Proof of the Arzelà–Ascoli Theorem

The Problem

In $\mathbb{R}^n$, the Bolzano–Weierstrass theorem says every bounded sequence has a convergent subsequence. Can we do the same for sequences of functions?

The naive answer is no. The sequence $f_n(x) = \sin(nx)$ on $[0,1]$ is uniformly bounded — $|f_n(x)| \leq 1$ for all $n$ and $x$ — but admits no uniformly convergent subsequence.

What additional structure do we need?

The Theorem

Equicontinuity Is the Key Condition

Uniform boundedness alone is not enough, $\sin(nx)$ shows this. Equicontinuity is the condition that the functions in $\mathcal{F}$ do not oscillate too rapidly, uniformly across the family.

Proof

Step 1: $X$ is separable. Let $n \in \mathbb{N}$ be arbitrary; the collection $\{B(x,\frac{1}{n})\}_{x \in X}$ is an open cover of $X$, and by compactness it has a finite subcover $\{B(x_i,\frac{1}{n})\}_{i \in I_n}$. We will show that the set

is dense in $X$. Let $y \in X$ and $\varepsilon > 0$; choose $n \in \mathbb{N}$ such that $\frac{1}{n} < \varepsilon$. Since $\{B(x_i,\frac{1}{n})\}_{i \in I_n}$ covers $X$, there exists $i \in I_n$ such that $d(x_i,y) < \frac{1}{n} < \varepsilon$ and $y \in B(x_i,\frac{1}{n})$. Then $x_i \in B(y,\frac{1}{n}) \subseteq B(y,\varepsilon)$. As we have

for all $y \in X$ and any $\varepsilon > 0$, $S$ is dense in $X$, and hence $X$ is separable.

Step 2: Recall the Bolzano–Weierstrass theorem:

Theorem. Let $(x_n) \subseteq \mathbb{R}$ be a bounded sequence. Then there exists a subsequence $(x_{n_k})$ that converges in $\mathbb{R}$.

We enumerate $S = \{x_1, x_2, \dots\}$ and define the subsequence $\{f_{n,k}\}$ inductively in $k$ as follows.

$k = 1$. Since $\{f_n(x_1)\}$ is bounded, Bolzano–Weierstrass gives a subsequence of functions $\{f_{n,1}\} \subseteq \{f_n\}$ such that $\{f_{n,1}(x_1)\}$ converges in $\mathbb{R}$.

$k \geq 2$. Suppose the subsequence $\{f_{n,k-1}\} \subseteq \{f_{n,k-2}\}$ is defined such that $\{f_{n,k-1}(x_{k-1})\}$ converges. Since $\{f_{n,k-1}(x_k)\}$ is bounded, there is a subsequence of functions $\{f_{n,k}\} \subseteq \{f_{n,k-1}\}$ such that $\{f_{n,k}(x_k)\}$ converges in $\mathbb{R}$.

Define $g_n = f_{n,n}$ for all $n \in \mathbb{N}$. We claim that $g_n(x_i)$ converges for all $x_i \in S$. Indeed, since $\{g_n(x_i)\}_{n \geq i} \subseteq \{f_{n,i}(x_i)\}$, and a subsequence of a convergent sequence converges, $g_n(x_i)$ converges in $\mathbb{R}$ and is moreover Cauchy.

Step 3: We show that $\{g_n\}$ converges uniformly on $X$. Let $\varepsilon > 0$ be arbitrary. Since $\{g_n\}$ is equicontinuous, there exists $\delta_\varepsilon > 0$ such that

whenever $d(x,y) < \delta_\varepsilon$ and for all $n \in \mathbb{N}$. Since $X$ is compact, there is a finite set $S_\varepsilon \subseteq S$ such that $\{B(x_i, \delta_\varepsilon)\}_{x_i \in S_\varepsilon}$ covers $X$. Since $\{g_n\}$ converges pointwise on $S$, for each $x_i \in S_\varepsilon$ there exists $N_i$ such that

for all $m, n \geq N_i$. Let $N_\varepsilon = \max\{N_i\}$. For any $x \in X$, one can choose $s \in S_\varepsilon$ such that $d(x,s) < \delta_\varepsilon$, so

Applying the triangle inequality,

for all $m, n \geq N_\varepsilon$ and all $x \in X$. Hence $\{g_n\}$ is uniformly Cauchy, and therefore converges uniformly. $\blacksquare$

Exercises

Exercise 1. Let $f_n : [0,1] \to \mathbb{R}$ be defined by $f_n(x) = x^n$. Show that $\{f_n\}$ is uniformly bounded but not equicontinuous, and identify the pointwise limit. Does $\{f_n\}$ have a uniformly convergent subsequence on $[0,1]$?

Exercise 2. Let $\{f_n\}$ be a sequence of differentiable functions on $[a,b]$ such that $|f_n(x)| \leq M$ and $|f_n'(x)| \leq M$ for all $n \in \mathbb{N}$ and all $x \in [a,b]$. Show that $\{f_n\}$ is equicontinuous, and conclude that it has a uniformly convergent subsequence.

Exercise 3. Let $\{f_n\}$ be a sequence of functions on $[0,1]$ defined by

Show that $\{f_n\}$ is uniformly bounded. Is it equicontinuous? Find the pointwise limit and determine whether any subsequence converges uniformly.

Exercise 4. Give an example of a sequence $\{f_n\} \subset C([0,1])$ that is equicontinuous but has no uniformly convergent subsequence. What hypothesis of the theorem does your example violate?

References

[1] T. Bühler and D. A. Salamon, Functional Analysis, ETH Zürich, 2017.