PREPRINT 2007:14. On Hankel Forms of Higher Weights, the Case of Hardy Spaces MARCUS SUNDHÄLL EDGAR TCHOUNDJA

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1 PREPRINT 2007:14 On Hankel orms of Higher Weights, the Case of Hardy Spaces MARCUS SUNDHÄLL EDGAR TCHOUNDJA Department of Mathematical Sciences Division of Mathematics CHALMERS UNIVERSITY O TECHNOLOGY GÖTEORG UNIVERSITY Göteborg Sweden 2007

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3 Preprint 2007:14 On Hankel orms of Higher Weights, the Case of Hardy Spaces Marcus Sundhäll and Edgar Tchoundja Department of Mathematical Sciences Division of Mathematics Chalmers University of Technology and Göteborg University SE Göteborg, Sweden Göteborg, May 2007

4 Preprint 2007:14 ISSN Matematiska vetenskaper Göteborg 2007

5 ON HANKEL ORMS O HIGHER WEIGHTS, THE CASE O HARDY SPACES MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA Abstract. In this paper we study bilinear Hankel forms of higher weights on Hardy spaces in several dimensions (see [10] and [11] for Hankel forms of higher weights on weighted ergman spaces). We get a full characterization of S p class Hankel forms, 2 p < (and 1 p < for the case of weight zero), in terms of the membership for the symbols to be in certain esov spaces. Also, the Hankel forms are bounded and compact if and only if the symbol satisfies certain Carleson measure criterion and vanishing Carleson measure criterion, respectively. 1. Introduction and main results Schatten-von Neumann class Hankel forms of higher weights on ergman spaces are characterized in [10] and [11]. In the same way, as for the case of ergman spaces, Hankel forms of higher weights on a Hardy space are explicit characterizations of irreducible components in the tensor product of Hardy spaces under the Möbius group; see [7]. Recall from [10] and [11] the case of weighted ergman spaces L 2 a(dι ν ) of holomorphic functions, square integrable with respect to the measure (1) dι ν (z) = c ν (1 z 2 ) ν (d+1) dm(z), where ν > d, c ν a normalization constant and dm(z) is the Lebesgue measure on the unit ball = {z C d : z < 1}. The bilinear Hankel Date: May 18, Mathematics Subject Classification. 32A25, 32A35, 32A37, Key words and phrases. Hankel forms, Schatten-von Neumann classes, ergman spaces, Hardy spaces, esov spaces, transvectants, unitary representations, Möbius group. The visit at the Silesian University in Opava, Czech Republic, for the second author was financially supported by the International Science Program (ISP), Uppsala, Sweden. 1

6 2 MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA forms of weight s = 0, 1, 2, are given in [10] by (2) H s (f 1, f 2 ) = T s (f 1, f 2 ), z (1 z 2 ) 2ν (d+1) dm(z). The transvectant, T s, is given by s ( ) s T s (f 1, f 2 )(z) = ( 1) s k k f 1 (z) s k f 2 (z), k (ν) k (ν) s k where s f(z) = k=0 d j 1,,j s=0 j1 js f(z) dz j1 dz js, and (ν) k = ν(ν +1) (ν +k 1) is the Pochammer symbol. Also, the Möbius invariant inner product, z is given in the following way; for u, v s ( C d), where the tangent space at z is identified with C d, (3) u, v z = s t (z, z)u, v s (C d ) where (z, z) = (1 z 2 )(I, z z) is the ergman operator on C d and t (z, z) is the dual operator acting on the dual space of C d. The tensor-valued holomorphic function is called the symbol corresponding to the Hankel form H s. Now, let be the boundary of the unit ball of C d. The irreducible components in the decomposition of tensor products of Hardy spaces H 2 ( ) in [7] can be given explicitly as bilinear Hankel forms of weight s on the Hardy space H 2 ( ) by (4) H s (f 1, f 2 ) = T s (f 1, f 2 ), z (1 z 2 ) d 1 dm(z), where the transvectant, T s, is here given by s ( ) s T s (f 1, f 2 )(z) = ( 1) s k k f 1 (z) s k f 2 (z), k (d) k (d) s k k=0 where, in fact, this is the limiting case ν = d of (2). The main results for Hankel forms, H s, defined by (4) are given below in Theorem A and Theorem. Theorem A. H s is (compact) bounded if and only if dµ (z) = 2 z(1 z 2 ) 2d 1 dm(z) is a (vanishing) Carleson measure on H 2 ( ), with equivalent norms.

7 HANKEL ORMS O HIGHER WEIGHTS ON HARDY SPACES 3 Remark. Note that 2 z =, z. Theorem. H s is of Schatten class S p, 2 p <, if and only if ( 1/p pd,s,p = p z(1 z 2 ) dm(z)) pd d 1 <, with equivalent norms. Remark. If s = 0 we rewrite the Schatten class criterion as (5) (R )(z) p (1 z 2 ) (p 1)(d+1) dm(z) <, where R = d i=1 z i z i is the radial derivative. Theorem is then extended to 1 p < where (5) is equivalent to pd,0,p < for 1 < p <. Remark. Janson and Peetre obtained Theorem A and Theorem in the case d = 1, by using paracommutator arguments; see [5]. Our approach extend their results and provides a different proof of the case d = 1 they have treated. Approach. In this paper we use different techniques to deal with the case of weight zero and the case of weight s = 1, 2,, and they are therefore treated separately in Section 3 and Section 4, respectively. In [10] the criteria for boundedness, compactness and Schatten-von Neumann class for higher weights, on weighted ergman spaces, are natural generalizations of the case of weight zero. or Hardy spaces, as the Example 4.2 shows, the transvectant of various weight does not behave as in the case of ergman spaces where the boundedness properties for the transvectant was necessary in order to generalize the weight zero case to arbitrary weights. This explains why we, in this paper, treat the weight zero and nonzero cases separately. The Hankel forms, on Hardy spaces, of weight zero can be rewritten, using the radial derivative, into the classical Hankel forms in [12] and then we use results from [13], [14] and [15] to get the right conditions for the symbols. We have results for Carleson measures which together with invariance properties give criteria for the boundedness and compactness for Hankel forms of nonzero weights. The Schatten class criteria are proved by using interpolation for analytic families of operators. or this purpose we need results about Hankel forms on

8 4 MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA ergman-sobolev-type spaces. The preliminaries in Section 2 gives the prerequisites we need. Notation. If 1 and 2 are two equivalent norms on a vector space X, then we write x 1 x 2, x X. Also, for two real-valued functions, f and g, on X we write f g if there is a constant C > 0, independent of the variables in questions, such that Cf(x) g(x). Acknowledgement. The main results in this paper are the outcome after the authors visited each other in Göteborg and Opava, respectively. We are therefore grateful to the Mathematical Sciences at Chalmers University of Technology and Göteborg University and to the Mathematical Institute at the Silesian University in Opava for their generous hospitality. We would also like to thank Professor Genkai Zhang in Göteborg and Professor Miroslav Engliš in Opava for useful encouragement and advise. 2. Preliminaries or α > d, let A 2 α be the ergman-sobolev-type space of holomorphic functions f : C with the property that f 2 α = c(m) 2 Γ(d + α)m! Γ(d + m + α) <, m N d where f(z) = m N d c(m)z m is the Taylor expansion of f. Then A 2 α is a Hilbert space with the inner product f 1, f 2 α = Γ(d + α)m! c 1 (m)c 2 (m) Γ(d + m + α), m N d where f i (z) = m N d c i (m)z m, i = 1, 2. A 2 α has a reproducing kernel, K α w for w, given by (6) K α w(z) = 1 (1 z, w ) α+d. If α > 0, then A 2 α is the weighted ergman space L 2 a(dι α+d ), where dι α+d is given by (1). Also, A 2 0 is the Hardy space H 2 ( ).

9 HANKEL ORMS O HIGHER WEIGHTS ON HARDY SPACES Decomposition of A 2 α A 2 β. Let G be the group of biholomorphic self-maps on. If g G with g(z) = 0, then there is a linear fractional map ϕ z on and a unitary map U U(d) such that g = Uϕ z. The fractional linear map, ϕ z, is given by (7) ϕ z (w) = z P zw (1 z 2 ) 1/2 Q z w 1 w, z where P z =, z z/ z 2 and Q z = I P z. The complex Jacobian is therefore given by J g = det U J ϕz, where (8) J ϕz (w) = ( 1) d (1 z 2 ) (d+1)/2 (1 z, w ) d+1. The group G acts unitarily on A 2 α via the following: (9) π ν (g)f(z) = f(g 1 (z))j g 1(z) ν/(d+1), where ν = α + d, and it gives an irreducible unitary (projective) representation of G. In addition, for β > d, the group G acts on the Hilbert space tensor product A 2 α A 2 β by, (10) π ν1 (g) π ν2 (g)(f 1 (z), f 2 (w)) = f 1 (g 1 (z))f 2 (g 1 (w))j g 1(z) ν 1/(d+1) J g 1(w) ν 2/(d+1), where ν 1 = α + d, ν 2 = β + d, and it gives an unitary (projective) representation of G. However this is not irreducible and the irreducible decomposition is given in [7]. In particular, if α + β > d s: (11) A 2 α A 2 β Hα+β+2d,s 2, s=0 where Hu,s, 2 u > d s, is the space of holomorphic functions : s (C d ) with the property that 2 z (1 z 2 ) u d 1 dm(z) <, where we recall that 2 z =, z = s t (z, z) (z), (z) s (C d ). The group G acts unitarily on H 2 u,s by, (12) π u,s (g 1 ) (z) = s dg(z) t (g(z))j g (z) u/(d+1),

10 6 MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA where dg(z) : T z () T g(z) () is the differential map, and gives an irreducible unitary (projective) representation of G. Via the transvectant, Ts α,β, defined on A 2 α A 2 β by s ( ) s Ts α,β (f 1, f 2 ) = ( 1) s k k f 1 (z) s k f 2 (z) (13), k (α + d) k (β + d) s k k=0 the irreducible components in the decomposition (11) are realized in [7] as Hankel forms of higher weights (order s): (14) H α,β,s (f 1, f 2 ) = Ts α,β (f 1, f 2 ), α+β+2d,s,2, where, u,s,2 is the H 2 u,s-pairing, and A 2 α+β+2d,s. Remark 2.1. or α = β = 0 in (14) we get the Hankel forms of weight s on Hardy spaces defined by (4). The transvectant Ts α,β intertwining property: (15) Hence, (16) T α,β s : A 2 α A 2 β A2 α+β+2d,s is onto and has an (π α+d (g)f 1, π β+d (g)f 2 ) = π α+β+2d,s (g)t α,β s (f 1, f 2 ). H α,β,s (π α+d (g)f 1, π β+d (g)f 2 ) = H α,β,s π α+β+2d,s (g 1 ) (f 1, f 2 ) Spaces of symbols and Schatten class Hankel forms. Let, for 1 p < and u > d ps, 2 Hp u,s be the space of all holomorphic functions : s (C d ) such that p u,s,p = p z(1 z 2 ) u d 1 dm(z) <. Also, for u s, let 2 H u,s be the space of holomorphic functions : s (C d ) such that u,s, = sup z (1 z 2 ) u <. z Then H p u,s, for 1 p, are anach spaces. In [10] and in [11] there are several results about H p pν,s for ν > d and we can use the same arguments as in [10] and [11] to generalize these results to H p u,s. Hence, the results below will be stated without proofs. The reader is referred to [10] and [11] for more details.

11 HANKEL ORMS O HIGHER WEIGHTS ON HARDY SPACES 7 Lemma 2.2. Let u > max(0, d s). Then the reproducing kernel of H 2 u,s is, up to a nonzero constant c, given by K u,s (w, z) = (1 w, z ) u s t (w, z) 1. Namely, for any x s (C d ) and any H 2 u,s, (z), x s (C d ) = c, K u,s(, z)x u,s,2 = c, K u,s (, z)x w (1 w 2 ) u d 1 dm(w). Lemma 2.3. Let 1 < p < and 1/p + 1/q = 1. or u > d ps and 2 v > d qs the following duality 2 (Hu,s) p = Hv,s q holds, with respect to the H 2 u p + v,s-pairing. That is, for any bounded q linear functional, l, on Hu,s p there exists an element G Hv,s q such that l( ) =, G u p + v q,s,2 for all Hu,s, p and l G v,s,q. Lemma 2.4. Let u > d s and v d s. If 2 < p < then 2 (H 2 u+2d,s, H v+d,s) [1 2 p ] = Hp (p 2)v+u+pd,s. Lemma 2.5. Let α, β > d with α + β > d s. Then there is a constant C(α, β, s, d) > 0 such that H α,β,s S2 (A 2 α,a 2 β ) = C(α, β, s, d) α+β+2d,s,2, for all holomorphic : s (C d ). Remark 2.6. y cumputing the norms for = s dz 1 we can see that C(α, β, s, d) is continuous in α and β, since for some C(d, s) > 0 we have (17) C(α, β, s, d) 2 = C(s, d) 2 s k=0 ( ) s k 1 (α + d) k (β + d) s k. Lemma 2.7. Let α, β > 0. Then H α,β,s is bounded on A 2 α A 2 β only if H 1 (α+β)+d,s, with equivalent norms. 2 or α, β 0 define an operator (18) T α,β s (A)(z) = s k=0 T α,β s ( ) s ( 1) s k k if and on S (A 2 α, A 2 β ) by ( ) w k s k ζ A(Kw, α K β ζ ) (z, z), (α + d) k (β + d) s k

12 8 MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA where K α w is the reproducing kernel for A 2 α given by (6). Remark 2.8. If A has rank one then by (13). T α,β s is the transvectant given In the following next two results in this subsection we get use of Lemma 2.4. Namely, to get the results we need to interpolate the spaces Hα+β+2d,s 2 and H 1 (α+β)+d,s, where α, β > 0. In fact, by 2 Lemma 2.4, (19) (H 2 α+β+2d,s, H 1 2 (α+β)+d,s) [1 2 p ] = Hp 1 2 p(α+β)+pd,s T α,β s Lemma 2.9. Let α, β 0 and 2 p. Then maps S p (A 2 α, A 2 β ) into Hp 1 boundedly, and if Hα,β,s p(α+β)+pd,s S p (A 2 α, A 2 β ), 2 for α, β > 0 and 2 p < or α = β = 0 and p = 2, then T s α,β (H α,β,s ) =. Using Lemma 2.5, Lemma 2.7 with (19) on one hand and Lemma 2.9 on the other hand we get the following theorem. Theorem Let α, β > 0 and 2 p. Then H α,β,s is in S p (A 2 α, A 2 β ) if and only if Hp 1 p(α+β)+pd,s, with equivalent norms. 2 Remark We want to extend this result to α, β > 1/p, to include the Hardy case, and need therefore the theory for families of analytic operators. We use the approach by ergh-janson-löfström- Peetre-Peller and Theorem 2.12, given below, can be found in [6]. Let X 0, X 1 be anach spaces continuously imbedded into a anach space X and Y 0, Y 1 be anach spaces continuously imbedded into a anach space Y. Theorem Let Γ be a bounded holomorphic function on the strip 0 < R(z) < 1, continuous on 0 R(z) 1 and taking values in the space of operators from X 0 X 1 to Y 0 + Y 1. Suppose that 1) or any y R the operator Γ(iy) can be extended to a bounded operator from X 0 to Y 0 and sup y R Γ(iy) X0 Y 0 = M 0 <. 2) or any y R the operator Γ(1 + iy) can be extended to a bounded operator from X 1 to Y 1 and sup y R Γ(1+iy) X1 Y 1 = M 1 <.

13 HANKEL ORMS O HIGHER WEIGHTS ON HARDY SPACES 9 Then for any θ (0, 1) the operator Γ(θ) can be extended to a bounded operator from X [θ] = (X 0, X 1 ) [θ] to Y [θ] = (Y 0, Y 1 ) [θ] and Γ(θ) X[θ] Y [θ] M 1 θ 0 M θ Hankel forms of weight zero To find the Schatten-von Neumann class Hankel forms of weight zero on Hardy spaces we shall rewrite H 0 in terms of the small Hankel operators studied in [12]. The problem then boils down to finding the relationship between the corresponding symbols. The Hankel form, H G, in [12] is given by (20) H G (f 1, f 2 ) = f 1 (w)f 2 (w)g(w) dσ(w), where dσ is the normalized Lebesgue measure on. Denote by R the radial derivative, defined as d f Rf(z) = z i (z), z i i=1 where f : C is holomorphic. If R d := (R + 2d 1)(R + 2d 2) (R + d), then for holomorphic functions f 1 and f 2 we have, by means of Taylor expansion, (21) f 1 (w)f 2 (w) dσ(w) = c(d) f 1 (z)r d f 2 (z)(1 z 2 ) d 1 dm(z). Lemma 3.1. Let H 0 be given by (4) and H G by (20). Then H 0 = H G if and only if Proof. Since H 0 (f 1, f 2 ) = and R d G(z) = c(d) (z). H G (f 1, f 2 ) = f 1 (z)f 2 (z) (z)(1 z 2 ) d 1 dm(z) f 1 (w)f 2 (w)g(w) dσ(w), then the result follows by applying (21) on f 1 = f 1 f 2 and f 2 = G.

14 10 MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA 3.1. Schatten-von Neumann class S p Hankel forms. In this subsection we present sufficient and necessary conditions for Hankel forms of weight zero to be in Schatten-von Neumann class S p, 1 p < (see Theorem 3.2). Theorem 3.2. The Hankel form H 0 is of Schatten-von Neumann class S p, for 1 p <, if and only R (z) p (1 z 2 ) (p 1)(d+1) dm(z) <. This theorem is a direct consequence of Lemma 3.1 and Theorem 1 in [12] (see also Theorem C in [4]): Theorem 3.3. Let α > 1 and 1 p <. Then the Hankel form H G, defined by (20), is of Schatten-von Neumann class S p if and only if R d+1 G(z) p (1 z 2 ) (p 1)(d+1) dm(z) < ounded and compact Hankel forms. In this subsection we present necessary and sufficient conditions for Hankel forms of weight zero to be bounded and compact; see Theorem irst we need some preliminaries, which basically can be found in [14] and [13]. We also remark that the one dimensional case of Lemma 3.14 is already proved (see Corollary 15 in [15]) but since we have not been able to find an explicit version of this result in several variables we prove this result. Once we have results for Carleson measures and MOA spaces, then the corresponding results for vanishing Carleson measures and VMOA spaces will be easily deduced. When necessary we will give brief proofs for the cases of vanishing Carleson measures and VMOA spaces, but in most cases these results will only be stated without proofs. Definition 3.4 (See [14]). Let ζ and r > 0 and let Q r (ζ) = {z : d(z, ζ) < r} where d(z, ζ) = 1 z, ζ 1/2 is the non-isotropic metric on. A positive orel measure µ in is called a Carleson measure if there exists a constant C > 0 such that µ(q r(ζ)) Cr d,

15 HANKEL ORMS O HIGHER WEIGHTS ON HARDY SPACES 11 for all ζ and r > 0, and called a vanishing Carleson measure if uniformly for ζ. µ(q r(ζ)) lim = 0 r 0 + r d Remark 3.5. y Lemma 3.9 below, the definition above concerns Carleson measures on Hardy spaces H 2 ( ). The Hardy space H 2 ( ) consists of all holomorphic functions f : C such that ( f H 2 = sup 0<r<1 sup z 1/2 f(rζ) dσ(ζ)) 2 <. Lemma 3.6 (Theorem 45 in [13]). A positive orel measure µ in is a Carleson measure if and only if, for each (or some) s > 0, (1 z 2 ) s dµ(w) <. 1 z, w d+s Remark 3.7. This is a generalization of Theorem 5.4 in [14]: A positive orel measure µ in is a Carleson measure if and only if P (z, w) dµ(w) <, where sup z P (z, w) = (1 z 2 ) d ; z, w. 1 z, w 2d Lemma 3.8 (See [13]). A positive orel measure µ in is a vanishing Carleson measure if and only if, for each (or some) s > 0, (1 z 2 ) s lim dµ(w) = 0. z 1 1 z, w d+s Lemma 3.9 (Theorem 5.9 in [14]). A positive orel measure µ in is a Carleson measure if and only if there exists a constant C > 0 such that f(z) 2 dµ(z) C f 2 H 2 ( ), for all f H 2 ( ). Lemma 3.10 (Theorem 5.10 in [14]). Let µ be a positive orel measure on. Then µ is a vanishing Carleson measure if and only if, for

16 12 MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA every sequence {f j } bounded in H 2 ( ) such that f j (z) 0 for every z, we have lim f j (z) 2 dµ(z) = 0. j Lemma 3.11 (Theorem 50 in [13]). Let µ be a positive orel measure in. Then the following conditions are equivalent (a) There is a constant C > 0 such that (Rf)(z) 2 dµ(z) C f 2 H 2 ( ), for all f H 2 ( ). (b) There is a constant C > 0 such that for all ζ and r > 0. µ(q r(ζ)) Cr (d+2), Lemma 3.12 (See [13]). Let µ be a positive orel measure in. Then the following conditions are equivalent (a) or every sequence {f j } bounded in H 2 ( ) and f j (z) 0 for every z, we have lim j (b) If r 0 +, then (Rf j )(z) 2 dµ(z) = 0. µ(q r(ζ)) r d+2 0 uniformly for ζ. Also, we need a result about the radial derivative, which can be obtained by using Taylor expansion. Lemma Let t > 1 and a > 0. Then (R + a)f(z) 2 (1 z 2 ) t+2 dm(z) f(z) 2 (1 z 2 ) t dm(z), for all holomorphic f : C. Lemma Let t > 1 and a 0. or any holomorphic function g : C, dµ 1 (z) = g(z) 2 (1 z 2 ) t dm(z) is a (vanishing) Carleson measure if and only if dµ 2 (z) = ((R + a)g)(z) 2 (1 z 2 ) t+2 dm(z) is a (vanishing) Carleson measure.

17 HANKEL ORMS O HIGHER WEIGHTS ON HARDY SPACES 13 Proof. We only prove equivalence for the Carleson measure case, the vanishing Carleson measure case then follows by using the same techniques. Also, we may assume that a > 0 since exactly the same arguments then can be applied on h = g g(0) instead of g (for estimating g g(0) ; see proof of Theorem 2.16 in [14]), and then the result follows by using the triangle inequality and the fact that g(0) 2 (1 z 2 ) t dm(z) is a Carleson measure. Assume first that dµ 1 is a Carleson measure. Then there is a constant C > 0 such that (1 z 2 ) 2 dµ 1 (z) 4r 2 Q r (ζ) Q r (ζ) dµ 1 (z) 4Cr (d+2), for all ζ and r > 0, so that (1 z 2 ) 2 dµ 1 (z) satisfies the condition (b) in Lemma Hence, by the triangle inequality, there is a constant C 1 > 0 such that (22) ( 1/2 ((R + a)f)(z) 2 (1 z 2 ) 2 dµ 1 (z)) C 1 f H 2 ( ), for all f H 2 ( ). y Lemma 3.13 and by the inequality (22), ) 1/2 ( f(z) 2 dµ 2 (z) ( 1/2 ((R + a)(fg))(z) 2 (1 z 2 ) dm(z)) t+2 + ( ((R + a)f)(z) 2 (1 z 2 ) 2 dµ 1 (z) ( 1/2 C 2 f(z) 2 dµ 1 (z)) + C 1 f H 2 ( ) C 3 f H 2 ( ), ) 1/2 for all f H 2 ( ), so that dµ 2 is a Carleson measure by Lemma 3.9.

18 14 MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA Assume now that dµ 2 (z) = ((R + a)g)(z) 2 (1 z ) t+2 dm(z) is a Carleson measure. Then, by Lemma 3.13, ( ) 1/2 f(z)g(z) 2 (1 z 2 ) t dm(z) ( ) 1/2 C ((R + a)(fg))(z) 2 (1 z 2 ) t+2 dm(z) ( ) 1/2 C ((R + a)f)(z) 2 g(z) 2 (1 z 2 ) t+2 dm(z) ( 1/2 + f(z) 2 dµ 2 (z)), for all f H 2 ( ). Since dµ 2 is a Carleson measure, then f(z) 2 dµ 2 (z) C f H 2, for all f H 2 ( ). y Lemma 3.11 it remains to prove that (23) g(z) 2 (1 z 2 ) t+2 dm(z) Cr d+2. Q r (ζ) y Lemma 3.6, (23) is equivalent to that for each (or some) s > 0 it holds that (1 w 2 ) s (24) sup w 1 ζ, w d+2+s dµ (ζ) < +, where dµ (ζ) = g(ζ) 2 (1 ζ 2 ) t+2 dm(ζ). Now, again by Lemma 3.6, since dµ 2 is a Carleson measure, then for each (or some) s > 0 there is a constant C > 0 such that (25) ((R + a)g)(ζ) 2 (1 ζ 2 ) t+2 1 ζ, w d+s dm(ζ) C (1 w 2 ) s, for all w. Let k = [(t + 1)/2] + 1. Then we have the reproducing property (R + a)g(ζ) (R + a)g(z) = c (1 z, ζ ) (1 d+1+k ζ 2 ) k dm(ζ). This implies that g(z) = c h( z, ζ )((R + a)g)(ζ) (1 z, ζ ) d+k (1 ζ 2 ) k dm(ζ),

19 HANKEL ORMS O HIGHER WEIGHTS ON HARDY SPACES 15 where h is a polynomial of degree 1. Hence ((R + a)g)(ζ) g(z) C 1 z, ζ (1 d+k ζ 2 ) k dm(ζ), so that (26) ( ) ( ) g(z) 2 C 2 dµ(ζ) (1 ζ 2 ) 2k (t+2) dm(ζ). 1 z, ζ d+s 1 z, ζ d+2k s Since t > 1 we can choose s such that 0 < s < t + 1. y (25) and Proposition in [9] we have g(z) 2 C (1 z 2 ) t+1. Thus, again by Proposition in [9], g(z) 2 (1 z 2 ) t+2 dm(z) 1 z, w d+2+s C 1 z 2 1 z, w dm(z) C d+2+s (1 w 2 ), s which proves (24). Definition 3.15 (See [14]). Let MOA denote the space of functions f H 2 ( ) such that f 2 MO = f(0) sup f(ξ) f Q(ζ,r) 2 dσ(ξ) <, Q(ζ,r) Q(ζ, r) Q(ζ,r) where, for any ζ and r > 0, Q(ζ, r) = { ξ : 1 ζ, ξ 1/2 < r }, and 1 f Q(ζ,r) = f(ξ) dσ(ξ). Q(ζ, r) Q(ζ,r) VMOA is the closure in MOA of the sets of polynomials, namely the space of functions f H 2 ( ) such that lim sup 1 f(ξ) f Q(ζ,r) 2 dσ(ξ) = 0. r 0 + Q(ζ, r) Q(ζ,r) Q(ζ,r) As a direct consequence of Lemma 3.14 we get a generalized version of (Theorem 5.19) Theorem 5.14 in [14].

20 16 MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA Lemma Let k be a positive integer, a 1,..., a k 0, and f be holomorphic on. Then the following properies are equivalent: (i) f (VMOA) MOA. (ii) ((R+a 1 ) (R+a k )f)(z) 2 (1 z 2 ) 2k 1 dm(z) is a (vanishing) Carleson measure. The classical Hankel form (small Hankel operator) H G on the Hardy space H 2 ( ), as in [12], is bounded if and only if G MOA and H G is compact if and only if G VMOA; see [2] and [3]. Then, as a consequence of Lemma 3.16 and Lemma 3.1, we have the following theorem. Theorem The Hankel form H 0 only if (z) 2 (1 z 2 ) 2d 1 dm(z) is a (vanishing) Carleson measure. 4. The case s = 1, 2, 3, is (compact) bounded if and In this section we study boundedness, compactness and the class S p properties, 2 p <, for the case s 1. As in [10] we have the esov characterization (see Lemma 4.1 below). However, this lemma does not hold for s = 0. Lemma 4.1. or any positive integer s, f(0) s 1 f(0) 2 + s f 2 z for all f H 2 ( ). dm(z) 1 z 2 f 2 H 2, The difficulty for the Hardy spaces are explained by this example, where it is shown that we can find f 1, f 2 H 2 ( ) such that T s (f 1, f 2 ) / Hd,s 1 : Example 4.2. This example is based on the proof of Theorem II in [8]. irst consider the case when s = 1 and d = 1. Let 1 f 1 (z) = k z2k and f 2 (z) = 1. k=1 Then f 1, f 2 H 2 ( D) and since the series f 1 (z) is lacunary then T 1 (f 1, f 2 ) 1,1,1 = f 1(z) dm(z) =. D

21 HANKEL ORMS O HIGHER WEIGHTS ON HARDY SPACES 17 This is a consequence of a result about lacunary series by Zygmund; see [8]. Namely, if n k+1 /n k > λ for some λ > 1, and if h(z) = k=0 c kz n k satisfies 1 0 h (re iθ ) dr < for some θ, then k=0 c k <. In the general case, d 1 and s = 1, 2,, we just change f 1 into f 1 (z) = k=1 1 k z2k 1, and still let f 2 (z) = 1. Then T s (f 1, f 2 ) d,s,1 = (1 z 1 2 ) s/2 (1 z 2 ) s/2 1 s f 1 (z) z1 s dm(z) (1 z 2 ) s 1 s f 1 (z) dm(z). z s 1 y Theorem 2.17 in [14] there is a constant C > 0 such that (1 z 2 ) s 1 s f 1 (z) dm(z) C f 1 (z) z 1 dm(z) z s 1 and the right hand side of the inequality above is infinite, as we can see in the initial case (s = 1, d = 1) oundedness and compactness. Criteria for boundedness and compactness are given in Theorem 4.5 and Theorem 4.7, respectively. To prove these theorems we need some lemmas. or holomorphic : s (C d ) we consider the norm CM given by (27) 2 CM = sup w (1 w 2 ) d 1 z, w 2d dµ (z), where dµ (z) = 2 z(1 z 2 ) 2d 1 dm(z). Lemma 4.3. Let : s (C d ) be holomorphic and let k be a nonnegative integer. If the measure dµ (z) = 2 z(1 z 2 ) 2d 1 dm(z) is a Carleson measure, then there is a constant C k > 0 such that k f 2 z dµ (z) C k 2 CM f 2 H, 2 for all f H 2 ( ).

22 18 MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA Proof. Clear if k = 0. Assume k is a positive integer. If f H 2 ( ) then k f Hd,s 2, by Lemma 4.1. Hence, by the reproducing property in Lemma 2.2 and by Lemma 7.1 in [10], k (1 z 2 ) s/2 (1 w 2 ) s/2 1 f z k f 1 z, w d+s w dm(w) Let 0 < ε < 1. Then, by Proposition in [9], k f 2 (1 w 2 ) ε 1 z 1 z, w d+ε k f 2 w dm(w), and hence, by Lemma 4.1, k f 2 z dµ (z) ( (1 w 2 ) ε 1 k f 2 w 2 CM k f 2 w ) dµ (z) dm(w) 1 z, w d+ε dm(w) 1 w 2 2 CM f 2 H 2. We need to consider subspaces of H 2 u,s, u > d s, namely 2 u,s which consists of elements = s f, where f : C is holomorphic and u,s,2 <. Lemma 4.4. Let { } X = S H3d,s 2 : S 3d,s,2 = sup s f d,s,2 =1 s f, S 2d,s,2. Then (d,s 2 ) X, with respect to the pairing s f, S 2d,s,2. That is, for any bounded linear functional, l, on d,s 2 there is an element S X such that l( s f) = s f, S 2d,s,2 and l S 3d,s,2. Proof. Let l (d,s 2 ). Extend l to l (Hd,s 2 ) with l = l and l( s f) = l( s f), by Hahn-anach Theorem. Then, by Lemma 2.3, there is an element S H3d,s 2 with l S 3d,s,2 so l S 3d,s,2. In this sense we can imbedd (d,s 2 ) continuously in H3d,s 2 and can therefore be viewed as a subspace of H3d,s 2 {. Hence, (d,s) 2 S H3d,s 2 : S 3d,s,2 = sup s f, S 2d,s,2 s f d,s,2 =1 },

23 HANKEL ORMS O HIGHER WEIGHTS ON HARDY SPACES 19 with respect to the pairing s f, S 2d,s,2. Now we can prove the criterion for boundedness. Theorem 4.5. The Hankel form H s is bounded if and only if dµ (z) = 2 z(1 z 2 ) 2d 1 dm(z) is a Carleson measure, with equivalent norms. Proof. irst assume that dµ is a Carleson measure. It suffices to prove that, for k > 0, k f 1 s k f 2, (1 z z 2 ) d 1 dm(z) CM f 1 H 2 f 2 H 2. This is a direct consequence of Lemma 4.1 and Lemma 4.3, since k f 1 s k f 2, (1 z z 2 ) d 1 dm(z) k f 1 z s k f 2 z z (1 z 2 ) d 1 dm(z) ( ) 1/2 ( ) 1/2 k f 1 2 dm(z) z s k f 1 z z dµ (z) C s,k CM f 1 H 2 f 2 H 2 for some constant C s,k > 0. Now assume that H s is bounded. Let G w = π 2d,s (ϕ w ) where the action g π 2d,s (g) is defined in (12), and the fractional linear map is defined in (7). Since ϕ 1 w = ϕ w, then by (16), (28) H s G w (f 1, f 2 ) = H s ( πd (ϕ w )f 1, π d (ϕ w )f 2 ), where g π d (g) is the unitary action on H 2 ( ) defined in (9). Since π 2d,s (ϕ w ) is unitary on H2d,s 2 (or even on L2 2d,s ; the space of measurable with 2d,s,2 < ) then π 2d,s (ϕ w ) 2 3d,s,2 = 2 z(1 ϕ w (z) 2 ) d (1 z 2 ) d 1 dm(z) = (1 w 2 ) d 1 z, w 2d 2 z(1 z 2 ) 2d 1 dm(z). Hence we can make the following reformulation of CM : (29) CM = sup G w 3d,s,2. w

24 20 MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA It follows by (28) that HG s w = H s and hence Hs G w is bounded for any w. Define, T Gw ( s f) = s f, G w 2d,s,2 on d,s 2. Then T Gw ( s f) = HG s w (f, 1), and by Lemma 4.1, T Gw ( s f) H s G w f H 2 H s G w s f d,s,2, so T Gw : d,s 2 C is a bounded linear functional on 2 d,s. Hence, by Lemma 4.4 and Lemma 4.1, so by (29), G w 3d,s,2 sup s f d,s,2 =1 s f, G w 2d,s,2 = sup H s Gw (f, 1) s f d,s,2 =1 sup HG s w f H 2 s f d,s,2 =1 H s G w = H s, CM H s. efore we can prove the criterion for compactness we need one more lemma. Lemma 4.6. Let : s (C d ) be holomorphic, and r (z) = (rz) for 0 < r < 1. If dµ (z) = 2 z(1 z 2 ) 2d 1 dm(z) is a vanishing Carleson measure, then r CM 0, as r 1. Proof. If dµ (z) is a vanishing Carleson measure, then (1 w 2 ) d lim w 1 1 z, w dµ 2d (z) = 0. Hence, this lemma is a direct consequence of the fact that (1 w 2 ) d 1 z, w dµ (1 rw 2 ) d 2d r (z) 1 z, rw dµ 2d (z) and dominated convergence. Theorem 4.7. The Hankel form H s is a vanishing Carleson measure. is compact if and only if dµ (z) = 2 z(1 z 2 ) 2d 1 dm(z)

25 HANKEL ORMS O HIGHER WEIGHTS ON HARDY SPACES 21 Proof. irst, assume that dµ (z) is a vanishing Carleson measure. Then, by Theorem 4.5 and Lemma 4.6, H s r H s r CM 0 as r 1. Hence, it suffices to prove that H s r is compact. ut since r can be approximated in Carleson norm by its Taylor polynomials P (r) N and Hs has finite rank, then H P (r) s r is clearly compact (see the N proof of the sufficiency in Theorem 1.1(b) in [10]). Now assume that H s is compact. As in the proof of Theorem 4.5, let G w = π 2d,s (ϕ w ). Then dµ (z) is a vanishing Carleson measure if and only if, for any sequence {w n } such that w n 1 as n, (30) lim G w n 3d,s,2 = 0. n Again, as in the proof of Theorem 4.5, by Lemma 4.4, G wn 3d,s,2 = sup H s Gwn (f, 1) s f d,s,2 =1 = sup ( H s πd (ϕ wn )f, π d (ϕ wn )1 ) s f d,s,2 =1 The action g π d (g) is unitary on H 2 ( ) and {π d (ϕ wn )1} is a sequence in H 2 ( ) converging weakly to 0. Since H s is compact, then there is a sequence {c n } of positive number converging to 0 such that ( H s πd (ϕ wn )f, π d (ϕ wn )1 ) c n f H 2. y Lemma 4.1, G wn 3d,s,2 c n 0 as n, which proves (30) Schatten-von Neumann class. In this subsection we prove Theorem for s 1. or this purpose we prove two more general results; Theorem 4.12 (valid for s 1) and Theorem 4.13 (valid for s 0), and then Theorem follows by letting α = β = 0. The main idea is to use the interpolation theorem for families of analytic operators. To do this we first need to rewrite Hankel forms on ergman- Sobolev-type spaces to forms on Hardy spaces.

26 22 MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA or t C, we define the radial fractional derivative of order t by (1 + R) t f(z) = m N d (1 + m ) t c(m)z m, where f(z) = m N d c(m)z m is the Taylor expansion of f. The following lemma follows by using Taylor expansion and Stirling s formula. Lemma 4.8. If 2R(t) + α > 1 then f 2 α (1 + R) t f(z) 2 (1 z 2 ) 2R(t)+α dm(z), for all holomorphic functions f : C. As a direct consequence of Lemma 4.8 we have the following lemma. Lemma 4.9. Let α > d. Then f H 2 (1 + R) α/2 f α, for all holomorphic functions f : C. y Lemma 4.9 the Hankel forms H α,β,s given by (14), defined on A 2 α A 2 β, can be regarded as forms defined on H2 ( ) H 2 ( ) via (31) H α,β,s (f 1, f 2 ) := H α,β,s ( (1 + R) α/2 f 1, (1 + R) β/2 f 2 ). Namely, as a direct consequence of Lemma 4.9, using (31), we have the following result. Lemma Let α, β > d and p {2, }. Then α,β,s H Sp(H 2,H 2 ) H α,β,s Sp(A 2 α,a2 β ). Remark We can extend (31) to complex numbers α and β. In this case, if R(α), R(β) > d then α,β,s R(α),R(β),s H Sp(H 2,H 2 ) = H Sp(H 2,H 2 ) for p {2, }, by unitary operators. Theorem Let 2 p < and α, β > 1/p. Then S p (H 2, H 2 ) if H p 1 and p(α+β)+pd,s 2 H α,β,s Sp(H 2,H 2 ) 1 2 p(α+β)+pd,s,p. H α,β,s

27 HANKEL ORMS O HIGHER WEIGHTS ON HARDY SPACES 23 Proof. Put α 1 = α p 2, β 2p 1 = β p 2, α 2p 2 = α + 1 and β p 2 = β + 1. p Clearly α 1, β 1 > 1/2 and α 2, β 2 > 0. We will use interpolation for the analytic families of operators. or this purpose consider, for αz,βz,s 0 R(z) 1, the forms H, given by (31), where α z = α 1 + z(α 2 α 1 ) and β z = β 1 + z(β 2 β 1 ). Now we can define the analytic family of operators, {Γ(z)}, on the strip 0 R(z) 1 into operators from the intersection Hα 2 1 +β 1 +2d,s H into S 1 2 (α 2+β 2 )+d,s 2 + S, where αz,βz,s Γ(z) = H. Consider R(z) = 0: y Remark 4.11, Lemma 4.10 and Lemma 2.5, if Hα 2 1 +α 2 +2d,s then H αz,βz,s S2 = H α 1,β 1,s S2 α1 +β 1 +2d,s,2. Consider R(z) = 1: y Remark 4.11, Lemma 4.10 and Lemma 2.7, if H 1 2 (α 2+β 2 )+d,s then H αz,βz,s S = H α 2,β 2,s S 1 2 (α 2+β 2 )+d,s,. Now we claim that there is a constant C(d, s) such that (32) Γ(z) S2 C(d, s) α1 +α 2 +2d,s,2 for 0 R(z) 1 and for all Hα 2 1 +α 2 +2d,s. Accepting temporarily the claim, since S 2 S continuously and since H 2 α 1 +β 1 +2d,s H 1 2 (α 2+β 2 )+d,s H2 α 1 +β 1 +2d,s continuously, then Γ is bounded on the strip 0 R(z) 1. Hence we can apply the interpolation theorem for the analytic families of operators (see Theorem 2.12). We obtain, for fixed 0 < θ < 1, that Γ(θ) is bounded from (Hα 2 1 +β 1 +2d,s, H 1 2 (α 2+β 2 )+d,s [θ] into (S 2, S ) [θ]. Put θ = (p 2)/p. Using Lemma 2.4 we get and hence ( H 2 α1 +β 1 +2d,s, H 1 2 (α 2+β 2 )+d,s ) [1 2 p ] = Hp 1 2 p(α+β)+pd,s, H α,β,s Sp 1 2 p(α+β)+pd,s,p, since α θ = α and β θ = β when θ = (p 2)/p. Now we go back to the claim (32). We may assume that z is real, and we therefore put z = θ [0, 1]. y Lemma 4.10, Γ(θ) S2 (H 2,H 2 ) = H α θ,β θ,s S2 (A 2 α θ,a 2 β θ ),

28 24 MARCUS SUNDHÄLL AND EDGAR TCHOUNDJA and since α θ > α 1 > 1/2, β θ > β 1 > 1/2 then H α θ,β θ,s S2 (A 2 α θ,a 2 β θ ) C(d, s) s! αθ +β θ +2d,s,2 C(d, s) α1 +β 1 +2d,s,2, by Lemma 2.5, where C(d, s) is the constant in (17). Theorem Let 2 p < and α, β 0. Then H p 1 2 p(α+β)+pd,s if H α,β,s S p (H 2, H 2 ) and 1 p(α+β)+pd,s,p Hα,β,s 2 Sp(H 2,H 2 ). T α,β s Proof. Consider defined by (13). y Lemma 2.9 it remains to 0,0 prove that T s (H 0,0,s ) = if H 0,0,s S p (H 2, H 2 ) for 2 p <. Let H 0,0,s S p and let r (z) = (rz), for r (0, 1). Since H 0,0,s is compact then 2 z(1 z 2 ) 2d 1 dm(z) is a vanishing Carleson measure, by Theorem 4.7, and hence r CM 0 as r 1, by Lemma 4.6. Then r pointwise and also, by Theorem 4.5, we have H 0,0,s r H 0,0,s S 0 as r 1. Hence, by Lemma 2.9, T 0,0 s (H 0,0,s ) = lim 0,0 T r 1 s (H 0,0,s r ) = lim r =. r 1 References [1]. eatrous, J. urbea, Holomorphic Sobolev spaces on the ball, Dissertationes Math. (Rozprawy Mat.), 276 (1989), 60 pp. [2] J. urbea, oundary behaviour of holomorphic functions in the ball, Pacific J. of Math., 127:1 (1987), [3] R. Coifman, R. Rochberg, G. Weiss, actorization theorems for Hardy spaces in several variables, Ann. Math. (2), 103 (1976), no. 3, [4] M. eldman, R. Rochberg, Singular value estimates for commutators and Hankel operators on the unit ball and the Heisenberg group, Analysis and partial differential equations, , Lecture Notes in Pure and Appl. Math., 122, Dekker, New York, (1990). [5] S. Janson and J. Peetre, A new generalization of Hankel operators (the case of higher weights), Math. Nachr., 132 (1987), [6] V. V. Peller, Vectorial Hankel operators, commutators and related operators of the Schatten-von Neumann class γ p, Integral Eq. Operator Theory, 5 (1982), no. 2, [7] L. Peng and G. Zhang, Tensor product of holomorphic representations and bilinear differential operators, J. unct. Anal., 210 (2004), no. 1,

29 HANKEL ORMS O HIGHER WEIGHTS ON HARDY SPACES 25 [8] W. Rudin, The radial variation of analytic functions, Duke Math. J., 22, (1955b), [9] W. Rudin, unction theory in the unit ball of C n, Springer-Verlag, [10] M. Sundhäll, Schatten-von Neumann properties of bilinear Hankel forms of higher weights, Math. Scand., 98 (2006), [11] M. Sundhäll, Trace class criteria for bilinear Hankel forms of higher weights, Proc. of AMS, 135 (2007), [12] G. Zhang, Hankel operators on Hardy spaces and Schatten classes. A Chinese summary appears in Chinese Ann. Math. Ser. A, 12 (1991), no. 3, 522. Chinese Ann. Math. Ser., 12 (1991), no. 3, [13] R. Zhao, K. Zhu, Theory of ergman spaces in the unit ball of C n, preprint, (2005). [14] K. Zhu, Spaces of holomorphic functions in the unit ball, GTM, 226, Springer- Verlag, New York, (2005). [15] K. Zhu, A class of Möbius invariant function spaces, Illinois J. Math., to appear. Marcus Sundhäll, Mathematical Institute, Silesian University in Opava, Na Rybnicku 1, CZ , Czech Republic address: Marcus.Sundhall@math.slu.cz Edgar Tchoundja, Department of Mathematical Sciences, Division of Mathematics, Chalmers University of Technology and Göteborg University, SE Göteborg, Sweden address: tchoundj@math.chalmers.se