I'm taking complex analysis course with the book 'Complex Variables with Applications' of Silverman and Ponusamy. I am tryng to solve Exercise 10.35.10 as follows.
Exercise 10.35.10 Formulate and solve a Dirichlet problem for a half-plane. More precisely, prove the following: If $F(x)$ is a continuous function on $\mathbb{R}$, bounded and $\lim_{x\to-\infty}F(x)=\lim_{x\to\infty}F(x)$, then show that the function $u(x, y)$ defined by $$u(x, y) = \frac{y}{\pi}\int_{-\infty}^{\infty}\frac{u(t, 0)}{(t-x)^2+y^2}dt$$ is a solution of the Dirichlet problem in the upper half-plane $\text{Im}(z) > 0$ with the boundary condition $u(x, 0) = F(x)$ for $x\in\mathbb{R}$.
And I tried this with analytic function. If $f(z)$ is an analytic function with real part $u$ on the upper half plane, then we have two equations $$2\pi if(z) = \int_C\frac{f(\zeta)}{\zeta-z}d\zeta, 0=\int_C\frac{f(\zeta)}{\zeta-\overline{z}}d\zeta$$ where $C$ is positively oriented contour consisting of $C_R = \{z:|z|=R, \text{Im}(z)\ge 0\}$ and line segment $[-R, R]$. If we subtract second from the first and do some simplification, we find $$f(z) = \frac{y}{\pi}\int_{-R}^{R}\frac{f(t)}{(t-x)^2+y^2}dt + \frac{1}{\pi}\int_{C_R}\frac{\text{Im}(z)f(\zeta)}{(\zeta-z)(\zeta-\overline{z})}d\zeta.$$ I found that the second integral converges to $0$ as $R\to\infty$ if we guarantee that $f$ is bounded on the upper half plane. This is again guaranteed if $u$ is bounded on the upper half plane, by Borel-Caratheodory Theorem. Now my question arises.
How can we say that $u$ should be bounded? Do we need to assume this? And how can we assume analyticity of $f$ on the region containing the plane $\{z:\text{Im}\ge 0\}$?
I thought of using maximum principle, but I failed since the region is not actually enclosed by the real line.
Besides, I know there is a way to obtain this by using LFT on the original Poisson integral formula. I think that method will have no problem of this kind. But is that the only way? Can I obtain the solution from this method?