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Consider $f : \Omega \subset R^n \rightarrow R $ a locally Lipschitz function with $\Omega$ a bounded and connected set. Then $f$ is Lipschitz?

My best is this :

For each $x \in \Omega$ exist a ball $B(x,r_x)$ centered in x such that

$$|f(a) - f(b)| \leq K_x || a-b||, \forall a,b, \in B(x,r_x) $$

where $K_x$ is a constant that depends on $x$. By the Heine Borel theorem we can take a covering $B(x_i , r_{x_i}) \ , i=1,...,n$ of $\Omega$.

I believe that from this i can conclude that $f$ is Lipschtz but i dont know how to do that . someone can give me a help to prove that f is Lipschitz? (or give a counter example)

thanks in advance

math student
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1 Answers1

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It's not true: Consider $f(x)=1/x$ on $\Omega=(0,1)\subset\mathbb{R}$. $f$ is $C^1$ and therefore locally Lipschitz, but not Lipschitz on $\Omega$, since the derivative $f^\prime(x)=-\frac{1}{x^2}$ is unbounded on $(0,1)$.

However: If you want the Lipschitz condition on a compact set $\Omega$ (which you seem to assume in your proof attempt), then you are fine:

Your proof is almost finished. The only thing left to do is pick $L=\max_{i=1,\dots,n} K_{x_i}$ and that is your Lipschitz constant.

You can find it also here.

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J.R.
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