I'm reading a proof from my teacher about the weak principle of the maximum that is,
$$\max_{\overline{\Omega}} u = \max_{\partial\Omega} u$$
when $u$ is harmonic ($\Delta u = 0$) in $\Omega$ and $u\in C^2(\Omega)\ (\overline\Omega)$, and $\Omega$ is an open bounded.
It goes like this:
Let's name $M = \max_{\overline{\Omega}} u$ and $m=\max_{\partial\Omega} u$. It is clear that $m\le M$ because $\partial\Omega\subset\overline\Omega$. Suppose that $m<M$. Let's pick a point $x_0\in\Omega$ such that $u(x_0) = M$
Then we can construct the following function:
$$v(x) = u(x) + \frac{|x-x_0|}{2d^2}(M-m)$$
where $d$ is the diameter of $\overline\Omega$
and we see that
$$v(x_0) = u(x_0) = M$$
For $x\in\partial\Omega$:
$$v(x) \le m+\frac{1}{2}(M-m) = \frac{1}{2}(M+m) < M$$
Therefore there exists $x^*\in\Omega$ such that $v(x^*) = \max_{\overline\Omega} v$. Since it's a maximum, $$\Delta v(x^*) \le 0 \tag{1}$$
The main question is: why $v(x^*)$ is a maximum for $v$ in $\overline\Omega$?. The only thing I can conclude is that $v$
Now we'll find a contradiction to $(1)$:
$$\Delta v = \Delta u + \frac{M-m}{2d^2}\Delta(|x-x_0|^2) = \frac{M-m}{2d^2}2N >0 \mbox{ in $\Omega$}$$
Q.E.D.
I think since $v$ in $\partial\Omega$ has value less than $M$, it's only possible place to attain a maximum is is in $\Omega$. There are still some details on why. For example, what if $v$ is concave downwards, therefore has no maximum in $\Omega$?
Best Answer
Here's a proof that seems simpler to me. Suppose that $M>m$. Pick $x_0$ with $u(x_0)=M$, such that $x_0$ is at minimal distance to the boundary $$d(x_0,\partial\Omega)=\min\{d(x,\partial\Omega):u(x)=M\}.$$
Now, assuming we're in $\Bbb R^2$ just to simplify the notation, if $0<r<d(x_0,\partial\Omega)$ then $$M=u(x_0)=\frac1{2\pi}\int_0^{2\pi}u(x_0+re^{it})\,dt<M,$$contradiction. (That average is less than $M$ because $u(x_0+re^{it})\le M$ for all $t$, but there is an interval $I$ such that $u(x_0+re^{it})<M$ for $t\in I$, since $d(x_0+re^{it},\partial\Omega)<d(x_0,\partial\Omega)$ for $t\in I$.)