Convergent sequence of functions and compact set contains the spectrum of the limit Let  $K\subseteq \mathbb{C}$ be a compact subset. Let $(x_n)$ be a convergent sequence of normal elements in a unital $C^*$-algebra $\mathcal{A}$, with limit $x$, such that $\sigma(x_n)\subseteq K$ for all $n$.
Problems: 
(i) Suppose that $\sigma(x)\subseteq K$. Show that for every $f\in C(K)$, one has $f(x_n)\to f(x)$ in $\mathcal{A}$.
(ii) Show that $\sigma(x)\subseteq K$.
Normally, I do not have any problems dealing with similar sequences. I do not know where to start in (i). Any hints would be helpful.
 A: Actually, I think one has to prove (ii) first because $f(x)$ is not well-defined if $\sigma(x)\not\subset K$.
(ii) Let $\lambda\in \mathbb{C}\setminus K$ and $f(z)=\frac 1{z-\lambda}$. Then
$$
\|(x_n-\lambda)^{-1}-(x_m-\lambda)^{-1}\|=\|(x_n-\lambda)^{-1}(x_m-x_n)(x_m-\lambda)^{-1}\|\leq \|f\|_{C(K)}^2\|x_m-x_n\|.
$$
Thus $(x_n-\lambda)^{-1}$ is a Cauchy sequence. Let's call its limit $y$. We have
\begin{align*}
\|y(x-\lambda)-1\|&=\lim_{n\to\infty}\|(x_n-\lambda)^{-1}(x-\lambda)-(x_n-\lambda)^{-1}(x_n-\lambda)\|\\
&=\lim_{n\to\infty}\|(x_n-\lambda)^{-1}(x-x_n)\|\leq \|f\|_{C(K)}\lim_{n\to\infty}\|x-x_n\|\\
&=0.
\end{align*}
Hence $y(x-\lambda)=1$. Similarly one shows $(x-\lambda)y=1$. Therefore $\lambda\in \rho(x)$.
(i) As established in the comments, if $p$ is a polynomial, then $p(x_n)\to p(x)$. This follows from the continuity of addition and multiplication in $\mathcal{A}$. Now let $f\in C(K)$ and $(p_k)$ a sequence of polynomials such that $p_k\to f$ uniformly on $K$. The existence of such a sequence is guaranteed by the Stone-Weierstrass theorem. Then
\begin{align*}
\|f(x)-f(x_n)\|&\leq \|f(x)-p_k(x)\|+\|p_k(x)-p_k(x_n)\|+\|p_k(x_n)-f(x_n)\|\\
&\leq 2\|p_k-f\|_{C(K)}+\|p_k(x)-p_k(x_n)\|.
\end{align*}
Letting first $n\to\infty$ and then $k\to \infty$ yields the desired result.
