An inequality equivalent to Hörmander's condition $\sup_{y\in\mathbb R^n}\int_{\{x: |x|>2|y|\}}|K(x-y)-K(x)|\,dx<\infty$ 
Let $K\in L_{\text{loc}}^1(\mathbb R^n\setminus\{0\})$. Prove that
$$\sup_{y\in\mathbb R^n}\int_{\{x: |x|>2|y|\}}|K(x-y)-K(x)|\,dx<\infty\label{1}\tag{1}$$
if and only if
$$\sup_{r>0}\frac1{r^n}\int_{B(0,r)}\int_{\{x: |x|>2r\}}|K(x-y)-K(x)|\,dx\,dy<\infty.\label{2}\tag{2}$$

This is an old exam problem on Harmonic Analysis. Formula \eqref{1} is called the Hörmander's condition for singular integrals. The proof of \eqref{1}$\Rightarrow$\eqref{2} is quite easy: assume
$$\int_{\{x: |x|>2|y|\}}|K(x-y)-K(x)|\,dx\leq M,\qquad \forall y\in\mathbb R^n,$$
then for $r>0$ and $y\in B(0,r)$ we have $\{x: |x|>2r\}\subset \{x: |x|>2|y|\}$, so
$$\int_{\{x: |x|>2r\}}|K(x-y)-K(x)|\,dx\leq \int_{\{x: |x|>2|y|\}}|K(x-y)-K(x)|\,dx\leq M,$$
hence
$$\frac1{r^n}\int_{B(0,r)}\int_{\{x: |x|>2r\}}|K(x-y)-K(x)|\,dx\,dy\leq \frac1{r^n}\int_{B(0,r)}M\,dy=M|B(0,1)|,\ \ \ \forall r>0.$$
This completes the proof of \eqref{1}$\Rightarrow$\eqref{2}.
However, for \eqref{2}$\Rightarrow$\eqref{1}, I don't know how to start.
Any help would be appreciated!
 A: Let $R=|y|, r=|y| / 2$, define
$$
\Omega=B(0,R)\cap B(y,r).
$$
Then $|\Omega| \sim r^n \sim R^n$,
$$
\begin{aligned}
&\int_{|x|>2|y|}|K(x-y)-K(x)| \mathrm{d} x \\
&\leq \int_{|x|>2|y|}\left|K\left(x-y^{\prime}\right)-K(x)\right| \mathrm{d} x+\int_{|x|>2|y|}\left|K\left(x-y^{\prime}\right)-K(x-y)\right| \mathrm{d} x
\end{aligned}
$$
Averaging with respect to $y^{\prime}$ on $\Omega$,
$$
\begin{aligned}
\int_{|x|>2|y|}|K(x-y)-K(x)| \mathrm{d} x \\
& \leq \frac{1}{|\Omega|} \int_{y^{\prime} \in \Omega} \mathrm{d} y^{\prime} \int_{|x|>2|y|}\left|K\left(x-y^{\prime}\right)-K(x)\right| \mathrm{d} x \\
&+\frac{1}{|\Omega|} \int_{y^{\prime} \in \Omega} \mathrm{d} y^{\prime} \int_{|x|>2|y|}\left|K\left(x-y^{\prime}\right)-K(x-y)\right| \mathrm{d} x
\end{aligned}
$$
While $R=|y|$ and $\Omega \subset B(0, R)$,
$$
\frac{1}{|\Omega|} \int_{y^{\prime} \in \Omega} \mathrm{d} y^{\prime} \int_{|x|>2|y|}\left|K\left(x-y^{\prime}\right)-K(x)\right| \mathrm{d} x \leq \frac{C}{R^n} \int_{\left|y^{\prime}\right|<R} \mathrm{~d} y^{\prime} \int_{|x|>2 R}\left|K\left(x-y^{\prime}\right)-K(x)\right| \mathrm{d} x
$$
On the other hand,
$$
\int_{y^{\prime} \in \Omega} \mathrm{d} y^{\prime} \int_{|x|>2|y|}\left|K\left(x-y^{\prime}\right)-K(x-y)\right| \mathrm{d} x
$$
$$
\begin{aligned}
&=\int_{y^{\prime} \in \Omega} \mathrm{d} y^{\prime} \int_{\left|x^{\prime}+y\right|>2|y|}\left|K\left(x^{\prime}+y-y^{\prime}\right)-K\left(x^{\prime}\right)\right| \mathrm{d}\left(x^{\prime}+y\right) \\
&\leq \int_{y^{\prime} \in \Omega} \mathrm{d} y^{\prime} \int_{\left|x^{\prime}\right|>|y|}\left|K\left(x^{\prime}+y-y^{\prime}\right)-K\left(x^{\prime}\right)\right| \mathrm{d} x^{\prime} \\
&\leq \int_{\left|y^{\prime}-y\right|<r} \mathrm{~d}\left(y^{\prime}-y\right) \int_{\left|x^{\prime}\right|>|y|}\left|K\left(x^{\prime}+y-y^{\prime}\right)-K\left(x^{\prime}\right)\right| \mathrm{d} x^{\prime} \\
&\leq \int_{\left|y^{\prime}-y\right|<r} \mathrm{~d}\left(y^{\prime}-y\right) \int_{\left|x^{\prime}\right|>2 r}\left|K\left(x^{\prime}-\left(y^{\prime}-y\right)\right)-K\left(x^{\prime}\right)\right| \mathrm{d} x^{\prime}
\end{aligned}
$$
Taking the supremum in $y'$ firstly and in $y$ secondly results the desired conclusion.
