[Math] In $\mathbb{R}^n$, locally lipschitz on compact set implies lipschitz

Question

I need to prove:

Let $A$ be open in $\mathbb{R}^m$, $g:A \longrightarrow \mathbb{R}^n$ a locally lipschitz function and $C$ a compact subset of $A$.

Show that $g$ is lipschitz on $C$.

Can anyone help me?

Answer 1

In general I prefer covering arguments (such as the one @WillieWong posted) to "choose a sequence and get a contradiction", but in this particular problem the second approach could be easier to implement (it avoids the technicalities pointed out by Willie).

Suppose $g$ is not Lipschitz on $C$: that is, there exists two sequences $x_n,y_n\in C$ such that $|g(x_n)-g(y_n)|/|x_n-y_n|\to \infty$. Since $g$ is bounded on $C$ (why?), we have $|x_n-y_n|\to 0$. Choose a convergent subsequence $x_{n_k}\to x$. Observe that the local Lipschitz-ness fails at $x$.