I have a question that let $1<p<\infty$, and f $\in L^{p}(\mathbb{R}^{n})$ . I want to prove that $\|Mf\|_{p}\leq2(3^{n}p')^{\frac{1}{p}}\|f\|_{p}$ whereas p' is given by $\frac{1}{p'}+\frac{1}{p}=1$

The question has a hint that use $\|Mf\|^{p}_{p}=p\int^{\infty}_{0}\lambda(\{x|Mf(x)>t\})t^{p-1}dt$

I have tried for several hours but the integration always seems diverge for me. Any suggestions or clues would be appreciated!

$\lambda(x)$ is the Lebesgue measure and $Mf(x)$ is the Hardy-Littlewood maximal function

Here is my effort:

\begin{align*} \|Mf\|^{p}_{p}&=p\int^{\infty}_{0}\lambda(\{x|Mf(x)>t\})t^{p-1}dt\\ &\leq p\int^{\infty}_{0}\frac{2\cdot3^{n}}{t}\int_{\{|f(x)|>\frac{t}{2}\}}|f(x)|dx\ t^{p-1}dt\\ &=p\cdot2\cdot3^{n}\int^{\infty}_{0}\int_{\{|f(x)|>\frac{t}{2}\}}|f(x)|\ t^{p-2}dx\ dt\\ \end{align*}

I am not sure how to proceed. Thank you very much again.

  • $\begingroup$ It's not obvious to me that $\int_0^{\epsilon} t^{p - 1} \, dt$ diverges for $p > 2$.... in fact, it only diverges for $p \le 0$. $\endgroup$ – T. Bongers Apr 19 '18 at 3:22
  • $\begingroup$ @user296602 Oh sorry, you are right. But I still have difficulty solving the inequality then. $\endgroup$ – user543972 Apr 19 '18 at 3:25

Let's start by changing the order of integration:

\begin{align*} \|Mf\|_p^p &\le p \cdot (2 \cdot 3^n) \int_{\mathbb{R}^n} \int_0^{2|f(x)|} |f(x)| t^{p - 2} \, dt \, dx \\ &= p \cdot (2 \cdot 3^n) \int_{\mathbb{R}^n}|f(x)| \cdot \frac{(2|f(x)|)^{p - 1}}{p - 1} \, dx \\ &= \frac{p}{p - 1} 2^p 3^n \int_{\mathbb{R}^n} |f(x)|^p \, dx \\ &= 2^p \cdot \left(\frac{1}{1-1/p} \cdot 3^n\right) \|f\|_p^p \\ &= 2^p \cdot (3^n p') \|f\|_p^p \end{align*}

which is the desired result.

  • $\begingroup$ So essentially you have assumed the weak type $(1,1)$ bound. $\endgroup$ – user284331 Apr 19 '18 at 3:36
  • $\begingroup$ @user284331 Yes, pretty much (although what I really assumed was that the asker's work was correct thus far :)). $\endgroup$ – T. Bongers Apr 19 '18 at 3:48

Not sure how to do it in an elementary way, but the proof can be found in many harmonic analysis books, for example, Classical Fourier Analysis, Loukas Grafakos:

First we need to show that the maximal operator is of weak type $(1,1)$.

Also, it maps $L^{\infty}$ to $L^{\infty}$ with norm $1$, then by Marcinkiewicz Interpolation Theorem, it maps $L^{p}$ to $L^{p}$ with the required norm.

The norm is by no means the best.

  • 1
    $\begingroup$ Although this answer is quite right, it's unfortunately maybe a bit circular. The proof of Marcinkiewicz (or at least one of the standard ones...) runs through a very similar argument involving integrating a distribution function like the OP is trying to. I'm sure there's a way around it, though. $\endgroup$ – T. Bongers Apr 19 '18 at 3:32

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