WEIGHTED BMO AND HANKEL OPERATORS BETWEEN BERGMAN SPACES

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1 WEIGHTED BMO AND HANKEL OPERATORS BETWEEN BERGMAN SPACES JORDI PAU, RUHAN ZHAO, AND KEHE ZHU Abstract. We introduce a family of weighted BMO saces in the Bergman metric on the unit ball of C n and use them to characterize comlex functions f such that the big Hankel oerators H f and H f are both bounded or comact from a weighted Bergman sace into a weighted Lesbegue sace with ossibly different exonents and different weights. As a consequence, when the symbol function f is holomorhic, we characterize bounded and comact Hankel oerators H f between weighted Bergman saces. In articular, this resolves two questions left oen in [7, 12]. 1. Introduction Let be the oen unit ball in C n and dv the usual Lebesgue volume measure on, normalized so that the volume of is one. Given a arameter α > 1 we write dv α (z) = c α (1 z 2 ) α dv(z), where c α is a ositive constant such that v α ( ) = 1. Denote by H( ) the sace of holomorhic functions on. For 0 < < the weighted Bergman sace A α := A α( ) consists of functions f H( ) that are in the Lebesgue sace L α := L (, dv α ). The corresonding norm is given by ( ) 1/ f,α = f(z) dv α (z). When = 2, the sace A 2 α is a reroducing kernel Hilbert sace: for each z there is a function Kz α A 2 α such that f(z) = f, Kz α α 2010 Mathematics Subject Classification. Primary 47B35; Secondary 32A36. Key words and hrases. Bergman saces, Hankel oerators, BMO, Bergman metric. This work started when the second named author visited the University of Barcelona in He thanks the suort given by the IMUB during his visit. The first author was suorted by DGICYT grant MTM C02-01 (MCyT/MEC) and the grant 2014SGR289 (Generalitat de Catalunya). 1

2 2 J. PAU, R. ZHAO, AND K. ZHU whenever f A 2 α. Here f, g α = fḡ dv α is the natural inner roduct in L 2 α. Kz α is called the reroducing kernel of the Bergman sace A 2 α. It is exlicitly given by the formula Kz α 1 (w) = (1 w, z ), z, w. n+1+α We also let k α z denote the normalized reroducing kernel at z. Thus k α z (w) = K α z (w)/ K α z (z) = (1 z 2 ) (n+1+α)/2 (1 w, z ) n+1+α. The orthogonal rojection P α : L 2 α A 2 α is an integral oerator given by f(w) dv α (w) P α f(z) = (1 z, w ), f n+1+α L2 (, dv α ). The (big) Hankel oerator H β f with symbol f is defined by H β f g = (I P β)(fg). We are interested in the maing roerties of H β f between different Lebesgue saces. Hankel oerators are closely related to Toelitz oerators and have been extensively studied by many authors in recent decades. For analytic f, Axler [3] first characterized the boundedness and comactness of H f on the unweighted Bergman sace of the unit disk. Later on, Axler s result was generalized in [1, 2] to weighted Bergman saces of the unit ball in C n. For general symbol functions, Zhu [15] first established the connection between size estimates of Hankel oerators and the mean oscillation of the symbols in the Bergman metric. This idea was further investigated in a series of aers [5], [6], and [4] in the context of bounded symmetric domains, and in [8, 9] in the context of strongly seudo convex domains. The main urose of this aer is to characterize real-valued functions f L q β such that Hβ f is bounded or comact from A α to L q β with 1 < q <. This is equivalent to characterizing comlex-valued functions f L q β such that both Hβ f and Hβ f are bounded or comact between the above saces. As a consequence, we will characterize holomorhic symbols f A 1 β such that is bounded or comact from Hβ f A α to L q β with 1 < q <. Our characterizations are based on a family of weighted BMO saces in the Bergman metric.

3 BMO AND HANKEL OPERATORS 3 Most revious results of this tye are for bounded and comact Hankel oerators from A α to L α. When f is holomorhic, Janson [7] and Wallstén [12] characterized bounded and comact Hankel oerators between weighted Bergman saces (in the Hilbert sace case) with different weights on the unit disk and the unit ball, resectively. Our results generalize theirs and solve two cases left oen by them. In the following, the notation A B means that there is a ositive constant C such that A CB, and the notation A B means that both A B and B A hold. 2. Preliminaries and auxiliary results In this section we collect some reliminary results that are needed for the roof of the main theorems. We begin with notation for the rest of the aer. For any two oints z = (z 1,..., z n ) and w = (w 1,..., w n ) in C n, we write z, w = z 1 w z n w n, and z = z, z = z z n 2. For any a with a 0 we denote by ϕ a (z) the Möbius transformation on that interchanges the oints 0 and a. It is known that ϕ a (z) = a P a(z) s a Q a (z), z, 1 z, a where s a = 1 a 2, P a is the orthogonal rojection from C n onto the one dimensional subsace [a] generated by a, and Q a is the orthogonal rojection from C n onto the orthogonal comlement of [a]. When a = 0, ϕ a (z) = z. It is known that ϕ a satisfies the following roerties: ϕ a ϕ a (z) = z, 1 ϕ a (z) 2 = (1 a 2 )(1 z 2 ) 1 z, a 2. (2.1) For z, w, the distance between z and w induced by the Bergman metric is given by β(z, w) = 1 2 log 1 + ϕ z(w) 1 ϕ z (w). For z and r > 0, the Bergman metric ball at z is given by D(z, r) = { w : β(z, w) < r }. We refer to [17] for more information about automorhisms and the Bergman metric on. A sequence {a k } of oints in is called a searated sequence (in the Bergman metric) if there exists a ositive constant δ > 0 such that

4 4 J. PAU, R. ZHAO, AND K. ZHU β(a i, a j ) > δ for any i j. The following result is Theorem 2.23 in [17]. Lemma 2.1. There exists a ositive integer N such that for any 0 < r < 1 we can find a sequence {a k } in with the following roerties: (i) = k D(a k, r). (ii) The sets D(a k, r/4) are mutually disjoint. (iii) Each oint z belongs to at most N of the sets D(a k, 4r). Any sequence {a k } satisfying the conditions of the above lemma will be called an r-lattice in the Bergman metric. Obviously any r-lattice is searated. The following integral estimate is well known and can be found in [17, Theorem 1.12] for examle. Lemma 2.2. Let t > 1 and s > 0. There is a ositive constant C such that (1 w 2 ) t dv(w) 1 z, w C (1 n+1+t+s z 2 ) s for all z. We also need a well-known variant of the revious lemma. Lemma 2.3. Let {z k } be a searated sequence in and let n < t < s. Then (1 z k 2 ) t 1 z, z k C (1 s z 2 ) t s, z. k=1 Lemma 2.3 above can be deduced from Lemma 2.2 after noticing that, if a sequence {z k } is searated, then there is a constant r > 0 such that the Bergman metric balls D(z k, r) are airwise disjoint. The following result is from [13]. Lemma 2.4. Given real numbers b and c, consider the integral oerator on defined by f(w)(1 w 2 ) b dv(w) S b,c f(z) =. 1 z, w c Let 1 < q <, α > 1, β > 1, and λ = n β q n α. Then the oerator S b,c is bounded from L α to L q β α + 1 < (b + 1), c n b + λ. if and only if

5 BMO AND HANKEL OPERATORS 5 We show that, under the same conditions, with an extra (unbounded) factor β(z, w) in the integrand, the modified oerator is still bounded from L α to L q β. Thus we consider the oerator f(w) β(z, w) T b,c f(z) = 1 z, w (1 c w 2 ) b dv(w). Proosition 2.5. Let b and c be real numbers. Let 1 < q <, α > 1, β > 1, and λ = n β q n α. If α + 1 < (b + 1) and c n b + λ, then T b,c is bounded from L α to L q β. Proof. Pick ε > 0 so that α + 1 < (b + 1 ε) and β qε > 1. Since β(z, w) grows logarithmically, we have β(z, w) = β(0, ϕ z (w)) C(1 ϕ z (w) 2 ) ε. It follows from (2.1) that T b,c f(z) C(1 z 2 ) ε (1 w 2 ) b ε f(w) dv(w). 1 z, w c 2ε Thus T b,c is bounded from L α to L q β if the oerator S b ε,c 2ε is bounded from L α to L q β εq. The desired result then follows from the revious lemma. In a similar manner, the following version of Lemma 2.2 can be obtained. The roof is left to the interested reader. Lemma 2.6. Let t > 1, s > 0, and d > 0. There is a ositive constant C such that (1 w 2 ) t β(z, w) d dv(w) C (1 z 2 ) s 1 z, w n+1+t+s for all z. The rest of this section is devoted to the roof of Theorem 2.8, which can be interreted as some sort of tangential maximum rincile. We begin with the following elementary fact. Lemma 2.7. Suose F and G are holomorhic functions on B 2. If F (z 1, z 2 ) = G(z 1, z 2 ), z k = (u/ 2)e iθ k, (2.2) where θ k are arbitrary real numbers and u is arbitrary from the unit disk D, then F = G on B 2.

6 6 J. PAU, R. ZHAO, AND K. ZHU Proof. Suose F (z 1, z 2 ) = a kl z1z k 2, l G(z 1, z 2 ) = k,l=0 Then we have ( ) k+l u 2 a kl e ikθ 1 e ilθ 2 = k,l=0 b kl z1z k 2. l k,l=0 ( ) k+l u 2 b kl e ikθ 1 e ilθ 2 k,l=0 for all u D and all real θ 1 and θ 2. By the uniqueness of Fourier coefficients on the torus, we must have a kl = b kl for all k and l. This shows that F = G on B 2. For n > 1, f H( ), and z {0} we will write { } f t f(z) = su u (z) : u = 1, u [z] and call it the comlex tangential gradient of f at z. Theorem 2.8. Let n > 1 and f H( ). If t f(z) 0 as z 1, then f is constant. Proof. We will first rove the case n = 2. In this case, the condition is equivalent to Let t f(z) 0, z 1, (2.3) ( ) f f lim z 2 (z) z 1 (z) = 0. z 1 z 1 z 2 z 1 = u 2 e iθ 1, z 2 = u 2 e iθ 2, where u D and θ k are arbitrary real numbers. Since z z 2 2 = u 2, we see that z 1 if and only if u 1. Thus condition (2.3) imlies that [ lim e iθ f ( ) 2 u u 1 z 1 2 eiθ 1 u, e iθ 2 e iθ f ( )] 1 u 2 z 2 2 eiθ 1 u, e iθ 2 = 0. 2 For any fixed θ k the exression inside the brackets above is an analytic function F (u) on the unit disk. By the classical maximum rincile for analytic functions on the unit disk, we conclude that the condition

7 BMO AND HANKEL OPERATORS 7 in (2.3) imlies that F (u) is identically zero on D. Therefore, the condition in (2.3) imlies that e iθ f ( ) 1 u z 1 2 eiθ 1 u, e iθ 2 = e iθ f ( ) 2 u 2 z 2 2 eiθ 1 u, e iθ 2 2 for all u D. Multily the above equation by u/ 2, we conclude that whenever z 1 f z 1 (z 1, z 2 ) = z 2 f z 2 (z 1, z 2 ) (2.4) z k = u 2 e iθ k, k = 1, 2. By Lemma 2.7, we see that the condition in (2.3) imlies that the identity in (2.4) must hold for all z = (z 1, z 2 ) in the unit ball. Write f(z) = a ij z1z i j 2 i,j=0 and assume that the identity in (2.4) holds for all z = (z 1, z 2 ) B 2. Then ia ij z1z i j 2 = ja ij z1z i 2. j i,j=0 i,j=0 This gives ia ij = ja ij for all i and j, which imlies that a ij = 0 whenever i j. Writing a j = a jj, we obtain f(z) = a j (z 1 z 2 ) j. In this case, we have j=0 z 2 f z 1 (z) z 1 f z 2 (z) = ( z 2 2 z 1 2 ) Consider the case in which ja j (z 1 z 2 ) j 1. j=1 z 1 = u sin θ, z 2 = u cos θ, where u D is arbitrary and 0 < θ < π/4 is fixed. Then f f z 2 (z) z 1 (z) = u 2 cos(2θ) ja j (u 2 sin θ cos θ) j 1. z 1 z 2 Let u 1 and aly the classical maximum rincile on the unit disk, we must have a j = 0 for all j 1, namely, f is constant. This comletes the roof of the theorem in the case n = 2. j=1

8 8 J. PAU, R. ZHAO, AND K. ZHU Next let us assume that n 3 and t f(z) 0 as z 1 for some f H( ). We want to show that f is constant. For any z = (z 1, z 2, z 3,, z n ), the vector (z 2, z 1, 0,, 0) is erendicular to z. Therefore, ( ) f f lim z 2 (z) z 1 (z) = 0. z 1 z 1 z 2 We roceed to show that this imlies that f is indeendent of the first two variables. Fix z = (z 1, z 2, z 3,, z n ) (we secifically mention that it is OK if some of the z k are 0) and write 1 r 2 = z z n 2, where 0 < r 1. Consider the function g(w 1, w 2 ) = f(rw 1, rw 2, z 3,, z n ), where w = (w 1, w 2 ) B 2. It is clear that w 1 if and only if So we also have rw rw z z n 2 1. [ ] g g lim w 2 (w) w 1 (w) = 0. w 1 w 1 w 2 By the n = 2 case that we have already roved, g must be constant. If we choose w B 2 such that rw k = z k for k = 1, 2. Then f(z 1, z 2, z 3,, z n ) = g(w 1, w 2 ) = g(0, 0) = f(0, 0, z 3,, z n ). Reeat the argument for the first and kth variable, where k 3, and let k run from 3 to n. The result is f(z 1, z 2, z 3,, z n ) = f(0, 0, 0,, 0). Since z = (z 1,, z n ) is arbitrary, we have shown that f is constant. For f H( ) we write ( f f(z) = (z),, f ) (z), z z 1 z n and call f(z) the comlex gradient of f at z. As a consequence of Theorem 2.8, we obtain the following maximum rincile in terms of the invariant gradient f(z) = (f ϕ z )(0). Corollary 2.9. Let n > 1 and f H( ). If (1 z 2 ) 1/2 f(z) 0 as z 1, then f is constant. Proof. This follows from [17, Theorem 7.22] and Theorem 2.8.

9 BMO AND HANKEL OPERATORS 9 3. A family of weighted BMO saces Let γ R. For any ositive radius r and every exonent with 1 <, the sace BMO r,γ consists of those functions f L loc () (the sace of locally L integrable functions on ) such that f BMO r,γ = su { (1 z 2 ) γ MO,r (f)(z) : z } <, where [ 1 MO,r (f)(z) = v σ (D(z, r)) ] 1/ f(w) f r (z) dv σ (w) is the -mean oscillation of f at z in the Bergman metric. Here 1 f r (z) = f(w) dv σ (w) v σ (D(z, r)) is the averaging function of f and dv σ (z) = (1 z 2 ) σ dv(z). At first glance, the function MO,r (f) seems to deend on the real arameter σ, but the weight factor (1 z 2 ) σ in dv σ is essentially canceled out by the extra factor (1 z 2 ) σ in v σ (D(z, r)) (1 z 2 ) n+1+σ. As a consequence, the sace BMO r,γ is actually indeendent of the weight arameter σ. In articular, this indeendence on σ is a consequence of the following lemma. Lemma 3.1. Let 1 <, f L loc (), and r > 0. Then f BMOr,γ if and only if there exists some constant C > 0 such that for any z, there is a constant λ z satisfying (1 z 2 ) γ v σ (D(z, r)) f(w) λ z dv σ (w) C. (3.1) Proof. If f BMOr,γ, then (3.1) holds with C = f BMO r,γ and λ z = f r (z). Conversely, if (3.1) is satisfied, then by the triangle inequality for the L -norm, MO,r (f)(z) is less than or equal to [ ] 1 1 f(w) λ z dv σ (w) + fr (z) λ z. v σ (D(z, r)) By Hölder s inequality, f r (z) λ z = 1 v σ (D(z, r)) [ 1 v σ (D(z, r)) (f(w) λ z ) dv σ (w) ] 1 f(w) λ z dv σ (w).

10 10 J. PAU, R. ZHAO, AND K. ZHU It follows that (1 z 2 ) γ MO,r (f)(z) 2C 1/ for all z, so that f BMOr,γ. For a continuous function f on let ω r (f)(z) = su { f(z) f(w) : w D(z, r) }. The function ω r (f)(z) is called the oscillation of f at the oint z in the Bergman metric. For any r > 0 and γ R, let BO r,γ denote the sace of continuous functions f on such that f BOr,γ = su z (1 z 2 ) γ ω r (f)(z) <. Lemma 3.2. Let r > 0, γ R, and f be a continuous function on. If γ 0, then f BO r,γ if and only if there is a constant C > 0 such that β(z, w) + 1 f(z) f(w) C min(1 z, 1 w ) γ for all z and w in. If γ < 0, then f BO r,γ if and only if there is a constant C > 0 such that β(z, w) + 1 f(z) f(w) C [ ] γ 1 z, w 2γ max(1 z, 1 w ) for all z and w in. Proof. Assume that f BO r,γ. If β(z, w) r, the result is clear, because then 1 z, w 1 z 1 w. Fix any z, w with β(z, w) > r. Let λ(t), 0 t 1, be the geodesic in the Bergman metric from z to w. Let N = [β(z, w)/r] + 1 and t i = i/n, 0 i N, where [x] denotes the largest integer less than or equal to x. Set z i = λ(t i ), 0 i N. Then Therefore, f(z) f(w) β(z i, z i+1 ) = β(z, w) N r. N f(z i 1 ) f(z i ) i=1 N f BOr,γ (1 z i ) γ. i=1 If γ 0, it follows from the obvious inequality (1 z i ) min(1 z, 1 w ) N ω r (f)(z i ) i=1

11 that BMO AND HANKEL OPERATORS 11 f(z) f(w) f BOr,γ N min(1 z, 1 w ) γ f BOr,γ β(z, w)/r + 1 min(1 z, 1 w ) γ max(1, 1/r) f BOr,γ β(z, w) + 1 min(1 z, 1 w ) γ. If γ < 0, the result is roved in the same way once the inequality 1 z i 2 1 z, w 2 max(1 z 2, 1 w 2 ) (3.2) is established. To rove this, simly note that the Möbius transformation ϕ z sends the geodesic joining z and w to the geodesic joining 0 and ϕ z (w). This gives 1 ϕ z (w) 2 1 ϕ z (z i ) 2. Develoing this inequality using (2.1), we get 1 w 2 1 z, w 2 1 z i 2 1 z, z i z, z i, which gives 2 1 z, w 2 1 z, z i. 1 w 2 Interchanging the roles of z and w, we get (3.2). The converse imlication is obvious. A consequence of the above lemma is that the sace BO r,γ is indeendent of the choice of r. So we will simly write BO γ = BO 1,γ and f BOγ = f BO1,γ = su(1 z 2 ) γ ω 1 (f)(z). z Let 0 < <, γ R, and r > 0. We say that f BA r,γ if f L loc () and ] 1/ f BA r,γ = su(1 z 2 ) [ f γ r (z) <. z We roceed to show that the sace BA r,γ is also indeendent of r. For σ > 1 and c > 0 the generalized Berezin transform B c,σ (ϕ) of a function ϕ L 1 (, dv σ ) is defined as B c,σ (ϕ)(z) = (1 z 2 ) c ϕ(w) 1 w, z dv σ(w). n+1+c+σ

12 12 J. PAU, R. ZHAO, AND K. ZHU In the case when c = n+1+σ, this coincides with the ordinary Berezin transform B σ ϕ(z) = ϕk σ z, k σ z σ. Lemma 3.3. Let 0 < q <, α, β > 1, and f L q loc (). Set γ = (n β)/q (n α)/, dµ f,β = f q dv β. The following conditions are equivalent: (i) The embedding i : A α L q (, dµ f,β ) is bounded. (ii) f BA q r,γ for some (or all) r > 0. (iii) (1 z 2 ) γq B c,σ ( f q ) L ( ) for all σ > 1 + γq and all c > max(0, γq). Proof. By [14, Theorem 50], condition (i) is equivalent to for some (or all) r > 0. Since µ f,β (D(z, r)) C(1 z 2 ) (n+1+α)q/ f q r(z) µ f,β(d(z, r)) (1 z 2 ) n+1+β, it follows that (i) and (ii) are equivalent. Since 1 z, w (1 z 2 ) for w D(z, r), we have B c,σ ( f q )(z) = (1 z 2 ) c f(w) q dv σ (w) 1 z, w n+1+c+σ (1 z 2 ) c f(w) q dv σ (w) 1 z, w n+1+c+σ f q r(z). This roves that (iii) imlies (ii). To finish the roof, let {a j } be an r-lattice. Then B c,σ ( f q )(z) = (1 z 2 ) c f(w) q dv σ (w) 1 z, w n+1+c+σ (1 z 2 ) c j=1 D(a j,r) By the estimate in (2.20) on age 63 of [17], we have Thus f(w) q dv σ (w) 1 z, w n+1+c+σ. 1 z, w 1 z, a j, w D(a j, r). B c,σ ( f q )(z) (1 z 2 ) c j=1 (1 a j 2 ) σ β 1 z, a j n+1+c+σ µ f,β(d(a j, r)).

13 So condition (ii) imlies that BMO AND HANKEL OPERATORS 13 B c,σ ( f q )(z) (1 z 2 ) c j=1 (1 a j 2 ) n+1+σ γq 1 z, a j n+1+c+σ. Since σ > 1 + γq, or n σ γq > n, an alication of Lemma 2.3 shows that condition (ii) imlies (iii). As a consequence of the revious result, we see that the sace BA r,γ is indeendent of the choice of r. Thus we will simly call it BA γ. Lemma 3.4. Let f L 1 loc () and r > 0. Then f r is continuous. Proof. The roof is elementary and we omit the details here. Theorem 3.5. Suose r > 0, γ R, 1 <, and f L loc (). The following conditions are equivalent: (a) f BMO r,γ. (b) f = f 1 + f 2 with f 1 BO γ and f 2 BA γ. (c) For some (or all) σ > max( 1, 1 + γ) and for each c > max(0, 2γ) we have su z Bn f(w) f r (z) (1 z 2 ) c+γ 1 z, w n+1+c+σ dv σ(w) <. (d) For some (or all) σ > max( 1, 1 + γ) and for each c > max(0, 2γ) there is a function λ z such that su z Bn f(w) λ z (1 z 2 ) c+γ 1 z, w n+1+c+σ dv σ(w) <. Proof. That (c) imlies (d) is obvious, and the imlication (d) (a) is a consequence of Lemma 3.1 and the inequality (1 z 2 ) γ f(w) λ z dv σ (w) v σ (D(z, r)) Bn f(w) λ z (1 z 2 ) c+γ 1 z, w n+1+c+σ dv σ(w), which follows from the well-known facts that 1 z, w (1 z 2 ), v σ (D(z, r)) (1 z 2 ) n+1+σ, for all z and w D(z, r). The roof of (a) (b) can be done as in [16, Theorem 5]. Indeed, since r is arbitrary, it suffices to show that BMO 2r,γ BO γ + BA γ.

14 14 J. PAU, R. ZHAO, AND K. ZHU Given f BMO 2r,γ and oints z, w with β(z, w) r, we have f r (z) f r (w) f r (z) f 2r (z) + f 2r (z) f r (w) 1 f(u) v σ (D(z, r)) f 2r (z) dv σ (u) 1 + f(u) v σ (D(w, r)) f 2r (z) dv σ (u). D(w,r) Since v σ (D(w, r)) v σ (D(z, r)) for w D(z, r), and since D(z, r) and D(w, r) are both contained in D(z, 2r), it follows from Hölder s inequality that the two integral summands above are both bounded by constant times (1 z 2 ) γ f BMO 2r,γ. This together with Lemma 3.4 roves that f r belongs to BO r,γ = BO γ. On the other hand, we can rove that the function g = f f r is in BA γ whenever f BMO2r,γ. In fact, it is rather easy to see that f BMO 2r,γ imlies that f BMOr,γ. By the triangle inequality for L integrals, [ g r (z) ] [ 1/ 1 v σ (D(z, r)) f(u) f r (z) dv σ (u) ] 1 + ωr ( f r )(z). Since f r BO r,γ and f BMOr,γ, we deduce that g belongs to BA γ. To show that (b) imlies (c), first observe that it follows from Lemma 2.2 that the integral aearing in art (c) is dominated by ( (1 z 2 ) γ B c,σ ( f )(z) + f r (z) ), and by Hölder s inequality, we have f r (z) f r(z). Thus Lemma 3.3 shows that f BA γ imlies condition (c). On the other hand, if f BO γ, we write f(w) f 1 ( ) r (z) = f(w) f(ζ) dvσ (ζ) v σ (D(z, r)) and use Lemma 3.2 and the triangle inequality to obtain and f(w) f r (z) C f BOγ β(z, w) + 1 min(1 z, 1 w ) γ, γ 0, f(w) f r (z) C f BOγ (β(z, w) + 1) 1 z, w 2γ (1 z ) γ, γ < 0. In both cases, the integral estimates in Lemmas 2.2 and 2.6 show that (c) holds if f BO γ. This shows that condition (b) imlies (c) and comletes the roof of the theorem.

15 BMO AND HANKEL OPERATORS 15 One of the consequences of the above result is that the sace BMO r,γ is also indeendent of r. So from now on it will simly be denoted by BMO γ. Next we are going to identify the sace of all holomorhic functions in BMO γ with certain Bloch-tye saces. Recall that for α 0, the Bloch tye sace B α = B α ( ) consists of holomorhic functions f in for which f B α = f(0) + su z (1 z 2 ) α f(z) <. Note that the comlex gradient f(z) can be relaced by the radial derivative Rf(z). We will simly obtain equivalent norms. When α > 1/2, a descrition can also be obtained using the invariant gradient f(z) = (f ϕ z )(0). That is, for α > 1/2, a holomorhic function f belongs to B α if and only if f(0) + su z (1 z 2 ) α 1 f(z) <, and this quantity defines an equivalent norm in B α. Note that when n = 1, this is true for all α 0, because in this case we actually have f(z) = (1 z 2 )f (z). When 0 < α < 1, the Bloch tye sace B α coincides (with equivalent norms) with the holomorhic Lischitz sace Λ 1 α = Λ 1 α ( ) consisting of all holomorhic functions f in such that { } f(z) f(w) f Λ1 α = f(0) + su : z, w B z w 1 α n, z w <. Note that when α = 0, the sace B α consists of holomorhic functions f with bounded artial derivatives. Equivalently, B 0 consists of all holomorhic functions f such that { } f(z) f(w) su : z w <. z w This sace is not what is usually called the Lischitz sace Λ 1. We refer to [17, Chater 7] for all these roerties of Bloch and Lischitz tye saces. Proosition 3.6. Let γ R, 1 <, and f H( ). Then f BMO γ if and only if f γ, = su z (1 z 2 ) γ f(z) <.

16 16 J. PAU, R. ZHAO, AND K. ZHU Proof. We will show that condition (d) of Theorem 3.5 is satisfied with λ z = f(z) if f γ, <. To this end, let σ > max( 1, 1 + γ) and c > max(0, γ). By [11, Lemma 7] and Lemma 2.2, we have f(w) f(z) Bn (1 z 2 ) c+γ 1 z, w dv σ(w) n+1+c+σ C Bn f(w) (1 z 2 ) c+γ 1 z, w dv σ(w) n+1+c+σ C f γ, (1 z 2 ) c+γ dv σ γ(w) 1 z, w n+1+c+σ C. Thus f γ, < imlies that f BMOγ. To rove the other imlication, we use the inequality f(z) 1 f(w) f(z) dv σ (w), v σ (D(z, r)) which aears on age 182 of [17]. By the triangle inequality for L saces, f(z) MO,r (f)(z) + f(z) f r (z). Alying Lemma 2.24 of [17] to the function g(w) = f(w) f r (z) and the oint z, we find a constant C such that f(z) f r (z) C MO,r (f)(z) for all z. Thus f BMO γ imlies f γ, <. Corollary 3.7. Let γ R and 1 <. Then (a) If n = 1, we have H( ) BMO γ = { B 1+γ, γ 1, C, γ < 1. (b) If n > 1, we have { H( ) BMOγ B 1+γ, γ > 1/2, = C, γ < 1/2. (c) If n > 1 and γ = 1, the sace H(B 2 n) BMOγ those holomorhic functions f on with su(1 z 2 ) 1/2 f(z) <. z consists of

17 BMO AND HANKEL OPERATORS 17 Proof. Parts (a) and (b) are consequence of Proosition 3.6, the remarks receding it, and Corollary 2.9. Part (c) is just a restatement of Proosition 3.6. According to [17, Theorem 7.2], the sace BMO contains nonconstant functions. In fact, any function in B α with 0 < α < 1/2 is in 1 2 BMO. However, this sace differs from B 1/2. See Corollary In the next theorem we record some characterizations obtained for Bloch tye saces, which are of some indeendent interest. Theorem 3.8. Suose r > 0, 1 <, σ > 1, and f H( ). Let γ 1 if n = 1; and γ > 1/2 if n > 1. Then the following statements are equivalent. (i) f B 1+γ. (ii) There is a constant C > 0 such that (1 z 2 ) γ f(w) f(z) dv σ (w) C v σ (D(z, r)) for all z. (iii) There is a constant C > 0 such that (1 z 2 ) γ f(w) v σ (D(z, r)) f r (z) dv σ (w) C for all z. (iv) For each z there exists a comlex number λ z such that (1 z 2 ) γ su f(w) λ z dv σ (w) <. z v σ (D(z, r)) (v) For some (or all) η > max( 1, 1 + γ) and for each c > max(0, 2γ) we have su z Bn f(w) f(z) (1 z 2 ) c+γ 1 z, w n+1+c+η dv η(w) <. Proof. In the roof of Proosition 3.6, we roved the imlications (i) (v) (ii) (i). The other equivalences are obtained from Theorem 3.5 and Lemma 3.1. When γ = 0, the equivalences of (i)-(iv) in Theorem 3.8 is just [17, Theorem 5.22], and the equivalence with (v) aears in [10].

18 18 J. PAU, R. ZHAO, AND K. ZHU 4. Bounded Hankel oerators The main result of this section is the following result, which characterizes bounded Hankel oerators induced by real-valued symbols between weighted Bergman saces. Theorem 4.1. Let 1 < q <, α > 1, β > 1, f L q β, and γ = n β q n α. Then H β f, Hβ f : A α L q β are both bounded if and only if f BMOq γ. We are going to rove Theorem 4.1 with a series of lemmas and roositions. For t 0 let K β,t z (w) = 1 (1 w, z ) n+1+β+t. Also, for f L q β and z, we consider the function MO β,q,t f defined by MO β,q,t f(z) = fh t z g z (z)h t z q,β, where, for z, the function g z (that deends on f and t) is given by g z (w) = P β( fh t z)(w), w B h t n, z(w) with the function h t z defined by h t z(w) = h α,β,t z (w) = Kβ,t z (w), w B Kz β,t n.,α Clearly, h t z,α = 1. When t = 0, it is easily seen that g z (z) = B β f(z) is the Berezin transform of f at the oint z. Also, since h t z(w) never vanishes on, the function g z is holomorhic on. Lemma 4.2. Let 1 < q < and α, β > 1. Let γ be as in Theorem 4.1 and t 0 such that n β + t > (n α)/. If MO β,q,t f L ( ), then f BMO q γ. Proof. Since n β + t > (n α)/, Lemma 2.2 gives us the estimate K β,t z,α (1 z 2 ) (n+1+α)/ (n+1+β+t).

19 BMO AND HANKEL OPERATORS 19 It follows that [ MO β,q,t f(z) ] q is comarable to (1 z 2 ) γq+(n+1+β)(q 1)+tq which dominates (1 z 2 ) γq D(z, r) f(w) g z (z) q 1 z, w (n+1+β+t)q dv β(w), f(w) g z (z) q dv(w). This shows that, if MO β,q,t f L ( ), the condition in Lemma 3.1 is satisfied with λ z = g z (z), so f BMOγ. q The following result gives the necessity in Theorem 4.1. It generalizes, in several directions, Proosition 8.19 in [18], where the method of roof is based on Hilbert sace techniques. A different method was used in [16] to deal with the case α = β and = q. Our method here is more flexible and allows us to obtain the result in much more generality. Proosition 4.3. Let 1 <, q <, α, β > 1, and t 0. Then for any f L q β we have MO β,q,t f(z) H β q,β + H β q,β. f ht z Proof. By the triangle inequality and the definition of Hankel oerators, we have MO β,q,t f(z) = fh t z g z (z) h t z q,β fh t z P β (fh t z) q,β + Pβ (fh t z) g z (z) h t z q,β = H β q,β + Pβ (fh t z) g z (z) h t z q,β. f ht z For any g A 1 β () it is easy to check that f ht z g(z)h t z = P β+t (ḡh t z). (4.1) This together with the boundedness of P β+t on L q β yields Pβ (fh t z) g z (z) h t z q,β = Pβ (fh t z) P β+t (g z h t z) q,β = ( Pβ+t Pβ (fh t z) g z h t z) ) q,β Pβ+t L q Pβ (fh t z) g z h t z β q,β.

20 20 J. PAU, R. ZHAO, AND K. ZHU Finally, Pβ (fh t z) g z h t z q,β fh t z P β (fh t z) q,β + fh t z g z h t z q,β = H β f ht z q,β + f h t z g z h t z q,β = H β f ht z q,β + f h t z P β ( f h t z) q,β = H β f ht z q,β + H β f ht z q,β. This roves the result with constant C = ( 1 + ) Pβ+t L q. β Note that the roosition above does not require q. The next two roositions, which require the condition q, will establish the sufficiency of Theorem 4.1. Proosition 4.4. Let 1 < q <, α > 1, β > 1, and If f BA q γ, then H β f : A α L q β γ = (n β)/q (n α)/. is bounded. Proof. Since q > 1, the Bergman rojection P β is bounded on L q β. Thus H β f g q,β fg q,β + P β (fg) q,β fg q,β = g L q (dµ f,β ). The result then follows from Lemma 3.3. We note that the roof of the revious roosition also works for 1 = < q <. In order to show that H β f is bounded if f BO γ with γ < 0, we need the following result. Lemma 4.5. Let s β > 1, 1 < q <, f L q β, and g H. Then H β f g q,β C H s f g q,β. Proof. Since g H, we have gf L q β. Also, H β f g q,β = (I Pβ )(gf) q,β (I P s )(gf) q,β + (P s P β )(gf) q,β = H s f g q,β + (Pβ P s )(gf) q,β. Since P s is bounded on L q β, the reroducing formula yields P βp s (gf) = P s (gf). Thus (P β P s )(gf) = (P β P β P s )(gf) = P β (I P s )(gf) = P β (H s fg). This gives (Pβ P s )(gf) Pβ q,β H s fg q,β, so we obtain the desired inequality with C = 1 + P β.

21 BMO AND HANKEL OPERATORS 21 Proosition 4.6. Let 1 < q <, α > 1, β > 1, and If f BO γ, then H β f : A α L q β γ = (n β)/q (n α)/. is bounded. Proof. We first consider the case γ 0. For g H, which is dense in A α, we have H β f g q q,β = H β f g(z) q dv β (z) B n q = (f(z) f(w)) g(w) (1 z, w ) dv β(w) n+1+β dv β (z) ( ) q f(z) f(w) g(w) dv 1 z, w n+1+β β (w) dv β (z). By Lemma 3.2, H β f g q q,β Write I 1 (g) = ( ( ) q (1 + β(z, w)) g(w) dv β (w) dv 1 z, w n+1+β min(1 z, 1 w ) γ β (z). ) q g(w) dv β (w) dv 1 z, w n+1+β min(1 z, 1 w ) γ β (z), and slit the inner integral in two arts, I 1,1 (g) over w z and I 1,2 (g) over w > z. Since min(1 z, 1 w ) γ = (1 z ) γ for w z, we have ( ) q g(w) dv β (w) I 1,1 (g) dv β qγ(z) 1 z, w n+1+β = Sb,c ( g )(z) q dv β qγ (z), where S b,c is the integral oerator aearing in Theorem 2.4 with b = β and c = n β. Notice that β qγ > 1 is equivalent to ( 1 n q 1 ) < 1 + α, which is automatically satisfied since q. Alying Theorem 2.4, we obtain I 1,1 (g) g q,α, rovided 1 + α < (1 + b) and c n b + λ. Since b = β and λ = n (β qγ) q n α = 0,

22 22 J. PAU, R. ZHAO, AND K. ZHU the condition c n b + λ is satisfied with equality. It remains to check that the condition 1 + α < (1 + b) = (1 + β) (4.2) is satisfied. Since γ 0 and q, we have 0 γ = (n β)/q (n α)/ (β α)/. This gives α β, so(4.2) holds since > 1. Similarly, we have I 1,2 (g) Sb,c ( g )(z) q dv β (z) with b = β γ, c = n β, and λ = γ. We want to aly Theorem 2.4 to estimate I 1,2 (g). In this case, the condition c n b + λ in Theorem 2.4 holds with equality. The other condition in Theorem 2.4 is α + 1 < (1 + β γ), which is equivalent to (n β) q < β + + n. If q is the conjugate exonent of q, the above condition is equivalent to ( 1 n q 1 ) < 1 + β, q which is automatically satisfied since q. Hence, by Theorem 2.4, we have I 1,2 (g) g q,α. This together with the revious estimate yields I 1 (g) g q,α. The remaining estimate I 2 (g) g q,α with ( ) q g(w) β(z, w) dv β (w) I 2 (g) := dv 1 z, w n+1+β min(1 z, 1 w ) γ β (z), can be roved in a similar manner, using Proosition 2.5 instead of Theorem 2.4. The roof of the case γ 0 is now comlete. If γ < 0 and g H, we use Lemma 4.5, with s β big enough so that (s + γ + 1) > α + 1, to obtain H β f g q H s q,β fg q q,β By Lemma 3.2, Therefore, H β f g q q,β [ f(z) f(w) g(w) dv 1 z, w n+1+s s (w) β(z, w) + 1 f(z) f(w) C (1 w ) 1 z, γ w 2γ. ( ] q dv β (z). ) q (β(z, w) + 1) g(w) 1 z, w dv s+γ(w) dv n+1+s+2γ β (z),

23 BMO AND HANKEL OPERATORS 23 and the boundedness of H β f : A α L q β follows from Theorem 2.4 and Proosition 2.5 again. The roof of Theorem 4.1 is now comlete: the necessity of the condition f BMO q γ follows from Lemma 4.2 and Proosition 4.3. Since f BMO q γ if and only if f BMO q γ, the sufficiency is a consequence of Theorem 3.5, Proosition 4.4, and Proosition 4.6. As an immediate consequence of Theorem 4.1 and Proosition 3.6, we obtain the following result that characterizes the boundedness of Hankel oerators with conjugate holomorhic symbols. Corollary 4.7. Let f A 1 β, 1 < q <, α > 1, β > 1, and γ = n β q (1) For n = 1 we have n α. (a) If γ 1, then H f : A α L q β is bounded if and only if f B 1+γ. (b) If γ < 1, then H f : A α L q β is bounded if and only if f is constant. (2) For n > 1 we have (a) If γ > 1/2, then H f : A α L q β is bounded if and only if f B 1+γ. (b) If γ < 1/2, then H f : A α L q β is bounded if and only if f is constant. (c) If γ = 1/2, then H f : A α L q β is bounded if and only if su(1 z 2 ) 1/2 f(z) <. z Proof. Since f A 1 β, the Hankel oerator H f is densely defined. If H f : A α L q β is bounded, by testing the boundedness on the function 1 we see that f A q β. Since B1+γ A q β, the result follows from Theorem 4.1 and Proosition 3.6. In the case q = = 2, this recovers the results of Janson and Wallstén [7, 12], where the case γ = 1 for n = 1 and the case γ = 1/2 for n > 1 were left oen. Thus we have resolved these oen cases. 5. Weighted VMO saces Let γ R. For any ositive radius r and every exonent with 1 <, the sace V MO r,γ consists of those functions f in BMO r,γ such that lim z 1 (1 z 2 ) γ MO,r (f)(z) = 0.

24 24 J. PAU, R. ZHAO, AND K. ZHU Again, the sace V MO r,γ is actually indeendent of the weight arameter σ. Similarly as before, for r > 0, we define V O r,γ as the sace of functions f in BO r,γ satisfying lim (1 z 1 z 2 ) γ ω r (f)(z) = 0, and V A r,γ as the sace of functions f in BA r,γ satisfying ] 1/ lim (1 z 1 z 2 ) [ f γ r (z) = 0. The following result shows that V A r,γ does not deend on r. Lemma 5.1. Let 0 < q <, α > 1, β > 1, f L q loc (), and γ = (n β)/q (n α)/, dµ f,β = f q dv β. The following are equivalent: (i) If {f k } is a bounded sequence in A α and f k 0 uniformly on every comact subset of, then lim f k (z) q dµ f,β (z) = 0. k (ii) f V A q r,γ for some (or all) r > 0. (iii) The condition lim (1 z 1 z 2 ) γq B c,σ ( f q )(z) = 0 holds for all σ > max( 1, 1 + γq) and all c > max(0, γq). Proof. By the corresonding little-oh result of Theorem 50 in [14], we know that (i) is equivalent to lim z 1 µ f,β (D(z, r)) (1 z 2 ) (n+1+α)q/ = 0 for some (or all) r > 0. The equivalence of (i) and (ii) is a consequence of this result and the fact that f q r(z) µ f,β(d(z, r)) (1 z 2 ) n+1+β. That (iii) imlies (ii) follows from the fact that f q r(z) B c,σ ( f q )(z), which has been shown in the roof of Lemma 3.3.

25 BMO AND HANKEL OPERATORS 25 It remains to rove that (ii) imlies (iii). Let f V A q r,γ. By definition, we have q f (1 z 2 ) γq B c,σ ( f q r)(z) = (1 z 2 ) c+γq r(w) dv σ (w) 1 z, w. n+1+c+σ For 0 < s < 1 let and I 1 (s) = (1 z 2 ) c+γq I 2 (s) = (1 z 2 ) c+γq w s s< w <1 f q r(w) dv σ (w) 1 z, w n+1+c+σ, f q r(w) dv σ (w) 1 z, w n+1+c+σ. Let ε be an arbitrary ositive number. By (ii), and then by Lemma 2.2, there exists an s > 0 such that I 2 (s) ε (1 z 2 ) c+γq (1 w 2 ) σ γq dv(w) ε. 1 z, w n+1+c+σ s< w <1 Since f V A q r,γ BA q r,γ, we know that f q r(w) (1 w 2 ) qγ. Since 1 z, w (1 w 2 ), we obtain I 1 (s) (1 z 2 ) c+γq dv σ (w) (1 w 2 ) n+1+c+σ+γq w s (1 z 2 ) c+γq (1 s 2 ) n+1+c+γq. Hence, we can find a δ (0, 1) such that I 1 (s) ε whenever 1 δ < z < 1. Combining the above two inequalities for I 1 (s) and I 2 (s) we deduce for 1 δ < z < 1 that Therefore, Let dµ f,σ = f q dv σ. Since (1 z 2 ) γq B c,σ ( f q r)(z) ε. lim (1 z 1 z 2 ) γq B c,σ ( f q r)(z) = 0. f q r(z) µ f,σ (z) := the above equation is equivalent to lim (1 z 1 z 2 ) γq B c,σ ( µ f,σ )(z) = 0. µ f,σ(d(z, r)), (5.1) (1 z 2 ) n+1+σ

26 26 J. PAU, R. ZHAO, AND K. ZHU By [14, Lemma 52], we have where Thus we obtain which is the same as B c,σ (µ f,σ )(z) B c,σ ( µ f,σ )(z), B c,σ (µ f,σ )(z) = (1 z 2 ) c dµ f,σ(w) 1 z, w n+1+c+σ. lim (1 z 1 z 2 ) γq B c,σ (µ f,σ )(z) = 0, lim (1 z 1 z 2 ) γq B c,σ ( f q )(z) = 0. This roves (ii) imlies (iii) and comletes the roof of the lemma. The next result shows that V O r,γ does not deend on r. Lemma 5.2. Let γ R and r 1, r 2 > 0. If f V O r1,γ, then f V O r2,γ. Proof. If r 1 > r 2, the result is obvious. So we assume that r 1 < r 2 and fix z. It follows from the continuity of f on that ω r2 (f)(z) = su{ f(z) f(ζ), ζ D(z, r 2 )}, and we can find w D(z, r 2 ) such that f(z) f(w) = ω r2 (f)(z). Let λ = λ(t), 0 t 1, be the geodesic in the Bergman metric from z to w. Then λ lies entirely in D(z, r 2 ). As in the roof of Lemma 3.2, we let N = [r 2 /r 1 ] + 2 and t i = i/n, 0 i N, where [x] denotes the largest integer less than or equal to x. Set z i = λ(t i ), 0 i N. Since N r 2 /r > r 2 /r 1, we have β(z, w) β(z i 1, z i ) = N r 2 N < r 1. Because z i is in the closure of D(z, r 2 ), there exists a constant K > 0, indeendent of i, such that 1 K (1 z 2 ) γ (1 z i 2 ) γ K(1 z 2 ) γ. Since f V O r1,γ, we know that lim (1 z i 2 ) γ ω r1 (f)(z i ) = 0. z i 1 But z i 1 as z 1. So for any ε > 0 there exists δ > 0 such that (1 z i 2 ) γ ω r1 (f)(z i ) < ε NK

27 whenever 1 z < δ. Thus Therefore, f(z) f(w) BMO AND HANKEL OPERATORS 27 = N f(z i 1 ) f(z i ) i=1 N ω r1 (f)(z i ) i=1 ε K(1 z i 2 ) ε γ K K (1 z 2 ) γ ε (1 z 2 ). γ ω r2 (f)(z) = f(z) f(w) for 1 δ < z < 1, which shows that ε (1 z 2 ) γ lim (1 z 2 ) γ ω r2 (f)(z) = 0, z 1 or f V O r2,γ. Because of Lemmas 5.1 and 5.2, we can denote V A r,γ and V O r,γ by V A γ and V O γ, resectively. Just as in the big-oh case, we have the following result for V MO r,γ. Theorem 5.3. Suose r > 0, γ R, 1 <, and f BMO γ. The following conditions are equivalent: (a) f V MO r,γ. (b) f = f 1 + f 2 with f 1 V O γ and f 2 V A γ. (c) For some (or all) σ > max( 1, 1 + γ) and for each c max(n σ, n σ 2γ), we have lim z 1 f(w) Bn f r (z) (1 z 2 ) c+γ 1 z, w dv σ(w) = 0. n+1+c+σ (d) For some ( or all) σ > max( 1, 1 + γ) and for each c max(n σ, n σ 2γ), there is a function λ z such that lim z 1 Bn f(w) λ z (1 z 2 ) c+γ 1 z, w n+1+c+σ dv σ(w) = 0. (e) For some (or all) σ > 1 there is a function λ z such that (1 z 2 ) γ lim f(w) λ z dv σ (w) = 0. z 1 v σ (D(z, r))

28 28 J. PAU, R. ZHAO, AND K. ZHU Proof. That (c) imlies (d) is obvious. That (d) imlies (e) follows from the simle inequality (1 z 2 ) γ f(w) λ z dv σ (w) v σ (D(z, r)) Bn f(w) λ z (1 z 2 ) c+γ 1 z, w n+1+c+σ dv σ(w). An easy modification of the roof of Lemma 3.1 shows that (e) imlies (a). That (a) imlies (b) follows easily from the roof of (a) imlying (b) in Theorem 3.5. Thus we only need to rove that (b) imlies (c). Suose that (b) holds. As in the roof of Theorem 3.5, from Lemma 5.1 it is not difficult to see that (c) is satisfied for f V A γ. Now, for f V O γ, it is obvious that f BO γ. Set I(z) = Bn f(w) f r (z) (1 z 2 ) c+γ 1 z, w n+1+c+σ dv σ(w). Making the change of variables w = ϕ z (ζ), we obtain I(z) = Bn f ϕ z (ζ) f r (z) (1 z 2 ) γ 1 z, ζ n+1+σ c dv σ(ζ). (5.2) In the case γ 0, it follows from the roof of Theorem 3.5 and the invariance of the Bergman metric that f ϕ z (ζ) f r (z) f BOγ β(ϕ z (ζ), z) + 1 min(1 z, 1 ϕ z (ζ) ) γ f BOγ β(ζ, 0) + 1 min(1 z, 1 ϕ z (ζ) ) γ. Let E = {ζ : ϕ z (ζ) z }. For ζ E we have 1 ϕ z (ζ) 2 1 z 2 and f ϕ z (ζ) f r (z) (1 z 2 ) γ (β(ζ, 0) + 1). (5.3) For ζ \ E we have 1 ϕ z (ζ) 2 1 z 2 and f ϕ z (ζ) f r (z) (1 z 2 ) γ (β(ζ, 0) + 1) (1 z 2 ) γ (1 ϕ z (ζ) 2 ) γ Since σ > 1 + γ 1, we have (β(ζ, 0) + 1) dv σ (ζ) <, (β(ζ, 0) + 1) (1 ζ 2 ) γ.

29 and Let BMO AND HANKEL OPERATORS 29 (β(ζ, 0) + 1) (1 ζ 2 ) γ dv σ (ζ) <. { (β(ζ, 0) + 1), H(ζ) = (β(ζ, 0) + 1) (1 ζ 2 ) γ ), ζ E ζ \ E The above argument shows that H(ζ) is in L 1 (, dv σ ) and, since c n σ, we have f ϕ z (ζ) f r (z) (1 z 2 ) γ 1 z, ζ c (n+1+σ) H(ζ) (5.4) for γ 0 and all z. If γ < 0, it follows from the roof of Theorem 3.5 again that f ϕ z (ζ) f r (z) f BOγ ( β(ϕz (ζ), z) + 1 ) (1 z 2 ) γ 1 z, ϕz (ζ) 2γ = f BOγ (β(ζ, 0) + 1)(1 ζ 2 ) γ (1 ϕ z (ζ) 2 ) γ Since c n σ 2γ, we also have = f BOγ (β(ζ, 0) + 1) 1 z, ζ 2γ (1 z 2 ) γ. f ϕ z (ζ) f r (z) (1 z 2 ) γ 1 z, ζ c (n+1+σ) G(ζ) (5.5) for all z, where G(ζ) = (β(ζ, 0) + 1) is in L 1 (, dv α ). Fix any ζ and let t = β(ζ, 0). Since β(ϕ z (ζ), z) = β(ζ, 0) = t and f V O γ, we get lim f ϕ z(ζ) f(z) (1 z 2 ) γ lim (1 z 2 ) γ ω t (f)(z) = 0. (5.6) z 1 z 1 On the other hand, we have f ϕ z (ζ) f r (z) f ϕ z (ζ) f(z) + f(z) f r (z) Therefore, f ϕ z (ζ) f(z) f(z) f(t) dv v σ (D(z, r)) σ (t) f ϕ z (ζ) f(z) + ω r (f)(z). f ϕ z (ζ) f r (z) (1 z 2 ) γ f ϕ z (ζ) f(z) (1 z 2 ) γ + ω r (f)(z) (1 z 2 ) γ,

30 30 J. PAU, R. ZHAO, AND K. ZHU which tends to zero as z 1, because f V O γ and (5.6). Since c n σ, we also have lim f ϕ z(ζ) f r (z) (1 z 2 ) γ 1 z, ζ c (n+1+σ) = 0. z 1 Thus in all cases, due to (5.4) and (5.5), we can aly Lebesgue s Dominated Convergence Theorem (bearing in mind the exression for I(z) given in (5.2)) to obtain I(z) 0 as z 1, which is (c). This comletes the roof of the theorem. Since condition (b) in the theorem above is indeendent of r, we see that the sace V MOγ,r is actually indeendent of r. Thus we will simly use the notation V MOγ. Notice that in (c) and (d) of Theorem 5.3 we require a somehow stronger condition c n σ than c > 0 in Theorem 3.5. It would be nice to know whether it is ossible to relace condition c n σ by c > 0 with c > γ in (c) and (d) here. Anyway, for our main urose here (to characterize comactness of Hankel oerators) condition c n σ is enough. For α > 0 let B0 α = B0 α ( ) denote the closure of the set of olynomials in B α. The sace B0 α consists exactly of those holomorhic functions f such that lim z 1 (1 z 2 ) α f(z) = 0. As before, the comlex gradient can be relaced by the radial derivative Rf. Furthermore, for α > 1/2, a function f is in B0 α if and only if the function (1 z 2 ) α 1 f(z) belongs to C 0 ( ). Again, we refer to [17, Chater 7] for all these facts. With minor modifications in the roof of Proosition 3.6 together with Corollary 2.9 we obtain the following result. Proosition 5.4. Let γ R and 1 <. Then H( ) V MOγ = B 1+γ 0 for n = 1 and γ > 1, or for n > 1 and γ > 1/2. In all other cases, the sace H( ) V MOγ consists of only constants. Proof. The details are left to the interested reader. 6. Comact Hankel oerators In this section we rove the following characterization of comact Hankel oerators between weighted Bergman saces. Theorem 6.1. Let 1 < q <, α, β > 1, f L q β, and γ = n β q n α.

31 BMO AND HANKEL OPERATORS 31 Then both H β f, Hβ f : A α L q β are comact if and only if f V MOq γ. Again, we are going to rove Theorem 6.1 with several lemmas. We begin with the necessity. Lemma 6.2. Let, q, α, β and γ be as in Theorem 6.1. If both H β f, H β f : A α L q β are comact, then f V MOq γ. Proof. Fix a nonnegative t such that n β + t > (n α)/. It is easy to see that h t z converges to zero uniformly on comact subsets of as z 1. Since each h t z is a unit vector in A α, we conclude that h t z 0 weakly in A α as z 1. It follows from the comactness of H β f that lim z 1 Hβ f ht z q,β = 0. The same is true if f is relaced by f. By Proosition 4.3 we have lim MO q,β,t(f)(z) = 0. z 1 In other words, we have lim z 1 (1 z 2 ) c+γq f(w) gz (z) q dv β (w) = 0, 1 z, w n+1+c+β where c = (q 1)(n+1+β)+tq. This imlies condition (e) in Theorem 5.3 with λ z = g z (z), so f V MOγ. q The sufficiency will follow from the next two results. Lemma 6.3. Let, q, α, β and γ be as in Theorem 6.1. If f V A q γ, then H β f : A α L q β is comact. Proof. Let {g n } be a bounded sequence in A α converging to zero uniformly on comact subsets of. We must rove that H β f g n q,β 0. Following the roof of Proosition 4.4, we know that H β q,β g n L q (dµ f,β ) f g n with dµ f,β = f q dv β. The desired result then follows from Lemma 5.1. Lemma 6.4. Let, q, α, β and γ be as in Theorem 6.1. If f V O γ, then H β f : A α L q β is comact. Proof. Let {g n } be a bounded sequence in A α converging to zero uniformly on comact subsets of. By Lemma 4.5 and the density of H in A α, for any δ β we have H β f g q,β C H δ f g q,β, g A α.

32 32 J. PAU, R. ZHAO, AND K. ZHU We will be done if we can rove that lim H δ f g q,β n = 0 n for some δ β. Since β > 1, we can find some η > 0 satisfying β η max(q, q ) > 1. We then choose some δ β large enough so that c = ηq + δ β satisfies c n σ + max(0, 2γq), with σ = β ηq. In fact, this is the same as δ n β + max(0, 2γq) 2ηq. So the choice δ = n β + max(0, 2γq) works. Let ε > 0 be arbitrary. Since V O γ V MOγ, q by art (c) of Theorem 5.3 and with the above c and σ, we may choose t 1 sufficiently close to 1 so that (1 w 2 ) (ηq+δ β)+γq f(z) f r (w) q 1 z, w dv β ηq(z) < ε (6.1) n+1+δ for all t 1 < w < 1. Fix r > 0. By the definition of V O γ, there exists t 2, 0 < t 2 < 1, such that ω r (f)(w) < ε(1 w 2 ) γ, w > t 2. We have H δ fg n q q,β ( ) q f(z) f(w) g n (w) dv 1 z, w n+1+δ δ (w) dv β (z). Let t = max(t 1, t 2 ) and slit the inner integral above in two arts: one for w t and the other for w > t. The integral on w t can be made as small as we want because of the uniform convergence to zero on comact subsets of g n. For the other, we will use our assumtion f V O γ. Since f(z) f(w) f(z) f r (w) + f(w) f r (w), we get two integrals. The first one involves the function f(w) I 1 (z) = f r (w) g n (w) dv 1 z, w n+1+δ δ (w). Since f(w) f r (w) = w >t 1 ( ) f(w) f(ζ) dvδ (ζ), v δ (D(w, r)) D(w,r)

33 we obtain for w > t that BMO AND HANKEL OPERATORS 33 f(w) f r (w) ω r (f)(w) < ε (1 w 2 ) γ. Therefore, I 1 := I 1 (z) q dv β (z) ε q = ε q S b,d ( g n )(z) q dv β (z), ( ) q g n (w) dv δ γ(w) dv 1 z, w n+1+δ β (z) with d = n+1+δ and b = δ γ. Now we want to aly Theorem 2.4 to show that S b,d : L α L q β is bounded. In the notation of Theorem 2.4 we have λ = γ and It remains to check the condition n b + λ = n δ = d. α + 1 < (b + 1) = (1 + δ γ), which is easily seen to be equivalent to ( 1 n q 1 ) < (1 + δ) 1 + β, (6.2) q By the roof of Proosition 4.6, we have ( 1 n q 1 ) < 1 + β = (1 + β) 1 + β, q q where q is the conjugate exonent of q. Since β δ, we see that (6.2) is indeed true. Therefore, by Theorem 2.4, we have I 1 ε q g n q,α C ε q. It remains to deal with I 2 := I 2 (z) q dv β (z), where I 2 (z) = w >t f(z) f r (w) g n (w) 1 z, w n+1+δ dv δ (w).

34 34 J. PAU, R. ZHAO, AND K. ZHU By Hölder s inequality and Lemma 2.2, [ I 2 (z) q f(z) f ] r (w) q g n (w) q dv w >t 1 z, w n+1+δ δ+ηq (w) [ ] q/q dv δ ηq (w) 1 z, w n+1+δ (1 z 2 ) ηq ( w >t Thus I 2 is dominated by [ g n (w) q (1 w 2 ) ηq w >t By (6.1), we get I 2 ε w >t f(z) f r (w) q g n (w) q 1 z, w n+1+δ dv δ+ηq (w) ) f(z) f ] r (w) q 1 z, w dv β ηq(z) dv n+1+δ δ (w). g n (w) q dv β γq (w) ε g n q A α. For the last inequality we used the fact that A α A q β γq, which can be obtained from Theorem 69 in [14]. Putting everything together we conclude that Hf δg n q,β 0 as n. This finishes the roof. To summarize, the necessity of Theorem 6.1 is roved by Lemma 6.2. Since f V MO q γ if and only if f V MO q γ, the sufficiency is a consequence of Theorem 5.3, Lemma 6.3, and Lemma 6.4. As a direct consequence of Theorem 6.1 and Proosition 5.4, we obtain the following characterization of comactness of Hankel oerators with conjugate holomorhic symbols. Corollary 6.5. Let f A 1 β, 1 < q <, α, β > 1, and γ = n β q n α. If n = 1 and γ > 1, or if n > 1 and γ > 1/2, then H β f : A α L q β is comact if and only if f B 1+γ 0. In all other cases, H β f : A α L q β is comact if and only if f is constant. References [1] J. Arazy, S. Fisher, and J. Peetre, Hankel oerators on weighted Bergman saces, Amer. J. Math. 110 (1988), [2] J. Arazy, S. Fisher, S. Janson, and J. Peetre, Membershi of Hankel oerators on the ball in unitary ideals, J. London Math. Soc. 43 (1991),

35 BMO AND HANKEL OPERATORS 35 [3] S. Axler, The Bergman sace, the Bloch sace, and commutators of multilication oerators, Duke Math. J. 53 (1986), [4] C. Békollé, C. Berger, L. Coburn, and K. Zhu, BMO in the Bergman metric on bounded symmetric domains, J. Funct. Anal. 93 (1990), [5] C. Berger, L. Coburn, and K. Zhu, BMO on the Bergman saces of the classical domains, Bull. Amer. Math. Soc. (N.S.) 17 (1987), [6] C. Berger, L. Coburn, and K. Zhu, Function Theory on Cartan Domains and the Berezin-Toelitz Symbol Calculus, Amer. J. Math. 110 (1988), [7] S. Janson, Hankel oerators between weighted Bergman saces, Ark. Mat. 26 (1988), [8] H. Li, BMO, VMO, and Hankel oerators on the Bergman sace of strongly seudo convex domains, J. Funct. Anal. 106 (1992), [9] H. Li, Hankel oerators on the Bergman sace of strongly seudo convex domains, Integr. Equ. Oer. Theory 19 (1994), [10] S. Li and H. Wulan, Some new characterizations of Bloch saces, Taiwanese J. Math. 14 (2010), [11] J. Pau and R. Zhao, Weak factorization and Hankel forms for weighted Bergman saces on the unit ball, Math. Ann. (to aear), available at htt://arxiv.org/abs/ [12] R. Wallstén, Hankel oerators between weighted Bergman saces in the ball, Ark. Mat. 28 (1990), [13] R. Zhao, Generalization of Schur s test and its alication to a class of integral oerators, Integr. Equ. Oer. Theory, doi: /s (2015). [14] R. Zhao and K. Zhu, Theory of Bergman saces in the unit ball of C n, Mem. Soc. Math. Fr. (N.S.) 115 (2008), vi+103. [15] K. Zhu, VMO, ESV, and Toelitz oerators on the Bergman sace, Trans. Amer. Math. Soc. 302 (1987), [16] K. Zhu, BMO and Hankel oerators on Bergman saces, Pacific J. Math. 155 (1992), [17] K. Zhu, Saces of Holomorhic Functions in the Unit Ball, Sringer-Verlag, New York, [18] K. Zhu, Oerator Theory in Function Saces, Second Edition, Math. Surveys and Monograhs 138, American Mathematical Society, Providence, Rhode Island, Jordi Pau, Deartament de Matemàtica Alicada i Analisi, Universitat de Barcelona, Barcelona, Catalonia, Sain address: jordi.au@ub.edu Ruhan Zhao, Deartment of Mathematics, State University of New York, Brockort, NY 14420, USA address: rzhao@brockort.edu Kehe Zhu, Deartment of Mathematics and Statistics, State University of New York, Albany, NY 12222, USA address: kzhu@math.albany.edu