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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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You might want to improve your accept rate (17% at time of commenting), or otherwise point what is missing in the answers given to your other questions. –  Asaf Karagila Jan 1 '12 at 11:11
What's the maximal function of $f$? –  Matt N. Jan 1 '12 at 11:14
Hello molan. By "improving your acceptance rate" they mean that you do accept answers to your previous questions by doing this: meta.math.stackexchange.com/a/3287/5798 –  Matt N. Jan 1 '12 at 11:18
@AsafKaragila I am so sorry for that. –  molan Jan 1 '12 at 11:29
@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

1 Answer 1

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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