The Arzelà–Ascoli Theorem Part 3

# The Arzelà–Ascoli Theorem Part 3

We are now going to look at a very famous theorem known as The Arzelà–Ascoli Theorem. We will prove the theorem in three parts on The Arzelà–Ascoli Theorem Part 1, The Arzelà–Ascoli Theorem Part 2, and **The Arzelà–Ascoli Theorem Part 3** pages.

First we review some important definitions.

- Recall from The Set of Real-Valued Continuous Functions on a Compact Metric Space X, C(X) that if $(X, d)$ is a compact metric space then the set of real-valued functions $f : X \to \mathbb{R}$ that are continuous is denoted $C(X)$.

- On the Boundedness of a Subset of C(X) page we saw that a collection of functions $\Gamma \subseteq C(X)$ is said to be bounded if there exists an $M \in \mathbb{R}$, $M > 0$ such that for all $f \in \Gamma$ and for all $x \in X$ we have that:

\begin{align} \quad \mid f(x) \mid \leq M \end{align}

- On the Equicontinuity of a Subset of C(X) page we saw that a collection of functions $\Gamma \subseteq C(X)$ is said to be equicontinuous on $X$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $f \in \Gamma$ and for all $x, y \in X$, if $d(x, y) < \delta$ then:

\begin{align} \quad \mid f(x) - f(y) \mid < \epsilon \end{align}

We will now finally complete the proof of the Arzelà–Ascoli theorem.

Theorem 1 (The Arzelà–Ascoli Theorem): Let $(X, d)$ be a compact metric space, and for each $n \in \mathbb{N}$ let $f_n : X \to \mathbb{R}$ be a continuous function. Then if the sequence of functions $(f_n(x))_{n=1}^{\infty}$ in $C(X)$ is bounded and equicontinuous then there exists a uniformly convergent subsequence. |

**Proof:**So far we have shown that there exists a countable dense subset $S$ and that there exists a subsequence $(f_{n, n}(x))_{n=1}^{\infty}$ of $(f_n(x))_{n=1}^{\infty}$ that converges pointwise to each $x_j \in S$ for $j \in \{1, 2, ..., n, ... \}$. We will now show that this sequence converges uniformly on $S$.

- Let $(g_n(x))_{n=1}^{\infty} = (f_{n, n}(x))_{n=1}^{\infty}$. Let $\epsilon > 0$ be given. Now, since the collection $\Gamma = \{ f_n(x) : n = 1, 2, ..., n, ... \} \subset C(X)$ is equicontinuous on $S$ we have that $\Gamma^*= \{ g_n(x) : n = 1, 2. ..., n, ... \} \subset \Gamma \subset C(X)$ is also equicontinuous on $S$. So for $\displaystyle{\epsilon_1 = \frac{\epsilon}{3} > 0}$ there exists a $\delta_1 > 0$ such that for all $g_n \in \Gamma^*$ and for all $x, y \in S$, if $d(x, y) < \delta_1$ then:

\begin{align} \quad \mid g_n(x) - g_n(y) \mid < \epsilon_1 = \frac{\epsilon}{3} \end{align}

- Now fix $M \in \mathbb{N}$ be such that $\frac{1}{M} < \delta_1$. Then $S_M = \{ x_1, x_2, ..., x_k \}$ is the collection of centers of the balls $B \left (x_1, \frac{1}{M} \right )$, $B \left ( x_2, \frac{1}{M} \right )$, …, $B \left (x_k, \frac{1}{M} \right )$ which cover all of $X$ from the first part of the proof. So since the subsequence $(g_n(x))_{n=1}^{\infty}$ converges pointwise at each $s \in S$ we have that the numercal sequence $(g_n(s))_{n=1}^{\infty}$ is Cauchy for each $s \in S$ and so there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:

\begin{align} \quad \mid g_n(s) - g_m(s) \mid < \epsilon_1 = \frac{\epsilon}{3} \end{align}

- So for each $x \in X$ we have that the distance from $x$ to an element in $s \in S_M$ is less than $\delta_1$ and so if $m, n \geq N$ then:

\begin{align} \quad \mid g_n(x) - g_m(x) \mid &= \mid g_n(x) - g_n(s) + g_n(s) - g_m(x) \mid \\ \quad & \leq \mid g_n(x) - g_n(s) \mid + \mid g_n(s) - g_m(x) \mid \\ \quad & \leq \mid g_n(x) - g_n(s) \mid + \mid g_n(s) - g_m(s) + g_m(s) - g_m(x) \mid \\ \quad & \leq \mid g_n(x) - g_n(s) \mid + \mid g_n(s) - g_m(s) \mid + \mid g_m(s) - g_m(x) \mid \end{align}

- Now $d(x, s) < \delta_1$ so the terms $\mid g_n(x) - g_m(s) \mid < \epsilon_1$ and $\mid g_m(s) - g_m(x) \mid < \epsilon_1$ by the equicontinuity of $\Gamma^*$, and $\mid g_n(s) - g_m(s) \mid < \epsilon_1$ by the choice of $N$, and so, for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ we have that for all $x \in S$ that:

\begin{align} \quad \mid g_n(x) - g_m(x) \mid < \epsilon_1 + \epsilon_1 + \epsilon_1 = \epsilon \end{align}

- Therefore the sequence $(g_n(x))_{n=1}^{\infty}$ is uniformly Cauchy and is hence uniformly convergent since for each $s \in S$ we have that $(g_n(s))_{n=1}^{\infty}$ is a numerical sequence in the complete metric space $\mathbb{R}$ and every Cauchy sequence in a complete metric space converges to that metric space by definition.

- So, if $(X, d)$ is a compact metric space and $(f_n(x))_{n=1}^{\infty}$ is a sequence of functions where for each $n \in \mathbb{N}$ we have $f_n : X \to \mathbb{R}$ then if the sequence of functions $(f_n(x))_{n=1}^{\infty}$ of $C(X)$ is bounded and equicontinuous then there exists a uniformly convergent subsequence. $\blacksquare$