Real Analysis – Lipschitz Continuous $\Rightarrow$ Uniformly Continuous

continuityreal-analysissolution-verification

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?

Best Answer

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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