What can be said about the Hilber transform of an entire function?

Question

Is the Hilbert transform of an entire function also entire? I can't find much information there, mostly due to confusing terminology of entire function = analytic function ≠ analytic signal...

Answer 1

Let $T(x) = pv(\frac{1}{x}) = \frac{d}{dx}\log |x|$ a distribution of order $\le 1$.

• If $f$ is entire and $L^1$ on the real axis and $$\forall y, \qquad \int_{-\infty}^\infty (|f'(x+iy)|+|f''(x+iy)|)| \log |x|\, | dx < \infty$$ then its Hilbert transform $$F(z) = \int_{-\infty}^\infty f(z-x) T(x)dx= \int_{-\infty}^\infty f'(z-x)\log |x| dx$$ is everywhere well-defined and holomorphic so it is entire.

• Assuming $F(z) = \int_{-\infty}^\infty f(z-x)T(x)dx$ converges for $|\Im(z)| < \epsilon$ this is a iff because if $F$ is entire then $F_n = \int_{-n}^n f(z-x)T(x)dx \to F$ locally uniformly.

• What means analytic signal ? If $h(x)$ is a signal write $$h(x) = \int_{-\infty}^\infty H(\omega) e^{i \omega x}d\omega$$ Assume $|H(\omega)|\le C$. Then $$h_a(z) = h \ast (iT+\pi \delta)(z)= \int_0^\infty H(\omega) e^{i \omega z}d\omega$$ is complex analytic for $\Im(z) > 0$ (and not entire). Note if $h$ is real then $h(x) = \frac{2}{\pi}\Re(h_a(x))$