Deposited on: 20 April 2010

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1 Pott, S. (2007) A sufficient condition for the boundedness of operatorweighted martingale transforms and Hilbert transform. Studia Mathematica, 182 (2). pp SSN Deposited on: 20 April 2010 Enlighten Research publications by members of the University of Glasgow

2 A SUFFCENT CONDTON FOR THE BOUNDEDNESS OF OPERATOR-EGHTED MARTNGAE TRANSFORMS AND HBERT TRANSFORM SANDRA POTT Abstract. et be an operator weight taking values almost everywhere in the bounded positive invertible linear operators on a separable Hilbert space H. e show that if and its inverse 1 both satisfy a matrix reverse Hölder property introduced in [2], then the weighted Hilbert transform H : 2 (R, H) 2 (R, H) and also all weighted dyadic martingale transforms T σ : 2 (R, H) 2 (R, H) are bounded. e also show that this condition is not necessary for the boundedness of the weighted Hilbert transform. 1. ntroduction The question of finding vector analogues to the celebrated Hunt-Muckenhoupt- heeden Theorem [8] has been studied intensively in recent years. S. Treil and A. Volberg showed in [12] that a weight function taking values almost everywhere in the positive invertible d d matrices satisfies the vector A 2 condition (1) sup 1/2 < C, R, interval if and only if the Hilbert transform H defines a bounded linear operator on the operator-weighted 2 -space { } 2 (R, C d ) = f : R C d : (t)f(t), f(t) dt < or equivalently, if and only if the weighted Hilbert transform M 1/2 1/2 HM : 2 (R, C d ) 2 (R, C d ) defines a bounded linear operator. Here, M 1/2 R denotes the densely defined multiplication operator with the matrix function 1/2. There exists also a dyadic version of this theorem [12]: et D denote the collection of all dyadic intervals in R. For each σ { 1, 1} D, let T σ denote the dyadic martingale transform on 2 (R, H) given by f D σ()h f Mathematical Subjects Classifications. Primary 42A50, 47B38 The author gratefully acknowledges support by the Nuffield Foundation and the European Network on Analysis and Operators. The author also thanks the Department of Pure Mathematics at the University Bordeaux, where part of this work was carried out, for their hospitality. 1

3 2 SANDRA POTT Then a matrix weight function taking values almost everywhere in the positive invertible d d matrices satisfies the vector A d 2 condition (2) sup 1/2 < C, D if and only if all weighted dyadic martingale transforms M 1/2 T σm 1/2 : 2 (R, C d ) 2 (R, C d ) are uniformly bounded. Characterisations for the boundedness of matrix-weighted Hilbert transforms on p (R, C d ) for 1 < p < were found in [9], [13]. A different approach via a matrix-weighted Hardy-ittlewood maximal function was suggested more recently by M. Christ and Goldstein in [2]. They introduce the matrix reverse Hölder property (3) 1/2 (x) r dx C ( R interval) and show that this implies the boundedness of a matrix-weighted maximal function on p (R, H) for p < r even in the infinite-dimensional situation, namely when replacing C d by a separable Hilbert space H, and considering a weight : R (H) taking values almost everywhere in the bounded positive invertible linear operators on H. t was shown in [7] that in the finite-dimensional situation, the boundedness of the weighted maximal function in turn implies the boundedness of the weighted Hilbert transform. n the finite-dimensional situation, the vector A 2 condition implies the matrix reverse Hölder property for some r > 2 [2], so that this approach provides a new proof of the above mentioned result in [12]. The characterisation of the boundedness of weighted Hilbert transform and martingale transforms in the infinite-dimensional setting has proved very difficult, even in the case p = 2. t was shown in [3], [4] and [5] that the operator versions of (1) and (2) are not sufficient for the boundedness of the weighted Hilbert transform and the weighted dyadic martingale transforms, respectively ((1) and (2) were shown to be necessary in [12]). n this paper, we show that also in the infinite-dimensional situation, the matrix reverse Hölder condition (3) for an operator weight and its inverse 1 for some r > 2 implies the uniform boundedness of all weighted dyadic martingale transforms and the boundedness of the weighted Hilbert transform on 2 (R, H). e use a slightly different route from [2], [7]. nstead of using the weighted maximal function, we show first that the weighted square function operator is bounded and bounded below by means of a stopping time argument from [10]. This gives us the uniform boundedness of the weighted dyadic martingale transforms. The case of the Hilbert transform then follows from a result in [11]. Using the theory of vector BMO functions, we also show that the reverse Hölder property is not necessary in the infinite-dimensional case, even with r = 2.

4 OPERATOR-EGHTED HBERT TRANSFORM 3 2. eights with reverse Hölder property and decaying stopping time Definition 2.1. [2] e say that an operator weight : R (H) has the matrix reverse Hölder property, if there exist constants C > 0 and r > 2 such that 1/2 (x) r dx C for all intervals R. Definition 2.2. e say that an operator weight : R (H) has the dyadic matrix reverse Hölder property, if there exist constants C > 0 and r > 2 such that 1/2 (x) r dx C for all dyadic intervals D. Such weights satisfies in particular the (dyadic) vector A 2 condition (1), (2). Furthermore, for each operator-valued weight and each interval R, one has the elementary inequality (4) 1/2 1 1/2 1 (see [12], [2]). For each D, let D() denote the collection of all D with. Given D and λ > 1, let λ,1 () denote the collection of all maximal dyadic subintervals λ of such that (5) 1 λ or (6) 1 λ (x) dx > λ λ λ 1/2 1 (x) 1/2 dx > λ. e write λ,0 () = {}, λ,k () = λ,k 1 () () for k 1, F() = ( ) F λ,1 () = D()\ () (), and F λ,k () = λ,k 1 () F() for k 1. Note that D() = k=1 F λ,k() for each D with this notation. Somewhat loosely, we will write () for the set () and () for (). emma 2.3. f and 1 both have the dyadic matrix reverse Hölder property, then for sufficiently large λ, there exists a constant c, 0 < c < 1, such that λ,k () c k for all D, k N. (i.e. λ () is a decaying stopping time in the sense of [10], Section 3.3.) Proof. The proof is an adaption of the proof of the eight emma 3.17 for scalar weights in [10]. e first introduce an auxiliary stopping time G. For D, let G() denote the collection of all maximal dyadic subintervals of such that (5) holds. e show that G is a decaying stopping time for sufficiently large λ. Note that (x) λ1 H

5 4 SANDRA POTT in the sense of an operator inequality almost everywhere on G = \ G(). Therefore, ( ) (7) (x)dx = 1 1/2 (x) dx 1 1/2 G G λ G 1 1 λ G. by (4). t is enough to show that there exists a constant α > 0 (independent of ) such that G α. Assume towards a contradiction that this is false. Then there exists an interval D such that G 2λ, and (x)dx 1 2. et G() = { λ,j } j. Then (8) (x)dx = λ,j However, one also has (9) j and therefore (10) j G (x)dx = 1 1/2 λ,j G (x)dx 1 2. (x) dx 1 1/2 j λ,j 2λ 1 1 λ,j 2λ λ,j. j λ,j 1 4λ. By the dyadic matrix reverse Hölder property, we have for some p > 1 (11) C (x) p dx (x) p dx j λ,j ( ) p 1 j λ,j p 1 (x) dx λ,j λ p. λ,j j Thus λ,j C λ p. j Choosing λ (4C) 1/p 1, we obtain a contradiction to (10). This proves that for λ (4C) 1/p 1, there exists a constant 0 < c < 1 such that G() c for all D. t then follows by induction that G k () c k for k N and D. Notice that for each interval G 2λ 2(), there exists Gλ 2 () with. To see this, let be the unique dyadic interval in G λ () such that. j j

6 OPERATOR-EGHTED HBERT TRANSFORM 5 riting (x) dx = 1 1/2 (x) dx 1 1/2, we obtain 2λ 2 < λ 1 (x) (x) dx (x) dx dx by (4), which yields 1 (x) dx > λ. Consequently, there exists such that λ () λ,2 (). Therefore G 2λ 2() Gλ 2() c 2. By iteration, it follows that by choosing λ sufficiently large, we can assume that G λ () 1 4 for D. e now define a second auxiliary stopping time G. For D, let G() be the collection of all maximal dyadic subintervals of such that (6) holds. Using now the reverse Hölder property of 1, we find that also G is a decaying stopping time for sufficiently large λ. Again, by choosing λ large enough, we can assume that G λ () 1 4 for D. This means that for λ large enough, λ () G λ () + G λ () 1/2 ( D). Thus λ is a decaying stopping time. 3. The weighted square function Theorem 3.1. et : R (H) be an operator weight such that and 1 both have the dyadic matrix reverse Hölder property. Then the weighted square function operator S : 2 (R, H) 2 (R, H), is bounded and invertible. Proof. e will first show that the operator T : 2 (R, H) 2 (R, H), f ( 1/2 f) h (x) D f D 1/2 (x) 1/2 f h (x) is bounded, following the steps of the proof of Th in [10].

7 6 SANDRA POTT e choose λ > 1 such that λ () is a decaying stopping time, and write just (). First note that for almost every x \ (), and therefore (12) (x) λ1 H 1/2 1 (x) 1/2 λ1 H, 1 λ 1 H (x) λ1 H by (4). et f 2 (R, H) have finite Haar expansion. e can assume without loss of generality that f is supported in [0, 1] and has mean 0. e write k for k ([0, 1]), and F k for F k ([0, 1]). For j N, write j f = h f = f h = f, F j j 1 and M j f = 1/2 f h = F j T j f = T j f = 1/2 (x)m j f = j 1 F() j 1 F() j 1 1/2 (x) 1/2 f h = M f j 1 F() 1/2 f h = j 1 1/2 (x)m f. Then j=1 jf = f and j=1 T jf = T f. e will show that the T j satisfy the conditions of Cotlar s emma. Each M f has support in, so T j f has support on the disjoint intervals in j 1. e write T j f 2 = T j f 2 dx = j 1 T j f 2 dx + j 1\ j T j f 2 dx j and estimate the terms separately. First note that (13) T j f 2 dx = j 1\ j = j 1 j 1 \ () \ () j 1 \ () T j f 2 dx 1/2 (x) 1 1/2 M f 2 dx 1/2 (x) 2 1 1/2 M f 2 dx λ j 1 \ () 1 1/2 M f 2 dx.

8 OPERATOR-EGHTED HBERT TRANSFORM 7 Note that for each j 1, (14) 1 1/2 M f 2 = 1 1/2 C F() F() 1/2 f h 2 1/2 2 f k 2 Cλ f 2, since for F(), one has 1/2 1 1/2 = 1 1/2 1 (x) 1/2 dx λ. Using the disjointness of the j 1, we obtain (15) j 1\ j T j f 2 dx Cλ 2 j f 2. Now we consider the other part: (16) j T j f 2 dx = = j 1 () j 1 () j 1 () 1/2 1/2 1/2 2 1/2 (x) F() 1/2 2λ 2 F() 1/2 f kh k 2 dx 1/2 f kh k 2 dx F() 1/2 f kh k 2 dx j 1 f 2 = 2λ 2 j f 2. Therefore, there exists a constant A > 0 such that T j f 2 A j f 2. e will now show that there exist 0 < d < 1 and a constant à > 0 such that for any k > j, (17) k 1 T j f 2 dx Ãdk j j f 2.

9 8 SANDRA POTT Note that M j f is contant on each j 1. Thus T j f 2 dx k 1 = 1/2 (x)m j f() 2 dx j k j 1 () = j k j 1 () = j k j 1 () = j k j 1 () j 1 ( ) j 1 ( ) 2λ 2λ 1/2 M j f() 2 dx 1/2 M j f() 2 M j f(), M j f() j 1 ( ) j 1 ( ) k j 1 () k j 1 () ( 2λC 1/p (c 1/p ) k j 1 = 2λC 1/p (c 1/p ) k j 1 k j 1 () k j 1 () 1 1/2 1/2 2 1/2 M jf() 2 j 1 ( ) j 1 C 1/p 4λ 3 (c 1/p ) k j 1 j f 2 1/2 1/2 2 1/2 M jf() 2 (x) dx 1/2 M jf() 2 1/2 (x) 2p dx ) 1/p c (k j 1)/p 1 1/p 1/2 M jf() 2 1/2 M jf() 2 ( ) T j f 2 dx by (16). Here, p = r/2 > 1, and 1/p + 1/p = 1. Choosing d = c 1/p, we obtain the statement. Boundedness of T now follows from Cotlar s emma (see e. g. [10], 2.4), since T k f is supported on k 1, and the Tk have orthogonal ranges. t remains to be shown that S is bounded. et M 1/2 be the densely defined multiplication operator with the operator function 1/2 on 2 (R, H), let D be the densely defined operator on 2 (R, H) which is given by D eh = 1/2 eh for each D and e H, let D 1 be the densely defined operator on 2 (R, H)

10 OPERATOR-EGHTED HBERT TRANSFORM 9 given by D 1 eh = 1/2 correspondingly. ith this notation, we have shown that T = M 1/2 D 1 eh for D, e H, and let D 1, D 1 be defined 1 extends to a bounded linear operator on 2 (R, H). Since 1 satisfies the vector A 2 condition, D 1 D 1 1 defines a bounded invertible linear operator on 2 (R, H). t follows that S = D 1M 1/2 is a bounded linear operator on 2 (R, H) and has a bounded inverse S 1 = M 1/2 1 D 1 1 = S 1 D 1 D Boundedness of the weighted dyadic martingale transforms and of the weighted Hilbert transform e can now prove our main result. Theorem 4.1. et : R (H) be an operator weight.. (1) Suppose that and 1 both have the dyadic matrix reverse Hölder property. Then for each σ { 1, 1} D, the weighted martingale transform M 1/2 T σm 1/2 defines a bounded invertible linear operator on 2 (R, H), and the norms of these operators are uniformly bounded. (2) Suppose that and 1 both have the matrix reverse Hölder property. Then the weighted Hilbert transform M 1/2 1/2 HM defines a bounded invertible linear operator on 2 (R, H). Proof. t suffices to show (1), since the case of the Hilbert transform then follows from [11] together with the fact that the matrix reverse Hölder property 2.1 implies the dyadic matrix reverse Hölder property 2.2 for all translations and dilations of and 1. By Theorem 3.1, M 1/2 T σm 1/2 = M 1/2 D 1 T 1 σ D 1M 1/2 = S 1 T σs extends to a bounded invertible operator on 2 (R, H) for each σ { 1, 1} D. 5. A counterexample to the necessity of the reverse Hölder property Theorem 5.1. There exist an infinite-dimensional Hilbert space and an operator weight : R () such that the weighted Hilbert transform M 1/2 1/2 HM : 2 (R, ) 2 (R, ) is bounded, but the matrix reverse Hölder condition 2.1 does not hold for. Proof. et H be an infinite-dimensional separable Hilbert space, and consider an operator weight of the form ( ) : R (H H), (t) = V (t) 1H B(t) V (t), where V (t) =, 0 1 H where B is a weakly integrable (H)-valued function. An easy calculation (see [6], [3]) shows that then the Hilbert tranform defines a bounded linear operator 2 (R, H H) 2 (R, H H), if and only if the commutator [H, B] defines a bounded linear operator 2 (R, H) 2 (R, H).

11 10 SANDRA POTT Now notice that for each interval R and each x, (18) (x) = ρ( 1 (x)) = ρ( V 1 V 1 V (x)v (x)) and = ρ(v (x) V 1 V 1 V (x)) = V (x) V 1 V 1 V (x) (19) V (x) V 1 V 1 V (x) ( ) ( ) ( ) 1H B(x) 1H + BB = B 1H H B 1 H B (x) 1 H ( ) B(x) B = B (x) B. 1 H n particular, (x) B(x) B. This means, the matrix reverse Hölder condition 2.1, even with r = 2, implies the norm BMO condition 1 (20) sup B(x) B dx <. R interval However, it is known that boundedness of the commutator [H, B] on 2 (R, H) does not imply that the norm BMO condition (20) for B holds, since (H) is not a UMD space (see [1]). 6. Acknowledgement The author thanks T.A. Gillespie for numerous fruitful discussions and Svetlana Roudenko for providing her with the unpublished manuscript [7]. References [1] O. Blasco, Hardy spaces of vector-valued functions:duality, Trans. Am. Math. Soc. 308(1988), no 2, [2] M. Christ and M. Goldberg, Vector A 2 weights and a Hardy-ittlewood maximal function, Trans. Amer. Math. Soc. 353 (2001), no. 5, [3] T. A. Gillespie, S. Pott, S. Treil and A. Volberg, The transfer method in estimates for vector Hankel operators Algebra i Analiz12(2000), no. 6, ; English version in St. Petersburg Math.. 12 (2001), no. 6, [4] T. A. Gillespie, S. Pott, S. Treil and A. Volberg, ogarithmic growth for matrix martingale transforms,. ondon Math. Soc. (2)64(2001), no. 3, [5] T. A. Gillespie, S. Pott, S. Treil and A. Volberg, ogarithmic growth for weighted Hilbert transform and vector Hankel operators, to appear in. Op. Th. [6] T. A. Gillespie, S. Pott and. ilson, Matrix BMO and matrix A 2 weights: A substitute for the classical exp-relationship, Preprint, submitted [7] M. Goldberg, unpublished manuscript [8] R. Hunt, B. Muckenhoupt and R. heeden, eighted norm inequalities for the conjugate function and Hilbert transform, Trans. Amer. Math. Soc. 176(1973), [9] F. Nazarov and S. Treil, The hunt for a Bellman function: applications to estimates for singular integral operators and to other classical problems of harmonic analysis, Algebra i Analiz 8(1996), [10] M. C. Pereyra, ecture Notes on Dyadic Harmonic Analysis, Contemporary Mathematics 289(2001), 1-60 [11] S. Petermichl and S. Pott, A version of Burkholder s Theorem for operator-weighted 2 - spaces, to appear in Proc. Amer. Math. Soc. 131 (2003), , electronically published on

12 OPERATOR-EGHTED HBERT TRANSFORM 11 [12] S. Treil and A. Volberg, avelets and the angle between past and future,. Funct. Anal. 143(1997), [13] A. Volberg, Matrix A p weights via S-functions,. Am. Math. Soc. 10(1997), address: sp23@york.ac.uk Department of Mathematics, University of York, York YO10 5DD, U