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I am working on the following problem from Stein and Shakarchi:

Let $f$ be an integral function on $\mathbb{R}^d$ such that $\|f\|_{L^1} = 1$ and let $f^*$ by the Hardy-Littlewood maximal function corresponding of $f$.

  • Prove that if $f$ is integrable on $\mathbb{R}^d$, and $f$ is not identically zero, then $f^*(x) \geq c/|x|^d $ for some $c > 0$ and all $|x| > 1$.
  • Conclude that $f^*$ is not integrable on $\mathbb{R}^d$.
  • Then, show that the weak type estimate $m(\{x:f^*(x) > \alpha \}) \geq c/\alpha$ for all $\alpha > 0$ whenever $\int|f| = 1$, is best possible in the following sense: if $f$ is supported in the unit ball with $\int|f| = 1$, then $m(\{x:f^*(x) > \alpha \}) \geq c'/\alpha$ for some $c' > 0$ and all sufficiently small $\alpha$.

[Hint: For the first part, use the fact that $\int_B|f| > 0$ for some ball $B$.]

I am not really sure where to begin with this. Any help would be greatly appreciated.

user49097
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1 Answers1

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Wlog, take $f$ nonnegative. Since $f \neq 0$, find an $R > 0$ large enough so that $\int_{B(0;R)} f > 0$. Fix any $x \in \mathbb{R}^d$, and take $S$ large enough so that $B(0;R) \subset B(x;S)$ (say $S = R + |x|$). So $f^*(x) \ge \frac{c}{S^d} \int_{B(x;s)} f \ge \frac{c}{(R+|x|)^d} \int_{B(0;R)} f$, where $c^{-1}$ is the volume of the unit ball in $\mathbb{R}^d$. It should be easy to obtain the last part of the estimate from here.

anonymous
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  • This doesn't establish the inequality for all $|x|\ge 1$, just for all $|x|$ sufficiently large, so this answer is incomplete. – Alex Ortiz Nov 15 '17 at 04:10
  • I don't see where an assumption is made on $|x|$, but it's also been like 5 years since I've thought about this answer. Can you elaborate? – anonymous Nov 18 '17 at 19:16
  • It's not true that the maximal function $f^*$ is greater than or equal to $c/|x|^d$ for all $x\in\Bbb R^d$. A simple example is to let $f$ be the indicator function of $[-1,1]$. This answer at best shows that the maximal function is greater than or equal to $c/|x|^d$ outside of some large ball, and at worst is just incomplete. You can actually prove it for all $|x|\ge 1$, not just $|x|\ge R$ for some large $R$. – Alex Ortiz Nov 18 '17 at 20:19
  • @AOritz - I don't think that really answered what I'm asking or have provided a valuable comment. You said the answer is incomplete because it assumes that $|x|$ is sufficiently large. I don't see where that assumption has been used. Can you elaborate and indicate specifically where this assumption is being implicitly used? – anonymous Nov 19 '17 at 14:15
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    Maybe you can start by showing exactly how $f^*(x)\ge c/|x|^d$ for all $|x|\ge 1$ based on your answer. – Alex Ortiz Nov 20 '17 at 18:06
  • To answer Alex's question, notice that $|x|/(|x|+R) > 1/(1+R)$ as $|x|>1$. Do I have to spell out the last bit? – Petra Axolotl May 06 '22 at 00:51