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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.

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2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Apr 25, 2021 at 5:30
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    $\begingroup$ 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. $\endgroup$
    – robjohn
    Commented Apr 25, 2021 at 5:57
  • $\begingroup$ Got it, thanks a lot! @robjohn $\endgroup$ Commented Apr 25, 2021 at 5:59
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I think your corollary is a restatement of the theorem, in fact baby Rudin proves your corollary as Arzelà Ascoli in theorem 7.25.

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