We have tempered distribution $K$ in $\mathbb{R}^n$ which coincides with a locally integrable function in $\mathbb{R}^n\setminus \{0\}$. We call the condition
$$\int\_{|x|>2|y|}|K(x-y)-K(x)|dx\leq B \hspace{1cm}\forall y\in \mathbb{R}^n$$
for some constant $B$, the Hörmander condition. 



> 
> **I want to show that if**
> $$|\nabla K(x)|\leq \frac{C}{|x|^{n+1}} \hspace{1cm}\forall x\neq 0$$ **then the Hörmander condition holds.**
> 
> 
> 


*For reference, it is from page 92 of the book 'Fourier Analysis' written by Javier.* 


**What I have tried is that**, by mean value theorem, for fixed $y$ and for each $x$, there is $t\in [0,1]$ with such that $z(x)=tx+(1-t)y$ and
$$\int\_{|x|>2|y|}|K(x-y)-K(x)|dx =\int\_{|x|>2|y|} |\nabla K(z(x))||y|dx\leq \int\_{|x|>2|y|}\frac{C}{|z(x)|^{n+1}}|y| dx$$ 


We are able to say $|z(x)|^{n+1}\geq |tx+(1-t)y|^{n+1}\geq (t|x|-(1-t)|y|)^{n+1}$.


But my problem here is that $t$ is dependent on $x$ so I currently cannot see how I can get further.


Thanks in advance for any helps!



Gradient condition implies Hörmander condition
I figured out proving it by using mean value theorem (thanks to @WillieWong). Observe that
\begin{align\*}
\int\_{|x|>2|y|}|K(x-y)-K(x)|dx &\leq \int\_{|x|>2|y|}|\nabla K(tx+(1-t)(x-y))||y| dx\\
&\leq \int\_{|x|>2|y|} \frac{C}{|x-(1-t)y|^{n+1}}|y|dx\\
&\leq \int\_{|x|>2|y|} \frac{C}{[|x|-(1-t)|y|]^{n+1}}|y|dx&\because |x|>|y|\\
&\leq \int\_{|x|>2|y|} \frac{C}{[|x|-|y|]^{n+1}}|y| dx&\because t\in (0,1)\\
&\leq \int\_{|x|>2|y|} \frac{C}{[|x|-\frac{|x|}{2}]^{n+1}}|y| dx &\because \frac{|x|}{2}>|y|\\
&\leq \int\_{2|y|}^{\infty} \frac{2^{n+1}C}{r^{n+1}}\sigma(n) r^{n-1}|y|dr\\
&=\frac{2^{n+1}C}{2|y|}\sigma(n)|y|\\
&={2^nC\sigma(n)}
\end{align\*}


where $\sigma(n)$ is the measure of unit sphere in $\mathbb{R}^{n}$. 


