Singular Integrals on Product Homogeneous Groups

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(1)Singular Integrals on Product Homogeneous Groups 著者 journal or publication title volume number page range year URL. Ding Yong, Sato Shuichi Integral Equations and Operator Theory 76 1 55-79 2013-01-01 http://hdl.handle.net/2297/34648 doi: 10.1007/s00020-013-2049-1.

(2) Singular integrals on product homogeneous groups Yong Ding and Shuichi Sato Abstract. We consider singular integral operators with rough kernels on the product space of homogeneous groups. We prove Lp boundedness of them for p 2 (1 1) under a sharp integrability condition of the kernels. Mathematics Subject Classication (2010). Primary 42B20. Keywords. Multiple singular integrals, homogeneous groups.. 1. Introduction. Let Rd be the d-dimensional Euclidean space, where d 2. We assume that Rd is equipped with multiplication given by a polynomial mapping which makes Rd a homogeneous group. This requires the existence of a dilation family fAt gt>0 on Rd such that each At is an automorphism of the group structure, where At is of the form At x = (ta1 x1 ta2 x2 : : : tad xd ) x = (x1 : : : xd ) with some real numbers a1 : : : ad satisfying 0 < a1 a2 ad (see. 12], 25] and 13, Section 2 of Chapter 1]). We also write Rd = H . Therefore, in addition to the Euclidean structure, H is equipped with a homogeneous nilpotent Lie group structure, where Lebesgue measure is bi-invariant Haar measure, the identity is the origin 0 and x;1 = ;x. We note that multiplication xy, x y 2 H , satises (1) At (xy) = (At x)(At y), x y 2 H , t > 0 (2) (ux)(vx) = ux + vx, x 2 H , u v 2 R (3) if z = xy, zk = Pk (x y), then P1 (x y) = x1 + y1 and Pk (x y) = xk + yk + Rk (x y) for k 2, where Rk (x y) is a polynomial depending only on x1 : : : xk;1 y1 : : : yk;1 . We have a norm function r(x) satisfying r(At x) = tr(x) for t > 0 and x 2 Rd . We may assume the following: The rst author is supported by NSFC (No.10931001)..

(3) 2. Yong Ding and Shuichi Sato. (4) the function r is continuous on Rd and smooth in Rd n f0g (5) r(x + y) B1 (r(x) + r(y)), r(xy) B2 (r(x) + r(y)) for some constants B1 B2 1 (6) r(x;1 ) = r(x) and if we denote by jxj the Euclidean norm for x 2 Rd , then c1 jxj1 r(x) c2 jxj2 if r(x) 1 c3 jxj1 r(x) c4 jxj2 if r(x) 1 for some positive constants c1 c2 c3 c4 1 2 1 and 2 (7) if we dene d = fx 2 Rd : r(x) = 1g, d coincides with S d;1 = fx 2 Rd : jxj = 1g (8) if = a1 + + ad (the homogeneous dimension of H ), then dx = t ;1 dSd dt, that is,. Z. f (x) dx = d. R. Z 1Z. d. 0. f (At )t ;1 dSd( ) dt. with dSd = ! d d , where ! is a strictly positive C 1 function on d and d d is the Lebesgue surface measure on d . If we dene a left invariant quasi-metric d by d(x y) = r(x;1 y), the space H with the quasi-metric d can be regarded as a space of homogeneous type. See. 3, 7, 12, 13, 16, 23, 24, 25] for more details about background materials. The convolution f g on H is dened by. f g(x) =. Z. H. f (y)g(y;1 x) dy:. Let be locally integrable in Rd n f0g. We assume that is homogeneous of degree 0 with respect to the dilation group fAt g, that is, (At x) = (x) for x 6= 0, t > 0 and that Z ( ) dSd ( ) = 0: d We dene the singular integral Z Tf (x) = p:v:f K (x) = p:v: f (y)K (y;1 x) dy (1.1) Rd. for appropriate functions f , where K (x) = (x0 )r(x); , x0 = Ar(x);1 x for x 6= 0. The following result was proved in 25]. Theorem A. Let Tf be as in (1.1). Suppose that 2 L log L(d). Then, T is bounded on Lp (Rd ) for all p 2 (1 1). The maximal singular integral operator is dened by. Z ; 1 Tf (x) = sup f (xy )K (y) dy : >0 r(y)>. (1.2). Then the following result is known (see 21]). Theorem B. Suppose that 2 L log L(d ). Let Tf be dened as in (1.2). Then, T is bounded on Lp (Rd ) for p 2 (1 1)..

(4) Singular integrals on product homogeneous groups. 3. See 4, 5, 6, 8, 14, 15, 16, 18] for relevant results and also 19, 22, 25] for weak (1 1) boundedness. An analogue of a theory of Duoandikoetxea and Rubio de Francia 10] for homogeneous groups was developed in 21], where the use of Fourier transform estimates was replaced by a variant of the L2 estimates given in T. Tao 25]. The theory enables us to prove Theorem B and to give a dierent proof of Theorem A. In this note we shall show that the theory extends to the case of product spaces of homogeneous groups. Consequently, we can obtain an analogue of Theorem A for multiple singular integrals with rough kernels. We consider the product space Rn = Rn1 Rn2 , where n = n1 + n2 and (2) n R 1 = H 1 , Rn2 = H 2 are homogeneous groups with dilations A(1) t , At and 1 norm functions r1 r2 , respectively. Let 2 L (n1 n2 ) satisfy. Z. n1. Z. (u v) dSn1 (u) =. n2. (u v) dSn2 (v) = 0. (1.3). for all (u v) 2 n1 n2 . Dene the singular integral. Tf (x y) = p:v: f K (x y) = p:v:. Z. Rn1 Rn2. f (xu;1 yv;1 )K (u v) du dv. (1.4) (2) 0 where K (u v) = r1 (u);1 r2 (v);2 (u0 v0 ), u0 = A(1) r1 (u);1 u, v = Ar2 (v);1 v , with 1 and 2 denoting the homogeneous dimensions of H 1 and H 2 , respectively. Then we shall prove the following. Theorem 1. Suppose that 2 L(log L)2 (n1 n2 ). Let T be as in (1.4). Then, T is bounded on Lp (H 1 H 2 ) for all p 2 (1 1). Also, we consider the maximal singular integral. Z T f (x y) = sup f (xu;1 yv;1 )K (u v) du dv : > rr uv > > > 1 0 2 0. 1( ) 2( ). 1 2. (1.5). We anticipate Lp boundedness of T under a condition of the kernels similar to the one in Theorem 1. See 1, 2, 9, 11] for previous works about singular integrals on product of Euclidean spaces. Our methods will give dierent proofs for some previous results, where singular integrals are dened by Euclidean convolution, since our proof of Theorem 1 will not use Fourier transform estimates explicitly. Theorem 1 is an extension of a result of 2] to the case of singular integrals on product homogeneous groups. The optimality of the kernel class L(log L)2 , in the case of Euclidean convolution, can be found in 2]. To prove Theorem 1, we apply extrapolation arguments via the following estimate. Proposition 1. Let 1 < p < 1, 1 < s 2 and 2 Ls (n1 n2 ). Then, there exists a constant Cp independent of s and such that kTf kp Cp (s ; 1);2 kkskf kp:.

(5) 4. Yong Ding and Shuichi Sato We can prove Theorem 1 P from Proposition 1 by decomposing 2. L(log L)2 (n1 n2 ) as = 1 k=1 ck k , where each k satises (1.3), supk1 kkPk1+1=k 1 and fck g is a sequence of non-negative real numbers 2 such that 1 k=1 k ck < 1 (see 20] 2and also 17, 18]). In Section 2 we prove a basic L -estimate (Lemma 1) by applying meth-. ods of Tao 25]. This enables us to adapt the theory of 10] for the case of multiple singular integrals on product homogeneous groups to prove Proposition 1 in Section 3. In Section 4 we prove an estimate for a certain maximal function which is closely related to the maximal singular integral in (1.5). In the proofs of the results, we deal with real valued functions only to simplify our arguments. In this note, the letter C , along with some others, will be used to denote non-negative constants which may be dierent in dierent occurrences.. 2. Orthogonality estimates in L2 via convolution. We write x = (x(1) x(2) ) 2 Rn , x(1) 2 Rn1 , x(2) 2 Rn2 , n = n1 + n2 . Let Bi (x(i) t) be the ball with center x(i) and radius t in Rni dened by Bi (x(i) t) = fy(i) 2 Rni : ri ((x(i) );1 y(i) ) < tg: Let

(6) (i) , i = 1 2, be a C 1R function on Rni with support in Bi (0 1) n Bi (0 1=2). We assume that

(7) (i) = 1,

(8) (i) =

(9) ~(i) ,

(10) (i) (x(i) ) 0 for all x(i) 2 Rni , where

(11) ~(i) (x(i) ) =

(12) (i) ((x(i) );1 ). Set (ki) = (ik);1

(13) (i) ; (ik)

(14) (i) k 2 Z where t(i)

(15) (i) (x(i) ) = t;i

(16) (i) ((A(ti) );1 x(i) ), 2 and Z denotes the set of integers. Dene s t = s(1) t(2) . Note that supp((ki) ) Bi (0 k ) n P Bi (0 k;1 =2), (ki) = ~ (ki) and k (ki) = (i) , where (i) is the delta function on Rni . Choose j 2 C01 (R), j 2 Z, satisfying supp(j ) ft 2 R : j t j+2 g j 0 X (log 2) j (t) = 1 for t 6= 0 j 2Z m j(d=dt) j (t)j cm jtj;m for m = 0 1 2 : : :. where cm is a constant independent of this is possible since we assume 2. We may assume that j (t) = 0 (;j t). Dene. Sj k F (x) =. Z 1Z 1 0. 0. j (s)k (t)s t F (x) dss dtt . (2.1). where F 2 L1 (Rn ), supp(F ) D0 , D0 = D0(1) D0(2) , D0(i) = fx(i) 2 Rni : 1 ri (x(i) ) 2g. Let K0 (x) = K P(x)D0 (x), where E denotes the characteristic function of a set E . Then j k2ZSj k K0 = K ..

(17) Singular integrals on product homogeneous groups. 5. R. Let (i) be a non-negative smooth function on Rni such that (i) (x(i) ) dx(i) = 1. We also assume that ~ (i) = (i) , supp((i) ) Bi (0 1) n Bi (0 1=2). For F 2 L1 (Rn ) with supp(F ) D0 , dene the operator U = U (F ) by X (2.2) U f = U (F )(f ) = j k f j k where. j k. (2) j k (x) = j k (F )(x) = Sj k F (x) ; (1) j k (x) ; j k (x) + j k (x) (1) (1) j k (x) = j k (F )(x) = (2) (2) j k (x) = j k (F )(x) =. Z Z. Z. . Sj k F (x) dx(1) (1)j (1) (x(1) ). . Sj k F (x) dx(2) (2)k (2) (x(2) ). . Sj k F (x) dx j k (x) = (1) (2) and R = f j k g is an arbitrary sequence such that j k = 1 or ;1. We note that j k (x) dx(i) = 0, Sj k K0 = j k (K0 ) and U (K0 )(f ) = Tf if j k = 1 for all j k. For s 1, let Ls (D0 ) denote the subspace of Ls (H 1 H 2 ) consisting of functions F supported in D0 . We prove the following L2 estimates. Lemma 1. Suppose that F 2 Ls (D0 ), s 2 (1 2]. Let j1 j2 = j1 j2 (F ), a(t) = min(1 ;t ). Then, for ji ki 2 Z, i = 1 2, we have j k (x) = j k (F )(x) =. kf j1 j2 k1 k2 k2 C (log ). 2. Y2 i=1. a((jji ; ki j ; c)=s0 ). !. kF ks kf k2. (2.3) for some positive constants C and c independent of , s and F , where (2) 0 k1 k2 = (1) k1 k2 and s = s=(s ; 1). Proof. It suces to prove Lemma 1 by assuming ji = 0. This can be seen by applying ;j1 ;j2 to f j1 j2 k1 k2 and noting that s t (f g) = (s t f ) (s t g), ;j1 ;j2 k1 k2 = k1 ;j1 k2 ;j2 . Let = 0 0 . If k1 k2 0, then from the cancellation condition for and the smoothness of k1 k2 we shall show. k k1 k2 k1 C (log )2 kF k1. Y2. i=1. a(ki ; ). (2.4). for some > 0, which implies the conclusion by Young's inequality. We need the following estimates. Lemma 2. Suppose that F 2 Lq (D0 ) for some q 2 1 2]. Put S = S0 0 F . Then kS kq C (log )2 kF kq where the constant C is independent of , q and F ..

(18) 6. R. Yong Ding and Shuichi Sato. Proof. Since 01 0 (t) dt=t (log 2);1 2 log , by Holder's inequality it follows that ZZZ q ; 1 2q=q 0 kS kq ((log 2) 2 log ) 0 (s)0 (t)js t F (x)jq ds dt dx. = ((log 2);1 2 log )2q=q. ZZZ 0. C (log )2q=q0 +2 kF kqq :. s t 0 (s)0 (t)s1 (1;q) t2 (1;q) dss dtt jF (x)jq dx. This implies the conclusion. To show (2.4), we rst note that k k1 k2 k1 C (log )2 kF k1 (2.5) by Lemma 2 with q = 1. Let t1 = k1 ;1 t2 = k2 ;1 . Then k1 k2 = t1 t2 1 1 . Since Z Z (1) dy = dy(2) = 0 we have k1 k2 (x) = = =. Z Y2 ; (i) (i) ti i 1 ((Ati );1 ((y(i) );1 x(i) )) ; (1i) ((A(tii) );1 x(i) ) (y) dy Z i=1 (1) (1) ; (1) (1) t1. Z. 1 ((At1 );1 ((y(1) );1 x(1) )) ; 1 ((At1 );1 x(1) ). 1. (2) ;1 (2) (2) k2 ((y ) x ) (y ) dy:. . . (2) ;1 (2) (2) ;1 (2) (2) ;1 (2) t;2 2 (2) 1 ((At2 ) ((y ) x )) ; 1 ((At2 ) x ) (1) ;1 (1) (1) k1 ((y ) x ) (y ) dy:. Therefore, if k1 k2 1, then it is not dicult to see that k k1 k2 k1 C kF k1;k1 ;k2 + (2.6) for some > 0 also, if k1 1, k k1 k2 k1 C;k1 + kF k1 (2.7) and if k2 1, k k1 k2 k1 C;k2 + kF k1 (2.8) (see the proof of (3.7) of 21]). By (2.5){(2.8), we have (2.4) for k1 k2 0. Let Z1 Sj(i) F (x(i) ) = j (s)s(i) F (x(i) ) dss : 0. (2) (2) (1) (1) Then (1) 0 0 (x) = S0 G(x ) (x ), where. Z. G(x(2) ) = S0(1) F (x) dx(1).

(19) Singular integrals on product homogeneous groups. 7. with S0(1) acting on x(1) variable only. Note that (1) (1) (2) (2) (1) f (1). S0(2)G) ((1) (2) 0 0 k1 k2 = f ( ) (k1 ) ( k2 ): By Lemma 1 of 21]. kg ((1) S0(2) G) ((1) (2) k2 )k2 C (log )a((jk2 j ; c)=s0 )kGks kgk2:. It is easy to see that. kGks C (log )kF ks :. Combining these estimates, we have kf (1) (2.9) 0 0 k1 k2 k2 (2) C (log )2 a((jk2 j ; c)=s0 )kF kskf ((1) (2) ) ((1) k1 )k2 C (log )2 a((jk2 j ; c)=s0 )kF kskf k2: R = 0, as in the proof of (2.6) we have If k1 ;1, since (1) 0 (1) (1) k k1 k1 = k((1);k1 (1) ) (1) 0 k1 Ca((jk1 j ; c)): From this, (2.9) and Young's inequality, it follows that. kf (0i )0 k1 k2 k2 C kF ks kf k2(log )2. Y2. j =1. a((jkj j ; c)=s0 ). (2.10). if ki ;1, for i = 1 the result R for i = 2 can be proved similarly. If k1 k2 ;1, since (0i) = 0 and k k1 k2 k1 = k;k1 ;k1 0 0 k1 , by the proof of (2.6) and Lemma 2 we have. kf 0 0 k1 k2 k2 C kF k1kf k2(log )2. Y2. i=1. a((jki j ; c)):. (2.11). 0 and k2 ;1, we write (2) 0 0 k1 k2 = (0 0 k1 k2 ; (2) 0 0 k1 k2 ) + 0 0 k1 k2 : We note that the rst term on the right hand side is equal to If k1. Z. Z. ( S dx) ; S dx (1). (2). . . . k1 (2) (2) k2 (1). where S = S0 0 F as above. Therefore, arguing as above, we see that Y ;(jkij;c) k0 0 k1 k2 ; (2) 2 : 0 0 k1 k2 k1 C kF k1 Combining this with Lemma 2, we have. i=1. 2 k0 0 k1 k2 ; (2) 0 0 k1 k2 k1 C kF k1 kf k2 (log ). Y2 i=1. a((jki j ; c)):.

(20) 8. Yong Ding and Shuichi Sato. By this and (2.10) with i = 2, we have (2.11) in the case k1 The case k1 ;1 and k2 0 can be handled similarly. We shall prove. kf S k1 k2 k2 C kF ks kf k2(log )2 for k1 k2 ;1,. Y2. i=1. 0 and k2 ;1.. a((jki j ; c)=s0 ). kf S k1 k2 k2 C kF ks (log )2 a((jk2 j ; c)=s0 )kf k2 for k1 0 k2 ;1, kf S k1 k2 k2 C kF ks (log )2 a((jk1 j ; c)=s0 )kf k2 for k1 ;1 k2 0.. (2.12) (2.13) (2.14). By (2.10), (2.11) and (2.12), we get the conclusion of Lemma 1 for. k1 k2 ;1. If k1 0 k2 ;1, by (2.9), (2.10) with i = 2, (2.11) and (2.13) we have. kf k1 k2 k2 C kF ks(log )2 a((jk2 j ; c)=s0 )kf k2 :. Combining this with (2.7) and Lemma 2, we get the desired inequality of Lemma 1 for k1 0 k2 ;1. If k1 ;1 k2 0, by an analogue of (2.9) for kf (2) 0 0 k1 k2 k2 , (2.10) with i = 1, (2.11) and (2.14) we have kf k1 k2 k2 C kF ks(log )2 a((jk1 j ; c)=s0 )kf k2 : By this and (2.8) with Lemma 2 we get the conclusion of Lemma 1 for k1 ;1 k2 0. If k1 k2 0, we can use (2.4) to prove the estimate in Lemma 1. This will complete the proof of Lemma 1. It remains to prove (2.12), (2.13) and (2.14). Let k1 k2 ;1. On account of Lemma 2 and the T T method, to prove (2.12) it suces to show that n f k1 k2 S~ S k1 k2 1 . 2. C (log )4n1 (k1 +k2 +c)=s0 kF k2sn1 kf k2 (2.15). for some c > 0, where fm denotes the convolution product of m factors of f and we may assume n1 n2 without loss of generality. This is deduced from Young's inequality and the L1 estimate n k1 k2 S~ S k1 k2 1 C (log )4n1 (k1+k2+c)=s0 kF k2sn1 : (2.16) . 1. R. 2 0 0 k1 k2 S~k1 C (log ) kF k1 and k1 k2 S~(x) = y (x)k1 k2 S~(y) dy, where y (x) is the delta function concentrated at y and 0k1 k2 is either k1 k2 or k1 k2 k1 k2 . Therefore, we get (2.16) if we prove S S C (log )2n1 (k1+k2 +c)=s0 kF kn1 w1 wn1 k1 k2 1 s. Note that k0.

(21) Singular integrals on product homogeneous groups. 9. uniformly for w1 : : : wn1 2 B1 (0 C2 ) B2 (0 C2 ), wj = (wj(1) wj(2) ). This is a consequence of S S g C (log )2n1 (k1+k2 +c)=s0 kF kn1 w1 wn1 k1 k2 s (2.17) uniformly in w1 : : : wn1 2 B1 (0 C2 ) B2 (0 C2 ), for all g in C 1 with compact support such that kgk1 1. The inner product on the left hand side of (2.17) can be written as. ZZZ. n Y 1. Y2. !. F (yu yu ) (tu ) dy dt" dx u=1 i=1 where (s) = 0 (s), t(i) = (t(1i) : : : t(ni1) ), y(i) = (y1(i) : : : yn(i1) ) 2 (D0(i) )n1 , dt" = dt" (1) dt" (2) , dt" (i) = (dt(1i) =t(1i) ) : : : (dt(ni1) =t(ni1) ), dy = dy(1) dy(2) , dy(i) = dy1(i) : : : dyn(i1) and k1 k2 (x)g. (1). (2). (i). g = g(H1 (y(1) t(1) )x(1) H2 (y(2) t(2) )x(2) ) (i). (i). Hi (y t ) =. n Y 1. j =1. wj(i) A(t(ii)) yj(i) = w1(i) A(t(ii)) y1(i) : : : wn(i1) A(t(nii)) yn(i1) : j. 1. 1. Let DHi (y(i) t(i) ) be the ni ni matrix whose j th column vector is L @t(i) Hi (y(i) t(i) ), 1 j ni : j (i). (i). DHi (y t. . . ) = @tL(i) Hi (y(i) t(i) ) : : : @tL(ni) Hi (y(i) t(i) ) 1 i. . where @tL(i) Hi is the left invariant derivative (see 25], 21]). Then, to obtain j (2.17) it suces to show that. ZZZ . k1 k2 (x)gGij. n Y 1. u=1. F (yu yu ) (1). (2). Y2 i=1. (i). (tu ). !. dy dt" dx. C (log )2n1

(22) (k1 +k2 +c)=s0 kF kns 1 (2.18). for i j = 1 2, with Gij = Gij (y(1) y(2) t(1) t(2) ) denoting. i ;k1 det(DH1 (y(1) t(1) )) j ;k2 det(DH2 (y(2) t(2) )) where 1 is a function in C01 (R) such that 0 1 1, supp(1 ) ;1 1], 1 (t) = 1 for t 2 ;1=2 1=2], 2 = 1 ; 1, and are small positive numbers. We prove (2.18) by considering three cases. Case 1: i = 1 and j = 1. We note that. Z. (D0(i) )n1. . . 0 1] ;ki det(DHi (y(i) t(i) )) dy(i) C

(23) (ki +c). (2.19).

(24) 10. Yong Ding and Shuichi Sato. uniformly in t(i) 2 1 2]n1 and w1(i) : : : wn(i1) 2 Bi (0 C2 ). If i = 1, or i = 2 and n2 = n1 , this was proved in Section 3 of 21]. To prove it for the case i = 2 and n2 < n1 , we note that the arguments of 21] easily implies that. Z. . (D0(2) )n2. . 0 1] ;k2 det(DH2 (y(2) t(2) )) dy1(2) : : : dyn(2)2 C

(25) (k2 +c). (2.20). uniformly in t(2) 2 1 2 ]n1 , w1(2) : : : wn(2)1 2 B2 (0 C2 ) and yn(2)2 +1 : : : yn(2)1 2 (D0(2) )n1 ;n2 . Thus, integrating (2.20) with respect to yn(2)2 +1 : : : yn(2)1 , we obtain (2.19) for i = 2. By (2.19), applying Holder's inequality, we obtain (2.18) for i j = 1. Case 2: i = 2 and j = 2. It suces to prove. ZZ . k k (x)gG22 (tu ) dt" dx C

(26) (k +k +c) i=1 u=1 n Y2 Y 1. 1. 2. (i). 1. 2. (2.21). uniformly in y(i) 2 (D0(i) )n1 , i = 1 2, and w1 : : : wn1 2 B1 (0 C2 ) B2 (0 C2 ). To prove (2.21) we need the following four lemmas. Lemma 3. Let f be a continuous function on Rni such that supp(f ) Bi (0 C1 ). Z. f (x(i) ) dx(i) = 0 kf k1 C2 :. Then we can nd functions f1 f2 : : : fni such that ni X. f (x(i) ) =. j =1. @x(ji) fj (x(i) ). supp(fj ) Bi (0 C10 ) kfj k1 C20 for j = 1 2 : : : ni . Lemma 4. There exist functions Fj(i) on Rni , j = 1 2 : : : ni , such that supp(Fj(i) ) Bi (0 Cki ), kFj(i) k1 Cki for some > 0 and (kii) (x(i) ) =. ni X j =1. @x(ji) Fj(i) (x(i) ):. Lemma 5. Dene DHi (y(i) t(i) )x(i) in the same way as DHi (y(i) t(i) ) with Hi (y(i) t(i) )x(i) in place of Hi (y(i) t(i) ). Suppose that det(DHi (y(i) t(i) )x(i) ) 6= 0. Then, for 1 j ni ,. @x(ji) g(Hi (y(i) t(i) )x(i) ).

(27). = rt(i) g(Hi (y(i) t(i) )x(i) ) (DHi (y(i) t(i) )x(i) );1 (@xL(i) (Hi (y(i) t(i) )x(i) )) j. where rt(i) g = (rt(1i) g : : : rt(nii) g)..

(28) Singular integrals on product homogeneous groups. 11. Lemma 6. Suppose that det(DHi (y(i) t(i) )x(i) ) 6= 0, for i = 1 2. Then. @x(1) @ (2) g(H1 (y(1) t(1) )x(1) H2 (y(2) t(2) )x(2) ) i xj.

(29). ;. . = rt(1) rt(2) g (DH2 (y(2) t(2) )x(2) );1 (@xL(2) (H2 (y(2) t(2) )x(2) )) j. (DH1 (y t )x );1 (@xL(1) (H1 (y(1) t(1) )x(1) )) (1). (1). (1). i. where rt(1) rt(2) g denotes the n1 n2 matrix whose (u v) component is (1) (1) (1) (2) (2) (2) @t(1) u @t(2) v g (H1 (y t )x H2 (y t )x ):. Lemma 3 is from Lemma 7.1 of 25]. Lemma 4 follows from Lemma 3. See Lemma 7.2 of 25] for Lemmas 5 and 6. To obtain (2.21), by Lemma 4 it suces to prove that. ZZ gb(t(1) t(2)) Y2 @ i Fj(i)(x(i) ) dt" dx C

(30) (k +k +c): i x i=1 ji 1. (). 2. (2.22). for all ji , 1 ji ni , where g is as in (2.21),. b(t(1) t(2) ) = G22. n Y2 Y 1. i=1 u=1. (t(ui) ):. We can deduce (2.22) from the estimate. Z @xj @xj gb(t(1) t(2)) dt" C;d(k +k ;c) (1) 1. 1. (2) 2. 2. (2.23). for all x 2 B1 (0 Ck1 ) B2 (0 Ck2 ) with a suciently small > 0 and a constant d > 0, if we apply integration by parts and use the L1 norm estimate for Fj(ii) in Lemma 4. By Lemma 6, we can replace @x(1) @ (2) g by the j1 xj2 expression involving rt(1) rt(2) g, and using integration by parts, we can get (2.23) from the estimate. Z i " (1) (1) (1) (2) (2) (2) (2) (1) (2) g@tu @tv hh(1) ( y t x ) h ( y t x ) b ( t t ) dt u v (1). where. (2). C;d(k1 +k2 ;c) (2.24). (DHi (y(i) t(i) )x(i) );1 (@xL(i) x(i) ) = (h(1i) : : : h(nii) ) ji. (note that @xL(i) (Hi (y(i) t(i) )x(i) ) = @xL(i) x(i) ). We see that ji ji (1) (2) jrt(1) rt(2) b(t t )j C;(k1 +k2 ) c. and. j det(DHi (y(i) t(i) )x(i) )j Cki .

(31) 12. Yong Ding and Shuichi Sato. on the support of b, since j det(DHi (y(i) t(i) )x(i) )j = j det C x(i) ] det DHi (y(i) t(i) )j C j det DHi (y(i) t(i) )j (see 25] and Section 2 of 21]). By these estimates along with Cramer's formula, we have the pointwise estimate. h (1) (1) (1) (2) (2) (2) (2) (1) (2) i ;d(k +k ;c) @tu @tv h(1) u (y t x )hv (y t x )b(t t ) C (1). 1. (2). 2. which implies (2.24). Case 3: i = 1 and j = 2. Using Lemmas 4 and 5 for i = 2 and arguing similarly to the proof of the case i = 2, j = 2, we have. ZZ . " dx k k (x)gG12 (F (yu yu ) (tu )) dt u=1 i=1 n Y C

(32) (k +c) 0 1] ;k det(DH1 (y(1) t(1) )) (jF (yu(1) yu(2) )j(t(1) u )) n Y 1. 1. 2. 2. (1). (2). Y2. (i). (i). (2). 1. 1. u=1. (i) 0. uniformly in y(i) 2 (D )n1 , t(1) 2 1 2]n1 and w1 : : : wn 2 B1 (0 C2 ) B2 (0 C2 ). Integrating this with respect to t(1) and y(i) and using Holder's inequality and (2.19), we have. ZZ . " dx k k (x)gG12 (F (yu(1) yu(2)) (t(ui) )) dy dt u=1 i=1 n Y. Y2. 1. 1. 2. C

(33) (k2 +c)

(34) (k1 +c)=s0 kF kns 1 :. This implies the desired estimate. If i = 2 and j = 1, (2.18) can be proved similarly. If k1 0 k2 ;1, to show (2.13) we apply arguments similar to those for k1 k2 ;1 above. Let. Gj = j ;k2 det(DH2 (y(2) t(2) )) . for j = 1 2. Then, arguing as in the proof of (2.18) with respect to y(2) t(2) x(2) variables, we have. ZZZ . k1 k2 (x)gGj. n Y 1. (F (yu yu ). u=1. (1). (2). Y2 i=1. . (tu )) dy dt" dx (i). C (log )2n1

(35) (k2 +c)=s0 kF kns 1 . for j = 1 2, where the function g is as above. This and Lemma 2 imply (2.13). We can prove (2.14) similarly. This completes the proof of Lemma 1..

(36) Singular integrals on product homogeneous groups. 13. 3. Proof of Proposition 1.. By a standard method (see Lemma 6 of 21] for the one-parameter case) we can prove the following Littlewood-Paley inequalities. (2) Lemma 7. Let 1 < p < 1 and k1 k2 = (1) k1 k2 (see Lemma 1). Then. 0 11=2 . X f C @ X jf j2A k k k k k k p k k k k. p p 0. 11=2 X @ jf k k j2 A Cpkf kp k k p 1. 1. 2. 1. 2. 1. 2. 1. 1. 1. 2. (3.1). 2. 2. (3.2). 2. where the constant Cp is independent of 2. Dene MF f (x) = sup jf Sj (jF j)(x)j j. where we write j = (j1 j2 ) 2 Z Z and Sj (F ) = Sj1 j2 (F ) is as in Section 2 (see (2.1)). Similar notation will be used in what follows. Let f = MF f . We prove the following result for along with Proposition 1. Lemma 8. Let p > 1 s 2 (1 2] = 2s0 and F 2 Ls (D0 ). Then, there exists a positive constant Cp independent of s and F such that k f kp Cp (s ; 1);2 kF kskf kp: P Proof. Put U = U (F ) with = 2s0 (see (2.2)) and write U f = k(1) k(2) Uk(1) k(2) f , where k(i) = (k1(i) k2(i) ) 2 Z2 and X Uk(1) k(2) f = j f k(1) +j j k(2) +j j = j (F ) j = (j1 j2 ): j. Fix k k 2 Z2. Using Lemma 1 with = 2s0 and duality, we have (1). (2). kf k j k2. Y C (s ; 1);2 kF ks kf k2 ((jki ; ji j ; c)) 2. i=1. where (t) = min(1 2;t). Applying this and Lemma 1, to j and ~j , and noting that. kk(2) +j k(2) +j0 k1 C. Y2. i=1. ((jji ; ji0 j ; c)) j 0 = (j10 j20 ). along with kk k1 C , we have f ( (1) ) ( (2) (2) 0 ) (~ 0 (1) 0 ) j k +j j k +j k +j k +j 2. CA2 kf k2. Y2. i=1. (2(jki(1) j ; c))((jji ; ji0 j ; c)) (3.3).

(37) 14. f k. Yong Ding and Shuichi Sato. j (j k(2) +j ) (k(2) +j 0 ~j 0 ) k(1) +j 0 2. (1) +. CA2 kf k2. Y2. i=1. (2(jki(2) j ; c)) (3.4). where A = (s ; 1);2kF ks . Taking the geometric mean, by (3.3) and (3.4) we have. f k. j j k(2) +j k(2) +j 0 ~j 0 k(1) +j 0 2. (1) +. CA kf k2 2. We can treat. Y2 Y2. i=1 m=1. !. (m). ((jki j ; c)) ((jji ; ji0 j ; c)=2):. kf k(2) +j0 ~j0 k(1) +j0 k(1) +j j k(2) +j k2. similarly. Thus, by the Cotlar-Knapp-Stein lemma it follows that. U k. (1). k f 2 CAkf k2 (2). Y2 Y2. m=1 i=1. ((jki(m) j ; c)=2). (3.5). uniformly in . From (3.5) we deduce that. kU f k2 . X. k(1) k(2). kUk(1) k(2) f k2 CAkf k2:. (3.6). Let 2 (0 1). Let fpj g1 1 be a sequence of positive numbers dened by p1 = 2 and 1=pj+1 = 1=2 + (1 ; )=(2pj ) for j 1. Then, 1=pj = (1 ; aj )=(1 + ), with a = (1 ; )=2. We note that fpj g is decreasing and converges to 1 + . We prove that kU f kpm Cm A kf kpm m 1 (3.7) for all F 2 Ls (D0 ), where Cm is a constant independent of , F and s. This follows from (3.6) for m = 1. We x m 1 and assume (3.7) for this m. If it is applied to U (jF j), via the Khintchine inequality, we see that kg(f )kpm CAkf kpm (3.8) where 0 1. g(f ) =. @X j. 1=2. jf j (jF j)j A 2. j = (j1 j2 ):. Let (f ) = supj jf jj jj, j = j (F ), and (f ) = supj jf j (jF j)j. Then (f ) (jf j) + C (jf j) + C1 (jf j) + C2 (jf j) (3.9) g(jf j) + C (jf j) + C1 (jf j) + C2 (jf j) where i f = sup jf (ji) (jF j)j: j.

(38) Singular integrals on product homogeneous groups It is easy to see that. (jf j) C (s ; 1);2 kF k1 Mf where M is the strong maximal function dened as. Mf (x x ) = sup t;1 1 t;2 2 (1). (2). t1 t2 >0. Z. B1 (x(1) t1 )B2 (x(2) t2 ). 15 (3.10). jf (y(1) y(2) )j dy(1) dy(2) :. By Lemma 7 of 21] and the Hardy-Littlewood maximal theorem (see 7, 12]), ki f kp C (s ; 1);2 kF ks kf kp p > 1: (3.11) p By the estimates (3.8){(3.11) and L boundedness of M it follows that k (f )kpm CAkf kpm : (3.12) By (3.12) and the estimate kj k1 CA we have. X. jgk k j21=2 CA X jgk j2 1=2 rm rm. (3.13). where 1=rm ; 1=2 = 1=(2pm) (see 10] and also 17, 18]). Thus, applying the Littlewood-Paley theory (Lemma 7) and (3.13) we have. kUk(1) k(2). 0 11=2 X f krm C @ jf k +j j j2 A j rm 0. 1 1=2 X. 2 CA @ jf k +j j A j rm (1). (3.14). (1). CAkf krm :. Since 1=pm+1 = (1 ; )=rm + =2, interpolating between (3.5) and (3.14), we get. kUk(1) k(2) f kpm+1 CAkf kpm+1 and hence. kU f kpm+1 . X k(1) k(2). Y2 Y2. `=1 i=1. ( (jki(`) j ; c)=2). (3.15). kUk(1) k(2) f kpm+1 CAkf kpm+1 . which proves (3.7) for all m by induction. For any p 2 (1 2], we take 2 (0 1) such that p 2 (1 + 2] and dene the sequence fpj g1 1 by using it. Then, we have pj+1 < p pj with some j . It follows that kU f kp CAkf kp (3.16) by interpolation between the estimates (3.7) with m = j and m = j + 1. By the estimate (3.16) we have kg(f )kp CAkf kp for p 2 (1 2], where g(f ) is as in (3.8). This estimate and (3.9), (3.10) with the strong maximal theorem, (3.11) imply Lemma 8 for p 2 (1 2]..

(39) 16. Yong Ding and Shuichi Sato. If p > 2, interpolating between the estimate for p = 2 of Lemma 8 and the estimate k (f )k1 C (log )2 kF k1kf k1 we get the desired result. This completes the proof of Lemma 8. Proof of Proposition 1. Since Tf = U (K0 )(f ) if j = 1 for all j = (j1 j2 ), by (3.16) we have kTf kp C (s ; 1);2 kkskf kp for p 2 (1 2] a duality argument will imply the conclusion for p 2 2 1).. 4. Lp estimates for certain maximal functions. We prove the following result, which may be useful in studying Lp boundedness of T in (1.5). Proposition 2. Let s be as in Proposition 1. We dene. 1 1 X X R(f )(x) = sup f Sj j K0(x) k k 2Zj =k j =k 1. 2. 1 2. 1. 1 2. 2. where Sj1 j2 K0 is as in Section 2. Let A = (s ; 1);2 kks and = 2s0 . Then, for p 2 (1 1) we have kR(f )kp Cp Akf kp with some positive constant Cp independent of s 2 (1 2] and 2 Ls .. P. (2) Proof. We write 'k1 k2 = m1 k1 +2 m2 k2 +2 m1 m2 = '(1) k1 'k2 , where (2) (i) (i) m1 m2 = (1) m1 m2 and 'ki = ki +1

(40) (i) . According to the decomposition. 1 X 1 X. j1 =k1 j2 =k2. f Sj1 j2 K0 (x) = T (f ) 'k1 k2. 0 k ;1 1 1 X X ;@ f Sj j K0 A 'k k j =;1 j =;1 0 1 k ;1 1 X X ;@ f Sj j K0 A 'k k j =;1 j =;1 0 k ;1 k ;1 1 X X +@ f Sj j K0 A 'k k j =;1 j =;1 01 1 1 X X +@ f Sj j K0 A ( ; 'k k ) 1. 1. 1 2. 1. 2. 1 2. 1. 2. 1 2. 1. 2. 2. 2. 1. 2. 1. 1. 2. 2. j1 =k1 j2 =k2. 1 2. 1. 2.

(41) Singular integrals on product homogeneous groups. 17. where is the delta function, we see that R(f ) sup jT (f ) 'k1 k2 j + M1 + M2 + M3 + M4. (4.1). k1 k2. with. 0 k ;1 1 1 X X M1 = sup @ f Sj j K0 A 'k k k k j =;1 j =;1 0 1 k ;1 1 X X M2 = sup @ f Sj j K0 A 'k k k k j =;1 j =;1 0 k ;1 k ;1 1 X X M3 = sup @ f Sj j K0 A 'k k k k j =;1 j =;1 0 1 1 1 X X M4 = sup @ f Sj j K0A ( ; 'k k ) : k k j =k j =k 1. 1. 2. 1. 1 2. 1. 2. 1 2. 1. 2. 1 2. 1. 2. 2. 2. 1. 2. 1. 2. 1. 1. 1. 2. 2. 2. 1. 2. 1 2. 1. 1 2. 1. 2. 2. Note that Proposition 1 and the strong maximal theorem imply that sup jT (f ) ' j C kM (Tf )k C Akf k p 2 (1 1): (4.2) k1 k2 p p p p k1 k2. Thus, to prove Proposition 2, it remains to give the Lp estimates of Mi (i = 1 2 3 4) by (4.1). Firstly, we consider M4 . Note that M4 Q1 + Q2 + Q3 , where. 0 1 1 1 X X Q1 = sup @ f Sj j K0 A ((1) ; '(1) ) (2) k k k j =k j =k 0 1 1 1 X X Q2 = sup @ f Sj j K0 A (1) ((2) ; '(2) ) k k k j =k j =k 0 1 1 1 X X Q3 = sup @ f Sj j K0 A ((1) ; '(1) ) ((2) ; '(2) ) : k k k k j =k j =k 1. 2. 1. 2. 1. 2. 1 2. 1. 1 2. 2. 1. 1 2. 2. 1 2. 2. 1. 1 2. 2. 1 2. 1. 1. 2. Now let us give the estimates of Q1 , Q2 and Q3 , one by one. Estimate of Q3 . We write j = (j1 j2 ) and k = (k1 k2 ), then it is easy to check that. Q3 where. . 1 X 1 X. j1 =0 j2 =0. . Nj (f ). . (2) (2) Nj (f ) = sup (f Sj+k K0) ((1) ; '(1) k1 ) ( ; 'k2 ) :. k1 k2.

(42) 18. Yong Ding and Shuichi Sato. Hence. kQ3 kp . 1 X 1 X j1 =0 j2 =0. kNj (f )kp :. Since Lemma 8 and the Hardy-Littlewood maximal theorem imply that kNj (f )ku Cu Akf ku for u > 1, (4.3) if we can show that there are c > 0 such that. kNj (f )k2 CAkf k2 then we may get. Y2. i=1. Y2. kNj (f )kp CAkf kp. i=1. ((ji ; c)). (4.4). ( (ji ; c)). for some 2 (0 1] by interpolating between (4.3) and (4.4). In fact, for p 2 (1 1), it is enough to take u 2 (1 1) and 2 (0 1] such that 1=p = (1 ; )=u + =2. Thus we have X kQ3kp kNj (f )kp CAkf kp : (4.5) j. So, it suces to verify (4.4) for estimating Lp norm of Q3 . (2) (2) Let Jk = ((1) ; '(1) k1 ) ( ; 'k2 ). Then. Nj (f ) Fix j and let. X. V f =. k. jf Sj+k K0 Jk j. X k. 2. !1=2. :. (4.6). k f Sj+k K0 Jk . where = f k g, k = 1 or ;1. If we can show that. kV f k2 CAkf k2. Y2. i=1. ((ji ; c)). (4.7). for some c > 0, uniformly in , then (4.4) follows from (4.6), (4.7) and Khintchine's inequality. To prove (4.7) we argue similarly to the proof of (3.5). By the CotlarKnapp-Stein lemma, it is easy to see that (4.7) can be deduced from the following two estimates: kf Sj+k(1) K0 Jk(1) Jk(2) Sj+k(2) K~0k2. CA2 kf k2. Y2. i=1. ((ji ; c))((jki(1) ; ki(2) j ; c)) (4.8).

(43) Singular integrals on product homogeneous groups. 19. and. kf Jk(2) Sj+k(2) K~0 Sj+k(1) K0 Jk(1) k2 CA kf k2 2. Y2. i=1. ((ji ; c))((jki(1) ; ki(2) j ; c)) (4.9). for some c > 0. Hence, to estimate Q3 it remains to give the estimates (4.8) and (4.9). P P Proof of (4.8). Note that (i) ;'(kii) = mi ki +1 (mi)i , Jk = mk+(1 1) m , where k = (k1 k2 ), m = (m1 m2 ) and m k means that m1 k1 and m2 k2 . Therefore,. kf Sj+k(1) K0 Jk(1) Jk(2) Sj+k(2) K~0k2 X kf Sj+k(1) K0 m(1) m(2) Sj+k(2) K~0 k2: (4.10) m(1) k(1) +(11) m(2) k(2) +(11). Lemma 1 implies that. kf (Sj+k(1) K0 m(1) ) (m(2) Sj+k(2) K~0 )k2 CA kf k2 2. Y2. i=1. (2) (2) ((jji + ki(1) ; m(1) i j ; c))((jji + ki ; mi j ; c)):. (4.11). Also, we see that. kf Sj+k(1) K0 (m(1) m(2) ) Sj+k(2) K~0 k2 CA2 kf k2. Y2. i=1. (2) ((jm(1) i ; mi j ; c)): (4.12). By the estimates (4.11) and (4.12) it follows that. kf Sj+k(1) K0 m(1) m(2) Sj+k(2) K~0 k2 CA2 kf k2. Y2. i=1. (2) (2) ((jji + ki(1) ; m(1) i j ; c)=2)((jji + ki ; mi j ; c)=2). By (4.10) and (4.13) we obtain (4.8).. (2) ((jm(1) i ; mi j ; c)=2): (4.13).

(44) 20. Yong Ding and Shuichi Sato. The proof of the estimate (4.9) is similar. In fact, kf Jk(2) Sj+k(2) K~0 Sj+k(1) K0 Jk(1) k2 X kf m(2) Sj+k(2) K~0 Sj+k(1) K0 m(1) k2. . m(1) k(1) +(11) m(2) k(2) +(11). X. X ` `. . m(1) m(2). k(1) +(11) (1) (2) k(2) +(11). (4.14). kf m(2) Sj+k(2) K~0 `(1) `(2) Sj+k(1) K0 m(1) k2 :. Lemma 1 implies. kf (m(2) Sj+k(2) K~0) (`(1) `(2) ) (Sj+k(1) K0 m(1) )k2. Y2. CA2 kf k2. i=1. (1) (1) ((jji + ki(2) ; m(2) i j ; c))((jji + ki ; mi j ; c)) (2) ((j`(1) i ; ì j ; c)) (4.15). kf m(2) (Sj+k(2) K~0 `(1) ) (`(2) Sj+k(1) K0 ) m(1) k2 CA2 kf k2. Y2. i=1. (1) (2) ((jji + ki(2) ; `(1) i j ; c))((jji + ki ; ì j ; c)): (4.16). By (4.15) and (4.16), taking the geometric mean, kf m(2) Sj+k(2) K~0 `(1) `(2) Sj+k(1) K0 m(1) k2. CA2 kf k2. Y2. i=1. (1) (1) ((jji + ki(2) ; m(2) i j ; c)=2)((jji + ki ; mi j ; c)=2). (2) (2) (1) ((j`(1) i ; ì j ; c)=2)((jji + ki ; ì j ; c)=2) ((jji + ki(1) ; `(2) i j ; c)=2): (4.17). Summing in `(1), `(2) in (4.17), we have kf m(2) Sj+k(2) K~0 Sj+k(1) K0 m(1) k2. CA kf k2 2. Y2. i=1. ((jki(1) ; ki(2) j ; c)). (1) (1) ((jji + ki(2) ; m(2) i j ; c)=2)((jji + ki ; mi j ; c)=2) (4.18). for some c > 0. By (4.14) and (4.18) we have (4.9). Estimate of Q1 . We decompose. 01 1 1 X X (2) @ f Sj j K0A ((1) ; '(1) = W1 (k) ; W2 (k) + W3 (k) k ) j1 =k1 j2 =k2. 1 2. 1.

(45) Singular integrals on product homogeneous groups. 21. where. 01 1 1 X X (2) W1 (k) = @ f Sj j K0 A ((1) ; '(1) ). ' k k j =k j =;1 0 1 k ;1 1 X X (2) f Sj j K0 A ((1) ; '(1) ). ' W2 (k) = @ k k j =k j =;1 01 1 1 X X (2) (2) W3 (k) = @ f Sj j K0 A ((1) ; '(1) ). ( ; ' ) : k k 1 2. 1. 2. 1 2. 1. 2. 1 2. 1. 2. 1 2. 1. 1 2. j1 =k1 j2 =k2. 1. 2. We note that supk jW3 (k)j = Q3 . It is easy to see that. 0 1 X sup jW1 (k)j sup @ Nj (f )A ((1) '(2) ) k k k j 0 1. 2. where. 2. 1. X Nj (f ) = sup (f Sk +j j K0 ) (((1) ; '(1) ) (2) ) : k k j 1. 1. We need to prove. 1. 1 2. 1. 2. kNj1 (f )k2 CA((j1 ; c))kf k2:. (4.19). Thus, by using (4.19) and by applying interpolation suitably, we have kNj1 (f )kp CA((j1 ; c))kf kp for some c > 0, and hence X k sup jW1 (k)jkp C kNj1 (f )kp CAkf kp : (4.20) k. j1. To prove (4.19), let. V 0 (f ) =. X k. (2) k1 (f S(j1 0)+k K0 ) (((1) ; '(1) k1 ) ) k = (k1 k2 ). where = f k1 g is a sequence such that k1 = 1 or k1 = ;1. Then as in the case of Nj (f ), it suces to show kV 0 (f )k2 CA((j1 ; c))kf k2: To prove this, as in the proof of (4.7) above, we need to estimate X kf S(j1 0)+k(1) K0 m(1) m(2) S(j1 0)+k(2) K~0 k2 (1) m(1) 1 k1 +1 (2) m(2) k 1 1 +1.

(46) 22. Yong Ding and Shuichi Sato. and. X (1) m(1) 1 k1 +1 (2) m(2) k 1 1 +1. kf m(2) S(j1 0)+k(2) K~0 S(j1 0)+k(1) K0 m(1) k2 :. However, it is easy to see that a computation similar to the one in the proof of (4.7) gives the desired results. As for the estimate of supk jW2 (k)j, we see that XX sup jW2 (k)j Pj k. j1 0 j2 1. where j = (j1 j2 ) and (2) Pj = sup (f Sj1 +k1 k2 ;j2 K0 ) ((1) ; '(1) k1 ) 'k2 : k1 k2. As in the arguments above, we need to show that. kPj1 j2 k2 CAkf k2. Y2. i=1. ((ji ; c)). (4.21). for some c > 0. Let j = (j1 ;j2 ). To prove this we use the following estimates: kf Sj +k(1) K0 m(1) m(2) Sj +k(2) K~0 k2. CA2 kf k2 and. Y2. i=1. ((j(j )i + ki(1) ; m(1) i j ; c)=2). (1) (2) ((j(j )i + ki(2) ; m(2) i j ; c)=2)((jmi ; mi j ; c)=2). kf m(2) Sj +k(2) K~0 Sj +k(1) K0 m(1) k2 CA2 kf k2. Y2. i=1 (2). ((jki(1) ; ki(2) j ; c)). ((j(j )i + ki ; mi j ; c)=2)((j(j )i + ki(1) ; m(1) i j ; c)=2) (1) (2) (2) (1) k2(1) +2 m(2) for m(1) m(2) with m(1) 1 k1 +1 m1 k1 +1 and m2 2 (2). k2(2) + 2. These estimates can be proved as (4.13) and (4.18). Using them as in the proof of (4.4), we can deduce that. . X (f S K ) ((1) ; '(1) ) '(2) 2!1=2 k k k j+k 0 1. 2. 2. CAkf k2 for some c > 0, which implies (4.21).. Y2 i=1. ((ji ; c)).

(47) Singular integrals on product homogeneous groups. 23. Interpolating between the estimates of (4.21) and Lemma 8, we have. kPj1 j2 (f )kp CAkf kp. Y2. i=1. ((ji ; c)). for some c > 0, and hence X k sup jW2 (k)jkp C kPj1 j2 (f )kp CAkf kp : k. j1 j2. By (4.5), (4.20) and (4.22). (4.22). kQi kp CAkf kp. (4.23) for i = 1. We can prove (4.23) for i = 2 in the same way. From (4.5) and (4.23) for i = 1 2, it follows that kM4kp CAkf kp: (4.24) Secondly, we give the estimate of M3 . Note that. M3 . where. . XX. j1 1 j2 1. Qj1 j2 . . . (2) Qj1 j2 = sup (f Sk1 ;j1 k2 ;j2 K0) '(1) k1 'k2 :. k1 k2. P Writing 'ki = mi ki +2 (mi)i and arguing as in the proof of (4.4), we can. prove. (i). kQj1 j2 k2 CAkf k2. Y2 i=1. ((ji ; c)). for some c > 0, which implies, in the same way as above, kM3kp CAkf kp 1 < p < 1: (4.25) Alternatively, we can handle M3 by noting M3 CAMf (see Lemma 10 of. 21]). Finally, let us estimate M1 and M2 . We observe that. 0 1 X M1 sup @ Nj0 (f )A ((1) '(2) ) k k j 1 X Nj0 (f ) = sup (f Sk ;j j K0 ) ('(1). (2) ) : k k j 1. 2. where. 1. 1. 2. 1. 1. 1 2. 1. 2. In the same way as in the estimate of Nj1 (f ), we can prove kNj01 (f )k2 CA((j1 ; c))kf k2: Using this, we can show kMi kp CAkf kp 1 < p < 1:. (4.26).

(48) 24. Yong Ding and Shuichi Sato. for i = 1. The estimate (4.26) for M2 can be proved similarly. By (4.1), (4.2), (4.24), (4.25) and (4.26) for i = 1 and 2, we obtain the conclusion of Proposition 2.. References. 1] H. Al-Qassem and Y. Pan, p boundedness for singular integrals with rough kernels on product domains, Hokkaido Math. J. 31 (2002), 555{613.. 2] A. Al-Salman, H. Al-Qassem and Y. Pan, Singular integrals on product domains, Indiana Univ. Math. J., 55 (2006), 369{387.. 3] A. P. Calderon and A. Torchinsky, Parabolic maximal functions associated with a distribution, Advances in Math. 16 (1975), 1{64.. 4] A. P. Calderon and A. Zygmund, On singular integrals, Amer. J. Math. 78 (1956), 289{309.. 5] Y. Chen, Y. Ding and D. Fan, A parabolic singular integral operator with rough kernel, J. Aust. Math. Soc. 84 (2008), 163{179.. 6] M. Christ, Hilbert transforms along curves I. Nilpotent groups, Ann. of Math. 122 (1985), 575{596.. 7] R. R. Coifman and G. Weiss, Analyse Harmonique Non-Commutative sur Certains Espaces Homogenes, Lecture Notes in Math. 242, Springer-Verlag, Berlin and New York, 1971.. 8] Y. Ding and X. Wu, Littlewood-Paley -functions with rough kernels on homogeneous groups, Studia Math. 195 (2009), 51{86.. 9] J. Duoandikoetxea, Multiple singular integrals and maximal functions along hypersurfaces, Ann. Inst. Fourier 36 (1986), 185{206.. 10] J. Duoandikoetxea and J. L. Rubio de Francia, Maximal and singular integral operators via Fourier transform estimates, Invent. Math. 84 (1986), 541{561.. 11] R. Feerman and E. M. Stein, Singular integrals on product spaces, Adv. in Math. 45 (1982), 117{143.. 12] G. B. Folland and E. M. Stein, Hardy spaces on homogeneous groups, Princeton Univ. Press, Princeton, N.J. 1982.. 13] A. Nagel and E. M. Stein, Lectures on pseudo-dierential operators, Mathematical Notes 24, Princeton University Press, Princeton, NJ, 1979.. 14] F. Ricci and E. M. Stein, Harmonic analysis on nilpotent groups and singular integrals, I. Oscillatory integrals, J. Func. Anal. 73 (1987), 179{194.. 15] F. Ricci and E. M. Stein, Harmonic analysis on nilpotent groups and singular integrals, II. Singular kernels supported on submanifolds, J. Func. Anal. 78 (1988), 56{84.. 16] N. Riviere, Singular integrals and multiplier operators, Ark. Mat. 9 (1971), 243{278.. 17] S. Sato, Estimates for singular integrals and extrapolation, Studia Math. 192 (2009), 219{233.. 18] S. Sato, Estimates for singular integrals along surfaces of revolution, J. Aust. Math. Soc. 86 (2009), 413{430.. 19] S. Sato, Weak type (1 1) estimates for parabolic singular integrals, Proc. Edinb. Math. Soc. 54 (2011), 221{247. L. g. .

(49) Singular integrals on product homogeneous groups. 25. 20] S. Sato, A note on p estimates for singular integrals, Sci. Math. Jpn. 71 (2010), 343{348.. 21] S. Sato, Estimates for singular integrals on homogeneous groups, J. Math. Anal. Appl. 400 (2013) 311{330.. 22] A. Seeger, Singular integral operators with rough convolution kernels, J. Amer. Math. Soc. 9 (1996), 95{105.. 23] E. M. Stein, Harmonic Analysis: Real-Variable Methods, Orthogonality and Oscillatory Integrals, Princeton University Press, Princeton, NJ, 1993.. 24] E. M. Stein and S. Wainger, Problems in harmonic analysis related to curvature, Bull. Amer. Math. Soc. 84 (1978), 1239{1295.. 25] T. Tao, The weak-type (1 1) of log homogeneous convolution operator, Indiana Univ. Math. J. 48 (1999), 1547{1584. L. . L. L. Yong Ding School of Mathematical Sciences, Laboratory of Mathematics and Complex Systems (BNU), Ministry of Education, Beijing Normal University, Beijing, 100875 P. R. of China e-mail: [email protected] Shuichi Sato Department of Mathematics, Faculty of Education, Kanazawa University, Kanazawa 920-1192, Japan e-mail: [email protected].

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