MATH 409 Lecture 12

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Lecture Slides

Uniform Continuity

A function $f:E\to \mathbb {R}$ defined on a set $E\subset \mathbb {R}$ is called uniformly continuous on $E$ if for every $e>0$ there exists $\delta =\delta (\epsilon )>0$ such that $\left|x-y\right|<\delta$ and $x,y\in E$ imply $\left|f(x)-f(y)\right|<\epsilon$.

First introduced by Cauchy with all his work on $\epsilon$'s and $\delta$'s.

Recall that $f$ is continuous at a point $y\in E$ if for every $\epsilon >0$ there exists $\delta =\delta (y,\epsilon )>0$ such that $\left|x-y\right|<\delta$ and $x\in E$ imply $\left|f(x)-f(y)\right|<\epsilon$.

Therefore the uniform continuity of $f$ is a stronger property than the continuity of $f$ on $E$.

Examples

Constant function. $f(x)=a$ is uniformly continuous

Indeed $\left|f(x)-f(y)\right|=0<\epsilon$ for any $\epsilon >0$ and $x,y\in \mathbb {R}$.

Identity function. $f(x)=x$

Since $f(x)-f(y)=x-y$, we choose $\delta =\epsilon$ and we have $\left|f(x)-f(y)\right|<\epsilon$ and $\left|x-y\right|<\epsilon$.

Sine function. $f(x)=\sin {x}$

It was shown in the previous lecture that $\left|\sin {x}-\sin {y}\right|\leq \left|x-y\right|$ for all $x,y\in \mathbb {R}$, therefore $\left|f(x)-f(y)\right|<\epsilon$ whenever $\left|x-y\right|<\epsilon$

Lipschitz Functions

A function $f:E\to \mathbb {R}$ is called a Lipschitz function if there exists a constant $L>0$ such that $\left|f(x)-f(y)\right|\leq L\left|x-y\right|$ for all $x,y\in E$.

Theorem. Any Lipschitz function is uniformly continuous.

Proof. Using notatino of the the definition, let $\delta =\epsilon /L$ for any $\epsilon >0$, then $\left|x-y\right|<\delta$ implies

$\left|f(x)-f(y)\right|\leq L\left|x-y\right|<L\delta =\epsilon \,\!$

for all $x,y\in E$.

quod erat demonstrandum

This is nice because we don't have to work with $\epsilon$'s and $\delta$'s to show that a function is uniformly continuous.

Theorem. The function $f(x)={\sqrt {x}}$ is uniformly continuous on $[0,\infty )$, but not Lipschitz.

Proof. For any $n\in \mathbb {N}$, we have $\left|f\left({\frac {1}{n}}\right)-f(0)\right|={\sqrt {\frac {1}{n}}}={\sqrt {n}}\,\left|{\frac {1}{n}}-0\right|$. It follows that $f$ is not Lipschitz.

Given $\epsilon >0$, let $\delta =\epsilon ^{2}$ In order to change the function by $\epsilon$, we can only choose a difference in $x$ and $y$ smaller than $\epsilon ^{2}<\epsilon$.

Suppose $\left|x-y\right|<\delta$, where $x,y\geq 0$. To estimate $\left|f(x)-f(y)\right|$, we consider two cases:

1. $x,y\in [0,\delta )$: we use the fact that $f$ is strictly increasing. Then $\left|f(x)-f(y)\right|<f(\delta )-f(0)={\sqrt {\delta }}=\epsilon$. Since the function ${\sqrt {x}}$ is strictly increasing, the farthest that $f(x)$ and $f(y)$ can be is at the extremes of an interval. Note that $\delta$ is not included in this interval, hence the strict "less than".
2. $x,y\not \in [0,\delta )$: we have $\delta \leq \max(x,y)$. Then $\left|{\sqrt {x}}-{\sqrt {y}}\right|=\left|{\frac {x-y}{{\sqrt {x}}+{\sqrt {y}}}}\right|\leq {\frac {\left|x-y\right|}{\sqrt {\max(x,y)}}}<{\frac {\delta }{\sqrt {\delta }}}={\sqrt {\delta }}=\epsilon$.

Thus $f$ is uniformly continuous.

quod erat demonstrandum

Theorem. $f(x)=x^{2}$ is not uniformly continuous on $\mathbb {R}$.

Let $\epsilon =2$ and an arbitrary $\delta >0$. Let $n_{\delta }$ be a natural number such that ${\frac {1}{n_{\delta }}}<\delta$. Further, let $x_{\delta }=n_{\delta }+{\frac {1}{n_{\delta }}}$ and $y_{\delta }=n_{\delta }$.

(The numbers are large, but the distance between them is small)

Then $\left|x_{\delta }-y_{\delta }\right|={\frac {1}{n_{\delta }}}<\delta$ while $f(x_{\delta })-f(y_{\delta })=\left(n_{\delta }+{\frac {1}{n_{\delta }}}\right)^{2}-{n_{\delta }}^{2}=2+{\frac {1}{{n_{\delta }}^{2}}}>\epsilon$.

We conclude that $f$ is not uniformly continuous.

quod erat demonstrandum

Theorem. The function $f(x)=x^{2}$ is Lipschitz (and hence uniformly continuous) on any bounded interval $[a,b]$.

Proof. For any $x,y\in [a,b]$ we obtain

${\begin{aligned}\left|x^{2}-y^{2}\right|&=\left|x+y\right|\,\left|x-y\right|\\&\leq \left(|x|+|y|\right)\,\left|x-y\right|\\&\leq 2\,\max(|a|,|b|)\,\left|x-y\right|\end{aligned}}$

quod erat demonstrandum

Theorem. Any function continuous on a closed bounded interval $[a,b]$ is also uniformly continuous on $[a,b]$.

Proof by contraposition. Assume that $f:[a,b]\to \mathbb {R}$ is not uniformly continuous on $[a,b]$. We have to show that $f$ is not continuous on $[a,b]$,

By assumption, there exists a $\epsilon >0$ tsuch that for any $\delta >0$, we can find two points $x,y\in [a,b]$ satisfying $\left|x-y\right|<\delta$ and $\left|f(x)-f(y)\right|\geq \epsilon$. In particular, for any $n\in \mathbb {N}$, there exist points $x_{n},y_{n}\in [a,b]$ such that $\left|x_{n}-y_{n}\right|<{\frac {1}{n}}$ while $\left|f(x_{n})-f(y_{n})\right|\geq \epsilon$.

By construction, $\left\{x_{n}\right\}$ is a bounded sequence. According to the Bolzano-Weierstrass theorem, there is a subsequence $\left\{x_{n_{k}}\right\}$ converging to a limit $c$. Moreover, $c$ belongs to $[a,b]$ as $\left\{x_{n}\right\}\subset [a,b]$.

Since $x_{n}-{\frac {1}{n}}<y_{n}<x_{n}+{\frac {1}{n}}$ for all $n\in \mathbb {N}$, the subsequence $\left\{y_{n_{k}}\right\}$ also converges to $c$. However, the inequalities $\left|f(x_{n_{k}})-f(x_{n_{k}})\right|\geq \epsilon$ imply that at least one of the sequences $\left\{f(x_{n_{k}})\right\}$ and $\left\{f(y_{n_{k}})\right\}$ is not converging to $f(c)$. It follows that the function $f$ is not continuous at $c$.

quod erat demonstrandum

Theorem. Suppose that a function $f:E\to \mathbb {R}$ is uniformly continuous on $E$. Then it maps Cauchy sequences to Cauchy sequences, that is, for any Cauchy sequence $\left\{x_{n}\right\}\subset E$, the sequence $\left\{f(x_{n})\right\}$ is also Cauchy.

Proof. Let $\left\{x_{n}\right\}\subset E$ be a Cauchy sequence. Since $f$ is uniformly continuous on $E$, for every $\epsilon >0$ there exists $\delta =\delta (\epsilon )$ such that $\left|x-y\right|<\delta$ and $x,y\in E$ imply $\left|f(x)-f(y)\right|<\epsilon$.

Since $\left\{x_{n}\right\}$ is a Cauchy sequence, there exists $N=N(\delta )\in \mathbb {N}$ (note that $\delta$ depends on $\epsilon$, so ultimately $N$ depends on $\epsilon$) such that $\left|x_{n}-x_{m}\right|<\delta$ for all $m,n\geq N$. Then $\left|f(x_{n})-f(x_{m})\right|<\epsilon$ for all $n,m\geq N$. We conclude that $\left\{f(x_{n})\right\}$ is a Cauchy sequence.

quod erat demonstrandum

Cauchy sequences are nice because every Cauchy sequence is a convergent sequence (and vice versa), so the image of a convergent sequence is also a convergent sequence that converges to the limit of the function.

Dense Subsets

Given a set $E\subset \mathbb {R}$ and its subset $E_{0}\subset E$, we say that $E_{0}$ is dense in $E$ if for any point $x\in E$ and any $\epsilon >0$ the interval $\left(x-\epsilon ,x+\epsilon \right)$ contains an element of $E_{0}$.

Examples

• An open bounded interval $(a,b)$ is dense in the closed interval $[a,b]$.
• The set $\mathbb {Q}$ of rational numbers is dense in $\mathbb {R}$ (previously shown without notion of density.

Theorem. A subset $E_{0}$ of the set $E\subset \mathbb {R}$ is dense in $E$ if and only if for any $c\in E$ there exists a sequence $\left\{x_{n}\right\}\subset E_{0}$ converging to $c$.

Proof. Suppose that for any point $c\in E$ there is a sequence $\left\{x_{n}\right\}\subset E_{0}$ converging to $c$. Then any $\epsilon$-neighborhood $(c-\epsilon ,c+\epsilon )$ of $c$ contains an element of that sequence.

Conversely, suppose that $E_{0}$ is dense in $E$. Then, given $c\in E$, for any $n\in \mathbb {N}$ there is a point $x_{n}\in \left(c-{\frac {1}{n}},c+{\frac {1}{n}}\right)\cap E_{0}$. Clearly $x_{n}\to c$ as $x\to \infty$.

quod erat demonstrandum

Continuous Extension

Main function of the day.

Theorem. Suppose that $E_{0}\subset E\subset \mathbb {R}$ is dense in $E$. Then any uniformly continuous function $f:E_{0}\to \mathbb {R}$ can be extended to a continuous function on $E$. Moreover, the extension is unique and uniformly continuous.

Proof. First let us show that a continuous extension of the function $f$ to the set $E$ is unique (assuming it exists). Suppose $g,h:E\to \mathbb {R}$ are two continuous extensions of $f$. Since the set $E_{0}$ is dense in $E$, for any $c\in E$ there is a sequence $\left\{x_{n}\right\}\subset E_{0}$ converging to $c$. Since $g$ and $h$ are continuous at $c$, we get $g(x_{n})\to g(c)$ and $h(x_{n})\to h(c)$ as $n\to \infty$ (by sequential characterization of continuity). However $g(x_{n})=h(x_{n})=f(x_{n})$ for all $n\in \mathbb {N}$. Hence $g(c)=h(c)$. This shows the uniqueness of the extension.

Now to prove existence. given $c\in E$, let $\left\{x_{n}\right\}$ be a sequence of elements of $E_{0}$ converging to $c$. The sequence $\left\{x_{n}\right\}$ is Cauchy. Since $f$ is uniformly continuous, it follows that the sequence $\left\{f(x_{n})\right\}$ is also Cauchy. Hence it converges to a limit $L$. We claim that the limit $L$ depends only on $c$ and does not depend on the choice of the sequence $\left\{x_{n}\right\}$.

Indeed, let $\left\{{\tilde {x}}_{n}\right\}\subset E_{0}$ be another sequence converging to $c$. Then a sequence $x_{1},{\tilde {x}}_{1},x_{2},{\tilde {x}}_{2},\ldots$ also converges to $c$. Consequently, $f(x_{1}),f({\tilde {x}}_{1}),f(x_{2}),f({\tilde {x}}_{2}),\ldots$ is convergent. The limit is $L$ since the subsequence $\left\{f(x_{n})\right\}$ converges to $L$. Another subsequence is $\left\{f({\tilde {x}}_{n})\right\}$, hence it converges to $L$ as well. Now we set $F(c)=L$, which defines a function $F:E\to \mathbb {R}$.

The continuity of the function $f$ implies that $F(c)=f(c)$ for $c\in E_{0}$, that is, $F$ is an extension of $f$.

Finally, we need to show that the extension $F$ is uniformly continuous.

Given $\epsilon >0$, let $\epsilon _{0}={\frac {\epsilon }{2}}$. Since $f$ is uniformly continuous, there is $\delta >0$ such that $\left|x-y\right|<\delta$ implies $\left|f(x)-f(y)\right|<\epsilon _{0}$ for all $x,y\in E_{0}$. For any $c,d\in E$ we can find sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ of elements of $E_{0}$ such that $x_{n}\to c$ and $y_{n}\to d$ as $n\to \infty$.

By construction of $F$, we have $f(x_{n})\to F(c)$ and $f(y_{n})\to F(d)$ as $n\to \infty$.

If $\left|c-d\right|<\delta$, then $\left|x_{n}-y_{n}\right|<\delta$ for all sufficiently large $n$. Consequently, $\left|f(x_{n})-f(y_{n})\right|<\epsilon _{0}$ for all sufficiently large $n$, which implies $\left|F(c)-F(d)\right|\leq \epsilon _{0}$ (by comparison theorem of taking limit of inequality). Thus $\left|F(c)-F(d)\right|\leq \epsilon _{0}<\epsilon$ and $F$ is uniformly continuous.

quod erat demonstrandum

The implication of this theorem is the definition of the exponential function. We will define it for rational exponents and then extend it to the real line.

Exponential Functions

Theorem. For any $a>0$, there exists a unique function $F_{a}:\mathbb {R} \to \mathbb {R}$ satisfying the following conditions:

1. $F_{a}(1)=a$
2. $F_{a}(x+y)=F_{a}(x)\,F_{a}(y)$ for all $x,y\in \mathbb {R}$
3. $F_{a}$ is continuous at $0$.

This function is denoted $F_{a}(x)=a^{x}$ and called the exponential function with base $a$.

Sketch of proof of uniqueness. We know how to prove this function is not negative: at any point $c$. Let's take $x={\frac {c}{2}}$ and $y={\frac {c}{2}}$. Thus $f(x+y)=f(c)=F(c/2)^{2}>0$

(nonzero)

If it's continuous at 0 and strictly positive, we can show it must be continuous at any point.

Sketch of proof of existence. Let $a^{0}=1$, $a^{1}=a$, and $a^{n+1}=a^{n}\,a$ for all $n\in \mathbb {N}$. Further, let $a^{-n}={\frac {1}{a^{n}}}$ for all $n\in \mathbb {N}$.

Lemma 1. a^{m+n} = a^m \, a^n and $a^{m\,n}=\left(a^{m}\right)^{n}$ for all $m,n\in \mathbb {Z}$.

Lemma 2. If $m_{1},m_{2}\in \mathbb {Z}$ and $n_{1},n_{2}\in \mathbb {N}$ satisfy ${\frac {m_{1}}{n_{1}}}={\frac {m_{2}}{n_{2}}}$, then ${\sqrt[{n_{1}}]{a^{m_{1}}}}={\sqrt[{n_{2}}]{a^{m_{2}}}}$.

For any $r\in \mathbb {Q}$, let $a^{r}={\sqrt[{n}]{a^{m}}}$, where $m\in \mathbb {Z}$ and $n\in \mathbb {N}$ are chosen so that $r={\frac {m}{n}}$.

Lemma 3. $a^{r+s}=a^{r}\,a^{s}$ and $a^{r\,s}=\left(a^{r}\right)^{s}$ for all $r,s\in \mathbb {Q}$.

Lemma 4. The function $f(r)=a^{r}$, $r\in \mathbb {Q}$, is monotone.

Lemma 5. $a^{\frac {1}{n}}\to 1$ as $n\to \infty$.

Lemma 6. The function $f(r)=a^{r}$ for $r\in \mathbb {Q}$ is uniformly continuous on $\left[b_{1},b_{2}\right]\cap \mathbb {Q}$ for any bounded interval $\left[b_{1},b_{2}\right]$.

quod erat demonstrandum