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. $\forall \ x, c \in \mathrm{Dom}\, (f)$ $$ |x - c| \le \delta \ \Rightarrow \ |f(x) - f(c)| \le \varepsilon$$


$f$ 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?


2 Answers 2


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: for $n\in\Bbb Z^+\setminus\{0\}$ we have

$$\begin{align*} \left|\frac{f\left(\frac1{2n\pi}\right)-f\left(\frac1{2n\pi+\frac{\pi}2}\right)}{\frac1{2n\pi}-\frac1{2n\pi+\frac{\pi}2}}\right|&=\left|\frac{\frac1{2n\pi+\frac{\pi}2}}{\frac1{2n\pi}-\frac1{2n\pi+\frac{\pi}2}}\right|\\ &=\left|\frac1{\frac{2n\pi+\frac{\pi}2}{2n\pi}-1}\right|\\ &=\left|\frac{2n\pi}{\pi/2}\right|\\ &=4|n|\,, \end{align*}$$

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


I suppose one small detail to add to Brian's answer is that in the argument we assumed $M\neq 0$. But the case for $M=0$ is trivial enough as that means $f$ is constant, which is clearly uniformly continuous (setting $\delta=\epsilon$).


Suppose $f:S\to\mathbb{R}$ is Lipchitz continuous, then we have some $M\geq 0$ such that for all $x,y\in S$, $$|f(x)-f(y)|\leq M|x-y|$$

Suppose $M>0$. Then, given some $\epsilon>0$, we set $\delta=\frac{\epsilon}{M}$. Then for all $x,y\in S$, we have $$|x-y|<\delta\implies |f(x)-f(y)|\leq M|x-y|<M\cdot \frac{\epsilon}{M}=\epsilon$$ Hence $f$ is uniformly continuous.

If $M=0$, then it is easy to see $f(x)=f(y)$ for all $x,y\in S$, i.e. $f$ is constant. One can check easily that this is uniformly continuous (setting $\delta=\epsilon$).

Note, however, that the converse is not true. For instance, one can check that the function $f:[0,\infty)\to\mathbb{R}$ given by $f(x)=\sqrt{x}$ is uniformly continuous but not Lipschitz continuous.

  • 1
    $\begingroup$ Since this doesn't fully answer the question, I would recommend you try to gain some reputation (you need 50), and then comment this under the relevant answer. Non-answers like this tend to be downvoted and/or deleted. $\endgroup$ Commented Jan 26, 2023 at 2:35
  • $\begingroup$ Yes sorry I did try to comment initially. How can I gain reputation? $\endgroup$
    – David
    Commented Jan 27, 2023 at 10:10
  • $\begingroup$ In short, you can gain reputation by asking or answering questions. Here is a quick and in-depth guide. I do understand if you don't want to go through all the trouble if all you wanted to do was post this comment, but that's just how the site happens to work... In any case, welcome to MSE. $\endgroup$ Commented Jan 27, 2023 at 10:36
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    $\begingroup$ @C-RAM Thank you for letting me know! I suppose I will just make this a full answer for now so that it doesn't get deleted :) $\endgroup$
    – David
    Commented Jan 29, 2023 at 23:17

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