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)=pv 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)=pv 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

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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=0Ò1ÒÒm, k 2 Z and fõsÒkg be a family of measures n + on R with õÃ 0Òk =0for every k 2 Z.Letfãsj : s =1Ò2ÒÒm and j =1Ò2g²R, + n n fës : s =1Ò2ÒÒmg²R nf1g,fNs :s=1Ò2ÒÒmg²N, and Ls: R ! R be linear transformations for s =1Ò2ÒÒm. Suppose for s =1Ò2ÒÒm (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.

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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 ÒÒ 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 ÒÒ 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). 2k Sn1 Also, we easily see

Z2k+1 Z 1 0 0 kõÃ msÒkkÄ jb(t)jt ja( y )j dõ( y ) dt Ä C 2k Sn1

for all k 2 Z and s =0Ò1ÒÒm1. n In the rest of this paper, for any non-zero ò =(ò1ÒÒòn) 2 R , we write òÛjòj = ò0 ò0ÒÒò0 2 n1 =( 1 n) S . Suppose n ½ 3anda()isan1-atom on Sn1 with supp(a) Â Sn1 \ B(ò0Òö), where B(ò0Òö) is the ball in Rn centered at ò0.Let Z  Á 0 2 (n3)Û2 2 1Û2 0 0 Fa(úÒò )=(1ú ) ü(1Ò1)(ú) a úÒ(1 ú ) y dõ(y) Sn2

Then, we have the following estimates for Fa when n ½ 3.

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0 LEMMA 4. Up to a constant multiplier independent of a(),Fa(úÒò ) is an 1-atom on R. More precisely, there is a constant C which is independent of a() such that  Á Â ò0 ò0 Òò0 ò0 supp(Fa) 1 2r( ) 1 +2r( ) ; 0 kF k1 ÄCÛr(ò); Za Fa(ú) dú =0Ò R

0 1 2 where r(ò )=jòj jAöòj and Aöò =(öò1Òöò2ÒÒöòn).

The function Fa can be similarly de®ned in the case n = 2. Suppose n =2anda()isan 1 1  ò0 ò0Òò0 2 1 -atom on S satisfying (4)±(6). The center of the support of a( )is =( 1 2) S . Let   Á  Áà 0 2 1Û2 2 1Û2 2 1Û2 fa(úÒò )=(1ú ) ü(1Ò1)(ú) a úÒ(1 ú ) +a úÒ(1 ú ) 

Similar to Lemma 4, we have 0 LEMMA 5. Up to a constant multiplier independent of a(),fa(úÒò ) is a q-atom on R, where q is any ®xed number in the interval (1Ò 2). The radius of its support is ò0 jòj1fö4ò2 ö2ò2g1Û2 ò0 r( )= 1 + 2 , and the center of its support is 1. The proofs of the above two Lemmas can be found in [6]. By inspecting the proof of Theorem 1 using Lemma 3, we only need to check that the family fõmsÒkg satis®es the following conditions: for s =0Ò1Ò2ÒÒm1 1 (ms)k (i) jõÃ msÒkjÄCj2 jAöòkåmsk 4(m s) and åms 6=0; (ms)k (ii) jõÃ Ò (ò) õÃ Ò (ò)jÄCj2 jAöòkå k; m s k m s 1 k m s (iii) sup 2 jõ Ò Ł f j Ä Ck f k p n k Z m s k Lp(Rn) L (R ) where C Ù 0 is independent of k 2 Z, ò2Rn, and the coef®cients of the polynomial Qms. We will only prove the case n Ù 2, since the proof for n = 2 is essentially the same (using Lemma 5 instead of Lemma 4) with a slight modi®cation. In fact for s =1Ò2ÒÒm1wehave þ þ þõÃmsÒk(ò)õÃms1Òk(ò)þ þZ k+1 P Z þ 2 m j 0 0 þ 1 i å t hx Òòi 0 iQ (t)hy Òòi 0 =þ b(t)t e j=ms+1 j 0 a( y )e m s dõ( y ) dt 2k Sn1 Z k+1 P Z þ 2 m j 0 0 þ 1 i å t hx Òòi 0 iQ (t)hy Òòi 0 þ b(t)t e j=ms j 0 a( y )e m s 1 dõ( y ) dtþ 2k Sn1

Let O be the rotation such that O(ò)=jòj1 and O1 be its inverse. Noting that a( y0)has support in B(1Òö), so by an argument of rotation Z Z 0 h 0Òòi 0 0 jòjh 0Ò i 0 iQms(t) y õ 1 iQms(t) y 1 õ a( y )e d ( y )= a(O y)e d ( y ) Sn 1 ZSn 1 0 jòjh 0Ò i 0 = A( y )e iQms(t) y 1 dõ( y ) Sn1

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 409 0 1 ò0Òö 0 úÒ 0 ÒÒ 0 where A( y )isan -atom supported in B( ). Thus, let y =( y2 yn), then by 0 the de®nition of Fa(úÒò ) and Lemma 4, we have

jõÃmsÒk(ò) õÃ ms1Òk(ò)j þZ k+1 P Z þ 2 m j 0 þ 1 i åjt hx Òòi 0 iQms(t)jòjú = þ b(t)t e j=m s+1 0 Fa(úÒò )e dú dt 2k R Z k+1 P Z þ 2 m j 0 þ 1 i å t hx Òòi 0 iQ (t)jòjú þ b(t)t e j=ms j 0 F (úÒò )e m s 1 dú dtþ k a þ 2 R Z k+1 P þ 2 m j 0 þ 1 i å t hx Òòi iQ (t)jòjú = þ b(t)t e j=ms j 0 e m s k 2 Z þ ms 0 þ 0 iåmst (jòjúhx Òòi) þ ð Fa(úÒò )fe 0 1g dú dtþ R þZ k+1 Z þ þ 2 ms 0 þ þ 1 0 iå t (jòjúhx Òòi) þ Ä þ jb(t)jt jF (úÒò )jje m s 0 1j dú dtþ k a þ 2 R þ þZ 2k+1 Z þ  Áþ þ Ä þ j j 1 j úÒò0 jþå ms jòjú h 0 Òòi þ ú þ þ b(t) t Fa( ) mst x0 d dtþ 2k R 0 Ò ÒÒ Since x0 = 1 =(1 0 0), then þ þ þ þ þ þ þjòjú hx0 Òòiþ = þjòjú h1Òòiþ=þjòjú jòjò0 þ 0 þ þ þ þ 1 þjòj úò0 þÄ þjòj ò0 þÒ ú2  = ( 1) C r( ) for all supp(Fa)

Thus þ þ þ þ þZ2k+1 Z þ þ þÁþ þ þ 1 0 þ ms þ 0 þ þ þ jõÃmsÒk(ò) õÃ ms1Òk(ò)jÄCþ jb(t)jt jFa(úÒò )jþåmst þjòjr(ò )þ þ dú dtþ þ2k R þ Z 2k+1 0 ms1 Ä Ckbk1jåmsj jòj jr(ò )j t dt 2k (ms)k Ä C2 jåmsjjAöòj

Similarly, we can prove that

mk jõÃmÒk(ò) õÃ m1Òk(ò)jÄC2 jåmjjAöòj

This proves (ii). Now

Z k+1 P Z 2 m j 0 1 i åjt hx Òòi 0 iQms(t)jòjú õÃ msÒk(ò)= b(t)t e j=m s+1 0 Fa(úÒò )e dú dt 2k R

By Lemma 4, withoutÁ loss of generality we may assume that Fa is a 2-atom with support ò0 Ò ò0 ò0 jòj1j òj ò ö2ò Òöò ÒÒöò in 2r( ) 2r( ) ,where r( )=Á Aö ,andAö =( 1 2 n). 0 0 0 Thus A(ú)=r(ò)Fa r(ò)úÒò is a 2-atom with support in the interval (1Ò 1). After change of variables we have

Z k+1 P Z 2 m j 0 0 1 i åjt hx Òòi iQms(t)r(ò )jòjú õÃ msÒk(ò)= b(t)t e j=m s+1 0 A(ú)e dú dt 2k R

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By Holder'sÈ inequality,

²Z k+1 þZ þ ¦ Û 2 þ 0 þ2 1 2 kÛ2 iQms(t)r(ò )jòjú jõÃ msÒk(ò)jÄCkbk12 þ A(ú)e dúþ dt 2k R ²Z þZ þ ¦ Û 2 k 0 1 2 kÛ2 kÛ2 þ iQ (2 t)r(ò )jòjú þ2 = Ckbk12 2 þ A(ú)e m s dúþ dt 1 R = Ckbk1Θk

1 To estimate Θk, we choose a function †2C (R) satisfying

†(t) Á 1forjtjÄ1Ò†(t)Á0forjtj½2

De®ne Tk by Z k 0 iQms(2 t)r(ò )jòjú (Tk f )(t)=ü(1Ò2)(t) e †(ú) f (ú) dú R Then Z Ł Òú ú úÒ TkTk f (t)= L(t )f( )d R where Z k k ú ò0 jòj iv(Qms(2 t) Qms(2 ))r( ) 2 L(tÒú)= e † (v) dvü(1Ò2)(t)ü(1Ò2)(ú) R We easily see that jL(tÒú)jÄCü(1Ò2)(t)ü(1Ò2)(ú) On the other hand, by Lemma 1, we have

n o k k 0 1 jL(tÒú)jÄC jQms(2 t) Qms(2 ú)jjr(ò)j jòj ü(1Ò2)(t)ü(1Ò2)(ú)

Thus

n o 1 k k 0 2(ms) jL(tÒú)jÄCjQms(2 t) Qms(2 ú)jjr(ò)j jòj ü(1Ò2)(t)ü(1Ò2)(ú)

By invoking Lemma 2 we have Z Z sup jL(tÒú)jdt ≤ sup jL(úÒt)jdú úÙ0 R tÙ0 R n o 1 Z n o 1 0 2(ms) k k 2(ms) Ä C jr(ò )j jòj sup jQms(2 ú) Qms(2 t)j dú tÙ0 R n o 1 1 0 2(ms) k Ä C jr(ò )j jòj j2 åmsj 2(m s) 

This shows þ þ 1 þ 0 k þ 4(ms) kTk f k2 Ä Cþr(ò )jòj2 jåmsjþ which leads to

þ þ 1 þ þ 1 þ 0 k þ 4(ms) þ k þ 4(ms) jõÃmsÒk(ò)jÄCþr(ò)jòj2 jåmsjþ = CþjAöòj2 jåmsjþ 

This shows, for s =0Ò1Ò2ÒÒm2,

þ þ 1 þ k þ 4(ms) jõÃmsÒk(ò)jÄþ2jåmsjjAöòjþ 

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For s = m 1, following the same argument above, we easily obtain

þ þ 1 þ k þ 4 jõÃ1Òk(ò)jÄþ2jå1jjAöòjþ 

This proves (i). Finally, we need to check (iii). For s =1Ò2ÒÒm1, we can write

Z Â  Xm ÁÃ n 0 0 j 0 õmsÒk Ł f (x)= b(jyj)jyj a( y ) f x Qms(jyj)y åjjyj x dy kÄj jÄ k+1 0 2 y 2 j=ms+1 P  Á j j 0 m å j jj 0 j j Ò j j ÒÒ j j Write Qms( y )y + j=ms+1 j y x0 = P1( y ) P2( y ) Pn(y) , where each Pj is a j j 0 0 å0 polynomial of y whose coef®cients depends on y Á, x0 and js. We denote PÄ (jyj)= P1(jyj)ÒP2(jyj)ÒÒPn(jyj) ,then Z  Á n 0 õmsÒk Łf(x)= b(jyj)jyj a( y ) f x PÄ (jyj) dy 2kÄjyjÄ2k+1 Thus Z þ  Áþ nk þ þ sup jõmsÒk Ł f (x)jÄsup kbk12 þa( y) f x PÄ(jyj) þ dy k Äj jÄ k+1 k2Z k2Z 2 y 2 Z Z þ  Áþ '¾ sup 2k ja( y0)j þ f x PÄ (t) þ dõ( y0) dt j jÄ k+1 n1 k2Z t 2 S So we have  Á Z Â Z þ  Áþ Ã p 1 rþ þ p 0 0 sup jõ Ò Ł f (x)j Ä sup þ f x PÄ (t) þ dt ja( y )j dõ( y ) m s k n1 k2Z S rÙ0 r 0 Therefore, we have

Z Z þ  Áþ p p 0 1 rþ þ 0 sup jõmsÒk Ł f j Ä ja( y )j sup þ f PÄ(t) þdt dõ( y ) Lp(Rn) n1 p n k2Z S rÙ0 r 0 L (R ) It was shown that

Z þ  Áþ p 1 rþ  Ä þ Ä k kp Ú Ú1 sup f P(t) dt C f Lp(Rn) for 1 p rÙ0 r 0 Lp(Rn) Ä 0 0 å0 with C independent of the coef®cients of P (thus independent of y , x0 and js)(see [8, pp. 476±478]). So we have for s =1Ò2ÒÒm1

sup jõ Ò Ł f j Ä Ck f k p n for 1 Ú p Ú1 m s k Lp(Rn) L (R ) k2Z Using the exact same argument, we can check that

sup jõ Ò Ł f j Ä Ck f k p n for 1 Ú p Ú1 m k Lp(Rn) L (R ) k2Z The theorem is proved.

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REFERENCES 1. L. Colzani, Hardy Spaces on Sphere. Ph.D. Thesis, Washington University, St. Louis, 1982. 2. L. Colzani, M. Taibleson and G. Weiss, Maximal estimates for CesÁaro and Riesz means on sphere. Indiana Univ. Math. J. (6) 33(1984), 873±889. 3. D. Fan, K. Guo and Y. Pan, Singular Integral along Submanifolds of Finite Type. Michigan Math. J. 45(1998), 135±142. 4. D. Fan and Y. Pan, Singular Integral Operators with Rough Kernels Supported by Subvarieties.Amer. J. Math. 119(1997), 799±839. 5. , L2-Boundedness of a singular integral operator. Publ. Mat. 41(1997), 317±333. 6. , A singular integral operator with rough kernel. Proc. Amer. Math. Soc. 125(1997), 3695±3703. 7. F. Ricci and E. M. Stein, Harmonic analysis on nilpotent groups and singular integrals I: Oscillatory integrals.J.Funct.Anal.73(1987), 56±84. 8. E. M. Stein, Harmonic Analysis: Real-Variable Methods, Orthogonality and Oscillatory Integrals. Princeton University Press, Princeton, NJ, 1993.

Department of Mathematical Sciences University of Wisconsin-Milwaukee Milwaukee, WI 53201 USA email: [email protected] [email protected]

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