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Suppose $f$ is a compactly supported measurable function (say in the interval $[-1,1]$) which is Hölder continuous of order $\alpha\in (0,1)$. I have read that the Hilbert transform $Hf$ of $f$ is also Hölder continuous of order $\alpha$, but that the result is not true at the endpoint $\alpha=1$. Is there an obvious counterexample? I am struggling to come up with one and would appreciate some suggestions.

Silvia Ghinassi
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Matt Rosenzweig
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1 Answers1

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Any "corner" will do, like $|x|$ multiplied by a smooth cutoff. The Hilbert transform commutes with taking distributional derivative (both are multiplication operators on the Fourier side). So, if you can find an $L^\infty$ function whose Hilbert transform is not in $L^\infty$, then its antiderivative gives the desired Lipschitz example.

Specifically: the Hilbert transform of the signum function has a logarithmic singularity at $0$, which can be seen either directly from the integral formula, or by observing that $\operatorname{Re}\log z$ and $\operatorname{Im}\log z$ are conjugate harmonic functions on the upper half-plane. Therefore, the Hilbert transform of a function that behaves like $|x|$ near $0$ behaves like $\int \log |x|\,dx$ near $0$, i.e., $x\log |x|$.

  • Perhaps you can help me with an even more basic question. It's easy to find a compactly supported $L^{\infty}$ function $f$ such that $Hf$ is not in $L^{\infty}$; take $f=\chi_{[a,b]}$. I want to find a $C_{c}(\mathbb{R})$ function $f$ such that the distribution $Hf$ does not coincide with a continuous function. It's easy to find a $C_{c}(\mathbb{R})$ function such that principal value integral diverges to $\infty$ at a point (e.g. $1/log|x|$ smoothly cutoff). However, this isn't what I want. Do you see a nice way to go about this without brushing a bunch of asymptotics under the rug? – Matt Rosenzweig Oct 27 '15 at 04:23
  • The failure of $C^0\to C^0$ is more subtle than the failure of $L^\infty\to L^\infty$. I usually reach for conformal maps as examples, but apparently this is not what you want. I suggest posting a separate question. That said, $f(x) = \operatorname{sign}x / \log|x|$ should work. –  Oct 27 '15 at 04:43