The Arzela-Ascoli theorem (as stated in Carothers' Real Analysis) says:
Theorem, Arzela-Ascoli: Let $X$ be a compact metric space, and let $\mathcal F$ be a subset of $C(X)$. Then, $\mathcal F$ is compact if and only if $\mathcal F$ is closed, uniformly bounded and equicontinuous.
Here is the corollary I want to prove:
Corollary: Let $X$ be a compact metric space. If $(f_n)_{n=1}^\infty$ is a uniformly bounded, equicontinuous sequence in $C(X)$, then some subsequence of $(f_n)_{n=1}^\infty$ converges uniformly on $X$.
I know that a set $S$ is compact in a metric space $(M,d)$ iff every sequence in $S$ has a subsequence that converges to a point in $S$. Since convergence and uniform convergence are identical in $(C(X), \|\cdot\|_\infty)$, it would suffice to prove that $(f_n)_{n=1}^\infty$ is a compact set in $(C(X), \|\cdot\|_\infty)$. In order to use Arzela-Ascoli's theorem, I need a set of functions which is (i) closed (ii) equicontinuous (iii) uniformly bounded. By the assumption in the corollary, $(f_n)_{n=1}^\infty$ is uniformly bounded and equicontinuous. If it were closed, I could have directly applied the Arzela-Ascoli theorem to get that $(f_n)_{n=1}^\infty$ is a compact set in $C(X)$ – as a result it has a convergent (uniformly) subsequence. Since this is not generally the case, I started thinking about the closure of $Y = \{f_n: n\ge 1\}$, i.e. $Z = \overline Y$.
If I am able to prove that $Z$ is uniformly bounded and equicontinuous (we already know that $Y$ is), then I can use Arzela-Ascoli's theorem to conclude that $Z$ is compact (since $Z$ is closed). $(f_n)_{n=1}^\infty$ is a sequence in $Z$ too, so if $Z$ is compact, $(f_n)_{n=1}^\infty$ must have a convergent (uniformly) subsequence. Done!
How do I prove that $Z$ is uniformly bounded and equicontinuous? Are there any other ways to prove this corollary? Thanks a lot!
Here, I found a related question. The answer states but doesn't prove that $Z$ is uniformly bounded and equicontinuous.
Best Answer
I assume that the metric on $C(X)$ is $\|f-g\|=\sup\limits_{x\in X}|f(x)-g(x)|$.
Suppose that $g\in\overline{Y}$. That means that for any $\epsilon\gt0$, there is an $f_n$ so that $\|f_n-g\|\le\epsilon$.
Since $\sup\limits_{x\in X}|f_n(x)|\le M$, we have $\sup\limits_{x\in X}|g(x)|\le M+\epsilon$. Thus $\overline{Y}$ is uniformly bounded (since we can choose $\epsilon$ as small as we want, the bound is the same).
Suppose that if $|x-y|\le\delta$, then $|f_n(x)-f_n(y)|\le\epsilon$. It follows that $$ \begin{align} |g(x)-g(y)| &\le|g(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-g(y)|\\[3pt] &\le3\epsilon \end{align} $$ So, with a slight change in the $\delta$ for a given $\epsilon$, we get that $\overline{Y}$ is equicontinuous.
Now you can proceed as you intended.