Research Blog

Nontangential Maximal Function

2020, November 10

Introduction

We are interested in the relations between a function $f$ defined on $\Rn$ and a corresponding function $F$ defined on $\RR{n+1}_+$. For instance, $F (\cdot, t) = f * \Phi _t$ for some convolution kernel $\Phi$. We define the nontangential maximal function

$$\begin{align*} F ^* (x) := \sup _{|x - y| < t} |F (y, t)| = \sup _{\Gamma (x)} |F|. \end{align*}$$

Here $\Gamma (x) = \set{(y, t): \abs{y - x} < t}$ is a cone centered at $x$ with aperture $1$. We may also define cone with aperture $a$ by $\Gamma _a (x) = \set{(y, t): \abs{y - x} < at}$, and corresponding $F ^* _a (x)$.

Carleson Measure

$\newcommand{\cN}{\mathcal N} \newcommand{\cC}{\mathcal C}$ For a ball $B = B _r (x _0) \subset \Rn$, let $T (B) \subset \RR{n+1}_+$ be the cone with base $B$:

$$\begin{align*} T (B) = \set{ (x, t): \abs{x - x _0} \le r - t } = \RR{n+1} _+ \setminus \bigcup _{x \notin B} \Gamma (x). \end{align*}$$

In general for an open set $O \subset \Rn$, we define

$$\begin{align*} T (O) = \RR{n+1} _+ \setminus \bigcup _{x \notin O} \Gamma (x). \end{align*}$$

It is easy to see

$$\begin{align*} A \subset B \Rightarrow B ^c \subset A ^c \Rightarrow \bigcup _{x \notin B} \Gamma (x) \subset \bigcup _{x \notin A} \Gamma (x) \Rightarrow T (A) \subset T (B) \end{align*}$$

and

$$\begin{align*} (x, t) \in T (A) \cap T (B) &\Leftrightarrow (x, t) \notin \bigcup _{x \notin B} \Gamma (x) \cup \bigcup _{x \notin A} \Gamma (x) \newline &\Leftrightarrow (x, t) \notin \bigcup _{x \notin A \cap B} \Gamma (x) \newline &\Leftrightarrow (x, t) \in T (A \cap B) \end{align*}$$

so $T (A) \cap T (B) = T (A \cap B)$. As for the union,

$$\begin{align*} (x, t) \in T (A) \cup T (B) &\Leftrightarrow (x, t) \notin \bigcup _{x \notin A} \Gamma (x) \cap \bigcup _{x \notin B} \Gamma (x) \newline &\Rightarrow (x, t) \notin \bigcup _{x \notin A \cup B} \Gamma (x) \newline &\Leftrightarrow (x, t) \in T (A \cup B). \end{align*}$$

so $T (A) \cap T (B) \subset T (A \cap B)$.

Consider the following two spaces:

$$\begin{align*} \cN &= \set{F: \RR{n+1} _+ \to \R \text{ measurable function with } F ^* \in L^1(\Rn)}, \newline \cC &= \set{\mu \text{ Borel measure on $\RR{n+1} _+$}: \mu (T (B)) \le C |B| \text{ for all $B$ for some $C < \infty$}}. \end{align*}$$

These two spaces equipped with norm

$$\begin{align*} \nor{F} _{\cN} &= \nor{F ^*} _{L ^1 (\Rn)}, \newline \nor{\mu} _{\cC} &= \sup _B \frac{\mu (T (B))}{|B|} \newline &= \sup _{x \in \Rn} \underbrace{\sup _{B \ni x} \frac{\mu (T (B))}{|B|}} _{:=C (\d\mu) (x)}. \end{align*}$$

are Banach spaces. $\cC$ is called the Carleson measure. The main theorem is the following.

Proof.

The proof is based on the following two observations:

1. $\set{\abs F > \alpha} \subset T \pth{ \set{ F ^* > \alpha}}$.
2. $\mu (T(B)) \le \abs B$ for all $B$ implies $\mu (T(O)) \le c \abs O$ for all $O \subset \Rn$.

Let’s see why they are true. Assume $F > 0$ for simplicity. For the first one,

$$\begin{align*} (x, t) \in \set{F > \alpha} &\Rightarrow \forall y \in B _t (x), F ^* (y) > \alpha \newline &\Rightarrow (x, t) \in T (B _t (x)) \subset T \pth{ \set{ F ^* > \alpha } }. \end{align*}$$

For the second one, pick $(x, t) \in T (O)$, then $B _t (x) \subset O$. Let $\seq Q k$ be a Whitney decomposition of $O$, such that $\diam (Q _k) \sim \dist (Q _k, O ^c)$. Let $Q _k$ be the cube that contains $x$, and let $B _k$ be a ball centered at $Q _k$ with radius $c \diam (Q _k)$. Then by choosing $c$ large enough, $B _t (x) \subset B _k$, and $(t, x) \in T (B _k)$. Therefore, $T (O) \subset \bigcup _k T (B _k)$, and

$$\begin{align*} \mu (T(O)) \le \sum _k \mu (T(B _k)) \end{align*}$$

Now for any $x \in Q _k \subset B _k$,

$$\begin{align*} \mu (T(B _k)) \le C (\d \mu) (x) |B _k|, \end{align*}$$

therefore

$$\begin{align*} \mu (T(B _k)) \le \frac1{|Q _k|} \int _{Q _k} C (\d \mu) (x) \d x |B _k| = c \int _{Q _k} C (\d \mu) (x) \d x, \end{align*}$$

thus

$$\begin{align*} \mu (T (O)) \le c \sum _k \int _{Q _k} C (\d \mu) (x) \dx = c \int _O C (\d \mu) (x) \d x. \end{align*}$$

After these two observations, we have

$$\begin{align*} \mu \ptset{F > \alpha} \le \mu( T( \set{F ^* > \alpha})) \le c \intset{F ^* > \alpha} C (\d\mu) (x) \d x. \end{align*}$$

Integrate with respect to $\alpha$,

$$\begin{align*} \int _{\RR{n+1}_+} F \d \mu \le c \intRn F ^* (x) C (\d\mu) (x) \d x. \end{align*}$$

The Effect of Aperture

We now show that the choice of aperture is not important.

Note that $F ^* _a \ge F ^* _b$ pointwisely, so the above inequalities are comparible.

Proof.

Denote $O = \set{F ^* _b > \alpha}$. For $x \in \set{F ^* _a > \alpha}$, there exists $(\bar x, t)$ with $F (\bar x, t) > \alpha$ and $x \in B _{at} (\bar x)$. For every $x \in B _{bt} (\bar x)$ we have $F ^* _b (x) > \alpha$, so $B _{bt} (\bar x) \subset O$. If we set $B = B _{(a + b) t} (x) \supset B _{bt} (\bar x)$, then

$$\begin{align*} \frac{|O \cap B|}{|B|} \ge \frac{B _{bt} (\bar x)}{|B|} \ge \pthf b{a+b} ^n. \end{align*}$$

This shows that

$$\begin{align*} \mm \ind O (x) \ge \pthf b{a+b} ^n. \end{align*}$$

Therefore,

$$\begin{align*} \abset{F ^* _a > \alpha} \le \abset{\mm \ind O \ge \pthf b{a+b} ^n} \le c_n \pthf{a+b}b ^n \nmL1{\ind O} = c _n \pthf{a+b}b ^n \abset{F ^* _b > \alpha}. \end{align*}$$

Atomic Decomposition

Proof.

We denote the dyadic level sets of $F ^*$ by $O ^j = \set{F ^* > 2 ^j}$, then $\supp F \subset \bigcup _{j \in \Z} T (O ^j)$. Moreover, we Whitney decompose $O ^j$ into $Q ^j _k$, so that

$$\begin{align*} T (O ^j) = \bigcup _k T (B ^j _k) \cap (Q ^j _k \times (0, \infty)). \end{align*}$$

Denote

$$\begin{align*} \Delta ^j _k = T (B ^j _k) \cap (Q ^j _k \times (0, \infty)) \cap (T (O ^j) \setminus T(O ^{j + 1})). \end{align*}$$

Then $\supp F$ is the disjoint union of $\Delta _k ^j$. Now set $\lambda ^j _k a ^j _k = F \ind{\Delta _k ^j}$, where $\lambda ^j _k = 2 ^{j + 1} \abs{B _k ^j}$. Then $a ^j _k$ is an atom and

$$\begin{align*} \sum _{j, k} \lambda _k ^j = \sum _{j, k} 2 ^{j + 1} |B _k ^j| = c \sum _{j, k} 2 ^{j + 1} |Q _k ^j| = c \sum _j 2 ^{j + 1} |O ^j| \le c \nmL1{F ^*} = c \nor F _{\cN}. \end{align*}$$

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