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.