Continuity in C$^*$-Algebras Let $f$ : $\mathbb{R}$ $\rightarrow$ $\mathbb{C}$ be a continuous function and $A$ be a C$^{*}$-algebra with unit. Show that the mapping $\phi$: $\{x\in A : x=x^*\}$ $\rightarrow$ $A$ given by $\phi(x)$ = $f(x)$ is continuous.
First of all, are both sets equipped with the norm topology? If so,
I am getting nowhere with this. Thanks for any help.
 A: Let's denote the set of self-adjoint elements of $A$ by $A_{sa}$.  Fix $x_0\in A_{sa}$ and $\varepsilon>0$.  Put $M=\|x_0\|+1$.  By Stone-Weierstrass, we can choose a polynomial $P$ such that $\|f-P\|_{C([-M,M])}<\frac{\varepsilon}{3}$.  Since addition and multiplication are continuous in $A$, $P$ is continuous on $A$, so there is some $\delta_0>0$ such that $\|P(x)-P(x_0)\|<\frac{\varepsilon}{3}$ whenever $\|x-x_0\|<\delta_0$.  Now put $\delta=\min\{\delta_0,1\}$.  If $x\in A_{sa}$ and $\|x-x_0\|<\delta$, then $\|x\|<\|x_0\|+\delta$, so that $\sigma(x),\sigma(x_0)\subset[-M,M]$.  Then we have 
\begin{align*}
\|f(x)-f(x_0)\|&\leq\|f(x)-P(x)\|+\|P(x)-P(x_0)\|+\|P(x_0)-f(x_0)\|\\
&\leq2\|f-P\|_{C([-M,M])}+\|P(x)-P(x_0)\|\\
&<\varepsilon,
\end{align*}
and therefore $f$ is continuous.
A: I think it makes sense to say both spaces are equipped with the norm topology; it seems like the right setting to me. I've provided part of a proof below.
Let $x,y \in A$. We know that there exists an isometric isomorphism $\tau_x : C(\sigma(x)) \rightarrow A_x$, where $\sigma(x)$ is the spectrum of $x$, $C(\sigma(x))$ is the $C^*$-algebra of continuous functions on $\sigma(x)$ equipped with the supremum norm, and $A_x$ is the $C^*$-algebra generated by $\{1,x\}$, where $1$ is the unit of $A$. 
By definition, we have $\tau_x(f) = f(x)$. Moreover, for any polynomial $P$, we have $\tau_x(P) = P(x)$, where $P(x)$ is just the usual polynomial expression of the element $x$. We have a similar function $\tau_y$.
Now, remark that we have:
\begin{align*}
\|f(x) - f(y)\|
&=
\|\tau_x(f) - \tau_x(P)+\tau_x(P)-\tau_y(P)+\tau_y(P)-\tau_y(f)\|
\\
&\leq
\|\tau_x(f) - \tau_x(P)\| + \|\tau_x(P)-\tau_y(P)\| + \|\tau_y(P)-\tau_y(f)\|
\\
&=
\|P(x)-P(y)\| + \|f-P\|_{C^{\infty}(\sigma(x))} + \|f-P\|_{C^{\infty}(\sigma(y))}
\end{align*}
Now, since $f$ is continuous and $\sigma(x), \sigma(y)$ are both compact, we can approximate $f$ by polynomials on these sets (Stone-Weierstrass theorem). So, we have reduced the problem to the following lemma:

If $P$ is a complex polynomial, then $A \rightarrow A : x \mapsto P(x)$ is continuous.

If this is true, then for any $\epsilon > 0$, we can pick a polynomial $P$ such that 
$\|f-P\|_{C^{\infty}(\sigma(x))} < \epsilon/3$ 
and
$\|f-P\|_{C^{\infty}(\sigma(y))} < \epsilon/3$,
and then pick $\delta>0$ such that $\|P(x)-P(y)\| < \epsilon/3$ whenever $\|x-y\|<\delta$. By the above inequality, this gives $\|f(x) - f(y)\| < \epsilon$, which finishes the proof.
I leave it to you to figure out the lemma. I think it could be proven the same way that we prove regular polynomials are continuous.
P.S. This proof feels unnecessarily lengthy to me; I'd be happy to see a more succinct and elegant one.
