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?
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1$\begingroup$ 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). $\endgroup$– Mateusz KwaśnickiCommented Apr 12, 2021 at 7:10
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$\begingroup$ 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? $\endgroup$– M. RahmatCommented Apr 12, 2021 at 19:45
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1$\begingroup$ 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 $u(x_n) \to 0$. $\endgroup$– Mateusz KwaśnickiCommented Apr 12, 2021 at 20:10
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