# If $f$ is locally Lipschitz, then for any compact set $K$, $f \mid_K$ is globally Lipschitz

This problem has been asked here before. For example, this question Prove that the restriction $f|_K$ of $f$ to $K$ is globally Lipschitz where $K$ is a compact set only treated the continuity of $f$ on $K$ while this question $f \in C^1$ defined on a compact set $K$ is Lipschitz? already assumed $f$ to be $C^1$ function. Although, the latter was close to answering my question but used terms difficult for me to understand. So, in both cases, they have not answered my question. So, here it is

If $f:O\subset \Bbb{R}^n\to\Bbb{R}^m$ is locally Lipschitz, then prove that for any compact set $K$ in $O,$ $f \mid_K$ is Lipschitz such that $\exists \;c\in [x,y]$ such that \begin{align}\Vert f(x)-f(y) \Vert\leq c\Vert x-y \Vert,\;\;\forall\;x,y\in K\end{align}

My efforts

Let $K$ be compact in $O$. Let $x,y\in K$, then by MVT, $\exists \;r\in [x,y]$ such that \begin{align}\Vert f(x)-f(y) \Vert\leq \sup\limits_{r\in [x,y]}\Vert f'(r) \Vert\Vert x-y \Vert\end{align} \begin{align}\qquad\qquad\qquad\leq \sup\limits_{r\in K}\Vert f'(r) \Vert\Vert x-y \Vert\end{align} Since $f$ is locally Lipschitz, then it is continuous and since $K$ is compact, then the maximum is reached. So, let \begin{align}c= \sup\limits_{r\in K}\Vert f'(r) \Vert\end{align} Thus, \begin{align}\Vert f(x)-f(y) \Vert\leq c\Vert x-y \Vert,\;\;\forall\;x,y\in K\end{align}

Please, can anyone help me check if my proof is correct? If no, alternative proofs will be highly regarded! Thanks!

• Your proof works only when $f$ is $C^1$. – Crostul Aug 22 '18 at 16:48
• @Crostul: But $f$ Lipschitz doesn't imply $f$ is $C^1$ – Omojola Micheal Aug 22 '18 at 16:51
• Well, the proof would work (since a Lipschitz function is ae. differentiable), but it needs a bit more finness & something like the Rademacher theorem. – copper.hat Aug 22 '18 at 17:05
• The incorrect assumption that $f$ must differentiable renders it invalid..... BTW "compact" is an intrinsic property of a space or sub-space. $K$ is compact in $O$ iff $K$ is compact in $\Bbb R^n$ iff $K$ is a compact sub$space$ of $\Bbb R^n$. – DanielWainfleet Aug 22 '18 at 18:08

## 2 Answers

While I am not a fan of proof by contradiction, it works efficiently here.

Suppose $S(x,y)={\|f(x)-f(y)| \over \|x-y\|}$ is unbounded for $x,y \in K, x \neq y$. Then we can find $x_k, y_k \in K$ such that $S(x_k,y_k) \to \infty$. Since $K$ is compact, we can assume that $x_k \to x, y_k \to y$. Since $f$ is bounded on $K$, we must have $x=y$ (otherwise $S(x_k,y_k)$ would not be unbounded). By assumption, $f$ is locally Lipschitz around $x$, hence $S(x_k,y_k) \le L$ for some (finite) $L$, which is a contradiction.

Here is a constructive proof:

Since $f$ is locally Lipschitz, for each $x$ there is some $r_x>0$ and $L_x$ such that $f$ is Lipschitz with rank $L_x$ on $B(x,r_x)$.

Then the sets $B(x, {1 \over 2} r_x)$, $x \in O$ form an open cover of $K$, so a finite number cover $K$. For convenience, denote these by $B(x_k, {1 \over 2} r_k)$ (instead of $r_{x_k}$).

Let $M= \sup_{x \in M} \|f(x)\|$, $r= {1 \over 2}\min r_k$, $L_0 = {2M \over r}$ and $L= \max (L_0, L_k)$. Then $L$ is a Lipschitz constant for $f$ on $K$.

To see this, pick $x,y \in K$. If $\|x-y\| \ge r$ then we see that ${ \|f(x)-f(y) \| \over \|x - y \|} \le {2M \over r} = L_0 \le L$. If $\|x-y\| < r$, then for some $x_k$ we have $x \in B(x_k, {1 \over 2} r_k)$. Then $y \in B(x_k, r_k)$ and so $\|f(x)-f(y) \| \le L_k \|x - y \| \le L \|x - y \|$.

• Thanks for the proof by contradiction. I'll go through the proof. Ask questions where needed and then, thick! – Omojola Micheal Aug 22 '18 at 17:07
• There is the "hidden", yet noteworthy, detail that $y_k\ne x_k$ for all $k$. – Saucy O'Path Aug 22 '18 at 17:09
• @Saucy O'Path: Thanks for the note! – Omojola Micheal Aug 22 '18 at 17:12
• There are many details omitted, ironically for clarity. Following Polonius, brevity is the soul of wit... – copper.hat Aug 22 '18 at 17:14
• @Mike: You are correct, but we can always choose a subsequence and renumber. This is a fairly standard trick to reduce word count :-). – copper.hat Aug 22 '18 at 17:50

By our assumption, $\forall x \in O$, there is a open neighbourhood of $x$, $U_x$, and $c \in \mathbb{R}$ such that $$||f(y) - f(z)|| \leq c_x ||y-z|| \quad \forall y,z \in U_x.$$

(Thanks to @copper.hat)

Let $K \subset O$ be compact, then it is the finite union of compact connected $J_ks$, and observe that $S = \{U_x\}_{x\in K}$ is an open cover for $K$, but since $K$ is compact, we can cover with a finite subset of $S$, namely $\{U_x\}_{x \in J}$ for some finite $J \subset K$.

Now, for each $k$, define

$$c_k = max\{c_x| x \in J_k\},$$ and $$c = max\{c_k\},$$ hence $f|_K$ is Lipschitz on $K$ with Lipschitz constant $c$.

• This is not quite correct. – copper.hat Aug 22 '18 at 19:35
• Take $f=0$ on $[0,1]$ and $f=1$ on $[2,3]$. Then $f$ is of local Lipschitz rank zero, but global rank one, hence taking the $\max$ is not sufficient. In addition, there is an implicit assumption in your proof that the domain is convex and can be joined by a straight line. It is straightforward to create an example (think horseshoe shape :-)) in which there is a small local rank but large global rank. A constructive proof requires a bit more work. – copper.hat Aug 22 '18 at 20:08
• @copper.hat Thanks for pointing out. What if we assume that $K$ has finite compact connected components, and take the max. of each component ? – onurcanbektas Aug 23 '18 at 5:49
• Sure, for example take a strictly decreasing, convergent sequence along with the limit. In any event, my example shows why the approach of taking the max fails. – copper.hat Aug 23 '18 at 6:55
• The set is closed and bounded. The open set that contains $0$ contains all but a finite number of the points. – copper.hat Aug 24 '18 at 18:23