Arzelà-Ascoli Theorem

Let’s conclude our contemplations on compactness with function spaces that will hopefully motivate future topological discussions on function spaces.

Theorem 1. $\mathcal C([0, 1], \mathbb R)$ equipped with the supremum metric is not compact.

Proof of Theorem 1. The sequence $\{ f_n : n \in \mathbb N \}$ defined by $f_n = n$ has no convergent subsequence.

Theorem 2. For any $n \in \mathbb N_0$ and $M > 0$ , define

$\mathcal D_{n,M} := \{f \in \mathcal C^{(n)}([0, 1], \mathbb R) : \| f^{(n)} \|_\infty \leq M\}.$

Then $\mathcal D_{0,M}$ and $\mathcal D_{1,M}$ are not compact, but $\mathcal D_{0,M} \cap \mathcal D_{1,M}$ is.

Partial Proof of Theorem 2. The sequences $\{ f_n \}$ and $\{g_n\}$ defined by $f_n(x) = M \cdot \sin(nx)$ and $g_n = n$ have no convergent subsequence. We delay the proof of the third claim.

The third claim is in fact one of the many key applications of the Arzelà-Ascoli theorem, which is the content of this post. Fix $\mathcal F \subseteq \mathcal C([0, 1], \mathbb R)$ , equipped with the uniform norm.

Lemma 1. Suppose $\mathcal F$ is compact. Then the following properties hold.

$\mathcal F$ is closed in the following sense: for any $\{f_n\} \subseteq \mathcal F$ , if $f_n \to f$ in the uniform metric, then $f \in \mathcal F$ .
$\mathcal F$ is bounded in the following sense: there exists $M > 0$ such that $f \in \mathcal D_{0,M}$ .
$\mathcal F$ is equicontinuous in the following sense: for any $\epsilon > 0$ , there exists $\delta > 0$ such that for any $x,y \in [0, 1]$ , $|x-y| < \delta$ implies that for any $f \in \mathcal F$ , $|f(x) - f(y)| < \epsilon$ .

Proof. Compactness implies closedness and boundedness by metric space arguments discussed before. For any open $U \subseteq [0, 1]$ , define

$\displaystyle \mathcal U(\epsilon, U) := \left\{f \in \mathcal F : \sup_{t \in U}f(t) - \inf_{t \in U} f(t) < \epsilon \right\}.$

Fix $x \in [0, 1]$ and $\epsilon > 0$ . For each $x \in [0, 1]$ , let $\mathcal U_x$ denote the collection of neighborhoods of $x$ . Then $\{U(\epsilon, V) : V \in \mathcal U_x\}$ forms an open cover for $\mathcal F$ . To verify this, since each $f \in \mathcal F$ is continuous at $x$ , there exists $\delta > 0$ such that for $y \in (x-\delta,x+\delta) =: V_\delta$ ,

$\displaystyle |f(x) - f(y)| < \frac{\epsilon}3 \quad \Rightarrow \quad \sup_{t \in V_\delta} f(t) - \inf_{t \in V_\delta} f(t) \leq \frac{ 2\epsilon }3< \epsilon,$

so that $f \in U(\epsilon, V_\delta)$ . Since $\mathcal F$ is compact, we can extract a finite sub cover $\{U(\epsilon, V_{\delta_i}) : i = 1,\dots,n\}$ . Selecting $\delta := \min\{\delta_1,\dots,\delta_n\}$ yields equicontinuity near $x$ .

Finally, we aim to prove uniform equicontinuity. Let $k_1,k_2,k_3 > 0$ be constants to be tuned. For each $x \in [0, 1]$ , use the pointwise equicontinuity of $\mathcal F$ to extract a constant $\delta_x > 0$ such that

$|x-y| < \delta_x \quad \Rightarrow \quad |f(x) - f(y)| < k_1 \cdot \epsilon,\quad f \in \mathcal F.$

Since $\{(x- k_2 \cdot \delta_x, x+k_2 \cdot \delta_x) : x \in [0, 1]\}$ yields an open cover for $[0, 1]$ , we can extract a finite sub-cover

$\{(x_i -k_2 \cdot \delta_{x_i}, x_i +k_2 \cdot \delta_{x_i}) : i = 1,\dots, n\}.$

Define $\delta := k_3 \cdot \min \{\delta_{x_1},\dots,\delta_{x_n}\}$ and $x, y \in [0, 1]$ . Suppose $|x-y | < \delta$ . Find $x_i$ such that $|x - x_i| < k_2 \cdot \delta_{x_i}$ , so that $|y - x_i| < (k_2 + k_3) \cdot \delta_{x_i}$ . Setting $k_2 = k_3 = 1/2$ ,

$|f(x) - f(y)| \leq |f(x) - f(x_i)| + |f(x_i) - f(y)| < 2k_1 \cdot \epsilon.$

Setting $k_1 = 1/2$ yields the desired result.

The simplest form of the Arzelà-Ascoli theorem claims that the converse holds.

Theorem 3 (Arzelà-Ascoli Theorem). $\mathcal F$ is compact if and only if it is closed, bounded, and equicontinuous.

Proof of Theorem 3. We have proven $(\Rightarrow)$ in Lemma 1, and turn our attention to $(\Leftarrow)$ .

Let $\mathcal F$ be closed, bounded, and equicontinuous in the sense of Lemma 1. Since $\mathcal F$ is a metric space, it suffices to prove that it is sequentially compact. Fix any sequence $\{f_n\}$ of $\mathcal F$ . We wish to extract a convergent subsequence $\{f_{n_k}\}$ .

Let $\mathbf x : \mathbb N \to \mathbb Q \cap [0, 1]$ be an enumeration of the rational numbers in $[0, 1]$ (so that $\mathbf x$ is bijective) and denote $x_i := \mathbf x(i)$ . Since $\mathcal F_0 := \mathcal F$ is bounded, $\{f(x_1) : f \in \mathcal F_0\}$ is bounded and hence contains a convergent subsequence $\{f_{1, i}(x_1) : i \in \mathbb N\}$ . Define $\mathcal F_1 := \{f_{1, i} : i \in \mathbb N\}$ .

Inductively, given each bounded collection $\mathcal F_k$ , $\{f(x_{k+1}) : f \in \mathcal F_k\}$ is bounded and hence contains a convergent subsequence $\{f_{k+1, i}(x_1) : i \in \mathbb N\}$ . Define $\mathcal F_{k+1} := \{f_{k+1, i} : i \in \mathbb N\}$ , so that inductively we can construct the decreasing chain of subsequences

$\mathcal F = \mathcal F_0 \supseteq \mathcal F_1 \supseteq \mathcal F_2 \supseteq \cdots.$

Now define the sub-sequence $\{g_n\}$ of $\{f_n\}$ by $g_k := f_{k,k}$ . We claim that $\{g_n\}$ converges. By our construction, it is clear that for any $x_i$ , $\{ g_n(x_i) : n \in \mathbb N\}$ converges and is thus Cauchy. Now, fix $\epsilon > 0$ and $x_i \in \mathbb Q \cap [0, 1]$ . For any $k_1 > 0$ , find $N \in \mathbb N$ such that for $m > n \geq N$ ,

$|g_m(x_i) - g_n(x_i)| < k_1 \cdot \epsilon.$

Since $\mathcal F$ is equicontinuous, for any $k_2 > 0$ , we have that for any $x \in [0, 1]$ , there exists a neighbourhood $U_x$ such that for any $y \in [0, 1]$ and $f \in \mathcal F$ ,

$y \in U_x \quad \Rightarrow \quad |f(x) - f(y)| < k_2 \cdot \epsilon.$

Since $\{U_x : x \in [0, 1]\}$ forms an open cover for $[0, 1]$ , we can extract a finite sub-cover $\{U_{x_1'},\dots, U_{x_J'}\}$ . For each $1 \leq j \leq J$ , find $x_{n_j} \in U_{x_j'}$ .

Now, for any $x \in [0, 1]$ , $x \in U_{x_j'}$ for some $j$ . Hence, for sufficiently large $m,n$ ,

$\begin{aligned} |g_m(x) - g_n(x)| &\leq |g_m(x) - g_m(x_{n_j})| + |g_m(x_{n_j}) - g_n(x_{n_j})| + |g_n(x_{n_j}) - g_n(x)| \\ &\leq k_2 \cdot \epsilon + k_1 \cdot \epsilon + k_2 \cdot \epsilon = (k_1 + 2k_2) \cdot \epsilon. \end{aligned}$

Setting $k_1 = 1/2$ , $k_2 = 1/4$ proves that $\{g_n(x)\}$ is Cauchy. A compactness argument shows that $\{g_n\}$ is Cauchy, and since $\mathcal C([0,1],\mathbb R)$ is complete, converges to some $g \in \mathcal C([0, 1],\mathbb R)$ . Since $\mathcal F$ is closed, $g \in \mathcal F$ , so that $g_n \to g$ , as required.

The Arzelà-Ascoli theorem is incredibly useful in real analysis. For now, we complete the proof of Theorem 2.

Complete Proof of Theorem 2. By the Arzelà-Ascoli theorem, we need to prove that $\mathcal F := \mathcal D_{0,M} \cap \mathcal D_{1,M}$ is closed, bounded, and equicontinuous. It is bounded since $\mathcal D_{0,M}$ is. For any $f \in \mathcal F$ , the mean value theorem yields the estimate

$|f(x) - f(y)| = |f'(t)| \cdot |x-y| \leq M \cdot |x-y|.$

Thus, given $\epsilon > 0$ , the choice $\delta := \epsilon/(2M)$ yields the desired equicontinuity. To check that $\mathcal F$ is closed, fix a sequence $\{f_n\} \subseteq \mathcal F$ such that $f_n \to f$ . Then

$\| f \|_\infty \leq \|f_n - f\|_\infty + M \quad \Rightarrow \quad f \in \mathcal D_{0, M}$

and by the differentiable limit theorem, $f$ is differentiable at any $x \in [0, 1]$ with a similar estimate

$\|f'\|_\infty \leq \|f_n' - f' \|_\infty + M \quad \Rightarrow \quad f \in \mathcal D_{1, M}.$

—Joel Kindiak, 29 May 2025, 1439H

KindiakMath

Arzelà-Ascoli Theorem

Leave a comment Cancel reply

Arzelà-Ascoli Theorem

Share this:

Leave a comment Cancel reply