Let $M$ denotes the Hardy-Littlewood maximal operator, $\chi_{B(x,r)}$ denotes the characteristic function of the open ball $B(x,r)$. Is the following inequality always true?

$$\chi_{B(x,2r)}(z) \le 2^n M\chi_{B(x,r)}(z), \text{ for all } z \in {\mathbb R}^n$$


Yes. In the following I assume the non-centered maximal operator. If $|z-x|>2\,r$ then $\chi_{B(x,2r)}(z)=0$. If $|z-x|<r$, then $ M\chi_{B(x,r)}(z)=\chi_{B(x,2r)}(z)=1$. Finally, if $r\le|z-x|\le2\,r$ consider the ball $B$ with center on the segment joining $x$ and $z$, containing $B(x,r)$ of radius $(|z-x|+r)/2$ and with $z$ on its boundary. Then $$ M\chi_{B(x,r)}(z)\ge\frac{|B(x,r)|}{|B|}=\frac{r^n}{\Bigl(\dfrac{|z-x|+r}{2}\Bigr)^2}\ge\Bigl(\frac{2}{3}\Bigr)^n. $$

  • $\begingroup$ I couldn't see how can we write $ M\chi_{B(x,r)}(z)=1$, when $z\in B(x,r)$. $\endgroup$ – bjk1806 Mar 28 '16 at 19:47
  • $\begingroup$ It is clear that $M\chi_{B(x,r)}(z)\le1$ for all $z$. If $|z-x|<r$ then there is $\delta>0$ such that $B(\delta,z)\subset B(r,x)$. Then $$\frac{1}{|B(\delta,z)|}\int_{B(\delta,z)}\chi_{B(\delta,z)}(y)\,dy=1.$$ $\endgroup$ – Julián Aguirre Mar 28 '16 at 21:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.