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While reading the paper "A geometric proof of the strong maximal theorem", by A. Cordoba and R. Fefferman -Annals of Mathematics Vol 102 no. 1, I got stuck trying to understand a main step in the proof.

The goal is to prove a inequality for the strong maximal operator: $$ Mf(x) = \sup_R \frac{1}{R}\int_R |f|, $$ where $R$ ranges over all rectangles with sides parallel to the coordinate axes which contain $x$.

In particular they state $$ \bigl|\{x:Mf(x)>\lambda\}\bigr| \leq A\int\frac{|f(x)|}{\lambda}\biggl( 1 + \log^+\frac{|f(x)|}{\lambda} \biggr)\,dx. $$ (for dimension 2).

To achieve this they prove that given a collection of rectangles $\{R_i\}$ there is a subcollection $\{\tilde{R}_j\}$ such that

i) $|\cup_i R_i| \leq C |\cup_j \tilde{R}_j|$

ii) $\|\exp(\sum \chi_{\tilde{R}_j})\| \leq C |\cup_i R_i|$

Then, they assert that the strong maximal theorem follows from this and their proof of the fact that the strong maximal operator if weak-type $(p,p)$ if and only if you have the $V_q$ property where $q$ is the dual exponent of $p$: Given any collection of rectangles $\{R\}$ there exists a subcollection $\{\tilde{R}\}$ such that

$(i)$ $|\cup_i R_i| \leq C |\cup_j \tilde{R}_j|$

$(ii)_q$ $\|\sum \chi_{\tilde{R}_j}\|_{q} \leq C |\cup_i R_i|^{1/q}$

I don't understand how their proof, which is based on Holder's inequality, extends to the case $p=1.$ The main obstacle for me is the fact that the right hand side of $(ii)_\infty$ should be just $1$ ($\text{something}^{1/\infty}$) but the exponent in their main result is instead $1$.

I tried using the inequality

$$ \int fg \leq C \|f\|_{L\log L} \|g\|_{e^L} $$ but I end up with $|\{x: Mf(x)>\lambda\}|$ multiplied on both sides of the equation...

How is this proved?

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I am not sure how Cordoba and Fefferman intended the argument to go, but the final argument you indicate basically works, as long as one adjusts things by an epsilon so that the factor of $|\bigcup_i R_i|$ on the RHS is dominated by that on the LHS. [This is an example of a more general principle, namely that one should not be disheartened by finding a "circularity" in one's argument, such as finding the same term on the RHS as on the LHS of an estimate, so long as the circularity is introduced in a non-trivial fashion, as there can be a chance that one can tweak the argument so that the argument becomes non-circular.]

More specifically, one can start with the Young type inequality $$ f g \leq C_\varepsilon f (1 + \log^+ f) + \varepsilon e^g \quad (1) $$ that is valid for any $f, g \geq 0$ and $\varepsilon > 0$, and some $C_\varepsilon$ depending on $\varepsilon$. If now $f \in L \log L$ and $\lambda > 0$, then taking a family $R_i$ of rectangles covering $\{ Mf \geq \lambda\}$ with $\int_{R_i} f \geq \lambda |R_i|$ and using the subcollection $\tilde R_j$ and (i) we have $$ \int \frac{|f|}{\lambda} \sum_j \chi_{\tilde R_j} \geq C^{-1} |\bigcup_i R_i|$$ while from (1) and (ii) one has $$ \int \frac{|f|}{\lambda} \sum_j \chi_{\tilde R_j} \leq C_\varepsilon \int \frac{|f|}{\lambda} (1 + \log_+ \frac{|f|}{\lambda}) + C\varepsilon |\bigcup_i R_i|.$$ Choosing $\varepsilon$ small enough we obtain $$ |\bigcup_i R_i| \leq C' \int \frac{|f|}{\lambda} (1 + \log_+ \frac{|f|}{\lambda})$$ which then gives the required bound on $|\{ Mf \geq \lambda \}|$.

(There is a technical issue that $|\bigcup_i R_i|$ may be infinite, so that one cannot immediately cancel the terms on both sides of the estimates, but one can first work for instance with finite subcollections of the $R_i$, then use a limiting argument to recover a bound for all of $|\bigcup_i R_i|$.)

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