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Some sharp inequalities for multilinear integral operators

Abstract

In this paper, some sharp inequalities for certain multilinear operators related to the Littlewood-Paley operator and the Marcinkiewicz operator are obtained. As an application, we obtain the ( L p , L q )-norm inequalities and Morrey spaces boundedness for the multilinear operators.

MSC:42B20, 42B25.

1 Introduction and results

In this paper, we study some multilinear operators related to some integral operators, whose definitions are as follows.

Fix n>δ0. We denote Γ(x)={(y,t) R + n + 1 :|xy|<t} and the characteristic function of Γ(x) by χ Γ ( x ) . Suppose that m j are the positive integers (j=1,,l), m 1 ++ m l =m and A j are the functions on R n (j=1,,l). Let

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

Definition 1 Let ε>0 and ψ be a fixed function which satisfies the following properties:

  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 + 1 + ε δ ) when 2|y|<|x|.

The multilinear Littlewood-Paley operator is defined by

S ψ A (f)(x)= [ Γ ( x ) | F t A ( f ) ( x , y ) | 2 d y d t t n + 1 ] 1 / 2 ,

where

F t A (f)(x,y)= R n j = 1 l R m j + 1 ( A j ; x , z ) | x z | m ψ t (yz)f(z)dz

and ψ t (x)= t n + δ ψ(x/t) for t>0. Set F t (f)(y)=f ψ t (y). We also define that

S ψ (f)(x)= ( Γ ( x ) | F t ( f ) ( y ) | 2 d y d t t n + 1 ) 1 / 2 ,

which is the Littlewood-Paley operator (see [1]).

Let H be the Hilbert space H={h:h= ( R + n + 1 | h ( y , t ) | 2 d y d t / t n + 1 ) 1 / 2 <}. Then for each fixed x R n , F t A (f)(x,y) may be viewed as a mapping from (0,+) to H, and it is clear that

S ψ A (f)(x)= χ Γ ( x ) F t A ( f ) ( x , y ) , S ψ (f)(x)= χ Γ ( x ) F t ( f ) ( y ) .

Definition 2 Let 0<γ1 and Ω be homogeneous of degree zero on R n with S n 1 Ω( x )dσ( x )=0. Assume that Ω Lip γ ( S n 1 ), that is, there exists a constant M>0 such that for any x,y S n 1 , |Ω(x)Ω(y)|M | x y | γ . The multilinear Marcinkiewicz operator is defined by

μ S A (f)(x)= [ Γ ( x ) | F t A ( f ) ( x , y ) | 2 d y d t t n + 3 ] 1 / 2 ,

where

F t A (f)(x,y)= | y z | t j = 1 l R m j + 1 ( A j ; x , z ) | x z | m Ω ( y z ) | y z | n 1 δ f(z)dz.

Set

F t (f)(y)= | y z | t Ω ( y z ) | y z | n 1 δ f(z)dz.

We also define that

μ S (f)(x)= ( Γ ( x ) | F t ( f ) ( y ) | 2 d y d t t n + 3 ) 1 / 2 ,

which is the Marcinkiewicz operator (see [2]).

Let H be the Hilbert space H={h:h= ( R + n + 1 | h ( y , t ) | 2 d y d t / t n + 3 ) 1 / 2 <}, then for each fixed x R n , F t A (f)(x,y) may be viewed as a mapping from (0,+) to H, and it is clear that

μ S A (f)(x)= χ Γ ( x ) F t A ( f ) ( x , y ) , μ S (f)(x)= χ Γ ( x ) F t ( f ) ( y ) .

Note that when m=0, S ψ A and μ S A are just the multilinear commutators (see [3, 4]). While when m>0, S ψ A and μ S A are non-trivial generalizations of the commutators. It is well known that multilinear operators are of great interest in harmonic analysis and have been widely studied by many authors (see [59]). In [10], Hu and Yang proved a variant sharp estimate for the multilinear singular integral operators. In [1113], authors proved a sharp estimate for the multilinear commutator. The main purpose of this paper is to prove the sharp inequalities for the multilinear integral operators S ψ A and μ S A when D α A j BMO( R n ) for all α with |α|= m j . As an application, we obtain the ( L p , L q )-norm inequalities and Morrey spaces boundedness for the multilinear operators.

First, let us introduce some notations. Throughout this paper, Q will denote a cube of R n with sides parallel to the axes. For any locally integrable function f, the sharp function of f is defined by

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

where, and in what follows, f Q = | Q | 1 Q f(x)dx. It is well-known that (see [14, 15])

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

We say that f belongs to BMO( R n ) if f # belongs to L ( R n ) and f B M O = f # L . For 1p< and 0δ<n, let

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

we write that M μ (f)= M n μ , 1 (f), which is the fractional maximal operator.

Fixed λ>0. For 1p<, let

f L p , λ = sup x R n , d > 0 ( 1 d λ B ( x , d ) | f ( y ) | p d y ) 1 / p ,

where B(x,d)={y R n :|xy|<d}. The Morrey spaces are defined by (see [1620])

L p , λ ( R n ) = { f L l o c 1 ( R n ) : f L p , λ < } .

As the Morrey space may be considered as an extension of the Lebesgue space, it is natural and important to study the boundedness of the multilinear integral operator on the Morrey space.

We shall prove the following theorems.

Theorem 1 Let D α A j BMO( R n ) for all α with |α|= m j and j=1,,l.

  1. (1)

    Then there exists a constant C>0 such that for any f C 0 ( R n ), 1<r<n/δ and x R n ,

    ( S ψ A ( f ) ) # (x)C j = 1 l ( | α j | = m j D α j A j B M O ) M δ , r (f)(x);
  2. (2)

    If 1<p<n/δ and 1/p1/q=δ/n, then S ψ A is bounded from L p ( R n ) to L q ( R n ), that is,

    S ψ A ( f ) L q C j = 1 l ( | α j | = m j D α j A j B M O ) f L p ;
  3. (3)

    If 1<p<n/δ, 0<λ<npδ, 1/q=1/pδ/(nλ), then S ψ A is bounded from L p , λ ( R n ) to L q , λ ( R n ), that is,

    S ψ A ( f ) L q , λ C j = 1 l ( | α j | = m j D α j A j B M O ) f L p , λ .

Theorem 2 Let D α A j BMO( R n ) for all α with |α|= m j and j=1,,l.

  1. (1)

    Then there exists a constant C>0 such that for any f C 0 ( R n ), 1<r<n/δ and x R n ,

    ( μ S A ( f ) ) # (x)C j = 1 l ( | α j | = m j D α j A j B M O ) M δ , r (f)(x);
  2. (2)

    If 1<p<n/δ and 1/p1/q=δ/n, then μ S A is bounded from L p ( R n ) to L q ( R n ), that is,

    μ S A ( f ) L q C j = 1 l ( | α j | = m j D α j A j B M O ) f L p ;
  3. (3)

    If 1<p<n/δ, 0<λ<npδ, 1/q=1/pδ/(nλ), then μ S A is bounded from L p , λ ( R n ) to L q , λ ( R n ), that is,

    μ S A ( f ) L q , λ C j = 1 l ( | α j | = m j D α j A j B M O ) f L p , λ .

Remark The conclusions of Theorems 1 and 2 are completely the same. Thus, they explain that the Littlewood-Paley and Marcinkiewicz operators have the many similar bondedness properties.

2 Proofs of theorems

To prove the theorems, we need the following lemmas.

Lemma 1 [7]

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

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

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

Lemma 2 [21]

Suppose that 1r<p<n/δ and 1/q=1/pδ/n. Then

M δ , r ( f ) L q C f L p .

Lemma 3 [16, 17]

Let 1<p< and 0<λ<n. Then the following estimates hold:

  1. (a)

    M ( f ) L p , λ C f # L p , λ ;

  2. (b)

    M μ ( f ) L q , λ C f L p , λ for 0<μ<(nλ)/np and 1/q=1/pnη/(nλ).

Lemma 4 Let 1<p<n/δ and 1/q=1/pδ/n. Then S ψ and μ S are all bounded from L p ( R n ) to L q ( R n ).

Proof For S ψ , by Minkowski inequality and the condition of ψ, we have

S ψ ( f ) ( x ) R n | f ( z ) | ( Γ ( x ) | ψ t ( y z ) | 2 d y d t t 1 + n ) 1 / 2 d z C R n | f ( z ) | ( 0 | x y | t t 2 n + 2 δ ( 1 + | y z | / t ) 2 n + 2 2 δ d y d t t 1 + n ) 1 / 2 d z C R n | f ( z ) | ( 0 | x y | t 2 2 n + 2 t 1 n ( 2 t + | y z | ) 2 n + 2 2 δ d y d t ) 1 / 2 d z ,

noting that 2t+|yz|2t+|xz||xy|t+|xz| when |xy|t and

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

we obtain

S ψ ( f ) ( x ) C R n | f ( z ) | ( 0 t d t ( t + | x z | ) 2 n + 2 2 δ ) 1 / 2 d z = C R n | f ( z ) | | x z | n δ d z .

For μ S , note that |xz|2t, |yz||xz|t|xz|3t when |xy|t, |yz|t, we have

μ S ( f ) ( x ) R n [ | x y | t ( | Ω ( y z ) f ( z ) | | y z | n 1 δ ) 2 χ Γ ( z ) ( y , t ) d y d t t n + 3 ] 1 / 2 d z C R n | f ( z ) | [ | x y | t χ Γ ( z ) ( y , t ) t n 3 ( | x z | 3 t ) 2 n 2 2 δ d y d t ] 1 / 2 d z C R n | f ( z ) | | x z | 3 / 2 [ | x z | / 2 d t ( | x z | 3 t ) 2 n 2 ] 1 / 2 d z C R n | f ( z ) | | x z | n δ d z .

Thus, the lemma follows from [21]. □

Proof of Theorem 1 (1) It suffices to prove for f C 0 ( R n ) and some constant C 0 , the following inequality holds:

1 | Q | Q | S ψ A ( f ) ( x ) C 0 | dxC j = 1 l ( | α j | = m j D α j A j B M O ) M δ , r (f)(x).

Without loss of generality, we may assume l=2. Fix a cube Q=Q( x 0 ,d) and x ˜ Q. Let Q ˜ =5 n Q and A ˜ j (x)= A j (x) | α | = m j 1 α ! ( D α A j ) Q ˜ x α , then R m j ( A j ;x,y)= R m j ( A ˜ j ;x,y) and D α A ˜ j = D α A j ( D α A j ) Q ˜ for |α|= m j . We write, for f 1 =f χ Q ˜ and f 2 =f χ R n Q ˜ ,

F t A ( f ) ( x , y ) = R n j = 1 2 R m j + 1 ( A ˜ j ; x , z ) | x z | m ψ t ( y z ) f ( z ) d z = R n j = 1 2 R m j + 1 ( A ˜ j ; x , z ) | x z | m ψ t ( y z ) f 2 ( z ) d z + R n j = 1 2 R m j ( A ˜ j ; x , z ) | x z | m ψ t ( y z ) f 1 ( z ) d z | α 1 | = m 1 1 α 1 ! R n R m 2 ( A ˜ 2 ; x , z ) ( x z ) α 1 | x z | m D α 1 A ˜ 1 ( z ) ψ t ( y z ) f 1 ( z ) d z | α 2 | = m 2 1 α 2 ! R n R m 1 ( A ˜ 1 ; x , z ) ( x z ) α 2 | x z | m D α 2 A ˜ 2 ( z ) ψ t ( y z ) f 1 ( z ) d z + | α 1 | = m 1 , | α 2 | = m 2 1 α 1 ! α 2 ! R n ( x z ) α 1 + α 2 D α 1 A ˜ 1 ( z ) D α 2 A ˜ 2 ( z ) | x z | m ψ t ( y z ) f 1 ( z ) d z ,

then

| S ψ A ( f ) ( x ) S ψ A ˜ ( f 2 ) ( x 0 ) | = | χ Γ ( x ) F t A ( f ) ( x , y ) χ Γ ( x 0 ) F t A ˜ ( f 2 ) ( x 0 , y ) | χ Γ ( x ) F t A ( f ) ( x , y ) χ Γ ( x 0 ) F t A ˜ ( f 2 ) ( x 0 , y ) χ Γ ( x ) R n j = 1 2 R m j ( A ˜ j ; x , z ) | x z | m ψ t ( y z ) f 1 ( z ) d z + χ Γ ( x ) | α 1 | = m 1 1 α 1 ! R n R m 2 ( A ˜ 2 ; x , z ) ( x z ) α 1 | x z | m D α 1 A ˜ 1 ( z ) ψ t ( y z ) f 1 ( z ) d z + χ Γ ( x ) | α 2 | = m 2 1 α 2 ! R n R m 1 ( A ˜ 1 ; x , z ) ( x z ) α 2 | x z | m D α 2 A ˜ 2 ( z ) ψ t ( y z ) f 1 ( z ) d z + χ Γ ( x ) | α 1 | = m 1 , | α 2 | = m 2 1 α 1 ! α 2 ! R n ( x z ) α 1 + α 2 D α 1 A ˜ 1 ( z ) D α 2 A ˜ 2 ( z ) | x z | m ψ t ( y z ) f 1 ( z ) d z + χ Γ ( x ) F t A ˜ ( f 2 ) ( x , y ) χ Γ ( x 0 ) F t A ˜ ( f 2 ) ( x 0 , y ) : = I 1 ( x ) + I 2 ( x ) + I 3 ( x ) + I 4 ( x ) + I 5 ( x ) ,

thus,

1 | Q | Q | S ψ A ( f ) ( x ) S ψ A ˜ ( f 2 ) ( x 0 ) | d x 1 | Q | Q I 1 ( x ) d x + C | Q | Q I 2 ( x ) d x + C | Q | Q I 3 ( x ) d x + C | Q | Q I 4 ( x ) d x + 1 | Q | Q I 5 ( x ) 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, for xQ and z Q ˜ , by Lemma 1, we get

R m j ( A ˜ j ;x,z)C | x y | m j | α j | = m j D α j A j B M O .

Thus, by the ( L r , L q )-boundedness of S ψ , for 1<r<n/δ and 1/q=1/rδ/n, we obtain

I 1 C j = 1 2 ( | α j | = m j D α j A j B M O ) 1 | Q | Q | S ψ ( f 1 ) ( x ) | d x C j = 1 2 ( | α j | = m j D α j A j B M O ) ( 1 | Q | Q | S ψ ( f 1 ) ( x ) | q d x ) 1 / q C j = 1 2 ( | α j | = m j D α j A j B M O ) | Q | 1 / q ( Q | f 1 ( x ) | r d x ) 1 / r C j = 1 2 ( | α j | = m j D α j A j B M O ) M δ , r ( f ) ( x ˜ ) .

For I 2 , denoting r=pq for 1<p<n/δ, q>1, 1/q+1/ q =1 and 1/s=1/pδ/n, we have, by Hölder’s inequality,

I 2 C | α 2 | = m 2 D α 2 A 2 B M O | α 1 | = m 1 1 | Q | Q | S ψ ( D α 1 A ˜ 1 f 1 ) ( x ) | d x C | α 2 | = m 2 D α 2 A 2 B M O | α 1 | = m 1 ( 1 | Q | R n | S ψ ( D α 1 A ˜ 1 f 1 ) ( x ) | s d x ) 1 / s C | α 2 | = m 2 D α 2 A 2 B M O | α 1 | = m 1 | Q | 1 / s ( R n | D α 1 A ˜ 1 ( x ) f 1 ( x ) | p d x ) 1 / p C | α 2 | = m 2 D α 2 A 2 B M O × | α 1 | = m 1 ( 1 | Q | Q ˜ | D α 1 A ˜ 1 ( x ) | p q d x ) 1 / p q ( 1 | Q | 1 r δ / n Q ˜ | f ( x ) | p q d x ) 1 / p q C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r ( f ) ( x ˜ ) .

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

I 3 C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r (f)( x ˜ ).

Similarly, for I 4 , denoting r=p q 3 for 1<p<n/δ, q 1 , q 2 , q 3 >1, 1/ q 1 +1/ q 2 +1/ q 3 =1 and 1/s=1/pδ/n, we obtain

I 4 C | α 1 | = m 1 , | α 2 | = m 2 1 | Q | Q | S ψ ( D α 1 A ˜ 1 D α 2 A ˜ 2 f 1 ) ( x ) | d x C | α 1 | = m 1 , | α 2 | = m 2 ( 1 | Q | R n | S ψ ( D α 1 A ˜ 1 D α 2 A ˜ 2 f 1 ) ( x ) | s d x ) 1 / s C | α 1 | = m 1 , | α 2 | = m 2 | Q | 1 / s ( R n | D α 1 A ˜ 1 ( x ) D α 2 A ˜ 2 ( x ) f 1 ( x ) | p d x ) 1 / p C | α 1 | = m 1 , | α 2 | = m 2 ( 1 | Q | Q ˜ | D α 1 A ˜ 1 ( x ) | p q 1 d x ) 1 / p q 1 ( 1 | Q | Q ˜ | D α 2 A ˜ 2 ( x ) | p q 2 d x ) 1 / p q 2 × ( 1 | Q | 1 r δ / n Q ˜ | f ( x ) | p q 3 d x ) 1 / p q 3 C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r ( f ) ( x ˜ ) .

For I 5 , we write

χ Γ ( x ) F t A ˜ ( f 2 ) ( x , y ) χ Γ ( x 0 ) F t A ˜ ( f 2 ) ( x 0 , y ) = R n ( χ Γ ( x ) χ Γ ( x 0 ) ) j = 1 2 R m j ( A ˜ j ; x , z ) | x z | m ψ t ( y z ) f 2 ( z ) d z + χ Γ ( x 0 ) R n ( 1 | x z | m 1 | x 0 z | m ) j = 1 2 R m j ( A ˜ j ; x , z ) ψ t ( y z ) f 2 ( z ) d z + χ Γ ( x 0 ) R n ( R m 1 ( A ˜ 1 ; x , z ) R m 1 ( A ˜ 1 ; x 0 , z ) ) R m 2 ( A ˜ 2 ; x , z ) | x 0 z | m ψ t ( y z ) f 2 ( z ) d z + χ Γ ( x 0 ) R n ( R m 2 ( A ˜ 2 ; x , z ) R m 2 ( A ˜ 2 ; x 0 , z ) ) R m 1 ( A ˜ 1 ; x 0 , z ) | x 0 z | m ψ t ( y z ) f 2 ( z ) d z | α 1 | = m 1 1 α 1 ! R n [ R m 2 ( A ˜ 2 ; x , z ) ( x z ) α 1 χ Γ ( x ) | x z | m R m 2 ( A ˜ 2 ; x 0 , z ) ( x 0 z ) α 1 χ Γ ( x 0 ) | x 0 z | m ] × D α 1 A ˜ 1 ( z ) ψ t ( y z ) f 2 ( z ) d z | α 2 | = m 2 1 α 2 ! R n [ R m 1 ( A ˜ 1 ; x , z ) ( x z ) α 2 χ Γ ( x ) | x z | m R m 1 ( A ˜ 1 ; x 0 , z ) ( x 0 z ) α 2 χ Γ ( x 0 ) | x 0 z | m ] × D α 2 A ˜ 2 ( z ) ψ t ( y z ) f 2 ( z ) d z + | α 1 | = m 1 , | α 2 | = m 2 1 α 1 ! α 2 ! R n [ ( x z ) α 1 + α 2 χ Γ ( x ) | x z | m ( x 0 z ) α 1 + α 2 χ Γ ( x 0 ) | x 0 z | m ] × D α 1 A ˜ 1 ( z ) D α 2 A ˜ 2 ( z ) ψ t ( y z ) f 2 ( z ) d z = I 5 ( 1 ) + I 5 ( 2 ) + I 5 ( 3 ) + I 5 ( 4 ) + I 5 ( 5 ) + I 5 ( 6 ) + I 5 ( 7 ) .

By Lemma 1 and the following inequality (see [15])

| b Q 1 b Q 2 |Clog ( | Q 2 | / | Q 1 | ) b B M O for  Q 1 Q 2 ,

we know that, for xQ and z 2 k + 1 Q ˜ 2 k Q ˜ ,

| R m ( A ˜ ; x , z ) | C | x z | m | α | = m ( D α A B M O + | ( D α A ) Q ˜ ( x , z ) ( D α A ) Q ˜ | ) C k | x z | m | α | = m D α A B M O .

Note that |xz|| x 0 z| for xQ and z R n Q ˜ , we obtain, similar to the proof of Lemma 4,

I 5 ( 1 ) R n ( R + n + 1 [ j = 1 2 | R m j ( A ˜ j ; x , z ) | | ψ t ( y z ) | | f 2 ( z ) | | x z | m I 5 ( 1 ) × | χ Γ ( x ) ( y , t ) χ Γ ( x 0 ) ( y , t ) | ] 2 d y d t t n + 1 ) 1 / 2 d z I 5 ( 1 ) C R n j = 1 2 | R m j ( A ˜ j ; x , z ) | | f 2 ( z ) | | x 0 z | m I 5 ( 1 ) × | Γ ( x ) t 1 n d y d t ( t + | y z | ) 2 n + 2 2 δ Γ ( x 0 ) t 1 n d y d t ( t + | y z | ) 2 n + 2 2 δ | 1 / 2 d z I 5 ( 1 ) C R n j = 1 2 | R m j ( A ˜ j ; x , z ) | | f 2 ( z ) | | x 0 z | m I 5 ( 1 ) × ( | y | 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 I 5 ( 1 ) C R n j = 1 2 | R m j ( A ˜ j ; x , z ) | | f 2 ( z ) | | x 0 z | m ( | y | t | x x 0 | t 1 n d y d t ( t + | x + y z | ) 2 n + 3 2 δ ) 1 / 2 d z I 5 ( 1 ) C R n j = 1 2 R m j ( A ˜ j ; x , z ) | f 2 ( z ) | | x x 0 | 1 / 2 | x 0 z | m + n + 1 / 2 δ d z I 5 ( 1 ) C j = 1 2 ( | α | = m j D α A j B M O ) k = 0 2 k + 1 Q ˜ 2 k Q ˜ k 2 | x x 0 | 1 / 2 | x 0 z | n + 1 / 2 δ | f ( z ) | d z I 5 ( 1 ) C j = 1 2 ( | α | = m j D α A j B M O ) k = 1 k 2 2 k / 2 1 | 2 k Q ˜ | 1 δ / n 2 k Q ˜ | f ( z ) | d z I 5 ( 1 ) C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r ( f ) ( x ˜ ) ; I 5 ( 2 ) C R n | x x 0 | | x 0 z | m + n + 1 δ j = 1 2 | R m j ( A ˜ j ; x , z ) | | f 2 ( z ) | d z I 5 ( 2 ) C j = 1 2 ( | α | = m j D α A j B M O ) k = 0 2 k + 1 Q ˜ 2 k Q ˜ k 2 | x x 0 | | x 0 z | n + 1 δ | f ( z ) | d z I 5 ( 2 ) C j = 1 2 ( | α | = m j D α A j B M O ) k = 1 k 2 2 k 1 | 2 k Q ˜ | 1 δ / n 2 k Q ˜ | f ( z ) | d z I 5 ( 2 ) C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r ( f ) ( x ˜ ) .

For I 5 ( 3 ) and I 5 ( 4 ) , by the formula (see [7])

R m ( A ˜ ;x,z) R m ( A ˜ ; x 0 ,z)= | β | < m 1 β ! R m | β | ( D β A ˜ ; x , x 0 ) ( x z ) β

and Lemma 1, we have

| R m ( A ˜ ; x , z ) R m ( A ˜ ; x 0 , z ) | C | β | < m | α | = m | x x 0 | m | β | | x z | | β | D α A B M O .

Thus, similar to the proof of Lemma 4,

I 5 ( 3 ) C j = 1 2 ( | α | = m j D α A j B M O ) k = 0 2 k + 1 Q ˜ 2 k Q ˜ k | x x 0 | | x 0 z | n + 1 δ | f ( y ) | d y I 5 ( 3 ) C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r ( f ) ( x ˜ ) ; I 5 ( 4 ) C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r ( f ) ( x ˜ ) .

Similarly, we get

I 5 ( 5 ) C | α 1 | = m 1 R n [ R m 2 ( A ˜ 2 ; x , z ) ( x z ) α 1 χ Γ ( x ) | x z | m R m 2 ( A ˜ 2 ; x 0 , z ) ( x 0 z ) α 1 χ Γ ( x 0 ) | x 0 z | m ] I 5 ( 5 ) × ψ t ( y z ) | D α 1 A ˜ 1 ( z ) | | f 2 ( z ) | d z I 5 ( 5 ) C | α | = m 2 D α A 2 B M O | α 1 | = m 1 k = 1 k ( 2 k / 2 + 2 k ) I 5 ( 5 ) × ( 1 | 2 k Q ˜ | 2 k Q ˜ | D α 1 A ˜ 1 ( y ) | r d y ) 1 / r ( 1 | 2 k Q ˜ | 1 r δ / n 2 k Q ˜ | f ( y ) | r d y ) 1 / r I 5 ( 5 ) C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r ( f ) ( x ˜ ) ; I 5 ( 6 ) C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r ( f ) ( x ˜ ) .

For I 5 ( 7 ) , taking q 1 , q 2 >1 such that 1/r+1/ q 1 +1/ q 2 =1, then

I 5 ( 7 ) C | α 1 | = m 1 , | α 2 | = m 2 R n [ ( x z ) α 1 + α 2 χ Γ ( x ) | x z | m ( x 0 z ) α 1 + α 2 χ Γ ( x 0 ) | x 0 z | m ] ψ t ( y z ) × | D α 1 A ˜ 1 ( z ) | | D α 2 A ˜ 2 ( z ) | | f 2 ( z ) | d z C | α 1 | = m 1 , | α 2 | = m 2 k = 1 k ( 2 k / 2 + 2 k ) ( 1 | 2 k Q ˜ | 1 p δ / n 2 k Q ˜ | f ( y ) | r d y ) 1 / r × ( 1 | 2 k Q ˜ | 2 k Q ˜ | D α 1 A ˜ 1 ( y ) | q 1 d y ) 1 / q 1 ( 1 | 2 k Q ˜ | 2 k Q ˜ | D α 2 A ˜ 2 ( y ) | q 2 d y ) 1 / q 2 C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r ( f ) ( x ˜ ) .

Thus

I 5 C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r (f)( x ˜ ).

We choose 1<r<p in (1), then (2) follows from Lemma 2. For (3), taking 1<r<min(p,(nλ)/pδ) in (1) and by Lemma 3, we obtain

S ψ A ( f ) L q , λ C M ( S ψ A ( f ) ) L q , λ C ( S ψ A ( f ) ) # L q , λ C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r ( f ) L q , λ C j = 1 2 ( | α | = m j D α A j B M O ) ( M r δ / n ( | f | r ) ) 1 / r L q , λ C j = 1 2 ( | α | = m j D α A j B M O ) M r δ / n ( | f | r ) L q / r , λ 1 / r C j = 1 2 ( | α | = m j D α A j B M O ) | f | r L p / r , λ 1 / r C j = 1 2 ( | α | = m j D α A j B M O ) f L p , λ .

This completes the proof of Theorem 1. □

Proof of Theorem 2 It is only to prove (1). Let Q, Q ˜ , A ˜ j (x), f 1 and f 2 be the same as the proof of Theorem 1. We write

F t A ( f ) ( x , y ) = R n j = 1 2 R m j + 1 ( A ˜ j ; x , z ) | x z | m Ω ( y z ) | y z | n 1 δ f 2 ( z ) d z + R n j = 1 2 R m j ( A ˜ j ; x , z ) | x z | m Ω ( y z ) | y z | n 1 δ f 1 ( z ) d z | α 1 | = m 1 1 α 1 ! R n R m 2 ( A ˜ 2 ; x , z ) ( x z ) α 1 D α 1 A ˜ 1 ( z ) | x z | m Ω ( y z ) | y z | n 1 δ f 1 ( z ) d z | α 2 | = m 2 1 α 2 ! R n R m 1 ( A ˜ 1 ; x , z ) ( x z ) α 2 D α 2 A ˜ 2 ( z ) | x z | m Ω ( y z ) | y z | n 1 δ f 1 ( z ) d z + | α 1 | = m 1 , | α 2 | = m 2 1 α 1 ! α 2 ! R n ( x z ) α 1 + α 2 D α 1 A ˜ 1 ( z ) D α 2 A ˜ 2 ( z ) | x z | m Ω ( y z ) | y z | n 1 δ f 1 ( z ) d z ,

then

1 | Q | Q | μ S A ( f ) ( x ) μ S A ˜ ( f 2 ) ( x 0 ) | d x 1 | Q | Q χ Γ ( x ) F t A ( f ) ( x , y ) χ Γ ( x 0 ) F t A ˜ ( f 2 ) ( x 0 , y ) d x 1 | Q | Q χ Γ ( x ) R n j = 1 2 R m j ( A ˜ j ; x , z ) | x z | m Ω ( y z ) | y z | n 1 δ f 1 ( z ) d z d x + 1 | Q | Q χ Γ ( x ) | α 1 | = m 1 1 α 1 ! × R n R m 2 ( A ˜ 2 ; x , z ) ( x z ) α 1 D α 1 A ˜ 1 ( z ) | x z | m Ω ( y z ) | y z | n 1 δ f 1 ( z ) d z d x + 1 | Q | Q χ Γ ( x ) | α 2 | = m 2 1 α 2 ! × R n R m 1 ( A ˜ 1 ; x , z ) ( x z ) α 2 D α 2 A ˜ 2 ( z ) | x z | m Ω ( y z ) | y z | n 1 δ f 1 ( z ) d z d x + 1 | Q | Q χ Γ ( x ) | α 1 | = m 1 , | α 2 | = m 2 1 α 1 ! α 2 ! × R n ( x z ) α 1 + α 2 D α 1 A ˜ 1 ( z ) D α 2 A ˜ 2 ( z ) | x z | m Ω ( y z ) | y z | n 1 δ f 1 ( z ) d z d x + 1 | Q | Q χ Γ ( x ) F t A ˜ ( f 2 ) ( x , y ) χ Γ ( x 0 ) F t A ˜ ( f 2 ) ( x 0 , y ) d x : = J 1 + J 2 + J 3 + J 4 + J 5 .

Similar to the proof of Theorem 1, we get

J 1 C j = 1 2 ( | α j | = m j D α j A j B M O ) 1 | Q | Q | μ S ( f 1 ) ( x ) | d x J 1 C j = 1 2 ( | α j | = m j D α j A j B M O ) ( 1 | Q | Q | μ S ( f 1 ) ( x ) | q d x ) 1 / q J 1 C j = 1 2 ( | α j | = m j D α j A j B M O ) M δ , r ( f ) ( x ˜ ) ; J 2 C | α 2 | = m 2 D α 2 A 2 B M O | α 1 | = m 1 1 | Q | Q | μ S ( D α 1 A ˜ 1 f 1 ) ( x ) | d x J 2 C | α 2 | = m 2 D α 2 A 2 B M O | α 1 | = m 1 ( 1 | Q | R n | μ S ( D α 1 A ˜ 1 f 1 ) ( x ) | s d x ) 1 / s J 2 C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r ( f ) ( x ˜ ) ; J 3 C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r ( f ) ( x ˜ ) ; J 4 C | α 1 | = m 1 , | α 2 | = m 2 1 | Q | Q | μ S ( D α 1 A ˜ 1 D α 2 A ˜ 2 f 1 ) ( x ) | d x J 4 C | α 1 | = m 1 , | α 2 | = m 2 ( 1 | Q | R n | μ S ( D α 1 A ˜ 1 D α 2 A ˜ 2 f 1 ) ( x ) | s d x ) 1 / s J 4 C j = 1 2 ( | α | = m j D α A j B M O ) M δ , p ( f ) ( x ˜ ) .

For J 5 , we write

χ Γ ( x ) F t A ˜ ( f 2 ) ( x , y ) χ Γ ( x 0 ) F t A ˜ ( f 2 ) ( x 0 , y ) = R n ( χ Γ ( x ) χ Γ ( x 0 ) ) j = 1 2 R m j ( A ˜ j ; x , z ) | x z | m Ω ( y z ) | y z | n 1 δ f 2 ( z ) d z + χ Γ ( x 0 ) R n ( 1 | x z | m 1 | x 0 z | m ) j = 1 2 R m j ( A ˜ j ; x , z ) Ω ( y z ) | y z | n 1 δ f 2 ( z ) d z + χ Γ ( x 0 ) R n ( R m 1 ( A ˜ 1 ; x , z ) R m 1 ( A ˜ 1 ; x 0 , z ) ) R m 2 ( A ˜ 2 ; x , z ) | x 0 z | m Ω ( y z ) | y z | n 1 δ f 2 ( z ) d z + χ Γ ( x 0 ) R n ( R m 2 ( A ˜ 2 ; x , z ) R m 2 ( A ˜ 2 ; x 0 , z ) ) R m 1 ( A ˜ 1 ; x 0 , z ) | x 0 z | m Ω ( y z ) | y z | n 1 δ f 2 ( z ) d z | α 1 | = m 1 1 α 1 ! R n [ R m 2 ( A ˜ 2 ; x , z ) ( x z ) α 1 χ Γ ( x ) | x z | m R m 2 ( A ˜ 2 ; x 0 , z ) ( x 0 z ) α 1 χ Γ ( x 0 ) | x 0 z | m ] × Ω ( y z ) | y z | n 1 δ D α 1 A ˜ 1 ( z ) f 2 ( z ) d z | α 2 | = m 2 1 α 2 ! R n [ R m 1 ( A ˜ 1 ; x , z ) ( x z ) α 2 χ Γ ( x ) | x z | m R m 1 ( A ˜ 1 ; x 0 , z ) ( x 0 z ) α 2 χ Γ ( x 0 ) | x 0 z | m ] × Ω ( y z ) | y z | n 1 δ D α 2 A ˜ 2 ( z ) f 2 ( z ) d z + | α 1 | = m 1 , | α 2 | = m 2 1 α 1 ! α 2 ! R n [ ( x z ) α 1 + α 2 χ Γ ( x ) | x z | m ( x 0 z ) α 1 + α 2 χ Γ ( x 0 ) | x 0 z | m ] × Ω ( y z ) | y z | n 1 δ D α 1 A ˜ 1 ( z ) D α 2 A ˜ 2 ( z ) f 2 ( z ) d z .

Then, similar to the proof of Lemma 4 and Theorem 1, we get

J 5 C j = 1 2 ( | α | = m j D α A j B M O ) M δ , r (f)( x ˜ ).

The same argument as the proof of Theorem 1 will give the proof of (2) and (3), we omit the details and finish the proof. □

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Lu, D. Some sharp inequalities for multilinear integral operators. J Inequal Appl 2013, 445 (2013). https://doi.org/10.1186/1029-242X-2013-445

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Keywords

  • multilinear operator
  • Littlewood-Paley operator
  • Marcinkiewicz operator
  • Morrey space
  • BMO