Estimative with kernels of the Riesz transform

Question

Let $x=(x_1,x_2)$ and $k(x)=\frac{x_1}{|x|^3}$, $|x|>0$.

Consider $x,y\in\mathbb{R^2},\xi=|x-y|, \tilde{x}=\frac{x+y}{2}$. Then

$|k(x-t)-k(y-t)|\leqslant C\frac{|x-y|}{|\tilde{x}-t|^3}$ when $|\tilde{x}-t|\geqslant\xi$?

Answer 1

First, by replacing $x$ by $x-t$ and $y$ by $y-t$, we can assume without loss of generality that $t=0$, and also $|x| \ge |y|$ (and so $|x| \ge |\tilde x|$).

Case 1: $|x| > 2|y|$. Then $|x-y| \approx |\tilde x| \approx |x|$, the inequality is easy.

Case 2: $|x| \ge |y| \ge |x|/2$. $$ \frac{x_1}{|x|^3} - \frac{y_1}{|y|^3} = \frac{|y|^3 x_1 - |x|^3 y_1}{|x|^3|y|^3} = \frac{|y|^3 x_1 - |y|^3 y_1}{|x|^3|y|^3} + \frac{x_1(|x|^3 - |y|^3)}{|x|^3|y|^3} $$ Now $$ \left|\frac{|y|^3 x_1 - |y|^3 y_1}{|x|^3|y|^3} \right| = \frac{|x_1-y_1|}{|x|^3} \le \frac{|x-y|}{|\tilde x|^3} ,$$ and $$ \left|\frac{x_1(|x|^3 - |y|^3)}{|x|^3|y|^3}\right| \le \frac{(|x|-|y|)(|x|^2 + |x||y|+|y|^2)}{|x|^2|y|^3} \le \frac{3(|x|-|y|)}{|y|^3} \le \frac{24|x-y|}{|\tilde x|^3} .$$ I have a feeling this argument is over complicated. But something like this will work.