# Why this set satisfies the interior sphere condition? [Solved]

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 $\partial B_r(\xi)\cap\overline\Omega=\{z\}$.

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What is the "interior sphere condition"? –  Henning Makholm Nov 22 '12 at 13:42
I edited the question with the definition! –  José Carlos Nov 22 '12 at 13:45
I am not sure that your condition is the standard interior sphere condition. –  Siminore Nov 22 '12 at 13:57
Yes, in this form, this condition seems not to hold for $\Omega$, even with constant $\varphi$.. anyway, what is $\xi$? –  Berci Nov 22 '12 at 14:09
Im sorry, I repaired the definition. $\xi$ is a point of $\Omega$. This condition means that for each point $z$ in the boundary, you can construct a ball insided $\Omega$, that the boundary of this ball intersect only the point $z$. –  José Carlos Nov 22 '12 at 14:17

## 1 Answer

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 $0$.

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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