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BULLETINof the Malaysian Mathematical Sciences Society

http://math.usm.my/bulletin

Bull. Malays. Math. Sci. Soc. (2)33(1) (2010), 93–103

Boundedness for Multilinear Singular Integral Operators on Morrey Spaces

Lanzhe Liu

College of Mathematics and Computer, Changsha University of Science and Technology Changsha 410077, P. R. of China

[email protected]

Abstract. In this paper, we prove the boundedness for some multilinear op- erators related to certain singular integral operators on the Morrey spaces by using a sharp inequality of the multilinear operators.

2000 Mathematics Subject Classification: 42B20, 42B25

Key words and phrases: Multilinear operator, singular integral operator, Mor- rey space, bounded mean oscillation (BMO).

1. Preliminaries and theorem

Fixλ≥0. For 1≤p <∞, set

||f||Lp,λ = sup

x∈Rn, d>0

1 dλ

Z

B(x,d)

|f(y)|pdy

!1/p ,

where B(x, d) = {y ∈Rn :|x−y|< d}. The Morrey spaces is defined by (see [14, 15])

Lp,λ(Rn) ={f ∈L1loc(Rn) :||f||Lp,λ <∞}.

In this paper, we shall study some singular integral operators as following.

Definition 1.1. Fix ε > 0 and 0 ≤ δ < n. Let S be the Schwartz space of test functions, and S0 be its dual, that is S0 is the set of all tempered distributions on S (see [19, p.88–89]). Suppose T :S → S0 be a linear operator and there exists a locally integrable functionK(x, y)onRn×Rn\ {(x, y)∈Rn×Rn:x=y} such that

T(f)(x) = Z

Rn

K(x, y)f(y)dy

for every bounded and compactly supported function f, where K satisfies:

|K(x, y)| ≤C|x−y|−n+δ

Communicated byLee See Keong.

Received:August 8, 2008;Revised: February 16, 2009.

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and

|K(y, x)−K(z, x)|+|K(x, y)−K(x, z)| ≤C|y−z|ε|x−z|−n−ε+δ

if2|y−z| ≤ |x−z|. Letmj be some positive integers(j= 1,· · ·, l),m1+· · ·+ml=m andAjbe some functions onRn (j= 1,· · ·, l), setA= (A1,· · ·, Al). The multilinear operator related to T is defined by

TA(f)(x) = Z

Rn

Ql

j=1Rmj+1(Aj;x, y)

|x−y|m K(x, y)f(y)dy, where

Rmj+1(Aj;x, y) =Aj(x)− X

|α|≤mj

1

α!DαAj(y)(x−y)α. Note that when m = 0, TA(f)(x) = R

Rn

Ql

j=1(Aj(x)−Aj(y))K(x, y)f(y)dy, that is TA is just the multilinear commutator of T and A (see [18]). While when m >0,TA is a non-trivial generalizations of the commutator. It is well-known that multilinear operators are of great interest in harmonic analysis and have been widely studied by many authors (see [3–5, 8]). Cohen and Gosselin (see [3–5]) obtained the Lp(p >1) boundedness of the multilinear singular integral operator. Hu and Yang (see [10]) proved a sharp estimate for the multilinear singular integral operators. In [16–18], the authors prove some sharp estimate for the multilinear singular integral commutator and obtained theLp (p >1) boundedness for the multilinear operator.

As the Morrey space may be considered as an extension of the Lebesgue space (Morrey space Lp,λ becomes Lebesgue space Lp when λ = 0), it is natural and important to study the boundedness of the multilinear singular integral operator on the Morrey Spaces Lp,λ with λ > 0. The purpose of this paper has twofold, first, we establish a sharp inequality for the multilinear singular integral operators, and second, we prove the boundedness for the multilinear singular integral operators on the Morrey spaces by using the sharp inequality.

First, let us introduce some notations. Throughout this paper,Q will denote a cube ofRnwith sides parallel to the axes. For any locally integrable functionf, the sharp function off is defined by

f#(x) = sup

Q3x

1

|Q|

Z

Q

|f(y)−fQ|dy, where, and in what follows,fQ=|Q|−1R

Qf(x)dx. It is well-known that (see [9]) f#(x)≈sup

Q3x c∈Cinf

1

|Q|

Z

Q

|f(y)−c|dy.

We say that f belongs to BM O(Rn) if f# belongs to L(Rn) and ||f||BM O =

||f#||L. For 0≤η <1, letMη be the fractional maximal operator defined by Mη(f)(x) = sup

Q3x

1

|Q|1−η Z

Q

|f(y)|dy.

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We denote byM(f)(x) =Mη(f)(x) ifη= 0, which is the Hardy-Littlewood maximal operator. For 1≤p <∞and 0≤δ < n, let

Mδ,p(f)(x) = sup

Q3x

1

|Q|1−pδ/n Z

Q

|f(y)|pdy 1/p

. We shall prove the following theorem.

Theorem 1.1. Let 1 < p < n/δ, 0 < λ < n−pδ, 1/q = 1/p−δ/(n−λ) and DαAj ∈ BM O(Rn) for all α with |α|=mj and j = 1,· · ·, l. Suppose that TA is the multilinear operator as Definition 1.1 such that T is bounded from Lu(Rn) to Lv(Rn)for any1< u < n/δ and1/v= 1/u−δ/n. Then, for any f ∈Lp,λ(Rn),

||TA(f)||Lq,λ≤C

l

Y

j=1

 X

j|=mj

||DαjAj||BM O

||f||Lp,λ.

2. Some lemmas

To prove the theorem, we need the following lemmas.

Lemma 2.1. [5] Let A be a function on Rn and DαA ∈ Lq(Rn) for all α with

|α|=mand some q > n. Then

|Rm(A;x, y)| ≤C|x−y|m X

|α|=m

1

|Q(x, y)|˜ Z

Q(x,y)˜

|DαA(z)|qdz

!1/q

,

whereQ˜ is the cube centered atxand having side length 5√

n|x−y|.

Lemma 2.2. [2, 6] Let 1 < p < ∞, 0 < λ < n and f ∈ Lp,λ(Rn). Then the following estimates hold:

(a) ||M(f)||Lp,λ ≤C||f#||Lp,λ;

(b) ||Mη(f)||Lq,λ ≤C||f||Lp,λ for0< η <(n−λ)/npand1/q= 1/p−nη/(n−λ).

Lemma 2.3. Let DαAj ∈ BM O(Rn) for all α with |α| = mj and j = 1,· · ·, l.

Suppose thatTA is the multilinear operator as Definition 1.1 such thatT is bounded fromLu(Rn)toLv(Rn)for any1< u < n/δand1/v= 1/u−δ/n. Then there exists a constant C >0 such that for anyf ∈Lp,λ(Rn),1< r < p < n/δ,0< λ < n−pδ anda.e. x˜∈Rn,

(TA(f))#(˜x)≤C

l

Y

j=1

 X

j|=mj

||DαjAj||BM O

Mδ,r(f)(˜x).

Proof. It suffices to prove for some constant C0 and f ∈ Lp,λ(Rn), the following inequality holds:

1

|Q|

Z

Q

|TA(f)(x)−C0|dx≤C

l

Y

j=1

 X

j|=mj

||DαjAj||BM O

Mδ,r(f)(˜x).

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Without loss of generality, we may assumel= 2. Fix a cubeQ=Q(x0, d) and ˜x∈Q.

Let ˜Q = 5√

nQ and ˜Aj(x) = Aj(x)− P

|α|=mj

1

α!(DαAj)Q˜xα, then Rmj(Aj;x, y) = Rmj( ˜Aj;x, y) andDαj =DαAj−(DαAj)Q˜ for|α|=mj. We write, forf1=f χQ˜

andf2=f χRn\Q˜,

TA(f)(x) = Z

Rn

Q2

j=1Rmj+1( ˜Aj;x, y)

|x−y|m K(x, y)f(y)dy

= Z

Rn

Q2

j=1Rmj+1( ˜Aj;x, y)

|x−y|m K(x, y)f2(y)dy +

Z

Rn

Q2

j=1Rmj( ˜Aj;x, y)

|x−y|m K(x, y)f1(y)dy

− X

1|=m1

1 α1!

Z

Rn

Rm2( ˜A2;x, y)(x−y)α1

|x−y|m Dα11(y)K(x, y)f1(y)dy

− X

2|=m2

1 α2!

Z

Rn

Rm1( ˜A1;x, y)(x−y)α2

|x−y|m Dα22(y)K(x, y)f1(y)dy

+ X

1|=m1,2|=m2

1 α12!

Z

Rn

(x−y)α12Dα11(y)Dα22(y)

|x−y|m K(x, y)f1(y)dy, then

TA(f)(x)−TA˜(f2)(x0)

≤ Z

Rn

Q2

j=1Rmj( ˜Aj;x, y)

|x−y|m K(x, y)f1(y)dy

+

X

1|=m1

1 α1!

Z

Rn

Rm2( ˜A2;x, y)(x−y)α1

|x−y|m Dα11(y)K(x, y)f1(y)dy

+

X

2|=m2

1 α2!

Z

Rn

Rm1( ˜A1;x, y)(x−y)α2

|x−y|m Dα22(y)K(x, y)f1(y)dy

+

X

1|=m1,2|=m2

1 α12!

Z

Rn

(x−y)α12Dα11(y)Dα22(y)

|x−y|m K(x, y)f1(y)dy +

TA˜(f2)(x)−TA˜(f2)(x0)

:=I1(x) +I2(x) +I3(x) +I4(x) +I5(x), thus,

1

|Q|

Z

Q

TA(f)(x)−TA˜(f2)(x0) dx

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≤ 1

|Q|

Z

Q

I1(x)dx+ 1

|Q|

Z

Q

I2(x)dx+ 1

|Q|

Z

Q

I3(x)dx+ 1

|Q|

Z

Q

I4(x)dx + 1

|Q|

Z

Q

I5(x)dx :=I1+I2+I3+I4+I5.

Now, let us estimate I1,I2, I3,I4 andI5, respectively. First, forx∈Qand y∈Q,˜ by Lemma 2.1, we get

Rm( ˜Aj;x, y)≤C|x−y|m X

j|=m

||DαjAj||BM O,

thus, by the (Lr, Lq)-boundedness ofT with 1< r < n/δand 1/q= 1/r−δ/n, we obtain

I1≤C

2

Y

j=1

 X

j|=mj

||DαjAj||BM O

 1

|Q|

Z

Q

|T(f1)(x)|dx

≤C

2

Y

j=1

 X

j|=mj

||DαjAj||BM O

 1

|Q|

Z

Rn

|T(f1)(x)|qdx 1/q

≤C

2

Y

j=1

 X

j|=mj

||DαjAj||BM O

|Q|−1/q Z

Rn

|f1(x)|rdx 1/r

≤C

2

Y

j=1

 X

j|=mj

||DαjAj||BM O

 1

|Q|1−rδ/n Z

Q˜

|f(x)|rdx 1/r

≤C

2

Y

j=1

 X

j|=mj

||DαjAj||BM O

Mδ,r(f)(˜x).

ForI2, denotingr=pqfor 1< p < n/δ,q >1, 1/q+ 1/q0 = 1 and 1/s= 1/p−δ/n, we have, by H¨older’s inequality,

I2≤C X

2|=m2

||Dα2A2||BM O

X

1|=m1

1

|Q|

Z

Q

|T(Dα11f1)(x)|dx

≤C X

2|=m2

||Dα2A2||BM O

X

1|=m1

1

|Q|

Z

Rn

|T(Dα11f1)(x)|sdx 1/s

≤C X

2|=m2

||Dα2A2||BM O

X

1|=m1

|Q|−1/s Z

Rn

|Dα11(x)f1(x)|pdx 1/p

≤C X

2|=m2

||Dα2A2||BM O

X

1|=m1

1

|Q|˜ Z

Q˜

|Dα11(x)|pq0dx 1/pq0

× 1

|Q|˜ 1−rδ/n Z

Q˜

|f(x)|pqdx 1/pq

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≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

Mδ,r(f)(˜x).

ForI3, similar to the proof ofI2, we get

I3≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

Mδ,r(f)(˜x).

Similarly, forI4, denotingr=pq3for 1< p < n/δ,q1, q2, q3>1, 1/q1+1/q2+1/q3= 1 and 1/s= 1/p−δ/n, we obtain

I4≤C X

1|=m1,|α2|=m2

1

|Q|

Z

Q

|T(Dα11Dα22f1)(x)|dx

≤C X

1|=m1,|α2|=m2

1

|Q|

Z

Rn

|T(Dα11Dα22f1)(x)|sdx 1/s

≤C X

1|=m1,|α2|=m2

|Q|−1/s Z

Rn

|Dα11(x)Dα22(x)f1(x)|pdx 1/p

≤C X

1|=m1,|α2|=m2

1

|Q|˜ Z

Q˜

|Dα11(x)|pq1dx

1/pq1 1

|Q|˜ Z

Q˜

|Dα22(x)|pq2dx 1/pq2

× 1

|Q|˜1−rδ/n Z

Q˜

|f(x)|pq3dx 1/pq3

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

Mδ,p(f)(˜x).

ForI5, we write

TA˜(f2)(x)−TA˜(f2)(x0)

= Z

Rn

K(x, y)

|x−y|m− K(x0, y)

|x0−y|m 2

Y

j=1

Rmj( ˜Aj;x, y)f2(y)dy

+ Z

Rn

Rm1( ˜A1;x, y)−Rm1( ˜A1;x0, y)Rm2( ˜A2;x, y)

|x0−y|m K(x0, y)f2(y)dy +

Z

Rn

Rm2( ˜A2;x, y)−Rm2( ˜A2;x0, y)Rm1( ˜A1;x0, y)

|x0−y|m K(x0, y)f2(y)dy

− X

1|=m1

1 α1!

Z

Rn

"

Rm2( ˜A2;x, y)(x−y)α1

|x−y|m K(x, y)

−Rm2( ˜A2;x0, y)(x0−y)α1

|x0−y|m K(x0, y)

#

Dα11(y)f2(y)dy

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− X

2|=m2

1 α2!

Z

Rn

"

Rm1( ˜A1;x, y)(x−y)α2

|x−y|m K(x, y)

−Rm1( ˜A1;x0, y)(x0−y)α2

|x0−y|m K(x0, y)

#

Dα22(y)f2(y)dy

+ X

1|=m1,2|=m2

1 α12!

Z

Rn

(x−y)α12

|x−y|m K(x, y)−(x0−y)α12

|x0−y|m K(x0, y)

×Dα11(y)Dα22(y)f2(y)dy

=I5(1)+I5(2)+I5(3)+I5(4)+I5(5)+I5(6).

By Lemma 2.1 and the following inequality (see [19])

|bQ1−bQ2| ≤Clog(|Q2|/|Q1|)||b||BM O for Q1⊂Q2, we know that, forx∈Qandy∈2k+1Q˜\2kQ,˜

|Rm( ˜A;x, y)| ≤C|x−y|m X

|α|=m

(||DαA||BM O+|(DαA)Q(x,y)˜ −(DαA)Q˜|)

≤Ck|x−y|m X

|α|=m

||DαA||BM O.

Note that|x−y| ∼ |x0−y|forx∈Qandy∈Rn\Q, we obtain, by the conditions˜ onK,

|I5(1)| ≤C Z

Rn

|x−x0|

|x0−y|m+n+1−δ + |x−x0|ε

|x0−y|m+n+ε−δ 2

Y

j=1

|Rmj( ˜Aj;x, y)||f2(y)|dy

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

X

k=0

Z

2k+1Q\2˜ kQ˜

k2

|x−x0|

|x0−y|n+1−δ + |x−x0|ε

|x0−y|n+ε−δ

|f(y)|dy

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

X

k=1

k2(2−k+ 2−εk) 1

|2kQ|˜ 1−δ/n Z

2kQ˜

|f(y)|dy

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

X

k=1

k2(2−k+ 2−εk)

×

1

|2kQ|˜1−rδ/n Z

2kQ˜

|f(y)|dy 1/r

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

Mδ,r(f)(˜x).

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ForI5(2), by the formula (see [5]):

Rm( ˜A;x, y)−Rm( ˜A;x0, y) = X

|β|<m

1

β!Rm−|β|(DβA;˜ x, x0)(x−y)β and Lemma 2.1, we have

|Rm( ˜A;x, y)−Rm( ˜A;x0, y)| ≤C X

|β|<m

X

|α|=m

|x−x0|m−|β||x−y||β|||DαA||BM O,

thus

|I5(2)| ≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

X

k=0

Z

2k+1Q\2˜ kQ˜

k |x−x0|

|x0−y|n+1−δ|f(y)|dy

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

Mδ,r(f)(˜x).

Similarly,

|I5(3)| ≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

Mδ,r(f)(˜x).

ForI5(4), we get

|I5(4)| ≤C X

1|=m1

Z

Rn

(x−y)α1K(x, y)

|x−y|m −(x0−y)α1K(x0, y)

|x0−y|m

× |Rm2( ˜A2;x, y)||Dα11(y)||f2(y)|dy

+C X

1|=m1

Z

Rn

|Rm2( ˜A2;x, y)−Rm2( ˜A2;x0, y)||(x0−y)α1K(x0, y)|

|x0−y|m

× |Dα11(y)||f2(y)|dy

≤C X

|α|=m2

||DαA2||BM O X

1|=m1

X

k=1

k(2−k+ 2−εk)

× 1

|2kQ|˜ Z

2kQ˜

|Dα11(y)|r0dy 1/r0

1

|2kQ|˜1−rδ/n Z

2kQ˜

|f(y)|rdy 1/r

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

Mδ,r(f)(˜x).

Similarly,

|I5(5)| ≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

Mδ,r(f)(˜x).

(9)

ForI5(6), takingq1, q2>1 such that 1/r+ 1/q1+ 1/q2= 1, then

|I5(6)| ≤C X

1|=m1,|α2|=m2

Z

Rn

(x−y)α12K(x, y)

|x−y|m −(x0−y)α12K(x0, y)

|x0−y|m

× |Dα11(y)||Dα22(y)||f2(y)|dy

≤C X

1|=m1,|α2|=m2

X

k=1

k(2−k+ 2−εk)

1

|2kQ|˜ 1−rδ/n Z

2kQ˜

|f(y)|rdy 1/r

× 1

|2kQ|˜ Z

2kQ˜

|Dα11(y)|q1dy

1/q1 1

|2kQ|˜ Z

2kQ˜

|Dα22(y)|q2dy 1/q2

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

Mδ,r(f)(˜x).

Thus

|I5| ≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

Mδ,r(f)(˜x).

This completes the proof of Lemma 2.3.

3. Proof of theorem

Proof of Theorem 1.1. Taking 1< r <min(p,(n−λ)/pδ) in Lemma 2.3, by Lemma 2.2, we obtain

||TA(f)||Lq,λ ≤C||M(TA(f))||Lq,λ

≤C||(TA(f))#||Lq,λ

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

||Mδ,r(f)||Lq,λ

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

||(Mrδ/n(|f|r))1/r||Lq,λ

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

||Mrδ/n(|f|r)||1/rLq/r,λ

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

|||f|r||1/rLp/r,λ

≤C

2

Y

j=1

 X

|α|=mj

||DαAj||BM O

||f||Lp,λ.

This completes the proof of the theorem.

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4. Applications

In this section, we shall apply the results of the paper to the Calder´on-Zygmund singular integral operator and fractional integral operator.

Application 1.Calder´on-Zygmund singular integral operator.

LetT be the Calder´on-Zygmund operator (see [19, 20]), then the kernelK ofT satisfies the conditions of Definition 1.1 with δ= 0, and T is bounded on Lp(Rn) for any 1< p <∞. The multilinear operator related toT is defined by

TA(f)(x) = Z

Rn

Ql

j=1Rmj+1(Aj;x, y)

|x−y|m K(x, y)f(y)dy.

Then Theorem 1.1 and Lemma 2.3 hold forTAwithδ= 0.

Application 2.Fractional integral operator with rough kernel.

For 0< δ < n, letTδ be the fractional integral operator with rough kernel defined by (see [1, 8])

Tδf(x) = Z

Rn

Ω(x−y)

|x−y|n−δf(y)dy.

The multilinear operator related toTδ is defined by TδA(f)(x) =

Z

Rn

Ql

j=1Rmj+1(Aj;x, y)

|x−y|m+n−δ Ω(x−y)f(y)dy, where Ω is homogeneous of degree zero on Rn, R

Sn−1Ω(x0)dσ(x0) = 0 and Ω ∈ Lipε(Sn−1) for some 0< ε≤1, that is there exists a constantM >0 such that for any x, y ∈Sn−1, |Ω(x)−Ω(y)| ≤ M|x−y|ε. In this time, the kernel satisfies the conditions of Definition 1.1. When Ω≡1,Tδ is the Riesz potentials. Then Theorem 1.1 and Lemma 2.3 hold forTδA.

Acknowledgement.The author would like to express his gratitude to the referees for the comments and suggestions.

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