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Inequality estimates for the boundedness of multilinear singular and fractional integral operators

Abstract

In this paper, the inequality of boundedness for the multilinear fractional singular integral operators associated to the weighted Lipschitz functions is estimated. The operators include the Calderón-Zygmund singular integral operator and the fractional integral operator.

MSC:42B20, 42B25.

1 Introduction

As the development of the singular integral operators, their commutators and multilinear operators have been well studied (see [15]). In [13, 57], the authors proved that the commutators and multilinear operators generated by the singular integral operators and BMO functions are bounded on L p ( R n ) for 1<p<. Chanillo (see [8]) proved a similar result when singular integral operators are replaced by the fractional integral operators. In [912], the boundedness for the commutators and multilinear operators generated by the singular integral operators and Lipschitz functions on L p ( R n ) (1<p<) and Triebel-Lizorkin spaces are obtained. In [13, 14], the weighted boundedness for the commutators generated by the singular integral operators and BMO or Lipschitz functions on L p ( R n ) (1<p<) spaces are obtained. The purpose of this paper is to study the weighted boundedness for some multilinear operators associated to the fractional singular integral operators and the weighted Lipschitz functions. As application, the weighted boundedness for the multilinear operators associated to the Calderón-Zygmund singular integral operator and the fractional integral operator is obtained.

2 Notations and theorems

In this paper, we are going to consider some multilinear operators as follows.

Let m j be positive integers (j=1,,k), m 1 ++ m k =m, and let b j be locally integrable functions on R n (j=1,,k). Set

R m j + 1 ( b j ;x,y)= b j (x) | α | m j 1 α ! D α b j (y) ( x y ) α .

Definition 1 Let T:S S be a linear operator, and there exists a locally integrable function K(x,y) on R n × R n such that

T(f)(x)= R n K(x,y)f(y)dy

for every bounded and compactly supported function f, where K satisfies: for fixed 0δ<n and 0<ε1,

| K ( x , y ) | C | x y | n + δ ,
(1)

and

| K ( y , x ) K ( z , x ) | C | y z | ε | x z | n + δ ε if 2|yz||xz|.
(2)

Given bounded and compactly supported functions f defined on R n , the multilinear operator associated to T is defined by

T b (f)(x)= R n j = 1 k R m j + 1 ( b j ; x , y ) | x y | m k K(x,y)f(y)dy.

Note that when m=0, T b is just a multilinear commutator of T and b (see [15, 16]), while when m>0, it is non-trivial generalization of the commutators; when η=0, T b is just a multilinear commutator of the singular integral operator, when 0η<n, T b is just a multilinear commutator of the fractional integral operator. It is well known that multilinear operators are of great interest in harmonic analysis and have been widely studied by many authors (see [14, 6, 7, 9, 10, 15, 16]). The purpose of this paper is to study the weighted boundedness properties for the multilinear operator.

Throughout this paper, Q will denote a cube of R n with sides parallel to the axes. For a cube Q and a locally integrable function f, let f(Q)= Q f(x)dx, f Q = | Q | 1 Q f(x)dx and

f # (x)= sup Q x 1 | Q | Q | f ( y ) f Q | dy.

It is well known that (see [17, 18])

f # (x) sup Q x inf c C 1 | Q | Q | f ( y ) c | dy.

For 1p< and 0η<n, let

M η , p (f)(x)= sup Q x ( 1 | Q | 1 p η / n Q | f ( y ) | p d y ) 1 / p ,

which is the Hardy-Littlewood maximal function when p=1 and η=0.

The A p weight is defined by (see [17])

A p = { w : sup Q ( 1 | Q | Q w ( x ) d x ) ( 1 | Q | Q w ( x ) 1 / ( p 1 ) d x ) p 1 < } , 1 < p < , A 1 = { w > 0 : M ( w ) ( x ) C w ( x ) , a.e. } ,

and A = p 1 A p . We know, for w A 1 , w satisfies the double condition, that is, for any cube Q,

w(2Q)Cw(Q).

The A(p,q) weight is defined by (see [19])

A ( p , q ) = { w > 0 : sup Q ( 1 | Q | Q w ( x ) q d x ) 1 / q ( 1 | Q | Q w ( x ) p / ( p 1 ) d x ) ( p 1 ) / p < } , 1 < p , q < .

Given a weight function w. For 1<p<, the weighted Lebesgue space L p (w) is the space of functions f such that

f L p ( w ) = ( R n | f ( x ) | p w ( x ) d x ) 1 / p <.

For β>0 and p>1, let F ˙ p β , (w) be the weighted homogeneous Triebel-Lizorkin space. For 0<β<1, the weighted Lipschitz space Lip β (w) is the space of functions f such that

f Lip β ( w ) = sup Q 1 w ( Q ) 1 + β / n Q | f ( y ) f Q | dy<.

We shall prove the following theorems in Section 3.

Theorem 1 Suppose that T b is the multilinear operator as Definition 1 such that T is bounded from L u ( R n ) to L v ( R n ) for any 0η<n, 1<u<n/η and 1/u1/v=η/n. Let 0<β1, β+δ<n, w A 1 and D α b j Lip β (w) for all α with |α|= m j and j=1,,k. Then T b is bounded from L p (w) to L q ( w 1 q ( k ( δ + k ) / n ) ) for any 1<p<n/(δ+k+kβ) and 1/p1/q=(δ+k+kβ)/n.

Theorem 2 Suppose that T b is the multilinear operator as Definition 1 such that T is bounded from L u ( R n ) to L v ( R n ) for any 0η<n, 1<u<n/η and 1/u1/v=η/n. Let 0<β<min(1/k,ε/k), β+δ<n, w A 1 and D α b j Lip β (w) for all α with |α|= m j and j=1,,k. Then T b is bounded from L p (w) to F ˙ q k β , ( w 1 q ( k ( δ + k k β ) / n ) ) for any 1<p<n/(δ+k) and 1/p1/q=(δ+k)/n.

3 Proofs of theorems

We begin with some preliminary lemmas.

Lemma 1 (see [1])

Let b be a function on R n and D α b L q ( R n ) for |α|=m and some q>n. Then

| R m ( b ; x , y ) | C | x y | m | α | = m ( 1 | Q ˜ ( x , y ) | Q ˜ ( x , y ) | D α b ( z ) | q d z ) 1 / q ,

where Q ˜ (x,y) is the cube centered at x and having side length 5 n |xy|.

Lemma 2 (see [11, 12])

For 0<β<1, 1<p< and w A , we have

f F ˙ p β , ( w ) sup Q 1 | Q | 1 + β / n Q | f ( x ) f Q | d x L p ( w ) sup Q inf c 1 | Q | 1 + β / n Q | f ( x ) c | d x L p ( w ) .

Lemma 3 (see [8])

Suppose that 1s<p<n/η, 1/q=1/pη/n and wA(p,q). Then

M η , s ( f ) L q ( w q ) C f L p ( w p ) .

Lemma 4 (see [17, 20])

For 0<β<1, 1p and w A 1 , we have

b Lip β ( w ) sup Q 1 w ( Q ) β / n ( 1 w ( Q ) Q | b ( x ) b Q | p w ( y ) 1 p d x ) 1 / p .

Lemma 5 (see [20])

For any cube Q, b Lip β (w), 0<β<1 and w A 1 , we have

sup x Q | b ( x ) b Q | C b Lip β ( w ) w ( Q ) 1 + β / n | Q | 1 .

To prove the theorems, we need the following lemmas.

Key Lemma 1 Suppose that T b is the multilinear operator as Definition 1 such that T is bounded from L u ( R n ) to L v ( R n ) for any 0η<n, 1<u<n/η and 1/u1/v=η/n. Let 0<β1, β+δ<n, w A 1 and D α b j Lip β (w) for all α with |α|= m j and j=1,,k. Then there exists a constant C 0 such that for every f C 0 ( R n ), 1<s<n/δ and x ˜ R n ,

sup Q x ˜ 1 | Q | Q | T b ( f ) ( x ) C 0 | d x C j = 1 k ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) k + k β / n M δ + k + k β , s ( f ) ( x ˜ ) .

Key Lemma 2 Suppose that T b is the multilinear operator as Definition 1 such that T is bounded from L u ( R n ) to L v ( R n ) for any 0η<n, 1<u<n/η and 1/u1/v=η/n. Let 0<β<min(1/k,ε/k), β+δ<n, w A 1 and D α b j Lip β (w) for all α with |α|= m j and j=1,,k. Then there exists a constant C 0 such that for every f C 0 ( R n ), 1<s<n/δ and x ˜ R n ,

sup Q x ˜ 1 | Q | 1 + k β / n Q | T b ( f ) ( x ) C 0 | d x C j = 1 k ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) k + k β / n M δ + k , s ( f ) ( x ˜ ) .

Proof of Key Lemma 1 Without loss of generality, we may assume k=2. Fix a cube Q=Q( x 0 ,d) with Q x ˜ . Let Q ˜ =5 n Q and b ˜ j (x)= b j (x) | α | = m 1 α ! ( D α b j ) Q ˜ x α , then R m + 1 ( b j ;x,y)= R m + 1 ( b ˜ j ;x,y) and D α b ˜ j = D α b j ( D α b j ) Q ˜ for |α|= m j . We split f=f χ Q ˜ +f χ R n Q ˜ = f 1 + f 2 . Write

T b ( f ) ( x ) = R n j = 1 2 R m j ( b ˜ j ; x , y ) | x y | m 2 K ( x , y ) f 1 ( y ) d y | α 1 | = m 1 1 α 1 ! R n R m 2 ( b ˜ 2 ; x , y ) ( x y ) α 1 | x y | m 2 D α 1 b ˜ 1 ( y ) K ( x , y ) f 1 ( y ) d y | α 2 | = m 2 1 α 2 ! R n R m 1 ( b ˜ 1 ; x , y ) ( x y ) α 2 | x y | m 2 D α 2 b ˜ 2 ( y ) K ( x , y ) f 1 ( y ) d y + | α 1 | = m 1 , | α 2 | = m 2 1 α 1 ! α 2 ! R n ( x y ) α 1 + α 2 D α 1 b ˜ 1 ( y ) D α 2 b ˜ 2 ( y ) | x y | m 2 K ( x , y ) f 1 ( y ) d y + R n j = 1 2 R m j + 1 ( b ˜ j ; x , y ) | x y | m 2 K ( x , y ) f 2 ( y ) d y ,

then

1 | Q | Q | T b ( f ) ( x ) T b ˜ ( f 2 ) ( x 0 ) | d x 1 | Q | Q | R n j = 1 2 R m j ( b ˜ j ; x , y ) | x y | m 2 K ( x , y ) f 1 ( y ) d y | d x + C | Q | Q | | α 1 | = m 1 R n R m 2 ( b ˜ 2 ; x , y ) ( x y ) α 1 | x y | m 2 D α 1 b ˜ 1 ( y ) K ( x , y ) f 1 ( y ) d y | d x + C | Q | Q | | α 2 | = m 2 R n R m 1 ( b ˜ 1 ; x , y ) ( x y ) α 2 | x y | m 2 D α 2 b ˜ 2 ( y ) K ( x , y ) f 1 ( y ) d y | d x + C | Q | Q | | α 1 | = m 1 , | α 2 | = m 2 R n ( x y ) α 1 + α 2 D α 1 b ˜ 1 ( y ) D α 2 b ˜ 2 ( y ) | x y | m 2 K ( x , y ) f 1 ( y ) d y | d x + 1 | Q | Q | T b ˜ ( f 2 ) ( x ) T b ˜ ( f 2 ) ( x 0 ) | d x : = I 1 + I 2 + I 3 + I 4 + I 5 .

Now, let us estimate I 1 , I 2 , I 3 , I 4 and I 5 , respectively. First, by Lemmas 1 and 5, we get

| R m ( b ˜ j ; x , y ) | C | x y | m | α | = m sup x Q ˜ | D α b j ( x ) ( D α b j ) Q ˜ | C | x y | m w ( Q ˜ ) 1 + β / n | Q | | α | = m D α b j Lip β ( w ) .

Thus, by the ( L s , L t )-boundedness of T with 1<s<n/δ and 1/t=1/sδ/n, we obtain

I 1 C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) ( | Q | 2 / n w ( Q ˜ ) 2 + 2 β / n | Q | 2 ) 1 | Q | Q | T ( f 1 ) ( x ) | d x C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) ( | Q | 2 / n w ( Q ˜ ) 2 + 2 β / n | Q | 2 ) ( 1 | Q | R n | T ( f 1 ) ( x ) | t d x ) 1 / t C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) ( | Q | 2 / n w ( Q ˜ ) 2 + 2 β / n | Q | 2 ) | Q | 1 / t ( R n | f 1 ( x ) | s d x ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) ( w ( Q ˜ ) | Q ˜ | ) 2 + 2 β / n ( 1 | Q ˜ | 1 s ( δ + 2 + 2 β ) / n Q ˜ | f ( y ) | s d y ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s ( f ) ( x ˜ ) .

For I 2 , by the ( L s , L t )-boundedness of T with 1<s<n/δ, 1/t=1/sδ/n and Lemma 5, we get

I 2 C | α 2 | = m 2 D α 2 b 2 Lip β ( w ) | Q | 2 / n w ( Q ˜ ) 1 + β / n | Q | | α 1 | = m 1 1 | Q | Q | T ( D α 1 b ˜ 1 f 1 ) ( x ) | d x C | α 2 | = m 2 D α 2 b 2 Lip β ( w ) | Q | 2 / n w ( Q ˜ ) 1 + β / n | Q | × | α 1 | = m 1 ( 1 | Q | R n | T ( D α 1 b ˜ 1 f 1 ) ( x ) | t d x ) 1 / t C | α 2 | = m 2 D α 2 b 2 Lip β ( w ) | Q | 2 / n w ( Q ˜ ) 1 + β / n | Q | | Q | 1 / t × | α 1 | = m 1 ( R n | D α 1 b 1 ( x ) ( D α b 1 ) Q ˜ | s | f 1 ( x ) | s d x ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) | Q | 2 / n w ( Q ˜ ) 2 + 2 β / n | Q | 2 | Q | 1 / t ( 1 | Q ˜ | Q ˜ | f ( y ) | s d y ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) ( w ( Q ˜ ) | Q ˜ | ) 2 + 2 β / n ( 1 | Q ˜ | 1 s ( δ + 2 + 2 β ) / n Q ˜ | f ( y ) | s d y ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s ( f ) ( x ˜ ) .

For I 3 , similar to the proof of I 2 , we get

I 3 C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s (f)( x ˜ ).

Similarly, for I 4 , we obtain, for 1<s<n/δ and 1/t=1/sδ/n,

I 4 C | α 1 | = m 1 , | α 2 | = m 2 | Q | 2 / n | Q | Q | T ( D α 1 b ˜ 1 D α 2 b ˜ 2 f 1 ) ( x ) | d x C | α 1 | = m 1 , | α 2 | = m 2 | Q | 2 / n ( 1 | Q | R n | T ( D α 1 b ˜ 1 D α 2 b ˜ 2 f 1 ) ( x ) | t d x ) 1 / t C | α 1 | = m 1 , | α 2 | = m 2 | Q | 2 / n | Q | 1 / t ( R n | D α 1 b ˜ 1 ( x ) D α 2 b ˜ 2 ( x ) | s | f 1 ( x ) | s d x ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) | Q | 2 / n w ( Q ˜ ) 2 + 2 β / n | Q | 2 | Q | 1 / t ( 1 | Q ˜ | Q ˜ | f ( y ) | s d y ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) ( w ( Q ˜ ) | Q ˜ | ) 2 + 2 β / n ( 1 | Q ˜ | 1 s ( δ + 2 + 2 β ) / n Q ˜ | f ( y ) | s d y ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s ( f ) ( x ˜ ) .

For I 5 , we write

T b ˜ ( f 2 ) ( x ) T b ˜ ( f 2 ) ( x 0 ) = R n ( K ( x , y ) | x y | m K ( x 0 , y ) | x 0 y | m ) j = 1 2 R m j ( b ˜ j ; x , y ) f 2 ( y ) d y + R n ( R m 1 ( b ˜ 1 ; x , y ) R m 1 ( b ˜ 1 ; x 0 , y ) ) R m 2 ( b ˜ 2 ; x , y ) | x 0 y | m K ( x 0 , y ) f 2 ( y ) d y + R n ( R m 2 ( b ˜ 2 ; x , y ) R m 2 ( b ˜ 2 ; x 0 , y ) ) R m 1 ( b ˜ 1 ; x 0 , y ) | x 0 y | m K ( x 0 , y ) f 2 ( y ) d y | α 1 | = m 1 1 α 1 ! R n D α 1 b ˜ 1 ( y ) f 2 ( y ) × [ R m 2 ( b ˜ 2 ; x , y ) ( x y ) α 1 | x y | m K ( x , y ) R m 2 ( b ˜ 2 ; x 0 , y ) ( x 0 y ) α 1 | x 0 y | m K ( x 0 , y ) ] d y | α 2 | = m 2 1 α 2 ! R n D α 2 b ˜ 2 ( y ) f 2 ( y ) × [ R m 1 ( b ˜ 1 ; x , y ) ( x y ) α 2 | x y | m K ( x , y ) R m 1 ( b ˜ 1 ; x 0 , y ) ( x 0 y ) α 2 | x 0 y | m K ( x 0 , y ) ] d y + | α 1 | = m 1 , | α 2 | = m 2 1 α 1 ! α 2 ! R n [ ( x y ) α 1 + α 2 | x y | m K ( x , y ) ( x 0 y ) α 1 + α 2 | x 0 y | m K ( x 0 , y ) ] × D α 1 b ˜ 1 ( y ) D α 2 b ˜ 2 ( y ) f 2 ( y ) d y = I 5 ( 1 ) + I 5 ( 2 ) + I 5 ( 3 ) + I 5 ( 4 ) + I 5 ( 5 ) + I 5 ( 6 ) .

Note that |xy|| x 0 y| for xQ and y R n Q ˜ = l = 0 ( 2 l + 1 Q ˜ 2 l Q ˜ ). By Lemmas 1 and 5, we get

| R m j ( b ˜ j ; x , y ) | C | α | = m j D α b j Lip β ( w ) ( w ( 2 l + 1 Q ˜ ) 1 + β / n | 2 l + 1 Q | ) | x y | m j .

Then, by the conditions on K, we obtain

| I 5 ( 1 ) | C R n Q ˜ ( | x x 0 | | x 0 y | m + n 1 δ + | x x 0 | ε | x 0 y | m + n + ε 2 δ ) j = 1 2 | R m j ( b ˜ j ; x , y ) | | f ( y ) | d y C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) l = 0 w ( 2 l + 1 Q ˜ ) 2 + 2 β / n | 2 l + 1 Q | 2 × 2 l + 1 Q ˜ 2 l Q ˜ ( | x x 0 | | x 0 y | n 1 δ + | x x 0 | ε | x 0 y | n + ε 2 δ ) | f ( y ) | d y C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) l = 0 w ( 2 l + 1 Q ˜ ) 2 + 2 β / n | 2 l + 1 Q | 2 × 2 l + 1 Q ˜ 2 l Q ˜ ( d ( 2 l d ) n 1 δ + d ε ( 2 l d ) n + ε 2 δ ) | f ( y ) | d y C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) l = 0 ( w ( 2 l + 1 Q ˜ ) | 2 l + 1 Q ˜ | ) 2 + 2 β / n ( 2 l + 2 l ε ) × 1 | 2 l + 1 Q ˜ | 1 ( δ + 2 + 2 β ) / n 2 l + 1 Q ˜ | f ( y ) | d y C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n l = 0 ( 2 l + 2 l ε ) × ( 1 | 2 l + 1 Q ˜ | 1 s ( δ + 2 + 2 β ) / n 2 l + 1 Q ˜ | f ( y ) | s d y ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s ( f ) ( x ˜ ) l = 1 ( 2 l + 2 l ε ) C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s ( f ) ( x ˜ ) .

For I 5 ( 2 ) , by the formula (see [1])

R m j ( b ˜ j ;x,y) R m j ( b ˜ j ; x 0 ,y)= | η | < m j 1 η ! R m j | η | ( D η b ˜ j ; x , x 0 ) ( x y ) η

and Lemma 5, we get

| I 5 ( 2 ) | C j = 1 2 ( | α | = m j D α b j Lip β ( w ) ) × l = 0 w ( 2 l + 1 Q ) 2 + 2 β / n | 2 l + 1 Q | 2 2 l + 1 Q ˜ 2 l Q ˜ | x x 0 | | x 0 y | n 1 δ | f ( y ) | d y C j = 1 2 ( | α | = m j D α b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s ( f ) ( x ˜ ) .

Similarly,

| I 5 ( 3 ) | C j = 1 2 ( | α | = m j D α b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s (f)( x ˜ ).

For I 5 ( 4 ) , similar to the estimates of I 5 ( 1 ) and I 5 ( 2 ) , and noticing that for b Lip β (w), w A 1 and xQ, by Lemma 4, we have

| b Q b 2 l Q |Clw(x)w ( 2 l Q ) β / n b Lip β ( w ) .

Thus, we obtain

| I 5 ( 4 ) | C | α 1 | = m 1 R n Q ˜ | ( x y ) α 1 K ( x , y ) | x y | m 2 ( x 0 y ) α 1 K ( x 0 , y ) | x 0 y | m 2 | × | R m 2 ( b ˜ 2 ; x , y ) | | D α 1 b ˜ 1 ( y ) | | f ( y ) | d y + C | α 1 | = m 1 R n Q ˜ | R m 2 ( b ˜ 2 ; x , y ) R m 2 ( b ˜ 2 ; x 0 , y ) | × | ( x 0 y ) α 1 K ( x 0 , y ) | | x 0 y | m 2 | D α 1 b ˜ 1 ( y ) f ( y ) | d y C | α 2 | = m 2 D α 2 b 2 Lip β ( w ) | α | = m 1 l = 0 2 l + 1 Q ˜ 2 l Q ˜ ( | x x 0 | | x 0 y | n 1 δ + | x x 0 | ε | x 0 y | n + ε 2 δ ) × w ( 2 l + 1 Q ˜ ) 1 + β / n | 2 l + 1 Q ˜ | ( | D α 1 b 1 ( y ) ( D α 1 b 1 ) 2 l + 1 Q ˜ | + | ( D α 1 b 1 ) 2 l + 1 Q ˜ ( D α 1 b 1 ) Q ˜ | ) | f ( y ) | d y C j = 1 2 ( | α | = m j D α b j Lip β ( w ) ) l = 0 2 l + 1 Q ˜ 2 l Q ˜ ( d ( 2 l d ) n 1 δ + d ε ( 2 l d ) n + ε 2 δ ) × w ( 2 l + 1 Q ˜ ) 1 + β / n | 2 l + 1 Q ˜ | [ w ( 2 l + 1 Q ˜ ) 1 + β / n | 2 l + 1 Q ˜ | + l w ( x ˜ ) w ( 2 l + 1 Q ˜ ) β / n ] | f ( y ) | d y C j = 1 2 ( | α | = m j D α b j Lip β ( w ) ) l = 0 w ( x ˜ ) ( w ( 2 l + 1 Q ˜ ) | 2 l + 1 Q ˜ | ) 1 + 2 β / n × ( 1 | 2 l + 1 Q ˜ | 1 s ( δ + 2 + 2 β ) / n 2 l + 1 Q ˜ | f ( y ) | s d y ) 1 / s l ( 2 l + 2 l ε ) C j = 1 2 ( | α | = m j D α b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s ( f ) ( x ˜ ) .

Similarly,

| I 5 ( 5 ) | C j = 1 2 ( | α | = m j D α b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s (f)( x ˜ ).

For I 5 ( 6 ) , we get

| I 5 ( 6 ) | C | α 1 | = m 1 , | α 2 | = m 2 R n Q ˜ | ( x y ) α 1 + α 2 K ( x , y ) | x y | m ( x 0 y ) α 1 + α 2 K ( x 0 , y ) | x 0 y | m | × ( | D α 1 b 1 ( y ) ( D α 1 b 1 ) 2 l + 1 Q ˜ | + | ( D α 1 b 1 ) 2 l + 1 Q ˜ ( D α 1 b 1 ) Q ˜ | ) × ( | D α 2 b 2 ( y ) ( D α 2 b 2 ) 2 l + 1 Q ˜ | + | ( D α 2 b 2 ) 2 l + 1 Q ˜ ( D α 2 b 2 ) Q ˜ | ) | f ( y ) | d y C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) l = 0 2 l + 1 Q ˜ 2 l Q ˜ ( | x x 0 | | x 0 y | n 1 δ + | x x 0 | ε | x 0 y | n + ε 2 δ ) × [ w ( 2 l + 1 Q ˜ ) 1 + β / n | 2 l + 1 Q ˜ | + l w ( x ˜ ) w ( 2 l + 1 Q ˜ ) β / n ] 2 | f ( y ) | d y C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) l = 0 [ w ( x ˜ ) ( w ( 2 l + 1 Q ˜ ) | 2 l + 1 Q ˜ | ) β / n ] 2 × ( 1 | 2 l + 1 Q ˜ | 1 s ( δ + 2 + 2 β ) / n 2 l + 1 Q ˜ | f ( y ) | s d y ) 1 / s l 2 ( 2 l + 2 l ε ) C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s ( f ) ( x ˜ ) .

Thus

| T b ˜ ( f 2 ) ( x ) T b ˜ ( f 2 ) ( x 0 ) | C j = 1 2 ( | α | = m j D α b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s (f)( x ˜ )

and

I 5 C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 + 2 β , s (f)( x ˜ ).

This completes the proof of the lemma. □

Proof of Key Lemma 2 Without loss of generality, we may assume k=2. By using the same argument as in the proof of Key Lemma 1, we have

1 | Q | 1 + 2 β / n Q | T b ( f ) ( x ) T b ˜ ( f 2 ) ( x 0 ) | d x 1 | Q | 1 + 2 β / n Q | R n j = 1 2 R m j ( b ˜ j ; x , y ) | x y | m 2 K ( x , y ) f 1 ( y ) d y | d x + C | Q | 1 + 2 β / n Q | | α 1 | = m 1 R n R m 2 ( b ˜ 2 ; x , y ) ( x y ) α 1 | x y | m 2 D α 1 b ˜ 1 ( y ) K ( x , y ) f 1 ( y ) d y | d x + C | Q | 1 + 2 β / n Q | | α 2 | = m 2 R n R m 1 ( b ˜ 1 ; x , y ) ( x y ) α 2 | x y | m 2 D α 2 b ˜ 2 ( y ) K ( x , y ) f 1 ( y ) d y | d x + C | Q | 1 + 2 β / n Q | | α 1 | = m 1 , | α 2 | = m 2 R n ( x y ) α 1 + α 2 D α 1 b ˜ 1 ( y ) D α 2 b ˜ 2 ( y ) | x y | m 2 K ( x , y ) f 1 ( y ) d y | d x + 1 | Q | 1 + 2 β / n Q | T b ˜ ( f 2 ) ( x ) T b ˜ ( f 2 ) ( x 0 ) | d x : = J 1 + J 2 + J 3 + J 4 + J 5 .

We obtain

J 1 C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) | Q | 2 / n ( w ( Q ˜ ) | Q | ) 2 + 2 β / n ( 1 | Q | R n | T ( f 1 ) ( x ) | t d x ) 1 / t C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) | Q | 2 / n ( w ( Q ˜ ) | Q | ) 2 + 2 β / n | Q | 1 / t ( R n | f 1 ( x ) | s d x ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) ( w ( Q ˜ ) | Q ˜ | ) 2 + 2 β / n ( 1 | Q ˜ | 1 s ( δ + 2 ) / n Q ˜ | f ( y ) | s d y ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 , s ( f ) ( x ˜ ) , J 2 C | α 2 | = m 2 D α 2 b 2 Lip β ( w ) | Q | 2 / n w ( Q ˜ ) 1 + β / n | Q | 1 + 2 β / n | α 1 | = m 1 ( 1 | Q | R n | T ( D α 1 b ˜ 1 f 1 ) ( x ) | t d x ) 1 / t C | α 2 | = m 2 D α 2 b 2 Lip β ( w ) | Q | 2 / n w ( Q ˜ ) 1 + β / n | Q | 1 + 2 β / n | Q | 1 / t | α 1 | = m 1 ( R n | D α 1 b ˜ 1 ( x ) | s | f 1 ( x ) | s d x ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) ( w ( Q ˜ ) | Q ˜ | ) 2 + 2 β / n ( 1 | Q ˜ | 1 s ( δ + 2 ) / n Q ˜ | f ( y ) | s d y ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 , s ( f ) ( x ˜ ) , J 3 C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 , s ( f ) ( x ˜ ) , J 4 C | α 1 | = m 1 , | α 2 | = m 2 | Q | 2 / n | Q | 2 β / n ( 1 | Q | R n | T ( D α 1 b ˜ 1 D α 2 b ˜ 2 f 1 ) ( x ) | t d x ) 1 / t C | α 1 | = m 1 , | α 2 | = m 2 | Q | 2 / n | Q | 2 β / n | Q | 1 / t ( R n | D α 1 b ˜ 1 ( x ) D α 2 b ˜ 2 ( x ) | s | f 1 ( x ) | s d x ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) ( w ( Q ˜ ) | Q ˜ | ) 2 + 2 β / n ( 1 | Q ˜ | 1 s ( δ + 2 ) / n Q ˜ | f ( y ) | s d y ) 1 / s C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 , s ( f ) ( x ˜ ) .

For J 5 , similar to the proof of I 5 in Key Lemma 1, we obtain

| T b ˜ ( f 2 ) ( x ) T b ˜ ( f 2 ) ( x 0 ) | C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) l = 0 w ( 2 l + 1 Q ˜ ) 2 + 2 β / n | 2 l + 1 Q | 2 × 2 l + 1 Q ˜ 2 l Q ˜ ( d ( 2 l d ) n 1 δ + d ε ( 2 l d ) n + ε 2 δ ) | f ( y ) | d y + C j = 1 2 ( | α | = m j D α b j Lip β ( w ) ) l = 0 w ( 2 l + 1 Q ) 2 + 2 β / n | 2 l + 1 Q | 2 2 l + 1 Q ˜ 2 l Q ˜ d ( 2 l d ) n 1 δ | f ( y ) | d y + C j = 1 2 ( | α | = m j D α b j Lip β ( w ) ) l = 0 2 l + 1 Q ˜ 2 l Q ˜ ( d ( 2 l d ) n 1 δ + d ε ( 2 l d ) n + ε 2 δ ) × w ( 2 l + 1 Q ˜ ) 1 + β / n | 2 l + 1 Q ˜ | [ w ( 2 l + 1 Q ˜ ) 1 + β / n | 2 l + 1 Q ˜ | + l w ( x ˜ ) w ( 2 l + 1 Q ˜ ) β / n ] | f ( y ) | d y + C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) l = 0 2 l + 1 Q ˜ 2 l Q ˜ ( d ( 2 l d ) n 1 δ + d ε ( 2 l d ) n + ε 2 δ ) × [ w ( 2 l + 1 Q ˜ ) 1 + β / n | 2 l + 1 Q ˜ | + l w ( x ˜ ) w ( 2 l + 1 Q ˜ ) β / n ] 2 | f ( y ) | d y C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) | Q | 2 β / n l = 0 ( w ( 2 l + 1 Q ˜ ) | 2 l + 1 Q | ) 2 + 2 β / n ( 2 l ( 2 β 1 ) + 2 l ( 2 β ε ) ) × ( 1 | 2 l + 1 Q ˜ | 1 s ( δ + 2 ) / n 2 l + 1 Q ˜ | f ( y ) | s d y ) 1 / s + C j = 1 2 ( | α | = m j D α b j Lip β ( w ) ) | Q | 2 β / n l = 0 w ( x ˜ ) ( w ( 2 l + 1 Q ˜ ) | 2 l + 1 Q ˜ | ) 1 + 2 β / n × ( 1 | 2 l + 1 Q ˜ | 1 s ( δ + 2 ) / n 2 l + 1 Q ˜ | f ( y ) | s d y ) 1 / s l ( 2 l ( 2 β 1 ) + 2 l ( 2 β ε ) ) + C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) | Q | 2 β / n l = 0 [ w ( x ˜ ) ( w ( 2 l + 1 Q ˜ ) | 2 l + 1 Q ˜ | ) β / n ] 2 × ( 1 | 2 l + 1 Q ˜ | 1 s ( δ + 2 ) / n 2 l + 1 Q ˜ | f ( y ) | s d y ) 1 / s l 2 ( 2 l ( 2 β 1 ) + 2 l ( 2 β ε ) ) C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) | Q | 2 β / n w ( x ˜ ) 2 + 2 β / n M δ + 2 , s ( f ) ( x ˜ ) .

Thus

J 5 C j = 1 2 ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) 2 + 2 β / n M δ + 2 , s (f)( x ˜ ).

This completes the proof of the lemma. □

Proof of Theorem 1 By Key Lemma 1, we get the sharp function estimate of T b as follows:

( T b ( f ) ) # ( x ˜ )C j = 1 k ( | α j | = m j D α j b j Lip β ( w ) ) w ( x ˜ ) k + k β / n M δ + k + k β , s (f)( x ˜ ).

Now, choose 1<s<p in Key Lemma 1, by using Lemma 3 and notice that w 1 q ( k ( δ + k ) / n ) A and w 1 / p A(p,q). We get

T b ( f ) L q ( w 1 q ( k ( δ + k ) / n ) ) M ( T b ( f ) ) L q ( w 1 q ( k ( δ + k ) / n ) ) C ( T b ( f ) ) # L q ( w 1 q ( k ( δ + k ) / n ) ) C j = 1 k ( | α j | = m j D α j b j Lip β ( w ) ) M δ + k + k β , s ( f ) w k + k β / n L q ( w 1 q ( k ( δ + k ) / n ) ) = C j = 1 k ( | α j | = m j D α j b j Lip β ( w ) ) M δ + k + k β , s ( f ) L q ( w q / p ) C j = 1 k ( | α j | = m j D α j b j Lip β ( w ) ) f L p ( w ) .

This completes the proof of the theorem. □

Proof of Theorem 2 Choose 1<s<p in Key Lemma 2, notice that w 1 q ( k ( δ + k k β ) / n ) A and w 1 / p A(p,q). By using Lemmas 2 and 3, we obtain

T b ( f ) F ˙ q k β , ( w 1 q ( k ( δ + k k β ) / n ) ) C j = 1 k ( | α j | = m j D α j b j Lip β ( w ) ) M δ + k , s ( f ) w k + k β / n L q ( w 1 q ( k ( δ + k k β ) / n ) ) = C j = 1 k ( | α j | = m j D α j b j Lip β ( w ) ) M δ + k , s ( f ) L q ( w q / p ) C j = 1 k ( | α j | = m j D α j b j Lip β ( w ) ) f L p ( w ) .

This completes the proof of the theorem. □

4 Applications

In this section we shall apply Theorems 1 and 2 of the paper to some particular operators such as the Calderón-Zygmund singular integral operator and the fractional integral operator.

Application 1 Calderón-Zygmund singular integral operator.

Let T be the Calderón-Zygmund operator (see [17, 18]). The multilinear operator related to T is defined by

T b (f)(x)= R m + 1 ( b ; x , y ) | x y | m k K(x,y)f(y)dy.

Then Theorem 1 holds for T b with δ=0.

Application 2 Fractional integral operator with rough kernel.

For 0<δ<n, let T δ be the fractional integral operator with rough kernel defined by (see [8, 17, 18])

T δ f(x)= R n Ω ( x y ) | x y | n δ f(y)dy,

which is the fractional integral operator when Ω=1 (see [8]). The multilinear operator related to T δ is defined by

T δ b f(x)= R n R m + 1 ( b ; x , y ) | x y | m + n k δ Ω(xy)f(y)dy,

where Ω is homogeneous of degree zero on R n , S n 1 Ω( x )dσ( x )=0 and Ω Lip ε ( S n 1 ) for some 0<ε1, that is, there exists a constant M>0 such that for any x,y S n 1 , |Ω(x)Ω(y)|M | x y | ε . Then Theorem 2 holds for T δ b .

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Acknowledgements

This work was jointly supported by the National Natural Science Foundation of China under Grant No. 11101053, the Key Project of Chinese Ministry of Education under Grant no. 211118, the Scientific Research Funds of Hunan Provincial Science and Technology Department of China no. 2013SK3143, the Hunan Provincial NSF no. 11JJ1001, the Scientific Research Fund of Hunan Provincial Education Department of China (no. 12C0028) and the Open Fund Project of Key Research Institute of Philosophies and Social Sciences in Hunan Universities (no. 12FEFM05).

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Correspondence to Chuangxia Huang.

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XZ carried out the Theorem 1 studies, participated in the disscussion of Theorem 2. CH is the corresponding author, designed the outline of this paper, carried out the Theorem 1 and Theorem 2 studies, participated in the disscussion of Applications. HH carried out the Theorem 2 studies, participated in the disscussion of Applications. LL participated in the disscussion of Applications and drafted the manuscript. All authors read and approved the final manuscript.

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Zhou, X., Huang, C., Hu, H. et al. Inequality estimates for the boundedness of multilinear singular and fractional integral operators. J Inequal Appl 2013, 303 (2013). https://doi.org/10.1186/1029-242X-2013-303

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