distance function is p- superharmonic?

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) ?

• I am doing this question beacause if the answer is "yes" then I understand a affirmation in a proof of a theorem that I am studying – math student Oct 11 '13 at 1:24
• What is the relation of your question with the paper you have cited? – Tomás Oct 11 '13 at 1:37
• Are you asking whether $\theta$ is p-harmonic, or p-superharmonic? And do you have a link to the proof you are studying? – Anthony Carapetis Oct 11 '13 at 2:19
• @Tomás . Eu citei o paper pois é meio que razoavel de se esperar que a funcao $\theta$ satisaça o que eu escrevi. No paper que estou lendo parece que o autor usa ( e nem menciona) que $\theta$ satisfaz o que escrevi na questao – math student Oct 11 '13 at 3:22
• For $p=2$ at least it seems the convexity of $A$ is necessary: jstor.org/stable/2045528. Since distance functions have unit gradient almost everywhere it feels like $p$ should not be too relevant - certainly in the smooth case harmonicity and $p$-harmonicity are equivalent when $|\nabla \theta| = 1$. I'm unsure if the weak case differs. – Anthony Carapetis Oct 11 '13 at 3:31

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.