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Need to demonstrate:

Let $A$ be open in $\mathbb{R}^m$; let $g:A \longrightarrow \mathbb{R}^n$ a function locally Lipschitz. Show that if $C$ is a compact subset of $A$, then $g$ satisfies the Lipschitz condition on $C$.

Someone can help me?

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Very similar: math.stackexchange.com/questions/154391/… So, possible duplicate? –  Weltschmerz Jun 6 '12 at 15:41
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@Weltschmerz The proof on $\mathbb R$ used the order structure which isn't available here. –  user31373 Jun 6 '12 at 16:45
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3 Answers

Hint: $C$ being compact means for any open cover it has a finite subcover. Let $x\in C$ and let $U_x$ be the corresponding open set on which $g:U_x\to\mathbb{R}^n$ is Lipschitz. This means that $C$ can be covered by finitely many of the $U_x$. Lastly use the fact that the among a finite set of positive numbers you can choose a maximum one.

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Some technical notes: (a) it may help to assume first that $C$ is connected (b) you may need the following fact: on a connected graph with $k$ nodes, between any two nodes there exists a path with at most $k-1$ edges. –  Willie Wong Jun 6 '12 at 15:55
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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 $x$.

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+1. That a contradiction argument is often easier to implement is a fact of life. There is sometimes a trade off, but in this case I agree with you that your argument is much cleaner to write. –  Willie Wong Jun 7 '12 at 8:54
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Maximize the continuous function $f(x,y)=\frac{|g(x)-g(y)|}{|x-y|}$ over the compact set $C\times C\cap\{|x-y| \geq \varepsilon\}$ with a sufficiently small $\varepsilon>0$. Locally Lipschitz condition is used to show that $f$ is bounded in $C\times C\cap\{|x-y|<\varepsilon\}$.

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thank you very much for the suggestions. –  Kenchin Haos Jun 7 '12 at 23:20
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