Let us setup some notation first: Let $\mathcal{A}$ be a unital $C^*$ -algebra and $z$ a normal element of $\mathcal{A}$ . Then

• $\sigma(z)$ denotes the spectrum of $z$ .
• $C(\sigma(z))$ denotes the $C^*$ -algebra of continuous functions $\sigma(z) \longrightarrow \mathbb{C}$ .
• If $f \in C(\sigma(z))$ then $f(z)$ is the element of $\mathcal{A}$ given by the continuous functional calculus.

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Theorem - Let $\mathcal{A}$ , $\mathcal{B}$ be unital $C^*$ -algebras and $\Phi :\mathcal{A} \longrightarrow \mathcal{B}$ a *-homomorphism. Let $x$ be a normal element in $\mathcal{A}$ . If $f \in C(\sigma(x))$ then

Proof: The identity elements of $\mathcal{A}$ and $\mathcal{B}$ will be both denoted by $e$ and it will be clear from the context which one we are referring to.

First, we need to check that $f(\Phi(x))$ is a well-defined element of $\mathcal{B}$ , i.e. that $\sigma(\Phi(x)) \subseteq \sigma(x)$ . This is clear since, if $x - \lambda e$ is invertible for some $\lambda \in \mathbb{C}$ , then $\Phi(x)-\lambda e = \Phi(x- \lambda e)$ is also invertible.

Let $\{p_n\}$ be sequence of polynomials in $C(\sigma(x))$ converging uniformly to $f$ . Then we have that

• $\Phi(p_n(x)) \longrightarrow \Phi(f(x))$ , by the continuity of $\Phi$ (see this entry) and the continuity of the continuous functional calculus mapping.

• $p_n(\Phi(x)) \longrightarrow f(\Phi(x))$ , by the continuity of the continuous functional calculus mapping.

It is easily checked that $\Phi(p_n(x)) = p_n(\Phi(x))$ (since $\Phi$ is an homomorphism). Hence we conclude that $\Phi(f(x)) = f(\Phi(x))$ as intended. $\square$