A note on a conjecture of Calderón

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1 A ote o a cojecture of Calderó Jiecheg CHEN & Xiagrog ZHU Dept of Math Xixi Campus), Zhejiag Uiversitry Abstract For f SR ) ad Ω L 1 S 1 ), S 1 Ωx )dx = 0, defie T Ω f)x) = lim ɛ 0+ x y ɛ Ωy/ y ) y fx y)dy I this paper, we shall prove that there are a class of fuctios i H 1 S 1 ) L l + LS 1 ) such that T Ω is weak type L 1 bouded Supported by 973 project ad NSFZJ 1

2 1 Itroductio For f SR d ) ad Ω L 1 S d 1 ), S d 1 Ωx )dx = 0, defie Ωy/ y ) T Ω f)x) = lim ɛ 0+ x y ɛ y d fx y)dy 1) I [1], Calderó ad Zygmud proved that if Ω L l + LS d 1 ), ie S d 1 Ωx ) l + Ωx ) )dx <, ) T Ω is L p bouded for 1 < p < I [7] ad [5], Ricci, Weiss ad idepedetly Coett proved that if Ω H 1 S d 1 ), T Ω is L p bouded for 1 < p < Also see [4] But, it remaied ope for log time if T Ω is weak type L 1 bouded uder the correspogdig coditios for Ω L l + LS d 1 ), it is a cojecture of Calderó) I [3], Christ ad Rubio de Fracia proved that for Ω L l + LS d 1 ) d 7), T Ω is weak type L 1 bouded Ad at the same time, i [6], Hofma idepedetly proved that for Ω L q S d 1 ) d =, q > 1), T Ω is weak type L 1 bouded Fially, i [8], Seeger geeralized the result to all d ad Ω L l + LS d 1 ) We kow that L l + LS d 1 ) H 1 S d 1 ) So, it is atural to ask a harder questio: for Ω H 1 S d 1 ), is T Ω weak type L 1 bouded? I [9], Stefaov proved that if Ω is a fiite sum of H 1 S 1 ) atoms with the additioal assumptio that the atoms are supported o almost disjoit arcs of comparable size ote that such a Ω must be L S 1 ) fuctio), T Ω L 1 W L 1 essetially depeds oly o Ω H1 Precisely, he proved Theorem 1 Let N, l be positive itegers satisfyig N π l c 0 where c 0 is suitably choose, ad I deote the arc i S 1 with ceter e ad satisfyig I l, I I m 1 mi I, I m ) for m Suppose Ω = N 1 a where > 0 ad a is a H 1 S 1 ) atom o S 1 satisfyig suppa ) I, a l ad S 1 a θ)dθ = 0 The, T Ω L 1 W L 1 C N =1 ) where C is idepedet of N ad l I this paper, we shall prove that there are a class of fuctios i H 1 S 1 ) L l + LS 1 ) such that T Ω is weak type L 1 bouded We have

3 Theorem Let I deote the arc i S 1 with ceter e ad legth ρ, disjoit mutually, ad A = sup θ S 1 :θ / I ρ < ; 3) θ e let Ω = 1 a where 1 < ad a is a H 1 S 1 ) atom o S 1 satisfyig i) suppa ) I, ii) S 1 a θ)dθ = 0, iii) a ρ 1, 4) The, {x : T Ω f)x) > } C A ) 1 f 1 5) for all f SR ) ad > 0, where C is idepedet of f, ad Ω Theorem 3 Suppose ρ 0, 1) ad ρ < There are {e } S 1, arcs I S 1 with ceter e ad legth ρ such that for ay sequece of H 1 atoms {a } satisfyig 4))ad Ω = a, {x : T Ω f)x) > } C ρ f 1 6) for all f SR ) ad > 0, where C is idepedet of f, ad Ω From Theorem 3, we have Corollary 4 For ay icreasig fuctio ϕ : [0, ) [0, ) with ϕ0+) = 0, if ϕt) lim =, 7) t t the, there must be a Ω H 1 ϕl) such that T Ω is weak type L 1 bouded, where Some lemmas ϕl) = { } Ω : ϕ Ωθ) )dθ < S 1 Without loss of geerality, we always assume that 1 = 1 ad > 0 3

4 For a rectagle Q = Qy, r) R ) with ceter y ad sides legth r = r 1, r ), let mq = Qy, mr), dq) = maxr 1, r ) Set ad A = Q : the loger side of Q is parallel to e, the shorter side legth is ρ time of the loger side legth 1 M f)x) = sup x Q A Q Q fz) dz It is easy to see that M is weak type L 1 bouded ad sup M L 1 W L 1 we first give a modified Whitey s decompositio, < Now, Lemma 5 Suppose E R is ope There is m R + such that for ay, there are mutually disjoit rectagles {Q,i } A satisfyig i) ii) E = i Q,i 4Q,i E iii) mq,i E c { } iv) dq,i ) k : k = 0, ±1, ±, This lemma ca be proved alog the idea of the proof of the Whitey s decompositio, see[10] For f L 1 R ) ad > 0, let E = { } x : M f)x) > E = {x : Mf)x) > } E ) where M is the Hardy-Littlewood maximal operator For ay, we shall make C Z decompositio of f based o the modified Whitey s decompositio of E ot E, key poit) By Lemma 5, we have that E = i Q,i where {Q,i } satisfy the coditios i Lemma 5 Take We have b Q,i = fx) 1 g x) = i 1 Q,i Q,i gx) = fx)χ E cx) B,j = i:dq,i )= j b Q,i Q,i fy)dy)χ Q,i x) Q,i fy)dyχ Q,i x) 4

5 Lemma 6 For ay, f = g + g + i b Q,i, ad i) ii) R b Q,i x)dx = 0 bq,i R x) dx C Q,i bq,i i Q,i x) dx f 1 g C f 1, g C f 1 iii) E E + {x : Mf)x) > } C 1 f 1 Lemma 7 Uder the hypothesis of Theorem, for all f SR ) ad > 0 a ) g C A f 1 8) Proof We first prove that a ) ) ξ) C A 9) By a well-kow computatio see [10] page 39), a ) ) 1 ξ) = a θ) l S 1 θ, ξ ξ + πi sig θ, ξ dθ 10) which is idepedet of ξ So, we may assume ξ = 1 Let ξ S 1 deote ayoe of the TWO uit vectors orthogoal to ξ If ±ξ / I, sig θ, ξ is costat for θ I, thus without loss of geerality, we may assume that e, ξ ) < π/, i this case, e, ξ e ξ ) ) a ) ξ) = S 1 a θ) l θ, ξ 1 l e, ξ 1) dθ S 1 a θ) l1 + θ e, ξ e, ξ 1 )dθ C sup θ I θ e e ξ C ρ e ξ 11) By 11), we have ote that A 1) ) a ) ξ) = + ) ) a ) ξ) :ξ I or ξ I :±ξ / I C1 + ρ ) C A e ξ 1) 5

6 Thus a ) g = = S 1 Lemma 7 is proved m S 1 m m a ) ) g ) a ) C A m ĝ m C A m m ) ) ) ξ) g ) am ξ) m ξ) g m ) ξ)dξ g m ) ξ) ) a ) ξ) dξ ) m m f 1 C A f 1 Take β C c 1, )) such that 0 β 1 ad j β j t) = 1 for all t R +, ψ Cc 1, 1)) such that 0 ψ 1 ad ψ [ 1, 1 ] = 1 ad ψ k) ± 1 ) = 0 for all k = 0, 1,, Let a,j x) = β j x ) ax/ x ) x ) ϕ s x) = ψ s 13) 6 ρ 1, e ) x) We have Lemma 8 Uder the hypothesis of Theorem, a,j ϕ s B,j s C A f 1 14) s>0 j Proof By a similar estimate i [9] see sectio 5 below for details), we have a,j B,j s C s f j 1 15) Now, we estimate ϕ s ξ) where s 0 If ϕ s ξ) = ψ s 6 ρ 1 ξ ξ, e ) > 0, we have s 6 ρ 1 ξ ξ, e 1, thus ξ ξ, e s 6 ρ 16) Note that ξ ξ, e ξ e for all satisfyig ±ξ / I ad e, ξ ) < π/, we have ξ s e C 6 ξ ρ, ad thus ρ 1 e C s 6 By Theorem 3), the umber of satisfyig 16) does ot exceed CA s 3, thus ϕ s ξ) CA s 3 17) 6

7 By 15) ad 17), ) a,j ϕ s B,j s s>0 j s>0 a,j ϕ s B,j s j ) ) = ϕ s>0 s â,j B,j s j s>0 ϕ s ξ) ) 1/ ) 1/ â,j B,j s j CA s 3 1/ a s>0,j B,j s j ) ) ) CA s 3 s 1/ f 1 = CA f 1 s>0 Lemma 8 is proved Lemma 9 Uder the hypothesis of Theorem, a,j δ ϕ s ) B,j s 1 C f 1 18) s>0 j Proof By a similar estimate i [9] see sectio 5 below for details), we have a,j δ ϕ s ) 1 C s 1 19) So Lemma 9 is proved a,j δ ϕ s ) B,j s 1 s>0 j C s 1 B,j s 1 = C f 1 s>0 j 3 Proof of Theorem For fixed f ad, T Ω f)x) = T Ω g)x) + a ) g x) + i a ) b Q,i x), 7

8 so, {x : T Ω f)x) > } I + II + III I = {x : T Ω g)x) > /3} { } II = x : a ) g x) > /3 { } III = x : i a ) b Q,i x) > /3 0) By L boudedess of T Ω, I C T Ω g) C Ω H 1 g C ) f 1 = C 1 f 1 1) Similarly, we have For III, we have II C a ) g x) CA 1 f 1 ) a ) b i Q,i x) = a,j B,j s x) s0 j + a,j B,j s x) s>0 j = a,j B,j s x) s0 j + a,j ϕ s B,j s x) s>0 j + a,j δ ϕ s ) B,j s x) s>0 j 3) If dq,i ) j, suppa,j b Q,i ) 4Q,i, so supp a,j B,j s ) i 4Q,i = E s0 j Noticig that E C 1 f 1, we have { } III E + x : a,j ϕ s B,j s x) s>0 j > /9 { } + x : a,j δ ϕ s ) B,j s x) > /9 s>0 = E + IV + V j 4) 8

9 By Lemma 8, IV C a,j ϕ s B,j s s>0 j CA 1 f 1 5) By Lemma 9, V C 1 a,j δ ϕ s ) B,j s CA 1 f 1 6) 1 s>0 j Combiig 0)-) ad 4)-6), we get {x : T Ω f)x) > } CA 1 f 1 Theorem is proved 4 Proof of Theorem 3 Without loss of geerality, we may assume that {ρ } is decreasig ad ρ < π/64 Let d m = 1 sup m m ρ We have Lemma 10 m d m 16 ρ Proof For m, ρ 1 i ρi ) im ρi ), thus d m m m m im ρi ) O the other had, m ρi ) m ρi ρj m im m jm ij = ρj ρi ) m j ij m j 4 j 4 j j 1 ρj ij ρi ) ρ j ) 1/ j 1 ρi ) ) 1/, j ij so d m m m m im ρi ) 16 j ρ j Lemma 10 is proved 9

10 By Lemma 10 ad the assumptio ρ < π/64, we get m d m < π/ So, we ca choose {e m } S 1, such that e m+1 e m = d m ad 0 < arg e m < arg e m+1 < π/ for all m I additio, by the fact d m ρ m + ρ m+1, {I m } are disjoit mutually We shall first apply iductio to prove that e m e for m > For m = + 1, e m e = d > m ) m ρ, we have m ) m ρ 7) +1 ρ = m ) m e m+1 e = e m+1 e m + e m e d m + m ) m ρ > m+1 ) m+1 ρ So, 7) holds for all m > Now, we shall prove that For θ S 1, we first cosider sup θ S 1 :θ / I m ρ Suppose e m e + m ) ) m ρ ρ < 8) θ e N + θ N θ N 0 θ def = { : 0 arg θ arg e π/} def = { : 0 arg θ arg e π/} def = { : arg θ arg e > π/} Label the elemets i N + θ by sub-idex such that < θ e < θ e 1 < θ e 0 The, N + θ = { < 1 < 0 } or N + θ = { K < < 1 < 0 } By 7), i the secod case, θ e l > e K e l > l K l ) K ρ l, thus : N + θ,θ / I ρ θ e i the fist case, θ e l > l ρ l, so 1 + K 1 K 1 l=0 l=1 ρ θ e : N + θ,θ / I 1 + l=1 ) l K l ) K l K l ) K ) C < ; l ) = C < 9) 10

11 Label the elemets i N θ by sub-idex such that θ e 0 < θ e 1 < θ e < The, N + θ = { 0 < 1 < } or N + θ = { 0 < 1 < < K } By 7), it is easy to show that I additio, : N 0 θ,θ / I : N θ,θ / I ρ C < 30) θ e ρ θ e C : N 0 θ,θ / I ρ = C < 31) From 9)-31), we get 8) By 8) ad Theorem, we get Theorem 3 Fially, we prove Corollary 4 By 7), we ca choose {t } such that 1 < t 1 < t < ad ϕt ) > t Set =, ρ = t 1, Ω = a where {a } are H 1 atoms satisfyig 4) ad a θ) = ρ 1 ie Ω / ϕl) for θ I The, Ω H 1 S 1 ), but S 1 ϕ Ωθ) )dθ = ρ ϕ ρ 1 ) =, 5 Appedix I the proofs of Lemmas 8-9, we apply the estimates 15) ad 19) without proofs I what follows, we shall give details of their proofs alog the ideas developped i [], [3], [6], [8] ad [9] Proof of 15) We have j a,j B,j s = a,j B,j s, a,i B,i s j i B,j s, ã,j a,i B,i s j ij B,j s 1 ã,j a,i B,i s j ij f 1 sup j ã,j a,i B,i s, ij 3) 11

12 where ã,j x) = a,j x) We first estimate i 3 ã,0 a,i B,i s 0) We have ã,0 a,i B,i s 0) = R B,i s y) R a,0 y + z)a,i z)dz)dy = B,i s y) + R S 1 def = S 1 a θ)t i,s a )θ)dθ 0 a y+tθ y+tθ ) β y+tθ ) β i t) y+tθ t dt)a θ)dθ)dy where So, by 4), T i,s a )θ) = R B,i s y)l y i a )θ)dy L y i a )θ) = + i 3 ã,0 a,i B,i s 0) 0 a y+tθ y+tθ ) β y+tθ ) β i t) y+tθ t dt i 3 S 1 a θ)t i,s a )θ)dθ sup θ i 3 T i,sa )θ) 33) For coveiece, let Q θ) def = Θ,i,θ def = { } y : 1 y, 4 y 3, θ 4ρ { y, t) : a y+tθ y+tθ ) β y+tθ ) β i t) y+tθ t 0 } 34) where θ be oe of the two uit vectors orthogoal to θ The suppl y i a )θ)) Q θ) 35) Actually, if L y i a )θ) 0, i 3 ad y, t) Θ,i,θ, we have y y + tθ + tθ + i+1 3, y y + tθ tθ 1 i+1 1 4, ad Let y, θ = y + tθ, θ y + tθ y+tθ y+tθ, θ y + tθ y+tθ y+tθ θ ) y+tθ y + tθ y+tθ e + e θ 4ρ Q s θ) = { } y : y 4 ad y, θ ρ s where s > 0, the i 3 T i,s a )θ) i 3 i 3 R B,i s y)l y i a )θ)dy + ) j:q,j Q s θ),dq,j )= i s j:q,j Q s θ)=,dq,j )= i s R b,j y)l y i a )θ)dy def = I + II 36) 1

13 For I, oticig that we have L y i a )θ) I + 0 ρ 1 β y + tθ ) β i t) y + tθ t Cρ 1 i 3 j:q,j Q s θ),dq,j )= i s Cρ 1 Cρ 1 j:q,j Q sθ),dq,j ) s Q,j ρ s = C s dt Cρ 1, Q,j To estimate II, for y Q,j Q θ) where dq,j ) = i s, y θ such that y y θ //θ, y θ y Q //θ where y Q is the ceter of Q,j, thus 37) y y θ = y y Q, θ i s y θ y Q = y y Q, θ C i s ρ yy θ Q s θ) = y θ y Q Q s θ) = By Lemma 11 below ad 38), we have L y i a )θ) L y θ i a )θ) i s sup θ Lz i a )θ) C s ρ 1 z yy θ L y θ i a )θ) L y Q i a )θ) i s ρ sup L z θ i a )θ) C s ρ 1 z y θ y Q 38) 39) So, II = i 3 R b,j y)l y i a )θ)dy j:q,j Q s θ)=,dq,j )= i s R b,j y) L y i a )θ) L y Q i a )θ) dy i 3 j:q,j Q sθ)=,dq,j )= i s C i 3 j:q,j Q s θ)=,q,j Q θ),dq,j )= i s sup y L y i a )θ) L y θ i a )θ) + L y θ i C s ρ 1 j:q,j Q θ),dq,j ) s Q,j C Q,j a )θ) L y ) Q a )θ) i s 40) By 33), 36), 38) ad 40), we get i 3 ã,0 a,i B,i s 0) C s 13

14 By traslatio argumets, ã,0 a,i B,i s By dilatio argumets, i 3 L R ) C s ã,j a,i B,i s C ij s 41) Combiig with 3), it gives 15) Lemma 11 For θ suppa ), y, θ ρ s where s > 0, y 4, we have i) ii) θ Ly i a )θ) i ρ 1 L y θ i a )θ) i+ s ρ Proof Without loss of geerality, we may assume θ = 1, 0) Let e 1 = 1, 0), e = 0, 1), the y = y 1 e 1 + y e, y ρ s Settig w = y+tθ y+tθ = y+te 1 y+te 1, we have Jw, y) = dw dt = y y + te 1 4) Note that y + te 1, w = 0, so t = y,w e 1,w Obviously, for y, t) Θ,i,θ where Θ,i,θ is defied by 34), we have t y 1 = 1 t y = e,w e 1,w = e 1,w e,w Cρ 1 s We first estimate y 1 L y i a )θ) By 43), y+te 1) y 1 = 0 = y+te 1 Thus, y 1 Jw, y) = y y 1 y 1 y + te 1 ) = 0, ad for y, t) Θ,i,θ, β y + te 1 ) β i t) y 1 y + te 1 ) t C β i t) ) y 1 t C i t + 1 t C i Therefore y 1 L y i a )θ) a w) + a w) y 1 β y+tθ ) β i t) dw y+tθ t ) Jw,y) β y+tθ ) β i t) Jw,y) y 1 y+tθ t Jw,y) dw Jw,y) Cρ 1 i w[ i 1, i+1 ]) Cρ 1 i i+1 i 1 dt = Cρ 1 i dw Jw,y) 43) 44) 14

15 Now, we estimate y L y i a )θ) At first, by 43), we have y+te 1 y = y+te 1, C y+te y 1 ) y+te 1 t y + 1) Cρ 1 s y y + te 1 ) 45) Thus, by 43) ad 45), for y, t) Θ,i,θ, y β i t) t ) C i t+1 t t y χ i 1 t i+1t) C i ρ 1 s y β y+te 1 ) ) y+te 1 C y+te 1 + y+te 1 y+te 1 ) y+te 1 4 y χ 1 y+te 1 Cρ 1 s, which meas that β y + te 1 ) β i t) y y + te 1 ) t C i ρ 1 s 46) I additio, for y, t) Θ,i,θ such that y, θ ρ s, we have thus, By 46) ad 47), we get Lemma 11 is proved Proof of 19) Jw,y) y = y y y+te 1 ) C1 + y ρ 1 ) y+te1 + y y y+te 1 ) y y+te 1 y ) Jw, y) Jw, y) y C1 + y ρ 1 y y L y i a )e 1 ) a w) + a w) C i+ s ρ 1 +Cρ 1 ) y β y+tθ ) β i t) dw y+tθ t ) Jw,y) β y+tθ ) β i t) Jw,y) y y+tθ t Jw,y) ρ 1 dw w[ i 1, i+1 ]) Jw,y) dw w[ i 1, i+1 ]) Jw,y) s ρ 1 C i+ s ρ Cρ 1 s 47) i+1 i 1 dt = C i+ s ρ dw Jw,y) For simplicity, we omit the subidex By the defiitio of ϕ s ie ϕ s, see 13)) ad ψ, we have δ ϕ s ) ) = 1 ψ s 6 ρ 1, e ) = φ s 6 m ρ 1, e ) m 0 15

16 where φ C0, 1 ) 1, )) is defied by φ {u: 1 < u 1} = 1 ψ, φ {u:1 u <} = ψ ), φ {u: 1 < u <}c = 0 which satisfies m 0 φ m u) = 1 ψu) for all u R 1 Take a oegative fuctio ς { } C0 1 < < ) such that + ς k u) = 1 for all u R 1, ad set L k = ς k )), the a j δ ϕ s ) = Cm s,k ) a j L k m 0 k 48) Cm s,k y) = ς k y)φ s 6 m ρ 1 y y, e ) where a j is just a,j defied by 13) with ceter e = e ad radius ρ = ρ We first estimate Cm s,k ) 1 Defie a iversible liear operator A km : R R by A k,m y = k+ s 6 +m ρy 1, k y ) where y 1 //e ad y e Let h = s 6 +m ρ, by Sobolev imbeddig theorem f C 1 α α fq )) where Q is a osigular liear trasform o R ) C s,k m ) 1 C = C I the supports of ς ad φ, we have which meas that so From 49)-51), we get C s,k m ) 1 C y α α y α α C s,k m A k,m y)) ςhy 1 + y )φ y 1,e hy 1 +y ) ) 1 hy 1 + y, 1 y 1 hy 1 + y, y 1 4, y, h hy 1 + y y 1 49), 50) ) y 1, e α y ςhy 1 + y )φ hy 1 + y ) C 51) α y α ) y 1, e ςhy 1 + y )φ hy 1 + y ) C 5) 16

17 Now, a j L k 1 = ay/ y )β j y ) y = S 1 0 R S 1 j+1 C C β j r) aθ) r sup r [ j 1, j+1 ],θ suppa) L k x y)dy dx L kx rθ)drdθ dx j 1 j aθ) L k x rθ) L k x re) drdθdx R L k x rθ) L k x re) dx sup r θ e L k 1 C k+j ρ r [ j 1, j+1 ],θ suppa) 53) By 5)-53), C s,k m ) a j L k 1 C k+j ρ 54) But, for larger k say, k + j 0), the estimate 54) is ot eough, we eed some other estimate Applyig Sobolev imbeddig thorem agai, we get by 50)-51) Write C s,k m ) a j L k 1 C C y α α α C s,k m A k,m y)â j A k,m y)) α y â j A k,m )) χ suppc s,k m A k,m )) 55) Note that y = y 1 e + y e suppc s,k m A k,m )), θ = θ 1 e + θ e suppa), θ 1 1, θ ρ ς k A k,m y) 0 = A k,m y k φ s 6 m ρ 1 A k,my A k,m y, e ) 0 = A k,my A k,m y, e s 6 +m ρ I additio, θ e < ρ ad s 6 + m > 0, so A k,my A k,m y, θ s 6 +m ρ A k,m y, θ s 6 +m+k ρ, 56) 57) ad ) α y A k,m y, θ = y α θ 1 s 6 +m+k ρy 1 + θ k y C s 6 +m+k ρ) α1 k ρ) α C s 6 +m+k ρ) α 58) 17

18 Now, by itegratio by parts, â j A k,m y) = aθ) β j r) r e ir Ak,my,θ drdθ = aθ) i A k,m y,θ ) 3 3 r β j r) r )e ir Ak,my,θ drdθ So, by 57)-58) ad the fact r 3 β j r) )r α 1 C j3 α +1), we have r y α â j A k,m y)) C aθ) ir) γ y γ A k,m y,θ i A 0γα k,m y,θ ) 3 3 r β j r) r )e ir Ak,my,θ drdθ C 0γα ρρ 1 s 6 +m+k ρ) 3 j3 γ +1) j s 6 +m+k ρ) γ = C s 6 +m+k+j ρ) 3+ γ 0γα 59) By 55), 50) ad 59), we have C s,k m ) a j L k 1 C α s 6 +m+k+j ρ) 3+ α = C 3 s 6 +m+k+j ρ) l 60) l=1 So, C s,k m ) a j L k m 0 k 1 C { } mi k+j 3 ρ, s 6 +m+k+j ρ) l m 0 k l=1 C k+j ρ m 0 k: k+j ρ 1 s m +C 3 s 6 +m+k+j ρ) l m 0 k: k+j ρ> 1 s m l=1 C s 1 m = C s m ) From 48), 61), we get 19) 18

19 Refereces [1] A P Calderó ad A Zygmud, O sigular itegrals, Amer Math Soc ), [] M Christ, Weak type bouds for rough kerels, A of Math ), 19-4 [3] M Christ ad J-L Rubio de Fracia, Weak type 1,1) bouds for rough operators II, Ivet Math ), 9-37 [4] R R Coifma ad G Weiss, Extesios of Hardy spaces ad their use i aalysis, Bull Amer Math Soc ), [5] W C Coett, Sigular itegrals ear L 1, Proc Sympos Pure Math of Amer Math Soc S Waiger ad G Weiss eds), Vol35 1) 1979), [6] S Hofma, Weak 1,1) boudedess of sigular itegrals with osmooth kerels, Proc Amer Math Soc ), [7] F Ricci ad G Weiss, A characterizatio of H 1 1 ), Proc Sympos Pure Math of Amer Math Soc S Waiger ad G Weiss eds), Vol35 1) 1979), [8] A Seeger, Sigular itegral operator with rough covolutio kerels, J Amer Math Soc 91996), [9] A Stefaov, Weak type estimates for certai Calderó-Zygmud sigular itegral operators, Studia Math 1471)001), 1-13 [10] E M Stei, Sigular itegrals ad differetiability properties of fuctios, Priceto Uiversity Press, Priceto N J,