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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 bBMO( 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 [1]) proved that the commutator [b,T] is bounded on L p ( R n ) (1<p<). In [24], 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<n, 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<r<, b j (j=1,,m) are the fixed locally integrable functions on R n . Set Γ(x)={(y,t) R + n + 1 :|xy|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 (yz)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<n, 1<p<n/δ. 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<r<, 0<δ<n, and b =( b 1 ,, b m ) for b j BMO, 1jm. Then | S ψ , δ b | r is bounded from L n / δ to BMO( R n ).

Theorem 2 Let 1<r<, 0<δ<n, 1<p<n/δ, and b =( b 1 ,, b m ), with b j BMO( R n ), for 1jm. 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<r<, 0<δ<n, 1<p<n/δ, 1/q=1/pδ/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 |dxC | 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<p<n/δ, 1/q=1/pδ/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<u<n/δ, 1/v=1/uδ/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+yz|2t+|xz||y|t+|xz|, and

0 t d t ( t + | x z | ) 2 n + 3 2 δ =C|xz | 2 n 1 + 2 δ ,

then, for xQ,

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)|dxC 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,1jm,

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<p<n/δ, 1/q=1/pδ/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<p<n/δ, 1/q=1/pδ/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<p<n/δ, 1/q=1/pδ/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)|dxC 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 |dxC 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, 1jm, 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/pδ/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<u<p<n/δ, 1/v=1/uδ/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<u<p<n/δ, 1/v=1/uδ/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)|dxC b BMO | f | r B p δ .

This completes the proof of Theorem 2. □

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Keywords

  • vector-valued multilinear commutator
  • Triebel-Lizorkin space
  • Lipschitz space
  • Lebesgue space
  • BMO( R n )