2 SPECTRUM OF AN ELEMENT OF A NORMED ALGEBRA 313 cannot be defined and analytic in the region |(| > r, because the function £H-»(£? - £x)~\ which is analytic and equal to the sum of the series ]T ({jc)n n for all sufficiently small |£|, would then be equal to the sum of this series for |£| r'"1. The number p(x) is called the spectral radius of x. We have (15.2.7.1) P(x)£ \\x\\, (15.2.7.2) p(xk) = (p(x))k for all integers A: ^ 0, by virtue of (15.2.4(iii)) and (15.2.3.1). (15.2.8) (i) Let A, B be two Banach algebras with nonzero unit elements e, ef respectively, and let u : A-»B be an algebra homomorphism such that u(e) = e'. Then SpB(w(*)) cz SpA(x)for all x e A. (ii) In particular, suppose that A is a closed subalgebra of B, having the same unit element. Then SpB(x) c SpA(x), and every frontier point of SpA(jt) belongs to SpB(x). Hence, z/SpA(,x) has empty interior, we have SpB(jc) = SpA(jc). (i) If £ e C is such that x < £e is invertible in A, then u(x — £e) = u(x) — £e' is invertible in B, whence the result. (ii) The first assertion is a particular case of (i). To prove the second assertion, it is enough to show that if A0 is a frontier point of SpA(x), then x — A0 e is not invertible in B. Now, by hypothesis there exists a sequence (AM) of regular values for x (in A) tending to A0 . For each n ^ 1, the inverse (x — kne)~l of x — kne in A exists, and is therefore also the inverse of x — "kne in B. If A0 <£ SpB(x), the sequence ((x — Xne)~l) would tend in B to the inverse y ofx — A0e in B (15.2.4(ii)). But since A is closed in B, we should havey e A, so that y would also be the inverse of x — 10 e in A, which is absurd. PROBLEMS 1. Let A be a normed algebra with no unit element. Then A can be considered as a closed two-sided ideal in the normed algebra A defined in Problem 5 of Section 15.1. For each x 6 A, the spectrum of x in A (denoted by Sp(*) or SpA(*)) is defined to be the spectrum of x in A, and p(x) is the spectral radius of x considered as an element of A. ' "ITTSBIJRGH, PEE'SYLVAIIA 152T3