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Let $f:\mathcal{R}^n\rightarrow \mathcal{R}$ be continuous. Let $K\subset \mathcal{R}^n$ be nonempty and compact. Suppose $f$ is defined as $f=f_1(x)$ $\forall x\in K$ and $f=f_2(x)$ $\forall x\notin K$. Suppose $f_1$ and $f_2$ are locally Lipschitz functions. Is $f$ locally Lipschitz? If so, where can I start in proving that? Any help will be greatly appreciated.

Remark: I can see it is differentiable almost everywhere but that is not enough to imply local Lipschitzness. How do I show it is locally Lipschitz on the boundary of $K$? Those are the key points, right?

Frederik vom Ende
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curiosity
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Consider $n=1$, $K=\{0\}$, $f(x)=\sqrt{\lvert x\rvert}$ and $f_1,f_2$ the restrictions. If you don't like $K$ having empty interior, consider $K=[-1,1]$ and $f(x)=\sqrt{\max\{x^2-1,0\}}$.

Sassatelli Giulio
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  • Is $f_{1}(x)=\sqrt{|x|}$ Lipschitz on $(0,1]$ ? – Medo Nov 06 '24 at 19:24
  • @Sassatelli Giulio Like Medo, I also fail to see why $\sqrt{|x|}$ is Lipschitz on $(0,1]$. – curiosity Nov 06 '24 at 19:37
  • @Sassatelli Giulio I am having difficulty seeing why $f_1$, $f_2$ restrictions are Lipschitz for both of the counter examples you provided. Could you please elaborate? Thank you so much for your help. – curiosity Nov 06 '24 at 20:00
  • @curiosity Locally Lipschitz on $\Bbb R\setminus{0}$. And it's because for all $x\in\Bbb R\setminus{0}$ there is some $\varepsilon>0$ such that there is some $h$ such that, for all $y,z\in (x-\varepsilon, x+\varepsilon)\cap(\Bbb R\setminus{0})$, the inequality $\lvert\sqrt{\lvert y\rvert}-\sqrt{\lvert z\rvert}\rvert\le h\lvert y-z\rvert$ holds. You may take $\varepsilon_x=\frac{\lvert x\rvert}2$ and $h_x=\sup_{v\in (x-\varepsilon_x,x+\varepsilon_x)}\frac1{2\sqrt{\lvert v\rvert}}$. – Sassatelli Giulio Nov 06 '24 at 20:23
  • @curiosity More of the same for my other example. – Sassatelli Giulio Nov 06 '24 at 20:26
  • @Sassatelli Giulio Just to be sure, so the statement I was trying to prove is false and these are the counter examples, right? – curiosity Nov 06 '24 at 20:28
  • @curiosity The way I read your statement, yes. – Sassatelli Giulio Nov 06 '24 at 20:30
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    @Sassatelli Giulio The second counter example is not Lipschitz at $x=\pm 1$ if the domain is $R$, even though it is Lipschitz on the restrictions. Is the right? Please bear with my incessant questions. I just want to make sure I get this. Thank you so much again. – curiosity Nov 06 '24 at 20:37
  • I agree that the statement is false for a general compact set $K$. – Medo Nov 06 '24 at 20:39
  • @SassatelliGiulio If I require that the extensions of $f_1$ and $f_2$ are locally Lipschitz on $\mathcal{R}^n$, would we be able to proof local Lipschitzness of $f$? Should I write it as a new question? – curiosity Nov 06 '24 at 20:57