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Let $f$ be in $L^{2}(\mathbb{R}^n)$ then $f^{*}\in L^{2}(\mathbb{R}^n)$?

Is there any weaker result like $f^* \in L^2_{loc}(\mathbb{R}^n)$ or $ff^*\in L^{1}((\mathbb{R}^n)$?

Notation:

$$f^*(x)=\sup_{x\in B}\int_{B}|f(y)|\;dy$$

where $B$ are ball containing $x$.

I want a proof of the weaker statements without the main result( which I don't know if is true). A perfect answer is: proof the main result or give it a counter example and proof the weaker statements or give counter examples to all of them.

Whatever I am happy if you can proof the weaker thesis.

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Your definition is not quite correct - here is the relevant information for those who wish to understand the question. Anyway, If $Mf$ is integrable, then certainly its also true that it is locally integrable. Are you asking if those weaker results hold, or for a proof of only the weaker results that you hope is simpler than the full proof? –  Ragib Zaman Feb 21 '12 at 15:21
    
The first weaker result was motivated by the $L^1$ case however the other are more subtle and may not be true. If possible I want counter examples. –  checkmath Feb 21 '12 at 15:48
    
If $f$ is integrable you may have that $f^*$ is not even locally integrable see the case: $f(x)=\frac{1}{|x|(\ln|x|^{-1})^2}$, if $-1/2\leq x\leq 1/2$ it is easy to check that $f^*(x)\geq \frac{c}{|x|\ln(|x|^{-1})}$ whenever $-1/2\leq x\leq 1/2$. –  checkmath Feb 23 '12 at 2:49

1 Answer 1

If $f \in L^2$ then by the Hardy-Littlewood maximal inequality $f^* \in L^2$ as well. Thus, by the Cauchy-Schwarz inequality $$ \left(\int_{\mathbb{R}^n} |f(x) f^*(x)| dx \right)^2 \leq \left( \int_{\mathbb{R}^n} |f(x) |^2 dx \right)\left(\int_{\mathbb{R}^n} |f^* (x)|^2 dx\right) < \infty $$ we see it is indeed true that $ ff^* \in L^1.$

And of course, $f^*$ is locally $L^2$ since it is in fact $L^2.$ So both the results are true.

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