A note on the multilinear oscillatory singular integral operators
Zefu Chu, Guoen Hu and Zhibo Lu
(Received December 17, 1999) (Revised November 13, 2000)
Abstract. In this paper, we consider theLp Rn boundedness for a class of multi- linear oscillatory singular integral operators with polynomial phases. We show that if the polynomial phases are non-trivial and the homogeneous kernels satisfy a certain minimum size condition, then the Lp Rn boundedness for the multilinear oscillatory singular integral operators can be deduced from the Lp Rn boundedness for the corresponding local multilinear singular integral operators.
1. Introduction
We will work on Rn nb2. Let P x;y be a real-valued polynomial on RnRn,W xbe homogeneous of degree zero which has a mean value zero on the unit sphere Snÿ1. De®ne the oscillatory singular integral operator
Tf x
RneiP x;yW xÿy
jxÿyjn f ydy:
1
It is well-known that the operators of this type have arisen in the study of Hilbert transforms along curves, singular integrals supported on lower- dimensional varieties and singular Radon transforms, etc. A celebrated result of Ricci and Stein [9] says that ifWALip1 Snÿ1, thenTis bounded onLp Rn for 1<p<y, with a bound depending only on n, p and degP (the total degree of P), not on the coe½cients of the polynomial. Chanillo and Christ [2] showed thatWALip1 Snÿ1is also su½cient forTto be a bounded mapping from L1 to weak L1, and the bound depends only on n and degP. Lu and Zhang [7] improved the result of Ricci and Stein, and proved that if WA 6q>1Lq Snÿ1, then T is bounded on Lp Rn with a bound C n;p;degP for 1<p<y.
In this paper, we will study the multilinear operators de®ned by TA1;...;Akf x
RneiP x;y W xÿy
jxÿyjnm Yk
j1
Rmj1 Aj;x;yf ydy;
2
2000Mathematics Subject Classi®cation. 42B20
Key words and phrases. multilinear operator, oscillatory singular integral, BMO.
The research was supported by the NSF of China (19701039) and the NSF of Henan Province.
wherekandmj j1;. . .;kare positive integers,mPk
j1mj,Aj j1;. . .;k
has derivatives of ordermj in BMO Rn,Rmj1 Aj;x;ydenotes the mj1-th Taylor series remainder of Aj at x about y, that is,
Rm1 Aj;x;y Aj x ÿ X
jajamj
1
a!DaAj y xÿya:
Operators of this type have been studied in [3], [4], [6] and many other works.
It is easy to see that the operator TA1;...;Ak is closely related to the oscillatory singular integral operator de®ned by (1) and the multilinear singular integral operator de®ned by
T~A1;...;Akf x
Rn
W xÿy
jxÿyjnm Yk
j1
Rmj1 Aj;;x;yf ydy:
3
Using good-l-inequality techniques, Cohen and Gosselin [5] showed that if W satis®es a certain vanishing moment and WALip1 Snÿ1, then for 1<p<y,
kT~A1;A2fkpaY2
j1
X
jajmj
kDaAjkBMO Rn 0
@
1 Akfkp:
In [3], Chen, Hu and Lu considered the Lp Rn boundedness for the operator TA1;A2 and proved that if WA6q>1Lq Snÿ1, and the polynomial P x;y is non-trivial, then the Lp Rn boundedness for TA1;A2 can be obtained from the Lp Rn boundedness for the local multilinear singular integral operator
SA1;A2f x
jxÿyja1
W xÿy
jxÿyjnm1m2 Y2
j1
Rmj1 Aj;;x;yf ydy;
(see [2, Theorem 2]). The purpose of this paper is to show that if WA L logLk1 Snÿ1, and P is non-trivial, then the Lp Rn boundedness for
TA1;...;Ak can be obtained from the Lp Rn boundedness for the local version of
the operator T~A1;...;Ak. Our main result in this paper can be stated as follows.
Theorem1. Let 1<p<y, k and mj j1;2;. . .;kbe positive integers, mPk
j1mj, Aj j1;2;. . .;kbe functions on Rn whose derivatives of order mj
are inBMO Rn. Suppose thatWis homogeneous of degree zero and belongs to the space L logLk1 Snÿ1, that is,
Snÿ1jW x0jlogk1 2 jW x0jdx0<y;
and the operator
SA1;...;Akf x
jxÿyja1
W xÿy
jxÿyjnm Yk
j1
Rmj1 Aj;x;yf ydy 4
is bounded on Lp Rn. Then for any real-valued non-trivial polynomial P x;y, the operator TAde®ned by(2)is also bounded on Lp Rn, with a bound depending on n, p, mj j1;. . .;k, Qk
j1 P
jajjmjkDajAjkBMO Rn and degP, not on the coe½cients of P.
2. Proof of Theorem 1
We begin with some preliminary lemmas.
Lemma1 (see [5]). Let b xbe a function onRn with derivatives of order m in Lq Rn for some n<qay. Then
jRm b;x;yjaCm;njxÿyjm X
jajm
1 jI x;~ yj
I x;~ yjDab zjqdz
!1=q
; where I x;~ y is the cube centered at x with diameter 5
pn
jxÿyj.
Lemma 2. Let 1<p<y, k and mj j1;2;. . .;k be positive integers, mPk
j1mj, Aj j1;2;. . .;kbe functions on Rn whose derivatives of order mj
are inBMO Rn. Suppose thatW~ is homogeneous of degree zero and belongs to the space Ly Snÿ1. Set
lW;k~ inf l>0:kWk~ 1
l logk 2kWk~ y
l
! a1
( )
:
Then for any r>0, the operator
UA1;...;Ak;rf x rÿnÿm
r=2<jxÿyjarjW x~ ÿyjYk
j1
jRmj1 Aj;x;yjjf yjdy 5
is bounded on Lp Rnwith a bound C n;m;plW;k~
Qk
j1 P
jajjmkDajAjkBMO Rn.
Proof. Note that for each t>0, ltW;~k inf l>0:ktWk~ 1
l logk 2ktWk~ y l
! a1
( )
inf tl~: ~l>0;ktWk~ 1
tl~ logk 2ktWk~ y
tl~
! a1
( )
tlW;k~ :
Thus we may assume that lW;k~ 1=2. Therefore, kWk~ 1logk 2 kWk~ ya1:
De®ne the operator E by Eh x
jxÿyja1jW x~ ÿyjh ydy:
Denote by E the adjoint operator of E, that is, Eh x
jxÿyja1jW y~ ÿxjh ydy:
Let b1;b2;. . .;bkABMO Rnand Q be a cube with side length 1. Denote by mQ bj the mean value of bj on Q. We claim that for 1<p<y, supph H10nQ and non-negative integer lak,
QjEh xjpYl
j1
jbj x ÿmQ bjjpdx 6
aClog ÿklp 2 kWk~ yYk
j1
kbjkBMO Rp nkhkpp; with the interpretation that when l0, Ql
j1jbj x ÿmQ bjj11. To prove (6), we can assume that khkp 1. Choose 1<rj <y such that Pk
j11=rj
1. By the well-known John-Nirenberg inequality, there is a positive constant Cj C p;rj;n such that
Qjbj x ÿmQ bjj2prjdx
1= 2rj
aCjkbjkBMO Rp n:
We may also assume thatkbjkBMO Rp n1=Cj for all 1ajak. We shall carry out our argument by induction onl. If l0, the Young inequality gives that
QjEh yjpdyaCkWk~ 1pkhkppaClogÿkp 2 kWk~ y:
Now let dakÿ1 be a non-negative integer and assume that the estimate (6) holds forld. We will show that (6) holds forld1. Observe thatF t
tlogp 2t is a Young function and its complementary Young function is C tAexpt1=p. By the general HoÈlder inequality, it follows that
QjEh xjpYd1
j1
jbj x ÿmQ bjjpdx
aCinf l>0:
Q
jEh xjp
l logp 2jEh xjp l
Yd
j1
jbj x ÿmQ bjjpdxa1
( )
inf l>0:
Qexp jbl1 x ÿmQ bl1j l1=p
Yd
j1
jbj x ÿmQ bjjpdxa2
( )
;
(see [1] or [8]). Applying the Young inequality again, we have kEhkyakWk~ ykhk1aCkWk~ ykhkpaCkWk~ y. Our induction assumption now gives that
QjEh xjplogp 2jEh xjp l
Yd
j1
jbj x ÿmQ bjjpdx
aClogp 2CkWk~ yp l
!
log ÿkdp 2 kWk~ y:
Set l0log ÿkd1p 2 kWk~ y. An easy computation then leads to that
QjEh xjplogp 2jEh xjp l0
Yd
j1
jbj x ÿmQ bjjpdxaCl0:
On the other hand, by the HoÈlder inequality,
Qexp jbl1 x ÿmQ bl1j l1=p
Yd
j1
jbj x ÿmQ bjjpdx
a
Qexp 2jbl1 xÿmQ bl1j l1=p
dx
1=2Yd
j1
Qjbj xÿmQ bjj2prjdx
1= 2rj
a
Qexp 2jbl1 x ÿmQ bl1j l1=p
dx
1=2
;
which together with the John-Nirenberg inequality implies that
inf l>0:
Qexp jbl1 x ÿmQ bl1j l1=p
Yd
j1
jbj x ÿmQ bjjpdxa2
( )
aC;
Therefore,
QjEh xjpYd1
j1
jbj x ÿmQ bjjpdxaClog ÿkd1p 2 kWk~ y:
We can now prove our Lemma 2. By dilation-invariance, it su½ces to consider the case r1. Write Rn6jIj, where each Ij is a cube having side length 1 and the cubes have disjoint interiors. Let wj be the characteristic function of Ij. Set fj fwj. Then
f x X
j
fj x; a:e: xARn:
Since the support of UA1;...;Ak;1fj is contained in a ®xed multiple of Ij, the supports of various terms fUA1;...;Ak;1fjg have bounded overlaps, and so we have
kUA1;...;Ak;1fkppaCX
j
kUA1;...;Ak;1fjkpp:
Thus we may assume that supp fHI for some cube Iwith side length 1. Set Aej y Aj y ÿ X
jajjmj
1
aj!mI DajAjya: A straightforward computation shows that for x;yARn,
Rmj1 Aj;x;y Rmj1 Aej;x;y.
Choose n<q<y Lemma 1 now tells us that jRmj Aej;x;yj
aCjxÿyjmj X
jajjmj
1 jI~ x;yj
I x;~ yjDajAj z ÿmI DajAjjqdz
!1=q
aCjxÿyjmj X
jajjmj
1 jI~ x;yj
I x;~ yjDajAj z ÿmI x;~ y DajAjjqdz
!1=q
Cjxÿyjmj X
jajjmj
jmI DajAj ÿmI x;y~ DajAjj
aCjxÿyjmj X
jajjmj
kDajAjkBMO Rn jmI DajAj ÿmI x;y~ DajAjj:
Note that if yAI and jxÿyja1, then I~ x;yH100nI. This in turn implies that for yAI and 1=2ajxÿyja1,
jmI DajAj ÿmI x;y~ DajAjjaCkDajAjkBMO Rn. Thus in this case, we have
jRmj Aej;x;yjaCjxÿyjmj X
jajjmj
kDajAjkBMO RnaC X
jajjmj
kDajAjkBMO Rn: Let
f y Yk
j1
X
jajjmj
kDajAjkBMO Rn jDajAj y ÿmI DajAjj
0
@
1 A: We can write
UA1;...;Ak;1f xaE jffj x:
A standard duality arguement and the HoÈlder inequality then show that
kUA1;...;Ak;1fkpa sup
supphH10nI;khkp0a1
E jffj xh xdx
sup
supphH10nI;khkp0a1
jEh yf yf yjdy aCkfkp sup
supphH10nI;khkp0a1kfEhkp0;
where p0 is the dual exponent of p, i.e. p0p= pÿ1. Invoking the estimate (6) for 0alak, we ®nally obtain
kUA1;...;Ak;1fkpaCYk
j1
X
jajjmj
kDajAjkBMO Rn 0
@
1 Akfkp:
This completes the proof of Lemma 2.
Proof ofTheorem1. Without loss of generality, we may assume that for
1ajak, X
jajjmj
kDajAjkBMO Rn1:
Letk0 be a positive integer and P x;ybe a real-valued non-trivial polynomial having degree k0 in x and degree l0 in y. Write
P x;y X
jmjk0;jnjl0
am;nxmynR x;y;
where R x;y is a real-valued polynomial which has degree less k0 in x. By dilation-invariance, we may assume thatP
jmjk0;jnjl0jamnj1. SpliteTA1;...;Ak as TA1;...;Akf x
jxÿyja1eiP x;y W xÿy
jxÿyjnm Yk
u1
Rmu1 Au;x;yf ydy
Xy
j1
2jÿ1<jxÿyja2jeiP x;y W xÿy
jxÿyjnm Yk
u1
Rmu1 Au;x;yf ydy
TA01;...;Akf x Xy
j1
TAj1;...;Akf x:
We ®rst consider the operator TAj1;A2;...;Ak for jb1. Let E0 fx0ASnÿ1, jW x0ja2g and El fx0ASnÿ1;2l<jW x0ja2l1g for positive integer l.
Let Wl be the restriction of W on El. De®ne the operator TAj1;...;Ak;l by
TAj1;...;Ak;lf x
2jÿ1<jxÿyja2jeiP x;y Wl xÿy
jxÿyjnm Yk
u1
Rmu1 Au;x;yf ydy:
To estimate the Lp Rn boundedness for TAj1;...;Ak;l, we will use the following lemma.
Lemma 3. Let the polynomial P x;y, k, mu and Au u1;. . .;k be the same as above, W~ be homogeneous of degree zero and belong to the space Ly Snÿ1. De®ne the operator
Vjf x
1<jxÿyja2eiP 2jx;2jy W x~ ÿy
jxÿyjnm Yk
u1
Rmu1 Au;x;yf ydy:
Then for1< p<y, there exists positive constants C anddwhich are depending only on n, p and degP such that
kVjfkpaCkWk~ y2ÿdjkfkp.
For the case of k1, this lemma was proved essentially in [3, page 43±46]. For general positive integer k, Lemma 3 can be proved by induction on k. We omit the details.
We now estimateTAj1;...;Ak;l. Note that forbABMO Rn and t>0, bt x
b tx also belongs to the space BMO Rn and kbtkBMO Rn kbkBMO Rn. Thus by dilation-invariance and Lemma 3,
kTAj1;...;Ak;lfkpaC2ÿdj2lkfkp: 7
On the other hand, Lemma 2 states that
kTAj1;...;Ak;lfkpaClWl;kkfkp: 8
Set llk lkkWlk12ÿl. A trivial computation gives that kWlk1
llk logk 2kWlky
llk
!
a kWlk1
lkkWlk1 logk 2kWlky
2ÿl
aC;
which in turn implies
lWl;kaC lkkWlk12ÿl:
9
Our hypothesis on Wnow says thatP
l>0lk1kWlk1<y. Let Nbe a positive integer such that N>2dÿ1. Combining the inequalities (7) and (8) yields that
X
jb1
X
lb0
TAj1;...;Ak;lf
paX
jb1
kTAj1;...;Ak;0fkpX
l>0
X
j>Nl
kTAj1;...;Ak;lfkp
X
l>0
X
1ajaNl
kTAj1;...;Ak;lfkp
aCX
jb1
2ÿdjkfkpCX
l>0
2l X
jbNl
2ÿdjkfkp
CX
l>0
llWl;kkfkpaCkfkp:
Now we turn our attention to the operatorTA01;...;Ak. The estimate for this term follows from the following lemma directly.
Lemma 4. Let 1<p<y, and SA1;...;Ak be de®ned by (4) with WA L logLk Snÿ1. Suppose that SA1;...;Ak is bounded on Lp Rn. Then for any real-valued polynomial P x;~ y, the operator
WA1;...;Akf x
jxÿyja1eiP x;y~ W xÿy
jxÿyjnm Yk
u1
Rmu1 Au;x;yf ydy;
is bounded on Lp Rn with a bound C n;m;p;degP.~
Proof. We follow along the same line as in the proof of Lemma 6 in [3].
We shall carry out the arguement by a double induction on the degree in x and yof the polynomial. Obviously, Lemma 4 holds if the polynomialP x;~ y
depends only onx or only on y. Let u andvbe two positive integers and the
polynomialP x;~ yhave degree u in x andv in y. We assume that Lemma 4 is known for all polynomials which are sums of monomials of degree less than u in x times monomials of any degree in y, together with monomials which are of degree u in x times monomials which are of degree less than v in y.
We can now write
P x;~ y X
jmju;jnjv
bmnxmynP0 x;y;
where P0 x;y satis®es the inductive assumption. We consider the following two cases.
Case I. P
jmju;jnjvjbmnja1. As in the proof of Lemma 2, we may assume that supp fHI for some cube I centered at x0 and having side length 1. By translation-invariance (note that our result is independent of the coe½cients of the polynomial), we may assume that supp fHI0, the cube centered at the origin and having side length 1. Set
P x;y P0 x;y X
jmju;jnjv
bmnymn:
Observe that if jxÿyja1 and yAI0, then
jeiP x;~ yÿeiP x;yjaCjxÿyj.
Thus,
jWA1;...;Akf xja
jxÿyja1eiP x;y W xÿy
jxÿyjnm Yk
j1
Rmj1 Aj;x;yf ydy
C
jxÿyja1
jW xÿyj jxÿyjnmÿ1
Yk
j1
jRmj1 Aj;x;yjjf yjdy
a
jxÿyja1eiP x;y W xÿy
jxÿyjnm Yk
j1
Rmj1 Aj;x;yf ydy
CXy
j0
2ÿjUA1;...;Ak;2ÿjf x;
where UA1;...;Ak;2ÿj is de®ned by (5). Set UAl1;...;Ak;2ÿjf x 2ÿj nm
2ÿjÿ1<jxÿyja2ÿjjWl xÿyjYk
u1
jRmu1 Au;x;yjjf yjdy:
It follows from Lemma 2 and the inequality (9) that Xy
j0
2ÿjkUA1;...;Ak;2ÿjfkpaCXy
j0
2ÿjX
lb0
kUAl1;...;Ak;2ÿjfkp
aCkfkpXy
j0
2ÿjX
lb1
lWl;kkfkpaCkfkp:
This via the induction hypothesis tells us that
kWA1;...;AkfkpaC n;m;p;degPk~ fkp.
Case II. P
jmju;jnjvjbmnj>1. Set J P
jmju;jnjvjbmnj1= uv. Let Q x;y X
jmju;jnjv
bmn
JuvxmynP0 x=J;y=J:
Then P x;~ y Q Jx;J y. De®ne the operator W~A1;...;Akf x
jxÿyjaJeiQ x;y W xÿy
jxÿyjnm Yk
u1
Rmu1 Au;x;yf ydy:
By dilation-invariance, it su½ces to prove that
kW~A1;...;AkfkpaC n;m;p;degPk~ fkp: 10
We splite the operator W~A1;...;Ak as W~A1;...;Akf x
jxÿyja1eiQ x;y W xÿy
jxÿyjnm Yk
u1
Rmu1 Au;x;yf ydy
Xj0
j1
2jÿ1<jxÿyja2jeiQ x;y W xÿy
jxÿyjnm Yk
u1
Rmu1 Au;x;yf ydy
2j0<jxÿyjaJeiQ x;y W xÿy
jxÿyjnm Yk
u1
Rmu1 Au;x;yf ydy
W~If x W~IIf x W~IIIf x;
where j0 is the positive integer such that 2j0 <Ja2j01. The conclusion of Case I applies to W~I, so
kW~IfkaC n;m;p;degPk~ fkp.
By the inequalities (7), (8) and (9) as in the estimate for P
jb1TAj1;A2;...;Ak, we can obtain that
kW~IIfkpaC n;m;p;degPk~ fkp.
On the other hand, it follows from Lemma 2 and the estimate (9) that kW~IIIfkpaC n;m;p;degPk~ fkp.
This leads to the estimate (10), and completes the proof of Lemma 4.
Acknowledgement
The authors would like to thank the referee for some valuable suggestions and corrections.
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Zefu Chu, Guoen Hu and Zhibo Lu Department of Applied Mathematics University of Information Engineering P.O. Box 1001-747, Zhengzhou 450002
People's Republic of China Z. Chu: [email protected] G. Hu: [email protected] Z. Lu: [email protected]