Pointwise Convergence and Uniform Convergence

Definition 8.1.1 (Pointwise Convergence)

A sequence of real valued functions $\{f_n\}$ defined on a set $E (\subset \mathbb{R})$ converges pointwise on $E$ if $\{f_n(x)\}$ converges for every $x\in E$. If this is the case, then $f$ defined by

\[\begin{align*} f(x) = \lim_{n\to\infty}f_n(x), \quad x \in E, \end{align*}\]

defines a function on $E$. The function $f$ is called the limit of the sequence $\{f_n\}$.

Remark

In terms of $\epsilon$ and $N$, the sequence $\{f_n\}$ converges pointwise to $f$ if for each $x\in E$ there is $f(x)\in\mathbb{R}$, given $\epsilon >0$, there exists $N=N(x,\epsilon)\in\mathbb{N}$ such that

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

for all $n\geq N$.

Remark

Let $\{f_n\}$ be a sequence of functions on $E$. We can associate the $\{S_n\}$ of $n$-th partial sums, where for each $n\in\mathbb{N}$, $S_n$ is the real valued function on $E$ defined by for all $x\in E$

\[\begin{align*} S_n(x) = (f_1 + \cdots + f_n)(x) = \sum_{k=1}^n f_k(x). \end{align*}\]

$\sum_{k=1}^\infty f_k$ converges pointwise if for each $x\in $, $\{S_n(x)\}$ converges.

Examples 8.1.2

(a) $E=[0,1], f_n(x) = x^n$ for each $x\in E$. Note that each $f_n$ is continuous on $E$.

\[\begin{align*} f(x) = \begin{cases} 0 & \text{if } x\in [0,1) \\ 1 & \text{if } x=1 \end{cases} \end{align*}\]

$\therefore f$ is not continuous at $x=1$.

(b) $f_k (x) = x^2 / (1+x^2)^k$, $x\in\mathbb{R}$.

\[\begin{align*} \sum_{k=0}^\infty f_k(x) = \sum_{k=0}^\infty x^2\left( \frac{1}{1+x^2}\right)^k=\begin{cases} 0 & \text{if } x=0 \\ 1+x^2 & \text{if } x\neq 0 \end{cases} \end{align*}\]

$\therefore f$ is not continuous at $x=0$.

(c) $\{x_k\}_{k=1}^\infty$ is an enumeration of $\mathbb{Q}$ in $[0,1]$.

\[\begin{align*} f_n(x) = \begin{cases} 0 & \text{if } x = x_k, 1\leq k \leq n \\ 1 & \text{otherwise} \end{cases} \end{align*}\]

Each $f_n$ is Riemann integrable on $[0,1]$.

Let $x\in \mathbb{Q}$ be given. Then there is $x_{n_0}$ such that $x=x_{n_0}$ for some $n_0\in\mathbb{N}$. Then $f_{n_0}(x)= 0$.

Let $\epsilon >0$ be given. If $n\geq n_0$, by the definition of $f_n$

\[\begin{align*} \lvert f_n (x)\rvert = 0 < \epsilon. \end{align*}\]

If $x \notin \mathbb{Q}, f_n(x)= 0$ for all $n\in\mathbb{N}$. Thus,

\[\begin{align*} f(x) = \lim_{n\to\infty} f_n(x) = \begin{cases} 0 & \text{if } x\in \mathbb{Q} \\ 1 & \text{if } x \notin \mathbb{Q} \end{cases} \end{align*}\]

$\therefore f$ is not Riemann integrable.

Examples

(a) $f_n:\mathbb{R}\to\mathbb{R}$.

\[\begin{align*} f_n(x) = \chi_{[n,n+1]}(x) = \begin{cases} 1 & \text{if } x \in [n,n+1] \\ 0 & \text{otherwise} \end{cases} \end{align*}\]

converges pointwise to $f(x)=0$.

(b)

\[\begin{align*} f_m (x) = \begin{cases} 1 & \text{if } m!x\in \mathbb{Z} ( \text{i.e., if } x= \frac{p}{m!} \text{ for some } p\in\mathbb{Z} \\ 0 & \text{otherwise} \end{cases} \end{align*}\]

If $x\notin \mathbb{Q}$, then $\lim_{m\to\infty}f_m(x)=0$.

If $x\in\mathbb{Q}$, we can write $x=\frac{p}{q}$. Then $f_q(x)=1$, i.e., $f_m(\frac{p}{q})=1$ for all $m\geq q$.

$\therefore \lim_{m\to\infty}f_m(x)=1$

$\therefore \lim_{m\to\infty}f_m(x) = \chi_\mathbb{Q}(x)$.

Definition 8.2.1 (Uniform Convergence)

A sequence of real-valued functions $\{f_n\}$ defined on a set $E(\subset \mathbb{R})$ converges uniformly to $f$ on $E$, if for every $\epsilon>0$, there exists a positive integer $\mathbb{N}\in\mathbb{N}$ such that

\(\begin{align*} \lvert f_n(x) - f(x)\rvert < \epsilon \end{align*}\) for all $x\in E$ and for all $n\geq N$. We write $f_n\rightrightarrows f.$ Similarly, a series $\sum_{k=1}^\infty f_k$ of real-valued functions converges uniformly on a set $E$ if and only if the sequences $\{S_n=\sum_{k=1}^n f_k\}$ of partial sums converges uniformly on $E$.

Remark

(a) If $f_n \rightrightarrows f$, then $f_n\rightarrow f$.

(b) Moreover, if $f_n\rightrightarrows f$ and $f_n\to g$, then. $f=g$

<Proof> (a) It is clear.

(b) Let $\epsilon >0$ be given and let $x\in E$ be given. Then there is $N_1\in\mathbb{N}$ such that

\[\begin{align*} n\geq N_1 \Rightarrow \lvert f_n(x) -f(x) \rvert < \frac{\epsilon}{2}. \end{align*}\]

Similarly, for a given $x\in E$, there is $N_2\in\mathbb{N}$ such that

\[\begin{align*} n\geq N_2 \Rightarrow \lvert f_n(x) -g(x)\rvert < \frac{\epsilon}{2}. \end{align*}\]

Take $N=\max\{N_1, N_2\}$. Then

\[\begin{align*} \lvert f(x)-g(x)\vert &\leq \lvert f_n(x)-f(x)\rvert + \lvert f_n(x)-g(x) \rvert \\ &< \frac{\epsilon}{2} +\frac{\epsilon}{2} \\ &=\epsilon \end{align*}\]

Since $\epsilon >0$ is arbitrary, $f(x)=g(x)$ for all $x\in E$.

$\therefore f=g$. \(\tag*{$\square$}\)

Examples 8.2.2

(a) $f_n(x) = x^n$ on $E=[0,1]$. Then $f_n \to f$ as $n\to\infty$ where

\[\begin{align*} f(x) = \begin{cases} 0 & \text{if } x \in [0, 1) \\ 1 & \text{if } x = 1. \end{cases} \end{align*}\]

We want to show that $f_n$ does not converge uniformly to $f$. Suppose that $f_n\rightrightarrows f$ . Given $\epsilon>0$, there is $n_0\in\mathbb{N}$ such that

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

for all $x\in E$ and for all $n \geq n_0$. If $x\in [0,1)$, then $x^{n_0}<\epsilon$. Take $0<\epsilon_0 <1-\epsilon$.

Since $f_{n_0}$ is continuous on $E$, there is a $\delta>0$ such that

\[\begin{align*} \lvert x-1 \rvert <\delta \text{ with } x\in E \Rightarrow \lvert x^{n_0}-1\rvert < \epsilon_0. \end{align*}\]

In other words,

\[\begin{align*} x\in (1-\delta, 1+\delta) \cap E \Rightarrow 1-\epsilon_0 < x^{n_0} < 1+\epsilon_0. \end{align*}\]

But it implies that $x^{n_0} < \epsilon < 1-\epsilon_0< x^{n_0}$, which is a contradiction.

$\therefore f_n$ does not converge uniformly.

On the other hand, if we set the domain $E=[0,a)$ with $0<a<1$, then $f_n\rightrightarrows f$ on $[0,a)$.

Let $\epsilon >0$ be given. Since $0<x<a, \lvert x^n\rvert <a^n$. Since $0<a<1$, there is $n_0\in\mathbb{N}$ such that

\[\begin{align*} n\geq n_0 \Rightarrow a^n <\epsilon. \end{align*}\]

Thus, for $n\geq n_0, x\in [0,a)$, $\lvert x^n\rvert <\epsilon$.

$\therefore f_n \rightrightarrows f$ on $[0,a)$.

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

Theorem 8.2.3 (Cauchy Criterion)

A sequence $\{f_n\}$ of real-valued functions defined on a set $E$ converges uniformly on $E$ if and only if for every $\epsilon>0$, there exists $n_0\in\mathbb{N}$ such that

\(\begin{align*} \lvert f_n(x) - f_m(x) \rvert <\epsilon \end{align*}\) for all $x\in E$ and all $n,m \geq n_0$.

<Proof>

$\Rightarrow$ Let $\epsilon >0$ be given. There is $N\in\mathbb{N}$ such that

\[\begin{align*} m,n\geq N, x\in E \Rightarrow \lvert f_n(x) - f_m(x) \rvert <\frac{\epsilon}{2}. \end{align*}\]

For $n,m\in\mathbb{N}$ with $m,n \geq N$,

\[\begin{align*} \lvert f_m(x) - f_n(x) \rvert &= \lvert f_m(x) - f(x) + f(x) - f_n(x) \rvert \\ &\leq \lvert f_m(x) - f(x) \rvert + \lvert f(x) - f_n(x)\rvert \\ &\leq \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ &=\epsilon. \end{align*}\]

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

\[\begin{align*} m,n \geq N, x\in E \Rightarrow \lvert f_m(x) - f_n(x)\rvert <\frac{\epsilon}{2}. \end{align*}\]

For each $x\in E, \{f_n(x)\}$ is Cauchy sequence and hence it converges. Let $f(x) = \lim_{n\to\infty}f_n(x)$ for every $x\in E$.

Fix $m\geq N$ and $x\in E$. Then,

\[\begin{align*} \lvert f(x) - f_m (x)\rvert &=\lvert \lim_{n\to\infty}f_n(x)-f_m(x)\rvert \\ &= \lim_{n\to\infty}\lvert f_n(x) - f_m(x) \rvert \quad (\because y\mapsto \lvert y-f_m(x)\rvert \text{ is continuous})\\ &\leq \frac{\epsilon}{2} \\ &< \epsilon \end{align*}\]

$\therefore f_n\rightrightarrows f$.

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

Corollary 8.2.4

The series $\sum_{k=1}^\infty f_k$ of real-valued functions on $E$ converges uniformly on $E$ if and only if given $\epsilon >0$, there exists a positive integer $n_0\in\mathbb{N}$ such that

\[\begin{align*} \left\lvert \sum_{k=n+1}^m f_k \right\rvert < \epsilon \end{align*}\]

for all $x\in E$ and for all integers $m>n\geq n_0$.

Theorem 8.2.5

Suppose the sequence $\{f_n\}$ of real-valued functions on the set $E$ converges pointwise to $f$ on $E$. For each $n\in\mathbb{N}$, set

\[\begin{align*} M_n = \sup_{x\in E}\lvert f_n(x) - f(x)\rvert. \end{align*}\]

Then $\{f_n\}$ converges uniformly to $f$ on $E$ if and only if $\lim_{n\to\infty} M_n =0$.

$\Rightarrow$ Let $\epsilon >0$ be given. Since $f_n \rightrightarrows f$, there exists a $N\in\mathbb{N}$ such that

\[\begin{align*} n\geq N, x\in E \Rightarrow \lvert f_n(x) - f(x) \rvert < \frac{\epsilon}{2}. \end{align*}\]

For $n\geq N, \frac{\epsilon}{2}$ is an upper bound of $\{\lvert f_n (x) -f(x) \rvert : x\in E\}$. Thus,

\[\begin{align*} M_n=\sup_{x\in E} \lvert f_n(x) - f(x) \rvert \leq \frac{\epsilon}{2} < \epsilon \end{align*}\]

for all $n\geq N$.

$\therefore \lim_{n\to\infty}M_n = 0.$

$\Leftarrow$ Let $\epsilon >0$ be given. Since $M_n\to 0$ as $n\to\infty$, there is $N\in\mathbb{N}$ such that

\[\begin{align*} n\geq N\Rightarrow \lvert M_n \rvert <\epsilon. \end{align*}\]

Thus, for $n\geq N$ and for all $x\in E$,

\[\begin{align*} \lvert f_n(x) -f(x) \rvert \leq \sup_{x\in E} \lvert f_n(x) -f(x)\rvert <\epsilon. \end{align*}\]

$\therefore f_n \rightrightarrows f$.

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

Theorem 8.2.7 (Weierstrass M-Test)

Suppose $\{f_k\}$ is a sequence of real-valued functions defined on a set $E$, and $\{M_k\}$ is a sequence of real numbers satisfying

\[\begin{align*} \lvert f_k(x) \rvert \leq M_k \quad \text{for all } x\in E \text{ and } k\in\mathbb{N}. \end{align*}\]

If $\sum_{k=1}^\infty M_k$ converges, then $\sum_{k=1}^\infty f_k(x)$ converges uniformly and absolutely on $E$.

<Proof>

Let $\epsilon >0$ be given and let $T_n = \sum_{k=1}^n M_k$. Since $T_n$ converges, $T_n$ is Cauchy sequence. Thus, for given $\epsilon >0$, there is $N\in\mathbb{N}$ such that

\[\begin{align*} n>m\geq N \Rightarrow \lvert T_n - T_m \rvert = \left\lvert \sum_{k=m+1}^n M_k \right\rvert < \epsilon. \end{align*}\]

Define $S_n(x) = \sum_{k=1}^n f_k(x)$. If $n>m\geq N$,

\[\begin{align*} \left \lvert S_n (x) - S_m(x)\right \rvert &= \left\lvert \sum_{k=m+1}^n f_k(x) \right\rvert \\ &\leq \sum_{k=m+1}^n\lvert f_k(x) \rvert \\ &\leq \sum_{k=m+1}^n M_k \\ &<\epsilon. \end{align*}\]

By Cauchy criterion, $\sum_{k=1}^\infty f_k$ converges uniformly on $E$.

\(\tag*{$\square$}\)

Reference

• Manfred Stoll, 『Introduction to Real Analysis』