# Singular integral of the composition of the Hilbert transform and fractional Laplacian

Given $$0, we can define the Fractional Laplacian by

$$\Lambda^{-s}f(x):=(-\Delta)^{-s/2}(x)=\int_{-\infty}^{+\infty}|x-y|^{-1+s}f(y)dy$$ or by means of Fourier transform as $$\widehat{\Lambda^{-s}f}(\xi)=c_s|\xi|^{-s}\widehat{f}(\xi)\;\mbox{for all}\;\xi\neq0.$$

The Hilbert transform is defined by $$Hf(x)=p.v. \int_{-\infty}^{+\infty}\frac{f(y)}{x-y}dy$$ or by means of Fourier transform as $$\widehat{Hf}(\xi)=-isgn(\xi)\widehat{f}(\xi)\;\mbox{for all}\;\xi\neq0.$$

Therefore, we can write $$$$\label{eq1}\widehat{\Lambda^{-s}Hf}(\xi)=C_ssgn(\xi)|\xi|^{-s}\widehat{f}(\xi)\;\mbox{for all}\;\xi\neq0.$$$$ My question is how can i define $$\Lambda^{-s}Hf$$ by singular integrals.

• Up to a constant, $\Lambda^{-s} H f(x) = \int_{-\infty}^\infty f(y) |y - x|^{1 - s} \operatorname{sign}(y - x) dy$. Informally: up to a constant, $H = \partial \Lambda^{-1}$, so $\Lambda^{-s} H = \partial \Lambda^{2-s}$. I am not sure where a rigorous derivation can be found; I would check Stein's Singular integrals and differentiability properties of functions first, and if not, Samko's Hypersingular integrals and their applications. – Mateusz Kwaśnicki Dec 20 '18 at 6:23
• Thanks @MateuszKwaśnicki. The details can be found in Stein and Weiss, Introduction to Fourier Analysis on Euclidean Spaces, p. 160. – Valter Moitinho Dec 20 '18 at 12:51