CAlgebraHomomorphismsPreserveContinuousFunctionalCalculus 2008-04-21 22:10:11 2008-04-21 23:21:41 Theorem continuous functional calculus % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let us setup some notation first: Let $\mathcal{A}$ be a unital \PMlinkname{$C^*$-algebra}{CAlgebra} and $z$ a normal element of $\mathcal{A}$. Then \begin{itemize} \item $\sigma(z)$ denotes the spectrum of $z$. \item $C(\sigma(z))$ denotes the $C^*$-algebra of continuous functions $\sigma(z) \longrightarrow \mathbb{C}$. \item If $f \in C(\sigma(z))$ then $f(z)$ is the element of $\mathcal{A}$ given by the continuous functional calculus. \end{itemize} $\,$ {\bf Theorem -} Let $\mathcal{A}$, $\mathcal{B}$ be unital \PMlinkname{$C^*$-algebras}{CAlgebra} 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 \begin{align*} \Phi(f(x)) = f(\Phi(x)) \end{align*} $\,$ {\bf \emph{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 \begin{itemize} \item $\Phi(p_n(x)) \longrightarrow \Phi(f(x))$, by the continuity of $\Phi$ (see \PMlinkname{this entry}{HomomorphismsOfCAlgebrasAreContinuous}) and the continuity of the continuous functional calculus mapping. \end{itemize} \begin{itemize} \item $p_n(\Phi(x)) \longrightarrow f(\Phi(x))$, by the continuity of the continuous functional calculus mapping. \end{itemize} 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. $\square$