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Commutators of multilinear strongly singular integrals on nonhomogeneous metric measure spaces

  • Hailian Wang and Rulong Xie EMAIL logo
Published/Copyright: December 31, 2021

Abstract

Let ( X , d , μ ) denote nonhomogeneous metric measure space satisfying geometrically doubling and the upper doubling measure conditions. In this paper, the boundedness in Lebesgue spaces for two kinds of commutators, which are iterated commutators and commutators in summation form, generated by multilinear strongly singular integral operators with RBMO ( μ ) function on nonhomogeneous metric measure spaces ( X , d , μ ) is obtained.

MSC 2010: 42B20; 42B25

1 Introduction

It is well known that Hytönen [1] introduced nonhomogeneous metric measure spaces ( X , d , μ ) , satisfying geometrically doubling and the upper doubling measure conditions (see Definitions 1.1 and 1.2). The nonhomogeneous metric measure space includes both the homogeneous spaces and nondoubling measure spaces as special cases. We recall that the homogeneous space is a space equipped with a nonnegative doubling measure μ , where μ is said to satisfy the doubling condition if there exists a constant C > 0 such that μ ( B ( x , 2 r ) ) C μ ( B ( x , r ) ) for all x supp μ and r > 0 . The nondoubling measure space is a space equipped with a nonnegative measure μ , where μ only needs to satisfy the polynomial growth condition, i.e., for all x R n and r > 0 , there exist a constant C 0 > 0 and k ( 0 , n ] , such that

(1.1) μ ( B ( x , r ) ) C 0 r k ,

where B ( x , r ) = { y R n : y x < r } . There are many important results in nondoubling measure spaces (see [2,3, 4,5,6, 7,8] and references therein). The analysis on nondoubling measures has important applications in solving the long-standing open Painlevé’s problem (see [5]).

Next, let us recall some results on nonhomogeneous metric measure spaces. Hytönen et al. [9] and Bui and Duong [10] independently introduced the atomic Hardy space H 1 ( μ ) and proved that the dual space of H 1 ( μ ) is RBMO ( μ ) . In [10], the authors also proved that Calderón-Zygmund operator and commutators generated by Calderón-Zygmund operators and RBMO function are bounded in L p ( μ ) ( 1 < p < + ) . Recently, some equivalent characterizations are established by Liu et al. [11] for the boundedness of Calderón-Zygmund operators on L p ( μ ) ( 1 < p < + ) . In [12], Fu et al. established boundedness of multilinear commutators of Calderón-Zygmund operators on Orlicz spaces on nonhomogeneous spaces. More results on nonhomogeneous metric measure spaces have been obtained (see [13,14,15,16, 17,18,19, 20,21,22, 23,24] and references therein).

The theory of multilinear singular integral operators has been considered by some researchers. In [25], Coifman and Meyer first established the theory of bilinear Calderón-Zygmund operators. Later, Grafakos and Torres [26,27] established the boundedness of multilinear singular integral on the product Lebesgue spaces and Hardy spaces. The boundedness of multilinear singular integrals and commutators on nondoubling measure spaces ( R n , μ ) was established by Xu [8,28]. Weighted norm inequalities for multilinear Calderón-Zygmund operators on nonhomogeneous metric measure spaces were also constructed in [29]. Boundedness for commutators of multilinear Calderón-Zygmund operators and multilinear fractional integral operators on nonhomogeneous metric measure spaces was also established in [30,16]. Zheng and Tao [31] established the boundedness for iterated commutators of multilinear singular integrals of Dini’s type on nonhomogeneous metric measure spaces.

The introduction of a strong singular integral operator is motivated by a class of multiplier operators whose symbol is given by e i ξ α / ξ β away from the origin, where 0 < α < 1 and β > 0 . Fefferman and Stein [32] enlarged the multiplier operators onto a class of convolution operators. Coifman [33] also considered a related class of operators for n = 1 . The strongly singular nonconvolution operators were introduced and researched by Alvarez and Milman [34,35], whose properties are similar to those of Calderón-Zygmund operators, but the kernel is more singular near the diagonal than those of the standard case. Furthermore, Lin and Lu [36,37, 38,39] obtained the boundedness for strongly singular integral and its commutators on Lebesgue spaces, Morrey spaces and Hardy spaces. Recently, we [40] established the boundedness for multilinear strongly singular integral operators on nonhomogeneous metric measure spaces.

In this paper, two kinds of commutators generated by multilinear strongly singular integral operators and RBMO ( μ ) function on nonhomogeneous metric measure spaces are introduced. We will prove that they are bounded in m -multiple Lebesgue spaces, provided that multilinear strongly singular integral is bounded from m -multiple L 1 ( μ ) × × L 1 ( μ ) to L 1 / m , ( μ ) , where L p ( μ ) and L p , ( μ ) denote the Lebesgue space and weak Lebesgue space, respectively.

Before stating the main results, let us first recall some notations and definitions.

Definition 1.1

[1] A metric space ( X , d ) is called geometrically doubling if there exists some N 0 N such that, for any ball B ( x , r ) X , there exists a finite ball covering { B ( x i , r / 2 ) } i of B ( x , r ) such that the cardinality of this covering is at most N 0 .

Definition 1.2

[1] A metric measure space ( X , d , μ ) is said to be upper doubling if μ is a Borel measure on X and there exists a dominating function λ : X × ( 0 , + ) ( 0 , + ) and a constant C λ > 0 such that for each x X , r λ ( x , r ) is nondecreasing, and for all x X , r > 0 ,

(1.2) μ ( B ( x , r ) ) λ ( x , r ) C λ λ ( x , r / 2 ) .

Remark 1.3

  1. A space of homogeneous type is a special case of upper doubling spaces, where one can take the dominating function λ ( x , r ) = μ ( B ( x , r ) ) . On the other hand, a metric space ( X , d , μ ) satisfying the polynomial growth condition (1.1) (in particular, ( X , d , μ ) = ( R n , , μ ) with μ satisfying (1.1) for some k ( 0 , n ) is also an upper doubling measure space if we take λ ( x , r ) = C r k .

  2. Let ( X , d , μ ) be an upper doubling space and λ be a dominating function on X × ( 0 , + ) as in Definition 1.2. In [10], it was shown that there exists another dominating function λ ˜ such that for all x , y X with d ( x , y ) r ,

    (1.3) λ ˜ ( x , r ) C ˜ λ ˜ ( y , r ) .

    Thus, we assume that λ always satisfies (1.3) in this paper.

Definition 1.4

[11] Let 1 < α , β < + . A ball B X is called ( α , β ) doubling if μ ( α B ) β μ ( B ) .

Remark 1.5

As pointed in Lemma 2.3 of [11], there exist plenty of doubling balls with small radii and with large radii. In the rest of this paper, unless α and β are specified otherwise, by an ( α , β ) doubling ball, we mean a ( 6 , β 0 ) doubling with a fixed number β 0 > max { C λ 3 log 2 6 , 6 n } , where n = log 2 N 0 is viewed as a geometric dimension of the space.

Definition 1.6

A kernel K ( , , ) L loc 1 ( ( X ) m + 1 \ { ( x , y 1 , , y j , , y m ) : x = y 1 = = y j = = y m } ) is called an m -linear strongly singular integral kernel if it satisfies:

  1. (1.4) K ( x , y 1 , , y j , , y m ) C j = 1 m λ ( x , d ( x , y j ) ) m

    for all ( x , y 1 , , y j , , y m ) ( X ) m + 1 with x y j for some j .

  2. There exist 0 < α < 1 and 0 < δ 1 such that

    (1.5) K ( x , y 1 , , y j , , y m ) K ( x , y 1 , , y j , , y m ) C d ( x , x ) δ j = 1 m d ( x , y j ) δ / α j = 1 m λ ( x , d ( x , y j ) ) m ,

    provided that C d ( x , x ) α max 1 j m d ( x , y j ) and for each j ,

    (1.6) K ( x , y 1 , , y j , , y m ) K ( x , y 1 , , y j , , y m ) C d ( y j , y j ) δ j = 1 m d ( x , y j ) δ / α j = 1 m λ ( x , d ( x , y j ) ) m ,

    provided that C d ( y j , y j ) α max 1 j m d ( x , y j ) , where C is a positive constant.

A multilinear operator T is called a multilinear strongly singular integral operator with the above kernel K satisfying (1.4)–(1.6) if, for f 1 , , f m are L functions with bounded support and x j = 1 m supp f j ,

(1.7) T ( f 1 , , f m ) ( x ) = X m K ( x , y 1 , , y m ) f 1 ( y 1 ) f m ( y m ) d μ ( y 1 ) d μ ( y m ) .

Definition 1.7

[11] For any two balls B Q , let N B , Q be the smallest integer satisfying 6 N B , Q r B r Q and denote

(1.8) K B , Q = 1 + k = 1 N B , Q μ ( 6 k B ) λ ( x B , 6 k r B ) ,

where x B and r B denote center and radius of ball B , respectively.

Definition 1.8

[11] Let ρ > 1 be some fixed constant. A function b L loc 1 ( μ ) is said to belong to RBMO ( μ ) if there exists a constant C > 0 such that for any ball B ,

(1.9) 1 μ ( ρ B ) B b ( x ) m B ˜ ( b ) d μ ( x ) C ,

and for any two doubling balls B Q ,

(1.10) m B ( b ) m Q ( b ) C K B , Q ,

where B ˜ is the smallest ( α , β ) -doubling ball of the form 6 k B with k N { 0 } , and m B ˜ ( b ) is the mean value of b on B ˜ , namely,

m B ˜ ( b ) = 1 μ ( B ˜ ) B ˜ b ( x ) d μ ( x ) .

The minimal constant C appearing in (1.9) and (1.10) is defined to be the RBMO ( μ ) norm of b and denoted by b .

Let us introduce the definitions of two kinds of commutators, one can refer to [41,42].

Definition 1.9

Iterated commutator generated by multilinear strongly singular integral T with b = ( b 1 , , b m ) RBMO ( μ ) m is defined by

(1.11) T b ( f 1 , , f m ) = [ b 1 , [ b 2 , [ b m 1 , [ b m , T ] m ] m 1 ] 2 ] 1 ( f 1 , , f m ) .

In particular, when m = 2 , we obtain

(1.12) [ b 1 , b 2 , T ] ( f 1 , f 2 ) ( x ) = b 1 ( x ) b 2 ( x ) T ( f 1 , f 2 ) ( x ) b 1 ( x ) T ( f 1 , b 2 f 2 ) ( x ) b 2 ( x ) T ( b 1 f 1 , f 2 ) ( x ) + T ( b 1 f 1 , b 2 f 2 ) ( x ) .

Also, we define [ b 1 , T ] and [ b 2 , T ] as follows, respectively.

(1.13) [ b 1 , T ] ( f 1 , f 2 ) ( x ) = b 1 ( x ) T ( f 1 , f 2 ) ( x ) T ( b 1 f 1 , f 2 ) ( x ) ,

(1.14) [ b 2 , T ] ( f 1 , f 2 ) ( x ) = b 2 ( x ) T ( f 1 , f 2 ) ( x ) T ( f 1 , b 2 f 2 ) ( x ) .

Definition 1.10

The commutator in the summation form generated by multilinear strongly singular integral T with b = ( b 1 , , b m ) RBMO ( μ ) m is defined by

(1.15) T b ( f 1 , , f m ) = j = 1 m T b j j ( f 1 , , f m ) ,

where

T b j j ( f 1 , , f m ) = b j T ( f 1 , , f j , , f m ) T ( f 1 , , b j f j , , f m ) .

For the sake of simplicity and without loss of generality, we only consider the case of m = 2 in this paper. Let us state the main results as follows.

Theorem 1.11

Suppose that μ is a Radon measure with μ = . Let [ b 1 , b 2 , T ] be defined by (1.12). Let 1 < p 1 , p 2 , q < + , f 1 L p 1 ( μ ) , f 2 L p 2 ( μ ) , b 1 RBMO ( μ ) and b 2 RBMO ( μ ) . If T is bounded from L 1 ( μ ) × L 1 ( μ ) to L 1 / 2 , ( μ ) , then there exists a constant C > 0 such that

(1.16) [ b 1 , b 2 , T ] ( f 1 , f 2 ) L q ( μ ) C f 1 L p 1 ( μ ) f 2 L p 2 ( μ ) ,

where 1 q = 1 p 1 + 1 p 2 .

Theorem 1.12

Suppose that μ is a Radon measure with μ = . Let T b be defined by (1.15) with m = 2 . Let 1 < p 1 , p 2 , q < + , f 1 L p 1 ( μ ) , f 2 L p 2 ( μ ) , b 1 RBMO ( μ ) and b 2 RBMO ( μ ) . If T is bounded from L 1 ( μ ) × L 1 ( μ ) to L 1 / 2 , ( μ ) , then there exists a constant C > 0 such that

(1.17) T b ( f 1 , f 2 ) L q ( μ ) C f 1 L p 1 ( μ ) f 2 L p 2 ( μ ) ,

where 1 q = 1 p 1 + 1 p 2 .

Remark 1.13

For μ < , by Lemma 2.1 in Section 2, Theorem 1.11 also holds if one assumes that X G ( f 1 , f 2 ) ( x ) d μ ( x ) = 0 with the operator G be replaced by T , [ b 1 , T ] , [ b 2 , T ] and [ b 1 , b 2 , T ] . Also Theorem 1.12 holds if one assumes that X G ( f 1 , f 2 ) ( x ) d μ ( x ) = 0 with the operator G be replaced by T , [ b 1 , T ] and [ b 2 , T ] .

Remark 1.14

Since the proof of Theorem 1.12 is similar to that of Theorem 1.11, we only give the proof of Theorem 1.11 in this paper.

Throughout the paper, χ E denotes the characteristic function of set E . p is the conjugate index of p , namely, 1 p + 1 p = 1 . C denotes a positive constant independent of the main parameters involved, but it may be different in different places.

2 Proof of main result

To prove Theorem 1.11, we first give some notations and lemmas.

Let f L loc 1 ( μ ) , the sharp maximal operator is defined by

M f ( x ) = sup B x 1 μ ( 6 B ) B f ( y ) m B ˜ ( f ) d μ ( y ) + sup ( B , Q ) Δ x m B ( f ) m Q ( f ) K B , Q ,

where Δ x { ( B , Q ) : x B Q and B , Q are doubling balls } , and the noncentered doubling maximal operator is defined by

N f ( x ) = sup B x , B doubling 1 μ ( B ) B f ( y ) d μ ( y ) .

For any 0 < δ < 1 , we also define that

M δ f ( x ) = { M ( f δ ) ( x ) } 1 / δ

and

N δ f ( x ) = { N ( f δ ) ( x ) } 1 / δ .

We can obtain that for any f L loc 1 ( μ ) ,

f ( x ) N δ f ( x )

for μ a . e . x X .

Let ρ > 1 , p ( 1 , ) and r ( 1 , p ) , the noncentered maximal operator M r , ( ρ ) f is defined by

M r , ( ρ ) f ( x ) = sup B x 1 μ ( ρ B ) B f ( y ) r d μ ( y ) 1 / r ,

when r = 1 , we simply write M 1 , ( ρ ) f ( x ) as M ( ρ ) f . If ρ 5 , then the operator M ( ρ ) f is bounded on L p ( μ ) for p > 1 and M r , ( ρ ) is bounded on L p ( μ ) for p > r (see [11]).

Lemma 2.1

[28] Let f L loc 1 ( μ ) with X f ( x ) d μ ( x ) = 0 if μ < . For 1 < p < + and 0 < δ < 1 , if inf ( 1 , N δ f ) L p ( μ ) , then there exists a constant C > 0 such that

N δ ( f ) L p ( μ ) C M δ ( f ) L p ( μ ) .

Lemma 2.2

[13] Let 1 p < + and 1 < ρ < + . Then, b R B M O ( μ ) if and only if for any ball B X ,

1 μ ( ρ B ) B b B m B ˜ ( b ) p d μ ( x ) 1 / p C ,

and for any two doubling balls B Q ,

m B ( b ) m Q ( b ) C K B , Q .

Lemma 2.3

[13]

m 6 j 6 5 B ˜ ( b ) m B ˜ ( b ) C j b .

Lemma 2.4

[13] K ( B , Q ) has the following properties:

  1. For all balls B R Q , K ( B , R ) 2 K ( B , Q ) .

  2. For any ρ [ 1 , + ) , there exists a positive constant C ( ρ ) , depending on ρ , such that, for all balls B Q with r Q ρ r B , K ( B , Q ) C ( ρ ) .

  3. There exists a positive constant C such that, for all balls B , K ( B , B ˜ ) C .

  4. There exists a positive constant C such that, for all balls B R Q , K ( B , Q ) K ( B , R ) + C K ( R , Q ) .

  5. There exists a positive constant C such that, for all balls B R Q , K ( R , Q ) C K ( B , Q ) .

Lemma 2.5

[40] Suppose that μ is a Radon measure with μ = . Let T be defined by (1.7). Let 1 < p 1 , p 2 , q < + , f 1 L p 1 ( μ ) and f 2 L p 2 ( μ ) . If T is bounded from L 1 ( μ ) × L 1 ( μ ) to L 1 / 2 , ( μ ) , then there exists a constant C > 0 such that

T ( f 1 , f 2 ) L q ( μ ) C f 1 L p 1 ( μ ) f 2 L p 2 ( μ ) ,

where 1 q = 1 p 1 + 1 p 2 .

Lemma 2.6

Suppose that [ b 1 , b 2 , T ] is defined by (1.12), 0 < δ < 1 / 2 , 1 < p 1 , p 2 , q < + , 1 < r < q and b 1 , b 2 RBMO ( μ ) . If T is bounded from L 1 ( μ ) × L 1 ( μ ) to L 1 / 2 , ( μ ) , then there exists a constant C > 0 such that for any x X , any ball B with B x , f 1 L p 1 ( μ ) and f 2 L p 2 ( μ ) , we have the following two cases:

  1. When l = l ( B ) sup x , y B d ( x , y ) 1 , assume that B ˜ is the smallest ( α , β ) doubling ball of the form 6 k B with k N { 0 } , it follows

M δ [ b 1 , b 2 , T ] ( f 1 , f 2 ) ( x ) C b 1 b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) + C b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 , f 2 ) ) ( x ) + C b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 , f 2 ) ) ( x ) + C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) + C b 1 b 2 M r , ( 6 ) ( T ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) ( x ) + C b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) ( x ) + C b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) ( x ) , (2.1)

(2.2) M δ [ b 1 , T ] ( f 1 , f 2 ) ( x ) C b 1 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) + C b 1 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) + C b 1 M r , ( 6 ) ( T ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) ( x )

and

(2.3) M δ [ b 2 , T ] ( f 1 , f 2 ) ( x ) C b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) + C b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) + C b 2 M r , ( 6 ) ( T ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) ( x ) .

  1. When 0 < l < 1 , assume that B 0 , Q 0 and B ˜ 0 are concentric with B , Q and B ˜ , respectively, and l ( B 0 ) = l ( B ) α , l ( Q 0 ) = l ( Q ) α , l ( B ˜ 0 ) = l ( B ˜ ) α , it follows

M δ [ b 1 , b 2 , T ] ( f 1 , f 2 ) ( x ) C b 1 b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) + C b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 , f 2 ) ) ( x ) + C b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 , f 2 ) ) ( x ) + C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) + C b 1 b 2 M r , ( 6 ) ( T ( f 1 χ 6 5 B ˜ 0 , f 2 χ 6 5 B ˜ 0 ) ) ( x ) + C b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 χ 6 5 B ˜ 0 , f 2 χ 6 5 B ˜ 0 ) ) ( x ) + C b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 χ 6 5 B ˜ 0 , f 2 χ 6 5 B ˜ 0 ) ) ( x ) , (2.4)

(2.5) M δ [ b 1 , T ] ( f 1 , f 2 ) ( x ) C b 1 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) + C b 1 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) + C b 1 M r , ( 6 ) ( T ( f 1 χ 6 5 B ˜ 0 , f 2 χ 6 5 B ˜ 0 ) ) ( x )

and

(2.6) M δ [ b 2 , T ] ( f 1 , f 2 ) ( x ) C b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) + C b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) + C b 2 M r , ( 6 ) ( T ( f 1 χ 6 5 B ˜ 0 , f 2 χ 6 5 B ˜ 0 ) ) ( x ) .

Proof

Because L ( μ ) with bounded support is dense in L p ( μ ) for 1 < p < + , we only consider the situation of f 1 , f 2 L ( μ ) with bounded support. Also, by Corollary 3.11 in [6], without loss of generality, we can assume that b 1 , b 2 L ( μ ) . Next, we divide two cases for proving the result.

Case 1: l ( B ) = l 1 . As in the proof of Theorem 9.1 in [6], to obtain (2.1), it suffices to show that

(2.7) 1 μ ( 6 B ) B [ b 1 , b 2 , T ] ( f 1 , f 2 ) ( z ) δ h B δ d μ ( z ) 1 / δ C b 1 b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) + C b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 , f 2 ) ) ( x ) + C b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 , f 2 ) ) ( x ) + C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) ,

holds for any x and ball B with x B , and

(2.8) h B h Q C K B , Q 4 [ b 1 b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) + b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 , f 2 ) ) ( x ) + b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 , f 2 ) ) ( x ) + b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) + b 1 b 2 M r , ( 6 ) ( T ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) ( x ) + b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) ( x ) + b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) ( x )

for all balls B Q with x B , where B is an arbitrary ball, Q is a doubling ball, B ˜ is the smallest ( α , β ) doubling ball of the form 6 k B with k N { 0 } . We denote

h B m B T ( b 1 m B ˜ ( b 1 ) ) f 1 χ 6 5 B , ( b 2 m B ˜ ( b 2 ) ) f 2 χ X \ 6 5 B + T ( b 1 m B ˜ ( b 1 ) ) f 1 χ X \ 6 5 B , ( b 2 m B ˜ ( b 2 ) ) f 2 χ 6 5 B + T ( b 1 m B ˜ ( b 1 ) ) f 1 χ X \ 6 5 B , ( b 2 m B ˜ ( b 2 ) ) f 2 χ X \ 6 5 B

and

h Q m Q T ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 Q , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 Q + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 Q , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 Q + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 Q , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 Q .

It is easy to see that

[ b 1 , b 2 , T ] = T ( ( b 1 b 1 ( z ) ) f 1 , ( b 2 b 2 ( z ) ) f 2 )

and

T ( ( b 1 m B ˜ ( b 1 ) ) f 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 ) = T ( ( b 1 b 1 ( z ) + b 1 ( z ) m B ˜ ( b 1 ) ) f 1 , ( b 2 b 2 ( z ) + b 2 ( z ) m B ˜ ( b 2 ) ) f 2 ) = ( b 1 ( z ) m B ˜ ( b 1 ) ) ( b 2 ( z ) m B ˜ ( b 2 ) ) T ( f 1 , f 2 ) + ( b 1 ( z ) m B ˜ ( b 1 ) ) T ( f 1 , ( b 2 b 2 ( z ) ) f 2 ) + ( b 2 ( z ) m B ˜ ( b 2 ) ) T ( ( b 1 b 1 ( z ) ) f 1 , f 2 ) + T ( ( b 1 b 1 ( z ) ) f 1 , ( b 2 b 2 ( z ) ) f 2 ) .

Then,

1 μ ( 6 B ) B [ b 1 , b 2 , T ] ( f 1 , f 2 ) ( z ) δ h B δ d μ ( z ) 1 / δ C 1 μ ( 6 B ) B [ b 1 , b 2 , T ] ( f 1 , f 2 ) ( z ) h B δ d μ ( z ) 1 / δ C 1 μ ( 6 B ) B ( b 1 ( z ) m B ˜ ( b 1 ) ) ( b 2 ( z ) m B ˜ ( b 2 ) ) T ( f 1 , f 2 ) ( z ) δ d μ ( z ) 1 / δ + C 1 μ ( 6 B ) B ( b 1 ( z ) m B ˜ ( b 1 ) ) T ( f 1 , ( b 2 b 2 ( z ) ) f 2 ) ( z ) δ d μ ( z ) 1 / δ + C 1 μ ( 6 B ) B ( b 2 ( z ) m B ˜ ( b 2 ) ) T ( ( b 1 b 1 ( z ) ) f 1 , f 2 ) ( z ) δ d μ ( z ) 1 / δ + C 1 μ ( 6 B ) B T ( ( b 1 m B ˜ ( b 1 ) ) f 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 ) ( z ) h B δ d μ ( z ) 1 / δ E 1 + E 2 + E 3 + E 4 .

We first estimate E 1 . Choosing r 1 , r 2 > 1 such that 1 r + 1 r 1 + 1 r 2 = 1 δ . By Hölder’s inequality and Lemma 2.2, it follows

(2.9) E 1 C 1 μ ( 6 B ) B b 1 ( z ) m B ˜ ( b 1 ) r 1 d μ ( z ) 1 / r 1 1 μ ( 6 B ) B b 2 ( z ) m B ˜ ( b 2 ) r 2 d μ ( z ) 1 / r 2 × 1 μ ( 6 B ) B T ( f 1 , f 2 ) r d μ ( z ) 1 / r C b 1 b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) .

For E 2 , choosing s > 1 such that 1 s + 1 r = 1 δ , by Hölder’s inequality and Lemma 2.2, we obtain

(2.10) E 2 C 1 μ ( 6 B ) B b 1 ( z ) m B ˜ ( b 1 ) s d μ ( z ) 1 / s 1 μ ( 6 B ) B [ b 2 , T ] ( f 1 , f 2 ) r d μ ( z ) 1 / r C b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 , f 2 ) ) ( x ) .

Similar to estimate E 2 , we immediately have

(2.11) E 3 C b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 , f 2 ) ) ( x ) .

Let us turn to estimate E 4 . Denote f j 1 = f j χ 6 5 B and f j 2 = f j f j 1 for j = 1 , 2 , we have

T ( ( b 1 m B ˜ ( b 1 ) ) f 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 ) ( z ) h B T ( ( b 1 m B ˜ ( b 1 ) ) f 1 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 1 ) ( z ) + T ( ( b 1 m B ˜ ( b 1 ) ) f 1 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ( z ) m B [ T ( ( b 1 m B ˜ ( b 1 ) ) f 1 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ]

+ T ( ( b 1 m B ˜ ( b 1 ) ) f 1 2 , ( b 2 m B ˜ ( b 2 ) ) f 2 1 ) ( z ) m B [ T ( ( b 1 m B ˜ ( b 1 ) ) f 1 2 , ( b 2 m B ˜ ( b 2 ) ) f 2 1 ) ] + T ( ( b 1 m B ˜ ( b 1 ) ) f 1 2 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ( z ) m B [ T ( ( b 1 m B ˜ ( b 1 ) ) f 1 2 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ] E 41 ( z ) + E 42 ( z ) + E 43 ( z ) + E 44 ( z ) .

Then,

E 4 C 1 μ ( 6 B ) B E 41 ( z ) δ d μ ( z ) 1 / δ + C 1 μ ( 6 B ) B E 42 ( z ) δ d μ ( z ) 1 / δ + C 1 μ ( 6 B ) B E 43 ( z ) δ d μ ( z ) 1 / δ + C 1 μ ( 6 B ) B E 44 ( z ) δ d μ ( z ) 1 / δ E 41 + E 42 + E 43 + E 44 .

To estimate E 41 , we need the classical Kolmogorov’s theorem: Let ( X , μ ) be a probability measure space and let 0 < p < q < , then there exists a constant C > 0 , such that f L p ( μ ) C f L q , ( μ ) for any measurable function f . Let p = δ and q = 1 / 2 such that 0 < δ < 1 / 2 . Using Kolmogorov’s theorem, Lemmas 2.2 and Hölder’s inequality, we obtain

E 41 C T ( ( b 1 m B ˜ ( b 1 ) ) f 1 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 1 ) L 1 / 2 , 6 5 B , d μ ( z ) μ ( 6 B ) C 1 μ ( 6 B ) 6 5 B ( b 1 ( z ) m B ˜ ( b 1 ) ) f 1 ( z ) d μ ( z ) 1 μ ( 6 B ) 6 5 B ( b 2 ( z ) m B ˜ ( b 2 ) ) f 2 ( z ) d μ ( z ) C 1 μ ( 6 B ) 6 5 B b 1 ( z ) m B ˜ ( b 1 ) p 1 d μ ( z ) 1 / p 1 1 μ ( 6 B ) 6 5 B f 1 ( z ) p 1 d μ ( z ) 1 / p 1 × 1 μ ( 6 B ) 6 5 B b 2 ( z ) m B ˜ ( b 2 ) p 2 d μ ( z ) 1 / p 2 1 μ ( 6 B ) 6 5 B f 2 ( z ) p 2 d μ ( z ) 1 / p 2 C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

To compute E 42 , let z , y B , y 1 6 5 B and y 2 X \ 6 5 B , then max 1 i 2 d ( z , y i ) d ( z , y 2 ) C l ( B ) C l ( B ) α C d ( z , y ) α . By Definition 1.6, Lemmas 2.2, 2.3, Hölder’s inequality and the properties of λ , it follows

T ( ( b 1 m B ˜ ( b 1 ) ) f 1 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ( z ) T ( ( b 1 m B ˜ ( b 1 ) ) f 1 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ( y ) C X \ 6 5 B 6 5 B K ( z , y 1 , y 2 ) K ( y , y 1 , y 2 ) i = 1 2 ( b i ( y i ) m B ˜ ( b i ) ) f i ( y i ) d μ ( y 1 ) d μ ( y 2 ) C X \ 6 5 B 6 5 B d ( z , y ) δ i = 1 2 d ( z , y i ) δ / α i = 1 2 λ ( z , d ( z , y i ) ) 2 i = 1 2 ( b i ( y i ) m B ˜ ( b i ) ) f i ( y i ) d μ ( y 1 ) d μ ( y 2 ) C 6 5 B b 1 ( y 1 ) m B ˜ ( b 1 ) f 1 ( y 1 ) λ ( z , d ( z , y 1 ) ) d μ ( y 1 ) X \ 6 5 B d ( z , y ) δ d ( z , y 2 ) δ / α b 2 ( y 2 ) m B ˜ ( b 2 ) f 2 ( y 2 ) λ ( z , d ( z , y 2 ) ) d μ ( y 2 ) C μ ( 6 B ) λ ( x B , 6 5 r B ) 1 μ ( 6 B ) 6 5 B b 1 ( y 1 ) m B ˜ ( b 1 ) f 1 ( y 1 ) d μ ( y 1 ) k = 1 6 k 6 5 B \ 6 k 1 6 5 B d ( z , y ) δ d ( z , y 2 ) δ / α b 2 ( y 2 ) m B ˜ ( b 2 ) f 2 ( y 2 ) λ ( z , d ( z , y 2 ) ) d μ ( y 2 )

C 1 μ ( 6 B ) 6 5 B b 1 ( y 1 ) m B ˜ ( b 1 ) p 1 d μ ( y 1 ) 1 / p 1 1 μ ( 6 B ) 6 5 B f 1 ( y 1 ) p 1 d μ ( y 1 ) 1 / p 1 × k = 1 1 λ ( x B , 6 k 1 6 5 r B ) 6 k δ / α l δ ( 1 1 / α ) 6 k 6 5 B b 2 ( y 2 ) m B ˜ ( b 2 ) f 2 ( y 2 ) d μ ( y 2 ) C b 1 M p 1 , ( 5 ) f 1 ( x ) k = 1 6 k δ / α μ 5 × 6 k 6 5 B λ x B , 6 k 1 6 5 r B 1 μ 5 × 6 k 6 5 B × 6 k 6 5 B b 2 ( y 2 ) m 6 k 6 5 B ˜ ( b 2 ) + m 6 k 6 5 B ˜ ( b 2 ) m B ˜ ( b 2 ) f 2 ( y 2 ) d μ ( y 2 ) C b 1 M p 1 , ( 5 ) f 1 ( x ) k = 1 6 k δ / α 1 μ 5 × 6 k 6 5 B 6 k 6 5 B b 2 ( y 2 ) m 6 k 6 5 B ˜ ( b 2 ) p 2 d μ ( y 2 ) 1 / p 2 × 1 μ 5 × 6 k 6 5 B 6 k 6 5 B f 2 ( y 2 ) p 2 d μ ( y 2 ) 1 / p 2 + C k b 2 1 μ 5 × 6 k 6 5 B 6 k 6 5 B f 2 ( y 2 ) d μ ( y 2 ) C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Taking the mean over y B , we have

E 42 ( z ) C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Therefore,

E 42 C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Similarly, we get

E 43 C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

For E 44 , by Definition 1.6, Lemma 2.2, Hölder’s inequality and the properties of λ , we obtain

T ( ( b 1 m B ˜ ( b 1 ) ) f 1 2 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ( z ) T ( ( b 1 m B ˜ ( b 1 ) ) f 1 2 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ( y ) C X \ 6 5 B X \ 6 5 B K ( z , y 1 , y 2 ) K ( y , y 1 , y 2 ) i = 1 2 ( b i ( y i ) m B ˜ ( b i ) ) f i ( y i ) d μ ( y 1 ) d μ ( y 2 ) C X \ 6 5 B X \ 6 5 B d ( z , y ) δ i = 1 2 ( b i ( y i ) m B ˜ ( b i ) ) f i ( y i ) d μ ( y 1 ) d μ ( y 2 ) ( d ( z , y 1 ) + d ( z , y 2 ) ) δ / α j = 1 2 λ ( z , d ( z , y j ) ) 2 C i = 1 2 X \ 6 5 B d ( z , y ) δ i b i ( y i ) m B ˜ ( b i ) f i ( y i ) d μ ( y i ) d ( z , y i ) δ i / α λ ( z , d ( z , y i ) ) C i = 1 2 k = 1 6 k δ i / α l δ ( 1 1 / α ) μ 5 × 6 k 6 5 B λ ( x B , 6 k 1 6 5 r B ) 1 μ 5 × 6 k 6 5 B 6 k 6 5 B b i ( y i ) m B ˜ ( b i ) f i ( y i ) d μ ( y i ) C i = 1 2 k = 1 6 k δ i / α 1 μ 5 × 6 k 6 5 B 6 k 6 5 B b i ( y i ) m B ˜ ( b i ) p i d μ ( y i ) 1 / p i 1 μ 5 × 6 k 6 5 B 6 k 6 5 B f i ( y i ) p i 1 / p i

C i = 1 2 k = 1 6 k δ i / α M p i , ( 5 ) f i ( x ) 1 μ 5 × 6 k 6 5 B 6 k 6 5 B b i ( y i ) m 6 k 6 5 B ˜ ( b i ) + m 6 k 6 5 B ˜ ( b i ) m B ˜ ( b i ) p i d μ ( y i ) 1 / p i C i = 1 2 k = 1 k 6 k δ i / α b i M p i , ( 5 ) f i ( x ) C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

where δ 1 , δ 2 > 0 and δ 1 + δ 2 = δ .

Taking the mean over y B , then

E 44 ( z ) C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Thus,

E 44 C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

So (2.7) can be obtained.

Next, we prove (2.8). Consider two balls B Q with x B , where B is an arbitrary ball and Q is a doubling ball. Recall that B ˜ is the smallest ( α , β ) doubling ball of the form 6 k B with k N { 0 } . Now we have the two cases: B Q B ˜ or B B ˜ Q . By Lemma 2.4, we can obtain the following facts. If B Q B ˜ , we have K Q , B ˜ C K B , B ˜ C . If B B ˜ Q , we have K B ˜ , Q C K B , Q . Without loss of generality, we only consider B B ˜ Q in this paper. The another case can be considered by the same method.

Let N = N B ˜ , Q + 1 . It is easy to see that

h B h Q h B h B ˜ + h B ˜ h Q .

Now we first consider h B ˜ h Q . We write

h B ˜ h Q m B ˜ T ( b 1 m B ˜ ( b 1 ) ) f 1 χ 6 5 B ˜ , ( b 2 m B ˜ ( b 2 ) ) f 2 χ X \ 6 5 B ˜ + T ( b 1 m B ˜ ( b 1 ) ) f 1 χ X \ 6 5 B ˜ , × ( b 2 m B ˜ ( b 2 ) ) f 2 χ 6 5 B ˜ + T ( b 1 m B ˜ ( b 1 ) ) f 1 χ X \ 6 5 B ˜ , ( b 2 m B ˜ ( b 2 ) ) f 2 χ X \ 6 5 B ˜ m B ˜ T ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 B ˜ + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 B ˜ + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 B ˜ + m B ˜ T ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 B ˜ + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 B ˜ + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 B ˜ m Q T ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 Q , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 Q + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 Q , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 Q + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 Q , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 Q F 1 + F 2 .

To estimate F 1 , recall the fact that

T ( f 1 , f 2 ) = T f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ + T f 1 χ 6 5 B ˜ , f 2 χ X \ 6 5 B ˜ + T f 1 χ X \ 6 5 B ˜ , f 2 χ 6 5 B ˜ + T f 1 χ X \ 6 5 B ˜ , f 2 χ X \ 6 5 B ˜ ,

then

F 1 m B ˜ T ( ( b 1 m B ˜ ( b 1 ) ) f 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 ) T ( ( b 1 m B ˜ ( b 1 ) ) f 1 χ 6 5 B ˜ , ( b 2 m B ˜ ( b 2 ) ) f 2 χ 6 5 B ˜ ) m B ˜ T ( ( b 1 m Q ( b 1 ) ) f 1 , ( b 2 m Q ( b 2 ) ) f 2 ) T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 B ˜ )

m B ˜ [ T ( ( b 1 m B ˜ ( b 1 ) ) f 1 , ( b 2 m B ˜ ˜ ( b 2 ) ) f 2 ) T ( ( b 1 m Q ( b 1 ) ) f 1 , ( b 2 m Q ( b 2 ) ) f 2 ) ] + m B ˜ T ( b 1 m B ˜ ( b 1 ) ) f 1 χ 6 5 B ˜ , ( b 2 m B ˜ ( b 2 ) ) f 2 χ 6 5 B ˜ T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 B ˜ ) F 11 + F 12 .

For F 11 , recall the fact that

T ( ( b 1 m B ˜ ( b 1 ) ) f 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 ) = T ( ( b 1 b 1 ( z ) + b 1 ( z ) m B ˜ ( b 1 ) ) f 1 , ( b 2 b 2 ( z ) + b 2 ( z ) m B ˜ ( b 2 ) ) f 2 ) = ( b 1 ( z ) m B ˜ ( b 1 ) ) ( b 2 ( z ) m B ˜ ( b 2 ) ) T ( f 1 , f 2 ) + ( b 1 ( z ) m B ˜ ( b 1 ) ) T ( f 1 , ( b 2 b 2 ( z ) ) f 2 ) + ( b 2 ( z ) m B ˜ ( b 2 ) ) T ( ( b 1 b 1 ( z ) ) f 1 , f 2 ) + T ( ( b 1 b 1 ( z ) ) f 1 , ( b 2 b 2 ( z ) ) f 2 ) = ( b 1 ( z ) m B ˜ ( b 1 ) ) ( b 2 ( z ) m B ˜ ( b 2 ) ) T ( f 1 , f 2 ) ( b 1 ( z ) m B ˜ ( b 1 ) ) [ b 2 , T ] ( f 1 , f 2 ) ( b 2 ( z ) m B ˜ ( b 2 ) ) [ b 1 , T ] ( f 1 , f 2 ) + [ b 1 , b 2 , T ] ( f 1 , f 2 ) .

Then,

F 11 m B ˜ [ ( b 1 ( z ) m B ˜ ( b 1 ) ) ( b 2 ( z ) m B ˜ ( b 2 ) ) T ( f 1 , f 2 ) ] + m B ˜ [ ( b 1 ( z ) m B ˜ ( b 1 ) ) [ b 2 , T ] ( f 1 , f 2 ) ] + m B ˜ [ ( b 2 ( z ) m B ˜ ( b 2 ) ) [ b 1 , T ] ( f 1 , f 2 ) ] + m B ˜ [ ( b 1 ( z ) m Q ( b 1 ) ) ( b 2 ( z ) m Q ( b 2 ) ) T ( f 1 , f 2 ) ] + m B ˜ [ ( b 1 ( z ) m Q ( b 1 ) ) [ b 2 , T ] ( f 1 , f 2 ) ] + m B ˜ [ ( b 2 ( z ) m Q ( b 2 ) ) [ b 1 , T ] ( f 1 , f 2 ) ] J 1 + J 2 + J 3 + J 4 + J 5 + J 6 .

Recall that K B ˜ , Q C K B , Q with B B ˜ Q . Note that B ˜ is a doubling ball, similar to estimate for (2.9), (2.10) and (2.11), respectively, we immediately obtain

J 1 + J 4 C K B , Q 2 b 1 b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) , J 2 + J 5 C K B , Q b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 , f 2 ) ) ( x ) , J 3 + J 6 C K B , Q b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 , f 2 ) ) ( x ) .

For F 12 , with the same method to estimate F 11 , we have

F 12 C K B , Q 2 b 1 b 2 M r , ( 6 ) ( T ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) ( x ) + C K B , Q b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) ( x ) + C K B , Q b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) ( x ) .

Combining the estimates of F 11 and F 12 , we complete the estimate for F 1 .

Now we turn to estimate F 2 . By decomposing the region of the integral, we have

F 2 m B ˜ { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ \ 6 5 B ˜ ) } + m B ˜ { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ \ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 B ˜ ) } + m B ˜ { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ \ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ \ 6 5 B ˜ ) } + m B ˜ { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 N B ˜ ) } m Q { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 N B ˜ ) } + m B ˜ { T ( ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 N B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ ) } m Q { T ( ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 N B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ ) }

+ m B ˜ { T ( ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 N B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 N B ˜ ) } m Q { T ( ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 N B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 N B ˜ ) } + m Q { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 Q , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ \ 6 5 Q ) } + m Q { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ \ 6 5 Q , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 Q ) } + m Q { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ \ 6 5 Q , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ \ 6 5 Q ) } i = 1 9 F 2 i .

For F 21 , we first compute T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ \ 6 5 B ˜ ) ( z ) .

T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ \ 6 5 B ˜ ) ( z ) C 6 N B ˜ \ 6 5 B ˜ 6 5 B ˜ ( b 1 ( y 1 ) m Q ( b 1 ) ) f 1 ( y 1 ) ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) i = 1 2 λ ( z , d ( z , y i ) ) 2 d μ ( y 1 ) d μ ( y 2 ) C 6 5 B ˜ ( b 1 ( y 1 ) m Q ( b 1 ) ) f 1 ( y 1 ) λ ( z , d ( z , y 1 ) ) d μ ( y 1 ) k = 1 N 1 6 k + 1 B ˜ \ 6 k B ˜ ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) λ ( z , d ( z , y 2 ) ) d μ ( y 2 ) + 6 B ˜ \ 6 5 B ˜ ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) λ ( z , d ( z , y 2 ) ) d μ ( y 2 ) C μ ( 6 B ˜ ) λ ( x B ˜ , 6 5 r B ˜ ) 1 μ ( 6 B ˜ ) 6 5 B ˜ ( b 1 ( y 1 ) m Q ( b 1 ) ) f 1 ( y 1 ) d μ ( y 1 ) k = 1 N 1 μ ( 5 × 6 k + 1 B ˜ ) λ ( x B ˜ , 6 k r B ˜ ) 1 μ ( 5 × 6 k + 1 B ˜ ) 6 k + 1 B ˜ ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) d μ ( y 2 ) + μ ( 5 × 6 B ˜ ) λ ( x B ˜ , 6 r B ˜ ) 1 μ ( 5 × 6 B ˜ ) 6 B ˜ ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) d μ ( y 2 ) C K B ˜ , Q 3 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) C K B , Q 3 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) ,

here we have used the property K B ˜ , Q C K B , Q with B B ˜ Q .

Taking the mean over z B ˜ , we obtain

F 21 C K B , Q 3 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Using the similar method to estimate F 21 , we immediately have

F 22 + F 27 + F 28 C K B , Q 3 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Now we estimate F 23 .

T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ \ 6 5 B ˜ , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ \ 6 5 B ˜ ) ( z ) C 6 N B ˜ \ 6 5 B ˜ 6 N B ˜ \ 6 5 B ˜ ( b 1 ( y 1 ) m Q ( b 1 ) ) f 1 ( y 1 ) ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) i = 1 2 λ ( z , d ( z , y i ) ) 2 d μ ( y 1 ) d μ ( y 2 ) C j = 1 N 1 k = 1 N 1 6 j + 1 B ˜ \ 6 j B ˜ 6 k + 1 B ˜ \ 6 k B ˜ ( b 1 ( y 1 ) m Q ( b 1 ) ) f 1 ( y 1 ) ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) i = 1 2 λ ( z , d ( z , y i ) ) 2 d μ ( y 1 ) d μ ( y 2 ) + 6 B ˜ \ 6 5 B ˜ 6 B ˜ \ 6 5 B ˜ ( b 1 ( y 1 ) m Q ( b 1 ) ) f 1 ( y 1 ) ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) i = 1 2 λ ( z , d ( z , y i ) ) 2 d μ ( y 1 ) d μ ( y 2 )

C k = 1 N 1 μ ( 5 × 6 k + 1 B ˜ ) λ ( x B ˜ , 6 k r B ˜ ) 1 μ ( 5 × 6 k + 1 B ˜ ) 6 k + 1 B ˜ ( b 1 ( y 1 ) m Q ( b 1 ) ) f 1 ( y 1 ) d μ ( y 1 ) × j = 1 N 1 μ ( 5 × 6 j + 1 B ˜ ) λ ( x B ˜ , 6 j r B ˜ ) 1 μ ( 5 × 6 j + 1 B ˜ ) 6 j + 1 B ˜ ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) d μ ( y 2 ) + μ ( 30 B ˜ ) λ ( x B ˜ , 6 r B ˜ ) 2 i = 1 2 1 μ ( 30 B ˜ ) 6 B ˜ ( b i ( y i ) m Q ( b i ) ) f i ( y i ) d μ ( y i ) C K B ˜ , Q 4 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) . C K B , Q 4 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Taking the mean over z B ˜ , we obtain

F 23 C K B , Q 4 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

For F 29 , with the similar method to estimate F 23 , we obtain

F 29 C K B , Q 4 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Using the similar method to estimate E 42 ( z ) , it follows

F 24 + F 25 C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Using the similar method to estimate E 44 ( z ) , we have

F 26 C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Next we consider h B h B ˜ . With the similar method to estimate F 2 , we easily obtain that

h B h B ˜ = m B T ( b 1 m B ˜ ( b 1 ) ) f 1 χ 6 5 B , ( b 2 m B ˜ ( b 2 ) ) f 2 χ X \ 6 5 B + T ( b 1 m B ˜ ( b 1 ) ) f 1 χ X \ 6 5 B , ( b 2 m B ˜ ( b 2 ) ) f 2 χ 6 5 B + T ( b 1 m B ˜ ( b 1 ) ) f 1 χ X \ 6 5 B , ( b 2 m B ˜ ( b 2 ) ) f 2 χ X \ 6 5 B m B ˜ T ( b 1 m B ˜ ( b 1 ) ) f 1 χ 6 5 B ˜ , ( b 2 m B ˜ ( b 2 ) ) f 2 χ X \ 6 5 B ˜ + T ( b 1 m B ˜ ( b 1 ) ) f 1 χ X \ 6 5 B ˜ , ( b 2 m B ˜ ( b 2 ) ) f 2 χ 6 5 B ˜ + T ( b 1 m B ˜ ( b 1 ) ) f 1 χ X \ 6 5 B ˜ , ( b 2 m B ˜ ( b 2 ) ) f 2 χ X \ 6 5 B ˜ C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Thus, (2.8) holds, and hence, (2.1) is proved. With the same method to prove (2.1), we can obtain that (2.2) and (2.3) also hold. Here, we omit the details. Thus, Lemma 2.6 in this case is proved.

Case 2: 0 < l ( B ) = l < 1 . Recall that B ˜ is the smallest ( α , β ) doubling ball of the form 6 k B with k N { 0 } . Assume that B 0 , Q 0 and B ˜ 0 are concentric with B , Q and B ˜ , respectively, and l ( B 0 ) = l ( B ) α , l ( Q 0 ) = l ( Q ) α , l ( B ˜ 0 ) = l ( B ˜ ) α , then B B 0 , Q Q 0 and B ˜ B ˜ 0 . Let ρ be a number such that 6 B 0 = ρ B . As in the proof of Theorem 9.1 in [6], to obtain (2.4), it suffices to show that

(2.12) 1 μ ( 6 B 0 ) B [ b 1 , b 2 , T ] ( f 1 , f 2 ) ( z ) δ h B δ d μ ( z ) 1 / δ C b 1 b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) + C b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 , f 2 ) ) ( x ) + C b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 , f 2 ) ) ( x ) + C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x )

holds for any x and ball B with x B , and

(2.13) h B h Q C K B , Q 4 [ b 1 b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) + b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 , f 2 ) ) ( x ) + b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 , f 2 ) ) ( x ) + b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) + b 1 b 2 M r , ( 6 ) ( T ( f 1 χ 6 5 B ˜ 0 , f 2 χ 6 5 B ˜ 0 ) ) ( x ) + b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 χ 6 5 B ˜ 0 , f 2 χ 6 5 B ˜ 0 ) ) ( x ) + b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 χ 6 5 B ˜ 0 , f 2 χ 6 5 B ˜ 0 ) ) ( x )

for all balls B Q with x B , where B is an arbitrary ball, Q is a doubling ball. For any ball B , we denote

h B m B T ( b 1 m B ˜ ( b 1 ) ) f 1 χ 6 5 B 0 , ( b 2 m B ˜ ( b 2 ) ) f 2 χ X \ 6 5 B 0 + T ( b 1 m B ˜ ( b 1 ) ) f 1 χ X \ 6 5 B 0 , ( b 2 m B ˜ ( b 2 ) ) f 2 χ 6 5 B 0 + T ( b 1 m B ˜ ( b 1 ) ) f 1 χ X \ 6 5 B 0 , ( b 2 m B ˜ ( b 2 ) ) f 2 χ X \ 6 5 B 0

and

h Q m Q T ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 Q 0 , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 Q 0 + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 Q 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 Q 0 + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 Q 0 , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 Q 0 .

Then,

1 μ ( 6 B 0 ) B [ b 1 , b 2 , T ] ( f 1 , f 2 ) ( z ) δ h B δ d μ ( z ) 1 / δ C 1 μ ( 6 B 0 ) B [ b 1 , b 2 , T ] ( f 1 , f 2 ) ( z ) h B δ d μ ( z ) 1 / δ C 1 μ ( 6 B 0 ) B ( b 1 ( z ) m B ˜ ( b 1 ) ) ( b 2 ( z ) m B ˜ ( b 2 ) ) T ( f 1 , f 2 ) ( z ) δ d μ ( z ) 1 / δ + C 1 μ ( 6 B 0 ) B ( b 1 ( z ) m B ˜ ( b 1 ) ) T ( f 1 , ( b 2 b 2 ( z ) ) f 2 ) ( z ) δ d μ ( z ) 1 / δ + C 1 μ ( 6 B 0 ) B ( b 2 ( z ) m B ˜ ( b 2 ) ) T ( ( b 1 b 1 ( z ) ) f 1 , f 2 ) ( z ) δ d μ ( z ) 1 / δ + C 1 μ ( 6 B 0 ) B T ( ( b 1 m B ˜ ( b 1 ) ) f 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 ) ( z ) h B δ d μ ( z ) 1 / δ G 1 + G 2 + G 3 + G 4 .

We first estimate G 1 . Choosing r 1 , r 2 > 1 such that 1 r + 1 r 1 + 1 r 2 = 1 δ . By Hölder’s inequality, Lemma 2.2 and the fact B B 0 , we have

G 1 C 1 μ ( 6 B 0 ) B b 1 ( z ) m B ˜ b 1 r 1 d μ ( z ) 1 / r 1 1 μ ( 6 B 0 ) B b 2 ( z ) m B ˜ b 2 r 2 d μ ( z ) 1 / r 2 × 1 μ ( 6 B 0 ) B 0 T ( f 1 , f 2 ) r d μ ( z ) 1 / r C b 1 b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) .

For G 2 , let s > 1 such that 1 s + 1 r = 1 δ , by Hölder’s inequality, Lemma 2.2 and the fact B B 0 , we deduce

G 2 C 1 μ ( 6 B 0 ) B b 1 ( z ) m B ˜ b 1 s d μ ( z ) 1 / s 1 μ ( 6 B 0 ) B 0 [ b 2 , T ] ( f 1 , f 2 ) r d μ ( z ) 1 / r C b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 , f 2 ) ) ( x ) .

Similar to estimate G 2 , we immediately obtain

G 3 C b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 , f 2 ) ) ( x ) .

Let us turn to estimate G 4 . Denote f j 1 = f j χ 6 5 B 0 and f j 2 = f j f j 1 for j = 1 , 2 , we have

T ( ( b 1 m B ˜ ( b 1 ) ) f 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 ) ( z ) h B T ( ( b 1 m B ˜ ( b 1 ) ) f 1 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 1 ) ( z ) + T ( ( b 1 m B ˜ ( b 1 ) ) f 1 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ( z ) m B [ T ( ( b 1 m B ˜ ( b 1 ) ) f 1 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ] + T ( ( b 1 m B ˜ ( b 1 ) ) f 1 2 , ( b 2 m B ˜ ( b 2 ) ) f 2 1 ) ( z ) m B [ T ( ( b 1 m B ˜ ( b 1 ) ) f 1 2 , ( b 2 m B ˜ ( b 2 ) ) f 2 1 ) ] + T ( ( b 1 m B ˜ ( b 1 ) ) f 1 2 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ( z ) m B [ T ( ( b 1 m B ˜ ( b 1 ) ) f 1 2 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ] G 41 ( z ) + G 42 ( z ) + G 43 ( z ) + G 44 ( z ) .

Then,

G 4 C 1 μ ( 6 B 0 ) B E 41 ( z ) δ d μ ( z ) 1 / δ + C 1 μ ( 6 B 0 ) B E 42 ( z ) δ d μ ( z ) 1 / δ + C 1 μ ( 6 B 0 ) B E 43 ( z ) δ d μ ( z ) 1 / δ + C 1 μ ( 6 B 0 ) B E 44 ( z ) δ d μ ( z ) 1 / δ G 41 + G 42 + G 43 + G 44 .

To estimate G 41 , using Kolmogorov’s theorem, Lemma 2.2 and Hölder’s inequality, we obtain

G 41 C T ( ( b 1 m B ˜ ( b 1 ) ) f 1 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 1 ) L 1 / 2 , ( 6 5 B 0 , d μ ( z ) μ ( 6 B 0 ) ) C 1 μ ( 6 B 0 ) 6 5 B 0 ( b 1 ( z ) m B ˜ ( b 1 ) ) f 1 ( z ) d μ ( z ) 1 μ ( 6 B 0 ) 6 5 B 0 ( b 2 ( z ) m B ˜ ( b 2 ) ) f 2 ( z ) d μ ( z ) C 1 μ ( 6 B 0 ) 6 5 B 0 b 1 ( z ) m B ˜ ( b 1 ) p 1 d μ ( z ) 1 / p 1 1 μ ( 6 B 0 ) 6 5 B 0 f 1 ( z ) p 1 d μ ( z ) 1 / p 1 × 1 μ ( 6 B 0 ) 6 5 B 0 b 2 ( z ) m B ˜ ( b 2 ) p 2 d μ ( z ) 1 / p 2 1 μ ( 6 B 0 ) 6 5 B 0 f 2 ( z ) p 2 d μ ( z ) 1 / p 2 C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

To compute G 42 , let z , y B , y 1 6 5 B 0 and y 2 X \ 6 5 B 0 , then max 1 i 2 d ( z , y i ) d ( z , y 2 ) C l ( B 0 ) = C l ( B ) α C d ( z , y ) α . Using Definition 1.6, Lemmas 2.2, 2.3, Hölder’s inequality and the properties of λ , we know

T ( ( b 1 m B ˜ ( b 1 ) ) f 1 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ( z ) T ( ( b 1 m B ˜ ( b 1 ) ) f 1 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ( y ) C X \ 6 5 B 0 6 5 B 0 K ( z , y 1 , y 2 ) K ( y , y 1 , y 2 ) i = 1 2 ( b i ( y i ) m B ˜ ( b i ) ) f i ( y i ) d μ ( y 1 ) d μ ( y 2 )

C X \ 6 5 B 0 6 5 B 0 d ( z , y ) δ i = 1 2 d ( z , y i ) δ / α i = 1 2 λ ( z , d ( z , y i ) ) 2 i = 1 2 ( b i ( y i ) m B ˜ ( b i ) ) f i ( y i ) d μ ( y 1 ) d μ ( y 2 ) C 6 5 B 0 b 1 ( y 1 ) m B ˜ ( b 1 ) f 1 ( y 1 ) λ ( z , d ( z , y 1 ) ) d μ ( y 1 ) X \ 6 5 B 0 d ( z , y ) δ d ( z , y 2 ) δ / α b 2 ( y 2 ) m B ˜ ( b 2 ) f 2 ( y 2 ) λ ( z , d ( z , y 2 ) ) d μ ( y 2 ) C μ ( 6 B 0 ) λ ( x B , 6 5 r B 0 ) 1 μ ( 6 B 0 ) 6 5 B 0 b 1 ( y 1 ) m B ˜ ( b 1 ) f 1 ( y 1 ) d μ ( y 1 ) k = 1 6 k 6 5 B 0 \ 6 k 1 6 5 B 0 d ( z , y ) δ d ( z , y 2 ) δ / α b 2 ( y 2 ) m B ˜ ( b 2 ) f 2 ( y 2 ) λ ( z , d ( z , y 2 ) ) d μ ( y 2 ) C 1 μ ( 6 B 0 ) 6 5 B 0 b 1 ( y 1 ) m B ˜ ( b 1 ) p 1 d μ ( y 1 ) 1 / p 1 1 μ ( 6 B 0 ) 6 5 B 0 f 1 ( y 1 ) p 1 d μ ( y 1 ) 1 / p 1 × k = 1 1 λ ( x B , 6 k 1 6 5 r B 0 ) 6 k δ / α 6 k 6 5 B 0 b 2 ( y 2 ) m B ˜ ( b 2 ) f 2 ( y 2 ) d μ ( y 2 ) C b 1 M p 1 , ( 5 ) f 1 ( x ) k = 1 6 k δ / α 1 μ ( 5 × 6 k 6 5 B 0 ) 6 k 6 5 B 0 b 2 ( y 2 ) m 6 k 6 5 B 0 ˜ ( b 2 ) p 2 d μ ( y 2 ) 1 / p 2 × 1 μ ( 5 × 6 k 6 5 B 0 ) 6 k 6 5 B 0 f 2 ( y 2 ) p 2 d μ ( y 2 ) 1 / p 2 + C k b 2 1 μ ( 5 × 6 k 6 5 B 0 ) 6 k 6 5 B 0 f 2 ( y 2 ) d μ ( y 2 ) C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Taking the mean over y B , we have

G 42 ( z ) C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Thus,

G 42 C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Similarly, we get

G 43 C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

For G 44 , by Definition 1.6, Lemmas 2.2, 2.3, Hölder’s inequality and the properties of λ , we obtain

T ( ( b 1 m B ˜ ( b 1 ) ) f 1 2 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ( z ) T ( ( b 1 m B ˜ ( b 1 ) ) f 1 2 , ( b 2 m B ˜ ( b 2 ) ) f 2 2 ) ( y ) C X \ 6 5 B 0 X \ 6 5 B 0 K ( z , y 1 , y 2 ) K ( y , y 1 , y 2 ) i = 1 2 ( b i ( y i ) m B ˜ b i ) f i ( y i ) d μ ( y 1 ) d μ ( y 2 ) C X \ 6 5 B 0 X \ 6 5 B 0 d ( z , y ) δ i = 1 2 ( b i ( y i ) m B ˜ b i ) f i ( y i ) d μ ( y 1 ) d μ ( y 2 ) ( d ( z , y 1 ) + d ( z , y 2 ) ) δ / α [ j = 1 2 λ ( z , d ( z , y j ) ) ] 2 C i = 1 2 X \ 6 5 B 0 d ( z , y ) δ i b i ( y i ) m B ˜ b i f i ( y i ) d μ ( y i ) d ( z , y i ) δ i / α λ ( x B , d ( z , y i ) ) C i = 1 2 k = 1 6 k 6 5 B 0 6 k δ i / α μ ( 5 × 6 k 6 5 B 0 ) λ ( x B , 5 × 6 k 6 5 r B 0 ) 1 μ ( 5 × 6 k 6 5 B 0 ) b i ( y i ) m B ˜ ( b i ) f i ( y i ) d μ ( y i )

C i = 1 2 k = 1 6 k δ i / α 1 μ ( 5 × 6 k 6 5 B 0 ) 6 k 6 5 B 0 b i ( y i ) m B ˜ ( b i ) p i d μ ( y i ) 1 / p i 1 μ ( 5 × 6 k 6 5 B 0 ) 6 k 6 5 B 0 f i ( y i ) p i 1 / p i C i = 1 2 k = 1 6 k δ i / α M p i , ( 5 ) f i ( x ) ( 1 μ ( 5 × 6 k 6 5 B 0 ) 6 k 6 5 B 0 b i ( y i ) m 6 k 6 5 B 0 ˜ ( b i ) + m 6 k 6 5 B 0 ˜ ( b i ) m B ˜ ( b i ) p i d μ ( y i ) ) 1 / p i C i = 1 2 k = 1 6 k δ i / α k b i M p i , ( 5 ) f i ( x ) C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) ,

where δ 1 , δ 2 > 0 and δ 1 + δ 2 = δ .

Taking the mean over y B , then

G 44 ( z ) C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Therefore,

G 44 C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

So (2.12) can be obtained.

Next, we prove (2.13). Consider two balls B Q with x B , where B is an arbitrary ball and Q is a doubling ball. Assume that B 0 , Q 0 and B ˜ 0 are concentric with B , Q and B ˜ , respectively, and l ( B 0 ) = l ( B ) α , l ( Q 0 ) = l ( Q ) α , l ( B ˜ 0 ) = l ( B ˜ ) α , then B B 0 , Q Q 0 and B ˜ B ˜ 0 . Thus, we have the four cases: B Q Q 0 B ˜ B ˜ 0 , B Q B ˜ Q 0 B ˜ 0 , B B ˜ B ˜ 0 Q Q 0 , B B ˜ Q B ˜ 0 Q 0 . Without loss of generality, we only consider B B ˜ B ˜ 0 Q Q 0 in this paper. The other cases can be considered by the same method.

Let N = N B ˜ 0 , Q 0 + 1 . It is easy to see that

h B h Q h B h B ˜ + h B ˜ h Q .

Now we first consider h B ˜ h Q . Write

h B ˜ h Q m B ˜ T ( b 1 m B ˜ ( b 1 ) ) f 1 χ 6 5 B ˜ 0 , ( b 2 m B ˜ ( b 2 ) ) f 2 χ X \ 6 5 B ˜ 0 + T ( b 1 m B ˜ ( b 1 ) ) f 1 χ X \ 6 5 B ˜ 0 , ( b 2 m B ˜ ( b 2 ) ) f 2 χ 6 5 B ˜ 0 + T ( b 1 m B ˜ ( b 1 ) ) f 1 χ X \ 6 5 B ˜ 0 , ( b 2 m B ˜ ( b 2 ) ) f 2 χ X \ 6 5 B ˜ 0 m B ˜ T ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 B ˜ 0 + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 B ˜ 0 + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 B ˜ 0 + m B ˜ T ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 B ˜ 0 + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 B ˜ 0 + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 B ˜ 0 m Q T ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 Q 0 , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 Q 0 + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 Q 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 Q 0 + T ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 5 Q 0 , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 5 Q 0 H 1 + H 2 .

To estimate H 1 , write

H 1 m B ˜ T ( ( b 1 m B ˜ ( b 1 ) ) f 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 ) T ( ( b 1 m B ˜ ( b 1 ) ) f 1 χ 6 5 B ˜ 0 , ( b 2 m B ˜ ( b 2 ) ) f 2 χ 6 5 B ˜ 0 ) m B ˜ T ( ( b 1 m Q ( b 1 ) ) f 1 , ( b 2 m Q ( b 2 ) ) f 2 ) T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 B ˜ 0 ) m B ˜ [ T ( ( b 1 m B ˜ ( b 1 ) ) f 1 , ( b 2 m B ˜ ( b 2 ) ) f 2 ) T ( ( b 1 m Q ( b 1 ) ) f 1 , ( b 2 m Q ( b 2 ) ) f 2 ) ] + m B ˜ T ( ( b 1 m B ˜ ( b 1 ) ) f 1 χ 6 5 B ˜ 0 , ( b 2 m B ˜ ( b 2 ) ) f 2 χ 6 5 B ˜ 0 ) T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 B ˜ 0 ) H 11 + H 12 .

For H 11 , since it equals to F 11 , we immediately have

H 11 C b 1 b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) ( x ) + C b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 , f 2 ) ) ( x ) + C b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 , f 2 ) ) ( x ) .

For H 12 , with the same method to estimate F 11 , we have

H 12 C K B , Q 2 b 1 b 2 M r , ( 6 ) ( T ( f 1 χ 6 5 B ˜ 0 , f 2 χ 6 5 B ˜ 0 ) ) ( x ) + C K B , Q b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 χ 6 5 B ˜ 0 , f 2 χ 6 5 B ˜ 0 ) ) ( x ) + C K B , Q b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 χ 6 5 B ˜ 0 , f 2 χ 6 5 B ˜ 0 ) ) ( x ) .

Combining the estimates of H 11 and H 12 , we complete the estimate for H 1 .

Now we turn to estimate H 2 . By decomposing the region of the integral, we have

H 2 m B ˜ { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ 0 \ 6 5 B ˜ 0 ) } + m B ˜ { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ 0 \ 6 5 B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 B ˜ 0 ) } + m B ˜ { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ 0 \ 6 5 B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ 0 \ 6 5 B ˜ 0 ) } + m B ˜ { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 N B ˜ 0 ) } m Q { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 N B ˜ 0 ) } + m B ˜ { T ( ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 N B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ 0 ) } m Q { T ( ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 N B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ 0 ) } + m B ˜ { T ( ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 N B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 N B ˜ 0 ) } m Q { T ( ( b 1 m Q ( b 1 ) ) f 1 χ X \ 6 N B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ X \ 6 N B ˜ 0 ) } + m Q { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 Q 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B 0 \ 6 5 Q 0 ) } + m Q { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ 0 \ 6 5 Q 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 5 Q 0 ) } + m Q { T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 N B ˜ 0 \ 6 5 Q 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ 0 \ 6 5 Q 0 ) } i = 1 9 H 2 i .

Similar to estimate F 2 i , 1 i 9 , we obtain that

H 21 + H 22 + H 27 + H 28 C K B , Q 3 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) , H 23 + H 29 C K B , Q 4 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) , H 24 + H 25 + H 26 C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

For simplicity, we only compute H 21 . We write

T ( ( b 1 m Q ( b 1 ) ) f 1 χ 6 5 B ˜ 0 , ( b 2 m Q ( b 2 ) ) f 2 χ 6 N B ˜ 0 \ 6 5 B ˜ 0 ) ( z ) C 6 N B ˜ 0 \ 6 5 B ˜ 0 6 5 B ˜ 0 ( b 1 ( y 1 ) m Q ( b 1 ) ) f 1 ( y 1 ) ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) i = 1 2 λ ( z , d ( z , y i ) ) 2 d μ ( y 1 ) d μ ( y 2 )

C 6 5 B ˜ 0 ( b 1 ( y 1 ) m Q ( b 1 ) ) f 1 ( y 1 ) λ ( z , d ( z , y 1 ) ) d μ ( y 1 ) k = 1 N 1 6 k + 1 B ˜ 0 \ 6 k B ˜ 0 ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) λ ( z , d ( z , y 2 ) ) d μ ( y 2 ) + 6 B ˜ 0 \ 6 5 B ˜ 0 ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) λ ( z , d ( z , y 2 ) ) d μ ( y 2 ) C μ ( 6 B ˜ 0 ) λ ( x B ˜ 0 , 6 5 r B ˜ 0 ) 1 μ ( 6 B ˜ 0 ) 6 5 B ˜ 0 ( b 1 ( y 1 ) m Q ( b 1 ) ) f 1 ( y 1 ) d μ ( y 1 ) × k = 1 N 1 μ ( 5 × 6 k + 1 B ˜ 0 ) λ ( x B ˜ 0 , 6 k r B ˜ 0 ) 1 μ ( 5 × 6 k + 1 B ˜ 0 ) 6 k + 1 B ˜ 0 ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) d μ ( y 2 ) + μ ( 5 × 6 B ˜ 0 ) λ ( x B ˜ 0 , 6 r B ˜ 0 ) 1 μ ( 5 × 6 B ˜ 0 ) 6 B ˜ 0 ( b 2 ( y 2 ) m Q ( b 2 ) ) f 2 ( y 2 ) d μ ( y 2 ) C K B ˜ 0 , Q 2 K B ˜ 0 , Q 0 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) C K B , Q 3 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Here, we have used the conclusions K B ˜ 0 , Q C K B , Q and K B ˜ 0 , Q 0 C K B , Q , which can be obtained by Lemma 2.4.

Taking the mean over z B ˜ , we obtain

H 21 C K B , Q 3 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Combining the estimate for H 2 i , 1 i 9 , then

H 2 C K B , Q 4 b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Next we consider h B h B ˜ . With the similar method to estimate H 2 , we easily obtain that

h B h B ˜ C b 1 b 2 M p 1 , ( 5 ) f 1 ( x ) M p 2 , ( 5 ) f 2 ( x ) .

Thus, (2.13) holds, and hence, (2.4) is proved. With the same method to prove (2.4), we can obtain that (2.5) and (2.6) also hold. Here, we omit the details. Thus, Lemma 2.6 has been proved.□

Proof of Theorem 1.11

Let 0 < δ < 1 / 2 , 1 < p 1 , p 2 , q < , 1 q = 1 p 1 + 1 p 2 , 1 < r < q , f 1 L p 1 ( μ ) , f 2 L p 2 ( μ ) , b 1 RBMO ( μ ) and b 2 RBMO ( μ ) .

When l ( B ) 1 , by f ( x ) N δ f ( x ) , Lemmas 2.1, 2.5, 2.6, Hölder’s inequality and the boundedness of M ( ρ ) and M r , ( ρ ) for ρ 5 and q > r , it follows

[ b 1 , b 2 , T ] ( f 1 , f 2 ) L q ( μ ) N δ ( [ b 1 , b 2 , T ] ( f 1 , f 2 ) ) L q ( μ ) C M δ ( [ b 1 , b 2 , T ] ( f 1 , f 2 ) ) L q ( μ ) C b 1 b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) L q ( μ ) + C b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 , f 2 ) ) L q ( μ ) + C b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 , f 2 ) ) L q ( μ ) + C b 1 b 2 M p 1 , ( 5 ) f 1 M p 2 , ( 5 ) f 2 L q ( μ ) + C b 1 b 2 M r , ( 6 ) ( T ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) L q ( μ ) + C b 1 M r , ( 6 ) ( [ b 2 , T ] ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) L q ( μ ) + C b 2 M r , ( 6 ) ( [ b 1 , T ] ( f 1 χ 6 5 B ˜ , f 2 χ 6 5 B ˜ ) ) L q ( μ ) C b 1 b 2 f 1 L p 1 ( μ ) f 2 L p 2 ( μ ) + C b 1 ( [ b 2 , T ] ( f 1 , f 2 ) ) L q ( μ ) + C b 2 ( [ b 1 , T ] ( f 1 , f 2 ) ) L q ( μ ) C b 1 b 2 f 1 L p 1 ( μ ) f 2 L p 2 ( μ ) + C b 1 M δ ( [ b 2 , T ] ( f 1 , f 2 ) ) L q ( μ ) + C b 2 M δ ( [ b 1 , T ] ( f 1 , f 2 ) ) L q ( μ ) C b 1 b 2 f 1 L p 1 ( μ ) f 2 L p 2 ( μ ) + C b 1 M r , ( 6 ) ( T ( f 1 , f 2 ) ) L q ( μ ) + C b 1 M p 1 , ( 5 ) f 1 M p 2 , ( 5 ) f 2 L q ( μ ) + C b 2 M r , ( 6 ) ( T ( f 1 , f 2 ) ) L q ( μ ) + C b 2 M p 1 , ( 5 ) f 1 M p 2 , ( 5 ) f 2 L q ( μ ) C b 1 b 2 f 1 L p 1 ( μ ) f 2 L p 2 ( μ ) + C M p 1 , ( 5 ) f 1 L p 1 ( μ ) M p 2 , ( 5 ) f 2 L p 2 ( μ ) C b 1 b 2 f 1 L p 1 ( μ ) f 2 L p 2 ( μ ) .

When 0 < l ( B ) < 1 , using the aforementioned method, we also obtain the results of Theorem 1.11. Thus, the proof of Theorem 1.11 has been completed.□

Acknowledgments

The authors would like to thank the referees for their very careful reading and many valuable suggestions. The authors thank Dr. Taotao Zheng for discussing a question with us.

  1. Funding information: This work was supported by Natural Science Foundation of China (No. 11971026), Natural Science Foundation of Education Committee of Anhui Province (Nos. KJ2016A506 and KJ2017A454) and Excellent Young Talents Foundation of Anhui Province (Nos. GXYQ2017070 and GXYQ2020049).

  2. Author contributions: The authors have the same contributions.

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

  4. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2021-02-25
Revised: 2021-08-01
Accepted: 2021-12-05
Published Online: 2021-12-31

© 2021 Hailian Wang and Rulong Xie, published by De Gruyter

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

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