# Question about the Ascoli-Arzelá Theorem proof

Ascoli-Arzelá Thoerem: Let $K$ be a compact space and $M$ be a metric space and $C(K,M)$ be the set of continuous functions from $K$ to $M$. $H \subset C(K,M)$ is relatively compact if and only if $H$ is equicontinuous and $H(x) : = \{ f(x): f\in H\}$ is relatively compact.

I want to understand the step in the implication ($\Leftarrow$). In order to show that $H$ is relatively compact I'd like to know how to prove that $\overline{H}$ is complete.

$\underline{Ideas:}$

First one.

Let $\overline{f}_n$ be a Cauchy sequence in $\overline{H}$. Then for any $n$, let $f_n \in H$ such that $$d(f_n,\overline{f}_n) < 1/n$$ As by hypothesis $\overline{H(x)}$ is compact and therefore complete, $(f_n(x))_n$ is a Cauchy sequence and we can define $f(x):= \lim_{n} f_n(x)$. There is a result that if $\overline{f}_n$ is equicontinuous and converges punctually to $f$, then $f$ is continuous and converges uniformly. I can see that $\overline{f}_n$ and converges punctually to $f$ but I can't see that $\overline{f}_n$ is equicontinuous. The place where I saw this Idea says that this is by construction. I can see by construction the party that I alredy said. How to see this?

Second one.

It suffices to show that $f_n \rightarrow f$ uniformly since $f$ will be continuous as uniform limit of continuous functions. As $\overline{f}_n$ is a Cauchy sequence for all $\varepsilon >0$ there is $n_0$ such that $n>n_0$ impllies $$d(\overline{f}_n,\overline{f}_m) : = \sup_{x \in K} d(f_n(x),f_m(x)) < \varepsilon.$$ Then, by triangle inequality and taking the limit as $m \rightarrow \infty$ in $$d(\overline{f}_n,f) \le d(\overline{f}_n,\overline{f}_m) + d(\overline{f}_m, f)$$ to obtain $$d(\overline{f}_n,f) \le \varepsilon + \lim_{m} d(\overline{f}_m, f).$$ Then we only need to prove that $$\lim_{m\rightarrow \infty} \sup_{x \in K} d(\overline{ f}_n(x),f(x)).$$ Note that we can use the same idea to show that $C(K,M)$ is complete (I do not know if this is true, for example is true if $M = \mathbb{R}$). Because if this is true $\overline{H}$ is complete because is the closure of a set contained in a in a complete space is compact.

Any help will be good. I will be very grateful.

I can see that $\overline{f}_n$ and converges punctually to $f$ but I can't see that $\overline{f}_n$ is equicontinuous.

The easiest way to see that is indirectly.

By construction, we have $f_n(x) \to f(x)$ for all $x\in K$. Now, $\{ f_n : n \in \mathbb{N}\}$ is a subsetof $H$, hence equicontinuous by assumption. Therefore, the convergence is in fact uniform. Hence $f$ is continuous. And

$$d(f,\overline{f}_n) \leqslant d(f,f_n) + d(f_n,\overline{f}_n) < d(f,f_n) + \frac{1}{n}.\tag{1}$$

That shows that $\overline{f}_n$ converges uniformly to $f$ (given $\varepsilon > 0$, choose $n_0 \geqslant 2/\varepsilon$ such that $d(f,f_n) < \varepsilon/2$ for all $n \geqslant n_0$, then $(1)$ shows $d(f,\overline{f}_n) < \varepsilon$ for $n \geqslant n_0$). But a uniformly convergent sequence of continuous functions on a compact space is equicontinuous.

Well, but the used theorems, that a pointwise convergent equicontinuous sequence of functions converges uniformly on compact sets, and that a uniformly convergent sequence of continuous functions is equicontinuous, may not be very familiar.

So let's show that if $H$ is equicontinuous, then $\overline{H}$ is also equicontinuous.

Let $x\in K$ arbitrary. For $\varepsilon > 0$, there is, by definition of equicontinuity, a neighbourhood $U_\varepsilon$ of $x$ such that

$$\bigl(\forall f\in H\bigr)\bigl(\forall y \in U_\varepsilon\bigr)\bigl(d_M(f(x),f(y)) \leqslant \varepsilon\bigr).$$

But then, for $\overline{f} \in \overline{H}$, and any $\delta > 0$, there is an $f\in H$ with $d(\overline{f},f) < \delta$, and hence we have

$$d_M(\overline{f}(x),\overline{f}(y)) \leqslant d_M(\overline{f}(x),f(x)) + d_M(f(x),f(y)) + d_M(f(y),\overline{f}(y)) \leqslant d(f(x),f(y)) + 2d(\overline{f},f) \leqslant \varepsilon + 2\delta$$

for all $y\in U_\varepsilon$. But since $\delta > 0$ was arbitrary, it follows that

$$d_M(\overline{f}(x),\overline{f}(y)) \leqslant \varepsilon$$

for all $\overline{f} \in \overline{H}$ and all $y\in U_\varepsilon$. Thus we have shown that $\overline{H}$ is equicontinuous when $H$ is.

Note: it is even true that $\operatorname{cl}_{pw}(H)$, the closure of $H$ in the topology of pointwise convergence, is equicontinuous when $H$ is equicontinuous.

Note that we can use the same idea to show that $C(K,M)$ is complete

No, we cannot do that, since $C(K,M)$ is complete only if $M$ is complete, and that was not assumed. Part of the hypotheses is that for every $x\in K$ the set $H(x)$ is contained in a compact and hence complete subset of $M$, and that can be the case even if $M$ itself is not complete.

• Why the fact that $\{f_n: n \in \mathbb{N}\} \subset H$ implies that $\{f_n: n \in \mathbb{N}\}$ is equicontinuous? – user29999 Nov 15 '14 at 22:59
• A subset of an equicontinuous set is a fortiori equicontinuous. The condition constrains fewer functions, so it is easier to be met. – Daniel Fischer Nov 15 '14 at 23:08