Our teaching assistant claimed the following. If you are in a let's say commutative $C^*$-algebras $X,Y$ and a linear $*$-morphism $\varphi:X\to Y$, let $x\in X$ and let $f\in C(\text{sp} (x),\mathbb{C})$. Then our teaching assistant claimed that $f(\varphi(x))=\varphi(f(x))$, where the action of $f$ on $x$ is defined by functional calculus. Since the algebras are commutative, one can define consider $X$ and $Y$ to be continuous functions of some compact set to $\mathbb{C}$ and $f(g)$ as $f\circ g$. I really don't see why $f$ and $\varphi$ should commute. I've tried looking for a counterexample, but didn't really find one.
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- 3$\begingroup$ First show it for polynomials and then approximate $\endgroup$Evangelopoulos Foivos– Evangelopoulos Foivos2024-12-24 17:51:18 +00:00Commented Dec 24, 2024 at 17:51
- $\begingroup$ @EvangelopoulosFoivos Thanks that makes sense! $\endgroup$Vincent Batens– Vincent Batens2024-12-25 19:40:30 +00:00Commented Dec 25, 2024 at 19:40
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1 Answer
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Posting a community wiki answer for archiving reasons.
Let $\pi\colon A\to B$ be a $^*$-homomorphism between $\mathrm{C}^\ast$-algebras and let $a\in A$ be normal. If $f\in C_0(\sigma(a)\setminus\{0\})$, then $\pi(f(a))=f(\pi(a))$.
By considering the forced unitizations $A^{(+)}, B^{(+)}$ and the extension $\pi^{(+)}\colon A^{(+)}\to B^{(+)}$, we can assume without loss of generality that both $A,B$ and $\pi$ are unital. Clearly then $\sigma(\pi(a))\subset\sigma(a)$ and $\pi(a)$ is normal, so $f(\pi(a))$ makes sense.
If $0\in\sigma(a)$, then $f$ can be extended to a continuous function on $\sigma(a)$ by setting $f(0)=0$. Otherwise, if $0\not\in\sigma(a)$, then $\sigma(a)\setminus\{0\}=\sigma(a)$, which is compact. In both cases, given $\varepsilon>0$, by Weierstrass approximation we can find a polynomial $p(z)$ such that $|f(z)-p(z)|<\varepsilon$ for all $z\in\sigma(a)$. Observe that
$$\|\pi(f(a))-f(\pi(a))\|\le\|\pi(f(a))-\pi(p(a))\|+\|\pi(p(a))-p(\pi(a))\|+\|p(\pi(a))-f(\pi(a))\|$$
$$\le2\sup_{z\in\sigma(a)}|p(z)-f(z)|+\|\pi(p(a))-p(\pi(a))\|<2\varepsilon+\|\pi(p(a))-p(\pi(a))\|$$
where we used that continuous functional calculus is isometric. Moreover, since $p$ is a polynomial and $\pi$ a $^*$-homomorphism, it is immediately seen that $\pi(p(a))=p(\pi(a))$, by mere algebraic calculations. We have thus shown that $\|\pi(f(a))-f(\pi(a))\|<2\varepsilon$. As $\varepsilon>0$ was arbitrary, we are done.