Arzela-Ascoli: Proof?

Problem

Given a compact domain.

Regard the function space: $$\mathcal{C}(\Omega):=\{f:\Omega\to\mathbb{C}:f\text{ continuous}\}$$

Consider a bounded family: $$\mathcal{F}\subseteq\mathcal{C}(\Omega):\quad\|f\|_{f\in\mathcal{F}}<\infty$$

Then Arzela-Ascoli states: $$\mathcal{F}\text{ precompact}\iff\mathcal{F}\text{ equicontinuous}$$ How to prove this from scratch?

Attempt

For a precompact family one finds: $$\mathcal{F}\subseteq\mathcal{B}_\delta(g_1)\cup\ldots\cup\mathcal{B}_\delta(g_I)$$ So one can always pick one close enough: $$f\in\mathcal{F}:\quad|f(x)-f(z)|\leq|f(x)-g_f(x)|+|g_f(x)-g_f(z)|+|g_f(z)-f(z)|<\varepsilon\quad(x\in B_\delta(z))$$

Conversely, prove for a sequence: $$f_n\in\mathcal{F}:\quad\|f_{m'}-f_{n'}\|\to0$$

For a compact domain one finds: $$\Omega\subseteq B_\delta(a_1)\cup\ldots\cup B_\delta(a_I)$$

Bolzano-Weierstrass gives a subsequence: $$|f_n(a_i)|_{n\in\mathbb{N}}<\infty:\quad|f_{m'}(a_i)-f_{n'}(a_i)|\to0$$

Take as threshold: $$m',n'\geq N':=\max_{i=1\ldots I}N'_i$$

So one can again always pick one close enough: $$x\in\Omega:\quad|f_{m'}(x)-f_{n'}(x)|\leq|f_{m'}(x)-f_{m'}(a_x)|+|f_{m'}(a_x)-f_{n'}(a_x)|+|f_{n'}(a_x)-f_{n'}(x)|<\varepsilon$$

Is this proof correct or do I miss something?

Discussion

Moreover, why does the usual proof exploit separability before?

(For example see wiki: Arzela-Ascoli: Proof)

Sure for a proposition on its own: $$\Omega\text{ separable}:\quad|f_n(x)-f(x)|\to0$$ $$\Omega\text{ precompact}:\quad\|f_n-f\|\to0$$

But why both together in a single proof?

• Speaking generally ... Cantor diagonalization allows you to take any sequence of complx functions $\{ f_{n} \}$ defined on a countable set $S$ for which $\{ f_{n}(x)\}$ is a bounded in $n$ for each fixed $x \in S$, and obtain a subsequence $\{ f_{n_{k}}\}_{k=1}^{\infty}$ which converges everywhere on $S$. This can be extended to functions on a space $\Omega$ with a countable dense subset by assuming an equicontinuity condition. Equicontinuity naturally arises when dealing with solutions of differential equations through integral operators. And that's where Fredholm theory came from originally. – DisintegratingByParts Feb 18 '15 at 21:11
• @T.A.E.: Yes I had that diagonal argument when I started writing this thread but that seems superfluos if one has a compact domain anyway, or? So it makes sense as a proposition on pointwise convergence on its own but within the proof of Arzela-Ascoli? – C-Star-W-Star Feb 18 '15 at 21:13
• Another example: If you start with a uniformly bounded sequence of non-decreasing real functions on $\mathbb{R}$, then you can get a subsequence which converges at the rationals, and you can show that there is a further subsequence that will converge pointwise everywhere; equicontinuity can be replaced by monotonicity instead. That, in some sense, becomes a compactness type of result for Borel measures on $\mathbb{R}$. – DisintegratingByParts Feb 18 '15 at 21:15
• By the way, the definition of a compact operator was an abstraction of the Arzela-Ascoli arguments used to study integral operators, especially their Fredholm properties and index. – DisintegratingByParts Feb 18 '15 at 21:31
• @T.A.E.: Aaaah that's why Arzela-Ascoli remains so important. – C-Star-W-Star Feb 18 '15 at 21:35

Consider any separable metric space $X$ and a complete metric space $Y$. Then endow $C(X,Y)$ with the compact convergence topology: a sequence $(f_n)$ converges to $f\in C(X,Y)$ if and only if for every compact subset $K$ of $X$, $f_n\mid K$ converges uniformly to $f\mid K$. You can show this is actually metrizable, so that compact and and sequentially compact are equivalent properties, and hence precompact and sequentially precompact are too.

What Arzela-Ascoli says is that a family $\mathscr F$ of $C(X,Y)$ is precompact if and only if it is locally uniformly equicontinuous -- every point $x\in X$ has a neighborhood $V$ where $\mathscr F\mid V$ is uniformly equicontinuous, and pointwise bounded, i.e. every set ${\rm ev}_x(\mathscr F)$ is bounded in $Y$ for each $x\in X$.

It seems to me that when one puts it in this way, the proof is very natural and easy to remember.

Consider a sequence $(f_n)$ in $\mathscr F$. First, take a dense countable subset $A=\{a_1,\ldots\}$. The general diagonal argument gives a subsequence $g_j=f_{i_j}$ such that $g_j(a_i)$ converges for each $a_i$ as $j\to\infty$. Now pick a compact set $K$, and take $\varepsilon >0$.

Since $\mathscr F$ is locally uniformly equicontinuous, each point $x$ in $K$ has a neighborhood $V_x$ where $\mathscr F \mid V_x$ is is uniformly equicontinuous. Since $K$ is compact, we can find finitely many $V_1,\cdots ,V_s$ that cover $K$, and where $z,y\in V_x$ implies $d(f(z),f(y))<\varepsilon$ for any $z,y\in V_x$ and $f\in\mathscr F$.

Since $A$ is dense, we can find $a'_i=a_{i_j}\in V_j$ for each $j$. I claim that $(g_j)$ converges uniformly in $K$. Indeed, pick any $z\in K$. If $z$ is in $V_i$ we have that

$$d(g_n(z),g_m(z))\leqslant d(g_n(z),g_n(a_i))+d(g_n(a_i),g_m(a_i))+d(g_m(a_i),g_m(z))<3\varepsilon$$

Since $a_i,z\in V_i$, we have that $d(g_n(z),g_n(a_i))$ and $d(g_m(a_i),g_m(z))$ are both $<\varepsilon$ and since $g_j(a_i)$ converges we can pick $N$ so that $n,m>N$ gives $d(g_n(a_i),g_m(a_i))<\varepsilon$. This means that if $n,m>N$ and $z\in K$, $$d(g_n(z),g_m(z))<3\varepsilon$$

Which means $(g_n)$ converges uniformly in $K$.

• Ah right so the natural topology in this situation would be that of uniform convergence on compact sets. So in that sense a separable space would do the job too. Thanks! – C-Star-W-Star Feb 18 '15 at 21:10
• @Freeze_S Yes, indeed. – Pedro Tamaroff Feb 18 '15 at 21:19
• Sorry for invoking an old one, but how do you show the metrizability of $C(X\rightarrow Y)$? – kazuki Dec 23 '17 at 19:43

The separability argument does not lose any generality. The reason is that a compact metric space is totally bounded (which you are using), and a totally bounded metric space is separable. So if you assume that the domain is compact then you get that it is separable for free.

Here's a proof of the totally bounded implies separable argument:

Suppose $X$ is a totally bounded metric space, and consider covering $X$ by $N_n$ balls of radius $1/n$ which are centered at $\{ x_{i,n} \}_{i=1}^{N_n}$. Then the set $\{ x_{i,n} : i \in \{ 1,\dots,N_n \},n \in \mathbb{N} \}$ is a countable dense subset of $X$.

I find the separability argument convenient because it constructs the limit: we extract a subsequence which converges pointwise to a function $f$ on the dense subset, and then the limit function is the unique continuous extension of $f$ to the whole space. Equicontinuity then guarantees that this extension actually exists and must be the uniform limit of the original sequence.

• Yes sure but that seems like an redundant argument when one has already a compact domain, or? – C-Star-W-Star Feb 18 '15 at 21:11
• @Freeze_S I don't really follow. I understood your question as asking about why you can use separability, not really why you should. I was trying to explain that if your domain is compact then you can automatically assume it is separable. – Ian Feb 18 '15 at 22:11
• Sure precompact implies separable. But building in the separability argument is superfluous. – C-Star-W-Star Feb 18 '15 at 22:15
• @Freeze_S I'm not talking about separability of the function space, I'm talking about separability of the domain. You assumed a priori that the domain is compact so a fortiori it is also separable. – Ian Feb 18 '15 at 22:19
• Ah to bad, I'm sorry I have stolen your time. :/ (See first line.) – C-Star-W-Star Feb 18 '15 at 22:21