Parr. R. Coifman, A. McIntosh, Y. Meyer. Translated by Ting Zhou. + C ϕ (x, y)f(y) dy (2) g(x) = v.p. g 2 C(1 + M) 9 f 2 (3)

Transcription

1 The L 2 Boundedness of he Cauchy Inegral over Lipschiz Curves (L inégrale de Cauchy défini un opéraeur borné sur L 2 pour les courbes lipschiziennes) Parr. R. Coifman, A. McInosh, Y. Meyer Translaed by Ting Zhou We propose o prove he following heorem. Theorem I. Le ϕ : R R be a Lipschiz funcion, namely, here exiss a consan M such ha ϕ(x) ϕ(y) M x y for all x, y R. Denoe by C ϕ (x, y) he kernel, defined for x y by 1 + iϕ (y) C ϕ (x, y) = x y + i(ϕ(x) ϕ(y)). (1) Then, for all f L 2 (R), he singular inegral g(x) = v.p. + C ϕ (x, y)f(y) dy (2) exiss almos everywhere and we have where C is an absolue consan and f 2 = g 2 C(1 + M) 9 f 2 (3) ( + f(x) 2 dx) 1/2. The exisence of he principle value in (2) in he general case (for arbirary M) and he inequaliy (3) in he special case where M is sufficienly small are due o A. Calderón [3]. The perurbaion mehod used by A. Calderón canno be generalized o he case we rea here. Moreover, our proof does no use he complex variables nor he conformal mapping. 1

2 Denoe by ψ : R C a Lipschiz funcion, of complex variables, such ha ψ 1 and le K n (x, y) be he singular kernel (ψ(x) ψ(y)) n /(x y) n+1. Then we have he resul specified as follows. Theorem II. For any f L 2 (R), he singular inegral v.p. + K n (x, y)f(y) dy exiss a.e. and defines a funcion g n (x) saisfying g n 2 C(1 + M) 4 f 2 ; C denoes an absolue consan. The Theorem I follows immediaely from Theorem II by renormalizaion. Indeed, suppose ψ δ < 1. Then i follows, wih he noaions from Theorem II, and hence g n 2 C(1 + n) 4 δ n f 2 ( 1) n g n 2 C /(1 δ) 5. To conclude, we se, for ϕ : R R saisfying ϕ M, ψ(x) = M 2 M 2 +1 x + i 1 M 2 +1 ϕ(x) so ha x + ψ(x) = x+iϕ(x) M 2 +1 and The choice of ψ ensures C ϕ (x, y) = L(x, y) 1 + iϕ (x) M ψ 2 M 2 + M M 2 + M 4 = M 2 < 1 and he muliplicaion operaor by (1 + iϕ (x))/(m 2 + 1) is bounded wih he norm no exceeding 1/(M 2 + 1), which proves Theorem I. The case n = 1 of Theorem II is due o A. Calderón [2]. I hen ook unil 1974 ha wo of he auhors of his aricle obained he case n = 2. In 1976, he wo auhors showed ha g n 2 C n f 2 if ψ 1 wihou being able o specify he growh of C n in n. The firs resul in his direcion is again due o A. Calderón who shows in 1977 ha we have C n C n wihou being able o deermine C > 1 [3]. A ha momen, Theorem II was presened as a corollary of he specific case of Theorem I where M δ (δ > sufficienly small). In fac, Theorem II can be generalized furher. 2

3 Theorem III. Le A 1, A 2,..., A n : R C be n Lipschiz funcions and K n (x, y) be he singular kernel K n (x, y) = (A 1(x) A 1 (y))(a 2 (x) A 2 (y))... (A n (x) A n (y)) (x y) n+1. (4) Then he operaor T n defined by T n [f](x) = v.p. is bounded over L 2 (R) and we have where C is an absolue consan. + K n (x, y)f(y) dy (5) T n [f] 2 C(1 + n) 4 A 1 A 2... A n f 2 (6) We will emporarily limi he lis of resuls and give successively he plan of he proof of Theorem III, hen he proof iself. Then we esablish he oher heorems (corollaries or generalizaions of Theorem III and he soluion of a problem by Kao). 1 Plan of he proof of Theorem III I is well known ha in he proof of he inequaliy (6), i suffices o resric o he case where A 1 = a 1 C (R),..., A n = a n C (R) since ha he passing o he limi ha allows o obain he general case is done hrough runcaing he kernel K n,ε (x, y) = K n (x, y)1 { x y ε}. 1 In fac, a famous heorem by A. Calderón, M. Colar and A. Zygmund (recalled in Appendix) allows o bound (from above) he norm of he operaor T n,ε : L 2 (R) L 2 (R), defined by he runcaed kernel K n,ε, uniformly in ε >, by he norm of T n : L 2 (R) L 2 (R) and by he hree numbers sup x y K n (x, y), sup x y 2 x y x y x K n(x, y), sup x y 2 x y y K n(x, y). 1 1 E is he characerisic funcion of se E. 3

4 For every fixed ε >, he passing o he limi is immediae for he runcaed kernel K n,ε (x, y). Indeed, a j L (R) is he limi a.e. of a sequence a j,ν C (R) also saisfying a j,ν a j ; using he primiives A j,ν of a j,ν we define he runcaed kernel K n,ε,ν and we have, uniformly in ν N, I follows ha, if f L 2 (R), we have K n,ε,ν (x, y) a 1... a n x y 1 1 {x y ε}. g ε (x) = + = lim ν + K n,ε (x, y)f(y) dy = + K n,ε,ν (x, y)f(y) dy. lim g ν,ε(x) ν + If we could esablish he inequaliy (6), wih C = 1 6, in he special case where A j C (R), we shall have g ν,ε 2 C 1 (1 + n) 4 a 1... a n f 2, hen by he heorem of Calderón, Colar and Zygmund, successively, also he upper bounds for g ε 2 and g 2 by applying wice he Faou Lemma. The a.e. exisence of lim ε g ε (x) = g(x) is proven by inducion over n: he case f C (R) resuls from he inducion hypohesis, by inegraion by pars, and he general case f L 2 (R) is obained hen by he mehod inroduced by A. Calderón and A. Zygmund (chaper IV of [5]). We hen assume, in all ha follows, A j = a j C (R). The primary ideniy (Proposiion 1) allows o express he operaor T n in erms of he basic operaors we will define now. We begin wih D = i(d/dx) : S(R) S(R) which allows, for all R, o define hree operaors bounded on L 2 (R), P = I I + 2 D 2, Q = D I + 2 D 2 e P iq = I I + id.2 On he oher hand M aj : L 2 (R) L 2 (R) is he poin-wise muliplicaion operaor by he funcion a j. We define, for b j C (R), he operaor L n(b 1,..., b n ) : L 2 (R) L 2 (R) by L n (b 1,..., b n ) = v.p. + (P iq )M b1 (P iq )M b2... (P iq )M bn (P iq ) d. (7) The principle value in he inegral operaor is defined as he limi of runcaed inegral... d/, ε >. ε 1/ε 2 The symbols associaed o P and Q are, respecively, 1/(1 + 2 ξ 2 ) and ξ/(1 + 2 ξ 2 ). 4

5 Then we have he following remarkable formula T n = 1 L n (a n! σ(1), a σ(2),..., a σ(n) ) (8) σ S n where he sum is over all he permuaions σ of {1, 2,..., n}. Therefore he proof of (6) reduces o proving he corresponding inequaliy concerning L n. Then he mehod of real variables comes ino play. We show ha he sudy of L n follows from ha of a Lilewood-Paley-Sein funcion associaed wih he problem; known as ( G k [f](x) = Q M a1 P M a2 P... M ak P [f](x) 2 d ) 1/2. (9) The boh hand sides of (9) are funcions of x and here, as in all ha follows, U V W denoes he composiion of corresponding operaors. The fundamenal esimae is G k [f] 2 C(1 + k) f 2 (1) where C > is an absolue consan. I is easy o show (6) from he esimae (1). The esimae (1) is obained due o a very precise descripion of he swiching rules beween operaors M a, P and Q, a C (R), and by inroducing he Carleson measure Q a 2 (dx d/) on R 2 + = R ], + [. 2 Proof of he represenaion inegral formula Le ε ], 1[ and H ε be an operaor defined by iπh ε = (D iy) 1 dy = 2 ε y 1/ε ε y 1/ε D(D 2 + y 2 ) 1 dy. Then i(π/2)h ε is he operaor associaed o he muliplicaion m ε (ξ) = Arcg(ξ/ε) Arcg(εξ). The kernel of H ε is k ε (x y) where k ε (x) = (1/πx)(e ε x e x /ε ). The operaor H ε is hen a varian of he Hilber ransform runcaed a he origin and a he infiniy. In paricular, in he sense of disribuion, lim ε k ε (x) = v.p.(1/πx). I follows ha wih D = i(d/ dx), we have, in he sense of disribuions, lim D n k ε (x) = i n!π 1 1 v.p. ε x n+1 (11) 5

6 where we define he disribuion v.p.(1/x n+1 ) by 1 + ] v.p. x n+1, g = v.p. [g(x) g() xn 1 dx (n 1)! g(n 1) () (he laer principle value has he usual sense for g C (R)). x n+1 Fixing a R and apply above consideraions o k ε (a x) insead of k ε (x) and o g n (x) = (A 1 (a) A 1 (x))... (A n (a) A n (x))f(x). We hen have g n (a) = g n(a) = = g n (n 1) (a) = and herefore + (A 1 (a) A 1 (x))... (A n (a) A n (x)) v.p. (a x) n+1 f(x) dx (12) uniformly wih respec o a R. = ( i) n π n! lim ε (Dn H ε )[g n ](a), Denoe by M Aj : S(R) S(R) he operaor of muliplicaion by A j and by δ j he derivaion of he algebra A = L(S(R), S(R)) defined by δ j (T ) = M Aj T T M Aj. Then, if g 1 (x) = (A 1 (a) A 1 (x))f(x), we have (T g 1 )(a) = (δ 1 T )[f](a) and, by inducion, i gives (δ 1 δ 2 δ n T )[f](a) = T [g n ](a) (13) and, in paricular, (D n H ε )[g n ](a) = (δ 1 δ 2 δ n D n H ε )[f](a). (14) I remains o calculae δ 1 δ 2 δ n (D n H ε ). This can be obained using he formal properies of a derivaion of an algebra ha we show easily. We have, for S, T, T 1,..., T n belonging o A = L(S(R), S(R)), and herefore δ j (ST ) = δ j (S)T + Sδ j (T ) (15) δ j (T 1... T n ) = δ j (T 1 )T 2... T n + T 1 δ j (T 2 )T 3... T n + + T 1 T 2... T n 1 δ j (T n ). (16) If S is inverible and if S 1 is is inverse, i saisfies and finally, because D = i(d/ dx), we have δ j (D) = im aj. δ j (S 1 ) = S 1 δ j (S)S 1 (17) 6

7 Since he muliplicaion operaors commue, and more generally, for k n 1, δ j2 δ j1 (D) =, (18) δ jn δ jn 1 δ j1 (D k ) =. (19) If y R and y, hen I/(D iyi) A and i saisfies ( ) I I δ 1 = D iyi D iyi δ I 1(D) D iyi. (2) Then apply δ 2 o (2), we obain, using (18) and (17) ( ) I I δ 2 δ 1 = D iyi D iyi δ I 2(D) D iyi δ I 1(D) D iyi + I D iyi δ I 1(D) D iyi δ I 2(D) D iyi. By inducion, we have ( ) I δ 1 δ 2 δ n = ( 1) n D iyi σ S n Furhermore, we have he following lemma. Lemma 1. For any ineger n 1 and all y, we have ( ) D n δ 1 δ 2 δ n D iyi The proof of his ideniy is immediae. Indeed, we have, D n (iy) n I D iyi I D iyi δ I σ(1)(d) D iyi δ I σ(2)(d)... D iyi δ I σ(n)(d) D iyi. = δ 1 δ n ( (iy) n I D iyi n 1 = (iy) k D n k 1 (21) ). (22) and i is sufficien o apply (19). We are now able o prove (8). We have, by (21) and (22) δ 1 δ 2 δ n D n H ε = i L n,ε (a π σ(1), a σ(2),..., a σ(n) ) (23) σ S n where L n,ε (a 1,..., a n ) = ε y 1/ε (D iyi) 1 M a1 (D iyi) 1... M an (D iyi) 1 y n dy. I suffices, a las, o perform a change of variable y = 1/ in he las inegral and o use (12) and (14). We obain he following resul. 7

8 Proposiion 1. Suppose, for A j = a j C (R), L n,ε (a 1,..., a n ) = (I + id) 1 M a1 (I + id) 1 M a2... M an (I + id) ε 1/ε 1 d. Then, for all funcions f C (R), we have, uniformly in x R, + (A 1 (x) A 1 (y))... (A n (x) A n (y)) v.p. (x y) n+1 f(y) dy = 1 n! σ S n lim ε L n,ε (a 1,..., a n )[f](x). 3 Remarkable ideniies saisfied by he operaors P and Q Le m : R C be a Lipschiz funcion and bounded and, for all R, le M be he operaor whose symbol is m(ξ) : M (e ixξ ) = m(ξ)e ixξ. Suppose ξm (ξ) L (R). Then he operaor ( / )M has he symbol m 1 (ξ) where m 1 (ξ) = ξm (ξ). Apply his remark o he case where m(ξ) = 1/(1 + ξ 2 ) corresponding o M = P. Then ξm (ξ) = 2ξ 2 /(1 + ξ 2 ) 2, ha is, ( / )P = 2Q 2. The calculaion is he same for ( / )Q. We firs have m(ξ) = ξ/(1 + ξ 2 ) and i saisfies Similarly, We can herefore conclude. ξm (ξ) = ξ/(1 + ξ 2 ) + 2ξ/(1 + ξ 2 ) 2 = m 1 (ξ). ξm 1(ξ) = ξ/(1 + ξ 2 ) 8ξ 3 /(1 + ξ 2 ) 3. Proposiion 2. We have ( / )P = 2Q 2 ( / )A = Q 8Q 3. while ( / )Q = Q + 2P Q = A and 4 Reducion of he problem o he Lilewood-Paley-Sein funcion If a j C (R) for j n, we se ( G k,n [f](x) = Q M ak+1 P M ak+2... P M an P [f](x) 2 d ) 1/2 (24) 8

9 when k n 1 and while, for 1 j n, we se ( G j[f](x) = ( G n,n [f](x) = Q [f](x) 2 d ) 1/2 Q Māj P Māj 1 P... P Mā1 P [f](x) 2 d ) 1/2 (25) wih ( G [f](x) = Q [f](x) 2 d ) 1/2. Naurally hese wo Lilewood-Paley-Sein funcions are of he same ype. The corresponding L 2 esimaes will be proved laer. Proposiion 3. For f and g in C (R), we have, if a j 1, 1 j n, [L n,ε (a 1,..., a n )f]ḡ dx 8 G k,n (f) 2 G j(g) f 2 g 2. (26) R Recall ha L n,ε = ε 1/ε j k n (P iq )M a1 (P iq )M a2... (P iq )M an (P iq ) d. To sudy his operaor, we expand he produc (P iq )M an (P iq )M a2... (P iq )M an (P iq ) ino 2 n+1 words of he form T M a1 T 1... T n 1 M an T n where T j {P, iq }. Among hese words, one is of he form P M a1 P... P M an P and i disappears afer inegraion since P = P. Then we find n + 1 words where Q appears only once. Se d L j = T M a1 T 1 M a2... T n 1 M an T n (27) ε 1/ε where T = P,..., T j 1 = P, T j = Q, T j+1 = = T n = P. Since Q = Q, he inegral of he erm where Q appears only once in he index j reads 2iL j. Finally we find he words where Q appears a leas wice. They are oo many o be reaed one by one and hence will be grouped ino packes. To do his we reform he sum of he 9

10 words such ha T =P,..., T j 1 = P, T j = Q, T j+1 {P, iq },..., T k 1 {P, iq }, T k = Q, T k+1 = = T n = P. This means we combine all he words in which he firs Q we found has index j and he las has index k. So we have, if j < k n, L j,k = P M a1 P... P M aj Q M aj+1 (P iq )M aj+2 (P iq ) ε 1/ε... M ak 1 (P iq )M ak Q M ak+1 P... M an P d. To simplify, we hen se U = P M a1... P M aj Q, V = M aj+1 (P iq )M aj+2 (P iq )... M ak 1 (P iq )M ak, and we have where W = Q M ak+1 P... M an P d L j,k = U V W ε 1/ε = L+ j,k L j,k L + j,k = 1/ε Finally L n,ε = n L j + j<k n L j,k. ε U V W d. Noe ha if a j 1, we have V op 1 where he operaor norm is ha of V : L 2 (R) L 2 (R). Again o simplify, we will se u v = R u(x) v(x) dx for u L2 (R), v L 2 (R) and denoe T : L 2 (R) L 2 (R) he adjoin of T. Then 1/ε L + j,k (f) g = U V W f g d ε 1/ε = V W f U g d ε 1/ε ε 1/ε W f 2 U d g 2 ε ( ) 1/2 ( 1/ε W f 2 d 1/ε 2 U g 2 2 G k,n f 2 G jg 2. ε ε V W f 2 U g 2 d ) 1/2 d 1

11 The same inequaliy applies o L j,k (f) g and we hen have L j,k (f) g 2 G k,n f 2 G jg 2. (28) j<k n The reamen of he operaor L j is more suble. j<k n We wrie Q = 8Q 3 + ( / )A where A = 2P Q Q = V Q = Q V wih V op = 1. Then we have L j = L 1 j + L2 j and he sudy of L 1 j = 8 1/ε ε P M a1 P... M aj 1 P M aj Q 3 M aj+1 P... P d is immediae by he mehod above. Because Q op = 1/2, i follows L 1 j f g 4 G j,nf 2 G j g 2. We consider L 2 j = 1/ε ε P M a1 P... M aj 1 P M aj ( A ) M aj+1 P... P d. An inegraion by pars gives wo erms whose operaor norms are obviously bounded from above by 1/2 and n erms where ( / )A is replaced by A while he one of P is replaced by 2Q 2. Each of hese erms is reaed by he mehod above and we have L 2 jf g f 2 g 2 + G k,n f 2 G jg 2. This complees he proof of Proposiion 3. j<k n 5 The L 2 esimae for he Lilewood-Paley-Sein funcion The wo definiions we have inroduced only appear differenly and i suffices o sudy, for a 1 C (R),..., a n C (R), We herefore have ( G n f = Q M a1 P M a2 P... M an P f 2 d ) 1/2. (29) 11

12 Proposiion 4. There exiss an absolue consan C > such ha, for all n N, if a j 1, 1 j n, G n f 2 C(1 + n) f 2 (3) The case n = is rivial and we have immediaely by he Plancherel formula ( Q f 2 d ) ( 1/2 ) 1/2 = Q f 2 dx d = 1 f The Proposiion 4 is proved by inducion over n. We sar wih he simple idea ha if, in he definiion of G n, we replace Q M a1 P M a2... by P M a1 Q M a2..., hen i can be verified ha, since P op = 1 and M a1 op 1, G n (f) 2 G n 1 (f) 2. Indeed his is he case because he error erms can be analyzed using he Carleson measures. The proof of Proposiion 4 will be given in secions 6 o 1. R The Commuaion Ideniies Proposiion 5. If b C (R), denoe respecively by M b, M b and M β he poin-wise muliplicaion operaors by b, b = P (b) and β = Q (b). Then we have where R = M β P Q M β Q. Q M b P = P M b Q + R (31) Considering he imporance of his ideniy, we provide wo proofs. To verify (31), i is sufficien o prove ha he wo side of (31) give he same resul when applied o he same characer e iξx, ξ R. By lineariy, we can also suppose ha b(x) = e iαx, α R. Therefore his is reduced o verify (α + ξ) (α + ξ) ξ 2 = (α + ξ) α 2 ξ ξ 2 + α α ξ 2 which doesn presen much difficuly. (α + ξ) (α + ξ) 2 α α 2 ξ ξ 2 The second proof is deeper since i allows o prove (31) for any derivaion D of an commuaive algebra A (here A = S(R), for example) such ha, for all real, I + id : A A are inverible. 12

13 Indeed, we have, for u A and v A, by he definiion of he derivaion. (I + id)(uv) = [(I + id)u]v + uidv Because I + id : A A is inverible, we have uv = (I + id) 1 {[(I + id)u]v} + (I + id) 1 [uidv] and finally, se f = (I + id)u, g = v, rewrien as [(I + id) 1 f]g = (I + id) 1 (fg) + (I + id) 1 {[(I + id) 1 f](idg)} (I + id) 1 (fg) = [(I + id) 1 f]g + r (32) where he error erm r = (I + id) 1 {[(I + id) 1 f](idg)}. Finally (31) is obained by aking he imaginary par of (32). Indeed, we have, from (32), (P iq )(fg) = [(P iq )f]g i(p iq ){[(P iq )f]dg} and, changing o, we also have (P + iq )(fg) = [(P + iq )f]g + i(p + iq ){[(P + iq )f]dg} Subracing boh sides of he wo ideniies above, we obain (31) in is equivalen form Q (fg) = (Q f)g + P {(P f)(dg)} Q {(Q f)(dg)}. 7 Recall he Carleson measure Le dµ(x, ) be a Radon measure on R 2 + = R [, + [. For any open inervals I =]a, b[, denoe by T (I) he riangle {(x, ) : d(x, I c )} in R 2 + where d(x, I c ) is he disance from x o he complemen of I. We say ha µ is a Carleson measure if here exiss a consan C such ha, for any open inerval, we have µ(t (I)) C I = C(b a) and he lower bound of he consans C will be denoed by C µ. Le g : R 2 + C be a coninuous funcion, vanishing a infiniy. We se, for x R, Mg(x ) = and hen we have he following resul [8]. sup g(x, ) (33) x x 13

14 Proposiion 6. Le µ be a Carleson measure wih consan C µ and g : R 2 + C be a coninuous funcion and vanishing a infiniy. Then we have, for any p >, + g(x, ) 2 dµ(x, ) C µ (Mg) p (x) dx. (34) R 2 + This resul is supplemened by he following classical lemma ([5] page 148). Lemma 2. If b L (R), hen Q b 2 (dx d/) = dµ(x, ) is a Carleson measure wih consan C µ 1 b 2. 8 The fundamenal inequaliy Le f(x, ) : R 2 + C be a coninuous funcion vanishing a infiniy. We wrie f(x, ) = f (x) where is regarded as a parameer and, in paricular, P f or Q f indicaes ha is fixed and ha he operaors P or Q ac on he funcion f of x. A las we se F (x) = sup > f (x). Wih above noaions we have Proposiion 7. Le f(x, ) : R 2 + C be a coninuous funcion vanishing a infiniy and b L (R) be a complex-valued funcion such ha b 1. Denoe by M b he poin-wise muliplicaion operaor by b(x) and by L 2 (R 2 +) he space L 2 (R 2 +, (dx d/)). Then we have where C is an absolue consan. Q M b P f L 2 (R 2 + ) Q f L 2 (R 2 + ) + C F L 2 (R) (35) Denoe by k(x) he funcion 1 2 e x and by h(x) he funcion Dk(x) = i(sign x)k(x). For all >, se k (x) = 1 ( x ) k h (x) = 1 ( x ) h. Thus P (f) = k f and Q (f) = h f which leads o P (f) f and he same for Q. Moreover P (f) 2 f 2 and Q f f 2. Reurning o Proposiion 7, we sar wih applying Proposiion 5. We have Q M b P f = P M b Q f + g where g = (Q b)(p f) Q [(Q b)(q f )]. 14

15 We can deduce ha if L 2 (R 2 +) = L 2 (R 2 +, (dx d/)), Q M b P f L 2 (R 2 + ) P M b Q f L 2 (R 2 + ) + (Q b)(p f ) L 2 (R 2 + ) + Q [(Q b)(q f ) L 2 (R 2 + ) To bound = A + B + C. A 2 = R dx d P M b Q f, we firs inegrae in x, fixing. The norm of P : L 2 (R) L 2 (R) is equal o 1. On he oher hand b = 1 leads o b = P (b) 1. Then we have A 2 2 dx d Q f = Q f 2 L 2 (R 2 ). + R 2 + The erms B and C are error erms. We have f (x) F (x) for all x R and P f = k f where k >. We now proceed o bound P f (x) for x x. We remark ha for x x, k (x y) ek (x y) leads o P f (x) ep F (x ). Denoe by F he Hardy and Lilewood maximal funcion of F. We have apparenly P F (x ) F (x ) and finally MP f (x ) ef (x ). (36) This inequaliy, ogeher wih Lemma 2 and Proposiion 6, gives B 2 = Q b 2 P f 2 dxd C F 2 dx C R 2 + R R F 2 dx. The reamen o C 2 is done in a similar way. We firs ge rid of he operaor Q by reasoning he same as he bound of A 2 and we hen remark ha Q f = h f k F = P F which brings us back o he previous case. 9 Maximal funcions ieraed We denoe by K he convex se of all he funcions K : R [, + ] ha are even, decreasing on [, + [ and saisfy R K(x) dx = 1. Then, as one can easily check, K 1 K and K 2 K imply K 1 K 2 K. On he oher hand, if K K and if f L 1 (R), hen K f f in which f always denoes he Hardy and Lilewood maximal funcion of f. 15

16 Denoe by b 1,..., b n n funcions in L (R) saisfying b j 1 and call M b1,..., M bn he corresponding muliplicaion operaors. Then we have Lemma 3. For all >, M b1 P M b2 P... M bn P f f. Indeed, P g = k g where k K and we hen have k,n = k k K which implies M b1 P M b2... M bn P f k,n f f. 1 The proof of Proposiion 4 We have already remarked ha he case n = is rivial. Suppose ha Proposiion 4 is esablished for n m. To pass o n = m + 1, we se f m, = M b1 P M b2 P... M bm P f and we hen have, for b L (R) and b 1, ( G m+1 (f) = Q M b P f m, 2 d and hence G m+1 (f) 2 = ( ( by Proposiion 7 and Lemma 3. Therefore we have R 2 + R 2 + ) 1/2 ) 1/2 2 dx d Q M b P f m, ) 1/2 2 dx d Q f m, + C f 2 G m+1 (f) 2 G m (f) 2 + C f 2 which complees he proof of Proposiion 4 and Theorem III. 11 Generalizaion of Theorem III Le m L (R) be an even funcion: m( ) = m() for all real. Using such a funcion m, we define a variaion of he Hilber ransform H m = v.p I I + id m()d

17 and H m : L 2 (R)] L 2 (R) is coninuous. So he kernel of H m is K m (x y) where K m (x) is he odd funcion whose resricion o ], + [ is ( ) 1 K m (x) = e xu m du. u We denoe by n N an ineger; A 1,..., A n : R C are n Lipschiz funcions and K m,n (x, y) is he singular kernel defined by n!k m,n (x, y) = (A 1 (x) A 1 (y))... (A n (x) A n (y))d n K m (x y). (37) Wih hese noaions, we prove he generalizaion of Theorem III as follows. Theorem IV. here exiss an absolue consan C > such ha, for any even funcion m L (R) and any sequence A j : R C, 1 j n, of n Lipschiz funcions, he norm of he operaor L m,n : L 2 (R) L 2 (R) defined by he singular kernel K m,,n does no exceed C(1 + n) 4 A 1... A n m. We resric ourselves o he proof of he special case where, in addiion, m is coninuous and m() = O( ε ) in he viciniy of he origin and m() = O( ε ) in he viciniy of he infiniy for a cerain ε >. The general case is obained by passing o he usual limis. To simplify he noaions, se q(x) = m ( 1 x) so ha + H m = i q(y) D iy dy, wih H m op + q(y) D iy dy < +. op We only consider he case where A j C (R) and we denoe by δ A 1,..., δ An he derivaions of he algebra L(L 2 (R), L 2 (R)) defined by δ Aj (T ) = M Aj T T M Aj ; M Aj : L 2 (R) L 2 (R) is he poin-wise muliplicaion operaor by A j. Firs, + q(y) H m = i D iy dy, we have if q vanishes ouside η y 1/η, η >, and herefore D n H m = i + δ A1 δ An D n H m = ( i) n+1 (iy) n D iy q(y) dy + α I + + α n 1 D n 1 σ S n + T σ (y)y n q(y) dy (38) 17

18 wih T σ (y) = I D iyi M I a σ(1) D iyi M I a σ(2)... D iyi M I a σ(n) D iyi. If q is no zero near and he infiniy bu saisfies q(y) = o( y ε ) near and q(y) = o( y ε ) in he viciniy of he infiniy, we obain (38) by simply passing o he limi. Finally, we se again y = 1/ and we obain where L m,n = 1 n! σ S n + R σ ()m() d I R σ () = I + id M I a σ(1) I + id M I a σ(2)... I + id M I a σ(n) I + id. We wrie (I + id) 1 = P iq which allows o re-apply he mehod from secion 4. The only difficul erms are L j = P M a1 P... M aj Q M aj+1 P... M an P m() d. We remark again ha (1/)Q = ( / )(2P Q Q ) + (1/)Q 3. The erm associaed o Q 3 reduces o he Lilewood-Paley-Sein funcion. Afer inegraion by pars, we have wo ypes of erms. Those where he derivaive P conains Q 2 and will be reaed by he mehod of secion 4. The oher erm is, assuming ha m () is coninuous on [, + [, L j = P M a1 P... M aj (Q 2P Q )M aj+1 P... M an P () m () d. We observe hen ha ( / )Q = Q + wp Q so ha we can inegrae L j by pars. Suppose ha m () = m 1 () has he following properies: m 1 () is coninuous on [, + [, m 1 () = o( ε ) near, m 1 () = o( ε ) near infiniy and m 2 () = m 1 () also possesses hese wo properies. Therefore, module he erms ha we can rea by he mehod in secion 4, we have L j This implies ha we can conrol L j L j m 2 () P M a1 P... M aj Q M aj+1 P... M an P d = L j. by he Lilewood-Paley-Sein funcion. Bu L j L j has he same srucure as L j, wih he difference ha m is replaced by m m 2 = r. 18

19 The soluion is hen quie simple. Given r L (R) wih r( ) = r(), r 1 and he hypohesis of addiional regulariy: r() = o( ε ) near, r() = o( ε ) near infiniy and r is coninuous on R. Using r, define he even funcion m : R C by m(x) = 1 ( x inf 2, ) r() d x = 1 2x We hen have m 1 (x) = xm (x) = 1 2x x x for x > so ha m 1 r and m r. r()d + x 2 r()d + x 2 x x r() d 2 (x > ). r() d 2 Furhermore m 2 (x) = xm 1 (x) = r(x) + m(x) as shown by immediae calculaion. The Theorem IV is proved. A variaion of Theorem IV is presened as follows. Proposiion 8. Le m L (], + [) and le k : R\{} C be an odd funcion whose resricion o ], + [ is k(x) = e x m() d. Denoe by A 1,..., A n n Lipschiz funcions such ha A j 1 and by K n (x, y) he kernel A 1 (x) A 1 (y)... A n(x) A n (y) k(x y). x y x y + Then he singular inegral g n (x) = v.p. K n(x, y)f(y)dy exiss a.e. when f L 2 (R) and we have g n 2 C(1 + n) 4 m f 2 where C is an absolue consan. This proposiion is a corollary of Theorem IV. Indeed define he disribuion S n = v.p. x n k(x) by ] S n, f = v.p. [f(x) f() xn 1 k(x) (n 1)! f (n 1) () dx (39) xn for f C (R); The principal value of he inegral has he usual sense. Then i is a simple exercise of he disribuion heory o verify he ideniy n!s n = D n k n where k n : R\{} C is he odd funcion defined by k n (x) = e xu m n (u) du when x > wih m n (u) = n!u n u (u )n 1 m() d. I is herefore reduced o Theorem IV. Theorem V. Le K C be a compac and convex se, F : U C be a holomorphic funcion in an open neighborhood of K and A : R C be a Lipschiz funcion such ha, for all pairs (x, y) of wo disinc real numbers, we have A(x) A(y) x y K. 19

20 Then for any odd kernel k : R\{} C as in he hypohesis of Proposiion 8, he singular kernel ( ) A(x) A(y) F k(x y) x y defines a bounded operaor on L 2 (R). Indeed, le Γ U be a closed curve, simple, recifiable, oriened and conained in he inerior of compac K. We wrie F (z) = 1 F (ζ) 2πi Γ ζ z dζ which reduces he proof of Theorem V o he special case where F (z) = 1/(ζ z) and where he disance from ζ o K exceeds ε >. Then we wan o obain a uniform esimae in ζ for he norm of he operaor defined by he kernel K ζ (x, y) = 1/(ζx A(x) (ζy A(y))). We can, since he compac se K is convex, find α C such ha α z R for all z K as well as α ζ R + ε/2; R doesn dependen on ζ since d(ζ, K) ε. 3 We se A(x) = αx + B(x). Then we have R and B(x) B(y) x y K ζ (x, y) = 1 (ζ α)(x y) (B(x) B(y)) = (ζ α) j 1 K j (x, y) j= where K j (x, y) = (B(x) B(y))j (x y) j+1. I is hen sufficien o apply Proposiion The generalized Hardy spaces by C. Kenig We denoe by ϕ : R R a Lipschiz funcion and by Γ = {(x, y) R 2 ; y = ϕ(x)} he graph of ϕ. Le Ω 1 and Ω 2 be open, bounded by Γ and defined respecively by y > ϕ(x) and y < ϕ(x). Following C. Kenig [1], we consider he wo generalized Hardy spaces H 2 (Ω 1 ) and H 2 (Ω 2 ) defined as follows: F H 2 (Ω 1 ) means ha F : Ω 1 C is holomorphic and F (x + iϕ(x) + iε) 2 dx < +. sup ε> R 3 Remark ha a compac convex se is an inersecion of compac discs. 2

21 We define similarly H 2 (Ω 2 ) and, for ϕ =, we rerieve he usual Hardy spaces. Le us se L 2 (Γ) = L 2 (Γ, ds) where ds is he arc lengh on Γ and we can [1] define a race operaor θ 1 : H 2 (Ω 1 ) L 2 (Γ) wih he following properies. If F (z) = 1/(z z 2 ) and z 2 Ω 2, hen θ 1 (F ) is he resricion of F o Γ. Furhermore θ 1 is linear, coninuous and he image θ 1 (H 2 (Ω 1 )) is closed in L 2 (Γ). In fac f θ 1 (H 2 (Ω 1 )) if and only if f L 2 dz (Γ) and f(z) = Γ z z 2 for all z 2 Ω 2. For simplified noaions, we wrie H 2 (Ω 1 ) Γ insead of θ 1 (H 2 (Ω 1 )). We define similarly he closed subspace H 2 (Ω 2 ) Γ of L 2 (Γ). Wih hese noaions we have Theorem VI. If Γ is he graph of a Lipschiz funcion, hen L 2 (Γ) is he direc sum of he closed subspaces H 2 (Ω 1 ) Γ and H 2 (Ω 2 ) Γ. The direc sum is orhogonal if ϕ(x) = ax + b. Suppose again ha Γ is he graph of he Lipschiz funcion ϕ and ha ϕ M. Le α > M and S α be he closed secor y α x of he complex plane. Denoe by Ω an open se of C 2 defined by Ω = {(z, w), z w / S α } and by K : Ω C a holomorphic funcion in Ω such ha K(z, w) C z w 1 for a cerain consan C and all (z, w) Ω. Using K(z, w) and Γ we define for any funcion f L 2 (Γ) a holomorphic funcion F 1 : Ω 1 C by F 1 (w) = K(z, w)f(z) dz. (4) We hen have [6]. Γ Theorem VII. For any graph Γ of a Lipschiz funcion and any holomorphic kernel K of above ype, F 1, defined by (4), belongs o H 2 (Ω 1 ) every ime ha f L 2 (Γ) and he operaor T : L 2 (Γ) H 2 (Ω 1 ) is well-defined and bounded. Naurally Theorem VI is only a paraphrase of Theorem I while Theorem VII is deduced from Theorem VI by he echnique from [6]. 13 Generalizaion of Theorem V by ransference Le (Ω, C, µ) be a space provided wih a ribe and a finie measure µ over C. Denoe by U a group of auomorphisms of (Ω, C, µ), indexed by R, preserving he measure µ and such 21

22 ha for E C and µ(e) < +, he measure of he symmeric difference beween U (E) and E ends o wih. We say ha a funcion A : Ω C is relaively Lipschiz o U if here exiss a consan C > such ha A(U (x)) A(x) C for all x Ω and all R. Theorem VIII. Le K be a compac and convex subse of he complex plane, U be an open se conaining K and F : U C be a holomorphic funcion in U. Le A : Ω C be a funcion such ha, for all real nonzero, and all x Ω, we have (A(U (x)) A(x))/ K. Then here exiss a consan C 1 such ha for all ε > and all N >, he norm T ε,n op of he operaor T ε,n : L 2 (Ω, C, dµ) L 2 (Ω, C, dµ) defined by ( ) A(U (x)) A(x) T ε,n f(x) = F f(u (x)) d doesn exceed C 1. ε N To show his, we se g(x) = T ε,n f(x) and we hen have, for all M >, g p L p (Ω) = g(u s(x)) p L p (Ω) = I = 1 M g(u s (x)) p ds dµ(x) 2M Ω M M ( ) A(U+s (x) A(U s (x)) = F f(u +s (x)) d Ω M ε N p ds 2M dµ(x). Denoe by θ x : R C he funcion defined by θ x () = f(u (x))1 { M+N} ; Muliplicaion by his indicaor funcion has no effec on he inegral I which we can herefore bound by ( ) ax ( + s) a x (s) J = F θ x () d p ds 2M dµ(x). Ω R ε N We have se a x () = A(U (x)) and we have (a x () a x (s))/( s) K for any pair (s, ) of wo disinc reals. Applying he Fubini heorem o calculae J, we are led o fixing x and o applying Theorem V. Then i follows ha J C p 1 θ x p dµ(x) M+N L p (ds) Ω 2M = Cp p dµ(x) ds 1 f(u s (x)) Ω M N 2M C p M + N 1 M f p L p (Ω) Cp 1 f p L p (Ω) (M + ). 22

23 14 Generalizaion o R n We say ha A : R n C is Lipschiz if here exiss a consan C such ha A(x) A(y) C x y for all x R n and all y R n. Theorem IX. Suppose ha A : R n R and ϕ : R n R are wo Lipschiz funcions. Then he kernel A(x) A(y) K(x, y) = [ x y 2 + (ϕ(x) ϕ(y)) 2 ] (n+1)/2 defines a bounded operaor on L 2 (R n ). To show his, we apply he following lemma. Lemma 4. Le K be a compac and convex subse of he complex plane. U be an open se conaining K, F : U C be a holomorphic funcion in u and A : R C a Lipschiz funcion such ha for all pairs (x, y) of wo disinc real numbers, we have (A(x) A(y))/(x y) K. Finally denoe by B : R C a second Lipschiz funcion and form he singular kernel N(x, y) = B(x) B(y) (x y) 2 F ( A(x) A(y) x y Then, for any funcion f L 2 (R) he singular inegral g(x) = v.p. N(x, y)f(y) dy exiss a.e. and we have g 2 C(K, U, F ) B f 2. The proof of his lemma is similar o ha of Theorem V and lef o he reader. Theorem IX is hen obained, from he lemma by A. Calderón s roaion mehod ha we can apply because, due o Theorem XI from he Appendix, he L 2 coninuiy of he operaor T defined by he kernel N(x, y) leads o, uniformly in ε >, he L 2 coninuiy of operaors T ε defined using he runcaed kernels N ε (x, y) = N(x, y)1 { x y ε}. ). 15 Domains of square roos of cerain accreive operaors Le H be a Hilber space over C, V H be a dense subspace and T : V H be a linear operaor. We say, following Kao [9], ha T is accreive-maximal if he following wo properies are saisfied: Re T f f for all f V ; (41) 23

24 Then we show he following properies: I + T : V H is surjecive. (42) For all λ C such ha Re λ >, λi + T : V H is surjecive, (λi + T ) 1 : H V is coninuous and he is norm is less equal han (Re λ) 1. There exiss an operaor and a unique S ha is accreive-maximal and saisfies S 2 = T. (43) (44) We hen wrie S = T 1/2 and we have, for all f V, T 1/2 f = 1 π λ 1/2 (T + λi) 1 T f dλ. (45) The convergence of his inegral a infiniy does no pose a problem due o (43) and we remark ha if f V, we have (T + λi) 1 T f = f (T + λi) 1 λf which leads o, for λ >, (T + λi) 1 T f 2 f ; The convergence a he origin of (45) is ensured. The problem arising is o deermine he domain of T 1/2. We will solve his problem in a special case. Le D = i(d/ dx) : H 1 L 2 (H 1 is he usual Sobolev space) and a(x) L (R) be a funcion such ha, for a cerain δ >, we have Re a(x) δ almos everywhere. In all wha follows he reference Hilber space H is L 2 (R). We hen have, for M a : H H he muliplicaion operaor by a(x), he following lemma. Lemma 5. The operaor DM a D is defined on a dense subse V H and is accreivemaximal. We sar wih specifying he domain of DM a D. Firs we consider bounded operaors L = D(I + D 2 ) 1/2 and L 1 = (I + D 2 ) 1/2 and remark ha, by he Plancherel formula, we have, for all f H, L (f) 2 + L 1 (f) 2 = f 2. Then se S = L L M a L. The condiion Re a(x) δ immediaely leads o Re Sf f ε f 2 where ε = inf(1, δ). I follows ha S : H H is an isomorphism. Then we se V = S 1 (H 1 ); V is a dense subspace in L 2 (R) and V = (I + D 2 ) 1/2 V is dense in H 1 (hence also in L 2 (R)). Finally we wrie I + DM a D = (I + D 2 ) 1/2 S(I + D 2 ) 1/2 which allows o verify immediaely ha I D M a D : V H is an isomorphism of vecor spaces. (46) 24

25 I follows ha, for f V H 1, M a Df H 1 and obviously we have Re DM a Df f δ Df 2. An equivalen poin of view is ha of Kao [9]. We sar wih he sesquilinear form J : H 1 H 1 C defined by J(f, g) = f g + M a Df Dg and we apply Theorem VI.2.1 of [9]. Theorem X. Le a L (R) such ha Re a(x) δ > and D = i(d/ dx) : H 1 L 2 (R). Then he domain of he accreive-maximal operaor (DM a D) 1/2 is H 1 (he usual Sobolev space) and, more precisely (DM a D) 1/2 + I : H 1 L 2 is an isomorphism beween his wo Hilber spaces. To prove he heorem, we recall he noaions from secion 1. We can obviously, by renormalizing he funcion a, assume ha Re a(x) 1 a.e.. We hen se b = 1 (1/a) and we have b 1 (1/M 2 ) < 1 (wih M = a ). Also we have he following ideniy. Lemma 6. For all, This can be verified by observing ha which is immediae since M b = I M 1 a. ( 2 DM a D + I) 1 DM a = Q (I M b P ) 1. ( 2 DM a D + I)Q = DM a (I M b P ) Reurning he proof of Theorem X. We make he change of variable λ = 1/ 2 in (45) and i verifies T 1/2 f = 2 ( 2 T + I) 1 T f d π or, in a way more explici, we have (DM a D) 1/2 = JD where J = 2 π = 2 π = 2 π ( 2 DM a D + I) 1 d DM a 1 d Q (I M b P ) Q (M b P ) k d. Here Q (M b P ) k (d/) = T k is he operaors sudied in he proof of Proposiion 3 (and noe L in his proof). We have T k op C(1 + k) 3 b k, which leads o he desired convergence as long as b < 1. 25

26 Finally we verify ha (DM a D) 1/2 + I : H 1 L 2 is an isomorphism beween wo Hilber spaces. Since (DM a D) 1/2 is accreive, i follows immediaely for all f H 1 (where = L 2 (R)). ((DM a D) 1/2 + I)f f I remains o verify ha (DM a D) 1/2 f + f ε Df for a cerain ε >. To show his, we se S = (DM a D) 1/2 and we have S = (DMāD) 1/2. Since Re ā = Re a δ, i follows S f C Df. Then he complex number ζ = (DM a D) 1/2 f+f (DMāD) 1/2 f saisfies ζ C (DM a D) 1/2 f+ f Df. On he oher hand, if ζ 1 = DM a Df f and ζ 2 = (DM a D) 1/2 f f we have Re ζ 1 δ Df 2 and ζ = ζ 1 + ζ 2. Then (DM a D) 1/2 is accreive and Re ζ 2. Finally Re ζ δ Df 2 which leads o he desired inequaliy wih ε = δ/c. 16 Appendix (Calderón-Zygmund s general singular kernel) Le Ω R n R n be an open se of pairs (x, y), x R n, y R n, such ha x y. Denoe by K : Ω C a funcion wih he following wo properies: There exiss a consan C such ha K(x, y) C x y n for all (x, y) Ω. (47) K is locally Lipschiz on Ω and, le x and y denoe he corresponding gradiens, we have x K(x, y) C x y n 1 and y K(x, y) C x y n 1 for all (x, y) Ω. (48) Using such a kernel K(x, y) we define he runcaed operaor T ε, for ε >, by T ε f(x) = K(x, y)f(y) dy (49) y x ε and his inegral is absoluely convergen for f L 2 (R n ). The fundamenal problem of he heory is hen o decide if T ε sends L 2 (R n ) coninuously o L 2 (R n ) and if here exiss a consan C 1 such ha, for all ε > and funcion f L 2 (R n ) 26

27 we have T ε f 2 C 1 f 2. (5) When (47), (48) and (5) are verified we have ha K(x, y) is a Calderón-Zygmund kernel. We remark ha his propery of K does no depend on he chosen euclidian norm on R n. Using he family T ε, ε >, we define he maximal operaor T by T f(x) = sup ε> T ε f(x). Wih hese noaions, we have he following heorem (A. Calderón, M. Colar and A. Zygmund). Theorem XI. For any Calderón-Zygmund kernel (in he sense we jus defined), he maximal operaor associaed o T exends, for 1 < p < +, o all L p (R n ) and we have T f p C p f p where C p does no depend on p, n, C and C 1 (consans defined by (47),(48) and (5)). Furhermore, le K : Ω C be a kernel saisfying (47) and (48) and le T : L 2 (R n ) L 2 (R n ) be a coninuous operaor such ha for any funcion f C (Rn ) and any x no belonging o he suppor of f we have T f(x) = K(x, y)f(y) dy. Then K is a Calderón- Zygmund kernel. A proof can be found in [5], Chaper IV. The heorem has he following corollary. Corollary 1. Le T j : L 2 (R n ) L 2 (R n ) be a bounded sequence of coninuous operaors. Suppose ha T j f(x) = K j (x, y)f(y) dy is as in he previous heorem and ha K j (x, y) saisfies, uniformly in j N, (47) and (48). If, in he sense of simple convergence on Ω, K j (x, y) K(x, y) Then K(x, y) is a Calderón- Zygmund kernel. Naurally all he kernels sudied in his work are Calderón-Zygmund kernels. References [1] A. P. Calderón, Algebra of Singular Inegral Operaors, Proc. Symp. Pure Mahemaics, X, A.M.S. (1967). [2] A. P. Calderón, Commuaors of singular inegrals operaors, Proc. Na. Acad. Sc. USA 53 (1965), [3] A. P. Calderón, Cauchy inegrals on Lipschiz curves and relaed operaors, Proc. Na. Acad. Sc. USA 74 (1977),

28 [4] A. P. Calderón, Commuaors, singular inegrals on Lipschiz curves and applicaions, Proc. I.C.M. Helsinki 1978, Vol. I, [5] R. Coifman and Y. Meyer, Au del des opraeurs pseudo-diffreniels, Asrisque 57, Soci Mahmaique de France (1978). [6] R. Coifman and Y. Meyer, Fourier analysis of mulilinear convoluions, Calderón s heorem and analysis on Lipschiz curves, Springer-Verlag, Lecure Noes in Mah. 779, [7] E. B. Fabes, M. Jodei, and N. M. Rivire, Poenial echniques for boundary value problems on C 1 -domains, Aca. Mah. 141 (1978), [8] C. Fefferman and E. M. Sein, H p -spaces of several variables, Aca. Mah. 129 (1972), [9] T. Kao, Perurbaion Theory for Linear Operaors, Springer-Verlag, New York, (1966). [1] C. Kenig, Weighed H p -spaces on Lipschiz domains. Amer. J. Mah. 12 (198),