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Let $K\subset \mathcal{R}^n$ be nonempty and closed. Let $f_1$ be locally Lipschitz on $\mathcal{R}^n$. Assume there is a finite $a=\inf_{x\in K^C}f_1(x)$. Suppose $f$ is defined as $f=a$ $\forall x\in K$ and $f=f_1(x)$ $\forall x \in K^C$. Also assume $f_1(x) \rightarrow a$ as $x\rightarrow bdry(K)$. Is $f$ locally Lipschitz on $\mathcal{R}^n$? If so, where can I start in proving that? Any help will be greatly appreciated.

Remark: Note that I am requiring $f_1$ to be locally Lipschitz on $\mathcal{R}^n$ not just on $K^C$. I asked a similar question here (see the link below) but my conditions are too relaxed so counter examples are provided. With these modified conditions I am hoping to prove local Lipschitzness of $f$ on $\mathcal{R}^n$. 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?

Is this piecewise defined function locally Lipschitz?

curiosity
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  • Why does $a = \text{inf}_{x \in K^C} f_1(x)$ exist? Is that an added assumption? Note that this assumption is not always true, for example if $f_1$ is the Lipschitz function $f_1(x)=x$ for $x \in \mathbb R$. – Lee Mosher Nov 07 '24 at 12:32
  • @Lee Mosher Thank you for pointing that out. Yes, that is an extra assumption. I have edited the question. I also added $f_1(x)\rightarrow a$ as $x\rightarrow bdry(K)$. Does that guarantee $f$ is at least continuous? – curiosity Nov 07 '24 at 12:52

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