Let's consider the usual Hilbert transform $H$ defined as $$Hf = P.V. (\frac{1}{x}*f).$$ A wellknown 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 zeropoint $x_0$ of infinite order, that is, $f^{(m)}(x_0) = Hf^{(m)}(x_0) = 0$ for any nonnegative integer $m$, then $f\equiv 0$? We can assume that $f$ is smooth to make the statement more rigorous.
1 Answer
No. Let $$u(z) = \exp((iz)^{1/2}(iz)^{1/2})$$ for $z$ in the closed upper complex halfplane, with the principal branch of the complex power. Then $u$ is a bounded holomorphic function in the open halfplane, 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 $x = 0$.

$\begingroup$ 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 LuOct 7, 2020 at 17:05

$\begingroup$ This seems to be quite standard. Wikipedia entry points to Titchmarsh and provides a (nonrigorous) explanation. $\endgroup$ Oct 7, 2020 at 20:52
