Some estimates for commutators of Riesz transforms associated with Schrödinger operators

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Some estimates for commutators of Riesz transforms associated with Schrödinger operators ✩ Yu Liu ∗ , Jielai Sheng School of Mathematics and Physics, University of Science and Technology Beijing, Beijing, 100083, China

a r t i c l e

i n f o

Article history: Received 6 December 2013 Available online xxxx Submitted by J. Xiao Keywords: Commutator Hardy space Campanato space Reverse Hölder inequality Riesz transform Schrödinger operator

a b s t r a c t We consider the Schrödinger operator L = −Δ + V on Rn , where the nonnegative potential V belongs to the reverse Hölder class Bq1 for some q1 ≥ n2 . Let q2 = 1 when q1 ≥ n and q1 = 1 − q1 + n1 when n2 < q1 < n. Set δ  = min{1, 2 − qn }. Let 2 1 1 p n HL (Rn ) be the Hardy space related to the Schrödinger operator L for n+δ  < p ≤ 1. The commutator [b, R] is generated by a function b ∈ Λθν , where Λθν is a function space which is larger than the classical Companato space, and the Riesz transform 1 . R = ∇(−Δ + V )− 2 . We show that the commutator [b, R] is bounded from Lp (Rn ) p ν and bounded from HL (Rn ) into Lq (Rn ) into Lq (Rn ) for 1 < p < q2 , where 1q = p1 − n for

n n+ν

< p ≤ 1, where n n+ν

1 q

=

1 p

−

ν . n

Moreover, we prove that the commutator [b, R]

n

maps HL (R ) continuously into weak L1 (Rn ). At last, we give a characterization for the boundedness of the commutator [b, R] in an extreme case. © 2014 Elsevier Inc. All rights reserved.

1. Introduction Let L = −Δ + V be the Schrödinger operator, where Δ is the Laplacian on Rn and the nonnegative potential V belongs to the reverse Hölder class Bq1 for some q1 ≥ n2 and n ≥ 3. In this paper we consider the Riesz transform associated with the Schrödinger operator L R = ∇(−Δ + V )− 2 . 1

(1)

Let b be a locally integrable function on Rn and T be a linear operator. For a suitable function f , the commutator is defined by [b, T ]f = bT (f ) − T (bf ). It is well known that Coifman, Rochberg and Weiss [3] proved that [b, T ] is a bounded operator on Lp for 1 < p < ∞ if and only if b ∈ BMO(Rn ), when T ✩ Supported by the National Natural Science Foundation of China under grant No. 10901018, the Fundamental Research Funds for the Central Universities under grant No. FRF-BR-13-002 and Program for New Century Excellent Talents in University under grant No. NCET-13-0664. * Corresponding author. E-mail addresses: [email protected] (Y. Liu), [email protected] (J. Sheng).

http://dx.doi.org/10.1016/j.jmaa.2014.04.053 0022-247X/© 2014 Elsevier Inc. All rights reserved.

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is a Calderón–Zygmund operator. In [9], Harboure, Segovia and Torrea found sufficient conditions on the function b in order to obtain the Hardy type and the BMO type boundedness of the commutator [b, T ], where T is an operator with a kernel satisfying some smoothness. Janson [10] proved that the commutator [b, T ] is bounded from Lp (Rn ) into Lq (Rn ), 1 < p < q < ∞, if and only if b ∈ Lipν (Rn ) with ν = ( p1 − 1q )n, where the Lipschitz space Lipν (Rn ) consists of the functions f satisfying f Lipν 

sup x,y∈Rn , x=y

|f (x) − f (y)| < ∞, |x − y|ν

0 < ν < 1.

Furthermore, Lu, Wu and Yang studied the boundedness properties of the commutator [b, T ] on the classical Hardy spaces when b ∈ Lipν (Rn ) in [18]. In recent years, the problems related to Schrödinger operators with nonnegative potentials have attracted much attention; see, for example [6,19,23,11,1,2,4,5,20–22,15,13] and their references. Moreover, a few scholars have investigated the boundedness of the commutators generated by a BMO function b and Riesz transforms associated with the Schrödinger operator L (cf. [8,12,2,14,16,17]). Based on their results we know that the boundedness of the commutators depends on the conditions of the nonnegative potential V . Motivated by the work in [18,8,12], the goal of this paper is to study the boundedness properties of the commutator [b, R] on the Lebesgue spaces and the Hardy spaces related to the Schrödinger operator L, where b belongs to the new function space which is larger than the classical Companato space. Note that a nonnegative locally Lq integrable function V on Rn is said to belong to the class Bq (1 < q < ∞) if there exists C > 0 such that the reverse Hölder inequality 

1 |B|

 q

 q1

V (x) dx B

C ≤ |B|

 V (x) dx

(2)

B

holds for every ball B in Rn . Obviously, Bq2 ⊂ Bq1 if q2 > q1 . But it is important that the Bq class has a property of “self-improvement”; that is, if V ∈ Bq , then V ∈ Bq+ε for some ε > 0 (cf. [19]). Assume V ∈ Bq1 for some q1 > n2 . Then the auxiliary function ρ(x, V ) = ρ(x) introduced by Shen in [19] is defined to be ρ(x) =

 1 1 . = sup r: n−2 m(x, V ) r>0 r



 V (y) dy ≤ 1 ,

x ∈ Rn .

B(x,r)

In order to obtain the estimates of Hardy spaces of the commutators in this paper, we also need to recall the Hardy space associated with the Schrödinger operator L which had been studied by Dziubański and n n 2 Zienkiewicz (cf. [4] and [5]). Because V ≥ 0 and V ∈ Lloc (R ), the Schrödinger operator L generates a (C0 ) contraction semigroup {TsL : s > 0} = {e−sL : s > 0}. The maximal function associated with {TsL : s > 0} is defined by M L f (x) = sups>0 |TsL f (x)|. We always denote δ = 2 − qn1 and δ  = min{1, δ} throughout this p n n paper. For n+δ  < p ≤ 1, the Hardy space HL (R ) associated with the Schrödinger operator L is defined as follows. Definition 1. We say that f is an element of HLp (Rn ) if the maximal function M L f belongs to Lp (Rn ). The quasi-norm of f is defined by f HLp (Rn ) = M L f Lp (Rn ) . p,q n 2 n Definition 2. Let n+δ  < p ≤ 1 ≤ q ≤ ∞. A function a ∈ L (R ) is called an HL -atom if r < ρ(x0 ) and the following conditions hold:

(i)

supp a ⊂ B(x0 , r),

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 1−1 (ii) aLq (Rn ) ≤ B(x0 , r) q p ,  ρ(x0 ) , then a(x) dx = 0. (iii) if r < 4 B(x0 ,r)

It follows from (ii) in Definition 2 that an HLp,∞ -atom is also an HLp,q -atom for 1 ≤ q < ∞. Dziubański and Zienkiewicz have characterized the Hardy space HLp (Rn ) by using HLp,∞ -atoms in [5]. We immediately obtain another following atomic characterization by their results.  p n n Proposition 1. Let n+δ  < p ≤ 1 ≤ q ≤ ∞. Then f ∈ HL (R ) if and only if f can be written as f = j λj aj ,  where aj are HLp,q -atoms, j |λj |p < ∞, and the sum converges in the HLp (Rn ) quasi-norm. Moreover, f HLp (Rn ) ∼ inf



|λj |p

 p1  ,

j

where the infimum is taken over all atomic decompositions of f into HLp,q -atoms. The above atomic decomposition of HLp (Rn ) implies that the Hardy space HLp (Rn ) is larger than the classical Hardy space H p (Rn ). Especially, the Hardy space HLp (Rn ) is exactly the local Hardy space hp (Rn ) introduced by Goldberg in [7] when the potential V is a positive constant. Then we introduce a class of new function spaces. Let θ > 0 and 0 < ν < 1, the Campanato class Λθν consists of the locally integrable functions b such that    b(y) − bB  dy ≤ C 1 +



1 |B(x, r)|1+ν/n

r ρ(x)

θ (3)

B(x,r)

holds true for all x ∈ Rn and r > 0, where bB = and is defined by [b]θν



sup

1 |B(x,r)|

1 ( |B(x,r)| 1+ν/n

B(x,r)



b(y) dy. A seminorm on Λθν is denoted by [b]θν

|b(y) − bB | dy)

B(x,r) r (1 + ρ(x) )θ

x∈Rn , r>0

< ∞.

(4)

Note that if θ = 0 in (3), Λθν is exactly the classical Campanato space; if ν = 0, Λθν is exactly the space BMO θ (cf. [2]); if θ = 0 and ν = 0, it is exactly the John–Nirenberg space BMO. Now we define the space Lipθν which consists of the functions f satisfying   |f (x) − f (y)| < ∞. f Lipθν  sup |x−y| |x−y| θ x,y∈Rn , x=y |x − y|ν (1 + ρ(x) + ρ(y) ) It is easy to see that this space is exactly the Lipschitz space when θ = 0. We will establish the relation between Lipθν and Λθν in Section 2. We are now in a position to state the main results in this paper. Theorem 1. Assume V ∈ Bq1 for some q1 > 1 = q2



n 2.

Let if q1 ≥ n,

1 1−

1 q1

+

1 n

if

n 2

< q1 < n.

= (−Δ + V )−1/2 ∇. Then for any b ∈ Λθν , 0 < ν < 1, the commutator Denote the adjoint operator of R by R is bounded from Lq (Rn ) into Lp (Rn ), 1 = 1 − ν , if q2 < p < ∞. [b, R] p q n

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We immediately deduce Corollary 1 by duality. Corollary 1. Assume V ∈ Bq1 for some q1 > 1 = q2



n 2.

Let if q1 ≥ n,

1 1−

1 q1

+

1 n

if

n 2

< q1 < n.

Then for any b ∈ Λθν , 0 < ν < 1, the commutator [b, R] is bounded from Lp (Rn ) into Lq (Rn ), 1 < p < q2 . Theorem 2. Assume V ∈ Bq1 for some q1 > n2 . Suppose b ∈ Λθν , 0 < ν < δ  . If then the commutator [b, R] is bounded from HLp (Rn ) into Lq (Rn ).

n n+ν

1 q

=

1 p

< p ≤ 1 and

1 q

=

− nν , if

1 p

− nν ,

Furthermore, we obtain the endpoint estimate for the commutator [b, R]. Theorem 3. Assume V ∈ Bq1 for some q1 > n n+ν

bounded from HL

n 2.

Suppose b ∈ Λθν , 0 < ν < δ  . Then the commutator [b, R] is

(Rn ) into weak L1 (Rn ).

Remark 1. When V ∈ Bq1 for q1 ≥ n, the Riesz transform R is exactly a Calderón–Zygmund operator. Because HLp (Rn ) is larger than the classical Hardy space H p (Rn ), Theorem 2 and Theorem 3 in this paper generalize Theorem 2.1 and Theorem 2.3 in [18], respectively. n

Theorem 2.3 in [18] implies [b, R] is not bounded from HLn+ν (Rn ) into L1 (Rn ) even if V ≡ 0. We need to impose stronger condition on the function b. To be precise, we have the following characterization. Theorem 4. Assume V ∈ Bq1 for some q1 > statements are equivalent.

n 2.

Suppose b ∈ Λθν , 0 < ν < δ  . Then the following two

n

(i) The commutator [b, R] is bounded from HLn+ν (Rn ) into L1 (Rn ). (ii) For any atom a supported in a ball B with center x0 and radius r < 

     K(x, u) b(y)a(y) dy  dx ≤ C,  

(2B)c

ρ(x0 ) 4 ,

for u ∈ B, (5)

B

where K(x, u) is the kernel of the operator R. This paper is organized as follows. In Section 2 we establish the relationship between Λθν and Lipθν . In Section 3, we recall some basic facts for the auxiliary function ρ(x) and give some estimates for the kernels In Section 4 we give some lemmas to prove the main of the Riesz transform R and its adjoint operator R. results. Section 5 gives the proofs of Theorem 1, Theorem 2, Theorem 3 and Theorem 4. Throughout this paper, unless otherwise indicated, C will be used to denote a positive constant that is not necessarily the same at each occurrence and it depends at most on the constant in (2) and the dimension n. A By A ∼ B, we mean that there exist constants C > 0 and c > 0 such that c ≤ B ≤ C. 2. The relationship between Λθν and Lipθν In this section, we shall investigate the relationship between Λθν and Lipθν . Firstly, we will introduce some lemmas.

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Lemma 1. For f ∈ Λθν , there exists a positive constant C such that

|fB(x,r) − fB(x,s) | ≤

C[f ]θν

rn+ν + sn+ν rn



s 1+ ρ(x)

θ ,

where x ∈ Rn and 0 < r < s. Proof. Assume f ∈ Λθν . Then |fB(x,r) − fB(x,s) | ≤ |fB(x,r) − f | + |f − fB(x,s) |      fB(x,r) − f (z) dz + ⇒ |fB(x,r) − fB(x,s) | dz ≤ B(x,r)

⇒

B(x,r)



  B(x, r)|fB(x,r) − fB(x,s) | ≤

  f (z) − fB(x,s)  dz

B(x,r)

  fB(x,r) − f (z) dz +

B(x,r)



  f (z) − fB(x,s)  dz.

B(x,s)

Since f ∈ Λθν , we have   B(x, r)|fB(x,r) − fB(x,s) | ≤



  fB(x,r) − f (z) dz +

B(x,r)

⇒ ⇒



  f (z) − fB(x,s)  dz

B(x,s)

 n θ n+ν r |fB(x,r) − fB(x,s) | ≤ C[f ]ν r 1+

 θ  r s n+ν 1+ +s ρ(x) ρ(x)

n+ν   θ s + sn+ν r 1+ . |fB(x,r) − fB(x,s) | ≤ C[f ]θν rn ρ(x)

Thus, we complete the proof of the lemma.

θ

2

Lemma 2. Suppose f ∈ Λθν . For any integer m ≥ 0, there exists a positive constant C such that  |fB(x,

t 2m+1

)

t − fB(x, 2m )| ≤

C[f ]θν

t 1+ ρ(x)

θ

tν , 2mν

where x ∈ Rn and t > 0. Proof. By Lemma 1, we have

  t ( 2m+1 )n+ν + ( 2tm )n+ν ( 2tm ) θ t t 1+ |fB(x, m+1 ) − fB(x, 2m )| ≤ t 2 ρ(x) ( 2m+1 )n  θ  (1+m)ν  mν−n  t 1 1 ≤ C[f ]θν 1 + tν + ρ(x) 2 2 

 −ν−n  θ ν t t 1 θ n 2 1+ ≤ C[f ]ν 1 + mν ρ(x) 2 2  θ ν t t . 2 ≤ C[f ]θν 1 + ρ(x) 2mν C[f ]θν

5

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Lemma 3. Suppose f ∈ Λθν . For any integer k > 0, there exists a positive constant C such that  |fB(x,t) − fB(x,

t 2k

)|

≤

C[f ]θν

t 1+ ρ(x)

 θ  k−1 ν −mν t 2 , m=0

where x ∈ Rn and t > 0. Proof. Because the constant C in Lemma 2 is independent of m, it is easy to check that Lemma 3 holds true by Lemma 2 and the triangle inequality. 2 ¯ 0 , t) for x0 ∈ Rn that Lemma 4. For all f ∈ Λθν , there exists a function F defined on the closed ball B(x equals f a.e. in the ball B(x0 , t) and there exists a positive constant C such that    fB(x,s) − F (x) ≤ C[f ]θν sν 1 +

s ρ(x)

θ ,

where x ∈ B(x0 , t) and s > 0. Proof. The existence of F is just the Lebesgue differentiation theorem: lims→0 fB(x,s) a.e. in B(x0 , t). ¯ 0 , t), we set Therefore, for each x ∈ B(x F (x)  lim fB(x, k→∞

t 2k

).

By Lemma 3, we have |fB(x,s) − fB(x,

s 2k

 θ ν 1+ | ≤ C[f ] s ) ν

s ρ(x)

θ k−1

2−mν .

m=0

Taking k → ∞, we get    fB(x,s) − F (x) ≤ C[f ]θν sν 1 +

s ρ(x)

θ 2

.

¯ 0 , t), x0 ∈ Rn , Lemma 5. There exists a positive constant C such that for all f ∈ Λθν and for any x, y ∈ B(x  |fB(x,s) − fB(y,s) | ≤ C[f ]θν 1 +

s s + ρ(x) ρ(y)

θ sν ,

where s = 2|x − y|. ¯ t). Then s = 2|x − y| implies that B(x, s) ∩ B(y, s) contains both B(x, s ) Proof. Let f ∈ Λθν and x, y ∈ B(x, 2 s and B(y, 2 ). Since 

 |fB(x,s) − fB(y,s) | dz ≤ B(x,s)∩B(y,s)



  fB(x,s) − f (z) dz +

B(x,s)∩B(y,s)

  f (z) − fB(y,s)  dz,

B(x,s)∩B(y,s)

then   B(x, s) ∩ B(y, s)|fB(x,s) − fB(y,s) | ≤

 B(x,s)

  fB(x,s) − f (z) dz +



B(y,s)

  f (z) − fB(y,s)  dz.

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Therefore, θ  θ  s s 1+ + 1+ ρ(x) ρ(y) θ  s s + . ≤ Csn+ν [f ]θν 1 + ρ(x) ρ(y)

  B(x, s) ∩ B(y, s)|fB(x,s) − fB(y,s) | ≤ sn+ν [f ]θν



Since B(x, 2s ) ⊂ B(x, s) ∩ B(y, s) and |B(x, 2s )| ≥ C( 2s )n , we conclude that there exists a positive constant C such that  |fB(x,s) − fB(y,s) | ≤ (l +1)θ

Theorem 5. Lipθν ⊆ Λν 0

C[f ]θν

s s 1+ + ρ(x) ρ(y)

θ sν .

2

and Λθν ⊆ Lipθν , l0 is the constant appearing in Lemma 9.

Proof. At first, for any f ∈ Lipθν , by Lemma 9, 1 |B(x, r)|1+ν/n



  f (y) − fB  dy

B(x,r)

1 ≤ |B(x, r)|2+ν/n



 

B(x,r)

≤

1 |B(x, r)|2+ν/n

B(x,r)

 





rν B(x,r)

1 ≤ |B(x, r)|2+ν/n

     f (y) − f (x) + f (x) − f (z) dz dy |x − y| |x − y| + ρ(x) ρ(y)

θ

θ   |x − z| |x − z| + + 1+ dz dy ρ(x) ρ(z)

B(x,r)

 

 r B(x,r)

1+ 

|x − y| 1+ ρ(x)

ν

(l0 +1)θ



|x − z| + 1+ ρ(x)

(l0 +1)θ

 dz dy

B(x,r)

 (l0 +1)θ 1+ν/n  r 1   B(x, r) 1+ ≤C ρ(x) |B(x, r)|1+ν/n (l0 +1)θ  r ≤C 1+ , ρ(x) (l +1)θ

(l +1)θ

then f ∈ Λν 0 . Therefore, Lipθν ⊆ Λν 0 . Secondly, we deal with the part that f ∈ Λθν ⇒ f ∈ Lipθν . For any x0 ∈ Rn , given a ball B(x0 , s) for s > 0. By Lemma 4, F (x) = limr→0 fB(x,r) and F = f a.e. in B(x0 , s). Let x, y ∈ B(x0 , s), t = 2|x − y|. Then we have       F (x) − F (y) ≤ F (x) − fB(x,t)  + |fB(x,t) − fB(y,t) | + fB(y,t) − F (y). By the last relation in the proof of Lemma 4, we obtain    fB(x,t) − F (x) ≤ C[f ]θν tν 1 +

t ρ(x)

θ

and    fB(y,t) − F (y) ≤ C[f ]θν tν 1 +

t ρ(y)

θ .

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Also, via Lemma 5, we get  |fB(x,t) − fB(y,t) | ≤

C[f ]θν

t t 1+ + ρ(x) ρ(y)

θ tν .

Putting everything together, we have        F (x) − F (y) ≤ F (x) − fB(x,t)  + |fB(x,t) − fB(y,t) | + fB(y,t) − F (y) ≤ C[f ]θν 1 + ⇒ ⇒

 θ t t |F (x) − F (y)| θ + ≤ C[f ]ν 1 + tν ρ(x) ρ(y)  θ t t |F (x) − F (y)| θ + ≤ C[f ]ν 1 + . |x − y|ν ρ(x) ρ(y)

t t + ρ(x) ρ(y)

θ tν

2

Therefore, we complete the proof of Theorem 5. 3. Estimates for the kernels of R and R

Firstly, we recall some lemmas of the auxiliary function m(x, V ) which have been proved by Shen in [19]. Throughout this section, we always assume V ∈ Bq1 for some q1 > n2 . Lemma 6. The measure V (x) dx satisfies the doubling condition, that is, there exists C > 0 such that 

 V (y) dy ≤ C B(x,2r)

V (y) dy B(x,r)

holds for all balls B(x, r) in Rn . Lemma 7. For 0 < r < R < ∞, 



1

R V (y) dy ≤ C r

rn−2

 qn −2 1



1

V (y) dy.

Rn−2

B(x,r)

B(x,R)

Lemma 8. If r = ρ(x), then 1

 V (y) dy = 1.

rn−2 B(x,r)

Moreover, 1

 V (y) dy ∼ 1

rn−2

if and only if

r ∼ ρ(x).

B(x,r)

Lemma 9. There exists l0 > 0 such that, for any x and y in Rn ,  −l0   l0 |x − y| |x − y| l0 +1 ρ(y) 1 1+ ≤C 1+ ≤ . C ρ(x) ρ(x) ρ(x) In particular, ρ(x) ∼ ρ(y) if |x − y| < Cρ(x).

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Lemma 10. There exists l1 > 0 such that  B(x,R)

l1  R V (y) dy ≤ C 1 + . ρ(x)



C V (y) dy ≤ n−2 n−2 |x − y| R

B(x,R)

For the proofs of Lemmas 6–10, we refer readers to [19]. The definition of fractional maximal operator Mγ on Rn is given as follows. Definition 3. Let f ∈ L1loc (Rn ). For 0 < γ < n, the fractional maximal operator is defined by Mγ f (x) = sup x∈B



1 γ |B|1− n

  f (y) dy,

x ∈ Rn ,

B

where the supremum on the right side is taken over all balls B ⊆ Rn . We also introduce the following maximal functions given in [2]. Definition 4. Given α > 0, the maximal functions for g ∈ L1loc (Rn ) and x ∈ Rn are defined by Mρ,α g(x) =

sup x∈B∈Bρ,α

1 |B|



  g(y) dy

B

and Mρ,α g(x) =

1 |B|

sup x∈B∈Bρ,α



  g(y) − gB  dy,

B

where Bρ,α = {B(y, r): y ∈ Rn , and r ≤ αρ(y)}. We recall the estimates of the fundamental solution of the operator −Δ + V + iτ and give the estimates of Let Γ (x, y, τ ) denote the fundamental solution kernels of the Riesz transform R and its adjoint operator R. for the operator −Δ + V + iτ , where τ ∈ R. Clearly, Γ (x, y, τ ) = Γ (y, x, −τ ). The following lemmas have been proved by Shen in [19]. Lemma 11. Let N be a positive integer. (1) Suppose V ∈ Bq1 for some q1 >   Γ (x, y, τ ) ≤ (2) Suppose V ∈ Bq1 for some q1 >

n 2.

Then there exists CN > 0 such that, for x = y, CN

(1 + |x − y||τ | n 2.

  ∇x Γ (x, y, τ ) ≤

1 2

)N (1

+ |x −

y|ρ(x)−1 )N

1 . |x − y|n−2

Then there exists CN > 0 such that, for x = y,

1 2

(1 + |x − y||τ |  1 × |x − y|n−2

CN + |x − y|ρ(x)−1 )N   1 V (z) dz . + |z − y|n−1 |x − y|n−1

)N (1

B(y,|x−y|)

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Doctopic: Real Analysis

[m3L; v 1.133; Prn:5/05/2014; 10:41] P.10 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

10

Especially, if V ∈ Bq1 for some q1 ≥ n, then there exists CN > 0 such that, for x = y,   ∇x Γ (x, y, τ ) ≤

CN (1 + |x − y||τ |

1 2

)N (1

+ |x −

y|ρ(x)−1 )N

1 . |x − y|n−1

By the functional calculus, we may write (−Δ + V )

− 12



1 =− 2π

(−iτ )− 2 (−Δ + V + τ )−1 dτ. 1

(6)

R

Let f ∈ C0∞ (Rn ). From (−Δ + V + τ )−1 f (x) =

Rn

Γ (x, y, τ )f (y) dy, it follows that

 Rf (x) =

K(x, y)f (y) dy,

(7)

Rn

where K(x, y) = −

1 2π



(−iτ )− 2 ∇x Γ (x, y, τ ) dτ . 1

(8)

R

And the adjoint operator of R is defined to be 

(x) = Rf

K(x, y)f (y) dy,

(9)

Rn

where 1 K(x, y) = − 2π



(−iτ )− 2 ∇y Γ (y, x, τ ) dτ . 1

(10)

R

can be obtained by Lemma 11 and (7)–(10). The following estimates of the kernels for R and R Lemma 12. Suppose V ∈ Bq1 for some q1 >   K(x, y) ≤

n 2.

CN (1 + |x − y|ρ(x)−1 )N

For any integer N > 0, there exists CN > 0 such that



1 |x − y|n−1



1 V (z) dz + |z − y|n−1 |x − y|n

 (11)

B(y,|x−y|)

and   K(x, y + h) − K(x, y) ≤

CN |h|δ (1 + |x − y|ρ(x)−1 )N |x − y|n−1+δ





1 V (z) dz + n−1 |z − y| |x − y|

 (12)

B(y,|x−y|)

for some δ > 0 and 0 < h < |x−y| 16 . If V ∈ Bq1 for some q1 ≥ n, then   K(x, y) ≤

CN 1 −1 N (1 + |x − y|ρ(x) ) |x − y|n

(13)

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Doctopic: Real Analysis

[m3L; v 1.133; Prn:5/05/2014; 10:41] P.11 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

11

and   K(x, y + h) − K(x, y) ≤ Lemma 13. Suppose V ∈ Bq1 for some q1 >   K(x, y) ≤

n 2.

CN (1 + |x − y|ρ(x)−1 )N

CN |h|δ . −1 N (1 + |x − y|ρ(x) ) |x − y|n+δ

(14)

For any integer N > 0, there exists CN > 0 such that





1 |x − y|n−1

1 V (z) dz + |z − y|n−1 |x − y|n

 (15)

B(y,|x−y|)

and   + h, y) − K(x, K(x y) ≤

CN |h|δ (1 + |x − y|ρ(x)−1 )N |x − y|n−1+δ





1 V (z) dz + n−1 |z − y| |x − y|

 (16)

B(y,|x−y|)

for some δ > 0 and 0 < h < |x−y| 16 . If V ∈ Bq1 for some q1 ≥ n, then   K(x, y) ≤

CN 1 (1 + |x − y|ρ(x)−1 )N |x − y|n

(17)

and   + h, y) − K(x, K(x y) ≤

CN |h|δ . −1 N (1 + |x − y|ρ(x) ) |x − y|n+δ

(18)

4. Technical lemmas n Proposition 2. There exists a sequence of points {xk }∞ k=1 in R , so that the family of critical balls Qk = B(xk , ρ(xk )), k ≥ 1, satisfies

 (i) k Qk = Rn ; (ii) There exists N such that for every k ∈ N, {j: 4Qj ∩ 4Qk = ∅} ≤ N . Proposition 3. Let θ > 0 and 1 ≤ s < ∞. If b ∈ Λθν , then there exists a positive constant C such that 

1 |B|



  b(y) − bB s dy



1/s ≤

C[b]θν rν

r 1+ ρ(x)

(l0 +1)θ

B

for all B = B(x, r) with x ∈ Rn and r > 0. Proof. By Theorem 5, b ∈ Λθν implies b ∈ Lipθν . Then 

1 |B(x, r)|



  b(y) − bB s dy

1/s

 ≤

1 |B(x, r)|

B(x,r)



  b(y) − b(x)s dy

1/s

B(x,r)

 +

1 |B(x, r)|

 B(x,r)

  b(x) − bB s dy

1/s

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[m3L; v 1.133; Prn:5/05/2014; 10:41] P.12 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

12







1 ≤2 |B(x, r)|

r

|x − y| |x − y| 1+ + ρ(x) ρ(y)

sν

sθ

1/s dy

B(x,r)

 ≤ C[b]θν rν 1 +

r ρ(x)

(l0 +1)θ ,

where we have used Lemma 9 in the last step. 2 Similar to the proof of Proposition 3, we immediately have Lemma 14. Let b ∈ Λθν , B = B(x, r) and s ≥ 1. Then there exists a positive constant C such that 

1 k |2 B|



  b(y) − bB s dy



1/s ≤

2k r 1+ ρ(x)

C[b]θν 2kν rν

(l0 +1)θ

2k B

for all k ∈ N. Lemma 15. (Cf. Lemma 2 in [2].) For 1 < p < ∞, there exist positive constants C, β and h such that if {Qk }∞ k=1 is a sequence of balls as in Proposition 2, then 

  Mρ,β g(x)p dx ≤ C

Rn

 Rn

p        1 M g(x)p dx + Σk |Qk | g(x) dx ρ,h |Qk | 2Qk

for all g ∈ L1loc (Rn ). Before we prove the main results, we need to give some necessary lemmas. Lemma 16. Let V ∈ Bq1 for q1 > n/2, and let 1 = q2



if q1 ≥ n,

1 1−

1 q1

+

1 n

if

n 2

< q1 < n

and b ∈ Λθν . Then, for q2 < m < ∞, there exists a positive constant C such that 1 |Q|



     1 (x) dx ≤ C[b]θ inf Mmν |f |m (y) m [b, R]f ν y∈Q

Q

n holds true for all f ∈ Lm loc (R ) and every ball Q = B(x0 , ρ(x0 )), where Mmν is a fractional maximal operator.

Proof. Throughout the proof of the lemma, we always assume B(x0 , ρ(x0 )). At first, we consider

n 2

< q1 < n. Let f ∈ Lp (Rn ) and Q =

  = (b − bQ )Rf −R f (b − bQ ) [b, R]f

(19)

and therefore, we need to deal with the average on Q for each term. By the Hölder inequality with m > q2 and Proposition 3,

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Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

1 |Q|



  (x) dx ≤ (b − bQ )Rf

Q



1 |Q|



    b(x) − bQ m dx

1/m 

Q

 ν ≤ C[b]θν ρ(x0 )





1 |Q|

1 |Q|

  (x)m dx Rf



13

  (x)m dx Rf

1/m

Q

1/m .

Q

is bounded on Lm (Rn ) with q2 < m, we If we write f = f1 + f2 with f1 = fχ2Q , then using the fact that R have 

ρ(x0 )

ν



1 |Q|



  1 (x)m dx Rf

1/m

 ν ≤ C ρ(x0 )

Q





1 |2Q|



  f (x)m dx

1/m

2Q



 1 ≤ C inf Mmν |f |m (y) m .

(20)

y∈Q

Now for x ∈ Q and using (15) in Lemma 13, we have     2 (x) =  Rf 

  K(x, z)f (z) dz 



|x0 −z|>2ρ(x0 )

  ≤ C I1 (x) + I2 (x) , where  I1 (x) = |x0 −z|>2ρ(x0 )

|f (z)| dz |x − z|n (1 + |x − z|/ρ(x))N

and  I2 (x) = |x0 −z|>2ρ(x0 )

|f (z)| |x − z|n−1 (1 + |x − z|/ρ(x))N

 B(z, |x−z| ) 4

V (u) du dz. |u − z|n−1

To deal with I1 (x), noting that ρ(x) ∼ ρ(x0 ) and |x − z| ∼ |x0 − z|, we split |x0 − z| > 2ρ(x0 ) into annuli to obtain 



ν  ν ρ(x0 ) I1 (x) = ρ(x0 )

|x0 −z|>2ρ(x0 )

 ν ≤ C ρ(x0 ) k≥1



≤ C inf Mmν y∈Q



|x −

2−N k k (2 ρ(x0 ))n

z|n (1

|f (z)| dz + |x − z|/ρ(x))N



  f (z) dz

|x0 −z|<2k ρ(x0 )

 1 |f |m (y) m .

(21)

Secondly, we consider the term I2 (x). We have, for x ∈ Q, 

ν  ν ρ(x0 ) I2 (x) = ρ(x0 )

 |x0 −z|>2ρ(x0 )

|f (z)| n−1 |x − z| (1 + |x − z|/ρ(x))N

 B(z, |x−z| ) 4

V (u) du dz |u − z|n−1

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Doctopic: Real Analysis

[m3L; v 1.133; Prn:5/05/2014; 10:41] P.14 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

14





ν

≤ C ρ(x0 )

|x0 −z|>2ρ(x0 )

ν



≤ C ρ(x0 )

k≥1

 ν ≤ C ρ(x0 ) k≥1



≤ C inf Mmν y∈Q



|f (z)| n−1 |x0 − z| (1 + |x0 − z|/ρ(x))N 

−N k

2 (2k ρ(x0 ))n−1 2−N k (2k ρ(x0 ))n−1



|x0

B(z,4|x0 −z|)



  f (z)

|x0 −z|<2k+1 ρ(x0 )



B(z,2k+3 |x0 −z|)

  f (z)I1 (V χB(x

−z|<2k+1 ρ(x

V (u) du dz |u − z|n−1

V (u) du dz |u − z|n−1

k 0 ,2 ρ(x0 ))

)(z) dz

0)

 1 |f |m (y) m ,

where I1 denotes the fractional integral operator and we have used the proof of I2 in Lemma 5 in [2] in the last inequality. mm ˜ To deal with the second term of (19), we also split f = f1 + f2 . Choose q2 < m ˜ < m and set t = m− m ˜. m ˜ n Using the boundedness of R on L (R ) and the Hölder inequality, we get 1 |Q|





  1 (b − bQ )(x) dx ≤ Rf

Q

1 |Q| 

≤C  ≤C



 ˜ 1 (b − bQ )(x)m Rf dx

1/m ˜

Q

1 |Q| 1 |Q|



   ˜ f (x) b(x) − bQ m dx

1/m ˜

2Q



  f (x)m dx

1/m 

2Q



≤ C[b]θν inf Mmν



y∈Q

1 |Q|



   b(x) − bQ t dx

1/t

2Q

 1 |f |m (y) m ,

where we have used Proposition 3 in the last inequality. Similarly, for x ∈ Q and using (15) in Lemma 13, we have     2 (x) =  Rf 

    z)f (z) b(z) − bQ dz  K(x, 



|x0 −z|>2ρ(x0 )

  ≤ C I˜1 (x) + I˜2 (x) , where  I˜1 (x) = |x0 −z|>2ρ(x0 )

|f (z)(b(z) − bQ )| dz |x − z|n (1 + |x − z|/ρ(x))N

and  I˜2 (x) = |x0 −z|>2ρ(x0 )

|f (z)(b(z) − bQ )| |x − z|n−1 (1 + |x − z|/ρ(x))N

We start by observing that for 1 ≤ m ˜ < m, t =  f (b − bQ )χB(x

k 0 ,2 ρ(x0 ))

 

m ˜

mm ˜ m−m ˜,

 B(z, |x−z| ) 4

V (u) du dz. |u − z|n−1

and by Lemma 14, we obtain

  ≤ f χB(x0 ,2k ρ(x0 )) m (b − bQ )χB(x0 ,2k ρ(x0 )) t

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[m3L; v 1.133; Prn:5/05/2014; 10:41] P.15 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••





  f (y)m dy

≤C

 m1

15

ν  n/t  (l0 +1)θ  k [b]θν 2k ρ(x0 ) 1 + 2k 2 ρ(x0 )

B(x0 ,2k ρ(x0 ))

    1 n ≤ C2k(l0 +1)θ [b]θν 2k ρ(x0 ) m˜ inf Mmν |f |m (y) m . y∈Q

(22)

For I˜1 (x), using (22) with m ˜ = 1, we have I˜1 (x) ≤ C



2−N k (2k ρ(x0 ))n

k≥1



   b(z) − bQ f (z) dz

|x0 −z|>2k ρ(x0 )

    1 k(−N +(l0 +1)θ) ≤ C[b]θν inf Mmν |f |m (y) m 2 y∈Q

k≥1



≤ C[b]θν inf Mmν



y∈Q

 1 |f |m (y) m

if we choose N sufficiently large. To deal with I˜2 (x), we discuss as in the estimate for I2 (x) with f (b − bQ ) instead of f and m ˜ and q˜1 1 1 1 instead of m and q1 , where m ˜  = q˜1 − n . Similar to the proof of (28) in [2] and also using (22), we have I˜2 (x) ≤ Cρ(x0 )−n+ q˜1 −1 n



   2k(−N +1+nμ−n−n/(˜q1 )) (b − bQ )f χB(x0 ,2k ρ(x0 )) m ˜

k≥1

  1 k(−N +(l0 +1)θ+2−n+nμ) ≤ C[b]θν inf Mmν |f |m (y) m 2 







y∈Q

k≥1

≤ C[b]θν inf Mmν y∈Q

 1 |f |m (y) m ,

where we choose N large enough to ensure that the above series converges. 2 which can be Now we recall the Lp –Lq estimates for the Riesz transform R and its adjoint operator R, deduced by Theorem 0.5 in [19]. Proposition 4. Assume V ∈ Bq1 for q1 > 1 = q2

n 2.



Let if q1 ≥ n,

1 1−

1 q1

+

1 n

if

n 2

< q1 < n.

Then there exists C > 0 such that for any f ∈ C0∞ (Rn ),   Rf (x)

Lp (Rn )

≤ Cf Lp (Rn ) ,

where 1 < p ≤ q2 . Proposition 5. Assume V ∈ Bq1 for some q1 > 1 = q2



n 2.

Let if q1 ≥ n,

1 1−

1 q1

+

1 n

Then there exists C > 0 such that for any f ∈ C0∞ (Rn ),

if

n 2

< q1 < n.

(23)

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Doctopic: Real Analysis

[m3L; v 1.133; Prn:5/05/2014; 10:41] P.16 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

16

Lp (Rn ) ≤ Cf Lp (Rn ) , Rf where q2 < p < ∞. = (−Δ + V )−1/2 ∇ be the adjoint operator of the Riesz transform R. Then there exists Lemma 17. Let R C > 0, such that for any f ∈ L1loc (Rn ) and b ∈ Λθν ,     1  (x) ≤ C[b]θ Mmν |f |m (x) m . Mρ,h [b, R]f ν

(24)

Proof. Let f ∈ L1loc (Rn ), x ∈ Rn and a ball I = B(x0 , r) with x ∈ I and r < hρ(x0 ), h > 0, we need to

1 (y) − c| dy by the right side of (24) for some constant c, which will be designated |[b, R]f control J = |I| I later. Let f = f1 + f2 , where f1 = f χ2I and f2 = f − f1 . Then = [b − b2I , R]f = (b − b2I )Rf − R(b − b2I )f1 − R(b − b2I )f2 [b, R]f . = A1 f + A2 f + A3 f. Take c =

|x0 −z|≥2r

˜ 0 , z)(b(z) − b2I )f (z) dz. Then we have K(x 1 J≤ |I|



  A1 f (y) dy + 1 |I|

I

 I

  A2 f (y) dy + 1 |I|



  A3 f (y) − c dy

I

. = J1 + J2 + J3 . At first, we consider J1 . Note that |x − x0 | ≤ r < hρ(x0 ) implies ρ(x) ∼ ρ(x0 ). By the Hölder inequality, we have    1 A1 f (y) dy J1 ≤ |I| I

1 = |I| C ≤ |I| C ≤ |I|



   (y) dy  b(z) − b2I Rf

I



  m b(z) − b2I  m−1 dy

I

 2I

  m b(z) − b2I  m−1 dy 

1 n



r C θ [b]ν 1 + ≤ |I| ρ(x0 ) ≤ C  [b]θν |I|− m + n 1

≤C for m > q2 .

[b]θν

  (y)m dy Rf

 m1

I

|2I| C θ [b]ν 1 + ≤ |I| ρ(x0 )



1− m1 



(l0 +1)θ

(l0 +1)θ

 I

Mmν |f |

m



ν n

ν n

  f (y)m dy  m1

m−1 m



 m1

  (y)m dy Rf

I

|I| |I|

(x)

  (y)m dy Rf

I

|I| |I|

ν



1− m1 

 m1

m−1 m

 I

  (y)m dy Rf

 m1  m1

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AID:18478 /FLA

Doctopic: Real Analysis

[m3L; v 1.133; Prn:5/05/2014; 10:41] P.17 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

17

For J2 , by the Hölder inequality and Proposition 5, J2 ≤

1 |I| 

≤



  A2 f (y) dy

I

1 |I|



1 ≤C |I| C 1 |I| m˜

 m1˜  

˜    b(y) − b2I f (y)m dy



  m b(y) − b2I  m−m˜ dy

 m1˜ − m1 

2I

  f (y)m dy

 m1

2I 1

|2I| n 1+ ρ(x0 )

ν C[b]θν |I| n

 ≤

 m1˜

2I

 ≤

 m1˜

I



≤

˜  A2 f (y)m dy

ν C[b]θν |I| n

r 1+ ρ(x0 )

≤ C  [b]θν |I| n − m ν

1



(l0 +1)θ 

1 |I|

(l0 +1)θ 

  f (y)m dy

1 |I|

 m1  

  f (y)m dy

 m1

2I

 m1  

  f (y)m dy

 m1

2I

 m1

2I

≤ C  [b]θν





1 |2I|1−

mν n

  f (y)m dy

 m1

2I

   1 ≤ C  [b]θν Mmν |f |m (x) m , where m > m ˜ > q2 . Finally, we consider J3 . Case of q 1 ≥ n: By Lemma 13 and Lemma 14, we have 



   z) − K(x 0 , z) m dz  K(y,

2k r≤|z−x0 |<2k+1 r

≤

CN rδ {1 + m(x0 , V )2k r}N

 m1 



   b(z) − b2I s dz

 1s

2k r≤|z−x0 |<2k+1 r





2k r≤|z−x0 |<2k+1 r

1 dz |z − x0 |m (n+δ)



1 m





   b(z) − b2I s dz

 1s

|z−x0 |<2k+1 r

 (l0 +1)θ  k  ns 2k r 1 rδ θ kν ν 1 + ≤ CN [b] 2 r 2 r n ν {1 + m(x0 , V )2k r}N (2k r)n− m +δ ρ(x0 ) ≤ CN [b]θν 2kν rν where

1 m

+

1 m

+

1 s

1 rδ , n k N −(l +1)θ k 0 {1 + m(x0 , V )2 r} (2 r) m +δ

= 1.

(25)

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Doctopic: Real Analysis

[m3L; v 1.133; Prn:5/05/2014; 10:41] P.18 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

18

Therefore, via the Hölder inequality,    2 A3 f (y) − c dy J3 ≤ |I| I

   1  ≤ |I| 



z) − K(x 0 , z) K(y,



   b(z) − b2I f (z) dz  dy

|z−x0 |>2r

I

1 ≤ |I|







    z) − K(x 0 , z) b(z) − b2I f (z) dz dy  K(y,

I |z−x0 |>2r

 ∞ 1 ≤ |I|

k=1

I



    z) − K(x 0 , z) b(z) − b2I f (z) dz dy  K(y,

2k r≤|z−x0 |<2k+1 r

 ∞ C[b]θν 1 rδ ≤ 2kν rν n n |I| {1 + m(x0 , V )2k r}N −(l0 +1)θ (2k r) m − s +δ k=1 I





  f (z)m dz

×

 m1

|z−x0 |<2k+1 r

≤ C[b]θν  ×

∞

CN

k=1

n  k ν  k+1  m 1 rδ −ν r 2 r 2 n +δ k N −(l +1)θ k 0 m {1 + m(x0 , V )2 r} (2 r)



1 k+1 (2 r)n−mν

  f (z)m dz

 m1

|z−x0 |<2k+1 r

≤ C[b]θν

∞

CN 2−kδ

k=1

   1 1 Mmν |f |m (x) m k N −(l +1)θ 0 {1 + m(x0 , V )2 r}

   1 ≤ C[b]θν Mmν |f |m (x) m if we choose N sufficiently large. Case of

n 2



< q 1 < n: By Lemma 10, Lemma 13 and (25), we have 

  z) − K(x 0 , z) p dz  K(y,

 p1 

2k r≤|z−x0 |<2k+1 r

≤ CN



×



 CN [b]θν 2kν rν ≤ {1 + m(x0 , V )2k r}N −(l0 +1)θ







2k r≤|z−x0 |<2k+1 r B(x0 ,2k+3 r)

   b(z) − b2I s dz

[b]θν 2kν rν {1 + m(x0 , V )2k r}N −(l0 +1)θ

   



 1s +

2k r≤|z−x0 |<2k+1 r

≤ CN

   b(z) − b2I t dz

 1t

2k r≤|z−x0 |<2k+1 r

rδ 1 k (2 r)n−1+δ {1 + m(x0 , V )2k r}N





[b]θν 2kν rν rδ n {1 + m(x0 , V )2k r}N −(l0 +1)θ (2k r) m +δ 

rδ

 V (z)q1 dz

n

(2k r)n−1− t +δ

rδ (2k r) −2 n (2k r)n−1− t +δ n q1

p  p1  V (y)  dz dy  n−1 |x0 − y|



 q1

1

+

B(x0 ,2k+3 r)





1 (2k+3 r)n−2

V (z) dz B(x0

,2k+3 r)

rδ n (2k r) m +δ



rδ + k n +δ (2 r) m



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[m3L; v 1.133; Prn:5/05/2014; 10:41] P.19 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••



 ≤ CN

[b]θν 2kν rν {1 + m(x0 , V )2k r}N −l1 −(l0 +1)θ

 ≤ CN

[b]θν 2kν rν rδ , n {1 + m(x0 , V )2k r}N −l1 −(l0 +1)θ (2k r) m +δ



rδ n−1− n t +δ

(2k r)

2k r

 qn −2 1

+

rδ n (2k r) m +δ

19



1 where m + p1 + 1t = 1 and q11 = p1 + n1 . Therefore, for m > q2 ,    2 A3 f (y) − c dy J3 ≤ |I| I

   1  ≤ |I| 

k=1

I

 [b]θν CCN

≤ C[b]θν ×

   b(z) − b2I f (z) dz  dy

    z) − K(x 0 , z) b(z) − b2I f (z) dz dy  K(y,

I |z−x0 |>2r





z) − K(x 0 , z) K(y,





 ∞ 1 ≤ |I| ≤



|z−x0 |>2r

I

1 ≤ |I|



    z) − K(x 0 , z) b(z) − b2I f (z) dz dy  K(y,

2k r≤|z−x0 |<2k+1 r

1 |I|

∞



 ∞ I

 CN

k=1

[b]θν 2kν rν rδ n {1 + m(x0 , V )2k r}N −l1 −(l0 +1)θ (2k r) m +δ k=1





  f (z)m dz

 m1

|z−x0 |<2k+1 r

n  k ν  k+1  m 1 rδ −ν 2 r 2 r n +δ k N −l −(l +1)θ k 1 0 m {1 + m(x0 , V )2 r} (2 r)



1 k+1 (2 r)n−mν

  f (z)m dz

 m1

|z−x0 |<2k+1 r  ≤ CCN [b]θν

∞

2−kδ

k=1

   1 1 Mmν |f |m (x) m k N −l −(l +1)θ 1 0 {1 + m(x0 , V )2 r}

   1 ≤ C  [b]θν Mmν |f |m (x) m if we choose N sufficiently large. 2 5. Proofs of the main results In this section we show that Theorems 1–4 hold true, respectively. Proof of Theorem 1. Suppose b ∈ Λθν . We choose m such that it satisfies q2 < m < p. We conclude from Proposition 2, Lemma 15, Lemma 16 and Lemma 17 that        p p n ≤ Mρ,β [b, R] (x)p dx [b, R]f L (R ) Rn



≤C Rn



≤C Rn

p        p 1   M  [b, R]f (x) dx |Qk | ρ,h [b, R]f (x) dx + |Qk | k

   p  θ p  M ρ,h [b, R]f (x) dx + [b]ν



k 2Q k

2Qk

    1  Mmν |f |m m (x)p dx

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Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

20

   1 p ≤ C[b]θν  Mmν |f |m m Lp (Rn ) ≤ C[b]θν f pLq (Rn ) , where p1 = 1q − nν . This completes the proof of Theorem 1.

2

Proof of Theorem 2. Choose ε such that 1 < ε < q2 , where q2 is given as in Theorem 1. In order to prove Theorem 2, we only need to show that for any HLp,ε -atom a,   [b, R]a q n ≤ C L (R ) holds true, where C is a constant independent of a. Suppose supp a ⊆ B = B(x0 , r) with r < ρ(x0 ). Then we have   [b, R]a

Lq





  [b, R]a(x)q dx

≤

 q1





  [b, R]a(x)q dx

+

|x−x0 |<2r

 q1

|x−x0 |≥2r

. = A1 + A 2 . Let

1 q3

=

1 ε

− nν . By Corollary 1 and the size condition of the atom a, we have 



  [b, R]a(x)q dx

A1 ≤

 q1

|x−x0 |<2r





  [b, R]a(x)q3 dx

≤

 q1

3

(2r) q − q3 n

n

|x−x0 |<2r





  a(x)ε dx

≤C

 1ε

(2r) q − q3 n

n

|x−x0 |<2r

≤ C(2r) ε − p (2r) q − q3 n

n

n

n

≤ C. For A2 , we consider two cases, that is,

ρ(x0 ) 4

≤ r < ρ(x0 ) and r <

ρ(x0 ) 4 .

Case I: At first, we note that Λθν ⊆ Lipθν . Therefore, b ∈ Λθν implies  θ   b(x) − b(x0 ) ≤ C[b]θν |x − x0 |ν 1 + |x − x0 | + |x − x0 | ρ(x) ρ(x0 )  (l0 +1)θ |x − x0 | ≤ C[b]θν |x − x0 |ν 1 + , ρ(x0 ) where we have used Lemma 9 in the last step. When r < ρ(x4 0 ) , by the vanishing condition of a, we have                [b, R]a(x) ≤ b(x) − b(x0 ) K(x, y) − K(x, x0 ) a(y) dy  +  K(x, y) b(y) − b(x0 ) a(y) dy      B

B

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[m3L; v 1.133; Prn:5/05/2014; 10:41] P.21 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

 ≤



|x − x0 | 1+ ρ(x0 )

|x − x0 |

C[b]θν

ν

  + C[b]θν rν 1 +

21

 (l0 +1)θ       K(x, y) − K(x, x0 ) a(y) dy    B

 (l0 +1)θ    r  K(x, y)a(y) dy  .   ρ(x0 ) B

By the Minkowski inequality,  A2 ≤

C[b]θν |x

− x0 |

ν



|x − x0 | 1+ ρ(x0 )

(l0 +1)θ 

q  q1       K(x, y) − K(x, x0 ) a(y) dy  dx  



|x−x0 |≥2r B

q  q1     K(x, y)a(y) dy  dx  



+ C[b]θν rν

|x−x0 |≥2r B

  (l +1)θ    |x − x0 | 0 a(y) ≤ C[b]θν |x − x0 |ν 1 + ρ(x0 )   a(y)

+ C[b]θν rν



  K(x, y) − K(x, x0 )q dx

 q1 dy

|x−x0 |≥2r

B







  K(x, y)q dx

 q1 dy

|x−x0 |≥2r

B

. = A21 + A22 . Step I(1). If q1 ≥ n, by Lemma 12, we have 



   K(x, y) − K(x, x0 ) q dx

 q1

2k r≤|x−x0 |<2k+1 r

1 ≤ CN r {1 + m(x0 , V )2k r}N





δ

≤ CN

2k r≤|x−x0 |<2k+1 r

1 dx |x − x0 |q(n+δ)

 q1

1 rδ , n {1 + m(x0 , V )2k r}N (2k r) q +δ

(26)

where 1q + q1 = 1. Note that 

  a(x) dx ≤ aLε |B|1− 1ε ≤ Crn− np .

B

Then by (26), we have  A21 ≤ C[b]θν

 (l0 +1)θ  ∞    k ν 2k r a(y) 1+ 2 r ρ(x0 ) k=1

B

≤ C  [b]θν



  a(y)

B

≤ C  [b]θν rn− p

n



  K(x, y) − K(x, x0 )q dx

2k r≤|x−x0 |<2k+1 r

∞

 k ν CN rδ 2 r n +δ {1 + m(x0 , V )2k r}N −(l0 +1)θ (2k r) q k=1

∞

 k ν CN rδ 2 r n +δ k N  {1 + m(x0 , V )2 r} (2k r) q k=1

 q1 dy

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[m3L; v 1.133; Prn:5/05/2014; 10:41] P.22 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

22

≤ C  [b]θν

∞

2−kδ+ p k−nk n

k=1

≤C



[b]θν

n if we choose N large enough. Moreover, since p > n+ν , then p > n If 2 ≤ q1 < n, by Lemma 10 and Lemma 12 we have





   K(x, y) − K(x, x0 ) q dx

≥

n n+δ 

n n+δ .

Therefore, −kδ + np k − nk < 0.

 q1

2k r≤|x−x0 |<2k+1 r

≤ CN

rδ 1 k (2 r)n−1+δ {1 + m(x0 , V )2k r}N



   





2k r≤|x−x0 |<2k+1 r B(x0 ,2k+3 r)

1 rδ n k N {1 + m(x0 , V )2 r} (2k r) q +δ   1 rδ ≤ CN {1 + m(x0 , V )2k r}N (2k r)n−1+δ

q  q1  V (z)  dx dz  n−1 |x0 − z|

+

≤

≤

 CN

 CN

 ≤ CN

1 {1 + m(x0 , V )2k r}N



V (z) dz

1 {1 + m(x0 , V )2k r}N −l1





+

 V (z) dz

B(x0

,2k+3 r)

 k  qn −2 r rδ 1 2 r + n (2k r)n−1+δ (2k r) q +δ δ

n q

(2k r)



1 k+3 (2 r)n−2



rδ

1

q1

B(x0 ,2k+3 r)

rδ (2k r)n−1+δ

 q1





+δ

k

2 r

 qn −2 1

+

rδ



n

(2k r) q +δ



1 rδ , n {1 + m(x0 , V )2k r}N −l1 (2k r) q +δ

(27)

where 1q + q1 = 1 and 1q = q11 − n1 . Then by (27), we similarly can arrive at the estimate of A21 . Step I(2). We are now in a position to deal with the term A22 . Note that |x − x0 | ≥ 2r and |y − x0 | < r. Therefore, 1 1 |x − y| ≥ |x − x0 | − |y − x0 | ≥ |x − x0 | − |x − x0 | = |x − x0 |. 2 2 By Lemma 9, 

   1   1 + m(x, V )|x − y| ≥ C 1 + m(x, V )|x − x0 | ≥ Cc 1 + m(x0 , V )|x − x0 | l0 +1 .

(28)

If q1 ≥ n, by (28) and Lemma 12 we have 



  K(x, y)q dx

 q1

≤ CN rδ

2k r≤|x−x0 |<2k+1 r

≤ CN where

1 q

+

1 q

= 1.



1



N

{1 + m(x0 , V )2k r} l0 +1

rδ

1 {1 + m(x0 , V )2k r}

2k r≤|x−x0 |<2k+1 r

N l0 +1

n

(2k r) q +δ

,

1 dx |x − x0 |q(n+δ)

 q1

(29)

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[m3L; v 1.133; Prn:5/05/2014; 10:41] P.23 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

23

Since |y − x0 | < r < ρ(x0 ), ρ(y) ∼ ρ(x0 ). Then by (29),  A22 ≤

∞   a(y)

C[b]θν rν

k=1

B



∞   a(y)

≤ C[b]θν rν

k=1

B n

≤ C[b]θν rν+n− p ∞



  K(x, y)q dx

 q1

2k r≤|x−x0 |<2k+1 r

CN {1 + m(x0 , V

∞

1 N

n

)2k r} l0 +1

CN

(2k r) q 1

{1 + m(x0 , V )2k r}

k=1

≤ C[b]θν



N l0 +1

n

(2k r) q

2−kν+ p k−nk n

k=1

≤

C[b]θν

n if we choose N sufficiently large and p > n+ν . n If 2 ≤ q1 < n, by (28), Lemma 10 and Lemma 12 again, we have





  K(x, y)q dx

 q1

2k r≤|x−x0 |<2k+1 r

≤ CN

1 (2k r)n−1





1

2k r≤|z−x0 |<2k+1 r

+

1 {1 + m(x0 , V

≤ CN

≤ CN  ≤ CN  ≤ CN

where

1 q

+

1 q





1 N

{1 + m(x0 , V )2k r} l0 +1

1 (2k r)n−1 1 (2k r)n−1 

{1 + m(x0 , V )2k r} l0 +1 −l1 1 N

{1 + m(x0 , V )2k r} −

1 n

N l0 +1 −l1

and

1 q1

+



V (y) dy 

,2k+3 r)



V (y) dy

1 n

n

(2k r) q

− qn

1

|x−x0 |≥2r

,



n

(2k r) q

= 1. Then by (30), we similarly deduce the estimate of A22 . 1 4.

The atom a does not satisfy the vanishing

q  q1   θ   r  1+ K(x, y)a(y) dy  dx  ρ(x0 )

|x−x0 |≥2r B



1

(30)

B



+



q  q1      b(x) − b(x0 )q  K(x, y)a(y) dy  dx  



A2 ≤

+ C[b]θν rν

2k r

 k n−2− qn 1 1 1 + 2 r n k n−1 (2 r) (2k r) q

Case II: When ρ(x4 0 ) ≤ r < ρ(x0 ), that is, rm(x0 , V ) ≥ condition. By the Minkowski inequality, 

 

+



1

1

B(x0 ,2k+3 r)

(2k r) q 1 q1

 q1

 q1

B(x0

1

1 q1

B(x0 ,2k+3 r)

n

N

=

0

q  q1  V (y)  dz dy  n−1 |x0 − y|

(2k r) q

{1 + m(x0 , V )2k r} l0 +1

1 q



1 N

)2k r} l0 +1

1

= 1,

{1 + m(x0 , V )2k r}

   N  l +1

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[m3L; v 1.133; Prn:5/05/2014; 10:41] P.24 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

24

 ≤

C[b]θν |x

− x0 |

ν

|x − x0 | 1+ ρ(x0 )

(l0 +1)θ 



+ C[b]θν rν



|x−x0 |≥2r

B

  K(x, y)q dx

 q1 dy

|x−x0 |≥2r

B

   a(y)





  a(y)

 q  q1 θ    1+ r  dx K(x, y) dy   ρ(x0 )

= A21 + A22 . Step II(1). Note that |x − y| ∼ |x − x0 |. If q1 ≥ n, by (29),  A21 ≤

C[b]θν

 (l0 +1)θ  ∞    k ν 2k r a(y) 1+ 2 r ρ(x0 ) k=1

B

 ≤ C[b]θν

∞   a(y) k=1

B



≤ C[b]θν

∞   a(y) k=1

B n

≤ C[b]θν rn− p ≤ CCN [b]θν

∞

CN

(2k r)

N l0 +1 −(l0 +1)θ

(2k r) q

{1 +

 q1 dy



1 n

(2k r) q

2k r

n

2k r

2k r

ν

ν

ν

1

n

2 p k−nk



1

{1 + m(x0 , V )2k r} N

n q

N l0 +1 −(l0 +1)θ

CN

2k } l0 +1 −(l0 +1)θ



1

{1 + m(x0 , V )2k r}

CN

k=1

  K(x, y)q dx

2k r≤|x−x0 |<2k+1 r

∞ k=1



{1 +

2k } l0 +1 −(l0 +1)θ N

≤ CCN [b]θν , where we choose N sufficiently large such that the above series converges. If n2 ≤ q1 < n, we also obtain the corresponding estimate of A22 via (30). Hence, we omit the detail of the proof. 2 n n ∞ ,2 Proof of Theorem 3. For f ∈ HLn+ν (Rn ), we write f = j=−∞ λj aj , where each aj is an HLn+ν -atom and ∞ n n+ν n . Suppose that supp a ⊆ B = B(x , r ) with r < ρ(x ). Write ( j=−∞ |λj | n+ν ) n ≤ 2f  n+ν j j j j j j

HL

[b, R]f (x) =

∞ j=−∞

j: rj ≥



+ j:



  λj b(x) − b(xj ) Raj (x)χ8Bj (x) + 



ρ(xj ) 4

  λj b(x) − b(xj ) Raj (x)χ(8Bj )c (x) 

∞

λj b(x) − b(xj ) Raj (x)χ(8Bj )c (x) − R

   λj b − b(xj ) aj (x)

j=−∞

ρ(x ) rj < 4 j

= A1 (x) + A2 (x) + A3 (x) + A4 (x). We show this case by a method similar to the one used in the proof of Theorem 4.1 in [12]. By Proposition 4, we obtain     b(x) − b(xj ) Raj (x)χ8B (x) 1 n ≤ C[b]θν j L (R )





|x−xj |≤8rj



8rj 1+ ρ(xj )

θ

 |x − xj | Raj (x) dx 

ν



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[m3L; v 1.133; Prn:5/05/2014; 10:41] P.25 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

≤C





 12 

 |x − xj |

[b]θν

2ν

dx

|x−xj |≤8rj ν+ n 2

≤ C  [b]θν rj





25

  Raj (x)2 dx

 12

|x−xj |≤8rj



  aj (x)2 dx

 12

|x−xj |≤8rj ν+ n 2

≤ C  [b]θν rj

n 2 −(n+ν)

rj

= C  [b]θν . Therefore,   A1 (x)

≤C

L1 (Rn )



∞

|λj | ≤ C

j=−∞

≤ Cf 

 n+ν n

∞

|λj |

n n+ν

j=−∞ n

HLn+ν

.

Secondly, we consider the term A2 (x). It is easy to see that |x − xj | ∼ |x − y| and 

     1 1 + m(x, V )|x − y| ≥ C 1 + m(x, V )|x − xj | ≥ Cc 1 + m(xj , V )|x − xj | l0 +1 .

Note that rj ≥ 

ρ(xj ) 4 .

If



n 2

< q1 < n, by the Hölder inequality, (11) and (8), we obtain, for some t > 1,

  K(x, y) dx



2k rj ≤|x−xj |<2k+1 rj





  K(x, y)t dx

≤

 1t

 k+1  n 2 rj t

2k rj ≤|x−xj |<2k+1 rj

≤ CCN

+

N

{1 + m(xj , V )2k rj } l0 +1 

1 N

{1 + m(xj , V )2k r} l0 +1

≤ CCN

1 t

=

 N

N

{1 + m(xj , V )2k rj } l0 +1

N

{1 + m(xj , V )2k rj } l0 +1 −l1 − n1 .

n

(2k+1 rj ) t (2k rj )n−1

{1 + m(xj , V )2k rj } l0 +1 −l1 1

1 q1



,



1 dx |x − xj |tn

 q1



1

t  1t  V (z)  dz dz  n−1 |xj − z|

 1t

 q1

V (z) dz B(xj ,2k+3 rj )

1



2k rj ≤|x−xj |<2k+1 rj B(xj ,2k+3 rj )

n



N

   





(2k+1 rj ) t (2k rj )n−1



1



≤ C CN

rj

 tn



2k rj ≤|x−xj |<2k+1 rj

{1 + m(xj , V )2k rj } l0 +1

≤ C CN

 ≤ CN

2

(2k+1 rj ) t (2k rj )n−1

k+1

1



where

n

1



(2k+3 rj )n



1

+1 



 n V (z) dz 2k rj q1 + 1

B(xj ,2k+3 rj )

 n (2k+1 rj ) t  k −2+ qn1 2 rj +1 (2k rj )n−1

(31)

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[m3L; v 1.133; Prn:5/05/2014; 10:41] P.26 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••

26

If q1 ≥ n, we similarly have 



  K(x, y) dx



2k rj ≤|x−xj |<2k+1 rj





  K(x, y)t dx

≤

 1t

 k+1  tn rj 2

2k rj ≤|x−xj |<2k+1 rj

≤ CCN

 ≤ CN



1 N

{1 + m(xj , V )2k r} l0 +1 1 N

{1 + m(xj , V )2k rj } l0 +1

2k+1 rj

 tn





2k rj ≤|x−xj |<2k+1 rj

1 dx |x − xj |tn

 1t

,

(32)

where 1t = q11 − n1 . Similar to the proof of A21 of Case II in Theorem 2 and by the estimates (31) and (32) of K(x, y),    b(x) − b(xj ) Raj (x)χ(8B  ≤

j)

c

 (x)L1 (Rn )



 |x − xj |

[b]θν

ν

|x−xj |≥8rj

 ≤

C[b]θν 

≤

  (l0 +1)θ     K(x, y)aj (y) dy  dx   Bj



∞    k ν |x − xj | aj (y) 1+ 2 rj ρ(xj )

(l0 +1)θ 

k=1

Bj

C[b]θν

|x − xj | 1+ ρ(xj )

∞   aj (y) k=1

Bj

≤ C[b]θν rj−ν  ≤ CCN [b]θν

   K(x, y) dx dy



2k rj ≤|x−xj |<2k+1 rj

 CN (2k rj )ν

{1 + m(xj , V )2k rj } l0 +1 −l1 −(l0 +1)θ N

∞

 CN (2k rj )ν

k=1

{1 + 2k } l0 +1 −l1 −(l0 +1)θ N

∞

2kν

k=1

1 {1 +

2k } l0 +1 −l1 −(l0 +1)θ N

 ≤ CCN [b]θν

(33)

if we choose N large enough. Therefore,

j:

n

HLn+ν

n 2

|λj | ≤ C

ρ(x ) rj ≥ 4 j

≤ Cf  Thirdly, we consider the term A3 . If t > 1,





  A2 (x) 1 n ≤ C L (R )

∞

 n+ν n |λj |

n n+ν

j=−∞

.

< q1 < n, by the Hölder inequality, (12) and (2) we get, for some

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Doctopic: Real Analysis

[m3L; v 1.133; Prn:5/05/2014; 10:41] P.27 (1-31)

Y. Liu, J. Sheng / J. Math. Anal. Appl. ••• (••••) •••–•••





  K(x, y) − K(x, xj ) dx

27



2k rj ≤|x−xj |<2k+1 rj





  K(x, y) − K(x, xj )t dx

≤

 1t

 k+1  tn 2 rj

2k rj ≤|x−xj |<2k+1 rj

≤ CCN

+

n

N

{1 + m(xj , V )2k rj } l0 +1 1

rjδ

N

{1 + m(xj , V )2k r} l0 +1

≤ CCN



2

k+1

 N

{1 + m(xj , V )2k rj } l0 +1

≤ C  CN

rj

 tn



   









n

rjδ (2k+1 rj ) t (2k rj )n−1+δ

t  1t  V (z) dz  dz |xj − z|n−1 

1 dx |x − xj |t(n+δ)

 V (z)q1 dz

 q1

1

 1t

+ 2−kδ



B(xj ,2k+3 rj )



1 N

{1 + m(xj , V )2k rj } l0 +1

n t

rjδ (2k+1 rj ) (2k rj )n−1+δ

1

rjδ

{1 + m(xj , V )2k rj } l0 +1 −l1 N

1 {1 + m(xj , V )2k rj } l0 +1 −l1







1

V (z) dz

(2k+3 rj )n B(xj ,2k+3 rj )



N



2k rj ≤|x−xj |<2k+1 rj B(xj ,2k+3 rj )

2k rj ≤|x−xj |<2k+1 rj

1

≤ C  CN

 ≤ CN

rjδ (2k+1 rj ) t (2k rj )n−1+δ

1

(2k+1 rj )  k −2+ qn1 2 rj + 2−kδ (2k rj )n−1+δ n t

 k  qn 2 rj 1 + 2−kδ





2−kδ ,

(34)

where 1t = q11 − n1 . If q1 > n, using the Hölder inequality, we have, for some t > 1, 



  K(x, y) − K(x, xj ) dx



2k rj ≤|x−xj |<2k+1 rj





  K(x, y) − K(x, xj )t dx

≤

 1t



2k+1 rj

 n t

2k rj ≤|x−xj |<2k+1 rj

≤ CCN  ≤ CN

1 N

{1 + m(xj , V )2k r} l0 +1 1

{1 + m(xj , V

)2k r

j}

N l0 +1

rjδ



2

k+1

rj

 tn





2k rj ≤|x−xj |<2k+1 rj

1 dx |x − xj |t(n+δ)

 1t

2−kδ ,

(35)

where 1t = q11 − n1 . Similar to the proof of A21 of Case I in Theorem 2, by the vanishing condition of aj and the estimates (34) and (35), then    b(x) − b(xj ) Raj (x)χ(8B  ≤

j)

c

 (x)L1 (Rn ) 

 |x − xj |

C[b]θν

ν

|x−xj |≥2rj

|x − xj | 1+ ρ(xj )

  (l0 +1)θ       K(x, y) − K(x, xj ) aj (y) dy  dx   Bj

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28

 ≤

C[b]θν

 (l0 +1)θ  ∞    k ν 2k r j aj (y) 1+ 2 rj ρ(xj ) k=1

Bj

 ≤ C[b]θν

2k rj ≤|x−xj |<2k+1 rj

∞   aj (y) dy k=1

Bj

≤ C[b]θν rj−ν  ≤ CCN [b]θν

   K(x, y) − K(x, xj ) dx dy



 CN (2k rj )ν 2−kδ

{1 + m(xj , V )2k rj } l0 +1 −l1 −(l0 +1)θ N

∞

 CN (2k rj )ν 2−kδ

k=1

{1 + m(xj , V )2k rj } l0 +1 −l1 −(l0 +1)θ N

∞

2kν−kδ

k=1

≤

 CCN [b]θν ,

(36)

where we use the fact that ν < δ. Therefore,   A3 (x)

L1 (Rn )

≤C





|λj | ≤ C

n

HLn+ν

 n+ν n |λj |

n n+ν

j=−∞

j: rj <ρ(xj )

≤ Cf 

∞

.

Note that     b − b(xj ) aj 

 L1

≤

C[b]θν rjν

rj 1+ ρ(xj )

(l0 +1)θ 

  aj (y) dy ≤ C  [b]θν rjν aj L2 |Bj | 12

Bj

≤C



−n−ν 2 [b]θν rjν rj2 rj n

n

≤C



[b]θν .

(37)

It follows from Theorem 2.7 in [12] that R is bounded from L1 (Rn ) into weak L1 (Rn ). Then     ∞         C λ   ≤   x ∈ Rn : A4 (x) > λ b − b(x ) a  j j j   4  λ j=−∞ ≤

L1

∞ C |λj |. λ j=−∞

Therefore,    4           x ∈ Rn : [b, R]f (x) > λ  ≤ C  x ∈ Rn : Ai (x) > λ     4 4  i=1 ∞ C C ≤ |λj | ≤ λ j=−∞ λ

≤

C n . f  n+ν HL λ



∞

 ν+n n |λj |

n n+ν

j=−∞

2

Proof of Theorem 4. Following from Proposition 1, we know that the commutator [b, R] is bounded from n

n

,2

HLn+ν (Rn ) into L1 (Rn ) if and only if [b, R]aL1 ≤ C holds for any HLn+ν -atom a, where C is independent of a. Suppose that supp a ⊆ B = B(x0 , r) with r < ρ(x).

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29

Then we have   [b, R]a 1 = L



    [b, R]a(x) dx +



|x−x0 |<2r



  [b, R]a(x) dx



|x−x0 |≥2r

. = B1 + B2 . Let

1 s

=

1 2

− nν . By Corollary 1 and the size condition of the atom a, we have 



  [b, R]a(x) dx

B1 =



|x−x0 |<2r





  [b, R]a(x)s dx

≤

 1s

n

(2r)n− s

|x−x0 |<2r





  a(x)2 dx

≤C

 12

n

(2r)n− s

|x−x0 |<2r

≤ C(2r) 2 −(n+ν) (2r)n− s n

n

≤ C. ρ(x0 ) 4

For B2 , we consider two cases, that is,

≤ r < ρ(x0 ) and r <

ρ(x0 ) 4 .

Case I: At this time, the atom a does not satisfy the vanishing condition. At first, we decompose [b, R]a(x)χ(2B)c (x) as follows:      [b, R]a(x)χ(2B)c (x) = b(x) − b(u) Ra(x)χ(2B)c (x) − R b − b(u) a (x)χ(2B)c (x), where u ∈ B(x0 , r). Then by using the proof of A2 in the proof of Theorem 3,     b(x) − b(u) Ra(x)χ(2B)c (x)

L1 (Rn )

≤ C.

Moreover, similar to the estimates of (31) and (33), we obtain      R b − b(u) a (x)χ(2B)c (x)

L1 (Rn )

 ≤ |x−x0 |≥2r



≤

     R b − b(u) a (x) dx        K(x, y) b(y) − b(u) a(y) dy  dx  

|x−x0 |≥2r B



≤

   b(y) − b(u)a(y) dy



  K(x, y) dx ≤ C.

|x−x0 |≥2r

B

Case II: At this time, the atom a satisfies the vanishing condition. Similar to (36), we immediately have     b(x) − b(x0 ) Ra(x)χ(2B)c (x)

L1 (Rn )

≤ C.

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30

Moreover, since (ii) holds true, similar to the estimates of (34) and (36), we obtain      R b − b(u) a (x)χ(2B)c (x) 1 n L (R )       R b − b(u) a (x) dx ≤ |x−x0 |≥2r



≤

        K(x, y) − K(x, u) b(y) − b(u) a(y) dy  dx +  

|x−x0 |≥2r B



   b(y) − b(u)a(y) dy

≤ 

   b(y) − b(u)a(y) dy

≤



+

≤

     K(x, u) b(y)a(y) dy  dx  



  K(x, y) − K(x, u) dx +

∞ j=1

B



|x−x0 |≥2r B



|x−x0 |≥2r

B

     K(x, u)b(y)a(y) dy  dx  



|x−x0 |≥2r



B

  K(x, y) − K(x, u) dx

2j r<|x−x0 |≤2j+1 r

     K(x, u) b(y)a(y) dy  dx  

|x−x0 |≥2r

B

∞     b(y) − b(u)a(y) dy CN

B

j=1

1 {1 + m(u, V )2k r} l0 +1 −l1 N

2

−kδ

 + |x−x0 |≥2r

     K(x, u) b(y)a(y) dy  dx   B

≤ C. By the above estimates, it is clear that [b, R]aL1 ≤ C if and only if  |x−x0 |≥2r

     K(x, u) b(y)a(y) dy  dx ≤ C.   B

Therefore, (i) is equivalent to (ii). This completes the proof. 2 Acknowledgments The authors are grateful to Jie Xiao and Lin Tang for their helpful advice on this subject. Also, we would like to express our gratitude to the referee for the valuable comments and suggestions. References [1] P. Auscher, B.B. Ali, Maximal inequalities and Riesz transform estimates on Lp spaces for Schrödinger operators with nonnegative potentials, Ann. Inst. Fourier (Grenoble) 57 (2007) 1975–2013. [2] B. Bongioanni, E. Harboure, O. Salinas, Commutators of Riesz transforms related to Schrödinger operators, J. Fourier Anal. Appl. 17 (2011) 115–134. [3] R. Coifman, R. Rochberg, G. Weiss, Factorization theorems for Hardy spaces in several variables, Ann. of Math. 103 (1976) 611–635. [4] J. Dziubański, J. Zienkiewicz, Hardy space H 1 associated to Schrödinger operator with potential satisfying reverse Hölder inequality, Rev. Mat. Iberoam. 15 (1999) 279–296. [5] J. Dziubański, J. Zienkiewicz, H p spaces for Schrödinger operators, in: Fourier Analysis and Related Topics, in: Banach Center Publ., vol. 56, Institute of Mathematics, Polish Academy of Sciences, Warszawa, 2002, pp. 45–53. [6] C. Fefferman, The uncertainty principle, Bull. Amer. Math. Soc. 9 (2) (1983) 129–206. [7] D. Goldberg, A local version of real Hardy spaces, Duke Math. J. 46 (1) (1979) 27–42. [8] Z.H. Guo, P.T. Li, L.Z. Peng, Lp boundedness of commutators of Riesz transform associated to Schrödinger operator, J. Math. Anal. Appl. 341 (2008) 421–432. [9] E. Harboure, C. Segovia, J. Torrea, Boundedness of commutators of fractional and singular integrals for the extreme values of p, Illinois J. Math. 41 (1997) 676–700.

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