Uniform Convergence and Continuity

Theorem 8.3.1

Suppose $\{f_n\}$ is a sequence of real-valued functions that converges uniformly to a function $f$ on a subset $E$ of $\mathbb{R}$. Let $p$ be a limit point of $E$ and suppose that for each $n\in\mathbb{N}$,

\[\begin{align*} \lim_{x\to p}f_n(x) = A_n. \end{align*}\]

Then the sequence $\{A_n\}$ converges and

\[\begin{align*} \lim_{x\to p} f(x) = \lim_{n\to\infty} A_n \end{align*}\]

<Proof>

Let $\epsilon >0$ be given. Since $f_n \rightrightarrows f$ on $E$, there is $N\in\mathbb{N}$ such that

\[\begin{align*} m,n\geq N, x\in E \Rightarrow \lvert f_n(x) - f_m(x) \rvert < \frac{\epsilon}{3}. \end{align*}\]

Since $f_n(x) \to A_n$ and $f_m(x) \to A_m$ as $x\to p$, there is a $\delta >0$ such that

\[\begin{align*} 0 < \lvert x-p \rvert < \delta, x\in E \Rightarrow \lvert f_n(x) - A_n \rvert < \frac{\epsilon}{3} \text{ and } \lvert f_m(x) - A_m \rvert < \frac{\epsilon}{3}. \end{align*}\]

Thus,

\[\begin{align*} \lvert A_m - A_n\rvert &= \lvert A_m - f_m(x) + f_m(x) - f_n(x) + f_n(x) - A_n\rvert \\ &\leq \lvert A_m - f_m(x) \rvert + \lvert f_m(x) -f_n(x) \rvert + \lvert f_n(x) -A_n \rvert \\ &< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{align*}\]

for any $m,n \geq N$ and $x\in N^\prime_\delta (p) \cap E$.

$\therefore \{A_n\}$ is Cauchy sequence.

Let $A=\lim_{n\to\infty}A_n$. We want to show that $\lim_{x\to p} f(x) = A$.

Let $\epsilon >0$ be given. Since $f_n \rightrightarrows f$ on $E$, there is $N_1\in\mathbb{N}$ such that

\[\begin{align*} m\geq N_1, x\in E \Rightarrow \lvert f(x) - f_m(x) \rvert < \frac{\epsilon}{3}. \end{align*}\]

Since $A_n \to A$, there is $N_2\in\mathbb{N}$ such that

\[\begin{align*} m\geq N_2 \Rightarrow \lvert A - A_m\rvert < \frac{\epsilon}{3}. \end{align*}\]

Since $\lim_{x\to p} f_m(x) = A_m$, there exists $\delta>0$ such that

\[\begin{align*} 0<\lvert x-p \rvert < \delta \Rightarrow \lvert f_m(x) -A_m\rvert < \frac{\epsilon}{3}. \end{align*}\]

Put $N=\max\{N_1, N_2\}\in\mathbb{N}$. If $0<\lvert x - p \rvert <\delta$ and $x\in E$, then

\[\begin{align*} \lvert f(x) - A \rvert &\leq \lvert f(x) - f_N(x) \rvert + \lvert f_N(x) - A_N\rvert + \lvert A_N -A \rvert \\ &< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon. \end{align*}\]

$\therefore \lim_{x\to p}f(x) = A= \lim_{n\to\infty}A_n$.

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

Remark

The last sentence can be written as

\[\begin{align*} \lim_{x\to p} \left(\lim_{n\to\infty}f_n(x) \right)= \lim_{n\to\infty}\left( \lim_{x\to p}f(x)\right) \end{align*}\]

Corollary 8.3.2

(a) If $\{f_n\}$ is a a sequence of continuous real-valued functions on $E$, and if $\{f_n\}$ converges uniformly to $f$ on $E$, then $f$ is continuous on $E$.

(b) If $\{f_n \}$ is a sequence of continuous real-valued functions on $E$, and if $\sum_{n=1}^\infty f_n$ converges uniformly on $E$, then

\[\begin{align*} S(x) = \sum_{n=1}^\infty f_n(x) \end{align*}\]

is continuous on $E$.

<Proof>

(a) If $p$ is isolated point, i.e., $N_{\delta_0}(p) \cap E=\{p\}$ for some $\delta_0>0$, then $f$ is continuous at $p$. If $p\in E$ is a limit point of $E$, then since $f_n$ is continuous for each $n\in\mathbb{N}$,

\[\begin{align*} \lim_{x\to p}f_n(x) = f_n(p). \end{align*}\]

Thus by the previous theorem,

\[\begin{align*} \lim_{x\to p} f(x) &= \lim_{x\to p}\left(\lim_{n\to\infty}f_n(x)\right) \\ &=\lim_{n\to\infty}\left(\lim_{x\to p}f_n(x) \right) \\ &= \lim_{n\to\infty}f_n(p) \\ &= f(p) \end{align*}\]

(b) Let

\[\begin{align*} S_n(x) = \sum_{k=1}^n f_k(x). \end{align*}\]

Then for each $n\in\mathbb{N}$, $S_n$ is continuous on $E$. Since $\{S_n\}$ converges uniformly to $S$ on $E$, by part (a) $S$ is also continuous on $E$.

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

Theorem 8.3.5 (Dini)

Suppose $K$ is a compact subset of $\mathbb{R}$ and $\{f_n\}$ is a a sequence of continuous real-valued functions on $K$ satisfying the following:

(a) $\{f_n\}$ converges pointwise on $K$ to a continuous function $f$, and

(b) $f_n(x) \geq f_{n+1}(x)$ for all $x\in K$ and $n\in\mathbb{N}$.

Then $\{f_n\}$ converges uniformly to $f$ on $K$.

<Proof> We want to show that for every $\epsilon>0$, there exists $N\in\mathbb{N}$ such that

\[\begin{align*} n\geq N, x\in K \Rightarrow \lvert f_n(x) - f(x)\rvert < \epsilon. \end{align*}\]

Observe that it suffices to show that there exists $N\in\mathbb{N}$ such that

\[\begin{align*} f_N(x) - f(x) < \epsilon \end{align*}\]

for all $x\in K$ since $\{f_n(x)\}$ is monotone decreasing for all $x\in K$. Let $g = f_n - f$ and let $K_n=\{x\in K: g(x) \geq \epsilon\}$ for each $n\in \mathbb{N}$. We want to show that $K_N=\emptyset$ for some $N\in\mathbb{N}$.

First, $K_n$ is closed set for all $n\in\mathbb{N}$. Let $p\in K^\prime_n$ be given. Then there is a sequence $\{x_k\}_{k=1}^\infty$ such that $x_k\to p$ with $x_k\neq p$ for all $k\in\mathbb{N}$.

Since $g_n$ is continuous and $g_n(x_k)\geq \epsilon$ for all $k\in\mathbb{N}$,

\[\begin{align*} g_n(p) = g_n\left(\lim_{k\to\infty}(x_k)\right) = \lim_{k\to\infty}\left(g_n(x_k)\right) \geq \epsilon. \end{align*}\]

$\because$ Suppose that $g_n(p) < \epsilon$. Let $\epsilon_0= \epsilon-g_n(p)$. Since $g_n(x_k)\to g(p)$ as $p\to\infty$, there exists $N\in\mathbb{N}$ such that

\[\begin{align*} k\geq N \Rightarrow \lvert g_n(x_k) -g(p) \rvert < \epsilon_0. \end{align*}\]

However,

\[\begin{align*} \lvert g_n(x_k) - g_n(p) \rvert &= g_n(x_k) - g_n(p) \\ &\geq \epsilon-g_n(p) \\ &=\epsilon_0, \end{align*}\]

which is a contradiction. Thus, $g_n(p) \geq \epsilon$, i.e., $p\in K_n$.

$\therefore K_n$ is closed.

Alternatively, we can show $K_n$ is closed by using the topological characterization of continuous function. Since $[\epsilon, \infty)^c = (-\infty, \epsilon)$ is open, $[\epsilon, \infty)$ is closed. Since $K_n=g^{-1}\left([\epsilon, \infty)\right)$ and $g$ is continuous, $K_n$ is closed.

Since $K_n$ is closed and $K_n \subset K$, $K_n$ is compact and $K_n\supset K_{n+1} (\because g_n(x) \geq g_{n+1}(x)).$

Now we want to show that $\bigcap_{n=1}^\infty K_n=\emptyset$. Suppose that $\bigcap_{n=1}^\infty K_n\neq \emptyset$ and let $x_0 \in \bigcap_{n=1}^\infty K_n$ be given. In other words,

\[\begin{align*} g_n(x_0) \geq \epsilon \text{ for all } n \in \mathbb{N}. \end{align*}\]

Since $g_n(x_0) \to 0$ as $n\to\infty$, for the given $\epsilon$, there is $N\in\mathbb{N}$ such that

\[\begin{align*} n\geq N\Rightarrow \lvert g_n(x_0)\rvert <\epsilon, \end{align*}\]

which is a contradiction to the assumption that $\lvert g_n(x_0)\rvert$ for all $n\in\mathbb{N}$.

$\therefore \bigcap_{n=1}^\infty K_n\neq \emptyset$.

By Theorem 3.2.7, there is a $K_{n_0}=\emptyset$ for some $n_0\in\mathbb{N}$. That is if $n\geq n_0$, then

\[\begin{align*} g_n(x) < \epsilon \end{align*}\]

for all $x\in K$.

$\therefore g_n \rightrightarrows 0$ on $K$, i.e., $f_n\rightrightarrows f$ on $K$.

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

The Space $\mathscr{C}[a,b]$

$\mathscr{C}[a,b]$ is the set of all continuous real-valued functions on $[a,b]$. For $f,g\in \mathscr{C}[a,b]$ and $c\in\mathbb{R}$, we can define binary operation and scalar multiplication as follows:

(a) $(f+g)(x) := f(x) + g(x)$

(b) $(cf)(x) :=c\cdot f(x)$

Then $(\mathscr{C}[a,b], +, \cdot)$ is $\mathbb{R}$-vector space.

Definition 8.3.7

For each $f\in \mathscr{C}[a,b]$, set

\[\begin{align*} \lVert f\rVert_u=\sup\{\lvert f(x)\rvert: x\in [a,b] \}. \end{align*}\]

The quantity $\lVert f\rVert_u$ is called the uniform norm of $f$ on $[a,b]$.

Theorem 8.3.8

A sequence $\{f_n\}$ in $\mathscr{C}[a,b]$ converges uniformly to $f\in\mathscr{C}[a,b]$ if and only if given $\epsilon>0$, there exists $n_0\in\mathbb{N}$ such that $\lVert f-f_n\rVert_u < \epsilon$ for all $n\geq n_0$.

<Proof>

Since

\[\begin{align*} \lVert f_n - f\rVert_u = \sup\{\lvert f_n(x) - f(x)\rvert: x\in [a,b]\}, \end{align*}\]

by Theorem 8.2.5, we are done.

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

Theorem 8.3.11

The metric space $(\mathscr{C}[a,b], \lVert \cdot \rVert_u)$ is complete.

<Proof>

Let $\{f_n\}$ be a Cauchy sequence in $\mathscr{C}[a,b]$. Given $\epsilon>0$ there is $N\in\mathbb{N}$ such that

\[\begin{align*} m,n \geq N \Rightarrow \lVert f_n - f_m \rVert_u < \epsilon. \end{align*}\]

By the definition of uniform norm,

\[\begin{align*} \lvert f_n(x) - f_m(x) \rvert \leq \sup_{x\in [a,b]}\{\lvert f_n(x) - f_m(x)\rvert\}=\lVert f_n -f_m\rVert_u \end{align*}\]

for all $x\in [a,b]$.

By Cauchy criterion, $\{f_n\}$ converges uniformly to $f$. Since $f_n$ is continuous for all $n\in\mathbb{N}$, by Corollary 8.3.2 $f$ is continuous on $[a,b]$, i.e., $f\in\mathscr{C}[a,b]$.

Since $f_n \rightrightarrows f$, $f\stackrel{\lVert \cdot \rVert_u}{\to}f$ by Theorem 8.3.8.

$\therefore (\mathscr{C}[a,b], \lVert \cdot \rVert_u)$ is complete.

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

Reference

• Manfred Stoll, 『Introduction to Real Analysis』