The Riesz transform kernel satisfies Hörmander's condition Define the kernel
$$K_i(x) = \frac{x_i}{\lvert x\rvert^{d+1}}$$
as a function $\mathbb{R}^d \setminus \{ 0 \} \to \mathbb{C}$ for $d \geq 3$ (say). I want to prove that this kernel satisfies Hörmander's condition:
$$ \int_{\lvert x \rvert > 2 \lvert y \rvert} \lvert K_i(x) - K_i(x-y) \rvert \, dx < B$$
for $y \ne 0$, where $B$ is an absolute constant.
I am having difficulty handling the coordinate variables in the integrand to obtain cancellation. If someone could help me find a tactic to obtain cancellation, I would appreciate it. My next task will be to use this hypothetical tactic on the double Riesz transform's kernel.
 A: EDIT: Comments on a standard generalization at the bottom...
First, the change of variables $x\to x/|y|$ shows that you can assume $|y|=1$ (you should work out the details; this depends on $K_j$ having exactly the homogeneity that it does).
And now you really just have to prove the integral is finite; any sane proof that it's finite for $|y|=1$ will actually prove it's bounded for $|y|=1$.
Now $$K_1(x)-K_1(x-y)=x_1\left(\frac1{|x|^{d+1}}-\frac1{|x-y|^{d+1}}\right)+\frac{y_1}{|x-y|^{d+1}}.$$The second term is certainly integrable over $|x|>2$, since $|y|=1$, so $$|x-y|\ge |x|/2.$$
For the first term, add the fractions:$$x_1\left(\frac1{|x|^{d+1}}-\frac1{|x-y|^{d+1}}\right)=x_1\frac{|x-y|^{d+1}-|x|^{d+1}}{|x|^{d+1}|x-y|^{d+1}}.$$Use some calculus or something to show $$|t^{d+1}-s^{d+1}|\le c t^d\quad(t\ge2,|t-s|\le1).$$ Note that $||x|-|x-y||\le1$. Noting again that $|x-y|\sim|x|$ and $|x_1|\le|x|$ you see that the above is no larger than $c/|x|^{d+1}$.

Well, that seemed a lot easier than the last time I worried about this stuff, a few decades ago. I just realized exactly the same argument works for the more general kernel $$K(x)=\frac{\Omega(x)}{|x|^d},$$where $\Omega$ is smooth away from the origin and is homogeneous of degree $0$ (that is, $\Omega(tx)=\Omega(x)$ for $t>0$). The very first step gets replaced by $$K(x)-K(x-y)=\Omega(x)\left(\frac1{|x|^d}-\frac1{|x-y|^d}\right)+(\Omega(x)-\Omega(x-y))\frac1{|x-y|^d}.$$Didn't look like it worked at first; what I was missing was $$\left|\Omega(x)-\Omega(x-y)\right|=\left|\Omega\left(\frac x{|x|}\right)-\Omega\left(\frac x{|x|}-\frac y{|x|}\right)\right|\le\frac c{|x|}\quad(|x|\ge2, |y|\le1).$$I don't recall what the double Riesz transform is; this may or may not be relevant there, but it certainly covers a lot of singular integrals.
