Prove $\int_\Omega G(x,y)f(y) dy \to 0$ as $x\to \partial \Omega$ Let $G(x,y)$ be the Green function for the Laplacian on a bounded domain $\Omega$. If $f$ is bounded and measurable, how can I show that
$$u(x)=\int_\Omega G(x,y)f(y)\text dy\to 0$$
as $x\to\partial\Omega$?
I found this question but I am afraid I am still confused. It does not work for the case $n=2$ since $G$ cannot be bounded by the fundamental solution in that case. I also fail to see how exactly are we showing that the limit vanishes.
This is problem 2.3 from Gilbarg and Trudinger's book on elliptic PDE, so I assume a simpler solution should be possible, one not relying on weak convergence or Mazur's lemma (which have not been covered in the book yet).
 A: Throughout my answer I'll use the notation adopted in Gilbarg and Trudinger.  Since $f$ is bounded $$ \vert u(x) \vert \leqslant \int_\Omega \vert G(x,y) \vert \cdot \vert f (y) \vert d y\leqslant \| f \|_{L^\infty(\Omega)}\int_\Omega \vert G(x,y) \vert  d y. $$ Note that, by definition, if $x_0 \in \partial \Omega $ then $G(x_0,y)=0$ for all $y \in \Omega$. Hence, $$\int_\Omega \vert G(x_0,y) \vert  d y =0. $$ Thus, all we need to show is that $$ \int_\Omega \vert G(x,y) \vert  d y \to \int_\Omega \vert G(x_0,y) \vert  d y\tag{$\ast$}$$ as $x \to x_0$. This can be achieved by the general dominated convergence theorem, see this. (If you're not a fan of this method I can give a second method). Indeed, for $n>2$, $$ \vert G(x,y) \vert \leqslant C \vert x-y \vert^{2-n}$$ (I can add more detail about this if you would like). Hence, \begin{align*}
\int_\Omega \vert G(x,y) \vert  d y &\leqslant C\int_\Omega \vert  x - y  \vert^{2-n}  d y.
\end{align*} Choose $R>0$ such that $\vert B_R(x) \vert = \vert \Omega\vert $. Then  \begin{align*}
\int_\Omega \vert  x - y  \vert^{2-n}  d y&\leqslant \int_{B_R(x)}\vert  x - y  \vert^{2-n}  d y \\
&= \int_0^R \int_{\partial B_r(x)}\vert  x - y  \vert^{2-n}ds_x d r \\
&= n \omega_n \int_0^R r  d r \\
&= \frac12 n \omega_n R^2 < \infty,
\end{align*} so $y\mapsto \vert x-y \vert^{2-n}$ is integrable. Clearly, $\vert x-y \vert^{2-n} \to \vert x_0-y \vert^{2-n} $ pointwise as $x \to x_0$. Finally, by the mean-value theorem and the estimates (2.14) in Gilbarg and Trudinger,\begin{align*}
\big \vert  \vert  x - y  \vert^{2-n} -\vert  x_0 - y  \vert^{2-n} \big \vert \leqslant C \vert x-x_0 \vert \vert \hat{x} - y \vert^{1-n}
\end{align*} for some $\hat{x}$ between $x$ and $x_0$. By the same method as before \begin{align*}
\int_\Omega  \vert \hat{x} - y \vert^{1-n} d y \leqslant C
\end{align*} for some constant independent of $\hat{x}$. Hence, $$\bigg \vert  \int_\Omega \vert  x - y  \vert^{2-n} -\vert  x_0 - y  \vert^{2-n} dy \bigg \vert \leqslant C \vert x-x_0\vert \to 0$$ as $x \to x_0$. Thus, $\vert x-y \vert^{2-n} \to \vert x_0-y \vert^{2-n} $ in $L^1(\Omega)$ and so ($\ast$) follows from the general dominated convergence theorem.
