In order to prove a result about Hilbert transforms, I need to show the complex function $F(z) = \arcsin(z)-i\operatorname{Log}(-iz)-i\log 2$ lies in $H^2(\mathbb{H})$, the Hardy Space for the upper half plane. In other words, I need to show that
\begin{equation} \sup_{y\in(0,\infty)}\int_{-\infty}^{\infty}|F(x+iy)|^2\,dx < \infty. \end{equation}
However, the nature of the function makes this very fiddly.
I've had two main thoughts so far.
(1) $\arcsin(z) = -i\operatorname{Log}(iz + \sqrt{1-z^2})$, so with some more algebra bashing, we can write $F(z) = i\operatorname{Log}\left(\frac{1}{2} + \frac{i}{2}\frac{\sqrt{1-z^2}}{z}\right)$, which is a significantly nicer form and highlights how $F(z) \to 0$ as $z \to \infty$ in $\mathbb{H}$. It doesn't seem nice enough to bound however.
(2) If you fix $x \in \mathbb{R}$, it seems as if $|F(x+iy)|$ increases as $y$ decreases to $0$ from above. If I could prove this, I'd be done, as then we could bound $\sup_{y\in(0,\infty)}\int_{-\infty}^{\infty}|F(x+iy)|^2\,dx$ by $\int_{-\infty}^{\infty}|F(x)|^2\,dx$ which certainly is finite. However, it's proving very difficult to work with $|F(x+iy)|$ as a function of $y$.
Supposedly, this result is "straightforward to prove" so I'm hoping someone has a nice way of proving this result.
