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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

3 Answers 3

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
    
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 $|f(x)-f(y)|\leq L|x-y|$? –  Jack Sep 26 at 1:03
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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 $V_z$. –  Willie Wong Sep 26 at 14:34
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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 $$|f(x) - f(y)| \leq k L´ R / r \cdot |x-y| = L|x-y|$$ –  Willie Wong Sep 26 at 15:40
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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 $U_{z_i} \cap U_{z_{i+1}} \neq \emptyset$. –  Willie Wong Sep 26 at 15:42

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

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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