The Arzelà–Ascoli theorem from the duality perspective

Introduction

The Arzelà–Ascoli theorem is an indispensable tool in mathematical analysis which characterizes compact sets of continuous functions in terms of equicontinuity. A “hard analysis way” of understanding this theorem is that, the only ways to “escape to infinity” for a collection of continuous functions are either:

1. values of the functions at a point blow up, or
2. regularities of the functions blow up.

In other words, if values at any given point are confined in a compact set and also the functions are “uniformly regular” (i.e., equicontinuous), then the collection must be relatively compact.

On the other hand, we can also view this statement in a more “soft analysis way”: it is a natural consequence of how compactness interplays with joint continuity of the pairing of functions and points. As an illustration of this viewpoint, I provided below a proof of the Arzelà–Ascoli theorem based on such an idea.

Setup

Let me describe the precise setup first. Let $X,Z$ be any sets and $\left\langle\,\cdot\,,\,\cdot\,\right\rangle\colon Z\times X \to Y$ be any function into a bounded metric space $(Y,d)$. Here are some remarks:

• The set $X$ is meant to be the “space”, i.e., the set of points.
• The set $Z$ is meant to be the set of $Y$-valued functions on $X$.
• The pairing $\left\langle z,x\right\rangle$ is meant to be the evaluation of $z$ at $x$.
• However, $X$ being the space and $Z$ being the functions is just the matter of interpretation, and we can freely reverse their roles at any time if we want. In fact, when we write $f(x)$, we can either regard $x$ as an object and $f$ as an operator acting on $x$, or $f$ as an object and $x$ as an operator acting on $f$. There really is no fundamental difference between the two, which can be considered the essence of many of the so-called “dualities” in mathematics. This is one of the key ingredients of the proof.
• $Y$ can (should) be in fact any uniform space. We chose it to be a bounded metric space solely because of possible readers’ convenience, and all proofs given below can be generalized to uniform spaces without any nontrivial issue. (In particular, the boundedness condition on $d$ is not really a restriction as we can always replace any metric by its truncation without changing the induced uniformity and topology.)

Given a set $A$, the set $Y^{A}$ of all functions $A\to Y$ can be either given with the pointwise convergence topology which is nothing but the product topology, or the uniform convergence topology which is the topology induced by the metric

Let us denote the pointwise convergence topology on $Y^{A}$ as $\mathscr{P}(A)$ and the uniform convergence topology on $Y^{A}$ as $\mathscr{U}(A)$.

When $A$ is a topological space, there is another topology we often consider, the compact convergence topology (a.k.a. the topology of uniform convergence on compacta), which is the coarsest (i.e., smallest) topology on $Y^{A}$ which makes all the restriction maps

continuous where $K$ runs over all compact subsets of $A$. In other words, this is the topology induced by the family \(\left\{(f,g)\mapsto\sup_{a\in K}d(f(a),g(a))\right\}_{K}\) of pseudometrics on $Y^{A}$. We denote this topology as $\mathscr{K}(A)$.

The theorem we want to prove is the following:

Theorem 1 (Arzelà–Ascoli theorem).

Suppose $X$ is a topological space and $\mathscr{F}\subseteq Y^{X}$ a collection of functions $X\to Y$ whose restrictions onto each compact subset $K\subseteq X$ are continuous. Then $\mathscr{F}$ is contained in a $\mathscr{K}(X)$-compact subset of $Y^{X}$ if and only if the following conditions hold:

1. for each $x\in X$, the set \(\left\{f(x)\in Y\colon f\in\mathscr{F}\right\}\) is contained in a compact subset of $Y$, and
2. for each compact subset $K\subseteq X$, \(\left\{f\vert_{K}\in Y^{K}\colon f\in\mathscr{F}\right\}\) is an equicontinuous collection of functions $K\to Y$.

(The reason why we say “contained in a compact subset” rather than “relatively compact” is because, in general, the latter is strictly stronger than the former when the ambient space is not preregular.)

Here, equicontinuous collection of functions means the following:

Definition 2 (Equicontinuity).

Suppose $X$ is a topological space. Then a collection $\mathscr{F}\subseteq Y^{X}$ of functions $X\to Y$ is said to be equicontinuous at $x\in X$ if for each $\epsilon>0$, there exists a neighborhood $U$ of $x$ such that $d(f(x’),f(x))\leq\epsilon$ holds for all $x’\in U$ and all $f\in\mathscr{F}$. Also, we call the collection $\mathscr{F}$ equicontinuous if it is equicontinuous at every point in $X$.

Applying the space-function duality, we can view $\mathscr{F}$ as a space where each $x\in X$ acts as a function, which allows us to interpret equicontinuity as follows: $\mathscr{F}$ is equicontinuous at $x$ if and only if the evaluation map

is continuous at $x$. Note that this is really just a paraphrase of the definition, and there is nothing deep here. In any case, inspired from this interpretation, we define the following:

Definition 3 (Equicontinuity with respect to the pairing).

Suppose $X$ is a topological space. Then the set $Z$ is said to be equicontinuous (with respect to the pairing $\left\langle\,\cdot\,,\,\cdot\,\right\rangle$) at $x\in X$ if the map

\[\begin{aligned} X &\to (Y^{Z},\mathscr{U}(Z)) \\ x' &\mapsto (z\mapsto \left\langle z,x'\right\rangle) \end{aligned}\]

is continuous at $x$, and $Z$ is said to be equicontinuous (with respect to the pairing $\left\langle\,\cdot\,,\,\cdot\,\right\rangle$) if it is equicontinuous at every $x\in X$. Similarly, suppose $Z$ is a topological space, then the set $X$ is said to be equicontinuous (with respect to the pairing $\left\langle\,\cdot\,,\,\cdot\,\right\rangle$) at $z\in Z$ if the map

\[\begin{aligned} Z &\to (Y^{X},\mathscr{U}(X)) \\ z' &\mapsto (x\mapsto \left\langle z',x\right\rangle) \end{aligned}\]

is continuous at $z$, and $X$ is said to be equicontinuous (with respect to the pairing $\left\langle\,\cdot\,,\,\cdot\,\right\rangle$) if it is equicontinuous at every $z\in Z$.

With these terminologies, Theorem 1 follows quite automatically by combining three simple lemmas. We state and prove them one by one.

Lemmas and proofs

Lemma 4.

Suppose $X$ is a topological space and $A\subseteq X$ a dense subspace. Then for any subset $\mathscr{F}\subseteq Y^{X}$ consisting of continuous functions, the restriction map $(\mathscr{F},\mathscr{U}(X))\to (Y^{A},\mathscr{U}(A))$ is a homeomorphism onto its image.

Proof. The map is clearly continuous, and injectivity follows immediately by continuity of each element in $\mathscr{F}$. Hence, it is enough to show that a net \(\left(f_{\alpha}\right)_{\alpha\in D}\) in $\mathscr{F}$ converges uniformly to a function $f\in\mathscr{F}$ whenever the corresponding net \(\left(f_{\alpha}\vert_{A}\right)_{\alpha\in D}\) in $Y^{A}$ converges uniformly to \(f\vert_{A}\). Indeed, for such a case, for any given $\epsilon>0$, $x\in X$ and $\alpha\in D$, continuity of $f$ and $f_{\alpha}$ ensures the existence of $a\in A$ such that

\[\begin{aligned} d(f_{\alpha}(x),f(x)) &\leq d(f_{\alpha}(x),f_{\alpha}(a)) + d(f_{\alpha}(a),f(a)) + d(f(a),f(x)) \\ &\leq \epsilon + d_{\infty}(f_{\alpha}\vert_{A}, f\vert_{A}) + \epsilon. \end{aligned}\]

Since $\epsilon>0$, $x\in X$ and $\alpha\in D$ are all arbitrary, the uniform convergence \(f_{\alpha}\to f\) follows.$\quad\blacksquare$

An immediate consequence of this lemma is that, when $X,Z$ are topological spaces and each $\left\langle \,\cdot\,,x\right\rangle\colon Z\to Y$ is continuous, then a subset $S\subseteq Z$ is equicontinuous if and only if $\overline{S}\subseteq Z$ is equicontinuous. Note that continuity of every $\left\langle \,\cdot\,,x\right\rangle\colon Z\to Y$ means nothing but that the topology on $Z$ is finer than (i.e. contains) the pointwise convergence topology inherited from $Y^{X}$, which shows:

Corollary 5. Suppose $X,Z$ are topological spaces and the map $Z\mapsto (Y^{X},\mathscr{P}(X))$ is continuous. Then a subset $S\subseteq Z$ is equicontinuous if and only if $\overline{S}\subseteq Z$ is equicontinuous.

The other two lemmas relate joint continuity of the pairing $\left\langle\,\cdot\,,\,\cdot\,\right\rangle$ to how the topology on $Z$ compares with topologies we can give on $Y^{X}$.

Lemma 6. Suppose $X,Z$ are topological spaces such that $\left\langle z,\,\cdot\,\right\rangle\colon X\to Y$ is continuous for each $z\in Z$. If for each $x\in X$, there exists a neighborhood $U\subseteq X$ of $x$ such that the map $Z\to (Y^{U},\mathscr{U}(U))$ is continuous, then the pairing $\left\langle\,\cdot\,,\,\cdot\,\right\rangle\colon Z\times X\to Y$ is jointly continuous.

That is, if $Z$ is consisting of continuous functions $X\to Y$ and if the topology on $Z$ is finer than the local uniform convergence topology inherited from $Y^{X}$, then the pairing must be jointly continuous. Note that this local uniform convergence topology coincides with the usual uniform convergence topology $\mathscr{U}(X)$ when $X$ is compact.

Proof. Take any net \(\left((z_{\alpha},x_{\alpha})\right)_{\alpha\in D}\) in $Z\times X$ convergent to some $(z,x)\in Z\times X$. Then we can find a neighborhood $U$ of $x$ such that the map $Z\to (Y^{U},\mathscr{U}(U))$ is continuous, so in particular \(\left(\left\langle z_{\alpha},\,\cdot\,\right\rangle\right)_{\alpha\in D}\) converges to $\left\langle z,\,\cdot\,\right\rangle$ uniformly on $U$. Without loss of generality, we can assume $x_{\alpha}\in U$ for all $\alpha\in D$. Then since $\left\langle z,\,\cdot\,\right\rangle\colon X\to Y$ is continuous, we obtain

\[\begin{aligned} d(\left\langle z_{\alpha},x_{\alpha}\right\rangle, \left\langle z,x\right\rangle) \leq d(\left\langle z_{\alpha},x_{\alpha}\right\rangle, \left\langle z,x_{\alpha}\right\rangle) + d(\left\langle z,x_{\alpha}\right\rangle, \left\langle z,x\right\rangle) \to 0 \end{aligned}\]

as $\alpha\to\infty$, which establishes the joint continuity of $\left\langle \,\cdot\,,\,\cdot\,\right\rangle$.$\quad\blacksquare$

The last lemma is an implication in the other direction:

Lemma 7. Suppose $X,Z$ are topological spaces. If the pairing $\left\langle\,\cdot\,,\,\cdot\,\right\rangle\colon Z\times X\to Y$ is jointly continuous, then the map $Z\to(Y^{X},\mathscr{K}(X))$ is continuous.

That is, if the pairing is jointly continuous, then the topology on $Z$ must be finer than the compact convergence topology inherited from $Y^{X}$. This topology is in general coarser than the local uniform convergence topology, but of course they coincide if $X$ is locally compact. Hence, for that case, the continuity of the map $Z\to(Y^{X},\mathscr{K}(X))$ together with continuity of each $\left\langle z,\,\cdot\,\right\rangle$ is equivalent to the joint continuity of the pairing.

Proof. Take any net \(\left(z_{\alpha}\right)_{\alpha\in D}\) in $Z$ convergent to some $z\in Z$ and fix a compact subset $K\subseteq X$. Suppose for sake of contradiction that the net \(\left(\left\langle z_{\alpha},\,\cdot\,\right\rangle\right)_{\alpha\in D}\) does not converge uniformly on $K$ to $\left\langle z,\,\cdot\,\right\rangle$. This means that, there exists $\epsilon>0$ such that for any $\alpha\in D$, there exists $h(\alpha)\geq \alpha$ and \(x_{\alpha}\in K\) satisfying

\[d\left(\left\langle z_{h(\alpha)},x_{\alpha}\right\rangle, \left\langle z,x_{\alpha}\right\rangle\right) \geq \epsilon.\]

Since $K$ is compact, passing to a subnet if necessary, we can assume that \(\left(x_{\alpha}\right)_{\alpha\in D}\) converges to some $x\in K$. However, since the nets \(\left((z_{h(\alpha)},x_{\alpha})\right)_{\alpha\in D}\) and \(\left((z,x_{\alpha})\right)_{\alpha\in D}\) in $Z\times X$ both converge to $(z,x)$ and the pairing $\left\langle\,\cdot\,,\,\cdot\,\right\rangle$ is jointly continuous, we have reached a contradiction, establishing the continuity of the map $Z\mapsto (Y^{X},\mathscr{K}(X))$.$\quad\blacksquare$

As noted earlier, we obtain the following corollary:

Corollary 8. Suppose $X,Z$ are topological spaces and $X$ is locally compact. Then the pairing $\left\langle\,\cdot\,,\,\cdot\,\right\rangle\colon Z\times X\to Y$ is jointly continuous if and only if the following conditions hold:

1. the map $Z\to (Y^{X},\mathscr{K}(X))$ is continuous, and
2. $\left\langle z,\,\cdot\,\right\rangle\colon X\to Y$ is continuous for each $z\in Z$.

Another corollary, which is a very useful statement in its own, is that, for an equicontinuous family of functions, the pointwise convergence topology and the compact convergence topology coincide:

Corollary 9. Suppose $X$ is a topological space and $Z$ is equicontinuous. Then the pointwise convergence topology (i.e., the topology inherited from $(Y^{X},\mathscr{P}(X))$) and the compact convergence topology (i.e., the topology inherited from $(Y^{X},\mathscr{K}(X))$) coincide on $Z$.

(By “inherited from” we mean the initial topology on $Z$ induced by the map $Z\to Y^{X}$.)

Proof. Let us denote the topologies on $Z$ inherited from $(Y^{X},\mathscr{P}(X))$ and $(Y^{X},\mathscr{K}(X))$ as $\mathscr{P}(X)$ and $\mathscr{K}(X)$, respectively. Clearly, $\mathscr{K}(X)$ is finer than (i.e. contains) $\mathscr{P}(X)$, so we only need to show that $(Z,\mathscr{P}(X)) \to (Z,\mathscr{K}(X))$ is continuous. Recall that $Z$ being equicontinuous means that the map $X\to (Y^{Z},\mathscr{U}(Z))$ is continuous. Then since $\left\langle\,\cdot\,,x\right\rangle\colon Z\to Y$ is continuous with respect to $\mathscr{P}(X)$ for each $x\in X$, applying Lemma 6 with $X\leftarrow (Z,\mathscr{P}(X))$ and $Z\leftarrow X$ shows that the pairing $\left\langle\,\cdot\,,\,\cdot\,\right\rangle$ is jointly continuous on $(Z,\mathscr{P}(X))\times X$, so that now applying Lemma 7 with $X\leftarrow X$ and $Z\leftarrow (Z,\mathscr{P}(X))$ shows that the map $(Z,\mathscr{P}(X))\to (Y^{X},\mathscr{K}(X))$ is continuous. Hence, continuity of $(Z,\mathscr{P}(X))\to (Z,\mathscr{K}(X))$ follows.$\quad\blacksquare$

Now, we prove our main theorem.

Proof of Theorem 1. $(\Rightarrow)$ Suppose that $\mathscr{F}$ is contained in a $\mathscr{K}(X)$-compact subset $\mathscr{B}\subseteq Y^{X}$. Define $\mathscr{L} \mathrel{\unicode{x2254}} \mathscr{B}\cap\overline{\mathscr{F}}$ where the closure is taken in $Y^{X}$ with respect to $\mathscr{K}(X)$, then $\mathscr{L}$ is $\mathscr{K}(X)$-compact and contains $\mathscr{F}$. Also, since uniform limits of continuous functions are continuous, we know that each $f\in \mathscr{L}$ is continuous when restricted to each compact subset $K\subseteq X$. Therefore, fix such $K$ and endow $\mathscr{L}$ with the uniform convergence topology on $K$, then we can apply Corollary 8 with $X\leftarrow K$ and $Z\leftarrow \mathscr{L}$ to conclude that the pairing $(f,x)\mapsto f(x)$ is jointly continuous on $\mathscr{L}\times K$. Then, since the evaluation map $f\mapsto f(x)$ is clearly continuous on $\mathscr{L}$ for each $x\in K$, we can again apply Corollary 8 with $X\leftarrow \mathscr{L}$ and $Z\leftarrow K$ to conclude that the map $K\to(Y^{\mathscr{L}},\mathscr{U}(K))$ is continuous, which means nothing but that the collection \(\left\{f\vert_{K}\in Y^{K}\colon f\in \mathscr{L}\right\}\) of functions $K\to Y$ is equicontinuous. Clearly, \(\left\{f\vert_{K}\in Y^{K}\colon f\in \mathscr{F}\right\}\) is equicontinuous as well.

On the other hand, the continuity of the evaluation map $f\mapsto f(x)$ on $(\mathscr{L},\mathscr{K}(X))$ for given $x\in X$ shows that the image \(\left\{f(x)\in Y\colon f\in \mathscr{L}\right\}\) of $\mathscr{L}$ under this map is compact. Therefore, \(\left\{f(x)\in Y\colon f\in \mathscr{F}\right\}\) is contained in a compact subset of $Y$, as claimed.

$(\Leftarrow)$ For each $x\in X$, let $B_{x}$ be a compact subset of $Y$ containing \(\left\{f(x)\in Y\colon f\in\mathscr{F}\right\}\). Define $\mathscr{B} \mathrel{\unicode{x2254}} \prod_{x\in X}B_{x}\subseteq Y^{X}$, then $\mathscr{B}$ is a $\mathscr{P}(X)$-compact subset of $Y^{X}$ by the Tychonoff’s theorem and it contains $\mathscr{F}$. Let $\mathscr{L} \mathrel{\unicode{x2254}} \mathscr{B}\cap \overline{\mathscr{F}}$ where the closure is taken in $Y^{X}$ with respect to $\mathscr{P}(X)$, then $\mathscr{L}$ is $\mathscr{P}(X)$-compact and contains $\mathscr{F}$.

Fix a compact subset $K\subseteq X$. Then the equicontinuity of \(\left\{f\vert_{K}\in Y^{K}\colon f\in \mathscr{F}\right\}\) implies the equicontinuity of \(\left\{f\vert_{K}\in Y^{K}\colon f\in \mathscr{L}\right\}\). Indeed, $\mathscr{L}$ is contained in the $\mathscr{P}(X)$-closure of $\mathscr{F}$, so we can apply Corollary 5 with $Z\leftarrow (\mathscr{L},\mathscr{P}(X))$ and $S\leftarrow \mathscr{F}$. Therefore, Corollary 9 shows that $\mathscr{P}(K)$ and $\mathscr{U}(K)$ coincide on $\mathscr{L}$, but since $K$ is arbitrary, we conclude that $\mathscr{P}(X)$ and $\mathscr{K}(X)$ coincide on $\mathscr{L}$, which in turn shows that $\mathscr{L}$ is $\mathscr{K}(X)$-compact.$\quad\blacksquare$

As you can see, proofs of both directions leverage the aforementioned duality of $X$ and $Z$, or more precisely, that they can switch their roles, using joint continuity of the pairing $(f,x)\mapsto f(x)$ as a pivot of the switch.

A final remark is that, in fact, the same proof still applies to a slightly more general setup which I find occasionally useful. The observation is that, even though we are heavily relying on compactness of the domain, we do not necessarily consider all compact subsets, maybe just some compact subsets. This means that we can in fact replace the topology $\mathscr{K}(X)$ by a coarser topology obtained by a smaller collection of compact sets. In such a case, we also have to modify the equicontinuity condition accordingly; that is, the functions in the collection needs to be equicontinuous only when restricted to those compact sets in consideration, but not necessarily when restricted to any compact sets. Such a generalization yields, as a simple corollary, a slight variation of Theorem 1 in terms of uniformly continuous functions and totally bounded sets instead of continuous functions and compact sets. We omit details in this post.