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Consider the subset $\Omega\subset\mathbf{R}^n$, $$\Omega=\{x=(x_1,...,x_n)\in\mathbf{R}^n;x_n>\varphi(x_1,...,x_n)\},$$ where $\varphi$ is a Lipschitz continuous function, that is, $\Omega$ is a unbounded set, bounded for a Lipschitz graph.

Why this set satisfies the interior sphere condtion?

Interior sphere condition means that for each $z\in\partial\Omega$, there is a ball $B_r(\xi)$ satisfying $\partial B_r(\xi)\cap\overline\Omega=\{z\}$.

José Carlos
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1 Answers1

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This set does certainly not satisfy the interior sphere (or interior ball) condition in general, as the example $\varphi (x) = |x_1| + \ldots + |x_{n-1}|$ shows. This functions is $1$-Lipschitz, but the condition is violated at $0$.

Lukas Geyer
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  • I need to use a Hopf Lemma in this set. You know how can i use this without the interior ball condition? – José Carlos Nov 22 '12 at 14:55
  • Maybe you just need it almost everywhere? Lipschitz functions are differentiable a.e., that should give you the condition a.e. At points of non-differentiability like $0$ in the example you already have problems writing down the statement of the Hopf lemma, since there is no well-defined normal vector at these points. – Lukas Geyer Nov 22 '12 at 19:54
  • Thank you very much Lukas Geyer. – José Carlos Nov 22 '12 at 23:51