# Lipschitz Continuous $\Rightarrow$ Uniformly Continuous

The Question: Prove that if a function $f$ defined on $S \subseteq \mathbb R$ is Lipschitz continuous then $f$ is uniformly continuous on $S$.

Definition. A function $f$ defined on a set $S \subseteq \mathbb R$ is said to be Lipschitz continuous on $S$ if there exists an $M$ so that $$\frac{|f(x) - f(c)|}{|x - c|} \le M$$ for all $x$ and $c$ in $S$ such that $x \ne c$.

My Heuristic Interpretation: if $f$ is Lipschitz continuous then the "absolute slope" of $f$ is never unbounded i.e. no asymptotes.

Definition. A continuous function $f$ defined on $\mathrm{Dom}\, (f)$ is said to be uniformly continuous if for each $\varepsilon > 0 \ \exists \ \delta > 0$ s.t. $\forall \ x, c \in \mathrm{Dom}\, (f)$ $$|x - c| \le \delta \ \Rightarrow \ |f(x) - f(c)| \le \varepsilon$$

Proof:

$f$ Lipschitz continuous $\Rightarrow$ $|f(x) - f(c)| \le M|x - c|$. Since we suppose $|x - c| \le \delta$ for uniform continuity, we have $x$ within $\delta$ of $c$, so $|x| \le |c| + \delta$. So taking $\delta = \varepsilon/M$ \begin{align*} |f(x) - f(c)| &\le M|x - c| \\ & \le M\delta \\ & = \varepsilon \end{align*}

My Question: Is my proof valid with the assumptions taken?

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It’s not very well organized, and it has some extraneous clutter, but it also has the core of the argument. You want to show that for each $\epsilon>0$ there is a $\delta>0$ such that $|f(x)-f(c)|<\epsilon$ whenever $x,c\in\operatorname{dom}f$ and $|x-c|<\delta$, so in a polished version of the argument your first step should be:

Suppose that $f$ is Lipschitz continuous on some set $S$ with Lipschitz constant $M$, and fix $\epsilon>0$.

You’ve already worked out that $\epsilon/M$ will work for $\delta$, so you can even start out with:

Fix $\epsilon>0$ and let $\delta=\frac\epsilon{M}$.

Now you want to show that this choice of $\delta$ does the job.

Clearly $\delta>0$. Suppose that $x,c\in S$ and $|x-c|<\delta$. Then by the Lipschitz continuity of $f$ we have $|f(x)-f(c)|\le M|x-c|<M\delta=\epsilon$, so $f$ is uniformly continuous on $S$. $\dashv$

Added: Your heuristic interpretation of Lipschitz continuity is inaccurate enough that it may well lead you astray at some point. Consider the function

$$f(x)=\begin{cases} x\sin\frac1x,&\text{if }x\ne0\\ 0,&\text{if }x=0\;. \end{cases}$$

This function has no vertical asymptotes, but it’s not Lipschitz continuous:

$$\frac{f\left(\frac1{2n\pi}\right)-f\left(\frac1{2n\pi+\frac{\pi}2}\right)}{\frac1{2n\pi}-\frac1{2n\pi+\frac{\pi}2}}=\frac{\frac1{2n\pi+\frac{\pi}2}}{\frac1{2n\pi}-\frac1{2n\pi+\frac{\pi}2}}=\frac1{\frac{2n\pi+\frac{\pi}2}{2n\pi}-1}=\frac{2n\pi}{\pi/2}=4n\;,$$

which can be made as large as you want. This function has very, very steep bits, but they’re also very, very short.

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