Let $R>0$ and $u:\mathbb{C}\setminus B(0,R)\rightarrow \mathbb{R}$ be a continuous function such that $u$ is harmonic on $\mathbb{C}\setminus\overline{B(0,R)}$. Assume that $u$ is bounded at $\infty$.
How do I prove that $u(z)= - \frac{1}{2\pi} \int_0^{2\pi } \frac{ R^2 - |z|^2 }{|Re^{i\theta}-z|^2} u(Re^{i\theta}) d\theta$?
I could prove it when $u$ vanishes at infinity, but I'm not sure how to prove this when $u$ is just bounded at infinity. (I think this is false now) If this is false, what would be a counterexample?
Hint: Assume $R=1$ for convenience. The singularity of the bounded harmonic $v(z) =u(1/ z)$ at $0$ is removable. Therefore $v$ equals its Poisson integral in the disc. Now map back.
In fact the problem can be stated as an "iff". Sticking with $R=1,$ suppose $u$ is not bounded at $\infty.$ Then the formula fails. That's because $u$ is bounded on $\{|z|=1\},$ hence the integral is bounded for, say, $|z|>2$ (This is an easy estimate.) Therefore the integral cannot equal $u.$