Continuity of functional calculus Let $\mathcal{A}$ be an unital C*-Algebra. $a,b$ be normal elements in $\mathcal{A}$. $X\subset \Bbb C$ is a compact subset. $f:X\rightarrow \Bbb C$ is continuous.  I need to show that for all $\epsilon >0$ there exists $\delta>0$ such that $||f(a)-f(b)||<\epsilon$ whenever $||a-b||<\delta$ with $\sigma (a), \sigma (b)\subset X$.
I have no clue on how to attack the problem as I know to define $f(a)$ by considering $C^*(a)$. Now that I have to work with both $f(a)$ and $f(b)$ I cannot use gelfand transformation directly. I tried to look at the definition of $f(a)$ and tried to write the map $f:\mathcal{A} \rightarrow \mathcal{A}$ explicitly that is giving all the identifications a name. But it got complicated and I could not work. Can you show me a way?
 A: Given $f$ and $\epsilon$, choose a polynomial $p$ with $\Vert f-p\Vert_{\infty,X}<\epsilon$ (where $\Vert\cdot\Vert_{\infty,X}$ is the supremum norm oin $X$). Now see the corresponding polynomial function in $\mathcal{A}$, $p:\mathcal{A}\to\mathcal{A}$. (Remember: the functional calculus respects this notation, i.e., $p(a)$, in the functional calculus, is obtained simply by plugging $a$ inside the polynomial $p$.) This is continuous, so, given $a\in\mathcal{A}$, there exists $\delta>0$ such that $\Vert b-a\Vert<\delta$ implies $\Vert p(b)-p(a)\Vert<\epsilon$.
Now let's use the fact that injective $*$-homomorphisms are isometric: If $\sigma(b)\subseteq $ and $\Vert b-a\Vert<\delta$, then
\begin{align*}
\Vert f(a)-f(b)\Vert&\leq\Vert f(a)-p(a)\Vert+\Vert p(a)-p(b)\Vert+\Vert p(b)-f(b)\Vert\\
&=\Vert f-p\Vert_{\infty,\sigma(a)}+\Vert p(a)-p(b)\Vert+\Vert f-p\Vert_{\infty,\sigma(b)}\\
&\leq 2\Vert f-p\Vert_{\infty,X}+\Vert p(a)-p(b)\Vert<3\epsilon,
\end{align*}
where $\Vert\cdot\Vert_{\infty,\sigma(a)}$ is the supremum norm on $\sigma(a)$ and similarly for $\sigma(b)$.
With this, you can then show that the map on two variables $C(X)\times\left\{a\in\mathcal{A}:\sigma(a)\subseteq X\right\}\to\mathcal{A}$, $(f,a)\mapsto f(a)$, is continuous.
