This problem has been asked in MSE, 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!