Maximal singular integrals on product
homogeneous groups
著者 Ding Yong, Sato Shuichi
journal or publication title Studia Mathematica volume 222 number 1 page range 41-49 year 2014-01-01 URL http://hdl.handle.net/2297/43048 doi: 10.4064/sm222-1-4
MAXIMAL SINGULAR INTEGRALS ON PRODUCT
HOMOGENEOUS GROUPS
YONG DING AND SHUICHI SATO
Abstract. We prove L
p boundedness for
p2 (11) of maximal
sin-gular integral operators with rough kernels on product homogeneous groups under a sharp integrability condition of the kernels.
1. Introduction
LetRd,d 2, be thed-dimensional Euclidean space. We assume thatRd is also equipped with a homogeneous group structure, where multiplication is given by a polynomial mapping the underlying manifold isRd itself. We also write Rd =H. Thus, H is associated with a dilation group fAtgt>
0 of automorphisms of the group structure such that
Atx= (ta1x 1t a2x 2:::t adxd) x= (x 1:::xd) 2H where real numbersa1:::ad satisfy 0< a1
a 2
ad (see 10], 21], 18], 6] and 11, Section 2 of Chapter 1]). So, we have, for each t >0,
At(xy) = (Atx)(Aty) xy2H:
Consequently, H is endowed with both the Euclidean structure and a homogeneous nilpotent Lie group structure. The group law ofH is given by a polynomial mapping which conforms to the Campbell-Hausdor formula in a corresponding Lie algebra via an exponential map and the action of the automorphism family fAtg. We note that the identity is the origin 0 and
x;1 =
;x furthermore, Lebesgue measure is bi-invariant Haar measure. Let us recall a norm function r(x) associated with fAtg. The function
r(x), which is non-negative and vanishes only at the origin, satises that
r(Atx) = tr(x) for t > 0 and x 2 Rd. We assume that r(x) is even, continuous on Rd and smooth in Rd nf0g, and also that the unit sphere d =fx 2Rd : r(x) = 1g dened by the norm function coincides with the unit sphere Sd;1 =
fx 2 Rd : jxj = 1g, where jxj denotes the Euclidean norm. Let =a1+
+ad (the homogeneous dimension of H). We shall
2010 Mathematics Subject Classication. Primary 42B20.
Key words and phrases. Multiple singular integrals, homogeneous groups, maximal
use the formula: Z Rd f(x)dx= Z 1 0 Z d f(At)t;1dS d()dt dSd =! d d
where ! is a strictly positive C1 function on
d and d d is the Lebesgue
surface measure on d. See 18, 6] and also 3, 10, 11, 14, 19, 20, 21] for
more details and related results.
We consider a function which is locally integrable in Rd n f0g and homogeneous of degree 0 with respect to the dilation group fAtg, that is, (Atx) = (x) forx6= 0, t >0. We assume the cancellation property:
Z d ()dSd() = 0: (1.1) Convolution onH is dened by f g(x) = Z H f(y)g(y;1x)dy: Let Tf(x) = p:v:f K(x) = p:v: Z H f(y)K(y;1x )dy (1.2)
for appropriate functions f, where K(x) = (x0)r(x);, x0 = A
r(x)
;1x for
x6= 0. We also dene the maximal singular integral operator
Tf(x) = sup >0 Z r(y)> f(xy;1 )K(y)dy : (1.3)
Then the following results are known.
Theorem A
( 21]).
If 2 LlogL(d) with (1.1) and Tf is as in (1.2), then T is bounded on Lp(H) forall p2(11).Theorem B
( 18]).
Let T fbe dened as in (1.3) with 2 LlogL(d) satisfying (1.1). Let p2(11). Then theoperator T
is bounded on Lp(H). We refer to 4, 12, 13, 14, 15, 16] for relevant results.
Part of a theory of Duoandikoetxea and Rubio de Francia 8] for singular integrals on the Euclidean spaces has been generalized to the case of ho-mogeneous groups by 18]. The arguments of 18] replace Fourier transform estimates by a variant of Tao'sL2 estimates via convolution (see 21]). As a result, 18] proved Theorem B and some weighted estimates, and also gave another proof of Theorem A.
Also, it has been shown in 6] that the theory of 18] extends to the case of product homogeneous groups to treat multiple singular integrals. Let Rn = Rn
1 Rn
2 be a product homogeneous group with Rn 1 = H 1, Rn 2 = H 2, where n = n1 +n2 and H 1, H
dilations At , At and norm functions r1r2, respectively. We consider a function in L1( n1 n 2) which satises Z n 1
(uv)dSn1(u) = 0 for allv 2n 2, (1.4) Z n 2
(uv)dSn2(v) = 0 for all u 2n 1. (1.5) Dene K(uv) =r1(u) ; 1r 2(v) ; 2(u 0v0 ) u0 =A(1) r1 (u) ;1uv 0 =A(2) r2 (v) ;1v where1 and2 are the homogeneous dimensions of
H 1 and
H
2, respectively. We consider the multiple singular integral
Tf(xy) = p:v:f K(xy) = p:v: Z H 1 H 2 f(xu;1yv;1 )K(uv)dudv: (1.6)
Then the following result is proved in 6].
Theorem C.
Let T be dened as in (1.6) with in L(logL) 2(n1 n
2) satisfying (1.4) and (1.5). Let 1< p <1. The operator T is then bounded on Lp(H
1 H
2) .
We can nd in 2] the optimality of theL(logL)2 integrability condition for multiple singular integrals with Euclidean convolution.
Let us recall that the maximal singular integral is dened by
Tf(xy) = sup 1> 0 2> 0 Z r1 (u)> 1 r2 (v)> 2 f(xu;1yv;1 )K(uv)dudv : (1.7)
In this note we shall prove the following.
Theorem 1.
LetTbedenedasin (1.7).SupposethatisinL(logL) 2(
n1 n2)
and satises (1.4), (1.5). Then T is bounded on Lp(H 1 H 2) for all p2(11).
Previous works concerning singular integrals on product of Euclidean spaces can be found in 1, 2, 7, 9]. Theorem 1 is an analogue of a result of 2] for multiple singular integrals on product homogeneous groups.
Similarly to the proof of Theorem C in 6], we use extrapolation argu-ments in proving Theorem 1 by applying the following result.
Theorem 2.
Let1< s2.Supposethatisin Ls(n 1n 2)
andsatises (1.4), (1.5). Then, for 1< p <1 we have
kT f
kp Cp(s;1) ;2
with a constant Cp independent of s and .
Let be as in Theorem 1. Then there exist a sequencefkgof functions inL1(
n1 n
2) and a sequence
fckgof non-negative real numbers such that each ksatises (1.4) and (1.5), supk1
kkk 1+1=k 1, P 1 k=1k 2c k<1, and = 1 X k=1 ckk:
Theorem 1 easily follows from this decomposition of and Theorem 2 (see 17, 15, 16]).
In Section 2 we recall some preliminary results from 6]. We shall prove Theorem 2 in Section 3 by using results of 6, 5, 18] similar arguments, via Fourier transform estimates, for singular integrals with Euclidean convolu-tion can be found in 1, 2].
2. Preliminaries
Let 2 and let j 2C 1 0 (
R), j 2Z (the set of integers), be such that supp( j)ft2R : j t j+2 g j 0 (log2)X j2Z j(t) = 1 for t >0 furthermore, j(d=dt) m j(t) jcmjtj ;m for m= 012:::
wherecm is a constant independent of we note that this is possible since
2.
Suppose thatF belongs toL1( H
1 H
2) with support inD0, whereD0 =
D(1) 0 D (2) 0 , D(1) 0 = fx2H 1 : 1 r 1(x) 2g D (2) 0 = fy 2H 2 : 1 r 2(y) 2g: Let (1) s f(x) = s; 1f((A (1) s );1x), (2) t g(y) = t; 2g((A (2) t );1y). Dene st = (1) s (2) t and let SjkF(xy) = Z 1 0 Z 1 0 j(s) k(t) stF(xy)dss dtt : (2.1) Then P jk2ZSjkK 0 =K, where K0(xy) = K(xy) (xy)2D 0, 0 otherwise. (2.2) Fors 1, let Ls(D
0) denote the subspace of L
s(H 1
H
2) consisting of functionsF with support in D0. Dene
MFf(xy) = supjk
2Z
for F 2Ls(D
0). The following result is Lemma 8 of 6].
Lemma 1.
Let p > 1. Suppose that s 2(12]= 2s 0 , s 0 = s=(s ;1), and F 2Ls(D 0) . Then, kMFfkp Cp(s;1) ;2 kFkskfkp with a positive constant Cp independent of s and F.Also, we need another result of 6].
Lemma 2.
Let 1 < s 2. Suppose that belongs to Ls(n 1n 2)
and satises (1.4) and (1.5). Let
Rf(xy) = sup `m2Z 1 X j=` 1 X k=m fSjkK 0(xy) where K 0 is dened by (2.2) and SjkK 0 is as in (2.1) with K 0 in place of F. Let = 2s 0
. Then, for p 2 (11) there exists a positive constant Cp independent of s2(12] and 2Ls such that
kRfkp CpA(s)kfkp where A(s) = (s;1) ;2 kks. This is Proposition 2 of 6]. 3. Proof of Theorem 2. We rst note that SjkK0(xy) =r1(x) ; 1r 2(y) ; 2(x 0y0 ) Z 1 1=2 j(sr1(x)) ds s Z 1 1=2 k(tr2(y)) dt t where x0 =A (1) r1(x) ;1x, y 0 =A (2) r2(y)
;1y. From this we easily see that supp(SjkK0) j r 1(x) 2 j+2 k r 2(y) 2 k+2 :
Thus, if`m2Zare determined by the conditions` +2
< `
+3, m+2
< m+3 and if f is a compactly supported smooth function, we have Z
r1 (u)>
r2 (v)>
f(xu;1yv;1)K(uv)dudv (3.1) = X j` km Z r1 (u)> r2 (v)> f(xu;1yv;1)S jkK0(uv)dudv
where Af(xy) = X j>`+3 k>m+3 Z H1H2 f(xu;1yv;1)S jkK0(uv)dudv = X j>`+3 k>m+3 fSjkK 0(xy) Bf(xy) = X `+3j` k>m+3 Z fr 1 (u)>gH 2 f(xu;1yv;1)S jkK0(uv)dudv Cf(xy) = X j>`+3 m+3km Z H 1 fr 2 (v)>g f(xu;1yv;1)S jkK0(uv)dudv Df(xy) = X `+3j` m+3km Z r1 (u)> r2 (v)> f(xu;1yv;1)S jkK0(uv)dudv: Let Af(xy) = sup >0 jAf(xy)j B f(xy) = sup >0 jBf(xy)j Cf(xy) = sup >0 jCf(xy)j D f(xy) = sup >0 jDf(xy)j: Then, (3.1) implies Tf(xy) A
f(xy) +Bf(xy) +Cf(xy) +Df(xy): (3.2) Let= 2s0 . Since Af Rf, by Lemma 2 we have kA f kp CpA(s)kfkp: (3.3) Also, sinceDf CMK 0( jfj), Lemma 1 implies kD f kp CkMK 0( jfj)kpCpA(s)kfkp: (3.4)
To estimate Bf, we note that jBf(xy)j X `+3j` Z `r 1 (u)2 ` +5 X k>m+3 Z H 2 f(xu;1yv;1)S jkK0(uv)dv du:
By changing variables with respect to u, we see that the right hand side is equal to X `+3j` Z 2 ` +5 ` Z n 1 X k>m+3 Fk(xys) j(s)dss dSn1() where Fk(xys) = Z H f(x(A(1) s );1yv;1 )(v0 )r2(v) ; 2 k(r2(v))dv
k(t) = 1=2 k(rt)dr=r. Thus, since 0 j(s)1, we have jBf(xy)jC Z 2 ` +5 ` Z n 1 X k>m+3 Fk(xys) ds s dSn1(): We write K(2) (v) =K0(v) S (2) k K(2) (v) = Z 1 0 k(t) (2) t K(2) (v)dtt : Then Fk(xys) =f(x(A(1) s );1 ) (2)S (2) k K(2) (y) where
(2) denotes the convolution on H 2. Consequently, (3.5) jBf(xy)j C Z 2 ` +5 ` Z n 1 X k>m+3 f(x(A(1) s );1 ) (2)S (2) k K(2) (y) ds s dSn1(): Let R(2) g(y) = supm 2Z X k>mg (2)S (2) k K(2) (y) forgonH
2. We writefx(y) =f(xy) when consideringf(xy) as a function onH
2 xing x similarly, we write fy(x) =f(xy). Dene
F(xy) =R(2)
fx(y):
Then, using (3.5), we have jB f(xy) jCsup `2Z Z 2 ` +5 ` Z n 1 F(x(A(1) s );1y)ds s dSn1() (3.6) Clog Z n 1 M(1) Fy(x)dSn1() where M(1) h(x) = supt> 0 1 t Z t 0 jh(x(A (1) s );1) jds for h on H
1. The last inequality of (3.6) can be seen as follows. Take a positive integer d such that 2d 2
5 <2d+1. Then Z 2 ` +5 ` jh(x(A (1) s );1 )j ds s d X i=0 Z 2i +1 ` 2i ` jh(x(A (1) s );1 )j ds s d X i=0 2M(1) h(x) = 2(d+ 1)M(1) h(x) C(log)M (1) h(x)
By M. Christ 5]M(1)
is bounded onLp,p >1, with a bound independent
of . Thus, using (3.6) and the Minkowski inequality we have kB f kp C(log) Z n 1 kF kpdSn 1(): (3.7) By Lemma 9 of 18] with = 2s0 , we have kR (2) gkp Cp(log) Z n 2 j(!)j sdSn 2(!) ! 1=s kgkp: Thus kFxkp Cp(log) Z n 2 j(!)jsdSn 2(!) ! 1=s kfxkp: Using this in (3.7) and notingkFkp = (
R kFxkppdx) 1=p, we have kB f kp Cp(log) 2 Z n 1 Z n 2 j(!)j sdSn 2(!) ! 1=s dSn1() kfkp (3.8) Cp(log) 2 kkskfkp
where the last inequality follows from H older's inequality. Similarly, we have kC f kp Cp(log) 2 kkskfkp: (3.9)
Combining (3.2), (3.3), (3.4), (3.8) and (3.9), we get the conclusion of The-orem 2.
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School of Mathematical Sciences, Laboratory of Mathematics and Com-plex Systems (BNU), Ministry of Education, Beijing Normal University, Beijing, 100875 P. R. of China
E-mail address: [email protected]
Department of Mathematics, Faculty of Education, Kanazawa Univer-sity, Kanazawa 920-1192, Japan
