I want to prove the following:
Let $\Omega \subset \mathbb{R}^n$ be a domain and $u:C^2(\Omega) \cap C(\overline{\Omega}) \rightarrow \mathbb{R}$ be a superharmonic function, i.e. $-\Delta u \geq 0$.
Then, it satisfies the minimum principle:
$\displaystyle\min_{\overline{\Omega}}u = \displaystyle\min_{\partial{\Omega}}u$.
I have previously proven the mean value inequality for $u$, which I think I'll have to use:
For all $x \in \Omega$ and $r>0$ such that $B(x,r) \subset \Omega$, it holds that: $$\displaystyle u(x) \geq \frac{1}{|\partial{B(x,r)}|}\int_{\partial{B(x,r)}} u \ dS $$
I tried to start with considering a point $x_0$ s.t. $u(x_0) = \displaystyle\min_{\overline{\Omega}}u = \mu $.
Then for any $0<r< dist(x_0, \partial{\Omega})$, we have:
$\mu = \displaystyle u(x_0) \geq \frac{1}{|\partial{B(x_0,r)|}}\int_{\partial{B(x_0,r)}} u(y) \ dS(y) \geq \mu $.
This inequality holds for any $0 < r' \leq r$ and hence the continuity of $u$ implies that $u(y) = u(x_0)= \mu$ for all $y \in B(x_0,r)$.
Can you give me a hint on how to proceed?
Best Answer
Firstly, we should note that $\Omega$ is a bounded domain in $\mathbb R^n$ which you omit there.
It’s the weak minimum principle, we have two different ways to prove it.
First way :
Firstly, we consider the case $-\Delta u>0$ in $\Omega $. If there is a $x_0\in\Omega$ such that $$u(x_0)=\min_{\bar\Omega}u(x).$$ By necessity for extreme values, we know that the hessian matrix at $x_0$ is semi-positive definite, which means $$-\Delta u(x_0)\leq 0.$$ It’s a contradiction. That is $$\displaystyle\min_{\bar{\Omega}}u = \displaystyle\min_{\partial{\Omega}}u.$$ Generally, for the case $-\Delta u\geq 0$ we define for $\forall \epsilon>0$ $$v(x)=u(x)-\epsilon|x|^2\quad in\,\Omega.$$ Hence, we obtain $$-\Delta v=-\Delta(u-\epsilon|x|^2)=-\Delta u+2n\epsilon>0.$$ Thus $$\displaystyle\min_{\bar{\Omega}}v= \displaystyle\min_{\partial{\Omega}}v.$$ Note that $$\min_{\bar\Omega}u\geq \min_{\bar\Omega}v=\min_{\partial\Omega}v\geq\min_{\partial\Omega}u -\epsilon \max_{\partial\Omega}|x|^2.$$ Let $\epsilon\to 0$ yields $$\min_{\bar\Omega}u\geq\min_{\partial\Omega}u$$ since $\Omega$ is bounded. And it’s clear that $$\min_{\bar\Omega}u\leq\min_{\partial\Omega}u$$ which gives us $$\min_{\bar\Omega}u =\min_{\partial\Omega}u.$$
Second way :
In this part, we will use mean value inequality.
Let $m:=\min_\limits{\bar\Omega}u$ and $D:=\{x\in\Omega: u(x)=m\}$. It’s clear that D is relative closed to $\Omega$. Next, we will prove D is relative open.
For $x_0\in D$, by mean value inequality we know $$m= \displaystyle u(x_0) \geq \frac{1}{|\partial{B(x_0,r)|}}\int_{\partial{B(x_0,r)}} u(y) \ dS(y) \geq m$$for any $0<r< dist(x_0, \partial{\Omega})$, which yields $u(y) = u(x_0)= m$ for all $y \in B(x_0,r)$.
Thus $B(x_0,r)\subset D$, that is D is relative open.And since $\Omega$ is connected, we get $D=\Omega$ or $D=\phi$.
Hence, we know $$\min_{\bar\Omega} u=\min_{\partial\Omega} u.$$
If you want to learn more about maximal principle, we refer you to Han Qing and Lin Fanghua’s elliptic pde.