# Let $v$ sub-harmonic s.t. $v$ has his maximum in $\Omega$. Then $v$ is constant.

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.

• 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/… – Thomas Nov 5 '17 at 11:36
• @Thomas : Yes (since $\Delta u$ is well defined ;-)). Thanks for the link :-) – user386627 Nov 5 '17 at 11:49