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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(yàn)
    關(guān)鍵詞:李翔

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

    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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