Pointwise and Uniform Convergence

Definition (Pointwise Convergence)

Let $E$ be a set (most of the time, a Metric Space). Consider

$$f_n:E\to \mathbb{C}n\ge 1$$

and $f:E\to \mathbb{C}$. We see the sequence $\{f_n{\}}_{n\ge 1}$ converges pointwise to $f$, denoted by $f_n\to f$ if

$$\underset{n\to \infty }{\lim }{f}_n(x)=f(x)\forall x\in E$$

Precisely, for any

$$(\forall x\in E)(\forall \epsilon >0)(\exists N>0)(\forall n>N)(|f_n(x)-f(x)|<\epsilon )$$

$N$ depends on both $\epsilon ,x$.

Remark (Pointwise Equivalence)

Pointwise convergence is equivalent to limits of sequences of numbers, indexed by $E$.

Example:

Let $E=[0,1]$. Let $f_n(x)=x^n$. For fixed $x\in [0,1]$, the values of

$$\underset{n\to \infty }{\lim }{f}_n(x)=\underset{n\to \infty }{\lim }{x}^n=\{\begin{array}{ll}0& 0\le x<1\\ 1& x=1\end{array}$$

So, $f_n\to f$ where

$$f(x)=\{\begin{array}{ll}0& 0\le x<1\\ 1& x=1\end{array}$$

giving us a point of simple discontinuity. So, if $f_n\to f$, and $f_n$ is continuous, then $f$ is not necessarily continuous.

Definition (Uniform Convergence)

Let $f_n:E\to \mathbb{C}$ and $f:E\to \mathbb{C}$ for $n\ge 1$. We say that $\{f_n{\}}_{n\ge 1}$ uniformly converges to $f$ if for any $\epsilon >0$, there exists $N>0$, such that for any $n>N$, for any $x\in E$, $|f_n(x)-f(x)|<\epsilon$. This denoted by

$$f_n\rightrightarrows f$$

precisely,

$$(\forall \epsilon >0)(\exists N>0)(\forall x\in E)(\forall n\ge N)(|f_n(x)-f(x)|<\epsilon )$$

Here, $N$ only relies on $\epsilon$. Thus, it is a stronger result than Definition (Pointwise Convergences), shown in Lemma (Uniform to Pointwise)

Lemma (Uniform to Pointwise)

If $f_n\rightrightarrows f$, then $f_n\to f$.

Example:

Let $\{x^n\}/\rightrightarrows f$ over $[0,1]$, where

$$f(x)=\{\begin{array}{ll}0& 0\le x<1\\ 1& x=1\end{array}$$

Proof: Suppose that $x^n\rightrightarrows f$ where $\epsilon =1/2$. Then $\exists N>0$ such that

$$|f_n(x)-f(x)|<1/2\forall x\in E,\forall n>N$$

Then take $n=N+1$, we have $f_n(1)=1$ which is continuous since it is constant. Then $\exists \delta$ such that if $x>1-\delta$, then $f_n(x)>1/2$. By assumption,

$$|f_n(x)-f(x)|<1/2f(x)=0\text{ if 1−δ<x<1}$$

which implies $|f_n(x)|<1/2$, a contradiction.

Criterion 1 (By Supremum)

Let $M_n={\sup }_{x\in E}|f_n(x)-f(x)|$ using the Supremum. Then

$$f_n\rightrightarrows f⟺\underset{n\to \infty }{\lim }{M}_n=0$$

This is just the definition, so the proof can be omitted.

Criterion 2 (Cauchy Criterion)

The sequence of functions $\{f_n\}$ defined on $E$, converges uniformly on $E$ if and only if for every $\epsilon >0$ there exists an integer $N$ such that $m\ge N,n\ge N,x\in E$ implies

$$|f_n(x)-f_m(x)|\le \epsilon$$

Proof: Suppose $\{f_n\}$ converges uniformly on $E$, and let $f$ be the limit function. Then there is an integer $N$ such that $n\ge N,x\in E$ and

$$|f_n(x)-f(x)|\le \epsilon /2$$

so that

$$|f_n(x)-f_m(x)\le |f_n(x)-f(x)|+|f(x)-f_m(x)|\le \epsilon$$

and the forward direction is done.

Conversely, if the Cauchy condition holds, then by theorem and that $E\subset {\mathbb{R}}^1,\{f_n(x)\}$ converges for every $x$. Let the limit be $f(x)$. Let $\epsilon >0$ be given, and choose $N$ such that

$$|f_n(x)-f_m(x)|\le \epsilon$$

holds. We fix some $n$, and let $m\to \infty$. Then as $f_m(x)\to f(x)$,

$$|f_n(x)-f(x)|\le \epsilon$$

for every $n\ge N$ and every $x\in E$.

Criterion 3 (Comparison Test)

Suppose $\{f_n\}$ is a sequence of functions defined on $E$, and suppose

$$|f_n(x)|\le M_{n}x\in E,n=1,2,3,\dots$$

Then $\sum f_n$ converges uniformly on $E$ if $\sum M_n$ converges.

The converse is not necessarily true.

Proof: If $\sum M_n$ converges, then for arbitrary $\epsilon >0$,

$$\left|\sum _{i=n}^m{f}_i(x)\right|\le \sum _{i=n}^m{M}_i\le \epsilon (x\in E)$$

provided $m,n$ are large enough. Then uniformly convergence is satisfied from Criterion 2 (Cauchy Criterion).

Theorem (Uniform Convergence and Continuity)

Suppose $f_n\to f$ uniformly on a set $E$ is a metric space. Let $x$ be a limit point of $E$, and suppose that

$$\underset{t\to x}{\lim }{f}_n(t)=A_n(n=1,2,3,\dots )$$

Then $\{A_n\}$ converges, and

$$\underset{t\to x}{\lim }f(t)=\underset{n\to \infty }{\lim }{A}_n$$

In other words,

$$\underset{t\to x}{\lim }\underset{n\to \infty }{\lim }{f}_n(t)=\underset{n\to \infty }{\lim }\underset{t\to x}{\lim }{f}_n(t)$$

Proof: Let $\epsilon >0$ be given. By uniform convergence of $\{f_n\}$, $\exists N$ such that $n,m\ge N,t\in E$ imply

$$|f_n(t)-f_m(t)|\le \epsilon$$

Letting $t\to x$, we obtain

$$|A_n-A_m|\le \epsilon$$

for $n,m\ge N$ so that $\{A_n\}$ is a Cauchy Sequence and therefore converges. Let $A$ denote the convergence point. Then,

$$|f(t)-A|\le |f(t)-f_n(t)|+|f_n(t)-A_n|+|A_n-A|$$

If we pick $n$ such that

$$|f(t)-f_n(t)|\le \epsilon /3$$

for all $t\in E$ (possible by uniform convergence), and such that

$$|A_n-A|\le \epsilon /3$$

Then for this $n$, we choose neighborhood $N_r(x)$ of $x$ such that

$$|f_n(t)-A_n|\le \epsilon /3$$

if $t\in V\cap E,t\ne x$. By substitution,

$$\begin{array}{rl}|f(t)-A|& \le |f(t)-f_n(t)|+|f_n(t)-A_n|+|A_n-A|\\ & \le \epsilon /3+\epsilon /3+\epsilon /3\\ & =\epsilon \end{array}$$

and we are done.

Corollary (Uniform Converges Retains Continuity)

If $\{f_n\}$ is a sequence of continuous functions on $E$, and if $f_n\to f$ uniformly on $E$, then $f$ is continuous on $E$.

Proof: $\forall p\in E$, WTS that ${\lim }_{x\to p}f(x)=f(p)$ or that

$$\forall \epsilon >0,\exists \delta >0,|x-p|<\delta ⟹|f(x)-f(p)|<\epsilon$$

so we have that

$$|f(x)-f(p)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(p)|+|f_n(p)-f(p)|$$

pick $N$ large enough so that

$$\begin{array}{rl}|f(x)-f_N(x)|& <\epsilon /3\\ |f(p)-f_N(p)|& <\epsilon /3\end{array}$$

since $f_n$ is continuous, then

$$\exists \delta >0,|x-p|<\delta ⟹|f_n(x)-f_n(p)|<\epsilon /3$$

and summing up all portions, we have that

$$|f(x)-f(p)|\le \epsilon$$

Definition (Complex Continuous Bounded Functions)

If $X$ is a Metric Space then let $\mathcal{C}(X)$ denote the set of all complex-valued continuous bounded functions with domain $X$.

• Boundedness is redundant if $X$ is compact, by Heine-Borel Theorem

We associate with $f\in \mathcal{C}(X)$ with a Supremum norm:

$$||f||=\underset{x\in X}{\sup }|f(x)|$$

• $f$ is bounded, so any norm is finite. It is obvious that $||f||=0$ only if $\forall x\in X,f(x)=0$.
• $h(x)=f(x)+g(x)⟹|h(x)|\le |f(x)|+|g(x)|+||f||+||g||$ for all $x\in X$ such that
• $||f+g||\le ||f||+||g||$
• By defining $d(f,g)=||f-g||$ , then we have that by Definition, $\mathcal{C}(X)$ is a Metric Space.

Theorem (Complete)

$\mathcal{C}(X)$ is also a complete metric space.

Proof: If $\{f_n\}\subset \mathcal{C}(\mathcal{X})$ is a Cauchy Sequence, then $\exists f\in \mathcal{C}(X)$ such that

$$\begin{array}{rl}\underset{n,m\ge N}{\sup }|f_n(x)-f_m(x)|& ⟺\underset{n\to \infty }{\lim }{f}_n=f\\ & ⟺\underset{n\to \infty }{\lim }d(f_n,f)=0\\ & ⟺\underset{n\to \infty }{\lim }\underset{x\in X}{\sup }d(f_n(x)-f(x))=0\\ & ⟺f_n\rightrightarrows f\end{array}$$

by Criterion 2 (Cauchy Criterion) and Definition (Uniform Convergence). Then by Corollary (Uniform Converges Retains Continuity), $f$ is continuous and since $f_n$ is bounded, $f$ is bounded. Thus $f\in \mathcal{C}(X)$.

Theorem (Countable is Pointwise Convergence)

We are guaranteed to have pointwise convergence on a subsequence if the domain $K$ is countable.

Proof: WLOG, let us assume that all the functions in the sequence are distinct. Since $K$ is countable, we can write

$$K=\{x_1,x_2,x_3,\dots \}$$

For $x_1$, since $\{f_n(x_1)\}$ is bounded by Bolzano-Weierstrass, then we have the subsequence

$$S_1:=f_{1,1},f_{1,2},f_{1,3},\cdots$$

converges. Now we evaluate on $x_2$. Using Bolzano-Weierstrass again, , we have the following subsequence

$$S_2:=f_{2,1},f_{2,2},f_{2,3},\cdots$$

which also converges for both $x_1,x_2$. By repeating this theorem for every element in $K$ (which we can do, since it is countable), then by Cantor’s Diagonal Argument, we have a pointwise convergent subsequence $S_{\infty }:=f_{k,k}$ for every point in $K$.

I believe we need diagonalization for a couple reasons.

1. Since for each time we use Bolzano-Weierstrass, there is a chance we may remove some elements that ensure a previous point (if we were at the $n$th step, but we accidentally made it no longer pointwise convergent on $n/2$th step) is still point wise convergent.
2. It ensures that this subsequence is convergent for all $x_i\in K$. Indeed, since $S_1$ ensures $f_{k,k}(x_1)$ is convergent, $S_2$ ensures $f_{k,k}(x_2)$ is convergent, and so on.