TITLE: Convexity of distance-to-boundary function
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: Here is another answer to the convexity (or more precisely concavity) part of the question, which I think is even more simple than the one Anton gave:
Let $p\in\Omega$, and $q$ be one of its closest points on $\partial\Omega$. Since $\Omega$ is convex, there exists a support hyperplane $H$ of $\Omega$ which passes through $q$. Furthermore $H$ is orthogonal to $pq$, since it supports the sphere $S_p$ centered at $p$ and passing through $q$. So $d_{\partial\Omega}(p)=d_H(p)$. Clearly $d_H(p)$ is not bigger than $d_{H'}(p)$ for any other support plane $H'$ of $\Omega$, because all these planes lie outside $S_p$. So 
$d_{\partial\Omega}$ is the infimum of the distance functions to support hyperplanes of $\Omega$. These functions are linear and therefore concave.  So $d_{\partial\Omega}$ is concave.