TITLE: The Hilbert transform of analytic function is still analytic.
QUESTION [5 upvotes]: If a function $f$ is analytic in the strip $\mathcal{D}_d = \left\{ z \in \mathbb{C} : |\Im(z)| < d \right\}$, how to show that the Hilbert transform of $f$, which is $\mathcal{H}f(x) = p.v. \int_{\mathbb{R}} \frac{f(t)}{x-t}dt$ ($p.v.$ means Cauchy principal value), is also analytic in this strip?

REPLY [3 votes]: I have enough reputation that I can see a deleted answer.  And that answer seems correct to me.
Of course, there is a problem of defining $\mathcal{H}f(x) = \text{p.v.} \int_{\mathbb{R}} \frac{f(t)}{x-t}\,dt$ for functions that are analytic on $\mathcal{D}_d = \left\{ z \in \mathbb{C} : |\Im(z)| < d \right\}$.  But if the function $f$ is sufficiently bounded (e.g. $|\hat f(\zeta)| \le C e^{-(d+\epsilon)|\zeta|}$ for all $\zeta \in \mathbb R$ and some $\epsilon>0$), so that the Hilbert transform is well defined, then one could argue like this:
$$ \frac d{dz} \mathcal{H}f(z) = \frac d{dz} \text{p.v.}\int_{\mathbb{R}} \frac{f(z-t)}{t}\,dt = \text{p.v.}\int_{\mathbb{R}} \frac{f'(z-t)}{t}\,dt .$$
Or if you feel squeamish about pulling the derivative inside of the integral, instead use a contour integral:
$$ \oint_\gamma \mathcal{H}f(z) \, dz = \text{p.v.}\int_{\mathbb{R}} \frac{\oint_\gamma f(z-t)\, dz}{t}\,dt = 0 $$
and then use Morea's Theorem.