If $a$ and $b$ commute in a $C^*$-algebra and $a$ is normal, then $f(a)$ and $b$ commute for any continuous $f$ 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.

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 $f(a)$.
