On the I. I. Privalov theorem on the Hilbert transform of Lipschitz functions
Žurnal matematičeskoj fiziki, analiza, geometrii, Tome 11 (2004) no. 4, pp. 380-407 Cet article a éte moissonné depuis la source Math-Net.Ru

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It is known that the Hilbert transform $h(f)$ of a bounded Lipschitz (order one) function $f$ on $\mathbb{R}$ is uniformly continuous ($h$ is understood as the singular integral operator with the Cauchy kernel regularized at infinity, so that $h$ is defined on the class of all functions summable on $\mathbb{R}$ w.r. to the Poisson measure). It is shown that the above theorem does not hold (in a very strong sense) for unbounded Lipschitz f's. Conditions sufficient (and “almost necessary”) for $h(f)$ to be Lipschitz are given. The results are motivated by some uniqueness problems of the Fourier analysis.
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Yu. S. Belov; V. P. Havin. On the I. I. Privalov theorem on the Hilbert transform of Lipschitz functions. Žurnal matematičeskoj fiziki, analiza, geometrii, Tome 11 (2004) no. 4, pp. 380-407. http://geodesic.mathdoc.fr/item/JMAG_2004_11_4_a1/