The function $u(x, y)$ satisfies Laplace's equation $$ u_{x x}(x, y)+u_{y y}(x, y)=0 \quad(x \in \mathbb{R}, y>0) $$ and satisfies the boundary conditions $$ u(x, 0)=f(x) $$ where $f$ is integrable, $u(x, y)$ remains bounded as $x^{2}+y^{2} \rightarrow \infty .$ Then, $$ u(x, y)=\frac{y}{\pi} \int_{-\infty}^{\infty} \frac{f(s) \,\mathrm{d} s}{y^{2}+(x-s)^{2}}. $$
In the official solution of this problem, it uses that $\mathcal{F}(u)$ remains bounded as $y \to \infty$, but why is this? That is, $$\mathcal{F}(u) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} u(x,y) e^{-i\omega x} \, \mathrm{d}x.$$ How do I know this is bounded? Just the real part of the integral is $\int_{-\infty}^{\infty} u(x,y) \cos (\omega x)\, \mathrm{d}x$, and if $u(x,y)$ is somehow equal to the sign of $\cos (\omega x)$ then the integral is not bounded. But then $u(x,y)$ depends on $\omega$, so this isn't necessarily a contradiction.
Also, a more common boundary condition is that $u(x,y) \to 0$ as $x^2 + y^2 \to \infty$. In the same spirit as above, how can I show that $\mathcal{F}(u)$ is bounded as $x^2 + y^2 \to \infty$ in this case? Surprisingly, it wasn't as easy as it had seemed at first. Any partial answers (answering only one of the questions) are very welcome! I will upvote them too.