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 x Qm(jyj)y dy m R is bounded in Lp(Rn)for1ÚpÚ1, and the bound is independent of the coef®cients of Qm(t).
Let Sn 1 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(Sn 1)and Z (1) Ω(x0)dõ(x0)=0Ò Sn 1 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 x Qm(jyj)y dy m Rn P 0 Ûj j2 n 1 j jΩ j j n 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(Sn 1) 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(Sn 1)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 Sn 1 is de®ned by
0 2 0 0 n Pry0 (x )=(1 r )Ûjry x j where 0 Ä r Ú 1andx0Òy0 2Sn 1. For any Ω 2 L1(Sn 1), 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 n 1 rx 0ÄrÚ1 S k +Ωk Ú1 Ω 1 n 1 1 If P L1(Sn 1) , we say that belongs to the Hardy space H (S ) with H -norm kΩk k +Ωk H1(Sn 1) = P L1(Sn 1). The space H1(Sn 1) was studied in [1] (see also [2]). In particular, it is known that
L1(Sn 1) Ã H1(Sn 1) Ã L Log+ L(Sn 1) Ã Lq(Sn 1)
for any q Ù 1. Another important property of H1(Sn 1) 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 2Sn 1Òjx0 x0jÚöfor some x0 2 Sn 1 and öÙ0g; Z 0 0 (5) a(ò0) dõ(ò0)=0Ò Sn 1 (n 1) (6) kak1 Äö Ω 2 1 n 1 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 n 1 C H1(Sn 1). 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 s 1Òk( ) 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 q 1 , 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 n 1 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(Sn 1) 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 x Qm(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 \ n 1 0 2 n 1 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 iQm s(t)hy Òòi 0 õÃ m sÒk(ò)= b(t)t e j=m s+1 0 a( y )e dõ( y ) dtÒ 2k Sn 1 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 Sn 1 then Z2k+1 Z 1 0 0 õÃ 0Òk(ò)= b(t)t a( y ) dõ( y ) dt =0Ò by (5). 2k Sn 1 Also, we easily see
Z2k+1 Z 1 0 0 kõÃ m sÒkkÄ jb(t)jt ja( y )j dõ( y ) dt Ä C 2k Sn 1
for all k 2 Z and s =0Ò1ÒÒm 1. n In the rest of this paper, for any non-zero ò =(ò1ÒÒòn) 2 R , we write òÛjòj = ò0 ò0ÒÒò0 2 n 1 =( 1 n) S . Suppose n ½ 3anda()isan1-atom on Sn 1 with supp(a) Â Sn 1 \ B(ò0Òö), where B(ò0Òö) is the ball in Rn centered at ò0.Let Z Á 0 2 (n 3)Û2 2 1Û2 0 0 Fa(úÒò )=(1 ú ) ü( 1Ò1)(ú) a úÒ(1 ú ) y dõ(y) Sn 2
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 =(öò1Òöò2ÒÒöòn).
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 )=(1 ú ) 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 jòj 1fö4ò2 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 fõm sÒkg satis®es the following conditions: for s =0Ò1Ò2ÒÒm 1 1 (m s)k (i) jõÃ m sÒkjÄCj2 jAöòkåm sk 4(m s) and åm s 6=0; (m s)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 Qm s. 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 =1Ò2ÒÒm 1wehave þ þ þõÃm sÒk(ò) õÃm s 1Òk(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=m s+1 j 0 a( y )e m s dõ( y ) dt 2k Sn 1 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=m s j 0 a( y )e m s 1 dõ( y ) dtþ 2k Sn 1
Let O be the rotation such that O(ò)=jòj1 and O 1 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 iQm s(t) y õ 1 iQm s(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 iQm s(t) y 1 dõ( y ) Sn 1
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 ÒÒ 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
jõÃm sÒk(ò) õÃ m s 1Òk(ò)j þZ k+1 P Z þ 2 m j 0 þ 1 i åjt hx Òòi 0 iQm s(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=m s 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=m s j 0 e m s k 2 Z þ m s 0 þ 0 iåm st (jòjú hx Òòi) þ ð Fa(úÒò )fe 0 1g dú dtþ R þZ k+1 Z þ þ 2 m s 0 þ þ 1 0 iå t (jòjú hx 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 m s jòjú h 0 Òòi þ ú þ þ b(t) t Fa( ) m st x0 d dtþ 2k R 0 Ò ÒÒ 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 ú ò0 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 m s þ 0 þ þ þ jõÃm sÒk(ò) õÃ m s 1Òk(ò)jÄCþ jb(t)jt jFa(úÒò )jþåm st þjòjr(ò )þ þ dú dtþ þ2k R þ Z 2k+1 0 m s 1 Ä Ckbk1jåm sj jòj jr(ò )j t dt 2k (m s)k Ä C2 jåm sjjAöòj
Similarly, we can prove that
mk jõÃmÒk(ò) õÃ m 1Òk(ò)jÄC2 jåmjjAöòj
This proves (ii). Now
Z k+1 P Z 2 m j 0 1 i åjt hx Òòi 0 iQm s(t)jòjú õÃ m sÒk(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 jòj 1j òj ò ö2ò Òöò ÒÒöò 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 iQm s(t)r(ò )jòjú õÃ m sÒk(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 iQm s(t)r(ò )jòjú jõÃ m sÒk(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 iQm s(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(Qm s(2 t) Qm s(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 jQm s(2 t) Qm s(2 ú)jjr(ò)j jòj ü(1Ò2)(t)ü(1Ò2)(ú)
Thus
n o 1 k k 0 2(m s) jL(tÒú)jÄCjQm s(2 t) Qm s(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(m s) k k 2(m s) Ä C jr(ò )j jòj sup jQm s(2 ú) Qm s(2 t)j dú tÙ0 R n o 1 1 0 2(m s) k Ä C jr(ò )j jòj j2 åm sj 2(m s)
This shows þ þ 1 þ 0 k þ 4(m s) kTk f k2 Ä Cþr(ò )jòj2 jåm sjþ which leads to
þ þ 1 þ þ 1 þ 0 k þ 4(m s) þ k þ 4(m s) jõÃm sÒk(ò)jÄCþr(ò)jòj2 jåm sjþ = CþjAöòj2 jåm sjþ
This shows, for s =0Ò1Ò2ÒÒm 2,
þ þ 1 þ k þ 4(m s) jõÃm sÒk(ò)jÄþ2jåm sjjAöòjþ
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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 =1Ò2ÒÒm 1, we can write
Z Â Xm ÁÃ n 0 0 j 0 õm sÒk Ł f (x)= b(jyj)jyj a( y ) f x Qm s(jyj)y åjjyj x dy kÄj jÄ k+1 0 2 y 2 j=m s+1 P Á j j 0 m å j jj 0 j j Ò j j ÒÒ j j Write Qm s( y )y + j=m s+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)ÒÒPn(jyj) ,then Z Á n 0 õm sÒk Łf(x)= b(jyj)jyj a( y ) f x PÄ (jyj) dy 2kÄjyjÄ2k+1 Thus Z þ Áþ nk þ þ sup jõm sÒk 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 2 k ja( y0)j þ f x PÄ (t) þ dõ( y0) dt j jÄ k+1 n 1 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 n 1 k2Z S rÙ0 r 0 Therefore, we have
Z Z þ Áþ p p 0 1 rþ þ 0 sup jõm sÒk Ł f j Ä ja( y )j sup þ f PÄ(t) þdt dõ( y ) Lp(Rn) n 1 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 =1Ò2ÒÒm 1
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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Department of Mathematical Sciences University of Wisconsin-Milwaukee Milwaukee, WI 53201 USA email: [email protected] [email protected]
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