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.

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.
