# Uniform Continuity

## Definition 4.3.1

Let $E$ be a subset of $\mathbb{R}$ and let $f:E\to\mathbb{R}$ be a function.
The function $f$ is uniformly continuous on $E$ if
$\forall \epsilon >0, \exists \delta >0$ such that \begin{align} |x-y|<\delta \text{ for all }x,y\in E \Rightarrow |f(x)-f(y)|<\epsilon. \end{align}

## Examples 4.3.2

(a) $E\subset \mathbb{R}$ is bounded and $f(x) = x^2 \Rightarrow f$ is uniformly continuous on $E$.

<proof> Since $E$ is bounded, there is $C>0$ such that $|x|\leq C$ for all $x\in E$.
Let $\epsilon >0$ be given. Then, if $x,y \in E$, \begin{align} \begin{split} |f(x)-f(y)| = |x^2 - y^2| &= |x+y|\cdot |x-y| \\ &\leq (|x|+|y|)\cdot |x-y| \\ &\leq 2C \cdot |x-y| \end{split} \end{align} Take $\delta := \epsilon / 2 >0$. If $|x-y| <\delta$ for all $x,y\in E$, then \begin{align} |f(x)-f(y)| \leq 2C|x-y| < 2C \cdot \frac{\epsilon}{2C} = \epsilon \end{align} $\tag*{\square}$

(b) $f(x) = \sin x$ is uniformly continuous.

<proof> \begin{align} \begin{split} |f(x) - f(y) | &= \left| 2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2}) \right| \\ &\leq 2|\sin(\frac{x-y}{2}) | \\ &\leq |x-y| \end{split} \end{align} Take $\delta :=\epsilon$, then we are done. $\tag*{\square}$

(c) $f(x) = \frac{1}{x}$. If $E = (0,\infty)$, then $f$ is not uniformly continuous on $E$. But if $E=[a,\infty] \text{for all }a>0$, $f$ is uniformly continuous on $E$.

<proof> Let $E=(0,\infty)$ and suppose that $f$ is uniformly continuous on $E$. Taking $\epsilon=1$, there exists a $\delta >0$ such that \begin{align} \text{for all } x,y \in E \text{ with } |x-y|<\delta \Rightarrow \left|\frac{1}{x}-\frac{1}{y}\right| < 1 \end{align} Choose $n_0 \in \mathbb{N}$ such that $\frac{1}{n_0}<\delta$ by Archimedean property and set a sequence $x_n = \frac{1}{n}$. Then \begin{align} |x_n-x_{n+1}| = \frac{1}{n}-\frac{1}{n+1} = \frac{1}{n} \leq \frac{1}{n_0} < \delta \end{align} whenever $n\geq n_0$. But $|f(x_n) -f(x_{n+1}) | = |n - (n+1)|=1$, which is a contradiction.

$\therefore f$ is not uniformly continuous on $E$.

Now, let $E=[a,\infty)$ with $a>0$ and $\epsilon >0$ be given. For $x,y \in E$, \begin{align} |f(x)-f(y) | = \left|\frac{1}{x}-\frac{1}{y} \right| = \left|\frac{x-y}{xy} \right| = \frac{1}{a^2}|x-y| \end{align} Take $\delta := a^2\epsilon >0$. If $|x-y| <\delta$ for all $x,y\in E$, then \begin{align} |f(x) - f(y) | \leq \frac{1}{a^2}|x-y| < \frac{1}{a^2}\cdot a^2\epsilon \end{align} $\therefore f$ is uniformly continuous on $E$. $\tag*{\square}$

## Definition (Lipschitz continuity)

Let $E\subset \mathbb{R}$ and let $f:E\to \mathbb{R}$ be a function satisfying a Lipschitz condition
if there is a $M>0$ such that \begin{align} x,y\in E \Rightarrow |f(x)- f(y)| \leq M |x-y| \end{align}

## Theorem 4.3.3

Let $f:E\to\mathbb{R}$ be a function satisfying Lipschitz condition on $E$. Then $f$ is uniformly continuous on $E$.

<proof> Since $f$ is Lipschitz function, there is a $M>0$ such that $x,y \in E \Rightarrow |f(x)-f(y)| < M|x-y|$ Let $\epsilon >0$ be given. Take $\delta := \epsilon / M$. Then for $x,y\in E$, $|f(x)-f(y)| \leq M|x-y| < M\cdot\frac{\epsilon}{M} = \epsilon$ $\therefore f$ is uniformly continuous on $E$. $\tag*{\square}$

## Theorem 4.3.4

Let $K\subset\mathbb{R}$ be a compact set and let $f:K\to\mathbb{R}$ be a continuous function on $K$. Then $f$ is uniformly continuous on $K$.

<proof> Let $\epsilon >0$ be given. Since $f$ is continuous on $K$, for every $p\in K$, there is a $\delta_p >0$ such that \begin{align} |x-p|<\delta_p \Rightarrow |f(x)-f(p)| <\frac{\epsilon}{2} \end{align} Now, $\{N_{\delta_p/2} (p): p\in K\}$ is an open cover of $K$. Since $K$ is compact, there are $p_1,\ldots, p_n \in K$ such that \begin{align} K \subset \bigcup_{i=1}^n N_{\delta_{p_i}/2} (p_i) \end{align} Take $\delta :=\min\{\delta_{p_1}/2,\ldots, \delta_{p_n}/2\} >0$. Suppose that $x,y\in K$with $|x-y|<\delta$. Then $x\in N_{\delta_{p_k}/2} (p_k)$ for some $1 \leq k\leq n$.

\begin{align} \begin{split} |p_k -y| &< |p_k -x| + |x-y| \\ &<\frac{\delta_{p_k}}{2} + \delta \\ &<\frac{\delta_{p_k}}{2} + \frac{\delta_{p_k}}{2} \\ &= \delta_{p_k} \end{split} \end{align}

That is $|x-y|<\delta < \delta_{p_i}$. So, $x,y\in N_{\delta_{p_i}}(p_i)$. Thus, \begin{align} |f(x) - f(y)| \leq |f(x) - f(p_i)| + |f(y)-f(p_i)| < \frac{\epsilon}{2} +\frac{\epsilon}{2} =\epsilon \end{align}

$\therefore f$ is uniformly continuous on $K$. $\tag*{\square}$

## Example

$A\subset \mathbb{R}$. Define $d(p,A) :=\inf\{|p-x|:x\in A\}$. Then, \begin{align} |d(x, A) - d(y, A)| \leq |x-y| \end{align} So, $d(\cdot, A)$ is Lipschitz function.

## Definition 4.4.1

$E\subset \mathbb{R}, f:E\to\mathbb{R}, p$ is a limit point of $E\cap (p,\infty)$. $f$ has a right limit at $p$ if there exists a $L\in\mathbb{R}$ such that $\forall \epsilon >0, \exists \delta >0$ such that \begin{align} p<x<p+\delta, x\in E \Rightarrow |f(x)-L| <\epsilon \end{align} We write \begin{align} f(p+) = \displaystyle{\lim_{x\to p+}f(x) = \lim_{x\to p, x> p}f(x)}. \end{align} Similarly, if $p$ is a limit point of $E\cap (-\infty, p)$, the left limit of $f$ at $p$, if it exists, is denoted by $f(p-)$, and we write \begin{align} f(p-) = \lim_{x\to p}f(x) = \lim_{x\to p, x<p} f(x). \end{align}

## Remark

\begin{align} \lim_{x\to p} f(x) = L \iff f(p+), f(p-) \text{ exist and } f(p+) = f(p-) = L \end{align} But $p\in E^\prime$ does not imply $p$ is a limit point of $E\cap (p, \infty)$ or $E\cap (-\infty, p)$. We can find an interval $I \subset E$ with $p\in I$. If $p\in \text{Int}(I)\neq \emptyset, N_{\delta_0} (p) \subset I$ for some $\delta_0 >0$.
Then for every $\delta >0, N^\prime_{\delta_0} (p) \cap N^\prime_\delta (p) \neq \emptyset$. Since $N^\prime_{\delta_0} (p) \cap (p,\infty) \neq \emptyset, N^\prime_{\delta_0} (p) \cap(I \cap (p,\infty)) \neq \emptyset.$
Thus, $p$ is a limit point of $I\cap (p,\infty)$. Similarly, $p$ is a limit point of $I\cap (-\infty,p)$.

If $p$ is the left end point of $I$, then the right limit of $f$ at $p$ coincides with the the limit of $f$ at $p$. Similar for $p$ is the right end point of $I$.

## Definition 4.4.2

$E\subset \mathbb{R}, f:E\to\mathbb{R}, p\in E$. The function $f$ is right continuous (left continuous) at $p$
if $\forall \epsilon >0, \exists \delta>0$ such that \begin{align} p\leq x <p+\delta (p-\delta < x \leq p) \Rightarrow |f(x)-f(p) |< \epsilon. \end{align}

## Definition 4.4.4

$f$ has a jump discontinuity at $p$ if $f(p+), f(p-)$ exist but $f(p+) \neq f(p-)$.

## Reference

• Manfred Stoll, Introduction to Real Analysis, Pearson

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