This problem has been asked in [MSE](https://math.stackexchange.com/questions/4548511/an-inequality-equivalent-to-h%c3%b6rmanders-condition-sup-y-in-mathbb-rn-int), but got no answers. I guess that this exam problem may be a small lemma in some research papers, so I post it here on MathOverflow.



> 
> 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!



An inequality equivalent to Hörmander's condition $\sup_{y\in\mathbb R^n}\int_{\{x: |x|>2|y|\}}|K(x-y)-K(x)|\,dx<\infty$
Condition \eqref{2} implies \eqref{1} with $5|y|$ instead of $2|y|$. Fix $y$, let $r=|y|$ and $I=\int\_{|x| >5|y| }|K(x-y)-K(x)|\, dx$. Then
$$I \leq \int\_{|x| >5r }|K(x-y)-K(x-z)|\, dx+\int\_{|x| >5r }|K(x-z)-K(x)|\, dx:=I\_1(z)+I\_2(z)
$$
for every $|z| \leq r$. If $K$ is the supremum in \eqref{2}, then $r^{-n} \int\_{B(0,r)} I\_2(z)\, dz \leq K$. In $I\_1$ we set $\xi=x-z$ so that $|\xi| \geq 4r$ and
$$I\_1(z) \leq \int\_{|\xi| \geq 4r} |K(\xi-(y-z))-K(\xi)|\, d\xi.$$ Since $|y-z| \leq 2r$, then
$$r^{-n}\int\_{B(0,r)} I\_1(z)\, dz \leq r^{-n} \int\_{B(0,2r)}|K(\xi-w)-K(\xi)|\, dw
\leq 2^n K.$$ The estimate of $I$ in terms of $K$ now follows by averaging the inequality $I \leq I\_1(z)+I\_2(z)$ over $B(0,r)$.


