# Locally Lipschitz implies continuity. Does the converse implication hold?

Let $$A$$ be open in $$\mathbb{R}^m$$; let $$g:A\rightarrow\mathbb{R}^n$$. If $$S\subseteq A$$, we say that $$S$$ satisfies the Lipschitz condition on $$S$$ if the function $$\lambda(x,y)=|g(x)-g(y)|/|x-y|$$ is bounded for $$x\neq y\in S$$. We say that $$g$$ is locally Lipschitz if each point of $$A$$ has a neighborhood on which $$g$$ satisfies the Lipschitz condition.

Show that if $$g$$ is locally Lipschitz, then $$g$$ is continuous. Does the converse hold?

For the first part, suppose $$g$$ is locally Lipschitz. So for each point $$r\in A$$, there exists a neighborhood for which $$|g(x)-g(y)|/|x-y|$$ is bounded. Suppose $$|g(x)-g(y)|/|x-y| in that neighborhood. Then $$|g(x)-g(r)| in that neighborhood of $$r$$. Therefore $$g(x)\rightarrow g(r)$$ as $$x\rightarrow r$$, and so $$g$$ is continuous at $$r$$. This means $$g$$ is continuous everywhere in $$A$$.

What about the converse? I don't think it holds, but can't come up with a counterexample.

Intuitively, a counterexample must be a function which is very steep without having a jump or other sort of discontinuity. Consider, for example, $$g(x) = x^{1/3}$$ at $0$. Then
$$\frac{|g(x) - g(0)|}{|x - 0|} = x^{-2/3}$$
This cannot be bounded in a neighborhood of $0$.
Or $x \mapsto \sqrt{|x|}$: ${}{}{}{}$