6

I'm trying to find a way to demonstrate the following:

Let $(A,*,\|\cdot\|)$ be a unital $C^*$-algebra. If $a,b\in A$ commute and $a\in A$ is normal (i.e. $a^*a=aa^*$), then for every continuous function $f:$Sp$(a)\to\mathbb{C}$, $f(a)$ and $b$ commute (where Sp$(a)$ denotes the spectrum of $a$ and $f(a)$ is given by functional calculus).

So far, I've been trying to show that $\|f(a)b-bf(a)\|=0$ knowing that $ab=ba$ or, equivalently, $\|ab-ba\|=0$, but I've got nowhere with this. Any hint/suggestion would be greatly appreciated.

Davide Giraudo
  • 181,608
  • 3
    Have you tried proving it for polynomials and then using Weierstrass's approximation theorem? – minimalrho Oct 11 '13 at 19:20
  • This does indeed make things simpler. Thanks for the idea. –  Oct 11 '13 at 19:36
  • 1
    @minimalrho, why don't you post this as answer? – Norbert Oct 11 '13 at 20:19
  • Note that $a^*$ need not be a limit of polynomials in $a$. – Jonas Meyer Oct 13 '13 at 18:59
  • @Norbert: It isn't an answer, or at least I don't understand it to be one. The problem is to show that $b$ commutes with everything in $C^(a)$, the main point of which is that $b$ commutes with $a^$. Thus it lies in Fuglede, not Weierstrass. – Jonas Meyer Oct 16 '13 at 12:46

1 Answers1

4

By Fuglede's theorem, $b$ also commutes with $a^*$. Therefore $b$ commutes with every element of the unital $*$-algebra generated by $a$, hence also with every element of its closure $C^*(1,a)$. For every continuous $f:\sigma(a)\to\mathbb C$, $f(a)$ is in $C^*(1,a)$, and therefore $b$ commutes with $f(a)$.

Jonas Meyer
  • 55,715