Pointwise and Uniform Convergence

Section 7.1 Pointwise and Uniform Convergence

Definition 7.1.

Let \(X\) be a set and let \(Y\) be a metric space. For functions \((f_n:X\rightarrow Y)_{n=1}^\infty\) and \(f:X\rightarrow Y\text{,}\) we say

\(f_n\) converges to \(f\) point-wise if for all \(\varepsilon>0\) and all \(x\in X\) there exists \(N\geq 1\) such that for all \(n\geq N\text{:}\)

\begin{equation*} d_Y(f(x),f_n(x))<\varepsilon. \end{equation*}
\(f_n\) converges to \(f\) uniformly if for all \(\varepsilon>0\) there exists \(N\geq 1\) such that fr all \(x\in X\) and \(n\geq N\text{:}\)

\begin{equation*} d_Y(f(x),f_n(x))<\varepsilon. \end{equation*}

In def of uniform convergence, \(N\) does not depend on \(X\text{.}\)

Exploration 7.2.

If \(f_n\rightarrow f\) uniformly, then \(f_n\rightarrow f\) point-wise.

A point-wise limit of continuous functions need not be continuous.

Example 7.3.

Let

\begin{equation*} \begin{aligned} f_n:[0,1]&\rightarrow\mathbb{R}\\ f_n(x) &=\begin{cases} 1-nx & 0\leq x\leq \frac{1}{n}\\ 0 & \frac{1}{n}<x\leq 1 \end{cases};\\ f:[0,1]&\rightarrow\mathbb{R}\\ f(x) &=\begin{cases} 1 & x=0\\ 0 &0<x\leq 1 \end{cases} \end{aligned} \end{equation*}

Solution.

Get \(\lim_{n\rightarrow\infty}f_n(0)=1=f(0)\) and if \(x\in(0,1]\) then for all \(n\in\mathbb{Z}_+\) with \(\frac{1}{n}<x\text{,}\) we have \(f_n(x)=0\text{.}\) So \(\lim_{n\rightarrow\infty}f_n(x)=0=f(x)\text{.}\) Then \(f_n\rightarrow f\) pointwise and \(f\) is not continuous.

Example 7.4.

Another example: \(g,g_n:[0,1]\rightarrow\mathbb{R}\text{,}\)

\begin{equation*} \begin{aligned} g_n(x)&=x^{\frac{1}{n}}\\ g(x)&=\begin{cases} 0 & x=0\\ 1 & 0<x\le 1. \end{cases} \end{aligned} \end{equation*}

Pointwise limits of functions often don't behave well with integrals.

Example 7.5.

Let \(f_n:[0,1]\rightarrow\mathbb{R}\) by \(f_n(x)=\begin{cases} n & 0<x<\frac{1}{n}\\ 0 & \text{otherwise} \end{cases}.\) Then

\begin{equation*} \lim_{n\rightarrow \infty}f_n(x)=0 \end{equation*}

for all \(x\in[0,1]\) and

\begin{equation*} \int_0^1f_n(x)\,dx=1 \end{equation*}

for all \(n\geq1\) so

\begin{equation*} 1=\lim_{n\rightarrow\infty}\int_0^1f_n(x)\,dx\neq \int_0^1\lim_{n\rightarrow\infty}f_n(x)\,dx=0. \end{equation*}

Theorem 7.6.

If \(X,Y\) are metric spaces and \((f_n:X\rightarrow Y)_{n=1}^\infty\) is a sequence of continuous functions, \(f:X\rightarrow Y\) is a function, and \(f_n\rightarrow f\) uniformly, then \(f\) is continuous. Further, if each \(f_n\) is uniformly continuous, then so is \(f\text{.}\)

Convention 7.7.

For a set \(X\text{,}\) let \(\mathcal{B}(X)\)\(=\{{\text{bounded functions }X\rightarrow\mathbb{C}}\}\text{.}\) (If necessary, we'll use \(\mathcal{B}(X,\mathbb{C})\) or \(\mathcal{B}(X,\mathbb{R})\) when it is important whether codomain is \(\mathbb{C}\) or \(\mathbb{R}\))

Definition 7.8.

Define \(\left|{\left|{{\cdot}}\right|}\right|_\infty\)\(:\mathcal{B}(X)\rightarrow[0,\infty)\) by \(\left|{\left|{{f}}\right|}\right|_\infty=\sup\{{\left|{f(x)}\right|}\,:\,{x\in X}\}\text{.}\) This is called the \(\infty\)-norm of \(f\) (aka supremum norm or uniform norm

Convention 7.9.

For \(f\in\mathcal{C}[a,b]\) (or more generally) and \(1\leq p<\infty\text{,}\) can define \(\left|{\left|{{f}}\right|}\right|_p\)\(=\left({\int_a^b\left|{f(x)}\right|^p\,dx}\right)^{1/p}\text{.}\) It turns out that \(\lim_{p\rightarrow\infty}\left|{\left|{{f}}\right|}\right|_p=\left|{\left|{{f}}\right|}\right|_\infty\) (this is reason for notation; \(\left|{\left|{{\cdot}}\right|}\right|_p\) makes sense for \(0<p<1\text{,}\) but it doesn't behave well)

Theorem 7.10.

\(\mathcal{B}(X)\) is a vector space
For \(f,g\in\mathcal{B}(X),\alpha\in\mathbb{C}\text{,}\)
1. \(\displaystyle \left|{\left|{{f}}\right|}\right|_\infty=0\iff f=0\)
2. \(\displaystyle \left|{\left|{{\alpha f}}\right|}\right|_\infty=\left|{\alpha}\right|\left|{\left|{{f}}\right|}\right|_\infty\)
3. \(\left|{\left|{{f+g}}\right|}\right|_\infty\leq \left|{\left|{{f}}\right|}\right|_\infty+\left|{\left|{{g}}\right|}\right|_\infty\) (triangle inequality)
4. \(\displaystyle \left|{\left|{{\overline{f}}}\right|}\right|_\infty=\left|{\left|{{f}}\right|}\right|_\infty\)
5. \(\left|{\left|{{fg}}\right|}\right|_\infty\leq \left|{\left|{{f}}\right|}\right|_\infty\left|{\left|{{g}}\right|}\right|_\infty\) (since \(\left|{fg(x)}\right|=\left|{f(x)}\right|\left|{g(x)}\right|\leq \left|{\left|{{f}}\right|}\right|_\infty\left|{\left|{{g}}\right|}\right|_\infty \forall x\in X\text{.}\))
The function \(d_\infty:\mathcal{B}(x)\times \mathcal{B}(x)\rightarrow[0,\infty)\) by \(d_\infty(f,g)=\left|{\left|{{f-g}}\right|}\right|_\infty\) is a metric on \(\mathcal{B}(X)\text{.}\)
if \((f_n)_{n=1}^\infty\subseteq\mathcal{B}(X)\) and \(f\in\mathcal{B}(X)\text{,}\) then \(f_n\rightarrow f\) in the metric \(d_\infty\iff f_n\rightarrow f\) uniformly on \(X\text{.}\) (both conditions mean

\begin{equation*} \lim_{n\rightarrow\infty}d_\infty(f_n,f)=\lim_{n\rightarrow\infty}\sup_{x\in X}\left|{f_n(x)-f(x)}\right|=0) \end{equation*}
Addition, multiplication, and conjugation are continuous on \(\mathcal{B}(X)\text{:}\) ie if \((f_n)_{n=1}^\infty,(g_n)_{n=1}^\infty\subseteq\mathcal{B}(X)\text{,}\) \(f_n\rightarrow f\) and \(g_n\rightarrow g\) uniformly then

\begin{equation*} \begin{aligned} f_n+g_n&\rightarrow f+g\\ f_ng_n&\rightarrow fg\\ \overline{f_n}&\rightarrow \overline{f} \end{aligned} \end{equation*}

all converge uniformly.

Theorem 7.11.

For any set \(X\text{,}\) \((\mathcal{B}(X),d_\infty)\) is complete, so every uniformly Cauchy sequence of bounded function converges uniformly.

In a few cases we have tests to check uniform convergence.

Theorem 7.12. Dini's Theorem.

Suppose \(X\) is a compact metric space \((f_n)_{n=1}^\infty\subseteq C(X,\mathbb{R}),f\in C(X,\mathbb{R})\text{,}\) \(f_n\rightarrow f\) pointwise and \(f_1\leq f_2\leq f_3\leq \dots\) pointwise. Then \(f_n\rightarrow f\) uniformly.

Exploration 7.13.

Dini's theorem holds for decreasing functions as well.

There are increasing sequences of continuous functions \(f_1\leq f_2\leq f_3\leq \dots\) which converge pointwise to a function \(f\text{,}\) but not uniformly. Dini's Thm tells us this can't happen when the limit of \(f\) is continuous though.

Example 7.14.

\(X=[0,1],f_n(x)=x^{1/n},f(x)=\begin{cases} 0 & x=0\\ 1 & x>0 \end{cases}.\)

Example 7.15.

The function \(f:[0,1]\rightarrow\mathbb{R}, f(t)=\sqrt{t}\) is a uniform limit of polynomials.

Mathematical Analysis

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