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.