# Proving $\sup_{K(0,1)}|u|\leq \frac{1}{4}\sup_{K(0,1)}|f|$ using Green function

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.

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I changed $\displaystyle sup_{K(0,1)}$ to $\displaystyle\sup_{K(0,1)}$. That is standard usage. The backslash in \sup not only prevents "$\sup$" from getting italicized, but also causes the subscript to appear directly under it (when in "display" style as opposed to "inline") and makes proper spacing appear before and after it. But I wonder: Is there some web site somewhere giving $\TeX$ advice that condones the use of \mbox{} as a substitute for \text{}? (I changed that too.) – Michael Hardy Jul 4 '12 at 16:54
I guess there is a typo in the first displayed equation. – Davide Giraudo Jul 4 '12 at 17:21
+1 for good title. What do you mean by $K(0,1)$? – leo Oct 6 '12 at 14:56

It is enough to prove than $G$ is non-positive. This can be done using the maximum principle and the fact that $$\lim_{(x,y)\to(x',y')}G(x,y,x',y')=-\infty,\quad (x',y')\in K.$$
Is that true for all Green functions? Now that you have told me the idea I have just explicitly calculated, that the Green function for the circle $K(0,1)$ is nonpositive. – user35058 Jul 4 '12 at 18:30