The below is the proof for the Arzela-Ascoli Theorem from Carothers' Real Analysis. I had a few questions regarding some steps in his proof which I have put in blue. If anyone could explain the blue lines, it would be appreciated.
Some definitions that I recently learned:
- Uniformly bounded means $\sup_{f\in\mathcal{F}}||f||_{\infty}<\infty$
- Equicontinuous means $\forall \epsilon>0 \exists \delta:d(x,y)<\delta\implies |f(x)-f(y)|<\epsilon \text{ for all } f\in\mathcal{F}$
Theorem (Arzela-Ascoli)
Let $X$ be a compact metric space, and let $\mathcal{F}\subset C(X)$. Then $\mathcal{F}$ compact if and only if $\mathcal{F}$ is closed, uniformly bounded, and equicontinuous.
(Note that $\mathcal{F}$ is a collection of continuous real-valued functions on $X$, and $C(X)$ is the space of all continuous functions $f:X\to\mathbb{R}$)
Proof.
(I understand the $\implies$ direction, so I omit that part of the proof.)
For $\impliedby$:
Suppose $\mathcal{F}$ is closed, uniformly bounded, and equicontinuous and let $(f_n)\in\mathcal{F}.$ $\color{blue}{\text{We need to show that $(f_n)$ has a uniformly convergent subsequence.}}$
$\color{blue}{\text{First note that $(f_n)$ is equicontinuous.}}$ Thus given $\epsilon>0, \exists \delta >0: d(x,y)<\delta\implies |f_n(x)-f_n(y)|<\epsilon/3, \forall n$.
Next since $X$ is totally bounded, $X$ has a finite $\delta$-net, i.e, there exists $x_1,\ldots, x_k\in X$ such that each $x\in X$ satisfies $d(x,x_i)<\delta$. $\color{blue}{\text{Now since $(f_n)$ is also uniformly bounded,}}$ each of the sequences $(f_n(x_i))_{n=1}^{\infty}$ is bounded in $\mathbb{R}$ for $i=1,2,\ldots, k$.
$\color{blue}{\text{Thus by passing to a subsequence of the $f_n$ (and relabeling), we may suppose that $(f_n(x_i))_{n=1}^{\infty}$ converges for each $i=1,2,\ldots, k$}}$.
In particular, we can find some $N$ such that $|f_m(x_i)-f_n(x_i)|<\epsilon/3$ for any $m,n\geq N$ and any $i=1,2,\ldots, k$.
And now we are done! Given $x\in X$, first find $i$ such that $d(x,x_i)<\delta$ and then whenever $m,n\geq N$, we will have
\begin{align*}
&|f_m(x)-f_n(x)|\\
&\leq|f_m(x)-f_m(x_i)|+|f_m(x_i)-f_n(x_i)|+|f_n(x_i)-f_n(x)|\\ &<\epsilon/3+\epsilon/3+\epsilon/3\\
&=\epsilon
\end{align*}
That is, $f_n$ is uniformy Cauchy, since our choice of $N$ does not depend on $x$. Since $\mathcal{F}$ is closed in $C(X)$ by assumption, it follows that $f_n(x)$ converges to some $f\in\mathcal{F}$, uniformly.
$\blacksquare$
More explictly, my questions are:
- For the first blue line, why are we trying to show that $(f_n)$ has a uniformly convergent subsequence?
- For the second blue line, how do we know $(f_n)$ is equicontinuous?
- For the third blue line, how can we tell that $(f_n)$ is uniformly bounded?
- For the fourth blue line, can someone elaborate on what Carothers is trying to do here? I'm having trouble understanding what is occurring at this step.
Thanks.
Best Answer
Because if you have a uniformly convergent subsequence and the space is closed your subsequence will converge in the space, then you have a that for any sequence you can extract a convergent subsequence in the space, which is one of the characterizations of a compact space.
If the space is equicontinuous any sequence inside it is equicontinuous.
Same as above. Or: $$\{f_n\} \subseteq \{f\in\mathcal{F}\}$$
so
$$\sup_{f_n}\||f_n||_{\infty} \leq \sup_{f\in\mathcal{F}}||f||_{\infty}<\infty $$
For each $x$ he has some subsequence $f_{n_i}(x_i)$ that converges pointwise, taking the intersection of all the $n_i$ and renaming as $n$ he gets a sequence that converges for every $x_i$