# 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.

## Author: mathtuition88

http://mathtuition88.com

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