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Let $f : \mathbb{R}^d \to \mathbb{R}^{\geq 0}$ be a probability density. Let

$$(Mf)(x) = \sup_{r>0} \frac{1}{V(B(x, r))} \int_{B(x, r)} f(y) \;\mathrm{d}y$$

be the Hardy-Littlewood maximal function corresponding to $f$, where $B(x, r)$ is an $r$-radius ball centered at $x$, and $V$ is for volume (Lebesgue measure). Then $Mf$ is measurable, and by the Hardy-Littlewood maximal inequality is finite almost everywhere. Does it follow that $\lVert Mf \rVert_{L^1} = \int_{\mathbb{R}^d} (Mf)(x) \;\mathrm{d}x < \infty$? If not, what would be some sufficient additional conditions for this to hold?

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In fact we always have $\int_{\mathbb R^d} Mf = \infty.$

Proof: Choose $R>0$ such that $\int_{B(0,R)} f >1/2.$ Note that for $|x|>R$ we have $B(0,R)\subset B(x,2|x|).$ Hence for such $x,$

$$Mf(x) \ge \frac{1}{V(B(x,2|x|))}\int_{B(x,2|x|)} f > \frac{1}{V(B(x,2|x|))}\cdot \frac{1}{2} > c\frac{1}{|x|^d}$$

for some constant $c>0$ independent of $x.$ Since

$$\int_{|x|>R} \frac{1}{|x|^d}\, dx = \infty,$$

the result is proved.

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  • $\begingroup$ This proof works when $d = 1$; great! However, for $d > 1$ the integral is finite, right (spherical coordinates)? $\endgroup$ – kaba Sep 9 '17 at 20:43
  • $\begingroup$ @Kaba No, the integral is infinite for all $d.$ $\endgroup$ – zhw. Sep 9 '17 at 21:32
  • $\begingroup$ Yes, I can see that now. $\endgroup$ – kaba Sep 9 '17 at 21:50

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