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An example of an application of Arzelà-Ascoli is that we can use it to prove that the following operator is compact: $$ T: C(X) \to C(Y), f \mapsto \int_X f(x) k(x,y)dx$$ where $f \in C(X), k \in C(X \times Y)$ and $X,Y$ are compact metric spaces.

To prove that $T$ is compact we can show that $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ is bounded and equicontinuous so that by Arzelà-Ascoli we get what we want. It's clear to me that if $TB_{\|\cdot\|_\infty} (0,1)$ is bounded then $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ is bounded too. What is not clear to me is why $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ is equicontinuous if $TB_{\|\cdot\|_\infty} (0, 1)$ is.

I think about it as follows: $TB_{\|\cdot\|_\infty} (0, 1)$ is dense in $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ with respect to $\|\cdot\|_\infty$ hence all $f$ in $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ are continuous (since they are uniform limits of continuous sequences). Since $Y$ is compact they are uniformly continuous. Now I don't know how to argue why I get equicontinuity from this. Thanks for your help.

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The closure for the topology of uniform convergence (or even simple convergence) of an equicontinuous set of functions is still equicontinuous. This is a standard fact you can find in almost every textbook. Uniform continuity has nothing to do with it. – Ahriman Aug 18 '12 at 8:22
Try to do it (what Ahriman says) yourself: Adapt the proof you give here to show that if $(f_k)_k$ is an equicontinuous sequence and $f_k \to f$ uniformly then $\{f\} \cup \{f_k\,:\,k \in \mathbb{N}\}$ is equicontinuous. Analyzing the resulting argument you should be able to deduce Ahriman's statement. – t.b. Aug 18 '12 at 8:37
up vote 4 down vote accepted

Take any $S\subseteq C(X)$ which is equicontinuous. Let $\epsilon>0$. We have some $\delta>0$ such that $$f\in S, \|x-x'\|<\delta\implies \|f(x)-f(x')\|<\epsilon.$$ For any $f\in \bar S$, we have a sequence $f_n\to f$ uniformly with each $f_n\in S$. Then $$\|x-x'\|<\delta\implies \|f(x)-f(x')\|\leq \|f(x)-f_n(x)\|+\|f_n(x)-f_n(x')\|+\|f_n(x')-f(x)\|$$ which becomes less than $\epsilon$ for sufficiently large $n$. Thus $\bar S$ is equicontinuous.

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+1 For a nice and simple answer. – user38268 Aug 18 '12 at 8:46
Thank you Alex. I wanted to do it on my own, so I posted an answer of my own. Of course I accept you, not my own answer. Thanks for your help. – Rudy the Reindeer Aug 18 '12 at 9:24

In this case, equicontinuity means that for all $f$ such that $\|f\|_\infty \leq 1$ and for all $\epsilon >0 $, there exists a $\delta>0$ such that if $d_Y(x,y) < \delta$, then $|Tf(x)-Tf(y)| \leq \epsilon$.

Now suppose there is a sequence $f_n$ (in the unit ball) such that $T f_n \to \phi$. Suppose $d_Y(x,y) < \delta$, then since $|Tf_n(x)-Tf_n(y)| \leq \epsilon$, for all $n$, it follows that $|\phi(x)-\phi(y)| \leq \epsilon$, as well. Hence any limit point is also equicontinuous (with the same modulus of continuity). Hence the closure of $TB_{\|\cdot\|_\infty} (0, 1)$ is equicontinuous.

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Recall that being an equicontinuous family implies that it is totally bounded with respect to the sup metric derived from the one on $Y$. Call the metric on $Y$ $d$. Now it is a general fact in real analysis that if a subset of a metric space is totally bounded then so is its closure. But then

$$\overline{TB_{\|\cdot\|_\infty} (0,1)}$$

totally bounded with respect to the sup metric derived from that on $Y$ implies that the family is equicontinuous under $f$.

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Parsing help: Some people write $B(x,r)$ for the ball of radius $r$ around $x$. The index $\lVert \cdot \rVert_\infty$ indicates that the ball is to be taken in the $\sup$-norm. Thus $TB_{\lVert \cdot\rVert_\infty}(0,1)$ is the image in $C(Y)$ of the unit ball of $C(X)$ with the sup norm under $T\colon C(X) \to C(Y)$. – t.b. Aug 18 '12 at 8:47
@t.b. Thanks for the heads up. I don't know functional analysis so that's nice to know. – user38268 Aug 18 '12 at 8:49

Following tb's comment:

Claim: If $\{f_n\}$ is equicontinuous and $f_n \to f$ uniformly then $\{f\} \cup \{f_n\}$ is equicontinuous.

Proof: Let $\varepsilon > 0$.

(i) Let $\delta^\prime$ be the delta that we get from equicontinuity of $\{f_n\}$ so that $d(x,y) < \delta^\prime$ implies $|f_n(x) - f_n(y)| < \varepsilon$ for all $n$.

(ii) Since $f_n \to f$ uniformly, $f$ is continuous and since $X$ is compact, $f$ is uniformly continuous so there is a $\delta^{\prime \prime}$ such that $d(x,y) < \delta^{\prime \prime}$ implies $|f(x) - f(y)| < \varepsilon$.

Now let $\delta = \min(\delta^\prime, \delta^{\prime \prime})$ then $d(x,y) < \delta$ implies $|g(x) - g(y)| < \varepsilon$ for all $g$ in $\{f\} \cup \{f_n\}$.

Edit What I wrote above is rubbish and doesn't lead anywhere. As pointed out in Ahriman's comment to the OP, we don't need continuity of $f$. We can bound $f$ as follows (in analogy to the proof of the uniform limit theorem): Let $\delta$ be such that $|f_n(x) - f_n(y)| < \varepsilon / 3$ for all $n$ and all $x,y$. Since $f$ is the uniform limit of $f_n$, for $x$ fixed, $f_n(x)$ is a Cauchy sequence converging to $f(x)$. Let $n$ be such that $\|f-f_n\|_\infty < \varepsilon / 3$. Then $$ |f(x) - f(y)| \leq |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f(y)-f_n (y)| < \varepsilon$$

Hence we may choose $\delta$ such that $|f(x) - f(y)| < \varepsilon / 3$ for all $f$ in $TB(0,1)$ to get that $\overline{TB(0,1)}$ is equicontinuous, too.

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Perhaps I should've mentioned it in the question. I'd thought of something like this, but since I don't have a finite number of $f$s in the closure, I am not sure how to pick the $\delta$ to bound all of them, even though I can do it for one sequence. – Rudy the Reindeer Aug 18 '12 at 9:00
If you look closely at the argument for continuity of $f$ you see that you can take $\delta'$ (see Alex's approach). No need to invoke compactness of $X$. That's one reason why I insisted in various comment threads that you should prove continuity yourself and not appeal to any theorem. – t.b. Aug 18 '12 at 9:02
I'm sorry. I did neither forget nor ignore these comments. I am simply too panicked to think. – Rudy the Reindeer Aug 18 '12 at 9:23

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