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# Endpoint estimates for vector-valued multilinear commutator of fractional area integral operator

## Abstract

In this paper, we prove the endpoint estimates for vector-valued multilinear commutator of fractional area integral operator.

MSC:42B20, 42B25.

## 1 Introduction

Let $b∈BMO( R n )$ and T be the Calderón-Zygmund operator, the commutator $[b,T]$ generated by b and T is defined by

$[b,T](f)(x)=b(x)T(f)(x)−T(bf)(x).$

A classical result of Coifman, Rochberb and Weiss (see ) proved that the commutator $[b,T]$ is bounded on $L p ( R n )$ ($1). In , the boundedness properties of the commutators for the extreme values of p are obtained. In this paper, we introduce vector-valued multilinear commutator of fractional area integral operator and prove the endpoint estimates for the commutator $| S ψ , δ b → | r$ generated by the fractional area integral operator $S ψ , δ$ and BMO functions.

## 2 Notations and results

We give the following definitions (see [2, 3, 57]).

Definition 1 Let $0<δ, a function ψ satisfies:

1. (1)

$∫ R n ψ(x)dx=0$;

2. (2)

$|ψ(x)|≤C ( 1 + | x | ) − ( n + 1 − δ )$;

3. (3)

$|ψ(x+y)−ψ(x)|≤C | y | ε ( 1 + | x | ) − ( n + 2 − δ )$, $2|y|<|x|$.

Suppose that $1, $b j$ ($j=1,…,m$) are the fixed locally integrable functions on $R n$. Set $Γ(x)={(y,t)∈ R + n + 1 :|x−y|≤t}$ and the eigenfunction by $χ Γ ( x )$. We define the vector-valued multilinear commutator of fractional area integral operator by

$| S ψ , δ b → (f)(x) | r = ( ∑ i = 1 ∞ ( S ψ , δ b → ( f i ) ( x ) ) r ) 1 / r ,$

where

$S ψ , δ b → (f)(x)= ( ∫ Γ ( x ) | F t b → ( f ) ( x , y ) | 2 d y d t t n + 1 ) 1 / 2$

and

$F t b ˜ (f)(x)= ∫ R n [ ∏ j = 1 m ( b j ( x ) − b j ( z ) ) ] ψ t (y−z)f(z)dz.$

Definition 2 We call a locally integrable function b in the central BMO space, namely $CMO( R n )$, if the function b satisfies

$∥ b ∥ CMO = sup r > 1 |Q(0,r) | − 1 ∫ Q |b(y)− b Q |dy<∞.$

We have

$∥ b ∥ CMO ≈ sup r > 1 inf c ∈ C |Q(0,r) | − 1 ∫ Q |b(y)−c|dy.$

Definition 3 Let $0<δ, $1. We call a locally integrable function b in $B p δ ( R n )$, if the function b satisfies

$∥ b ∥ B p δ = sup r > 1 r − n ( 1 / p − δ / n ) ∥ b χ Q ( 0 , r ) ∥ L p <∞.$

Now we state our theorems as follows.

Theorem 1 Suppose $1, $0<δ, and $b → =( b 1 ,…, b m )$ for $b j ∈BMO$, $1≤j≤m$. Then $| S ψ , δ b → | r$ is bounded from $L n / δ$ to $BMO( R n )$.

Theorem 2 Let $1, $0<δ, $1, and $b → =( b 1 ,…, b m )$, with $b j ∈BMO( R n )$, for $1≤j≤m$. Then $| S ψ , δ b → | r$ is bounded from $B p δ ( R n )$ to $CMO( R n )$.

## 3 Proofs of theorems

We begin with a preliminaries lemma.

Lemma 1 (see [3, 4])

Let $1, $0<δ, $1, $1/q=1/p−δ/n$. Then $| S ψ , δ | r$ is bounded from $L p ( R n )$ to $L q ( R n )$.

Proof of Theorem 1 It is only to prove that there exists a constant $C Q$, the following inequality holds:

$1 | Q | ∫ Q || S ψ , δ b → (f)(x) | r − C Q |dx≤C ∥ | f | r ∥ L n / δ .$

Fix a cube $Q=Q( x 0 ,r)$, let $f=g+h={ g i }+{ h i }$ for $g i = f i χ Q$, $h i = f i χ ( Q ) c$.

When $m=1$, set $( b 1 ) Q = | Q | − 1 ∫ Q b 1 (y)dy$, then

$F t b 1 ( f i )(x,y)= ( b 1 ( x ) − ( b 1 ) Q ) F t ( f i )(y)− F t ( ( b 1 − ( b 1 ) Q ) g i ) (y)− F t ( ( b 1 − ( b 1 ) Q ) h i ) (y),$

so

$| S ψ , δ b 1 ( f ) ( x ) | r − | S ψ , δ ( ( ( b 1 ) 2 Q − b 1 ) h ) ( x 0 ) r | ≤ ( ∑ i = 1 ∞ ∥ χ Γ ( x ) ( b 1 ( x ) − ( b 1 ) Q ) F t ( f i ) ( y ) ∥ r ) 1 / r + ( ∑ i = 1 ∞ ∥ χ Γ ( x ) F t ( ( ( b 1 ) Q − b 1 ) g i ) ( y ) ∥ r ) 1 / r + ∥ χ Γ ( x ) F t ( ( b 1 − ( b 1 ) Q ) f 2 ) ( y ) − χ Γ ( x 0 ) F t ( ( b 1 − ( b 1 ) Q ) h ) ( y ) ∥ r = A ( x ) + B ( x ) + C ( x ) .$

For $A(x)$, suppose $1, $1/q=1/p−δ/n$ and $1/q+1/ q ′ =1$, by the Hölder inequality, then

$1 | Q | ∫ Q | A ( x ) | d x = 1 | Q | ∫ Q | b 1 ( x ) − ( b 1 ) Q | | S ψ , δ ( f ) ( x ) | r d x ≤ ( 1 | Q | ∫ Q | b 1 ( x ) − ( b 1 ) Q | q ′ d x ) 1 / q ′ × ( 1 | Q | ∫ R n | S ψ , δ ( f ) ( x ) | r q χ Q ( x ) d x ) 1 / q ≤ C ∥ b 1 ∥ BMO | Q | − 1 / q ( ∫ R n | f ( x ) | r p χ Q ( x ) d x ) 1 / p ≤ C ∥ b 1 ∥ BMO | Q | − 1 / q × [ ( ∫ R n | f ( x ) | r n / δ d x ) δ p / n ( ∫ Q χ Q ( x ) d x ) 1 − δ p / n ] 1 / p ≤ C ∥ b 1 ∥ BMO | Q | − 1 / q ∥ | f | r ∥ L n / δ | Q | ( 1 − δ p / n ) / p ≤ C ∥ b 1 ∥ BMO ∥ | f | r ∥ L n / δ .$

For $B(x)$, fix $1, $1/v=1/u−δ/n$, by the Hölder inequality, then

$1 | Q | ∫ Q | B ( x ) | d x = 1 | Q | ∫ Q | S ψ , δ ( ( b 1 − ( b 1 ) Q ) g ) ( x ) | r d x ≤ ( 1 | Q | ∫ R n | S ψ , δ ( ( b 1 − ( b 1 ) Q ) g ) ( x ) ) v d x | r ) 1 / v ≤ C | Q | − 1 / v ( ∫ R n | b 1 ( x ) − ( b 1 ) Q | u | f ( x ) | r u χ Q ( x ) d x ) 1 / u ≤ C ( 1 | Q | ∫ Q | b 1 ( x ) − ( b 1 ) Q | u d x ) 1 / u ∥ | f | r ∥ L n / δ ≤ C ∥ b 1 ∥ BMO ∥ | f | r ∥ L n / δ .$

For $C(x)$, we have

$C ( x ) = ∥ χ Γ ( x ) F t ( ( b 1 − ( b 1 ) Q ) f 2 ) ( y ) − χ Γ ( x 0 ) F t ( ( b 1 − ( b 1 ) Q ) h ) ( y ) ∥ r ≤ [ ∫ ∫ R + n + 1 ( ∫ Q c | χ Γ ( x ) − χ Γ ( x 0 ) ∥ b 1 ( z ) − ( b 1 ) Q ∥ ψ t ( y − z ) | | f ( z ) | r d z ) 2 d y d t t n + 1 ] 1 / 2 ≤ C ∫ Q c | b 1 ( z ) − ( b 1 ) Q | | f ( z ) | r × | ∫ ∫ | x − y | ≤ t t 1 − n d y d t ( t + | y − z | ) 2 n + 2 − 2 δ − ∫ ∫ | x 0 − y | ≤ t t 1 − n d y d t ( t + | y − z | ) 2 n + 2 − 2 δ | 1 / 2 d z ≤ C ∫ Q c | b 1 ( z ) − ( b 1 ) Q | | f ( z ) | r × ( ∫ ∫ | y | ≤ t , | x + y − z | ≤ t | 1 ( t + | x + y − z | ) 2 n + 2 − 2 δ − 1 ( t + | x 0 + y − z | ) 2 n + 2 − 2 δ | d y d t t n − 1 ) 1 / 2 d z ≤ ∫ Q c | b 1 ( z ) − ( b 1 ) Q | | f ( z ) | r ( ∫ ∫ | y | ≤ t , | x + y − z | ≤ t | x − x 0 | t 1 − n ( t + | x + y − z | ) 2 n + 3 − 2 δ d y d t ) 1 / 2 d z .$

Notice that when $|y|≤t$, $2t+|x+y−z|≥2t+|x−z|−|y|≥t+|x−z|$, and

$∫ 0 ∞ t d t ( t + | x − z | ) 2 n + 3 − 2 δ =C|x−z | − 2 n − 1 + 2 δ ,$

then, for $x∈Q$,

$C ( x ) ≤ ∫ Q c | b 1 ( z ) − ( b 1 ) Q | | f ( z ) | r ( ∫ ∫ | y | ≤ t 2 2 n + 3 − 2 δ | x 0 − x | t 1 − n d y d t ( 2 t + 2 | x + y − z | ) 2 n + 3 − 2 δ ) 1 / 2 d z ≤ C ∫ Q c | b 1 ( z ) − ( b 1 ) Q | | f ( z ) | r | x − x 0 | 1 / 2 ( ∫ ∫ | y | ≤ t t 1 − n d y d t ( t + | x − z | ) 2 n + 3 − 2 δ ) 1 / 2 d z ≤ C ∫ Q c | b 1 ( z ) − ( b 1 ) Q | | f ( z ) | r | x − x 0 | 1 / 2 ( ∫ 0 ∞ t d t ( t + | x − z | ) 2 n + 3 − 2 δ ) 1 / 2 d z ≤ C ∫ Q c | b 1 ( z ) − ( b 1 ) Q | | f ( z ) | r | x 0 − x | 1 / 2 | x 0 − z | n + 1 / 2 − δ d z ≤ C ∑ k = 1 ∞ ∫ 2 k + 1 Q ∖ 2 k Q | b 1 ( z ) − ( b 1 ) Q | | f ( z ) | r | x 0 − x | 1 / 2 | x 0 − z | n + 1 / 2 − δ d z ≤ C ∑ k = 1 ∞ 2 − k / 2 | 2 k + 1 Q | − 1 + δ / n ∫ 2 k + 1 Q | b 1 ( z ) − ( b 1 ) Q | | f ( z ) | r d z ≤ C ∥ b 1 ∥ BMO ∑ k = 1 ∞ k 2 − k / 2 ∥ | f | r ∥ L n / δ ≤ C ∥ b 1 ∥ BMO ∥ | f | r ∥ L n / δ ,$

so that

$1 | Q | ∫ Q |C(x)|dx≤C ∥ b 1 ∥ BMO ∥ | f | r ∥ L n / δ .$

When $m>1$, let $b → Q =( ( b 1 ) Q ,…, ( b m ) Q )∈ R n$, where

$( b j ) Q =|Q | − 1 ∫ Q b j (y)dy,1≤j≤m,$

let $f=g+h={ g i }+{ h i }$ for $g i = f i χ Q$, $h i = f i χ ( Q ) c$. We have

$F t b → ( f i ) ( x , y ) = ∫ R n [ ∏ j = 1 m ( b 1 ( x ) − b 1 ( z ) ) ] ψ t ( y − z ) f i ( z ) d z = ( b 1 ( x ) − ( b 1 ) Q ) ⋯ ( b m ( x ) − ( b m ) Q ) F t ( f i ) ( y ) + ( − 1 ) m F t ( ( b 1 − ( b 1 ) Q ) ⋯ ( b m − ( b m ) Q ) f i ) ( y ) + ∑ j = 1 m − 1 ∑ σ ∈ C j m ( − 1 ) m − j ( b → ( x ) − b → Q ) σ × ∫ R n ( b → ( z ) − b → Q ) σ c ψ t ( y − z ) f i ( z ) d z = ( b 1 ( x ) − ( b 1 ) Q ) ⋯ ( b m ( x ) − ( b m ) Q ) F t ( f i ) ( y ) + ( − 1 ) m F t ( ( b 1 − ( b 1 ) Q ) ⋯ ( b m − ( b m ) Q ) g i ) ( y ) + ( − 1 ) m F t ( ( b 1 − ( b 1 ) Q ) ⋯ ( b m − ( b m ) Q ) h i ) ( y ) + ∑ j = 1 m − 1 ∑ σ ∈ C j m ( − 1 ) m − j ( b → ( x ) − b → Q ) σ F t ( ( b → − b → Q ) σ c f i ) ( x , y ) ,$

by the Minkowski inequality, we have

$| | S ψ , δ b → ( f ) ( x ) | r − | S ψ , δ ( ( ( b 1 ) Q − b 1 ) ⋯ ( ( b m ) Q − b m ) h ) ( x 0 ) | r | ≤ ∥ χ Γ ( x ) ( b 1 ( x ) − ( b 1 ) Q ) ⋯ ( b m ( x ) − ( b m ) Q ) F t ( f ) ( y ) ∥ r + ∑ j = 1 m − 1 ∑ σ ∈ C j m ∥ χ Γ ( x ) ( b → ( x ) − b → Q ) σ F t ( ( b → − b → Q ) σ c f ) ( x , y ) ∥ r + ∥ χ Γ ( x ) F t ( ( b 1 − ( b 1 ) Q ) ⋯ ( b m − ( b m ) Q ) g ) ( y ) ∥ r + ∥ χ Γ ( x ) F t ( ∏ j = 1 m ( b j − ( b j ) Q ) h ) ( y ) − χ Γ ( x 0 ) F t ( ∏ j = 1 m ( b j − ( b j ) Q ) h ) ( y ) ∥ r = M 1 ( x ) + M 2 ( x ) + M 3 ( x ) + M 4 ( x ) .$

For $M 1 (x)$, similar to the proof of $m=1$, we take $1, $1/q=1/p−δ/n$, by the Hölder inequality and Lemma 1, we have

$1 | Q | ∫ Q M 1 ( x ) d x ≤ ( 1 | Q | ∫ Q | ∏ j = 1 m ( b j ( x ) − ( b j ) Q ) | q ′ d x ) 1 / q ′ ( 1 | Q | ∫ Q | S ψ , δ ( f ) ( x ) | r q d x ) 1 / q ≤ C ∥ b → ∥ BMO | Q | − 1 / q ( ∫ Q | f ( x ) | r p d x ) 1 / p ≤ C ∥ b → ∥ BMO | Q | − 1 / q ( ∫ Q | f ( x ) | r n / δ d x ) δ / n | Q | ( 1 − ( δ p / n ) ) / p ≤ C ∥ b → ∥ BMO ∥ | f | r ∥ L n / δ .$

For $M 2 (x)$, taking $1, $1/q=1/p−δ/n$, we get

$1 | Q | ∫ Q M 2 ( x ) d x ≤ C ∑ j = 1 m − 1 ∑ σ ∈ C j m ∥ b → σ ∥ BMO | Q | − 1 / q ( ∫ R n | ( b ( x ) − b Q ) σ c f ( x ) | r p χ Q ( x ) d x ) 1 / p ≤ C ∑ j = 1 m − 1 ∑ σ ∈ C j m ∥ b → σ ∥ BMO ( 1 | Q | ∫ Q | ( b ( x ) − b Q ) σ c | q d x ) 1 / q ∥ | f | r ∥ L n / δ ≤ C ∑ j = 1 m − 1 ∑ σ ∈ C j m ∥ b → σ ∥ BMO ∥ b → σ c ∥ BMO ∥ f ∥ L n / δ ≤ C ∥ b → ∥ BMO ∥ | f | r ∥ L n / δ .$

For $M 3 (x)$, taking $1, $1/q=1/p−δ/n$, we obtain

$1 | Q | ∫ Q M 3 ( x ) d x ≤ ( 1 | Q | ∫ Q | S ψ , δ ( ( b 1 − ( b 1 ) Q ) ⋯ ( b m − ( b m ) Q ) g ) ( x ) | r q d x ) 1 / q ≤ C | Q | − 1 / q ∥ ( b 1 ( x ) − ( b 1 ) Q ) ⋯ ( b m ( x ) − ( b m ) Q ) | g ( x ) | r ∥ L p ≤ C ( 1 | Q | ∫ Q | ( b 1 ( x ) − ( b 1 ) Q ) ⋯ ( b m ( x ) − ( b m ) Q ) | q d x ) 1 / q ∥ | f | r ∥ L n / δ ≤ C ∥ b → ∥ BMO ∥ | f | r ∥ L n / δ .$

For $M 4 (x)$, we have

$M 4 ( x ) ≤ C ∫ Q c | x 0 − x | 1 / 2 | x 0 − z | − ( n + 1 / 2 − δ ) | ∏ j = 1 m ( b j ( z ) − ( b j ) Q ) | | f ( z ) | r d z ≤ C ∑ k = 1 ∞ ∫ 2 k Q ∖ 2 k − 1 Q | x 0 − x | 1 / 2 | x 0 − z | − ( n + 1 / 2 − δ ) | ∏ j = 1 m ( b j ( z ) − ( b j ) Q ) | | f ( z ) | r d z ≤ C ∑ k = 1 ∞ ∫ 2 k Q ∖ 2 k − 1 Q | x 0 − x | 1 / 2 | x 0 − z | n + 1 / 2 − δ | ∏ j = 1 m ( b j ( z ) − ( b j ) Q ) | | f ( z ) | r d z ≤ C ∑ k = 1 ∞ 2 − k / 2 1 | 2 k Q | 1 − δ / n ∫ 2 k Q | ∏ j = 1 m ( b j ( z ) − ( b j ) Q ) | | f ( z ) | r d z ≤ C ∑ k = 1 ∞ 2 − k / 2 ( ∫ 2 k Q | f ( z ) | r n / δ d z ) δ / n × ( 1 | 2 k Q | ∫ 2 k Q | ∏ j = 1 m ( b j ( z ) − ( b j ) Q ) | n / ( n − δ ) d z ) ( n − δ ) / n ≤ C ∑ k = 1 ∞ k m 2 − k / 2 ∏ j = 1 m ∥ b j ∥ BMO ∥ | f | r ∥ L n / δ ≤ C ∥ b → ∥ BMO ∥ | f | r ∥ L n / δ ,$

so

$1 | Q | ∫ Q | M 4 (x)|dx≤C ∥ b → ∥ BMO ∥ | f | r ∥ L n / δ .$

This completes the proof of Theorem 1. □

Proof of Theorem 2 It is only to prove that there exists a constant $C Q$, for any of the cubes $Q=Q(0,d)$ ($d>1$), the following inequality holds:

$1 | Q | ∫ Q || S ψ , δ b → (f)(x) | r − C Q |dx≤C ∥ f ∥ B p δ .$

Fix a cube $Q=Q(0,d)$ ($d>1$). Let $f=g+h={ g i }+{ h i }$, where $g i = f i χ Q$, $h i = f i χ ( Q ) c$ and $b → Q =( ( b 1 ) Q ,…, ( b m ) Q )$. For $( b j ) Q =|Q | − 1 ∫ Q | b j (y)|dy$, $1≤j≤m$, we have

$F t b → ( f i ) ( x , y ) = ∫ R n [ ∏ j = 1 m ( b 1 ( x ) − b 1 ( z ) ) ] ψ t ( y − z ) f i ( z ) d z = ( b 1 ( x ) − ( b 1 ) Q ) ⋯ ( b m ( x ) − ( b m ) Q ) F t ( f i ) ( y ) + ( − 1 ) m F t ( ( b 1 − ( b 1 ) Q ) ⋯ ( b m − ( b m ) Q ) g i ) ( y ) + ( − 1 ) m F t ( ( b 1 − ( b 1 ) Q ) ⋯ ( b m − ( b m ) Q ) h i ) ( y ) + ∑ j = 1 m − 1 ∑ σ ∈ C j m ( − 1 ) m − j ( b → ( x ) − b → Q ) σ F t ( ( b → − b → Q ) σ c f i ) ( x , y ) .$

By the Minkowski inequality, we have

$| | S ψ , δ b → ( f ) ( x ) | r − | S ψ , δ ( ( ( b 1 ) Q − b 1 ) ⋯ ( ( b m ) Q − b m ) h ) ( x 0 ) | r | ≤ ∥ χ Γ ( x ) ( b 1 ( x ) − ( b 1 ) Q ) ⋯ ( b m ( x ) − ( b m ) Q ) F t ( f ) ( y ) ∥ r + ∑ j = 1 m − 1 ∑ σ ∈ C j m ∥ χ Γ ( x ) ( b → ( x ) − b → Q ) σ F t ( ( b → − b → Q ) σ c f ) ( x , y ) ∥ r + ∥ χ Γ ( x ) F t ( ( b 1 − ( b 1 ) Q ) ⋯ ( b m − ( b m ) Q ) g ) ( y ) ∥ r + ∥ χ Γ ( x ) F t ( ∏ j = 1 m ( b j − ( b j ) Q ) h ) ( y ) − χ Γ ( x 0 ) F t ( ∏ j = 1 m ( b j − ( b j ) Q ) h ) ( y ) ∥ r = H 1 ( x ) + H 2 ( x ) + H 3 ( x ) + H 4 ( x ) .$

For $H 1 (x)$, take $1/q=1/p−δ/n$, by the Hölder inequality and Lemma 1, we have

$1 | Q | ∫ Q H 1 ( x ) d x ≤ ( 1 | Q | ∫ Q | ∏ j = 1 m ( b j ( x ) − ( b j ) Q ) | q ′ d x ) 1 / q ′ ( 1 | Q | ∫ Q | S ψ , δ ( f ) ( x ) | r q d x ) 1 / q ≤ C ∥ b → ∥ BMO | Q | − 1 / q ( ∫ R n | f ( x ) | r p χ Q ( x ) d x ) 1 / p ≤ C ∥ b → ∥ BMO d − n ( 1 / p − δ / n ) ∥ | f | r χ Q ∥ L p ≤ C ∥ b → ∥ BMO ∥ | f | r ∥ B p δ .$

For $H 2 (x)$, taking $1, $1/v=1/u−δ/n$, we get

$1 | Q | ∫ Q H 2 ( x ) d x ≤ C ∑ j = 1 m − 1 ∑ σ ∈ C j m ( 1 | Q | ∫ Q | ( b ( x ) − b Q ) σ | v ′ d x ) 1 / v ′ × ( 1 | Q | ∫ Q | S ψ , δ ( ( b − b Q ) σ c f ) ( x ) | r u d x ) 1 / u ≤ C ∑ j = 1 m − 1 ∑ σ ∈ C j m ∥ b → σ ∥ BMO | Q | − 1 / v ( ∫ R n | ( b ( x ) − b Q ) σ c f ( x ) | r u χ Q ( x ) d x ) 1 / u ≤ C ∑ j = 1 m − 1 ∑ σ ∈ C j m ∥ b → σ ∥ BMO | Q | ( δ / n − 1 / p ) ∥ | f | r χ Q ∥ L p × ( 1 | Q | ∫ Q | ( b ( x ) − b Q ) σ c | p r / ( p − r ) d x ) ( p − u ) / p u ≤ C ∑ j = 1 m − 1 ∑ σ ∈ C j m ∥ b → σ ∥ BMO ∥ b → σ c ∥ BMO d − n ( 1 / p − δ / n ) ∥ | f | r χ Q ∥ L p ≤ C ∥ b → ∥ BMO ∥ | f | r ∥ B p δ .$

For $H 3 (x)$, taking $1, $1/v=1/u−δ/n$, we obtain

$1 | Q | ∫ Q H 3 ( x ) d x ≤ ( 1 | Q | ∫ Q | S ψ , δ ( ( b 1 − ( b 1 ) Q ) ⋯ ( b m − ( b m ) Q ) g ) ( x ) | r v d x ) 1 / v ≤ C | Q | − 1 / v ( ∫ Q | ( b 1 ( x ) − ( b 1 ) Q ) ⋯ ( b m ( x ) − ( b m ) Q ) g ( x ) | r u d x ) 1 / u ≤ C | Q | − 1 / v ∥ ( b 1 ( x ) − ( b 1 ) Q ) ⋯ ( b m ( x ) − ( b m ) Q ) | f | r χ Q ∥ L v ≤ C ∥ b → ∥ BMO ∥ | f | r ∥ B p δ .$

For $H 4 (x)$, we have

$I 4 ( x ) ≤ [ ∫ ∫ R + n + 1 ( ∫ Q c | χ Γ ( x ) − χ Γ ( x 0 ) | ∏ j = 1 m | b j ( z ) − ( b j ) Q | | ψ t ( y − z ) | | f ( z ) | r d z ) 2 d y d t t n + 1 ] 1 / 2 ≤ C ∑ k = 0 ∞ ∫ 2 k + 1 Q ∖ 2 k Q | x 0 − x | 1 / 2 | x 0 − z | − ( n + 1 / 2 − 2 δ ) | ∏ j = 1 m ( b j ( z ) − ( b j ) Q ) | | f ( z ) | r d z ≤ C ∑ k = 1 ∞ 2 − k / 2 | 2 k + 1 Q | − 1 + δ / n ∫ 2 k + 1 Q | ∏ j = 1 m ( b j ( z ) − ( b j ) Q ) | | f ( z ) | r d z ≤ C ∑ k = 1 ∞ k m 2 − k / 2 | 2 k Q | − ( 1 / p − δ / n ) ∥ b → ∥ BMO ∥ | f | r χ 2 k Q ∥ L p ≤ C ∥ b → ∥ BMO ∥ | f | r ∥ B p δ ,$

so

$1 | Q | ∫ Q | H 4 (x)|dx≤C ∥ b → ∥ BMO ∥ | f | r ∥ B p δ .$

This completes the proof of Theorem 2. □

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Correspondence to Weiping Kuang.

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• $BMO( R n )$ 