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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 $|g(x)-g(y)|/|x-y|<M$ in that neighborhood. Then $|g(x)-g(r)|<M|x-r|$ 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.

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2 Answers 2

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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 $0$.

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  • $\begingroup$ Yes, your intuitive description summarizes it well. Thanks T. Bongers. $\endgroup$
    – Mika H.
    Commented Oct 18, 2013 at 4:55
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Or $x \mapsto \sqrt{|x|}$: ${}{}{}{}$

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