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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?

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

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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 $$ M\chi_{B(x,r)}(z)\ge\frac{|B(x,r)|}{|B|}=\frac{r^n}{\Bigl(\dfrac{|z-x|+r}{2}\Bigr)^2}\ge\Bigl(\frac{2}{3}\Bigr)^n. $$

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  • $\begingroup$ 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

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