An Equivalent Characterization of CMO(Rn)with Ap Weights

Minghui Zhong and Xianming Hou

School of Mathematical Sciences,Xiamen University,Xiamen 361005,China.

Abstract.Let 1＜p＜∞and ω ∈Ap.The space CMO(Rn)is the closure in BMO(Rn)of the set of (Rn).In this paper,an equivalent characterization of CMO(Rn)with Ap weights is established.

Key words:BMOω(Rn),CMO(Rn),Ap,John-Nirenberg inequality.

1 Introduction

The goal of this paper is to provide an equivalent characterization ofCMO(Rn),which is useful in the study of compactness of commutators of singular integral operator and fractional integral operator.

The spaceBMO(Rn)is defined by the set of functions(Rn)such that

where

The spaceCMO(Rn)is the closure inBMO(Rn)of the set of(Rn),which is a proper subspace ofBMO(Rn).

In fact,it is known thatCMO(Rn)=VMO0(Rn),whereVMO0(Rn)is the closure ofC0(Rn)inBMO(Rn),see[2,3,9].HereC0(Rn)is the set of continuous functions on Rnwhich vanish at infinity.Neri[8]gave a characterization ofCMO(Rn)by Riesz transforms.Meanwhile,Neri proposed the following characterization ofCMO(Rn)and its proof was established by Uchiyama in his remarkable work[11].

Theorem 1.1.Let f ∈BMO(Rn).Then f ∈CMO(Rn)if and only if f satisfies the following three conditions

Recently,Guo,Wu and Yang[6]established an equivalent characterization of spaceCMO(Rn)by local mean oscillations.Lots of works about spaceCMO(Rn)have been studied,see[4]for example. Muckenhoupt and Wheeden[7,Theorem 5]showed the norm ofBMOω(Rn)(see Definition 1.2)is equivalent to the norm ofBMO(Rn),where the weight functionωis MuckenhouptApweight.So it is natural to consider equivalent characterizations ofCMO(Rn)associated toApweights.

To state our main results,we first recall some relevant notions and notations.

The following class ofApwas introduced in[1,5].

Definition 1.1.Let ω(x)≥0and ω(x)(Rn).For1＜p ＜∞,we say that ω(x)∈Ap if there exists a constant C＞0such that for any cube Q,

Also,for p=1,we say that ω(x)∈A1if there is a constant C＞0such that

where M is the Hardy-Littlewood maximal operator.For p ≥1,the smallest constant appearing in(1.1)and(1.2)is called the Ap characteristic constant of ω and is denoted by[ω]Ap.

Definition 1.2.Let ω∈Ap.For a cube Q inRn,we say a function f ∈(Rn)is in BMOω(Rn)if f satisfies

where

Letω∈Ap(p≥1),q＞1,ThenBMOω,q(Rn)is defined by

where

Now,we can formulate our main results as follows.

Theorem 1.2.Let p ≥1,1＜q ＜∞.Suppose f ∈BMO(Rn)and ω ∈Ap.Then the following conditions are equivalent:

(1)f ∈CMO(Rn);

(2)f satisfies the following three conditions:

(3)f satisfies the following three conditions:

Throughout this paper,the letterC,will stand for positive constants,not necessarily the same one at each occurrence,but independent of the essential variables.Iff ≤Cg,we writefgorgf;and iffgf,we writef ～g.A dyadic cubeQon Rnis a cube of the form

Rjmeans{x ∈Rn:|xi|＜2j,i=1,2,···,n}.Forλ ＞0,λQdenotes the cube with the same center asQand side-lengthλtimes the side-length ofQ.

2 The proof of Theorem 1.2

This section is devoted to the proof of Theorem 1.2.To do this,we firstly recall some auxiliary lemmas.Note that[7,Theorem 3]impiles the following weighted John-Nirenberg inequalities,also see[1,10].

Lemma 2.1.(John-Nirenberg)Let p∈[1,∞),ω∈Ap and f ∈BMOω(Rn).For every α＞0and cube Q,there exist constants c1and c2such that

Next,we recall some useful properties ofApweights.

Lemma 2.2([5]).Let ω∈Ap and1≤p＜∞.

1.There exist0＜δ＜1and C＞0that depending only on the dimension n,p,and[ω]Ap such that for any cube Q and any measurable subset S of Q we have

2.There exist constants C and γ ＞0that depending only on the dimension n,p,and[ω]Ap such that for every cube Q we have

3.For all λ＞1,and all cubes Q,

Now,we are in position to prove the Theorem 1.2.

Proof.To prove(1)?(2)in Theorem 1.2.Assume thatf ∈CMO(Rn).If(Rn),then(i)?(iii)hold.It is obvious that(i)holds for uniformly continuous functionsf.Without loss of generality,we assumesupp(f)?Q0.Then for eachQ?Rn,there existsh∈Rn,for|x|＞|h|,we haveQ0∩(Q+x)=?,(iii)holds.

Note that

For(Rn),we have

On the other hand,Q(0,r)denotes the closed cube centered at 0 with side-lengthr.For anyx0∈Q(0,r),there exists a cubeQcentered atx0such thatQ(0,r)?Q,by(2.1),we get

which tends to 0 as|Q|tends to+∞,(ii)holds.

Iffor any givenε ＞0,there existsfε ∈(Rn)satisfying(i)?(iii)andThen by Lemma 2.1 and(2.2),forω∈Ap,1＜p＜∞,it is easy to see

The detailed proof of(2.4)also can be found in[1,7].By(2.4)and the triangle inequality,we deduce that(i)?(iii)hold forf.

The proof of(2)?(3).By the H?lder inequality,we get

where 1/q+1/q′=1.

It follows from(2.5)that iffsatisfies(i)?(iii)thenfsatisfies(i′)?(iii′).

The proof of(3)?(1).Now we show that iffsatisfies(i′)?(iii′)then for allε＞0,there existsgε ∈BMO(Rn)such that

We prove(2.6)and(2.7)by the following two steps.

Step IBy(i′)and(ii′),there existiεandkεsuch that

By(iii′),there existsjε ＞kεsuch that

Now for eachx ∈Rjε,we take dyadic cubeQxwith side-length 2iεcontainingx;ifmeans a dyadic cube of side-length 2iε+m?jε.Set=m(f,Qx)ω,by(ii′),there existsmε ＞jεsuch that

To see this,by(ii′),letmε＞jε+kε?iεbe large enough such that when

for some positive constantC1.

For,it is obvious that

This together with(2.12)and(2.3)imply that

Sinceby(2.3)and(2.12),we have

By(2.13),(2.14)and(2.12),we conclude that for anyQxwith

For anyby(2.15),we get

Step IIDefinewhenx∈RmεandwhenNotice that

By the definition ofiε,jεandmε,iforx,,there existsC2＞0 such that

In fact,assume that|x|＜|y|.Firstly,we show that ifx,,then(2.17)holds.By noting thatx,,we get

and(2.17)holds.Next,if,we deduce from(2.11)that

Thirdly,ifand,(2.15)indicates that

by(2.8),(2.17)holds.Similarly,ifandby(2.8),(2.17)also holds.Ifnotice thatQ∩Rjε=?and by(2.10),

Similarly,we have

Hence

Combining these cases,(2.17)holds.

We turn to prove thatgεsatisfies(2.6).Set

By the definition ofgε,we get

Selectt＜2iε,then

where in the second inequality we make the change of variabley=ut.

Sinceu∈B(0,1)andt＜2iε,?x∈Rn,

By(2.17),ifx,x?tu∈Rmε,,hence

If one ofxandx?tuin,the other must be in,we also have

Therefore

We obtain that(2.6)holds.

Now we prove(2.7).By the definitioniεandjεagain,we obtain that for anyx∈Rmε,

Indeed,

IfQx∩Rjε=?,by(2.10),(2.18)holds.Ifusing(2.8),(2.18)holds.

LetQbe a arbitrary cube in Rn.In order to prove(2.7)holds,it suffices to show

We consider the following four cases:

Case(i):Q ?Rmεand max{diam Qx:Qx∩Q?}＞4diam Q,by(2.16),the number ofis finite.Ifand,by(2.17),

Moreover,if,then,by(2.8),we haveM(f,Q)ω ＜ε;ifQ∩Rjε=?,by(2.10),we also obtainM(f,Q)ω ＜ε.Hence

Case(ii):Q?Rmεanddiam Q,we have

Invoking(2.18),we get

Case(iii):,thenQ∩Rjε=?andM(f,Q)ω ＜ε.Using(2.17),

Hence

Case(iv):andLetPQbe a smallest positive number such thatQ?RPQ.Then

Moreover,

On the one hand,by(2.18),we have

On the other hand,it is easy to prove that

Combining with(2.9)and(2.18)and the fact thatgε(x)=gε(y)for any,we obtain

This implies(2.7)and completes the proof of Theorem 1.2.