Riesz transform after linear transformation

Question

I am encountering the term $\partial_x \mathcal{R}_x(f(x,y))$. I needed to do the following linear transformation

$$x' = a x+ by,\,\,\,\,\, y'=ax-by,\,\,\, and \,\,f(x,y)=g(x',y') $$

I ended up with the term $a (\partial_{x'} + \partial_{y'}) \mathcal{R}_{\frac{1}{2a}(x'+y')}(g(x',y'))$.

Given that $\mathcal{F}[\partial_x \mathcal{R}_x(f(x,y))]{(\xi,\eta)}=\frac{- \xi^2}{|(\xi,\eta)|} \hat{f}(\xi,\eta)$, I am nut sure how to express the transform Riesz, in particular, I doubt if the Riesz transform can be distributed among subscript $\frac{1}{2a} (x'+y')$. Any hint appreciated.

Answer 1

The Riesz transform of the function $f(x,y)$ of two variables reads, $$ \mathcal{R}_x(f(x,y))=\frac{1}{2\pi}\iint \frac{(x-u)f(u,v)}{[(x-u)^2+(y-v)^2]^{3/2}}\,dudv.$$ Let me define $F(x,y)=\partial_x\mathcal{R}_xf(x,y)$. The transformed functions $\tilde{f}(x',y')$ and $\tilde{F}(x',y')$ are defined by the coordinate transformation $$x=-\frac{-x'-y'}{2 a},y=-\frac{y'-x'}{2 b},$$ with Jacobian $2ab$. Upon substitution you end up with $$\tilde{F}(x',y')=\frac{2}{\pi}\iint \left(a^2(u'-v'-x'+y')^2-2b^2 (u'+v'-x'-y')^2\right)$$ $$\qquad{}\times\left(b^2(u'+v'-x'-y')^2+a^2(u'-v'-x'+y')^2\right)^{-5/2}\tilde{f}(u',v')\,du'dv'.$$ Any reason why you would want to go through this transformation?