Distribution of Hilbert transforms of measures
Let \(\eta\) be a real Borel measure on the real line with finite total variation. Its Hilbert transform \[ g(x)=\lim_{\delta\to 0+}{1\over \pi} \int_{|t-x|\geq\delta} {d\eta(t) \over t-x}={1\over \pi}PV\int {d\eta (t)\over t-x} \] is a measurable, almost everywhere finite but in general not locally integrable function. Its signed distribution function is \[ \nu_g(s)= \begin{cases}\lambda \bigl\{x:g (x)\geq s\bigr\}, & s>0\\ -\lambda\bigl\{ x:g(x)\leq s \bigr\}, & s<0 \end{cases} \] (where \(\lambda\) denotes the Lebesgue measure). Now consider the inverse Hilbert transform of \(\nu_g\) \[ r(t)= -\lim_{\varepsilon\to 0+} {1\over\pi} PV\int_{|s|\geq \varepsilon}{\nu_g(s) \over s-t}ds. \] The main result in this paper gives bounds for the integral over \(r^+\), \(r^-\) and \(r\) in terms of integrals over \(\eta\) resp. its absolutely continuous and singular parts, which have the following consequences: \(\|r\|_{L_1}\leq 2\int|d\eta|\), whereas the inverse Hilbert transform \(q\) of the nonincreasing rearrangement \(g^\#\) of \(g\) satisfies \(\|q\|_{L_1}\leq 4\int|d\eta|\). Furthermore it follows that \(\eta\) is absolutely continuous iff \(g^\#\) is the Hilbert transform of some \(L_1\)-function. The proof uses the special case of Hilbert transforms of test functions and a certain class of subharmonic functions, which is discussed as well.
scientific article; zbMATH DE number 1443219