Assume is locally Lipschitz on , that is, for any , there exists (depending on ) such that for all .

Then, for any compact set , there exists a constant (depending on ) such that for all . That is, is Lipschitz on .

## Proof

Suppose to the contrary is not Lipschitz on , so that for all , there exists such that

Then there exists two sequences such that

Since is locally Lipschitz implies is continuous, so is bounded on by Extreme Value Theorem. Hence .

By sequential compactness of , there exists a convergent subsequence , and thus .

Then for any , there exists such that but which contradicts that is locally Lipschitz.