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In solving this question, I come across below result.

Let $(E, d)$ be a compact metric space and $f:E \to \mathbb R$ locally Lipschitz. Then $f$ is Lipschitz.

Could you have a check if my attempt is fine?

Because $f$ is locally Lipschitz, it is continuous. Let $M := \max_{x\in E} |f(x)|$. Suppose that $f$ is not Lipschitz on $E$.

  • For each $n \in \mathbb N^*$, there exist $x_n, y_n \in E$ such that $2M \ge |f(x_n)-f(y_n)| > n d(x_n, y_n)$. It follows that $d(x_n, y_n) \to 0$.
  • By compactness of $E$, we can assume $x_n \to a$ and $y_n \to a$ for some $a\in E$. Because $f$ is locally Lipschitz, there are $r,L>0$ such that $f$ is $L$-Lipschitz on the open ball $B(a,r)$.
  • For $n$ large enough, we get $|f(x_n)-f(y_n)| \le Ld(x_n,y_n)$. It follows that $Ld(x_n,y_n)> n d(x_n, y_n)$ for $n$ large enough. This is a contradiction. This completes the proof.
Akira
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    For a compact space, local is already global. Meaning, the definition of local Lipschitz continuity guarantees a Lipschitz constant for any compact subset. But the whole space is already a compact subset. // Or following your approach, $E$ has a finite diameter $d$. Thus $B(a,d)=E$ for any point $a$. – Lutz Lehmann May 25 '22 at 09:12
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    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

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