$x \mapsto \sup_{f \in K}{f(x)}$ is continuous Let $K \subset C[0, 1]$ be a compact subspace, then consider the following map $\varphi: x \mapsto \sup_{f \in K}{f(x)}$ How to prove that $\varphi$ is continuous?
Since, $K$ is compact, then, by Arzela-Ascoli theorem, $K$ is pointwise bounded, i.e. $|f_{n}(x)| \leq M$, $\forall f_{n} \in K$ and equicontinuity also holds, i.e. $\forall \epsilon > 0, \exists \delta >0: |x-y| < \delta => |f(x)-f(y)| < \epsilon$ for each $f \in K$. Then, the second condition seems to be extremely useful, since i need to prove that if $x_{n} \rightarrow x$, then $\sup_{f \in K}{f(x_{n})} \rightarrow \sup_{f \in K}{f(x)}$ by i can't make a good use of it. Are there any hints that might help?
 A: Let us show that $\varphi$ is continuous in $x$. Hence, we pick an arbitrary $\varepsilon > 0$ and choose the corresponding $\delta > 0$.
Now, let $y$ be given with $| x - y | \le \delta$. For any $f \in K$ we have
$$f(x) - \varepsilon \le f(y)\le f(x) + \varepsilon.$$
You should try to finish the proof.
A: Suppose $x_n \to x$, then $f(x_n) \to f(x)$ for all $f \in K$ and
since $\phi(x_n) \ge f(x_n)$ for all $f \in K$ we see that
$\liminf_n \phi(x_n) \ge \liminf_n f(x_n) = f(x)$ and so 
$\liminf_n \phi(x_n) \ge \phi(x)$. This is true regardless of the nature of $K$.
Now choose $f_n \in K$ such that $f_n(x_n) = \phi(x_n)$. Choose a subsequence
$x_{n_k}$ such that $\limsup_n \phi(x_n) = \lim_k \phi(x_{n_k})$. Then $f_{n_k}$ has a further subsequence such that $f_{n_{k_j}} \to f \in K$,
and hence we have $f_{n_{k_j}}(x_{n_{k_j}}) \to f(x)$ and so
\begin{eqnarray*}
\limsup_n \phi(x_n) &=& \lim_k \phi(x_{n_k})
= \lim_j \phi(x_{n_{k_j}}) \\
&=& \lim_j f_{n_{k_j}}(x_{n_{k_j}})
= f(x) \le \phi(x)
\end{eqnarray*}
Hence $\phi$ is continuous.
A: It's possible to complete a proof in the following way:
Since $K \subset C([0, 1], \mathbb{R})$ is compact, then by Arzela-Ascoli theorem $\forall \epsilon > 0, \exists \delta>0$, so that $|x-y| < \delta$ implies $|f(x)-f(y)|<\epsilon, \forall f \in K$. 
Let $|x_{n} - x| \to 0$, then we need to check that $|\sup_{f \in K}{f(x_{n})}-\sup_{f \in K}{f(x)}| \to 0$. But $|\sup_{f \in K}{f(x_{n})}-\sup_{f \in K}{f(x)}| \leq \sup_{f \in K}{|f(x_{n})-f(x)|} < \epsilon, \forall \epsilon > 0$ (by equicontinuity). 
