$C^{*}$-algebra homomorphisms preserve continuous functional calculus

Let us setup some notation first: Let $𝒜$ be a unital $C^{*}$-algebra (http://planetmath.org/CAlgebra) and $z$ a normal element of $𝒜$. Then

• $\sigma (z)$ denotes the spectrum of $z$.

• $C(\sigma (z))$ denotes the $C^{*}$-algebra of continuous functions $\sigma (z)⟶ℂ$.

• If $f\in C(\sigma (z))$ then $f(z)$ is the element of $𝒜$ given by the continuous functional calculus.

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Theorem - Let $𝒜$, $ℬ$ be unital $C^{*}$-algebras (http://planetmath.org/CAlgebra) and $\mathrm{\Phi }:𝒜⟶ℬ$ a *-homomorphism. Let $x$ be a normal element in $𝒜$. If $f\in C(\sigma (x))$ then

$\mathrm{\Phi }(f(x))=f(\mathrm{\Phi }(x))$

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Proof: The identity elements of $𝒜$ and $ℬ$ 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 $f(\mathrm{\Phi }(x))$ is a well-defined element of $ℬ$, i.e. that $\sigma (\mathrm{\Phi }(x))\subseteq \sigma (x)$. This is clear since, if $x-\lambda e$ is invertible for some $\lambda \in ℂ$, then $\mathrm{\Phi }(x)-\lambda e=\mathrm{\Phi }(x-\lambda e)$ is also invertible.

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

• $\mathrm{\Phi }(p_n(x))⟶\mathrm{\Phi }(f(x))$, by the continuity of $\mathrm{\Phi }$ (see this entry (http://planetmath.org/HomomorphismsOfCAlgebrasAreContinuous)) and the continuity of the continuous functional calculus mapping.

• $p_n(\mathrm{\Phi }(x))⟶f(\mathrm{\Phi }(x))$, by the continuity of the continuous functional calculus mapping.

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