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. $\blacksquare$