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Let $f: K \rightarrow \mathbb R$ be a locally Lipschitz mapping on a compact subset $K$ of $\mathbb R^n$. Is it then $f$ Lipschitz?

user172903
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    What do you expect, and why? – Daniel Fischer Oct 04 '14 at 15:40
  • I mean: Do you expect that $f$ will be Lipschitz, or don't you? Why do you expect what you expect? – Daniel Fischer Oct 04 '14 at 16:12
  • Each point $x\in K$ has a neighbourhood $U_x$ on which $f$ is Lipschitz. Such sets cover $K$. By compactness a finite numbers $U_{x_i}$ covers $K$.On each $U_{x_i}$ $f$ is Lipschitz. Maybe $f$ is Lipschitz on their union? – user172903 Oct 04 '14 at 16:20
  • @user172903 Yes, that works. You need to fill in details, of course. Another way of proving it is given in one of the answers of the duplicate. – Daniel Fischer Oct 04 '14 at 17:27

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