Lp boundedness of commutator operator associated with schrödinger operators on heisenberg group

Acta Mathematica Scientia 2012,32B(2):568–578 http://actams.wipm.ac.cn

Lp BOUNDEDNESS OF COMMUTATOR OPERATOR ¨ ASSOCIATED WITH SCHRODINGER OPERATORS ON HEISENBERG GROUP∗

)

Li Pengtao (

Department of Mathematics, Shantou University, Shantou 515063, China E-mail: [email protected]


)

Peng Lizhong (

LMAM School of Mathematical Sciences, Peking University, Beijing 100871, China E-mail: [email protected]

Abstract Let L = −H n +V be a Schr¨ odinger operator on Heisenberg group H n , where H n is the sublaplacian and the nonnegative potential V belongs to the reverse H¨ older class BQ/2 , where Q is the homogeneous dimension of H n . Let T1 = (−H n +V )−1 V , T2 = (−H n +V )−1/2 V 1/2 , and T3 = (−H n +V )−1/2 ∇H n , then we verify that [b, Ti ], i = 1, 2, 3 are bounded on some Lp (H n ), where b ∈ BMO(H n ). Note that the kernel of Ti , i = 1, 2, 3 has no smoothness. Key words Commutator; BMO; Heisenberg group; boundedness; Riesz transforms associated to Schr¨ odinger operators 2000 MR Subject Classification

1

47B32; 47A75; 42C40; 94A40

Introduction

Let L = −H n + V be a Schr¨ odinger differential operator on Heisenberg group H n . Throughout this article, we will assume that V (x) is a nonzero, nonnegative potential, and belongs to Bq for some q > Q/2, where Q = 2n + 2 is the homogeneous dimension of H n . Let T1 = (−H n + V )−1 V , T2 = (−H n + V )−1/2 V 1/2 , and T3 = (−H n + V )−1/2 ∇H n , then Lp boundedness of the operator Tj was studied by Z. Shen in [6]. In [4], the authors considered the Lp boundedness of [b, Ti ], i = 1, 2, 3, on Rn . In this article, we will discuss the Lp boundedness of the commutator operators [b, Ti ] = bTi − Ti b, where b ∈ BMO(H n ). This article is organized as follows. In the rest of this section, we state some knowledge, notations, and terminologies to be used throughout this article. In Section 2, we give some estimates of the kernels of the operators Ti , i = 1, 2, 3. In Section 3, we will prove our main results. ∗ Received

September 30, 2009; revised July 15, 2010. The first author was supported by NSFC 11171203, S2011040004131, and STU Scientific Research Foundation for Talents TNF 10026. The second author was supported by NSFC No. 10990012, 10926179, and RFDP of China No.200800010009.

No.2

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The (2n + 1)-dimensional Heisenberg group H n is a nilpotent Lie group with underlying manifold R2n × R. The group structure is given by   n  (x, t)(y, s) = x + y, t + s + 2 (xn+j yj − xj yn+j ) .

(1.1)

j=1

The Lie algebra of left-invariant vector fields on H n is spanned by ∂ ∂ ∂ ∂ + 2xn+j , Xn+j = − 2xj ∂xj ∂t ∂xn+j ∂t ∂ X2n+1 = . ∂t

Xj =

j = 1, 2 · · · n,

(1.2) (1.3)

All nontrivial commutation relations are given by [Xj , Xn+j ] = −4X2n+1 , j = 1, 2, · · · n. The sub-Laplacian H n and the gradient ∇H n are defined, respectively, by 

Hn

=

2n 

Xj2 and ∇H n = (X1 , X2 , · · · X2n ).

(1.4)

j=1

The dilations on H n have the form δr (x, t) = (rx, r2 t) t > 0.

(1.5)

The Haar measure on H n coincides with the Lebesgue measure on R2n × R. The measure of any measurable set E is denoted by |E|, we define a homogeneous norm on H n by |g| = (|x|4 + |t|2 )1/4 ,

|g

−1

g = (x, t) ∈ H n .

(1.6)

This norm satisfies the triangular inequality and leads to a left-invariant distance d(g, h) = h|. the ball of radius r and centered at g is denoted by B(g, r) = {h ∈ H n , |g −1 h| < r}

(1.7)

whose volume is given by |B(g, r)| = cn rQ , cn = |B(0, 1)| =

2π n+1/2 Γ(n/2) , (n + 1)Γ(n)Γ( n+1 2 )

(1.8)

where Q = 2n + 2 is the homogeneous dimension of H n . Now, we turn to the Schr¨ odinger operator L on Heisenberg group. Definition 1 A nonnegative locally Lq integrable function V on H n is said to belong to Bq (1 < q < ∞), if there exists C > 0 such that the reverse H¨ older inequality 

1 |B|

1/q

 V (g)q dg B

 ≤C

1 |B|



 V (g)dg

(1.9)

B

holds for every ball B in H n . Remark 1 Obviously, Bq1 ⊂ Bq2 if q1 > q2 . 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. In this article, we always assume that 0 = V ∈ BQ/2 and then V ∈ Bq0 for some q0 > Q/2. Of course, we may

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assume that q0 < Q. It is also well known that if V ∈ Bq , q > 1, then, V (g)dg is a doubling measure, namely,   V (g)dg ≤ C0 V (g)dg. (1.10) B(g0 ,2r)

B(g0 ,r)

The following theorem is a collection of properties of the potential V satisfying the reverse H¨older inequality on Heisenberg group H n . We first introduce an auxiliary function. Definition 2

For g ∈ H n , the function m(g, V ) is defined by    1 1 = sup r > 0, Q−2 V (h)dh ≤ 1 . m(g, V ) r B(g,r)

(1.11)

Theorem A There exist C > 0, c > 0, and k0 > 0 such that, for g, h ∈ H n , 0 < m(g, V ) < ∞, for every g ∈ Rn ;  1 1 , then Q−2 V (h)dh = 1; (b) If r = m(g, V ) r B(g,r) (a)

c ; m(g, V )

(1.12) (1.13)

(c)

m(g, V ) ∼ m(h, V ), if |gh−1 | ≤

(d)

 k0 m(h, V ) ≤ C 1 + |gh−1 |m(g, V ) m(g, V );

(1.15)

(e)

 −k0 /(1+k0 ) m(h, V ) ≥ Cm(g, V ) 1 + |gh−1 |m(g, V ) ;

(1.16)

(f)

 1 c 1 + |gh−1 |m(h, V ) k0 +1 ≤ 1 + |gh−1 |m(g, V ) ≤ C{1 + |gh−1 |m(h, V )}k0 +1 . (1.17)

(1.14)

Now, we turn to the estimate of the kernel. Let Γ(g, h, t) denote the fundamental solution of the operator −ΔH n + iτ , where τ ∈ R. In [5], C. Lin, H. Liu and Y. Liu proved that, for any k > 0, there exists a constant Cl > 0, such that |Γ(g, h, τ )| ≤

Cl 1 , (1 + |gh−1 ||τ |1/2 )l |gh−1 |Q−2

|∇H n ,g Γ(g, h, τ )| ≤

1 Cl , −1 −1 1/2 l (1 + |gh ||τ | ) |gh |Q−1

(1.18)

(1.19)

where ∇H n ,g denotes the gradient operator for g (See [5] for details). The above estimate still holds for ∇H n ,h instead of ∇H n ,g and can be easily reduced from the corresponding estimates of the heat kernel. We remark that the explicit expression of Γ(g, h) = Γ(g, h, 0) was obtained by Folland: 2n−2 γ(n/2)2 1 Γ(g, h) = . (1.20) π n+1 |gh−1 |Q−2 Let ΓL (g, h, τ ) denote the fundamental solution for the operator L + iτ , where τ ∈ R. For any k > 0, there exists a constant Cl > 0, such that |ΓL (g, h, τ )| ≤

Cl 1 . (1 + |gh−1 ||τ |1/2 )l (1 + |gh−1 |(m(g, V ) + m(h, V )))k |gh−1 |Q−2

(1.21)

No.2

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The Estimate of the Kernel of Ti, i = 1, 2, 3

Lemma 1 Suppose V ∈ Bq0 , q0 > 1. Assume that −ΔH n u + (V + iτ )u = 0 in B(g0 , 2R), then, (1) For x ∈ B(g0 , R),   V (h) C |∇H n u(g)| ≤ C sup |u| dh + |u(h)|dh; (2.1) −1 | RQ+1 B(g0 ,2R) B(g0 ,2R) B(g0 ,2R) |gh (2) If Q/2 < q0 < Q, let

1 t

=



1 q0

− n1 , k0 > log2 C0 + 1, where C0 is the constant in (1.10),

1/t |∇H n u|t dg

≤ CRQ/q0 −2 {1 + Rm(g0 , V )}k0

B(g0 ,R)

sup

|u|.

(2.2)

B(g0 ,2R)

Lemma 2 Suppose V ∈ Bq . For some q > Q 2 , let N > log2 C0 + 1, where C0 is the n constant in Remark 1, (1.10). Then, for any g0 ∈ H , R > 0,  1 V (h)dh ≤ CRQ−2 . (2.3) {1 + m(g0 , V )R}N B(g0 ,R) Proof There exists an integer j0 ∈ Z, such that 2j0 R ≤ m(g10 ,V ) < 2j0 +1 R. Case 1: j0 < 0. By the doubling property of V (g), Lemma 1 and (b) of Theorem A, we have 1 {1 + m(g0 , V )R}N

 V (h)dh ≤ B(g0 ,R)

1



{1 + 2−(j0 +1) R−1 R}N B(g0 ,R)  1 V (h)dh ≤ −j0 N (2 ) B(g0 ,R) C ≤ −j0 N C0−j0 (2j0 R)Q−2 (2 )

V (h)dh

≤ RQ−2 , (N > log2 C0 ).

(2.4)

Case 2: j0 ≥ 0. By (b) and (g) of Theorem A, we obtain   1 V (h)dh ≤ V (h)dh {1 + m(g0 , V )R}N B(g0 ,R) B(g0 ,R)  1 V (h)dh ≤ RQ−2 Q−2 R B(g0 ,R) ≤ RQ−2 .

(2.5)

Now, we estimate the kernels Ki (g, h) of Ti , i = 1, 2, 3. Lemma 3 Suppose V ∈ Bq for some q > Q/2, then there exists δ > 0 such that, for any integer k > 0 and 0 < |ω| < |gh−1 |/16, |K1 (g, h)| ≤

1 Ck V (h); {1 + m(g, V )|gh−1 |}k |gh−1 |Q−2

|K1 (gω, h) − K1 (g, h)| ≤

|ω|δ Ck V (h)1/2 . −1 k −1 {1 + m(g, V )|gh |} |gh |Q−2+δ

(2.6)

(2.7)

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Proof Because T1 = (−ΔH n + V )−1 V , then, we can see that K1 (g, h) = ΓL (g, h, 0)V (h). By the estimate of fundamental solution, we have |K1 (g, h)| ≤

1 Ck V (h). −1 k −1 {1 + m(g, V )|gh |} |gh |Q−2

(2.8) −1

For (2.7), taking g, h ∈ H n and Q/2 < q0 < min(Q, q), we know V ∈ Bq0 . Let R = |gh8 | , 1 1 1 R t = q0 − Q , then δ = 1 − Q/t > 0. For any 0 < h < 2 , it follows from the imbedding theorem of Morrey and Lemma 1, |K1 (gω, h) − K1 (g, h)| ≤ |ΓL (gω, h, 0) − ΓL (g, h, 0)|V (h)  1/t |∇g ΓL (u, h, 0)|t du V (h) ≤ C|ω|1−Q/t B(g,R) 1−Q/t

≤ C|ω| 

|ω| ≤C R ≤C

R

Q/q0 −2

{1 + Rm(g, v)}k0

sup

|ΓL (u, h, 0)|V (h)

u∈B(g,2R)

1−Q/t {1 + Rm(g, v)}k0

sup

|ΓL (u, h, 0)|V (h)

u∈B(g,2R)

Ck1 1 |ω|δ {1 + Rm(g, v)}k0 sup V (h) −1 |}k1 |uh−1 |Q−2 Rδ {1 + m(h, V )|uh u∈B(g,2R)

≤ Ck

|ω|δ 1 1 V (h). −1 δ −1 k −1 |gh | {1 + m(g, V )|gh |} |gh |Q−2

(2.9)

Lemma 4 Suppose V ∈ Bq for some q > Q/2. Then, there exists δ > 0 and for any integer k > 0, 0 < |ω| < |gh−1 |/16, |K2 (g, h)| ≤

1 Ck V (h)1/2 ; {1 + m(g, V )|gh−1 |}k |gh−1 |Q−1

|ω|δ Ck V (h)1/2 . −1 k −1 {1 + m(h, V )|gh |} |gh |Q−1+δ Proof By functional calculus, we write  1 (−iτ )−1/2 (−ΔH n + V + iτ )−1 dτ, (−ΔH n + V )−1/2 = − 2π R

1 −1/2 then, K2 (g, h) = − 2π Γ(g, h, τ )dτ V (h)1/2 . R (−iτ ) We claim that, for k > 2,  Ck . |τ |−1/2 {1 + |τ |1/2 |gh−1 |}−k dτ ≤ |gh−1 | R |K2 (gω, h) − K2 (g, h)| ≤

(2.10) (2.11)

(2.12)

(2.13)

In fact, we have 

|τ |−1/2 {1 + |τ |1/2 |gh−1 |}−k dτ R   ≤ |τ |−1/2 {1 + |τ |1/2 |gh−1 |}−k dτ + |τ |≤|gh−1 |−2 |τ |>|gh−1 |−2   |τ |−1/2 dτ + |τ |(−k−1)/2 |gh−1 |dτ ≤ |τ |≤|gh−1 |−2

≤

Ck . |gh−1 |

|τ |>|gh−1 |−2

(2.14)

P.T. Li & L.Z. Peng: Lp BOUNDEDNESS OF COMMUTATOR OPERATOR

No.2

573

We first prove (2.10). We obtain |K2 (g, h)|  1 |τ |−1/2 |ΓL (g, h, τ )|dτ V (h)1/2 ≤ 2π R  Ck 1 1 1 ≤ |τ |−1/2 dτ V (h)1/2 2π R (1 + |gh−1 |m(g, V ))k {1 + |gh−1 ||τ |1/2 }k |gh−1 |Q−2  Ck 1 1 1 1/2 ≤ V (h) |τ |−1/2 dτ −1 Q−2 −1 −1 1/2 k 2π |gh | (1 + |gh |m(g, V ))k {1 + |gh ||τ | } R 1 Ck ≤ V (h)1/2 . −1 k −1 {1 + m(g, V )|gh |} |gh |Q−1

(2.15)

For (2.11), fix g, h ∈ H n and Q/2 < q0 < min(Q, q), then, we know V ∈ Bq0 . Let R = |gh−1 |/8 and 1/t = 1/q0 − 1/Q, then, δ = 1 − Q/t > 0. For any 0 < |ω| < R2 , we have  1 |τ |1/2 |ΓL (gω, h, τ ) − ΓL (g, h, τ )|dτ V (y)1/2 . (2.16) |K2 (gω, h) − K2 (g, h)| ≤ 2π R By Morrey’s imbedding theorem, |K2 (gω, h) − K2 (g, h)|  1/t  C 1/2 1−Q/t L t |τ | |ω| |∇H n ,g Γ (u, h, τ )| dt dτ V (h)1/2 ≤ 2π R B(g,R)  C ≤ |τ |1/2 |ω|1−Q/t RQ/q0 −2 {1 + Rm(g, V )}k0 sup |ΓL (u, h, τ )|dτ V (h)1/2 2π R u∈B(g,2R)  δ  |ω| ≤C {1 + Rm(g, V )}k0 |τ |1/2 sup |ΓL (u, h, τ )|dτ V (h)1/2 . (2.17) R u∈B(g,2R) R Because |gu−1 | ≤ 2R = |gh−1 |/4, then, |uh−1 | ≥ c(|gh−1 | − |gu−1 |) ≥ c|gh−1 |. So, we obtain sup

|ΓL (u, h, τ )|

u∈B(g,2R)

≤

Ck 1 1 −1 ||τ |1/2 }k {1 + m(h, V )|uh−1 |}k |uh−1 |Q−2 {1 + |uh u∈B(g,2R)

≤

1 1 Ck . {1 + |gh−1 ||τ |1/2 }k {1 + m(h, V )|gh−1 |}k |gh−1 |Q−2

sup

(2.18)

Finally, we have |K2 (gω, h) − K2 (g, h)| δ   1 1 Ck |ω| |τ |1/2 dτ V (h)1/2 ≤C −1 −1 k −1 Q−2 |gh | {1 + m(h, V )|gh |} |gh | {1 + |gh−1 ||τ |1/2 }k R ≤

|ω|δ Ck V (h)1/2 . −1 k −1 {1 + m(h, V )|gh |} |gh |Q−1+δ

(2.19)

Lemma 5 Suppose V ∈ Bq for some Q/2 < q < Q. Then, there exists δ > 0 such that, for any integer k > 0, 0 < |ω| < |gh−1 |/16, |K3 (g, h)| 1 Ck ≤ −1 k −1 {1 + m(g, V )|gh |} |gh |Q−1

 B(h,|gh−1 |)

 V (u) 1 ; du + |hu−1 |Q−1 |gh−1 |

(2.20)

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|K3 (gω, h) − K3 (g, h)|   V (u) δ 1 Ck . du + ≤ −1 |Q−1 {1 + m(g, V )|gh−1 |}k |gh−1 |Q−1+δ |gh−1 | B(h,|gh−1 |) |hu

(2.21)

Proof By partial integral, we know that  1 K3 (g, h) = (−iτ )−1/2 ∇H n ,2 ΓL (g, h, τ )dτ. 2π R Fix g, h ∈ H n , let R = |gh−1 |/8, 1/t = 1/q − 1/Q and δ = Q/q − 2 > 0. For any 0 < |ω| < R/2, we have  1 |K3 (gω, h) − K3 (g, h)| ≤ |τ |−1/2 |∇H n ,2 ΓL (gω, h, τ ) − ∇H n ,2 ΓL (g, h, τ )|dτ. (2.22) 2π R By Morrey’s imbedding theorem, we have |∇H n ,2 ΓL (gω, h, τ ) − ∇H n ,2 ΓL (g, h, τ )|  1/t ≤ C|h|1−Q/t |∇H n ,1 ∇H n ,2 ΓL (u, h, τ )|t du B(g,R) 1−Q/t

≤ C|h|

R

Q/q−2

{1 + Rm(g, V )}k0

sup

|∇H n ,2 ΓL (u, h, τ )|.

(2.23)

u∈B(g,2R)

As ΓL (u, h, τ ) = ΓL (u, h, u, −τ ), then, ∇H n ,2 ΓL (u, h, τ ) = ∇H n ,1 ΓL (h, u, τ ). So, we obtain |∇H n ,2 ΓL (u, h, τ )|

sup u∈B(g,2R)

≤

|∇H n ,1 ΓL (h, u, −τ )|

sup u∈B(g,2R)

≤





sup

sup

u∈B(g,2R)

η∈B(h,|hu−1 |/4)

+

c |hu−1 |Q+1

|ΓL (η, u, −τ )|

B(h,|hu−1 |/2)





Γ (v, u, −τ )dv . L

B(h,|hu−1 |/2)

V (v) dv |hv −1 |Q−1 (2.24)

Using the fact that |ηu−1 | ∼ |hu−1 |, |vu−1 | ∼ |hu−1 |, |gh−1 | ∼ |hu−1 |, and choosing k1 large enough, we have sup

|∇H n ,2 ΓL (u, h, τ )|

u∈B(g,2R)

Ck 1 1/2 |hu−1 |}k1 {1 + m(u, V )|hu−1 |}k1 |hu−1 |Q−2 {1 + |τ | u∈B(g,2R)  Ck V (v) 1 × dv + −1 Q−1 −1 1/2 −1 k −1 k 1 1 | {1 + |τ | |hu |} {1 + m(u, V )|hu |} |hu |Q−1 B(h,|gh−1 |) |hv Ck ≤ 1/2 −1 k 1 {1 + |τ | |gh |} {1 + m(g, V )|gh−1 |}k1    V (v) 1 1 . (2.25) × dv + |gh−1 |Q−2 B(h,|gh−1 |) |hv −1 |Q−1 |gh−1 |Q−1

≤

sup

So, we obtain |∇H n ,2 ΓL (gω, h, τ ) − ∇H n ,2 ΓL (g, h, τ )|

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P.T. Li & L.Z. Peng: Lp BOUNDEDNESS OF COMMUTATOR OPERATOR

|ω|δ 1 1 ≤ Ck −1 δ −1 k −1 |gh | {1 + m(g, V )|gh |} {1 + |gh ||τ |1/2 }k    V (v) 1 1 . × dv + |gh−1 |Q−2 B(h,|gh−1 |) |hv −1 |Q−1 |gh−1 |Q−1

575

(2.26)

Finally, we have |h|δ 1 |K3 (gω, h) − K3 (g, h)| ≤ Ck |gh−1 |δ {1 + m(g, V )|gh−1 |}k    V (v) 1 1 . × dv + |gh−1 |Q−1 B(h,|gh−1 |) |hv −1 |Q−1 |gh−1 |Q

3

(2.27)

The Main Theorems

In this section, we prove the main theorems. We first introduce a new H¨ omander condition n on Heisenberg group H . Definition 3 K(g, h) is said to satisfy condition H(m) for some m ≥ 1, if there exists a constant C > 0 such that for ∀l > 0, g, g0 ∈ H n with gg0−1 ≤ l, then,  1/m ∞  k Q/m m k(2 l) |K(g, h) − K(g0 , h)| dh ≤ C, (3.1) k=5

2k l≤|hg0−1 |<2k+1 l

1 . where m1 = 1 − m Proposition 1 Let m > 1, and suppose T is bounded on Lp (H n ) for p ∈ (m , ∞) and K satisfies H(m), then, ∀b ∈ BMO(H n ), [b, T ] is bounded on Lp (H n ) for every p ∈ (m , ∞), and

[b, T ]f p ≤ Cp b BMO f p .

(3.2)

Proof Proposition 1 is an easy corollary of the following lemma. Recall that the sharp function of Fefferman-Stein is defined by  1  M f (g) = sup |f (h) − fB |dh, (3.3) x∈B |B| B

1 f (h)dh and the supremum is taken on all balls B with g ∈ B. where fB = |B| B Lemma 6 Let T satisfy the same condition in Proposition 1. Then, for ∀s > m , there exists a constant Cs > 0, such that for ∀f ∈ L1loc , b ∈ BMO(H n ), M  ([b, T ]f )(g) ≤ Cs b BMO{Ms (T f )(g) + Ms (f )(g)},

(3.4)

where Ms (f ) = (M (|f |s ))1/s and M is the Hardy-Littlewood maximal function. Proof Fix s > m , f ∈ L1loc , g ∈ H n and fix a ball I = B(g0 , l) with g ∈ I. We only need

1 to control J = |I| |[b, T ]f (h) − ([b, T ]f )I |dh by the right-hand side of (3.4). Let f = f1 + f2 , I where f1 = f χ32I , f2 = f − f1 . Then, [b, T ]f = (b − bI )T f − T (b − bI )f1 − T (b − bI )f2  A1 f + A2 f + A3 f.

(3.5)

Then, we have    1 1 1 |A1 f (h) − (A1 f )I |dh + |A2 f (h) − (A2 f )I |dh + |A3 f (h) − (A3 f )I |dh J≤ |I| I |I| I |I| I  J1 + J2 + J3 .

(3.6)

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We estimate J1 , J2 , and J3 , separately. For J1 , we have   2 2 |A1 f (h)|dh = |(b − bI )T f (h)|dh J1 ≤ |I| I |I| I 1/s   1/s   1 1 s s |b − bI | dh |T f (h)| dh ≤2 |I| I |I| I ≤ 2 b BMO(H n ) Ms (T f )(g).

(3.7)

For J2 , by the Lp boundedness of T , we have   1/s1  2 1 |A2 f (h)|dh ≤ 2 |A2 f (h)|s1 dh |I| I |I| I   1/s1 1 s1 ≤2 |A2 f (h)| dh |I| 32I  1/s2  1/s   1 1 ≤ CQ |b − bI |s2 dh |f (h)|s dh |32I| 32I |32I| 32I ≤ CQ b BMO Ms (f )(g).

J2 ≤

At last, we start to estimate J3 . Set CI = have

 2 J3 ≤ |I| I  2 ≤ |I| I  2 ≤ |I| I  2 ≤ |I| I  ×

|ug0−1 |>32l ∞ 

−1 k k+1 l k=5 2 l≤|ug0 |<2 ∞  

k=5

  ∞

k≥5

K(g0 , u)(b(u) − bI )f (u)du, then, we

|K(h, u) − K(g0 , u)||b(u) − bI ||f (u)|dudh

2k l≤|g0−1 u|<2k+1 l

k

k(2 l)

1/m |K(h, u) − K(g0 , u)|m du

|(b(u) − bI )f (u)| 1

Q/m

I k=5

≤ C sup

|ug0−1 |>32l

(3.8)

|A3 f (h) − CI |dh  |K(h, u) − K(g0 , u)||b(u) − bI ||f (u)|dudh

2k l≤|g0−1 u|<2k+1 l

2 ≤ |I|



(2k l)Q/m k

m

1/m du

 2k l≤|g0−1 u|<2k+1 l

|(b(u) − bI )f (u)|

m

1/m du

k+2

b BMO Ms f (g) ≤ C b BMO Ms f (g). k

(3.9)

Theorem 1 (i) Suppose V ∈ Bq , q ≥ Q/2, and let b ∈ BMO. Then, for q  ≤ p < ∞,

[b, T1 ]f p ≤ Cp b BMO f p .

(3.10)

(ii) Suppose V ∈ Bq , q ≥ Q/2, and let b ∈ BMO, then, for (2q) ≤ p < ∞,

[b, T2 ]f p ≤ Cp b BMO f p . (iii) Suppose V ∈ Bq and

Q 2

(3.11)

≤ q < Q, and let b ∈ BMO. Then, for p0 ≤ p < ∞,

[b, T3 ]f p ≤ Cp b BMO f p , where 1/p0 = 1/q − 1/Q.

(3.12)

P.T. Li & L.Z. Peng: Lp BOUNDEDNESS OF COMMUTATOR OPERATOR

No.2

577

Proof By use of Proposition 1, we only need to prove the kernels of Ti , i = 1, 2, 3 satisfy the condition H(m) for some m in Definition 3. (i) We only need to prove that the kernel K1 satisfies H(q). By Lemma 3 and V ∈ Bq , we have  1/q q |K1 (g, h) − K1 (g0 , h)| dy 2k l≤|hg0−1 |<2k+1 l

≤ CN

lδ 1 k Q−2+δ (2 l) {1 + m(g0 , V )2k l}N



1/q V q (h)dh

B(g0 ,2k+3 l)

δ

 l 1 ≤ CN k Q−2+δ (2k l)−Q/q (2 l) {1 + m(g0 , V )2k l}N

≤ CN Then, we have  ∞   k(2k l)Q /q k=5

lδ (2k l)Q−2+δ

(2k l)Q/q−2 ≤ C

lδ (2k l)(Q/q )+δ

V (h)dh B(g0 ,2k l)

.

(3.13)

1/q

q

2k l≤|g0−1 h|<2k+1 l



|K1 (g, h) − K1 (g0 , h)| dh

≤C

∞  k ≤ C. 2kδ

(3.14)

k=5

For the proof of (ii), we prove that K2 satisfies H(m) for m = 2q. Form Lemma 4,  1/2q 2q |K2 (g, h) − K2 (g0 , h)| dh 2k l≤|g0−1 |<2k+1 l



1/2q

≤ CN

lδ 1 k Q−1+δ (2 l) {1 + m(g0 , V )2k l}N

≤ CN

 lδ 1 (2k l)−Q/(2q ) k Q−1+δ k N (2 l) {1 + m(g0 , V )2 l}

≤ CN

δ

V q (h)dh B(g0 ,2k+3 l)



1/2 V (h)dh

B(g0 ,2k l)

δ

  l l (2k l)−Q/(2q )+(Q−2)/2 ≤ C k δ (2k l)−Q/(2q ) . (2k l)Q−1+δ (2 l)

(3.15)

Hence, ∞ 

k

Q/(2q)



k(2 l)

2q

2k l≤|g0−1 h|<2k+1 l

k=5

|K2 (g, h) − K2 (g0 , h)| dh

1/2q ≤C

∞  k ≤ C. (3.16) 2kδ k=5

For the proof of (iii), we prove that K3 satisfies H(p0 ). From Lemma 5,  1/p0 |K3 (g, h) − K3 (g0 , h)|p0 dh 2k l≤|g0−1 |<2k+1 l

    lδ 1 V (u) lδ   ≤ CN k Q−1+δ du +  (2 l) {1 + m(g0 , V )2k l}N  B(g0 ,2k+3 l) |uh−1 |Q−1 Lp0 (dh) (2k l)Q/p0  1/q lδ 1 lδ q ≤ CN k Q−1+δ V (h)dh + k (Q/p )+δ k N 0 (2 l) {1 + m(g0 , V )2 l} (2 l) B(g0 ,2k+3 l)   δ l 1 lδ k −Q/q ≤ CN k Q−1+δ (2 l) V (h)dh +  (2 l) {1 + m(g0 , V )2k l}N (2k l)Q/p0 +δ B(g0 ,2k l) ≤C

lδ  (2k l)Q/p0 +δ

.

(3.17)

578

ACTA MATHEMATICA SCIENTIA

Vol.32 Ser.B

Therefore, we have ∞ 





k(2k l)n/p0

k=5 ∞ 

≤C

k=5

2k l≤|g0−1 h|<2k+1 l

|K3 (g, h) − K3 (g0 , h)|p0 dy

1/p0

k ≤ C. 2kδ

(3.18)

By duality, we can get the following theorem. Theorem 2 For the operators T1∗ = V (−H n + V )−1 , T2∗ = V 1/2 (−H n + V )−1/2 , and T3∗ = −∇H n (−H n + V )−1/2 , we have

[b, T1∗]f p ≤ Cp b BMO f p , 1 < p ≤ q;

(3.19)

[b, T2∗]f p

≤ Cp b BMO f p , 1 < p ≤ 2q;

(3.20)

[b, T3∗]f p

≤ Cp b BMO f p , 1 < p ≤ p0 , where 1/p0 = 1/q − 1/Q.

(3.21)

From Theorem 1 (i), we can get the result concerning second order Riesz transform. Let T4 = (−H n + V )−1 ∇2H n and T4∗ = ∇2H n (−H n + V )−1 , we have Theorem 3 Suppose V ∈ Bq , q ≥ Q/2, then, we have

[b, T4 ]f p ≤ Cp b BMO f p , q  ≤ p < ∞;

(3.22)

[b, T4∗]f p

(3.23)

≤ Cp b BMO f p , 1 < p ≤ q.

Proof We only need to prove [b, T4 ]. Because 2 T4 = (−H n + V )−1 ∇2 = (−H n + V )−1 H n −1 H n H n

= (I − (−H n + V )V ) ∇2

∇2Hn ∇2 = (I − T1 ) Hn , H n H n

(3.24)

∇2

n n we can get [b, T4 ] = [b, I − T1 ] H − (I − T1 )[b, H ]. Hence, the boundedness of [b, T4 ] follows Hn Hn from that of [b, I − T1 ] and (I − T1 ).

References [1] Dziubanski J, Zienkiewicz J. Hardy space H 1 associated to Schr¨ odinger operator with potential satisfying reverse H¨ older inequality. Revista Matematica Iberoamericana, 1999, 15(2): 279–296 [2] Dziubanski J, Garrig´ os G, Martinez T, Torrea J L, Zienkiewicz J. BMO spaces related to Schr¨ odinger operators with potentials satisfying a reverse H¨ older inequality. Mathematische Zeitschrift, 2005, 249: 329–356 [3] Dziubanski J, Zienkiewicz J. H p spaces associated with Schr¨ odinger operators with potentials from reverse H¨ older class. Colloquium Mathematicum, 2003, 98(1): 5–38 [4] Guo Z H, Li P T, Peng L Z. Lp boundedness of commutator of Riesz transform associated to Schrodinger operator. Journal of Mathematical Analysis and Application, 2008, 341: 421–432 1 associated with Schr¨ [5] Lin C C, Liu H, Liu Y. The Hardy space HL odinger operator on the Heisenberg group (preprint) [6] Shen Z. Lp estimates for Schr¨ odinger operators with certain potentials. Ann Inst Fourier (Grenoble), 1995, 45(2): 513–546 [7] Stein E M. Harmonic Analysis: Real Variable Methods, Orthogonality and Oscillatory Integrals//Princeton Math Series 43. Princeton, NJ: Princeton University, 1993