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Boundedness for Riesz transform associated with Schrödinger operators and its commutator on weighted Morrey spaces related to certain nonnegative potentials

Abstract

Let L=Δ+V be a Schrödinger operator, where Δ is the Laplacian on R n and the nonnegative potential V belongs to the reverse Hölder class B q for qn/2. The Riesz transform associated with the operator L is denoted by T= ( Δ + V ) 1 2 and the dual Riesz transform is denoted by T = ( Δ + V ) 1 2 . In this paper, we establish the boundedness for the operator T and its commutator on the weighted Morrey spaces L α , V , ω p , λ ( R n ) related to certain nonnegative potentials belonging to the reverse Hölder class B q for n/2q<n, where p 0 <p< and 1 p 0 = 1 q 1 n .

1 Introduction

In this paper, we consider the Schrödinger operator

L=Δ+V(x)on  R n ,n3,

where V(x) is a nonnegative potential belonging to the reverse Hölder class B q for qn/2. The Riesz transform associated with the Schrödinger operator L is defined by T= L 1 2 and the commutator operator

[b,T](f)(x)=T(bf)(x)b(x)Tf(x),x R n ,
(1)

where f is a suitable integral function. Also, the dual Riesz transform associated with the Schrödinger operator L is defined by T = L 1 2 and the commutator operator

[ b , T ] (f)(x)= T (bf)(x)b(x) T f(x),x R n .
(2)

First, Tang and Dong established the boundedness of some Schrödinger type operators on the Morrey spaces related to the nonnegative potential V belonging to the reverse Hölder class in [1]. Furthermore, Liu and Wang investigated the boundedness of the dual Riesz transforms and its commutators on the Morrey spaces related to the nonnegative potential V belonging to the reverse Hölder class in [2]. Recently, Pan and Tang established the boundedness of some Schrödinger type operators on weighted Morrey spaces related to the nonnegative potential V belonging to the reverse Hölder class in [3]. Motivated by [3], our aim is to establish the boundedness for the dual Riesz transform associated with Schrödinger operators and its commutators on weighted Morrey spaces related to the certain nonnegative potentials, where the condition on the potential is weaker than that in [3]. Our result is a nontrivial generalization of the main results in [3].

A nonnegative locally L q integrable function V(x) on R n is said to belong to B q (1<q<) if there exists C>0 such that the reverse Hölder inequality

( 1 | B | B V ( x ) q d x ) 1 / q C ( 1 | B | B V ( x ) d x )
(3)

holds for every ball B in R n .

It is important that the B q class has a property of ‘self improvement’; that is, if V B q , then V B q + ε for some ε>0 (see [4]).

We assume the potential V B q for qn/2 throughout the paper. We introduce the auxiliary function ρ(x,V)=ρ(x) defined by

ρ(x)= sup r > 0 { r : 1 r n 2 B ( x , r ) V ( y ) d y 1 } ,x R n .

It is well known that 0<ρ(x)< for any x R n (cf. Lemma 1 in Section 2).

A kind of new Morrey spaces is established by Tang and Dong in [1]. Furthermore, the weighted Morrey space is introduced by Pan and Tang in [3]. Let p[1,), α(,), and λ[0,1). For f L loc p ( R n ) and V B q (q>1), we say f L α , V , ω p , λ ( R n ) (weighted Morrey spaces related to the nonnegative potential V) provided that

f L α , V , ω p , λ ( R n ) p = sup B ( x 0 , r ) R n ( 1 + r ρ ( x 0 ) ) α ω ( B ( x 0 , 2 r ) ) λ B ( x 0 , r ) |f(x) | p ω(x)dx<,

where B=B( x 0 ,r) denotes a ball with centered at x 0 and radius r, and the weight functions ω A p ρ , (see Section 2).

Now we are in a position to give the main results in this paper.

Theorem 1 Suppose V B q for n/2q<n, α(,), λ(0,1), and 1/ p 0 =1/q1/n. Then, for p 0 p< and ω A p / p 0 ρ , ,

T f L α , V , ω p , λ ( R n ) C f L α , V , ω p , λ ( R n ) ,

where C is independent of f.

Theorem 2 Suppose V B q for n/2q<n, b BMO ρ , α(,), λ(0,1), and p 0 so that 1/ p 0 =1/q1/n. Then, for p 0 p< and ω A p / p 0 ρ , ,

[ b , T ] f L α , V , ω p , λ ( R n ) C f L α , V , ω p , λ ( R n ) ,

where C is independent of f.

We will use C to denote a positive constant, which is not necessarily same at each occurrence and even is different in the same line, and may depend on the dimension n and the constant in (3). By AB, we mean that there exists a constant C such that 1/CA/BC.

2 Some lemmas

In this section, we collect some known results proved in [4] in order to prove the main results in this paper.

Lemma 1 There exist constants C, k 0 >0 such that

1 C ( 1 + | x y | ρ ( x ) ) k 0 ρ ( y ) ρ ( x ) C ( 1 + | x y | ρ ( x ) ) k 0 / ( k 0 + 1 ) .

In particular, ρ(y)ρ(x) if |xy|<Cρ(x).

Lemma 2 (1) For 0<r<R<,

1 r n 2 B ( x , r ) V(y)dyC ( r R ) 2 n q 1 R n 2 B ( x , R ) V(y)dy

and

1 r n 2 B ( x , r ) V(y)dy1if and only ifrρ(x).
  1. (2)

    There exist C>0 and l 0 >0 such that

    1 R n 2 B ( x , R ) V(y)dyC ( 1 + R ρ ( x ) ) l 0 .

Let be the kernel of T and K be the kernel of T .

Lemma 3 If V B q for qn/2, then for every N there exists a constant C N >0 such that

| K (x,z)| C N ( 1 + | x z | ρ ( x ) ) N 1 | x z | n 1 ( B ( z , | x z | / 4 ) V ( u ) | u z | n 1 d u + 1 | x z | ) .
(4)

Moreover, the last inequality also holds with ρ(x) replaced by ρ(z).

In this paper, we always write Ψ θ (B)= ( 1 + r / ρ ( x 0 ) ) θ , where θ>0; x 0 and r denote the center and radius of B, respectively.

A weight will always mean a nonnegative function which is locally integrable. As in [5], we say that a weight ω belongs to the class A p ρ , θ for 1<p<, if there is a positive constant C such that for the whole ball B=B(x,r)

( 1 Ψ θ ( B ) | B | B ω ( y ) d y ) ( 1 Ψ θ ( B ) | B | B ω 1 p 1 ( y ) d y ) p 1 C.

We also say that a nonnegative function ω satisfies the A 1 ρ , θ condition if there exists a positive constant C, for all balls B

M V θ (ω)(x)Cω(x),a.e. x R n ,

where

M V θ f(x)= sup x B 1 Ψ θ ( B ) | B | B |f(y)|dy.

Since Ψ θ (B)1, obviously, A p A p ρ , θ for 1p<, where A p denote the classical Muckenhoupt weights (see [6]). It follows from [7] that A p A p ρ , θ for 1p<. For convenience, we always assume that Ψ(B) denotes Ψ θ (B), A p ρ , = θ > 0 A p ρ , θ , and A ρ , = p 1 A p ρ , .

Lemma 4 ([7])

Let 0<θ<, then:

  1. (i)

    If 1 p 1 < p 2 <, then A p 1 ρ , θ A p 2 ρ , θ .

  2. (ii)

    ω A p ρ , θ if and only if ω 1 p 1 A p ρ , θ , where 1/p+1/ p =1.

  3. (iii)

    If ω A p ρ , θ for 1p<, then there exists a constant C>0 such that for any λ>1

    ω ( λ B ( x 0 , r ) ) C ( 1 + λ r ρ ( x 0 ) ) ( k 0 + 1 ) θ ω ( B ( x 0 , r ) ) .

Lemma 5 ([8])

Let 0<θ<, 1p<. If ω A p ρ , θ , then there exist positive constants δ, η, and C such that

( 1 | B | B ω ( y ) 1 + δ d y ) 1 / ( 1 + δ ) C 1 | B | B ω(y)dy ( 1 + r ρ ( x 0 ) ) η

for all ball B( x 0 ,r).

As a consequence of Lemma 5, we have the following result.

Corollary 1 ([8])

Let 0<θ<, 1p<. If ω A p ρ , θ , then there exist positive constants q>1, η, and C such that

ω ( E ) ω ( B ) C ( | E | | B | ) 1 / q ( 1 + r ρ ( x 0 ) ) η

for any measurable subset E of a ball B( x 0 ,r).

Bongioanni et al. [9] introduced a new space BMO θ (ρ) defined by

f BMO θ ( ρ ) = sup B R n 1 Ψ θ ( B ) | B | B |f(x) f B |dx<,

where f B = 1 | B | B f(y)dy and Ψ θ (B)= ( 1 + r / ρ ( x 0 ) ) θ , B=B( x 0 ,r), and θ>0.

In particularly, Bongioanni et al. [9] proved the following results for BMO θ (ρ).

Proposition 1 Let θ>0 and 1s<. If b BMO θ (ρ), then

( 1 | B | B | b ( x ) b B | s d x ) 1 / s C b BMO θ ( ρ ) ( 1 + r ρ ( x 0 ) ) θ

holds for all B=B( x 0 ,r), with x 0 R n and r>0, where θ =( k 0 +1)θ and k 0 is the constant appearing in Lemma  1.

Proposition 2 Let b BMO θ (ρ), B=B( x 0 ,r), and 1s<. Then

( 1 | 2 k B | 2 k B | b ( y ) b B | s d y ) 1 s C b BMO θ ( ρ ) k ( 1 + 2 k r ρ ( x 0 ) ) θ
(5)

for all kN, with θ =( k 0 +1)θ and the constant k 0 is given as in Proposition  1.

Obviously, the classical BMO space is properly contained in BMO θ (ρ); for more examples please see [9]. For convenience, we let BMO ρ = θ > 0 BMO θ (ρ).

From Corollary 2.2 in [3], the following result holds true.

Corollary 2 If b BMO ρ and ω A ρ , , then there exist positive constants C and η such that for every ball B=B(x,r), we have

1 ω ( B ) B |b(x) b B | p ω(x)dx ( 1 + r ρ ( x ) ) η b BMO ρ p ,

where b B = 1 | B | B b(y)dy.

3 The proof of our main results

Proof of Theorem 1 Without loss of generality, we may assume that α<0 and ω A p / p 0 ρ , θ . Pick any ball B=B( x 0 ,r), and write

f(x)= f 1 (x)+ f 2 (x),

where f 1 = χ B ( x 0 , 2 r ) f. Hence, we have

( B ( x 0 , r ) | T f ( x ) | p ω ( x ) d x ) 1 / p ( B ( x 0 , r ) | T f 1 ( x ) | p ω ( x ) d x ) 1 / p + ( B ( x 0 , r ) | T f 2 ( x ) | p ω ( x ) d x ) 1 / p .
(6)

By the L ω p boundedness of T (see Theorem 3 in [5]), we obtain

B ( x 0 , r ) | T f 1 (x) | p ω(x)dxC ( 1 + r ρ ( x 0 ) ) α ω ( B ( x 0 , 2 r ) ) λ f L α , V , ω p , λ ( R n ) p .
(7)

Now, for xB( x 0 ,r) and using Lemma 3, we have

| T f 2 ( x ) | = | | x 0 z | > 2 r K ( x , z ) f ( z ) d z | I 1 ( x ) + I 2 ( x ) ,
(8)

where

I 1 (x)= C N | x 0 z | > 2 r | f ( z ) | | x z | n ( 1 + | x z | ρ ( x ) ) N dz

and

I 2 (x)= C N | x 0 z | > 2 r | f ( z ) | | x z | n 1 ( 1 + | x z | ρ ( x ) ) N B ( z , | x z | / 4 ) V ( u ) | u z | n 1 dudz.

Then

( B ( x 0 , r ) | T f 2 ( x ) | p ω ( x ) d x ) 1 / p ( B ( x 0 , r ) ( I 1 ( x ) ) p ω ( x ) d x ) 1 / p + ( B ( x 0 , r ) ( I 2 ( x ) ) p ω ( x ) d x ) 1 / p .
(9)

By the proof of Theorem 1.1 in [3], we have

B ( x 0 , r ) ( I 1 ( x ) ) p ω(x)dxC ( 1 + r ρ ( x 0 ) ) α ω ( B ( x 0 , 2 r ) ) λ f L α , V , ω p , λ ( R n ) p .

Next we deal with I 2 (x). For xB( x 0 ,r), | x 0 z | 2 |xz| 3 | x 0 z | 2 . We get

B ( x 0 , r ) ( I 2 ( x ) ) p ω ( x ) d x = C N B ( x 0 , r ) ( | x 0 z | > 2 r | f ( z ) | | x z | n 1 ( 1 + | x z | ρ ( x ) ) N B ( z , | x z | / 4 ) V ( u ) | u z | n 1 d u d z ) p ω ( x ) d x i = 1 C N B ( x 0 , r ) ( B ( x 0 , 2 i + 1 r ) B ( x 0 , 2 i r ) | f ( z ) | | x 0 z | n 1 ( 1 + | x 0 z | ρ ( x ) ) N × B ( x 0 , 2 i + 3 r ) V ( u ) | u z | n 1 d u d z ) p ω ( x ) d x i = 1 C N B ( x 0 , r ) 1 ( 1 + 2 i r ρ ( x ) ) N p ( 2 i r ) ( n 1 ) p ( B ( x 0 , 2 i r ) | f ( z ) | I 1 ( V χ B ( x 0 , 2 i r ) ) d z ) p ω ( x ) d x .

Let p 0 p<. By simple computation, p p 0 =1+ p v . By the definition of A p / p 0 ρ , θ ,

( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) ω v / p ( y ) d y ) 1 / v = ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) ω 1 p v + 1 1 ( y ) d y ) ( p v + 1 1 ) 1 v ( p v + 1 1 ) C ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) ω ( y ) d y ) 1 p ,
(10)

where 1/q=1/s+1/n and 1/p+1/v+1/s=1.

Using Hölder’s inequality, (10), and the boundedness of the fractional integral I 1 : L q L s with 1/q=1/s+1/n, for 1/p+1/v+1/s=1, we have

B ( x 0 , 2 i r ) | f ( x ) | I 1 ( V χ B ( x 0 , 2 i r ) ) d x = B ( x 0 , 2 i r ) | f ( x ) | ω 1 / p ω 1 / p I 1 ( V χ B ( x 0 , 2 i r ) ) d x ( Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | ) 1 v ( B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p × ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) ω ( x ) v / p d x ) 1 / v × ( B ( x 0 , 2 i r ) ( I 1 ( V χ B ( x 0 , 2 i r ) ) ) s d x ) 1 / s C ( Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | ) 1 v ( B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p × ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) ω ( x ) d x ) 1 / p × ( B ( x 0 , 2 i r ) ( I 1 ( V χ B ( x 0 , 2 i r ) ) ) s d x ) 1 / s C ( Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | ) 1 / p + 1 / v ( B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p × ω ( B ( x 0 , 2 i r ) ) 1 / p I 1 ( V χ B ( x 0 , 2 i r ) ) s C ( Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | ) 1 / p + 1 / v ( B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p × ω ( B ( x 0 , 2 i r ) ) 1 / p V χ B ( x 0 , 2 i r ) q .
(11)

For V B q , using Lemma 2, we get

V χ B ( x 0 , 2 i r ) q C ( 2 i r ) n / q B ( x 0 , 2 i r ) V ( x ) d x C ( 2 i r ) n / q + n 2 ( 2 i r ) n + 2 B ( x 0 , 2 i r ) V ( x ) d x C ( 2 i r ) n / q + n 2 ( 1 + 2 i r ρ ( x 0 ) ) l 0 .
(12)

It is easy to check that (n1)p p n q +(n2)p+n+ p n v =0. Furthermore, using Corollary 1, we have

ω ( B ( x 0 , 2 r ) ) ω ( B ( x 0 , 2 i + 1 r ) ) C ( 2 i ) n q ( 1 + 2 i r ρ ( x 0 ) ) η .
(13)

Therefore, by (13),

B ( x 0 , r ) ( I 2 ( x ) ) p ω ( x ) d x = C N B ( x 0 , r ) ( | x 0 z | > 2 r | f ( z ) | | x z | n 1 ( 1 + | x z | ρ ( x ) ) N B ( z , | x z | / 4 ) V ( u ) | u z | n 1 d u d z ) p ω ( x ) d x i = 1 C N ( 2 i r ) ( n 1 ) p p n q + ( n 2 ) p + n + p n v × B ( x 0 , r ) ( 1 + 2 i r ρ ( x 0 ) ) ( 1 + p v ) θ + l 0 p ω ( B ( x 0 , 2 i r ) ) ( 1 + 2 i r ρ ( x ) ) N p B ( x 0 , 2 i r ) | f ( y ) | p ω ( y ) d y ω ( x ) d x i = 1 C N ω ( B ( x 0 , 2 i + 1 r ) ) λ 1 f L α , V , ω p , λ ( R n ) p B ( x 0 , r ) ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p v ) θ + l 0 p ( 1 + 2 i r ρ ( x ) ) N p ω ( x ) d x i = 1 C N ω ( B ( x 0 , 2 i + 1 r ) ) λ 1 ω ( B ( x 0 , r ) ) ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p v ) θ + l 0 p ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) f L α , V , ω p , λ ( R n ) p i = 1 C N ( ω ( B ( x 0 , 2 r ) ) ω ( B ( x 0 , 2 i + 1 r ) ) ) 1 λ ω ( B ( x 0 , 2 r ) ) λ ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p v ) θ + l 0 p ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) f L α , V , ω p , λ ( R n ) p i = 1 C N 2 i n ( 1 λ ) / q ω ( B ( x 0 , 2 r ) ) λ ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p v ) θ + l 0 p + η ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) f L α , V , ω p , λ ( R n ) p C f L α , V , ω p , λ ( R n ) p ,
(14)

where we choose N large enough so that the above series converges.

From (6)-(14), we obtain

T f L α , V , ω p , λ ( R n ) C f L α , V , ω p , λ ( R n ) .

Thus, Theorem 1 is proved. □

Proof of Theorem 2 During the proof of Theorem 2, we always denote θ =( k 0 +1)θ. Without loss of generality, we may assume that α<0, b BMO θ (ρ), and ω A p / p 0 ρ , θ . Pick any ball B=B( x 0 ,r), and write

f(x)= f 1 (x)+ f 2 (x),

where f 1 = χ B ( x 0 , 2 r ) f. Hence, we have

( B ( x 0 , r ) | [ b , T ] f ( x ) | p ω ( x ) d x ) 1 / p ( B ( x 0 , r ) | [ b , T ] f 1 ( x ) | p ω ( x ) d x ) 1 / p + ( B ( x 0 , r ) | [ b , T ] f 2 ( x ) | p ω ( x ) d x ) 1 / p .
(15)

By the L ω p boundedness of [b, T ] (see Theorem 2 in [8]), we obtain

B ( x 0 , r ) | [ b , T ] f 1 (x) | p ω(x)dxC ( 1 + r ρ ( x 0 ) ) α ω ( B ( x 0 , 2 r ) ) λ f L α , V , ω p , λ ( R n ) p .
(16)

Set b B = 1 | B ( x 0 , r ) | B ( x 0 , r ) b(x)dx. Write [b, T ] f 2 =(b b B ) T f 2 T ( f 2 (b b B )). Then

( B ( x 0 , r ) | [ b , T ] f 2 ( x ) | p ω ( x ) d x ) 1 / p ( B ( x 0 , r ) | ( b b B ) T f 2 | p ω ( x ) d x ) 1 / p + ( B ( x 0 , r ) | T ( f 2 ( b b B ) ) | p ω ( x ) d x ) 1 / p .
(17)

By (8) in the proof of Theorem 1, we obtain

B ( x 0 , r ) | ( b b B ) T f 2 | p ω ( x ) d x 2 p 1 ( B ( x 0 , r ) | b b B | p ( I 1 ( x ) ) p ω ( x ) d x + B ( x 0 , r ) | b b B | p ( I 2 ( x ) ) p ω ( x ) d x ) .

Let p 0 p<. By simple computation, p p 0 <1+ p p . By Lemma 4, A p / p 0 ρ , θ A 1 + p / p ρ , θ . Then

( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) ω p / p ( y ) d y ) p / p = ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) ω 1 1 + p p 1 ( y ) d y ) ( 1 + p p 1 ) C ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) ω ( y ) d y ) 1 .
(18)

By Lemma 1 and Corollary 2, as well as Lemma 3, we have

B ( x 0 , r ) | b b B | p ( I 1 ( x ) ) p ω ( x ) d x C N B ( x 0 , r ) | b b B | p ( | x 0 z | > 2 r | f ( z ) | | x z | n ( 1 + | x z | ρ ( x ) ) N d z ) p ω ( x ) d x i = 1 C N B ( x 0 , r ) | b b B | p ( B ( x 0 , 2 i + 1 r ) B ( x 0 , 2 i r ) | f ( z ) | | x 0 z | n ( 1 + | x 0 z | ρ ( x ) ) N d z ) p ω ( x ) d x i = 1 C N B ( x 0 , r ) | b b B | p 1 ( 1 + 2 i r ρ ( x ) ) N p ( 2 i r ) n p ( B ( x 0 , 2 i r ) | f ( z ) | d z ) p ω ( x ) d x i = 1 C N B ( x 0 , r ) | b b B | p 1 ( 1 + 2 i r ρ ( x ) ) N p ( 2 i r ) n p × ( B ( x 0 , 2 i r ) | f ( z ) | ω ( z ) 1 / p ω ( z ) 1 / p d z ) p ω ( x ) d x i = 1 C N B ( x 0 , r ) | b b B | p 1 ( 1 + 2 i r ρ ( x ) ) N p ( 2 i r ) n p ( B ( x 0 , 2 i r ) | f ( z ) | p ω ( z ) d z ) × ( B ( x 0 , 2 i r ) ω ( z ) p p d z ) p p ω ( x ) d x i = 1 C N B ( x 0 , r ) | b b B | p 1 ( 1 + 2 i r ρ ( x ) ) N p ( 2 i r ) n p ( B ( x 0 , 2 i r ) | f ( z ) | p ω ( z ) d z ) × ( 1 + 2 i r ρ ( x 0 ) ) ( 1 + p p ) θ | B ( x 0 , 2 i r ) | 1 + p p ω ( B ( x 0 , 2 i r ) ) 1 ω ( x ) d x i = 1 C N B ( x 0 , r ) | b b B | p ( 1 + 2 i r ρ ( x 0 ) ) p θ ( 1 + 2 i r ρ ( x ) ) N p ω ( B ( x 0 , 2 i r ) ) 1 × B ( x 0 , 2 i r ) | f ( z ) | p ω ( z ) d z ω ( x ) d x i = 1 C N ω ( B ( x 0 , 2 i + 1 r ) ) λ 1 f L α , V , ω p , λ ( R n ) p B ( x 0 , r ) | b b B | p ( 1 + 2 i r ρ ( x 0 ) ) α + p θ ( 1 + 2 i r ρ ( x ) ) N p ω ( x ) d x i = 1 C N ω ( B ( x 0 , 2 i + 1 r ) ) λ 1 ω ( B ( x 0 , r ) ) ( 1 + 2 i r ρ ( x 0 ) ) α + p θ + η ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) b BMO ρ p f L α , V , ω p , λ ( R n ) p i = 1 C N ( ω ( B ( x 0 , 2 r ) ) ω ( B ( x 0 , 2 i + 1 r ) ) ) 1 λ ω ( B ( x 0 , 2 r ) ) λ × ( 1 + 2 i r ρ ( x 0 ) ) α + p θ + η ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) b BMO ρ p f L α , V , ω p , λ ( R n ) p i = 1 C N 2 i n ( 1 λ ) / q ω ( B ( x 0 , 2 r ) ) λ ( 1 + 2 i r ρ ( x 0 ) ) α + p θ + 2 η ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) b BMO ρ p f L α , V , ω p , λ ( R n ) p ,
(19)

where we choose N large enough so that the above series converges.

For I 2 (x), we assume n/2<q<n due to Lemma 3. Then, since xB( x 0 ,r), we also have | x 0 z | 2 |xz| 3 | x 0 z | 2 . Then

B ( x 0 , r ) | b b B | p ( I 2 ( x ) ) p ω ( x ) d x C N B ( x 0 , r ) | b b B | p ( | x 0 z | > 2 r | f ( z ) | | x z | n 1 ( 1 + | x z | ρ ( x ) ) N × B ( z , | x z | / 4 ) V ( u ) | u z | n 1 d u d z ) p ω ( x ) d x i = 1 C N B ( x 0 , r ) | b b B | p ( B ( x 0 , 2 i + 1 r ) B ( x 0 , 2 i r ) | f ( z ) | | x 0 z | n 1 ( 1 + | x 0 z | ρ ( x ) ) N × B ( x 0 , 2 i + 3 r ) V ( u ) | u z | n 1 d u d z ) p ω ( x ) d x i = 1 C N B ( x 0 , r ) 1 ( 1 + 2 i r ρ ( x ) ) N p ( 2 i r ) ( n 1 ) p | b b B | p × ( B ( x 0 , 2 i r ) | f ( z ) | I 1 ( V χ B ( x 0 , 2 i r ) ) d z ) p ω ( x ) d x .

By (11) and (12) in the proof of Theorem 1, we obtain

B ( x 0 , r ) | b b B | p ( I 2 ( x ) ) p ω ( x ) d x C N B ( x 0 , r ) | b b B | p ( | x 0 z | > 2 r | f ( z ) | | x z | n 1 ( 1 + | x z | ρ ( x ) ) N × B ( z , | x z | / 4 ) V ( u ) | u z | n 1 d u d z ) p ω ( x ) d x i = 1 C N ω ( B ( x 0 , 2 i + 1 r ) ) λ 1 f L α , V , ω p , λ ( R n ) p × B ( x 0 , r ) | b b B | p ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p / v ) θ + l 0 p ( 1 + 2 i r ρ ( x ) ) N p ω ( x ) d x i = 1 C N ω ( B ( x 0 , 2 i + 1 r ) ) λ 1 ω ( B ( x 0 , r ) ) × ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p / v ) θ + l 0 p + η ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) b BMO ρ p f L α , V , ω p , λ ( R n ) p i = 1 C N ( ω ( B ( x 0 , 2 r ) ) ω ( B ( x 0 , 2 i + 1 r ) ) ) 1 λ ω ( B ( x 0 , 2 r ) ) λ × ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p / v ) θ + l 0 p + η ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) b BMO ρ p f L α , V , ω p , λ ( R n ) p i = 1 C N 2 i n ( 1 λ ) / q ω ( B ( x 0 , 2 r ) ) λ × ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p / v ) θ + l 0 p + 2 η ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) b BMO ρ p f L α , V , ω p , λ ( R n ) p
(20)

if we choose N large enough.

Now, for xB( x 0 ,r) and using Lemma 3, we have

| T ( f 2 ( b b B ) ) | = | | x 0 z | > 2 r K ( x , z ) f ( z ) ( b b B ) d z | I ˜ 1 ( x ) + I ˜ 2 ( x ) ,
(21)

where

I ˜ 1 (x)= C N | x 0 z | > 2 r | f ( z ) ( b b B ) | | x z | n ( 1 + | x z | ρ ( x ) ) N dz

and

I ˜ 2 (x)= C N | x 0 z | > 2 r | f ( z ) ( b b B ) | | x z | n 1 ( 1 + | x z | ρ ( x ) ) N B ( z , | x z | / 4 ) V ( u ) | u z | n 1 dudz.

Then,

( B ( x 0 , r ) | T ( f 2 ( b b B ) ) | p ω ( x ) d x ) 1 / p ( B ( x 0 , r ) ( I ˜ 1 ( x ) ) p ω ( x ) d x ) 1 / p + ( B ( x 0 , r ) ( I ˜ 2 ( x ) ) p ω ( x ) d x ) 1 / p .
(22)

Firstly, we consider I ˜ 1 (x). By Proposition 2 and (10), for 1/p+1/v+1/s=1, we have

B ( x 0 , 2 i r ) | f ( x ) ( b ( x ) b B ) | d x B ( x 0 , 2 i r ) | f ( x ) | ω 1 / p ω 1 / p | b ( x ) b B | d x Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | × ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p × ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) ω ( x ) v / p d x ) 1 / v × ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) | b ( x ) b B | s d x ) 1 / s C Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | × ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p × ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) ω ( x ) d x ) 1 / p × ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) | b ( x ) b B | s d x ) 1 / s C ( Ψ ( B ( x 0 , 2 i r ) ) ) 1 / p + 1 / v | B ( x 0 , 2 i r ) | × ( B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p ω ( B ( x 0 , 2 i r ) ) 1 / p × ( 1 | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) | b ( x ) b B | s d x ) 1 / s C i ( 1 + 2 i r ρ ( x 0 ) ) ( 1 / p + 1 / v ) θ + θ | B ( x 0 , 2 i r ) | ( B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p × ω ( B ( x 0 , 2 i r ) ) 1 / p b BMO ρ C i ( 1 + 2 i r ρ ( x 0 ) ) ( 1 / p + 1 / v ) θ + θ ( 2 i r ) n ( B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p × ω ( B ( x 0 , 2 i r ) ) 1 / p b BMO ρ .
(23)

Then we get

B ( x 0 , r ) ( I ˜ 1 ( x ) ) p ω ( x ) d x = C N B ( x 0 , r ) ( | x 0 z | > 2 r | f ( z ) ( b b B ) | | x z | n ( 1 + | x z | ρ ( x ) ) N d z ) p ω ( x ) d x i = 1 C N B ( x 0 , r ) ( B ( x 0 , 2 i + 1 r ) B ( x 0 , 2 i r ) | f ( z ) ( b b B ) | | x 0 z | n ( 1 + | x 0 z | ρ ( x ) ) N d z ) p ω ( x ) d x i = 1 C N B ( x 0 , r ) 1 ( 1 + 2 i r ρ ( x ) ) N p ( 2 i r ) n p ( B ( x 0 , 2 i r ) | f ( z ) ( b b B ) | d z ) p ω ( x ) d x i = 1 C N i p B ( x 0 , r ) ( 1 + 2 i r ρ ( x 0 ) ) ( 1 + p / v ) θ + p θ ( 1 + 2 i r ρ ( x ) ) N p ( B ( x 0 , 2 i r ) | f ( z ) | p ω ( z ) d z ) × ω ( B ( x 0 , 2 i r ) ) 1 b BMO ρ p ω ( x ) d x i = 1 C N i p ω ( B ( x 0 , 2 i + 1 r ) ) λ 1 f L α , V , ω p , λ ( R n ) p b BMO ρ p × B ( x 0 , r ) ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p / v ) θ + p θ ( 1 + 2 i r ρ ( x ) ) N p ω ( x ) d x i = 1 C N i p ω ( B ( x 0 , 2 i + 1 r ) ) λ 1 ω ( B ( x 0 , r ) ) × ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p / v ) θ + p θ ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) b BMO ρ p f L α , V , ω p , λ ( R n ) p i = 1 C N i p ( ω ( B ( x 0 , 2 r ) ) ω ( B ( x 0 , 2 i + 1 r ) ) ) 1 λ ω ( B ( x 0 , 2 r ) ) λ × ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p / v ) θ + p θ ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) b BMO ρ p f L α , V , ω p , λ ( R n ) p i = 1 C N i p 2 i n ( 1 λ ) / q ω ( B ( x 0 , 2 r ) ) λ × ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p / v ) θ + p θ + η ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) b BMO ρ p f L α , V , ω p , λ ( R n ) p ,
(24)

where we choose N large enough so that the above series converges.

For V B q , then V B q + ε for ε>0. Using Lemma 2, we get

V χ B ( x 0 , 2 i r ) q + ε C ( 2 i r ) n / ( q + ε ) B ( x 0 , 2 i r ) V ( x ) d x C ( 2 i r ) n / ( q + ε ) + n 2 ( 2 i r ) n + 2 B ( x 0 , 2 i r ) V ( x ) d x C ( 2 i r ) n / ( q + ε ) + n 2 ( 1 + 2 i r ρ ( x 0 ) ) l 0 .
(25)

Let p 0 p<. We choose u such that u= q ( q + ε ) ε and 1/p+1/v+1/u+1/s=1.

Let 1/(q+ε)=1/s+1/n. By simple computation,

p p 0 = p ( 1 1 q + 1 n ) = p ( 1 1 q + 1 n + 1 q + ε 1 q + ε ) = p ( 1 1 q ( q + ε ) 1 s ) = 1 + p v .

Finally, we deal with I ˜ 2 (x). Using Hölder’s inequality, (10), and the boundedness of the fractional integral I 1 : L q + ε L s , for 1/p+1/v+1/u+1/s=1, we have

B ( x 0 , 2 i r ) | f ( x ) ( b ( x ) b B ) | I 1 ( V χ B ( x 0 , 2 i r ) ) d x B ( x 0 , 2 i r ) | f ( x ) | ω 1 / p ω 1 / p | b ( x ) b B | I 1 ( V χ B ( x 0 , 2 i r ) ) d x ( Ψ ( B ( x 0 , 2 i r ) ) ) 1 / v ( | B ( x 0 , 2 i r ) | ) 1 / v + 1 / u ( B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p × ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) ω ( x ) v / p d x ) 1 / v × ( 1 | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) | b ( x ) b B | u d x ) 1 / u ( B ( x 0 , 2 i r ) ( I 1 ( V χ B ( x 0 , 2 i r ) ) ) s d x ) 1 / s C ( Ψ ( B ( x 0 , 2 i r ) ) ) 1 / v ( | B ( x 0 , 2 i r ) | ) 1 / v + 1 / u ( B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p × ( 1 Ψ ( B ( x 0 , 2 i r ) ) | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) ω ( x ) d x ) 1 / p × ( 1 | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) | b ( x ) b B | u d x ) 1 / u ( B ( x 0 , 2 i r ) ( I 1 ( V χ B ( x 0 , 2 i r ) ) ) s d x ) 1 / s C ( Ψ ( B ( x 0 , 2 i r ) ) ) 1 / p + 1 / v | B ( x 0 , 2 i r ) | 1 / p + 1 / v + 1 / u × ( B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p ω ( B ( x 0 , 2 i r ) ) 1 / p × ( 1 | B ( x 0 , 2 i r ) | B ( x 0 , 2 i r ) | b ( x ) b B | u d x ) 1 / u I 1 ( V χ B ( x 0 , 2 i r ) ) s C i ( 1 + 2 i r ρ ( x 0 ) ) ( 1 / p + 1 / v ) θ + θ | B ( x 0 , 2 i r ) | 1 / p + 1 / v + 1 / u ( B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p × ω ( B ( x 0 , 2 i r ) ) 1 / p b BMO ρ V χ B ( x 0 , 2 i r ) q + ε C i ( 1 + 2 i r ρ ( x 0 ) ) ( 1 / p + 1 / v ) θ + θ + l 0 ( 2 i r ) ( n 1 ) × ω ( B ( x 0 , 2 i r ) ) 1 / p ( B ( x 0 , 2 i r ) | f ( x ) | p ω ( x ) d x ) 1 / p b BMO ρ .
(26)

Then

B ( x 0 , r ) ( I ˜ 2 ( x ) ) p ω ( x ) d x = C N B ( x 0 , r ) ( | x 0 z | > 2 r | f ( z ) ( b b B ) | | x z | n 1 ( 1 + | x z | ρ ( x ) ) N B ( z , | x z | / 4 ) V ( u ) | u z | n 1 d u d z ) p ω ( x ) d x i = 1 C N B ( x 0 , r ) ( B ( x 0 , 2 i + 1 r ) B ( x 0 , 2 i r ) | f ( z ) ( b b B ) | | x 0 z | n 1 ( 1 + | x 0 z | ρ ( x ) ) N × B ( x 0 , 2 i + 3 r ) V ( u ) | u z | n 1 d u d z ) p ω ( x ) d x i = 1 C N B ( x 0 , r ) 1 ( 1 + 2 i r ρ ( x ) ) N p ( 2 i r ) ( n 1 ) p × ( B ( x 0 , 2 i r ) | f ( z ) ( b b B ) | I 1 ( V χ B ( x 0 , 2 i r ) ) d z ) p ω ( x ) d x i = 1 C N i p ω ( B ( x 0 , 2 i + 1 r ) ) λ 1 f L α , V , ω p , λ ( R n ) p b BMO ρ p × B ( x 0 , r ) ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p / v ) θ + p θ + l 0 p ( 1 + 2 i r ρ ( x ) ) N p ω ( x ) d x i = 1 C N i p ω ( B ( x 0 , 2 i + 1 r ) ) λ 1 ω ( B ( x 0 , r ) ) × ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p / v ) θ + p θ + l 0 p ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) b BMO ρ p f L α , V , ω p , λ ( R n ) p i = 1 C N i p ( ω ( B ( x 0 , 2 r ) ) ω ( B ( x 0 , 2 i + 1 r ) ) ) 1 λ ω ( B ( x 0 , 2 r ) ) λ × ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p / v ) θ + p θ + l 0 p ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) b BMO ρ p f L α , V , ω p , λ ( R n ) p i = 1 C N i p 2 i n ( 1 λ ) / q ω ( B ( x 0 , 2 r ) ) λ × ( 1 + 2 i r ρ ( x 0 ) ) α + ( 1 + p / v ) θ + p θ + l 0 p + η ( 1 + 2 i r ρ ( x 0 ) ) N p / ( k 0 + 1 ) b BMO ρ p f L α , V , ω p , λ ( R n ) p ,
(27)

where we choose N large enough so that the above series converges.

From (15)-(27), we obtain

[ b , T ] f L α , V , ω p , λ ( R n ) C f L α , V , ω p , λ ( R n ) .

Thus, we complete the proof of Theorem 2. □

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