Problem 1

Exercise 3.1 (Prove Arzelà-Ascoli)   Let $K$ be a compact metric space and let $A$ be a subset of $C(K)$. Prove that $A$ is compact if and only if $A$ is closed, bounded, and equicontinuous.

Proof. This exercise is asking us to prove the Arzelà-Ascoli Theorem.

First we present a diagonal argument: Let $\{f_n\}_{n=1}^\infty$ be a sequence in $\overline{A}$. Construct a diagonal subsequence as follows. Select a countable dense subset $\{x_k\}_{k=1}^\infty\subseteq K$ and select nested subsequences

$$f_n=f_{1,n}\supseteq\cdots\supseteq f_{k,n} \supseteq f_{k+1,n}\supseteq\cdots$$ (3.1)

in such a way that $\lim_{n\to\infty}f_{k,n}(x_k)$ exists, by the completeness of $\mathbb{R}$ and the boundedness of $\overline{A}$. It will be proven that $f_{n,n}$ is Cauchy.

Let $\epsilon>0$ be given. Select $\delta>0$ so that $\vert x-y\vert<\delta$ implies $\vert f(x)-f(y)\vert<\epsilon/3$. Cover $K$ and extract a finite subcover

$$K \subseteq \bigcup_{x\in K} B_\delta(x) \implies K\subseteq\bigcup_{i=1}^n B_\delta(x_i)$$ (3.2)

For each $x_i$, select $N_i$ for which $n,m\geq N_i$ implies $\vert f_{i,n}(x_i) - f_{i,m}(x_i)\vert < \epsilon/3$. Set $N=\max\{N_i\}$.

Now we are in a position to prove that if $n,m>N$ we have $\Vert f_m-f_n\Vert < \epsilon$. For each $x\in K$, there exists $\vert x-x_i\vert<\delta$. Then we may write

$$\Vert f_m-f_n\Vert = \sup_{x\in K}\vert f_m(x)-f_m(x_i)+f_m(x_i)-f_n(x_i) +f_n(x_i)-f_n(x)\vert$$ (3.3)

$$\leq \sup\vert f_m(x)-f_m(x_i)\vert + \sup\vert f_m(x_i)-f_n(x_i)\vert + \sup\vert f_n(x_i)-f_n(x)\vert$$ (3.4)

The first and last suprema are bounded by $\epsilon/3$ due to the equicontinuity estimate. The middle supremum equals $\max_i \vert f_m(x_i)-f_n(x_i)\vert$, and the selection of $N$ guarantees that this quantity is bounded by $\epsilon/3$, completing the proof.

An alternative proof which may be considered more explicit is given as follows, which rests on the equivalence that

$$\iff$$ (3.5)

in a metric space.

Let $\epsilon>0$ be given. Our goal is to determine $\{f_1,\dots,f_n\}\subseteq \overline{A}$ such that

$$\overline{A}\subseteq \bigcup_{i=1}^n B_\epsilon(f_i)$$ (3.6)

To proceed, we represent each function in $\overline{A}$ by a bounded “step” function, found by exploiting the compact domain $K$. Since $\overline{A}$ is bounded and equicontinuous, there exists $M>0$ and $\delta>0$ such that for all $h\in\overline{A}$ we have $\Vert h\Vert<M$ and that $\vert x-y\vert<\delta$ implies $\vert h(x)-h(y)\vert<\epsilon$.

To exploit the compact domain $K$, extract a finite subcover as follows

$$K \subseteq \bigcup_{x\in K} B_\delta(x)$$ (3.7)

$$\subseteq \bigcup_{i=1}^L B_\delta(x_i)$$ (3.8)

Define a collection of functions

$$G = \left\{ g : \bigcup_{i=1}^L B_\delta(x_i)\cap K \to \mathbb{R... ..._\delta(x_i))=\epsilon y_i, \vert\epsilon y_i\vert<M, y_i\in\mathbb{Z} \right\}$$ (3.9)

For each $f\in\overline{A}$, select $\{y_1, \dots, y_L\}$ satisfying

$$\epsilon y_i \leq f(x_i) \leq \epsilon(y_i+1)$$ (3.10)

and $\vert\epsilon y_i\vert<M$, so that we may define

$$g_f(B_\delta(x_i)) = \epsilon y_i$$ (3.11)

Then $\Vert f-g_f\Vert<\epsilon$. Consider this collection of functions $\overline{G}=\{g_f\}_f=\{g_1, \dots, g_N\}$. Realize that

$$\overline{A} \subseteq \bigcup_{i=1}^N B_\epsilon(g_i)$$ (3.12)

We are now so close, because we just need to invert each $g_i\to f_i$ where $f_i$ is simply an element of $\overline{A}$ such that $\Vert f_i-g_i\Vert<\epsilon$ as described above, so that an appropriate $\epsilon$-net is

$$\overline{A}\subseteq \bigcup_{i=1}^N B_\epsilon(f_i)$$ (3.13)

$\qedsymbol$