L 2 -BOUNDEDNESS OF THE CAUCHY INTEGRAL OPERATOR FOR CONTINUOUS MEASURES

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1 Vol. 98, No. 2 DUKE MATHEMATICAL JOURNAL 999 L 2 -BOUNDEDNESS OF THE CAUCHY INTEGRAL OPERATOR FOR CONTINUOUS MEASURES XAVIER TOLSA. Introduction. Let µ be a continuous i.e., without atoms positive Radon measure on the complex plane. The truncated Cauchy integral of a compactly supported function f in L p µ, p +, is defined by ε fz= ξ z >ε fξ ξ z dµξ, z C,ε>0. In this paper, we consider the problem of describing in geometric terms those measures µ for which ε f 2 dµ C f 2 dµ, for all compactly supported functions f L 2 µ and some constant C independent of ε>0. If holds, then we say, following David and Semmes [DS2, pp. 7 8], that the Cauchy integral is bounded on L 2 µ. A special instance to which classical methods apply occurs when µ satisfies the doubling condition µ2 Cµ, for all discs centered at some point of sptµ, where 2 is the disc concentric with of double radius. In this case, standard Calderón-Zygmund theory shows that is equivalent to f 2 dµ C f 2 dµ, 2 where fz= sup ε fz. ε>0 If, moreover, one can find a dense subset of L 2 µ for which fz= lim ε 0 ε fz 3 Received 3 May 997. Revision received 3 November Mathematics Subject Classification. Primary 30E20; Secondary 42B20, 30E25, 30C85. Author partially supported by Direccion General de Investigacion Cientifica y Tecnica grant number PB

2 270 XAVIER TOLSA exists a.e. µ i.e., almost everywhere with respect to µ, then 2 implies the a.e. µ existence of 3, for any f L 2 µ, and f 2 dµ C f 2 dµ, for any function f L 2 µ and some constant C. For a general µ, we do not know if the limit in 3 exists for f L 2 µ and almost all µ z C. This is why we emphasize the role of the truncated operators ε. Proving for particular choices of µ has been a relevant theme in classical analysis in the last thirty years. Calderón s paper [Ca] is devoted to the proof of when µ is the arc length on a Lipschitz graph with small Lipschitz constant. The result for a general Lipschitz graph was obtained by Coifman, McIntosh, and Meyer in 982 in the celebrated paper [CMM]. The rectifiable curves Ɣ, for which holds for the arc length measure µ on the curve, were characterized by David [D] as those satisfying µ z,r Cr, z Ɣ, r > 0, 4 where z,r is the closed disc centered at z of radius r. It has been shown in [MMV] that if µ satisfies the Ahlfors-David regularity condition C r µ z,r Cr, z E, 0 <r<diame, where E is the support of µ, then is equivalent to E being a subset of a rectifiable curve satisfying 4. A necessary condition for is the linear growth condition µ z,r C 0 r, z sptµ, r > 0, 5 as shown, for example, in [D2, p. 56]. To find another relevant necessary condition, we need to introduce a new object. The Menger curvature of three pairwise different points x,y,z C is cx,y,z = Rx,y,z, where Rx,y,z is the radius of the circumference passing through x,y,z with Rx,y,z = and cx,y,z = 0, if x,y,z lie on the same line. If two among the points x,y,z coincide, we let cx,y,z = 0. The relation between the Cauchy kernel and Menger curvature was found by Melnikov in [Me2]. It turns out that a necessary condition for is see [MV] and [MMV] cx,y,z 2 dµxdµydµz C µ, 6 for all discs. The main result of this paper is that, conversely, 5 and 6 are also sufficient for. It is not difficult to realize that 6 can be rewritten as

3 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 27 BMOµ, when µ is doubling and satisfies 5. Therefore, our result can be understood as a T-theorem for the Cauchy kernel with an underlying measure not necessarily doubling. In fact, the absence of a doubling condition is the greatest problem we must confront. We overcome the difficulty thanks to the fact that the operators to be estimated have positive kernels. Following an idea of Sawyer, we resort to an appropriate good λ-inequality to obtain a preliminary weak form of the L 2 -estimate. In a second step, we use an inequality of Melnikov [Me2] relating analytic capacity to Menger curvature to prove the weak,-estimate for ε, uniform in ε>0. It is worthwhile to mention that this part of the argument involves complex analysis in an essential way and no real variables proof is known to the author. From the weak,-estimate, we get the restricted weak-type 2,2 of ε. By interpolation, one obtains the strong-type p,p, for <p<2, and then by duality, one obtains the strong-type p,p, for 2 <p<. One more appeal to interpolation finally gives the strong-type 2,2. For other applications of the notion of Menger curvature, see [L] and [Ma2]. We now proceed to introduce some notation and terminology to state a more formal and complete version of our main result. We say that µ satisfies the local curvature condition if there is a constant C such that 6 holds for any disc centered at some point of sptµ. We say that the Cauchy integral is bounded on L p µ whenever the operators ε are bounded on L p µ uniformly on ε. Let MC be the set of all finite complex Radon measures on the plane. If ν MC, then we set ε νz = ξ z >ε ξ z dνξ. We say that the Cauchy integral is bounded from MC to L, µ, the usual space of weak L -functions with respect to µ, whenever the operators ε are bounded from MC to L, µ uniformly on ε. We can now state our main result. Theorem.. Let µ be a continuous positive Radon measure on C. Then the following statements are equivalent. µ has linear growth and satisfies the local curvature condition. 2 The Cauchy integral is bounded on L 2 µ. 3 The Cauchy integral is bounded from MC to L, µ. 4 The Cauchy integral is bounded from L µ to L, µ. Notice that if any of the statements, 2, 3, or 4 of Theorem. holds, then the Cauchy integral is bounded on L p µ, for <p<, by interpolation and duality. Conversely, if there exists p, such that the Cauchy integral is bounded on L p µ, then the Cauchy integral is bounded on L 2 µ by duality and interpolation.

4 272 XAVIER TOLSA Using the results of Theorem., in the final part of the paper we give a geometric characterization of the analytic capacity γ +, and we show that γ + is semiadditive for sets of area zero. The paper is organized as follows. In Section 2 we define the curvature operator K, and we prove that if µ has linear growth and satisfies the weak local curvature condition, then K is bounded on L p µ, for all p,. As a consequence, we get that for each µ-measurable subset A C, cx,y,z 2 dµxdµydµz CµA. A A A In Section 3 we explore the relation between the Cauchy integral, analytic capacity, and curvature. In Section 4 we complete the proof of Theorem.. Finally, in Section 5 we study the analytic capacity γ +. A constant with a subscript, such as C 0, retains its value throughout the paper, while constants denoted by the letter C may change in different occurrences. Acknowledgments. I would like to thank Mark Melnikov for introducing me to this subject and for his valuable advice. Also, I wish to express my thanks to Joan Verdera for many helpful suggestions and comments. 2. The curvature operator. Throughout the paper, µ is a positive continuous Radon measure on the complex plane. Also, if A C is µ-measurable, we set c 2 x,y,a= cx,y,z 2 dµz, x,y C, A and, if A,B,C C are µ-measurable, then c 2 x,a,b= cx,y,z 2 dµydµz, x C, A B and c 2 A,B,C= cx,y,z 2 dµxdµydµz. A B C The total curvature of A with respect to µ is defined as c 2 A = cx,y,z 2 dµxdµydµz. A A A Also, we define the curvature operator K as Kf x = kx,yf ydµy, x C,f Ł loc µ, where kx,y is the kernel kx,y = cx,y,z 2 dµz = c 2 x,y,c, x,y C.

5 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 273 For a µ-measurable A C, weset K A f x = c 2 x,y,af ydµy, x C,f Ł loc µ. Thus, Kf = K A f +K C\A f. We say that µ satisfies the weak local curvature condition if there are constants 0 <α and C 2 such that for each disc centered at some point of sptµ, there exists a compact subset S such that µs α µ and c 2 S C 2 µs. 7 In this section we prove the following result. Theorem 2.. Let µ be a positive Radon measure with linear growth that satisfies the weak local curvature condition. Then K is bounded from L p µ to L p µ, < p<, and from MC to L, µ. Corollary 2.2. Let µ be a positive Radon measure with linear growth that satisfies the weak local curvature condition. Then there exists a constant C such that for all µ-measurable sets A,B C, In particular, c 2 A,B,C C µaµb. c 2 A CµA. Proof. Since K is of strong-type 2,2, c 2 A,B,C = Kχ B dµ A χ A L 2 µ Kχ B L 2 µ C µaµb. From Corollary 2.2 it follows that if µ has linear growth, the local and the weak local curvature conditions are equivalent. Some remarks about Theorem 2. are in order. The proof of the L p -boundedness of the curvature operator is based on a good λ-inequality. The fact that K is a positive operator seems to be essential to proving the L p -boundedness of K without assuming that µ is a doubling measure. Recall that in [S], [S2], and [SW] the boundedness of some positive operators in L p µ is studied without assuming that µ is doubling, too. Our proof is inspired by these papers. We consider the centered Hardy-Littlewood maximal operator M µ f x = sup r>0 µ x,r f dµ. x,r

6 274 XAVIER TOLSA As is well known, M µ is bounded on L p µ for all p, and from MC to L, µ. This follows from the usual argument, by the Besicovitch covering lemma see [Ma, p. 40]. We now prove some lemmas for the proof of Theorem 2.. Lemma 2.3. If µ has linear growth with constant C 0 and f L loc µ, then for all x C and d>0we have x y 2 fy dµy 4C 0 d M µfx. x y d In particular, x y d x y 2 dµy 4C 0 d. The proof of this lemma is straightforward. One has only to integrate on annuli centered at x. See [D, Lemma 3], for example. The next lemma shows how the Menger curvature of three points changes as one of these points moves. Before stating the lemma, let us remark that if x,y,z C are three pairwise different points, then elementary geometry shows that cx,y,z = 2dx,L yz x y x z, where dx,l yz stands for the distance from x to the straight line L yz passing through y,z. Lemma 2.4. such that Let x,y,z C be three pairwise different points, and let x C be C x y x y C x y, 8 where C>0is some constant. Then cx,y,z c x,y,z x x 4+2C x y x z. 9 Proof. Since x = y, wehavex = y by 8. If x = z, then cx,y,z= 0. In this case, 9 is straightforward: cx,y,z c x,y,z = cx,y,z 2 x x y = 2 x x y x z.

7 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 275 For x = y and x = z, wehave cx,y,z c x,y,z 2dx,L yz = x y x z 2d x,l yz x y x z = 2 dx,l yz x y x z dx,l yz x y x z x y x z x y x z 2 dx,l yz dx,l yz x y x z x y x z x y x z +2d x,l yz x y x z x y x z x y x z x y x z = A+B. 0 To estimate the term A, notice that dx,l yz dx,l yz x x, and so x x A 2 x y x z. We turn now to the term B in 0. We have x y x z x y x z = x y x y x z + x z x z x y x x x z + x x x y. Thus, using that dx,l yz x y and dx,l yz x z, we obtain B 2 x x d x,l yz x z x y x z x y x z + d x,l yz x y x y x z x y x z x x x x 2 x y x z +2 x y x z x x 2+2C x y x z. Now, adding the inequalities obtained for A and B, we get 9. The following lemma is a kind of maximum principle for Menger curvature, which is essential in the proof of Theorem 2.. Lemma 2.5. Let us suppose that µ has linear growth with constant C 0. Let A,B C be µ-measurable, and assume that for some β we have c 2 x,a,b β, x A. Then there exists a constant β depending on C 0 and β such that c 2 x,a,b β, x C.

8 276 XAVIER TOLSA Proof. It is enough to prove the lemma assuming that A is compact. Otherwise, we can consider an increasing sequence of compact sets A n A such that µa\ n= A n = 0. Then we have c 2 x,a n,b c 2 x,a,b β, for all x A n. Applying the lemma to A n, it follows that c 2 x,a n,b β, for all x C. Hence, by monotone convergence, we conclude that c 2 x,a,b= lim n c 2 x,a n,b β. Therefore, we assume that A is compact. We only have to prove inequality when x A. Let r>0be the distance from x to A. Then we split c 2 x,a,b as c 2 x,a,b= c 2 x,a x,4r,b x,4r +c 2 x,a x,4r,b\ x,4r +c 2 x,a\ x,4r,b x,4r 2 +c 2 x,a\ x,4r,b\ x,4r. We estimate each term on the right-hand side of 2 separately. For the first, using cx,y,z 2 y x 4r,wehave c 2 x,a x,4r,b x,4r 4 r 2 dµydµz y x 4r z x 4r 4 r 2 C 04r 2 = 64C0 2. For the second term in 2, we use Lemma 2.3: c 2 x,a x,4r,b\ x,4r 4 y A x,4r z x >4r z x 2 dµydµz C y A x,4r r dµy C. The third term is estimated like the second one with z replaced by y. Finally, we consider the last term in 2. By the definition of r, there exists a point x A x,2r. Then, if y A \ x,4r and z B \ x,4r,wehaveby Lemma 2.4: cx,y,z c x,y,z r +C y x z x. Hence, c 2 x,a\ x,4r,b\ x,4r C +2 y A\ x,4r y A\ x,4r z B\ x,4r z B\ x,4r r 2 y x 2 z x 2 dµydµz cx,y,z 2 dµydµz

9 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 277 Cr 2 y x >4r y x 2 dµy z x >4r z x 2 dµz +2c 2 x,a,b C +2β, where the last inequality follows from the fact that x A. The proof is complete. Lemma 2.6. Let µ be a positive Radon measure that has linear growth with constant C 0 and satisfies the weak local curvature condition 7 with constants 0 <α and C 2. Then there exists some constant β depending on C 0, C 2, and α such that for each disc centered at some point of sptµ there exists a µ-measurable subset S such that µs α 4 µ and c2 x,s, β, x C. Proof. Let be a disc centered at some point of sptµ. Because of 7 there exists a subset S 0 such that µs 0 αµ and c 2 S 0 C 2 µs 0. By Chebyshev, there exists a subset S S 0 such that µs 2 µs 0 α 2 µ and c 2 x,s 0,S 0 2C 2, x S. Then, for all x S,wehavec 2 x,s,s 2C 2. Applying Lemma 2.5, we conclude that there is some constant C 2 such that c2 x,s,s C 2, for all x C. Therefore, c 2 S,S, C 2 µ 2C 2 α µs. Applying Chebyshev again, we find a subset S S such that and µs 2 µs α 4 µ c 2 x,s, 4C 2 α, x S. Thus, for all x S, wehavec 2 x,s, 4C 2 /α, and by Lemma 2.5 we get some constant β such that c 2 x,s, β, for all x C.

10 278 XAVIER TOLSA Lemma 2.7. Suppose µ has linear growth with constant C 0. Let Q C be a square and f L loc µ, f 0. Set f = fχ 2Q and f 2 = f f. Let ε>0 and R>2 be given. Then there exists some constant C R,ε, depending on R, ε, and C 0, such that for all x,x 0 Q and for all ω RQ\2Q we have Kf 2 x +εkf ω+c R,ε M µ f x 0. Proof. We split Kf 2 x as Kf 2 x = y 2Q = y 2Q + y 2Q z C z 3RQ cx,y,z 2 fydµydµz z 3RQ cx,y,z 2 fydµydµz cx,y,z 2 f ydµydµz. 3 To estimate the first integral on the right-hand side of 3, we apply Lemma 2.3: cx,y,z 2 fydµydµz y 2Q z 3RQ C C z 3RQ z 3RQ y x 0 > lq/2 y x 0 2 f ydµy dµz lq M µfx 0 dµz CM µ fx 0, 4 where lq is the side length of Q. We split the second integral on the right-hand side of 3 into two parts: cx,y,z 2 fydµydµz y 2Q z 3RQ = cx,y,z 2 fydµydµz y 3RQ\2Q z 3RQ + cx,y,z 2 f ydµydµz. 5 y 3RQ z 3RQ

11 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 279 To estimate the first integral on the right-hand side of 5, we apply Lemma 2.3 again: cx,y,z 2 fydµydµz y 3RQ\2Q z 3RQ C fy y 3RQ\2Q z x 0 >2lQ z x 0 2 dµz dµy C fy y 3RQ\2Q lq dµy CM µ fx 0. 6 Finally, we only have to estimate the second integral on the right-hand side of 5. If y,z 3RQ, since x,x 0,ω RQ, we can apply Lemma 2.4 and obtain lq cx,y,z cω,y,z C y x 0 z x 0. Then cx,y,z 2 lq 2 cω,y,z+c y x 0 z x 0 +εcω,y,z 2 +C +ε lq 2. y x 0 z x 0 Therefore, applying Lemma 2.3 twice, cx,y,z 2 fydµydµz y 3RQ z 3RQ +ε cω,y,z 2 fydµydµz y 3RQ z 3RQ +C +ε lq 2 y 3RQ z 3RQ y x 0 2 z x 0 2 fydµydµz +εkf ω+c +ε lq 2 y x 0 >2lQ y x 0 2 f ydµy z x 0 >2lQ z x 0 2 dµz +εkf ω+c +ε M µ fx 0. 7 Adding inequalities 4, 6, and 7, the lemma is proved. It is easy to check that if f 0, then Kf is lower-semicontinuous. That is, the set λ := {x C : Kf x > λ} is open for each λ. Our proof of Theorem 2. uses Whitney s decomposition of this open set. In the next lemma, we state the precise version of the decomposition we need.

12 280 XAVIER TOLSA Lemma 2.8. If C is open, = C, then can be decomposed as = Q k, k= where Q k are dyadic closed squares with disjoint interiors such that for some constants R>20 and D, the following hold: i 20Q k ; ii RQ k c = ; iii for each square Q k, there are at most D squares Q j such that 0Q k 0Q j =. Moreover, if µ is a positive Radon measure on C and µ < +, we can choose some subfamily {Q si } i of {Q k } k such that 0Q si 0Q sj =,ifi = j, and µ Q si D µ. i= Proof. Whitney s decomposition for closed squares satisfying i, ii, and iii is a well-known result. See, for example, [St, pp ] or [S2]. Let us prove that we can choose some subfamily {Q si } i {Q k } k as stated in the lemma. We assume that the squares {Q k } k are ordered in such a way that µq m µq n,ifm<n. We take Q s = Q. By induction, if Q s,...,q sn have been chosen, then we define Q sn+ as the square Q k such that 0Q k 0Q s = =0Q k 0Q sn = 8 and k is minimal with this property. In other words, Q sn+ is the square satisfying 8 of maximal µ-measure. Observe that, by condition iii, there are at most n D squares Q k such that 0Q k intersects some of the squares 0Q s,...,0q sn. So at least one of the squares 0Q,...,0Q n D+ does not intersect any of the squares 0Q s,...,0q sn. Then s n must satisfy s n n D +. 9 Now, since {µq k } k is nonincreasing, by 9 we have nd µ Dµ Q n D+ Dµ Qsn. Therefore, Q k k=n D+ µ Q i µ i= n= nd Q k k=n D+ D µ Q sn. n=

13 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 28 Proof of Theorem 2.. We prove that there exists 0 <η< such that for all ε>0 there is some δ>0 for which the following good λ-inequality holds: µ{x : Kf x > +ελ,m µ fx δλ} ηµ{x : Kf x > λ}, 20 for f L p µ,f 0. Let λ ={x C : Kfx>λ}. Then λ is open. Decompose λ as λ = Q k, k= according to Lemma 2.8. Choose a subfamily {Q i } i I of {Q k } k in such a way that 0Q i 0Q j =,ifi,j I, i = j, and µ Q i D µ λ. i I We denote by I the set of indices i I such that there exists some x i Q i spt µ so that M µ fx i δλ. Also, we denote I 2 = I \ I. For each i I, we define i = x i,3lq i. Then we have Q i 3Q i i 0Q i λ. On the other hand, since the weak local curvature condition holds, by Lemma 2.6 we know that there is some constant β such that for each disc i,i I, there exists a µ-measurable subset S i i such that µs i α 4 µ i, c 2 x,s i, i β, for all x C, 2 where β depends on C 2, C 0, and α. Fori I 2, we define i = S i = Q i. Since i j =,ifi,j I, i = j, wehave Therefore, µ S i α µ i 4 i I i I α µq i 4 i I α 4D µ λ. µ λ \ i I S i α µ λ. 22 4D

14 282 XAVIER TOLSA We prove below that for each i I we have µ S i { x : Kfx>+ελ } α 8D µ i, 23 if δ = δε,α is small enough. Then, by 22 and 23, µ{kf > +ελ,m µ f δλ} µ λ \ S i i I +µ S i {Kf > +ελ,m µ f δλ} i I α 4D α 4D = α 4D + α 8D α 8D µ λ + i I µ S i { Kf > +ελ } µ λ + α 8D µ λ. µ λ i I µ i Then, taking η = α/8d, 20 follows. We now prove 23. Observe that, due to 2, for i I,wehave K i χsi x = c 2 x,s i, i β, for all x C. Then, if i I,wehave µ { } 4 x S i : K i fχ i x>ε/4λ K i fχ i dµ ελ S i = 4 fχ i K i χsi dµ ελ 4β fdµ ελ i 4β ελ µ im µ fx i 4βδ ε µ i α 8D µ i, 24 where the last inequality holds provided we choose δ εα/32dβ. In fact, we set { } ε εα δ = δε,α = min, 4C R,ε/4 32Dβ, ε, 4C 3

15 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 283 where C R,ε/4 is the constant given by Lemma 2.7 and C 3 is some constant, which we define later. Due to the properties of Whitney s decomposition, there exists a point ω i in RQ i \ 2Q i c λ, and if moreover i I, then by Lemma 2.7, K fχ C\2Qi x Kf ω i +C R,ε/4 M µ fx i + ε 4 + ε 4 λ+c R,ε/4 δλ + ε 2 Consequently, if x Q i, i I, and Kfx>+ελ, then K ε fχ 2Qi x > λ We prove that ε K C\ i fχ2qi x 4 λ, x Q i,i I. 26 Since K = K i +K C\ i, 25 and 26 give ε K i fχ i x K i fχ2qi x > λ, 4 27 provided x Q i,i I, and Kfx>+ελ. Now 23 is a consequence of 24. Let us check that 26 holds: K C\ i fχ2qi x C fy y 2Q i z i y z 2 dµz dµy C fy y 2Q i y z lq i /2 y z 2 dµz dµy because 3Q i i C f ydµy by Lemma 2.3 lq i y 2Q i C 3 M µ fx i defining C 3 appropriately λ. ε 4 λ by the choice of δ. Hence 20 follows. It is well known see [St2, p. 52] or [D2, p. 60] that from 20, one gets, for p,+, provided f 0 and Kf L p µ C p M µ f L p µ C p f L p µ, 28 inf,kf L p µ. 29

16 284 XAVIER TOLSA Let us see that if f has compact support and is bounded by some constant A, then 29 holds. Suppose that sptf 0,R. Then Kf A Kχ 0,R.Forx C\ 0,2R, with d = x, wehave K χ 0,R x = c 2 x, 0,R,C = c 2 x, 0,R, 0,2d +c 2 x, 0,R,C\ 0,2d. 30 Now we estimate the first term on the right-hand side: c 2 x, 0,R, 0,2d y 0,R z 0,2d C y 0,R z 0,2d C dµy d y 0,R = C χ 0,R ydµy d y x,2d CM µ χ 0,R x. 4 y x 2 dµydµz d 2 dµydµz The second term on the right-hand side of 30 is estimated as c 2 x, 0,R,C\ 0,2d 4 y 0,R z x >d z x 2 dµydµz C dµy d y 0,R = C χ 0,R ydµy d y x,2d CM µ χ 0,R x. Thus, inf,kf inf,a Kχ 0,R χ 0,2R +A CM µ χ 0,R, and so inf,kf L p µ. Now, 28 holds for f 0 bounded with compact support, and a routine argument shows that K is bounded on L p µ. The boundedness of K from MC into L, µ can be proved as follows. One checks easily that 20 also holds for a positive finite measure ν MC: µ { x : Kνx > +ελ,m µ νx δλ } ηµ{x : Kνx > λ}, where Kν and M µ ν are defined in the obvious way. As before, this inequality yields the desired weak,-estimate.

17 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 285 The curvature operator K behaves like a Calderón-Zygmund operator. Moreover, it enjoys the very nice property of being a positive operator. If µ is a doubling measure with linear growth that satisfies the local curvature condition, it can be checked that K is bounded from H at µ to L µ and from L µ to BMOµ. Then, by interpolation, K is bounded on L p µ, for p,. 3. Relation of curvature with the Cauchy integral and analytic capacity. The relation between the Cauchy integral and curvature stems from the following formula [Me2], whose proof is a simple calculation see [MV], for example. Lemma 3.. cx,y,z 2 = 2Re Let x,y,z C be three pairwise different points. Then y xz x + z yx y + x zy z For any µ-measurable set A C, weset cε 2 A = x,y,z Acx,y,z 2 dµxdµydµz. x y >ε x z >ε y z >ε The following lemma is a generalization of an identity proved in [MV].. 3 Lemma 3.2. Suppose that µ has linear growth, and consider h : C [0,] to be µ-measurable. Then, for any µ-measurable set A C, we have 2 ε χ A 2 hdµ = cx,y,z 2 χ A xχ A yhzdµxdµydµz 4Re x y >ε x z >ε y z >ε A ε χ A ε hdµ+o µa 32 and c 2 ε A = 6 A ε χ A 2 dµ+o µa. 33 Proof. By 3, x y >εcx,y,z 2 χ A xχ A yhzdµxdµydµz y z >ε z x >ε

18 286 XAVIER TOLSA = 2 x y >εχ A xχ A yhz y z >ε z x >ε Re y xz x + z yx y + dµx dµy dµz x zy z = I +II+III. 34 We first estimate the integral I: I = 2Re χ x y >ε A xχ A yhz y xz x dµxdµydµz z x >ε 2Re χ A xχ A yhz y xz x dµxdµydµz = I +I 2. We clearly have [ I = 2Re A x y >ε z x >ε y z ε z x >ε ] hz z x dµz χ A y x y >ε y x dµy dµx = 2Re ε hx ε χ A xdµx. 35 A Now we proceed to estimate the integral I 2. Notice that if y z ε< z x, then y xz x y x 2 = y z y x 2 z x y x 2, and so y xz x 2 y x 2. Thus, I 2 4 χ A xχ A yhz Cε x y >ε z x >ε y z ε χ A x x y >ε y x 2 dµxdµydµz y x 2 dµy dµx CµA. 36 Interchanging the roles of x and y, it easily follows that II = I. 37

19 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 287 We turn now our attention to III: III = 2Re χ y z >ε A xχ A yhz x zy z dµxdµydµz z x >ε 2Re χ A xχ A yhz x zy z dµxdµydµz = III +III 2. We have [ III = 2Re = 2Re y z >ε z x >ε x y ε z x >ε ] χ A x x z dµx χ A y y z >ε y z dµy hz dµz ε χ A z 2 hzdµz. 38 We need an estimate for III 2.If x y ε< y z, then x zy z x z 2 = x y x z 2 y z x z 2, and so Therefore, III 2 4 Cε x zy z 2 x z 2. y z >ε z x >ε x y ε χ A x χ A xχ A yhz x z 2 dµxdµydµz z x >ε x z 2 dµz dµx CµA. 39 Because of 35, 36, 37, 38, and 39, we get I +II+III = 4Re ε hx ε χ A xdµx A +2 ε χ A z 2 hzdµz+o µa. From this equation and 34, we finally obtain 32. Identity 33 follows readily from 32 by taking h = χ A.

20 288 XAVIER TOLSA The analytic capacity γ of a compact set E C is γe= sup f, f where the supremum is taken over all analytic functions f : C \ E C such that f, with the notation f = lim z zf z f. Melnikov proved an inequality that relates curvature to analytic capacity, which we proceed to describe. Let E C be compact, and assume that E supports a positive Radon measure µ that has linear growth with constant C 0 and such that c 2 E,µ cx,y,z 2 dµxdµydµz < +. Then Melnikov s inequality [Me2] is E E E γe C 4 µe 3/2 µe+c 2 E,µ /2, 40 where C 4 > 0 is some constant depending only on C 0. For our purposes, we need a slightly improved version of 40, which we state as Theorem 3.3. Given a compactly supported measure ν MC, we denote by ν the locally integrable function /z ν. Theorem 3.3. Let E C be compact, and let µ be a positive Radon measure supported on E that has linear growth with constant C 0 and such that c 2 E,µ <. Then there exists a complex finite measure ν supported on E such that ν µe, νz < for all z E c, νz for almost all with respect to Lebesgue measure z C, and dν C µe 3/2 4 µe+c 2 E,µ /2, where C 4 is some positive constant depending on C 0. Proof. Given a positive integer n, set δ = /n. In [Me2], Melnikov shows that there exist finitely many discs n j, pairwise disjoint, such that E n j n j V δ E where V δ E is the δ-neighborhood of E, 2π j radius n j µe, and γe n C 4 µe 3/2 µe+c 2 E,µ /2. Let h n be the Ahlfors function of E n. That is, h n is analytic on the complement of E n, fz, z C \ E n, and f = γe n. Set ν n = h n zdz, where dz = dz Ɣn, Ɣ n = E n a finite union of disjoint circumferences, and h n z,z Ɣ n, are the

21 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 289 boundary values of h n. Then ν n z = h n z,z En c, and ν nz = 0,z En. Hence, νz, for almost all with respect to Lebesgue measure z C. Moreover, ν n µe, for each n. Passing to a subsequence if necessary, we can assume that ν n ν in the weak -topology of MC, for some ν MC, and we also assume that ν n h in the weak -topology of L C with respect to Lebesgue measure, for some h L C. Since ν n converges in the sense of distributions to ν, weget ν = h. We clearly have dν = lim dν n = lim γe n, n n and so ν fulfills the required conditions. 4. Boundedness of the Cauchy integral. We state now some known results, which we use in the proof of Theorem.. The following proposition is basic for the study of the boundedness of the Cauchy integral from MC to L, µ. Proposition 4.. Let X be a locally compact Hausdorff space, µ be a positive Radon measure on X, and T be a linear operator bounded from the space MX of complex finite Radon measures on X into 0 X, the space of continuous functions on X vanishing at. Suppose, furthermore, that T, the adjoint operator of T, boundedly sends MX into 0 X. Then the following statements are equivalent. a There exists a constant C such that µ { x : T νx >λ } C ν λ, λ>0, for all ν MX. b There exists a constant C such that, for each Borel set E X, we have { µe 2sup hdµ: hµ-measurable, } spth E,0 h, T hdµx C, x X. c There exists a constant C such that, for each compact set E X, we have µe { C sup dν }. : ν MX,sptν E, ν µe, T νx, x X Moreover, the least constants in a, b, c, which we denote by C a, C b, and C c, satisfy A C a C b AC a, A Ca /3 C c AC a +, for some constant A>0.

22 290 XAVIER TOLSA The implication a b is proved in [Ch, p. 07]. It is trivial that b c, and c a can be proved as in [Mu, pp ] or [Ve2]. We give a sketch of the argument for c a. Proof. Suppose that c holds. Let ρ be the constant that appears in c. For λ>0 and θ [0,2π, let E λ,θ = { x X : T νx λe iθ,ρλ/4 }. Covering C by disks λe iθ,ρλ/4, one can see as in [Mu, pp ] or [Ve2] that a follows if, for all λ>0 and θ [0,2π, µe λ,θ C ν λ. 4 Let us see that 4 holds. There is a compact set F λ,θ E λ,θ such that µf λ,θ µe λ,θ /2. Let η MX be such that sptη F λ,θ, η µf λ,θ, T ηx for all x X, and µf λ,θ 2ρ dη. Then ρ 2 λµf λ,θ λe iθ dη T λe νdη + iθ T ν dη Tηdν + ρ 4 λ η ν + ρ 4 λµf λ,θ. Therefore, µe λ,θ 2µF λ,θ 8 ν /ρλ. See also [Ve] for an interesting application of Proposition 4.. Remark 4.2. To apply the lemma, we use a standard technique. We replace ε by the regularized operator ε, defined as ε νx = r ε x ydνy, where ν is a complex finite measure and where r ε is the kernel r ε z = z, if z >ε, z ε 2, if z ε. Then ε ν is the convolution of the complex measure ν with the uniformly continuous kernel r ε, and so ε ν is a continuous function.

23 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 29 Also, we have r ε z = z χ ε πε 2, where χ ε is the characteristic function of 0,ε.Ifν is a compactly supported Radon measure, we have the identity ε ν = z χ ε πε 2 ν = χ ε πε 2 ν. Assume now that ν A a.e. with respect to Lebesgue measure. Since χ ε L πε 2 =, C we obtain ε νz A, for all z C. Also, notice that ε νx ε νx = ε 2 y x <ε y xdνy C 0M µ νx. 42 Hence, if µ has linear growth, ε is of strong-type p,p uniformly in ε if and only if ε is of strong-type p,p uniformly in ε. Also, ε is bounded from MC into L, µ uniformly in ε if and only if the same holds for ε. For the proof of Theorem., the following proposition is also necessary. Proposition 4.3. Let µ be a positive continuous Radon measure on C. If the Cauchy integral is bounded on L p µ, for some p,, or if it is bounded from L µ into L, µ, then µ has linear growth. The proof of this result can be found in [D2, p. 56]. Let us remark that, in fact, in [D2, p. 56] David states only that if the Cauchy integral operator is of strong-type p,p, for some p,, then µ has linear growth. With some minor changes in the proof, one can check that if the Cauchy integral is of weak-type,, then µ has linear growth also. 4.. Proof of the equivalence between the L 2 -boundedness of the Cauchy integral and the linear growth condition plus the local curvature condition of µ. If the Cauchy integral operator is of strong-type 2,2, µ has linear growth by Proposition 4.3 and µ satisfies the local curvature condition by 33 of Lemma 3.2. Assume now that µ has linear growth and satisfies the local curvature condition. First we show that the Cauchy integral operator is bounded from MC into L, µ. That is, we prove the implication 3 of Theorem.. We see that the operator ε, the adjoint of ε, is bounded from MC into L, µ uniformly in ε, which is equivalent to the uniform boundedness of ε : MC L, µ. To do so, we apply Proposition 4. to the operator ε, taking X = C. Notice

24 292 XAVIER TOLSA that ε and ε are bounded from MC into 0 C, with the norm depending on ε. We show that statement c of Proposition 4. holds; that is, for each compact set E C and each ε>0, there exists some complex finite measure ν MC such that µe C dν with C independent of ε, sptν E, ν µe, and ε νx for all x C. Since µ has linear growth and satisfies the local curvature condition, the curvature operator is bounded on L 2 µ. Hence, c 2 E CµE. If we apply Theorem 3.3 to the compact set E and the measure µ E, we conclude that there exist a constant C 4 > 0 and a complex measure ν MC such that sptν E, ν µe, νx < for all x E c, ν a.e. with respect to Lebesgue measure, and dν C µe 3/2 4 µe+c 2 E,µ /2 C 5µE, with C 5 > 0. By Remark 4.2 we have ε νx, for all x C. The measure ν has all the required properties, and hence the Cauchy integral is bounded from MC into L, µ. Now, we prove that the Cauchy integral operator is of restricted weak-type 2,2. That is, for each µ-measurable set A C and all λ>0, µ { x C : ε χ A x >λ } C µa λ 2, where C is some constant independent of ε. Set E λ = { x C : ε χ A x >λ }. Since ε is bounded from MC into L, µ uniformly in ε, by b of Proposition 4. there exists a constant C such that for all ε>0 there is a function h : C [0,] such that hx = 0, if x E λ, µe λ 2 hdµ, E λ and ε hx C, for all x C. By 42 we get ε hx C +C 0, for all x C. Applying 32 we obtain µe λ 2 hdµ 2 E λ λ 2 ε χ A 2 hdµ

25 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 293 λ 2 x y >ε cx,y,z 2 χ A xχ A yhzdµxdµydµz x z >ε y z >ε + 4 λ 2 ε χ A ε hdµ + C A λ 2 µa C KχA λ 2,χ A + ε λ 2 χ A dµ+ C A λ 2 µa C ε λ 2 χ A dµ+ C A λ 2 µa. The only task left is to estimate the integral A εχ A dµ. By 33, since K is of strong-type 2,2, we obtain ε χa dµ µa /2 /2 ε χa 2 dµ A A CµA /2 c 2 A+µA /2 CµA. Therefore, ε is of restricted weak-type 2,2 uniformly on ε. Finally, since the Cauchy integral is of weak-type, and of restricted weaktype 2,2, by interpolation see [Gu, p. 59] or [StW, p. 97] we conclude that it is of strong-type p,p, for <p<2. By duality, it is of strong-type p,p, for 2 <p<, and again by interpolation it is of strong-type 2, Proof of the remaining implications in Theorem.. We have proved 2 and 3. On the other hand, it is obvious that 3 4. So if we show that 4, the proof of Theorem. will be complete. Suppose that the Cauchy integral operator boundedly sends L µ into L, µ. By Proposition 4.3 we know that µ has linear growth. To prove that µ satisfies the local curvature condition, we need the following lemma. Lemma 4.4. Let µ be a positive Radon measure on C. Suppose that ε is bounded from L µ into L, µ uniformly in ε; that is, there exists some constant C such that µ { x : ε fx >λ } C f L µ, λ for all λ>0, f L µ, and ε>0. Then there exists a constant C depending on C such that for any compact set E C there is a µ-measurable function h : C [0,] independent of ε such that, for all ε>0, hx = 0, for all x E, 43

26 294 XAVIER TOLSA µe 8 hdµ, 44 hx C, for all x E and a.e. 2 x C, 45 and, for all ε>0, ε hx C +C 0, for all x C. 46 Proof. Observe that 46 follows from 45, since for each ε>0wehave ε hx = πε 2 χ ε h x C 6, for all x C, and by 42, ε hx ε h L C +C 0 h L µ C 6 +C 0, for all x C. So, given E C compact, we must show that there exists a function h : C [0, ] satisfying 43, 44, and 45. By the Lebesgue-Radon-Nikodym theorem, we have dµ E = gd 2 E +dσ, where 2 stands for the 2-dimensional Hausdorff measure, g L E 2, g 0, and σ is a positive finite Radon measure that is singular with respect to 2. So there is a Borel set E 0 E such that 2 E 0 = 0 and σe= σe 0. Let us consider the case σe 0 µe/2. Since µe = E\E 0 gd 2 + σe 0, we have E\E 0 gd 2 µe/2. Take N N and F E \ E 0 compact such that gx N, for all x F, and µf = F gd 2 µe/4. Since ε is bounded from L µ into L, µ uniformly in ε, we can apply Proposition 4. to the space X = sptµ. We conclude that there exists some function h ε : F [0,] so that µf 2 h ε dµ and ε h ε x C, for all x sptµ. Also, by 42 we have ε h ε x ε h ε L µ +C 0 h ε L µ C +C 0 = C, for all x F. 47 There is a sequence {ε k } k tending to zero such that h εk k converges weak in L µ to some function h L µ. Then it is straightforward to check that µf 2 hdµ, hx = 0, for all x F,

27 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 295 and Also, we have h L µ. lim ε k h εk x = hx, 48 k for all x C, since εk h εk x hx εk h εk x h εk x + hεk x hx = I +II. 49 The term I tends to zero as k, since I = h εk y y x ε k y x dµy y x ε k y x Nd 2 y = 2πNε k. 50 Now we consider the term II in 49: II = hεk y hy dµy y x. F Notice that, for any fixed x C, the function χ F y/y x belongs to L µ, as dµ F = gd 2, with g bounded. Therefore, II tends to zero as k. Hence, 48 holds. By 47 and 48 we get that hx C, for all x F. It is easily checked that h is continuous on C. Therefore, by the maximum principle, hx C, for all x C. Thus h satisfies 43, 44, and 45. Suppose now that σe 0 > µe/2. Let G E 0 be compact with µg µe/4. By Proposition 4. there exists some function h ε : G [0,] so that µg 2 h ε dµ and ε h ε x C, for all x sptµ. As above, by 42 we obtain ε h ε x C, for all x G. 5 We show that there is some constant C 7 such that, for all w C satisfying dw,g = 2ε, 52 we have ε h ε w C7. 53 Since ε h ε w = h ε w by 52, we get that h ε x C 7, for all x C, such that dx,g > 2ε, by the maximum principle. Now, we can take a sequence {ε k } k tending to zero such that h εk k converges weak in L µ to some function h L µ satisfying µg 2 hdµ,

28 296 XAVIER TOLSA and Also, for all x G, wehave hx = 0, for all x G, h L µ. lim h ε k x = hx. k As 2 G = 0, 43, 44, and 45 hold. Let us prove 53 for w satisfying dw,g = 2ε. Let x 0 G be such that w x 0 = 2ε. We denote h ε, = h ε χ x0,4ε and h ε,2 = h ε h ε,. Then, by 5, ε h ε w ε h ε x 0 + ε h ε w ε h ε x 0 C + ε h ε, w + ε h ε, x 0 + ε h ε,2 w ε h ε,2 x 0. Now, since h ε, it is easily checked that ε h ε, w C, ε h ε, x 0 C, and ε h ε,2 w ε h ε,2 x 0 C for some constant C; 53 follows. We can now continue the proof of 4. By Corollary 2.2 we only have to prove that µ satisfies the weak local curvature condition. Given a disc C, there exists a function h : C [0,] independent of ε such that µ 8 hdµ, hx = 0, for all x, and ε hx C, for all x C, with C not depending on the disc. Consider the set { S = z : hz }. 6 Then µ 8 hdµ+ S \S hdµ 8 µs+ 6 µ \S = 8 µ 5 6 µ \S.

29 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 297 Therefore, and so µ \S 4 5 µ, µs µ By equation 33 applied to the measure hdµ,wehave cε 2,hdµ = 6 ε h 2 hdµ+o µ Cµ, for all ε>0, and since hx /6, for x S, we obtain cε 2 S 63 cε 2 S,hdµ 63 cε 2,hdµ Cµ CµS, for all ε>0, and consequently c 2 S CµS. Therefore, by 54, µ satisfies the weak local curvature condition. Remark 4.5. After this paper was written, Nazarov, Treil, and Volberg [NTV] obtained some results that are related to the ones proved here. In particular, they proved that if µ is a positive Radon measure on C not doubling, in general with linear growth and if T is a Calderón-Zygmund operator such that Tχ Q 2 dµ CµQ, for all squares Q C, then T is bounded on L 2 µ. They also have shown that the L 2 -boundedness of T implies the weak,-boundedness. 5. A geometric characterization of the analytic capacity γ +. The analytic capacity γ + or capacity γ + of a compact set E C is defined as γ + E = sup f, where the supremum is taken over all analytic functions f : C\E C, with f on C \ E, which are the Cauchy transforms of some positive Radon measure µ supported on E. Obviously, γe γ + E. The analytic capacity γ was first introduced by Ahlfors [Ah] in order to study removable singularities of bounded analytic functions. He showed that a compact set is removable for all bounded analytic functions if and only if it has zero analytic capacity. However, this did not solve the problem of characterizing these sets in a geometric way this is known as Painlevé s problem, because of the lack of a geometric or metric characterization of analytic capacity. In fact, it is not even known

30 298 XAVIER TOLSA if analytic capacity as a set function is semiadditive; that is, if there is some absolute constant C such that γe F C γe+γf, for all compact sets E,F C see [Me], [Su], [Vi], and [VM], for example. On the other hand, as far as we know, the capacity γ + was introduced by Murai [Mu, pp. 7 72]. He introduced this notion only for sets supported on rectifiable curves, and he obtained some estimates involving γ + about the weak,-boundedness of the Cauchy transform on these curves. Also, until now, no characterization of γ + in geometric or metric terms has been known. In the following theorem, we obtain a more precise version of inequality 40, and we get a geometric and metric characterization of γ + for compact sets with area zero. Theorem 5.. If E C is compact, then µ 3/2 γ + E C sup µ +c 2 µ /2, 55 where C>0 is some absolute constant and the supremum is taken over all positive Radon measures µ supported on E such that µ x,r r, for all x C, r>0. If, moreover, 2 E = 0, then where the supremum is taken as above. µ 3/2 γ + E sup µ +c 2 µ /2, 56 The notation a b in 56 means that there is some positive absolute constant C such that C a b Ca. Using Theorem 5., we get the semiadditivity of γ +. Theorem 5.2. where C is some absolute constant. Let E,F C be compact with 2 E = 2 F = 0. Then γ + E F C γ + E+γ + F, Proof of Theorem 5.. First we show that 55 holds. Let µ be a positive finite Radon measure supported on E with linear growth with constant and such that c 2 µ <. Suppose that k = c 2 µ/ µ >. Then we set σ = µ k /2. Notice that σ has linear growth with constant less than or equal to and c 2 σ = c2 µ k 3/2 = k µ = σ. 57 k3/2

31 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 299 Furthermore, σ 3/2 σ +c 2 σ /2 = σ = µ. 2/2 2k /2 Thus, by the definition of k, µ 3/2 µ +c 2 µ /2 = µ +k /2 2k /2 σ 3/2 = +k σ +c 2 σ /2 2 /2 σ 3/2 σ +c 2 σ /2. Therefore, taking 57 into account, to prove 55 we can assume that the supremum is taken only over measures µ, supported on E, having linear growth with constant such that c 2 µ µ. Let us remark that this fact was already noticed in [Me2]. So, if µ is a positive finite measure supported on E with linear growth with constant and such that we have to show that c 2 µ µ, 58 γ + E C 8 µ, 59 with C 8 > 0, since µ 3/2 µ +c 2 µ /2 µ, because of 58. By Chebyshev, from 58 we get µ { x C : c 2 µ x > 2} 2 c2 µ 2 µ. Here we use the notation c 2 µ x instead of c2 x. So if we set σ = µ {c 2 µ x 2}, then Also, σ 2 µ. cσ 2 x 2, for σ -almost all x C.

32 300 XAVIER TOLSA Notice that σ has linear growth with constant and satisfies c 2 σ F 2σF, 60 for any Borel set F. Using Theorem., we get that the Cauchy transform is bounded on L 2 σ and is bounded also from L σ into L, σ, with the norm bounded by some absolute constant. Therefore, by Lemma 4.4, there exists some absolute constant C 9 such that for any compact set F C, there is a σ -measurable function h : F [0,] such that σf 8 hdσ and hdσ x C 9, for all x F.In particular, if we choose F = E, weget σ 8 hdσ, for some σ -measurable function h : E [0,], and hx C 9, for all x E. Thus 59 holds. For the second part of the theorem, assuming 2 E = 0, we only have to show that there exists some constant C such that µ 3/2 γ + E C sup µ +c 2 µ /2. 6 Let ν be some positive measure supported on E such that νx < for all x E and γ + E 2 ν. Let us check that ν has linear growth. Let x,r C be some closed disc. Recall that if z x =r z y dνyd z <, 62 where stands for the -dimensional Hausdorff measure, then ν x,r = νzdz. 63 2πi z x =r See [Ga, p. 40], for example. Notice that for all x C, 62 holds for a.e. r>0 this follows by Fubini. So we get that for all x C, 63 holds for a.e. r>0. Since 2 E = 0, we also get that for each fixed x, x,r E = 0, for -almost all r>0. Therefore, by 63, for all x C, wehave ν x,r r, 64 for -almost all r>0, and by approximation this can be extended to all r>0.

33 L 2 -BOUNDEDNESS OF THE CAUCHY OPERATOR 30 Now, by Remark 4.2, ε νx C, for all x C and for some constant C not depending on ε, so ε νx C +, for all x C by 42. Using the identity 33, we obtain c 2 ν C ν. Therefore, ν 3/2 γ + E 2 ν C ν +c 2 ν /2, and 6 follows. We do not know if the second part of Theorem 5. holds for sets with positive area. However, we have the following nonquantitative result. Corollary 5.3. Let E C be compact. Then γ + E > 0 if and only if E supports some positive finite Radon measure µ with linear growth such that c 2 µ <. Proof. If 2 E > 0, the result follows, choosing µ = E 2.If 2 E = 0, we apply 56 of Theorem 5.. Proof of Theorem 5.2. The semiadditivity of γ + for sets of area zero follows from 56 of Theorem 5. and the fact that the quantity µ 3/2 sup µ +c 2 µ /2 65 is semiadditive. Arguing as in the proof of Theorem 5., we know that there is a positive finite Radon measure µ with linear growth with constant, µ supported on E F, such that γ + E F C 0 µ and cµ 2 x 2, for µ-almost all x C. Then c 2 µ E 2µE and c 2 µ F 2µF. Thus, by Theorem 5., γ + E C µe and γ + F C µf, where C > 0 is some absolute constant. Therefore, γ + E F C 0 µe+µf C γ+ E+γ + F. With some minor changes in the proof of Theorem 5.2, one can check that, in fact, γ + is countably semiadditive on compact sets with area zero. Remark 5.4. Let us define the capacity γ +. Given a compact E C, weset γ + E = sup µ,

34 302 XAVIER TOLSA where the supremum is taken over all positive Radon measures µ supported on E such that µ a.e. 2 inc. Obviously, if 2 E = 0, then γ + E = γ + E. However, we do not know if γ + E = γ + E or γ + E γ + E holds for compact sets E with positive area. Arguing as in Theorem 5., it is easily seen that µ 3/2 γ + E sup µ +c 2 µ /2, with the supremum taken as in Theorem 5., for any compact set E C. Also, as in Theorem 5.2, γ + E F C γ + E+ γ + F, for all compact sets E,F C. So the notion of γ + seems to be more natural than the notion of γ +. On the other hand, observe that if we showed that γe γ + E, 66 for all sets E with 2 E = 0, then by Theorem 5.2 we could obtain easily that analytic capacity is a semiadditive function on compact sets. Of course, proving 66 seems difficult. In fact, 66 clearly implies the conjecture of Melnikov stating that γe>0 if and only if E supports some positive finite measure with linear growth and finite curvature. Remark 5.5. Nazarov, Treil, and Volberg [NTV3] have shown that γ + E > 0 if and only if γ c E > 0, where γ c E stands for γ c E = sup f, with the supremum taken over all functions f analytic on C\E, with f, which are the Cauchy transforms of some complex measure. However, from their estimates, one cannot derive that γ + E γ c E. It is not difficult to see that this would imply 66. Nazarov, Treil, and Volberg informed the author that they know how to obtain Theorem 5. with different arguments, using the Tbobtained in [NTV2]. References [Ah] L. Ahlfors, Bounded analytic functions, Duke Math. J ,. [Ca] A. P. Calderón, Cauchy integrals on Lipschitz curves and related operators, Proc. Nat. Acad. Sci. U.S.A , [Ch] M. Christ, Lectures on Singular Integral Operators, CBMS Regional Conf. Ser. in Math. 77, Amer. Math. Soc., Providence, 990.

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