard
cells, is called a standard enclosure.
2. For each e > 0, £fore are standard e-enclosures of any limited
m*.
For let (£ = jenj be any ry-enclosure of SI. Then
2en-i<)?. (2
Each en being metric, may be enclosed in the cells of a finite
standard outer enclosure Fn , such that
Fn-tn<-n/ln , ™ = l, 2,...
Then g = \Fn\ is an enclosure of SI, and
<i+27;, by 2).
But the enclosure .F can be replaced by a non-overlapping
standard enclosure © = 5gn|, as in 334, 3. But © < ^Fn.
Hence if 2 77 is taken < e,
and © is an e-enclosure.
3. Let £=!emj, 8=SW
be two non-overlapping enclosures of the same or of different
sets. Let emn = Dv(tm, fn).
360 MEASURE
^m==\^m, !"> ^m.21 ^wi, 3 " -V ~^~ ^m"> C"
then 0OT is measurable. By this process the metric or measurable
cell tm falls into an enumerable set of non-overlapping measur
able cells, as indicated in 3). If we suppose this decomposition to
take place for each cell of (§, we shall say we have superimposed g
on (£.
364. (IF". H. Young.} Let (£ be any complete set in limited 21.
2%era
| = Max <£. (1
For let 8 lie within a cube 2ft, and let ^ = 90? - 51, (7= 2ft - (£
be as usual the complementary sets.
Let 53= |bB} be a border set of (£ [328]. It is also a non-
overlapping enclosure of (7; we may suppose it is a standard en
closure of C. Let E be a standard e-enclosure of A. Let us
superimpose E on 53, getting a measurable enclosure A of both 0
and A. Then
(7= (7A>^A.
Hence
& = m-C=m- (7A<9^-^A.
Thus
<£ = (£, by 338
< Meas (2ft - ^4A)
!A, by 352, 2
Hence
(£<«,
and thus
Maxg<|- (2
On the other hand, it is easy to show that
Max§>2t. (3
For let AD be an e-outer enclosure of A, formed of standard
non-overlapping cells all of which, after having discarded certain
parts, lie in 2ft.
LOWER MEASURE 361
Let $ = 9ft - AD + ft, (4
where g denotes the frontier points of AD lying in 21. Obviously
$ is complete. Since each face of D is a null set, g is a null set.
Thus each set on the right of 4) is measurable, hence
I = m - AD + §
= m-ID
= 2JJ-2-e' , 0<e'<e
Thus Max (£>$ =>|-e,
from which follows 3), since e is small at pleasure.
365. 1. If 21 is complete, it is measurable, and
8 = 8.
For by 364,
| = 2I.
On the other hand,
S=l, by 338.
2. Let 33 be any measurable set in the limited set 21. Then
| = Max 5. (1
For |>33 = §.
Hence, 21 > Max 5. (2
But the class of measurable components of 21 embraces the
class of complete components (£, since each (£ is measurable by 1.
Thus Max 5 > Max £ (3
From 2), 3) we have 1), on using 364.
366. Van Week Sets. Let (E denote the unit interval (0, 1),
whose middle point call M. Let 3 denote the irrational points of
(5. Let the division Dn, w = 1, 2, ••• divide (g into equal intervals
$n of length l/2n.
3(32 MEASURE
We throw the points 3 into two classes 51 = \a j, 53 = \b\ having
the following properties :
1° To each a corresponds a point b symmetrical with respect
to M, and conversely.
2° If a falls in the segment & of J9n, each of the other seg
ments B' of Dn shall contain a point a' of 51 such that a' is situated
in S' as a is situated in B.
3° Each S of Dn shall contain a point a' of 51 such that it is
situated in 8, as any given point a of 51 is situated in (g.
4° 51 shall contain a point a situated in (£ as any given point
a' of 51 is in any 8n.
The 1° condition states that 51 goes over into 53 on rotating &
about M. The 2° condition states that 51 falls into n = 1, 2, 22,
23, ••• congruent subsets. The 3° condition states that the subset
2ln of 51 in Bn goes over into 51 on stretching it in the ratio 2n : 1.
The condition 4° states that 51 goes over into 5ln on contracting it
in the ratio 1 : 2n.
We show now that 51, and therefore 53 are not measurable. In
the first place, we note that
1-8,
by 1°. As 3 = 51 + 53, if 51 or 53 were measurable, the other would
be, and ^ ^
«=*-$.
Thus if we show 51 or 53 = 1, neither 51 nor 53 is measurable.
We show this by proving that if 21 = «< 1, then 53 is a measurable
set, and S3 = 1. But when 53 is measurable, $8 = |- as we saw, and
we are led to a contradiction.
Let e = el + e2 -}- ••• be a positive term series whose sum e is
small at pleasure. Let ^ = \en\ be a non-overlapping ej-enclosure
of 51, lying in <g. Then
i1 = 2en = « + e; = «1 , 0
Let 53! = 3 - (gj ; then 53j < 53, and
= 1 — « — €j>l — « —
. LOWER MEASURE 363
Each interval en contains one or more intervals 7;nl, ?/n2, ... of
some Z>,, such that
f*?nm = en-0-n , 0 < (7n
where v
& = Zcrn
may be taken small at pleasure.
Now each rjnm has a subset 5lnTO of 21 entirely similar to 21.
Hence there exists an enclosure (£nm of 5lnra, whose measure «nm is
such that
But @2 = {^n>«S is a non-overlapping enclosure of 21, whose
measure v~ ^^ .
«2 = «!Sr7nW = «l^ 'On - ^n)
if <r is taken sufficiently small.
Let 932 denote the irrational points in (^ — C?2. It is a part of
53, and 332 has no point in common with ^&l. We have
In this way we may continue. Thus 53 contains the measurable
component ^ ^
whose measure is
>!-€.
As e is small at pleasure, 53 = 1.
367. (If. H. Young.) Let
«!,**, V" C1
6e an infinite enumerable set of point sets whose union 51 is limited.
Let 5ln>a>0 , n = 1, 2 «•• Then there exists a set of points each
of which belongs to an infinity of the sets 1) and of lower measure > a.
364 MEASURE
For by 365, 2, there exists in the sets 1), measurable sets
G, , <52 , 6,— . (2
each of whose measures (£n > a. Let us consider the first n of
these sets, viz.: ^ > ^ _ ^ (g
The points common to any two of the sets 3) form a measurable
set £5^ by 358, 1. Hence the union (£ln = { $)l(C j is measurable, by
359. The difference of one of the sets 3), as (^ and Dv(&v (Sln),
is a measurable set cx which contains no point in common with the
remaining sets of 3). Moreover
/». <K.
cx > a - (Slfl.
In the same way we may reason with the other sets (£2, (E3 •••
of 3). Thus 51 contains w measurable sets Cj, C2 ••• cn no two of
which have a common point.
Hence c = Cl + ... + c.
is a measurable set and
The first and last members give
£,„>«- la.
n
Thus however small a > 0 may be, there exists a p such that
Si,, l-«. • , (4
Let us now group the sets 2) in sets of /a. These sets give rise
to a sequence of measurable sets
such that the points of each set in 5) belong to at least two of the-
sets 1) and such that the measure of each is > the right side of 4).
We may now reason on the sets 5) as we did on those in 2).
We would thus be led to a sequence of measurable sets
ASSOCIATE SETS 365
such that the points of each set in 6) lie in at least two of the sets
5), and hence in at least 22 of the sets 1), and such that their
measures are.
In this way we may continue indefinitely. Let now ^3l be the
union of all the points of 51, common to at least two of the sets 1).
Let 332 be the union of the points of 51 common to at least 22 of
the sets 1), etc. In this way we get the sequence
«!>«2> •••
each of which contains a measurable set whose measure is
We have now only to apply 25 and 364.
368. As corollaries of 367 we have:
1. Let Cj, Q2 ••• be an infinite enumerable set of non-overlapping
cubes whose union is limited. Let each Qn > a > 0. Then there
exists a set of points b whose cardinal number is c, lying in an infin
ity of the On and such that b > a.
2. (Arzeld.) Let ^ , ?/2 ••• =77. On each line yn there exists an
enumerable set of intervals of length £„. Should the number of inter
vals vn on the lines yn be finite, let vn = cc. In any case Bn > a > 0,
n = 1, 2, ... and the projections of these intervals lie in 51 = (a, 5).
Then there exists at least one point x = f in 51, such that the ordinate
through £ is cut by an infinity of these intervals.
Associate Sets
369. 1. Let e^e^eg ••• =0. (1
Let (gn be a standard en-enclosure of 2ln. If the cells of Q?n+1 lie in
<gn, we write @1>@2>-" (2
and call 2) a standard sequence of enclosures belonging to 1).
Obviously such sequences exist. The set
%e = Vvl®n\
is called an outer associated set of 51. Obviously
51<5L.
366 MEASURE
2. Each outer associated set 5le is measurable, and
f=ie=limin.
n=x>
For each (gn is measurable; hence 5le is measurable by 362, and
= , asen=0.
370. 1. Let A be the complement of 51 with respect to some
cube O containing 5(. Let ,4.e be an outer associated set of A.
Then a.= o-^.
is called aw inner associated set of 51. Obviously
2. The inner associated set 5It is measurable, and
§1=2.
For Ae is measurable by 369, 2. Hence 5lt = G — ^4.e is meas
urable. But
Ae = A
by 369, 2. Hence
Separated Sets
371. Let H, 53 be two limited point sets. If there exist
measurable enclosures (£, g °f H, 53 such that 3) = Dv(@, g) is a
null set, we say 51, 53 are separated.
If we superimpose g on (§, we get an enclosure of (£ = (51, 53)
such that those cells containing points of both 51, 53 form a null
set, since these cells are precisely £). We shall call such an en
closure of (£ a null enclosure.
Let 5l = {5ln} ; we shall call this a separated division of 51 into
the subsets 5ln, if each pair 5lm, 5ln is separated. We shall also
say the 5ln are separated.
SEPARATED SETS 367
372. For 21, 53 to be separated, it is necessary and sufficient that
2) = ZH2le, 53e)
is a null set.
It is sufficient. For let
Then G = (a,b,5»
is a measurable enclosure of (£, consisting of three measurable
cells. Of these only £) contains points of both 21, 53. But by
hypothesis £) is a null set. Hence 21, 53 are separated.
It is necessary. For let 9D? be a null distinct enclosure of (E,
such that those of its cells 9?, containing points of 21, $8 form a
null set. Let us superimpose 9# on the enclosure (£ above, get
ting an enclosure g of 21.
The cells of g arising from a contain no point of 53 ; similarly
the cells arising from b contain no point of 21. On the other
hand, the cells arising from £), split up into three classes
The first contains no point of 53, the second no point of 21, the
cells of the last contain both points of 21, 53. As £)a,6< W,
$a.6 = 0. (1
On the other hand,
He = a + 2>?l;
hence «+*>.+5>*>a.
Thus a + §0>S, (2
byl). Also g.=a + £=I by 369, 2.
This with 2) gives ^ ^ ^ ^
a+ :Ta^>a + :I).
Hence 3^^ (3
But l).>f}a + §6.
This with 3) gives £>6 = 0.
In a similar manner we find that £)a = 0. Hence 3) is a null
set by 3).
368 MEASURE
373. 1. J/ 51, $ are separated, then £) = Dv($l, $8) is a null set.
For S)e = Dv 23 is a nul1 set b 3?2- But £) < 2).
2. Let 51, 23 be the Van Vleck sets in 366. We saw there that
1 = 8 = 1. Then by 369, 2, §c = 5e = 1. The divisor of 5Ie, 23e is
not a null set. Hence by 372, 51, 33 are not separated. Thus the
condition that ^ be a null set is necessary, but not sufficient.
374. 1. Let J51J, \y$n\ oe separated divisions of 51. Let
(£IK = Zhj(5It, $3*). Then \&IK\ is a separated division 0/"5l also.
We have to show there exists a null enclosure of any two of the
sets <£„, £mn. Now <£„ lies in 51 and 53. ; also £mn lies in 5Tm, 23n.
By hypothesis there exists a null enclosure & of 5lt, 5lm; and a null
enclosure g of 23*, 23n. Then © = Dv(&, g) is a null enclosure of
5It, 5Im and of 58., 33n. Thus those cells of ®, call them ®a, con
taining points of both 5It, 5Tm form a null set; and those of its cells
©6, containing points of both 53K, 23n also form a null set.
Let Gr— \g\ denote the cells of © that contain points of both
(£„, (Smn. Then a cell g contains points of 5ft 5TTO 53K 53n. Thus g
lies in @a or ©6. Thus in either case G- is a null set. Hence \&lK\
form a separated division of 51.
2. Let D be a separated division of 51 into the cells d^ d2 -••
Let E be another separated division of 51 into the cells e1, e2---
We have seen that F = \flK\ where fllt = Dv(dt, eK~) is also a sepa
rated division of 51. We shall say that F is obtained by superim
posing E on D or D on ^, and write F '= D + E= E+ D.
3. Let E be a separated division of the separated component 53
of 51, while D is a separated division of 51. If dt is a cell of D, eK
a cell of E, and c?t(( = Dv(dL, eK), then
Thus superposing E on D causes each cell c?t to fall into sepa
rated cells e?tl, c?tl ••• St. The union of all these cells, arising from
different dt, gives a separated division of 5( which we also denote
by D + E.
375. Let J5tn| be a separated division of 51. Let 23 < 51, and let
23n denote the points of 23 in 5(n. TAew J23nj ^s a separated division
of*.
SEPARATED SETS 369
For let $) be a null enclosure of 5lm, 5In. Let S)^ denote the
cells of & containing points of both 5im, 5ln. Let (£ denote the
cells of 2) containing points of $8 ; let (£a)6 denote the cells con
taining points of both 53m, 53n. Then
As $Da& is a null set, so is (£«&.
376. 1. Letn = (53, <£) fo a separated division of 51.
i = i + I . (i
For let e1 > €2 > ••• = 0. There exist en-measurable enclosures
of 5t, 53, <£ ; call them respectively An, Bn, Cn. Then &n = An +
Bn + (7n is an en-enclosure of 51, 33, (S simultaneously.
Since 53, (S are separated, there exist enclosures B, 0 of 53, &
such that those cells of D = B + 0 containing points of both 53
and (E form a null set. Let us now superpose D on ($n getting
an en-enclosure En=\ena\ of 51, 53, & simultaneously. Let ebn
denote the cells of En containing points of 53 alone ; ecn those
cells containing only points of (£ ; and ebc those cells containing
points of both 53, (£. Then
2"ns = 2?fen + Zecn + tebc . (2
s
As 2e6c = 0, we see that as n == oo,
Hence passing to the limit w= oo, in 2) we get 1).
2. Let 51 = f53nj fo a separated division of limited 51. Then
l = 2g.. (1
For in the first place, the series
B = 2§n (2
is convergent. In fact let 5In = (53X , 532 • • - 53n).
Then 5IW < 5T, and hence In < I.
370 MEASURE
On the other hand, by 1
in=i1+ ... + §„=£„,
the sum of the first n terms of the series 2). Thus
^n<I,
and hence B is convergent by 80, 4. Thus
^<i.
On the other hand, by 339,
£>i.
The last two relations give 1).
CHAPTER XII
LEBESGUE INTEGRALS
General Theory
377. In the foregoing chapters we have developed a theory of
integration which rests on the notion of content. In this chapter
we propose to develop a theory of integration due to Lebesgue,
which rests on the notion of measure. The presentation here
given differs considerably from that of Lebesgue. As the reader
will see, the theory of Lebesgue integrals as here presented differs
from that of the theory of ordinary integrals only in employing
an infinite number of cells instead of a finite number.
378. In the following we shall suppose the field of integration
SI to be limited, as also the integrand SI lies in Sftm and for brevity
we set/(#) =f(xl >•• xm). Let us effect a separated division of
SI into cells Sj, S2 • ••. If each cell 8t lies in a cube of side d, we
shall say D is a separated division of norm d.
As before, let
, &>t = Osc/= M, - m, in 8t.
Then SD = I,Mfg , ^,= 20* A,
the summation extending over all the cells of SI, are called the
upper and lower sums off over SI with respect to D.
The sum
is called the oscillatory sum with respect to D.
379. If m = Min f,M= Max / in SI, then
m < mt <M^<M.
371
372 LEBESGUE INTEGRALS
Hence
Thus
t <SD<8j><
But 2lt = l,
by 376, 2.
380. 1. Since /is limited in 51,
Max $0 , Min Sj)
with respect to the class of all separated divisions D of 31, are
finite. We call them respectively the lower and upper Lebesgue
integrals of /over the field 51, and write
=U*xSD ; /=Min^.
In order to distinguish these new integrals from the old ones,
we have slightly modified the old symbol ( to resemble somewhat
script L, or I , in honor of the author of these integrals.
J^i JL^K
we say /is L-integrable over 51, and denote the common value by
which we call the L-integral.
The integrals treated of in Vol. I we will call R-integrals, i.e.
integrals in the sense of Riemann.
2. Letf be limited over the null set 51. Then f is L-integrable in
51, and
This is obvious from 379.
381. Let 51 be metric or complete. Then
GENERAL THEORY 373
For let dj, c?2 ••• be an unmixed metric or complete division of
31 of norm d. Let each cell dk be split up into the separated cells
Then since d^ is complete or metric,
(* =: (I = 2-.(1.
i t ue
Hence using the customary notation,
Thus summing over K,
»,5. <
Summing over A gives
2wt3t <
Thus by definition,
Letting now (^=0, we get 1).
2. Ze£ H i« metric or complete. If f is R-integrable in 31, it is
L-integrable and
(2
3. In case that H is not metric or complete, the relations 1), 2)
may not hold.
Example 1. Let 31 denote the rational points in the interval
(0, 1).
Let
/= 1, for x = — , n even
n
= 2, when n is odd.
Then
f/-l , (T/-2;
Ja Jr
while
since 31 is a null set. Thus 1) does not hold.
374 LEBESGUE INTEGRALS
Example 2. Let/= 1 at the rational points 51 in (0, 1). Then
=° ' ^Lf<&- (3
Let g= — 1 in 51. Then
Thus in 3) the i-integral is less than the ^-integral, while in
4) it is greater.
Example 3. Let /= 1 at the irrational points 51 in (0, 1).
Then
although 51 is neither metric nor complete.
382. Let 2), A be separated divisions of 51. Let
E=D + &= \et\.
For any cell d, of D splits up into c?tl, dl9 ••• on superimposing
A, and = =
But = * =
and = =
Thus v < v > s
383. 1. Extremal Sequences. There exists a sequence of sepa
rated divisions n r> r> (\
**\ •> ^2 ' ^3 "
each Dn+1 being obtained from Dn by superposition, such that
(2
GENERAL THEORY 375
For let el > e2 > ••• =0. For each en, there exists a division
En such that
Let JSi + D! = Da , ^3+D2 =
and for uniformity set El = Dr Then by 382,
sDn^<sDn , SD^<SE^
Hence
Letting n = oo we get 2).
Thus there exists a sequence \D'n} of the type 1) for 2), and a
sequence \D'i \ of the same type for 3). Let now Dn = D'n + #'„'•
Obviously 2), 3) hold simultaneously for the sequence \Dn\.
2. The sequence 1) is called an extremal sequence.
3. Let \Dn} be an extremal sequence, and E any separated divi
sion of 2(. Let En = Dn + E. Then E^ E^--- is an extremal
sequence also.
384. Let f be L-integrable in 31. Then for any extremal sequence
MU,
inhere dt are the cells of Dn, and £ any point o/8( in
F°r
Passing to the limit we get 1).
385. 1. Let m = Min/, M= Max/ tw 3[.
Jfl.
This follows at once from 379 and 383, 1.
< Af<
376 LEBESGUE INTEGRALS
2. Let F = Max \f\in 21, then
//
<
This follows from 1.
386. In order that f be L-integrable in 21, it is necessary that, for
each extremal sequence \ Dn \ ,
lim flDf= 0;
^s sufficient if there exists a sequence of superimposed separated
divisions \ En \ , such that
lim £1EJ = 0.
n=ao
It is necessary. For
/T
lim iS I = lim
As /is X-integrable,
0= f- r=lirn(^n-^n) = limflz)n
JL% 5^21
It is sufficient. For
Both \SEn\, \&sn\ are limited monotone sequences. Their
limits therefore exist. Hence
0 = lim £1E = lim SE — lim SE .
n H . H
Thus
oC'ffi ow2l
387. J^i order that f be L-integrable, it is necessary and sufficient
that for each e > 0, there exists a separated division D of H, for
which
It is necessary. For by 386, there exists an extremal sequence
Dn\, such that
0 <_ fl/> /< e , for any n > some m.
Thus we may take Dm for D.
GENERAL THEORY 377
It is sufficient. For let e1>e2> ••• = 0. Let \Dn\ be an
extremal sequence for which
Let Aj = D1, A2 = Ax + Z>2, A3 = A2 + D3 ••• Then JAnj is a
set of superimposed separated divisions, and obviously
Hence / is l/-iiitegrable by 386.
388. In order that f be L-integrable, it is necessary and sufficient
that, for each pair of positive numbers o>, cr there exists a separated
division D of 21, such that if 77^ ?;2, ••• are those cells in which
Osc/> o>, then
2f. < <r. (1
It is necessary. For by 387 there exists a separated division
D = JSJ for which
flDf = SwA < G)(T. (2
If ^, ^2 ... denote the cells of D in which Osc/ <.&),
This in 2) gives 1).
It is sufficient. For taking e > 0 small at pleasure, let us then
take
r-A . — ^ (4
2^1
where II = Osc /in 51.
From 1), 3), and 4) we have, since COL < H,
o-n H- aM = e.
We now apply 387.
389. 1. Iff is L-integrable in H, it is in 53 < H.
For let \Dn\ be an extremal sequence of /relative to SI. Then
by 386,
378 LEBESGUE INTEGRALS
But the sequence \Dn\ defines a sequence of superposed sepa
rated divisions of 33, which we denote by \En\. Obviously
nej< nDj.
Hence by 1),
<W=o,
and / is .L-integrable in 23 by 386.
2. If f is L-integrable in 51, so is |/|.
The proof is analogous to I, 507, using an extremal sequence
for /.
390. 1. Let j5lnj be a separated division of 51 into a finite or in
finite number of subsets. Letf be limited in 51. Then
//=//+//+- a
<£% 4^ ^2I2
For let us 1° suppose that the subsets 5^ ••• 5(r are finite in num
ber. Let \Dn\ be an extremal sequence of/ relative to 51, and
]Dmn\ an extremal sequence relative to 5Im. Let
En=Dn + Dln+ ... +Drn.
Then \En\ is an extremal sequence of /relative to 51, and also
relative to each 51TO.
Now - -
*.JL-f*»Wt •" +^K^n-
Letting 71 = 00, we get 1), for this case.
Let now r be infinite. We have
i=flm. (2
Let «„=(?!! ...?in) , en=?r-93n-
Then 53n, Sn form a separated division of 51, and
S =€„ + !„.
If v is taken large enough, 2) shows that
^n<~ , rc>" , jlf=Max|/| in5T.
GENERAL THEORY 379
Th as by case 1°,
'
(3
2In
where by 385, 2
|e'|<Mn<e , n>v.
Thus 1) follows from 3) in this case.
2. Let l^inl be a separated division 0/51. Then
iffis L integrable in 51, or if it is in each 2In, and limited i
391. 1. Letf = ^ i/i 51 except at the points of a null set
in
:21
Forlefc
//=/,. a
^21 4:21
=8+ ft. Then
f/= f/+f/=//. (2
4:21 ilJ 4SJ? 4^8
Similarly //=J/ (3
But/ = ^ in 83. Thus 2), 3) give 1).
392. 1. 7/00;
If e < 0 ; Jc/ = c J/, £cf = e jf.
The proof is similar to 3, 3, using extremal sequences.
2. Iffis L- integrable in 51, so is cf, and
^s a constant.
380 LEBESGUE INTEGRALS
393. 1. Let F(x) = fl(x) + ••• +/„<», each fm being limited
in 2i. Then
For let jDnj be an extremal sequence common to F^fv •••/„. In
each cell
dn\ » ^n2 ' ' '
of Dn we have
2 Min/m < Min F < Max F < 2 Max/m .
Multiplying by dns, summing over s and then letting w = oo,
gives 1).
2. Iff^x), •••/„(«) are each L-integrable in 31, so u
and
394. 1.
For using the notation of 393,
Min (/+#) < Min/+
in each cell dns of Dn.
2. If g is L-integrable in 51,
f (/+<
^21
Reasoning similar to 3, 4, using extremal sequences.
3- f(f-g)<ff-fg.
^21 <X2l ^2t
ff- f
JLw JLw,
g<
GENERAL THEORY 381
f (/-*) < f / + /'(-</) < ff- fg;
<^2l <=L>y( s^2l «^2l «^2l
etc.
4. Iff, g are L-integrable in Si, so isf — g, and
L(f~9)=Lf~
^ . . ^ .
395. -/f/, # are L-integrable in SI, so is./- (/.
J.?so ^eir quotient f/g is L-integrable provided it is limited in 51.
The proof of the first part of the theorem is analogous to I,
505, using extremal sequences common to both / and g. The
proof of the second half is obvious and is left to the reader.
396. 1. Let f, g be limited in SI, and f <^g, except possibly in a
null set yi. Then -^ -^
I f< I g- (i
Xsi ^,21
Let us suppose first that/<: # everywhere in SI.
Let \Dr\ be an extremal sequence common to both / and g.
Then „ f^ u ,
b/)nj <-&Dng.
Letting n = oo , we get 1).
We consider now the general case. Let SI = 53 4- $1- Then
since
//=// , /,=/
^21 ^33 <^2l =^33
lf-fc-*
But in 53, f<g without exception. We may therefore use the
result of case 1°.
2. Letf>Qin%. Then
382 LEBESGUE INTEGRALS
397. The relations of 4 also hold for L-integrals, viz. :
Iff </!/!• (i
l^fca £%.
f/< f/- (2
^51 <^2l
< f/< fl/l- (3
«L2l ^21
(4
The proof is analogous to that employed for the 72-integrals,
using extremal sequences.
398. Let 2l = 08u, @«) t>e a separated division for each u = 0.
Let ltt = 0. Then
lim
«=0
For by 390, l,
im f / = f /
=0 ^33M 4^21
But by 385, 2, the last integral = 0, since (£„ = 0, and since /is
limited.
399. Xe^ f be limited and continuous in 51, except possibly at the
points of a null set $ft. Then f is L-integrable in 5(.
Let us first take 91 = 0. Then/ is continuous in 51. Let 51 lie
in a standard cube £}. If Osc/ is not < e in 51, let us divide Q
into 2n cubes. If in one of these cubes
Osc/< e, (1
let us call it a black cube. A cube in which 1) does not hold we
will call white. Each white cube we now divide in 2n cubes.
These we call black or white according as 1) holds for them or
does not. In this way we continue until we reach a stage where
all cubes are black, or if not we continue indefinitely. In the
latter case, we get an infinite enumerable set of cubes
<h> <fe' ^3 •" (2
GENERAL THEORY 383
Each point a of 51 lies in at least one cube 2). For since / is
continuous at x = a,
!/CO-/OOI<«/2 , x in r,(«).
Thus when the process of division has been carried so far that
the diagonals of the corresponding cubes are < 8, the inequality
1) holds for a cube containing a. This cube is a black cube.
Thus, in either case, each point of 51 lies in a black cube.
Now the cubes 2) effect a separated division D of 51, and in
each of its cells 1) holds. Hence /is .L-integrable in 5T.
Let us now suppose %l > 0. We set
5l = £ + ft.
Then /is Z-integrable in (£ by case 1°. It is L-integrable in $1
by 380, 2. Then it is L-integrable in 51 by 390, 1.
2. If / is L-integrable in 51, we cannot say that the points of
discontinuity of/ form a null set.
Example. Let/= 1 at the irrational points $, in 51 = (0, 1) ;
= 0 at the other points 9?, in 51.
Then each point of 51 is a point of discontinuity. But here
since tit is a null set. Thus /is L-integrable.
400. Iff(xl ••• xni) has limited variation in 51, it is L-integrable.
For let D be a cubical division of space of norm d. Then by I,
709, there exists a fixed number V, such that
2a>(dm-l< V
for any D. Let o>, cr be any pair of positive numbers. We take
d such that
. a
Let d( denote those cells in which Osc/>&>, and let the number
of these cells be v. Let rj, denote the points of 51 in d{ . Then
l<^,(odm-l< V.
384 LEBESGUE INTEGRALS
Hence
v<
Thus
ox
2),
•• <Va<* . by*)- '
Hence /is .L-integrable by 388.
401. Let <t> =/, in 51 < 33 ;
= 0, in A = 33 - 5t.
f /- /V -
<X2t els
if 1°, <f) is L-integrable in 33 ; or 2°, / is L-integrable in 51,
are separated parts of 33.
On the 1° hypothesis let J(SS| be an extremal sequence of </>.
Let the cells of @8 be e^, ez ••• They effect a separated division
of 51 into cells d^ d% ••• Let mt, M~t be the extremes of /in c?t and
wt, JV[ the extremes of (/> in et. Then for those cells containing at
least a point of 5(,
is obviously true when et = d^. Let dk < et. If mt < 0,
nt^t <^ mtc?t , since mi = ni. (3
If mt > 0, wt = 0, and 3) holds.
If Mt <_ 0, Jftc?t < Nfc , since ^Vt = 0. (4
If M^ > 0, 4) still holds, since ML = Nt.
Thus 2) holds in all these cases. Summing 2) gives
for the division ($a, since in a cell e of (§, containing no point of
<t> = 0. Letting s = oo, we get 1), since the end members
INTEGRAND SETS 385
On the 2° hypothesis,
f </>= r*+ r*= r*= r /,
<X«B <^2l <^J. •&* el/t
since 0 being = 0 in A, is .L-integrable, and we can apply 390.
402. 1. If
/=o,
we call /a null function in 51.
2. Iff> 0 is a null function in 51, the points ty where f > 0 form
o
For let 51 = 3 + ?, so that/= 0 in 3.
By 401, . ^
= //=//• a
«z« x*
Let €j > e2 > ••• = 0. Let ^3n denote the points of $ where
/>€n. Then
%f = 0, byl).
Each <>pn is a null set. For
Hence ^n = 0.
Then * = J?nJ=Gi+C2+-
where ^ = ^, $2=^2-^, $3=^33-$3...
As each §n is a null set, $ is a null set.
Integrand Sets
403. Let 51 be a limited point set lying in an w-way space $ftm.
Let f (xl ••• xm) be a limited function denned over 51. Any
point of 21 may be represented by
a = (flj ... «m).
386 LEBESGUE INTEGRALS
The point x = ^ ••• amxm+l)
lies in an m + 1 way space $ftm+1. The set of points J x\ in which
xm+1 ranges from — oo to +00 is called an ordinate through a. If
xm+1 is restricted by Q < ^ ^ ^
we shall call the ordinate a positive ordinate of length I ; if it is re
stricted by _ l<Xm+l<0,
it is a negative ordinate. The set of ordinates through all the
points a of H, each having a length =/(«), and taken positively
or negatively, as /(a) is ^ 0, form a point set $ in 9?m+1 which
we call an integrand set. The points of $ f°r which xm+1 has a
fixed value xm+1 = e form a section of 3, and is denoted by 3(<?) or
by a-
404. Z0£ 21= j#J fo a limited point set in 9?TO. Through each
point «, let us erect a positive ordinate of constant length /, getting a
set £), in Wm+1 . Then g = ;fj (\
For let Q?! > (£2 > ••• form a standard sequence of enclosures of
O, such that j^ ^ ^ ^
Let us project each section of (£n corresponding to a given value
of zm+ on $m, and let 5ln be their divisor. Then 5ln > 31. Thus
Letting n = oo , and using 2), we get
6 = 1 • i.
To prove the rest of 1), let 0 be the complement of D with re
spect to some standard cube Q in 9^TO+1, of base Q in 9?m.
Then, as just shown,
0 =IA , where .4 = Q - 51.
Hence
INTEGRAND SETS 387
405. Letf >0 be L-integralle in 21. Then
where $ is the integrand set corresponding to f.
For let J£tj be a separated division .Z) of 21. On each cell
erect a cylinder (£t of height Mt = Max/ in St. Then by 404,
Let (£= {(SJ ; the (£t are separated. Hence, e>0 being small
at pleasure,
for a properly chosen D. Thus
§< f/. (2
^21
Similarly we find
f/<5. (3
^2t
From 2), 3) follows 1).
406. Letf>0 be L-integrable over the measurable field 21. Then
the corresponding integrand set 3 is measurable, and
For by 2) in 405,
§< f /•
x«
Using the notation of 405, let cn be a cylinder erected on 8n of
height mn = Min/ in 8B. Let c = JcJ. Then c < 3, and hence
c<3- (2
But 21 being measurable, each cn is measurable, by 404. Hence
c is by 359. Thus 2) gives
c<3- (3
Now for a properly chosen Z>,
-*+ r/<27wtst=T.
388 LEBESGUE INTEGRALS
Hence
.
as e is arbitrarily small. From 2), 3), 4)
from which follows 1).
Measurable Functions
407. Let/Oj ••• xm) be limited in the limited measurable set 51.
Let 2(v denote the points of 51 at which
If each 5lxM is measurable, we say/z's measurable in 5(.
We should bear in mind that when /is measurable in 5(, neces
sarily 51 itself is measurable, by hypothesis.
408. 1. Iff is measurable in 51, the points & of 51, at which f — 0,
form a measurable set.
For let 5ln denote the points where
where €l>62>...=o.
Then by hypothesis, 5ln is measurable. But ^ —
Hence (£ is measurable by 361.
2. Iffis measurable in 51, the set of points where
is measurable, and conversely.
Follows from 1, and 407.
3. If the points 5IA in 51 where f>\ form a measurable set for
each X, / is measurable in 51.
For 51^ having the same meaning as in 407,
5lx, = 5Tx-5TM.
Each set on the right being measurable, so is 5lxM •
MEASURABLE FUNCTIONS 389
409. 1. Iffis measurable in 21, it is L-integrable.
For setting m = Min /, M = Max / in 21, let us effect a division
D of the interval g = (w, M) of norm c?, by interpolating a finite
number of points
7W1<7W2<Wg< -••
Let us call the resulting segments, as well as their lengths,
dl, d%, d3 "-
Let 2lt denote the points of 21 in which
ml-i<f<ml , i = l, 2, ... ; mQ = m.
We now form the sums
SD = 2wt-iit , */, = 2wtit.
Obviously
But t/>-t^ = m151 + «A
= 5(i(mi - ™) 4-
= 0 , asrf = 0. (2
We may now apply 387.
2. Iff is measurable in 21
** A
/=lim
L
using the notation in 1.
This follows from 1), 2) in 1.
3. The relation 3) is taken by Lebesgue as definition of his
integrals. His theory is restricted to measurable fields and to
measurable functions. For Lebesgue's own development of his
theory the reader is referred to his paper, IntSgrale, Longueur •,
Aire, Annali di Mat., Ser. 3, vol. 7 (1902) ; and to his book,
Lemons sur I1 Integration. Paris, 1904. He may also consult the
excellent account of it in Nobsons book, The Theory of Functions
of a Real Variable. Cambridge, England, 1907.
390 LEBESGUE INTEGRALS
Semi-Divisors and Quasi- Divisors
410. 1. The convergence of infinite series leads to the two
following classes of point sets.
T of
Let
F= 2/.C*, ... xj = 2/; + 2/. = Fn
1 n+1
each/t being defined in 21.
Let us take e > 0 small at pleasure, and then fix it.
Let us denote by 2ln the points of H at which
- e < Fn(x) < e. (2
Of course 5ln may not exist. We are thus led in general to the
Sets or or or /•*
*»| ? <^2 » ^3 v*3
The complementary set An = 51 — 5ln will denote the points
where !>«. ..• (4
If now F is convergent at #, there exists a i> such that this point
lies in or or or sz
<*v » ^+1 ^ -u-v+2 \°
The totality of the points of convergence forms a set which has
this property : corresponding to each of its points #, there exists
a v such that x lies in the set 5). A set having this property is
called the semi-divisor of the sets 3), and is denoted by
Sdv J2IJ.
Suppose now, on the other hand, that 1) does not converge at
the point x in 51. Then there exists an infinite set of indices
^i < n% < ••• = oo,
such that =
I -F»aO) I > €.
Thus, the point x lies in an infinity of the sets
Al , A2 , A3 ••• (6
The totality of points such that each lies in an infinity of the
sets 6) is called the quasi-divisor of 6) and is denoted by
QdrM,!.
Obviously,
HnS + QdvJ^nS = 3l. (7
SEMI-DIVISORS AND QUASI-DIVISORS 891
We may generalize these remarks at once. Since F(x) is
nothing but
we can apply these notions to the case that the functions/,^ ••• xm)
are defined in 51, and that
2. We may go still farther and proceed in the following abstract
manner.
The divisor £) of the point sets
HI , v- • a
is the set of points lying in all the sets 1).
The totality of points each of which lies in an infinity of the sets
1) is called the quasi-divisor and is denoted by
(2
The totality of points a, to each of which correspond an index m
such that a lies in
forms a set called the semi-divisor of 1), and is denoted by
Sdv {'«.}. (3
If we denote 2), 3) by G and @ respectively, we have, obviously,
£) < @ < O. (4
3. In the special case that 2Ij >212 > ••• we have
Q = @ = 5). (5
For denoting the complementary sets by the corresponding
Roman letters, we have
But Q has precisely the same expression.
Thus G = S), and hence by 4), @ = £).
392 LEBESGUE INTEGRALS
4. Z^5ln + A» = 93> ra = l,2,... Then
For each point b of 93 lies
either 1° only in a finite number of 5ln, or in none at all,
or 2° in an infinite number of 5ln.
In the 1° case, b does not lie in 518, 5ls+1 ••• ; hence it lies in
A,, Aa+1 ••• In the 2° case b lies obviously in Qdv [51J.
5. If 5^, 5f2 ••• are measurable, and their union is limited,
n = Qdvj5u , e = sdvj5u
are measurable.
For let S)n = Dw(3Cn, 5In+1 •••) . Then @ = j$nj .
But @ is measurable, as each £)n is. Thus Sdv J J.nJ is measur
able, and hence Q is by 4.
6. £e£ O = Qdv 55ln| , c«(?A Hn 6ezw^ measurable, and their union
limited. If there are an infinity of the 5ln, say
measure is > a,
Q>«. (6
For let 53n = (5(ln, 5lln+1 .••), then
Let 53 = Dw{«»} . As «n > »B+1,
S = limSn>« (7
by 362. As O>93 we have 6) at once, from 7).
Limit Functions
411. Let
as x ranges over 51, r finite or infinite. Let f be measurable in 51
and numerically <M,for each t near r. Then <j> is measurable in
51 also.
To prove this we show that the points 93 of 51 where
LIMIT FUNCTIONS 393
form a measurable set for each X, p. For simplicity let T be finite.
Let £1? t2~. =T; also let €1>e2> • •• =0. Let (£„,, denote the
points of 21 where
.) </* + «»• (2
Then for each point x of 33, there is an «0 such that 2) holds for
any*,, if s> v Let £n = Sdv f(£n,J . Then 8 < <£B . But the &M
being measurable, (£ft is by 410, 5. Finally 23 = Dv J(£J , and hence
53 is measurable.
412. Let
for x in 51, and r finite or infinite. Let tf, t" • •• =T. Let each
fs =/(#, t(9)) be measurable, and numerically < M. Let <f>=fg + g8.
Let ©8 denote the points where
9.\>*-
Then for each e > 0, Um ^ = Q> (1
*=*>
For by 411, <^> is measurable, hence gt is measurable in 51, hence
®, is measurable.
Suppose now that 1) does not hold. Then
Then there are an infinity of the ®8, as ®,t, ®a2--- whose
measures are >X>0. Then by 410, 6, the measure of
is > X. But this is not so, since/, = (/>, at each point of SI.
413. 1. Let v
for x in 51, and T finite or infinite.
Ifeachf,=f(x, #a)) is measurable, and numerically <M in
each sequence 1), then
/-» /-a
0- (2
394 LEBESGUE INTEGRALS
For set , /.
£=/.+#n
and let \ \ ^ -\r -in
|&|:$-y , «=!, 2 ...
Then as in 412, <£ and #8 are measurable in 51. Then by 409,
they are Z-integrable, and
(3
Let $3a denote the points of 51, at which
and let $88 + B8 — 51. Then 238, B8 are measurable, since ga is.
Thus by 390, ~ „ „
J ^=2 gs+l/8'
Hence ., _
i/8 -
ovii
By 412, 58 = 0. Thus
Hence passing to the limit in 3), we get 2), for the sequence
1). Since we can do this for every sequence of points t which
= T, the relation 2) holds.
it*
converge in 51. If each term /t is measurable, and each
then F is L-integrable, and
Iterated Integrals
414. In Vol. I, 732, seq. we have seen that the relation,
holds when /is 72-integrable in the metric field 51. This result
was extended to iterable fields in 14 of the present volume. We
ITERATED INTEGRALS 395
wish now to generalize still further to the case that f is j
grable in the measurable field 51. The method employed is due to
Dr. W. A. Wilson,* and is essentially simpler than that employed
by Lebesgue.
1. Let x = (Zj ••• z,) denote a point in s-way space Sft,, s = m + n.
If we denote the first m coordinates by x1 ••• xm, and the remaining
coordinates by ^ ••• #„, we have
The points
*=
range over an ra-way space $Rm, when 2 ranges over 9?8. We call
x the projection of z on $TO .
Let z range over a point set 51 lying in $Ra, then x will range
over a set 53 in $m, called the projection of 51 o?i 9?m. The points
of 51 whose projection is x is called the section of 51 corresponding
to x. We may denote it by
5l(z), or more shortly by (£.
We write « = ».<£
to denote that 51 is conceived of as formed of the sections (E, cor
responding to the different points of its projection 53.
2. Let O denote a standard cube containing 51, .let q denote its
projection on Sftm. Then 53 <.q. Suppose each section 5I(z) is
measurable. It will be convenient to let 5l(X) denote a function
of x defined over q such that
/«, /«>
5l(V) = Meas 5l(z) = S when # lies in 53,
= 0 when x lies in q — 53.
This function therefore is equal to the measure of the section of
51 corresponding to the point z, when such a section exists ; and
when not, the function = 0.
When each section 5l(#) is not measurable, we can introduce
the functions
* Dr. Wilson's results were obtained in August, 1909, and were presented by me
in the course of an address which I had the honor to give at the Second Decennial
Celebration of Clark University, September, 1909.
396 LEBESGUE INTEGRALS
Here the first = (£ when a section exists, otherwise it = 0, in q.
A similar definition holds for the other function.
3. Let us note that the sections
where He, 2L are the outer and inner associated sets belonging to 31,
are always measurable.
For 2Ie = Dv \ (£„ \ , where each G?n is a standard enclosure, each
of whose cells tnm is rectangular. But the sections enm(V) are
also rectangular. Hence
being the divisor of measurable sets, is measurable.
415. Let $te be an outer associated set of tyt,both lying in the stand-
0t
ard cube O. Then 2le(V) is L-integrable in q, and
= f ioo- a
«4/
For let j@nj be a sequence of standard enclosures of 51, and
@n=5enni;. Then
£n = 2enm (2
m
and «.(*) = 2e»(*). (3
TO
Now enm being a standard cell, enm(^) nas a constant value > 0
for all x contained in the projection of enm on q. It is thus con
tinuous in q except for a discrete set. It thus has an ^-integral,
and
This in 2) gives
enm<», by 413, 2,
f €,0),
oLa
(4
by 3).
ITERATED INTEGRALS 397
On the other hand, (S(z) is a measurable function by 411. Also
1 = Se = lim <fn
= Aim in(z), by 413, 1. (5
<X
Thus this in 5) gives 1).
416. Let 51 lie in the standard cube O. Let Hi 50 a/i inner asso
ciated set. Then §t(V) is L-integraUe in q, and
For 0 = 21. + A-
Thus flCr) = SCO - A.(x).
Hence H^x) is i-integrable in q, and
f*{x) = fSO) - p.(*)
oLq 0Lq — q
= S-A , by 415,
= St = a by 370, 2.
417. ie£ measurable 21 fo'e iw #Ae standard cube O.
For a.
Hence ^ = T^^) < Tf^) < fi/^) = f, (2
~ <Zq iq~ «^q
using 396, 1, and 415, 416. From 2) we conclude 1) at once.
418. Let H = $8 • (S ^ measurable. Then S are L-integrable in
398 LEBESGUE INTEGRALS
For by 41 7, ~
by 401.
419. If 51 = 53 • (£ is measurable, the points of 53 at which & is not
measurable form a null set $1.
For by 418, g= f = = f
i i
Hence
= f (f-<£).
<X$
Thus
is a null function in 53, and by 402, 2, points where </> > 0 form a
null set.
420. Let 51 = 53 • (S £>e measurable. Let b denote the points of 53
/or which the corresponding sections (£ are measurable. Then
=
For by 419, 53=b +
and 9? is a null set. Hence by 418,
=/s.
<Xb
421. Let f>0 in 51. .Zf £Ae integrand set 3s? corresponding to f
be measurable, then f is L-integrable in 51,
3-J/
For the points of Q lying in an m -f- 1 way space 9?m+1 may be
denoted by x = (v •- y z}
where y = (yl - • • ?/m) ranges over 9?m, in which 51 lies. Thus 51
may be regarded as the projection of Q on 9?m. To each point y
ITERATED INTEGRALS 399
of 21 corresponds a section 3G/), which for brevity may be denoted
by $. Thus we may write
$»*-*.
As ft is nothing but an ordinate through y of length /Q/), we
have by 419, ^ /»=
§«/I
at*
422. Ze£ / 6e L-integraUe over the measurable field 21 = 53 • (£.
ie£ b denote those points of 33, /or which f is L-integralle over the
corresponding sections (£. 2%era
f /=///. a
^21 otb^S
Moreover $ft = 55 — b is a null set.
Let us 1° suppose f> 0. Then by 406, 3 is measurable and
§=r/. (2
•£«
Let 0 denote the points of 23 for which 3(z) is measurable.
Then by 420,
5 = f§(*). (3
Jip
By 419, the points
^ = 53 - /3 (4
form a null set.
On the other hand, $(x) is the integrand set of/, for SI (V) = 6.
Hence by 421, for any z in /3,
, (5
and /3 < b. (6
From 2), 3), 5) we have
w. Jlp J*&
From 6) we have
^=33-b<23-/3 = ^,
a null set by 4). Let us set
b = J3 + n.
400 LEBESGUE INTEGRALS
Then tt lying in the null set $, is a null set. Hence
JLp JL& JLn JL(s, JL\) cXg
This with 7) gives 1).
Let f be now unrestricted as to sign. We take 0 > 0, such
that the auxiliary function
ff=f+C>0, in St.
Then /, g are simultaneously Z-integrable over any section (£.
Thus by case 1°
f (/+<?)= c fcf+o). (8
e£/2l oC'b ^S
Now r /»/»./»-
(9
(10
By 418, (£ is Z-integrable in ^3, and hence in b. Thus
/* /* /» /» /»=
Li^f+c^=Llf+0l^" (11
As b differs from 33 by a null set,
ri=ri=s, (12
by 418. From 8), 9), 10), 11), 12) we have 1).
423. If f is L-integrable over the measurable set 51 = 53 • (£, £Aew
/* /* /T
J /=///. a
Xx «X^3 oC$
For by 422,
/» /» /»
(2
As ^3 — b = yt is a null set,
"/=o
ITERATED INTEGRALS 401
may be added to the right side of 2) without altering its value.
Thus
JLn JLbJ& °LftcL(§. JLsQcLd
424. 1. (TF. A. Wilson.) If f (^ • - - xm~) is L-integrable in
measurable 21, f is measurable in 21.
Let us first suppose that/> 0. We begin by showing that the
set of points 21A of 21 at which / > X, is measurable. Then by
408, 3, /is measurable in 21.
Now / being Z-integrable in 21, its integrand set 3> is measur
able by 406. Let QA be the section of 3 corresponding to xm+l= \.
Then the projection of 3A on 3tm is 21A. Since 3 is measurable, the
sections 3?A are measurable, except at most over a null set L of
values of X, by 419. Thus there exists a sequence
Xj < \2< ... =X
none of whose terms lies in L. Hence each 3xM is measurable, and
hence 2lAjt is also.
As 2lAn+1 < 2lAn, each point of 21A lies in
so that ' «A<5>. (2
On the other hand, each point d of 3) lies in 21A. For if not,
f(d)<\.
There thus exists an s such that
< X. < X. (3
But then d does not lie in 2IA,, for otherwise f (d) > X,, which
contradicts 3). But not lying in 2IA,, d cannot lie in £), and this
contradicts our hypothesis. Thus
£)<21A. (4
From 2), 4) we have
$«*A<
But then from 1), 21A is measurable.
Let the sign off be now unrestricted.
402 IMPROPER L-INTEGRALS
Since f is limited, we may choose the constant C, such that
Then g is Z-integrable, and hence, by case 1°, g is measurable.
Hence/, differing only by a constant from g, is also measurable.
2. Let 51 be measurable. Iff *'* L-integrable in 51, i£ i* measur
able in 51, a/ic? conversely.
This follows from 1 and 409, l.
3. From 2 and 409, 3, we have at once the theorem :
When the field of integration is measurable, an L-integrable func
tion is integrable in Lebesgue's sense, and conversely; moreover, both
have the same value.
Remark. In the theory which has been developed in the fore
going pages, the reader will note that neither the field of integra
tion nor the integrand needs to be measurable. This is not so in
Lebesgue's theory. In removing this restriction, we have been
able to develop a theory entirely analogous to Riemann's theory of
integration, and to extend this to a theory of upper and lower in
tegration. We have thus a perfect counterpart of the theory
developed in Chapter XIII of vol. I.
4. Let 51 be metric or complete. If f \xl ••• xni) is limited and
R-integrable, it is a measurable function in 51.
For by 381, 2, it is .L-integrable. Also since 51 is metric or
complete, 51 is measurable. We now apply 1.
IMPROPER L-INTEGRALS
Upper and Lower Integrals
425. 1. We propose now to consider the case that the integrand
/(#!••• #w) is not limited in the limited field of integration 51- In
chapter II we have treated this case for ^-integrals. To extend
the definitions and theorems there given to X-integrals, we have
in general only to replace metric or complete sets by measurable
sets; discrete sets by null sets; unmixed sets by separated sets;
UPPER AND LOWER INTEGRALS 403
finite divisions by separated divisions ; sequences of superposed
cubical divisions by extremal sequences; etc.
As in 28 we may define an improper .//-integral in any of the
three ways there given, making such changes as just indicated.
In the following we shall employ only the 3° Type of definition.
To be explicit we define as follows :
Let/ (2^ ... xm) be defined for each point of the limited set 51.
Let 5la/3 denote the points of 21 at which
The limits
„ /-
lim / / , lim / / (2
a, 0=* cL^p a, 0= * o^SJa/S
in case they exist, we call the lower and upper (improper) L-in-
tegrals, and denote them by
In case the two limits 2) exist and are equal, we denote their
common value by
and say/ is (improperly) L-integrable in 51, etc.
2. In order to use the demonstrations of Chapter II without too
much trouble, we introduce the term separated function. A func
tion / is such a function when the fields 5la0 defined by 1) are
separated parts of 51.
We have defined measurable functions in 407 in the case that
/ is limited in 51. We may extend it to unlimited functions by
requiring that the fields 5ltt/3 are measurable however large a, ft are
taken.
This being so, we see that measurable functions are special cases
of separated functions.
In case the field 51 of integration is measurable, 5la0 is a meas
urable part of 51, if it is a separated part. From this follows the
important result :
Iff is a separated function in the measurable field 51, it is L-in
tegrable in each 5la/3.
404 IMPROPER L-INTEGRALS
From this follows also the theorem:
Letf be a separated function in the measurable field 51. If either
the lower or upper integral off over 51 is convergent, f is L-integrable
in 51, and /» /»
/ /= lim / /.
JLK «, 0-=°^2la/3
426. To illustrate how the theorems on improper .72-integrals
give rise to analogous theorems on improper .//-integrals, which
may be demonstrated along the same lines as used in Chapter II,
let us consider the analogue of 38, 2, viz. :
£7*
f converges, so do I f.
cly
Let \En\ be an extremal sequence common to both
f , r /3'>/3-
<^2la/3' ^2la/3
Let e denote the cells of En containing a point of ^ ; e* those
cells containing a point of typ ; 8 those cells containing a point of
5Iaj3 but none of ^-. Then
L
= lim {2JC • e + 1MI • e'
In this manner we may continue using the proof of 38, and so
establish our theorem.
427. As another illustration let us prove the theorem analogous
to 46, viz. :
Let Sfj, 512, ••• 5ln form a separated division of 5(. If f is a
separated function m 51, then
//=//+-+//,
4^21 4/2li <?k2l»
provided the integral on the left exists, or all the integrals on the
right exist.
For let 51,, aft denote the points of 5lap in 51,. Then by 390, l,
,a/3 oSLn, aft
In this way we continue with the reasoning of 46.
L-INTEGRALS 405
428. In this way we can proceed with the other theorems ; in
each case the requisite modification is quite obvious, by a con
sideration of the demonstration of the corresponding theorem in
.R-integrals given in Chapter II.
This is also true when we come to treat of iterated integrals
along the lines of 70-78. We have seen, in 425, 2, that if 31 is
measurable, upper and lower integrals of separated functions do
not exist as such ; they reduce to .//-integrals. We may still
have a theory analogous to iterated 72-integrals, by extending the
notion of iterable fields, using the notion of upper measure. To
this end we define :
A limited point set at 31 = 33 • (£ is submeasurable with respect
to 33, when
1= f I.
f
We do not care to urge this point at present, but prefer to pass
on at once to the much more interesting case of ^-integrals over
measurable fields.
Lrlntegrals
429. These we may define for our purpose as follows :
Let/(a:1 ••• xm) be defined over the limited measurable set 31.
As usual let 3L0 denote the points of 31 at which
-«</<& «, £>0.
Let each 3la/3 be measurable, and let / have a proper Z-integral
in each 3L0- Then the improper integral of f over 31 is
f/ = Km C f, (1
Ji% «.0=*<X2U/3
when this limit exists. We shall also say that the integral on
the left of 1) is convergent.
On this hypothesis, the reader will note at once that the dem
onstrations of Chapter II admit ready adaptation ; in fact some
of the theorems require no demonstration, as they follow easily
from results already obtained.
406 IMPROPER L-INTEGRALS^
430. Let us group together for reference the following theo
rems, analogous to those on improper .R-integrals.
1. Iff is (improperly') L-integrable in H, it is in any measurable
part of 21.
2. If g, h denote as usual the non-negative functions associated
withf, then
3. If I f is convergent, so is I \f\, and conversely.
eW$l o*^2l
4. When convergent,
Lf^Lf- (2
5. If I f is convergent, then
e > 0, <r > 0,
for any measurable 53 < 51, such that $8 < cr.
6. Let $[ = ($(!, H2 ••• 5In) be a separated division of 21, each 5lt
being measurable. Then
f/= r /+- + r/, (3
alf[ JL*l JL%n
provided the integral on the left exists, or all the integrals on the
right exist.
7. Let 51 = \tyin\ be a separated division of 51, into an enumerable
infinite set of measurable sets 5ln. Then
- (4
212
provided the integral on the left exists.
8. Iff<9 in ^ except possibly at a null set, then
ff< Cff, (5
,, ., ^51 ei*
when convergent.
L-INTEGRALS 407
431. 1. To show how simple the proofs run in the present
case, let us consider, in the first place, the theorem analogous to
38, 2, viz. :
If I f converges, so do I f and I f.
JLyi J* JLm
The rather difficult proof of 38, 2 can be replaced by the follow
ing simpler one. Since
is a separated division of 2la/s> we have
/=/+/•
<**<# JL JL^
Hence
\f L =\f-f
\£%ap aU^r \^ «tf^
But the left side is < e, for a sufficiently large a, and /3, /?' >
some y£0. This shows that / is convergent. Similarly we show
^
the other integral converges.
2. This form of proof could not be used in 38, 2, since 1) in
general is not an unmixed division of 2fa/3.
3. In a similar manner we may establish the theorem analo
gous to 39, viz. :
If i f and I f converge, so does I f.
Jiy JL& JL*
4. Let us look at the demonstration of the theorem analogous
to 43, 1, viz. :
provided the integral on either side of these equations converges.
408 IMPROPER L-INTEGRALS
Let us prove the first relation. Let 53^ denote the points of
at which f</3. Then
8* «'R + ft
is a separated division of $8^, and hence
= Cf, etc.
cL
5. It is now obvious that the analogue of 44, l is the relation 1)
in 430.
6. The analogue of 46 is the relation 3) in 430. Its demon
stration is precisely similar to that in 46.
7. We now establish 430, 7. Let
Q>(ftt!fc»<lri>.
Then 2t = $m + ^m
is a separated division of 21, and we may take m so large that
Bm < <r, an arbitrarily small positive number. Hence by 430, 5,
we may take m so large that
f
J»Bm
f/=//+/
«t2l J*%m J*B
r*
< €.
From this our theorem follows at once.
Iterated Integrals
432. 1. Let us see how the reasoning of Chapter II may be
extended to this case. We will of coarse suppose that the field
of integration 51 = 93 • & is measurable. Then by 419, the points
of 58 for which the sections are not measurable form a null set.
Since the integral of any function over a null set is zero, we may
therefore in our reasoning suppose that every (£ is measurable.
Since 21 is measurable, there exists a sequence of complete com
ponents Am= BmOm in H, such that the measure of A = \Am\ is S.
ITERATED INTEGRALS 409
Since An is complete, its projection Bn is complete, by I, 717, 4.
The points of Bm for which the corresponding sections Om are not
measurable form a null set vm. Hence the union \vm\ is a null
set. Thus we may suppose, without loss of generality in our
demonstrations, that 21 is such that every section in each Am is
measurable.
Now from
we see that those points of 53 where (£ > C form a null set. We
may therefore suppose that (£ = 0 everywhere. Then (£ — C is a
null set at each point ; we may thus adjoin them to C. Thus we
may suppose that & = C at each point of 53, and that 53 = B is the
union of an enumerable set of complete sets Bm.
As we shall suppose that
/»
is convergent, let
«! < «2 < • • • = oo ,
= 00.
Let us look at the sets 5lan, 53^, which we shall denote by 2In.
These are measurable by 429. Moreover, the reasoning of 72, 2
shows that without loss of generality we may suppose that 21 is
such that 53n = 53. We may also suppose that each (£„ is measur
able, as above.
2. Let us finally consider the integrals
'/• (i
These may not exist at every point of 53, because / does not
admit a proper or an improper integral at this point. It will
suffice for our purpose to suppose that 1) does not exist at a null
set in 53. Then without loss of generality we may suppose in our
demonstrations that 1) converges at each point of 53.
On these assumptions let us see how the theorems 73, 74, 75,
and 76 are to be modified, in order that the proofs there given
may be adapted to the present case.
410 IMPROPER L-INTEGRALS
433. 1 The first of these may be replaced by this :
Let B^ n denote the points of $8 at which Tn > a. Then
21= ,
For by 419,
as by hypothesis the sections (£ are measurable. Moreover, by
hypothesis
is a separated division of (£, each set on the right being measur
able. Thus the proof in 73 applies at once.
2. The theorem of 74 becomes :
Let the integrals
, />0
be limited in the complete set 53. Let (gn denote the points of 53 at
which
Then
HmS^B.
n=ao
The proof is analogous to that in 74. Instead of a cubical
division of the space 9?p, we use a standard enclosure. The sets
33n are now measurable, and thus
is measurable. Thus bn = IT. The rest of the proof is as in 74.
3. The theorem of 75 becomes :
Let the integral ^
f , />0
XG
be limited in complete 53. Then
lim T //=0.
n=
ITERATED INTEGRALS 411
The proof is entirely similar to that in 75, except that we use
extremal sequences, instead of cubical divisions.
4. As a corollary of 3 we have
Let the integral
be limited and L-integrable in 53. Let ^3 = \Sm\ the union of an
enumerable set of complete sets. Then
lira f f/=0.
•"•JUftatc.
For if 53m = (#!, £2 ... Bm), and % = 53m 4- £)„, we have
/
But for w sufficiently large, £)m is small at pleasure. Hence
We have now only to apply 3.
434. 1. We are now in position to prove the analogue of
76, viz. :
Let 51 = 53 • (£ be measurable. Let I f be convergent. Let the
/» 0^21
integrals I f converge in 53, except possibly at a null set. Then
JL*
ff=f!f> (1
«C21 vL$&°L($
provided the integral on the right is convergent.
We follow along the line of proof in 76, and begin by taking
/ > 0 in a. By 423, we have
• • U-LU--
hence ,, ,, ,,
/=limj if. (2
I n=QOoL33 =£(£„
412 IMPROPER L-INTEGRALS
Now e > 0 being small at pleasure,
-e+f (/<f Cf , for a > some
<£% aid aL^G JL*i
Since we have seen that we may regard 33 as the union of an
enumerable set of complete sets, we see that the last term on the
right = 0, as n = oo, by 433, 4. Thus
c r <iim c r = r, (3
JL<$> 0&C JL^& JL&n JUK
by 2). On the other hand,
From 3) and 4) we have 1), when/> 0.
The general case is now obviously true. For
where f> 0 in ^P, and < 0 in ft. Here ^ and ft are measurable.
We have therefore only to use 1) for each of these fields and add
the results.
2. The theorem 1 states that if
1 /* /*
/ , / If,
loth converge, they are equal. Hobson* in a remarkable paper on
Lebesgue Integrals has shown that it is only necessary to assume
the convergence of the first integral ; the convergence of the second
follows then as a necessary consequence.
* Proceedings of the London Mathematical Society, Ser. 2, vol. 8 (1909),
p. 31.
ITERATED INTEGRALS 413
435. We close this chapter by proving a theorem due to
Lebesgue, which is of fundamental importance in the theory of
Fourier's Series.
Letf(x) be properly or improperly L-integrable in the interval
2l = a<5. Then
im r\
=0 JLa
For in the first place,
*)l«fc+ r\f\dx<^r\f\dx. (2
JLa JLa
Next we note that
Hence
Or f—g<f-g'
From 2), 3) we have
J,<J. + 2£\f-g\d*. (4
Let now ^=/ fo"r |/| < (?,
= 0 for |/| >#.
Then by 4), r
where e' is small at pleasure, for Cr sufficiently large. Thus the
theorem is established, if we prove it for a limited function,
\g(*)\<&.
Let us therefore effect a division of the interval F = (— 6r, 6r),
of norm d, by interpolating the points
- a< cl< cz< ... <a,
causing F to fall into the intervals
7i< 72' 7s ••'
414 IMPROPER L-INTEGRALS
Let hm = cm for those values of x for which g(x) falls in the in
terval 7m, and = 0 elsewhere in 21. Then
< 2 J^t + e', e' small at pleasure,
for d sufficiently small.
Thus we have reduced the demonstration of our theorem to a
function h(x) which takes on but two values in 21, say 0 and 7.
Let (£ be a <r/4 enclosure of the points where h = 7, while g may
denote a finite number of intervals of (§ such that g — ($ < o-/4.
Let <£ = 7 in (§:, and elsewhere = 0 ; let i/r = 7 in g, and else
where = 0. Thus using 4),
since h=</)in (a, /3), except at points of measure < <r/4. Similarly
^ < J^ rf I 7.
Thus Jh < J^ + 0-7 < J^ + e,
for <r sufficiently small.
Thus the demonstration is reduced to proving it for a i/r which
is continuous, except at a finite number of points. But for such a
function, it is obviously true.
CHAPTER XIII
FOURIER'S SERIES
Preliminary Remarks
436. 1. Let us suppose that the limited function f(x) can be
developed into a series of the type
f(x) = a0 + #! cos x 4- 02 cos 2 x 4- 03 cos 3 # -f- •••
4- bl sin a: 4- 62 sin 2 z + Z>3 sin 3 x 4- ••• (1
which is valid in the interval 51 = (— TT, TT). If it is also known
that this series can be integrated termwise, the coefficients an, bn
can be found at once as follows. By hypothesis
cos
r fdx=a,Tdx+al r
JL-K JL-* JL-*
sin xdx 4- • • •
r
As the terms on the right all vanish except the first, we have
I
Let us now multiply 1) by cos nx and integrate.
I /(V) cos nxdx = «0 / cos rz^Ja; + al I cos ^ cos nxdx 4-
/^ff
4- ^! i si n 2; cos wa;4- •••
/ cos ma: cos nxdx =0 , m=£n,
r.
£-
cos2 nzcfo =
sin mx cosna:=0.
415
416 FOURIER'S SERIES
Thus all the terms on the right of the last series vanish except
the one containing an. Hence
a — -
/(V)cos nxdx. (2
Finally multiplying 1) by sin 712:, integrating, and using the
relations „*
i sin mx sin nxdx = 0 , m^n,
fsi:
•7T
sin2 nxdx — TT,
fJU—
we get
bn = - T V (a;) sin nxdx. (2
TToC'-TT
Thus under our present hypothesis,
1 /*"• 1 « /*"'
/W = o~ l f (u)du + — -% cos nx I f(u) cc
^^JL-rr TT 1 ^-'r
'"
1 oo
H — 2 sin
7T l
f»
/ » V* N. * T /" Q
/^ ^
The series on the right is known as Fourier's series ; the coeffi
cients 2) are called Fourier's coefficients or constants. When the
relation 3) holds for a set of points 33, we say/(#) can be de
veloped in a Fourier's series in 33, or Fourier's development is valid
in $.
2. Fourier thought that every continuous function in 51 could
be developed into a trigonometric series of the type 3). The
demonstration he gave is not rigorous. Later Dirichlet showed
that such a development is possible, provided the continuous
function has only a finite number of oscillations in 51. The func
tion still regarded as limited may also have a finite number of
discontinuities of the first kind, i.e. where
/(a+O) , /(a-0) (4
exist, but one at least is =£/(#).
At such a point a, Fourier's series converges to
PRELIMINARY REMARKS 417
Jordan has extended Dirichlet's results to functions having
limited variation in 51. Thus Fourier's development is valid in
certain cases when / has an infinite number of oscillations or
points of discontinuity. Fourier's development is also valid in
certain cases when/ is not limited in 51, as we shall see in the
following sections.
We have supposed that /(#) is given in the interval
31 = ( — TT, TT). This restriction was made only for convenience.
For if/(V) is given in the interval 3 = (a < b), we have only to
change the variable by means of the relation
b-a
Then when x ranges over 3k u will range over H.
Suppose f is an even function in 51; its development in Fourier's
series will contain only cosine terms. For
00
f(x) — 2(an cos nx + bn sin nx),
o
GO
/(— x) = 2(an cos nx— bn sin nx).
o
Adding and remembering that f(x) =/( — x) in 51, we get
GO
f(x) = J2#n cos nx, f even.
" o
Similarly if f is odd, its development in Fourier's series" will
contain only sine terms ;
po
f(x) = -J-S5n sin nx, f odd.
" i
Let us note that if f(x) is given only in 53 = (0, TT), and has
limited variation in 53, we may develop / either as a sine or a
cosine series in 53. For let
g(.x)=f(x) » a: in 93
Then g is an even function in 51 and has limited variation.
Using Jordan's result, we see g can be developed in a cosine
series valid in 51. Hence / can be developed in a cosine series
valid in 53.
418 FOURIER'S SERIES
In a similar manner, let
h(x)=f(x) , x in $
Then h is an odd function in 21, and Fourier's development
contains only sine terms.
Unless /(0)= 0, the Fourier series will not converge to /(O)
but to 0, on account of the discontinuity at x = 0. The same is
true for x = TT.
If /can be developed in Fourier's series valid in 5l = (— ?r, TT),
the series 8) will converge for all #, since its terms admit the
period 2 TT. Thus 3) will represent f(x) in H, but will not
represent it unless f also admits the period 2 TT. The series 3)
defines a periodic function admitting 2 TT as a period.
EXAMPLES
437. We give now some examples. They may be verified by
the reader under the assumption made in 436. Their justifica
tion will be given later
Example 1. f(x)~x , for —7r<x<7r.
Then
I sin x sin 2 x sin 3 x \
t 1 ~2~ ~^~ "J
If we set x = — , we get Leibnitz s formula,
_ _
4 1 35 7
Example 2. f(x)—x , 0<z<?r
= -X , -7T<Z<0.
Then
* s - TT 4 f cos x cos 3 rr cos 5 a;
If we set x = 0, we get
8 1* 8* 5*
PRELIMINARY REMARKS 419
Example 3. f (V) = 1 • • , 0 < x < TT
= 0 , x = 0, ± TT
= -1 , -7T<Z<0.
Then
/. , N _ 4 f sin a; , sin 3 a: sin 5 a; 1
/ \x) — — I ~~q I 5 I r r " ' f*
7T I 1 3 O J
Example 4. f (x) = x , 0 < a; < ^
= 7T — X
By defining /as an odd function, it can be developed in a sine
series, valid in (0, TT). We find
/Y N_4 fsin a: sin 3 x sin 5 x _ 1 . ;r
Examples. /(^) = 1 , 0 .<#'<.---
= -1 , !<*<7r.
By defining / as an even function, we get a development in
cosines,
/, . ^ _ 4 f cos x cos 3 a: cos 5 x _
W iw (0, TT).
Example 6. /(«) = K71" ~ *) ' O<Z<TT.
By defining / as an odd function we get a development in
sines,
/ (x) = sin # + J sin 2 # + J sin 3 a; -f • • •
zc? in (— TT, TT).
> s"! i
A. v
Example 7. Let f(x) = - , 0 < a: < ^ -•; .
3 3
2Z ^)\
3
^ 2?r
3 ' 3
420 FOURIER'S SERIES
Developing/ as a sine series, we get
valid in (0, TT).
Example 8. /CO = e* » in (— TT, TT),
We find
valid for — TT < x < TT.
Example 9. We find
2 /i . f 1 cos :r cos 2 a: cos 3 x , ]
cos /iz = - £ sin 7r/i — - - — — - + — — — -- - - - + ...
TT 1 2 /A2 /i2 — 1 /A2 — 22 /i2 — 3a J
Let us set x = TT, and replace /* by a; ; we get
1 | 1 |
o "^ o 10 "^ o OQ '
s „ -i o o 10 " o OQ
2a; 2^ ir2— I2 r* — 22 ar2 — 32
a decomposition of cot irx into partial fractions, a result already
found in 216.
Example 10. We find
2 2 cos 2 ^ 2 cos 4 # 2 cos 6 a;
valid for Q < x <TT.
Summation of Fourier's Series
438. In order to justify the development of f(x) in Fourier's
series F, we will actually sum the F series and show that it con
verges to /(of) in certain cases. To this end let us suppose that
f(x) is given in the interval 5l = (— TT, TT), and let us extend/ by
giving it the period 2 TT. Moreover, at the points of discontinuity
of the first kind, let us suppose
SUMMATION OF FOURIER'S SERIES 4'2\
Then the function
<K«) = /O + 2 u) 4- /Or - 2 *) - 2/(*)
is continuous at u = 0, and has the value 0, at points of continuity,
and at points of discontinuity of 1° kind of/. Finally let us sup
pose that / is (properly or improperly) Z-integrable in 21 ; this
last condition being necessary, in order to make the Fourier co
efficients aB, bn have a sense.
Let
F=F(x) = \ 004- aa cos x 4- 03 cos 2 x 4- —
+ ^sin .r + 52sin 2 2- 4- ••• (1
X
= \ a0 4- 2(am.cos war + 6m sin na:),
where we will now write
1 /***•»
«« = - / /O)eosw2*Za?, (2'
£w = ^ r'/C^) sin «w^- (2; '
Since /(z) is periodic, the coefficients an, 5m have the same value
however c is chosen. If we make c = — TT, these integrals reduce
to those given in 436.
We may write
F= — I f (f)tfr \ I + 2(cos nx cos nt -h sin nx sin nt) \
2 COS »(« - 2") \f (t)dt. (3
Thus F _i/v.f(0(ft
where p.-i~*&»m(t-z).
c-t
1
Provided . ,
sin J(f — 2-) =£0,
(5
we may write
L- sin 2 C ^ — "^) ^
-f 2 I sin ^-^ (f — 2-) — sin ^^ (f — a:) J L
422 FOURIER'S SERIES
ThUS
sn - ,fi
if 5) holds. Let us see what happens when 5) does not hold.
In this case %(t — x) is a multiple of TT. As both £ and x lie in
(<?, c + 2 TT), this is only possible for three singular values :
t = x ; t = c, x = c -f- 2 TT ; £ = c + 2 TT, a; = c.
For these singular values 4) gives
n
As Pn is a continuous function of £, #, the expression on the
right of 6) must converge to the value 7) as #, t converge to these
singular values. We will therefore assign to the expression on
the right of 6) the value 7), for the above singular values. Then
in all cases
1 p*glPy2n + lXt-aO /(Q(ft.
irJic 2 sin ^(t — x)
Let us set 2 -I- 1 — t — —
Then
Fn^-\ f(x + 2uY-^^du.
sin
Let us choose c so that
c — x = — TT,
then ~f
<X_JT J,_7T
2 2
Replacing w by — u in the first integral on the right, it becomes
-, o ^ sin vu -,
f(x — 2 u) du.
„ . " sin u
Thus we get
7T
TT^io gin W
Let us now introduce the term — 2/(z) under the sign of inte
gration in order to replace the brace by <£(w). To this end let us
SUMMATION OF FOURIER'S SERIES 423
give x an arbitrary but fixed value and consider the Fourier's
series for the function
f(x), a constant.
If we denote the Fourier series corresponding to the g function
by = r
4- h1 sin t + h% sin 2 t +
we have
cos ntdt = 0,
f(x\ /^c+2w
-^L
7T ^^
sin ntdt = 0.
7T
Thus the sum of the first n -f 1 terms of the Fourier series
belonging to g(f) reduces to
<?. =/(*)• (9
But this sum is also given by 8), if we replace
by <?(*42M;
since g is a constant. We get thus
Sin W
Let us therefore subtract f(x) from both sides of 8), using 9),
10). We get
IT
F.(x) -f(x) = -T {/(* + 2 «) +f(x - 2 t.) - 2/O) j b
sin
Settin& AW = Trl^W -/(*)}, (11
We have thus the theorem :
For the Fourier Series to converge to f(x) at the point x, it is
necessary and sufficient that Dn(x) =0, as n == oo.
424 FOURIER'S SERIES
Validity of Fourier's Development*
439. The integral on the right side of 438, 12), on which the
validity of Fourier's development at the point x depends, is a
special case of the integral
In fact Jn goes over into Dn, if we set
,
sin u 2
To evaluate Jn let us break 23 up into the intervals
-.r=ar
n
These intervals are equal except the last, which is shorter than
the others unless b — a is a multiple of ir/n. We have thus
If we set
v = M + -
71
we see that while v ranges over 3325, u ranges over 332*-i- This
substitution enables us to replace the integrals over 332<s by those
over SBfc-n since
I g (v) sin nvdv — — I g(u + — ) sin nudu.
JL%2S £®2s-l * U'
Hence grouping the integrals in pairs, we get
Jn= I g(u) sin nudu + 2 I 1 9(.u) ~~ 9\u + ~ ) r s^n ww^w
f O3 ^^^ f c*j I \ M. / I
e^*50 * oL202«-l L \ ft/ J
/-»
+ I g (u) sin nudu,
JLw
* The presentation given in 439-448 is due in the main to Lebesgue. Cf. his
classic paper, Mathematische Annalen, vol. 61 (1905), p. 251. Also his Lemons sur
les Series Trigonometriques, Paris, 1906.
VALIDITY OF FOURIER'S DEVELOPMENT 425
where W is 33r or $,._! -f $r, depending on the parity of r. Now
I, (2
du
du.
IsC, {'<->-'("+i)J™
n
\L
Thus Jn = 0, if the three integrals 2), 3), 4) = 0. Moreover,
if these three integrals are uniformly evanescent with respect to
some point set (£ < 33, Jn is also uniformly evanescent in (£. In
particular we note the theorem
Jn = 0, if g is L-integrable in 53.
We are now in a position to draw some important conclusions
with respect to Fourier's series.
440. 1. Let f(x) be L-integraUe in (c, C + ^TT). Then the
Fourier constants an, bn = 0, as n = oo.
For
-^ />c+2jr
an = - f
irJic
cos nxdx
is a special case of the Jn integral. As /is L-integrable, we need
only apply the theorem at the close of the last article. Similar
reasoning applies to bn.
2 . For a given value of x in 51 = ( — TT, TT) let
sn u
IT
be L-integrable in 33 = f 0, — j. Then Fourier s development is valid
at the point x.
426 FOURIER'S SERIES
For by 438, Fourier's series = /(#) at the point x, if Dn(x) = 0.
But Dn is a special case of Jn for which the g function is in-
tegrable. We thus need only apply 439.
3. For a given x in 21 = (— vr, TT), let
u
/ \
be L-integrable in 33 = f 0, — j. Then Fourier s development is valid
at the point x.
^For let S>0, then :
sm
u
= 0 , as S = 0 , by hypothesis.
4. For a given x in 51 = (— TT, TT), Ze£
)-/C») (8
fte L-integrable in 51. 2TAe?^ Fourier s development is valid at the
point x. . . ,
f(x-2u)-f(x)
x
J u
f 7T\
Thus x ig -Zy-integrable in f 0, — j, as it is the difference of two
integrable functions.
441. (Lebesgue). For a given x in 5l = (— TT, TT) let
7T
/»«
«=« oto
P*
2° Km I |^(w + o) — ty(u) \du = 0
*-«aW
/br some y such that
VALIDITY OF FOURIER'S DEVELOPMENT
427
Then Fourier's development is valid at the point x.
For as we have seen,
\X)\<
sin M
it_
f
X.
du
du<D'
\srnu
where &, is a certain number which = » as n = oo.
us first consider Dr. Since 0<w<— , we have
Hence
z/%4
sin
sin
,
u h T —
6 24
|t, 0«r, T<
= r-
4 7
= V
'u -. TM \ ~> . a
~6\Ty ~*?
<z^, provided s>t.
But this is indeed so. For
« _ vau ^ -. _ TT
Hence
Thus
Df < v I |0| C?M = 0, by hypothesis.
eio
w0?# <i«rw ^o D'r. We have
428 FOURIER'S SERIES
Now / being L-integrable,
*(•+?)-*<•>
is £-integrable in f 77, 1L \ . Thus
lim = 0.
But by condition 2°,
lim f" =
•^•oLir
Thus lim IX ' = 0.
s=o
Finally we consider Dfff. But the integrand is an integrable
function in f /3, ^ J . Thus it = 0 as n = GO .
442. 1. 7^0 validity of Fourier's development at the point x de
pends only on the nature off in a vicinity of x, of norm $ as small as
we please.
For the conditions of the theorem in 441 depend only on the
value of /in such a vicinity.
2. Let us call a point x at which the function
is continuous at u = 0, and has the value 0, a regular point.
In 438, we saw that if # is a point of discontinuity of the first
kind for /(#), then # is a regular point.
3. Fourier's development is valid at a regular point x, provided
for some ij
lim
6=0
For at a regular point x, $(u) is continuous at u = 0, and = 0
for u = 0. Now
lim 1 /**! <K«) 1 du=
A=o hJio
LIMITED VARIATION 429
7T 77
Thus
du = TT • —
Hence condition 1° of 441 is satisfied.
Limited Variation
443. 1. Before going farther we must introduce a few notions
relative to the variation of a function f(x) defined over an interval
$1 =(#<£>). Let us effect a division D of 21 into subintervals,
by interpolating a finite number of points al < a2 < ••• The sum
F,- 2 |/(«.) -/(«.+!) I (1
is called the variation of fin 51 for the division D. If
Max VD (2
is finite with respect to the class of all finite divisions of 21, we say
/ 'has finite variation in 21. When 2) is finite, we denote its value by
Var/, or Vf, or V
and call it the variation of f in 31.
We shall show in 5 that finite variation means the same thing
as limited variation introduced in I, 509. We use the term finite
variation in sections 1 to 4 only for clearness.
2. A most important property of functions having finite vari
ation is brought out by the following geometric consideration.
Let us take two monotone increasing curves A, B such that one
of them crosses the other a finite or infinite number of times. If
f(x), g(x) are the continuous functions having these curves as
graphs, it is obvious that
is a continuous function which changes its sign, when the curves
A, B cross each other. Thus we can construct functions in infinite
variety, which oscillate infinitely often in a given interval, and
which are the difference of two monotone increasing functions.
430 FOURIER'S SERIES
For simplicity we have taken the curves A, B continuous. A
moment's reflection will show that this is not necessary.
Since d(x) is the difference of two monotone increasing functions,
its variation is obviously finite. Jordan has proved the following
fundamental theorem.
3. If f(x) has finite variation in the interval 51 = (a < 5), there
exists an infinity of limited monotone increasing functions g(x), h(x}
such that f—o_h (\
For let D be a finite division of St. Let
PD= sum of terms J/OWj)— /(<OS which are > 0,
Then VB = 2 l/Ow,) -/(«-) \=P0 + ND. (2
Also
) -/(«)! + !/<X)-/(«i)i + - + \f(V) -f(an)\=PD-ND.
On the left the sum is telescopic, hence
f(V)-f (.<*)= PD-ND. (3
From 2), 3) we have
VD = 2 PD +/(a)-/(J) = 2 Na +/(6) -/(«). (4
Let now MaxPB = P , Max^ = ^
with respect to the class of finite divisions D.
We call them the positive and negative variation of f(x) in 51.
Then 4) shows that
/(a). (5
Adding these, we get jr= p + jy. (6
From 5) we have
-/(«) = P-JV. (7
Instead of the interval 21 = («<&), let us take the interval
(a < x), where x lies in St. Replacing 5 by a: in 7), we have
/(*)=/(«) + P(x)-JIT(*). (8
LIMITED VARIATION 431
Obviously P(x), -ZV(#) are monotone increasing functions.
Let p(x) be a monotone increasing function in 21. If we set
(9
we get 1) from 8) at once.
4. From 8) we have
I <
5. We can now show that when f(x) has finite variation in the
interval 51 = (a < 6) it has limited variation and conversely.
For if / has finite variation in 21 we can set
/CO = *(*) -*(*),
where <£, ty are monotone increasing in 21. Then if 21 is divided
into the intervals Sx, S2 ••• we have
Osc/<
Osc 4> = A<£ , Osc ^ = A^ , in St
since these functions are monotone. Hence summing over all the
intervals S,, vrx * ^ <** ^ A
2 Osc/< 2A0 H- SA-v/r
< {0(5)- ^(a)} + {^(ft) - t(a) j
< some $f, for any division.
Hence / has limited variation.
If / has limited variation in 21,
| A/I <0sc/ , in£t.
Hence 2 | A/ 1 < 2 Osc/ < some M.
Hence /has finite variation.
6. If f(x) has limited variation in the interval 21, its points of
continuity form a pant act ic set in 21.
This follows from 5, and I, 508.
432 FOURIER'S SERIES
7. Let a<b< c ; then iff has finite variation in (a, c),
JW+ v»,J= va,j, (ii
^>6 means the variation of f in the interval (a, 5), efa.
r«/= Max VBf
with respect to the class of all finite divisions D of (a, <?). The
divisions D fall into two classes :
1° those divisions E containing the point 5,
2° the divisions F which do not.
Let A be a division obtained by interpolating one or more
points in the interval. Obviously
Let now Gr be obtained from a division F by adding the point
Then
Hence Max Fi>Max VF.
E F
Hence to find F^fC/, we may consider only the class E. Let
now E^ be a division of (a, 5), and E% a division of (5, <?). Then
^ + EI is a division of class E. Conversely each division of class
Ogives a division of (a, 5), (5, c). Now
From this 11) follows at once.
444. We establish now a few simple relations concerning the
variation of two functions in an interval 51 = (a < ft).
F°r
where for brevity we set /. _ ff .
/i — J\aJ'
|iy. (2
LIMITED VARIATION 433
3. Letf, g be monotone increasing functions in 21. Then
n/+<7) = Vf+ Vg. (3
For
4. .For any two functions f, g having limited variation,
Vf+Vg. (4
5. Letf,^ have limited variation in 21 = («, £).
7W
«=!/(«)! , «i=l/iO)|.
^(//i) <(« + fy)(«! + vfo (5
For by 443, 8) we have
where
Thus
Hence by 2, 4,
< ^PA + ^y^ + VPAl + ...
-) , by 3
But Vf=P + N , hence, etc.
445. Fourier's development is valid at the regular point x, if there
exists a 0 < f <~, such that in (0, f) £Ae variation V(u) of i/r(V)
-j
m awy (u, f ) zs limited^ and such that u V(u) = 0, u = 0.
By 442, we have only to show that
2
is evanescent with B.
434 FOURIER'S SERIES
Let us first suppose that ty(u) is monotone in some (0, £), say
monotone increasing. Similar reasoning will apply, if it is mono
tone decreasing. Then, taking 0 < rj + B < £ ,
^= r\^u + §)-^(u)\du = ( ^(u+S)du- I ^(u)du.
In the second integral from the end, set v = u + S.
Xrj /*17+5
^(u + tydu = / ty(v)dv.
aW
Hence,
r
ltl<^
We will consider the integrals on the right separately. Let
^m = Max |^|, in
Then
Hence, -, -i
1 - < some M.
= - + o-u ,
sin u u
Thus,
= 0 , as 8 = 0 , since </>(w) = 0,
as a: is a regular point.
TTe iwrw wow to %. In (17, 77 + 8), 3, ?? sufficiently small,
sin u > M - ^3 > ri(I - ??2).
LIMITED VARIATION 435
Thus, if <#v = Max | 0'| in (77, 77 + S), >>*
with S.
Thus, when ty is monotone in some (0, f), Fourier's develop
ment is valid. But obviously when -ty is monotone, the condition
that uV(u)=0 is satisfied. Our theorem is thus established in
this case.
Let us now consider the case that the variation r(v) of i/r is
limited in (u, f).
From 443, 10), we have
As before we have
By hypothesis there exists for each € > 0, a S0 > 0, such that
u V(u) < e , for any 0 < u < 80 .
Hence,
PX*)^1 '
u
s turn now to ^2. Since V(u) is the sum of two limited
monotone decreasing functions P, Nin. (u, f)' ^ ig integrable.
Thus,
/»7
I /
ot'T?
is evanescent with
436 FOURIER'S SERIES
446. 1. Fourier's development is valid at the regular point x, if
^>(u) has limited variation in some interval (0 < f ), f < — .
For let 0 < u < 7 < ?, then
Now T/r=<f>(X) • -
sm u
Hence Vuy+ < j V^ +
But sin w b.eing monotone,
__ ___ _
MY sin w sin w sin 7
Thus
sm
Similarly' r*<S**±.
vfr— sin 7 2*
Now
0<-^-<^f , in (0*, n.
sin w
The theorem now follows by 445. For we may take 7 so small
that T-r , €
Thus for any u < 7,
On the other hand, 3)^ being sufficiently large, and 7 chosen as
in 1) and then fixed,
Thus
for w<some 5r. Hence
for 0 < u < some B.
2. (Jordan. ) Fourier's development is valid at the regular point
x, *ff(v) has limited variation in some domain of x.
OTHER CRITERIA 437
For <K«) = !/(* + 2 «)-/O)!+ {/(*- 2 «)-/(»)}
has limited variation also.
3. Fourier's development is valid at every point of 21 = (0, 2 TT),
iff is limited and has only a finite number of oscillations in 21.
Other Criteria
447. Let X=
.Zf X = 0 rts S = 0, so does "^, awe? conversely.
For , ^x xA , <xx sinfw 4-
U + 0
• /- N > sin
Obviously X and "^ are simultaneously evanescent with
provided
R = = 0 , as 8 = 0.
Let
r*, v Sin U
Z(u) = -
u
Then . f N < r*s . ^^
Now
v cos v — sin v
tf
Thus
438
Hence
As
FOURIER'S SERIES
sin u
-
!J>|iO ,
with 8.
448. (Lipschitz-Dini.) At the regular point x, Fourier's devel
opment is valid, if for each e > 0, there exists a S0 > 0, swcA that for
each 0 < 8 < S0 ,
For
l .
log 5
n
,
Now a: being a regular point, there exists an rjf such that
I $00 I < €> f°r w in any (^» ^')'
Thus taking ^ ,
^7 > o, V,
<e
Thus
, log B | e VS
< 2 e, for any 8 < ?;.
X = 0, as 8 = 0.
I _1\
o V
Uniqueness of Fourier's Development
449. Suppose /(V) can be developed in Fourier's series
1 /1|r
an= - I /(a?)
T^cX-TT
cos
sn
, (2
UNIQUENESS OF FOURIER'S DEVELOPMENT 439
valid in 51 = (— TT,' TT). We ask can/" (V) be developed in a simi-
/ (x) = I a0' + 2(an' cos n# + £„' sin 712:), (3
also valid in 51, where the coefficients are not Fourier's coefficients,
at least not all of them.
Suppose this were true. Subtracting 1), 3) we get
0 = £ O0 - O + 2 { (an - <) cos war + (&„ - &i) sin r?:r J = 0,
?0 + 2 jtfn cos tt# -f c?n sin /i^j = 0, in 51. (4
Thus it would be possible for a trigonometric series of the type
4) to vanish without all the coefficients cm, dm vanishing.
For a power series
to vanish in an interval about the origin, however small, we know
that all the coefficients pm in 5) must => 0.
We propose to show now that a similar theorem holds for a
trigonometric series. In fact we shall prove the fundamental
Theorem 1. Suppose it is known that the series 4) converges to 0
for all the points of 51 = (— TT, TT), except at a reducible set 9?.
Then the coefficients cm, dm are all 0, and the series 4) = 0 at all the
points o/5l.
From this we deduce at once as corollaries :
TJieorem 2. Let $1 be a reducible set in 51. Let the series
«0 + 2 \an cos nx + @n sin nx\ (6
converge in 51, except possibly at the points Si. Then 6) defines a
function l?(x) in 51 - 3t
If the series , v< , Ql .
«0 + 2 \ «„ cos nx + fin sin nx \
converges to F(x) in 51 — 9?, its coefficients are respectively equal to
those in 6).
Theorem 3. If f(^x) admits a development in Fourier1 % series for
the set 51 — 9?, any other development of f(x) of the type 6), valid in
51 — 9£ is necessarily Fourier's series, i.e. the coefficients «TO, fim have
the values given in 2).
440 FOURIER'S SERIES
In order to establish the fundamental theorem, we shall make
use of some results due to Riemann, Gr. Cantor, Harnack and
Schwarz as extended by later writers. Before doing this let us
prove the easy
Theorem 4- If /(#) admits a development in Fourier s series
which is uniformly convergent in 21 = ( — TT, TT), it admits no other
development of the type 3), which is also uniformly convergent in 51.
For then the corresponding series 4) is uniformly convergent
in 21, and may be integrated termwise. Thus making use of the
method employed in 436, we see that all the coefficients in 4)
vanish.
450. 1. Before attempting to prove the fundamental theorem
which states that the coefficients #n, bn are 0, we will first show
that the coefficients of any trigonometric series which converges
in 21, except possibly at a point set of a certain type, must be such
that they = 0, as n = oo. We have already seen, in 440, 1, that
this is indeed so in the case of Fourier's series, whether it con
verges or not. It is not the case with every trigonometric series
as the following example shows, viz. :
§ sin n \ x. (1
i
When x = r^ all the terms, beginning with the r !th, vanish,
r I
and hence 1) is convergent at such points. Thus 1) is conver
gent at a pantactic set of points. In this series the coefficients an
of the cosine terms are all 0, while the coefficients of the sine
terms 6n, are 0 or 1. Thus bn does not = 0, as n = oo.
2. Before enunciating the theorem on the convergence of the
coefficients of a trigonometric series to 0, we need the notion of
divergence of a series due to Harnack.
Let A=a1 4- a2+ ••• (2
be a series of real terms. Let gn, Grn be the minimum and maxi
mum of all the terms
-^n+l » -An+2 ,
where as usual An is the sum of the first n terms of 2). Obviously
UNIQUENESS OF FOURIER'S DEVELOPMENT 441
Thus the two sequences \gn\, [Grn\ are monotone, and if limited,
their terms converge to fixed values. Let us say
The difference
b = #-<7
is called the divergence of the series 2).
3. For the series 2) to converge it is necessary and sufficient that
its divergence b = 0.
For if A is convergent,
Thus -e + A<gn<G-n<A + €.
Thus the limits Gr, g exist, and
G-g<2e ; or G- = g,
as e > 0 is small at pleasure.
Suppose now b = 0. Then by hypothesis, Gr, g exist and are
equal. There exists, therefore, an n, such that
<7-*<<7n<#n<# + e,
or #n-<7n<2e.
Thus \An+p-An\<2e , p=l,2-.
and A is convergent.
451. Let the series
GO
2 (an cos nx + bn sin nx)
o
be such that for each 8 > 0, there exists a subinterval of
Sl = (-ir, TT)
fa divergence b < S. Men an, 6n = 0, as
= QO.
For, as in 450, there exists for each x an wz, such that
£
| 0n cos nx + 5n sin nx \ < - , n> mx (1
442 FOURIER'S SERIES
for any point x in some interval $3 of 21. Thus if b is an inner
point of 23, x=-b + @ will lie in 33, if 0 lies in some interval
B = (p, q). Now
an cos ft (b + /3) + 6n sin n(b + j3)
= (an cos 716 + &„ sin w&) cos n/3 — (an sin nb — bn cos ^6) sin n/3.
an cos n(b — (3} + bn sin w (6 — /3)
v;^ r= (.ancosnb + &„ sin w5) cos w/3 + («n sin nb - 5n cos ^6) sin nj3.
Adding and subtracting these equations, and using 1) we have
• 8
. , |. (an cos nb + ;5n sin nb) cos n/5 | < -,
^
| (an sin nb — ln cos w5) sin n@ < -,
for all n>mx. Let us multiply the first of these inequalities by
cos nb sin nf$, and the second by sin nb cos nft, and add. • We get
| an sin w£J < 8 , /^ = 2 £ , n > mx. (2
Again if we multiply the first inequality by sin nb sin n/3, and
the second by cos nb cos n/3, and subtract, we get
bnsmn/3l\ <S \ n>mx. (3
From 2), 3), we can infer that for any e > 0
| an'\ < e ' y ;,| bn | < e. , n > some ?w, (4
or what is the same, that an, Jn = 0.
For suppose that the first inequality of 4) did not hold. Then
there exists a sequence
n\ < n2 < ... = oo (5
such that on setting
i«^1 = «+«ir , e-S = S'
we will have
«., > «'• (6
If this be so, we can show that there exists a sequence
J:, V.l < V2 < '•• = °°
in 5), such that for some ft' in B,
\aVrsmvr{3'\ >B, (7
UNIQUENESS OF FOURIER'S DEVELOPMENT 443
which contradicts 2). To this end we note that 70 > 0 may be
chosen so small that for any r and any | 7 | < 70 ,
| aVr | cos 7 > (B -f £') cos 70 > B. (8
Let us take the integer vl so that
Vl>7L±2h. (9
9-f
Then 2^^ N n N -^ r>
-Oi (?-?)- 270)^2-
7T
Thus at least one odd integer lies in the interval determined by
the two numbers
9 9
Let ml be such an integer. Then
9 9
j(M' + W<% < ^Oi - 70)-
If we set
we see that the interval Bl=(pr q^ lies in B. The length of
Bl is 270/i/j. Then for any /3 in .B^
"i/3 = mi^ + f>'i ' |7il<70-
Thus by 8),
| avi sin i/j/3 j = 1^1 cos j1 > 8. (12
But we may reason on Bl as we have on B. We determine z>2
by 9), replacing jt?, q by JPX, ^j. We determine the odd integer m2
by 10), replacing^, q, vl by p1, q1, v2. The relation 11) deter
mines the new interval B2 = (jt?2, ^2), on replacing m^ vl by w2, z^2.
The length of B2 is 2jQ/v2, and 52 lies in ^j. For this relation
of v2 , and for any /3 in B.2 we have, similar to 12),
| avt sin z/2/3 | > B.
In this way we may continue indefinitely. The intervals
Bl > ^2 > ... = to a point /3', and obviously for this (S\ the rela-
444 FOURIER'S SERIES
tion 7) holds for any x. In a similar manner we see that if bn does
not = 0, the relation 3) cannot hold.
452. As corollaries of the last theorem we have :
1. Let the series
00
bn sin nx) (1
be such that for each & > 0, the points in 5l = ( — TT, TT) at which
the divergence of 1) is >8, form an apantactic set in 51. Then
an, bn = 0, as n = oo .
2. Let the series 1) converge in 51, except possibly at the points of
a reducible set ^R. Then an, bn = 0.
For 9t being reducible [318, 6], there exists in 51 an interval 33
in which 1) converges at every point. We now apply 451.
453. Let
at the points of 51 = (— TT, TT), where the series is convergent. At the
other points of 51, let F(x) have an arbitrarily assigned value, lying
between the two limits of indetermination g, Gr of the series. If F is
R-integrable in 51, the coefficients an, bn = 0.
For there exists a division of 51, such that the sum of those in
tervals in which Osc F> o> is < cr. There is therefore an interval
3 in which Osc F < o>. If & is an inner interval of 3, the di
vergence of the above series is < w at each point of $. We now
apply 451.
454. Riemanrfs Theorem.
Let F(x) = ^ a0 + 2(an cos nx + bn sin nx) = 2J.n converge at
each point of 51 = (— TT, TT), except possibly at the points of a redu
cible set 9?. The series obtained by integrating Ms series termwise,
we denote by
a(x) = - a^ - JJ — (an cos nx + bn sin nx) = - AQx2 - % -f-
~t -i n M •« n
Then Q- is continuous in 51.
UNIQUENESS OF FOURIER'S DEVELOPMENT 445
Let <b(u) = G-(x + 2 u) + #(z - 2 w) - 2 G(x). (1
£ eacA point of 33 = 21 — 9?,
lira ^^ = ^0) ; (2
«=o 4 IT
edch point oj 21,
lim^^ = 0. (3
t*=o w
For, in the first place, since 9? is a reducible set, an, bn = 0. The
series 6r is therefore uniformly convergent in 21, and is thus a
continuous function.
Let us now compute <l>. We have
an cos n(x + 2 w) + an cos /i(a: — 2 w) — 2 an cos wa;
= 2 an cos no: (cos 2 nu — 1)
= — 4 an cos w# sin2 nu.
Also 6n sin n(> + 2 w) -f 6n sin w(z - 2 ^) - 2 on sin wa;
= '2bn sin /ix(cos 2 ww — 1)
= — 4 bn sin nx sin2 ww.
Thus
4 u2 o \ nu
if we agree to give the coefficient of A^ the value 1. Let us
give x an arbitrary but fixed value in 53. Then for each e > 0,
there exists an w such that •-
en , €B<€, n>i
Thus ^
(4
446 FOURIER'S SERIES
The index m being determined as above, let us take u such that
u < — , so that m < — ;
m u
and break jS into three parts
1 m+l K+I
where K is the greatest integer < TT/U, and then consider each sum
separately, as u = 0.
Obviously lim S1 = 0.
u=Q
As to the second sum, the number of its terms increases indefin
itely as u = 0.
For any u,
(i— • —19 i— • — 19 x
p^>-?_r~7
mu J KU J J
<,r!»2!.P.<«
WW
since each term in the brace is positive. In fact
sin v
v
is a decreasing function of v as v ranges from 0 to TT, and
nu < KU <_TT , n OB we, tn -f- 1, ... *.
Finally we consider Sy We may write the general term as
follows :
[rsin(n-l>"|2 _rsi
€n lL (n-l)w J " L
jTsin (n - l)^"]2 fsin nu~V\
' €n IL~" nu J L W J J
sin2 (yi — 1)u — s\ii2nu _ — sin (2n — l)u sin w , ^ , 1
UNIQUENESS OF FOURIER'S DEVELOPMENT 447
Thus
<^-T(
O-l)2 n*\ u
6 €
since
But K >_ — — 1 , or KU _> TT — w.
M
Thus I . y < c f 1 i 1 I -
1 (TT — w)2 TT — u)
Hence e o „ ^ - n . , • n
o = Oj -(- o2 + A33 — u, as u = u,
which proves the limit 2), on using 4).
To prove the limit 3), we have
Let us give wa definite value and break ^Tinto three sums.
m
T± = 2,
where m is chosen so that
\A n I < e , n > m ;
TO+l
where \ is the greatest integer such that
\U<_1',
and
A+l
Obviously for some M,
| Z\| <.w^f.
Also
since
448 FOURIER'S SERIES
As to the last sum,
u A+1.v
Thus T ^ Q as M = 0.
455. Schwarz-Luroth Theorem. •
In ty, = (a < b} let the continuous function f (x) be such that
except possibly at an enumerable set & in 51. At the points ($, let
uS(x, u) = 0 as u = 0. (2
Then f is a linear function in 51.
Let us first suppose with Schwarz that (g = 0. We introduce
the auxiliary function,
where _
?; = ± 1, and c is an arbitrary constant.
The function g (V) is continuous in 51, and <? (a) = </ (5) = 0.
Moreover , >. , , N 0 /• -\
flr(^4-1*)4-Q'(^ — tt J — !Z<7(2?). -n
e_v ! : — ! — ii_i . — L £_! — L = <?, as U = U.
Thus for all 0<u< some 8,
(3
From this follows that g(x)<0 in 51. For if #(V)>0, at any
point in 51, it takes on its maximum value at some point f within 51.
Thus
for 0 < w < 8, 8 being sufficiently small. Adding these two in
equalities gives 6r <_0, which contradicts 3). Thus #.< 0 in 51.
Let us now suppose L=fcO for some x in 51. We take c so small
that T r
sgn# = sgn ?;£ = 77 sgn .L.
UNIQUENESS OF FOURIER'S DEVELOPMENT 449
But 77 is at pleasure ± 1, hence the supposition that L =£ 0 is
not admissible. Hence L = 0 in 21, or
/<X)=/00-f5|j/(i)-/00! (4
is indeed a linear function.
Let us now suppose with Liiroth that (g>0. We introduce the
auxiliary continuous function.
Thus A(a)=0 ,
Suppose at some inner point £ of
This leads to a contradiction, as we proceed to show. For then
provided
We shall take c so that this inequality is satisfied, i.e. c lies in
the interval® = (0*, (7*). Thus
Hence A(V) takes on its maximum value at some inner point e
of 31. Hence for B > 0 sufficiently small,
<S. (6
(7
Now if e is a point of H — (S,
«=o
But this contradicts 7), which requires that
\\mH(e, w)<0.
450 FOURIER'S SERIES
Hence e is a point of (£. Hence by 2),
h(e + u)-h(e) + AQ - M) - &(e) ^ Q as w ^ Q.
w w
By 6), both terms have the same sign. Hence each term = 0.
Thus f or u > 0
0 = lim h(e ±M^~ h^ = lim f(e
±u ±u b — a
+ 2*0-a).
Thus to each c in the interval (£, corresponds an e in (£, at which
point the derivative of f(x) exists and has the value given on the
right of 8). On the other hand, two different <?'s, say c and c' , in
(£ cannot correspond to the same e in (S.
For then 8) shows that
c (e — a) = c'(e — #),
or as > , _
Thus there is a uniform correspondence between (£ whose cardi
nal number is c, and (£ whose cardinal number is e, which is absurd.
Thus the supposition 5) is impossible. In a similar manner, the
assumption that L < 0 at some point in 51, leads to a contradiction.
Hence L = 0 in 51, and 4) again holds, which proves the theorem.
456. Cantor s Theorem. Let
J a0 + 2)(#n cos nx + bn sin nx) (1
i
converge to 0 in 5l = (— TT, TT), except possibly at a reducible set 9?,
where nothing is asserted regarding its convergence. Then it con
verges to 0 at every point in 51, and all its coefficients
For by 452, 2, an, 5n = 0. Then Riemann's function
^v N i V 1 A *
/ (V) = -J- aQx2 — V — (an cos w# H- 6n sin 712:)
^ n^
UNIQUENESS OF FOURIER'S DEVELOPMENT 451
satisfies the conditions of the Schwarz-Liiroth theorem, 455, since 9?
is enumerable. Thus f(x) is a linear function of x in 51, and has
the form a + fix. Hence
« 1
a + fix — | aQx2 = — 2, -g («B cos W2: + ^n sin 712:) . (2
The right side admits the period 2 TT, and is therefore periodic.
Its period o> must be 0. For if &> > 0, the left side has this
period, which is absurd. Hence co = 0, and the left side reduces
to a constant, which gives /3=0, a0= 0. But in 51 — 9?, the right
side of 1) has the sum 0. Hence «= 0. Thus the right side of
2) vanishes in 51. As it converges uniformly in 51, we may deter
mine its coefficients as in 436. This gives
CHAPTER XIV
DISCONTINUOUS FUNCTIONS
Properties of Continuous Functions
457. 1. In Chapter VII of Volume I we have discussed some
of the elementary properties of continuous and discontinuous
functions. In the present chapter further developments will be
given, paying particular attention to discontinuous functions.
Here the results of Baire * are of foremost importance. Le-
besgue f has shown how some of these may be obtained by sim
pler considerations, and we have accordingly adopted them.
2. Let us begin by observing that the definition of a continu
ous function given in I, 339, may be extended to sets having iso
lated points, if we use I, 339, 2 as definition.
Let therefore/0^ ••• xm) be defined over 51, being either limited
or unlimited. Let a be any point of 21. If for each e > 0, there
exists a 8 > 0, such that
I/O) -/O) | < e, for any x in F5(a),
we say / is continuous at a.
By the definition it follows at once that / is continuous at each
isolated point of 21. Moreover, when a is a proper limiting point
of 21, the definition here given coincides with that given in I, 339.
If /is continuous at each point of 21, we say it is continuous in 21.
The definition of discontinuity given in I, 347, shall still hold,
except that we must regard isolated points as points of con
tinuity.
* " Sur les Functions de Variables reeles," Annali di Mat., Ser. 3, vol. 3
(1899).
Also his Lemons sur les Functions Discontinues. Paris, 1905.
t Bulletin de la Societe Mathematique de France, vol. 32 (1904), p. 229.
452
PROPERTIES OF CONTINUOUS FUNCTIONS 453
3. The reader will observe that the theorems I, 350 to 354
inclusive, are valid not only for limited perfect domains, but also
for limited complete sets.
458. 1. If f(x^ ••• #m) is continuous in the limited set 51, and its
values are known at the points of 53 < 51, then f is known at all
points of 53' lying in 51.
For let 5j, b2, b3 ••• be points of 53, whose limiting point b lies
in 5i. Then
2. If f is known for a dense set 53 in 5(, and is continuous in 51,
/ is known throughout 51.
For g, > 3,
3. If f(x1 -•- xm) is continuous in the complete set 51, the points
53 in 51 where f=c,a constant, form a complete set. If 51 is an
interval, there is a first and a last point 0/53.
For/= c at x= «1? «2 ••• which = a; we have therefore
/(a) = lim/(an) = c.
459. The points of continuity (£ of /(a:1---^m) in 51 lie in a
deleted enclosure $. Ifty. is complete, ft = (£.
For let ex > e2 > ••• = 0. For each en, and for each point of
continuity c in 51, there exists a cube Q whose center is c, such that
Osc/<en, in O.
Thus the points of continuity of / lie in an enumerable non-
overlapping set of complete metric cells, in each of which
Osc/< en. Let Qn be the inner points of this enclosure. Then
each point of the deleted enclosure
« = I* {CU
which lies in 51 is a point of continuity of /. For such a point c
lies within each Qn.
Hence r\ * ^ - rr s \
Osc/< e, in F6O),
for S > 0 sufficiently small and n sufficiently great.
454 DISCONTINUOUS FUNCTIONS
Oscillation
460' Let n
This is a monotone decreasing function of S. Hence if o>5 is
finite, for some 8 > 0,
&) = lim o>5
6=0
exists. We call co the oscillation off at x = a, and write
T W v> »^t»
Should o)s = -f oo, however small & > 0 is taken, we say o> = -f <x>.
A I v\"\A^^nen w = ^/ ig continuous at x = a, if a is a point in the domain
of definition of/. When o> > 0, /is discontinuous at this point.
It is a measure of the discontinuity off at x = a ; we write
461. 1.
at x = a. Then
\d-e\< Disc (/ ± ^) < d + e.
x—a
For in F5 (a),
|Osc/- Osc^l < Osc(/±^) < OSG/+
2. Tf/ is continuous at x = a, while Disc g = d, then
Disc (/ + 9) = d.
j:=a
For /being continuous at a, Disc/ = 0.
Hence Disc ^ < Disc (/ + g) < Disc g = d.
3. If c is a constant,
Disc (c/) = | c | Disc/ , at x = a.
For Osc (c/) = ; c | Osc/ , in any F«(a).
4. When the limits
OSCILLATION 455
exist and at least one of them is different from /"(a;), the point x
is a discontinuity of the first kind, as we have already said.
When at least one of the above limits does not exist, the point x
is a point of discontinuity of the second kind.
462. 1. The points of infinite discontinuity 3 of f, defined over
a limited set tyi,form a complete set.
For let i], t2 ••• be points of 3, having k as limiting point.
Then in any V(k) there are an infinity of the points in and hence
in any V(lc), Osc/= + oo. The point k does not of course
need to lie in 51.
2. We cannot say, however, that the points of discontinuity of
a function form a complete set as is shown by the following
Example. In 51 = (0, 1), let f(x) = x when x is irrational, and
= 0 when x is rational. Then each point of 51 is a point of dis
continuity except the point x — 0. Hence the points of disconti
nuity of /do not form a complete set.
3. Let f be limited or unlimited in the limited complete set 51.
The points ^of^at ivhich Oscf>^kform a complete set.
For let ar «2 ••• be points of $ which = a. However small
8 >0 is taken, there are an infinity of the an lying in F«(a). But
at any one of these points, Oscf~>_k. Hence Qscf>_k in V& (a)
and thus a lies in $.
4. Letf(x^ -• :rm) be limited and R-integrable in the limited set 51.
The points ft at which Oscf>^kform a discrete set.
For let D be a rectangular division of space. Let us suppose
$D > some constant c > 0, however D is chosen. In each cell 8
of D,
Osc/>fc.
Hence the sum of the cells in which the oscillation is _> k can
not be made small at pleasure, since this sum is &#. But this
contradicts I, 700, 5.
5. Let f(xl ••• zm) be limited in the complete set 51. If the points
$ in 51 at which Osc / > k form a discrete set, for each k, then f is
R-integrable in 51.
456 DISCONTINUOUS FUNCTIONS
For about each point of 51 — $ as center, we can describe a cube
& of varying size, such that Osc/< 2 k in (£. Let D be a cubical
division of space of norm d. We may take d so small that
&D = 2c?t is as small as we please. The points of 51 lie now within
the cubes (£ and the set formed of the cubes d,. By Borel's
theorem there are a finite number of cubes, say
such that all the points of 51 lie within these TJ'S. If we prolong
the faces of these ?;'s, we effect a rectangular division such that
the sum of those cells in which the oscillation is > 2 k is as small
as we choose, since this sum is obviously < ®D. Hence by I, 700,
5, / is 72-integrable.
6. Letf(x^ ••• xm*) be limited in 51; let its points of discontinuity
in 51 be £). If f is R-integrable, 2) is a null set. If 51 is complete
and 2) is a null set, f is R-integrable.
Let / be jR-integrable. Then £) is a null set. For let e: > e2
> ... = 0. Let $)„ denote the points at which Osc/> en. Then
$) = {£)„}. But since/ is 72-integrable, each £)n is discrete by 4.
Hence £) is a null set.
Let 51 be complete and £) a null set. Then each £)n is complete
by 3. Hence by 365, £n = £)„. As £) = 0, we see S)nis discrete.
Hence by 5, /is JZ-integrable.
If 51 is not complete, / does not need to be .R-integrable when
£) is a null set.
Example. Let 5IX = l , n = 1, 2 ... ; m < 2».
Let 51 = 51} + 5^ .
Let /(V) = — , at x = ^
= 1 in 512 .
Then each point of 51 is a point of discontinuity, and 51
But 5lj, 5Lj are null sets, hence 51 is a null set.
POINTWISE AND TOTAL DISCONTINUITY 457
On the other hand,
J/=l , f/=0,
^31 —21
and / is not 72-integrable in 51.
Pointwise and Total Discontinuity
463. Let/^j ••• 2?m) be defined over 51. If each point of 31 is a
point of discontinuity, we say /is totally discontinuous in 51.
We say / is pointwise discontinuous in 51, if / is not continuous
in 51= \a\, but has in any V(a) a point of continuity. If/ is
continuous or pointwise discontinuous, we may say it is at most
pointwise discontinuous.
Example 1. A function/^ ••• zm) having only a finite number
of points of discontinuity in 51 is pointwise discontinuous in 51.
Example 2. Let
f(x) = 0 , for irrational x in 51 = (0, 1)
1 £. m
= - , for x — —
n n
= 1 , for x = 0, 1.
Obviously / is continuous at each irrational x, and discontinuous
at the other points of 51. Hence / is pointwise discontinuous
in 51.
Example 3. Let £) be a Harnack set in the unit interval
51 = (0, 1). In the associate set of intervals, end points included,
let/(z)=l. At the other points of 51, let /= 0. As S) is
apantactic in 51, / is pointwise discontinuous.
Example 4. In Ex. 3, let $) = (g + g, where (§ is the set of end
points of the associate set of intervals. Let /= I/ft at the end
points of these intervals belonging to the nih stage. Let/= 0 in
g. Here / is defined only over £). The points g are points of
continuity in £). Hence/ is pointwise discontinuous in S).
Example 5. Let/(V) be^irichlfit!s_function, i.e. /= 0, for the
irrational points 3 in 51 = (0, 1), and = 1 for the rational points.
458 • DISCONTINUOUS FUNCTIONS
As each point of 21 is a point of discontinuity,/ is totally discon
tinuous in 21. Let us remove the rational points in 21 ; the deleted
domain is 3. In this domain/ is continuous. Thus on removing
certain points, a discontinuous function becomes a continuous
function in the remaining point set.
This is not always the case. For if in Ex. 4 we remove the
points g, retaining only the points (£, we get a function which is
totally discontinuous in (£, whereas before / was only pointwise
discontinuous.
464. 1. Iff(x1 ••• #m) is totally discontinuous in the infinite com
plete set 21, then the points b^ where
Disc/ > co , a) > 0,
form an infinite set, if co is taken sufficiently small.
For suppose bw were finite however small co is taken. Let
o)1>ft)2>... =0. Let Dj, D2, ••• be a sequence of superposed
cubical divisions of space of norms dn == 0. We shall only con
sider cells containing points of 21. Then if d1 is taken sufficiently
small, D1 contains a cell Sj, containing an infinite number of
points of 21, but no point at which Disc/>ft>r If d2 is taken
sufficiently small, D2 contains a cell S2<81, containing no point
at which Disc/><»2. In this way we get a sequence of cells,
which == a point p. As 21 is complete, p lies in 21. But / is
obviously continuous at p. Hence / is not totally discontinuous
in 21.
2. If 21 is not complete, bw .does not need to be infinite for
any co > 0.
Example. Let 21 = , n = 1, 2, ••• and m odd and <2". At
— , let</= — • Then each point of 21 is a point of discontinuity.
But bo, is finite, however small co > 0 is taken.
3. We cannot say /is not pointwise discontinuous in complete
21, when bo> is infinite.
EXAMPLES OF DISCONTINUOUS FUNCTIONS 459
Example. At the points | - \ = % let / = 0 ; at the other
(n)
points of 51 = (0, 1), let/=l.
Obviously / is pointwise discontinuous in 21. But bw is an
infinite set for &> < 1, as in this case it is formed of $ft, and the
point 0.
Examples of Discontinuous Functions
465. In volume I, 330 seq. and 348 seq., we have given ex
amples of discontinuous functions. We shall now consider a few
more.
Example 1. Riemanns Function.
Let (x) be the difference between x and the nearest integer;
and when x has the form n + J, let (x) = 0. Obviously (V) has
the period 1.
It can be represented by Fourier's series thus :
sin 2 TTX sin 2 • 2 TTX . sin 3 • 2 TTX
Riemann s function is now
This series is obviously uniformly convergent in 21 = (— oo, oo).
Since (x) has the period 1 and is continuous within ( — J, |),
we see that (nx) has the period -, and is continuous within
n
{ — - — , — j. The points of discontinuity of (nx) are thus
®n= \^- + -} , « = 0, ±1, ±2,...
( z n n )
Let (g= 5(gn{. Then at any a; not in (£, each term of 2) is a con
tinuous function of x. Hence F(x) is continuous at this point.
On the other hand, F is discontinuous at any point e of (£. For
F being uniformly convergent,
R lim F(x) = ZR lim (3
w
i lim ^(a;) = 2Z lim ^=^ (4
x=€ ar=e H2
460 DISCONTINUOUS FUNCTIONS
We show now that 3) has the value
jFO)--^ for e=2* + 1, e irreducible. (5
16 n2 2n
and 4) the value
||; ' F^+w <6
Hence 2
-OO- (7
To this end let us see when two of the numbers
1 +£, and -L + i
2mm 2 n n
are equal. If equal, we have
Thus if we take 2 s + 1 relatively prime to n, no two of the num
bers in (§„ are equal. Let us do this for each n. Then no two of
the numbers in (g are equal.
Let now x = e = h - • Then (mx) is continuous at this point,
2 n n
unless 8) holds; i.e. unless m is a multiple of n, say m= In. In
this case, 8) gives
Thus I must be odd ; I = 1, 3, 5 ••• In this case (mx) = 0 at e,
while .Rlim (mx) = — J. When m is not an odd multiple of w,
obviously R lim (mx) — (me).
Thus when m = In, I odd,
7?r (W) 1 1 (map 1 1 1
« m2 : 2 W~ m2 • 2 7*2 ' p'
When 77i is not a multiple of n,
m
EXAMPLES OF DISCONTINUOUS FUNCTIONS 461
Hence
This establishes 5). Similarly we prove 6). Thus F(x) is
discontinuous at each point of (§. As
F is limited. As the points (S form an enumerable set, F is
^-integrable in any finite interval.
466. Example 2. Let/(V)=0 at the points of a Cantor set
(7= m • ^#2 ••• ; m = 0, or a positive or negative integer, and the
a's = 0 or 2. Let f(x) = 1 elsewhere. Since / (x) admits the
period I,/ (3 tt#) admits the period -- Let 01 be the points of
3 n
C which fall in 21 =(0, 1). Let Dl be the corresponding set of
intervals. Let <72 = (7X + Fx, where. Tl is obtained by putting a
Cl set in each interval of D1 . Let Dz be the intervals correspond
ing to <72. Let <73 = (72 + F2 where F2 is obtained by putting a <72
set in each interval of .Z>2, etc.
The zeros of/(3nz) are obviously the points of Cn. Let
The zeros of F are the points of (S = \ Cn\. Since each (7n is a null
set, (S is also a null set. Let A = 21 — (L The points J., (£ are
each pantactic in 51. Obviously F converges uniformly in 51,
since 0</(3 nx) <1. Since fn(x) is continuous at each point a
of -4, F is continuous at a, and
462 DISCONTINUOUS FUNCTIONS
We show now that F is discontinuous at each point of (£. For
let em be an end point of one of the intervals of Dm+1 but not of
Dm. Then
--
7/1
Hence
As the points ^4. are pantactic in 21, there exists a sequence in
A which = e. For this sequence F = H. Hence
Similarly, if rjm is not an end point of the intervals Dm+1, but a
limiting point of such end points,
Disc = 5^
*=>?m
The function F is jft-integrable in 21 since its points of discon
tinuity (£ form a null set.
467. Let & = S«tl...la! be an enumerable set of points lying in the
limited or unlimited set 21, which lies in $ftm. For any x in 21 and
any eL in (£, let x— e^ lie in $8. Let g(x1 ••• #m) be limited in 53 and
continuous, except at x = 0, where
Disc #(V)= b.
Let (7=2cti...la converge absolutely. Then
is continuous in A = 21 — (£, and at x= e^
For when # ranges over 21, # — et remains in $8, and g is limited
in 33. Hence F is uniformly and absolutely convergent in 21.
Now each g(x — ej) is continuous in A ; hence ^ is continuous
in A by 147, 2.
EXAMPLES OF DISCONTINUOUS FUNCTIONS 463
On the other hand, F is discontinuous at x = eK. For
where If is the series F after removing the term on the right of
the last equation. But H, as has just been shown, is continuous
at x = eK .
468. Example 1. Let (& = \en\ denote the rational numbers.
•JO
= 0 , x=0.
Then rv \ V* 1 / N -i
^0*0=2,— #O-O > /*>!
?*M
is continuous, except at the points (g. At x = en,
Disc ^ = -^.
n*
Example 2. Let @ = \en\ denote the rational numbers.
Let "-
= 0 ,
which we considered in I, 331.
Then
is continuous, except at the rational points, and at x = em,
Disc F(x) = — .
m !
469. In the foregoing g(x) is limited. This restriction may be
removed in many cases, as the reader will see from the following
theorem, given as an example.
LetJ2=\eil...ls\ be an enumerable apantactic set in $(. Let (§ =
(J£, E'}. For any x in 51, and any et in E, let x — et lie within a
cube 53. Let g^ ••• xm) be continuous in 58 except at x=Q, where
9 == + QO, as x= 0. Let 2ctl...t< be a positive term convergent series.
464 DISCONTINUOUS FUNCTIONS
Then •
is continuous in A = H — (§. On the other hand, each point of ($ is a
point of infinite discontinuity.
For any given point x = a of A lies at a distance > 0 from (g.
Thus
as x ranges over some F^(a), and et over E.
Hence |0(s- Ol< some JK;
for a: in V^(a), and et in E. Thus .F is uniformly convergent at
x = a. As each g(x — et) is continuous at # = a, F is continuous
at a.
.Z/e£ next x = eK. Then there exists a sequence
x\ x" ... = eK (1
whose points lie in A. Thus the term g(x — eK) == + oo as x ranges
over 1). Hence a fortiori F = + oo. Thus each point of E is a
point of infinite discontinuity.
Finally any limit point of E is a point of infinite discontinuity,
by 462, 1.
470. Example. Let g(x) = - , an = -- , • a > 1.
x an
n
Then
is a continuous function, except at the points
:': ' " : o,-i,-^-v- : ••'••
a a* a6
which are points of infinite discontinuity.
471. Let us show how to construct functions by limiting
processes, whose points of discontinuity are any given complete
limited apantactic set (£ in an m-way space 9?m.
EXAMPLES OF DISCONTINUOUS FUNCTIONS 465
1. Let us for simplicity take m = 2, and call the coordinates of
a point #, y.
Let Q denote the square wlxose center is the origin, and one of
whose vertices is the point (1, 0).
The edge of Q is given by the points x, y satisfying
|*|+|y| = l. (1
Thus ., f J, on the edge
-
LO, outside
of the square Q. Hence
L ( „) = ifl - lim "tt-M-lyli ] = ( i, on the edge, g
2L -=.l+njl-|* -[y|d 10, off the edge.
edge.
Thus ff(*y)=0(*,y) + J(*,y) = {;'in.e'
10, without ().
2. We next show how to construct a function g which shall = 0
on one or more of the edges of Q. Let us call these sides e^ <?2, £3, e±,
as we go around the edge of Q beginning with the first quadrant.
If 6r = 0 on et, let us denote it by Grl ; if G- = 0 on «t, eK let us
denote it by 6rl(C, etc. We begin by constructing 6rr We observe
that
=oo 1 + w|< ! 10, for tj= 0.
Now the equation of a right line I may be given the form
x cos a + y sin a = p
where 0<a<27r, jt?>0. Hence
= 1 - lim n I g COS a + y sin a ~ p I = j1' on Z'
n=x 1 + w | a; cos a + ?/ sin a — p \ ( 0, off I.
If now we make I coincide with ev we see that
y) =
Hence
- ^(^ y) = ' n ,
0, on ex and without Q.
466 DISCONTINUOUS FUNCTIONS
In the same way,
By introducing a constant factor we can replace Q by a square
Qc whose sides are in the ratio c : 1 to those of Q.
Tnus i H , on the edge of ft,
Q^ y^ _ iim Trn \~^~\\^ = 1 1» inside,
1 + ( c + c) l°' °utside'
We can replace the square Q by a similar square whose center
is #, b on replacing |#|, \y\ by \x — a , \y — b\.
We have thus this result : by a limiting process, we can con
struct a function g(x, y) having the value 1 inside $, and on any
of its edges, and = 0 outside ft and on the remaining edges.
Q has any point a, b as center, its edges have any length, and its
sides are tipped at an angle of 45° to the axes.
We may take them parallel to the axes, if we wish, by replacing
x
, | y | in our fundamental relation 1) by
Finally let us remark that we may pass to w-way space, by re
placing 1) by
| xl | 4- |*2 I + - + |*m| = l.
3. Let now Q = jqn| be a border set [328], of non-overlapping
squares belonging to the complete apantactic set (£, such that
d-f- (£ = 9? the whole plane. We mark these squares in the
plane and note which sides qn has in common with the preceding
q's. We take the gn(xy) function so that it is = l in qn, except
on these sides, and there 0. Then
has for each point only one term ^= 0, if #, y lies in O, and no
term ^= 0 if it lies in (£.
Hence
_ { 1» f°r each point of O,
" I 0, for each point of '<£.
EXAMPLES OF DISCONTINUOUS FUNCTIONS 467
Since & is apantactic, each point of (S is a point of disconti
nuity of the 2° kind ; each point of Q is a point of continuity.
4. Let/(2#) be a function denned over 51 which contains the
complete apantactic set (£.
Then
472. 1. Let 51 = (0, 1), $B= the points 2 ™ + l in 51.
Then 53n, $„ have no points in common.
Let/n(» = 1 in »n, and = 0 in Bn = 51 - 23n.
Let«=[SBnJ. Then
The function J7 is totally discontinuous in 33, oscillating be
tween 0 and 1. The series F does not converge uniformly in
any subinterval of 51.
2. Keeping the notation in 1, let
At each point of 23n, G- = -, while Gr = 0 in ^.
ft
The function 6r is discontinuous at the points of 93, but con
tinuous at the points B. The series G- converges uniformly in
51, yet an infinity of terms are discontinuous in any interval in 51.
473. Let the limitecj. set 51 be the union of an enumerable set
of complete sets £5ln5. We show how to construct a function/,
which is discontinuous at the points of 51, but continuous else
where in an w-way space.
Let us suppose first that 51 consists of but one set and is com
plete. A point all of whose coordinates are rational, let us call
rational, the other points of space we will call non-rational. If 51
has an inner rational point, let /= 1 at this point, on the frontier
of 51 let /= 1 also ; at all other points of space let /= 0. Then
each point a of 51 is a point of discontinuity. For if x is a fron-
468 DISCONTINUOUS FUNCTIONS
tier or an inner rational point of 91, f(x) = 1, while in any V(x)
there are points where /= 0. If x is not in 91, all the points of
some D(x) are also not in 91. At these points /= 0. Hence /is
continuous at such points.
We turn now to the general case. We have
where A1 = 9^, Az — points of 912 not in 9lx, etc. Let/j = 1 at the
rational inner points of A^ and at the frontier points of 9lj ; at all
other points let /x = 0. Let /2 = 1 at the rational inner points of
A2, and at the frontier points of A% not in Al ; at all other points H
let /2 = 0. At similar points of Az let/3 = J, and elsewhere = 0,
etc.
Consider now TJ
Let x = a be a point of 91. If it is an inner point of some A,,
it is obviously a point of discontinuity of F. If not, it is a proper
frontier point of one of the A's. Then in any D (#) there are points
of space not in 91, or there are points of an infinite number of the
As. In either case a is a point of discontinuity. Similarly we
see F is continuous at a point not in 91.
2. We can obviously generalize the preceding problem by sup
posing 91 to lie in a complete set 33, such that each frontier point
of 91 is a limit point of A = 33 — 91.
For we have only to replace our w-way space by 33.
Functions of Class 1
474. 1. Baire has introduced an important classification of
functions as follows :
Let /(#!•••#,„) be defined over 91 ; /and 91 limited or unlimited.
If /is continuous in 91, we say its class is 0 in 91, and write
Class /=0 , orCl/=0 , Mod 91.
If ^.
each/n being of class 0 in 91, we say its class is 1, if/ does not lie
in class 0, mod 91.
FUNCTIONS OF CLASS 1 469
2. Let the series F(x) = ?,fn(x)
converge in 51, each term/n being continuous in 51. Since
we see F is of class 0, or class 1, according as F is continuous, or
not continuous in 51. A similar remark holds for infinite prod-
UCts G(x)=Hgn(x).
3. The derivatives of a function f(x) give rise to functions of
class 0 or 1. For let f(x) have a unilateral differential coeffi
cient g (x) at each point of 51. Both / and 51 may be unlimited.
To fix the ideas, suppose the right-hand differential coefficient
exists. Let h1>hz> • •• = (). Then
is a continuous function of x in 51. But
0O)=lim?.<»
n=oo
exists at each x in 51 by hypothesis.
A similar remark applies to the partial derivatives
<tf_ . V_
3*1 dxm
of a function f(xl •«• #n).
4-
each/n being of class 1 in 51. Then we say, Cl/= 2 if /does not
lie in a lower class. In this way we may continue. It is of
course necessary to show that such functions actually exist.
475. Example 1.
for x > 0,
for x = 0.
, = Um nx = f 1,
«=» 1 + nx \ 0,
This function was considered in I, 331. In any interval
51 = (0 < b) containing the origin x = 0, Cl jf = 1 ; in any inter
val (a < 6), a > 0, not containing the origin, Cl/= 0.
470 DISCONTINUOUS FUNCTIONS
Example 2.
n=oo enx
The class of f(x) is 0 in 51. Although each fn is limited in 51,
the graphs of fn have peaks near x = 0 which = oo, as n = oo.
Example S. If we combine the two functions in Ex. 1, 2, we
get f art^um J-JL l-lna^l1'*01**0'
"" n=™ \ 1 + nx enx* J 1 0, for x = 0.
Hence Cl/(z) = 1 for any set 53 embracing the origin ; = 0
for any other set.
Example 4-
Let /<» = liin.™*+» , in 21 = (0,1).
n=oo
Then /(aO=0 , for ^ = 0
i
= 2;^* , for 2; > 0.
We see thus that / is continuous in (0*, 1), and has a point of
infinite discontinuity at x = 0.
Hence Class / (a;) =1, in 51
= 0, in(OM).
Example 5.
Let j ^ = Um __! in ^ = (0? ^^
a: -
Then f(x)=^ , for 2: > 0
•2/
= -foo , for x — 0.
Here lim/*^)
n=co
does not exist at x = 0. We cannot therefore speak of the class
of /(a?) in 51 since it is not defined at the point x = 0. It is
defined in 53 = (0*, oo), and its class is obviously 0, mod 53.
FUNCTIONS OF CLASS 1
471
Example 6.
Let
f(x) = sin - , for x =£ 0
x
= a constant c , for x = 0.
We show that Cl/= 1 in 51 = (- 0, oo). For let
nx ( 1
712:
sin
= <7n 0*0
n
Now by Ex. 1,
0, for a; > 0,
e, for x = 0 ;
lim
sin -, for x > 0,
x
while nmnnW=}
0, for 2=0.
As each fn is continuous in 51, and
1* •£ f \ -/?/" *\ * OY
•/ TI V. y ~~" •/ V x "*^^
we see its class is < 1. As / is discontinuous at x = 0, its class
is not 0 in 51.
Example 7. Let ... ,. 1 .1
f(x) = lim - • sin -•
..— n
Here the functions fn(x) under the limit sign are not defined
for x= 0. Thus /is not defined at this point. We cannot there
fore speak of the class of /with respect to any set embracing the
point x— 0. For any set 58 not containing this point, Cl /= 0,
since f(x) = 0 in 53.
Let us set
Let
<j>(x) = sin - , for x =£ 0
x
= a constant c
g(x) = lim -
lim <f>n(x).
472 DISCONTINUOUS FUNCTIONS
Here g is a continuous function in 51 = (— oo, oo). Its class is
thus 0 in 51. On the other hand, the functions <£n are each of
class 1 in 5(.
Example 8.
is defined at all the points of (—00, oo) except 0, — 1, — 2, ...
These latter are points of infinite discontinuity. In its domain
of definition, F is a continuous function. Hence Cl F(V) = 0
with respect to this domain.
476. 1. If 51, limited or unlimited, is the union of an enumerable
set of complete sets, we say 51 is liy per complete.
Example 1. The points S* within a unit sphere $, form a
hypercomplete set. For let Sr have the same center as S, and
radius r<l. Obviously each 2r is complete, while J2r5 = A7*, r
ranging over r^ < r2 < - • • = 1 .
Example 2. An enumerable set of points al, a2 ••• form a hyper-
complete set. For each an may be regarded as a complete set,
embracing but a single point.
2. If 5lx, 5I2"* are limited hypercomplete sets, so is their union
For each 5lm is the union of an enumerable set of complete sets
,n. Thus 51= 151^ „( m, n = 1, 2 •-• is hypercomplete.
Let 51 be complete. If $8 is a complete part of 51, A = 51 — 53 is
hypercomplete.
For let Q= JqJ be a border set of 53, as in 328. The points
An of A in each qn are complete, since 51 is complete. Thus
A — \An\, and A is hypercomplete.
Let 51= J5lnj be hypercomplete, each 5In being complete. If 53 i*
a complete part of 51, ^. = 51 — 53 is hypercomplete.
For let An denote the points of 5In not in 53. Then as above,
An is hypercomplete. As A — \An\, A is also hypercomplete.
FUNCTIONS OF CLASS 1 473
477. 1. (Se Sets. If the limited or unlimited set 51 is the union
of an enumerable set of limited complete sets, in each of which
Osc/<e, we shall say 51 is an (ie set. If, however small e>0 is
taken, 51 is an (£€ set, we shall say 51 is an @€ set, e = 0, which we
may also express by (£e=o-
2. Let f(x± • • • xm) be continuous in the limited complete set 51.
Then 51 is an (ge set, e = 0.
For let €>0 be taken small at pleasure and fixed. By I, 353,
there exists a cubical division of space 7), such that if 5ln denote
the points of 51 in one of the cells of D, Osc/< e in 5ln. As 5ln is
complete, since 51 is, 51 is an (§€ set.
3. An enumerable set of points 51= \an\ is an (£e=y) 8e^-
For each an may be regarded as a complete set, embracing but
a single point. But in a set embracing but one point, Osc/= 0.
4. The union of an enumerable set of (§e sets 51 = \ 51™ j is an ^ set.
For each 51™ is the union of an enumerable set of limited sets
2lm = S5lm,nj, ns=l> 2> - and Osc/<e in each 5lmn.
Thus a = {8LJ , f», n =1,2,...
But an enumerable set of enumerable sets is an enumerable set.
Hence 51 is an (£e set.
5. Letf(x^ "- xm) be continuous in the complete set 51, except at the
points £) = c?j, dz ... dt. Then 51 is an (£^o set.
For let e>0 be taken small at pleasure and fixed. About each
point of £) we describe a sphere of radius p. Let 5lp denote the
points of 51 not within one of these spheres. Obviously 5lp is com
plete. Let p range over r1 > r2 > ••• = 0. If we set 5( = A -f- 2),
obviously A = \$[rn}. As/ is continuous in 5lfn, it is an (§e set.
Hence 51, being the union of A and £), is an (£e set.
478. 1. Let 51 be an @e set. The points £) of 51 common to the
limited complete set $8 form an (£e set.
For 51 is the union of the complete sets 5(n, in each of which
Osc/<e. But the points of 5ln in 33 form a complete set An, and
of course Osc/< e in An. As 2) = J An\, it is an (ge set.
474 DISCONTINUOUS FUNCTIONS
2. Let 51 be a limited (ge set. Let 23 be a complete part of 51.
Then A = % - S3 is an (ge se£.
For 51 is the union of the complete sets 5ln, in each of which
Osc /<e. The points of 5ln not in 33 form a set An, such that
Osc /<e in An also. But J. = \An\, and each ^4n being hyper-
complete, is an @e set.
3. Let/(^ ... xm} be defined over 51, either/ or 51 being limited
or unlimited. The points of 51 at which
<*</<£ (1
may be denoted by
(«</<£). (2
If in 1) one of the equality signs is missing, it will of course be
dropped in 2).
479. 1. Let^ ,/2, ••• be continuous in the limited complete set H.
If at each point of 51, Urn fn exists, 51 is an (£e^0 set and so is any
n=oo
complete 53 < 51.
For let lim fn (^ ••• zm) =f(xl • •• xm) in 51. Let us effect a
division of norm e/2 of the interval ( — 00, oo ) by interpolating
the points • • • m_2 , m_ ^ , m0 = 0, m^ , m2 • • •
Let 5lt = (wt </< wl+2), then 51 = j 51 J .
Next let ~ -Dvlm+-<f<m --}
Then ^=\^n,Pl » w, jt? = l, 2-» (1
For let a be a point of 5lt , and say /(#) = a. Then
But a — €</5(a) <«+ e ,
and we may take e and n so that
Hence a is in £)ni2? .
Conversely, let a be a point of 5£)w>pj. Then a lies in some
n^p. Hence,
i -^ ^ J? /* N ^^" -^
FUNCTIONS OF CLASS 1 475
But as/n(a) ==/(«), we have
Hence if e is sufficiently small,
and thus a is in 5lt .
Thus 1) is established. But £)np is a divisor of complete sets,
and is therefore complete. Thus 51 is the union of an enumerable
set of complete sets j53t5i in each of which Osc/<e, e small at
pleasure.
Let now 53 be any complete part of 51. Let at = Dv J53, 53tj.
Then at is complete, and Osc/<e, in at. Moreover, 53 = \^\.
Hence 53 is an (S€^ set-
2. If Class /< 1 in limited complete 51, / limited or unlimited,
51 is an (Se set.
This is an obvious result from 1.
3. Let /(#!••• #m) be a totally discontinuous function in the non-
enumerable set 51. Then Class f is not 0 or 1 in 51, (f b = Disc/a*
0a£^ point is < & > 0.
For in any subset 53 of 51 containing the point re, Osc f>k.
Hence Osc /is not <e, in any part of 51, if e < k. Thus 51 cannot
be an (Se set.
4. If Class /(«!«•• zm);< 1 m fAe limited complete set 5(, ^e se£
53 = («</< 5) is a hyper complete set, a, b being arbitrary numbers.
For we have only to take a = mt,b = mi+2. Then 53 = 5It, which,
as in 1, is hypercomplete.
480. (Lebesgue.) Let the limited or unlimited f unction f^ ••• zm)
be defined over the limited set 51. // 51 may be regarded as an
(£e-o set with respect to f, the class of f is < 1.
For let col> o>2 > ... = 0. By hypothesis 51 is the union of a
sequence of complete sets
5ln , 5I12 , 5I13... (^
in each of which Osc / <. Wj • 51 is also the union of a sequence
of complete sets
»„ , »« , »W" (1
476 DISCONTINUOUS FUNCTIONS
in each of which Osc/< o>2. If we superpose the division 1) of
51 on the division A^, each 5llK will fall into an enumerable set
of complete sets, and together they will form an enumerable
sequence
in each of wThich Osc^^o^- Continuing in this way we see that
21 is the union of the complete sets
such that in each set of Sn, Osc/< o>n, and such that each set lies
in some set of the preceding sequence ASrn_1.
With each 2ln>s we associate a constant (7n.s, such that
<». , in a.., (2
and call <7n, the corresponding field constant.
We show now how to define a sequence of continuous functions
/i 1/2 "* which =/. To this end we effect a sequence of super
imposed divisions of space Dj, D2 ••• of norms = 0. The vertices
of the cubes of Dn we call the lattice points Ln . The cells of Dn
containing a given lattice point I of Ln form a cube Q. Let 5Ilti
be the first set of Sl containing a point of Q. Let 5l2i2 be the first
set of iS72 containing a point of jQ lying in 2lltl. Continuing in
this way we get
%H >*W^ •'•>*•%•
To 5Inln belongs the field constant (7n,n ; this we associate with
the lattice point I and call it the corresponding lattice constant.
Let now (S be a cell of Dn containing a point of 31. It has 2n
vertices or lattice points. Let P8 denote any product of 8 differ
ent factors 2rn, #r2, ••• xrg. We consider the polynomial
4> = APn + 25Pn_j + 2 OP,_ + .- + S^Pj + i,
the summation in each case extending over all the distinct
products of that type. The number of terms in <f> is, by I, 96,
FUNCTIONS OF CLASS 1 477
We can thus determine the 2n coefficients of <£ so that the values
of </> at the lattice points of @ are the corresponding lattice con
stants. Thus c/> is a continuous function in Q, whose greatest and
least values are the greatest and least lattice constants belonging
to (£. Each cube (5 containing a point of 31 has associated with it
a <f> function.
We now define fn(%i •••%„) by stating that its value in any
cube (£ of Dn, containing a point of 21, is that of the correspond
ing 0 function. Since $ is linear in each variable, two <£'s belong
ing to adjacent cubes have the same values along their common
points.
We show now that/n(V) =f(x) at any point x of 51, or that
e>0, v, |/ (*)-/»<>) I <* , n >v. (3
Let a)e < e/8. Let Hltl be the first set in iS^ containing the point #,
H2ll the first set of Ss lying in 5Ilti and containing x. Continuing
we get ^ > ^ > ^ £ ... > 2lete.
Let tye be the union of the sets in $j preceding 5Iit ; of the sets in
$2 preceding 2I2l2 and lying in 2llti, and so on, finally the sets of
Se preceding 2lcte, and lying in $He_^ ^^ Their number being
finite, 8= Dist (5let<, ^e) is obviously > 0. We may therefore
take v > e so large that cubes of Dv about the point x lie wholly
in -/>,(», if < S.
Consider now/n(z), n > v, and let us suppose first that x is not
a lattice point of Dn. Let it lie within the cell (£ of Dn. Then
fn(x) is a mean of the values of
where I is any one of the 2n vertices of (E, and Cnjn is the corre
sponding lattice constant, which we know is associated with the
Bet**..
We observe now that each of the
^<3te.... (4
For each set in 8n is a part of some set in any of the preceding
sequences. Now 2ln7n cannot be a part of 2I1A:, k < ^, for none of
478 DISCONTINUOUS FUNCTIONS
these points lie in D^x). Hence H«yB is a part of 5114. For the
same reason it is a part of 5l2i2, etc., which establishes 4).
Let now x' be a point of 5I7y-J( . Then
I Cnin ~ C\f | < | Oa/n -/(*') +
<fi>n +*>.<! , by 2). (5
From this follows, since fn(x) is a mean of these Onjn, that
But now
I f (<r\ f (r\ I < I f (r\ f1 I J- I C1 f ( <r\ (7
\J yGJ ~~Jn\J'S I — \J \^J ^njn | T | ^njn J»\XJ • \ i
As x lies in 5lete,
I/ GO - °«i, I < /GO - C«. I + I ^,. - cm. I
<»«+|<|. (8
by 2), 5). From 6), 8) we have 3) for the present case.
The case that # is a lattice point for some division and hence
for all following, has really been established by the foregoing
reasoning.
481. 1. Let f be defined over the limited set 51. If for arbitrary
a, 5, the sets 53 = (« </< V) are hyper complete, then Class /< 1.
For let us effect a division of norm e/2 of (—00, GO) as in
479,1. Then 5T=J5U, where as before 5It = (mt < / < wl+2) .
But as Osc/<e in 5lt, and as each 5lt is hypercomplete by
hypothesis, our theorem is a corollary of 480.
2. For /(#! ••• #m) t° t>e °f class < 1 in the limited complete set
51, it is necessary and sufficient that the sets (a </< £>) are hyper-
complete, a, b being arbitrary.
This follows from 1 and 479, 2.
3. Let limited 51 be the union of an enumerable set of complete sets
|5lnJ, such that Cl/< 1 in each 5In, Mew Cl/< 1 in 51.
FUNCTIONS OF CLASS 1 479
For by 479, 1, 5(n is the union of an enumerable set of complete
sets in each of which Oscf < e. Thus 51 is also such a set, i.e. an
l§e set. We now apply 480, l.
4. If Class/ <1 in the limited complete set 51, its class is<~L,
in any complete part $8 of 51.
This follows from 479, l and 480, 1.
482. 1. Let f(xl "• xm) be defined over the complete set 51, and
have only an enumerable set (£ of points of discontinuity in 51.
Then Class/ = 1 in 51.
For the points E of 51 at which Osc/ > e/2 form a complete
part of 51, by 462, 3. But E, being a part of @, is enumerable
and is hence an (ge set by 477, 3. Let us turn to 23 = 51 — E. For
each of its points 5, there exists a 8 > 0, such that Osc/< e in
the set b of points of $8 lying in -A(6). As 51 is complete, so is b.
As E is complete, there is an enumerable set of these b, call them
bj, b2 •••, such that 33 = fbj. As 51 = 53+ E, it is the union of
an enumerable set of complete sets, in each of which Osc/ < e.
This is true however small e> 0 is taken. We apply now 480, l.
2. We can now construct functions of class 2.
Example. Let fn(x± ••• xm) — 1 at the rational points in the
unit cube Q, whose coordinates have denominators < n. Else
where let/n = 0. Since fn has only a finite number of discontinu
ities in O, Cl/n = 1 in Q. Let now
At a non-rational point, each fn = 0, .-. /=0. At a rational
point, fn = 1 for all n > some s. Hence at such a point /= 1.
Thus each point of O is a point of discontinuity and Disc/= 1.
Hence Cl/is not 1. As / is the limit of functions of class 1, its
class is 2.
483. Let f(x± ••• #m) be continuous with respect to each #t, at each
point of a limited set 51, each of whose points is an inner point.
Then Class /<!.
480 DISCONTINUOUS FUNCTIONS
For let 21 lie within a cube Q. Then A = Q — H is complete.
We may therefore regard 21 as a border set of A ; that is, a set of
non-overlapping cubes jqnj. We show now that C1/<1 in any
one of these cubes as q. To this end we show that the points $8m
of q at which
m m
form a complete set. For let 6j, 52 ••• be points of 33m, which = ft.
We wish to show that ft lies in 23m. Suppose first that 6,, £>,+1 •••
have all their coordinates except one, say a?, the same as the coordi
nates of ft. Since
1 1
w~~ m'
therefore -j -.
As /is continuous in x1, and as only the coordinate xl varies in
ba+p, we have
Hence /3 lies in 53m.
We suppose next that 68, 68+1 ••• have all their coordinates the
same as /? except two, say xl , #2 .
We may place each bn at the center of an interval t of length 8,
parallel to the xl axis, such that
since /is uniformly continuous in a^, by I, 352. These intervals
cut an ordinate in the a^, a;2 plane through y8, in a set of points
ca+p which = /3. Then as before,
m
As e is small at pleasure, ft lies in 33m. In this way we may
continue.
As Cl/< 1 in each qn, it is in 51, by 481, 3.
FUNCTIONS OF CLASS 1 481
484. (Volterra.) Letf^f^ ••• be at most po bitwise discontinuous
in the limited complete set 51. Then there exists a point of 51 at
which all thefn are continuous.
For if 51 contains an isolated point, the theorem is obviously
true, since every function is continuous at an isolated point. Let
us therefore suppose that 51 is perfect.
Let e1>e2>---=0. Let a1 be a point of continuity of /r
Then Osc/! < e , in some 5l: = F5l(«i)-
In 5lj there is a point b of continuity of /r Hence Osc/j < e2
in some F^), and we may take b so that F7?(6)<511. But in
1^(6) there is a point a2 at which /2 is continuous. Hence
OSC/L < e2 , Osc/2 < ej , in some 512 = F«2(a2),
and we may take az such that 512 < 5^ . Similarly there exists a
point a3 in 512, such that
e3 , OsC/2 < 62 ' °SC/3 < el > in S0me ^3= ^3(^3)'
and we may take a3 so that 5I3 < 512 .
In this way we may continue. As the sets 5ln are obviously
complete, Dv\y[n\ contains at least one point a of 51. But at this
point each/m is continuous.
485. 1. Let 51 = ^8 + £ be complete, let 53, £ be pantactic with
reference to 51. Then there exists no pair of functions f, g defined
over 51, such that if 53 are the points of discontinuity of f in 51, then
53 shall be the points of continuity of g in 51.
This is a corollary of Volterra's theorem. For in any Ffi(a) of
a point of 51, there are points of 53 and of (£. Hence there are
points of continuity of /and g. Hence/, g are at most pointwise
discontinuous in 51. Then by 484, there is a point in 51 where/
and g are both continuous, which contradicts the hypothesis.
2. Let 51= 53 +(£ be complete, and let 53, £ each be pantactic with
reference to 51. If 53 is hypercomplete, (£ is not.
For if 53, @ were the union of an enumerable set of complete
sets, 473 shows that there exists a function / defined over 51
which has 53 as its points of discontinuity ; and also a function g
482 DISCONTINUOUS FUNCTIONS
which has (£ as its points of discontinuity. But no such pair of
functions can exist by 1.
3. The non-rational points 3 in any cube O cannot be hyper-
complete.
For the rational points in O are hypercomplete.
4. As an application of 2 we can state :
The limited function f(xl--xm) which is < 0 at the irrational
points of a cube G, and > 0 at the other points 3 of £}, cannot be
of class 0 or 1 in O.
For if Cl/ < 1, the points of O where/ > 0 must form a hyper-
complete set, by 479, 4. But these are the points 3.
486. 1. (Baire.^) If the class off(^xl"-xm) is 1 in the com
plete set 51, it is at most pointwise discontinuous in any complete
8<H,
If Cl/ = 1 in 51, it is < 1 in any complete $ < 51 by 481, 4 ; we
may therefore take 33 = 51. Let a be any point of 51. We shall
show that in any V — V^a) there is a point c of continuity of /.
Let el > e2 > ••• = 0. Using the notation of 479, 1, we saw that
the sets 5lt = (wt </< mt+2) are hypercomplete. By 473, we can
construct a function </>t(^i ••• zm), defined over the w-way space
$ftOT which is discontinuous at the points 5ft, and continuous else
where in 9?m. These functions c^, </>2 ••• are not all at most point-
wise discontinuous in V. For then, by 484, there exists in V a
point of continuity #, common to all the <£'s. This point b must
lie in some 5lt, whose points are points of discontinuity of $t.
Let us therefore suppose that fa is not at most pointwise dis
continuous in V. Then there exists a point cl in V, and an rjl
such that Vl = J^C^i) contains no point of continuity of </>y .
Thus F1<.5ly. But in 5Iy and hence in Fa, Osc f<e1. The
same reasoning shows that in V\ there exists a F^= F^^), such
that Osc/< e2 in F^. As 51 is complete, V^> V^> ••• defines a
point c in Fat which /is continuous.
2. If the class of f(x^ ••• #m) is 1 in the complete set 51, its points
of discontinuity 3) form a set of the first category.
FUNCTIONS OF CLASS 1 483
For by 462, 3, the points £)„ of 3) at which Osc/> - form a
n
complete set. Each Dn is apantactic, since / is at most pointwise
discontinuous, and £)„ is complete. Hence 3) = jOn( is the union
of an enumerable set of apantactic sets, and is therefore of the 1°
category.
487. 1. Let f be defined over the limited complete set 31. If
Class f is not < 1, there exists a perfect set 3) in 31, such that f is
totally discontinuous in 3).
For if Cl/ is not <1 there exists, by 480, an e such that for
this e, 31 is not an (£e set. Let now c be a point of 31 such that
the points a of 31 which lie within some cube q, whose center is <?,
form an @€ set. Let 53 = {aj, (£ = \c\.
Then $8 = (£. For obviously (£ < 53, since each c is in some
a. On the other hand, 33 < (£. For any point b of 53 lies within
some q. Thus b is the center of a cube q' within q. Obviously
the points of 31 within q' form an (ge set.
By Borel's theorem, each point c lies within an enumerable set
of cubes Jcn|, such that each c lies within some q. Thus the
points an of 31 in cn, form an (ge set. As (£ = 5an|, (£ is an (5e set.
Let £) = 31 - 6. If 3) were 0, 31 = S and 31 would be an (g. set
contrary to hypothesis. Thus 3) > 0.
3) ^s complete. For if Z were a limiting point of 3) in (£, Z must
lie in some c. But every point of 31 in c is a point of (£ as we saw.
Thus I cannot lie in (£.
We show finally that at any point d of £),
Osc/>€, with respect to £).
If not, Osc/<e with respect to the points b of £) within
some cube q whose center is d. Then b is an (£e set. Also the
points e of & in q form an (5e set. Thus the points b -f c, that is,
the points of 31 in q form an (5e set. Hence d belongs to (£, and
not to £). As Osc/>e at each point of 3), each point of 3) is a
point of discontinuity with respect to 3). Thus/ is totally discon
tinuous in £).
This shows that 3) can .contain no isolated points. Hence 3) is
perfect.
484 DISCONTINUOUS FUNCTIONS
2. Let f be defined over the limited complete set SI. .If f is at
most pointwise discontinuous in any perfect 58 < SI, its class is < 1
This is a corollary of 1. For if Class / were not 0, or 1, there
exists a perfect set 2) such that /is totally discontinuous in £).
488. If the class of f, g < 1 in the limited complete set SI, the class
of their sum, difference, or product is < 1 . If f > 0 in SI, the class
of <f>
For example, let us consider the product h =fg. If Cl h is not
< 1, there exists a perfect set £) in SI, as we saw in 487, l, such
that h is totally discontinuous in 2). But/, # being of class <, 1,
are at most pointwise discontinuous in £) by 486. Then by 484,
there exists a point of £) at which/, g are both continuous. Then
h is continuous at this point, and is therefore not totally discon-
tinous in $).
Let us consider now the quotient <f>. If Cl <£ is not < 1, 0 is
totally discontinuous in some perfect set $) in SI. But since /> 0
in £), / must also be totally discontinuous in £). This contradicts
486.
489. 1. Let F = ^f^...,t(xl ••• #m) converge uniformly in the com
plete set SI. Let the class of each termft be < 1, then Class J?< 1
mSI.
For setting as usual [117],
1=1^ + ^ (1
there exists for each e > 0, a fixed rectangular cell 72A, such that
| JA | < e, as x ranges over SI. (2
As the class of each term in FK is < 1, Cl J^ < 1 in SI. Hence
SI is an (ge set with respect to FK.
From 1), 2) it follows that SI is an (ge set with respect to F.
2. .Letf ^ = II/^...^^ ••• #m) converge uniformly in the complete
set SI. /f the class of eachf^ is < 1, then Cl ^< 1 in SI.
SEMICONTINUOUS FUNCTIONS 485
Semicontinuous Functions
490. Let f(xl ••- XM) be defined over 21. If a is a point of 51,
Max/ in F"8(a) exists, finite or infinite, and may be regarded as a
function of 8. When finite, it is a monotone decreasing function
of 8. Thus its limit as 8 = 0 exists, finite or infinite. We call
this limit the maximum off at x= a, and we denote it by
Max/.
x=a
Similar remarks apply to the minimum of /in Fa(a). Its limit,
finite or infinite, as 8 == 0, we call the minimum of f at x = a, and
we denote it by
Min/
x=a
The maximum and minimum of /in Fg(a) niay be denoted by
Max/ , Min/
a, S a, S
Obviously, Max (-/)=- Min/,
x=a x=a
Min (-/)=- Max/.
x=a x=a
491. Example 1. -,
/O) = ±in(-l, 1) , for
*c
= 0 , forz=0.
Then Max/=+sc , Min/=-oo.
^=0 r=o
Example 2. -,
/(V) = sin - in (— 1, 1) , for x =£ 0
^c/
= 0 , for x = 0.
Then Max/=l , Min/=-l.
;r=o x=Q
Example 3. /(„) = l in (_ !, l) , f or
= 2 , for x = 0.
Then Max/ =2 . Min/=l.
"
486 DISCONTINUOUS FUNCTIONS
We observe that in Exs. 1 and 2,
Iim/=Max/ , lim/=Min/;
x=Q x=Q x=Q z=o
while in Ex. 3,
lim/= 1 , and hence Max/ > lim/.
X=Q X=Q X=Q
Also lim/ = Min/.
Example 4- -i
f(x) = (z2 + 1) sin - in (- 1, 1) , for x* 0
x
= - 2 , for x = 0.
Here Max/= 1 , Min/= - 2,
x=0 x=0
lim/=l , lim/== — 1.
X=Q #=0
Examples. Let f^ = x , for rational rr in (0, 1)
= 1 , for irrational x.
Here Max/=l , Min/=0,
z=0 x=0
lhn/=l.
2^=0
492. 1. For M to be the maximum of f at x = a, it is necessary
and sufficient that
1° e > 0, 8 > 0, /(z) < M + e, for any x in V8(a) ;
2° there exists for each e > 0, and in any Fg(a), a point a such
that
These conditions are necessary. For M is the limit of Max/
in Fs(a), as B = 0. Hence
e>0, o>0,
a,fi
But for any x in Fs(a),
f(x) < Max/.
SEMICONTINUOUS FUNCTIONS 487
Hence f^ <M+e ? x in y^
which is condition 1°.
As to 2°, we remark that for each e > 0, and in any Fi(a),
there is a point a, such that
-e + Max/ </(«).
a, 6
But M < Max/.
Hence
which is 2°.
These conditions are sufficient. For from 1° we have
a, 8
and hence letting £ = 0,
since e > 0 is small at pleasure.
From 2° we have
Max/> M-e,
a, 5
and hence letting S = 0,
Max/ > Jtf. (2
x=a
From 1), 2) we have M= Max/.
j^=a
2. For m to be the minimum of f at x = a, it is necessary and
sufficient that
1° e > 0, 8 > 0, m - e </O), for any x in V&(a) ;
2° that there exists for each e > 0, and in any VI (a), a point a
such that
/(«) < m + e.
493. When Max/ = /(#), we say / is supracontinuous at x = a.
x=a
When Min/ = /(a), we say / is infracontinuous at a. When/ is
x=a
supra (infra) continuous at each point of 51, we say / is supra
(infra) continuous in 51. When / is either supra or infracontinu
ous at a and we do not care to specify which, we say it is semi-
continuous at a.
488 DISCONTINUOUS FUNCTIONS
The function which is equal to Max /at each point x of 51 we
call the maximal function of/, and denote it by a dash above, viz.
f(x). Similarly the minimal f unction /(V) is defined as the value
of Min / at each point of 21.
Obviously Qsc/= Max/_ M;n/= D.sc/
x=a x=a x=a - x—a
We call -
<*)=/<>)-/<>)
the oscillatory function.
We have at once the theorem :
For f to be continuous at x — a, it is necessary and sufficient that
/00= 700 =/(«).
For Min / < f(a) < Max /.
a, 6 a, 8
Passing to the limit x = a, we have
Min /</<» < Max/,
x=a x=a
or
/(«)</(«)</(«).
But for / to be continuous at x — a, it is necessary and suffi
cient that
«(V) = Osc/= 0.
x=a
494. 1. For f to be supracontinuous at x = a, it is necessary and
sufficient that for each e > 0, there exists a S > 0, such that
/O) < /O) + € » for any x in V^a). (1
Similarly the condition for infracontinuity is
f(a) - e < /(» , for any x in V^a). (2
Let us prove 1). It is necessary. For when /is supracontinu
ous at a,
Then by 492, 1,
€>0 , S>0 , /(aO</(<0 + € , for any x in
which is 1).
SEMICONTINUOUS FUNCTIONS 489
It is sufficient. For 1) is condition 1° of 492, l. The condition
2° is satisfied, since for a we may take the point a.
2. The maximal function f(x) in supracontinuous ; the minimal
function f(x) is infracontinuous, in 31.
To prove that /is supracontinuous we use 1, showing that
f(x) < /(«) + e , for any x in some V^d).
Now by 492, l,
e' > 0, 8 > 0 , f(x) < /(a) + e' , for any x in F«(a).
Thus if e' < e
f(x) < /(a) + e , f or any x in F, (a) , 17 = |.
3. jT/ie swm (>/ ^wo supra (infra) continuous functions in 31 is a
supra (infra) continuous function in 31.
For let/, g be supracontinuous in 31 ; let/-f g = h. Then by 1,
for any a; in some Fs(a) ; hence
This, by 1, shows that h is supracontinuous at x= a.
4. If f(x) is supra (infra) continuous at z = a, g(jz)=—f(%)
is infra (supra) continuous.
Let us suppose that /is supracontinuous. Then by 1,
/(V)</(a)-|- e , for any x in some Fs(a).
Hence -/(«)-,€<-/(*), , .
g(a) - e < ^(2:) , for any x in F«(a).
Thus by 1, # is infracontinuous at a.
490 DISCONTINUOUS FUNCTIONS
495. If /(#! ••• Xm) i8 supracontinuous in the limited complete
set 51, the points 33 of 51 at which /> c an arbitrary constant form a
complete set.
For let f > c? at 6j, bz ••• which == b ; we wish to show that b lies
in $.
Since/ is supracontinuous, by 494, l,
/(V) </(£>) + e , for any x in some F6(5)= V.
But <?</(&„), by hypothesis ; and bn lies in Fi for n> some m.
Hence
*-«</(a).
As e > 0 is small at pleasure,
and b lies in 33.
496. 1. The oscillatory function oy(x) is supracontinuous.
For by 493, „(*)= Max/- Min/
= Max/+Max(-/).
But these two maximal functions are supracontinuous by 494, 2.
Hence by 494, 3, their sum a> is supracontinuous.
2. The oscillatory function o> is not necessarily infracon-
tinuous, as is shown by the following
Example. /= 1 in (-1, 1), except for z = 0, where /= 2.
Then co(x) = 0, except at x = 0, where o» = 1. Thus
Min &>(>) = 0 , while o>(0) = 1.
Hence &>(a?) is not infracontinuous at x = 0.
3. Let o>(V) be the oscillatory function of f(xl ••• xm) in 51. For
f to be at most pointwise discontinuous in 51, it is necessary that
Min &) = 0 at each point of 51. If 51 is complete, this condition is
sufficient.
SEMICONTINUOUS FUNCTIONS 491
It is necessary. For let a be a point of H. As / is at most
pointwise discontinuous, there exists a point of continuity in any
Ffi(a). Hence Min o>(V) = 0, in F6(a). Hence Min o>(V) = 0.
x=a
It is sufficient. For let €1>e2> ••• =0. Since Min 0(2:) = 0,
jr=a
there exists in any Fs(a) a point al such that &>(«!) <|e1.
Hence o>(V) < el in some ^(aj) < F6. In VSl there exists a point
«2 such that o>O) < e2 in some F52(«) < F6l, etc. Since 21 is com
plete and since we may let Sn = 0,
V^ > Ffi2 > ••• = a point « of 51,
at which f is obviously continuous. Thus in each Fs(tf) is a point
of continuity of/. Hence /is at most pointwise discontinuous.
497. 1. At each point x of 31,
<#> = Min \f(x) -/(*)}, and t = Min j/<» -/(*){
= 0.
Let us show that $ = 0 at an arbitrary point a of 51. By 494,
2, /(V) is supracontinuous ; hence by 494, 1,
f(x) </(«) + e , for any a: in some Fs(a) = F". (1
Also there exists a point a in Fsuch that
_€+/(«) </(„). (2
Also by definition
/(«) </(«)• (3
If in 1) we replace # by a we get
/(«)</(«) + e. (4
From 2), 3), 4) we have
-e +/(a) </(«) </(«) </(«) + e,
or
0 </(«)-/(«)< 2 €.
As e > 0 is small at pleasure, this gives
</><» = o.
492 DISCONTINUOUS FUNCTIONS
2. Iff is semicontinuous in the complete set 51, it is at most point-
wise discontinuous in 51.
' »CO=/CO-/CO
= [/CO -/CO] + [/co -/CO] (1
To fix the ideas let / be supracontinuous. Then <£ == 0 in 51.
Hence 1) gives
Min o>(V) = Min ^(z) = 0, by 1.
Thus by 496, 3, / is at most pointwise discontinuous in 51.
CHAPTER XV
DERIVATES, EXTREMES, VARIATION
Derivates
498. Suppose we have given a one-valued continuous function
f(x) spread over an interval 51 = (a<6). We can state various
properties which it enjoys. For example, it is limited, it takes
on its extreme values, it is integrable. On the other hand, we
do not know 1° how it oscillates in 51, or 2° if it has a differ
ential coefficient at each point of 51. In this chapter we wish to
study the behavior of continuous functions with reference to these
last two properties. In Chapters VIII and XI of volume I this
subject was touched upon ; we wish here to develop it farther.
499. In I, 363, 364, we have defined the terms difference quo
tient, differential coefficient, derivative, right- and left-hand dif
ferential coefficients and derivatives, unilateral differential coeffi
cients and derivatives. The corresponding symbols are
, /'(a) ,
Lf(a) ,
The unilateral differential coefficient and derivative may be de
noted by
Uf'(a) , Uf'(x). (1
When . ,
does not exist, finite or infinite, we may introduce its upper and
lower limits. Thus
A=O
(2
always exist, finite or infinite. We call them the upper and lower
differential coefficients at the point x = a. The aggregate of values
493
494 DERIVATES, EXTREMES, VARIATION
that 2) take on define the upper and lower derivatives of /(#), as
in I, 363.
In a similar manner we introduce the upper and lower right-
and left-hand differential coefficients and derivatives,
Rf> , Rf , Lf , Lf. (3
Thus, for example,
finite or infinite. Cf. I, 336 seq.
If f(x) is defined only in 51 = (a < /3), the points a, a + A must
lie in 51. Thus there is no upper or lower right-hand differential
coefficient at x = fB ; also no upper or lower left-hand differential
coefficient at x = a. This fact must be borne in mind. We call
the functions 3) derivates to distinguish them from the deriva
tives Rf, Lf. When Rf(a)=Ef'(a)i finite or infinite,
Rf(cC) exists also finite or infinite, and has the same value. A
similar remark applies to the left-hand differential coefficient.
To avoid such repetition as just made, it is convenient to in
troduce the terms upper and lower unilateral differential coeffi
cients and derivatives, which may be denoted by
Uf> , Uf. (4
The symbol U should of course refer to the same side, if it is
used more than once in an investigation.
When no ambiguity can arise, we may abbreviate the symbols
3), 4) thus:
R , R , L , L , U , U.
The value of one of these derivates as R at a point x — a may
similarly be denoted by
5(«).
The difference quotient
a-b
may be denoted by
A(a, b).
DERIVATES 495
1
Example 1. /(V) = zsin- , x=£ 0 in (— 1, 1)
= 0 , x=0.
li sin -
Here f or x = 0, -^ = — . — = sin r-
Hence -
"i
Z/'(0) =
Example 2. f(x) = z* sin - , a; ^ 0 in ( - 1,
x
= 0 , a: = 0.
. 1
A/ n 4
Here for x = 0 , — = — 5 —
Hence
+ QO , Z/'(0)=:-QO,
/'(O) = + oo ,
Example 8. f(x) = a; sin - , f or 0 < x < 1
= x* sin - , for — 1 < x < 0
x
= 0 , for x = 0.
Here
500. 1. Before taking up the general theory it will be well
for the reader to have a few examples in mind to show him how
complicated matters may get. In I, 367 seq., we have exhibited
functions which oscillate infinitely often about the points of a set
496 DERIVATES, EXTREMES, VARIATION
of the 1° species, and which may or may not have differential co
efficients at these points.
The following theorem enables us to construct functions which
do not possess a differential coefficient at the points of an enumer
able set.
2. Let @ = \en\ be an enumerable set lying in the interval SI. For
each x in SI, and en in (£, let x — en lie in an interval 33 containing
the origin. Let g(x) be continuous in 33. Let g\x) exist and be
numerically <_ M in 33, except at x = 0, where the difference quotients
are numerically <. M. Let A = San converge absolutely. Then
is a continuous function in SI, having a derivative in (£ = SI — G£.
At the points of (§, the difference quotient of F behaves essentially as
that of g at the origin.
For g(x) being continuous in 33, it is numerically < some con
stant in SI. Thus F converges uniformly in SI. As each term
g(x — #n) is continuous in SI, F is continuous in SI.
Let us consider its differential coefficient at a point x of (£.
Since g'(x— en) exists and is numerically < M,
^'(20=2«n</(*-O , by 156, 2.
Let now x = em, a point of (£,
W(x) = amg(x - em) + 2*a^C* - *») £* i *% M*$
-
The summation in 2* extends over 'dl\n=^m. Hence by what
has just been shown, G- has a differential coefficient at x = em.
Thus — — behaves at x = em, essentially as — at x = 0. Hence
501. Example 1. Let
^r(V) = a# , a: > 0
b < 0 < a.
= bx
DERIVATES 497
Then
is continuous in any interval SI, and has a derivative
-$r'O-en)
at the points of SI not in (£. At the point em,
r> TT/ /„ \ . >T^* J- is \
H± (x) = ama + > —og(em-en),
Let & denote the rational points in SI. The graph of F(x) is a
continuous curve having tangents at a pantactic set of points ;
and at another pantactic set, viz. the set (S, angular points (I, 366).
A simple example of a g function is
Example 2. Let g(x) = x2sin- , x =£ 0
= 0 , z=0.
This function has a derivative
#'(z) = 2zsin--7rcos- ,
X X
= 0 , z = 0.
Thus if 2cn is an absolutely convergent series, and Q?= \en\ an
enumerable set in the interval 31 = (0, 1),
is a continuous function whose derivative in SI is
Thus F has a derivative which is continuous in SI — (S, and at
the point x = em
Disc F' = 2 <?m7r,
since -'(*)= 2- Mtf. 451 '
498 DERIVATES, EXTREMES, VARIATION
If (£ is the set of rational points in 31, the graph of F(x) is a
continuous curve having at each point of 31 a tangent which does
not turn continuously as the point of contact ranges over the
curve; indeed the points of abrupt change in the direction of the
tangent are pantactic in 31.
Example 3. Let g(x) = x sin log x2 , x =£ 0
= 0 , x=0.
Then g '(x) = sin log x1 + 2 cos log x2 , x j= 0.
At x = 0, ^ = sin log h2
which oscillates infinitely often between ± 1, as h = A# = 0. Let
(g = \en\ denote the rational points in an interval 31. The series
- *%• /%P ^ ^ ein Inrp (v _ p ^2
== / ( •C "r, I olll H.^ii I Js "n I
satisfies the condition of our theorem. Hence F(x) is a continu
ous function in 31 which has a derivative in 31 — (§. At £ = em,
Thus the graph of ^ is a continuous curve which has tangents at
a pantactic set of points in 31, and at another pantactic set it has
neither right- nor left-hand tangents.
502. Weierstrass' Function. For a long time mathematicians
thought that a continuous function of x must have a derivative, at
least after removing certain points. The examples just given
show that these exceptional points may be pantactic. Weierstrass
called attention to a continuous function which has at no point a
differential coefficient. This celebrated function is defined by the
series
F (x) = 2 an cos bnirx = cos TTX -f- a cos ITTX + a2 cos WTTX + - • • (1
0
where 0 < a < 1 ; b is an odd integer so chosen that
a5l + 7r. (2
DERIVATES 499
The series F converges absolutely and uniformly in any interval
Hence .F is a continuous function in 2(. Let us now consider
the series obtained by differentiating 1) termwise,
Q(x) = - 7r2(«6)nsin bnirx.
If ab < 1, this series also converges absolutely and uniformly,
and F'(x) = <?O),
by 155, 1. In this case the function has a finite derivative in 21.
Let us suppose, however, that the condition 2) holds. We have
v ^ , c <«-hA) - cos bn7rx\ = Qm+ Qm. (3
A* 7 A
Now Tn_1 n
§m = Vs — 5 cos bn7r(x + 7t) — cos ftn7T2:|
<*4 A
0
» /*JT+/i
- I sin bn7rudu.
^ A *
Since
I fx+h I /*x+h
I sin bn7rudu < I I
I <M < ' W—- -' if
Consider now * n
§ro = 2J ^ jcos bnrjr(x + h) — cos l>nTrx\.
m ft
Up to the present we have taken h arbitrary. Let us now
take it as follows ; the reason for this choice will be evident in a
moment.
Let *-*=i.+f.,
where im is the nearest integer to bmx. Thus
-*<£.<!•
Then 6-(* + A) = *. + f » + Ai- = .» + ij. .
500 DERIVATES, EXTREMES, VARIATION
We choose h so that
Vm = £ m 4- hbm is ± 1, at pleasure.
Then _*
h = ^m ^m = 0, as m=cc ;
om
moreover , and „_ - . < .
This established, we note that
cos bnTr(x + h) = cos bn~m7r - bm(x + h) = cos 6n~m(tm -f- rjm)7r
= cos (tm + ^m)7r , since 6 is odd
= ( — l)l»+1 , since T;TO is odd.
Also cos bn7rx = cos bn~m(tm + f m)?r
bn~m^m7T.
Thus
where :••; «« = (-i>+'.
Now each 5 j > 0 and in particular the first is > 0. Thus
sgn Qm = sgn e-f = sgn emrjm,
h
and , ,xm
i^i>f=^->iW"
» ^m — ^m
Thus if 2) holds, | ^TO | > | Qm |. Hence from 3),
sgn Q = sgn Qm = sgn ^m?;m,
and
Let now m = QO . Since ?;m = ± 1 at pleasure, we can make
Q = -j-oo, or to — oo , or oscillate between ± oo, without becoming
definitely infinite. Thus F (x) has at no point a finite or infinite
differential coefficient. This does not say that the graph of F does
not have tangents; but when they exist, they must be cuspidal tangents.
DERIVATES 501
503. 1. Vblterra's Function.
In the interval 51 = (0, 1), let <£> = \r)l be a Harnack set of
measure 0<A<1. Let A = |8nj be the associate set of black
intervals. In each of the intervals £n = (a < /8), we define an
auxiliary function fn as follows :
/„(*) = (*- «)2 sin- , in («*, 7), (1
where 7 is the largest value of x corresponding to a maximum of
the function on the right of 1), such that 7 lies to the left of the
middle point /z of Sn. If the value of /„(#) at 7 is g, we now
Finally fn(a)= 0. This defines fn (x) for one half of the inter
val Sn. We define /„(#) for the other half of Sn by saying that if
x<x' are two points of Sn at equal distances from the middle
point ft then .
With Volterra we now define a function f(x) in 51 as follows:
f(x) = fn(x) , inSn , 7i = l,2,...
= 0 , in $.
Obviously/^) is continuous in 51.
At a point x of 51 not in &,f(x) behaves as
1 1
2 x sin -- cos-,
x x
as is seen from 1). Thus as x converges in 8n toward one of its
end points a, /3, we see that f(x) oscillates infinitely often be
tween limits which = ± 1. Thus
similar limits exist for the points $.
Let us now consider the differential coefficient at a point 77 of
&. We have
M = /^ + *)-/W = /0> + *) , 8ince/(,)=0.
502 DERIVATES, EXTREMES, VARIATION
If rj + k is a point of §,/(?? + &) = 0. If not, rj -f k lies in some
interval Sm . Let a; = e be the end point of 8m nearest rj + k.
Then
1*1
Thus /'(??)= 0. Hence Volterra's f unction f(x) has a differen
tial coefficient at each point of 51; moreover/' (x) is limited in 51.
Each point rj of & is a point of discontinuity of/' (V), and
Disc/ <» > 2.
Hence/' (V) is not 72-integrable, as & = h>0.
We have seen, in I, 549, that not every limited J2-integrable
function has a primitive. Volterra's function illustrates con
versely the remarkable fact that Not every limited derivative is
R-integrable.
2. It is easy to show, however, that The derivative of Volterra's
function is L-integrable.
For let 51A denote the points of 51 at which f (x) > X. Then
when X>l/m, m = 1, 2, ••• 51A consists of an enumerable set of
intervals. Hence in this case 51A is measurable. Hence 5(A, X>0,
is measurable. Now 51 , X>0, differs from the foregoing by add
ing the points 3n in ea°h &n at which/' (x) — 0, and the points £>.
But each 3n is enumerable, and hence a null set, and § is measur
able, as it is perfect. Thus 51A, X>0, is measurable. In the
same way we see 51A is measurable when X is negative. Thus 51A
is measurable for any X, and hence .L-integrable.
504. 1. We turn now to general considerations and begin by
considering the upper and lower limits of the sum, difference, prod
uct, and quotient of two functions at a point x = a.
Let us note first the following theorem :
Letf(xl -•• XM) be limited or not in 51 which has x = a as a limiting
point. Let <£fi = Max/ (f>& — Min/m Ffi*(a). Then
This follows at once from I, 338.
DERIVATES 503
2. Letf(x^ "• xm~), g(x^ -• xm) be limited or not in 21 which has
x = a as limiting point.
Let limf=a , limg = 0
\imf=A , Km g = B
as x= a. Then, these limits being finite,
, (1
« - B < Hm (/ - g) < A - 0. (2
For in any F^*(a),
Min / + Min g < Min (/ -f y) < Max (/+#)< Max / + Max g.
Letting S = 0, we get 1).
Also in
Min / — Max g < Min (/ — g) < Max (/ — g) < Max /— Min g.
Letting 8 = 0, we get 2).
3* If
a/3 < lim/#< AS. (3
If
(4
/O)>Q , ^)>*>0,
l^^|- (5
a<Q<A , 2;>^>0,
The relations 3), 4), 5), 6) may be proved as in 2. For exam
ple, to prove 5), we observe that in Fi*(«),
g g Min g
504 DERIVATES, EXTREMES, VARIATION
5. « + £<lim(/ + 0)<a + £. (7
a + Jg<Ihn~(/ + #) <.! + £. (8
a -.£< lim (/-#)<«-£. (9
A-£<Km(f-g)<A-0. (10
J/ /O)>o , <?(*)> o,
•
a/3 < limfg < *B, (11
(12
(13
6. T/" lira/ exists,
= lira / + lira #, (15
m^. (16
If g(x) > & > 0,
11 m ^ exists,
Baifl (17
lim5r. (18
/(») > 0, ^r(^) > 0. ie^ lim ^ ftw*«. Then
. lim ^, (19
lim fg — lira / • lim g. (20
g(x) > k > 0,
lim //^ = lim //lim ^, (21
Em f/g = En //lim #. (22
505. The preceding results can be used to obtain relations be
tween the derivates of the sum, difference, product, and quotient
of two functions as in I, 373 seq.
DERIVATES 505
1. Let w2;=tt+va. Tnen
Aw _ Aw Av ,-,
Az Az A# '
Thus from 504, 1), we get the theorem:
Uu' + v'U<Uw'< Uu' + W. (2
7/ M has a unilateral derivative Uu',
Uw' = Uu1 + Uv', (3
Uw' = CTw' + Uv'. (4
We get 3), 4) from 1), using 504, 15), 16).
2. In the interval 51, w, v arg continuous, u is monotone increasing,
v is > 0, and v' exists. Then, if w = wv,
Uw1 =uv' +vUuf. (2
For from A^ A A^
— - = (M + AM) — -h v — ,
A^ Ao: Ao:
wehave
/ , rri. w
= wv -|- v U lim —
— Aa:
which gives 1). Similarly we establish 2).
506. 1. We show now how we may generalize the Law of the
Mean, I, 393.
Let f(x) be continuous in 2I=(a<£>). Let m, M be the mini
mum and maximum of one of the four derivates of f in 51. Then for
— a
To fix the ideas let us take Rf'(x) as our derivate. Suppose
now there exists a pair of points a</3 in SI, such that
/(«>= M +e , e>0.
506 DERIVATES, EXTREMES, VARIATION
We introduce the auxiliary function
</>(*) =/<X>-(^+<0*, (2
where 0<c<e = c+S.
Then -K/3) - »(«) = /Q8) -/(«) ( Jf ; . _ ^
(3 — a /3 — a
Henoe
Consider now the equation
<K/3) -<£(*) = >?•
It is satisfied for x = a. If it is satisfied for any other x in the
interval (a/3), there is a last point, say x = 7, where it is satisfied,
by 458, 3.
Thus for *> 7, <K*0 is >(/>(<*)•
Hence
(3
Now from 2) we have
Hence M is not the maximum of Rf'(x) in 51. Similarly the
other half of 1) is established. The case that m or M is infinite
is obviously true.
2. Letf(x) be defined over 51 = (# < b). Let a1 < «2 < ••• < an lie
in 21. Let m and M denote the minimum and maximum of the dif
ference quotients
A(ara2) , A(a2, aB) , ••• A(an_1, aj.
Then m<^^an)<M. (1
For let us first take three points a < @ < 7 in 2(. We have iden
tically * a _
7).
Now the coefficients of A on the right lie between 0 and 1.
Hence 1) is true in this case. The general case is now obvious.
DERIVATES 507
507. 1. Let f(x) be continuous in H = (a < 6). The four deri-
vates off have the same extremes in 21.
To fix the ideas let
Min L = m , Min R = n, in 21.
We wish to show that m = p. To this end we first show that
For there exists an a in 21, such that
L(a) < m + e.
There exists therefore a 0 < a in 21, such that
a — /3
Now by 506, l,
ft = Min R<q.
Hence ^ < w,
as e>0 is small at pleasure.
TFe s^o^ wow ^Aa^ m< u, (2
For there exists an a in H, such that
R(a) < fji + €.
There exists therefore a /3 > « in 21, such that
a— /3
Thus by 506, l,
m= Min ^
Hence as before m<p. From 1), 2) we have m = /i.
2. In 499, we emphasized the fact that the left-hand derivates
are not defined at the left-hand end point of an interval, and the
right-hand derivates at the right-hand end point of an interval
for which we are considering the values of a function. The fol
lowing example shows that our theorems may be at fault if this
fact is overlooked.
508 DERIVATES, EXTREMES, VARIATION
Example. Let/(V) = j x \.
If we restrict x to lie in 21 = (0, 1), the four derivates = 1 when
they are defined. Thus the theorem 1 holds in this case. If,
however, we regarded the left-hand derivates as denned at x = 0,
and to have the value
i/'(0) = - 1,
as they would have if we considered values of / to the left of 21,
the theorem 1 would no longer be true.
For then Min £ = - 1 , Min ]R = + 1,
and the four derivates do not have the same minimum in SI.
3. Let f (x) be continuous about the point x= c. If one of its
four derivates is continuous at x = c, all the derivates defined at this
point are continuous, and all are equal.
For their extremes in any Fs(tf) are the same. If now R is
continuous at x = c,
R(c) - e < R(x) < R(c) + e,
for any x in some F5(c).
4. Let f(x) be continuous about the point x = c. If one of its
four derivates is continuous at x = c, the derivative exists at this
point.
This follows at once from 3.
.Remark. We must guard against supposing that the derivative
is continuous at x = c, or even exists in the vicinity of this point.
Example. Let F(x) be as in 501, Ex. 1. Let
21 = (0, 1) and <g = | -
Let
Then RH'(x) = 2 ^(a;) + x*RF'(x),
LH'(x) = 2 xF(x) + x*LF'(x~).
Obviously both 725"' and 1/5' are continuous at x = 0 and
5'(0) = 0. But H' does not exist at the points of (g, and lience
DERIVATES 509
does not exist in any vicinity (0, 8) of the origin, however small
S > 0 is taken.
5. If one of the derivates of the continuous function f(x) is
continuous in an interval El, the derivative fr(&) exists, and is con
tinuous in El.
This follows from 3.
6. If one of the four derivates of the continuous function f \x) is
= 0 in an interval El, f(x) = const in El.
This follows from 3.
508. 1. If one of the derivates of the continuous function f(x) is
> 0 in El = (a < £>), f(x) is monotone increasing in El.
For then m = Min Rf > 0, in (a < z). Thus by 506, i,
2. If one of the derivates of the continuous function f(x) is >_ 0
in El, f(%) is monotone decreasing.
3. If one of the derivates of the continuous function f(x) is > 0
in El, ivithout being constantly 0 in any little interval of El, f(x) is
an increasing function in El. Similarly f is a decreasing function
in El, if one of the derivates is <^ 0, without being constantly 0 in any
little interval of El.
The proof is analogous to I, 403.
509. 1. Letf(x) be continuous in the interval El, and have a deriv
ative, finite or infinite, within El. Then the points where the deriva
tive is finite form a pantactic set in El.
For let a < ft be two points of El. Then by the Law of the
Mean,
— a
As the right side has a definite value, the left side must have.
Thus in any interval (a, ft) in El, there is a point 7 where the
differential coefficient is finite.
510 DERIVATES, EXTREMES, VARIATION
2. Let f(x) be continuous in the interval 2l = O<6). Then
Uf (x) cannot be constantly -f oo, or constantly — oo in 51.
For consider
which is continuous, and vanishes for x = a, x = b. We observe
that <£O) differs from f(x) only by a linear function. If now
£/f'(V)=-f-cc constantly, obviously U<f>'(x) = + oo also. Thus </>
is a univariant function in 51. This is not possible, since <£ has
the same value at a and b.
3. ie£ /(z) fo continuous in 51 = (a < 6), awe? Aave a derivative,
finite or infinite, in 51= (a*, 6).
For the Law of the Mean holds, hence
A
Letting now A = 0, we get the theorem.
Remark. This theorem answers the question : Can a continu
ous curve have a vertical tangent at a point x = a, if the deriva
tives remain < M in V*(a) ? The answer is, No.
4. Let f(x) be continuous in 5t = (a < 5), and have a derivative,
finite or infinite, in &* = («*, b). Iff (a) exists, finite or infinite,
there exists a sequence «j > «2 > • • • = a in 51, such that
/'(«) = lim/'(«n)- (1
»=»
For /(a+AWfq), , „<„,<, + ,. (2
Let now A range over ^ > A2 > ••• = 0. If we set «n= «//n , the
relation 1) follows at once from 2), since /'(a) exists by
hypothesis.
510. 1. A right-hand derivate of a continuous function f(x)
cannot have a discontinuity of the 1° kind on the right. A similar
statement holds for the other derivates.
DERIVATES 511
For let R(x} be one of the right-hand derivates. It it has a
discontinuity of the 1° kind on the right at x — a, there exists a
number I such that
/ —e<_R(x)<^ I -f e , in some (a < a -f S) .
Then by 50G, 1,
h
Hence R(a)=l,
and R(jc) is continuous on the right at x = a, which is contrary
to hypothesis.
2. It can, however, have a discontinuity of the 1° kind on the
left, as is shown by the following
Example. Let/(ar) = | x = + Vx2 , in2I = (-l,l).
Here E(x) = + 1 , f or x > 0 in 21
= - 1 , for x < 0.
Thus at x = 0, R is continuous on the right, but has a discon
tinuity of the 1° kind on the left.
3. Let f(x) be continuous in 31 = («, 5), and have a derivative,
finite or infinite, in H* =(a*, £*). Then the discontinuities of f (x)
in 51, if any exist, must be of the second kind.
This follows from 1.
Example. /0c) = aa8inl , for x* 0 in 51 = (0, 1)
= 0 , for x = 0.
Then -, -,
f(x) = 2 a: sin - - cos - , x=^0
x x
= 0 , a?=0.
The discontinuity of /'(#) at x = 0, is in fact of the 2° kind.
4. Let f(x) be continuous in 2l = (a<6), except at x= a, which
is a point of discontinuity of the 2° kind. Let f (x) exist, finite or
infinite, in (a*, 6). Then x = a is a point of infinite discontinuity
*//'(*>
512 DERIVATES, EXTREMES, VARIATION
For if
p = R lim/ O) , q = R lim/ (»,
x==a *=a
there exists a sequence of points «1>«2 >•••=«, such that
/(an)=jt>; and another sequence /^ >y£2 > ••• = «, such that
/(/3n) = 9. We may suppose
«n>/3» , oran<^n , rc=l, 2, ...
Then the Law of the Mean gives
where 7n lies between «n, /3n. Now the numerator =p — q, while
the denominator = 0. Hence Qn = + oo , or — oo , as we choose.
5. Let f (a?) have a finite unilateral differential coefficient U at
each point of the interval 31. Then U is at most pointwise discon
tinuous in 31.
For by 474, 3, U is a function of class 1. Hence, by 486, l, it is
at most pointwise discontinuous in 31.
511. Let f (x) be continuous in the interval (a < £»). Let R(x)
denote one of the right-hand derivates of f(x). If R is not con
tinuous on the right at a, then
Q/JnPYP
I = R lim R(x) , m — R lim R(x) , x = a.
To fix the ideas let R be the upper right-hand derivate. Let us
suppose that a = Rf'(a) were >m. Let us choose 77, and c such
that
<a. (2
We introduce the auxiliary function
<£O)= cx-f(x).
1 ) = c-3f'(x). (3
Now if 8 > 0 is sufficiently small,
Rf'(x)<m + r? , for any x in 31* = (a*, a + 8).
DERIVATES 513
Thus 2), 3), show that
R<t>'(x)>(T , <r>0.
Hence <$>(x) is an increasing function in 51*. But, on the other
hand' 4/" oo =£/"(«),
since a > m. Hence
'a=c - «< 0.
Hence <j> is a decreasing function at x = a. This is impossible
since </> is continuous at a. Thus a<^m.
Similarly we may show that I < a.
512. 1. Let f(x) be continuous in 51 = (a < £>), and have a
derivative, finite or infinite. //"«=/'(«), £=/'(£), then f (x)
takes on all values between a, yS, as a: ranges over 51.
For let « < 7 < /3, and let
h>0.
We can take h so small that
<?O, A)<7 , and
Now $(6, -»). Q(b-h,h).
Hence ^7 z, IN .
§(6 -h,h)> 7.
If now we fix A, () (#, ^) is a continuous function of #. As Q
is < 7, for x = a, and > 7, for z = b — h, it takes on the value 7
for some a;, say for # = £, between a, b — h. Thus
But by the Law of the Mean,
<K6 *)=
where , ,
«<f<^<? + ^<^.
Thus/f (x) = 7, at x = 77 in 51.
2. ief /(#) 3e continuous in the interval 51, awe? admit a deriva
tive, finite or infinite. If /'(x) = 0 in 51, except possibly at an
enumerable set (£, ^£71 /' = 0 a/so t'w (S.
514 DER1VATES, EXTREMES, VARIATION
For if /'(«) = 0, and /'(£) = b * 0, then f (x) ranges over all
values in (0, b), as x passes from a to ft. But this set of values
has the cardinal number c. Hence there is a set of values in
(a, ft) whose cardinal number is c, where f'(x) =£ 0. This is
contrary to the hypothesis.
3. Let f(x), g(x) be continuous and have derivatives, finite or
infinite, in the interval 51. // in 51 there is an a for which
/'«> </(<*)'
and a ft for which
then there is a 7 for which
provided gw=/(20
has a derivative, finite or infinite.
For by hypothesis
8'(«)>0 ,
Hence by 1 there is a point where 8' = 0.
513. 1. If one of the four derivates of the continuous function
f(x) is limited in the interval 51, all four are, and they, have the
same upper and lower R-integrals.
The first part of the theorem is obvious from 507, 1. Let us
effect a division of 51 of norm d. Then
Cli = lim 2M& , M, = Max ^t in d,.
J% <M)
But the maximum of the three other derivates in d, is also M, by
507, 1. Hence the last part of the theorem.
2. Let f(x) be continuous and have a limited unilateral derivate
m=&. Then
f *&**</<?) -/(«) <
J.a
For let a < al «*2 < ••• < b determine a division of 51, of norm d.
DERIVATES 515
Then by 506, i,
in the interval (am, am+1) = dm.
Hence
2dm Min -B<6 -« < 2 Max ^.
Letting <#= 0, we get 1).
3. If f(x) is continuous, and Uf is limited and R-integrable in
JTI7 =
514. 1. Letf(x) be limited in 51 = (a < 5),
F(x)= ffdx , a<2:<
«/a
67 lira /< Z/JP'(M) < ^ lira/, (1
*=« «=M
/br a?i^ u within 51.
To fix the ideas let us take a right-hand derivate atx=u. Then
£u+k
fdx < h Max / , in (11 *, u + A), 7i > 0.
Thus
Letting h = 0, we get
x=u *=u '
which is 1) for this case.
2. Let f(x) be limited in the interval 51 = (a < 6). /f/O + 0)
R derivative Cfdx = f(x + 0) ;
*/a
and iff(x — 0) exists, a<x<b
L derivative I fdx =/(# — 0).
•Ja
516 DERIVATES, EXTREMES, VARIATION
3. Let f(x) be limited and R-integrable in 31 =(«<#). The
points where
F(x= dx , a<x<b
does not have a differential coefficient in tyi/orm a null set.
For F<x) =/O) by I, 537, 1,
when/ is continuous at x. But by 462, 6, the points where/ is
not continuous form a null set*.
515. In I, 400, we proved the theorem :
Let f(x) be continuous in 31 = (a < £>), and let its derivative
= 0 within 31. Then /is a constant in 31. This theorem we have'
extended in 507, 6, to a derivate of f(x). It can be extended still
farther as follows :
1. (L. Scheefer}. If f(x) is continuous in 31 = (a< 6), and if
one of its derivates = 0 in 31 except possibly at the points of an
enumerable set (g, then f = constant in 31.
If /is a constant, the theorem is of course true. We show that
the contrary case leads to an absurdity, by showing that Card (5
would = c, the cardinal number of an interval.
For if / is not a constant, there is a point c in 31 where
p=f(c)—f(a) is =#=0. To fix the ideas let jt?>0; also let us
suppose the given derivate is R = Rf'(x).
Let g(x,f)=f(x)-f<ji)-t(x-a) , t>0.
Obviously g \ is the distance/ is above or below the secant line,
y = t(x- «)+/(«).
Thus in particular for any £,
g(a, £)=0 , g(c>t) = p-t(c-a).
Let q > 0 be an arbitrary but fixed number < p. Then
g(c, t)-q=p -q-t(c-a)
if t < T, where
T=
c—a
DERIVATES 517
Hence
g(c, 0 > q
for any t in the interval £ = (T, T), 0 < T < T7. We note that
Card £ = c.
Since for any £ in £, #(a, £) = 0, and g(c, t) > q, let x = et be
the maximum of the points < c where g(x, t) = q. Then e < <?,
and for any A such that e + h lies in (e, c),
Hence
Thus for any £ in £, et lies in (g. As t ranges over !£, let 0*
range over (^ < (g. To each point e of (^ corresponds but one
point t of X. For
0 =g(e, f) - g(e, *') = (« - ^')(« ~ «)•
Hence .
f = r , as e > a.
Thus Card £ = Card @x < Card <g,
which is absurd.
2. Let f (x) be continuous in 5( = (a<5). Let § denote the
points of 51 wAere owe o/ tfAe derivates has one sign. If £ exists,
Card (E = c, ZAtf cardinal number of the continuum.
The proof is entirely similar to that in 1. For let c be a point
of (£. Then there exists a d > c such that
We now introduce the function
£(s»0=/(*)-/(<0-*<>-<0 , *>o,
and reason on this as we did on the corresponding g in 1, using
here the interval (c, c?) instead of (a, 5). We get
Card ^ = Card £ = c.
3. Letf(oi), g(x) be continuous in the interval 21. Ze£ a jt?az> of
corresponding derivates as Rf, Rg' be finite and equal, except pos
sibly at an enumerable set (g. Then f — g + C, in 21, where C is a
constant.
518 DER1VATES, EXTREMES, VARIATION
For let ^=f_g f ^_/B
Then in A = %-&
R(f>f > Rf-Rg' = o , Jfy' > o.
But if R(f>' < 0 at one point in 21, it is < 0 at a set of points 33
whose cardinal number is c. But $8 lies in (g. Hence R<f> is
never < 0, in 21. The same holds for ty. Hence, by 508, </> and
i/r are both monotone increasing. This is impossible unless
$ == a constant.
516. The preceding theorem states that the continuous function
/(#) in the interval 21 is known in 21, aside from a constant, when
/' (x) is finite and known in 21, aside from an enumerable set.
Thus f(x) is known in 21 when /' is finite and known at each
irrational point of 21.
This is not the case when/7 is finite and known at each rational
point only in 21.
For the rational points in 21 being enumerable, let them be
PI* »2' rs'~ 0-
Let z = z1 + /2+z3+...
be a positive term series whose sum I is < 21. Let us place rx
within an interval ^ of length < ^ . Let r be the first number
in 1) not in Bv Let us place it within a non-overlapping interval
S2 of length < Z2, etc.
We now define a f unction /(V) in 21 such that the value of /at
any x is the length of all the intervals and part of an interval
lying to the left of x. Obviously f(x) is a continuous function of
x in 21. At each rational point /' (x) = 1. But f(x) is not de
termined aside from a constant. For 2£n < I. Therefore when
I is small enough we may vary the position and lengths of the
S-intervals, so that the resulting /'s do not differ from each other
only by a constant.
517. 1. Let f(x) be continuous in 21 = (a < 5) and have a finite
derivate, say Rf', at each point of 21. Let S denote the points of 21
where R has one sign, say > 0. If (S exists, it cannot be a null set.
DERIVATES 519
For let c be a point of (£, then there exists a point d > c such
that
Let (£n denote the points of (£ where
n _ 1 < Rf' < n. (2
Then (£ = ^ + (£2 4- ••• Let 0 < <? < p. We take the positive
constants qv q2'" such that
If now (£ is a null set, each (£m is also. Hence the points of (£m
can be inclosed within a set of intervals Smn such that 2Smn < <?„».
n
Let now qm (a:) be the sum of the intervals and parts of intervals
$m, m w = 1, 2 ••• which lie in the interval (a < x). Let
Obviously Q(x) is a monotone increasing function, and
0 <<?<»<?• (3
Consider now
p<v>=/O) -/(«)- «(^).
We have at a point of 51 — (S,
Hence at such a point
EP'
But at a point x of (£, jRP' < 0 also. For x must lie in some
(£m, and hence within some Smn. Thus qm(%) increases by at least
Az when x is increased to x + Az. Hence mqm(x), and thus
(?(V) is increased at least wAz. Thus
' <Rf< -m<Q, by 2),
520 DERIVATES, EXTREMES, VARIATION
since x lies in (£m. Thus RP' < 0 at any point of 51. Thus P is
a monotone decreasing function in 51, by 508, 2. Hence
P(c)-P(d) > 0.
Hence /<*) - /(<*)- i « (0 - Q W S > 0,
or using 1), 3)
p - q < o,
which is not so, as p is > <?.
2. (Lebesgue.) Let f(x), g (x) be continuous^in the interval 5(,
and have a pair of corresponding derivates as Rf, Rg' which are
finite at each point of 51, and also equal, the equality holding except
possibly at a null set. Thenf(x) — g(x) = constant in 51.
The proof is entirely similar to that of 515, 3, the enumerable
set (£ being here replaced by a null set. We then make use of 1.
518. Letf(x) be continuous in some interval A = (u — S, u -f- 8).
Letf'(x) exist, finite or infinite, in A, but be finite at the point x = u.
Then
where
Let us first suppose that/'(V) = 0. We have for 0<h<rj<B,
f(u-h)-f(ju)\
-h }
, u<x'<u + h , u-h<x"<u
h
where |e'|, | e"| are < e/2 for ?; sufficiently small.
Now ^ - u ^ -, I x" - u | < -,
~T ~h~~
while /"(^) = 0 , by hypothesis.
Hence \Qf\<€ » for
and 1) holds in this case.
MAXIMA AND MINIMA 521
Suppose now that f" (u) = a ^ 0. Let
</(V) =/(V) — q(x) , where q(x) = -J- ax2 + bx + c.
Since q"(u) = a , #"(»=0.
Thus we are in the preceding case, and lim Qg = 0.
Bat Qg=Qf-Qq-
Hence lim Qf= a.
Maxima and Minima
519. 1. In I, 466 and 476, we have denned the terms f(x) as
a maximum or a minimum at a point. Let us extend these terms
as follows. Let/^j ••• #m) be defined over 51, and let x= a be an
inner point of 21.
We say f has a maximum at x = a if I0,/ (a) — f(x) > 0, for any
x in some V(a), and 2°, f(a) —f(x) > 0 for some x in any F(a).
If the sign > can be replaced by > in 1°, we will say / has a
proper maximum at a, when we wish to emphasize this fact; and
when > cannot be replaced by >, we will -say / has an improper
maximum. A similar extension of the old definition holds for
the minimum. A common term for maximum and minimum is
extreme.
2. If f(x) is a constant in some segment 53, lying in the inter
val 51, 53 is called a segment of invariability, or a constant segment
of /in «.
Example. Let f(x) be continuous in 51 = (0, 1*).
Lefc x= • flj^ag ••• (1
be the expression of a point of 51 in the normal form in the dyadic
system. Let -
be expressed in the triadic system, where an=an, when an — 0,
and =2 when an = l. The points (£ = S£5 form a Cantor set,
I, 272. Let j3nj be the adjoint set of intervals. We associate
522 DERIVATES, EXTREMES, VARIATION
now the point 1) with the point 2), which we indicate as usual by
x~%. We define now a function g(x) as follows :
#(O=/O) , when a;- f.
This defines g for all the points of (£. In the interval 3n, let g
have a constant value. Obviously g is continuous, and has a
pantactic set of intervals in each of which g is constant.
3. We have given criteria for maxima and minima in I, 468
seq., to which we may add the following :
Let f(x) be continuous in (a— 8, a + £). If Rf (a) > 0 and
Lf] '(#)< 0, finite or infinite, f(x) has a minimum at x = a.
If Ef'(a)<0 and Lf'(a)> 0, finite or infinite, f(x) has a maxi
mum at x = a.
For on the 1° hypothesis, let us take a such that R — a > 0.
Then there exists a 8' > 0 such that
Similarly if fi is chosen so that L + /5< 0, there exists a S" > 0,
suchthat
Hence ^ « + ^ in (a _ 5^, a»).
520. Example 1. Let /(a?) oscillate between the #-axis and the
two lines y = x and y = — x, similar to
y
• 7T
zsm —
In any interval about the origin, y oscillates infinitely often, hav
ing an infinite number of proper maxima and minima. At the
point # = 0,/has an improper minimum.
Example 2. Let us take two parabolas Pl, P2 defined by y — x*,
y = 2x2. Through the points x=±\, ±\-~ let us erect ordi-
nates, and join the points of intersection with Px, P2, alternately
by straight lines, getting a broken line oscillating between the
MAXIMA AND MINIMA 523
parabolas Pl , P2 . The resulting graph defines a continuous func
tion f(x) which has proper extremes at the points (£ = j ± - \ •
However, unlike Ex. 1, the limit point x= 0 of these extremes is
also a point at which f(x) has a proper extreme.
Example 3. Let \S\ be a set of intervals which determine a
Harnack set ^ lying in 51 = (0, 1). Over each interval S = (a, /3)
belonging to the nih stage, let us erect a curve, like a segment of
a sine curve, of height hn = 0, as n = oo, and having horizontal
tangents at a, /3, and at 7, the middle point of the interval S. At
the points jf j of 51 not in any interval 8, let/(V) = 0. The func
tion/ is now defined in 51 and is obviously continuous. At the
points \y[,f has a proper maximum ; at points of the type a, y8,
f, /has an improper minimum. These latter points form the set
& whose cardinal number is c. The function is increasing in each
interval (a, 7), and decreasing in each (7, /3). It oscillates in
finitely often in the vicinity of any point of &.
We note that while the points where / has a proper extreme
form an enumerable set, the points of improper extreme may form
a set whose cardinal number is c.
Example 4- We use the same set of intervals \&\ but change
the curve over S, so that it has a constant segment 77 = (\, JJL) in its
middle portion. As before /=0, at the points £ not in the
intervals 8.
The function f (x) has now no proper extremes. At the points
of ^p, / has an improper minimum ; at the points of the type \, /*, it
has an improper maximum.
Example 5. Weierstrass Function. Let (5 denote the points in
an interval 51 of the type
x = — , r, s, positive integers.
0s
For such an x we have, using the notation of 502,
bmx = im + f m = b»-'r.
Hence fm = 0 , formes.
Thus e= -l+i= _i+i.
524 DERIVATES, EXTREMES, VARIATION
A Ti1
Hence sgn — = sgn Q = sgn emr)m = sgn ( -
Thus sgn
if r is even, and reversed if r is odd. Thus at the points @, the
curve has a vertical cusp. By 519, 3, F has a maximum at the
points (£, when r is odd, and a minimum when r is even. The
points (£ are pantactic in 51.
Weierstrass' function has no constant segment 8, for then
f'(x) = 0 in B. But ^' does not exist at any point.
521. 1. Let f(xl "- xm) be continuous in the limited or unlimited
set 51. Let & denote the points of 51 where f has a proper extreme.
Then d: is enumerable.
Let us first suppose that 51 is limited. Let S > 0 be a fixed
positive number. There can be but a finite number of points a in
51 such that
For if there were an infinity of such points, let ft be a limiting
point and 77 < J S. Then in F^(/3) there exist points «', a" such
that F^(a'), Fg(«") overlap. Thus in one case
and in the other
which contradicts the first.
Let now Sj > S2 > -•• =0. There are but a finite number of
points a for which 1) holds for S = Sj, only a finite number for
£ = &,,, etc. Hence d: is enumerable. The case that 51 is unlim
ited follows now easily.
2. We have seen that Weierstrass' function has a pantactic set
of proper extremes. However, according to 1, they must be
enumerable. In Ex. 3, the function has a minimum at each point
of the non-enumerable set ^p; but these minima are improper. On
the other hand, the function has a proper maximum at the points
57}, but these form an enumerable set.
MAXIMA AND MINIMA 525
522. 1. Let f(x) be continuous in the interval 51. Letf have a
proper maximum at x = a, and x = ft in 51. Then there is a point 7
between a, ft where f has a minimum, which need not however be a
proper minimum.
For say a < ft. In the vicinity of a, f(x) is </(«) ; also in
the vicinity of ft, f(x) is </(/8). Thus there are points 53 in
(a, yS) where /is < either /(a) or/(y£). Let /it be the minimum
of the values otf(x)i as # ranges over 53. There is a least value
of # in (a, /3) for which f(x) = /z. We may take this as the
point in question. Obviously 7 is neither a nor ft.
2. That at the point 7, / does not need to have a proper mini
mum is illustrated by Exs. 1, or 3.
3. In 51 = (#, 6) let f(x) exist, finite or infinite. The points
within 51 at which f has an extreme proper or improper, lie among
the zeros of f (x).
This follows from the proof used in I, 468, 2, if we replace there
< 0, by < 0, and > 0, by > 0.
4. Let f (x) be continuous in the interval 51, and let f(x) have
no constant segments in 51. The points & of 51 ivhere f has an ex
treme, form an apantactic set in 51. Let 3 denote the zeros of f (x)
in 51. If 53 = fbnj is the border set of intervals lying in 51 corre
sponding to &f(x) is univariant in each bn.
For by 3, the points (g lie in 3- As f'(x) is continuous, S *s
complete and determines the border set 53- Within each bn,
f (x) has one sign. Hence f(x) is univariant in bn.
5. Letf(x) be a continuous function having no constant segment
in the interval 51. If the points Q, ivhere f has an extreme form a
pantactic set in 51, then the points 53 ivhere f(x) does not exist or is
discontinuous, form also a pantactic set in 51.
For if 53 is not pantactic in 51, there is an interval (£ in 51
containing no point of 53. Thus f'(x) is continuous in (E. But
the points of @ in (£ form an apantactic set in (5 by 4. This,
however, contradicts our hypothesis.
Example. Weierstrass' function satisfies the condition of the
theorem 5. Hence the points where F' (x) does not exist or is
526 DERIVATES, EXTREMES, VARIATION
discontinuous form a pantactic set. This is indeed true, since
F' exists at no point.
6. Let f(x) be continuous and have no constant segment in the
interval 51. Let /'(#) exist, finite or infinite. The points where
f'(x) is finite and is =^0form a pantactic set in 51.
For let a < ft be any two points in 51. If /(«) =/(/3), there is
a point a < 7 < ft such that /(«) T^/Oy), since / has no constant
segment in 51. Then the Law of the Mean gives
a-y
Thus in the arbitrary interval (a, ft) there is a point f, where
f (x) exists and is =£ 0.
7. Let f(x) be continuous in the interval 51. Then any interval
53 in 51 which is not a constant segment contains a segment (E in which
f is univariant.
For since / is not constant in 53, there are two points a, b in 53
at which /has different values. Then by the Law of the Mean
/(«)-/(*)=(«- ») AO . cin»-
Hence f (c) =£ 0. As f'(x) is continuous, it keeps its sign in
some interval (c — 8, c -f 8), and /is therefore univariant.
523. Letf(x) be continuous in the interval 51, and have in any in
terval in 51 a constant segment or a point at which f has an extreme.
If f (x) exists, finite or infinite, it is discontinuous infinitely often in
any interval in 51, not a constant segment. At a point of continuity
of the derivative, f (x) = 0.
For if f (x) were continuous in an interval 53, not a constant
segment, / would be univariant in some interval (£:<53, by 522, 7.
But this contradicts the hypothesis, which requires that any inter
val as (£ has a constant segment. Hence f'(x) is discontinuous
in any interval, however small.
Let now x = c be a point of continuity. Then if c lies in a con
stant segment, /'(V) = 0 obviously. If not, there is a sequence of
points e1, ez ••• = e such that f(x) has an extreme at en. But then
/'(en)=0, by 522,3. As/'<» is continuous at z = <?,/'(»= 0
also.
MAXIMA AND MINIMA ,527
524. (Konig.) Letf^x) be continuous in 51 and have a pantactic
set of cuspidal points (L Then for any interval $8 of $[, there exists
a ft such that f(x) = ft at an infinite set of points in 33. Moreover,
there is a pantactic set of points \%\ in 33, such that k being taken at
For among the points & there is an infinite pantactic set c of
proper maxima, or of proper minima. To fix the ideas, suppose
the former. Let x = c be one of these points within 33. Then
there exists an interval b <.33, containing c, such that
/(<?) >/(#) , for any x in b.
Let fji = Min/(V), in b.
Then there is a point x where / takes on this minimum value.
The point c divides the interval b into two intervals. Let ( be
that one of these intervals which contains #, the other interval we
denote by m. Within m let us take a point c^ of c. Then in I
there is a point c[ such that
The point c1 determines an interval bx , just as c determined b.
Obviously bj^tn, and bx falls into two segments Ix , n^ as before
b did. Within ml we take a point of c. Then in ( there is a
point c%, and in (x a point c'%, such that
In this way we may continue indefinitely. Let
be the points obtained in this way which fall in (. Let c1 be a
limit point of this set. Let
be the points obtained above which fall in (j, and let cn be a limit
point of this set. Continuing in this way we get a sequence of
limiting points ^ cu cnr ... (2
lying respectively in (, (x, I2 ...
528 DERIVATES, EXTREMES, VARIATION
Since /is continuous,
- /OO =/(«")=/(<"')=- (3
Thus if we set /(<?') = fi we see that /(#) takes on the value (S at
the infinite set of points 2), which lie in 33.
Let 7j, 72 ••• be a set of points in 2) which = 7.
/(y) ~
p.
7 - 7i 7 - 72
Thus if /'(#) exists at # = 7, the equations 3) show that
= 0. If f does not exist at 7, they show that
Let now k be taken at pleasure. Then
g(x)=f(x)-kx
is constituted as/, and
?O) =/(*)-*•
This gives 1).
525. 1. Lineo- Oscillating Functions. The oscillations of a con
tinuous function fall into two widely different classes, accord
ing as f(x) becomes monotone on adding a linear function
l(x) = ax + 6, or does not.
The former are called lineo-oscillating functions. A continu
ous function which does not oscillate in 2f, or if it does is lineo-
oscillating, we say is at most a lineo-oscillating function.
Example 1. Let /(_e) = 8ina! , Z(>) = :r.
Ifwesefc y =/(*)+*(*)
and plot the graph, we see at once that y is an increasing function.
At the point x = TT, the slope of the tangent to f(x) = sin x is
greatest negatively, i.e. sinx is decreasing here fastest. But the
angle that the tangent to sin x makes at this point is — 45°, while
the slope of the line l(x) is constantly 45°. Thus at x = TT, y has
a point of inflection with horizontal tangent.
If we take l(x) = ax, a > 1, y is an increasing function, increas
ing still faster than before.
MAXIMA AND MINIMA 529
All this can be verified by analysis. For setting
y = sin x + ax , a > 1,
we sret /
y = a -f- cos #,
and , Q
Thus y is a lineo-oscillating function in any interval.
Example 2. f(x) = z2 sin - , x 3= 0
= 0 , x=0.
l(x) = ax + b , y =f(x) + l(x).
Then ^ 1
ni ' — . "/ />« QI r-j _- ("*(~)G | ft sy* —t-~ O
X X
= a , x = 0.
Hence, if a > 1 -f 2 TT, z/ is an increasing function in 51 = ( — TT, TT).
The function /oscillates infinitely often in 51, but is a lineo-oscil
lating function.
Example 8. /(#) = # sin - , x =£ 0
x
= 0 , x = 0.
Here 111
y' = sin cos — f- a , # ^= 0.
For x=Q, yr does not exist, finitely or infinitely.
Obviously, however great a is taken, y has an infinity of oscilla
tions in any interval about x = 0. Hence /is not a lineo-oscillat
ing function in such an interval.
2. If on e of the four derivates of the continuous function /(#) is
limited in the interval 51, f(x) is at most lineo-oscillating in 51.
For say Rf > - a in 51. Let 0 < a < £,
530 DERIVATES, EXTREMES, VARIATION
Then
Hence g is monotone increasing by 508, 1.
3. Letf^x) be at most lineo-oscillating in the interval 31. If Uf
does not exist finitely at a point x in 31, it is definitely infinite at the
point. Moreover, the sign of the oo is the same throughout 31.
For if / is monotone in 31, the theorem is obviously true. If
be monotone. Then
Uf = W - a,
and this case is reduced to the preceding.
Remark. This shows that no continuous function whose graph
has a vertical cusp can be lineo-oscillating. All its vertical tan
gents correspond to points of inflection, as in
Variation
526. 1. Letf(x) be continuous in the interval 31, and have limited
variation. Let D be a division of 31 of norm d. Then using the no
tation of 443,
\imVDf=Vf , \imPDf=Pf , YimNDf=Nf. (1
For there exists a division A such that
where for brevity we have dropped / after the symbol V. Let
now A divide 31 into v segments whose minimum length call X.
Let D be a division of 31 of norm d<d0<\. Then not more
than one point of A, say aK, can lie in any interval as (#t, «l+1) of
D. Let E = D + A, the division obtained by superposing A on D.
Then ft denoting some integer < i>,
V- r-i
VARIATION 531
If now dQ is taken sufficiently small, Osc/ in any interval of D
is as small as we choose, say < — . Then
6 v
But since E is got by superposing A on Z>,
Hence for any D of norm < c?0,
which proves the first relation in 1. The other two follow at
once now from 443.
527. // f(x) is continuous and has limited variation in the in
terval 31 = (a < 6), then
P(x) , N(x) , V(x)
are also continuous functions of x in 31.
Let us show that V(x) is continuous ; the rest of the theorem
follows at once by 443.
By 526, there exists a c?0, such that for any division D of norm
d<dQ, F"(6) = Fi(6) + e' , 0<e'<e/3.
Then a fortiori, for any x < b in 31,
V(X) = VD(X} + 6j , 0<€!<€/3. (1
In the division Z>, we may take x as one of the end points of
interval, and x + h as the other end point. Then
On the other hand, if dQ is taken sufficiently small,
\ffx + A) -f(x) I < ^ , f or 0 < h < B. (3
o
From 1), 2), 3) we have
0 < V(x + A) - V(x) < e , f or any 0 < h < 8. (4
an
532 DERIVATES, EXTREMES, VARIATION
But in the division .Z), x is the right-hand end point of some in
terval as (x —k,x). The same reasoning shows that
\V(x-K)-V(x)\<€ , foranyO<&<£. (5
From 4), 5) we see V(x) is continuous.
528. 1. If one of the derivates of the continuous function f (V) is
numerically < M in the interval 51, the variation Voffis <
For by definition y= Max y
with respect to all divisions D=\di\ of 5L Here
Now by 506, 1,
or
Hence ^ < M^ <
2. Letf(x) be limited and R-integrable in 51 =
has limited variation in 51.
For let D be a division of 51 into the intervals dt = (at,
Then
a»+i) -F(a,) I =2 | I /(fe
Thus Max VD - F
and F has limited variation.
529. 1. If f (x) has limited variation in the interval 51, the
points $ where Osc/> &, are finite in number.
For suppose they were not. Then however large G- is taken,
we may take n so large that nk > G-. There exists a division D
VARIATION 533
of H, such that there are at least n intervals, each containing a
point of $ within it. Thus for the division D,
Thus the variation of / is large at pleasure, and therefore is not
limited.
2. If f has limited variation in the interval 51, its points of dis
continuity form an enumerable set.
This follows at once from 1.
530. 1. Let Dj, D2 ••• be a sequence of superposed divisions, of
norms dn = 0, of the interval 51. Let £lDn be the sum of the oscilla
tions of f in the intervals of Dn. If Max flDn is finite, f(x) has
limited variation in 51.
For suppose / does not have limited variation in 51. Then
there exists a sequence of divisions E^, E^ «•• such that if £lEn is
the sum of the oscillations of /in the intervals of En, then
n^<n^< ... = +00. (i
Let us take v so large that no interval of Dv contains more than
one interval of En or at most parts of two En intervals. Let
Fn= En + Dv. Then an interval 8 of Dv is split up into at most
two intervals 8', 8" in Fn. Let o>, a/, co" denote the oscillation of
/in 8, £', 8". Then the term co in Dv goes over into
a' + co' ' < 2 co
in £lFn. Hence if Max £lDn = M,
n£n < 2 QDr < 2 M,
which contradicts 1).
2. Let VDn = 2 |/(X) — /(at+1) | , the summation extended
over the intervals (at, al+1) of the division Dn. If Max VDn is
n
finite with respect to a sequence of superposed divisions \Dn\, we
cannot say that /has limited variation.
Example. For let / (z) = 0, at the rational points in the inter
val 51 = (0, 1), and = 1, at the irrational points. Let Dn be
534 DERIVATES, EXTREMES, VARIATION
f) i -|
obtained by interpolating the points — — — in 21. Then /=0
at the end points a,, al+l of the intervals of Dn. Hence VDn = 0.
On the other hand, f(x) has not limited variation in 21 as is
obvious.
531. Let F(x) = lim/(>, t), r finite or infinite, for x in the
interval 21. Let Var f(x, t) < M for each t near r.
Then F(x) has limited variation in 21.
To fix the ideas let T be finite. Let
Then for a division D of 21,
VDF< VDf+ VDg.
But
) - g (a
where (#m, am+l) are the intervals of D.
But for some t = t' near r, each
where « is the number of intervals in the division D.
Tlms .
Hence
and F has limited variation.
532. Let f(x), g(x) have limited variation in the interval 21, then
their sum, difference, and product have limited variation.
If also l#l>7>° i in^I
thenf/g has limited variation.
Let us show, for example, that h=fg has limited variation.
Forlet Min/=m , Min<7 = n
in the interval d,.
Osc/= (o , Osc g = r
VARIATION 535
rhen / = m + «&> , g = n + £r , in rft,
0<«<1 , 0<£<1.
Thus fg = mn + ra/3r -f wao> -f a/Scwr.
Now
Tim — Til r —
a> — car < < < mn + m
G>T.
Hence 77 = Osc A < 2 jr | m | -f *> n\ + a>Tj.
Bufc |wi|, |w|, r<.some JST.
Thus TA < 4 7f 2a> + 2
< some 6r,
and A has limited variation.
533. 1. Let us see what change will be introduced if we
replace the finite divisions D employed up to the present by
divisions JE, which divide the interval 31 = (a < £>) into an infinite
enumerable set of intervals («t, al+i).
and
for the class of finite or infinite enumerable divisions
Obviously TT> F;
hence if IT is finite, so is F.
We show that if F is finite, so is W. For suppose W were
infinite. Then for any Q- > 0, there exists a division E, and an
w, such that the sum of the first n terms in 1) is > 6r, or
(2
Let now D be the finite division determined by the points #j
az '" an+i which figure in 2).
Then VB>a,
hence F= ao, which is contrary to our hypothesis.
536
DERIVATES, EXTREMES, VARIATION
We show now that V and W are equal, when finite. For let
E be so chosen that
W — € ^ W ^ W
~ \ rr f \ V r «
Now
6/2
if n is sufficiently large.
Let D correspond to the points a1 a2 ••• in WE^n. Then
and hence Y A- ' > W -4- ' — W
Hence W-Vj,<e.
We may therefore state the theorem :
2. Iff has limited variation in the interval 21 with respect to the
class of finite divisions Z>, it has with respect to the class of enumer
able divisions E, and conversely. Moreover
Max VD = Max VE.
534. Let us show that Weierstrass' function F, considered in
502, does not have limited variation in any interval 2( = (a < /3)
when ab > 1. Since F is periodic, we may suppose a. > 0. Let
be the fractions of denominator bm which lie in 21.
These points effect a division Dm of 21, and
,/=0
bm
-F(k-±L
\ ¥
If I is the minimum of the terms Fi under the S sign,
Now
Hence
bm
- 2.
NON-INTUITIONAL CURVES 537
On the other hand, using the notation and results of 502,
and also ( x
h
Let us now take
Then r_£+/ h _. 1
"W -&
Hence from 3), -^ > mf%_ ^
V3 ^6 - 1
ThUS r/>« ^ am(~ — )(&"5 - 2) , by 1), 2).
As a < 1, and «£> > 1, we see that
VDm = + GO, as w = oo .
Non-intuitional Curves
535. 1. Let /(#) be continuous in the interval 21. The graph
of / is a continuous curve 0. If / has only a finite number of os
cillations in 21, and has a tangent at each point, we would call 0 an
ordinary or intuitional curve. It might even have a finite num
ber of angle points, i.e. points where the right-hand tangent is
different from the left-hand one [cf. I, 366]. But if there were
an infinity of such points, or an infinity of points in the vicinity
of each of which / oscillates infinitely often, the curve grows less
and less clear to the intuition as these singularities increase in
number and complexity. Just where the dividing point lies be
tween curves whose peculiarities can be clearly seen by the intui
tion, and those which cannot, is hard to say. Probably different
persons would set this point at different places.
For example, one might ask : Is it possible for a continuous
curve to have tangents at a pantactic set of points, and no tangent
at another pantactic set? If one were asked to picture such a
curve to the imagination, it would probably prove an impossibility.
538
DERIVATES, EXTREMES, VARIATION
Yet such curves exist, as Ex. 3 in 501 shows. Such curves might
properly be called non-intuitional.
Again we might ask of our intuition : Is it possible for a con
tinuous curve to have a tangent at every point of an interval SI,
which moreover turns abruptly at a pantactic set of points ? Again
the answer would not be forthcoming. Such curves exist, how
ever, as was shown in Ex. 2 in 501.
We wish now to give other examples of non-intuitional curve's.
Since their singularity depends on their derivatives or the nature
of their oscillations, they may be considered in this chapter.
Let us first show how to define curves, which, like Weierstrass'
curve, have a pantactic set of cusps. To effect this we will extend
the theorem of 500, 2, so as to allow g(x) to have a cusp at x— 0.
536. Let (£ = \en\ denote the rational points in the interval
51= (—a, a). Let g(x) be continuous in 53 = ( — 2 a, 2 a), and
= 0, at x = 0. Let 53* denote the interval 53 after removing the
point x = 0. Let g have a derivative in 53*, such that
M
\g'(*)\<
Then
a> 0.
(1
0
is a continuous function in SI, and - - behaves at x = em essentially
A#
as —9- does at the origin.*
A#
To simplify matters, let us suppose that (5 does not contain the
origin. Having established this case, it is easy to dispose of the
general case. We begin by ordering the en as in 233. Then
obviously if
en = - t q > 0 , p positive or negative,
we have
Let
p r
q s
qs
mn
(2
* Cf. Dini, Theorie der Functional, etc., p. 192 seq. Leipzig, 1892.
NON-INTUITIONAL CURVES 539
Let E(x) be the F series after deleting the mth term. Then
We show that E has a differential coefficient at x = em, obtained
by differentiating E termwise. To this end we show that as h = 0,
(e- + ~grf*-) ,'•'»*• (3
converges to ff = 2a^'(O , m*n. < ' . '- (4
That is, we show
e>0 , rj>0 , |D(A)-# <e , 0<|A|<i;. (5
Let us break up the sums 3), 4) which figure in 5), into three
parts r s
2 = 2 + 2 + 2. (6
1 1 r+l 5+1
\D-a\< Dr-ar\ + \Dr,8-ar,a\ + \Dt-aa\ (7
< A + B + C.
Since g'(emn) exists, the first term may be made as small as we
choose for an arbitrary but fixed r ; thus
Let us noiv turn to B. We have
B<\Dr,
j(0 ,(
provided g1 (x) exists in the interval (emn, emn +
But by 2),
mn ms
if
Thus by 1),
| ^'(^mn 4- A') | < 2aMm«n« < M^* , ^ a constant.
540
Hence a fortiori,
Now the sum
DERIVATES, EXTREMES, VARIATION
, <
(9
converges if n > 0. Hence Rp> q and Hp may be made as small as
we choose, by taking p sufficiently large. Let us note that by 91,
#,<--. GO
pf>
Thus if /* = Min (a, /3),
B< I A.I + Iff,. I
for a sufficiently large r.
We consider finally O. We have
h s+i
From 9) we see that
<7
8 . -,
for s sufficiently large. Since g(x) is continuous in
Hence
1
JV
2f
if s_>-r— , on using 10).
I h\
Taking s still larger if necessary, we can make
Thus
NON-IXTUITIOXAL CURVES 541
The reader now sees why we broke the sum 6) into three parts.
As h = 0, the middle term contains an increasing number of terms.
But whatever given value h has, * has a finite value.
Thus as J., B, C are each < e/3, the relation 5) is established.
Hence E has a differential coefficient at x = em, and as
A(0)
r- — am - 7 -- -- T~>
h h h
our theorem is established.
537. Example 1. Let
g(x) = Vz2.
Then for x * 0, g' (x) = — . Here «
Thus
is a continuous function, and at the rational points em in the in
terval 31,
RF' (a;) = + oo , X^' (a;) = - oo.
Hence the graph of F has a pantactic set of cuspidal tangents
in 31. The curve is not monotone in any interval of 31, however
small.
^
g (x) = x sin - , x =£ 0
Example 2. Let
Then 1 , -j
#'<» = sin- - -cos- ,
XX X
Here a = 1. For x = 0,
542 DERIVATES, EXTREMES, VARIATION
Then 1 1
*(*)- 2 (*-«.) sin -J- , /3>0
is a continuous function in 51, and at the rational point e
w
where E is the series obtained from F by deleting the rath term.
538. Pompeiu Curves.* Let us now show the existence of
curves which have a tangent at each point, and a pantactic set of
vertical inflectional tangents.
We first prove the theorem (Borel) :
Let Ja
where (£ = \en\ is an enumerable set in the interval 51, and
is convergent. Then B converges absolutely and uniformly in a set
53 < 51, and S is as near 51 as we choose.
The points £) where adjoint B is divergent form a null set.
For let us enclose each point en in an interval Sn of length
with en as center.
The sum of these intervals is
f or k > 0 sufficiently large. Let now k be fixed. A point x of 51
will not lie in any Sn if
r — I r - P \ -> ^an
'n — \ •£ Vn | ^> ~T-'
Then at such a point,
k
Adjoint B < 2an—=. = £2 Van = kA.
Van
* Math. Annalen, v. 03 (1907), p. 326.
NON-INTUITIONAL CURVES 543
As S > 51 — e, the points £) where B does not converge ab
solutely form a null set.
539. 1. We now consider the function
F(x) = I an(x - erf = Z/»O) (1
i
where (5 = \en\ is an enumerable pantactic set in an interval 51, and
A = 2«n (2
is a convergent positive term series.
Then F is a continuous function of x in 51. For | x — en |* is <
some M in 51.
Let us note that each f n(x) is an increasing function and the
curve corresponding to it has a vertical inflectional tangent at the
point x = en .
We next show that F (x) is an increasing function in 51. For let
x' < x". Then
/.(*')</.(*")•
Hence
Hence
2. Let us now consider the convergence of
obtained by differentiating F termwise at the points of 21 — @.
Let 55 denote the points in 31 where
?» — (4
diverges. We have seen £) is a null set if
(5
544 DERIVATES, EXTREMES, VARIATION
is convergent. Let 51 = £) + (£. Let # be a point of (£, i.e. a
point where 4) is convergent. We break 3) into two parts
such that in Dr each fn < 1. Then D2 is obviously convergent,
since each of its terms
where =
x— e,
tf
and the series 2) is convergent.
The series Dl is also convergent. For as fn< 1, the term
M<l^
£n
and the series 4) converges by hypothesis, at a point x in (£.
Hence D(x) is convergent at any point in (£, and S, = 51 when 5) is
convergent.
3. Let (7 denote the points in 51 where 3) converges. Let
51= 0+ A.
J7e Tieatf sAow #A«£ F'(x) = D(x), for x in 0. For taking a; at
pleasure in (7 but fixed,
We now apply 156, 2, showing that Q is uniformly convergent
in (0*, ?;). By direct multiplication we find that
Thus 6) gives
.OXTX _
Let us set
Then
NOX-INTUITIONAL CURVES 545
for 0 < | h <_?/, 77 sufficiently small. As the series on the right is
independent of A, Q converges uniformly in (0*, 77). Thus
by 156, 2
F1 = D , for any x in C.
4. Let now x be a point of A, not in Q. At such a point we show
that
.F'(X)= + oo, (8
and thus the curve F has a vertical inflectional tangent. For as
D is divergent at #, there exists for each M an w, such that
A»>2 M.
But the middle term in 7) shows that for | h \ < some rjr each
term in Qm is > \ the corresponding term in Dm. Thus
M , 0<|A|<y.
Since each term of Q is > 0, as 7) shows,
Q(h) > M.
Hence 8) is established.
5. Let us finally consider the points x= em. If <1> denotes the
series obtained from F by deleting the mth term, we have
As J7 is increasing, the last term is > 0.
Hence J"(:r)=+oo , ing.
As a result ive see the curve F has at each point a tangent. At an
enumerable pantactic set V, it has points of inflection with vertical
tangents.
7. Let us now consider the inverse of the function F, which we
denote by
x=G(t). (9
As x in 1) ranges over the interval 51, t =F(x) will range over
an interval 53, and by I, 381, the inverse function 9) is a one-
valued continuous function of t in 53 which has a tangent at each
546 DERIVATES, EXTREMES, VARIATION
point of 53. If TTare the points in 53 which correspond to the
points V in 51, then the tangent is parallel to the £-axis at the
points W, or Gr' (£) = 0, at these points. The points W are pan-
tactic in 53.
Let Z denote the points of 53 at which Gr' (t) = 0. We show
that Z is of the 2° category, and therefore
Card£=c.
For G-1 '(£) being of class < 1 in 53? its points of discontinuity 5
form a set of the 1° category, by 486, 2. On the other hand, the
points of continuity of Gr' form precisely the set Z, since the
points W are pantactic in 53 and Gr' = 0 in W. In passing let us
note that the points Z in 53 correspond 1-1 to a set of points 3 afc
which the series 3) diverges. For at these points the tangent to
F is vertical. But at any point of convergence of 3), we saw in
2 that the tangent is not vertical.
Finally we observe that 3) shows that
MinD(aO>- • — 2an , in 51.
3 gf
Hence Q ^2
Max
Summing up, we have this result :
8. Let the positive term series 2V«n converge. Let (& = \en\ be
an enumerable pantactic set in the interval 51. The Pompeiu curves
defined by
F(x)=Zan(x-Crf
have a tangent at each point in 51, whose slope is given by
when this series is convergent, i.e. for all x in 51 except a null set.
At a point set 3 of the 2° category which embraces (5, the tangents
are vertical. The ordinates of the curve F increase with x.
540. 1. Faber Curves.* Let F(x) be continuous in the interval
${ == (0, 1). Its graph AVC denote by F. For simplicity let
* Math. Annalen, v. 66 (1908), p. 81.
NON-INTUITIONAL CURVES 547
i = 0, .F(l) = /0. We proceed to construct a sequence of
broken lines or polygons,
which converge to the curve F as follows :
As first line LQ we take the segment joining the end points of
F. Let us now divide 21 into n^ equal intervals
of length 1
and having
as end points. As second line L^ we take the broken line or
polygon joining the points on F whose abscissae are the points 3).
We now divide each of the intervals 2) into n2 equal intervals,
getting the n^nz intervals
of length
and having
(5
as end points. In this way we proceed on indefinitely. Let us
call the points
4-K.l
terminal points. The number of intervals in the rth division is
vr = nl- n2.- nr.
If Lm(x) denote the one-valued continuous function in 51 whose
value is the ordinate of a point on Lm, we have
, (6
since the vertices of Lm lie on the curve F.
2. For each x in 51,
m(x) = F(x). ' (7
m=x>
For if x is a terminal point, 7) is true by 6).
548 DERIVATES, EXTREMES, VARIATION
If x is not a terminal point, it lies in a sequence of intervals
£i>s2> •••
belonging to the 1°, 2° ••• division of 21.
Let 5> ,
0»= (am,m «m,tt+l)-
Since F(x) is continuous, there exists an «, such that
| F(x) -F(am, „) | <|, m>s (8
for any a: in Sm. As Lm(x) is monotone in Sm,
| Zm(20 - Lm(amn~) \ < | Lm(amn) - Xm(aw,n+1) |
.' ^| > by 8). -
|^(*)-^(«*n)|<|- (9
Hence from 8), 9),
which is 7).
3. Tfe m^ ^r^g 1) as a telescopic series. For
etc. Hence
.F
If we set
) = lim Z.(a:) = i0<» + 2
we have F(x) = %fn(x) ,
o .
and
The function /n(V), as 10) shows, is the difference between the
ordinates of two successive polygons Ln_^ Ln at the point x. It
may be positive or negative. In any case its graph is a polygon
NON-INT UITIOXAL CURVES 549
fn which has a vertex on the z-axis at the end point of each
interval 8n_lt Let In8 be the value of fn(x) at the point x = ana,
that is, at a point corresponding to one of the vertices of fn. We
call l^ the vertex differences of the polygon Ln.
Let
p. -Mini/.. | , ?n = Max|?n.|.
* 5
Then l/nOOl<9n , in 51. (13
In the foregoing we have supposed F(x) given. Obviously if
the vertex differences were given, the polygons 1) could be con
structed successively.
We now show :
2<?n (14
is consent,
is uniformly convergent in 51, and is a continuous function in 5T.
For by 13), 14), F converges uniformly in 51. As each fn(x)
is continuous, F is continuous in 51.
The functions so defined may be called Faber functions.
541. 1. We now investigate the derivatives of Faber's functions,
and begin by proving the theorem :
If 2n1 — n.q.=*I,v.qt (1
converge, the unilateral derivatives of F(x) exist in 51 = (0, 1) . More
over they are equal, except possibly at the terminal points A= \amn\.
For let x be a point not in A. Let x', x" lie in V= V*(x) ;
lQtxf-x=h',x" -x=h".
Let Q-F(x')-F^
V h"
Then F'(x) exists at x, if
e>0 , 7;>0 , \Q\<e , for any x', x" in V. (2
550
Now
1 Q\ <
DER1VATES, EXTREMES, VARIATION
Fm(x')-Fm(x) Fm(x")-Fm(x) Fm(x*)-Fm(x)
h' h" '
+
h1
h"
But
Hence
Similarly
/.<>') -/.(*)
x'-x
00 _
< 2 2zy?«< - •> m sufficiently large.
Finally, if 77 is taken sufficiently small, x, x1 ', x" will correspond
to the side of the polygon Lm . Hence using 540, 12), we see
that Ql = 0. Thus 2) holds, and F' (x) exists at x.
If x is a terminal point #mn, and the two points x', x" are taken
on the same side of amn , the same reasoning shows that the uni
lateral derivatives exist at amn . They may, however, be different.
2. Let n^ = n2 = ••• = 2. For the differential coefficient F'(x) to
exist at the terminal point x, it is necessary that
Um2n^n<oo. (3
*f Em 2npn = oo, (4
the points where the differential coefficient does not exist form a
pantactic set in SI.
Let us first prove 3). Let 6< a< c be terminal points. Then
they belong to every division after a certain stage. We will
therefore suppose that 5, c are consecutive points in the nth
division, and a is a point of the n + 1st division falling in the
interval Sn = (5, c). If a differential coefficient is to exist at a,
(5
a — b
must be numerically less than some J/, as n = oo, and hence their
sum Q remains numerically <2M.
NON-INTUITIONAL CURVES 551
Now
.P (a) =£„+,(«) ,
Thus Q = 2»+i S2 in+1O) - [Z.(6) +
or | § | = 4 • 2nZn)8 , supposing a = ana.
Hence 2n?n < JJf,
which establishes 3).
Let us noiv consider 4). By hypothesis there exists a sequence
< ••• = oc, such that
(r being large at pleasure. Hence at least one of the difference
quotients 5) belonging to this sequence of divisions is numerically
large at pleasure.
3' If X = 2?m. (1
i« absolutely convergent, the functions F(x) have limited variation in
&
For/m(#) is monotone in each interval B^. Hence in 8^
Var/m -|C- ^,a+i I < Km. | + | ?m,.+i I-
Hence in 51, Var /„(*)< 2 2^.
Hence „
2\ , in 51.
771=1 5
We apply now 531.
552
DERIVATES, EXTREMES, VARIATION
542. Faber Functions without Finite or Infinite Derivatives.
To simplify matters let us consider the following example.
The method employed admits easy generalization
and gives a class of functions of this type. We
use the notation of the preceding sections.
Let /0(#) have as graph Fig. 1. We next
divide 21 = (0, 1) into 21! equal parts 8n, 812 and
take f^(x) as in Fig. 2. We now divide 21 into
22! equal parts £21, S22, S23, S24 and take /2 (x) as
The height of the peaks is 12 =
in Fig. 3.
In the mth division 21 falls into 2w! equal parts
FIG. 1
in
FIG. 2
one of which may be denoted by
Its length may be denoted by the same letter,
thus
FIG. 3
AAA/^
FIG. 4
In Fig. 4, Sm is an interval of the m — 1st
division.
The maximum ordinate of /m(V) is L = = -. The
10TO 2 10m
part of the curve whose points have an ordinate < J lm have been
marked more heavily. The x of such points, form class 1. The
other o;'s make up class 2. With each x in class 1, we associate
the points am < /3m corresponding to the peaks of fm adjacent to x.
Thus am<x<{3m. If # is in class 2, the points «m, /3m are the
adjacent valley points, where fm = 0.
Let now x be a point of class 1. The numerators in
(1
«.-*
am- x
have like signs, while their denominators are of opposite sign.
Thus the signs of the quotients 1) are different. Similarly if x
belongs to class 2, the signs of 1) are opposite. Hence for any #,
NON-INTUITIONAL CURVES
553
the signs of 1) are opposite. It will be convenient to let em denote
either c^ or /3m. We have
Hence,
m!
On the other hand, for any x^x' in 8TO,
JU As
Hence setting x' = en, and letting n > w,
1 2m! 1 2n"1!
— 1 Am ' On! 1 i\m On!
(2
(3
10
For if Iog2 a be the logarithm of a with the base 2,
_ n > n I0g2 10 , for n sufficiently large.
n-1
Hence
Thus
On!
and this establishes 4).
Let us now extend the definition of the functions fn(x) by giv
ing them the period 1. The corresponding Faber function F(x)
defined by 540, 12) will admit 1 as period. We have now
From 2) we have
554 DERIVATES, EXTREMES, VARIATION
As to T2, we have, using 4) and taking n sufficiently large,
Similarly
\T9\< i
-v_^ v_ __
* * " ™ ' »
< 2 2
— n-fl
Thus finally
As
Thus
sgn
- X
il>l*il +
F(x^
= sgn
en — x
- F(x)
en — x
As gn may be at pleasure «n or ^n, and as the signs of 1) are
opposite, we see that
F(x) has neither a finite nor an infinite differential coefficient
at any point.
CHAPTER XVI
SUB- AND INFRA-UNIFORM CONVERGENCE
Continuity
543. In many places in the preceding pages we have seen how
important the notion of uniform convergence is when dealing
with iterated limits. We wish in this chapter to treat a kind of
uniform convergence first introduced by Arzeld, and which we
will call subuniform. By its aid we shall be able to give condi
tions for integrating and differentiating series termwise much
more general than those in Chapter V.
We refer the reader to Arzela's two papers, " Sulle Serie di
Funzioni," R. Accad. di Bologna, ser. V, vol. 8 (1899). Also
to a fundamental paper by Osgood, Am. Journ. of Math., vol. 19
(1897), and to another b}^ Hobson, Proc. Lond. Math. Soc., ser. 2,
vol. 1 (1904).
544. 1. Let/^j ••« xm, ^ ••• tn)=f(x, t) be a function of two
sets of variables. Let x = (xl>~xm} range over Hi in an w-way
space, and t = (tl •«• £n) range over £ in an rc-way space. As x
ranges over Hi and t over £, the point (xl ••• ^ ...) = (#, £) will
range over a set 51 lying in a space 9?p , p = m + n.
Let r, finite or infinite, be a limiting point of £.
Let v f( t t^ = 6C "> in £
t=T
Let the point x range over 53 < X, while t remains fixed, then
the point (x, £) will range over a layer of ordinate t, which we
will denote by %t. We say x belongs to or is associated with this
layer.
We say now that/== <£, subuniformly in H when for each e>0,
555
556 SUB- AND INFRA-UNIFORM CONVERGENCE
1° There exists a finite number of layers £t whose ordinates t
lie in F;*(T).
2° Each point x of £ is associated with one or more of these
layers. Moreover if x = a belongs to the layer £,, all the points
x in some V^a also belon to ?,.
while (#, t) ranges over any one of the layers %t. When ra= 1,
that is when there is but a single variable # which ranges over an
interval,0 the layers reduce to segments. For this reason Arzela
calls the convergence uniform in segments.
2. In case that subuniform convergence is applied to the series
convergent in 51, we may state the definition as follows :
F converges subimiformly in 51 when
1° For each e > 0, and for each v there exists a finite set of
layers of ordinates > v, call them
8,, «,- (2
such that each point x of 51 belongs to one or more of them, and if
x = a belongs to £m, then all the points of 51 near a also belong
as the point (#, w) ranges over any one of the layers 2).
545. Example. Let
EVjA _ y f MS (n-l> ] in9J-r
' -^iTT^~l+(n-l)VJ
Here
The series converges uniformly in 51, except at x = 0. The
convergence is therefore not uniform in 51 ; it is, however, sub-
uniform. For
CONTINUITY 557
Hence taking m at pleasure and fixed,
Fm\<e , a; in «! = (-$, 8),
sufficiently small. On the other hand,
Thus for w sufficiently large,
Hence we need only three segments s1, s2, s3 to get subuniform
convergence.
546. 1. Let /(^ -~xmi V-'O^^Oi '••*«) m £, as ? = T,
finite or infinite. Let /(#, t) be continuous in Hi for each t near r.
For <f> to be continuous at the point x = a in £, it is necessary that
for each e > 0, there exists an 77 > 0, and a dt for each t in Fn*(r)
such that
for each t in V^ and for any x in Vd(a).
It is sufficient if there exists a single t=/3 in T/rT/*(r) for ivhich
the inequality 1) holds for any x in some V^a).
It is necessary. For since </> is continuous at x = a,
| $(x) — <£(«) | < | , for any x in some F^(«).
o
Also since /= c/>,
I/O Q - <£(» | < I , for any f in some r/(r).
Finally, since /is continuous in x for any £ near r,
I/O> 0 -/O» 0|< jj ^ for any ^ in some 1\(a).
Adding these three inequalities we get 1), on taking
558 SUB- AND INFRA-UNIFORM CONVERGENCE
It is sufficient. For by hypothesis
and hence in particular.
» n
o
Also since /O, £) *s continuous in a;,
I/O, £)—/(«, £)| < , for any a; in some F6,,(».
6
Thus if 8 < 8', 8", these unequalities hold simultaneously. Add-
ing them we get
— <Ka)l < e > f°r any ^ i
and thus <£ is continuous at x = a.
2. As a corollary we get :
Let ^O)=2/tl...lnOi-^)
converge in 21, each term being continuous in 51. For F(x) to be con
tinuous at the point x = a in 21, it is necessary that for each e > 0,
and for any cell R^_ > some J2A, there exists a S^ such that
^8 sufficient if there exists an R^ and a 8 > 0
| ^O)l < e -> for anV x in V^(
547. 1. Let limf^xl"-xm, ^ ••• £n) = <^>Oi "' ^m) ^ %, r finite
x^
or infinite. Letf(x, t) be continuous in Hfor each t near r.
1° Iff= </> subuniformly in £, <#> i* continuous in 3E-
2° ^ 3E ^8 complete, and <f> is continuous in £,/== <t> subuniformly
in X.
To prove 1°. Let x = a be a point of £. Let e > 0 be taken at
pleasure and fixed. Then there is a layer %ft to which the point
a belongs and such that
CONTINUITY 559
when (x, t) ranges over the points of 8^. But then 1) holds for
t = fi and x in some F^(a). Thus the condition of 546, l is satis
fied.
To prove 2°. Since (f> is continuous at z=a, the relation 1)
holds by 546, l, for each t in Vf(j) and for any x in Vdt(a).
With the point a let us associate a cube Ca^ lying in D^a) and
having a as center. Then each point of £ lies within a cube.
Hence by Borel's theorem there exists a finite number of these
cubes (7, such that each point of £ lies within one of them, say
But the cubes 2) determine a set of layers
^ , V (3
such that 1) holds as (x, £) ranges over the points of SI in each
layer of 3). Thus the convergence of /to <f> is subuniform in £.
2. As a corollary we have the theorem :
Let ps • . a? ^ = 2f (x ••• x }
converge in £, each / fteiVigr continuous in H. If F converges sub-
uniformly in £, F is continuous in £. If £ is complete and F is
continuous in X, F converges subuniformly in H.
548. 1. Let F, , _ 2 - (x . . x ^
* \%) — "HMi—Si VX1 ^m/
converge in SI.
i^^ ^Ag convergence be uniform in SI except possibly for the points
of a complete discrete set 53 = 55|. .For each b, let there exist a \0
swcA that for any X > X0,
.F converges subuniformly in SI.
For let D be a cubical division of norm d of the space 9?TO in
which SI lies. We may take d so small that 33^ is small at
pleasure. Let BD denote the cells of D containing points of SI
but none of 33. Then by hypothesis ^converges uniformly in BD.
Thus there exists a /x0 such that for any ft > /x0,
| Fp (x) | < e , for any z of SI in BD.
560 SUB- AND INFRA-UNIFORM CONVERGENCE
At a point b of $8, there exists by hypothesis a Fs(ft) and a X0
such that for each X > X0
| JFx(aO ! < € , for any x in F«(i).
Let (7ftjX be a cube lying in D«(5), having b as center. Since 33
is complete there exists a finite number of these cubes
such that each point of 23 lies within one of them.
Moreover —
l^.(*)l«i
for any x of 21 lying in tlie /cth cube of 1).
As 1?^ embraces but a finite number of cubes, and as the same
is true of 1), there is a finite set of layers 8 such that
\Fv(x)\<e , in each 8.
The convergence is thus subuniform, as X, p are arbitrarily large.
2. The reasoning of the preceding section gives us also the
theorem :
in £, r finite or infinite. Let the convergence be uniform in £ except
possibly for the points of a complete discrete set (§ = \e\. For each
point e< let there exist an rj such that setting e(#, t) =/(^, 0 — </>(^)»
lim e(x, t) = 0 , for any t in F^*(T).
x=e
Thenf= <£ subuniformly in %.
3. As a special case of 1 we have the theorem :
Let *(*>
converge in H, and converge uniformly in 51, except at x = «1? ••• a: = a,.
^4.^ a; = at Z^^ fA^re g^z's^ « j/t
lim J*ni(a;) =0 , nt > i/t , i = 1, 2 ... s.
X=aL
Then F converges subuniformly in 51.
CONTINUITY 561
4. When v ~, ,, . , x
lim/O, 0 = £(*)
<=T
we will often set
/(avO~*(*)4XM)i
and call e the residual function.
549. Example 1.
/(^n)«^=£(a?)=0 , for 7i = oo in 21 = (0 «z),
gW^.r"
«, /3, X > 0 , /I > 0.
The convergence is suburiiform in 81. For 2; = 0 is the only
possible point of non-uniform convergence, and for any m,
/- ~\ i H)\ OC • /\ • /\
= 0 , as x = 0.
Example 2. f(x, ri) = -- - = 6 (x) = 0 , as n = oo,
c + n^xP f
a; in 31 = (0 < a) , a, /S, X, /* > 0 , ^>X , <? > 0.
The convergence is uniform in iB = (e < a), where ^ > 0. For
" c
a» 7^
" e^ n^
< e , for n > some m.
Thus the convergence is uniform in 81, except possibly at x = 0.
The convergence is subuniform in 51. For obviously for a given n
lim/(#, n) = 0.
-r=0
550. 1. Let Hm/O^ ••• xm tl ... fn) = ^(^ .-• arm) «» Hi, r finite
t=T
or infinite.
Let the convergence be uniform in % except at the points
85 =(*„*,, - J,).
562 SUB- AND INFRA-UNIFORM CONVERGENCE
For the convergence to be sub-uniform in £, it is necessary that for
each b in 33, and for each e > 0, there exists a t= fi near r, such that
x=b
For if the convergence is subuniform, there exists for each
and 77 > 0 a finite set of layers £t, t in FV"(r) such that
| e(#, f) | < e , re in %t.
Now the point x= b lies in one of these layers, say in fy
Then
| e(#, /3) | < e , for all x in some V*(b).
But then 1) holds.
"' J^jOT CL/YYL'Y)i r I^Pt"
0
This is the series considered in 140, Ex. 2.
F converges uniformly in 21 = ( — 1, 1), except at x = 1.
As y ^ = _ ^
we see that y XT /^ \ _ -i
Hence ^ is not subuniformly convergent in 51.
Integrdbility
551. 1. Infra-uniform Convergence. It often happens that
subuniformly in 36 except possibly at certain points (£= \e] form
ing a discrete set. To be more specific, let A be a cubical divi
sion of $ftm in which X lies, of norm 5. Let JTA denote those cells
containing points of X, but none of (£. Since (§, is discrete,
JTA = 36. Suppose now /"=(/> subuniformly in any ^TA ; we shall
say the convergence is infra-uniform in £. When there are no
exceptional points, infra-uniform convergence goes over into sub-
uniform convergence.
INTEGRABILITY 563
This kind of convergence Arzela calls uniform convergence by
segments, in general.
2. We can make the above definition independent of the set (g,
and this is desirable at times.
Let £ = (^T, £) be an unmixed division of X such that £ may be
taken small at pleasure. If /=</> subuniformly in each X, we
say the convergence is infra-uniform in %.
3. Then to each e, 77 >0, and a given Jf, there exists a set of
layers (j, I2 ••• , t in FV"(r), such that the residual function e(z, f)
is numerically < e for each of these layers. As the projections of
these layers I do not in general embrace all the points of £, we
call them deleted layers.
4. The points £ we shall call the residual points.
5. Example 1. ^ y*
This series was studied in 150. We saw that it converges uni
formly in 21 = (0, 1), except at x— 0.
As .
nx
and as this = 1 as x = 0 for an arbitrary but fixed w, F does not
converge subuniformly in 51, by 550. The series converges infra-
uniformly in 21, obviously.
6. Example X. !•= L.»(I _*).
0
This series was considered in 550, 2. Although it does not
converge subuniformly in an interval containing the point x = 1,
the convergence is obviously infra-uniform.
552. 1. Let lira / (^ ••• xm ^ ... *n) = ^(^ ••• xm) be limited in X,
x=r
T finite or infinite. For each t near r, let f be limited and R-integrable
in H. For </> to be R-integrable in X, it is sufficient thatf = <f> infra-
uniformly in Hi. If Hi is complete, this condition is necessary.
564 SUB- AND INFRA-UNIFORM CONVERGENCE
It is sufficient. We show that for each e, co > 0 there exists a
division D of 9?m such that the cells in which
Osc <f> > co (1
have a volume < <r. For setting as usual
we have in any point set,
Oscc
Using the notation of 551,
in the finite set of deleted layers (x, (2 ••• corresponding to
t= t^ t2 -•• For each of these ordinates £t,/(#, Q is integrable
in H. There exists, therefore, a rectangular division D of 9?m,
such that those cells in which
have a content < f , whichever ordinate tL is used. Let JS be a
division of 9?m such that the cells containing points of the residual
set £ have a content < <r/2. Let F = D + E. Then those cells
of F in which
Osc/C*,O>!< or Osc ' €(s, *0 I >f
t = l, 2 ... have a content < <r. Hence those cells in which 1)
holds have a content < cr.
It is necessary, if £ is complete. For let
<1, *2 V- = T'
Since <£ and/ (z, fn) are integrable, the points of discontinuity of
<£O) and of /(&, «n) are null sets by 462, 6. Hence if (£, ^ denote
the points of continuity of <£(#) and /(#, 0 in X,
since 96 is measurable, as it is complete.
INTEGKABILITY
Let ©=QdvJ<£|},
then 5 = f
by 410, 6.
Let £) = Di»(6, ®),
then f) = S,
as we proceed to show. For if G = X —
6T = S)
565
(1
But ^ is a null set. Hence Meas Dy((S, 6r) = 0, and thus
S = £ = J), which is 1).
Let now f be a point of 5), let it lie in (£/,, (£/2 ••• where ^, ^2 ...
form a monotone sequence = r. Then since
there is an m such that
< § , for any n > w.
o
But f lying in 2), it lies in (£ and (£, .
Thus «
for any a: in
Now
Hence
n
(3
Hence from 2), 8),
<e , for any a: in
Thus associated with the point f, there is a cube F lying in DS(£),
having % as center. As D = £ — 5) is a null set, each of its points
can be enclosed within cubes (7, such that the resulting1 enclosure
566 SUB- AND INFRA-UNIFORM CONVERGENCE
(£ has a measure < <r, small at pleasure. Thus each point of £ lies
within a cube. By Borel's theorem there exists a finite set of
these cubes
1' 2 '" .f ' 1' 2 '" «'
such that each point of £ lies within one of them. But corre
sponding to the F's, are layers
81, 8a, - 8r
such that in each of them
|€O,01<€.
Thus/ = <t> subuniformly in X = (I\, T2~- Fr). Let £ be the
residual set. Obviously ji < <r. Thus the convergence is infra-
uniform.
2. As a corollary we have :
Let ^0^ = 2/tl...lnOi"-O
converge in 31. Xe^ JP be limited, and each f, be limited and R-in-
tegrable in 31. For F to be R-integrable in 31, it is sufficient that F
converges infra-uniformly in 31.
If 31 is complete, this condition is necessary.
553. Infinite Peaks. 1. Let lira f (x { "'Xmt1"-tn) = $(x) in £,
t=T
T finite or infinite. Although /(a;, t) is limited in £ for each t
near T, and although </>(» is also limited in X, we cannot say that
I/O, 01 < some ^ 0-
for any x in X and an}^ f near T, as is shown by the following
Example. Let/O> 0 = -^ = ^O) = °> as ^^^ for * in
^ = (-00,0)).
It is easy to see that the peak of / becomes infinitely high as
flfssOQ.
In fact, for x = —-/=_. Thus the peak is at least as high
V£ e
Vz , . T
as — , which = QO .
e
INTEGRABILITY 567
The origin is thus a point in whose vicinity the peaks of the
family of curves f (x, £) are infinitely high. In general, if the
Peaksof /CV^V'-O
in the vicinity Vs of x = f become infinitely high as t = 'T, however
small 8 is taken, we say f is a point with infinite peaks.
On the other hand, if the relation 1) holds for all x and t in
volved, we shall say/(#, £) is uniformly limited.
2. If lim /CV^ *1»-*») = <Kai •"*»)> and if ffr 0 «
<=T
uniformly limited in £, £fo?w $ z* limited in Hi.
For 2: being taken at pleasure in 36 and fixed, $(x) is a limit
point of the points /(z, £) as t = r. But all these points lie in
some interval (—#,#) independent of z. Hence <j> lies in this
interval.
3. IfH is complete, the points $ in Hi with infinite peaks also form
a complete set. If these points & are enumerable, they are discrete.
That $ is complete is obvious. But then $ = $ = 0, as $ is
enumerable.
554. 1. Let lim/0^ ••• xm ^ ••• fn) = ^(^ •••«„) in I, metric or
complete. Let f (x, t) be uniformly limited in X, and R-integrable
for each t near T. For the relation
Km
fo hold, it is sufficient thatf=(j> infra- uniformly in I. If X is
for each t complete, this condition is necessary.
For by 552, <f> is JZ-integrable if /= £ iufra-uniformly, and when
X is complete, this condition is necessary. By 424, 4, each f(x, t)
is measurable. Thus we may apply 381, 2 and 413, 2.
2. As a corollary we have the theorem :
Let m F(x) = -Sfli ..... Or,-*.)
converge in the complete or metric field 51. Let the partial sums J\ be
uniformly limited in 51- Let each term /t fo limited and R-integrable
in 21. Then for the relation
568 SUB- AND INFRA-UNIFORM CONVERGENCE
to hold it is sufficient that F is infra -uniformly convergent in 51. If
51 is complete, this condition is necessary.
555. Example 1. Let us reconsider the example of 150,
We saw that we may integrate termwise in 51 = (0, 1), al
though F does not converge uniformly in 51. The only point of
non-uniform convergence is x ~ 0. In 551, 5, we saw that it con
verges, however, infra-uniformly in 51. As
I ^n(X) I < 1 •> f°r anv x m ^ and for every n,
all the conditions of 554 are satisfied and we can integrate the
series termwise, in accordance with the result already obtained
in 150.
** (n ~*x
Example 2. Let F(x) = V - ~ = 0.
^ J ^4 [ enx* e(n-l)x* j
Then ^ ,^ _ nx
^w-p-
We considered this series in 152, i. We saw there that this
series cannot be integrated termwise in 51 = (0 < «). It is, how
ever, subuniformly convergent in 51 as we saw in 549, Ex. 1. We
cannot apply 554, however, as Fn is not uniformly limited. In
fact we saw in 152, l, that x = 0 is a point with an infinite peak.
Example 3. F(x) = 2^(1 - x).
o
We saw in 551, 6, that F converges infra-uniformly in 51 = (0, 1).
Here Fn (V) = | 1 - x* | < some M,
for any x in 51 = (0 < w), u <^ 1, and any n. Thus the Fn are
uniformly limited in 51.
We may therefore integrate termwise by 554, 2. We may
verify this at once. For
= 0
INTEGRABILITY 569
Hence C*F(x)dai = u. (1
On the other hand,
Xu lln+^
Fndx = u -- -- = u , as n = GO. (2
71+1
+
From 1), 2) we have
0 , n + 1 n H- 2
556. 1. // l°/Oi ••> a?» «! - O = *(«! — *.) infra-uniformly
in the metric or complete field 36, as f = r, r ^wite or infinite ;
2° /(#, 0 ^s uniformly limited in H and R-integrable for each t
near r;
uniformly with respect to the set of measurable fields H in X-
If 36 is complete, condition 1° wa^ be replaced by 3° <£(#) is
R-integrable in X.
For by 552, l, when 3° holds, 1° holds ; and when 1° holds, <f>
is 72-integrable in 36.
Now the points <&t where
| e O, tn) > e
are such that ^
lim @t = 0 , by 412.
Let £ = £,+ £«. Then
Ce(x,t)= /*€(«, 0+ f «C
«4/;e -t^fc otft
I f
«
But lim I, = 0,
?=T
which establishes the theorem.
2. As a corollary we have :
If 1° F(x)= ^fli...in(^x1 ••• ^m) converges infra-uniformly, and
each of its terms f, is R-integrable in the metric or complete field 51;
570 SUB- AND INFRA-UNIFORM CONVERGENCE
2° F>,(x) is uniformly limited in 51;
Tken r *•(*)= 2 r/.,
J« Jy
and the series on the right converges uniformly with respect to all
measurable SB < 51.
3. Jf 1° lim/(z, ^ ... fn) = <£(X) i« R-integrable in the interval
t=T
5( = (« < 6), T finite or infinite ;
2°f(x, £) is uniformly limited, and R-integr able for each t near r;
Thpw Cx fx
lim I/O, 0^=1 <KzVz = <£<>\
*=T •/« •/«
uniformly in 51, anc? <J>(a:) is continuous in 51.
termfL are R-integrable in the interval 21 = (a< 5);
2° jPx(^) *'* uniformly limited in 51;
2%eft xy, N v r^-, N 7
^(a;) = 2 I f,(x)dx , z tw 51
•OB
continuous.
For 6r is a uniformly convergent series in 51, each of whose terms
/.*•
%x u
i» « continuous function of x.
Differentiability
557. 1. J/ 1° lhn/(s, <!-«„)=: ^(«) m 51 = (a < 6), r finite or
infinite ;
%°fx(.xi 0 *'* R-integrable for each t near T, a/ic? uniformly limited
3° /j(a;, «)= i/r(V) infra- uniformly in 5(, a« £ = T;
^ a^ a point x of continuity of i/r in 51
^(aO-^OO. (1
or ttwa£ is ^e same
-f/(*,0- (2
dx
DIFFERENTIABILITY 571
For by 554,
lim f/J(z, f)dx = F+(x)dx (3
t=r "a •/•
= lim[/0r, 0-/O, 0] » by I, 538
/=T
= <£<»-<»<» , byl°.
Now by I, 537, at a point of continuity of i/r,
(*)** -*co- (4
From 3), 4), we have 1), or what is the same 2).
2. /ft the interval 51, if
1° F(x)= 2/tl...lB(a;) converges; (1
2° EacKf[(x) is limited and R-integrable ;
3° .FA' (a;) t* uniformly limited ;
4° G(x)= 2// i« infra- uniformly convergent;
Then at a point of continuity of 6r(#) m 51, t^e ??i(z?/ differentiate
the series 1) termivise, or F'(x)= G(x).
3. J/i the interval 51, if
1° f(x, t1 •- tn) = (f>(x) as t = r, r finite or infinite ;
2° /(#, f) is uniformly limited, and a continuous function of x ;
3° ^(a:) = limy^a;, £) z*8 continuous;
nen . *'(*)=*(*), a
or w^a^ zs ^6 same
= lim- /(«, 0- (2
For by 547, l, condition 3° requires that /' = i|r subuniformly
in K. But then the conditions of 1 are satisfied and 1) and 2)
hold.
4. In the interval 51 let us suppose that
1° F(x) = 2/ti...tn<» converges ; (1
572 SUB- AND INFRA-UNIFORM CONVERGENCE
2° Each termf, is continuous;
3° Fl(x) is uniformly limited ;
4° G(x) = 2/[(a;) is continuous ;
Then we may differentiate 1) termtvise, or F'(x) = 6r(#),
558. Example 1. We saw in 555, Ex. 3 that
' *a-«U). a
The series got by differentiating termwise is
#00 = 2zn(l-20=l » Q<z<l
o (2
= 0 , x = 0.
Thus by 557, 4, ^^ = ^^ in (Q#? 1) = 5p> (g
The relation 3) does not hold for x = 0.
Example 2.
H-
l J
1
Here ^(a:) = arctg x, for any z. (1
Hence (r(^) is continuous in any interval 51, not containing
x = 0. Thus we should have by 557, 4,
Ff(x)=G-(x), zinH. (3
This relation is verified by 1), 2). The relation 3) does not
hold for x = 0, since
DIFFERENTIABILITY 573
Example 3.
= 1 log (1 -f x2) , for any x.
In any interval 51, all the conditions of 557, 4, hold.
Hence
, for
In case we did not know the value of the sums 1), 2) we could
still assert that 3) holds. For by 545, G- is subuniformly con
vergent in 51, and hence is continuous.
Example 4-
F( ^ — V I ^ "*" nx — 1 +(n + 1)^ } _ 1 + # s-t
Here
*'(*)= -5. (2
The series obtained by differentiating jF termwise is
* (3
•7 I gi»- enx
and hence
/?•£ z>WX
The peaks of the residual function
e(>, n)=^x,
are of height = \/e. The convergence of G- is not uniform at
x = 0. The conditions of 557, 4, are satisfied and we can differ
entiate 1) termwise. This is verified by 2), 3).
574 SUB- AND INFRA-UNIFORM CONVERGENCE
559. 1. Ifl° lim/O, t1"- tn) = <$>(x) is limited and R-integraUe
t = T
in the interval 51 = (a < ft) ;
2° f(x, t) is limited, and R-integrable in 51, for each t near r ;
3° ^(x) = lim P/O, 0 = lim g(x, t)
t=r */a t=r
is a continuous function in 51;
4° The points (g in 51 in whose vicinity the peaks of f(x, t) as
t = T are infinitely high form an enumerable set ;
Then Cx Cx
6(x) = I ^(a;) = lim ( /(*, t)dx = ^O), (1
«/« <=T «^a
°^" fx rx
lim I /(a;, t)dx = I lim/(^, <)c?a;,
t=T J<* t/a <=T
ant? the set (S i* complete and discrete.
For (g is discrete by 553, 3.
Let a be a point of A = 51 — (g. Then in an interval a about a,
|/(#, t) | < some M , a; in a, any £ near r. (2
Now by 556, 3, taking e > 0 small at pleasure, there exists an
?? > 0 such that
for any x in a, and t in F^*(T). If we set x = a 4- h, we have
(3
h
Also by 556, 3, we have
P>O, 0^ = f*<Kx)dx + e" , €" | < e
*^a ^a
for an?/ 2: in a, and t in FT,*(T). Thus
(4
A h Ax h
From 3), 4) we have
i;=^+^ K|,|e"|<e.
h Ax h
DIFFERENTIABILITY 575
Now e may be made small at pleasure, and that independent of
h. Thus the last relation gives
forarinA.
z Az
As this holds however small h = Az is taken, we have
*± = M for x in A.
dx dx
Hence by 515, 3,
-^(x) = 6(x) + const , in 21.
For x = a,
and thus
, in 21.
2. J.8 a corollary we have :
If 1° F(x)= 2/4...tn(z) is limited and R integrable in the inter-
2° 1\(^) ^s limited and each termf, is R-integrable;
3° &(x)= Xf^ is continuous;
4° The points @ i/i 21 wi ^Aos^ vicinity the peaks of F^(x) are in
finitely high form an enumerable set;
Then
or we may integrate the F series termivise.
560. 1. If 1° lira /(a, «j — *») = *(») *w 51 = (« < J> T/w*fe or
<=T
infinite ;
2° /£(z, «) i8 limited and R-integrable for each t near r;
3° The points (§. of 21 m w^o86 vicinity f'x(x, t) has infinite peaks
as t = T form an enumerable set;
4° <£(V) is continuous at the points (g;
5° -^(a:) = lim/J(ir, 0 z* limited and R-integrable in 21;
576 SUB- AND INFRA-UNIFORM CONVERGENCE
Then at a point of continuity of ^(V) in 51
$(x) = ^o), (l
or what is the same
-j- lim/(^0 = lim-f/<>, 0-
ax t=r t=T ax
For let S = (a < /3) be an interval in 51 containing no point of
($. Then for any x in 8
X*CMfe:-/&Q~/C<Q * b72°-
m^ i*
Hence
lim I f'x(x, t)dx = lim\f(x, £) — ./(«, £)J
= c£(V) - </>(«) , by 1°. (2
By 556, 3, <f>(x) is continuous in S. Thus <£(#) is continuous
at any point not in (g. Hence by 4° it is continuous in 51.
We may thus apply 559, 1, replacing therein /(#, £) by /,£(#, £).
We get
/*x /*x /*x
lim I f'x(x,£)dx= I lim/^(a;, 0^ = I ^(x)dx. (3
/=T c/a »/a ^/a
Since 2) obviously holds when we replace a by a, this relation
with 3) gives
At a point of continuity, this gives 1) on differentiating.
2. Ifl° F(x) = 2/ti ... tn(#) converges in the interval 51;
2° 6r(:r) = 2/[(a;) aw^ ^ac^ o/ ^s ^erws ar^ limited and R-
integrable in 51;
3° TAe points of 51 zVi whose vicinity Gr\(x) has infinite peaks as
X = GO, form an enumerable set at which F(x) is continuous;
Then at a point of continuity of G-(x) we have
or what is the same
DIFFERENTIABILITY 577
561. Example. ^ / n** _ (n+1)^ 1 _ z2
TU"*2 ~ *'
Hence o
The series obtained by differentiating F termwise is
1
Here ~ _
Hence 2 ^ 9 ^
is a continuous function of x.
The convergence of the Gr series is not uniform at x = 0. For
set an = \/n. Then
I en en J
To get the peaks of the residual function we consider the
points of extreme of
y = — • f 3
««•
We find n(l-5nx*+OnW^
yr = 2 .
7jj;2
Thus y' = 0 when
2 nV - 5 wa^ + 1 = 0,
or when 2; = — — or - — = , a, a constants.
V/i
Putting these values in 3), we find that y has the form
Hence x = 0 is the only point where the residual function has
an infinite peak. Thus the conditions of 560, 2, are satisfied, and
we should have F' (JK) = G-(x) for any x. This is indeed so, as 1),
2) show.
CHAPTER XVII
GEOMETRIC NOTIONS
Plane Curves
562. In this chapter we propose to examine the notions of
curve and surface together with other allied geometric concepts.
Like most of our notions, we shall see that they are vague and
uncertain as soon as we pass the confines of our daily experience.
In studying some of their complexities and even paradoxical
properties, the reader will see how impossible it is to rely on his
unschooled intuition. HB will also learn that the demonstration
of a theorem in analysis which rests on the evidence of our
geometric intuition cannot be regarded as binding until the
geometric notions employed have been clarified and placed on a
sound basis.
Let us begin by investigating our ideas of a plane curve.
563. Without attempting to define a curve we would say on
looking over those curves most familiar to us that a plane curve
has the following properties :
1° It can be generated by the motion of a point.
2° It is formed by the intersection of two surfaces.
3° It is continuous.
4° It has a tangent at each point.
5° The arc between any two of its points has a length.
6° A curve is not superficial.
7° Its equations can be written in any one of the forms
(2
(3
and conversely such equations define curves.
678
PLANE CURVES 579
8° When closed it forms the complete boundary of a region.
9° This region has an area.
Of all these properties the first is the most conspicuous and
characteristic to the naive intuition. Indeed many employ this
as the definition of a curve. Let us therefore look at our ideas
of motion.
564. Motion. In this notion, two properties seem to be essen
tial. 1° motion is continuous, 2° it takes place at each instant in
a definite direction and with a definite speed. The direction of
motion, we agree, shall be given by dy/dx, its speed by ds/dt.
We see that the notion of motion involves properties 4°, 5°, and 7°.
Waiving this point, let us notice a few peculiarities which may
arise.
Suppose the curve along which the motion takes place has an
angle point or a cusp as in I, 366. What is the direction of
motion at such a point? Evidently we must say that motion is
impossible along such a curve, or admit that the ordinary idea of
motion is imperfect and must be extended in accordance with the
notion of right-hand and left-hand derivatives.
Similarly ds/dt may also give two speeds, a posterior and an
anterior speed, at a point where the two derivatives of s == <£(£)
are different.
Again we will admit that at any point of the path of motion,
motion may begin and take place in either direction. Consider
what happens for a path defined by the continuous function in
I, 367. This curve has no tangent at the origin. We ask how
does the point move as it passes this point, or to make the ques
tion still more embarassing, suppose the point at the origin. In
what direction does it start to move? We will admit that no
such motion is possible, or at least it is not the motion given us
by our intuition. Still more complicated paths of this nature are
given in I, 369, 371, and in Chapter XV of the present volume.
It thus appears that to define a curve as the path of a moving
point, is to define an unknown term by another unknown term,
equally if not more obscure.
565. 2° Property. Intersection of Two Surfaces. This property
has also been used as the definition of a curve. As the notion
580 GEOMETRIC NOTIONS
of a surface is vastly more complicated than that of a curve, it
hardly seems advisable to define a complicated notion by one still
more complicated and vague.
566. 3° Property. Continuity. Over this knotty concept philos
ophers have quarreled since the days of Democritus and Aristotle.
As far as our senses go, we say a magnitude is continuous when
it can pass from one state to another by imperceptible gradations.
The minute hand of a clock appears to move continuously, although
in reality it moves by little jerks corresponding to the beats of the
pendulum. Its velocity to our senses appears to be continuous.
We not only say that the magnitude shall pass from one state
to another by gradations imperceptible to our senses, but we also
demand that between any two states another state exists and so
without end. Is such a magnitude continuous ? No less a mathe
matician than Bolzano admitted this in his philosophical tract
Paradoxien des Unendlichen. No one admits it, however, to-day.
The different states of such a magnitude are pantactic, but their
ensemble is not a continuum.
But we are not so much interested in what constitutes a con
tinuum in the abstract, as in what constitutes a continuous curve
or even a continuous straight line or segment. The answer we
have adopted to these questions is given in the theory of irra
tional numbers created by Cantor and Dedekind [see Vol. I,
Chap. II], and in the notion of a continuous function due to
Cauchy and Weierstrass [see Vol. I, Chap. VII].
These definitions of continuity are analytical. With them we
can reason with the utmost precision and rigor. The consequences
we deduce from them are sufficiently in accord with our intuition
to justify their employment. We can show by purely analytic
methods that a continuous function /(V) does attain its extreme
values [I, 354], that if such a function takes on the value a at the
point P, and the value b at the point Q, then it takes on all inter
mediary values between a, b, as x ranges from P to Q [I, 357].
We can also show that a closed curve without double point does
form the boundary of a complete region [cf. 576 seq.].
567. 4° Property. Tangents. To begin with, what is a tangent ?
Euclid defines a tangent to a circle as a straight line which meets
PLANE CURVES 581
the circle and being produced does not cut it again. In com
menting on tliis definition Casey says, " In modern geometry a
curve is made up of an infinite number of points which are
placed in order along the curve, and then the secant through two
consecutive points is a tangent." If the points on a curve were
like beads on a string, we might speak of consecutive points. As,
however, there are always an infinite number of points between any
two points on a continuous curve, this definition is quite illusory.
The definition we have chosen is given in I, 365. That property
3° does not hold at each point of a continuous curve was brought
out in the discussion of property 1°. Not only is it not necessary
that a curve has a tangent at each of its points, but a curve does
not need to have a tangent at a pantactic set of points, as we saw
in Chapter XV.
For a long time it was supposed that every curve has a tangent
at each point, or if not at each point, at least in general. Analytic
ally, this property would go over into the following : every con
tinuous function has a derivative. A celebrated attempt to prove
this was made by Ampere.
Mathematicians were greatly surprised when Weierstrass ex
hibited the function we have studied in 502 and which has no
derivative.
Weierstrass* himself remarks: " Bis auf die neueste Zeit hat
man allgemein angenommen, dass eine eindeutige und continuir-
liche Function einer reellen Verandeiiichen auch stets eine erste
Ableitung habe, deren Werth nur an einzelnen Stellenunbestimmt
oder unendlich gross werden konne. Selbst in den Schriften von
Gauss, Cauchy, Dirichlet findet sich meines Wissens keine
Ausserung, aus der unzweifelhaft hervorginge, dass diese Mathe-
matiker, welche in ihrer Wissenschaft die strengste Kritik iiberall
zu iiben gewohnt waren, anderer Ansicht gewesen seien."
568. Property 5°. Length. We think of a curve as having
length. Indeed we read as the definition of a curve in Euclid's
Elements : a line is length without breadth. When we see two
simple curves we can often compare one with the other in regard
to length without consciously having established a way to measure
* Werke, vol. 2, p. 71.
582 GEOMETRIC NOTIONS
them. Perhaps we unconsciously suppose them described at a
uniform rate and estimate the time it takes. It may be that we
regard them as inextensible strings whose length is got by
straightening them out. A less obvious way to measure their
lengths would be to roll a straightedge over them and measure
the distance on the edge between the initial and final points of
contact.
We ask how shall we formulate arithmetically our intuitional
ideas regarding the length of a curve ? The intuitionist says, a
curve or the arc of a curve has length. This length is expressed
by a number L which is obtained by taking a number of points
Pj, P2, P3--* on the curve between the end points P, P', and
forming the sum
The limit of this sum as the points became pantactic is the
length L of the arc PP'.
Our point of view is different. We would say : Whatever
arithmetic formulation we choose we have no a priori assurance
that it adequately represents our intuitional ideas of length.
With the intuitionist we will, however, form the sum 1) and see if
it has a limit, however the points Pt are chosen. If it has, we will
investigate this number used as a definition of length and see if it
leads to consequences which are in harmony with our intuition.
This we now proceed to do.
569. 1. Let Z
be one-valued continuous functions of t in the interval 21 = (a < 5).
As t ranges over 51 the point x, y will describe a curve or an arc
of a curve O. We might agree to call such curves analytic, in
distinction to those given by our intuition. The interval 51 is
the interval corresponding to O.
Let D be a finite division of 21 of norm c?, defined by
To these values of t will correspond points
p,p,,p,...e (2
PLANE CURVES 583
on (7, which may be used to define a polygon PD whose vertices
are 2).
Let (m, TTi + 1) denote the side PmPm+1, as well as its length.
If we denote the length of PD by the same letter, we have
1 ^ PD (B
exists, it is called the length of the arc (7, and C is rectifiable.
2. (Jordan.) For the arc PQ to be rectifiable, it is necessary and
sufficient that the functions c/>, ^ in 1) have limited variation in 51.
For
But the sum on the right is the variation of <f> for the division D.
If now <f> does not have limited variation in 51, the limit 3) does
not exist. The same holds for ty. Hence limited variation is a
necessary condition.
The condition is sufficient. For
PD< 2 | Az | + 2 | Ay | = Var <£ + Var ^.
D D
As <f>, T/T have limited variation, this shows that
P
is finite. We show now that
P0 = Max PD
D
limPZ) = P0. (4
<f=0
For there exists a division A such that
Let A cause 51 to fall into v intervals, the smallest of which has
the length X. Let D be a division of 51 of norm d<dQ<\.
Then no interval of D contains more than one point of A.
Let E=D + A.
Obviously PE>P
or
584 GEOMETRIC NOTIONS
Suppose that the point tK of A falls in the interval (£t, £t+1) 01
D. Then the chord (t, i -f- 1) in PD is replaced by the two chords
(t, *), (K, i + 1) in PE- Hence
where ».£ C^) + (<• + 1^ - <•, V+ 1) .
Obviously as <£, -^ are continuous we may take dQ so small that
each
G-K<£- , for any d<d0.
'- V
Hence ' Pg-PD<e-. (6
From 5), 6) we have
PQ-PD<€ , for any d < d0,
which gives 4).
3. If the arc PQ is rectifiable, any arc contained in PQ is also
rectifiable.
For (/>, i/r having limited variation in interval 51, have a fortiori
limited variation in any segment of 51.
4. Let the rectifiable arc C fall into two arcs C1 , C2 . If s, *j , s2
are the lengths of (7, C\, <72, tf/ew,
S = *j + 8y (T
For we saw that (7X, (72 are rectifiable since (7 is. Let ^ , 5I2
be the intervals in 51 corresponding to 6\, (72. Let Dv D2 be
divisions of ^, ?12 °^ norm ^- Then
«! = lim PD. , s2 = limPn.
d=o rf=o
But Dj, D2 effect a division of 21, and since
s = lim PA, (8
e=0
with respect to the class of all divisions of 51, the limit 8) is the
same when E is restricted to range over divisions of the type of D.
Now
Pj>-Pft+*V
Passing to the limit, we get 7).
PLANE CURVES 585
The preceding reasoning also shows that if Cl , (72 are rectifiable
curves, then C is, and 7) holds again.
5. If 1) define a rectifiable curve, its length s is a continuous func
tion s(f) of t.
For (f>, ^r having limited variation,
where the functions on the right are continuous monotone increas
ing functions of t in the interval 51 = (« < ^).
For a division D of norm d of the interval A21 = (t, t + A) we
have
PD = 2
4-
where B^ = c/>1(^ + h) — <£(0> an(i similarly for the other func
tions. As c^j is continuous, 8^ = 0, etc., as h = 0. We may
therefore take 77 > 0 so small that S^, S(£2, 8-v^j, 3^2 < e/4, if A < 77.
Hence As = s(t + k) - s(t) < ^lax P^ < e , if 0 < h > 77.
Thus s is continuous.
6. The length s of the rectifiable arc C corresponding to the inter
val (a < t) is a monotone increasing function of t.
This follows from 4.
7. If x, y do not have simultaneous intervals of invariability, s(t)
is an increasing function of t. The inverse function is one-valued
and increasing and the coordinates x, y are one- valued functions of s.
That the inverse function t (s) is one-valued follows from I, 214.
We can thus express t in terms of s, and so eliminate t in 1).
570. 1. If <$>' , ty' are continuous in the interval 51,
8= fdt^^ + ^'t. (1
s = Km 2Ac£2-f- A^2. (2
586
Now
GEOMETRIC NOTIONS
where ^, £" lie in the interval Af*.
As (//, -*// are continuous they are uniformly continuous. Hence
for any division D of norm < some c?0,
where | «K , | /3K | < some 77, small at pleasure, for any K. Thus
and we may take
= lim
Hence
j;
<
which establishes 1).
For simplicity we have assumed <£', ^' to be continuous in
This is not necessary, as the following shows.
2. Let ar «.. aB, b^ -- bn>0 but not all = 0.
Then ,
For
= 1, 2
Hence
Vaf + V5J
But
This in 5) gives 4).
PLANE CURVES 587
Let us apply 4) to prove the following theorem, more general
than 1.
3. (Baire.) If <£', -^ are limited and R integrable, then
8
For by 4),
2 <
4>K _ yK = r
where ^j, ??" are numerically <1. Thus
! S^K^AC — ^-&K^K \ = ^^/e7?* Osc (£>' + 28K7^^' Osc ijr1 . (6
As <£', >/r' are integrable, the right side = 0, as d = 0. Now
lira
Thus passing to the limit in 6), we have
Km SA^V(£'(^)2 + A/r'(t')2= f-
^21
This with 2), 3) gives 1) at once.
571. Volterra's Curve. It is interesting to note that there are
rectifiaUe curves for which <£'(£)> V^CO are not both R-integr able.
Such a curve is Volterra's curve, discussed in 503. Let its equa
tion be 2/=/(z). Then f (x) behaves as
2 x sin -- cos -
X X
in the vicinity of a non null set in 31 = (0, 1). Hence f'(x) is
not ^-integrable in 51. But then it is easy to show that
does not exist. For suppose that
VI
588 GEOMETRIC NOTIONS
were .R-integrable. Then #2 = 1 +/'(V)2 is ,R-integrable, and
hence /'(^)2 also. But the points of discontinuity of f'2 in 51 do
not form a null set. Hence /'2 is not .K-integrable.
On the other hand, Volterra's curve is rectifiable by 569, 2, and
528, i.
572. Taking the definition of length given in 569, 1, we saw
that the coordinates
must have limited variation for the curve to be rectifiable. But we
have had many examples of functions not having limited variation
in an interval 21. Thus the curve defined by
y =x sin- , x^ 0
(4
does not have a length in 21 = (— 1, 1) ; while
y = x2 sin - , x =£ 0
x (5
=0 , x=0
does.
It certainly astonishes the na'ive intuition to learn that the
curve 4) has no length in any interval 8 about the origin how
ever small, or if we like, that this length is infinite, however small
8 is taken. For the same reason we see that
No arc of Weierstrass' curve has a length (or its length is infinite)
however near the end points are taken to each other, ivhen ab>l.
573. 1. 6° Property. Space-filling Curves. We wish now to
exhibit a curve whlcli passes through every point of a square, i.e.
which completely fills a square. Having seen how to define one
such curve, it is easy to construct such curves in great variety, not
only for the plane but for space. The first to show how this may
be done was Peano in 1890. The curve we wish now to define is
due to Hilbert.
We start with a unit interval 51 = (0, 1) over which t ranges,
and a unit square 33 over which the point x, y ranges. We define
PLANE CURVES
589
as one-valued continuous functions of t in 51 so that xy ranges over
53 as t ranges over 51. The analytic curve O defined by 1) thus
completely fills the square 53.
We do this as follows. We effect a division of 51 into four
equal segments SJ, B'2, S3, 8£, and of 53 into equal squares 77^, 77^,
773, 774, as in Fig. 1.
We call this the first division or D^. The corre
spondence between 51 and 53 is given in first
approximation by saying that to each point P in
B[ shall correspond some point Q in v[ .
We now effect a second division D2 by dividing
each interval and square of J)1 into four equal
parts.
We number them as in Fig. 2,
X" %" W
01 » °2 •" *M
As to the numbering of the rfs we observe the
following two principles : 1° we may pass over the
squares 1 to 16 continuously without passing the
same square twice, and 2° in doing this we pass
over the squares of Dl in the same order as in FIG. 2.
Fig. 1. The correspondence between 5( and 53 is
given in second approximation by saying that to each point P in
B[f shall correspond some point Q in rj{'. In this way we continue
indefinitely.
To find the point Q in 53 corresponding to P in 51 we observe
that P lies in a sequence of intervals
8' >S" >B"r > ••• =0, (2
to which correspond uniquely a sequence of squares
V >T," >r>'" >..- =0. (3
The sequence 3) determines uniquely a point whose coordinates
are one-valued functions of £, viz. the functions given in 1).
The functions 1) are continuous in 51.
For let t' be a point near t ; it either lies in the same interval as
t in Dn or in the adjacent interval. Thus the point Qf corre-
590 GEOMETRIC NOTIONS
spending to t' either lies in the same square of Dn as the point Q
corresponding to £, or in an adjacent square. But the diagonal
of the squares = 0, as n = oo. Thus
Thus
both = 0, as t' = t.
As t ranges over 51, the point x, y ranges over every point in the
square 53.
For let Q be a given point of 53. It lies in a sequence of
squares as 3). If Q lies on a side or at a vertex of one of the rj
squares, there is more than one such sequence. But having taken
such a sequence, the corresponding sequence 2) is uniquely de
termined. Thus to each Q corresponds at least one P. A more
careful analysis shows that to a given Q never more than four
points P can correspond.
2. The method we have used here may obviously be extended
to space. By passing median planes through a unit cube we
divide it into 23 equal cubes. Thus to get our correspondence
each division Dn should divide each interval and cube of the pre
ceding division Dn_1 into 23 equal parts. The cubes of each divi
sion should be numbered according to the 1° and 2° principles of
enumeration mentioned in 1.
By this process we define
as one-valued continuous functions of t such that as t ranges over
the unit interval (0, 1), the point a?, #, z ranges over the unit
cube.
574. 1. Hilbert's Curve. We wish now to study in detail the
correspondence between the unit interval 51 and the unit square
53 afforded by Hilbert's curve denned in 573. A number of inter
esting facts will reward our labor. We begin by seeking the
points P in 51 which correspond to a given Q in 53-
To this end let us note how P enters and leaves an TJ square.
Let B be a square of Dn. In the next division B falls into four
PLANE CURVES 591
squares Bl ••• B± and in the n 4- 2d division in 16 squares StJ.
Of these last, four lie at the vertices of B ; we call them vertex
squares. The other 12 are median squares. A simple considera
tion shows that the rj squares of Dn+2 are so numbered that we
always enter a square B belonging to Dn, and also leave it by a
vertex square.
Since this is true of every division, we see on passing to the
limit that the point Q enters and leaves any 77 square at the ver
tices of ?;. We call this the vertex law.
Let us now classify the points P, Q.
If P is an end point of some division Dn > we call it a terminal
point, otherwise an inner point, because it lies within a sequence
of 8 intervals B' > B" > ••• = 0.
The points Q we divide into four classes :
1° vertex points, when Q is a vertex of some division.
2° inner points, when Q lies within a sequence of squares
y>7/'>... =o.
3° lateral points, when Q lies on a side of some 77 square but
never at a vertex.
4° points lying on the edge of the original square 33. Points
of this class also lie in 1°, 3°.
We now seek the points P corresponding to a Q lying in one of
these four classes.
Class 1°. Q a Vertex Point. Let Dn be the first division such
that Q is at a vertex. Then Q lies in four squares TJL, ?/,•, TJK, ^ of
!>,.
There are 5 cases :
a) ij k I are consecutive.
yS) ij k are consecutive, but not I.
7) ij are consecutive, but not k I.
8) ij, also k I, are consecutive,
e) no two are consecutive.
A simple analysis shows that a), /3) are not permanent in the
following divisions ; 7), 8) may or may not be permanent ; e) is
permanent.
592 GEOMETRIC NOTIONS
Now, whenever a case is permanent, we can enclose Q in a se
quence of 7] squares whose sides = 0. To this sequence corre
sponds uniquely a sequence cf & intervals of lengths = 0. Thus
to two consecutive squares will correspond two consecutive inter
vals which converge to a single point P in 21. If the squares are
not consecutive, the corresponding intervals converge to two dis
tinct points in 31. Thus we see that when 7) is permanent, to Q
correspond three points P. When 8) is permanent, to Q corre
spond two points P. While when Q belongs to e), four points P
correspond to it.
Class 2°. Q an Inner Point. Obviously to each Q corresponds
one point P and only one.
Glass 3°. Q a Lateral Point. To fix the ideas let Q lie on a ver
tical side of one of the T/'S. Let it lie between ??t, r;;- of Dn. There
are two cases :
a) j = i + 1.
We see easily that a) is not permanent, while of course ft) is.
Thus to each Q in class 3°, there correspond two points P.
Class 4°. Q lies on the edge of 33. If Q is a vertex point, to it
may correspond one or two points P. If Q is not a vertex point,
only one point P corresponds to it.
To sum up we may say :
To each inner point Q corresponds one inner point P.
To each lateral point Q correspond two points P.
To each edge point Q correspond one or two points P.
To each vertex point Q, correspond two, three, or four points P.
2. As a result of the preceding investigation we show easily
that :
To the points on a line parallel to one of the sides of 23 correspond
in 51 an apantactic perfect set.
3. Let us now consider the tangents to Hilbert's curve which
we denote by H.
PLANE CURVES
593
$
7.
ft+i
Q
%+i
Let Q be a vertex point. We saw there were three permanent
cases 7), S), e).
In cases 7), 8) we saw that to two consecutive B intervals cor
respond permanently two contiguous ver
tical or horizontal squares.
Thus as t ranges over ' — = - ' - = — ' o
Ot 0<+i
ot, ot+1, the point #, y ranges
over these squares, and the secant line
joining Q and this variable point #, y oscillates through 180°.
There is thus no tangent at Q. In case e) we see similarly that
the secant line ranges through 90°. Again there is no tangent
at Q.
In the same way we may treat the three other classes. We find
that the secant line never converges to a fixed position, and may
oscillate through 360°, viz. when Q is an inner point. As a result
we see that Hubert's curve has at no point a tangent, nor even a
unilateral tangent.
4. Associated with Hilbert's curve ^Tare two other curves,
a; =0(0 , and ^ = ^(0.
The functions <£, T/T being one-valued and continuous in H, these
curves are continuous and they do not have a multiple point. A
very simple consideration shows that they do not have even a
unilateral tangent at a pantactic set of points in 31.
Pro£$rty 7°. Equations of a Curve. As already remarked,
it is commonly thought that the equation of a curve may be
written in any one of the three forms
y =/O),
(i
(2
and if these functions are continuous, these equations define con
tinuous curves.
Let us look at the Hilbert curve H. We saw its equation
could be expressed in the form 3). H cuts an ordinate at every
point of it for which 0 < y < 1. Thus if we tried to define H by
594 GEOMETRIC NOTIONS
an equation of the type 1), /(V) would have to take on every
value between 0 and 1 for each value of x in 51 = (0, 1). No such
. — ^ functions are considered in analysis.
WVXf u\ Again, we saw that to any value x = a in 21 corresponds a perfect
apantactic set of values \ta\ having the cardinal number c. Thus
the inverse function of x = <£(£) is a many-valued function of x
whose different values form a set whose cardinal number is c.
^ — -A Such functions have not yet been studied in analysis.
How is it possible in the light of such facts to say that we may
pass from 3) to 1) or 2) by eliminating t from 3). And if we
cannot, how can we say a curve can be represented equally well
by any of the above three equations, or if the curve is given by
one of these three equations, we may suppose it replaced by one
of the other two whenever convenient. Yet this is often done.
In this connection we may call attention to the loose way
elimination is treated. Suppose we have a set of equations
We often see it stated that one can eliminate ^ ••• tn and obtain
a relation involving the #'s alone. Any reasoning based on such
a procedure must be regarded as highly unsatisfactory, in view of
what we have just seen, until this elimination process has been
established.
576. Property 8». Closed Curves. A circle, a rectangle, an
ellipse are examples of closed curves. Our intuition tells us that
it is impossible to pass from the inside to the outside without
crossing the curve itself. If we adopt the definition of a closed
curve without multiple point given in I, 362, we find it no easy
matter to establish this property which is so obvious for the simple
closed curves of our daily experience. The first to effect the
demonstration was Jordan in 1892. We give here * a proof due
to de la ValUe-Poussin.-\
Let us call for brevity a continuous curve without double point
* The reader is referred to a second proof due to Brouwer and given in 598 seq.
t Cours tf Analyse, Paris, 1903, Vol. 1, p. 307.
PLANE CURVES 595
a Jordan curve. A continuous closed curve without double point
will then be a closed Jordan curve. Cf. I, 362.
577. Let 0 be a closed Jordan curve. However small <r > 0 is
taken, there exists a polygonal ring R containing C and such that
1° Each point of R is at a distance < a from 0.
2° Each point of C is at a distance < a- from the edges of R.
For let x = 0(0 , y = i/r(0 (1
be continuous one-valued functions of t in T =(#<£>) defining C.
Let D = (a, ax, <z2 ••• 6) be a division of T of norm d. Let
a, «j, «2 ••• be points of 0 corresponding to a, al ••• If d is suffi
ciently small, the distance between two points on the arc
<7t = (at_i, at) is <e', small at pleasure. Let A be a quadrate
division of the x, y plane of norm 8. Let us shade all cells con
taining a point of (7t. These form a connected domain since C, is
continuous. We can thus go around its outer edge without a
break.* If this shaded domain contains unshaded cells, let us
shade these too. We call the result a link A, . It has only one
edge En and the distance between any two points of Et is ob
viously < e' + 2 V2 8. We can choose d, & so small that
e' + 2V2 8 < 0-, arbitrarily small. (1
Then the distance between any two points of A, is < er. Let e"
be the least distance between non-consecutive arcs CL. We take
8 so small that we also have
- (2
Then two non-consecutive links AL, Aj have no point in common.
For then their edges would have a. common point P. As P lies
on EL its distance from OL is < V2 5. Its distance from Cy- is also
< V2 8. Thus there is a point Pt on (7t, and a point Pj on Cf such
that
* Here and in the following, intuitional properties of polygons are assumed as
known.
596 GEOMETRIC NOTIONS
But by hypothesis e" <_ rj. Hence
e"<2V2S,
which contradicts 2).
Thus the union of these links form a ring R whose edges are
polygons without double point. One of the edges, say Gt, lies
within the other, which we call Gr€ . The curve 0 lies within R.
The inner polygon 6rt must exist, since non-consecutive links have
no point in common.
578. 1. Interior and Exterior Points. Let <r1 > <r2 > ••• = 0.
Let Rl, R2~- be the corresponding rings, and let
be their inner and outer edges. A point P of the plane not on
0 which lies inside some Gr, we call an interior or inner point of C.
HP lies outside some Gre, we call it an exterior or outer point of C.
Each point P not on C must belong to one of these two classes.
For let p = Dist (P, (7); then p is > some <rn. It therefore lies
within 6r(tn) or without 6r<n), and is thus an inner or an outer point.
Obviously this definition is independent of the sequence of rings
\Rn\ employed. The points of the curve Care interior to each
G-™ and exterior to each G-(n).
Inner points must exist, since the inner polygons exist as al
ready observed. Let us denote the inner points by 3 and the
outer points by O. Then the frontiers of 3 and £) are the curve C.
2. We show now that
1° Two inner points can be joined by a broken line Ll lying in $.
2° Two outer points can be joined by a broken line Le lying in £).
3° Any continuous curve ® joining an inner point i and an outer
point e has a point in common with C.
To prove 3°, let
be the equations of $, the variable t ranging over an interval
, t=a corresponding to i and t=@ to e. Let t' be
T=
PLANE CURVES 597
such that a<t<t' gives inner points, while t = tf does not give an
inner point. Thus the point corresponding to t = t' is a frontier
point of 3 and hence a point of O.
To prove 1°. If A, B are inner points, they lie within some 6rt .
We may join A, B, 6rt by broken lines Za, Z& meeting (7t at the
points A', B', say. Let G-^ be the part of 6rt lying between A',
B1. Then
La + aab + Lb
is a broken line joining A to B.
The proof of 2° is similar.
579. 1. Let P', P" correspond to t = tr, t = t", on the curve 0
defined by 577, l). If t' <t", we say P' precedes P" and write
P'<P".
Any set of points on C corresponding to an increasing set of
values of t is called an increasing set.
As t ranges from a to £>, the point P ranges over C in a direct
sense.
We may thus consider a Jordan curve as an ordered set, in the
sense of 265.
2. (De la Vallee-Poussin.) On each arc Ot of the curve C, there
exists at least one point Pj such that
Jp1<p2<p3<... a
may be regarded as the vertices of a closed polygon without double
point and whose sides are all < e.
For in the first place we may take S > 0 so small that no square
of A contains a point lying on non-consecutive arcs (7t of C. Let
us also take A so that the point a corresponding to t = a lies
within a square, call it S^ of A. As t increases from t = #, there
is a last point Pl on C where the curve leaves Sr The point P^
lies in another square of A, call it S2< containing other points of
C. Let P2 be the last point of C in S2 . In this way we may
continue, getting a sequence 1).
There exists at least one point of 1) on each arc C, . For other
wise a square of A would contain points lying on non-consecutive
arcs OK . The polygon determined by 1) cannot have a double
598 GEOMETRIC NOTIONS
point, since each side of it lies in one square. The sides are < e,
provided we take SV2 < e, since the diagonal is the longest line
we can draw in a square of side S.
580. Existence of Inner Points. To show that the links form a
ring with inner points, Schonfliess* has given a proof which may
be rendered as follows :
Let us take the number of links to be even, and call them Z^,
L2,--L2n. Then L^ L3, L5~- lie entirely outside each other.
Since L^ L2 overlap, let P be an inner common point. Simi
larly let Q be an inner common point of Z2, L3. Then P, Q
lying within L2 may be joined by a finite broken line b lying
within L2. Let 62 be that part of it lying between the last point
of leaving L^ and the following point of meeting L3. In this
way the pairs of links
T T T T
X/j^g , ^3^5 I ••'
define finite broken lines
No two of these can have a common point, since they lie in
non-consecutive links. The union of the points in the sets
we call a ring, and denote it by $. The points of the plane not
in $ fall into two parts, separated by $. Let 2 denote the part
which is limited, together with its frontier. We call X the inte
rior of 9t That X has inner points is regarded as obvious since
it is defined by the links
which pairwise have no point in common, and by the broken lines
fi2 , Z>4 , #6 •••
each of which latter lies entirely within a link.
Let £ = l>v ( £) •• m = 1, 2, ...
* Die Entwickelung der Lehre von den Punktmannigfaltigkeiten. Leipzig, 1908,
Part 2, p. 170.
PLANE CURVES 599
Then these £ have pairwise no point in common since the L2m
have not.
Let £ = ?a + ?4+- +82n + fl.
Then $ > 0. For let us adjoin L2 to Sft, getting a ring 9?2 whose
interior call £2. That £2 aas inner points follows from the fact
that it contains ?4, £6 ••• Let us continue adjoining the links
L±< L6 •- Finally we reach Z2n, to which corresponds the
ring $R2n, whose interior, if it exists, is £2n. If £2n does not exist,
^2n-2 contains only £2n. This is not so, for on the edge of L^
bounding <£, is a point P, such that some Dp(P) contains points
of no L except Lr In fact there is a point P on the edge of Ll
not in either L2 or Z2n, as otherwise these would have a point in
common. Now, if however small p > 0 is taken, 7>P(P) contains
points of some L other than L^ the point P must lie in LK which
is absurd, since L^ has only points in common with £2, L2n, and
P is not in either of these. Thus the adjunction of L2, L±, •••
L2n produces a ring 9?2n whose interior £2n does not reduce to 0 ;
it has inner points.
581. Prggerti^ 9°. Area. That a figure defined by a closed
curve without double point, i.e. the interior of a Jordan curve,
has an area, has long been an accepted fact in intuitional geometry.
Thus Lindemann, Vorlesungen uber G-eometrie, vol. 2, p. 557, says
" einer allseitig umgrenzten Figur kommt ein bestimmter Flachen-
inhalt zu." The truth of such a statement rests of course on
the definition of the term area. In I, 487, 702 we have given a
definition of area for any limited plane point set 51 which reduces
to the ordinary definition when 51 becomes an ordinary plane figure.
In our language 51 has an area when its frontier points form a
discrete set. Let
define a Jordan curve (£, as t ranges over T=(a<b). The
figure 51 defined by this curve has the curve as frontier. In I,
708, 710, we gave various cases in which (£ is discrete. The
reasoning of I, 710, gives us also this important case :
If one of the continuous functions ^>, -fy defining the Jordan curve
(£, has limited variation in T, then (£ is discrete.
600
GEOMETRIC NOTIONS
It was not known whether (£ would remain discrete if the con
dition* of limited variation was removed from both coordinates,
until Osgood * exhibited a Jordan curve which is not discrete.
This we will now discuss.
. 582. 1. Osgood' s Curve. We start with a unit segment
T ' = (0, 1) on the t axis, and a unit square S in the xy plane.
We divide jTinto 17 equal parts
.12345 1C 17 rjj F7J rn s~\
d
and the square $ into 9 equal
squares
by drawing 4 bands J?x which
are shaded in the figure. On
these bands we take 8 segments,
marked heavy in the figure.
Then as t is ranging from left
to right over the even or black
intervals T2, T4, ••• Tu marked heavy in the figure, the point x, y
on Osgood's curve, call it £), shall range univariantly over the
segments 3).
While t is ranging over the odd or white intervals 2\, T3 ••• T17
the point xy on £) shall range over the squares 2) as determined
below.
Each of the odd intervals 1) we will now divide into 17 equal
intervals T^ and in each of the squares 2) we will construct
horizontal and vertical bands E2 as we did in the original square
$. Thus each square 2) gives rise to 8 new segments on £)
corresponding to the new black intervals in T, and 9 new squares
jSLJ> corresponding to the white intervals. In this way we may
continue indefinitely.
The points which finally get in a black interval call /3, the
others are limit points of the /3's and we call them X. The point
* Trans. Am. Math. Soc., vol. 4 (1903), p. 107.
PLANE CURVES 601
on D corresponding to a (3 point has been defined. The point of
O corresponding to a point X is defined to be the point lying in
the sequence of squares, one inside the other, corresponding to the
sequence of white intervals, one inside the other, in which X falls,
in the successive divisions of T.
Thus to each t in T corresponds a single point #, y in 8. The
aggregate of these points constitutes Osgood's curve. Obviously
the #, y of one of its points are one-valued functions of t in T, say
(4
The curve D has no double point. This is obvious for points of
£) lying in black segments. Any other point falls in a sequence
of squares
S,>S^>S^^
to which correspond intervals
in which the corresponding fs lie. But only one point t is thus
determined.
The functions 4) are continuous. This is obvious for points j3
lying within the black intervals of T. It is true for the points X.
For X lies within a sequence of white intervals, and while t ranges
over one of these, the point on £) ranges in a square. But these
squares shut down to a point as the intervals do. Thus $, i/r are
continuous at t = X. In a similar manner we show they are con
tinuous at the end points of the black intervals.
We note that to t = 0 corresponds the upper left-hand corner
of S, and to t = 1, the diagonally opposite point.
2. Up to the present we have said nothing as to the width of
the shaded bands „ D
-E>1 •> B\ "
introduced in the successive steps. Let
A = a1 + a2 +
be a convergent positive term series whose sum A<1. We
choose Bl so that its area is a1^ B^ so that its area is #2, etc.
Then c-> A cS 1 i /£
£) = 0,£j = l — -A, (o
602 GEOMETRIC NOTIONS
as we now show. For £) has obviously only frontier points ;
hence Q = 0. Since £) is complete, it is measurable and
5 = 6.
Let 0= #-O, and B= \Bn\. Then 0 < B. For any point
which does not lie in some Bn lies in a sequence of convergent
squares ^ > /Srt/ > ••• which converge to a point of £). Now
S=S1+S2+ ... =A.
On the other hand, B contains a null set of points of O, viz. the
black segments. Thus
0 = S = A , and hence 5 = 1 — A
and 5) is established.
Thus OsgoocTs curve is continuous, has no double point, and its
upper content is 1 — A.
3. To get a continuous closed curve C without double point
we have merely to join the two end points a, ft of Osgood's curve
by a broken line which does not cut itself or have a point in com
mon with the square S except of course the end points a, ft.
Then C bounds a figure g whose frontier is not discrete, and g
does not have an area. Let us call such curves closed Osgood
curves.
Thus we see that there exist regions bounded by Jordan curves
which do not have area in the sense current since the Greek
geometers down to the present day.
Suppose, however, we discard this traditional definition, and
employ as definition of area its measure. Then we can say :
A figure g formed of a closed Jordan curve J and its interior ^
has an area, viz. Meas g.
For Front g = J. Hence g is complete, and is therefore rneas-
ureable.
We note that & T , ^
tf = «/ + J-
We have seen there are Jordan curves such that
J>0.
DETACHED AND CONNECTED SETS 603
We now have a definition of area which is in accordance with the
promptings of our geometric intuition. It must be remembered,
however, that this definition has been only recently discovered,
and that the definition which for centuries has been accepted leads
to results which flatly contradict our intuition, which leads us to
say that a figure bounded by a continuous closed curve has an
area.
583. At this point we will break off our discussion of the
relation between our intuitional notion of a curve, and the con
figuration determined by the equations
where $, ty are one-valued continuous functions of t in an interval
T. Let us look back at the list of properties of an intuitional
curve drawn up in 563. We have seen that the analytic curve
1) does not need to have tangents at a pantactic set of points on
it; no arc on it needs have a finite length; it may completely fill
the interior of a square ; its equations cannot always be brought
in the forms y=f(x) or F(xy)=Q, if we restrict ourselves to
functions /or F employed in analysis up to the present; it does
not need to have an area as that term is ordinarily understood.
On the other hand, it is continuous, and when closed and with
out double point it forms the complete boundary of a region.
Enough in any case has been said to justify the thesis that
geometric reasoning in analysis must be used with the greatest
circumspection.
•
Detached and Connected Sets
584. In the foregoing sections we have studied in detail some
of the properties of curves defined by the equations
Now the notion of a curve, like many other geometric notions, is
independent of an analytic representation. We wish in the fol
lowing sections to consider some of these notions from this point
of view.
604 GEOMETRIC NOTIONS
585. 1. Let 51, 53 be point sets in w-way space 9?m. If
we say 51, 53 are detached. If 51 cannot be split up into two parts
53, (£ such that they are detached, we say 51 has no detached parts.
If 51 = 53 + (£ and Dist (53, £)> 0, we say 53, (£ are detached parts
of a.
Let the set of points, finite or infinite,
a, ax, «2, ••• 6 (1
be such that the distance between two successive ones is < e. We
call 1) an e-sequence between a, b ; or a sequence with segments
(«t, «t+1) of length < e. We suppose the segments ordered so
that we can pass continuously from a to b over the segments without
retracing. If 1) is a finite set, the sequence is finite, otherwise
infinite.
2. Let 51 have no detached parts. Let a, b be two of its points.
For each e > 0, there exists a finite e-sequence between a, b, and lying
• c\v
in 51.
For about a describe a sphere of radius e. About each point of
5( in this sphere describe a sphere of radius e. About each point
of 51 in each of these spheres describe a sphere of radius e. Let
this process be repeated indefinitely. Let 53 denote the points of
51 made use of in this procedure. If 53 < 51, let (£ = 51 — 53. Then
Dist (53, £)>€, an(^ 51 has detached parts, which is contrary to
hypothesis. Thus there are sets of e-spheres in 51 joining a and b.
Among these sets there are finite ones. For let g denote the
set of points in 51 which may be joined to a by finite sequences ; let
© = 51 - g. Then Dist (g, ®) > e. For if < e, there is a point /
in g, and a point g in ® whose distance is < e. Then a and g can
be joined by a finite e-sequence, which is contrary to hypothesis.
3. If 51 has no detached parts, it is dense.
For if not dense, it must have at least one isolated point a.
But then a, and 51 — a are detached parts of 51, which contradicts
the hypothesis.
4. Let 51, 53, S be complete and 51 = (53, 6). If 51 has no de
tached parts, 53, (5 ^ave at least one common point.
IMAGES 605
For if 23, (£ have no common point, 8 = Dist (23, d) is > 0.
But 8 cannot > 0, -since 23, (£ would then be detached parts of 21.
Since & = 0 and since 23, (£ are complete, they have a point in
common.
5. If 21 i« sw<?A £A«£ any two of its points may be joined by an
e-sequence lying in 21, where e > 0 is small at pleasure, 21 has no
detached parts.
For if 21 had 23, <£ as detached parts, let Dist (23, <£) = 8. Then
8 > 0. Hence there is no sequence joining a point of 23 with a
point of (£ with segments < 8.
6. If 21 is complete and has no detached parts, it is said to be
connected. We also call 21 a connex.
As a special case, a point may be regarded as a connex.
7. If 21 is connected, it is perfect.
For by 3 it is dense, and by definition it is complete.
8. If 21 is a rectilinear connex, it has a first point a and a last
point /3, and contains every point in the interval (a, /3).
For being limited and complete its minimum and maximum
lie in 21 and these are respectively a and 0. Let now
There exists an ersequence Cl between a, 0. Each segment has
an e2-sequence (72. Each segment of (72 has an e3-sequence (73,
etc. Let O be the union of all these sequences. It is pantactic
in (a, yS). As 21 is complete,
Images
586. Let ^./W-U ••• *-n=/n(*i-O (1
be one-valued functions of t in the point set £. As t ranges over
X, the point x = (xl ••• xn) will range over a set 21 in an w-way
space 9JB. We have called 21 the image of X. Cf. I, 238j 3. ^ |4
If the functions / are not one-valued, to a point t may correspond
several images xr, x" ••• finite or infinite in number. Conversely
606 GEOMETRIC NOTIONS
to the point x may correspond several values of t. If to each
point t correspond in general r values of #, and to each x in
general s values of £, we say the correspondence between £, 31 is
r to s. If r = s = 1 the correspondence is 1 to 1 or unifold ; if
r > 1, it is manifold. If r = 1, 31 is a simple image of £, other
wise it is a multiple image. If the functions 1) are one- valued
and continuous in £, we say 31 is a continuous image of £.
587. Transformations of the Plane. Example 1. Let
u = x sin y , v = x cos y. (1
We have in the first place
u^ + v2 = x2.
This shows that the image of a line x = a, a =£ 0, parallel to the
?/-axis is a circle whose center is the origin in the u, v plane, and
whose radius is a. To the ^/-axis in the a?, y plane corresponds
the origin in the u, v plane.
From 1) we have, secondly,
- = tan y.
v
This shows that the image of a line y = 6, is a line through the
origin in the u, v plane.
From 1) we have finally that w, v are periodic in y, having the
period 2 TT. Thus as x, y ranges in the band B, formed by the
two parallels y= ± TT, or — TT < y < TT, the point u, v ranges over
the entire u, v plane once and once only.
The correspondence between B and the u, v plane is unifold,
except, as is obvious, to the origin in the it, v plane corresponds
the points on the y-axis.
Let us apply the theorem of I, 441, on implicit functions. The
determinant A is here
— — x.
sin ?/, cos y
x cos y, — x sin y
As this is •=£ 0 when x, y is not on the «/-axis, we see that the
correspondence between the domain of any such point and its
image is 1 to 1. This accords with what we have found above.
IMAGES 607
It is, however, a much more restricted result than we have found ;
for we have seen that the correspondence between any limited
point set 21 in B which does not contain a point of the y-axis and
its image is unifold.
588. Example 2. Let
u
the radical having the positive sign. Let us find the image of the
first quadrant Q in the #, y plane.
From 1) we have at once
Hence the image of Q is a band B parallel to the v-axis.
From 1) we get secondly
y = uv , x = wVl — u2. (2
Hence
Thus the image of a circle in Q whose center is the origin and
whose radius is a is a segment of a right line v = a.
When x = y = 0, the equations 1) do not define the correspond
ing point in the u, v plane. If we use 2) to define the corre
spondence, we may say that to the line v = 0 in B corresponds the
origin in the x, y plane. With this exception the correspondence
between Q and B is uniform, as 1), 2) show.
The determinant A of 1) is, setting
r =
d(w, v) _ r3 r3 — x
for any point x, y different from the origin.
589. Example 3. Reciprocal Radii. Let 0 be the origin in the
x, y plane and H the origin in the u, v plane. To any point
P = (#, y) in the x, y plane different from the origin shall cor
respond a point Q = (u, v) in the w, v plane such that n Q has
608 GEOMETRIC NOTIONS
the same direction as OP and such that OP • £IQ = 1. Analyti
cally we have
x=\y , u = \v , \ > 0,
and
From these equations we get
u v f+
x = ~2 2 ' y = ~TT~2 v
and also
The correspondence between the two planes is obviously unifold
except that no point in either plane corresponds to the origin in
the other plane. We find for any point a?, y different from the
origin that a(M, v} J
d(x,y)' -(J + fyf
Obviously from the definition, to a line through the origin in
the #, y plane corresponds a similar line in the w, v plane. As xy
moves toward the origin, w, v moves toward infinity.
Let x, y move on the line x = a =£ 0. Then 1) shows that w, v
moves along the circle
a (>2 + v2) - u = 0
which passes through the origin. A similar remark holds when
z, y moves along the line y = b =£ 0.
590. Such relations between two point sets 51, 33 as defined in
586 may be formulated independently of the functions /. In fact
with each point a of H we may associate one or more points 51? bz •••
of 55 according to some law. Then 33 may be regarded as the
image of SI. We may now define the terms simple, manifold, etc.,
as in 586. When b corresponds to a we may write b ~ a.
We shall call 33 a continuous image of 51 when the following con
ditions are satisfied. 1° To each a in 51 shall correspond but one
b in 33, that is, 33 is a simple image of H. 2° Let b ~ a, let at, «2 •••
be any sequence of points in 51 which = #. Let bn ~ an. Then
bn must = Z> however the sequence \an\ is chosen.
IMAGES 009
When 53 is a simple image of 51, the law which determines
which b of 53 is associated with a point a of 51 determines obviously
n one-valued functions as in 586, 1), where ^ ••• tm are the m co
ordinates of a, and xl ••• xn are the n coordinates of b. We call these
functions 1) the associated functions. Obviously when 53 is a
continuous image, the associated functions are continuous in 51.
591. 1. Let 53 be a simple continuous image of the limited complete
set 5(. Then 1° 53 is limited and complete. If 2° 51 is perfect and
only a finite number of points of 51 correspond to any point 0/53, then
53 is perfect. -5^3° 51 is a connex, so is 53.
To prove 1°. The case that 53 is finite requires no proof. Let
ftj, £2 "• be points of 53 which = /9. We wish to show that 0 lies
in 53. To each bn will correspond one or more points in 51; call
the union of all these points a. Since 53 is a simple image, a is an
infinite set. Let av #2 "' be a set of points in a which = «, a
limiting point of 51. As 51 is complete, a lies in 51. Let b ~ a.
Let b,n ~ an. As an = a, bln = /3. But 53 being continuous, b^
must = b. Thus /3 lies in 53. That 53 is limited follows from the
fact that the associated functions are continuous in the limited
complete set 51. To prove 2°. Suppose that 53 had an isolated
point b. Let b ~ a. Since 51 is perfect, let a^ a2>~ = a. Let
bn ~ an. Then as 53 is continuous, bn = 5, and b is not an isolated
point. To prove 3°. We have only to show that there exists
an e-sequence between any two points a, /3 of 53, e small at pleasure.
Let a ~ a, @ ~ b. Since 51 is connected there exists an ^-sequence
between a, b. Also the associated functions are uniformly con
tinuous in 51, and hence 77 may be taken so small that each segment
of the corresponding sequence in 53 is > e.
2. Let f{t1 ••• £TO) be one-valued and continuous in the connex 51,
then the image of 51 is an interval including its end points.
This follows from the above and from 585, 8.
3. Let the correspondence between 51, 53 be unifold. If 53 is a
continuous image of 51, then 51 is a continuous image of 53.
For let \bn\ be a set of points in 53 which = b. Let an ~ 5n,
a ~ b. We have only to show that an = a. For suppose that it
does not, suppose in fact that there is a sequence ati, ati ••• which
610 GEOMETRIC NOTIONS
= a 3= a. Let {3 ~ a. Then 5tl, ba ••• = /3. But any partial se
quence of jfinj must = b. Tims & = /3, hence a = «, hence an = a.
4. w4 Jordan curve J is a unifold continuous image of an interval
T. Conversely if J is a unifold continuous image of an interval T,
there exist two one-valued continuous functions
such that as t ranges over T, the point x, y ranges over J. In case
J is closed it may be regarded as the image of a circle F.
All but the last part of the theorem has been already established.
To prove the last sentence we have only to remark that if we set
x = r cos t , y = r sin t
we have a unifold continuous correspondence between the points
of the interval (0, 2 TT*) and the points of a circle.
5. The first part of 4 may be regarded as a geometrical definition
of a Jordan curve. The image of a segment of the interval T or
of the circle F, will be called an arc of J.
592. Side Lights on Jordan Curves. These curves have been
defined by means of the equations
As t ranges over the interval T = (a < 5), the point P = (x, y)
ranges over the curve J. This curve is a certain point set in the
#, y plane. We may now propose this problem : We have given
a point set (£ in the a?, y plane ; may it be regarded as a Jordan
curve ? That is, do there exist two continuous one-valued func
tions 1) such that as t ranges over some interval T, the point P
ranges over the given set (£ without returning on itself, except
possibly for t = a, t = b, when the curve would be closed?
Let us look at a number of point sets from this point of view.
593. Example 1.
L Let y = sin - , x in the interval % = (- 1, 1), but ^ 0
x
= 0 for x = Q.
IMAGES 611
Is this point set (£ a Jordan curve ? The answer is, No. For a
Jordan curve is a continuous image of an interval 51. By 591, 1,
it is complete. But (E is not complete, as all the points on the
y axis, — 1 < y < 1 are limiting points of (£, and only one of them
belongs to (£, viz. the origin.
2. Let us modify (£ by adjoining to it all these missing limiting
points, and call the resulting point set (7. Is C a Jordan curve ?
The answer is again, No. For if it were, we can divide the inter
val T into intervals 8 so small that the oscillation of c£, ^ in any
one of them is < &>. To the intervals 8, will correspond arcs Ot on
the curve, and two non-consecutive arcs Ct are distant from each
other by an amount > some e, small at pleasure. This shows that
one of these arcs, say CK, must contain the segment on the y-axis
~~ 1 ^ V ^ !• But then Osc ^r = 2 as t ranges over the correspond
ing 8K interval. Thus the oscillation of i/r cannot be made < e,
however small £K is taken.
3. Let us return to the set (£ defined in 1. Let A, B be the
two end points corresponding to x = — 1, x = 1. Let us join them
by an ordinary curve, a polygon if we please, which does not cut
itself or (£. The resulting point set $ divides all the other points
of the plane into two parts which cannot be joined by a contin
uous curve without crossing $. For this point of view $ must be
regarded as a closed configuration. Yet $ is obviously not complete.
On the other hand, let us look at the curve formed by removing
the points on a circle between two given points a, b on it. The
remaining arc % including the end points a, b is a complete set, but
as it does not divide the other points of the plane into two sepa
rated parts, we cannot say £ is a closed configuration.
We mention this circumstance because many English writers
use the term closed set where we have used the term complete.
Cantor, who first introduced this notion, called such sets abge-
schlossen, which is quite different from geschlossen = closed.
_!
594. Example 2. Let p = e e ', for 6 in the interval 31 = (0, 1)
except 6 = 0, where p = 0. These polar coordinates may easily be
replaced by Cartesian coordinates
-i -1
e 'cos 6 , = e 9 sin d , in 51,
612 GEOMETRIC NOTIONS
except 6 = 0, when x, y both = 0. The curve thus denned is a
Jordan curve.
Let us take a second Jordan curve
with p = 0 for 6 = 0. If we join the two end points on these
curves corresponding to 6 = 1 by a straight line, we get a closed
Jordan curve J", which has an interior <J, and an exterior O.
The peculiarity of this curve J is the fact that one point of it,
viz. the origin # = ;z/ = 0, cannot be joined to an arbitrary point
°f 3s ky a finite broken line lying entirely in $ ; nor can it be
joined to an arbitrary point in £) by such a line lying in £)•
595. 1. It will be convenient to introduce the following terms.
Let 31 be a limited or unlimited point set in the plane. A set
of distinct points in 21
«i » «2 » <V" (1
determine a broken line. In case 1) is an infinite sequence, let an
converge to a fixed point. If this line has no double point, we call
it a chain, and the segments of the line links. In case not only the
points 1) but also the links lie in 21, we call the chain a path. If
the chain or path has but a finite number of links, it is called
finite.
Let us call a precinct a region, i.e. a set all of whose points are
inner points, limited or unlimited, such than any two of its points
may be joined by a finite path.
2. Using the results of 578, we may say that, —
A closed Jordan curve J divides the other points of the plane into
two precincts, an inner $ and an outer £)• Moreover, they have a
common frontier which is J.
3. The closed Jordan curve considered in 594 shows that not
every point of such a closed Jordan curve can always be joined to
an arbitrary point of 3 or O by a finite path.
Obviously it can by an infinite path. For about this point, call
it P, we can describe a sequence of circles of radii r = 0. Between
any two of these circles there lie points of Q and of £), if r is suf-
IMAGES 613
liciently small. In this way we may get a sequence of points in 3,
viz. /j, /2 ... = P. Any two of these Im, Im+1 may be joined by a
path which does not cut the path joining /j to Im. For if a loop
were formed, it could be omitted.
4. Any arc & of a closed Jordan curve J can be joined by a path
to an arbitrary point of the interior or exterior, which call 51.
For let «7= $ -f- £• Let & be a point of $ not an end point.
Let S = Dist(&, £), let a be a point of 51 such that Dist(a, &)
< ^ B. Then ~. ox 1 .
??= Dist(a, £) > Jo.
Hence the link I = (a, &) has no point in common with £. Let
b be the first point of I in common with $. Then the link
m = (a, £) lies in 31. If now a is any point of 21, it may be joined
to a by a path p. Then p + m is a path in 51 joining the arbi
trary point a to a point b on the arc $.
596. Example 3. For d in 51 = (0*, 1) let
and -O^iX
p = a(H-« v <").
These equations in polar coordinates define two non-intersecting
spirals jS^ Sz which coil about p = a as an asymptotic circle F.
Let us join the end points of the spirals corresponding to 0 = 1
by a straight line L. Let § denote the figure formed by the
spirals S^ $2, the segment L and the asymptotic circle F. Is £
a closed Jordan curve ? The answer is, No. This may be seen
in many ways. For example, § does not divide the other points
into two precincts, but into three, one of which is formed of points
within F.
Another way is to employ the reasoning of 593, 2. Here the
circle F takes the place of the segment on the y-axis which figures
there.
Still another way is to observe that no point on F can be joined
to a point within S by a path.
597. Example 4- Let d be formed of the edge & of a unit
square, together with the ordinates o erected at the points
614 GEOMETRIC NOTIONS
x = ™ , of length ^, ft = 1, 2 - Although (£ divides the other
points of the plane into two precincts 3 and D, we can say that
($ is not a closed Jordan curve.
For if it were, 3 and £) would have to have (£ as a common
frontier. But the frontier of £) is (§, while that of 3 is (£.
That (£ is not a Jordan curve is seen in other ways. For
example, let 7 be an inner segment of one of the ordinates o.
Obviously it cannot be reached by a path in £).
Brouwers Proof of Jordan's Theorem
598. We have already given one proof of this theorem in 577
seq., based on the fact that the coordinates of the closed curve are
expressed as one-valued continuous functions
Brouwer's proof * is entirely geometrical in nature and rests
on the definition of a closed Jordan curve as the unifold continu
ous image of a circle, cf. 591, 5.
If 51, 53, ••• are point sets in the plane, it will be convenient to
denote their frontiers by g^, g^ ••• so that
%^ = Front H , etc.
We admit that any closed polygon ty having a finite number of
sides, without double point, divides the other points of the plane
into an inner and an outer precinct $t, tye respectively. In the
following sections we shall call such a polygon simple, and usu
ally denote it by ty.
We shall denote the whole plane by (§.
Then <g= ?+$.+?..
Let 51 be complete. The complementary set A is formed, as
we saw in 328, of an enumerable set of precincts, say A = \An\.
* Math. Annalen, vol. 69 (1910), p. 169.
BROUWER'S PROOF OF JORDAN'S THEOREM 615
599. 1. If a precinct 31 and its complement* A each contain a
point of the connex (£, then g^ contains a point of (L
For in the contrary case c = Dv(2I, (£) is complete. In fact
33 = 21 + Sgi is complete. As (£ is complete, Dv($8, (S) is com
plete. But if §21 does not contain a point of Q, c = Dv(33, (£).
Thus on this hypothesis, c is complete. Now c = Dv(A, (E) is
complete in any case. Thus (£ = c + <?, which contradicts 585, 4.
2. If ^^ ^$e, the interior and exterior of a simple polygon ty each
contain a point of a connex (£, then ty contains a point of (L
3. Let $ be complete and not connected. There exists a simple
polygon $ such that no point of $ lies on $, while a part of $ lies in
$t and another part in tye.
For let $j, $2 be two non-connected parts of $ whose distance
from each other is p > 0. Let A be a quadrate division of the
plane of norm 8, so small that no cell contains a point of $j and
$2. Let A! denote the cells of A containing points of $j. It is
complete, and the complementary set A2 = @ — Aj is formed of one
or more precincts. No point of ^l lies in A2 or on its frontier.
Let Pj, P2 be points in $j, $2 respectively. Let D be that
precinct containing P2. Then g^ embraces a simple polygon ^3
which separates Pl and P2.
4. Let ®1, $2 be two detached connexes. There exists a simple
polygon $ which separates them. One of them is in ^ the other in
$e, and no point of either connex lies on ^3.
For the previous theorem shows that there is a simple polygon
$ which separates a point Pl in ®1 from a point P2 in $2 and no
point of $j or f 2 lies on $. Call this fact F.
Let now Pl lie in $t . Then every point of $x lies in % . For
otherwise ^ and $e each contain a point of the connex $j . Then
2 shows that a point of $j lies on $, which contradicts F.
5. Let 53 be a precinct determined by the connex (L Then
b = Front 53 is a connex.
* Since the initial sets are all limited, their complements may be taken with ref
erence to a sufficiently large square O ; and when dealing with frontier points, points
on the edge of O may be neglected.
616 GEOMETRIC NOTIONS
For suppose b is not a connex. Then by 3, there exists a simple
polygon ty which contains a part of b in ^ and another in ^3e,
while no point of b lies on ty. Hence a point fi' of b lies in $t,
and another point /3" in $„. As 53 is a precinct, let us join /3',
fi'f by a path v in 53- Thus ty contains at least one point of v,
that is, a point of 53 lies on $. As b and ty have no point in
common, and as one point of ty lies in 53> all the points of ^ lie
in 53. Hence ^(^ e) = 0> (1
As b is a part of (£ and hence some of the points of (£ are in $e
and some in ^^ it follows from 2 that a part of $ lies in (£. This
contradicts 1).
6. Let $1? $2 be two connexes without double point. By 3
there exists a simple polygon ty which separates them and has
one connex inside, the other outside ^3.
Now $ = $j + $2 is complete and defines one or more precincts.
One of these precincts contains ty.
For say $ lay in two of these precincts as 51 and 53. Then the
precinct 51 and its complement (in which 53 lies) each contain a
point of the connex $. Thus g^ contains a point of $. But g^
is a part of $, and no point of $ lies on $.
That precinct in Coinp $ which contains ^3 we call the inter
mediate precinct determined by $x, $2, or more shortly the pre
cinct between $x, $2 and denote it by Inter ($r $2).
7. Let $j, $2 be two detached connexes, and let ! = Inter ($r $2).
Then $!, $2 can I e joined by a path lying in I, except its end points
which lie on the frontiers of $v $2 respectively.
For by hypothesis p = Dist($j, $2)>0. Let P1 be a point of
5^ such that some domain b of Pl contains only points of $j and
of I. Let Q! be a point of f in b. Join Px, Ql by a right line, let
it cut g^j first at the point P' . In a similar way we may reason
on $2, obtaining the points P", Q2. Then P' Q^P" is the path
in question. If we denote it by v, we may let v* denote this
path after removing its two end points.
8. Let $j, $2 t>e ^wo detached connexes. A path v joining $x,
$2 and lying in 1= Inter ($j, $2), end points excepted, determines
one and only one precinct in f .
BROUWER'S PROOF OF JORDAN'S THEOREM 617
For from an arbitrary point P in !, let us draw all possible
paths to v. Those paths ending on the same side (left or right)
of v certainly lie in one and the same precinct fr or !j in f. Then
since one end point of v is inside, the other end point outside ^)3,
there must be a part of ty which is not met by v and which joins
the right and left sides of v. We take this as an evident property
of finite broken lines and polygons without double points.
Thus !j and !r are not detached ; they are parts of one precinct.
9. Two paths v^ v2 without common point, lying in I and joining
$j, $2, split t into two precincts.
Let i = ! — vl ; this we have just seen is a precinct. From any
point of it let us draw paths to i>2. Those paths ending on the
same side of v2 determine precincts tj, tr which may be identical.
Suppose they are. Then the two sides of v2 can be joined by a
path lying in I, which does not touch v2 (end points excepted),
has no point in common with v1, and together with a segment of
v% forms a simple polygon ty which has one end point of vl in ^Pt,
the other end point in tye. Thus by 2, $ contains a point of the
connex vr This is contrary to hypothesis.
Similar reasoning shows that
10. The n paths v^ ••• vn pair wise without common point, lying in
f, and joining the connexes $j, $2 split ! into n precincts.
Let us finally note that the reasoning of 595, 4, being independ
ent of an analytic representation of a Jordan curve, enables us to
use the geometric definition of 591, 5, and we have therefore the
theorem
11. Let 51 be a precinct whose frontier g is a Jordan curve. Then
there exists a path in $t joining an arbitrary point of 5( with any arc
0/8-
Having established these preliminary theorems, we may now
take up the body of the proof.
600. 1. Let 51 be a precinct determined by a closed Jordan curve
J. Then g = Front 51 is identical with J.
If J determines but one precinct 51 which is pantactic in (£, we
have obviously 5 = J,
618 GEOMETRIC NOTIONS
Suppose then that 51 is a precinct, not pantactic in (£. Let 53
be a precinct =£ 21 determined by g. Let b = Front 53. Then
b <_ g <^J. Suppose now b < J. As J" is a connex by 591, 1, g is a
connex by 599, 5. Similarly since g is a connex, b is a connex.
Since b < </, let b ~ b on the circle F whose image is J. We
divide b into three arcs 5X, £>2, 63 to which ~ b1? b2, b3 in b.
Let /3= Inter (bx, b3).
Then by 599, 11, we can join bx, b3 by a path vl in 51, and by a
path vz in 53. By 599, 9, these paths split /2 into two precincts
ffii A- We can join v1? #2 by a path wx lying in /3X, and by a
path w2 lying in /32 .
Now the precinct 53 and its complement each contain a point of
the connex Wj. Hence by 599, l, b contains a point of ur Simi
larly b contains a point of u2. Thus u^ u2 cut b, and as they
do not cut bx, b3 by hypothesis, they cut b2. Thus at least one
point of fit and one point of 02 lie in b2 .
Let p be a point of /^ lying in b2, let p ~p on the circle. Let
b' be an arc of b2 containing p. Let b' ~ b' . As the connex b'
has no point in common with Front /3X, b' must lie entirely in /31
by 599, 1. This is independent of the choice of b', hence the
connex b2, except its end points, lies in @v Thus /32 can contain
no point of b2, which contradicts the result in italics above.
Thus the supposition that b < J is impossible. Hence b = J,
and therefore g = J.
As a corollary we have :
2. A Jordan curve is apantactic in (§.
3. A closed Jordan curve J cannot determine more than two
precincts.
For suppose there were more than two precincts
"a,, a,, a,- (i
Let us divide the circle T into four arcs whose images call «7j, J%,
Ji,/4-
Then by 1, the frontier of each of the precincts 1) is J. Thus
by 599, 9, there is a path in each of the precincts 2lx, ^ ••• join
ing Jj and J"3. These paths split
DIMENSIONAL INVARIANCE 619
! = Inter (Jv <73)
into precincts fj, I2 ••'•
Now as in 1, we show on the one hand that each lt must contain
a point of «72 or <74, and on the other hand neither J2 nor J~4 can
lie in more than one !t.
4. A dosed Jordan curve J must determine at least tivo precincts.
Suppose that J determines but a single precinct 21. From a
point a of 21 we may draw two non-intersecting paths M15 w2 to
points 6j, #2 of J.
Since the point a may be regarded as a connex, a and J are two
detached connexes. Hence by 599, 9, the paths u^ uz split 21 into
two precincts 2lx, 2Lj- Let j = (w19 w2, J"). The points 6X, b2
divide J"into two arcs «7X, J"2, and
are closed Jordan curves. Regarding a and Jj as two detached
connexes, we see y\ determines two precincts, «j, o^. By 599, l, a
path which joins a point ax of ax with a point az of Og must cut jl
and hence y. It cannot thus lie altogether in ^ or in 212 . Thus
both «j, a2 do not lie in 2lj, nor both in 212. Let us therefore
say for example that 2lj lies in «x, and 2^ in «2. Hence by 2,
2lj is pantactic in «j, and 2(2 in «2- -^7 ^' eac^ P°int of jl is com
mon to the frontiers of ctj and of Og, and hence of 2IX and of 212,
as these are pantactic.
Let P be a point of J2 . It lies either in «j or ct^. Suppose it
lies in Oj. Then it lies neither in c^ nor on Front a^ and hence
neither in 212 nor on Front 2I2.. But every point of /2 and also
every point of jl lies on Front 212. We are thus brought to a
contradiction. Hence the supposition that J determines but a
single precinct is untenable.
Dimensional Invariance
601. 1. In 247 we have seen that the points of a unit interval
/, and of a unit square S may be put in one to one correspondence.
This fact, due to Cantor, caused great astonishment in the mathe
matical world, as it seemed to contradict our intuitional views
620 GEOMETRIC NOTIONS
regarding the number of dimensions necessary to define a figure.
Thus it was thought that a curve required one variable to define
it, a surface two, and a solid three.
The correspondence set up by Cantor is not continuous. On
the other hand the curves invented by Peano, Hilbert, and others
(cf. 573) establish a continuous correspondence between /and S,
but this correspondence is not one to one. Various mathemati
cians have attempted to prove that a continuous one to one corre
spondence between spaces of m and n dimensions cannot exist.
We give a very simple proof due to Lebesgue.*
It rests on the following theorem :
2. Let 51 be a point set in 9?m . Let O < 21 be a standard cube
aL<'2<T , 4=1, 2-~m.
Let (Sj, ^-"be a finite number of complete sets so small that each
lies in a standard cube of edge a. If each point of 51 lies in one of
the (S's, there is a point of 51 ivhich lies in at least m + 1 of them.
Suppose first that each (£t is the union of a finite number of
standard cubes. Let (^ denote those (Ts containing a point of
the face \l of Q lying in the plane xl = ar The frontier gx of (^
is formed of a part of the faces of the GTs. Let Fl denote that
part of gj which is parallel to fj. Let Ox = i)y(O, .Fj). Any
point of it lies in at least two (Ts.
Let @2 denote those of the GTs not lying altogether in @x and
containing a point of the face f2 of Q determined by x% = a2. Let
JP2 denote that part of Front @2 which is parallel to f2. Let
O2 = Dt^Qj, F^). Any point of it' lies in at least three of the (Ts.
In this way we may continue, arriving finally at Om, any point
of which lies in at least m -\- 1 of the GTs.
Let us consider now the general case. We effect a cubical
division of space of norm d<a. Let O, denote those cells of D
which contain a point of (£t. Then by the foregoing, there is a
point of 51 which lies in at least m + 1 of the (7's. As this is true,
however small d is taken, and as the (£' s are complete, there is at
least one point of 51 which lies in m + 1 of the GTs.
* Math Annalen, vol. 70 (1911), p. 166.
DIMENSIONAL INVARIANCE 621
3. We now note that the space $lm may be divided into congruent
cells so that no point is in more than m -h 1 cells.
For m = 1 it is obvious. For m = 2 we may
use a hexagonal pattern. We may also use
a quadrate division of norm B of the plane.
These squares may be grouped in horizontal
bands. Let every other band be slid a distance
^ 8 to the right. Then no point lies in more
than 3 squares. For m = 3 we may use a
cubical division of space, etc.
In each case no point of space is in more than m + 1 cells.
Let us call such a division a reticulation of 9?m .
4. Let 51 be a point set in $?OT having an inner point a. There is
no continuous unifold image $8 of 51 in 9?n, n^=m, such that b~a is
an inner point of $8.
For let n > m. Let us effect a reticulation R of 9?m of norm p.
If B > 0 is taken sufficiently small A = Z>26(a) lies in 51. Let
E = D^d) ; if p is taken sufficiently small, the cells
o^o.-.o. (i
of R which contain points of E, lie in A. Let the image of E be
(£, and that of the cells 1) be
SpG.-C.. (2
These are complete. Each point of (5 lies in one of the sets 2).
Hence by 2, they contain a point /3 which lies in n + 1 of them.
Then a~/3 lies in n 4- 1 of the cells 1). But these, being part of
the reticulation R, are such that no point lies in more than m + 1
of them. Hence the contradiction.
602. 1. Schotifliesi Theorem. Let
u =/O, #) , v = g(x, y*) (1
be one-valued and continuous in a unit square A whose center is
the origin. These equations define a transformation T. If T is
regular, we have seen in I, 742, that the domain Z>P(P) of a point
P = (x, y} within A goes over into a set E such that if Q~P
then -Z>a((?) lies in E, if cr >0 is sufficiently small.
622 GEOMETRIC NOTIONS
These conditions on /, g which make T regular are sufficient,
but they are much more than necessary as the following theorem
due to Schonfliess * shows.
2. Let A = B -f- c be a unit square in the x, y plane, whose center
is the origin and whose frontier is c.
u =/(?* y) , v = g(x, y)
be one-valued continuous functions in A. As (x, y} ranges over A,
let (w, v) range over 21 = 53 + c where c ~ c. Let the correspondence
between A and 51 be uniform. Then c is a closed Jordan curve and
the interior c, of c is identical with 53.
That c is a closed Jordan curve follows from 576 seq., or 598
seq. Obviously if one point of 53 lies in C0 all do. For if &, j3e
are points of 53, one within c and the other without, let Jt~&,
be~@e- Then 5t, be lying in B can be joined by a path in B
which has no point in common with c. The image of this path is
a continuous curve which has no point in common with c, which
contradicts 578, 2.
Let
be the equation of c in polar coordinates.
If 0 < ft < 1, the equation
P = 14(0)
defines a square, call it CM, concentric with c and whose sides are
in the ratio /* : 1 with those of c. The equations of CM ~ c^ are
v=g\- .•-....... j= 00, 0).
These CM curves have now the following property :
If a point (p, q) is exterior {interior) to cMo, it is exterior (in
terior) to CM , for all fA such that
I /* — MO I ^ some e > 0.
For let /3M be the distance of (/?, <?) from a point (w, v) on CM.
Then
*Goettingen Nachrichten, 1899. The demonstration here given is due to Osgood,
Goett. Nachr., 1900.
AREA OF CURVED SURFACES 623
is a continuous function of 6, p which does not vanish for //, = /*0,
when 0 < 6 < 2 TT. But being continuous, it is uniformly con
tinuous. It therefore does not vanish in the rectangle
- e + /i0 < AC < AIO + e , 0 < 0 < 2 TT.
We can now show that if 53<c0 it is identical with ct. To this
end we need only to show that any point ft of ct lies on some CM.
In fact, as /x = 0, CM contracts to a point. Thus ft is an outer point
of some CM, and an inner point of others. Let /u0 be the maximum
of the values of n such that ft is exterior to all cu, if ft < /v
Then ft lies on cMo. For if not, ft is exterior to cMo+«, by what we
have just shown, and /x0 is not the maximum of ft.
Let us suppose that 23 lay without c. We show this leads to a
contradiction. For let us invert with respect to a circle f, lying
in ct. Then c goes over into a curve f, and 51 goes over into
3) = (S -}- f . Then & lies inside f. Let f , 77 be coordinates of a
point of £). Obviously they are continuous functions of z, y in
A, and <TN « •£ i
A~x) , c ~ f , uniformly.
By what we have just proved, (£ must fill all the interior of f.
This is impossible unless 51 is unlimited.
3. We may obviously extend the theorem 2 to the case
u\ =/iOi •••'O "'um = /«(*!•••*„)
and A is a cube in m-way space 9?m, provided we assume that c, the
image of the boundary of J., divides space into two precincts
whose frontier is c.
Area of Curved Surfaces
603. 1. TJie Inner Definition. It is natural to define the area of a
curved surface in a manner analogous to that employed to define
the length of a plane curve, viz. by inscribing and circumscrib
ing the surface with a system of polyhedra, the area of whose
faces converges to 0. It is natural to expect that the limits of
the area of these two systems will be identical, and this common
limit would then forthwith serve as the definition of the area of
the surface. The consideration of the inner and the outer sys-
GEOMETRIC NOTIONS
terns of polyhedra afford thus two types of definitions, which
may be styled the inner and the outer definitions. Let us look
first at the inner definition.
Let the equations of the surface S under consideration be
x =
v)
y = ^r(w, v)
z = xO, v),
(1
the parameters ranging over a complete metric set 21, and a?, y, z
being one-valued and continuous in 21.
Let us effect a rectangular division D of norm d of the u, v
plane. The rectangles fall into triangles tK on drawing the
diagonals. Such a division of the plane we call quasi rectangular.
Let P0=(«0,«0) , P, = Oo + M) , P2 = O0, fo + 'O
be the vertices of tK. To these points in the u, v plane corre
spond three points $«. = (#t, y^ «t), t=l, 2, 3, of £ which form the
vertices of one of the triangular faces TK of the inscribed polyhe
dron 11^ corresponding to the division D. Here, as in the follow
ing sections, we consider only triangles lying in 21. We may do
this since 21 is metric.
Let XK, YK, ZK be the projections of TK on the coordinate planes.
Then, as is shown in analytic geometry,
where
2/2
2/2-2/0
A'y , A'z
and similar expressions for y"K, ZK.
Thus the area of II ^ is
the summation extending over all the triangles tK lying in the
set 21.
Let x, y, z have continuous first derivatives in 21- Then
dx
AREA OF CURVED SURFACES
62;')
with similar expressions for the other increments. Let
A =
du
*y
dv
dz
du
dz
dv
Then
X=(^K + «K)^
dz
du
dz
~dv
dz_
du
dz
dv
dz
du
dx_
~dv
du
dv
YK
where aK /3K yK are uniformly evanescent with d in 3(. Thus if
^., .6, C do not simultaneously vanish at any point of 21, we have
as area of the surface S
2. An objection which at once arises to this definition lies in
the fact that we have taken the faces of our inscribed polyhedra
in a very restricted manner. We cannot help asking, Would we
get the same area for S if we had chosen a different system of
polyhedra ?
To lessen the force of this objection we observe that by replac
ing the parameters u, v by two new parameters u', v' we may
replace the above quasi rectangular divisions which correspond to
the family of right lines u = constant, v = constant by the infinitely
richer system of divisions corresponding to the family of curves
u' = constant, v' = constant. In fact, by subjecting u', v' to cer
tain very general conditions, we may transform the integral 3)
to the new variables u', v' without altering its value.
But even this does not exhaust all possible ways of dividing 51
into a system of triangles with evanescent sides. Let us there
fore take at pleasure a system of points in the u, v plane having
no limiting points, and join them in such a way as to cover the
plane without overlapping with a set of triangles tK. If each
triangle lies in a square of side d, we may call this a triangular
division of norm d. We may now inquire if SD still converges
to the limit 3). as d = 0, for this more general system of divisions.
It was generally believed that such was the case, and standard
treatises even contained demonstrations to this effect. These
demonstrations are wrong ; for Schwarz * has shown that by
* Werke, vol. 2, p. 309.
626
GEOMETRIC NOTIONS
properly choosing the triangular divisions D, it is possible to
make SD converge to a value large at pleasure, for an extensive
class of simple surfaces.
604. 1. Schwarzs Example. Let Q be a right circular cylin-
der of radius 1 and height 1. A set of planes parallel to the base
at a distance - apart cuts out a system of circles rx, F2
Let
us divide each of these circles into m equal
arcs, in such a way that the end points of
the arcs on Fx, F3, F6 ••• lie on the same
vertical generators, while the end points of
F2, F4, F6 ••• lie on generators halfway
between those of the first set. We now
inscribe a polyhedron so that the base of
one of the triangular facets lies on one
circle while the vertex lies on the next circle above or below, as
in the figure.
The area t of one of these facets is
t = 1 bh
Thus
m
= \ A:+ 1
cos
m
m ' w* 2m
There are 2 m such triangles in each layer, and there are n
layers. Hence the area of the polyhedron corresponding to this
triangular division D is
84 = 2^ = 2 mn sin — \ — h 4 sin4 — .
m * M? 2m
Since the integers w, n are independent of each other, let us
consider various relations which may be placed on them.
Case 1°. Let n = \m. Then
S0 =
2
2
<2
9-v • 7T / 1
7W A. ^sl 11 A /
+ 4 sin4 —
2 m
• 7T
Sill —
2 TT m
7T
sin
2m
7T
.
m TT ,
'X2W2 1 ^w4
m V
TT , as m = QO
2m
AREA OF CURVED SURFACES
627
Case 2°. Let n = \m*. Then
+ 4^f-4
sin
7T
= 2 Try/1 + — X2 , as w = x.
Case 3°. Let w = Xw3. Then
sin -
m
/ .,-..•
sin^
TT
l + ^m\
7T
m J
V
'2m
, as m = x.
= +x>
2. Thus only in the first case does SD converge to 2 TT, which
is the area of the cylinder C as universally understood. In the
2° and 3° cases the ratio h/b = 0. As equations of C we may
take
x = cos u
= sin
z = v.
Then to a triangular facet of the inscribed polyhedron will cor
respond a triangle in the u, v plane. In cases 2° and 3° this tri
angle has an angle which converges to TT as m = oo. This is not
so in case 1°. Triangular divisions of this latter type are of great
importance. Let us call then a triangular division of the u, v
plane such that no angle of any of its triangles is greater than
7T — e, where e > 0 is small at pleasure but fixed, positive triangu
lar divisions. We employ this term since the sine of one of the
angles is > some fixed positive number.
605. The Outer Definition. Having seen one of the serious diffi
culties which arise from the inner definition, let us consider briefly
the outer definition. We begin with the simplest case in which
the equation of the surface S is
=/(*,
(1
/ being one-valued and having continuous first derivatives. Let
us effect a metric division A of the #, y plane of norm S, and on
628 GEOMETRIC NOTIONS
each cell dK as base, we erect a right cylinder (7, which cuts out an
element of surface 8K from S. Let *$„ be an arbitrary point of SK
and !£„ the tangent plane at this point. The cylinder O cuts out
of XK an element ASK . Let VK be the angle that the normal to £K
makes with the z-axis. Then
1
C°S "* = I 7]Tx2— -f^i
^(IHD:
and
COS VK
The area of & is now defined to be
lim 2A#K (2
5=0
when this limit exists. The derivatives being continuous, we have
at once that this limit is
which agrees with the result obtained by the inner definition in
603, 3).
The advantages of this form of definition are obvious. In the
first place, the nature of the divisions A is quite arbitrary ; however
they are chosen, one and the same limit exists. Secondly, the most
general type of division is as easy to treat as the most narrow, viz.
when the cells dK are squares.
Let us look at its disadvantages. In the first place, the elements
A/S^ do not form a circumscribing polyhedron of S. On the con
trary, they are little patches attached to S at the points tyK, and
having in general no contact with one another. Secondly, let us
suppose that S has tangent planes parallel to the z-axis. The de
rivatives which enter the integral 603, 3) are no longer continuous,
and the reasoning employed to establish the existence of the limit
2) breaks down. Thirdly, we have the case that z is not one-
valued, or that the tangent planes to S do not turn continuously,
or do not even exist at certain points.
AREA OF CURVED SURFACES 629
To get rid of these disadvantages various other forms of outer
definitions have been proposed. One of these is given by G-oursat
in his Cours d* Analyse. Instead of projecting an arbitrary
element of surface on a fixed plane, the xy plane, it is projected on
one of the tangent planes belonging to that element. Hereby the
more general type of surfaces defined by 603, 1) instead of those
defined by 1) above is considered. The restriction is, however,
made that the normals to the tangent planes cut the elements of
surface but once, also the first derivatives of the coordinates are
assumed to be continuous in 21. Under these conditions we get
the same value for the area as that given in 603, 3).
When the first derivatives of #, y, z are not continuous or do
not exist, this definition breaks down. To obviate this difficulty
de la ValUe-Poussin has proposed a third form of definition in his
Cours d Analyse, vol. 2, p. 30 seq. Instead of projecting the
element of surface on a tangent plane, let us project it on a plane
for which the projection is a maximum. In case that S has a con
tinuously turning tangent plane nowhere parallel to the 2-axis, de
la Vallee-Poussin shows that this definition leads to the same
value of the area of S as before. He does not consider other cases
in detail.
Before leaving this section let us note that Jordan in his Cours
employs the form of outer definition first noted, using the paramet
ric form of the equations of S. In the preface to this treatise the
author avows that the notion of area is still somewhat obscure, and
that he has not been able "a definir d'une maniere satisfaisante
1'aire d'une surface gauche que dans le cas ou la surface a un plan
tangent variant suivant une loi continue."
606. 1. Regular Surfaces. Let us return to the inner definition
considered in 603. We have seen in 604 that not every system of
triangular divisions can be employed. Let us see, however, if we
cannot employ divisions much more general than the quasi rec
tangular. We suppose the given surface is defined by
x = </>(>, v) , y = \lr(u, v) , z = %(>, v) (1
the functions </>, i/r, ^ being one- valued, totally differentiate func
tions of the parameters u, v which latter range over the complete
630
GEOMETRIC NOTIONS
metric set SI. Surfaces characterized by these conditions we
shall call regular. Let
be the vertices of one of the triangles tK, of a triangular division
2> of norm ^ of SI. As before let $0, ^x, $2 be the corresponding
points on the surface S. Then
and similar expressions hold for the other increments. Also
du dv
+ 2 jr.',
where X* denotes the sum of several determinants, involving the
infinitesimals
«i , < , ft , ft.
Similar expressions hold for J^, ZK. We get thus
XK = AJK+X'K , YK = £&+¥! , Z^C^ + Z't
where ^., 5, (7 are the determinants 2) in 603. Then the area of
the inscribed polyhedron corresponding to this division D is
Let us suppose that
as w, v ranges over SI. Also let us assume that
VI V? *7f
^< *K ^K
AREA OF CURVED SURFACES
631
remain numerically < e for any division D of norm d< c?0, e small
at pleasure, except in the vicinity of a discrete set of points, that
is, let 3) be in general uniformly evanescent in 51, as d = 0. Then
BI
where in general
If now A, B, Q are limited and ^-integrable in 2(, we have at
once
lim8D=fdudvJA*
~ ^a
as in 603.
2. We ask now under what conditions are the expressions 3)
in general uniformly evanescent in 51 ? The answer is pretty evi
dent from the example given by Schwarz. In fact the equation
of the tangent plane X at ^30 is
A(x - a-0) + B(y - #0) + C(z - z0) = 0.
On the other hand the equation of the plane T= ($0, $j, <$%)
is - y
^o v» z(
or finally
Thus for 3) to converge in general uniformly to zero, it is nec
essary and sufficient that the secant planes T converge in general
uniformly to tangent planes. Let us call divisions such that the
faces of the corresponding inscribed polyhedra converge in general
uniformly to tangent planes uniform triangular divisions. For
such divisions the expressions 3) are in general uniformly evanes
cent, as c?= 0. We have therefore the following theorem :
3. Let 51 be a limited complete metric set. Let the coordinates
x, y, 2 be one-valued totally differentiable functions of the parame-
632 GEOMETRIC NOTIONS
ters u, v in 51, such that A2 + IP + O2 is greater than some positive
constant, and is limited and H-integrable in 51. Then
tf = lim SD = V^2+^2+ C*dudv,
rf=o ^21
D denoting the class of uniform triangular divisions of norms d.
This limit we shall call the area of S. From this definition we
have at once a number of its properties. We mention only the
following :
4. Let 5Ij, ••• 5lw be unmixed metric sets whose union is 51. Let
$!, ••• Sm be the pieces of S corresponding to them. Then each SK
has an area and their sum is S.
5. Let 51A be a metric part of 51, depending on a parameter X= 0,
such that SA = S. Then
6. The area of S remains unaltered when S is subjected to a dis
placement or a transformation of the parameters as in I, 744 seq.
607. 1. Irregular Surfaces. We consider now surfaces which
do not have tangent planes at every point, that is, surfaces for
which one or more of the first derivatives of the coordinates #, y, z
do not exist, and which may be styled irregular surfaces. We
prove now the theorem :
Let the coordinates x, y, z be one-valued functions of w, v having
limited total difference quotients in the metric set 51. Let D be a
positive triangular division of norm d<dQ. Then
Max *S^
is finite and evanescent with 51.
For let the difference quotients remain </JL. We have
But
cosec
AREA OF CURVED SURFACES 633
where 0K is the angle made by the sides P$PI, ^V^V As D is a
positive division, one of the angles of tK is such that cosec 6K is
numerically less than some positive number M. Thus
where /-i, M are independent of K and d. Similar relations hold
for Y, , ZK. Hence
77)
where 77 > 0 is small at pleasure, for dQ sufficiently small.
2. Let 31 and x, y, z be as in 606, 3, except at certain points form
ing a discrete set a, the first partial derivatives do not exist. Let
their total difference quotients be limited in 31. Then
= C
J
ivhere D denotes a positive triajiyular division of norm d.
Let us first show that the limit on the left exists. We may
choose a metric part 55 of 51 such that (§ = 51 — 53 is complete and
exterior to 31 and such that 53 is as small as we please. Let /Sg
denote the area of the surface corresponding to (5. The triangles
tK fall into two groups : G1 containing points of 53 ; 6r2 containing
only points of (£• Then
SD = 2 V X2 + 17 + Z 2 = v + 2 .
«, ff»
But 53 may be chosen so small that the first sum is < e/4 for
any d<dQ. Moreover by taking d0 still smaller if necessary, we
have
|2-SC <e/4.
0,
HenCC |^-^|<6/2 , d<d0. (1
Similarly for any other division D' of norm d',
\SD,-Sg>\<€/2 , d'<d0
decreasing dQ still farther if necessary. Thus
|^-^|<€ , d,d'<dQ.
634 GEOMETRIC NOTIONS
Hence lim SD exists, call it S. Since S exists we may take dQ
so small that
\S-SD\<€/2
This with 1) gives
|tf
that is,
S = lim &k = lim f V^ + ^
«/S
= C ^/A? + & + (Fdudv
c/2l
by I, 724.
608. 1. The preceding theorem takes care of a large class of
irregular surfaces whose total difference quotients are limited.
In case they are not limited we may treat certain cases as follows:
Let us effect a quadrate division of the u, v plane of norm d,
and take the triangles tK so that for any triangular division D
associated with c?, no square contains more than n triangles, and
no triangle lies in more than v squares ; n, v being arbitrarily
large constants independent of d. Such a division we call a
quasi quadrate division of norm d. If we replace the quadrate by
a rectangular division, we get a quasi rectangular division.
We shall also need to introduce a new classification of functions
according to their variation in 51, or along lines parallel to the
u, v axes. Let D be a quadrate division of the w, v plane of norm
d < dQ . Let
COK = Osc/(w, v) , in the cell dK.
Then Max ^a)Kd
is the variation of / in 51. If this is not only finite, but evanes
cent with 51, we say/ has limited fluctuation in 51. Obviously this
may be extended to any limited point set in m-way space.
Let us now restrict ourselves to the plane. Let a denote the
points of 51 on a line parallel to the w-axis. Let us effect a divi
sion D' of norm df. Let co'K = Osc/(w, v) in one of the intervals
of D'. Then
?7a = Max 2o>^
is the variation of /in a-
AREA OF CURVED SURFACES 635
Let us now consider all the sets a lying on lines parallel to the
M-axis, and let
a <_ a- , (7 = 0.
If now there exists a constant G- independent of a such that
1«<*#,
that is, if ?7a is uniformly evanescent with <r, we say that/(w, v)
has limited fluctuation in 51 with respect to u.
With the aid of these notions we may state the theorems :
2. Let the coordinates x, y, z be one-valued limited functions in
the limited complete set 51. Let x, y have limited total difference
quotients, while z has limited variation in 51. Let D denote a quasi
quadratic division of norm d<dQ. Then
Max SD
D
is finite.
For, as before,
But p denoting a sufficiently large constant,
| A; '., | A;' | are < pd.
Let ft>t=Osc2 in the square st. If the triangle tK lies in the
squares *tl, ••• stjfc,
I A; i, i A» i<«.,+ - +«...
Thus, n denoting a sufficiently large constant,
the summation extending over those squares containing a triangle
of D. But z having limited variation,
2&)tc? < some M.
Hence *\XK\ , 2 | YK \ are <
Finally, as in 607,
2 |Z | <some M' .
The theorem is thus established.
636 GEOMETRIC NOTIONS
3. The coordinates x, y, z, being as in 2, except that z has limited
fluctuation in 51, and D denoting a quasi quadrate division of
norm d < ^
D
is finite and evanescent with 51.
The reasoning is the same as in 2 except that now M, M' are
evanescent with 51.
4. Let the coordinates x, y, z have limited total difference quo
tients in 51, while the variation of z along any line parallel to the u
or v axis is < M. Let 51 lie in a square of side s = 0. Then
Max/S^ < sG-,
D
where G is some constant independent of s, and D is a quasi rectan
gular division of norm d < dQ .
For here
2 | A»y • | A'z |
where M' denotes a sufficiently large constant; du, dv denote the
length of the sides of one of the triangles tK parallel respectively
to the u, v axes, and a)u, a)v the oscillation of z along these sides.
Since the variation is < M in both directions,
Ms.
Similarly
2o>A < M8.
The rest of the proof follows as before.
5. The symbols having the same meaning as before, except that z
has limited fluctuation with respect to u, v,
The demonstration is similar to the foregoing. Following the
line of proof used in establishing 607, 2 and employing the
theorems just given, we readily prove the following theorems :
AREA OF CURVED SURFACES 637
6. Let 51 be a metric set containing the discrete set a. Let b be
a metric part of 31, containing a such that 33 = 51 — b is exterior to a,
and t> = 0. Let the coordinates x, y, z be one-valued totally differ-
entiable functions in 33, such that A2 + IP + C2 never sinks below a
positive constant in any 33, is properly R-integrable in any 53, and
improperly integrable in 51. Let x, y have limited total difference
quotients, and z limited fluctuation in b. Then
lim S# = I V^i2 + B2 + 02dudv
d=0 J%
where A, B, C are the determinants in 603, 2), and D is any quasi
quadrate division of norm d.
7. Let the symbols have the same meaning as in 6, except that
1° a reduces to a finite set.
2° z has limited variation along any line parallel to the u, v axes.
3° D denotes a uniform quasi rectangular division. Then
\\mSD= VA2 + 7>2 + CPdudv.
d=o %/*
8. The symbols having the same meaning as in 6, except that
1° z has limited fluctuation with respect to w, v in b.
2° D denotes a uniform quasi rectangular division. Then
lim SD = I V 'A* + B2 -f C2dudv.
9. If we call the limits in theorems 6, 7, 8, area, the theorems
606, 3, 4, 5 still hold.
INDEX
(Numbers refer to pages)
-x
•
tf
AbeVs identity, 87
series, 87
Absolutely convergent integrals, 31
series, 79
products, 247
Addition of cardinals, 292
ordinals, 312
series, 128
Adherence, 340
Adjoint product, 247
series, 77, 139
set of intervals, 337
Aggregates, cardinal number, 278
definition, 276
distribution, 295
enumerable, 280
equivalence, 276
eutactic, 304
exponents, 294
ordered, 302
power or potency, 278
sections, 307
similar, 303
transfinite, 278
uniform or 1-1 correspondence, 2"
Alternate series, 83
Analytical curve, 582
Apantactic, 325
Area of curve, 599, 602
surface, 623
Arzela, 365, 555
Associated simple series, 144
products, 247
multiple series, 145
normal series, 245
logarithmic series, 243
inner sets, 365
Associated, outer sets, 365
non-negative functions, 41
Baire, 326, 452, 482, 587
Bernovillian numbers, 265
Bertram's test, 104
Bessel functions, 238
Beta functions, 267
Binomial series, 110
Bocher, 165
Bonnet's test, 121
Borel, 324, 542
Brouwer, 614
Cohen's test, 340
Cantors 1° and 2° principle, 316
theorem, 450
Category of a set, 326
Cauchy's function, 214
integral test, 99
radical test, 98
theorem, 90
Cell of convergence, 144
standard rectangular, 359
Chain, 612
Class of a function, 468, 469
Conjugate functions, 238
series, 147
products, 249
Connex, 605
Connected sets, 605
Contiguous functions, 231
Continuity, 452
infra, 487
semi, 487
supra, 487
Continuous image, 608
639
640
INDEX
Contraction, 287
Convergence, infra-uniform, 562
monotone, 176
uniform, 156
at a point, 157
in segments, 556
sub-uniform, 555
Co-product, 242
Curves, analytical, 582
area, 599, 602
Faber, 546
Jordan, 595, 610
Hilbert, 590
length, 579
non-intuitional, 537
Osgood, 600
Pompeiu, 542
rectifiable, 583
space-filling, 588
D'Alembert, 96
Deleted series, 139
Derivates, 494
Derivative of a set, 330
order of, 331
Detached sets, 604
Dilation, 287
Dim, 176, 185, 438, 538
series, 86
Discontinuity, 452
at a point, 454
of 1° kind, 416
of 2° kind, 455
point wise, 457
total, 457
Displacement, 286
Distribution, 295
Divergence of a series, 440
Division, complete, 30
separated, 366, 371
unmixed, 2
of series, 196
of products, 253
Divisor of a set, 23
quasi, 390
Divisor, semi, 390
Du Bois Reymond, 103
473
Elimination, 594
Enclosures, complementary e-, 355
deleted, 452
distinct, 344
divisor of, 344
e, 355
measurable, 356
non-overlapping, 344
null, 366
outer, 343
standard, 359
Enumerable, 280
Equivalent, 276
Essentially positive series, 78
negative series, 78
Euler's constant, 269
Eutactic, 304
Exponents, 294
Exponential series, 96
Extremal sequence, 374
Faber curves, 546
Fluctuation, 634, 635
Fourier's coefficient, 416
constants, 416
series, 416
Function, associated non-negative func
tions, 41
Bessel's, 238
Beta, 267
Cauchy's, 214
class of, 468, 469
conjugate, 233
contiguous, 231
continuous, 452
infra, 487
semi, 487
supra, 487
discontinuous, 452
of 1° kind, 416
of 2° kind, 455
INDEX
641
Function, Gamma, 267
Gauss' HO), 238
hyperbolic, 228
hypergeometric, 228
lineo-oscillating, 528
maximal, 488
measurable, 338
minimal, 488
monotone, 137
null, 385
oscillatory, 488
pointwise discontinuous, 457
residual, 561
Riemann's, 459
totally discontinuous, 457
truncated, 27
uniformly limited, 160, 567
Volterra's, 501, 583
Weierstrass', 498, 523, 581, 588
Gamma function, 267
Gauss' function !!(», 238
test, 109
Geometric series, 81, 139
Harnack, divergence of series, 440
sets, 354
Hermite, 300
Httbert's curves, 590
Hobson, 389, 412, 555
Hyperbolic functions, 228
Hypercomplete sets, 472
Hypergeometric functions, 229
series, 112
Images, simple, multiple, 606
unifold, manifold, 606
continuous, 606, 608
Integrals, absolutely convergent, 31
L- or Lebesgue, proper, 372
improper, 403, 405
improper, author's, 32
classical, 26
de la Vallee-Poussin, 27
inner, 20
Integrals, R- or Riemannian, 372
Integrand set, 385
Intervals, of convergence, 90
adjoint set of, 337
set of, belonging to, 337
Inversion, geometric, 287
of a series, 204
Iterable sets, 14
Iterated products, 251
series, 149
Jordan curves, 595, 610
variation, 430
theorem, 436
Kdnig, 527
Rummer's test, 106, 124
Lattice points, 137
system, 137
Law of Mean, generalized, 505
Layers, 555
deleted, 563
Lebesque or L- integrals, 372
theorems, 413, 424, 426, 452, 475,
520, 619
Leibnitz's formula, 226
Length of curve, 579 ,
Lindermann, 300, 599
Lineo-oscillating functions, 528
Link, 612
Liouville numbers, 301
Lipschitz, 438
Logarithmic series, 97
Liiroth, 448
Maclaurin's series, 206
Maximal, minimal functions, 488
Maximum, minimum, 521
at a point, 485
Measure, 348 —
lower, 348
upper, 343
Mertens, 130
Metric sets, 1
642
INDEX
Monotone convergence, 176
functions, 137
Moore-Osgood theorem, 170
Motion, 579
Multiplication of series, 129
cardinals, 293
ordinals, 314
infinite products, 253
Normal form of infinite product, 245
Null functions, 385
sets, 348
Numbers, Bernouillian, 265
cardinal, 278
class of ordinal numbers, 318
limitary, 314.
Liouville, 301
ordinal, 310
rank of limitary numbers, 331
Ordered sets, 302
Order of derivative of a set, 331
Oscillation at a point, 454
Oscillatory function, 488
Osyood curves, 600
-Moore theorem, 170
theorems, etc., 178, 555, 622
Pantactic, 325
Path, 612
Peaks, 179
infinite, 566
Poly ant, 153
Point sets, adherence, 340
adjoint set of intervals, 337
apantactic, 325
associated inner set, 365
outer set, 365
Baire sets, 326
category 1° and 2°, 326
coherence, 340
conjugate, 51
connected, 605
convex, 605
detached, 604
Point sets, divisor, 23
Ilarnack sets, 354
hypercomplete, 472
images, 605, 606
integrand sets, 385
iterable, 14
measurable, 343, 348
metric, 1
negative component, 37
null, 318
-^ pantactic, 325
positive component, 37
potency or power, 278
projection, 10
quasidivisor, 390
reducible, 335
reticulation, 621
semidivisor, 390
separated intervals, 337
sum, 22
transfinite derivatives, 330
union, 27
well-ordered, 304
Polntwise discontinuity, 457
Pompeiu, curves, 542
Potency or power of a set, 278
Power series, 89, 144, 187, 191
Precinct, 612
Pringsheim, theory of convergence, 113
theorems, etc., 141, 215, 216, 217,
220, 273
Projection, 10
Products, absolute convergence, 247
adjoint, 247
associate simple, 247
conjugate, 249
co-product, 242
iterated, 251
normal form, 245
Quasidivisor, 390
Raabe's test, 107
Rank of limitary numbers, 331
INDEX
643
Rate of convergence or divergence, 102
Ratio test, 96
Reducible sets, 335
Remainder series, 77
of Taylor's series, 209, 210
Rectifiable curves, 583
Regular points, 428
Residual function, 561
Reticulation, 621
Richardson, 32
Riemann's function, 459
theorem, 444
R- or Riemann integrals, 372
Rotation, 286
Scheefer, theorem, 516
Schonfliess, theorems, 598, 621
Schicarz, theorem, etc., 448, 626
Section of an aggregate, 307
Segment, constant, or of invariability,
521
Semidii-isor, 390
Separated divisions, 366, 371
functions, 403
sets, 366
of intervals, 337
Sequence, extremal, 374
m-tuple, 137
Series, Abel's, 87
absolute convergent, 79
adjoint, 77, 139
alternate, 83
associate logarithmic, 243
normal, 245
simple, 144
multiple, 144
Bessels, 238
binomial, 110
cell of convergence, 144
conjugate, 147
deleted, 139
Dini's, 86
divergence of, 440
essentially positive or negative, 78
exponential, 96
Series, Fourier's, 416
geometric, 81, 139
harmonic, 82
general of exponent /A, 82
hypergeometric, 112
interval of convergence, 90
inverse, 204
iterated, 149
logarithmic, 97
Maclaurin's, 206
power, 89, 114, 187
rate of convergence or divergence,
102
remainder, 77
simple convergence, 80
Taylor's, 206
tests of convergence, see Tests
telescopic, 85
trigonometric, 88
two-way, 133
Similar sets, 303
Similitude, 287
Simple convergence of series, 80
Singular points, 26
Space-filling curves, 588
Steady convergence, 176
Submeasurable, 405
Sum of sets, 22
Surface, area, 623
irregular, 632
regular, 629
Taylor's series, 206
Telescopic series, 85
Tests of convergence, Bertram, 104
Bonnet, 121
Cahen, 108
Cauchy, 98, 99
d'Alembert, 96
Gauss, 109
Rummer, 106, 124
Pringsheim, 123
Raabe, 107
radical, 98
ratio, 96
644 INDEX
Tests of convergence, tests of 1° and 2° Uniformly limited function, 160, 567
kind, 120 Union of sets, 22
Weierstrass, 120
Theta functions, 135, 184, 256 Vallee-Poussin (de la), 27, 594
Total discontinuity, 457 Van Vleck sets, 361
Transjinite cardinals, 278 Variation, limited or finite, 429, 530
derivatives, 330 positive and negative, 430
Translation, 286 Volterra curves, 501, 587
Trigonometric series, 88
Truncated function, 27 WaWs tonnolo, 26°
Two-way series, 133 Weierstrats' function, 498, 523, 588
test, 120
Undetermined coefficients, 197 Well-ordered sets, 304
Unifold image, 606 Wilson, W. A., vii, 395, 401
Uniform convergence, 156 youn^ w H theoremSj 360> 363
at a point, 157
correspondence, 276 Zeros of power series, 191
SYMBOLS EMPLOYED IN VOLUME II
(Numbers refer to pages)
Front 31, 1. Fa, 614 Xr K2 ••• , 318, 323
2o Z,, Zg-,318
gj(«) - 2jw? 330 ; 9i<a) - s^a? 331
fl, 1 1 = Meas 2t, 343 ; § = Meas 5t, 348
U>{ I'22 S = Meas a, 348
£,,22 r r r
Ad,J,81 /' J, J, 372, 403, 405
/A,M'31 Sdv, Qdv, 390
2laij3, 32. 3(/,afp, 34 j/^ 429; Var/= F/, 429
31^, s#_a, 34 QSC j- _ oscillation in a given set.
A „, A n, Ad j A , 77. A „, p, 78 QSC y; 454
f> _ 139 Disc/, 454
51 ~ 53, 276 ; 31 a* 53, 303 (S€, @t±0» 473
Card 51, 278 J(x) y^) 433
e = K 280; c.287 /'(^./'(x), 493
j/i^, jyu
Ord 31, 311 .ff(^), 494
w, 311 ; «, 318 A(a, ft), 494
INDEX
645
The following symbols are defined in Volume I and are repeated here for
the convenience of the reader.
Dist(a, x) is the distance between
a and x
Z^(a), called the domain of the point
a of norm 8 is the set of points a:,
such that Dist (a, x) < 8
Fg(rt), called the vicinity of the point
a of norm 8, refers to some set 31,
and is the set of points in D^(a)
which lie in 31
Z>6*(tt), V$*(a) are the same as the
above sets, omitting a. They are
called deleted domains, deleted vi
cinities
an = a means an converges to a
f(x) = a, means /(z) converges to a
A line of symbols as :
c < 0, wz, | a - an | < e, n > m
is of constant occurrence, and is to
be read : for each e > 0, there exists
an index m, such that | a — an \ < e,
for every n > m
Similarly a line of symbols as :
00, S>0, !/(*) - a|<£, x in F«*(«)
is to be read : for each e > 0, there
exists a 8 > 0, such that
I /GO -«|<€,
for every x in Pr5*(a)
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