Example of a compactly supported Lipschitz function with non-Lipschitz Hilbert transform

Question

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.

Answer 1

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., $x\log |x|$.