I have a question to the following problem.
Let $f\in C(K(0,1))\subset \mathbb{R}^2.$ Prove that $\sup_{K(0,1)}|u|\leq \frac{1}{4}\sup_{K(0,1)}|f|,$ if $u(x,y)$ is a solution to the problem:
$\Delta u=f, \ \mbox{on} \ K(0,1),$ $u(x,y)=0 \ \mbox{the boundary of} \ K(0,1).$
Let $G(x,y,x',y')$ be the appropriate Green function for $K(0,1).$
I tried to solve the problem, but came to a problem. My solution goes like this: $|u(x,y)|=|\int_{K(0,1)}G(x,y,x',y')f(x',y')~dx'~dy'|\leq \sup_{K(0,1)}|f|\int_{K(0,1)}|G(x,y,x',y')|~dx'~dy'.$
But now I have a problem. How to get rid of the absolute value inside the integral. Because, if it were not there, then $\int_{K(0,1)}G(x,y,x',y')~dx'~dy'=\frac{x^2+y^2-1}{4},$ which is the unique solution to the problem: $\Delta u=1, \ \mbox{on} \ K(0,1),$ $u(x,y)=0 \text{ the boundary of } K(0,1).$ Then it is easy to establish, that $|\frac{x^2+y^2-1}{4}|\leq \frac{1}{4}$ on $K(0,1).$
Can someone please tell me how to write a fully correct solution to this problem.