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Let $\Omega\subset \mathbb R^n $ a bounded open domain and $v:\Omega \longrightarrow \mathbb R$ sub-harmonic i.e. $\Delta v\geq 0$. Show that if $v$ has his maximum in the interior of $\Omega $, then $v$ is constant in $\Omega $.

Proof

Let $x_0\in \Omega $ s.t. $v(x_0)\geq v(x)$ for all $x\in \Omega $. I'm trying to show that $v(x)=v(x_0)$ for all $x\in \Omega $. I know that $\Delta v(x_0)=0$ and $\nabla v(x_0)=0$ but I cant conclude. I also know that $v(x)\leq \frac{1}{|B(x_0,r)|}\int_{B(x_0,r)}u(t)dt$ for all $B(x_0,r)\subset \subset \Omega $, but I can't conclude.

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  • $\begingroup$ Are you assuming that $v$ is continuous and do you know the corresponding result for harmonic functions? If yes then your question has already been answered here: math.stackexchange.com/questions/1489107/… $\endgroup$ – Thomas Nov 5 '17 at 11:36
  • $\begingroup$ @Thomas : Yes (since $\Delta u$ is well defined ;-)). Thanks for the link :-) $\endgroup$ – user386627 Nov 5 '17 at 11:49

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