Equicontinuous Families of Functions

Definition 7.19

Let $\{f_n\}$ be a sequence of functions defined on a set $E$. We say that $\{f_n\}$ is pointwise bounded on $E$ if the sequence $\{f_n(x)\}$ is bounded for every $x\in E$, that is, if there exists a finite valued function $\phi: E\to\mathbb{R}$ such that

\[\begin{align*} \lvert f_n(x) \rvert < \phi(x) \quad (x\in E, n\in\mathbb{N}). \end{align*}\]

We say that $\{f_n\}$ is uniformly bounded on $E$ if there exists a number $M$ such that

\[\begin{align*} \lvert f_n(x) \rvert < M \quad (x\in E, n\in\mathbb{N}). \end{align*}\]

Remark

Even if $\{f_n\}$ is a uniformly bounded sequence of continuous functions on a compact set $E$, there need not exist a subsequence which converges pointwise on $E$.

Let $f_n(x) = \sin (nx), x\in [0,2\pi], n\in\mathbb{N}$. Suppose that there exists a sequence $\{n_k\}$ such that $\{\sin(n_kx)\}$ converges for every $x\in [0,2\pi]$. In that case we must have

\[\begin{align*} \lim_{k\to\infty}(\sin(n_kx)-\sin(n_{k+1}x))=0 \quad (0\leq x \leq 2\pi), \end{align*}\]

which implies that

\[\begin{align*} \lim_{k\to\infty}(\sin (n_kx)-\sin(n_{k+1}x)^2=0 \quad (0\leq x \leq 2\pi). \end{align*}\]

By Lebesque dominated convergence theorem,

\[\begin{align*} \lim_{k\to\infty}\int_0^{2\pi}\left( \sin(n_kx)-\sin(n_{k+1}x)\right)^2 = \int_0^{2\pi}\lim_{k\to\infty} \left(\sin(n_kx)-\sin(n_{k+1}x) \right)^2) =0. \end{align*}\]

However

\[\begin{align*} \int_0^{2\pi}(\sin (n_kx)-\sin(n_{k+1}x))^2=2\pi, \end{align*}\]

which is a contradiction.

\[\tag*{$\square$}\]

Theorem 7.23

If $\{f_n\}$ is a pointwise bounded sequence of complex functions on a countable set $E$, then $\{f_n\}$ has a subsequence $\{f_{n_k}\}$ such that $\{f_{n_k}(x)\}$ converges for every $x\in E$.

Note that if $E=\{x_1, x_2\}$, by Bolzano-Weierstrass Theorem we can get a convergent subsequence $\{f_{n_k}(x_1)\}$ on $x_1$. Then consider the sequence

\[\begin{align*} \{f_{n_k}(x_2)\}_{k=1}^\infty. \end{align*}\]

Again by Bolzano-Weierstrass Theorem, we can choose a convergent subsequence $\{f_{n_{k_l}}(x_2)\}$ on $x_2$.

$\therefore \{f_{n_{k_l}}\}$ converges on $E=\{x_1, x_2\}$.

<Proof>

Let $E=\{x_i\}$. Since \(\begin{align*} \{f_n(x_1)\}_{n=1}^\infty \end{align*}\)

is bounded, there is a convergent subsequence

\[\begin{align*} \{f_{n_k}(x_1)\}_{k=1}^\infty. \end{align*}\]

We denote $\{f_{n_k}\}$, which converges to $\lim_{k\to\infty}f_{n_k}(x_1)$ by $\{f_{1,k}\}$.

Iterate the following procedure. Since $\{f_{1,n}(x_2)\}$ is bounded, there is a convergent subsequence

\[\begin{align*} \\{f_{1, n_k}(x_2)\\}_{k=1}^\infty. \end{align*}\]

Notate $f_{2,k}=f_{1,n_k}$. Now we consider sequences $S_1, S_2, S_3,\ldots$, which represent by the array

\[\begin{align*} &S_1: \: f_{1,1} \quad f_{1,2} \quad f_{1,3} \quad f_{1,4} \quad \cdots \\ &S_2: \: f_{2,1} \quad f_{2,2} \quad f_{2,3} \quad f_{2,4} \quad \cdots \\ &S_3: \:f_{3,1} \quad f_{3,2} \quad f_{3,3} \quad f_{3,4} \quad \cdots \\ \end{align*}\]

and which have the following properties:

(1) Each row is a subsequence of the earlier row

(2) $n$-th row converges on $\{x_1, \ldots, x_n\}$.

(3) $k$-th element of the $n$-th row occurs in the greater or than equal to $k$-th place in the previous row.

The sequence $S=\{f_{n,n}\}$ (except possibly its first $n-1$ terms) is a subsequence of $S_n$ for all $n\in\mathbb{N}$. Hence (2) implies that $\{f_{n,n}(x_i)\}$ converges as $n\to\infty$, for every $x_i\in E$.

Specifically, let $x_i$ be given and let $\epsilon >0$ be given. Then

\[\begin{align*} \{f_{n,n}\}_{n=i}^\infty \end{align*}\]

is a subsequence of $S_i=\{f_{i,n}\}$. Since $S_i$ converges pointwise on $\{x_1, \ldots, x_i\}$, there is $N_1\in\mathbb{N}$ such that

\[\begin{align*} n\geq N_1 \Rightarrow \lvert f_{i,n}(x_i) - \lim_{k\to\infty}f_{i,k}(x_i)\rvert < \epsilon. \end{align*}\]

Since $\{f_{n,n}(x_i)\}$ is a subsequence of

\(\begin{align*} \{f_{i,n}(x_i)\}_{n=1}^\infty \end{align*}\),

\[\begin{align*} n\geq N:=\max\{i, N_1\}\Rightarrow \lvert f_{n,n}(x_i) - \lim_{k\to\infty} f_{i,k}(x_i) \rvert <\epsilon. \end{align*}\]

$\therefore \{f_{n,n}(x_i)\}$ converges on $x_i$.

\[\tag*{$\square$}\]

Definition 7.22

A family $\mathscr{F}$ of complex functions $f$ defined on a set $E$ in a metric space $X$ to be equicontinuous on $E$ if for every $\epsilon>0$ there exists a $\delta>0$ such that

\[\begin{align*} d(x,y) <\delta , x,y\in E\Rightarrow \lvert f(x) - f(y) \rvert <\epsilon \text{ for all }f\in\mathscr{F}. \end{align*}\]

Remark

IS every finite collection of uniformly continuous functions are equicontinuous? Yes.

Let $\{f_n\}_{n=1}^k$ is a collection of uniformly continuous functions. Let $\epsilon > 0$ be given. Then for each $f_n\in\mathscr{F}$, there is a $\delta_n>0$ such that

\[\begin{align*} d(x,y) < \delta_n \Rightarrow \lvert f_n(x) - f_n(y)\rvert <\epsilon. \end{align*}\]

Take $\delta =\min\{\delta_1, \ldots, \delta_k\}$. Then $\mathscr{F}$ is equicontinuous.

Theorem 7.24

If $K$ is a compact metric space, if $\{f_n\}$ is a sequence of continuous functions on $K$, and if $\{f_n\}$ converges uniformly on $K$, then $\{f_n\}$ is equicontinuous.

<Proof>

Let $\epsilon >0$ be given. By uniform convergence, there is $N\in\mathbb{N}$ such that

\[\begin{align*} n > N \Rightarrow \lvert f_n(x) - f_N(x) \rvert < \frac{\epsilon}{3} \end{align*}\]

for all $x\in K$. Since $f_n$ is continuous on the compact set $K$, all $f_n$ are uniformly continuous on $K$ by Theorem 4.3.4. By previous remark $\{f_1, \ldots, f_N\}$ is equicontinuous. Thus, there exists $\delta>0$ such that

\[\begin{align*} d(x,y) <\delta \Rightarrow \lvert f_i(x) - f(y) \rvert <\frac{\epsilon}{3} \end{align*}\]

for all $i=1,\ldots, N$.

For $n> N$, if $d(x,y)<\delta$, then

\[\begin{align*} \lvert f_n(x) -f_n(y)\rvert &\leq \lvert f_n(x) - f_N(x) \rvert + \lvert f_N(x) - f_N(y) \rvert + \lvert f_N(y) - f_n(y)\rvert \\ &< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3}\\ &=\epsilon. \end{align*}\] \[\tag*{$\square$}\]

Reference

• Walter Rudin, 『Principles of Mathematical Analysis』