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.