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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.

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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 $x = 0$.

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  • $\begingroup$ 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! $\endgroup$
    – Jacob Lu
    Oct 7, 2020 at 17:05
  • $\begingroup$ This seems to be quite standard. Wikipedia entry points to Titchmarsh and provides a (non-rigorous) explanation. $\endgroup$ Oct 7, 2020 at 20:52
  • $\begingroup$ I see. Thank you very much! $\endgroup$
    – Jacob Lu
    Oct 8, 2020 at 0:45

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