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Find the Hardy-Littlewood maximal function $Mf$ of the $\Bbb R^\Bbb R$ function $f=\chi_{[-1,1]}$.

How do we find $Mf(x)$ for $|x| > 1$?
I see that it should decrease like $1/x$, but I can't find the formula.

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For future reference, it might be valuable to specify that this is the Hardy-Littlewood maximal function. There are actually many maximal functions that are studied in harmonic analysis. –  Christopher A. Wong Nov 13 '12 at 3:41
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up vote 3 down vote accepted

Since you're on $\mathbb{R}$, the maximal function looks like:

$$ Mf(x)=\sup_{B(x,\epsilon)}\frac{1}{2\epsilon}\int_{B(x,\epsilon)}\vert f\vert dx\\ =\sup_{\epsilon}\frac{1}{2\epsilon}\lambda(B(x,\epsilon)\cap [-1,1]) $$ where $\lambda$ is the Lebesgue measure.

If $\vert x\vert>1$, the sup should (though this would require an argument) occur when you take your ball to include all of $[-1,1]$ and nothing more - i.e. take $\epsilon=\vert x\vert+1$. Your maximal function will then be $Mf(x)=\frac{1}{\vert x\vert+1}$.

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