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Consider the function on $\mathbb R$ defined by

$$f(x)=\begin{cases}\frac{1}{|x|\left(\log\frac{1}{|x|}\right)^2} & |x|\le \frac{1}{2}\\ 0 & \text{otherwise}\end{cases}$$

Now suppose $f^*$ is the maximal function of $f$, then I want to show the inequality $f^*(x)\ge \frac{c}{|x|\left(\log\frac{1}{|x|}\right)}$ holds for some $c>0$ and all $|x|\le \frac{1}{2}$.

But I don't know how to prove it. Can anyone give me some hints?

Thanks very much.

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What's the maximal function of $f$? – Rudy the Reindeer Jan 1 '12 at 11:14
@AsafKaragila I am so sorry for that. – molan Jan 1 '12 at 11:29
@Matt The maximal function of $f$ is defined as follows.$f^*(x)=sup_{x\in B}\frac{1}{m(B)}\int_B|f(y)|dy$ where supremum is taken over all balls containing the point $x$ and $m(B)$ is the Lebesgue measure of $B$. – molan Jan 1 '12 at 11:30
@Matt No thanks. I am so sorry that I didn't note this before and thank you for your help. – molan Jan 1 '12 at 11:46
up vote 2 down vote accepted

By definition $$ f^*(x)=\sup_{B\in \text{Balls}(x)}\frac{1}{\mu(B)}\int\limits_B |f(y)|d\mu(y)\qquad(1) $$ where $\text{Balls}(x)$ the set of all closed balls containing $x$. We can express $f^*$ in another form $$ f^*(x)=\sup_{\alpha\leq x\leq\beta}\frac{1}{\beta-\alpha}\int\limits_{\alpha}^{\beta} |f(y)|d\mu(y) $$ Consider $0< x\leq1/2$. Obviously $$ f^*(x)=\sup_{\alpha\leq x\leq\beta}\frac{1}{\beta-\alpha}\int\limits_{\alpha}^{\beta} |f(y)|d\mu(y)\geq\frac{1}{x-0}\int\limits_{0}^{x}|f(y)|d\mu(y)=-\frac{1}{x\log x} $$ Thus $f^*(x)\geq\frac{1}{x\log\left(\frac{1}{x}\right)}$ for $0<x\leq 1/2$. Since $f$ is even then does $f^*$, hence inequality $$ f^*(x)\geq\frac{1}{|x|\log \left(\frac{1}{|x|}\right)} $$ holds for all $-1/2\leq x\leq1/2$

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