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The Riesz transform of the function $f(x,y)$ of two variables ($d=2$) 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 $$x\to -\frac{-x'-y'}{2 a},y\to -\frac{y'-x'}{2 b},$$ with Jacobian $2ab$. You thus 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?

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?

Riesz transform of the function $f(x,y)$ of two variables ($d=2$), $$ \mathcal{R}_x(f(x,y))=\frac{1}{2\pi}\iint \frac{(x-u)f(u,v)}{[(x-u)^2+(y-v)^2]^{3/2}}\,dudv.$$ Take the derivative with respect to $x$, make the desired substitutions $$x\to -\frac{-x'-y'}{2 a},y\to -\frac{y'-x'}{2 b},$$ $$u\to -\frac{-u'-v'}{2 a},v\to -\frac{v'-u'}{2 b},$$ with Jacobian $2ab$, and you end up with $$[\partial_x\mathcal{R}_xf](x',y')=\frac{2}{\pi}\iint \left(a^2(u'-v'-x'+y')^2-2b^2 (u'+v'-x'-y')^2\right)$$ $$\qquad{}\times\left(b^2(u'+v'-x'-y')^2+a^2(u'-v'-x'+y')^2\right)^{-5/2}f(u',v')\,du'dv'.$$ Any reason why you would want to go through this transformation?

The Riesz transform of the function $f(x,y)$ of two variables ($d=2$) 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 $$x\to -\frac{-x'-y'}{2 a},y\to -\frac{y'-x'}{2 b},$$ with Jacobian $2ab$. You thus 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?

Riesz transform of the function $f(x,y)$ of two variables ($d=2$), $$ \mathcal{R}_x(f(x,y))=\frac{1}{2\pi}\iint \frac{(x-u)f(u,v)}{[(x-u)^2+(y-v)^2]^{3/2}}\,dudv,$$ Take the derivative with respect to $x$, make the desired substitution $$x\to -\frac{-x'-y'}{2 a},y\to -\frac{y'-x'}{2 b},$$ and you end up with $$[\partial_x\mathcal{R}_xf](x',y')=\frac{4}{\pi}\iint \left(\frac{(2 b v-x'+y')^2}{b^2}-\frac{2 (-2 a u+x'+y')^2}{a^2}\right)$$ $$\qquad{}\times\left(\frac{(-2 a u+x'+y')^2}{a^2}+\frac{(2 b v-x'+y')^2}{b^2}\right)^{-5/2}f(u,v)\,dudv.$$ Any reason why you would want to go through this transformation?

Riesz transform of the function $f(x,y)$ of two variables ($d=2$), $$ \mathcal{R}_x(f(x,y))=\frac{1}{2\pi}\iint \frac{(x-u)f(u,v)}{[(x-u)^2+(y-v)^2]^{3/2}}\,dudv.$$ Take the derivative with respect to $x$, make the desired substitutions $$x\to -\frac{-x'-y'}{2 a},y\to -\frac{y'-x'}{2 b},$$ $$u\to -\frac{-u'-v'}{2 a},v\to -\frac{v'-u'}{2 b},$$ with Jacobian $2ab$, and you end up with $$[\partial_x\mathcal{R}_xf](x',y')=\frac{2}{\pi}\iint \left(a^2(u'-v'-x'+y')^2-2b^2 (u'+v'-x'-y')^2\right)$$ $$\qquad{}\times\left(b^2(u'+v'-x'-y')^2+a^2(u'-v'-x'+y')^2\right)^{-5/2}f(u',v')\,du'dv'.$$ Any reason why you would want to go through this transformation?