Let $\Omega$ open bounded be given and $G(x,y)$ denote the Green function in Evans setting. That is, we have $$ \Delta_yG(x,y)=\delta_{x-y} $$ and $$ G(x,y)=0 $$ if $y\in\partial\Omega$.
Now for $u$ we define $$ u(x):=\int_\Omega G(x,y)f(y)dy $$ where $f(y)$ is bounded and integrable. I want to prove that $$ u(x)\to0 \,\,\,\text{ as }x\to \partial\Omega$$
This is intuitively right but I can not prove it in details... Please help me with that. Thank you!
Here's an outline
First of all note that, by the maximum principle, there is a constant $c=c_n$ (the normalization constant in the definition of the fundamental solution for the laplacian), so that $|G(x,y)|\leq c|x-y|^{2-n}$. Therefore, by a variant of the dominated convergence theorem, it's enough to show that $$ \int_\Omega \frac{|f(y)|}{|x-y|^{n-2}} dy \to \int_\Omega \frac{|f(y)|}{|z-y|^{n-2}}dy, \qquad \text{ when } x\to z\in \partial \Omega. $$ This is easy to see: Clearly the integrands converge pointwise a.e. in $\Omega$, and moreover the functions $f_x(y)=|x-y|^{2-n}$ are uniformly bounded (in $x$) in $L^p(\Omega)$ for $p<n/(n-2)$, and thus they're weakly convergent as $x\to z$ by uniqueness of the pointwise and weak limits. But since $f\in L^\infty(\Omega) \subset L^{p'}(\Omega)$, we conclude the desired limit.