# Is this true? $C^1$ functions are Lipschitz on bounded sets

This is very simple and I'm probably missing something obvious but I can't seem to spot the mistake.

We know that $C^1$ functions on compact sets are Lipschitz.

Let $U \subset \mathbb R^n$ be open, $S \subset U$ be compact, and $f: U \to \mathbb R^m$ be $C^1$.

$f$ is Lipschitz,so $\exists M \in \mathbb R$ so that $|f(x)-f(y)| \leq M|x-y|$ for all $x,y \in S$.

But this also means that it's true for all $x,y$ in the interior of $S$. So $f$ is also Lipschitz with constant $M$ in the interior of a compact set. Compacts are closed and bounded, so only criterion for being the interior of a compact set is being bounded.

The conclusion is that $f$ is Lipschitz on bounded sets, as we can close it to get a compact set, there it will be Lipschitz with some constant, so it also must be Lipschitz with that same constant inside.

Is this correct? I don't see anything in the literature or online about it. Everything is just about compactness. Where is the mistake here?

Very important edit

I meant that $f$ is $C^1$ on an open set $U$ which contains compact set $S$, then $f$ is Lipschitz on interior of $S$.

• Sounds true to me. Why are you so sure it's wrong – Dalamar Apr 18 '18 at 7:45
• Lack of references and google hits, "too good to be true" feel to it. – Oria Gruber Apr 18 '18 at 7:48
• The Lipschitz constant $M$ depends in general on the compact set $S \subset U$. – user539887 Apr 18 '18 at 7:49
• After your edit, yes, it is true. – user539887 Apr 18 '18 at 8:04

Note that $\sqrt{x}$ is $C^1$ in $(0,1]$ (which is not compact) but this function is not Lipschitz in any bounded set $(0,a]$ with $0<a\leq 1$: if $(x_n)_n$ and $(y_n)_n$ are two sequences which go to $0^+$ with $x_n>y_n$ then $$\lim_{n\to +\infty} \frac{\sqrt{x_n}-\sqrt{y_n}}{x_n-y_n}= \lim_{n\to +\infty} \frac{1}{\sqrt{x_n}+\sqrt{y_n}}=+\infty.$$

On the other hand, if $f$ is $C^1$ in an open set $U$ which contains a compact set $S$, then $|f'|$ is continuous in the compact set $S$ and in $S$ it attains a maximum value $M$. Hence $f$ is Lipschitz in $S$ (and therefore also in its interior) with constant $M$.

• Yes but notice that I required that $f$ to be $C^1$ on $U$ which contains the compact set $S$. In your case, $\sqrt{x}$ is not $C^1$ on an open set that contains $[0,1]$ so it fails this property. – Oria Gruber Apr 18 '18 at 7:47
• Do you mind to rephrase your question clearly please. – Robert Z Apr 18 '18 at 7:52
• I'm terribly sorry for the confusion. What I meant is: $f$ is $C^1$ in open set $U$ which contains compact set $S$, then $f$ is lipschitz in the interior of $S$. Infact, $f$ is lipschitz in any bounded set $A \subset U$. So for instance if $f$ is $C^1$ in $\mathbb R^n$, then it will be Lipschitz in every bounded subset of $\mathbb R^n$, even those which are not closed. – Oria Gruber Apr 18 '18 at 7:53
• Please see my P.S. – Robert Z Apr 18 '18 at 7:56
• @OriaGruber Even the two statements in that comment of yours are not equivalent, since if $A \subseteq U$ is bounded, it doesn't mean that there is a compact set $S$ such that $A \subseteq S \subseteq U.$ – Adayah Apr 18 '18 at 7:58

Let $U=(0,1)$ and consider $f(x)=x^{-1}$ on $U$? We can choose by Mean Value Theorem that $\cdots<x_{2}<\eta_{2}<y_{2}<x_{1}<\eta_{1}<y_{1}<1$ such that $\left|-\dfrac{1}{\eta_{i}^{2}}\right|\leq M$ for $i=1,2,...$, and $\eta_{i}\rightarrow 0$ as $i\rightarrow\infty$.

• So the inequality blows up. – user284331 Apr 18 '18 at 7:51
• I think I understand, thank you. – Oria Gruber Apr 18 '18 at 7:52