Somebody can to help me in the following question?

Let $f$ be measurable in $\mathbb{R}^n$ and different from zero in some set of positive measure. Show that there is a positive constant с such that $f^*(x) \geq c |x|^{-n}$ for $|x| \geq 1$

Note: $f^*(x)$ is the Hardy-Littlewood maximal function, i.e.,

$\displaystyle{f^*(x) = \sup_{r>0} \; \frac{1}{|B_{r}(x)|}\; \int_{B_{r}(x)} \;|f(y)|\;dy}.$


From the given conditions it follows that there exists a ball $B=B_M(0)$ (wlog M>1) such that $$ \int_B |f| = c>0. $$

If $|x|\ge M$, then $B_{2|x|}(x)\supset B$ so that \begin{align} f^*(x) &\ge \frac{1}{|B_{2|x|}(x)|}\int_{B_{2|x|}(x)} |f| \\ &= \frac{1}{v_n2^n|x|^n}\int_{B_{2|x|}(x)} |f| \\ &\ge \frac{1}{v_n2^n|x|^n}c \\ \end{align}

If $1\le|x|\le M$, then $B_{2M|x|}\supset B$

\begin{align} f^*(x) &\ge \frac{1}{|B_{2M|x|}(x)|}\int_{B_{2M|x|}(x)} |f| \\ &= \frac{1}{v_n(2M)^n|x|^n}\int_{B_{2M|x|}(x)} |f| \\ &\ge \frac{1}{v_n(2M)^n|x|^n}c \\ \end{align}

here $v_n$ is the volume of the ball with unit radius in $\mathbb{R}^n$.

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  • $\begingroup$ To get $c$ we just factor out $\frac{1}{|x|^n}$ and take the minimum of the respectives co-effecients? $\endgroup$ – Brofessor Jun 21 '19 at 14:32

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