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        Commutators of Lipschitz Functions and Singular IntegralswithNon-Smooth KernelsonEuclideanSpaces

        2016-10-24 02:18:05HungVietLe
        Analysis in Theory and Applications 2016年2期

        Hung Viet Le

        APUInternationalSchool,286LanhBinhThang,Ward 11,District11,HoChiMinh City,Viet Nam

        ?

        Commutators of Lipschitz Functions and Singular IntegralswithNon-Smooth KernelsonEuclideanSpaces

        Hung Viet Le?

        APUInternationalSchool,286LanhBinhThang,Ward 11,District11,HoChiMinh City,Viet Nam

        .In this article,we obtain the Lp-boundedness of commutators of Lipschitz functions and singular integrals with non-smooth kernels on Euclidean spaces.

        Commutators,singularintegrals,maximalfunctions,sharpmaximalfunctions,muckenhoupt weights,Lipschitz spaces.

        AMS Subject Classifications:42B20,42B25,42B35

        1 Introduction

        Consider the singular integral operator T defined by

        where f is a continuous function with compact support,x/∈suppf;and the kernel K(x,y)is a measurable functiondefinedon(Rn×Rn)?with?={(x,x):x∈Rn}.If b∈BMO(Rn),then the commutator[b,T]of a BMO function b and the singular integral operator T is defined by

        The Lp-boundedness(1<p<∞)of T and Tbare well known in the Euclidean setting,provided that the kernel K(x,y)of the operator T satisfies H¨ormander's conditions(see[1,15-17]among many other good references).In 1999,Duong and McIntosh[3]obtained the Lp-boundedness of T,under the assumption that the kernel K(x,y)satisfies some conditions which are weaker than H¨ormander's integral conditions.The boundedness of the operator T with non-smooth kernel on Lp(w)(w∈Ap(Rn),1<p<∞)was proved by Martell[12].Moreover,Duong and Yan[4]obtained the Lp-boundedness ofthe commutator Tbunder some conditions which are weaker than H¨ormander's pointwise conditions.Lin and Jiang[11]also obtained the Lp-boundedness of Tb,but with b∈Lipα,w(Rn).See also[8,9,13,18]for additional results on these topics.

        The purpose of this paper is to extend the results in[11].That is,we would like to obtain the Lp-boundedness(1<p<∞)of the operator T→b,where

        2 Background

        2.1Apweights

        For a ball B in Rn,let|B|denote the measure of the ball B.A weight w is said to belong to the Muckenhoupt class Ap(Rn),1<p<∞,if there exists a positive constant C such that

        for all balls B in Rn.The smallest constant C for which the above inequality holds is the Apbound of w.The class A1(Rn)consists of non-negative functions w such that

        for all balls B in Rn.It is well-known that(see[7,17]for instance)if w∈Ap(Rn)for some p∈[1,∞),then for any measurable subset E?B,there exist positive constants γ and C such that

        Inequality(2.1)indeed holds with γ∈(0,1).This will be used in the estimate of(3.3)below.Furthermore,if w∈Ap(Rn)(1≤p≤∞),thenit satisfies the reverse H¨olderinequality. That is,there exist s′>1 and c>0(both depending on w)so that

        A weight w is said to belong to the class Ap,q(Rn),1<p,q<∞,if there exists a positive constant C such that

        for all balls B?Rn.Observe that

        When p=1 and q>1,we say that w∈A1,q(Rn)if there exists a positive constant C such that for all balls B?Rn,

        It follows from H¨older's inequality that for 1<q1<q2<∞,

        and

        The following lemma is necessary for the proof of our theorem.

        2.2Approximation of the identity

        We assume that there exists a class of operators At(t>0)which can be represented by the kernels at(x,y)in the sense that

        Moreover,the kernels at(x,y)satisfy the following conditions

        where

        Here s is a positive,bounded,decreasing function satisfying

        where k appears in(1.2).

        Remark 2.1.The functions htabove satisfy the following properties(see[4,5]): i)There exist positive constants C1and C2such that

        ii)There exists a positive constant C such that

        where M is the Hardy-Littlewood maximal operator.

        The class of operators Atplays the role of approximation to the identity.The existence of such a class of operators Atwas verified in[3].

        Now consider the operators T and T→bgiven in(1.1)and(1.2)respectively.Let Atand Bt(t>0)be two classes of operators which satisfy(2.3),(2.4)and(2.5).Denote by K(x,y)-Kt(x,y)the kernels ofthe operators(T-TBt),and K(x,y)-Kt(x,y)as the kernels of(T-AtT).We state below some assumptions which are necessary for our theorem.

        (a)T is a bounded linear operator from Lr(Rn)to Lr(Rn)for some r∈(1,∞);

        (b)There exist positive constants c1and CAsuch that

        (c)There exist positive constants c2,c3and β>nk(k appears in(1.2))such that

        In the sequel,the letter C will denotea constant,which may vary at different occurrences. However,it is independent of any essential variable.

        3 Main theorem

        Let τ=max{τ1,τ2,τ3}.Assume that w∈A1,τ(Rn),and bi∈Lipαi,w(Rn)for 1≤i≤k.

        Let T,given by(1.1),satisfy assumptions(a),(b)and(c).Then there exists a constant C>0,independent of f,such that

        where

        Remark 3.1.Observe that for the case k=1,w is only required to be in A1,τ1(Rn).

        Proof.First,we show that there exist r1,r2,r3>1 such that 1<rs<q0,where r:=r1r2r3.For the case k=1,since 1<s<q0,there exists an r>1 such that 1<rs<q0.We then choose some numbers r1,r2,r3>1 such that r=r1r2r3.Now suppose k>1.Since s<√q0,there exists an r3such that 1<s<r3<√q0.Then sr3<q0.Pick a number t1∈(sr3,q0),and let t=t1/sr3>1.We choose a number r2∈(1,t)and let r1=t/r2,r:=r1r2r3.Then we have r1,r2,r3>1,and 1<rs<q0.For the rest of the proof,we denote t=r1r2and r=r1r2r3=tr3. Let→λ=(λ1,···,λk)=((b1)B,···,(bk)B),where

        Then,the above equation implies that

        Note that Eqs.(3.1)and(3.2b)imply that for 1≤j≤k-1,

        We have the following lemmas.

        Proof.Observe that

        By H¨older's inequality,we have

        Thus,

        Let

        where the last inequality follows from(3.3).Combining(3.5)and(3.6)yields

        So,we complete the proof of the lemma.

        Proof.Take a ball B which contains x.We have

        Thus,by H¨older's inequality and Lemma 3.1,we see that

        On the other hand,y∈B and z∈2j+1B2jB imply that|y-z|≥2j-1rB.So,

        Hence,an application of Lemma 3.1 yields

        provided that

        for some ?>0.Combining the estimates of J3(x)and J4(x)and taking the supremum over all balls B containing x yields the conclusion.

        Lemma 3.3.It holds

        where ck,iare constants depending on k and i.

        Proof.For an arbitrary fixed x∈Rn,choose a ball B which contains x.Following[16],we split f=fχ2B+fχ(2B)c≡f1+f2,and write

        By H¨older's inequality,Lemma 3.1 and Lemma 3.2 respectively,we have

        and

        where

        Note that

        Therefore,by Theorem 5.3[12]and Lemma 3.1,

        Observe that

        The third and last inequalities are due to Theorem 5.3[12]and Lemma 3.1 respectively.

        It remains to estimate K7(x).By hypothesis and Lemma 3.1,we have

        provided that β>nk.Finally,the result follows from combining all of the estimates above and taking the supremum over all balls B containing x.

        Lemma 3.4.If w∈A1,s′(Rn)for some s′>1,then there exists a constant C>0 such that

        Proof.Let B be a ball which contains x.By H¨older's inequality,

        Taking the supremum over all balls B which contain x yields the desired result.

        By Eqs.(3.2a)and(3.2b),we have that,for 1≤j≤k,

        Thus for 1≤j≤k,

        where

        Therefore,we may apply Lemma 2.1,Lemma 3.4,Theorem 5.3[12],together with equation(3.2a)to conclude that

        and

        Since 1<rs<q0,it follows that for 1≤j≤k-1,

        Hence

        which implies that

        or equivalently,

        Again,by Lemma 2.1,Lemma 3.4 and Eq.(3.1),we infer that for 1≤i≤k-1,

        provided that 1<rs<q0<qk-i<n/νifor 1≤i≤k-1.Consequently,by Theorems 4.2,5.3[12],Lemma 3.3,inequalities(3.7)-(3.9),and induction argument,we conclude that

        Thus,we complete the proof.

        Remark 3.2.We could apply the reverse H¨older inequality to prove Lemma 3.4,thereby eliminating the assumption that w∈A1,s′(Rn).However,the exponent s′appearing in the reverse H¨older inequality(see(2.2))depends on w and may be very close to 1,which means that its conjugate exponent s could be very large(see[7,17]).This limits the value of q0for which the theorem holds,since q0is necessarily greater than s.

        Remark 3.3.Let ? be a non-decreasing positive function on R+.Denote by ?(f,B),the mean oscillation of a function f on a ball B?Rn,as|B|-1RB|f(x)-fB|dx.Define BMO?as the space of all functions f satisfying ?(f,B)≤C?(r),whenever B is a ball with radius r(see[10]).Note that when ?≡1,then BMO?=BMO,the space of all functions of bounded mean oscillation.Let Λα,0<α≤1,be the space of Lipschitz continuous functions,Λα={f:|f(x)-f(y)|≤C|x-y|α,?x,y∈Rn}.A function ψ:[0,∞)→[0,∞)is said to be a Young function if it is continuous,convex,increasing and satisfying ψ(0)=0 and limt→∞ψ(t)=∞.The Orlicz space Lψis defined as the space of all functions f such thatRψ(λ|f|)<∞,for some λ>0.

        Now consider the singular integral Tf and the commutator Tbf(as defined in the introduction),but with the convolution kernel

        With this type of kernel,Janson[10]proved that b belongs to BMO?if and only if Tbmaps Lp(1<p<∞)boundedly into Lψ,where ? and ψ are related by the equation ?(r)= rn/qψ-1(r-n),or equivalently,ψ-1(t)=t1/p?(t-1/n).When ?(t)=tα(0<α<1),ψ(t)=tqwith 1/q=1/p-α/n,then it is evident that Lipα(Rn)=BMOtα(Rn)and Lψ(Rn)=Lq(Rn). In this particular case,Janson's Theorem says that b belongs to Lipα(Rn)if and only if Tbmaps Lp(Rn)(1<p<∞)boundedly into Lq(Rn),where 1/q=1/p-α/n.It is interesting to note that the above necessary condition is the same as in Theorem 3.1,when k=1 and w≡1,but with different kernel K.

        Acknowledgments

        The author would like to express his gratitude toward the referees for their thorough review and precious suggestions.

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        .Email address:hvle2008@yahoo.com(H.V.Le)

        25 October 2015;Accepted(in revised version)11 April 2016

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