The Neumann Problem on a Half-space when dimension is $2$ Take $\Omega:=\\{x=(x_1,x_2):\,-\infty<x_1<\infty,\,x_2>0\\}$, i.e., the half-space, and I am interested in the Neumann problem \begin{cases} \Delta u=0&x\in \Omega\\\ \partial_2u(x_1,0)=f(x_1),&-\infty<x_1<\infty \end{cases} where $f$ is continuous. I am able to work out the Green's function for this problem and I have $$ u(x)=\frac{1}{2\pi}\int_{-\infty}^\infty \log |(y_1-x_1)^2+x_2^2|\cdot f(y_1)dy_1 $$ Now I am trying to prove that $u$ is bounded if and only if $\int_{-\infty}^\infty f(x_1)dx_1=0$. I can not do both directions... The problem is for unbounded domain, we can not apply Gauss-Green's formula directly, o.w. the only if part would be easy. But I highly suspect in this case I do can use Gauss-Green's formula, or the divergence formula... But I can not prove it. Also for if part, I don't really have an idea... Any help is really welcome!

If $f$ just bounded the integral can diverge (e.g. for $f\equiv 1$). If $f$ has compact support then passing to the polar coordinates, $x_1=r\cos\varphi$ etc. and expanding the kernel on the powers of $r=(x_1^2+x_2^2)^{1/2}$ we have $$ \log |(y_1-x_1)^2+x_2^2|=2 \log r-\frac{2 y_1 \cos \varphi }{r}+O\left(\frac{1}{r^2}\right),\quad r\to\infty, $$ and $$ u(x)=2 \log r\int_{\mathbb R}f(y_1)\,dy_1+O(r^{-1}). $$

**Update**

May be the requirement of compact support can be replaced by a suitable vanishing of $f$, say as $|x_1|^{-2}$. Perhaps to get a better understanding of necessary and suffitient conditions on $f$ one can map conformly the half-plane on the unit circle. The integral of $f$ will turn into in an integral with some weight on the unit circumference.