# Unique continuation of the Hilbert transform

Let's consider the usual Hilbert transform $$H$$ defined as $$Hf = P.V. (\frac{1}{x}*f).$$ A well-known unique continuation principle states that if $$Hf = f =0$$ on some interval $$I$$, then $$f \equiv 0$$. My question is whether the argument is still true if we replace the interval $$I$$ with a point $$x_0$$. More specifically, can we prove that if both the function $$f$$ and its Hilbert transform $$Hf$$ have a zero-point $$x_0$$ of infinite order, that is, $$f^{(m)}(x_0) = Hf^{(m)}(x_0) = 0$$ for any non-negative integer $$m$$, then $$f\equiv 0$$? We can assume that $$f$$ is smooth to make the statement more rigorous.

No. Let $$u(z) = \exp(-(-iz)^{1/2}-(-iz)^{-1/2})$$ for $$z$$ in the closed upper complex half-plane, with the principal branch of the complex power. Then $$u$$ is a bounded holomorphic function in the open half-plane, continuous up to the boundary, and vanishing sufficiently fast at complex infinity. Thus, the Hilbert transform of $$f(x) = \Re u(x) = \Re \exp(-e^{-i \pi/4 \operatorname{sign} x} |2x|^{1/2} - e^{i \pi/4 \operatorname{sign} x} |2x|^{-1/2})$$ is given by $$Hf(x) = \Im u(x) = \Im \exp(-e^{-i \pi/4 \operatorname{sign} x} |2x|^{1/2} - e^{i \pi/4 \operatorname{sign} x} |2x|^{-1/2}).$$ A standard argument shows that both $$f$$ and $$Hf$$ are smooth and have a zero of infinite order at $$x = 0$$.
• Many thanks for your answer! I understand that now $f$ and $Hf$ vanish of infinite order at zero. But I am not very familiar with complex analysis, may I ask you why the Hilbert transform of $f$ is given by the imaginary part of $u(x)$? Is there a reference for the general theory of it? Thanks a lot! Oct 7, 2020 at 17:05