Mathematical Analysis

Leture 10

5.3 Uniform continuity

Definition 5.3.1. Let $S \subseteq \R$, and let $f \colon S \to \R$ be a function. Suppose for every $\epsilon \gt 0$ there exists a $\delta > 0$ such that whenever $x, c \in S$ and $\abs{x-c} \lt \delta$, then $\abs{\,f(x)-f(c)} \lt \epsilon$. Then we say $f$ is uniformly continuous.






5.3 Uniform continuity

Uniform vs Nonuniform continuity




5.3 Uniform continuity

Uniform vs Nonuniform continuity




Example 5.3.1.

$f \colon [0,1] \to \R$ defined by $f(x) := x^2$ is uniformly continuous.

Proof. Note that $0 \leq x,c \leq 1$. Then

$\sabs{x^2-c^2}$ $= \sabs{x+c}\sabs{x-c}\qquad \qquad \qquad \qquad \qquad $
$ \quad\leq \bigl(\sabs{x}+\sabs{c}\bigr) \sabs{x-c}$ $\leq (1+1)\sabs{x-c} .$

Therefore, given $\epsilon > 0$, let $\delta := \dfrac{\epsilon}{2}$. If $\sabs{x-c} \lt \delta$, then $\sabs{x^2-c^2} \lt \epsilon$.


Example 5.3.1.

On the other hand, $g \colon \R \to \R$ defined by $g(x):=x^2$ is not uniformly continuous.

Proof. Suppose it is uniformly continuous, then for every $\epsilon > 0$, there would exist a $\delta > 0$ such that if $\sabs{x-c} \lt \delta$, then $\sabs{x^2 -c^2} \lt \epsilon$. So consider $\epsilon =1.$ If such $\delta$ existed and $c = x+ \delta/2$ $\;\Ra \ds\sabs{x^2 - \left(x+ \frac{\delta}{2}\right)^2} \lt 1.$

However $ \;\ds \abs{x\delta + \dfrac{\delta^2}{4}} \lt 1$ ❗

which is a contradiction, since we can choose $x$ large. $\;\bs$


Continuous functions

Comparison between continuity and uniform continuity

Let $S \subseteq \R$ and let $f \colon S \to \R$ be a function.

Continuous function

We say that
$f$ is continuous:

if for every $c\in S$ and every $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that whenever $x \in S$ and $\abs{x-c} \lt \delta$,
then $\abs{\,f(x)-f(c)} \lt \epsilon.$

Uniform continuous

We say that
$f$ is uniformly continuous:

if for every $\epsilon \gt 0$
there exists a $\delta > 0$ such that
whenever $x, c \in S$ and $\abs{x-c} \lt \delta$,
then $\abs{\,f(x)-f(c)} \lt \epsilon.$



Example 5.3.2.

The function $f \colon (0,1) \to \R$ defined by $f(x) := \dfrac{1}{x}$ is not uniformly continuous.

Proof. Choose $\epsilon=2$ and $0 \lt \delta.$ Set $\delta_0 = \min\{\delta/2, 1/4\}$, $x= \delta_0$, and $y = 2 \delta_0.$ Then $x,y\in(0,1)$ and $\abs{x-y}= \delta_0 \lt \delta$ but

$\ds\abs{\frac{1}{x}-\frac{1}{y}}$ $=\ds\abs{\frac{y-x}{xy}} $ $ \ds= \abs{\frac{\delta_0}{2\delta_0^2} }$ $=\ds\abs{ \frac{1}{2\delta_0}}$ $\geq 2 $ $= \epsilon. $

Hence $f$ is not uniformly continuous. $\; \bs$



5.3 Uniform continuity

Nonuniform continuity criterion

Theorem 5.3.1. Let $S\subseteq \R$ and let $f \colon S \to \R$. Then the following statements are equivalent:

  1. $f$ is not uniformly continuous on $S$.
  2. There exists an $\epsilon_0>0$ such that for every $\delta>0$ there are points $x_{\delta}$, $y_{\delta}$ in $S$ such that $|x_{\delta} - y_{\delta}|\lt \delta$ and $|\,f(x_{\delta}) - f(y_{\delta})|\geq\epsilon_0$.
  3. There exists an $\epsilon_0\gt 0$ and two sequences $\{x_n\}$ and $\{y_n\}$ in $S$ such that $\lim (x_n-y_n)=0$ and $|\,f(x_{n}) - f(y_{n})|\geq\epsilon_0$ for all $n\in \N$.


5.3 Uniform continuity

Nonuniform continuity criterion

Example 5.3.3. We can apply the previous result to show that $f(x)=\dfrac{1}{x}$ is not uniformly continuous on $(0,\infty)$.

Consider $x_n=\dfrac{1}{n}$ and $y_n=\dfrac{1}{n+1}$ in $(0,\infty)$. Then

$ \ds\lim_{n\ra \infty} (x_n-y_n) $ $=\ds\lim_{n\ra \infty} \left(\frac{1}{n}-\frac{1}{n+1}\right) $ $ =0, $

but $\,\abs{\,f(x_n)- f(y_n)}=1\,$ for all $\,n\in \N$.



5.3 Uniform continuity

Theorem 5.3.2. Let $f \colon [a,b] \to \R$ be a continuous function. Then $f$ is uniformly continuous.

👀 Complementary reading 📖

5.3.1 Lipschitz continuous functions

Definition 5.3.2. A function $f \colon S\subseteq \R \to \R$ is Lipschitz continuous, if there exists a $K > 0$, such that \begin{equation*} \abs{\,f(x)-f(y)} \leq K \abs{x-y} \;\;\text{for all } x \text{ and } y \text{ in } S. \end{equation*}





5.3.1 Lipschitz continuous functions

Theorem 5.3.3. A Lipschitz continuous function is uniformly continuous.








5.3.1 Lipschitz continuous functions

Theorem 5.3.3. A Lipschitz continuous function is uniformly continuous.

Proof. Let $f \colon S \to \R$ be a function and let $K$ be a constant such that $\abs{\,f(x)-f(y)} \leq K \abs{x-y}$ for all $x, y$ in $S$. For $\epsilon > 0$ be given. Take $\delta := \dfrac{\epsilon}{K}$. For all $x$ and $y$ in $S$ such that $\abs{x-y} \lt \delta$,

$ \ds\abs{\,f(x)-f(y)} $ $\ds \leq K \abs{x-y} $ $\ds\lt K \delta $ $\ds = K \frac{\epsilon}{K} $ $\ds = \epsilon .$

Hence $f$ is uniformly continuous. $\; \bs$




Example 5.3.4.

Not every uniformly continuous function is Lipschitz continuous.

For example $f(x):=\sqrt{x}$ defined on $I=[0,2]$.

$f$ is uniformly continuous on $I$ but there is no number $K>0$ such that $$|\,f(x)-f(0)|\leq K|x-0|$$ for all $x\in I.$



5.4 Monotone functions and continuity

Definition 5.4.1. Let $S \subseteq \R$. We say $f \colon S \to \R$ is increasing (resp. strictly increasing) if $x,y \in S$ with $x \lt y$ implies $f(x) \leq f(y)$ (resp. $f(x) \lt f(y)$).

We define decreasing and strictly decreasing in the same way by switching the inequalities for $f$.

If a function is either increasing or decreasing, we say it is monotone. If it is strictly increasing or strictly decreasing, we say it is strictly monotone.



5.4 Monotone functions and continuity

One-sided limits for monotone functions are computed by computing infima and suprema.

Let $S \subseteq \R$, $c \in \R$, $f \colon S \to \R$ be increasing, and $g \colon S \to \R$ be decreasing.

If $c$ is a cluster point of $S \cap (-\infty,c)$, then

$\ds \lim_{x \to c^-} f(x) = \sup \{ \,f(x) : x \lt c, x \in S \}$

$\ds \lim_{x \to c^-} g(x) = \inf \{ g(x) : x \lt c, x \in S \} .$



5.4 Monotone functions and continuity

One-sided limits for monotone functions are computed by computing infima and suprema.

Let $S \subseteq \R$, $c \in \R$, $f \colon S \to \R$ be increasing, and $g \colon S \to \R$ be decreasing.

If $c$ is a cluster point of $S \cap (c, \infty)$, then

$\ds \lim_{x \to c^+} f(x) = \inf \{ \,f(x) : x \gt c, x \in S \}$

$\ds \lim_{x \to c^+} g(x) = \sup \{ g(x) : x \gt c, x \in S \} .$



5.4 Monotone functions and continuity

$\ds \lim_{x \to c^+} f(x) = \inf \{ \,f(x) : x \gt c, x \in S \}$


5.4 Monotone functions and continuity

$\ds \lim_{x \to c^-} f(x) = \sup \{ \,f(x) : x \lt c, x \in S \}$


5.4 Monotone functions and continuity

Theorem 5.4.1. If $I \subseteq \R$ is an interval and $f \colon I \to \R$ is monotone and not constant, then $f(I)$ is an interval if and only if $f$ is continuous.


Theorem 5.4.2. Let $I \subseteq \R$ be an interval and $f \colon I \to \R$ be monotone. Then $f$ has at most countably many discontinuities.



5.4 Monotone functions and continuity

Theorem 5.4.3. If $I \subseteq \R$ is an interval and $f \colon I \to \R$ is strictly monotone, then the inverse $f^{-1} \colon f(I) \to I$ is continuous.








Example 5.4.1.

Theorem 5.4.3 does not require $\,f\,$ itself to be continuous.

Let $f \colon \R \to \R$ be defined by $\ds f(x):= \begin{cases} x & \text{if } x \lt 0, \\ x+1 & \text{if } x \geq 0. \\ \end{cases} $
The function $f$ is not continuous at $0$. The image of $I = \R$ is the set $(-\infty,0)\cup [1,\infty)$, not an interval. Then $f^{-1} \colon (-\infty,0)\cup [1,\infty) \to \R$ can be written as \begin{equation*} f^{-1}(y) = \begin{cases} y & \text{if } y \lt 0, \\ y-1 & \text{if } y \geq 1. \end{cases} \end{equation*} It is not difficult to see that $f^{-1}$ is a continuous function.


Example 5.4.1.

$\ds f(x):= \begin{cases} x & \text{if } x \lt 0, \\ x+1 & \text{if } x \geq 0. \\ \end{cases} $
$\ds f^{-1}(y) = \begin{cases} y & \text{if } y \lt 0, \\ y-1 & \text{if } y \geq 1. \end{cases} $


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