Concavity/convexity of distance-to-boundary function

Question

For $\Omega$ a bounded open set of $\mathbf{R}^d$, denote $\mathrm{d}_\Omega:x\mapsto \mathrm{d}(x,\partial\Omega)$ the distance-to-boundary function.

If $\Omega$ is convex, a short argument recalled here by Anton Petrunin proves that $\mathrm{d}_\Omega$ is a concave function inside $\Omega$. In particular, if $\Omega$ is smooth then in a neighborhood of $\partial\Omega$ and inside $\Omega$, the hessian matrix of $\mathrm{d}_\Omega$ is non-positive.

I would like to know if outside $\Omega$ (but still in a neighborhood of $\partial\Omega$), this hessian matrix still has a sign or equivalently if $\mathrm{d}_\Omega$ is convex or concave in $\mathbf{R}^d\setminus \Omega$.

I am a bit disturbed because on the one hand the short argument above does not seem to apply outside $\Omega$ (in a way or another) but at the same time I would expect that the convexity of the "enclosed" volume defined by a closed surface (here $\partial\Omega$) could be guessed directly from it (or using its distance function).

Answer 1

The signed distance function $f=\pm\textrm{dist}_{\partial \Omega}$ is concave everywhere; here $f=\textrm{dist}_{\partial \Omega}$ inside and $f=-\textrm{dist}_{\partial \Omega}$ outside.

The proof is straightforward, for inside you know it, for outside it is very similar:

Assume $B(x,r_x)\cap \Omega\ni p$ and $B(y,r_y)\cap \Omega\ni q$. It is sufficient to show that $$B(\tfrac{x+y}2,\tfrac{r_x+r_y}2)\cap \Omega\ne \varnothing.$$ The latter follows since $\tfrac{p+q}2\in B(\tfrac{x+y}2,\tfrac{r_x+r_y}2)$.