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PRELIMINARY

Let $\Omega \subseteq \mathbb{R}^n$ be an open set. We adopt the following notation: \begin{align} (x',x_n) &\equiv x \in \mathbb{R}^n \\ Q &\equiv \{(x',x_n) \,\vert\, |x'|<1, |x_n|<1 \} \\ Q_+ &\equiv Q \cap \{(x',x_n) \,\vert\, x_n>0\} \\ Q_0 &\equiv Q \cap \{(x',x_n) \,\vert\, x_n=0\} \end{align}

We say that $\Omega$ has a $C^{k,\alpha}$-boundary if for all $x_0 \in \partial \Omega \,$ there exist a neighborhood $U$ of $x_0$ and a function $H:U \to Q$ such that $H$ is of class $C^{k,\alpha}$, has an inverse $H^{-1}$ also of class $C^{k,\alpha}$ and the following relations hold: $$ H(U \cap \Omega) = Q_+ \hspace{5mm} H(U \cap \partial \Omega) = Q_0 $$ I am mostly interested in the cases $C^{1,0}$ ($\mathbf{C^1}$ boundary) and $C^{0,1}$ (Lipschitz boundary). Note that the former implies the latter (we just need to restrict to compact sets).

PROBLEM

Let $d(x), x \in \Omega$ denote the distance function of $x$ from the boundary of $\Omega$ (ie: $d(x) \equiv \text{dist}(x,\partial \Omega)$). I would like to prove the following statement:

If $\Omega$ has Lipschitz boundary then there exists $R>0$ such that $d(x)$ is differentiable ae in $\{x \in \Omega \,\vert\, d(x)<R\}$.

MY ATTEMPT

So far I tried proving the statement for a $C^1$ boundary following the sketch found in the answer to this question. Unfortunately I can't prove an intuitively immediate fact:

If $x_0 \in \partial \Omega$ is the closest point of the boundary to $x \in \Omega$ (assuming they are both contained in the domain of a local chart $H$) then the line connecting them has the same direction as the normal versor at $x_0$.

Suppose that I manage to prove this claim, then I should be able to follow the sketch. Finally, in order to weaken the proof in the Lipschitz Boundary case, can I invoke some form of Rademacher's Theorem (about the differentiability of Lipschitz functions)?

MOTIVATION

I need this statement for proving some claims made (without proofs) by my professors in a Calculus of Variations and a PDE courses, concerning the solutions of problems on domains with Lipschitz boundary.

Matteo Menghini
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    The distance function is always Lipschitz https://math.stackexchange.com/questions/3365754/distance-from-a-point-in-a-set-to-a-subset-of-that-set-is-lipschitz so it is differentiable a.e. By Rademacher – JackT Apr 24 '25 at 07:32
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    @JackT Thank you for the answer. You are absolutely right. I feel stupid for not using Rademacher immediately. Also, I don't understand why my professor told me that the Lipschitz condition was essential for this property... I will leave the question up only because I would like to know if someone could also prove the statement below about the normal versor. – Matteo Menghini Apr 24 '25 at 07:39
  • It’s not stupid at all dw :) Possibly your professor means it is Lipschitz up to the boundary - as you’ve written it, you only ask for the distance to be differentiable a.e. in the interior of $\Omega$ not $\overline \Omega$ – JackT Apr 24 '25 at 07:46
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    Perhaps, you can be interested by this old question of mine and its answer by @Andrew Hwang – Jean Marie Apr 24 '25 at 07:48

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