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Let $\Omega\subseteq\mathbb{R}^N$ be an open and bounded set with uniform Lipschitz boundary. Consider the signed distance function:

$d:\mathbb{R}^N\to\mathbb{R},\ d(x)=\begin{cases} \mathrm{dist}(x,\partial\Omega),\ x\notin\Omega \\ 0,\ x\in\partial\Omega \\ -\mathrm{dist}(x,\partial\Omega),\ x\in\Omega\end{cases}$

It is well-known that $d$ is almost everywhere differentiable (being 1-Lipschitz) and $|\nabla d(x)|\leq 1$. Also it is known that $|\nabla d(x)|=1$ in a neighbourhood of $\partial\Omega$ if $\Omega$ has a $C^2$ boundary (see Gilbard&Trudinger, page 355). Of course we know that $\partial\Omega$ is a compact set with finite perimeter and with null Lebesgue measure.

My question is the following: Is it true that the set $\{x\in \mathbb{R}^N\ |\ \exists y_1\neq y_2\in\partial\Omega\ \text{with}\ d(x)=|x-y_1|=|x-y_2|, x\notin\Omega \ \text{and} \ d(x)=-|x-y_1|=-|x-y_2|, x\in\Omega\}$ has a null Lebesgue measure?

This fact would guarantee that $|\nabla d(x)|=1$ almost everywhere for $\Omega$ a Lipschitz domain. Is this true?

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    $\begingroup$ I think something stronger might hold. Does the discussion at the following link answer your question? mathoverflow.net/questions/377613/… $\endgroup$
    – Leo Moos
    Dec 28, 2020 at 15:52
  • $\begingroup$ Wow! Thanks a lot!!! I did not beleived that this is so general. $\endgroup$
    – Bogdan
    Dec 28, 2020 at 16:26

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