If $f$ is locally Lipschitz on $X$ and $X$ is compact, then $f$ is Lipschitz on $X$.

Question

My proof: Since $f$ is locally Lipschitz on $X$, for each $x ∈ X$ there exists an open $B_x$ containing $x$ such that $f$ is Lipschitz on $B_x$. Consider the collection of all such $B_x$. This collection forms an open cover of $X$ and so there is a finite sub-collection $\{B_1, B_2, . . . , B_n\}$ which also covers $X$, since $X$ is compact. Since $f$ is Lipschitz on each $B_i$, there is an $B_i$ such that $d(f(x_i), f(y_i)) \leq M_i d(x_i, y_i)$ for all $x_i, y_i \in Wi$, for $i ∈ \{1, 2, . . . , n\}$. Then taking $B = max\{B_1, B_2, . . . , B_n\}$, we see that $f$ is Lipschitz on $X$.

Is my solution correct?

Can we use the fact of compact metric space that every sequence in $X$ has a convergent sub-sequence to prove the same fact??

I think there's a small problem with your proof. What if I pick $x,y$ belonging to two different balls $B_i$ and $B_j$? — TorsionSquid, Feb 16 '15 at 02:56
For those looking for a reference: while this result can be found in many books (but only as an exercise), on page 93 of Kumaresan's "Topology of Metric Spaces" (2005) the author gives a sufficiently detailed proof sketch. — Frederik vom Ende, Feb 05 '23 at 11:36
Could you please tell me some other books that mentioned this result (exercise is OK) — Dat Ba Tran, Mar 26 '23 at 15:05

score 6 · Answer 1 · answered Feb 17 '15 at 19:31

First note that locally Lipschitz implies continuous. This said, the argument suggested can be completed as follows. For every $x\in X$ there is an open ball $B_x$ centered at $x$ of radius $\varepsilon_x$ and a constant $M_x$ such that $d(f(x),f(y))\le M_xd(x,y)$ for $x,y\in B_x$. Now consider the open convering consisting of the open balls $B'_x$ of center $x$ and radius $\varepsilon_x/2$, and by compactness extract a finite subcovering $B'_{x_1},\dots,B'_{x_r}$. In this situation: $$ \delta=\min\big\{\frac{\varepsilon_{x_i}}{2}: i=1,\dots,r\big\}>0\quad\text{and}\quad K=\max\{d(f(x),f(y)): x,y\in X\} $$ exists by compactness and continuity. Then pick a positive constant $$ L\ge \frac{K}{\delta}, M_{x_1},\dots,M_{x_r}. $$ We see that $f$ is $L$-Lipschitz. Indeed,

(1) if $d(x,y)<\delta$, as $x\in B'_{x_i}$ for some $i$ we have $$ d(y,x_i)\le d(y,x)+d(x,x_i)<\delta+\frac{\varepsilon_{x_i}}{2}\le\varepsilon_{x_i}, $$ hence $x,y$ are both in $B_i$ and $$ d(f(x),f(y))<M_id(x,y)\le L\,d(x,y), $$

(2) if $d(x,y)\ge\delta$ we have $$ d(f(x),f(y))\le K=\frac{K}{\delta}\delta\le L\,d(x,y). $$ We are done.

I'm not sure that the open sets $B'_x$ are a cover of $X$ since $\frac{\epsilon_x}2<\epsilon_x$, what can you say about? — Antonio Maria Di Mauro, Jul 23 '20 at 20:07
@AntonioMariaDiMauro: They are definitely a cover of $X$, since for each $x\in X$ we have $x\in B_x'$. — Brian M. Scott, Jul 23 '20 at 20:28
@BrianM.Scott Right! It was simple: excuse me if I ask a trivial thing but here is late and I am a bit tired. Thanks too much for your assistance!!! — Antonio Maria Di Mauro, Jul 23 '20 at 20:32

If $f$ is locally Lipschitz on $X$ and $X$ is compact, then $f$ is Lipschitz on $X$.

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