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Weighted estimates for commutators of some singular integrals related to Schrödinger operators
Accepted Manuscript Weighted estimates for commutators of some singular integrals related to Schrödinger operators The Anh Bui
PII: DOI: Reference:
S0007-4497(13)00078-X 10.1016/j.bulsci.2013.06.007 BULSCI 2574
To appear in:
Bulletin des Sciences Mathématiques
Received date: 10 September 2012
Please cite this article in press as: T.A. Bui, Weighted estimates for commutators of some singular integrals related to Schrödinger operators, Bull. Sci. math. (2013), http://dx.doi.org/10.1016/j.bulsci.2013.06.007
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Weighted estimates for commutators of some singular integrals related to Schr¨odinger operators The Anh Bui∗
Abstract Let L = −Δ + V be a Schr¨odinger operator with non-negative potential V satisfying some appropriate reverse H¨older inequality. In this paper, we study the boundedness of the commutators of some singular integrals associated to L such as Riesz transforms and fractional integrals with the new BMO functions introduced in [BHS1] on the weighted spaces Lp (w) where w belongs to the new classes of weights introduced by [BHS2].
1
Introduction
Let L = −Δ + V be the Schr¨odinger operators on Rn with n ≥ 3 where the potential V is in the reverse H¨older class RHq for some q > n/2, i.e., V satisfies the reverse H¨older inequality ˆ 1/q 1 ˆ C V (y)q dy ≤ V (y)dy |B| B |B| B for all ball B ⊂ Rn . In this paper, we consider the following singular integrals associated to L: (i) Riesz transforms R = ∇L−1/2 and their adjoint R∗ = L−1/2 ∇; (ii) Fractional integrals Iα f (x) = L−α/2 f (x) for 0 < α < n. In the classical case when V = 0, it has been shown that Riesz transforms R and their commutators Rb with BMO functions b is bounded on Lp (w) for all 1 < p < ∞ and w in the Muckenhoupt classes Ap , see for example [St]. Also, the classical fractional integrals and their commutators with BMO functions b are bounded from Lp (wp ) to Lq (wq ) for all 1 < p < n/α, 1/p − 1/q = α/n and w ∈ A1+1/p ∩ RHq , or equivalently wq ∈ A1+ q , where Ap is the p
Muckenhoupt class of weights, see for example [MW, ST]. Recall that a non-negative and locally integrable function w is said to be in the Muckenhoupt Ap classes with 1 ≤ p < ∞, if the following inequality holds for all balls B ⊂ Rn ˆ 1/p 1/p ˆ − 1 w w p−1 ≤ C|B|. (1) B
B
∗
Department of Mathematics, University of Pedagogy, HoChiMinh City, Vietnam and Department of Mathematics, Macquarie University, NSW 2109, Australia. Email: bt [email protected] and [email protected] 2000 Mathematics Subject Classification: 42B35, 35J10. Key words: weight, Riesz transform, fractional integral, Schr¨ odinger operator, commutator, BMO.
1
Recently, in [BHS2], a new class of weights associated to Schr¨odinger operators L has been L,θ introduced. According to [BHS2], the authors defined the new classes of weights AL p = ∪θ>0 Ap for p ≥ 1, where AL,θ is the set of those weights satisfying p 1/p ˆ 1/p ˆ r θ − 1 w w p−1 ≤ C|B| 1 + (2) ρ(x) B B L for all ball B = B(x, r). We denote AL ∞ = ∪p≥1 Ap where the critical radius function ρ(·) is defined by ˆ 1 V ≤ 1 , x ∈ Rn , (3) ρ(x) = sup r > 0 : n−2 r B(x,r)
see [Sh]. It is easy to see that in certain circumstances the new class AL p is larger than the Muckenhoupt , see [BHS2, Proposition 5]. class Ap . The following properties hold for new classes AL p Proposition 1.1 The following statements hold: L i) AL p ⊂ Aq for 1 ≤ p ≤ q < ∞. L ii) If w ∈ AL p with p > 1 then there exists > 0 such that w ∈ Ap− . Consequently, L AL p = ∪q
1
Lp (w) for 1 < p < s with w p−1 ∈ AL p /s . (ii) If q ≥ n, the Riesz transforms R∗ and R are bounded on Lp (w) for 1 < p < ∞ and w ∈ AL p. (b) Let V ∈ RHq with q > n/2. Then Iα is bounded from Lp (wp ) to Lq (wq ) for all 1 < p < . n/α, 1/p − 1/q = α/n and wq ∈ AL 1+ q p
For the proof we refer to Theorem 3 and Theorem 4 in [BHS2]. Now we consider the commutators of the Riesz transforms R and R∗ with the BMO functions b. It was proved in [GLP] that the commutators Rb and Rb∗ are bounded on Lp here the range of p depends on the potential V . Then the authors in [BHS1] extended the classes of BMO functions to the new class BM OLθ with θ > 0 for the boundedness of the commutators Rb and Rb∗ . We would like to give a brief overview of the results in [BHS1]. According to [BHS1], the new BMO space BM OLθ with θ > 0 is defined as a set of all locally integrable functions b satisfying ˆ 1 r θ |b(y) − bB |dy ≤ C 1 + (4) |B| B ρ(x)
2
where B = B(x, r) and bB =
1 |B|
´ B
b. A norm for b ∈ BM OLθ , denoted by b θ , is given
θ2 by the infimum of the constants satisfying (4). Clearly BM OLθ1 ⊂ BM OL for θ1 ≤ θ2 and 0 ∞ θ BM OL = BM O. We define BM OL = ∪θ>0 BM OL . The following result can be considered to be a variant of John-Nirenberg inequality for the spaces BM OLθ , see [BHS1, pp.119-120]. θ then for all B = (x , r) Proposition 1.3 Let θ > 0, s ≥ 1. If b ∈ BM OL 0 i) 1 ˆ 1/s r θ |b(y) − bB |s dx b θ 1 + |B| B ρ(x0 )
where θ = (N0 + 1)θ and N0 is a constant in (5). ii) 1/s 1 ˆ 2k r θ |b(y) − b |dx
b
k 1 + B θ ρ(x0 ) |2k B| 2k B for all k ∈ N. ∞ we consider the commutators Let T be either R or R∗ . For b ∈ BM OL
Tb f (x) = T (bf )(x) − b(x)T f (x) and
Iαb f (x) = Iα (bf )(x) − b(x)Iα f (x).
It was proved in [BHS1, Theorem 1] that θ with θ > 0 and V ∈ RH . Theorem 1.4 Let b ∈ BM OL q (i) If n/2 < q < n and s is such that 1/s = 1/q − 1/n, the commutators Rb∗ are bounded on Lp for s < p < ∞ and hence by duality Rb is bounded on Lp for 1 < p < s. (ii) If q ≥ n, the commutators Rb∗ and Rb are bounded on Lp for 1 < p < ∞..
The aim of this paper is investigating the boundedness of the commutators Rb∗ , Rb and Iαb with b ∈ BM OL∞ on new weighted spaces Lp (w) with w ∈ AL p. The organization of this paper is as follows. In Section 2, we recall some basic properties of the critical radius function ρ(·) and consider weighted estimates for some localized operators. Section 3 is devoted to prove the main results on weighted estimates of the commutators Rb∗ , Rb and Iαb . Finally, we make some conventions. Throughout the whole paper, we denote by C a positive constant which is independent of the main parameters, but it may vary from line to line. The symbol X Y means that there exists a positive constant C such that X ≤ CY . Recently, we have learned that the AL p weighted norm inequalities for the commutators of the Riesz transforms was obtained independently in [BHS3]. However, the approach in our paper is different from that in [BHS3]. Moreover, the weighted norm inequalities for the commutators of fractional integrals L−α/2 is unique.
3
2
Weighted estimates for some localized operators
We would like to recall some important properties concerning the critical radius function which will play an important role to obtain the main results in the sequel, see [Sh, DZ1] respectively. Proposition 2.1 If V ∈ RHn/2 , there exist c0 and N0 ≥ 1 such that c−1 0 ρ(x)
N0 |x − y| −N0 |x − y| N0 +1 1+ ≤ ρ(y) ≤ c0 ρ(x) 1 + ρ(x) ρ(x)
(5)
for all x, y ∈ Rn . A ball of the form B(x, ρ(x)) is called a critical ball. From the inequality (5), we can imply that for x, y ∈ σQ where Q is a critical ball and σ > 0, then ρ(x) ≤ cσ ρ(y)
(6)
2N0 +1
where cσ = c20 (1 + σ) N0 +1 . Proposition 2.2 There exists a sequence of points xj , j ≥ 1 in Rn so that the family Qj := B(xj , ρ(xj )) satisfies (i) ∪j Qj = Rn . (ii) For every σ ≥ 1 there exist constants C and N1 such that j χσQj ≤ Cσ N1 . Following [BHS1], we introduce the following maximal functions for g ∈ L1loc (Rn ) and x ∈ Rn ˆ 1 Mρ,α g(x) = sup |g|, x∈B∈Bρ,α |B| B ˆ 1 g(x) = sup |g − gB |, Mρ,α x∈B∈Bρ,α |B| B where Bρ,α = {B(y, r) : y ∈ Rn and r ≤ αρ(y)}. Also, given a ball Q, we define the following maximal functions for g ∈ L1loc (Rn ) and x ∈ Q ˆ 1 MQ g(x) = sup |g|, x∈B∈F (Q) |B ∩ Q| B∩Q ˆ 1 g(x) = sup |g − gB∩Q |, MQ x∈B∈F (Q) |B ∩ Q| B∩Q where F(Q) = {B(y, r) : y ∈ Q, r > 0}. We have the following lemma. Lemma 2.3 For 1 < p < ∞, then there exist β and γ such that if {Qk }k is a sequence of balls as in Proposition 2.2 then ˆ ˆ 1 ˆ p |Mρ,β g(x)|p w(x)dx |Mρ,γ g(x)|p w(x)dx + w(Qk ) |g| |2Qk | 2Qk Rn Rn k
for all g ∈ L1loc (Rn ) and w ∈ AL ∞. 4
Proof: Note that the unweighted estimate of Lemma 2.3 was obtained [BHS1, Lemma 2]. Now we adapt some ideas in [BHS1, Lemma 2] to our present setting. According to [BHS1, p. 121], there exists β > 0 so that for all critical balls Q and x ∈ Q, we have Mρ,β g(x) ≤ M2Q (gχ2Q )(x), and for x ∈ 2Q,
(gχ2Q )(x) ≤ Mρ,2 g(x). M2Q
are HardyNote that since for each k, gχ2Qk is supported in 2Qk , operators M2Qk and M2Q k Littlewood and sharp maximal functions defined in 2Qk viewed as a space of homogeneous type with the Euclidean metric and the Lebesgues measure restricted to 2Qk . Moreover, by definition ∞ of A∞ ∞ , if w ∈ A∞ then w ∈ A∞ (2Qk ), where A∞ (2Qk ) = ∪p≥1 Ap (2Qk ) and Ap (2Qk ) is the class of Muckenhoupt weights on the spaces of homogeneous type 2Qk . Moreover, due to [BHS2, Lemma 5], and [PS, Proposition 3.4] gives ˆ ˆ p |Mρ,β g(x)| w(x)dx ≤ |Mρ,β g(x)|p w(x)dx Rn
Qk
k
≤
ˆ
Qk
k
|M2Q (gχ2Q )(x)|p w(x)dx
ˆ p 1 |g| |2Qk | 2Qk 2Qk k k ˆ ˆ p 1 |Mρ,2 g(x)|p w(x)dx + w(Qk ) |g| |2Qk | 2Qk 2Qk k ˆ ˆk p 1 |Mρ,2 g(x)|p w(x)dx + w(Qk ) |g| . |2Qk | 2Qk Rn
ˆ
|M2Q g(x)|p w(x)dx + k
w(Qk )
k
This completes our proof. The following maximal functions will be useful in the sequel. For N > 0, κ ≥ 1, 0 < α < n and 1 ≤ s < n/α, we define the following functions for g ∈ L1loc (Rn ) and x ∈ Rn GN κ,α,s f (x)
=
sup
∞
Qx;Q is critical k=0
and N f (x) = Hκ,s
sup
−N k
2
∞
Qx;Q is critical k=0
ˆ
1 1−αs/n |2k Q|
2−N k
1 |2k Q|
ˆ 2k Q
2k Q
|f (z)|s dz
|f (z)|s dz
1/s
1/s
= κQ. where Q N N N When κ = 1, we write GN α,s and Hs instead of G1,α,s and H1,s , respectively. We are now in N position to establish the weighted estimates for GN κ,α,s and Hκ,s .
5
Proposition 2.4 (i) Let wq ∈ AL,θ
q/s
1+ (p/s)
then we have
and N > N0 θ( 1s − αn ). If p > s and 1/p − 1/q = α/n,
GN κ,α,s f Lq (wq ) f Lp (wp )
(ii) Let w ∈ AL,θ p/s and N >
N0 θ s .
If p > s, then we have
N
Hκ,s f Lp (w) f Lp (w) .
Proof: (i) Without of loss of generality, we can assume that κ = 1. Assume that Q = B(x0 , ρ(x0 )). For x ∈ Q, the inequalities (5) tells us that C0−1 ρ(x0 ) ≤ ρ(x) ≤ C0 ρ(x0 ). This implies |B(x, ρ(x))| ≈ |Q| and Q ⊂ 2C0 B(x, ρ(x)). Therefore, GN α,s f (x)
∞
−N k
2
k=0
1 k |2 B(x, ρ(x))|1−αs/n
ˆ Bk (x,ρ(x))
|f (z)|s dz
1/s
where Bk (x, ρ(x)) = 2C0 × 2k B(x, ρ(x)). Let {Qj }j be the family of critical balls as in Proposition 2.2. By (5), C0−1 ρ(x) ≤ ρ(xj ) ≤ C0 ρ(x) for all x ∈ Qj . So, |B(x, ρ(x))| ≈ |Qj | and Bk (x, ρ(x)) ⊂ Qkj where Qkj = 2C0 × 2k Qj . For wq ∈ A1+ 1 , using H¨older inequalities, we obtain (p/s)
GN α,s f Lq (wq )
ˆ q/s 1/q 1 s q 2 |f (z)| dz w (x)dx 1−αs/n k Qj |2 B(x, ρ(x))| Bk (x,ρ(x)) j k=0 ˆ ˆ ∞ q/s 1/q 1 s q 2−N k |f (z)| dz w (x)dx 1−αs/n k Qj |2 Qj | Qkj j k=0 ˆ ∞ q/s 1/q wq (Qj ) s 2−N k |f (z)| dz |2k Qj |q/s−αq/n Qkj j k=0 ∞
∞
−N k
2−N k
ˆ
j
k=0
× wq (Qkj ) Since wq ∈ AL
1 |2k Qj |q/s−αq/n
ˆ Qkj
(wq )
− qs ( ps )
q/s (p/s)
ˆ Qkj
|f (z)|p wp (z)dz
q/p 1/q
. (7)
q/s
1+ (p/s)
w
q
, we have, by (5), (Qkj )
ˆ Qkj
(wq )
− qs ( ps )
q/s (p/s)
≤ C|Qkj |1/s−α/n 2kN0 θ×(1/s−α/n)
for some θ > 0. 6
This in combination with (7) gives
GN α,s f Lq (wq )
2−(N −N0 θ×(1/s−α/n))k
j
k
ˆ
2−(N −N0 θ×(1/s−α/n))k
ˆ
k
Rn
Qkj
|f (z)|p wp (z)dz
|f (z)|p wp (z)dz
1/p
1/p
C f Lp (wp ) where in the last inequality we used (ii) in Proposition 2.2. (ii) The proof of (ii) is just a direct consequence of (i) when α = 0. This completes our proof. For 0 ≤ α < n, let Mα be the fractional maximal function defined by ˆ 1 |f (y)|dy. Mα f (x) = sup 1−α/n Bx |B| B For s ≥ 1, we define
Mα,s f = sup
.
Bx
ˆ
1 |B|1−sα/n
B
|f (y)|s dy
1/s
α,s as follows For a family of balls {Qk }k given by Proposition 2.2, we define the operator M α,s f = χQk Mα,s (f χQk ) (8) M k
j = 4(2C 2 + 1)γQj and γ is a constant in Proposition 2.3. where Q 0 Remark 2.5 (i) For s < p < ∞ and 1/p − 1/q = α/n, it was proved in [MW] that Mα,s is bounded from Lp (wp ) to Lq (wq ) with wq ∈ A1+ (q/s) . This together with [BHS2, Proposition 4] (p/s)
α,s is bounded from Lp (wp ) to Lq (wq ) with wq ∈ AL shows that M
(q/s)
1+ (p/s)
here s < p < ∞ and
1/p − 1/q = α/n. 0,s . The similar argument as in (i) also shows s instead of M (ii) When α = 0, we write M L that for p > s and w ∈ Ap/s , Ms is bounded on Lp (w).
3
Weighted estimates for commutators of singular integrals
3.1 3.1.1
Riesz transforms Kernel estimates of Riesz transforms
In the sequel, let us remind that for the number N , we shall mean that N is a sufficiently large number and different from line to line. Let K and K ∗ be the vector valued kernels of R and R∗ respectively. The following propositions give some estimates on the kernels of R and R∗ , see for example [Sh, GLP]. 7
Proposition 3.1 a) If V ∈ RHq with q > n/2 then we have (i) For every N there exists a constant C such that ∗
|K (x, y)| ≤ C
|x−z| −N ρ(x) ) − z|n−1
(1 + |x
ˆ B(z,|x−z|/4)
V (u) 1 . + |u − z|n−1 |x − z|
(9)
(ii) For every N and 0 < δ < min{1, 2 − n/q} there exists a constant C such that ∗
∗
|K (x, z) − K (y, z)| ≤ C
|x−z| −N ρ(x) ) z|n−1+δ
|x − y|δ (1 + |x −
ˆ B(z,|x−z|/4)
V (u) 1 (10) + |u − z|n−1 |x − z|
whenever |x − y| < 23 |x − z|. b) If V ∈ RHq with q ≥ n then we have (i) For every N there exists a constant C such that |K(x, y)| ≤ C
|x−y| −N ρ(x) ) |x − y|n
(1 +
.
(11)
(ii) For every N and 0 < δ < min{1, 1 − d/q} there exists a constant C such that ∗
∗
|K (x, z) − K (y, z)| ≤ C
|x−z| −N ρ(x) ) n+δ z|
|x − y|δ (1 + |x −
(12)
whenever |x − y| < 23 |x − z|. 3.1.2
Commutators of Riesz transforms
The main result concerning the weighted estimates for Rb∗ and Rb is formulated by the following theorem. θ with θ > 0 and V ∈ RH . Theorem 3.2 Let b ∈ BM OL q (i) If n/2 < q < n and s is such that 1/s = 1/q − 1/n, the commutator Rb∗ is bounded on p Lp (w) for s < p < ∞ and w ∈ AL p/s and hence by duality Rb is bounded on L (w) for 1 < p < s −
1
with w p−1 ∈ AL p /s . (ii) If q ≥ n, the commutators Rb∗ and Rb are bounded on Lp (w) for 1 < p < ∞ and w ∈ AL p. Proof: (i) To prove (i), we exploit the strategy in [BHS1]. For any s < p < ∞ and w ∈ AL p/s , we have by Lemma 2.3 ˆ
Rb∗ f pLp (w) ≤ |Mρ,β (Rb∗ f )(x)|p w(x)dx n ˆR 1 ˆ p |Mρ,γ (Rb∗ f )(x)|p w(x)dx + w(Qk ) |Rb∗ f | |2Qk | 2Qk Rn k
8
where {Qk } is a family of critical balls given in Lemma 2.3. Now let N > 0 be a number which will be fixed later. For the sake of simplicity, we assume that the number N can change from´ each of its p appearances. 1 ∗ 1. Estimate k w(Qk ) |2Qk | 2Qk |Rb f |
Let s < p0 < p and Q = B(x0 , ρ(x0 )) so that w ∈ AL p/p0 . We borrow some ideas in [BHS1, Lemma 5] to write Rb∗ f = (b − bQ )R∗ f − R∗ ((b − bQ )f ).
So, we have ˆ ˆ ˆ 1 1 1 |Rb∗ f |dx ≤ |(b − bQ )R∗ f |dx + |R∗ ((b − bQ )f )|dx := I1 + I2 . |2Q| 2Q |2Q| 2Q |2Q| 2Q Let us estimate I1 first. By H¨older inequality, we can write 1 ˆ 1/p0 I1 b θ |R∗ f |p0 |2Q| 2Q 1 ˆ 1/p0 1 ˆ 1/p0 ∗ p0 b θ |R f1 | + |R∗ f2 |p0 |2Q| 2Q |2Q| 2Q := I11 + I12 where f = f1 + f2 with f1 = f χ4Q . Due to Lp0 -boundedness of R∗ , one has 1 ˆ 1/p0 I11 |f |p0 inf HpN0 f (z). z∈Q |2Q| 4Q To estimate I12 , for x ∈ 2Q, due to (9), we have ˆ ∗ |K ∗ (x, y)f (y)|dy R f2 (x) ≤ ˆ
Rn \4Q
|x−y| −N ρ(x) ) |x − y|n
(1 + Rn \4Q
ˆ
+
|f (y)|dy
|x−y| −N ρ(x) ) − y|n−1
(1 +
Rn \4Q
|x
ˆ B(y,|x−y|/4)
V (u) |f (y)|dudy |u − y|n−1
:= A1 (x) + A2 (x). To take care A1 , note that ρ(x) ≈ ρ(x0 ) and |x − y| ≈ |x0 − y| for all x ∈ 2Q and y ∈ Rn \4Q. So, decomposing Rn \4Q into annuli 2k+1 \2k Q, we have 2−kN ˆ A1 (x) |f (y)|dy |2k Q| 2k Q k≥2 1 ˆ 1/p0 p0 2−kN |f (y)| dy |2k Q| 2k Q k≥2
inf HpN0 f (z) z∈Q
9
for all x ∈ 2Q. For the term A2 , by decomposing Rn \4Q into annuli 2k+1 \2k Q, we get that ˆ ˆ 2−kN V (u) |f (y)| dudy A2 (x) n−1 1−1/n k |2 Q| 2k Q 2k+2 Q |u − y| k≥2 ˆ 2−kN |f (y)|I1 (V χ2k+2 Q )(y)dy |2k Q|1−1/n 2k Q k≥2 where I1 = (−Δ)−1/2 . Let us remind that I1 is Lq0 − Lp0 boundedness with 1/p0 = 1/q0 − 1/n. This together with H¨ older inequality gives ˆ 2−kN A2 (x) |(f χ2k Q )(y)|I1 (V χ2k+2 Q )(y)dy |2k Q|1−1/n Rn k≥2 1/p0 2−kN ˆ p0 |f |
I1 (V χ2k+2 Q ) Lp0 |2k Q|1−1/n 2k Q k≥2 1 ˆ 1/p0 2−kN p0 |f |
V χ2k+2 Q Lq0 |2k Q|1−1/n−1/p0 |2k Q| 2k Q k≥2
Noting that if we choose p0 to be close enough to s then V ∈ RHq0 . This in combination with the fact that V is a doubling measure gives ˆ 1
V χ2k+2 Q Lq0 k 1/q V |2 Q| 0 2k+2 Q ˆ 2kκ k 1/q V for some κ > 0 |2 Q| 0 Q 2kκ k 1/q −1+2/n . |2 Q| 0 Hence,
1 2−kN A2 (x) |2k Q|1/q0 −1+2/n |2k Q|1−1/n−1/p0 |2k Q| k≥2 1 ˆ 1/p0 −k(N −κ) p0 2 |f | |2k Q| 2k Q
2kκ
ˆ 2k Q
|f |p0
1/p0
k≥2
inf HpN0 f (z) z∈Q
for all x ∈ 2Q. From the estimates of A1 and A2 , we obtain I12 inf z∈Q HpN0 f (z). The term I2 can be estimated in the same line with I1 . Using the decomposition f = f1 + f2 again, one gets that 1 ˆ 1 ˆ ∗ I2 ≤ |R ((b − bQ )f1 )(y)|dy + |R∗ ((b − bQ )f2 )(y)|dy := I21 + I22 . |2Q| 2Q |2Q| 2Q 10
Choose s < r < p0 . Using H¨older inequality and Lr -boundedness of R∗ , we have 1 ˆ 1/r I21 |R∗ ((b − bQ )f1 )(y)|r dy |2Q| 2Q 1 ˆ 1/r |((b − bQ )f1 )(y)|r dy |2Q| 4Q 1 ˆ 1/p0 1 ˆ 1/ν p0 |f1 (y)| dy (b − bQ )ν dy for some ν > r |2Q| 4Q |2Q| 4Q b θ inf HpN0 f (z). z∈Q
The estimate of I22 b θ inf z∈Q HpN0 f (z) can be taken care similarly to ones of I12 and I21 . So we omit the details here. To sum up, it had proved that for any critical ball Q, we have ˆ 1 |R∗ f |dx b θ inf HpN0 f (z). (13) z∈Q |2Q| 2Q b p ´ 1 ∗ f | . Due to (19) and the fact that Return to the estimate of w(Q ) |R k k b |2Qk | 2Qk w ∈ AL p/p0 , we have
k
1 w(Qk ) |2Qk |
ˆ 2Qk
k
k
|Rb∗ f |
p
p
b pθ w(Qk ) inf HpN0 f (z)
b pθ
b pθ
ˆ Rn
ˆ Qk
z∈Q
|HpN0 f (z)|p w(z)dz
|HpN0 f (z)|p w(z)dz.
Due to Proposition 2.4, since we can take N sufficiently large so that HpN0 f Lp (w) f Lp (w) . Hence 1 ˆ p w(Qk ) |Rb∗ f | b pθ f pLp (w) . |2Qk | 2Qk k ´ 2. Estimate Rn |Mρ,γ (Rb∗ f )(x)|p w(x)dx For any ball B(x0 , r) with r ≤ γρ(x0 ) and x ∈ B, we write ˆ 1 |R∗ f (x) − (Rb∗ f )B |dx |B| B b ˆ ˆ 2 2 ∗ |(b − bB )R f (x)|dx + |R∗ ((b − bB )f1 )(x)|dx ≤ |B| B |B| B ˆ 1 |R∗ ((b − bB )f2 )(x) − (R∗ ((b − bB )f2 ))B |dx + |B| B := E1 + E2 + E3 . where f = f1 + f2 with f1 = f χ2B . 11
Let s < p0 < p so that w ∈ AL older inequality and Proposition 1.3 show that p/p0 . The H¨ 1 ˆ 1/p0 1 ˆ 1/p0 |b − bB |p0 |R∗ f |p0 |B| B |B| B 1 ˆ 1/p0 b θ |R∗ f |p0 . |B| B
E1
j . This yields For any critical ball Qj such that x ∈ Qj ∩ B. It can be verified that B ⊂ Q that p (R∗ f )(y). E1 b θ × inf M 0 y∈B
older inequality and and Proposition 1.3 again tell us that For some s < r < p0 < p, H¨ 1/r 1 ˆ |R∗ ((b − bB )f1 )|r E2 |B| B 1/r 1 ˆ |(b − bB )f1 |r |B| 2B 1/ν 1 ˆ 1/p0 1 ˆ |(b − bB )|ν |f |p0 for some ν > r |B| 2B |B| 2B p (f )(y). b θ × inf M y∈B
0
To estimate E3 , we need the to show that ˆ N p (f )(u)) |K ∗ (x, z) − K ∗ (y, z)||b(z) − bB ||f (z)|dz b θ ( inf Hγ,p f (u) + inf M 0 0 u∈B
Rn \2B
u∈B
(14)
for all f and x, y ∈ B. If this holds, then we have ˆ ˆ ˆ 1 |K ∗ (u, z) − K ∗ (y, z)||b(z) − bB ||f (z)|dz dydu E3 2 |B| B B Rn \2B N p (f )(x)). f (x) + M b θ (Hγ,p 0 0
These three estimates of E1 , E2 and E3 give N p (R∗ f )(x) + Hγ,p p (f )(x)). (Rb∗ )(x) b θ (M (x) + M Mρ,γ 0 0 0
This implies p (R∗ f ) Lp (w) + H N f Lp (w) + M p (f ) Lp (w) . (Rb∗ ) Lp (w) b θ ( M
Mρ,γ γ,p0 0 0 N is bounded on Lp (w). This together with Note that we can choose N large enough so that Hγ,p 0 p and R∗ f are bounded on Lp (w) for w ∈ AL gives the desired estimate. the fact that M 0 p/p0
Proof of (14): We adapt some ideas of [BHS1, Lemma 6] to our present situation. Setting Q = B(x0 , γρ(x0 )), due to (10) and the fact that ρ(x) ≈ ρ(x0 ) and |x − z| ≈ |x0 − z|, we get ˆ |K ∗ (x, z) − K ∗ (y, z)||b(z) − bB ||f (z)|dz K1 + K2 + K3 + K4 Rn \2B
12
where
ˆ
|f (z)(b(z) − bB )| dz, |x0 − z|n+δ Q\2B ˆ |f (z)(b(z) − bB )| K2 = rδ ρ(x0 )N dz, n+δ+N Qc |x0 − z| ˆ ˆ |f (z)(b(z) − bB )| V (u) dudz, K3 = r δ n−1+δ |u − z|n−1 |x − z| 0 Q\2B B(x0 ,4|x0 −z|)
K1 = r
δ
and δ
K4 = r ρ(x0 )
N
ˆ Qc
|f (z)(b(z) − bB )| |x0 − z|n−1+δ+N
Let j0 be the smallest integer so that k = 1, . . . , j0 , we obtain
2j 0 r
ˆ B(x0 ,4|x0 −z|)
V (u) dudz. |u − z|n−1
≥ γρ(x0 ). Splitting Q\2B into annuli 2k+1 B\2k B for
ˆ j0 2−kδ K1 |f (z)(b(z) − bB )|dz. |2k B| 2k B k=1
By H¨older inequality and Proposition 1.3, K1
j0
−kδ
k2
b θ
1 k |2 B|
k=1
ˆ 2k B
|f (z)|p0 dz
1/p0
j where Qj and Q
j are balls in (8). Therefore, Remind that if x ∈ B ∩ Qj then 2k B ⊂ Q K1
j0
p f (u). k2−kδ b θ inf M 0 u∈B
k=1
Splitting Qc into annuli and then applying H¨ older inequality and Proposition 1.3 again, we obtain ˆ ∞ ρ(x ) N 2−kδ−(k−j0 )N 0 |f (z)(b(z) − bB )|dz K2 2j0 r |2k−j0 +1 2j0 −1 B| 2k B k=j0 −1 ˆ ∞ ρ(x ) N 1/p0 1 0 −kδ−(k−j0 )(N −θ ) p0 k2 |f (z)| dz . b θ 2j0 r |2k−j0 +1 2j0 −1 B| 2k B k=j0 −1
Since 2j0 r ≥ γρ(x0 ) ≥ 2j0 −1 r, we get that K2 b θ b θ
∞
−(k−j0 )(N −θ )
2
k=j0 −1 ∞ −l(N −θ )
2
l=1
1 |2l Q|
1
ˆ
|2k−j0 +1 Q|
ˆ 2l Q
|f (z)|p0 dz
2k−j0 +1 Q
1/p0
N b θ inf Hγ,p f (u) since Q = γB(x0 , ρ(x0 )). 0 u∈B
13
|f (z)|p0 dz
1/p0
It can be verified that K3
j0
Let s < q0 < p0 , ν = Proposition 1.3 K3 b θ
2−kδ k |2 B|1−1/n k=2 p 0 q0 p0 −q0
ˆ 2k B
|f (z)(b(z) − bB )|I1 (V χ2k+2 B )(z)dz
and r such that 1/r = 1/q0 + 1/n then by H¨older inequality and
1 k2−kδ |2k B|−1/n+1/q0 |2k B| k=2
j0
1 k2−kδ b θ |2k B|−1/n+1/q0 |2k B| k=2 j0
b θ
ˆ 2j B
ˆ 2j B
|f |p0 |f |p0
1/p0 1/p0
I1 (V χ2k+2 B ) Lq0
V χ2k+2 B Lr
j0
k2−kδ p f (u)
V χ2k+2 B Lr inf M 0 k B|−1/n+1/q0 u∈B |2 k=2
Noting that we can choose q0 so that V ∈ RHr , then we have ˆ ˆ 1/r 1
V χ2k+2 B Lr V (z)r dz V (z)dz |Q|1−1/r Q Q 1 1 k 2/n−1/r |Q|2/n−1/r |2 B| for all k = 2, . . . , j0 . So, K3 b θ
j0
p f (x) := C b θ inf M p f (u). k2−kδ M 0 0 u∈B
k=2
The similar arguments to ones used to obtain the estimates K2 and K3 gives N K4 b θ inf Hγ,p f (u). 0 u∈B
This completes the proof of (i). (ii) The proof of part (ii) is completely analogous to that of (i). The main difference here is that in this situation we can work directly on R intead of R∗ via the pointwise estimates of the kernels K(x, y) of the Riesz transforms R, see for example [BHS1, Lemma 4]. Hence, we would like to omit the details here leave the proof to the interested readers.
3.2 3.2.1
Fractional integrals Kernel estimates of fractional integrals
Let Kα be the kernel of Iα . The following results give the estimates on the kernel Kα (x, y). Proposition 3.3 If V ∈ RHq with q > n/2 then we have 14
(i) For every N there exists a constant C such that |Kα (x, y)| ≤ C
|x−y| −N ρ(x) ) − y|n−α
(1 + |x
.
(15)
(ii) There is a number δ > 0 such that for every N there exists a constant C such that |Kα (x, y) − Kα (x, z)| ≤ C
|x−y| −N ρ(x) ) n+δ−α y|
|y − z|δ (1 + |x −
(16)
whenever |y − z| < 14 |x − y|. To prove Proposition 3.3, we need the following estimates of the heat kernels of e−tL , see [DZ2, p.12] Proposition 3.4 Let pt (x, y) be the kernels associated to the semigroups {e−tL }t>0 . If V ∈ RHq with q > n/2 then we have (i) For every N > 0 there exists a constant C such that √ √ t t −N C |x − y|2 ; (17) 1+ + |pt (x, y)| ≤ n/2 exp − c t ρ(x) ρ(y) t (ii) There is a number δ > 0 such that for every N there exists a constant C such that |pt (x, y) − pt (x, z)| + |pt (y, x) − pt (z, x)| √ √ t t −N |x − y|2 C |y − z| δ √ exp − c 1+ + ≤ n/2 t ρ(x) ρ(y) t t whenever |y − z| < 12 |x − y|. Proof of Proposition 3.3: (i) We have, by (17), ˆ ∞ dt |tα/2 pt (x, y)| |Kα (x, y)| ≤ t 0 √ √ ˆ |x−y|2 α/2 t t t −N dt |x − y|2 exp − c 1+ + n/2 t ρ(x) ρ(y) t t 0 √ √ ˆ ∞ α/2 2 −N t t t dt |x − y| + exp − c 1+ + n/2 t ρ(x) ρ(y) t 2 t |x−y| = I1 + I 2 . Let us estimate I2 first. Since t > |x − y|2 , we have, for > 0 so that n > α + , ˆ ∞ tα/2 |x − y|2 |x − y| −N dt exp − c I2 1 + n/2 t ρ(x) t |x−y|2 t ˆ ∞ α/2 n/2−α/2− −N t dt t |x − y| ( 1+ n/2 |x − y|2 ρ(x) t 2 t |x−y| |x − y| −N 1 1+ . n−α |x − y| ρ(x) 15
(18)
For I1 , we have ˆ I1 ˆ
0
0
ˆ
|x−y|2
√ n/2+N/2 t −N dt t 1+ 2 n/2 |x − y| ρ(x) t t √ √ α/2 N t t t + ρ(x) −N dt
|x−y|2
|x − y|n |x − y| ρ(x) t √ √ tα/2 t + ρ(x) N t + ρ(x) −N dt
|x−y|2 α/2 t
|x − y|n |x − y| + ρ(x) 1 |x − y| −N . 1+ n−α |x − y| ρ(x)
ρ(x)
0
t
This completes (i). (ii) For |y − z| < 14 |x − y|, using (18) gives
√ √ − z| δ t t −N dt |x − y|2 √ 1+ exp − c + |Kα (x, y) − Kα (x, z)| n/2 t ρ(x) ρ(y) t t t 0 √ √ |y − z| δ ˆ ∞ tα/2 2 −N |x − y| t t dt exp − c 1+ + n/2 |x − y| t ρ(x) ρ(y) t t 0 ˆ ∞ |y − z| δ ˆ |x−y|2 ... + ... |x − y| |x−y|2 0 |y − z| δ (II1 + II2 ). = |x − y| ˆ
∞ α/2 |y t
At this stage, repeating the arguments in (i), we obtain (ii). 3.2.2
Commutators of fractional integrals
We are now in position to state the result concerning the weighted estimates for Iαb . Theorem 3.5 Let b ∈ BM OLθ with θ > 0 and V ∈ RHq with q > n/2. Then the commutator . Iαb is bounded from Lp (wp ) to Lq (wq ) for 1 < p < ∞, 1/p − 1/q = α/n and wq ∈ AL 1+ q p
Proof: The strategy of the proof for Theorem 3.5 is similar to that of Theorem 3.2. , we have by Lemma 2.3 For any 1 < p < ∞, 1/p − 1/q = α/n and wq ∈ AL 1+ q
Iαb f qLq (wq ) ≤
ˆ ˆ
Rn Rn
p
|Mρ,β (Iαb f )(x)|q wq (x)dx |Mρ,γ (Iαb f )(x)|q wq (x)dx +
k
wq (Qk )
1 |2Qk |
ˆ 2Qk
|Iαb f |
q
where {Qk } is a family of critical balls given in Proposition 2.3. ´ So, to obtain the weighted estimates for Iαb , we need only to consider Rn |Mρ,γ (Iαb f )(x)|q wq (x)dx q ´ q and k w (Qk ) |2Q1 k | 2Qk |Iαb f | . 16
Now let N > 0 be a number which will be fixed later. For the sake of simplicity, we assume that the number N can change fromeach of its appearances. q ´ Step 1. Estimate k wq (Qk ) |2Q1 k | 2Qk |Iαb f | Let 1 < s < p, 1/s − 1/υ = α/n and Q = B(x0 , ρ(x0 )) so that wq ∈ AL
(q/s)
1+ (p/s)
. We have
Iαb f = (b − bQ )Iα f − Iα ((b − bQ )f ). Hence, 1 |2Q|
ˆ 2Q
|Iαb f |dx ≤
1 |2Q|
ˆ 2Q
|(b − bQ )Iα f |dx +
1 |2Q|
ˆ 2Q
|Iα ((b − bQ )f )|dx := I1 + I2 .
To take care I1 , using H¨older inequality and Proposition 1.3, we get that 1 ˆ 1/υ I1 b θ |Iα f |υ |2Q| 2Q 1 ˆ 1/υ 1 ˆ 1/υ υ b θ |Iα f1 | + |Iα f2 |υ |2Q| 2Q |2Q| 2Q := I11 + I12 where f = f1 + f2 with f1 = f χ4Q . Since Iα is Ls − Lυ bounded, one has I11
1
ˆ
|2Q|1−αs/n
4Q
|f |s
1/s
inf GN α,s f (z). z∈Q
To estimate I12 , for x ∈ 2Q, (15) implies that ˆ |Kα (x, y)f (y)|dy Iα f2 (x) ≤ ˆ
Rn \4Q
|x−y| −N ρ(x) ) − y|n−α
(1 + Rn \4Q
|x
|f (y)|dy.
In this situation, we have ρ(x) ≈ ρ(x0 ) and |x − y| ≈ |x0 − y| for all x ∈ 2Q and y ∈ Rn \4Q. So, decomposing Rn \4Q into annuli 2k+1 \2k Q, we have, by H¨older inequality, ˆ 2−kN |f (y)|dy k Q|1−α/n kQ |2 2 k≥2 ˆ 1/s 1 −kN s 2 |f (y)| dy |2k Q|1−αs/n 2k Q k≥2
Iα f2 (x)
inf GN α,s f (z) z∈Q
for all x ∈ 2Q. Hence I12 inf z∈Q GN α,s f (z).
17
The estimate for I2 can be proceeded in the same line with one of I1 . The decomposition f = f1 + f2 gives 1 ˆ 1 ˆ I2 ≤ |Iα ((b − bQ )f1 )(y)|dy + |Iα ((b − bQ )f2 )(y)|dy := I21 + I22 . |2Q| 2Q |2Q| 2Q Choose 1 < r < s < p and 1/r − 1/r0 = α/n. Using H¨older inequality, Proposition 1.3 and Lr − Lr0 -boundedness of Iα , we have 1 ˆ 1/r0 |I − α((b − bQ )f1 )(y)|r0 dy I21 |2Q| 2Q 1 ˆ 1/r 1 r |((b − b )f )(y)| dy 1 Q |2Q|−α/n |2Q| 4Q 1 ˆ 1/s 1 ˆ 1/ν 1 s ν |f (y)| dy (b − b ) dy for some ν > r 1 Q |2Q| 4Q |2Q|−α/n |2Q| 4Q b θ inf GN α,s f (z). z∈Q
The estimate I22 b θ inf z∈Q Gα,s f (z) can be obtained by the similar approach to ones of I12 and I21 . So we omit the details here. To sum up, for any critical ball Q, we have ˆ 1 |I b f |dx b θ inf GN (19) α,s f (z). z∈Q |2Q| 2Q α Return to the estimate of AL
(q/s)
1+ (p/s)
k
wq (Qk )
ˆ
1 |2Qk |
´
|Iαb |
2Qk
q
. Due to (19) and the fact that wq ∈
, we have k
1 w (Qk ) |2Qk | q
2Qk
k
k
|Iαb f |
q
q
b qθ wq (Qk ) inf GN f (z) α,s
b qθ
b qθ
ˆ Rn
ˆ Qk
z∈Q
q q |GN α,s f (z)| w (z)dz
q q |GN α,s f (z)| w (z)dz
b qθ f qLp (wp ) , q q where in the last in equality, we can choose N large enough so that GN α,s is bounded on L (w ). ´ b q q Step 2. Estimate Rn |Mρ,γ (Iα f )(x)| w (x)dx
18
For any ball B(x0 , r) with r ≤ γρ(x0 ) and x ∈ B, we write ˆ 1 |I b f (x) − (Iαb f )B |dx |B| B α ˆ ˆ 2 2 |(b − bB )Iα f (x)|dx + |Iα ((b − bB )f1 )(x)|dx ≤ |B| B |B| B ˆ 1 |Iα ((b − bB )f2 )(x) − (Iα ((b − bB )f2 ))B |dx + |B| B := E1 + E2 + E3 . where f = f1 + f2 with f1 = f χ2B . Applying H¨ older inequality and Proposition 1.3, we have 1 ˆ 1/s 1 ˆ 1/s s E1 |b − bB | |Iα f |s |B| B |B| B 1 ˆ 1/s b θ |Iα f |s . |B| B
j . Therefore, For any critical ball Qj such that x ∈ Qj ∩ B, it is easy to see that B ⊂ Q s (Iα f )(u). E1 b θ × inf M u∈B
For some 1 < r < s < p and 1/r −1/r0 = α/n so that wq ∈ AL
(q/s)
1+ (p/s)
. By using H¨older inequality
and Proposition 1.3, we get that 1/r0 1 ˆ |Iα ((b − bB )f1 )|r0 E2 |B| B 1 ˆ 1/r 1 r |(b − b )f | 1 B |B|−α/n |B| 2B 1 ˆ 1/ν 1 ˆ 1/s 1 ν s |(b − b )| |f | for some ν > r B |B| 2B |B|−α/n |B| 2B α,s (f )(u). b θ × inf M u∈B
ˆ
Before taking care E3 , we need the to show that
Rn \2B
|Kα (x, z) − Kα (y, z)||b(z) − bB ||f (z)|dz b θ ( inf GN γ,α,s f (u) + inf Mα,s (f )(u)) (20) u∈B
u∈B
for all f and x, y ∈ B. If this holds, then we have ˆ ˆ ˆ 1 E3 |K (u, z) − K (y, z)||b(z) − b ||f (z)|dz dydu α α B |B|2 B B Rn \2B b θ (GN γ,α,s (x) + Mα,s (f )(x)).
19
These three estimates of E1 , E2 and E3 give that α,s (f )(x) + GN f (x) + M s (Iα f )(x)). Mρ,γ (Iαb )(x) b θ (M γ,α,s
This implies α,s (f ) Lq (wq ) + GN
Mρ,γ (Iαb ) Lq (wq ) b θ ( M γ,α,s f Lq (wq ) + Ms (Iα f ) Lq (wq ) ). p p q q q L With sufficiently large N , GN γ,α,s is bounded form L (w ) to L (w ), with w ∈ A1+(q/s)/(p/s) . α,s is also bounded form Lp (wp ) to Lq (wq ) for all 1 < p < n/α, 1/p − 1/q = α/n Moreover, M and wq ∈ AL 1+(q/s)/(p/s) . Hence,
α,s (f ) Lq (wq ) + GN
M γ,α,s f Lq (wq ) f Lp (wp ) . s (see Remark 2.5) s (Iα f ) Lq (wq ) , from the weighted estimates of Iα and M For the last term M L L and the fact that A (q/s) ⊂ Aq/s , one gets that 1+ (p/s)
s (Iα f ) Lq (wq ) Iα f Lq (wq ) f Lp (wp ) .
M Hence,
Mρ,γ (Iαb ) Lq (wq ) b θ f Lp (wp ) .
This completes our proof. Proof of (20): Setting Q = B(x0 , γρ(x0 )), due to (16) and the fact that ρ(x) ≈ ρ(x0 ) and |x − z| ≈ |x0 − z|, we get ˆ |Kα (x, z) − Kα (y, z)||b(z) − bB ||f (z)|dz K1 + K2 Rn \2B
where K1 = r δ
ˆ
|f (z)(b(z) − bB )| dz, |x0 − z|n+δ−α
Q\2B
and K2 = rδ ρ(x0 )N
ˆ
|f (z)(b(z) − bB )| dz. |x0 − z|n+δ+N −α
Qc
Let j0 be the smallest integer so that 2j0 r ≥ γρ(x0 ). Splitting Q\2B into annuli 2k+1 B\2k B for k = 1, . . . , j0 , we obtain j0
2−kδ K1 |2k B|1−α/n k=1
ˆ 2k B
|f (z)(b(z) − bB )|dz.
Using H¨older inequality and Proposition 1.3, we obtain K1
j0 k=1
k2−kδ b θ
1 k |2 B|1−αs/n 20
ˆ 2k B
|f (z)|s dz
1/s
j where Qj and Q
j are balls in (8). Therefore, Note that if x ∈ B ∩ Qj then 2k B ⊂ Q j0
K1
α,s f (u). k2−kδ b θ inf M u∈B
k=1
Splitting Qc into annuli and then applying H¨ older inequality and Proposition 1.3 again, we obtain ˆ ∞ ρ(x ) N 2−kδ−(k−j0 )N 0 |f (z)(b(z) − bB )|dz K2 2j0 r |2k−j0 +1 2j0 −1 B|1−α/n 2k B k=j0 −1 ˆ ∞ ρ(x ) N 1/s 1 0 −kδ−(k−j0 )(N −θ ) s k2 |f (z)| dz . b θ 2j0 r |2k−j0 +1 2j0 −1 B|1−αs/n 2k B k=j −1 0
Since 2j0 r ≥ γρ(x0 ) ≥ 2j0 −1 r, we get that K2 b θ b θ
∞
2
k=j0 −1 ∞ −l(N −θ )
2
l=1
b θ inf
u∈B
ˆ 1/s 1 s |f (z)| dz |2k−j0 +1 Q|1−αs/n 2k−j0 +1 Q ˆ 1/s 1 s |f (z)| dz 1−αs/n
−(k−j0 )(N −θ )
|2l Q|
GN γ,α,s f (u)
2l Q
(since Q = γB(x0 , ρ(x0 ))).
This completes our proof. Acknowledgements The author is supported by a grant from the Australia Research Council and by NAFOSTED under the project “Some optimization and control problems for dynamical systems under uncerrtainties”. He would like to thank Prof. X. T. Duong for helpful comments and suggestions.
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22
PII: DOI: Reference:
S0007-4497(13)00078-X 10.1016/j.bulsci.2013.06.007 BULSCI 2574
To appear in:
Bulletin des Sciences Mathématiques
Received date: 10 September 2012
Please cite this article in press as: T.A. Bui, Weighted estimates for commutators of some singular integrals related to Schrödinger operators, Bull. Sci. math. (2013), http://dx.doi.org/10.1016/j.bulsci.2013.06.007
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Weighted estimates for commutators of some singular integrals related to Schr¨odinger operators The Anh Bui∗
Abstract Let L = −Δ + V be a Schr¨odinger operator with non-negative potential V satisfying some appropriate reverse H¨older inequality. In this paper, we study the boundedness of the commutators of some singular integrals associated to L such as Riesz transforms and fractional integrals with the new BMO functions introduced in [BHS1] on the weighted spaces Lp (w) where w belongs to the new classes of weights introduced by [BHS2].
1
Introduction
Let L = −Δ + V be the Schr¨odinger operators on Rn with n ≥ 3 where the potential V is in the reverse H¨older class RHq for some q > n/2, i.e., V satisfies the reverse H¨older inequality ˆ 1/q 1 ˆ C V (y)q dy ≤ V (y)dy |B| B |B| B for all ball B ⊂ Rn . In this paper, we consider the following singular integrals associated to L: (i) Riesz transforms R = ∇L−1/2 and their adjoint R∗ = L−1/2 ∇; (ii) Fractional integrals Iα f (x) = L−α/2 f (x) for 0 < α < n. In the classical case when V = 0, it has been shown that Riesz transforms R and their commutators Rb with BMO functions b is bounded on Lp (w) for all 1 < p < ∞ and w in the Muckenhoupt classes Ap , see for example [St]. Also, the classical fractional integrals and their commutators with BMO functions b are bounded from Lp (wp ) to Lq (wq ) for all 1 < p < n/α, 1/p − 1/q = α/n and w ∈ A1+1/p ∩ RHq , or equivalently wq ∈ A1+ q , where Ap is the p
Muckenhoupt class of weights, see for example [MW, ST]. Recall that a non-negative and locally integrable function w is said to be in the Muckenhoupt Ap classes with 1 ≤ p < ∞, if the following inequality holds for all balls B ⊂ Rn ˆ 1/p 1/p ˆ − 1 w w p−1 ≤ C|B|. (1) B
B
∗
Department of Mathematics, University of Pedagogy, HoChiMinh City, Vietnam and Department of Mathematics, Macquarie University, NSW 2109, Australia. Email: bt [email protected] and [email protected] 2000 Mathematics Subject Classification: 42B35, 35J10. Key words: weight, Riesz transform, fractional integral, Schr¨ odinger operator, commutator, BMO.
1
Recently, in [BHS2], a new class of weights associated to Schr¨odinger operators L has been L,θ introduced. According to [BHS2], the authors defined the new classes of weights AL p = ∪θ>0 Ap for p ≥ 1, where AL,θ is the set of those weights satisfying p 1/p ˆ 1/p ˆ r θ − 1 w w p−1 ≤ C|B| 1 + (2) ρ(x) B B L for all ball B = B(x, r). We denote AL ∞ = ∪p≥1 Ap where the critical radius function ρ(·) is defined by ˆ 1 V ≤ 1 , x ∈ Rn , (3) ρ(x) = sup r > 0 : n−2 r B(x,r)
see [Sh]. It is easy to see that in certain circumstances the new class AL p is larger than the Muckenhoupt , see [BHS2, Proposition 5]. class Ap . The following properties hold for new classes AL p Proposition 1.1 The following statements hold: L i) AL p ⊂ Aq for 1 ≤ p ≤ q < ∞. L ii) If w ∈ AL p with p > 1 then there exists > 0 such that w ∈ Ap− . Consequently, L AL p = ∪q
1
Lp (w) for 1 < p < s with w p−1 ∈ AL p /s . (ii) If q ≥ n, the Riesz transforms R∗ and R are bounded on Lp (w) for 1 < p < ∞ and w ∈ AL p. (b) Let V ∈ RHq with q > n/2. Then Iα is bounded from Lp (wp ) to Lq (wq ) for all 1 < p < . n/α, 1/p − 1/q = α/n and wq ∈ AL 1+ q p
For the proof we refer to Theorem 3 and Theorem 4 in [BHS2]. Now we consider the commutators of the Riesz transforms R and R∗ with the BMO functions b. It was proved in [GLP] that the commutators Rb and Rb∗ are bounded on Lp here the range of p depends on the potential V . Then the authors in [BHS1] extended the classes of BMO functions to the new class BM OLθ with θ > 0 for the boundedness of the commutators Rb and Rb∗ . We would like to give a brief overview of the results in [BHS1]. According to [BHS1], the new BMO space BM OLθ with θ > 0 is defined as a set of all locally integrable functions b satisfying ˆ 1 r θ |b(y) − bB |dy ≤ C 1 + (4) |B| B ρ(x)
2
where B = B(x, r) and bB =
1 |B|
´ B
b. A norm for b ∈ BM OLθ , denoted by b θ , is given
θ2 by the infimum of the constants satisfying (4). Clearly BM OLθ1 ⊂ BM OL for θ1 ≤ θ2 and 0 ∞ θ BM OL = BM O. We define BM OL = ∪θ>0 BM OL . The following result can be considered to be a variant of John-Nirenberg inequality for the spaces BM OLθ , see [BHS1, pp.119-120]. θ then for all B = (x , r) Proposition 1.3 Let θ > 0, s ≥ 1. If b ∈ BM OL 0 i) 1 ˆ 1/s r θ |b(y) − bB |s dx b θ 1 + |B| B ρ(x0 )
where θ = (N0 + 1)θ and N0 is a constant in (5). ii) 1/s 1 ˆ 2k r θ |b(y) − b |dx
b
k 1 + B θ ρ(x0 ) |2k B| 2k B for all k ∈ N. ∞ we consider the commutators Let T be either R or R∗ . For b ∈ BM OL
Tb f (x) = T (bf )(x) − b(x)T f (x) and
Iαb f (x) = Iα (bf )(x) − b(x)Iα f (x).
It was proved in [BHS1, Theorem 1] that θ with θ > 0 and V ∈ RH . Theorem 1.4 Let b ∈ BM OL q (i) If n/2 < q < n and s is such that 1/s = 1/q − 1/n, the commutators Rb∗ are bounded on Lp for s < p < ∞ and hence by duality Rb is bounded on Lp for 1 < p < s. (ii) If q ≥ n, the commutators Rb∗ and Rb are bounded on Lp for 1 < p < ∞..
The aim of this paper is investigating the boundedness of the commutators Rb∗ , Rb and Iαb with b ∈ BM OL∞ on new weighted spaces Lp (w) with w ∈ AL p. The organization of this paper is as follows. In Section 2, we recall some basic properties of the critical radius function ρ(·) and consider weighted estimates for some localized operators. Section 3 is devoted to prove the main results on weighted estimates of the commutators Rb∗ , Rb and Iαb . Finally, we make some conventions. Throughout the whole paper, we denote by C a positive constant which is independent of the main parameters, but it may vary from line to line. The symbol X Y means that there exists a positive constant C such that X ≤ CY . Recently, we have learned that the AL p weighted norm inequalities for the commutators of the Riesz transforms was obtained independently in [BHS3]. However, the approach in our paper is different from that in [BHS3]. Moreover, the weighted norm inequalities for the commutators of fractional integrals L−α/2 is unique.
3
2
Weighted estimates for some localized operators
We would like to recall some important properties concerning the critical radius function which will play an important role to obtain the main results in the sequel, see [Sh, DZ1] respectively. Proposition 2.1 If V ∈ RHn/2 , there exist c0 and N0 ≥ 1 such that c−1 0 ρ(x)
N0 |x − y| −N0 |x − y| N0 +1 1+ ≤ ρ(y) ≤ c0 ρ(x) 1 + ρ(x) ρ(x)
(5)
for all x, y ∈ Rn . A ball of the form B(x, ρ(x)) is called a critical ball. From the inequality (5), we can imply that for x, y ∈ σQ where Q is a critical ball and σ > 0, then ρ(x) ≤ cσ ρ(y)
(6)
2N0 +1
where cσ = c20 (1 + σ) N0 +1 . Proposition 2.2 There exists a sequence of points xj , j ≥ 1 in Rn so that the family Qj := B(xj , ρ(xj )) satisfies (i) ∪j Qj = Rn . (ii) For every σ ≥ 1 there exist constants C and N1 such that j χσQj ≤ Cσ N1 . Following [BHS1], we introduce the following maximal functions for g ∈ L1loc (Rn ) and x ∈ Rn ˆ 1 Mρ,α g(x) = sup |g|, x∈B∈Bρ,α |B| B ˆ 1 g(x) = sup |g − gB |, Mρ,α x∈B∈Bρ,α |B| B where Bρ,α = {B(y, r) : y ∈ Rn and r ≤ αρ(y)}. Also, given a ball Q, we define the following maximal functions for g ∈ L1loc (Rn ) and x ∈ Q ˆ 1 MQ g(x) = sup |g|, x∈B∈F (Q) |B ∩ Q| B∩Q ˆ 1 g(x) = sup |g − gB∩Q |, MQ x∈B∈F (Q) |B ∩ Q| B∩Q where F(Q) = {B(y, r) : y ∈ Q, r > 0}. We have the following lemma. Lemma 2.3 For 1 < p < ∞, then there exist β and γ such that if {Qk }k is a sequence of balls as in Proposition 2.2 then ˆ ˆ 1 ˆ p |Mρ,β g(x)|p w(x)dx |Mρ,γ g(x)|p w(x)dx + w(Qk ) |g| |2Qk | 2Qk Rn Rn k
for all g ∈ L1loc (Rn ) and w ∈ AL ∞. 4
Proof: Note that the unweighted estimate of Lemma 2.3 was obtained [BHS1, Lemma 2]. Now we adapt some ideas in [BHS1, Lemma 2] to our present setting. According to [BHS1, p. 121], there exists β > 0 so that for all critical balls Q and x ∈ Q, we have Mρ,β g(x) ≤ M2Q (gχ2Q )(x), and for x ∈ 2Q,
(gχ2Q )(x) ≤ Mρ,2 g(x). M2Q
are HardyNote that since for each k, gχ2Qk is supported in 2Qk , operators M2Qk and M2Q k Littlewood and sharp maximal functions defined in 2Qk viewed as a space of homogeneous type with the Euclidean metric and the Lebesgues measure restricted to 2Qk . Moreover, by definition ∞ of A∞ ∞ , if w ∈ A∞ then w ∈ A∞ (2Qk ), where A∞ (2Qk ) = ∪p≥1 Ap (2Qk ) and Ap (2Qk ) is the class of Muckenhoupt weights on the spaces of homogeneous type 2Qk . Moreover, due to [BHS2, Lemma 5], and [PS, Proposition 3.4] gives ˆ ˆ p |Mρ,β g(x)| w(x)dx ≤ |Mρ,β g(x)|p w(x)dx Rn
Qk
k
≤
ˆ
Qk
k
|M2Q (gχ2Q )(x)|p w(x)dx
ˆ p 1 |g| |2Qk | 2Qk 2Qk k k ˆ ˆ p 1 |Mρ,2 g(x)|p w(x)dx + w(Qk ) |g| |2Qk | 2Qk 2Qk k ˆ ˆk p 1 |Mρ,2 g(x)|p w(x)dx + w(Qk ) |g| . |2Qk | 2Qk Rn
ˆ
|M2Q g(x)|p w(x)dx + k
w(Qk )
k
This completes our proof. The following maximal functions will be useful in the sequel. For N > 0, κ ≥ 1, 0 < α < n and 1 ≤ s < n/α, we define the following functions for g ∈ L1loc (Rn ) and x ∈ Rn GN κ,α,s f (x)
=
sup
∞
Qx;Q is critical k=0
and N f (x) = Hκ,s
sup
−N k
2
∞
Qx;Q is critical k=0
ˆ
1 1−αs/n |2k Q|
2−N k
1 |2k Q|
ˆ 2k Q
2k Q
|f (z)|s dz
|f (z)|s dz
1/s
1/s
= κQ. where Q N N N When κ = 1, we write GN α,s and Hs instead of G1,α,s and H1,s , respectively. We are now in N position to establish the weighted estimates for GN κ,α,s and Hκ,s .
5
Proposition 2.4 (i) Let wq ∈ AL,θ
q/s
1+ (p/s)
then we have
and N > N0 θ( 1s − αn ). If p > s and 1/p − 1/q = α/n,
GN κ,α,s f Lq (wq ) f Lp (wp )
(ii) Let w ∈ AL,θ p/s and N >
N0 θ s .
If p > s, then we have
N
Hκ,s f Lp (w) f Lp (w) .
Proof: (i) Without of loss of generality, we can assume that κ = 1. Assume that Q = B(x0 , ρ(x0 )). For x ∈ Q, the inequalities (5) tells us that C0−1 ρ(x0 ) ≤ ρ(x) ≤ C0 ρ(x0 ). This implies |B(x, ρ(x))| ≈ |Q| and Q ⊂ 2C0 B(x, ρ(x)). Therefore, GN α,s f (x)
∞
−N k
2
k=0
1 k |2 B(x, ρ(x))|1−αs/n
ˆ Bk (x,ρ(x))
|f (z)|s dz
1/s
where Bk (x, ρ(x)) = 2C0 × 2k B(x, ρ(x)). Let {Qj }j be the family of critical balls as in Proposition 2.2. By (5), C0−1 ρ(x) ≤ ρ(xj ) ≤ C0 ρ(x) for all x ∈ Qj . So, |B(x, ρ(x))| ≈ |Qj | and Bk (x, ρ(x)) ⊂ Qkj where Qkj = 2C0 × 2k Qj . For wq ∈ A1+ 1 , using H¨older inequalities, we obtain (p/s)
GN α,s f Lq (wq )
ˆ q/s 1/q 1 s q 2 |f (z)| dz w (x)dx 1−αs/n k Qj |2 B(x, ρ(x))| Bk (x,ρ(x)) j k=0 ˆ ˆ ∞ q/s 1/q 1 s q 2−N k |f (z)| dz w (x)dx 1−αs/n k Qj |2 Qj | Qkj j k=0 ˆ ∞ q/s 1/q wq (Qj ) s 2−N k |f (z)| dz |2k Qj |q/s−αq/n Qkj j k=0 ∞
∞
−N k
2−N k
ˆ
j
k=0
× wq (Qkj ) Since wq ∈ AL
1 |2k Qj |q/s−αq/n
ˆ Qkj
(wq )
− qs ( ps )
q/s (p/s)
ˆ Qkj
|f (z)|p wp (z)dz
q/p 1/q
. (7)
q/s
1+ (p/s)
w
q
, we have, by (5), (Qkj )
ˆ Qkj
(wq )
− qs ( ps )
q/s (p/s)
≤ C|Qkj |1/s−α/n 2kN0 θ×(1/s−α/n)
for some θ > 0. 6
This in combination with (7) gives
GN α,s f Lq (wq )
2−(N −N0 θ×(1/s−α/n))k
j
k
ˆ
2−(N −N0 θ×(1/s−α/n))k
ˆ
k
Rn
Qkj
|f (z)|p wp (z)dz
|f (z)|p wp (z)dz
1/p
1/p
C f Lp (wp ) where in the last inequality we used (ii) in Proposition 2.2. (ii) The proof of (ii) is just a direct consequence of (i) when α = 0. This completes our proof. For 0 ≤ α < n, let Mα be the fractional maximal function defined by ˆ 1 |f (y)|dy. Mα f (x) = sup 1−α/n Bx |B| B For s ≥ 1, we define
Mα,s f = sup
.
Bx
ˆ
1 |B|1−sα/n
B
|f (y)|s dy
1/s
α,s as follows For a family of balls {Qk }k given by Proposition 2.2, we define the operator M α,s f = χQk Mα,s (f χQk ) (8) M k
j = 4(2C 2 + 1)γQj and γ is a constant in Proposition 2.3. where Q 0 Remark 2.5 (i) For s < p < ∞ and 1/p − 1/q = α/n, it was proved in [MW] that Mα,s is bounded from Lp (wp ) to Lq (wq ) with wq ∈ A1+ (q/s) . This together with [BHS2, Proposition 4] (p/s)
α,s is bounded from Lp (wp ) to Lq (wq ) with wq ∈ AL shows that M
(q/s)
1+ (p/s)
here s < p < ∞ and
1/p − 1/q = α/n. 0,s . The similar argument as in (i) also shows s instead of M (ii) When α = 0, we write M L that for p > s and w ∈ Ap/s , Ms is bounded on Lp (w).
3
Weighted estimates for commutators of singular integrals
3.1 3.1.1
Riesz transforms Kernel estimates of Riesz transforms
In the sequel, let us remind that for the number N , we shall mean that N is a sufficiently large number and different from line to line. Let K and K ∗ be the vector valued kernels of R and R∗ respectively. The following propositions give some estimates on the kernels of R and R∗ , see for example [Sh, GLP]. 7
Proposition 3.1 a) If V ∈ RHq with q > n/2 then we have (i) For every N there exists a constant C such that ∗
|K (x, y)| ≤ C
|x−z| −N ρ(x) ) − z|n−1
(1 + |x
ˆ B(z,|x−z|/4)
V (u) 1 . + |u − z|n−1 |x − z|
(9)
(ii) For every N and 0 < δ < min{1, 2 − n/q} there exists a constant C such that ∗
∗
|K (x, z) − K (y, z)| ≤ C
|x−z| −N ρ(x) ) z|n−1+δ
|x − y|δ (1 + |x −
ˆ B(z,|x−z|/4)
V (u) 1 (10) + |u − z|n−1 |x − z|
whenever |x − y| < 23 |x − z|. b) If V ∈ RHq with q ≥ n then we have (i) For every N there exists a constant C such that |K(x, y)| ≤ C
|x−y| −N ρ(x) ) |x − y|n
(1 +
.
(11)
(ii) For every N and 0 < δ < min{1, 1 − d/q} there exists a constant C such that ∗
∗
|K (x, z) − K (y, z)| ≤ C
|x−z| −N ρ(x) ) n+δ z|
|x − y|δ (1 + |x −
(12)
whenever |x − y| < 23 |x − z|. 3.1.2
Commutators of Riesz transforms
The main result concerning the weighted estimates for Rb∗ and Rb is formulated by the following theorem. θ with θ > 0 and V ∈ RH . Theorem 3.2 Let b ∈ BM OL q (i) If n/2 < q < n and s is such that 1/s = 1/q − 1/n, the commutator Rb∗ is bounded on p Lp (w) for s < p < ∞ and w ∈ AL p/s and hence by duality Rb is bounded on L (w) for 1 < p < s −
1
with w p−1 ∈ AL p /s . (ii) If q ≥ n, the commutators Rb∗ and Rb are bounded on Lp (w) for 1 < p < ∞ and w ∈ AL p. Proof: (i) To prove (i), we exploit the strategy in [BHS1]. For any s < p < ∞ and w ∈ AL p/s , we have by Lemma 2.3 ˆ
Rb∗ f pLp (w) ≤ |Mρ,β (Rb∗ f )(x)|p w(x)dx n ˆR 1 ˆ p |Mρ,γ (Rb∗ f )(x)|p w(x)dx + w(Qk ) |Rb∗ f | |2Qk | 2Qk Rn k
8
where {Qk } is a family of critical balls given in Lemma 2.3. Now let N > 0 be a number which will be fixed later. For the sake of simplicity, we assume that the number N can change from´ each of its p appearances. 1 ∗ 1. Estimate k w(Qk ) |2Qk | 2Qk |Rb f |
Let s < p0 < p and Q = B(x0 , ρ(x0 )) so that w ∈ AL p/p0 . We borrow some ideas in [BHS1, Lemma 5] to write Rb∗ f = (b − bQ )R∗ f − R∗ ((b − bQ )f ).
So, we have ˆ ˆ ˆ 1 1 1 |Rb∗ f |dx ≤ |(b − bQ )R∗ f |dx + |R∗ ((b − bQ )f )|dx := I1 + I2 . |2Q| 2Q |2Q| 2Q |2Q| 2Q Let us estimate I1 first. By H¨older inequality, we can write 1 ˆ 1/p0 I1 b θ |R∗ f |p0 |2Q| 2Q 1 ˆ 1/p0 1 ˆ 1/p0 ∗ p0 b θ |R f1 | + |R∗ f2 |p0 |2Q| 2Q |2Q| 2Q := I11 + I12 where f = f1 + f2 with f1 = f χ4Q . Due to Lp0 -boundedness of R∗ , one has 1 ˆ 1/p0 I11 |f |p0 inf HpN0 f (z). z∈Q |2Q| 4Q To estimate I12 , for x ∈ 2Q, due to (9), we have ˆ ∗ |K ∗ (x, y)f (y)|dy R f2 (x) ≤ ˆ
Rn \4Q
|x−y| −N ρ(x) ) |x − y|n
(1 + Rn \4Q
ˆ
+
|f (y)|dy
|x−y| −N ρ(x) ) − y|n−1
(1 +
Rn \4Q
|x
ˆ B(y,|x−y|/4)
V (u) |f (y)|dudy |u − y|n−1
:= A1 (x) + A2 (x). To take care A1 , note that ρ(x) ≈ ρ(x0 ) and |x − y| ≈ |x0 − y| for all x ∈ 2Q and y ∈ Rn \4Q. So, decomposing Rn \4Q into annuli 2k+1 \2k Q, we have 2−kN ˆ A1 (x) |f (y)|dy |2k Q| 2k Q k≥2 1 ˆ 1/p0 p0 2−kN |f (y)| dy |2k Q| 2k Q k≥2
inf HpN0 f (z) z∈Q
9
for all x ∈ 2Q. For the term A2 , by decomposing Rn \4Q into annuli 2k+1 \2k Q, we get that ˆ ˆ 2−kN V (u) |f (y)| dudy A2 (x) n−1 1−1/n k |2 Q| 2k Q 2k+2 Q |u − y| k≥2 ˆ 2−kN |f (y)|I1 (V χ2k+2 Q )(y)dy |2k Q|1−1/n 2k Q k≥2 where I1 = (−Δ)−1/2 . Let us remind that I1 is Lq0 − Lp0 boundedness with 1/p0 = 1/q0 − 1/n. This together with H¨ older inequality gives ˆ 2−kN A2 (x) |(f χ2k Q )(y)|I1 (V χ2k+2 Q )(y)dy |2k Q|1−1/n Rn k≥2 1/p0 2−kN ˆ p0 |f |
I1 (V χ2k+2 Q ) Lp0 |2k Q|1−1/n 2k Q k≥2 1 ˆ 1/p0 2−kN p0 |f |
V χ2k+2 Q Lq0 |2k Q|1−1/n−1/p0 |2k Q| 2k Q k≥2
Noting that if we choose p0 to be close enough to s then V ∈ RHq0 . This in combination with the fact that V is a doubling measure gives ˆ 1
V χ2k+2 Q Lq0 k 1/q V |2 Q| 0 2k+2 Q ˆ 2kκ k 1/q V for some κ > 0 |2 Q| 0 Q 2kκ k 1/q −1+2/n . |2 Q| 0 Hence,
1 2−kN A2 (x) |2k Q|1/q0 −1+2/n |2k Q|1−1/n−1/p0 |2k Q| k≥2 1 ˆ 1/p0 −k(N −κ) p0 2 |f | |2k Q| 2k Q
2kκ
ˆ 2k Q
|f |p0
1/p0
k≥2
inf HpN0 f (z) z∈Q
for all x ∈ 2Q. From the estimates of A1 and A2 , we obtain I12 inf z∈Q HpN0 f (z). The term I2 can be estimated in the same line with I1 . Using the decomposition f = f1 + f2 again, one gets that 1 ˆ 1 ˆ ∗ I2 ≤ |R ((b − bQ )f1 )(y)|dy + |R∗ ((b − bQ )f2 )(y)|dy := I21 + I22 . |2Q| 2Q |2Q| 2Q 10
Choose s < r < p0 . Using H¨older inequality and Lr -boundedness of R∗ , we have 1 ˆ 1/r I21 |R∗ ((b − bQ )f1 )(y)|r dy |2Q| 2Q 1 ˆ 1/r |((b − bQ )f1 )(y)|r dy |2Q| 4Q 1 ˆ 1/p0 1 ˆ 1/ν p0 |f1 (y)| dy (b − bQ )ν dy for some ν > r |2Q| 4Q |2Q| 4Q b θ inf HpN0 f (z). z∈Q
The estimate of I22 b θ inf z∈Q HpN0 f (z) can be taken care similarly to ones of I12 and I21 . So we omit the details here. To sum up, it had proved that for any critical ball Q, we have ˆ 1 |R∗ f |dx b θ inf HpN0 f (z). (13) z∈Q |2Q| 2Q b p ´ 1 ∗ f | . Due to (19) and the fact that Return to the estimate of w(Q ) |R k k b |2Qk | 2Qk w ∈ AL p/p0 , we have
k
1 w(Qk ) |2Qk |
ˆ 2Qk
k
k
|Rb∗ f |
p
p
b pθ w(Qk ) inf HpN0 f (z)
b pθ
b pθ
ˆ Rn
ˆ Qk
z∈Q
|HpN0 f (z)|p w(z)dz
|HpN0 f (z)|p w(z)dz.
Due to Proposition 2.4, since we can take N sufficiently large so that HpN0 f Lp (w) f Lp (w) . Hence 1 ˆ p w(Qk ) |Rb∗ f | b pθ f pLp (w) . |2Qk | 2Qk k ´ 2. Estimate Rn |Mρ,γ (Rb∗ f )(x)|p w(x)dx For any ball B(x0 , r) with r ≤ γρ(x0 ) and x ∈ B, we write ˆ 1 |R∗ f (x) − (Rb∗ f )B |dx |B| B b ˆ ˆ 2 2 ∗ |(b − bB )R f (x)|dx + |R∗ ((b − bB )f1 )(x)|dx ≤ |B| B |B| B ˆ 1 |R∗ ((b − bB )f2 )(x) − (R∗ ((b − bB )f2 ))B |dx + |B| B := E1 + E2 + E3 . where f = f1 + f2 with f1 = f χ2B . 11
Let s < p0 < p so that w ∈ AL older inequality and Proposition 1.3 show that p/p0 . The H¨ 1 ˆ 1/p0 1 ˆ 1/p0 |b − bB |p0 |R∗ f |p0 |B| B |B| B 1 ˆ 1/p0 b θ |R∗ f |p0 . |B| B
E1
j . This yields For any critical ball Qj such that x ∈ Qj ∩ B. It can be verified that B ⊂ Q that p (R∗ f )(y). E1 b θ × inf M 0 y∈B
older inequality and and Proposition 1.3 again tell us that For some s < r < p0 < p, H¨ 1/r 1 ˆ |R∗ ((b − bB )f1 )|r E2 |B| B 1/r 1 ˆ |(b − bB )f1 |r |B| 2B 1/ν 1 ˆ 1/p0 1 ˆ |(b − bB )|ν |f |p0 for some ν > r |B| 2B |B| 2B p (f )(y). b θ × inf M y∈B
0
To estimate E3 , we need the to show that ˆ N p (f )(u)) |K ∗ (x, z) − K ∗ (y, z)||b(z) − bB ||f (z)|dz b θ ( inf Hγ,p f (u) + inf M 0 0 u∈B
Rn \2B
u∈B
(14)
for all f and x, y ∈ B. If this holds, then we have ˆ ˆ ˆ 1 |K ∗ (u, z) − K ∗ (y, z)||b(z) − bB ||f (z)|dz dydu E3 2 |B| B B Rn \2B N p (f )(x)). f (x) + M b θ (Hγ,p 0 0
These three estimates of E1 , E2 and E3 give N p (R∗ f )(x) + Hγ,p p (f )(x)). (Rb∗ )(x) b θ (M (x) + M Mρ,γ 0 0 0
This implies p (R∗ f ) Lp (w) + H N f Lp (w) + M p (f ) Lp (w) . (Rb∗ ) Lp (w) b θ ( M
Mρ,γ γ,p0 0 0 N is bounded on Lp (w). This together with Note that we can choose N large enough so that Hγ,p 0 p and R∗ f are bounded on Lp (w) for w ∈ AL gives the desired estimate. the fact that M 0 p/p0
Proof of (14): We adapt some ideas of [BHS1, Lemma 6] to our present situation. Setting Q = B(x0 , γρ(x0 )), due to (10) and the fact that ρ(x) ≈ ρ(x0 ) and |x − z| ≈ |x0 − z|, we get ˆ |K ∗ (x, z) − K ∗ (y, z)||b(z) − bB ||f (z)|dz K1 + K2 + K3 + K4 Rn \2B
12
where
ˆ
|f (z)(b(z) − bB )| dz, |x0 − z|n+δ Q\2B ˆ |f (z)(b(z) − bB )| K2 = rδ ρ(x0 )N dz, n+δ+N Qc |x0 − z| ˆ ˆ |f (z)(b(z) − bB )| V (u) dudz, K3 = r δ n−1+δ |u − z|n−1 |x − z| 0 Q\2B B(x0 ,4|x0 −z|)
K1 = r
δ
and δ
K4 = r ρ(x0 )
N
ˆ Qc
|f (z)(b(z) − bB )| |x0 − z|n−1+δ+N
Let j0 be the smallest integer so that k = 1, . . . , j0 , we obtain
2j 0 r
ˆ B(x0 ,4|x0 −z|)
V (u) dudz. |u − z|n−1
≥ γρ(x0 ). Splitting Q\2B into annuli 2k+1 B\2k B for
ˆ j0 2−kδ K1 |f (z)(b(z) − bB )|dz. |2k B| 2k B k=1
By H¨older inequality and Proposition 1.3, K1
j0
−kδ
k2
b θ
1 k |2 B|
k=1
ˆ 2k B
|f (z)|p0 dz
1/p0
j where Qj and Q
j are balls in (8). Therefore, Remind that if x ∈ B ∩ Qj then 2k B ⊂ Q K1
j0
p f (u). k2−kδ b θ inf M 0 u∈B
k=1
Splitting Qc into annuli and then applying H¨ older inequality and Proposition 1.3 again, we obtain ˆ ∞ ρ(x ) N 2−kδ−(k−j0 )N 0 |f (z)(b(z) − bB )|dz K2 2j0 r |2k−j0 +1 2j0 −1 B| 2k B k=j0 −1 ˆ ∞ ρ(x ) N 1/p0 1 0 −kδ−(k−j0 )(N −θ ) p0 k2 |f (z)| dz . b θ 2j0 r |2k−j0 +1 2j0 −1 B| 2k B k=j0 −1
Since 2j0 r ≥ γρ(x0 ) ≥ 2j0 −1 r, we get that K2 b θ b θ
∞
−(k−j0 )(N −θ )
2
k=j0 −1 ∞ −l(N −θ )
2
l=1
1 |2l Q|
1
ˆ
|2k−j0 +1 Q|
ˆ 2l Q
|f (z)|p0 dz
2k−j0 +1 Q
1/p0
N b θ inf Hγ,p f (u) since Q = γB(x0 , ρ(x0 )). 0 u∈B
13
|f (z)|p0 dz
1/p0
It can be verified that K3
j0
Let s < q0 < p0 , ν = Proposition 1.3 K3 b θ
2−kδ k |2 B|1−1/n k=2 p 0 q0 p0 −q0
ˆ 2k B
|f (z)(b(z) − bB )|I1 (V χ2k+2 B )(z)dz
and r such that 1/r = 1/q0 + 1/n then by H¨older inequality and
1 k2−kδ |2k B|−1/n+1/q0 |2k B| k=2
j0
1 k2−kδ b θ |2k B|−1/n+1/q0 |2k B| k=2 j0
b θ
ˆ 2j B
ˆ 2j B
|f |p0 |f |p0
1/p0 1/p0
I1 (V χ2k+2 B ) Lq0
V χ2k+2 B Lr
j0
k2−kδ p f (u)
V χ2k+2 B Lr inf M 0 k B|−1/n+1/q0 u∈B |2 k=2
Noting that we can choose q0 so that V ∈ RHr , then we have ˆ ˆ 1/r 1
V χ2k+2 B Lr V (z)r dz V (z)dz |Q|1−1/r Q Q 1 1 k 2/n−1/r |Q|2/n−1/r |2 B| for all k = 2, . . . , j0 . So, K3 b θ
j0
p f (x) := C b θ inf M p f (u). k2−kδ M 0 0 u∈B
k=2
The similar arguments to ones used to obtain the estimates K2 and K3 gives N K4 b θ inf Hγ,p f (u). 0 u∈B
This completes the proof of (i). (ii) The proof of part (ii) is completely analogous to that of (i). The main difference here is that in this situation we can work directly on R intead of R∗ via the pointwise estimates of the kernels K(x, y) of the Riesz transforms R, see for example [BHS1, Lemma 4]. Hence, we would like to omit the details here leave the proof to the interested readers.
3.2 3.2.1
Fractional integrals Kernel estimates of fractional integrals
Let Kα be the kernel of Iα . The following results give the estimates on the kernel Kα (x, y). Proposition 3.3 If V ∈ RHq with q > n/2 then we have 14
(i) For every N there exists a constant C such that |Kα (x, y)| ≤ C
|x−y| −N ρ(x) ) − y|n−α
(1 + |x
.
(15)
(ii) There is a number δ > 0 such that for every N there exists a constant C such that |Kα (x, y) − Kα (x, z)| ≤ C
|x−y| −N ρ(x) ) n+δ−α y|
|y − z|δ (1 + |x −
(16)
whenever |y − z| < 14 |x − y|. To prove Proposition 3.3, we need the following estimates of the heat kernels of e−tL , see [DZ2, p.12] Proposition 3.4 Let pt (x, y) be the kernels associated to the semigroups {e−tL }t>0 . If V ∈ RHq with q > n/2 then we have (i) For every N > 0 there exists a constant C such that √ √ t t −N C |x − y|2 ; (17) 1+ + |pt (x, y)| ≤ n/2 exp − c t ρ(x) ρ(y) t (ii) There is a number δ > 0 such that for every N there exists a constant C such that |pt (x, y) − pt (x, z)| + |pt (y, x) − pt (z, x)| √ √ t t −N |x − y|2 C |y − z| δ √ exp − c 1+ + ≤ n/2 t ρ(x) ρ(y) t t whenever |y − z| < 12 |x − y|. Proof of Proposition 3.3: (i) We have, by (17), ˆ ∞ dt |tα/2 pt (x, y)| |Kα (x, y)| ≤ t 0 √ √ ˆ |x−y|2 α/2 t t t −N dt |x − y|2 exp − c 1+ + n/2 t ρ(x) ρ(y) t t 0 √ √ ˆ ∞ α/2 2 −N t t t dt |x − y| + exp − c 1+ + n/2 t ρ(x) ρ(y) t 2 t |x−y| = I1 + I 2 . Let us estimate I2 first. Since t > |x − y|2 , we have, for > 0 so that n > α + , ˆ ∞ tα/2 |x − y|2 |x − y| −N dt exp − c I2 1 + n/2 t ρ(x) t |x−y|2 t ˆ ∞ α/2 n/2−α/2− −N t dt t |x − y| ( 1+ n/2 |x − y|2 ρ(x) t 2 t |x−y| |x − y| −N 1 1+ . n−α |x − y| ρ(x) 15
(18)
For I1 , we have ˆ I1 ˆ
0
0
ˆ
|x−y|2
√ n/2+N/2 t −N dt t 1+ 2 n/2 |x − y| ρ(x) t t √ √ α/2 N t t t + ρ(x) −N dt
|x−y|2
|x − y|n |x − y| ρ(x) t √ √ tα/2 t + ρ(x) N t + ρ(x) −N dt
|x−y|2 α/2 t
|x − y|n |x − y| + ρ(x) 1 |x − y| −N . 1+ n−α |x − y| ρ(x)
ρ(x)
0
t
This completes (i). (ii) For |y − z| < 14 |x − y|, using (18) gives
√ √ − z| δ t t −N dt |x − y|2 √ 1+ exp − c + |Kα (x, y) − Kα (x, z)| n/2 t ρ(x) ρ(y) t t t 0 √ √ |y − z| δ ˆ ∞ tα/2 2 −N |x − y| t t dt exp − c 1+ + n/2 |x − y| t ρ(x) ρ(y) t t 0 ˆ ∞ |y − z| δ ˆ |x−y|2 ... + ... |x − y| |x−y|2 0 |y − z| δ (II1 + II2 ). = |x − y| ˆ
∞ α/2 |y t
At this stage, repeating the arguments in (i), we obtain (ii). 3.2.2
Commutators of fractional integrals
We are now in position to state the result concerning the weighted estimates for Iαb . Theorem 3.5 Let b ∈ BM OLθ with θ > 0 and V ∈ RHq with q > n/2. Then the commutator . Iαb is bounded from Lp (wp ) to Lq (wq ) for 1 < p < ∞, 1/p − 1/q = α/n and wq ∈ AL 1+ q p
Proof: The strategy of the proof for Theorem 3.5 is similar to that of Theorem 3.2. , we have by Lemma 2.3 For any 1 < p < ∞, 1/p − 1/q = α/n and wq ∈ AL 1+ q
Iαb f qLq (wq ) ≤
ˆ ˆ
Rn Rn
p
|Mρ,β (Iαb f )(x)|q wq (x)dx |Mρ,γ (Iαb f )(x)|q wq (x)dx +
k
wq (Qk )
1 |2Qk |
ˆ 2Qk
|Iαb f |
q
where {Qk } is a family of critical balls given in Proposition 2.3. ´ So, to obtain the weighted estimates for Iαb , we need only to consider Rn |Mρ,γ (Iαb f )(x)|q wq (x)dx q ´ q and k w (Qk ) |2Q1 k | 2Qk |Iαb f | . 16
Now let N > 0 be a number which will be fixed later. For the sake of simplicity, we assume that the number N can change fromeach of its appearances. q ´ Step 1. Estimate k wq (Qk ) |2Q1 k | 2Qk |Iαb f | Let 1 < s < p, 1/s − 1/υ = α/n and Q = B(x0 , ρ(x0 )) so that wq ∈ AL
(q/s)
1+ (p/s)
. We have
Iαb f = (b − bQ )Iα f − Iα ((b − bQ )f ). Hence, 1 |2Q|
ˆ 2Q
|Iαb f |dx ≤
1 |2Q|
ˆ 2Q
|(b − bQ )Iα f |dx +
1 |2Q|
ˆ 2Q
|Iα ((b − bQ )f )|dx := I1 + I2 .
To take care I1 , using H¨older inequality and Proposition 1.3, we get that 1 ˆ 1/υ I1 b θ |Iα f |υ |2Q| 2Q 1 ˆ 1/υ 1 ˆ 1/υ υ b θ |Iα f1 | + |Iα f2 |υ |2Q| 2Q |2Q| 2Q := I11 + I12 where f = f1 + f2 with f1 = f χ4Q . Since Iα is Ls − Lυ bounded, one has I11
1
ˆ
|2Q|1−αs/n
4Q
|f |s
1/s
inf GN α,s f (z). z∈Q
To estimate I12 , for x ∈ 2Q, (15) implies that ˆ |Kα (x, y)f (y)|dy Iα f2 (x) ≤ ˆ
Rn \4Q
|x−y| −N ρ(x) ) − y|n−α
(1 + Rn \4Q
|x
|f (y)|dy.
In this situation, we have ρ(x) ≈ ρ(x0 ) and |x − y| ≈ |x0 − y| for all x ∈ 2Q and y ∈ Rn \4Q. So, decomposing Rn \4Q into annuli 2k+1 \2k Q, we have, by H¨older inequality, ˆ 2−kN |f (y)|dy k Q|1−α/n kQ |2 2 k≥2 ˆ 1/s 1 −kN s 2 |f (y)| dy |2k Q|1−αs/n 2k Q k≥2
Iα f2 (x)
inf GN α,s f (z) z∈Q
for all x ∈ 2Q. Hence I12 inf z∈Q GN α,s f (z).
17
The estimate for I2 can be proceeded in the same line with one of I1 . The decomposition f = f1 + f2 gives 1 ˆ 1 ˆ I2 ≤ |Iα ((b − bQ )f1 )(y)|dy + |Iα ((b − bQ )f2 )(y)|dy := I21 + I22 . |2Q| 2Q |2Q| 2Q Choose 1 < r < s < p and 1/r − 1/r0 = α/n. Using H¨older inequality, Proposition 1.3 and Lr − Lr0 -boundedness of Iα , we have 1 ˆ 1/r0 |I − α((b − bQ )f1 )(y)|r0 dy I21 |2Q| 2Q 1 ˆ 1/r 1 r |((b − b )f )(y)| dy 1 Q |2Q|−α/n |2Q| 4Q 1 ˆ 1/s 1 ˆ 1/ν 1 s ν |f (y)| dy (b − b ) dy for some ν > r 1 Q |2Q| 4Q |2Q|−α/n |2Q| 4Q b θ inf GN α,s f (z). z∈Q
The estimate I22 b θ inf z∈Q Gα,s f (z) can be obtained by the similar approach to ones of I12 and I21 . So we omit the details here. To sum up, for any critical ball Q, we have ˆ 1 |I b f |dx b θ inf GN (19) α,s f (z). z∈Q |2Q| 2Q α Return to the estimate of AL
(q/s)
1+ (p/s)
k
wq (Qk )
ˆ
1 |2Qk |
´
|Iαb |
2Qk
q
. Due to (19) and the fact that wq ∈
, we have k
1 w (Qk ) |2Qk | q
2Qk
k
k
|Iαb f |
q
q
b qθ wq (Qk ) inf GN f (z) α,s
b qθ
b qθ
ˆ Rn
ˆ Qk
z∈Q
q q |GN α,s f (z)| w (z)dz
q q |GN α,s f (z)| w (z)dz
b qθ f qLp (wp ) , q q where in the last in equality, we can choose N large enough so that GN α,s is bounded on L (w ). ´ b q q Step 2. Estimate Rn |Mρ,γ (Iα f )(x)| w (x)dx
18
For any ball B(x0 , r) with r ≤ γρ(x0 ) and x ∈ B, we write ˆ 1 |I b f (x) − (Iαb f )B |dx |B| B α ˆ ˆ 2 2 |(b − bB )Iα f (x)|dx + |Iα ((b − bB )f1 )(x)|dx ≤ |B| B |B| B ˆ 1 |Iα ((b − bB )f2 )(x) − (Iα ((b − bB )f2 ))B |dx + |B| B := E1 + E2 + E3 . where f = f1 + f2 with f1 = f χ2B . Applying H¨ older inequality and Proposition 1.3, we have 1 ˆ 1/s 1 ˆ 1/s s E1 |b − bB | |Iα f |s |B| B |B| B 1 ˆ 1/s b θ |Iα f |s . |B| B
j . Therefore, For any critical ball Qj such that x ∈ Qj ∩ B, it is easy to see that B ⊂ Q s (Iα f )(u). E1 b θ × inf M u∈B
For some 1 < r < s < p and 1/r −1/r0 = α/n so that wq ∈ AL
(q/s)
1+ (p/s)
. By using H¨older inequality
and Proposition 1.3, we get that 1/r0 1 ˆ |Iα ((b − bB )f1 )|r0 E2 |B| B 1 ˆ 1/r 1 r |(b − b )f | 1 B |B|−α/n |B| 2B 1 ˆ 1/ν 1 ˆ 1/s 1 ν s |(b − b )| |f | for some ν > r B |B| 2B |B|−α/n |B| 2B α,s (f )(u). b θ × inf M u∈B
ˆ
Before taking care E3 , we need the to show that
Rn \2B
|Kα (x, z) − Kα (y, z)||b(z) − bB ||f (z)|dz b θ ( inf GN γ,α,s f (u) + inf Mα,s (f )(u)) (20) u∈B
u∈B
for all f and x, y ∈ B. If this holds, then we have ˆ ˆ ˆ 1 E3 |K (u, z) − K (y, z)||b(z) − b ||f (z)|dz dydu α α B |B|2 B B Rn \2B b θ (GN γ,α,s (x) + Mα,s (f )(x)).
19
These three estimates of E1 , E2 and E3 give that α,s (f )(x) + GN f (x) + M s (Iα f )(x)). Mρ,γ (Iαb )(x) b θ (M γ,α,s
This implies α,s (f ) Lq (wq ) + GN
Mρ,γ (Iαb ) Lq (wq ) b θ ( M γ,α,s f Lq (wq ) + Ms (Iα f ) Lq (wq ) ). p p q q q L With sufficiently large N , GN γ,α,s is bounded form L (w ) to L (w ), with w ∈ A1+(q/s)/(p/s) . α,s is also bounded form Lp (wp ) to Lq (wq ) for all 1 < p < n/α, 1/p − 1/q = α/n Moreover, M and wq ∈ AL 1+(q/s)/(p/s) . Hence,
α,s (f ) Lq (wq ) + GN
M γ,α,s f Lq (wq ) f Lp (wp ) . s (see Remark 2.5) s (Iα f ) Lq (wq ) , from the weighted estimates of Iα and M For the last term M L L and the fact that A (q/s) ⊂ Aq/s , one gets that 1+ (p/s)
s (Iα f ) Lq (wq ) Iα f Lq (wq ) f Lp (wp ) .
M Hence,
Mρ,γ (Iαb ) Lq (wq ) b θ f Lp (wp ) .
This completes our proof. Proof of (20): Setting Q = B(x0 , γρ(x0 )), due to (16) and the fact that ρ(x) ≈ ρ(x0 ) and |x − z| ≈ |x0 − z|, we get ˆ |Kα (x, z) − Kα (y, z)||b(z) − bB ||f (z)|dz K1 + K2 Rn \2B
where K1 = r δ
ˆ
|f (z)(b(z) − bB )| dz, |x0 − z|n+δ−α
Q\2B
and K2 = rδ ρ(x0 )N
ˆ
|f (z)(b(z) − bB )| dz. |x0 − z|n+δ+N −α
Qc
Let j0 be the smallest integer so that 2j0 r ≥ γρ(x0 ). Splitting Q\2B into annuli 2k+1 B\2k B for k = 1, . . . , j0 , we obtain j0
2−kδ K1 |2k B|1−α/n k=1
ˆ 2k B
|f (z)(b(z) − bB )|dz.
Using H¨older inequality and Proposition 1.3, we obtain K1
j0 k=1
k2−kδ b θ
1 k |2 B|1−αs/n 20
ˆ 2k B
|f (z)|s dz
1/s
j where Qj and Q
j are balls in (8). Therefore, Note that if x ∈ B ∩ Qj then 2k B ⊂ Q j0
K1
α,s f (u). k2−kδ b θ inf M u∈B
k=1
Splitting Qc into annuli and then applying H¨ older inequality and Proposition 1.3 again, we obtain ˆ ∞ ρ(x ) N 2−kδ−(k−j0 )N 0 |f (z)(b(z) − bB )|dz K2 2j0 r |2k−j0 +1 2j0 −1 B|1−α/n 2k B k=j0 −1 ˆ ∞ ρ(x ) N 1/s 1 0 −kδ−(k−j0 )(N −θ ) s k2 |f (z)| dz . b θ 2j0 r |2k−j0 +1 2j0 −1 B|1−αs/n 2k B k=j −1 0
Since 2j0 r ≥ γρ(x0 ) ≥ 2j0 −1 r, we get that K2 b θ b θ
∞
2
k=j0 −1 ∞ −l(N −θ )
2
l=1
b θ inf
u∈B
ˆ 1/s 1 s |f (z)| dz |2k−j0 +1 Q|1−αs/n 2k−j0 +1 Q ˆ 1/s 1 s |f (z)| dz 1−αs/n
−(k−j0 )(N −θ )
|2l Q|
GN γ,α,s f (u)
2l Q
(since Q = γB(x0 , ρ(x0 ))).
This completes our proof. Acknowledgements The author is supported by a grant from the Australia Research Council and by NAFOSTED under the project “Some optimization and control problems for dynamical systems under uncerrtainties”. He would like to thank Prof. X. T. Duong for helpful comments and suggestions.
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22