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\section{Equicontinuity}

\subsection{Existence of Convergent Subsequences}

The following type of problem frequently arises in analysis: Given a
family of functions, $\mathcal{F}\subset C[0,1]$, can we find a function
$f_0 \in C[0,1]$ ($f_0\in \mathcal{F}$?) that minimizes some quantity
$E(f)$ over all $f\in \mathcal{F}$.

For example, let $E(f) = \int_0^1|f(x)|^2dx$.  Since $E(f) \ge 0$,
$\forall\, f\in\mathcal{F}$, $m = \inf_{f\in\mathcal{F}} E(f)$ exists and
so $\exists\, \{f_n\}\subset\mathcal{F}$ such that $E(f_n)\to m$. Now if
we could find a subsequence $\{f_{n'}\}$ that converged \emph{uniformly} to
$f_0$, then $f_0$ would be continuous by Theorem~\ref{02:UNIFORMLIMITS}
and $E(f_{n'}) \to E(f_0) = m$.

Not all sequences $\{f_n\}\subset C[a,b]$ have convergent subsequences.

\emph{Example:} $f_n(x) = n$.

This problem does not occur if we can put a uniform bound on every
function in the sequence.

\begin{definition}
A family of functions, $\mathcal{F}$, is \emph{uniformly bounded} if
$\exists\, M$ such that $|f(x)| \le M$, $\forall\, x\in[a,b]$ and
$\forall\, f\in \mathcal{F}$.
\end{definition}

Not all \emph{bounded} sequences $\{f_n\}\subset C[a,b]$ have convergent
subsequences.

\emph{Example:} $f_n(x) = \sin(nx)$.

This problem does not occur if the oscillations (or the ``continuity'')
of the functions in the sequence can be controlled in a uniform way.

\begin{definition}
A family of functions, $\mathcal{F}$, is \emph{uniformly equicontinuous}
if $\forall\, 1/m$, $\exists\, 1/n$ (depending only on $1/m$) such that
\[ |x - y| < \frac{1}{n} \Rightarrow |f(x) - f(y)| < \frac{1}{m},\quad
\forall\, f\in \mathcal{F} \]
\end{definition}

\subsection{Arzela-Ascoli Theorem}

\begin{theorem}\label{05:ASCOLI}
If $\{f_n\}\subset C[a,b]$ is uniformly bounded and uniformly
equicontinuous, then there is a subsequence $f_{n'}$ converging uniformly
to $f\in C[a,b]$.
\end{theorem}

\proof Let $x_1, x_2, x_3, \dots$ be a countable dense subset of $[a,b]$,
e.g., $[a,b]\cap\Q$. Then for each fixed $k$, $f_1(x_k), f_2(x_k),
f_3(x_k),\dots$ is a bounded sequence ($|f_n(x)| \le M, \forall\, x\in[a,b],
\forall\, n$). So, for each $k$, there is a convergent subsequence,
$f_{n'}(x_k)\to c_k$.

From such subsequences we want to extract a subsequence that converges
simultaneously~$\forall\, k$. We do this as follows:
\begin{eqnarray*}
\{f_{1i}\}\subset\{f_n\} &:& \fbox{$f_{11}(x_1)$}, f_{12}(x_1), f_{13}(x_1),
\dots \to c_1 \\
\{f_{2i}\}\subset\{f_{1i}\} &:& f_{21}(x_2), \fbox{$f_{22}(x_2)$}, f_{23}(x_2),
\dots \to c_2 \\
\{f_{3i}\}\subset\{f_{2i}\} &:& f_{31}(x_3), f_{32}(x_3), \fbox{$f_{33}(x_3)$},
\dots \to c_3
\end{eqnarray*}
We take the diagonal subsequence of functions, $f_{11}$, $f_{22}$,
$f_{33}$, \dots. Except for the first $k$ functions,  $\{f_{ii}\}$ is a
subsequence of the $k$th row, $\{f_{ki}\}$.  Therefore, $\{f_{ii}\}$
converges simultaneously on all $x_n$.

We now show $\{f_{ii}\}$ converges uniformly on $[a,b]$ (the limit
function is automatically continuous by Theorem~\ref{02:UNIFORMLIMITS}).
We will prove the \emph{uniform Cauchy criterion} holds:
\begin{quote}
Given $1/m$, $\exists\, N$ such that $|f_{kk}(x) - f_{jj}(x)| <
1/m$, $\forall\, x\in[a,b]$ and $\forall\, j,k \ge N$.
\end{quote}
The assumption of uniform equicontinuity  implies $\exists\, 1/n$ such that
\begin{equation}\label{05:A}
|x - y| < \frac{1}{n} \Rightarrow
|f_j(x) - f_j(y)| < \frac{1}{3m}\quad \forall\, j
\end{equation}

Since $\bigcup(x_k - 1/n, x_k + 1/n)$ covers the compact interval $[a,b]$,
there is a finite subcover. To keep the notation simple, we re-index the
$x_k$'s so that 
\[ [a,b] \subset (x_1 -\frac{1}{n_1}, x_1+\frac{1}{n_1})\cup\dots\cup
(x_r -\frac{1}{n_r}, x_r+\frac{1}{n_r}) \]

We know $f_{ii}(x_p)$ converges for $p=1,\dots,r$. So $\exists\, N$ such that
\begin{equation}\label{05:B}
|f_{kk}(x_p) - f_{jj}(x_p)| < \frac{1}{3m}, \forall\, j,k \ge N, p=1,\dots,r
\end{equation}

\emph{Conclusion:} If $j,k \ge N$ and $x\in[a,b]$, then $\exists\, x_p$ such
that $|x - x_p| < 1/n$. Therefore, applying (\ref{05:A}) and~(\ref{05:B}),
\begin{eqnarray*} 
\lefteqn{|f_{kk}(x)-f_{jj}(x)|}&& \\
&\le&
|f_{kk}(x)-f_{kk}(x_p)|+|f_{kk}(x_p)-f_{jj}(x_p)|+|f_{jj}(x_p)-f_{jj}(x)|\\
&<& \frac{1}{3m} + \frac{1}{3m} + \frac{1}{3m} = \frac{1}{m}
\end{eqnarray*}
\qed

\subsection{Criterion for Equicontinuity}

\begin{theorem}
If $\{f_k\}\subset C^1[a,b]$ and $|f_k'(x)| \le M$, $\forall\, k$, and
$x\in[a,b]$, then $\{f_k\}$ is uniformly equicontinuous.
\end{theorem}

\proof Given $1/m$, let $n = m\cdot M$. Then $\forall\, x,y\in[a,b]$,
the Mean Value Theorem implies there is a $z$ between $x$ and $y$ such that
\[ f_k'(z) = \frac{f_k(x) - f_k(y)}{x-y} \]
Therefore,
\[ |f_k(x) - f_k(y)| = |f_k'(z)||x-y| \le M|x - y| < \frac{1}{m} \]
whenever $|x - y| < 1/n$. \qed

\begin{corollary}
Suppose $\{f_k\}\subset C^1[a,b]$, and $\exists\, M$ such that $|f_k(x)|
\le M$ and $|f_k'(x)| \le M$, $\forall\, k$ and $x\in[a,b]$. Then there is
a subsequence $\{f_{k'}\}$ converging uniformly $f_{k'}\to f\in C[a,b]$.
\end{corollary}

\emph{Remember:} \emph{uniform} means ``for all $x$'';
\emph{equi-} means ``for all $k$''
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