Superharmonicity of the distance function

Suppose $$V$$ is a convex open proper subset of $$\mathbb{R}^m$$ ($$m\geq2$$). It is known that the function $$u(x)=$$dist$$(x,\partial V)$$ is superharmonic on $$V$$. Is there a similar result without $$V$$ being convex? That is a positive superharmonic function $$u$$ on $$V$$ that vanishes at each point of the boundary of $$V$$ and goes to $$\infty$$ at infinity, in case $$V$$ is unbounded?

• If you replace "at each point of the boundary" by "at each regular point of the boundary", then yes, of course: just take the Green potential of an arbitrary finite, positive measure. It cannot go to infinity in the usual sense, though, unless the complement of $V$ is compact (in particular, if $V$ is an unbounded convex set). Apr 12 '21 at 7:10
• Thanks. If I am not mistaken, you are saying that if $V=\mathbb{R}^m\setminus K$ with $K$ compact, then the Green potential of $V$ goes to infinity at infinity (correct?). But it seems to me that the Green function of $V$ goes to zero at infinity; so how come the Green potential goes to infinity? Apr 12 '21 at 19:45
• No, this is not what I intended to say. I just wanted to point out that if $V$ is not the complement of a compact set, then there is a sequence $x_n$ such that $|x_n| \to \infty$, but $x_n$ is close enough to the boundary of $V$ so that $u(x_n) \to 0$. Apr 12 '21 at 20:10