Canad. Math. Bull. Vol. 41 (4), 1998 pp. 404±412 Lp-BOUNDEDNESS OF A SINGULAR INTEGRAL OPERATOR ABDELNASER J. AL-HASAN AND DASHAN FAN 1 0 ABSTRACT.Letb(t)beanL function on R, Ω( y )beanH1 function on the unit sphere satisfying the mean zero property (1) and Qm(t) be a real polynomial on R of degree m satisfying Qm(0) = 0. We prove that the singular integral operator Zn Ð n 0 TQ Òb( f )(x)=pv b(jyj)Ω(y)jyj f xQm(jyj)y dy m R is bounded in Lp(Rn)for1ÚpÚ1, and the bound is independent of the coef®cients of Qm(t). Let Sn1 be the unit sphere in Rn, n ½ 2, with normalized Lebesgue measure dõ = dõ(x0).LetΩ(x) be a homogeneous function of degree zero, with Ω 2 L1(Sn1)and Z (1) Ω(x0)dõ(x0)=0Ò Sn1 where x0 = xÛjxj for any x 6=0. Suppose b(t)isanL1(R) function. In this paper we investigate the Lp boundedness of the singular integral operator Z Á 0 (2) TQ Òb( f )(x)=pv K(y)f xQm(jyj)y dy m Rn P 0 Ûj j2 n1 j jΩ j jn m å k where y = y y S ,K(y)=b(y) (y)y and Qm(t)= k=1 kt is real. The study of such kind of operators has a long history. Readers can see [5] for more detailed background. In particular, the following theorem can be found in [5]. Ω 2 THEOREM A. Let TQmÒb be the singular integral operator de®ned in (2). If H1(Sn1) satis®es (1), then this operator is bounded in L2(Rn). The proof of Theorem A is based on the Plancherel theorem, so that the authors in [5] were only able to obtain the L2 boundedness. The main purpose of this note is to extend Theorem A to the following Lp boundedness theorem. Ú Ú THEOREM 1. Let TQmÒb be the singular integral operator de®ned by (2) and 1 p 1.IfΩ2H1(Sn1)satis®es (1) then this operator is bounded in Lp(Rn). More precisely, we have k k Ä kΩk k k (3) TQmÒb( f ) p C H1(Sn 1) f p Received by the editors June 4, 1997; revised May 6, 1998. AMS subject classi®cation: 42B20. Key words and phrases: singular integral, rough kernel, Hardy space. c Canadian Mathematical Society 1998. 404 Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 23:02:31, subject to the Cambridge Core terms of use. A SINGULAR INTEGRAL OPERATOR 405 where C is a constant independent of Ω, f and the coef®cients of Qm. The proof of Theorem 1 is based on a lemma in [4] by Fan and Pan (see Lemma 3 below). Throughout this paper, the letter C will denote a positive constant that may vary at each occurrence but is independent of the essential variables. To prove Theorem 1, we recall the de®nition of Hardy spaces on the unit sphere, as well as some lemmas. The Poisson kernel on Sn1 is de®ned by 0 2 0 0 n Pry0 (x )=(1r )Ûjry x j where 0 Ä r Ú 1andx0Òy0 2Sn1. For any Ω 2 L1(Sn1), we de®ne the radial maximal function P+Ω(x0)by þZ þ + 0 þ 0 0 0 þ P Ω(x)= supþ Ω(y)P 0 ( y ) dõ( y )þ n1 rx 0ÄrÚ1 S k +Ωk Ú1 Ω 1 n1 1 If P L1(Sn1) , we say that belongs to the Hardy space H (S ) with H -norm kΩk k +Ωk H1(Sn1) = P L1(Sn1). The space H1(Sn1) was studied in [1] (see also [2]). In particular, it is known that L1(Sn1) Ã H1(Sn1) Ã L Log+ L(Sn1) Ã Lq(Sn1) for any q Ù 1. Another important property of H1(Sn1) is the atomic decomposition, which will be reviewed below. 1 An exceptional atom is an L function E(x) satisfying kEk1 Ä 1. An 1-atom is an L1 function a() that satis®es (4) supp(a) ²fx0 2Sn1Òjx0 x0jÚöfor some x0 2 Sn1 and öÙ0g; Z 0 0 (5) a(ò0) dõ(ò0)=0Ò Sn1 (n1) (6) kak1 Äö Ω 2 1 n1 FromP [1] or [2], we ®nd that any H (S ) has an atomic decompositionP Ω = ïjaj,wheretheaj's are either exceptional atoms or 1-atoms and jïjjÄ kΩk Ω 2 1 n1 C H1(Sn1). In particular, if H (S ) has the mean zero property (1) then all the atoms aj in the atomic decomposition can be chosen to be 1-atoms. LEMMA 1[VAN DER CORPUT [8]]. Suppose û and h are smooth functions on [aÒ b] and û is real-valued. If jû(k)(x)j½1for x 2 [aÒ b] then þ þ Z b þ h i þ iïû(t) þ 1 0 þ e h(t) dtþ Ä Ckjïj k khk1 + kh k1 a holds when (i) k ½ 2 (ii) or k =1, if in addition it is assumed that û0(t) is monotonic. Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 23:02:31, subject to the Cambridge Core terms of use. 406 ABDELNASER J. AL-HASAN AND DASHAN FAN P ã n LEMMA 2[SEE [7]]. Suppose P( y)= jãjÄm aãy is a polynomial of degree m in R and ¢Ú 1.Then m Z X Á ¢ ¢ jP( y)j dy Ä A¢ jaãj jyjÄ1 jãjÄm The bound Aè may depend on ¢, m, and the dimension n, but it is independent of the coef®cients faãg. We shall need the following results from [4] (see also [3]). LEMMA 3. Let m 2 N,s=0Ò1ÒÒm, k 2 Z and fõsÒkg be a family of measures n + on R with õÃ 0Òk =0for every k 2 Z.Letfãsj : s =1Ò2ÒÒm and j =1Ò2g²R, + n n fës : s =1Ò2ÒÒmg²R nf1g,fNs :s=1Ò2ÒÒmg²N, and Ls: R ! R be linear transformations for s =1Ò2ÒÒm. Suppose for s =1Ò2ÒÒm (a) kõÃ sÒkkÄ1for k 2 Z; ã jõ ò jÄ ëkj òj s2 ò2 n 2 (b) Ã sÒk ( ) C( s Ls ) for R and k Z; ã jõ ò õ ò jÄ ëkj òj s1 ò2 n 2 (c) Ã sÒk ( ) Ã s1Òk( ) C( s Ls ) for R and k Z; (d) For some q Ù 1 there exists Aq Ù 0 such that þ þ þ þ supþjõsÒkjŁfþ ÄAqkfkLq(Rn) k2Z Lq(Rn) for all f 2 Lq(RnÂ). Ã 2 2q Ò 2q Then for every p q+1 q1 , there exists a positive constant Cp such that X (7) õ Ò Ł f Ä C k f k p n m k Lp(Rn) p L (R ) k2Z and X Á 1 2 2 (8) jõ Ò Ł f j Ä C k f k p n m k Lp(Rn) p L (R ) k2Z p n holds for all f 2 L (R ). The constant Cp is independent of the linear transformations f gm Ls s=1. Now we are in the position to prove Theorem 1. Note that TQmÒb( f ) is equal to Z Á n 0 0 (9) jyj b(jyj)Ω( y ) f x Qm(jyj)y dy Rn P Ω 2 1 n1 Ω ï where P H (S ) satis®es the mean zero property (1). We can write = jaj, jï jÄ kΩk 1 where j C H1(Sn1) and each aj is an -atom. So X k k Ä jï jk k TQmÒb( f ) p C j Bj( f ) p where Z Á n 0 0 Bj( f )(x)= b(jyj)jyj aj(y)f xQm(jyj)y dy Rn with aj being an 1-atom. Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 23:02:31, subject to the Cambridge Core terms of use. A SINGULAR INTEGRAL OPERATOR 407 Therefore, it suf®ces to show (10) kBj f kp Ä Ck f kp where C is independent of the coef®cients of the polynomial Qm(t) and the atoms aj(). For simplicity, we denote aj()bya()andBj(f)byB(f). Noting that supp(a)isin 0 Òö \ n1 0 2 n1 the ball B(x0 ) S for some x0 S , without loss of generality we may assume 0 Ò ÒÒ x0 = 1 =(1 0 0). k k+1 Let Ik =(2Ò2 ), then B( f )(x) is equal to Z X1 Á X1 n 0 0 b(jyj)jyj a( y ) ü (jyj) f x Q (jyj)y dy = õ Ò Ł f (x) n Ik m m k R k=1 k=1 where Z 0 n 0 iQm(jyj)hy Òòi õÃ mÒk(ò)= b(jyj)jyj a( y )e dy 2kÄjyjÄ2k+1 with m being the degree of the polynomial Qm. De®ne Z k+1 P Z 2 m j 0 0 1 i åjt hx Òòi 0 iQms(t)hy Òòi 0 õÃ msÒk(ò)= b(t)t e j=m s+1 0 a( y )e dõ( y ) dtÒ 2k Sn1 P Ò ÒÒ m å k s =1 2 m 1. Noting that Qm(t)= k=1 kt ,wehaveQ0(t)=0. So we de®ne k+1 Z2 Z 0 1 0 iQ0(t)hy Òòi 0 õÃ 0Òk(ò)= b(t)t a( y )e dõ( y ) dt 2k Sn1 then Z2k+1 Z 1 0 0 õÃ 0Òk(ò)= b(t)t a( y ) dõ( y ) dt =0Ò by (5).