A smoothness problem for an integral of Cauchy type
Let L be a rectifiable Jordan curve separating the complex plane into two domains \(D:=int L\) and \(G:=ext L\). Given a continuous function f on L, put \(\tilde f:=f\) on L and \[ \tilde f(z):=(1/2\pi i)\int_{L}(f(\zeta))/(\zeta -z)d\zeta,\quad z\in D. \] We say that L satisfies the Plemelj-Privalov theorem (shortly: PPT), if for every \(f\in H^{\alpha}(L)\) \((0<\alpha <1)\) the corresponding function \(\tilde f\in H^{\alpha}(\bar D)\) (Hölder classes). It is known [\textit{P. M. Tamrazov}, Smoothness and polynomial approximations (1975; Zbl 0351.41004)] that L satisfies PPT, if \[ \theta_ L:=\sup_{z\in L}mes\{\zeta \in L;\quad | \zeta -z| \leq \delta \}\approx \delta,\quad 0<\delta \leq diam L. \] Moreover, in the absence of this condition, the smoothness of \(\tilde f\) may be smaller than that of f. The present author shows that for every \(\mu\) with \(0<\mu <1\) there exists a quasiconformal Jordan curve L with \(\theta_ L(\delta)\geq C\delta^{\mu}\) such that for every \(f\in H^{\alpha}(L)\) \((0<\alpha <1)\) the corresponding \(\tilde f\) belongs to \(H^{\alpha}(\bar D)\). This shows that the applicability of PPT depends on subtler geometric characteristics of L than the rate of convergence to 0 of \(\theta_ L(\delta)\) as \(\delta\) tends to 0.
scientific article; zbMATH DE number 3979479
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