Locally Lipschitz implies Lipschitz on Compact Set Proof

Assume \phi is locally Lipschitz on \mathbb{R}^n, that is, for any x\in \mathbb{R}^n, there exists \delta, L>0 (depending on x) such that |\phi(z)-\phi(y)|\leq L|z-y| for all z,y\in B_\delta(x)=\{t\in\mathbb{R}^n: |x-t|<\delta\}.

Then, for any compact set K\subset\mathbb{R}^n, there exists a constant M>0 (depending on K) such that |\phi(x)-\phi(y)|\leq M|x-y| for all x,y\in K. That is, \phi is Lipschitz on K.

Proof

Suppose to the contrary \phi is not Lipschitz on K, so that for all M>0, there exists x,y\in K such that \displaystyle \frac{|\phi(x)-\phi(y)|}{|x-y|}>M.

Then there exists two sequences x_n, y_n\in K such that \displaystyle \frac{|\phi(x_n)-\phi(y_n)|}{|x_n-y_n|}\to\infty.

Since \phi is locally Lipschitz implies \phi is continuous, so \phi is bounded on K by Extreme Value Theorem. Hence |x_n-y_n|\to 0.

By sequential compactness of K, there exists a convergent subsequence x_{n_k}\to x, and thus y_{n_k}\to x.

Then for any L>0, there exists k such that x_{n_k},y_{n_k}\in B_\delta(x) but \displaystyle \frac{|\phi(x_{n_k})-\phi(y_{n_k})|}{|x_{n_k}-y_{n_k}|}>L which contradicts that \phi is locally Lipschitz.

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