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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?

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    Very similar: http://math.stackexchange.com/questions/154391/locally-lipschitz-implies-lipschitz So, possible duplicate?2012-06-06
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    @Weltschmerz The proof on $\mathbb R$ used the order structure which isn't available here.2012-06-06

3 Answers 3

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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](http://mathoverflow.net/questions/12342/reductio-ad-absurdum-or-the-contrapositive/12427#12427), but in this case I agree with you that your argument is _much_ cleaner to write.2012-06-07
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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.2012-06-06
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    I don't quite follow your hint. In every ball $U_x$ one has a Lipschitz constant and since there are only finitely many balls, one can choose a maximum one. What if two points in $C$ say $x,y\in C$, are contained in different balls? How can one find the estimate $|f(x)-f(y)|\leq L|x-y|$?2014-09-26
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    In your comment, do you mean "path-connected" instead of "connected"?2014-09-26
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    @Jack: for your second comment: sure, if it helps you. The assumption is not necessary in any case, it may just help the reader see the picture better. // For your first comment, you can be a bit clever: take $U_x$ to be balls and $V_x$ be the same balls with half the radius. Make finite subcover from $V_x$. If $x,y\in U_z$ you are done. If $x,y$ cannot be found within the same $U_z$, let $z$ be such that $V_z$ contains $x$. By assumption $U_z$ does not contain $y$. So $|x-y| \geq $ radius of $V_z$.2014-09-26
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    I'm trying to follow you. So $$|f(x)-f(y)|\leq |f(x)-f(z)|+|f(z)-f(y)|\leq L|x-z|+|f(z)-f(y)|$L$ is the Lipschitz constant on $V_z$ and $R$ is its radius. Are you saying that one needs to keep doing this trick for $|f(z)-f(y)|$? – 2014-09-26
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    Suppose for the moment that $C$ is connected, which implies that if $x,y$ not belong to the same $U_z$, there is a sequence of $z_i$ such that you can build a path. Let $k$ be the number of $V_z$ in the subcover. Let $L´$ be the largest of the Lip. constants of the corresponding $U_z$. Let $R$ be the largest diameter of the $U_z$ and $r$ the smallest radius of the $V_z$. Then $$ |f(x) - f(y)| \leq k \cdot L´ \cdot R $$ and $|x-y| \geq r$. So when ever $x,y$ cannot be found in the same $U_z$ we have that $$|f(x) - f(y)| \leq k L´ R / r \cdot |x-y| = L|x-y|$$2014-09-26
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    For $C$ disconnected replace $r$ by the lesser of the $r$ defined above and the minimum distance between the (finitely many) connected components of $C$. // Maybe I should clarify that in my previous comment, a "path" refers to a sequence of $z_i$ such that $U_{z_i} \cap U_{z_{i+1}} \neq \emptyset$.2014-09-26
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    Thank you for your elaboration! I though I need to assume path-connectedness in order get the "ball path" defined in your comment.(I don't see why compactness imply the the existence of the "path"...) What if there are infinitely many connected components of $C$? (It might be trivial for you to see that it must be finite, but I don't see why...)2014-09-26
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    I should have checked the topology basics for those questions. Just found something related(http://math.stackexchange.com/q/326259/9464). Thank you again!2014-09-28
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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 $C\times C\cap\{|x-y|<\varepsilon\}$.