The Arzelà–Ascoli Theorem Part 2

# The Arzelà–Ascoli Theorem Part 2

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 continue 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. |

*Be very careful when reading this section of the proof. We talk about both function sequences AND numerical sequences! Understand that plugging in a specific value of $x$ into a function sequence will give us a numerical sequence.*

**Proof:**So far we have proven that $(X, d)$ contains a countable and dense subset $S$. We will now show that there exists a subsequence of $(f_n(x))_{n=1}^{\infty}$ that converges pointwise on $S$.

- Since $S$ is countable we can write this set as:

\begin{align} \quad S = \{ x_1, x_2, ..., x_n, ... \} \end{align}

- Consider the following numerical sequence:

\begin{align} \quad (f_n(x_1))_{n=1}^{\infty} = (f_1(x_1), f_2(x_1), ..., f_n(x_1), ...) \end{align}

- Since $\Gamma = \{ f_n(x) : n = 1, 2, ... \} \subset C(X)$ is given to be bounded, we have that for each $f_n \in \Gamma$ and for all $x \in X$ that there exists an $M \in \mathbb{R}$, $M > 0$ such that $\mid f_n(x) \mid \leq M$. In particular, the sequence above is bounded. By The Bolzano-Weierstrass Theorem there then exists a convergent subsequence of $(f_n(x_1))_{n=1}^{\infty}$, call it:

\begin{align} \quad (f_{1, n}(x_1))_{n=1}^{\infty} = (f_{1, 1}(x_1), f_{1, 2}(x_1), ..., f_{1, n}(x_1), ...) \end{align}

- So the function subsequence $(f_{1, n}(x))_{n=1}^{\infty}$ converges pointwise at $x_1$. Now consider the following sequence:

\begin{align} \quad (f_{1, n}(x_2))_{n=1}^{\infty} = (f_{1, 1}(x_2), f_{1, 2}(x_2), ..., f_{1, n}(x_2), ...) \end{align}

- Once again, we have that this sequence is bounded and by the Bolzano-Weierstrass theorem there then exists a convergent subsequence of $(f_{1, n}(x_2))_{n=1}^{\infty}$, call it:

\begin{align} \quad (f_{2, n}(x_2))_{n=1}^{\infty} = (f_{2, 1}(x_2), f_{2, 2}(x_2), ..., f_{2, n}(x_2), ...) \end{align}

- So the function subsequence $(f_{2, n}(x))_{n=1}^{\infty}$ converges pointwise at $x_2$.

- We continue on in this fashion and obtain a countable collection of function subsequences:

\begin{align} \quad (f_{1, n}(x))_{n=1}^{\infty} = (f_{1, 1}(x), f_{1, 2}(x), f_{1, 3}(x), ..., f_{1, n}(x), ...) \\ \quad (f_{2, n}(x))_{n=1}^{\infty} = (f_{2, 1}(x), f_{2, 2}(x), f_{2, 3}(x), ..., f_{2, n}(x), ...) \\ \quad (f_{3, n}(x))_{n=1}^{\infty} = (f_{3, 1}(x), f_{3, 2}(x), f_{3, 3}(x), ..., f_{3, n}(x), ...) \\ \quad \quad \vdots \quad \quad \quad \quad \\ \quad (f_{m, n}(x))_{n=1}^{\infty} = (f_{m, 1}(x), f_{m, 2}(x), f_{m, 3}(x), ..., f_{m, n}(x), ... ) \\ \quad \quad \vdots \quad \quad \quad \quad \end{align}

- Each of these function sequences are subsequences of the function sequences above, and are all function subsequences of the original function sequence $(f_n(x))_{n=1}^{\infty}$. Furthermore, the $m^{\mathrm{th}}$ subsequence converges pointwise to the points $x_1, x_2, ..., x_m \in S$. Consider the diagonal sequence:

\begin{align} \quad (f_{n, n}(x))_{n=1}^{\infty} = (f_{1, 1}(x), f_{2, 2}(x), f_{3, 3}(x), ..., f_{n, n}(x), ...) \end{align}

- Then this sequence must be pointwise convergent to each $x_j \in S$ for all $j \in \{1, 2, ... \}$ because the $j$-tail of the sequence $(f_{n, n}(x_j))_{n=j}^{\infty} = (f_{j, j}(x_j), f_{j+1, j+1}(x_j), ...)$ is a subsequence of the numerical sequence $(f_{j, n}(x_j))_{n=1}^{\infty}$ which converges.