Understanding kernel of Riesz transforms $K(r,\zeta)=r^{-2}\Omega(\zeta)$

Question

In the appendix of the paper (https://arxiv.org/abs/math/0604185) the authors said "The Riesz transform are singular integral operators with kernels $K(r,\zeta)=r^{-2}\Omega(\zeta)$, where $(r,\zeta)$ are the polar coordinates. The function $\Omega$ is smooth and $\int_{S^1} \Omega(\zeta)d\sigma(\zeta)=0$." Can anyone explain a little bit more about this statement. I don't quite understand how this is related to the definition of Riesz transform by using Fourier multiplier method. Here the authors consider in $\mathbb R^2$.

Answer 1

We define the Riesz transform, say, in the first coordinate by \begin{align*} R_{1}(f)(x)=(f\ast W_{1})(x), \end{align*} where \begin{align*} \left<W_{1},\varphi\right>=\lim_{\epsilon\rightarrow 0}\int_{|y|\geq\epsilon}\dfrac{y_{1}}{|y|^{3}}\varphi(y)dy, \end{align*} so \begin{align*} (f\ast W_{1})(x)=\left<W_{1},f(x-\cdot)\right>=\lim_{\epsilon\rightarrow 0}\int_{|x-y|\geq\epsilon}\dfrac{x_{1}-y_{1}}{|x-y|^{3}}f(y)dy. \end{align*} However, the expression can be written as \begin{align*} (f\ast W_{1})(x)=\lim_{\epsilon\rightarrow 0}\int_{|x-y|\geq\epsilon}\dfrac{\Omega((x-y)/|x-y|)}{|x-y|^{2}}f(y)dy, \end{align*} where $\Omega$ is defined on the sphere ${\bf{S}}^{1}$, $\Omega(\theta)=\theta_{1}$, we now let $K(r,\theta)=r^{-2}\Omega(\theta)$, then \begin{align*} (f\ast W_{1})(x)=\lim_{\epsilon\rightarrow 0}\int_{{\bf{S}}^{1}}\int_{\epsilon}^{\infty}\dfrac{\Omega(\theta)}{r^{2}}f(x-r\theta)rdrd\theta=\lim_{\epsilon\rightarrow 0}\int_{{\bf{S}}^{1}}\int_{\epsilon}^{\infty}K(r,\theta)f(x-r\theta)rdrd\theta. \end{align*} On the other hand, we have \begin{align*} \int_{{\bf{S}}^{1}}\Omega(\theta)d\theta=\int_{{\bf{S}}^{1}}\theta_{1}d\theta=0 \end{align*} since $\theta\rightarrow\theta_{1}$ is odd.