equicontinuous functions in $C([a,b])$ Consider sequence of equicontinuous and bounded functions called $E$ in $C([a,b])$.
We define $g(x)=\sup\{f(x):f \in E\}$.
Now my question is how we can show that $g(x)$ is continuous function of $x$.
I write something.
Fix x that $g(x)=f_n(x)$.
For arbitrary $y$, $g(y)$ is one of the elements of $E$ called $f_m(x)$.
My problem is if we take $y$ close enough to $x$ I can't show that we can small $f_n(y)-f_m(y)$ as we want.
 A: Equicontinuity is the statement that for every $x\in[a,b]$ and $\epsilon>0$, there is a $\delta>0$ such that for every $f\in E$, $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$. The important thing here is that the same $\delta$ works for every $f\in E$.
Going forward, we'll need one simple to verify fact: if $E$ is equicontinuous and $b\in E$ is fixed, then the family $\{\max\{b,f\}:f\in E\}$ is also equicontinuous.
Now, fix $x\in[a,b]$, $\epsilon>0$ and find $f_x\in E$ with $0\leq g(x)-f_x(x)<\epsilon$, which can be done as $E$ is uniformly bounded. Also, for any $y\in[a,b]$ and find $f_y\in E$ with $0\leq g(y)-f_y(y)<\epsilon$ and define $f_y^*=\max\{f_x,f_y\}$. Notice that $f_y^*$ satisfies $0\leq g(x)-f_y^*(x),g(y)-f_y^*(y)<\epsilon$; thus
$$
|g(x)-g(y)|< 2\epsilon+|f_y^*(x)-f_y^*(y)|.
$$
Now, by the above mentioned fact, the family $\{f_y^*:y\in[a,b]\}$ is equicontinuous, so for $\delta$ sufficiently small, we also have $|f_y^*(x)-f_y^*(y)|<\epsilon$ whenever $|x-y|<\delta$, so we know that $g$ must be a continuous function.
A: Here are sketches to two different proofs that are common for such a problem: 
Call the elements of the sequence $f_n$, and let $g_n(x)=\max\{f_1(x),\cdots, f_n(x)\}.$ Note each $g_n$ is continuous. As you said, $g(x)=\sup_n\{f_n(x)\}$ is defined for all $x\in [a,b],$ given the family is uniformly bounded. Also, we can see that $g_n(x)\leq g_{n+1}(x)$ for all $x\in [0,1],$ so that $g_n\nearrow g$ pointwise. Now, note that the family $\{g_n:\ n\in\mathbb{N}\}$ is uniformly bounded (as $E$ is) and equicontinuous (check this). By the Arzela-Ascoli theorem, there exists a uniformly convergent subsequence. But, if a monotonic sequence has a uniformly convergent subsequence, then it converges uniformly. Since the pointwise limit must match the uniform limit, $g_n\rightarrow g$ uniformly, and hence $g$ is continuous.
You can also do it directly. We know that for all $\epsilon>0$, there exists $\delta>0$ so that for all $n\in\mathbb{N}$ and $x,y\in [a,b],$ $$|x-y|<\delta\implies |f_n(x)-f_n(y)|<\epsilon/2.$$ We want to find a $\delta'$ that gives continuity for $g$. If we take $\delta'=\delta$, then for some $n\in\mathbb{N}$ (given by the definition of sup), $$g(x)\leq f_n(x)+\epsilon/2<f_n(y)+\epsilon/2+\epsilon/2=f_n(y)+\epsilon\leq g(y)+\epsilon,$$ using continuity. If you reverse $x$ and $y$ and do the same argument, you can conclude that $|g(x)-g(y)|<\epsilon,$ provided $|x-y|<\delta.$
