Weak factorization and Hankel forms for weighted Bergman spaces on the unit ball

We establish weak factorizations for a weighted Bergman space $A^p_{\a}$, with $1<p<\infty$, into two weighted Bergman spaces on the unit ball of $\C^n$. To obtain this result, we characterize bounded Hankel forms on weighted Bergman spaces on the unit ball of $\C^n$.


INTRODUCTION
A classical theorem of Riesz asserts that any function in the Hardy space H p on the unit disk can be factored as f = Bg with f H p = g H p , where B is a Blaschke product and g is an H p -function with no zeros on the unit disk. An immediate consequence of that result is that any function in the Hardy space H p admits a "strong" factorization f = f 1 f 2 with f 1 ∈ H p1 , f 2 ∈ H p2 and f H p 1 · f H p 2 = f H p , for any p 1 and p 2 determined by the condition 1/p = 1/p 1 + 1/p 2 . In [12], C. Horowitz obtained strong factorizations of functions in a weighted Bergman space on the unit disk into functions of two weighted Bergman spaces with the same weight. These strong factorization results are no longer possible to obtain [11] in the setting of Hardy and Bergman spaces in the unit ball of the complex euclidian space C n of dimension n when n ≥ 2, but it is still possible to obtain some "weak" factorizations for functions in these spaces.
For two Banach spaces (or F -spaces) of functions, A and B, defined on the same domain, the weakly factored space A ⊙ B is defined as the completion of finite sums with the following norm: When 0 < p ≤ 1, weak factorizations for the Hardy spaces H p and the weighted Bergman spaces A p α on the unit ball of C n are well known (see [6] and [9] for Hardy spaces; and [5], [20] or [25,Corollary 2.33] for Bergman spaces). However, when 1 < p < ∞, even for unweighted Bergman spaces the problem is still open (see, for example [4]).
In this paper we completely solve the above problem for Bergman spaces by establishing weak factorizations for a weighted Bergman space A q β , with 1 < q < ∞ and β > −1, into two weighted Bergman spaces with non necessarily the same weight, on the unit ball B n of C n . The following is our main result. Theorem 1. Let 0 < q < ∞ and β > −1. Then A q β (B n ) = A p1 α1 (B n ) ⊙ A p2 α2 (B n ) for any p 1 , p 2 > 0 and α 1 , α 2 > −1 satisfying In this context, by "=" we mean equality of the function spaces and equivalence of the norms. We mention here that the case 0 < q ≤ 1 is well known, and follows easily from the atomic decomposition for Bergman spaces. Our contribution here is the case q > 1.
Now we are going to recall the definition of the weighted Bergman spaces. First we need some notations. For any two points z = (z 1 , ..., z n ) and w = (w 1 , ..., w n ) in C n , we use z, w = z 1w1 + · · · + z nwn to denote the inner product of z and w, and to denote the norm of z in C n . Let B n = {z ∈ C n : |z| < 1} be the unit ball in C n and S n = {z ∈ C n : |z| = 1} be the unit sphere in C n . Let H(B n ) be the space of all analytic functions on B n . We use dv to denote the normalized volume measure on B n and dσ to denote the normalized area measure on S n . For −1 < α < ∞, we let dv α (z) = c α (1 − |z| 2 ) α dv(z) denote the normalized weighted volume measure on B n , where c α = Γ(n + α + 1)/[n!Γ(α + 1)].
For 0 < p < ∞ and −1 < α < ∞, let L p (B n , dv α ) be the weighted Lebesgue space which contains measurable functions f on B n such that Denote by A p α = L p (B n , dv α ) ∩ H(B n ), the weighted Bergman space on B n , with the same norm as above. If α = 0, we simply write them as L p (B n , dv) and A p respectively and f p for the norm of f in these spaces.
It is a well-known fact that to obtain weak factorization results is equivalent to give a "good" description of the boundedness of certain Hankel forms. A Hankel form is a bilinear form B on a space of analytic functions such that for any f and g, B(f, g) is a linear function of f g. These forms have been extensively studied on Hardy spaces and on Bergman spaces. For the case of the Hardy space on the unit disk, a classical result by Nehari [18] says that the Hankel form (under the usual integral pair for Hardy spaces) with an analytic symbol b is bounded on H 2 × H 2 if and only if b ∈ BM OA, the space of analytic functions of bounded mean oscillation. The proof used the fact that a function in H 1 can be factored into product of two functions in H 2 . Unfortunately, such strong factorization is not possible (see [11]) for Hardy spaces in the unit ball B n of C n . However, Coifman, Rochberg and Weiss [6] were able to generalize Nehari's result to the unit ball B n by using a weak factorization of H 1 . Namely, they proved that Our approach to the problem for weighted Bergman spaces on the unit ball is the opposite to the one of Coifman, Rochberg and Weiss in [6]. We first characterize boundedness of the Hankel forms on weighted Bergman spaces, and with this result the weak factorization easily follows.
Given α > −1 and a holomorphic symbol function b we define the associated Hankel type bilinear form T α b for polynomials f and g by Since the polynomials are dense in the weighted Bergman spaces, the Hankel form T α b is densely defined on A p1 α1 × A p2 α2 for any p 1 , p 2 > 0 and any α 1 , The next result characterizes boundedness of the Hankel form T α b acting on A p1 α1 × A p2 α2 . We will see in Section 3 that this implies the weak factorization in Theorem 1.
Theorem 2. Let 1 < p 1 , p 2 < ∞, and α, α 1 , α 2 > −1 satisfy where q and β are real numbers satisfying (1), and q ′ and β ′ are determined by the condition Remarks. Note that, condition (3) guarantees that q > 1 and β ′ > −1. When q and β satisfy condition (1), automatically we would have β > −1 (to see this, simply add two equations in (1) together). By a general duality theorem for weighted Bergman spaces (see Theorem A in Section 2), the condition b ∈ A q ′ β ′ means that the symbol b belongs to the dual space of A q β under the pairing given by (2).
It turns out that boundedness of the Hankel form T α b is equivalent to boundedness of a (small) Hankel operator, which we are going to introduce in a moment. Let α > −1. It is well-known that, the integral operator is the orthogonal projection from L 2 (B n , dv α ) onto the weighted Bergman space A 2 α . The above formula can be used to extend P α to a linear operator from L 1 (B n , dv α ) into H(B n ). For 1 < p < ∞, P α is a bounded operator from L p (B n , dv α ) onto A p α . Denote by A p α the conjugate analytic functions f on B n that are in L p (B n , dv α ). Clearly, Let Q α denote the orthogonal projection from L 2 (B n , dv α ) onto A 2 α . Clearly one has Given f ∈ L 1 (B n , dv α ) and a polynomial g, the weighted (small) Hankel operator is defined by h α f g = Q α (f g). Due to the density of polynomials, the small Hankel operator h α f is densely defined on the weighted Bergman space A p α for 1 ≤ p < ∞. We will study boundedness of the small Hankel operator with conjugate analytic symbols, that is, Let f ∈ H(B n ) and α such that Then (5) is true (see [23,Theorem 70]). Hence the above result concerns the boundedness of h α f from a smaller space to a larger space. Also, by [25, Theorem 2.11], condition (6) means that the integral operator P α is a bounded projection from L p2 (B n , dv α2 ) onto A p2 α2 .
If one considers the operator , and the norms of h α f and S α f are equivalent. Now, if g ∈ A p1 α1 and h ∈ A p2 α2 , by Fubini's theorem we easily obtain , with equivalent norms. Here, the numbers α ′ 2 and p ′ 2 are defined by the relation Comparing Theorem 2 with Theorem 3, notice that the first inequality in (3) is equivalent to condition 1 < p ′ 2 < p 1 < ∞. Also, when p 2 and α 2 are replaced by p ′ 2 and α ′ 2 , condition (6) turns out to be equivalent to α 2 > −1, and therefore is always satisfied; and the second inequality in (3) is equivalent to condition (5). Therefore, Theorem 3 implies Theorem 2.
The paper is organized as follows: in Section 2 we give some necessary concepts and recall some key results which are needed in our proof of the main result. In Section 3 we give in detail the connection between weak factorizations and Hankel forms. The proof of Theorem 3 is given in Section 4.
In the following, the notation A B means that there is a positive constant C such that A ≤ CB, and the notation A ≍ B means that both A B and B A hold.

PRELIMINARIES
We need the following duality theorem. In this generality the result is due to Luecking [16] (see also, Theorem 2.12 in [25]).
(with equivalent norms) under the integral pair , α given by (2), where We need the following well known integral estimate that can be found, for example, in [25, Theorem 1.12].
Lemma B. Let t > −1 and s > 0. There is a positive constant C such that For any a ∈ B n with a = 0, we denote by ϕ a (z) the Möbius transformation on B n that exchanges 0 and a. It is known that, for any z ∈ B n where s a = 1 − |a| 2 , P a is the orthogonal projection from C n onto the one dimensional subspace [a] generated by a, and Q a is the orthogonal projection from C n onto the orthogonal complement of [a]. When a = 0, ϕ a (z) = −z. ϕ a has the following properties: For z, w ∈ B n , the pseudo-hyperbolic distance between z and w is defined by and the hyperbolic distance on B n between z and w induced by the Bergman metric is given by β(z, w) = tanh ρ(z, w).
For z ∈ B n and r > 0, the Bergman metric ball at z is given by It is known that, for a fixed r > 0, the weighted volume We refer to [25] for all of the above facts. A sequence {a k } of points in B n is a separated sequence (in Bergman metric) if there exists a positive constant δ > 0 such that β(z i , z j ) > δ for any i = j. We need a wellknown result on decomposition of the unit ball B n . The following version is Theorem 2.23 in [25] Lemma C. There exists a positive integer N such that for any 0 < r < 1 we can find a sequence {a k } in B n with the following properties: Any sequence {a k } satisfying the conditions of the above lemma is called a lattice (or an r-lattice if one wants to stress the dependence on r) in the Bergman metric. Obviously any r-lattice is separated.
For convenience, we will denote by D k = D(a k , r) andD k = D(a k , 4r). Then Lemma C says that B n = ∪ ∞ k=1 D k and there is an positive integer N such that every point z in B n belongs to at most N of setsD k .
We also need the following atomic decomposition theorem for weighted Bergman spaces. This turns out to be a powerful theorem in the theory of Bergman spaces. The result is basically due to Coifman and Rochberg [5], and can be found in Chapter 2 of [25].
Theorem D. Suppose p > 0, α > −1, and b > n max 1, Then we have (i) For any separated sequence {a k } in B n and any sequence λ = {λ k } ∈ ℓ p , the function and In the proof given in [25], part (i) requires that the sequence {a k } is an r-lattice for some r ∈ (0, 1], but it is well known that only the separation of the sequence {a k } is needed. As was said before, the fact that, for 0 < q ≤ 1, any function in the Bergman space A q β admits a weak factorization follows easily from part (ii) of Theorem D. Since in [25] it is only considered the case α 1 = α 2 = β, we give the details. Indeed, let f ∈ A q β and take with b big enough. Then f = k f k g k and by Lemma B we have f k p1,α1 |λ k | and g k p2,α2 ≤ C. Since 0 < q ≤ 1, we have Using the embedding of Hardy spaces into Bergman spaces together with the atomic decomposition for Bergman spaces, the proof of the weak factorization for H p with 0 < p < 1 is surprisingly simple, so that we suspect that it must be known to the experts, but since we couldn't find it in the literature it is included here. We refer to the books [21] and [25] for the theory of Hardy spaces in the unit ball.
Theorem 4. Let 0 < p < 1, and r, s such that 1 p = 1 r + 1 s . If f ∈ H p , then there are functions f j ∈ H r and g j ∈ H s such that Proof. By Corollary 4.49 in [25], for α = n p − (n + 1) we have H p ⊂ A 1 α with f 1,α ≤ C f H p . Using the atomic decomposition for the Bergman space A 1 α we see that there is a sequence of points {a k } ⊂ B n such that any f ∈ H p admits the decomposition It is clear that f = k f k g k . Since f k H r = |λ k | and g k H s = 1 we are done.

WEAK FACTORIZATIONS AND HANKEL FORMS
The equivalence between boundedness of T α b and weak factorization can be formulated as the following result. The proof here basically follows the argument of Corollary 1.2 in [1]. Proposition 5. Let 1 < q < ∞ and α, β > −1. Let p 1 , p 2 and α 1 , α 2 satisfy (3) and (1), and let q ′ and β ′ satisfy (4). The following are equivalent: Assume that (ii) holds, we prove that with duality under the pairing , α . Since A q β is reflexive and a Banach space X is reflexive if and only if X * is reflexive [22, p.31], this would give (i). Let F ∈ (A q β ) * . By Theorem A, there is a function b ∈ A q ′ β ′ such that F (ϕ) = ϕ, b α for any ϕ ∈ A q β . By (ii), Then, for any ε > 0, we can find two sequences {g k } ∈ A p1 α1 and {h k } ∈ A p2 α2 such that f = k g k h k , and Hence Since ε > 0 was arbitrary, we get that Hence F ∈ (A p2 α2 ) * , and so, by Theorem A, there is an unique function α2 , where p ′ 2 and α ′ 2 satisfy Now we let f = gh with g ∈ A p1 α1 and h ∈ A p2 α2 . Then f ∈ A p1 α1 ⊙ A p2 α2 . Since F ∈ (A p1 α1 ⊙ A p2 α2 ) * , we know that which shows that T α b extends to a continuous linear functional on A p1 Combining the above arguments we know that (7) is true, and so (i) is true. Next let us assume that (i) holds, and prove (ii). First, assume that b ∈ A q ′ β ′ . A simple application of Hölder's inequality clearly shows that f q,β . Hence, for any ε > 0, we can find two sequences {g k } ∈ A p1 α1 and {h k } ∈ A p2 α2 such that f = k g k h k , and Since ε was arbitrary we get that The proof is complete.

PROOF OF THEOREM 3
In this section we prove Theorem 3, from which Theorem 2 follows. We first prove an auxiliary result which may be of independent interest. Let b ≥ 0 and α > −1. For a measurable function f on B n , we define the following integral operator The operator R α,b is the same as the one appearing in [25,Section 1.4], and is the unique continuous linear operator on H(B n ), satisfying for all w ∈ B n . Lemma 6. Let b ≥ 0 be any fixed number. Let α, β > −1, q > 1 satisfy Let f ∈ H(B n ). If there is a constant M > 0 such that for any 0 < r ≤ 1 and any r-lattice then f ∈ A q β and f q,β M. Proof. Suppose (9) holds. Choose β ′ = q ′ (α − β/q) (which means α = β/q + β ′ /q ′ ). Note that condition 1 + α > (1 + β)/q in the lemma guarantees that β ′ > −1 by the following computation: we know that Hence, from Theorem D we know that, there exists a sequence {µ k } ∈ ℓ q ′ with {µ k } ℓ q ′ h q ′ ,β ′ and an r-lattice {a k } in B n such that Hence Since by Theorem A and Hölder's inequality we obtain and so f ∈ A q β . The proof is complete. Now we are ready to prove Theorem 3.
Proof of Theorem 3. As we noticed before, we just need to prove that S α f : α1 . If p 2 > 1 then P α : L p2 (B n , dv α2 ) → A p2 α2 is bounded, and then from Hölder's inequality the result follows. Indeed, Conversely, suppose S α f : A p1 α1 → A p2 α2 is bounded, we are going to show that f ∈ A q β . We begin with using an argument of Luecking (see, e.g., [17]). Let r k (t) be a sequence of Rademacher functions (see [8,Appendix A]). Let b be large enough so that (10) b > n + 1 + α 1 p 1 .
Fix any r > 0, and let {a k } be an r-lattice and {D k } be the associated sets in Lemma C. By Theorem D, we know that, for any sequence of real numbers {λ k } ∈ ℓ p1 , the function belongs to A p1 α1 with g t p1,α1 {λ k } ℓ p 1 for almost every t in (0, 1). Denote by Since S α f : A p1 α1 → A p2 α2 is bounded, we get that for almost every t in (0, 1). Integrating both sides with respect to t from 0 to 1, and using Fubini's Theorem and Khinchine's inequality (see [22, p.12]), we get If p 2 ≥ 2, then 2/p 2 ≤ 1, and from (12) we have If 1 < p 2 < 2, then 2/p 2 > 1, from (12), by Hölder's inequality we get since each z ∈ B n belongs to at most N of the setsD k . Combining the above two inequalities, and applying (11) we obtain By subharmonicity we know that, From this we obtain Let R α,b be the integral operator defined in (8). Then the equation (14) is the same as Since {λ k } was an arbitrary sequence in ℓ p1 , we know that {λ p2 k } is an arbitrary sequence in ℓ p1/p2 . Since the conjugate exponent of p 1 /p 2 is (p 1 /p 2 ) ′ = p 1 /(p 1 − p 2 ), by duality we obtain that Note that condition (6) guarantees that 1 + α > (1 + β)/q. Since (16) is true for any 0 < r ≤ 1 and any r-lattice {a k } in B n , we can apply Lemma 6 to obtain that f ∈ A q β and f q,β S α f . This finishes the proof.

Compactness.
Under the assumptions of Theorem 3, actually one has that the small Hankel operator h α f : A p1 α1 → A p2 α2 is bounded if and only if it is compact. This is from a general result of Banach space theory. It is known that, for 0 < p 2 < p 1 < ∞, every bounded operator from ℓ p1 to ℓ p2 is compact (see, for example Theorem I.2.7, p.31 in [15]). Since the weighted Bergman space A p α is isomorphic to ℓ p (see, Theorem 11, p.89 in [22], note that the same proof there works for weighted Bergman spaces on the unit ball B n ), we get directly the above result.

Small Hankel operators with the same weights.
In order to clarify the result on small Hankel operators between Bergman spaces given in Theorem 3, we consider the case of the same weights, that is, when α 1 = α 2 = β = α. In that case, all the restrictions in Theorem 3 reduces to p 2 > 1. We isolate this case here.
This proves a conjecture in [4]. Concerning the boundedness of h α f : A p1 α → A p2 α for all possible choices of 0 < p 1 , p 2 < ∞, we mention here that the case p 1 = p 2 > 1 is by now classical (see [13], [24] and [4]), and in this case the boundedness is equivalent to the symbol f being in the Bloch space B, that consists of those holomorphic functions f on Here, Rf denotes the radial derivative of f , that is, The Bloch space also admits an equivalent norm in terms of the invariant gradient ∇f (z) := ∇(f • ϕ z )(0) as follows The case 0 < p 1 ≤ p 2 is completely settled in [4] (actually the results are stated for the unweighted Bergman spaces A p , but the proofs works also for the weighted case). The description for the case p 1 = p 2 = 1 is that f must belong to the so called logarithmic Bloch space, a result that goes back to the one dimensional case obtained by Attele [2]. Concerning estimates with loss, in [4] Bonami and Luo obtained a description for the case 0 < p 2 < p 1 with p 2 < 1 (again the result in [4] is stated for the unweighted Bergman spaces). Thus, in view of our result, to complete the picture it remains to deal with the case p 1 > p 2 = 1 (this problem is also open for the unit disk). In that case, also in [4], some partial results are obtained (again for the unweighted case). Mainly, they provide a pointwise estimate that is necessary for the small Hankel operator to be bounded, and they show that the condition is sufficient. Moreover, they conjecture that the previous condition is also necessary. We have not been able to prove the conjecture, but we are going to shed some light on that problem. Before going to the proof we need first some preparation. First of all, recall that the Bloch space is the dual of A 1 α under the integral pairing , α (see [25,Theorem 3.17]). We also need the following lemma, whose one dimensional analogue is essentially proved in [3]. Lemma 9. Let 1 < p < ∞, σ > −1, and n + 1 + σ < b. Then for any f ∈ H(B n ) and a ∈ B n .
Therefore we obtain Due to our condition b > n + 1 + σ, we have t = 2(n + 1 + σ) − b < n + 1 + σ and we can apply (18) to get Proof. Part (i) follows directly from Theorem 8 and the pointwise estimate for Bloch functions |g(z)| ≤ g B log 2 1 − |z| 2 . To prove part (ii), for each z ∈ B n , the function