I am in the middle of the proof of the fact in the title and have had trouble understanding a certain intermediate estimate (I'm looking at the proof of theorem 7.11 in Duoandikoetxea's Fourier Analysis).
Let $f$ be a bounded function with compact support (say $\text{supp}(f) \subset B(0,R))$. If $|x| > 2R$, I should be able to see that $|Tf(x)| \leq \frac{C||f||_\infty}{|x|^d}$, for some constant $C$, and where $x \in \mathbb{R}^d$, $T$ is a CZ operator.
The closest I can get is showing that $|Tf(x)| \leq \frac{D||f||_\infty}{R^d}$, where $D$ comes from the bound of the standard kernel of $T$ and the Lebesgue measure of $B(0,R)$ (this ends up not being sufficient for the remainder of the proof).
How should I go about obtaining the desired bound above? I'm convinced I'm missing something extremely simple but have thought about it for a good amount of time and still don't see it. Thanks in advance for any help.
If the kernel of $T$ is $K$, then, for $|x|>2R$, $$|Tf(x)|\leq\int_{B(0,R)}|K(x,y)f(y)|\,dy\leq C\|f\|_{\infty}\int_{B(0,R)}\frac{1}{|x-y|^d}\,dy.$$ But, if $y\in B(0,R)$, then $$|x-y|\geq |x|-|y|=\frac{|x|}{2}+\frac{|x|}{2}-|y|>\frac{|x|}{2}+R-|y|\geq\frac{|x|}{2},$$ therefore $$|Tf(x)|\leq C\|f\|_{\infty}\int_{B(0,R)}\frac{2^d}{|x|^d}\,dy=\frac{CR^d}{|x|^d}\|f\|_{\infty}.$$