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](https://mathoverflow.net/questions/291308/convexity-of-distance-to-boundary-function) 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).



Concavity/convexity of distance-to-boundary function
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 $\tfrac{p+q}2\in B(\tfrac{x+y}2,\tfrac{r\_x+r\_y}2)$.


