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A CHARATERIZATION OF COMMUTATORS FOR PARABOLIC SINGULAR INTEGRALS

YANPING CHEN and YONG DING

Abstract

In this paper, the authors give a characterization of theLp-boundedness of the commutators for the parabolic singular integrals. More precisely, the authors prove that ifbBMOϕ(Rn, ρ), then the commutator [b, T] is a bounded operator fromLp(Rn)to the Orlicz spaceLψ(Rn), where the kernel functionhas no any smoothness on the unit sphereSn−1. Conversely, if assuming on a slight smoothness onSn−1, then the boundedness of [b, T] fromLp(Rn)toLψ(Rn)implies thatb BMOϕ(Rn, ρ). The results in this paper improve essentially and extend some known conclusions.

1. Introduction

Suppose thatSn1= {xRn:|x| =1}is the unit sphere inRnequipped with the Lebesgue measuredσ, where| · |denotes the Euclidean norm inRn. Let bLloc(Rn), then the commutator of the classical singular integral is defined by

[b, T]f (x)=p.v.

Rn

(xy)

|xy|n (b(x)b(y))f (y) dy, whereis homogeneous function of degree zero onRn\ {0}, that is, (1.1) (μx)=(x) for any μ >0 and xRn\ {0}. Moreover,L1(Sn1)satisfying the following cancellation condition:

(1.2)

Sn−1

(x) dσ (x)=0.

In 1976, Coifman, Rochberg and Weiss gave the following result:

TheoremA ([2]). If ∈ Lip1(Sn1) satisfies(1.1)and (1.2), andb ∈ BMO, then[b, T]is bounded onLp(Rn) (1< p <). Conversely, if [b, Rj]

This work was supported by NSF of China (Grants No. 10931001, 10901017) and SRFDPHE of China (Grant No. 20090003110018).

Received 17 October 2008.

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is bounded onLp(Rn)for some p,1 < p <and allj = 1, . . . , n, then b∈BMO, whereRj is the j’th Riesz transform.

In 1978, Janson [6] extended Theorem A. To state the result in [6], we give some notations and definitions.

Let ψ be a non-decreasing convex function on R+ with ψ (0) = 0, let ψ1denote the inverse function ofψ. The Orlicz spaceLψ(Rn)is the set of functionsf satisfying

(f, ψ )=

Rn

ψ (λ|f (y)|) dy <∞ for someλ >0. The norm inLψ is defined by

fLψ =inf

λ>0

1 λ

1+

Rn

ψ (λ|f (y)|) dy

= sup

gLψ

(g,ψ)1

Rn

f (y)g(y) dy ,

whereψis the complementary Young function ofψ, which is given by ψ(s)= sup

0t <[st−ψ (t )], 0≤s <.

The following generalized Hölder’s inequality holds in the Orlicz spaceLψ(Rn) (see [9] for its proof):

(1.3)

Rn

f (y)g(y) dy

fLψgLψ∗. Now we give the definition of BMOϕas follows. Denote

M(b, Q)= 1

|Q|

Q

|b(x)bQ|dx, wherebQ = |Q1|

Qb(x) dxandQis a cube inRn. Letϕbe a positive function, then

BMOϕ(Rn)=

bLloc(Rn):bBMOϕ = sup

xRn r>0

M(b, Q(x, r)) ϕ(r) <

whereQ(x, r)denotes the cube centered atxand with diameterr. With the above notations, Janson got the following conclusion:

TheoremB ([6]). Suppose thatC(Sn1)satisfies(1.1)and(1.2) and1< p <. Letϕandψbe two non-decreasing positive functions onR+

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connected by the relationϕ(r) = rn/pψ1(rn)(or equivalentlyψ1(t ) = t1/pϕ(t1/n)). Ifψ is convex,ψ (0)=0andψ (2t )Cψ (t ), thenbbelongs toBMOϕif and only if [b, T]mapsLp(Rn)boundedly intoLψ(Rn).

Notice that in Theorem B,was assumed on a very stronger smoothness condition onSn1. Hence, it is natural to ask if the conclusion of Theorem B holds still under a weaker condition of. In the paper, we will give a positive answer to this question. In fact, we will improve and extend Theorem B under the parabolic cases. Before stating our results, let us recall some definitions and some known results.

In 1966, Fabes and Rivière [5] introduced the parabolic singular integral.

Letα1, . . . , αnbe fixed real numbers withαi ≥1. It is easy to see that for each fixedxRn, the function

F (x, ρ)=

n

i=1

xi2 ρi

is a strictly decreasing function ofρ > 0. Therefore, there exists an unique ρ = ρ(x)such thatF (x, ρ) = 1. It was proved in [5] thatρ is a metric on Rnand the metric space is denoted by(Rn, ρ). Forμ > 0 andxRn, if we denote by

δμ:(x1, x2, . . . , xn)−→α1x1, μα2x2, . . . , μαnxn)

a dilation onRn, then we have the polar decompositionx = δρx withxSn1,ρ=ρ(x)anddx=ρα1J (x)dρ dσ (x), where

J (x)=α1x12+ · · · +αnxn2 and α=

n

i=1

αi.

Suppose thatis a real valued and measurable function defined onRn\{0}. It is said thatis homogeneous of degree zero with respect toδμ, if for any μ >0 andxRn\ {0}

(1.4) μx)=(x).

Moreover, we also assume thatsatisfies the following cancellation condition:

(1.5)

Sn−1

(x)J (x) dσ (x)=0.

In [5], Fabes and Rivière proved that ifC1(Sn1)satisfying (1.4) and (1.5), then the parabolic singular integral operatorT is bounded onLp(Rn)for

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1< p <∞, whereT is defined by

(1.6) Tf (x)=p.v.

Rn

(y)

ρ(y)αf (xy) dy.

Later, the above result was improved by Nagel, Riviere and Wainger [7]

and the regularity condition assumed onwas removed.

TheoremC ([7]). IfLlog+L(Sn1)satisfies(1.4)and(1.5), then the operatorT is bounded onLp(Rn)for1< p <.

Recently, the condition ofin Theorem C was weakened further by Chen, Ding and Fan:

Theorem D ([4]). If H1(Sn1) satisfies (1.4) and (1.5), where H1(Sn1) denotes the Hardy space on the unit sphereSn1, then the para- bolic singular integral operatorT is bounded onLp(Rn)for1< p <.

Let us now turn to the definitions of the BMO space and the Lipschitz space on(Rn, ρ). Denote byE(x, r) = {y : ρ(xy) < r}the ellipsoid centered atx and with radiusr. Forj > 0, jE denotes thej-times extension of the ellipsoidE with the same center. Moreover,|E(x, r)|is the Lebesgue measure ofE(x, r), which is comparable torα. LetEcdenote the complement ofE.

For a positive functionϕ, the parabolic BMOϕ(Rn, ρ)space is defined by BMOϕ(Rn, ρ)=

bLloc(Rn):bBMOϕ = sup

xRn r>0

M(b,E(x, r)) ϕ(r) <

,

where M(b,E) = |E1|

E |b(x)bE|dx, and bE = |E|1

Eb(x) dx. When ϕ ≡ 1, then we denote simply BMO1(Rn, ρ)by BMO(Rn, ρ)andbBMO1

bybBMO,ρ.

For 0< β≤1, the definition of parabolic Lipschitz spaceβ(Rn, ρ)is the following:

β(Rn, ρ)=

b:bβ = sup

x,yRn

|b(x)b(y)| ρ(xy)β <

.

ForbLloc(Rn), the commutator [b, T] of parabolic singular integral operator is defined by

(1.7) [b, T]f (x)=p.v.

Rn

(xy)

ρ(xy)α(b(x)b(y))f (y) dy.

The commutator of the parabolic singular integral operator arises naturally in the theory of parabolic PDE. In 1996, Bramanti and Cerutti [1] proved the

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commutator of parabolic singular integral operator is bounded onLp(Rn)and gave some applications of the commutator in the theory of parabolic PDE.

TheoremE ([1]). Suppose thatb∈BMO(Rn, ρ)and that∈Lipβ(Sn1) (0 < β ≤ 1) satisfies (1.4) and (1.5), then [b, T] is bounded on Lp(Rn) (1< p <).

In 2004, Palagachev and Softova [8] gave the boundedness of commutator of parabolic singular integral with variable kernel on the Morrey space and gave some applications of the commutators in studying parabolic PDE.

Recently, we improved Theorem E and removed the regularity condition assumed onin Theorem E.

TheoremF ([3]). SupposeL(log+L)2(Sn1)satisfies(1.4)and(1.5).

Ifb ∈ BMO(Rn, ρ), then the commutator[b, T]defined in(1.7)is bounded onLp(Rn)for1< p <.

The main results in this paper are as follows.

Theorem 1.1. Let ψ be a non-decreasing convex function on R+ with ψ (0)=0andψ (2t )Cψ (t ). For1< p <, denoteϕ(r)=rα/pψ1(rα) (or equivalently, ψ1(t ) = t1/pϕ(t1/α)). Suppose that0 < β ≤ 1satisfies 1 < p < α/β and Lα−βα (Sn1) satisfying (1.4)and (1.5). Then for b∈BMOϕ(Rn, ρ),[b, T]is bounded operator fromLp(Rn)toLψ(Rn).

The following result can be seen a reverse of Theorem 1.1.

Theorem1.2. Suppose thatsatisfies(1.4)and(1.5)and there are con- stantsC1>0andγ >1such that for anyx, ySn1

(1.8) |(x)(y)| ≤ C1

logρ(x2y)

γ.

If[b, T] for some1 < p <mapsLp(Rn)boundedly intoLψ(Rn), then b∈BMOϕ(Rn, ρ), whereψ andϕare both as in Theorem 1.1.

In particular, by taking ϕ(r) ≡ 1 and ψ (t ) = tp for 1 < p < ∞ in Theorem 1.1 and Theorem 1.2, we get the following corollary of Theorem 1.1 and Theorem 1.2:

Corollary1.3. Ifsatisfies(1.4),(1.5)and(1.8), then the commutator [b, T] is a bounded operator onLp(Rn) (1 < p <) if and only ifb ∈ BMO(Rn, ρ).

For 1 < p < q < ∞, if we takeψ (t ) = tq, thenϕ(r) = rα/prα/q in Theorem 1.1 and Theorem 1.2. Notice that BMOtβ(Rn, ρ)= β(Rn, ρ)(see

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Lemma 3.4 in Section 3), we therefore get another corollary of Theorem 1.1 and Theorem 1.2:

Corollary1.4. Suppose that1< p < q <with1/p−1/q≤ 1/α.

If satisfies(1.4), (1.5)and (1.8), then [b, T] is a bounded operator from Lp(Rn)toLq(Rn)if and only ifbα(1

p1q)(Rn, ρ).

Remark1.5. Notice that the regularity condition assumed onhas been removed in Theorem 1.1, hence the conclusion of Theorem 1.1 is a substantial improvement of Theorem B and Theorem E. Moreover, it is easy to see that the condition (1.8) is weaker than the Lipshitz condition Lipβ(Sn1)(0< β ≤1).

So Theorem 1.2 is also a substantial improvement and extension of Theorem B and Uchiyama’s result in [11].

In the proof of Theorem 1.1, we will need to apply the(Lp, Lq)-bounded- ness of the parabolic fractional integral T, which is an extension of the Hardy-Littlewood-Sobolev theorem for the Riesz potential (see [10]). So, these results have also itself independent interest. Here the parabolic fractional in- tegral is defined by

Tf (x)=

Rn

(y)

ρ(y)αβf (xy) dy, where 0< β < αwithLα−βα (Sn1)satisfying (1.4).

Theorem1.6. Suppose that0< β < α, and thatLα−βα (Sn1)satisfies (1.4).

(i)T is of weak type 1,ααβ

. That is, there exists a constantC > 0 such that for anyfL1(Rn)andλ >0,

{xRn:|Tf (x)|> λ}≤ C

λf1

α−βα . (ii)For1< p < αβ and 1q = p1βα,T is of type(p, q).

Corollary 1.7. Suppose that Lα−βα (Sn1) (0 < β ≤ 1)satisfies (1.4)and (1.5), letbβ(Rn, ρ). Then the commutator[b, T] defined in (1.7)is a bounded operator fromLp(Rn)to Lq(Rn)for1 < p < α/β and 1/q = 1/p−β/α. More precisely, we have[b, T]fqCbβfp, where the constantCis independent ofbandf.

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2. Proofs of Theorem 1.1 and Theorem 1.6 Let us begin with the proof of Theorem 1.6.

Proof of Theorem1.6. First we prove that for 1≤p < αβand1q = p1βα, T is of weak type(p, q). Set K(x) = ρ(x)(x)α−β and E(s) = {xRn :

|K(x)|> s}fors >0. Then

(2.1) |E(s)| ≤Asα−βα ,

whereAdepends only onα,βand. In fact, by (1.4) and|J (x)| ≤ α, we have that

|E(s)| ≤ 1 s

E(s)

|(x)| ρ(x)αβ dx

≤ 1 s

Sn−1|J (x)||(x)|

|(xs)|α−β1

0

rβ1dr dσ (x)

Asα−βα .

Obviously, we may takeA= αβα−βα

L

α−βα (Sn−1). Now for fixedμ >0, let K1(x)=sgn(K(x))(|K(x)| −μ)χE(μ)(x)

and K2(x) = K(x)K1(x). It is easy to see that K2μ. Thus for 1< p < αβ, from (2.1) we get

Rn

|K2(x)|pdx =p μ

0

sp1|E(s)|dspA μ

0

sp1α−βα ds

= αβ

α Aqμα−βα p

q. Hence, when 1≤p < αβ, we obtain that

(2.2) K2p

αβ α Aq

p1 μ(α−β)qα . So, forfLp(Rn), Hölder’s inequality implies that

K2f

αβ α Aq

p1

μ(α−β)qα fp.

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For anyλ >0, setμto satisfy αβ

α Aq p1

μ(α−β)qα fp= λ 2.

Then

xRn:|K2f (x)|> λ 2

=0.

Thus

(2.3)

{xRn:|Tf (x)|> λ}≤

xRn:|K1f (x)|> λ 2

≤ 2

λK1fp

p

. It follows from (2.1) that

(2.4)

Rn

|K1(x)|dx =

E(μ)

(|K(x)| −μ) dx

0

|E(t+μ)|dt

A

μ

tα−βα dt = βA

αβμα−ββ . For anyfL(Rn)andxRn, by (2.4), we conclude that (2.5) |K1f (x)| ≤ f

Rn

|K1(x)|dxβA αβμ

β

α−βf. For anyfL1(Rn), it also implies that

(2.6) K1f1

Rn

Rn

|K1(xy)||f (y)|dy dxβA

αβμα−ββ f1. Thus (2.5) and (2.6) show thatT1 : fK1f is of type(,)and of type(1,1). The Riesz-Thörin theorem leads to thatT1is also of(p, p)type for 1< p <∞, and

(2.7) T1(p,p)βA

αβμα−ββ . Combining (2.3) with (2.7) yields that

(2.8)

{xRn:|Tf (x)|> λ}≤ 2

λ βA

αβμα−ββ fp

p

=C 1

λfp

q

.

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whereCis independent ofλandf.

Therefore, (2.8) tells us thatT is of weak type(p, q) for 1 ≤ p < αβ and 1q = 1pβα. For 1 < p < βα, takep0such thatp < p0 < αβ and letq0

satisfy 1/q0=1/p0β/α. By (2.8) and applying Marcinkiewicz interpolation theorem, we obtain thatT is of type(p, q), where 1/q=1/p−β/α. Thus we complete the proof of Theorem 1.6.

Proof of Corollary1.7. For anyfLp(Rn)andxRn, by the defin- ition of the parabolic fractional integralT, we have the following pointwise relationship between the commutator [b, T](f )(x)andT||(|f|)(x):

|[b, T](f )(x)| ≤

Rn

|b(x)b(y)||(xy)|

ρ(xy)α |f (y)|dy

Cbβ

Rn

|(xy)|

ρ(xy)αβ|f (y)|dy

=CbβT||(|f|)(x).

Then by Theorem 1.6 we get

[b, T](f )qCbβT||(|f|)qCbβfp.

Before proving Theorem 1.1, let us give two lemmas. Suppose thatηC0(Rn)with supp(η)⊂ {x :ρ(x) < 1}and

Rnη(y) dy = 1. Forr >0 and b∈BMOϕ(Rn, ρ), we denote

(2.9) br(x)=

Rn

b(xδry)η(y) dy.

Lemma 2.1 ([6]).Suppose that b ∈ BMOϕ(Rn, ρ) withbBMOϕ ≤ 1.

Then for anyr >0,bbr ∈BMOϕ(Rn, ρ)andbbrBMOϕ,ρCϕ(r).

Lemma2.2.Let0< β ≤1. Ifb∈BMOϕ(Rn, ρ), thenbrβ(Rn, ρ)and (2.10) brβCrβϕ(r)bBMOϕ.

Proof. By the definition (2.9) ofbr, we get

|br(x)br(y)|

= rα

Rn

b(z)η(δr−1(xz)) dzrα

Rn

b(z)η(δr−1(yz)) dz . Note that

rα

Rn

η(δr−1y) dy=

Rn

η(y) dy =1,

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for anyν, independent ofz, we have

|br(x)br(y)|

=rα

Rn

(b(z)ν)η(δr−1(xz)) dz

Rn

(b(z)ν)η(δr−1(yz)) dz

rα

Rn

|b(z)ν||η(δr−1(xz))η(δr−1(yz))|dz.

Ifρ(xry) ≤1, byηC, we get

|br(x)br(y)|

Crα

Rn

|b(z)ν||η(δr−1(xz))η(δr−1(yz))|βdz

Crα|δr−1(xy)|β

Rn

|b(z)ν||∇η(δr−1(ιx+(1ι)yz))|βdz, where 0< ι <1. LetE = {z:ρ(ιx+(1ι)yz) < r}andν= bE. Then by|(xy)i| ≤ ρ(xy)αi, supp(∇η)⊂ {x :ρ(x) < 1}and∇ηC, we have

|br(x)br(y)|

C|δr−1(xy)|βrα

E|b(z)bE|dz

Cϕ(r)bBMOϕ

r1(x1y1)2+ · · · +rn(xnyn)2β/2

Cϕ(r)bBMOϕ

ρ(xy) r

1

+ · · · +

ρ(xy) r

nβ/2

Cϕ(r)bBMOϕ

ρ(xy) r

β

. Ifρ(xry) ≥1, we get

|br(x)br(y)|

Crα

Rn

|b(z)ν||η(δr−1(xz))η(δr−1(yz))|β/max{αi}dz

Crα|δr−1(xy)|β/max{αi}

Rn

|b(z)a|

× |∇η(δr−1(ιx+(1ι)yz))|β/max{αi}dz.

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LetE = {z : ρ(ιx+(1ι)yz) < r}andν = bE, then by|(xy)i| ≤ ρ(xy)αi, supp(∇η)⊂ {x:ρ(x) <1}, and∇ηC, we get

|br(x)br(y)|

C|δr−1(xy)|β/max{αi}rα

E|b(z)bE|dz

Cϕ(r)bBMOϕ

ρ(xy) r

1

+ · · · +

ρ(xy) r

n2 max{β

αi}

Cϕ(r)bBMOϕ

ρ(xy) r

β

.

Thus (2.10) follows from the above estimates in two cases.

For a measurable functionf, denote bymf the distribution function off, i.e.,

mf(t )={x :|f (x)|> t} for t >0.

Lemma 2.3 ([6]). Suppose that 1 ≤ p2 < p < p1 <, θ is a non- increasing function onR+,Lis a linear operator such that

(2.11)

⎧⎪

⎪⎨

⎪⎪

mLg(t1/p1·θ (t ))C

t , if gp1 ≤1, mLg(t1/p2·θ (t ))C

t , if gp2 ≤1.

Then

0

mLg(2t1/p·θ (t )) dtC if gp(p/p1)1/p.

Proof of Theorem1.1. We will apply some idea taken from [6] to prove Theorem 1.1. We may assume thatbBMOϕ≤1, then by (2.10)brβCrβϕ(r)forr >0. Choosepi (i = 1,2)such that 1< p2 < p < p1 < αβ and denote 1/qi =1/piβ/α. Then by Corollary 1.7, forfLpi (i=1,2) withfpi ≤1

(2.12) [br, T]fqiCrβϕ(r).

On the other hand, by Lemmas 2.1 we know thatbbr ∈BMOϕ(Rn, ρ)with bbrBMOϕ,ρCϕ(r). Note that Lα−βα (Sn1)L(log+L)2(Sn1), applying Theorem F fori=1,2 andfLpi withfpi ≤1, we have (2.13) [b−br, T]fpiCϕ(r).

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We now take r = t1/α, by (2.12) and (2.13) we obtain the following weak estimate:

(2.14) m[b,T]f

t1/piϕ(t1/α)

2Cϕ(r) t1/piϕ(r)

pi

+

2Crβϕ(r) t1/piϕ(r)

qi

= C t . If setθ (t )=ϕ(t1/α), then (2.14) is just (2.11). Hence, by Lemma 2.3, when fp(p/p1)1/pwe have

Rn

ψ

|[b, T]f (x)| 2

dx =

0

m[b,T]f(2ψ1(t )) dtC.

This shows that[b, T]fLψCand Theorem 1.1 is proved.

3. Some lemmas

In this section, we give some lemmas which will be used in the proof of Theorem 1.2.

Lemma3.1. Ifρ(x)≥4ρ(y), then

ρ((xy)x)≤3ρ(y) ρ(x) wherex = x1

ρ(x)α1, . . . , ρ(x)xnαn

Sn1. Proof. Write

(3.1)

ρ((xy)x)=ρ

x1y1

ρ(xy)α1x1

ρ(x)α1, . . . , xnyn

ρ(xy)αnxn ρ(x)αn

. Denoteϕi(x)= ρ(x)xiαi, then

(3.2) ∂ϕi

∂xi(x)=ρ(x)αixiαiρ(x)αi1∂ρ(x)

∂xi . ByF (x, ρ)=1, we get

∂ρ(x)

∂xi = xiρ(x)i n

j=1αjxj2ρ(x)j1. This together with (3.2) shows that

(3.3)

∂ϕi

∂xi(x)=ρ(x)αixiαiρ(x)αi1 xiρ(x)i n

j=1αjxj2ρ(x)j1

=ρ(x)αi

1− αixi2ρ(x)i n

j=1αjxj2ρ(x)j

.

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Sinceρ(x)|xi|αi ≤1 and

(3.4) min{αi} ≤

n

j=1

αjxj2ρ(x)j ≤max{αi},

we get by (3.3) and (3.4)

(3.5)

∂ϕi

∂xi(x)

1+ αi min{αj}

ρ(x)αi. Applying (3.5) and the mean value theorem toϕi(x), we have (3.6)

xiyi

ρ(xy)αixi ρ(x)αi

∂ϕi

∂xi

(ξ )

|yi| ≤(1+αi)ρ(ξ )αi|yi|, whereξ =t (xy)+(1t )xfor somet(0,1). Then we have

ρ(x)ρ(ξ )+ρ(xξ )=ρ(ξ )+ρ(ty)ρ(ξ )+ρ(y).

Bearing in mindρ(x)≥4ρ(y), soρ(ξ )≥3ρ(y)andρ(ξ )≥3/4ρ(x). This together (3.6) yields

(3.7)

xiyi

ρ(xy)αixi ρ(x)αi

(1+αi)(4/3)αiρ(x)αi|yi|

≤2αi(4/3)αiρ(x)αi|yi|.

Notice thatρ(δμx)=μρ(x)forμ >0, it follows from (3.1) and (3.7) ρ((xy)x)ρ

1

4 3ρ(x)

α1

|y1|, . . . ,n

4 3ρ(x)

αn

|yn|

≤ 4

3ρ(x)1max{(2αi)1/αi}ρ(|y1|, . . . ,|yn|)

= 4

3ρ(x)1max{(2αi)1/αi}ρ(y)

≤3ρ(x)1ρ(y).

Lemma3.2. Ifsatisfies the conditions(1.4)and(1.8), then forρ(x)

4ρ(y),

(xy)

ρ(xy)α(x) ρ(x)α

C ρ(x)α

logρ(x)ρ(y)γ.

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Proof. Sinceρ(x)≥4ρ(y), then by (1.4), (1.8) and Lemma 3.1 we get

(3.8) |(xy)(x)| ≤ C

logρ(x)ρ(y)γ. Now we claim that

(3.9) |ρ(x)αρ(xy)α| ≤C ρ(y) ρ(x)α+1.

In fact, we know 3/4≤ ρ(xρ(x)y) ≤ 5/4 byρ(x) ≥4ρ(y). On the other hand, using the convexity oftα, it is easy to check that

1−(1/2)α

1−1/2 ≤ 1−tα

1−t for t >1/2 and t =1.

Hence

|1−tα| ≤2 2α−1

|1−t| for t >1/2.

Thus, forρ(x)≥4ρ(y)we have

|ρ(x)αρ(xy)α| ≤2

2α−1 ρ(y) ρ(x)α+1.

Hence we get (3.9). Applying (3.8), (3.9) and notice thatL(Sn1), we have

(xy)

ρ(xy)α(x) ρ(x)α

≤ |(xy)(x)|

ρ(x)α + |(xy)||ρ(xy)αρ(x)α|

C

ρ(x)α

logρ(x)ρ(y)γ +C ρ(y) ρ(x)α+1

C

ρ(x)α

logρ(x)ρ(y)γ.

Lemma3.3 ([9]). Let|E|be a set of finite measure. Then χELψ∗ = |E|ψ1(|E|1).

Lemma3.4 ([6]). If 0 < β ≤ 1and ϕ(t ) = tβ, then BMOϕ(Rn, ρ) = β(Rn, ρ).

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4. Proof of Theorem 1.2

In the proof of Theorem 1.2, we will use some idea taken from [11]. Suppose that [b, T] is a bounded operator from Lp(Rn) to Lψ(Rn), we are going to prove thatb∈BMOϕ(Rn, ρ). Below,Bj(j =1, . . . ,15)denotes the positive constant depending only on,p,α,γ andBi (1i < j ).

Without loss of generality, we assume [b, T]LpLψ = 1. We wish to prove that there exists a constantB:=B(, p, α, γ )such that for anyx0Rn andrR+,

(4.1) N := 1

|E(x0, r)|ϕ(r)

E(x0,r)

|b(y)a0|dyB, wherea0= |E(x0, r)|1

E(x0,r)b(y) dy. If we denote T f (x) =p.v.

Rn

(xy)

ρ(xy)αf (y) dy,

where(x)= −(x), then it is easy to see that [b,T] is the adjoint operator of [b, T]. Hence[b,T]Lψ∗Lp =1. Notice that [b−a0,T]=[b,T], thus we may assumea0=0. Let

(4.2) f (y)=[sgn(b(y))−c0E(x0,r)(y), wherec0 = |E(x10,r)|

E(x0,r)sgn(b(y)) dy. By |E(x1

0,r)|

E(x0,r)b(y) dy = a0 = 0, it is easy to check that|c0|<1. Moreover, the following properties off are obvious:

(4.3) f ≤2,

(4.4) suppfE(x0, r),

(4.5)

Rn

f (y) dy =0,

(4.6) f (y)b(y) >0,

(4.7) 1

|E(x0, r)|ϕ(r)

Rn

f (y)b(y) dy=N.

Sincesatisfies (1.5) and (1.8), then there exists a positive numberB1 < 1 such that

(4.8) σ ()=σ

xSn1:(x)≥ 2C1

logB2

1

γ

>0,

(16)

whereσis the measure onSn1which is induced from the Lebesgue measure onRn. Let

=

xSn1:(x)≥ 2C1

logB2

1

γ

>0,

then is a closed set. If denote = {xSn1 : −x = (x)}, thenσ () = σ () > 0. We claim that ifxandySn1 satisfying ρ(y(x))B1, then

(4.9) (y)C1

logB2

1

γ.

In fact, since

|((x))(y)| ≤ C1

logρ(y2(x))

γC1

logB2

1

γ

and note that((x))≥2 C1 logB2

1

γ, we have(y) C1

logB2

1

γ. Now denote G=

xRn:ρ(xx0) > B2r and (xx0) , whereB2=3B11+1. Then forxG,

(4.10)

|[b,T]f (x)|

≥ |T (bf )(x) | − |b(x)||T f (x)|

=

Rn

((yx))

ρ(yx)α b(y)f (y) dy

− |b(x)|

Rn

((yx))

ρ(yx)α f (y) dy :=I1I2.

We first give the estimate ofI1. Ifρ(yx0) < r, then ρ(xx0) > B2ρ(yx0) >4ρ(y−x0).

By Lemma 3.1, we see that

ρ((yx)(x0x))≤3ρ(yx0) ρ(xx0)B1. Then((yx)) C1

logB2

1

γ by (4.9). Thus, it follows from (4.4), (4.6) and

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