In this article in page 3:
http://arxiv.org/pdf/0904.1332.pdf
says:
If I consider $\Omega$ a open bounded and convex set, then the function $\delta (x) = \displaystyle\min_{y \in \partial \Omega } |x-y|$ ($x \in \Omega$) is p - superharmonic with $p >1$.
Someone know if the function $\theta (x) = d (x ,A)$ ($x \in \Omega - A$) (d denotes the set distance) where $A \subset \Omega$ with $\overline{A} \subset \Omega$ is p-superharmonic in the weak sense in $\Omega - A$(p>1) ?
Let $\Omega\subset\mathbb{R}^N$, $N\geq 2$, be an open set containing the origin and $A=\{0\}$. Consider the function $$u(x)=\operatorname{d}(x,A)=\|x\|$$
Note that $u$ is subharmonic for every open set in $\mathbb{R}^N$ not containing the origin, hence, $u$ is weakly subharmonic in every open set not containing the origin. This implies that for any $\varphi\geq 0$ such that $\varphi\in C_0^\infty(U)$, with $U$ open and $\overline{U}\subset\Omega\setminus A$, we have that $$\int_{\Omega\setminus A} \nabla u\cdot\nabla\varphi dx=\int_U\nabla u\cdot\nabla\varphi\geq 0$$
Therefore the function $u$ cannot be weakly superharmonic in $\Omega\setminus A$, however, is its weakly subharmonic.