Consider the following the theorem in the classical PDE book of Evans (chapter 2):
(Part of the strong maximum principle) Let $U$ a open set in $R^n$ and $u \in C^2 (U) \cap C(\overline{U})$, with $\Delta u = 0$ in $U$.
If $U$ is connected and there exists a point $x_0 \in U$ such that
$$ u(x_0) = \displaystyle\max_{\overline{U}} u$$
then $u$ is constant within $U$.
Part of the proof:
Suppose there exists a point $x_0 \in U$ with $u(x_0) = M = \displaystyle\max_{\overline{U}} u . $ Then for $0 < r < \mbox{dist} (x_0 , \partial U)$, the mean value property asserts
$$ M = u(x_0) = \displaystyle\frac{\displaystyle1}{|B(x_0, r)|}\int_{B(x_0,r) } u \ dy \leq M.$$Then
$$u = M\quad\text{in}\quad B(x_0 , r)\tag{$*$}.$$
I dont understand the equality in $(*)$. If I be non rigorous, for me is clear to see the equality in $(*)$. But I dont know how to prove the equality… Someone can give me a hint about how to prove the equality in $(*)$?
Thanks in advance!
Best Answer
We can split the ball $B=B(x_0,r)=E_1\cup E_2\cup E_3$, where $$E_1=\{x:u(x)<M\},\qquad E_1=\{x:u(x)=M\},\qquad E_3=\{x:u(x)>M\}$$ these sets are disjoint so $$\int_Budx=\int_{E_1}udx+\int_{E_2}udx+\int_{E_3}udx$$ By the assumption on $x_0$ we must have $E_3=\emptyset$, but then $E_1$ must be empty too, and hence $B=E_2$.