Home Mean oscillation and boundedness of multilinear operator related to multiplier operator
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Mean oscillation and boundedness of multilinear operator related to multiplier operator

  • Qiaozhen Zhao and Dejian Huang EMAIL logo
Published/Copyright: August 5, 2021

Abstract

In this paper, the boundedness of certain multilinear operator related to the multiplier operator from Lebesgue spaces to Orlicz spaces is obtained.

MSC 2010: 42B20; 42B25

1 Introduction and preliminaries

Let b be a locally integrable function on R n and T be an integral operator. For a suitable function f , the commutator generated by b and T is defined by [ b , T ] ( f ) = T ( b f ) b T ( f ) . The investigation of the commutator begins with Coifman-Rochberg-Weiss pioneering study and classical result [1]. If T is the Calderón-Zygmund singular integral operator, a well known result in [1] states that the commutator [ b , T ] (where b B M O ) is bounded on L p ( R n ) ( 1 < p < ) . In fact, they even proved that this property characterizes B M O functions. There are two major reasons for considering the problem of commutators, one is that the boundedness of commutator can produce some characterizations of function space, and the other one is that the theory of commutator plays an important role in studying the regularity of solutions of the second order elliptic and parabolic partial differential equations. With the development of singular integral operators [2,3], their commutators have been well studied. In [1,4,5], the authors proved that the commutators generated by the singular integral operators and B M O functions are bounded on L p ( R n ) for 1 < p < . Chanillo [6] proved a similar result when singular integral operators are replaced by the fractional integral operators. In [7], Janson proved the boundedness of the commutators generated by the singular integral operators and B M O functions from Lebesgue spaces to Orlicz spaces; he pointed out that one can also characterize those functions for which the above commutator is bounded from L p to L q (for q p ). As it turned out, this is equivalent to the functions being in a certain Lipschitz space. In [8,9,10, 11,12], the authors studied the algebras of singular integral operators. Calderón [13] introduced the multilinear singular integral operator as follows:

T b ( f ) ( x ) = R n R m + 1 ( b ; x , y ) x y m K ( x y ) f ( y ) d y ,

where R m + 1 ( b ; x , y ) denotes the ( m + 1 )th remainder of Taylor series of the function b at x about y , which is a non-trivial and natural extension of the commutator if m = 0 .

In this paper, we will study a more complicated expression, which extends naturally the formula of the above commutator, and prove the analogue of Janson’s theorem for these “multilinear expressions,” following closely Janson’s method, that is, we will prove the boundedness of the multilinear operator associated with the multiplier operator from Lebesgue space to Orlicz space.

Throughout this paper, Q denotes 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 d y ,

where and in what follows, f Q = Q 1 Q f ( x ) d x . It is well known [2,3] that

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

Let M be the Hardy-Littlewood maximal operator defined by

M ( f ) ( x ) = sup Q x 1 Q Q f ( y ) d y .

We denote M p f = ( M ( f p ) ) 1 / p for 0 < p < . For 1 r < and 0 < β < n , let

M β , r ( f ) ( x ) = sup Q x 1 Q 1 r β / n Q f ( y ) r d y 1 / r .

We say that f belongs to B M O ( R n ) if f # belongs to L ( R n ) and f B M O = f # L . More generally, let ρ be a non-decreasing positive function on [ 0 , + ) and define B M O ρ ( R n ) as the space of all functions f such that

1 Q ( x , r ) Q ( x , r ) f ( y ) f Q d y C ρ ( r ) .

For β > 0 , the Lipschitz space Lip β ( R n ) is the space of functions f such that

f Lip β = sup x y f ( x ) f ( y ) / x y β < .

Let m f denote the distribution function of f , that is, m f ( t ) = { x R n : f ( x ) > t } .

Let ρ be a non-decreasing convex function on [ 0 , + ) with ρ ( 0 ) = 0 . ρ 1 denotes the inverse function of ρ . The Orlicz space L ρ ( R n ) is defined by the set of functions f such that R n ρ ( λ f ( x ) ) d x < for some λ > 0 , and the norm is given by

f L ρ = inf λ > 0 λ 1 1 + R n ρ ( λ f ( x ) ) d x .

2 Main results

In this paper, we will study the multilinear operator associated with the multiplier operator [14].

A bounded measurable function k defined on R n { 0 } is called a multiplier. The multiplier operator T associated with k is defined by

T ( f ) ( x ) = k ( x ) f ˆ ( x ) , f S ( R n ) ,

where f ˆ denotes the Fourier transform of f and S ( R n ) is the Schwartz test function class. We recall the definition of the class M ( s , l ) . Denote by x t the fact that the value of x lies in the annulus { x R n : a t < x < b t } , where 0 < a 1 < b < are values specified in each instance.

Definition 1

Let l 0 be a real number and 1 s 2 , then the multiplier k satisfies the condition M ( s , l ) , if

ξ R D α k ( ξ ) s d ξ 1 s < C R n / s α

for all R > 0 and multi-indices α with α l , when l is a positive integer. In addition, if

ξ R D α k ( ξ ) D α k ( ξ z ) s d ξ 1 s C z R γ R n s α

for all z < R / 2 and all multi-indices α with α = [ l ] , the integer part of l when l is not an integer.

Denote D ( R n ) = { ϕ S ( R n ) : supp ( ϕ ) is compact } and D ˆ 0 ( R n ) = { ϕ S ( R n ) : ϕ ˆ D ( R n ) with ϕ ˆ vanishes in a neightborhood of the origin } . The boundedness of T on L p ( R n ) is proved by [14].

Definition 2

For a real number l ˜ 0 and 1 s ˜ < , then K verifies the condition M ˜ ( s ˜ , l ˜ ) , and K M ˜ ( s ˜ , l ˜ ) if

x R D α ˜ K ( x ) s ˜ d x 1 s ˜ C R n / s ˜ n α ˜ , R > 0 ,

for all multi-indices α ˜ l ˜ . In addition, if

x R D α ˜ K ( x ) D α ˜ K ( x z ) s ˜ d x 1 s ˜ C z R v R n s ˜ n u , if 0 < v < 1 ,

x R D α ˜ K ( x ) D α ˜ K ( x z ) s ˜ d x 1 s ˜ C z R log R z R n s ˜ n u , if v = 1 ,

for all z < R 2 , R > 0 , and all multi-indices α ˜ with α ˜ = u , where u denotes the largest integer strictly less than l ˜ with l ˜ = u + v .

Lemma 1

[14] Let k M ( s , l ) , 1 s 2 , and l > n s . Then the associated mapping T , defined a priori for f D ˆ 0 ( R n ) , T ( f ) ( x ) = ( f K ) ( x ) , extends to a bounded mapping from L p ( R n ) into itself for 1 < p < and K ( x ) = k ˇ ( x ) .

Lemma 2

[14] Suppose k M ( s , l ) , 1 s 2 . Given 1 s ˜ < , let r 1 be such that 1 r = max 1 s , 1 1 s ˜ . Then K M ˜ ( s ˜ , l ˜ ) , where l ˜ = l n r .

Lemma 3

[14] Let 1 s < , suppose that l is a positive real number with l > n / r , 1 / r = max { 1 / s , 1 1 / s ˜ } , and k M ( s , l ) . Then there is a positive constant a , such that

Q k K ( x z ) K ( x Q z ) s ˜ d z 1 / s ˜ C 2 k a ( 2 k h ) n / s ˜ .

Now we define the multilinear operator associated with the multiplier operator as follows.

Definition 3

Let l and m j be the positive integers ( j = 1 , , l ), m 1 + + m l = m and b j be the functions on R n ( j = 1 , , l ). For 1 j l , 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 .

Let T be the multiplier operator. By Lemma 1, T ( f ) ( x ) = ( K f ) ( x ) for K ( x ) = k ˇ ( x ) . The multilinear operator associated with T is defined by

T b ( f ) ( x ) = R n j = 1 l R m j + 1 ( b j ; x , y ) x y m K ( x y ) f ( y ) d y .

Note that T b is just the multilinear commutators of T and b when m = 0 [4,5]. It is well known that multilinear operator, as a non-trivial extension of commutator, is of great importance in harmonic analysis and has been widely studied by many authors [1,3,4, 5,6,7,9,10,11, 12,15,16, 17,18,19, 20,21,22, 23,24]. The main purpose of this paper is to prove the boundedness of the multilinear operator T b from Lebesgue space to Orlicz space.

We shall prove the following main theorem in Section 3.

Theorem

Let 0 < β 1 , 1 < p < n / l β , and φ , ψ be two non-decreasing positive functions on [ 0 , + ) with ( ψ l ) 1 ( t ) = t 1 / p φ l ( t 1 / n ) . Suppose that ψ is convex, ψ ( 0 ) = 0 , ψ ( 2 t ) C ψ ( t ) . Let T be the multiplier operator as Definition 3. Then T b is bounded from L p ( R n ) to L ψ l ( R n ) if D α b j B M O ( R n ) for all α with α = m j and j = 1 , , l .

Remark

  1. If l = 1 and ψ 1 ( t ) = t 1 / p φ ( t 1 / n ) , then T b and S b are all bounded from L p ( R n ) to L ψ ( R n ) under the conditions of Lemmas 1 and 2.

  2. If l = 1 , φ ( t ) 1 and ψ ( t ) = t p for 1 < p < , then T b is bounded on L p ( R n ) if D α b B M O φ ( R n ) for all α with α = m .

  3. If l = 1 , ψ ( t ) = t q and φ ( t ) = t n ( 1 / p 1 / q ) for 1 < p < q < , then, by B M O t β ( R n ) = Lip β ( R n ) [11, Lemma 4], T b is bounded from L p ( R n ) to L q ( R n ) if D α b Lip n ( 1 / p 1 / q ) ( R n ) for all α with α = m .

3 Proof of Theorem

We begin with the following preliminary lemmas.

Lemma 4

[14] Let T be the multiplier operator as Definition 3. Then T is bounded on L p ( R n ) for 1 < p < .

Lemma 5

[7] Let ρ be a non-decreasing positive function on [ 0 , + ) and η be an infinitely differentiable function on R n with compact support such that R n η ( x ) d x = 1 . Denote that b t ( x ) = R n b ( x t y ) η ( y ) d y . Then b b t B M O C ρ ( t ) b B M O ρ .

Lemma 6

[7] Let 0 < β < 1 or β = 1 and ρ be a non-decreasing positive function on [ 0 , + ) . Then b t Lip β C t β ρ ( t ) b B M O ρ .

Lemma 7

[7] Suppose 1 p 2 < p < p 1 < , ρ is a non-increasing function on [ 0 , + ) , B is a linear operator such that m B ( f ) ( t 1 / p 1 ρ ( t ) ) C t 1 if f L p 1 1 and m B ( f ) ( t 1 / p 2 ρ ( t ) ) C t 1 if f L p 2 1 . Then 0 m B ( f ) ( t 1 / p ρ ( t ) ) d t C if f L p ( p / p 1 ) 1 / p .

Lemma 8

[6] Suppose that 0 < β < n , 1 r < p < n / β , and 1 / q = 1 / p β / n . Then M β , r ( f ) L q C f L p .

Lemma 9

[11] Let b 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 ( 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 ˜ is the cube centered at x and having side length 5 n x y .

In order to prove the theorem of the paper, we need the following Key Lemma.

Key Lemma. Let T be the multiplier operator as Definition 3. Suppose that Q = Q ( x 0 , d ) is a cube with supp f ( 2 Q ) c and x , x ˜ Q .

  1. If D α b j B M O ( R n ) for all α with α = m j , j = 1 , , l , then

    T b ( f ) ( x ) T b ( f ) ( x 0 ) C j = 1 l α j = m j D α j b j B M O M r ( f ) ( x ˜ ) for any r > 1 ;

  2. If 0 < β 1 and D α b j Lip β ( R n ) for all α with α = m j and j = 1 , , l , then

    T b ( f ) ( x ) T b ( f ) ( x 0 ) C j = 1 l α j = m j D α j b j Lip β M l β , r ( f ) ( x ˜ ) for any r > 1 .

Proof

Without loss of generality, we may assume l = 2 . Let Q ˜ = 5 n Q and b ˜ ( x ) = b ( x ) α = m 1 α ! ( D α b ) Q ˜ x α , then R m ( b ; x , y ) = R m ( b ˜ ; x , y ) and D α b ˜ = D α b ( D α b ) Q ˜ for α = m . For supp f ( 2 Q ) c and x , x ˜ Q , we write

T b ( f ) ( x ) T b ( f ) ( 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 ( 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 ( 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 ( 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 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 α 1 b ˜ 1 ( y ) f ( 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 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 α 2 b ˜ 2 ( y ) f ( 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 ( y ) d y = I 1 + I 2 + I 3 + I 4 + I 5 + I 6 .

  1. By Lemma 9 and the following inequality [3], for b B M O ( R n ) ,

    b Q 1 b Q 2 C log ( Q 2 / Q 1 ) b B M O for Q 1 Q 2 .

Note that x y x 0 y for x Q and y R n Q ˜ , then, for x Q and y 2 k + 1 Q 2 k Q with k 1 ,

R m ( b ˜ ; x , y ) C x y m α = m ( D α b B M O + ( D α b ) Q ˜ ( x , y ) ( D α b ) Q ˜ ) C k x y m α = m D α b B M O .

For I 1 , by the conditions on K , Lemmas 2 and 3, we obtain

I 1 k = 1 2 k + 1 Q 2 k Q 1 x y m 1 x 0 y m j = 1 2 R m j ( b ˜ j ; x , y ) f ( y ) d y + k = 1 2 k + 1 Q 2 k Q K ( x y ) K ( x 0 y ) x 0 y m 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 B M O k = 1 k 2 2 k + 1 Q 2 k Q x x 0 x 0 y K ( x y ) f ( y ) d y + C j = 1 2 α j = m j D α j b j B M O k = 1 k 2 2 k + 1 Q ˜ f ( y ) r d y 1 / r 2 k + 1 Q 2 k Q K ( x y ) K ( x 0 y ) r d y 1 / r C j = 1 2 α j = m j D α j b j B M O k = 1 k 2 ( 2 k + 2 k a ) 1 2 k + 1 Q 2 k + 1 Q f ( y ) r d y 1 / r C j = 1 2 α j = m j D α j b j B M O M r ( f ) ( x ˜ ) .

For I 2 , by the formula [11]

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

and Lemma 9, we have

R m ( b ˜ ; x , y ) R m ( b ˜ ; x 0 , y ) C γ < m α = m x x 0 m γ x y γ D α b B M O ,

thus,

I 2 C j = 1 2 α j = m j D α j b j B M O k = 1 2 k + 1 Q 2 k Q k x x 0 x 0 y K ( x 0 y ) f ( y ) d y C j = 1 2 α j = m j D α j b j B M O k = 1 k 2 k 1 2 k + 1 Q 2 k + 1 Q f ( y ) r d y 1 / r C j = 1 2 α j = m j D α j b j B M O M r ( f ) ( x ˜ ) .

Similarly,

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

For I 4 , similar to the proof of I 1 and I 2 , we get

I 4 C α 1 = m 1 R n 2 Q ( x y ) α 1 x y m ( x 0 y ) α 1 x 0 y m K ( x y ) R m 2 ( b ˜ 2 ; x , y ) D α 1 b ˜ 1 ( y ) f ( y ) d y + C α 1 = m 1 R n 2 Q R m 2 ( b ˜ 2 ; x , y ) R m 2 ( b ˜ 2 ; x 0 , y ) ( x 0 y ) α 1 K ( x y ) x 0 y m D α 1 b ˜ 1 ( y ) f ( y ) d y

+ C α 1 = m 1 R n 2 Q K ( x y ) K ( x 0 y ) ( x 0 y ) α 1 x 0 y m R m 2 ( b ˜ 2 ; x 0 , y ) D α 1 b ˜ 1 ( y ) f ( y ) d y C j = 1 2 α j = m j D α j b j B M O k = 1 k 2 ( 2 k + 2 k a ) 1 2 k + 1 Q 2 k + 1 Q f ( y ) r d y 1 / r C j = 1 2 α j = m j D α j b j B M O M r ( f ) ( x ˜ ) .

Similarly,

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

For I 6 , taking 1 < s 1 , s 2 < such that 1 / r + 1 / s 1 + 1 / s 2 = 1 , then

I 6 C α 1 = m 1 , α 2 = m 2 R n 2 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 2 b ˜ 2 ( y ) f ( y ) d y C α 1 = m 1 , α 2 = m 2 k = 1 ( 2 k + 2 k a ) 1 2 k + 1 Q 2 k + 1 Q f ( y ) r d y 1 / r × 1 2 k + 1 Q 2 k + 1 Q D α 1 b ˜ 1 ( y ) s 1 d y 1 / s 1 1 2 k + 1 Q 2 k + 1 Q D α 2 b ˜ 2 ( y ) s 2 d y 1 / s 2 C j = 1 2 α j = m j D α j b j B M O k = 1 k 2 ( 2 k + 2 k a ) M r ( f ) ( x ˜ ) C j = 1 2 α j = m j D α j b j B M O M r ( f ) ( x ˜ ) .

Thus,

T b ( f ) ( x ) T b ( f ) ( x 0 ) C j = 1 2 α j = m j D α j b j B M O M r ( f ) ( x ˜ ) .

  1. For b Lip β ( R n ) , by Lemma 9 and the following inequality:

b ( x ) b Q 1 Q Q b Lip β x y β d y b Lip β ( x x 0 + d ) β ,

we get

R m ( b ˜ ; x , y ) C α = m D α b Lip β ( x y + d ) m + β

and

R m ( b ˜ ; x , y ) R m ( b ˜ ; x 0 , y ) C α = m D α b Lip β ( x y + d ) m + β ,

then, similar to the proof of (I),

I 1 k = 1 2 k + 1 Q 2 k Q 1 x y m 1 x 0 y m K ( x y ) j = 1 2 R m j ( b ˜ j ; x , y ) f ( y ) d y + k = 1 2 k + 1 Q 2 k Q K ( x y ) K ( x 0 y ) x 0 y m 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 k = 1 ( 2 k + 2 k a ) 1 2 k + 1 Q 1 2 β r / n 2 k + 1 Q f ( y ) r d y 1 / r C j = 1 2 α j = m j D α j b j Lip β M 2 β , r ( f ) ( x ˜ ) , I 2 + I 3 C j = 1 2 α j = m j D α j b j Lip β k = 1 ( 2 k + 2 k a ) 1 2 k + 1 Q 1 2 β r / n 2 k + 1 Q f ( y ) r d y 1 / r C j = 1 2 α j = m j D α j b j Lip β M 2 β , r ( f ) ( x ˜ ) , I 4 C α 1 = m 1 R n 2 Q ( x y ) α 1 x y m ( x 0 y ) α 1 x 0 y m K ( x , x y ) R m 2 ( b ˜ 2 ; x , y ) D α 1 b ˜ 1 ( y ) f ( y ) d y + C α 1 = m 1 R n 2 Q R m 2 ( b ˜ 2 ; x , y ) R m 2 ( b ˜ 2 ; x 0 , y ) ( x 0 y ) α 1 K ( x , x y ) x 0 y m D α 1 b ˜ 1 ( y ) f ( y ) d y + C α 1 = m 1 R n 2 Q K ( x , x y ) K ( x 0 , x 0 y ) ( x 0 y ) α 1 x 0 y m R m 2 ( b ˜ 2 ; x 0 , y ) D α 1 b ˜ 1 ( y ) f ( y ) d y C j = 1 2 α j = m j D α j b j Lip β k = 1 ( 2 k + 2 k a ) 1 2 k + 1 Q 1 2 β r / n 2 k + 1 Q f ( y ) r d y 1 / r C j = 1 2 α j = m j D α j b j Lip β M 2 β , r ( f ) ( x ˜ ) , I 5 C j = 1 2 α j = m j D α j b j Lip β M 2 β , r ( f ) ( x ˜ ) ,

I 6 C α 1 = m 1 , α 2 = m 2 R n 2 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 2 b ˜ 2 ( y ) f ( y ) d y C j = 1 2 α j = m j D α j b j Lip β k = 1 ( 2 k + 2 k a ) 1 2 k + 1 Q 1 2 β r / n 2 k + 1 Q f ( y ) r d y 1 / r C j = 1 2 α j = m j D α j b j Lip β M 2 β , r ( f ) ( x ˜ ) .

Thus,

T b ( f ) ( x ) T b ( f ) ( x 0 ) C j = 1 2 α j = m j D α j b j Lip β M β , r ( f ) ( x ˜ ) .

Now we are in a position to prove the main theorem.

Proof of Theorem

Without loss of generality, we assume l = 2 . We prove the theorem in several steps. First, if D α b j B M O ( R n ) for all α with α = m j and j = 1 , , l , we prove

(1) ( T b ( f ) ) # C j = 1 2 α j = m j D α j b j B M O M r ( f )

for any 1 < r < . Fix a cube Q = Q ( x 0 , d ) and x ˜ Q . 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 ( b j ; x , y ) = R m ( b ˜ j ; x , y ) and D α b ˜ j = D α b j ( D α b j ) Q ˜ for α = m j . For f 1 = f χ Q ˜ and f 2 = f χ R n Q ˜ , we write

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

then

T b ( f ) ( x ) T b ( f 2 ) ( x 0 ) T j = 1 2 R m j ( b ˜ j ; x , ) x m f 1 + T α 1 = m 1 1 α 1 ! R m 2 ( b ˜ 2 ; x , ) ( x ) α 1 D α 1 b ˜ 1 x m f 1 + T α 2 = m 2 1 α 2 ! R m 1 ( b ˜ 1 ; x , ) ( x ) α 2 D α 2 b ˜ 2 x m f 1 + T α 1 = m 1 , α 2 = m 2 1 α 1 ! α 2 ! ( x ) α 1 + α 2 D α 1 b ˜ 1 D α 2 b ˜ 2 x m f 1 + T b ( f 2 ) ( x ) T b ( f 2 ) ( x 0 ) = L 1 ( x ) + L 2 ( x ) + L 3 ( x ) + L 4 ( x ) + L 5 ( x )

and

1 Q Q T b ( f ) ( x ) T b ( f 2 ) ( x 0 ) d x 1 Q Q L 1 ( x ) d x + 1 Q Q L 2 ( x ) d x + 1 Q Q L 3 ( x ) d x + 1 Q Q L 4 ( x ) d x + 1 Q Q L 5 ( x ) d x = L 1 + L 2 + L 3 + L 4 + L 5 .

Now, for L 1 , if x Q and y 2 Q , by Lemma 9, we get

R m ( b ˜ ; x , y ) C x y m α = m D α b B M O ,

thus, by the L r boundedness of T (see Lemma 1) and Hölder’s inequality, we obtain

L 1 C j = 1 2 α j = m j D α j b j B M O 1 Q Q T ( f 1 ) ( x ) d x C j = 1 2 α j = m j D α j b j B M O 1 Q R n T ( f 1 ) ( x ) r d x 1 / r C j = 1 2 α j = m j D α j b j B M O 1 Q R n f 1 ( x ) r d x 1 / r C j = 1 2 α j = m j D α j b j B M O 1 Q ˜ Q ˜ f ( x ) r d x 1 / r C j = 1 2 α j = m j D α j b j B M O M r ( f ) ( x ˜ ) .

For L 2 , denoting r = u v for 1 < u , v < and 1 / v + 1 / v = 1 , we have

L 2 C α 2 = m 2 D α 2 b 2 B M O α 1 = m 1 α = m 1 Q Q T ( D α 1 b ˜ 1 f 1 ) ( x ) d x C α 2 = m 2 D α 2 b 2 B M O α 1 = m 1 1 Q R n T ( D α 1 b ˜ 1 f 1 ) ( x ) u d x 1 / u C α 2 = m 2 D α 2 b 2 B M O α 1 = m 1 1 Q R n D α 1 b ˜ 1 ( x ) f 1 ( x ) u d x 1 / u C α 2 = m 2 D α 2 b 2 B M O 1 Q ˜ Q ˜ f ( x ) u v d x 1 / u v α 1 = m 1 1 Q ˜ Q ˜ D α 1 b ˜ 1 ( x ) ( D α 1 b ˜ 1 ) Q ˜ u v d x 1 / u v C j = 1 2 α j = m j D α j b j B M O M r ( f ) ( x ˜ ) .

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

L 3 C j = 1 2 α j = m j D α j b j B M O M r ( f ) ( x ˜ ) .

Similarly, for L 4 , denoting r = u w for 1 < u , v 1 , v 2 , w < and 1 / v 1 + 1 / v 2 + 1 / w = 1 , by Holder’s inequality, we obtain

L 4 C α 1 = m 1 , α 2 = m 2 1 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 1 Q R n T ( D α 1 b ˜ 1 D α 2 b ˜ 2 f 1 ) ( x ) u d x 1 / u C α 1 = m 1 , α 2 = m 2 Q 1 / u R n D α 1 b ˜ 1 ( x ) D α 2 b ˜ 2 ( x ) f 1 ( x ) u d x 1 / u C α 1 = m 1 , α 2 = m 2 1 Q ˜ Q ˜ D α 1 b ˜ 1 ( x ) u v 1 d x 1 / u v 1 1 Q ˜ Q ˜ D α 2 b ˜ 2 ( x ) u v 2 d x 1 / u v 2 1 Q ˜ Q ˜ f ( x ) u w d x 1 / u w C j = 1 2 α j = m j D α j b j B M O M r ( f ) ( x ˜ ) .

For L 5 , by Key Lemma, we have

L 3 C j = 1 2 α j = m j D α j b j B M O M r ( f ) ( x ˜ ) .

Put these estimates together and take the supremum over all Q such that x ˜ Q , we have

( T b ( f ) ) # ( x ˜ ) C j = 1 2 α j = m j D α j b j B M O M r ( f ) ( x ˜ ) .

Thus, taking r such that 1 < r < p , we obtain

(2) T b ( f ) L p C ( T b ( f ) ) # L p C j = 1 2 α j = m j D α j b j B M O M r ( f ) L p C j = 1 2 α j = m j D α j b j B M O f L p .

Second, if D α b j Lip β ( R n ) for all α with α = m j and j = 1 , , l , we prove

(3) ( T b ( f ) ) # C j = 1 2 α j = m j D α j b j Lip β M 2 β , r ( f )

for any r with 1 < r < n / 2 β . In fact, for x Q and y 2 Q , by Lemma 9, we have

R m ( b ˜ ; x , y ) C x y m α = m sup z 2 Q D α b ( z ) ( D α b ) Q C x y m Q β / n α = m D α b Lip β .

Similar to the proof of (1) and by Key Lemma, we obtain

1 Q Q T b ( f ) ( x ) T b ( f ) ( x 0 ) d x 1 Q Q T j = 1 2 R m j ( b ˜ j ; x , ) x m f 1 d x + C 1 Q Q T α 1 = m 1 1 α 1 ! R m 2 ( b ˜ 2 ; x , ) ( x ) α 1 D α 1 b ˜ 1 x m f 1 d x

+ C 1 Q Q T α 2 = m 2 1 α 2 ! R m 1 ( b ˜ 1 ; x , ) ( x ) α 2 D α 2 b ˜ 2 x m f 1 d x + C 1 Q Q T α 1 = m 1 , α 2 = m 2 1 α 1 ! α 2 ! ( x ) α 1 + α 2 D α 1 b ˜ 1 D α 2 b ˜ 2 x m f 1 d x + 1 Q Q T b ( f 2 ) ( x ) T b ( f 2 ) ( x 0 ) d x C j = 1 2 α j = m j D α j b j Lip β 1 Q 1 / r 2 β / n Q ˜ f ( x ) r d x 1 / r + 1 Q Q T b ( f 2 ) ( x ) T b ( f 2 ) ( x 0 ) d x C j = 1 2 α j = m j D α j b j Lip β M 2 β , r ( f ) ( x ˜ ) .

Thus, (3) holds. Taking 1 < r < p < n / 2 β , 1 / q = 1 / p 2 β / n , by Lemma 8 we obtain

(4) T b ( f ) L q C ( T b ( f ) ) # L q C j = 1 2 α j = m j D α j b j Lip β M 2 β , r ( f ) L q C j = 1 2 α j = m j D α j b j Lip β f L p .

Now we verify that T b satisfies the conditions of Lemma 6. In fact, for any 1 < p i < n / 2 β , 1 / q i = 1 / p i 2 β / n ( i = 1 , 2 ) , and f L p i 1 , note that T b ( f ) ( x ) = T b b s ( f ) ( x ) + T b s ( f ) ( x ) and D α ( b s ) = ( D α b ) s with D α ( b j b j s ) B M O ( R n ) and D α b j s Lip β ( R n ) , by (2) and Lemma 5, we obtain

T b b s ( f ) L p i C j = 1 2 α j = m j D α j ( b j b j s ) B M O f L p i C j = 1 2 α j = m j D α j b j ( D α j b j ) s B M O C j = 1 2 α j = m j D α j b j B M O φ φ 2 ( s ) ,

and by (4) and Lemma 6, we obtain

T b s ( f ) L q i C j = 1 2 α j = m j D α j b j Lip β f L p i C s 2 β φ 2 ( s ) j = 1 2 α j = m j D α j b j B M O φ .

Thus, for s = t 1 / n ,

m T b ( f ) ( ( ψ 2 ) 1 ( t ) ) m T b ( f ) ( t 1 / p i φ 2 ( t 1 / n ) ) m T b b s ( f ) ( t 1 / p i φ 2 ( t 1 / n ) / 2 ) + m T b s ( f ) ( t 1 / p i φ 2 ( t 1 / n ) / 2 ) C φ 2 ( s ) t 1 / p i φ 2 ( s ) p i + s 2 β φ 2 ( s ) t 1 / p i φ 2 ( s ) q i = C t 1 .

For f L p ( p / p 1 ) 1 / p , taking 1 < p 2 < p < p 1 < n / 2 β and by Lemma 7, we obtain

R n ψ 2 ( T b ( f ) ( x ) ) d x = 0 m T b ( f ) ( ( ψ 2 ) 1 ( t ) ) d t C ,

then, T b ( f ) L ψ 2 C .□

Acknowledgments

The authors express their gratitude to the anonymous referees for their valuable comments that largely polish this paper.

  1. Funding information: This work was supported by the School-level Scientific Research Projects of Jiaxing Nanyang Vocational and Technical College (Grant: p30018ky013).

  2. Conflict of interest: The authors state no conflict of interest.

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Received: 2020-06-28
Accepted: 2021-06-03
Published Online: 2021-08-05

© 2021 Qiaozhen Zhao and Dejian Huang, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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