Proving a Corollary of Arzela-Ascoli's Theorem

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.

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.

• Could you explain the part where you say: "with a slight change in $\delta$ for given $\epsilon$, we get..."? For equicontinuity, we want a $\delta > 0$ for given $\epsilon > 0$ such that $$|x-y| < \delta \implies |f(x) - f(y)| < \epsilon\ \ \text{for every } f\in \overline Y$$ How are we getting this? Commented Apr 25, 2021 at 5:30
• In the computation given above, we take the $\delta$ that gives a certain $\epsilon$ for $Y$ which gives $3\epsilon$ for $\overline{Y}$. Thus, to get $\epsilon$ for $\overline{Y}$, we would use the $\delta$ that gives $\epsilon/3$ for $Y$. Actually, by using a little more care, we can get the same $\delta$ and $\epsilon$, but that argument is a bit more complicated.
– robjohn
Commented Apr 25, 2021 at 5:57
• Got it, thanks a lot! @robjohn Commented Apr 25, 2021 at 5:59

I think your corollary is a restatement of the theorem, in fact baby Rudin proves your corollary as Arzelà Ascoli in theorem 7.25.