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        BOUNDEDNESS OF MULTILINEAR LITTLEWOOD-PALEY OPERATORS ON AMALGAM-CAMPANATO SPACES?

        2020-04-27 08:18:38XiangLI李翔QianjunHE何騫君DunyanYAN燕敦驗
        關(guān)鍵詞:李翔

        Xiang LI(李翔) Qianjun HE(何騫君) Dunyan YAN(燕敦驗)

        1.School of Mathematical Sciences,University of Chinese Academy of Sciences,Beijing 100049,China

        2.School of Applied Science,Beijing Information Science and Technology University,Beijing 100192,China

        E-mail:lixiang162@mails.ucas.ac.cn;heqianjun16@mails.ucas.ac.cn;ydunyan@ucas.ac.cn

        Abstract In this paper,we consider the boundedness of multilinear Littlewood-Paley operators which include multilinear g-function,multilinear Lusin’s area integral and multilinear Littlewood-Paley -function.Furthermore,norm inequalities of the above operators hold on the corresponding Amalgam-Campanato spaces.

        Key wordsmultilinear Littlewood-Paley g-function;multilinear -function;Amalgam-Campanato spaces

        1 Introduction

        Since the pioneering work of Coifman and Meyer[4,5],multilinear theory attracted much attention,including Christ and Journ′e[3],Fu et al.[8],Kenig and Stein[16],Grafakos and Torrers[9,10],Hu[15]and Lerner et al.[18].As a multilinearization of Littlewood-Paley’s function,Coifman and Meyer[6]introduced the following bilinear operator

        Recently,in[2,23,29,30],the authors de fined and studied the multilinear Littlewood-Paley operators,including multilinear g-function,Marcinkiewicz integral and-function.To state some known results,we first recall some de finitions.

        De finition 1.1A function K(x,y1,···,ym)de fined away from the diagonal x=y1=···=ymin(Rn)m+1is said to be a multilinear non-convolutiontype kernel,if for all(y1,···,ym)∈(Rn)m,there exists a positive constant C,such that K satis fies the following three conditions Z

        and

        Then the multilinear Littlewood-Paley g-function,multilinear Lusin’s area integral and multilinear Littlewood-Paley-function with non-convolution type kernels are de fined by

        and

        We need the following two results given by Xue and Yan[30].

        Theorem AAssume that 1

        Theorem BAssume that λ >2m,0< γ

        To state our results,we now recall the de finition of Amalgam-Campanato space which appears in[28].

        De finition 1.2(Amalgam-Campanato space)Let n<β<∞and 0

        with

        where fB(y,r)denotes the average of f over ball B(y,r)with radii r and centered at y,that is,

        Remark 1.3It is easy to see that the spaces goes back to the classical Campanato spacewhen p=∞.

        In this paper,we only consider the bilinear case,the multilinear case m>2 can be obtained in the similar way.Our main results are as follows.

        Theorem 1.4Let n<βj<∞,1

        Theorem 1.5Let λ>4,n<βj<∞,qj≤ αj≤pj≤ ∞ with j=1,2.Assume that

        Suppose that

        If g?λ(f1,f2)(x)is finite on a set of positive measure,then these exists a positive constant C independent of f1,f2such that

        Remark 1.6Notice that S(f1,f2)(x)≤Cg?λ(f1,f2)(x)point-wisely holds[24],and thus the above result is also true for multilinear Lusin’s area integral S.

        This paper is organized as follows.Section 2 will prepare some lemmas.The proof of Theorem 1.4 will be show in Section 3.Section 4 will be devoted to prove Theorem 1.5.

        Throughout this paper,we use the following notation.For 1≤ p≤ ∞,p′is the conjugate exponent of p,that is,1/p+1/p′=1.The notation A.B stand for A ≤ CB,for some positive constant C independent of A and B.The set B=B(y,r)denotes an open ball with center at y and radius r,and 4B denote the ball with the same center as B and with radius length 4r.The letter C will denote a constant which may be different in each occasion but is independent of the essential variables.

        2Some Lemmas

        In order to prove the main results,we need some preliminaries and lemmas.

        Lemma 2.1Assume that f ∈ (Cq,Lp)α,β,1 ≤ p,q ≤ ∞.If γ >0 and δ>0,then for any B=B(y,r),we have

        ProofUsing H?lder’s inequality,we obtain that

        The following lemma plays a key role in our proof.

        Lemma 2.2Assume that f ∈ (Cq,Lp)α,β,n< β < ∞,1≤ q≤ α ≤ p≤ ∞,If ε>0 and,then for any B=B(y,r),we have

        ProofWe first split the following integral into two parts as follows,

        We first estimate of I.It implies from H?lder’s inequality that

        Taking Lpnorm of the first part of(2.1),we obtain that

        To estimate II,we need the following estimate

        Since

        we have

        By(2.5)and Minkowski’s inequality,we obtain

        Next,we will deal with f1and f2,the programing is following[14].We decompose fi,i=1,2,as follows

        Then one gets g(f11,f2j)=g(f1j,f21)=0 by condition(1.2)for j=1,2,3.We shall use the following notations.For any nonnegative integer k,denote

        De fine

        and

        We need the following Lemma for.

        Lemma 2.3Assume that α,β,p and q as in Theorem 1.5.For any r>0,we have

        ProofTo prove inequality(2.8),by condition(1.3),we have

        As t

        Next,we give the proof of(2.10).Note that(f13,f23)can be controlled by a constant times of

        Since t<4r,y1,y2∈(4B)c,z∈B(y,),we have

        Thus,taking

        By H?lder’s inequality and Lemma 2.2,we obtain

        where we used the fact

        and

        Prove of inequality(2.10),when(z,t)∈J(k),we havet2kr≤1,and

        if λ1> λ2>4.Therefore,choosing λ ∈ (4,5)and using(1.3),we get

        In order to deal with the integral that contains f2,we first split the integral interval(4B)cinto(4B)c∩(2k+1B)cand(4B)c∩(2k+1B),then we magnify each of the two intervals to(4B)c∩(2k+1B)cand 2k+1B,respectively, fi nally,we get

        For y2∈ (2k+1B)c,z ∈ 2k?2B2k?3B and 2k?3r ≤ t<2k?2r,we obtain that t+|z? y2|~2kr+|y?y2|.Thus

        Using Minkowski’s inequality and Lemma 2.2,we obtain

        It remains to control II.In fact,it is easy to get

        Note that the first integral of(2.12)can be estimated

        Thus,we have

        To prove inequality(2.11),using similar way of(f12,f23)(x),we can conclude that

        Combining(2.12)with(2.13)and(2.14),it yields that

        We also need the following lemma.

        Lemma 2.4Assume that α,β,p and q as in Theorem 1.5.For any r>0,we have the following estimate

        ProofUsing the triangle inequality,we have

        Utilizing Lemma 2.3,we obtain

        Again by(1.3)of K and the mean value theorem,we have

        Similar to the estimate for Lemma 2.3,we conclude that

        where

        Similar to the proof of(2.10),we get

        By means of Minkowski’s inequality and Lemma 2.2,we get

        Similar to the proof of(2.10)for II,we obtain

        Take advantage of Lemma 2.2 and Minkowski’s inequality,we get

        Using symmetry,we also have

        Now,we will dealing with IV.By Minkowski’s inequality,we have

        By the Cauchy-Schwartz inequality,we have

        Hence,we get

        Combining(2.13)with(2.15)and by Minkowski’s inequality,we obtain

        where we used

        3 Proof of Theorem 1.4

        Proof of Theorem 1.4Using Minkowski’s inequality,we obtain the following estimate

        For the second term,by Minkowski’s inequality,we obtain

        According to(1.4)of K,for any x∈B,z1∈4B,z2∈(4B)c,we conclude that

        Hence,

        As a consequence,by H?lder’s inequality and Lemma 2.1,we obtain

        Similarly,we can conclude that

        Next we estimate the fourth term.Since,we can take ε satisfy condition

        Thus,using the estimated for the second term and H?lder’s inequality,we have

        The above inequality gives

        Taking Lp-norm on both side yields

        De fine

        Similar to the estimate of I1(y,r),it follows that

        Taking Lp-norm on both side inequality above,we obtain

        Let

        Similar to I2(y,r),we have

        and

        Taking Lp-norm on both side gives

        Using Minkowski’s inequality,we get

        This completes the proof of Theorem 1.4.

        4 Proof of Theorem 1.5

        Proof of Theorem 1.5Let fi=fi1+fi2+fi3,i=1,2 as in(2.7).Notice that(f11,f2j)=(f1j,f21)=0 for j=1,2,3,and by Minkowski’s inequality,we have

        Utilizing Lemma 2.3,we see that

        By symmetry,we also have

        Combining(4.1),(4.2)with(4.3)and Lemma 2.4,it yields that

        This completes the proof of Theorem 1.5.

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