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Let $\mu$ be a Lebesgue measure, $\mathcal{M}$ be the Hardy-Littlewood Centered Maximal Function, and the rest as in the title. Then to this end I define $f_{R}(x):=f(x)\chi_{|x|\le R}$ where $\chi$ is the characteristic function and $R>0$. Then we have $$\begin{aligned}&\mathcal{M}f_{R}(x)\ge\frac{\|f_{R}\|_{L^{1}}}{v_{n}}\cdot\frac{1}{(|x|+R)^{n}},\qquad \text{for } |x|\ge R \text{ and } v_{n} \text{ the volume of the unit ball in } \mathbb{R}^{n} \\ \iff& \sup_{R>0}\frac{1}{\mu(B(x,R))}\int_{B(x,R)}|f_{R}|d\mu\ge\frac{1}{\mu(B(x,|x|+R))}\int_{\mathbb{R}^{n}}|f(x)\chi_{|x|\le R}|dx \\ =&\frac{1}{\mu(B(x,|x|+R))}\int_{B(x,R)}|f(x)|dx \end{aligned} $$

I'm not sure that all my computations are correct. And how can I conclude from here that $f_{R}=0$ a.e.?

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  • $\begingroup$ What is $\mathcal M$? $\endgroup$ – gerw Feb 20 '16 at 13:12
  • $\begingroup$ @gerw $\mathcal{M}$ is the Hardy-Littlewood centered maximal function. You can see its expansion in my computation. $\endgroup$ – Jason Born Feb 20 '16 at 13:13
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Suppose $f$ does not vanish almost everywhere. Then there must be some $R>0$ such that $$ \int_{B(0,R)}|f|dm > 0. $$ For all $x \in \mathbb{R}^n$ with $|x|>R$, \begin{align} (Mf)(x) &\geq \frac{1}{m\left(B(x,2|x|)\right)}\int_{B(x,2|x|)}|f|dm \\ &\geq \frac{c}{|x|^n} \int_{B(0,R)}|f|dm \end{align} where $c$ is a constant. But $1/|x|^n \notin L^1(\mathbb{R}^n)$, hence a contradiction.

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  • $\begingroup$ Why did you choose the radius of your ball $2|x|$? And what does your $c$ depend on? Different from my proposed method, but nonetheless I like it. $\endgroup$ – Jason Born Feb 20 '16 at 14:15
  • $\begingroup$ @user3482534 $2|x|$ so the ball can cover $B(0,R)$. $c = [2^nm(B(x,1))]^{-1}$. $\endgroup$ – Qiyu Wen Feb 20 '16 at 14:41
  • $\begingroup$ I doubt if your approach will be good enough. Even if $f$ does not vanish a.e., $\int_{B(x,R)} |f|dm$ can really vanish in a large portion of the space. Then your task would be to find the small portion of the space where the lower bounds can show that $Mf$ is not integrable. Sounds hard for me. And you know $Mf$ can probably do better than that. $\endgroup$ – Qiyu Wen Feb 20 '16 at 14:51
  • $\begingroup$ This approach is correct. If the integrals of $|f|$ over $B(0, R)$ vanish for all $R$ then the integral of $|f|$ vanishes on $\mathbb{R}^n$ by the monotone convergence theorem, so $|f|$ and hence $f$ is a.e. zero. $\endgroup$ – Unit Feb 29 '16 at 17:29

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