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Let $\mu$ be the Hardy Littlewood centered maximal operator in $\mathbb{R}^n$

$\mu (f)(x) = \sup_{r>0} \frac{1}{|B_r(x)|} \int_{B_r(x)} |f(y)|dy.$

If $g(x)=|x|^{-\eta}$, com $\eta \in (0,n)$, how to prove that $\mu(g)(x)=Cg(x)$, for some $C$ constant?

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First observation: there is a close relation between $\mu(f)$ and $\mu(f\circ T)$ where $T$ is a linear operator that is either orthogonal or a multiple of identity. You should have $\mu(f\circ T)=\mu(f)\circ T$ in both cases.

Second observation to make: the function $g(x)=|x|^{-\eta}$ satisfies $g\circ T=g$ when $T$ is orthogonal, and $g\circ T=\lambda^{-\eta}g$ when $Tx=\lambda x$.

Then combine the 1st and 2nd observations to show that $\mu(g)$ has the same symmetry/scaling as $g$.

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Hint: Use the definition to show that $Mg(rx) = Mg(x)$ for any rotation $r$, and that $Mg(tx) = t^{-\eta}Mg(x)$ for any $t > 0$.

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    the advantage of writing 2 liners :)2012-06-19