Hi, there. Bellow is my attempt. I don't know if I have gone in the wrong way and I am stuck.
My attempt:
Using Green's representation formula, $u(y)=\int_{\partial \Omega}u \frac{\partial G}{\partial v}ds+\int_{\Omega}G\triangle udx=\int_{\partial B_1(0)}g \frac{\partial G}{\partial v}ds+\int_{B_1(0)}Gfdx$
$\le max_{\partial B_1(0)}|g| \int_{\partial B_1(0)}\frac{\partial G }{\partial v}ds+max_{B_1(0)}|f| \int_{B_1(0)}Gdx$.
Then I need to prove exist a $C$ depending only on $n$, such that $\int_{B_1(0)}Gdx \le C$ and $\int_{\partial B_1(0)}\frac{\partial G }{\partial v}ds \le C$.
I already know the functions in a ball:
$\int_{\partial B_1(0)}\frac{\partial G }{\partial v}ds=\frac{1}{nw_n}\int_{\partial B_1(0)}\frac{1-|y|^2}{|x-y|^n}ds(x)$
$G(x,y)=\Gamma(|x-y|)-\frac{R}{|y|^n-2} \Gamma(x-\overline y)$
But how to prove the above two $\le C$, where $C$ depends only on $n$?
Thanks so much!
Use Green's representation on the constant function $1$ to see that $$ \int_{\partial\Omega} \frac{\partial G}{\partial \nu} (x,y) d\sigma(y) =1, \quad \forall x\in B(0,1). $$ On the other hand, by the maximum principle we have $$ G(x,y)\leq C(n)|x-y|^{2-n}, \quad \forall x,y\in \Omega. $$ Now, for every $x\in B(0,1)$ we get $$ \int_{B(0,1)} |x-y|^{2-n} dy \leq \int_{B(x,2)} |x-y|^{2-n} dy = n \omega_n \int_0^2 t dt =C(n), $$ where the first inequality is because $B(0,1)\subset B(x,2)$ and the first equality is a change of variables followed by the formula for the integral of radial functions.