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).
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.