Let $\Omega\subset\mathbb{R}^{n}$ be an open,
bounded convex domain. Denote $d_{\Omega}:\Omega\rightarrow\mathbb{R}$
the distance-to-boundary function, that is,
$$
d_{\Omega}\left(x\right):=\inf\left\{ \left|x-y\right|:y\in\partial\Omega\right\} .
$$
It is well-known that if $\partial\Omega$ is smooth then $d_{\Omega}$
is convex in a neighborhood of $\partial\Omega$ and $\left|\nabla d_{\Omega}\right|=1$
in this neighborhood.
I would like to reduce the smoothness assumption of the boundary.
My question (suggestion) is: if $\partial\Omega$ is only Lipschitz, is it true
that $d_{\Omega}$ is convex in a neighborhood of $\partial\Omega$
and $\left|\nabla d_{\Omega}\right|=1$ almost everywhere in this
neighborhood?
Any reference is welcome.
Best Answer
The proof does not require smoothness.
Assume $B(x,r_x)\subset \Omega$, $B(y,r_y)\subset \Omega$. It is sufficient to show that $$B(\tfrac{x+y}2,\tfrac{r_x+r_y}2)\subset \Omega.$$ The latter follows from convexity of $\Omega$. Indeed, the ball $B(\tfrac{x+y}2,\tfrac{r_x+r_y}2)$ is the set of midpoints between $B(x,r_x)$ and $B(y,r_y)$.
The second condition, $|\nabla \textrm{dist}_{\partial\Omega}|\stackrel{ae}{=}1$ holds for all open domains --- nothing it needed. Indeed, $\textrm{dist}_{\partial\Omega}$ is $1$-Lipschitz. Note that if $\textrm{dist}_{\partial\Omega}$ is differentiable at $x$, then $|\nabla_x \textrm{dist}_{\partial\Omega}|\le 1$. Now, by Rademacher's theorem, $\textrm{dist}_{\partial\Omega}$ is differentiable almost everywhere. Assume that $v$ be the unit vector direction in the direction of a shortest way from $x$ to the boundary. Then $(d\textrm{dist}_{\partial\Omega})(v)=-1$ therefore $|\nabla_x \textrm{dist}_{\partial\Omega}|\ge 1$.