TITLE: an integral estimate from Stein's book, Singular Integral
QUESTION [6 upvotes]: I am reading the Stein's book Singular Integrals and Differentiability Properties of Functions. In the text (page 40), he states that
$$
\int_{|x|\geq 2|y|}\Big|\frac{1}{|x-y|^n}-\frac{1}{|x|^n}\Big| dx\leq C
$$
where $x,y\in \mathbb{R^n}$ and $C$ is a constant only depending on the dimension $n$.
Could you please explain me how to get this estimate?
Thanks in advance.

REPLY [6 votes]: The estimate $\left|x\right|\geq2\left|y\right|$ implies
$$
\left|x-y\right|\geq\left|x\right|-\left|y\right|\geq\frac{\left|x\right|}{2}
$$
as well as $\left|\left|x\right|-\left|x-y\right|\right|\leq\left|y\right|$
and by the mean value theorem, we have
$$
\left|x\right|^{n}-\left|x-y\right|^{n}=n\cdot\left[\left|x\right|-\left|x-y\right|\right]\cdot\xi^{n-1}
$$
for some $\xi$ between $\left|x-y\right|$ and $\left|x\right|$.
In particular, $$0\leq\xi\leq\max\left\{ \left|x-y\right|,\left|x\right|\right\} \leq\left|x\right|+\left|y\right|\leq2\left|x\right|.$$
Thus,
$$
\left|\left|x\right|^{n}-\left|x-y\right|^{n}\right|\leq n\cdot\left(2\left|x\right|\right)^{n-1}\cdot\left|y\right|.
$$
This leads to
$$
\left|\frac{1}{\left|x-y\right|^{n}}-\frac{1}{\left|x\right|^{n}}\right|=\frac{\left|\left|x\right|^{n}-\left|x-y\right|^{n}\right|}{\left|x\right|^{n}\left|x-y\right|^{n}}\leq2^{n}n\cdot\left|y\right|\frac{\left|x\right|^{n-1}}{\left|x\right|^{2n}}=2^{n}n\cdot\left|y\right|\cdot\left|x\right|^{-n-1}.
$$
Now calculate (using integration in polar coordinates)
\begin{eqnarray*}
\int_{\left|x\right|\geq2\left|y\right|}\left|x\right|^{-n-1}\, dx & = & \int_{2\left|y\right|}^{\infty}r^{n-1}\int_{S^{n-1}}\left|r\xi\right|^{-n-1}\, d\xi\, dr\\
 & = & C\cdot\int_{2\left|y\right|}^{\infty}r^{-2}\, dr\\
 & = & \frac{C'}{\left|y\right|}
\end{eqnarray*}
and hence
\begin{eqnarray*}
\int_{\left|x\right|\geq2\left|y\right|}\left|\frac{1}{\left|x-y\right|^{n}}-\frac{1}{\left|x\right|^{n}}\right|\, dx & \leq & 2^{n}n\cdot\left|y\right|\cdot\int_{\left|x\right|\geq2\left|y\right|}\left|x\right|^{-n-1}\, dx\\
 & \leq & \frac{2^{n}n\cdot C'}{\left|y\right|}\cdot\left|y\right|=C''.
\end{eqnarray*}