Proposition 6.4.14.label Let $K\suf\com$ be compact and $A_{K}$ denote the set of all normal elements $x$ with $\spec{A}{x}\suf K$. If $f$ is a continuous function on $K$ then the functional calculus $x\in A_{K}\mto f(x)\in A$ is continuous.
Proof. By Stone-Weierstrass theorem, for any $\e>0$ there exists a polynomial $p$ of $\lam$ and $\cl{\lam}$ such that $\SUP{\lam\in K}\abs{p(\lam,\cl{\lam})-f(\lam)}<\e$. Since $p$ is a polynomial there exists $\de>0$ such that $\norm{p(x,x^*)-p(y,y^*)}{}<\e$ for $\norm{x-y}{}<\de$ and $\norm{x}{},\norm{y}{}\leq M$ for some constant $M>0$ that doesn’t depend on $x,y$. let $M=\SUP{\lam\in K}\abs{\lam}$. If $\norm{x-y}{}<\de$ and $x,y\in A_{K}$ then
\begin{align*}\norm{f(x)-f(y)}{}\leq\norm{f(x)-p(x,x^*)}{}+\norm{p(x,x^*)-p(y,y^*)}{}+\norm{p(y,y^*)-f(y)}{}<3\e\end{align*}
$\square$