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I'm studying harmonic analysis and found that we can bound non-tangential maximal function by Hardy-Littlewood maximal function. Most books don't give the proof of it. How can I see that? Is there a proof of it using dyadic decomposition?


It appears the OP is asking the following question. Given $\phi$ a bounded positive integrable function on $\mathbb{R}^n$ that is radial and radially decreasing. For $t> 0$ let $\phi_t = t^{-n}\phi(x/t)$. Define the maximal operator $$ M_\phi f(x) = \sup \{ |\phi_t\ast f(y)| : |x-y| < t\} $$

Why does the following hold: $$ M_\phi f(x) \leq C \mathcal{M} f(x) $$ where $\mathcal{M}$ is the Hardy-Littlewood maximal operator.

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What do you mean by your question? In particular, what do you mean by "bound non-tangential maximal function by Hardy-Littlewood maximal function"? Classically the definition of the HL maximal operator $\mathcal{M}$ takes input a function defined on $\mathbb{R}^n$, whereas the non-tangential maximal operator $\mathcal{F}$ takes input a function defined on the upper-half-space $\mathbb{R}^n\times\mathbb{R}_+$. The two operators do not even operate on the same domain, how do you want to compare the two? –  Willie Wong Dec 18 '12 at 13:03
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It would help greatly if you state where you "found" this fact. –  Willie Wong Dec 18 '12 at 13:03
    
I mean that the non tangential maximal function is less than constant * maximal function. I mean that when the tangential function can be written as a convolution like sup (psi * f) and * here is convolution and psi< const(1+x)^n. Do you know something like that? –  Danny Dec 18 '12 at 17:01
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Do you mean non-tangential maximal function of harmonic extension or of a similar convolution? Aside: you are not a new user; it's about time to take up simple LaTeX like $(1+x)^{n}$. Had you used some notation in your question, chances are the meaning would be easier to understand. –  user53153 Dec 19 '12 at 4:29
    
Please either give a source for the claim (so we can check definitions or notations) or better yet include in your questions statement how you see the Hardy-Littlewood maximal function and the non-tangential maximal function defined. I suspect @Pavel is on the mark, but dude you are making it extremely hard for us to help you. –  Willie Wong Dec 19 '12 at 9:21

2 Answers 2

up vote 4 down vote accepted

Why insist on dyadic decomposition of cubes when you don't need it?

Observe that the maximal function $\mathcal{M}f = \mathcal{M} |f|$ by definition. And observe that since $\phi$ is positive, $|\phi_t\ast f| \leq \phi_t \ast |f|$. Hence we can assume without loss of generality that $f$ is positive.

Let $\lambda_\phi(s)$ for $s \geq 0$ be the set $\{ \phi(x) \geq s\}$. We have that $\phi(x) = \int_0^P \chi_{\lambda_{\phi}(s)}(x) \mathrm{d}s$, where $P = \sup \phi$. Note that by assumption $\lambda_\phi(s)$ for a decreasing family of balls around the origin. Let $\lambda^*\phi(s)$ be the ball whose radius is 1 more than the radius of $\lambda_\phi(s)$. Now, let $|y-x| < 1$. We have that

$$ \int_{\lambda_\phi(s)} f(y-z) \mathrm{d}z \leq \int_{\lambda^*_\phi(s)} f(x - z) \mathrm{d}z $$

since $f$ is positive and $x + \lambda^*_\phi(s) \supset y + \lambda_\phi(s)$.

Let $P'$ be the smallest number such that $\lambda_\phi(P')$ has radius at most 1. By assumption (that $\phi$ is a bounded integrable function) we have that $\lambda_\phi(P')$ has positive radius $R'$.

$$\begin{align} \phi\ast f(y) &= \int_0^{P'} \int_{\lambda_\phi(s)} f(y-z) \mathrm{d}z \mathrm{d}s + \int_{P'}^P \int_{\lambda_\phi(s)} f(y-z) \mathrm{d}z \mathrm{d}s \\ &\leq \int_0^{P'} \int_{\lambda^*_\phi(s)} f(x-z) \mathrm{d}z \mathrm{d}s + \int_{P'}^P\int_{\lambda^*_\phi(P')} f(x-z) \mathrm{d}z \mathrm{d}s \\ & \leq \int_0^{P'} |\lambda_\phi^*(s)| \mathcal{M}f(x) \mathrm{d}s + |P' - P||\lambda_\phi^*(P')|\mathcal{M}f(x) \end{align}$$

Now using that for $s \leq P'$ we have that $|\lambda_\phi^*(s)| \leq c^n |\lambda_\phi(s)|$ where $c = 1 + 1/R'$ we have

$$ \leq \mathcal{M}f(x)\left( c^n \int_0^P |\lambda_\phi(s)| \mathrm{d}s + |P-P'| |\lambda_\phi^*(P')|\right)\tag{*}$$

The first term inside the parenthesis gives $\int \phi(z) \mathrm{d}z$ from the distributional function characterisation of Lebesgue spaces. The second term is quite obviously a finite constant depending on $\phi$.


Now, if we replace $\phi$ by $\phi_t$ in the above argument, then $P \mapsto t^{-n} P$. We will need to consider $|y-x| < t$ and we let $\lambda^*_{\phi_t}(s)$ to have radius $t$ more than its counterpart without the star. We also let $P'$ be such that the corresponding ball has radius at least $t$: by the scaling property of $\phi_t$ we see that $P' \mapsto t^{-n} P'$, and $\lambda_\phi(P') \to t \lambda_\phi(P')$. So the above analysis goes to show that the constant derived above (the term inside the parentheses in (*)) does not depend on the scaling $t$. Hence we get the desired inequality.

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I certainly will read it carefully...many thanks... –  Danny Dec 19 '12 at 15:22

There are at least 3 books that cover this questin - Stein, Torchinsky, and Grafakos. In the context of the intervl, also Zygmund's book.

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