Research Blog

Vector-Valued Maximal Function

2020, November 07

For $f \in L ^1 _\loc (\Rn)$, we can define

$$\begin{align*} \mm f (x) := \sup _{r > 0} \fint _{B _x (r)} |f (y)| \d y. \end{align*}$$

$\newcommand{\vf}{\boldsymbol f} \newcommand{\vg}{\boldsymbol g} \newcommand{\vb}{\boldsymbol b} \newcommand{\vM}{\boldsymbol{\mathcal M}}$

Consider the space of function sequence $\vf = \set{f _j} \cnt j1\infty \in \pth{L^p (\Rn)} _\tensor$, define

$$\begin{align*} | \vf (x) | = \pth{ \SUM j1i |f _j (x)| ^2 } ^\half = \nor{\set{f _j (x)} \cnt j1\infty} _{\ell^2} \end{align*}$$

We say $\vf \in L ^p$ if $\abs{\vf} \in L ^p (\Rn)$, or $\nor{\vf} _{L ^p} = \nor{f _j (x)} _{L ^p _x \ell ^2 _j}$.

Vector-valued maximal operator $\vM$ is defined by

$$\begin{align*} \vM \vf (x) = \pth{ \SUM j1\infty |\mm f _j (x)| ^2 } ^\half = \nor{\set{\mm f _j (x)} \cnt j1\infty} _{\ell^2} . \end{align*}$$

Then:

We will prove this theorem first for $p = 2$, then for $p = 1$, then for $p \in (1, 2)$, then for $p \in (2, \infty)$.

Proof for $p = 2$

Case $p = 2$ is natural, because $\mm$ is bounded on $L^2$, and $L ^2 _x \ell ^2 _j = L ^2_{x, j}$.

Proof for $p = 1$

When $p = 1$, we want to prove for all $\alpha > 0$,

$$\begin{align*} \abset{ \vM \vf > \alpha } \le \frac A\alpha \nmL1\vf. \end{align*}$$

Assume $\vf \ge 0$. We have Calder'on-Zygmund decomposition, that is, we have a collection $\set{Q _k}$ disjoint cubes, $\vf = \vg + \vb = \vg + \sum _k \vb _k$, such that

1. $\abs{\vg} = \abs{\vf} \le \alpha$ in $\Rn \setminus \bigcup _k Q _k$, and $\vg = 0$ in $Q _k$.
2. $\vb _k$ is supported in $Q _k$, and $\fint _{Q _k} \abs{\vb _k} \le A \alpha$.
3. $\sum _k \abs{Q _k} \le \frac A\alpha \nor\vf _{L ^1}$.

Because $\abs{\vg} \le \min \set{\alpha, \abs{\vf}}$,

$$\begin{align*} \abset{ \vM \vg > \frac\alpha2 } \le \frac4{\alpha ^2} \nmL2{\vM \vg} ^2 \le \frac A{\alpha ^2} \nmL2\vg ^2 \le \frac A\alpha \nmL1\vf. \end{align*}$$

it suffices to show

$$\begin{align*} \abset{ \vM \vb > \frac\alpha2 } \le \frac A\alpha \nmL1\vf. \end{align*}$$

Denote $\bar \vb _k = \fint _{Q _k} \vb _k \le A\alpha$, and $\bar \vb = \sum _k \bar \vb _k \ind{Q _k}$. Then

$$\begin{align*} \abset{ \vM \bar \vb > \frac\alpha2 } \le \frac4{\alpha ^2} \nmL2{\vM \bar \vb} ^2 \le \frac A{\alpha ^2} \nmL2{\bar \vb} ^2 \le A ^3 \sum _k |Q _k| \le \frac A\alpha \nmL1\vf. \end{align*}$$

Similarly

$$\begin{align*} \abset{ c \vM \bar \vb > \frac\alpha2 } \le \frac A\alpha \nmL1\vf. \end{align*}$$

We want to show that

$$\begin{align*} \vM \vb (x) \le c \vM \bar \vb (x) \qquad x \notin \bigcup _k Q _k ^* \end{align*}$$

where $Q _k ^* = 2 Q _k$. This would end the proof because

$$\begin{align*} \abset{ \vM \vb (x) > \frac\alpha2 } &\le \abset{ c \vM \bar \vb > \frac\alpha2 } + \abset{ \vM \vb (x) > c \vM \bar \vb } \\&\le \frac A\alpha \nmL1\vf + \sum _k |Q _k ^*| \\&\le \frac A\alpha \nmL1\vf. \end{align*}$$

To see why $c \vM \bar \vb$ bound $\vM \vb$ away from $Q _k ^*$, it suffices to prove for each coordinate,

$$\begin{align*} \mm b _j (x) \le c \mm \bar b _j (x) \qquad x \notin \bigcup _k Q _k ^* \end{align*}$$

then the $\ell ^2$ norm are controlled. Let $B = B (x, r)$, and consider

$$\begin{align*} \fint _B b _j (x) \d x = \frac1{|B|} \sum _k \int _{B \cap Q _k} b _j (x) \d x. \end{align*}$$

If $x \notin Q _k ^*, B \cap Q _k \neq \varnothing$, then $Q _k \subset 3 B$, so

$$\begin{align*} \fint _B b _j \d x \le \frac1{|B|} \int _{3B} b _j \d x = \frac1{|B|} \int _{3B} \bar b _j \d x \le 3 ^n M \bar b _j (x). \end{align*}$$

Proof for $1 < p < 2$

By Marcinkiewicz interpolation, we have all the strong type bound for $1 < p \le 2$.

Proof for $2 < p < \infty$

First we introduce the weighted maximal inequality.

Proof.

The case $q = \infty$ is trivial, and the case $1 < q < \infty$ is by Marcinkiewicz interpolation.

For $x \in \set{\mm f (x) > \alpha}$, there exists $B _x$ such that

$$\begin{align*} |B _x| < \frac1\alpha \int _{B _x} |f (y)| \d y. \end{align*}$$

Then $$\begin{align*} \mu (5 B _x) = \int _{5 B _x} \omega \d x &\le A |B _x| \mm \omega (y), \qquad \forall y \in B _x \\& \le \frac A\alpha \int _{B _x} |f (y)| \mm \omega (y) \d y. \end{align*}$$

Fincally, we choose a disjoint subselection of $B _{x _k}$ such that $5 B _{x _k}$ is a covering of $\set{\mm f (x) > \alpha}$.

Back to the case $\vf = \set{f _j} \cnt j1i$, by proposition case $q = 2$ we have

$$\begin{align*} \nor{\mm f _j} _{L ^2(\omega \d x)} \le A _2 \nor{f _j} _{L ^2 (\mm \omega \d x)}. \end{align*}$$

Take $\ell ^2$ norm,

$$\begin{align*} \nor{\vM \vf} _{L ^2(\omega \dx)} \le A _2 \nor{\vf} _{L ^2 (\mm \omega \d x)}, \end{align*}$$

that is

$$\begin{align*} \int (\vM \vf) ^2 \omega \d x \le A \int |\vf| ^2 (\mm \omega) \d x. \end{align*}$$

Then by duality, set $q = \pthf p2 ^*$,

$$\begin{align*} \nmL p{\vM \vf} &= \nmL{\frac p2}{(\vM \vf) ^2} \\&= \sup _{\nmL q{\omega} = 1} \int (\vM \vf) ^2 \omega \d x \\& \le \sup _{\nmL q{\omega} = 1} A \int |\vf| ^2 (\mm \omega) \d x \\& \le A \nmL{\frac p2}{\vf ^2} \sup _{\nmL q{\omega} = 1} \nmL{q}{\mm \omega} \\& \le A \nmL p \vf. \end{align*}$$

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