Characterization of Schatten class Hankel operators on weighted Bergman spaces

We completely characterize the simultaneous membership in the Schatten ideals $S_ p$, $0<p<\infty$ of the Hankel operators $H_ f$ and $H_{\bar{f}}$ on the Bergman space, in terms of the behaviour of a local mean oscillation function, proving a conjecture of Kehe Zhu from 1991.


Main results
Problem: Describe the simultaneous membership in the Schatten ideals S p of the Hankel operators H f and H f acting on weighted Bergman spaces.
The answer given below is the main result of the paper.
Here dλ n (z) = dv(z) (1 − |z| 2 ) n+1 is the Möbius invariant volume measure on B n , and MO r (f ) is a certain type of local mean oscillation function to be defined next after we discuss briefly the history of the problem.
When f is holomorphic on B n one has H f = 0, and the membership of H f in S p is described by f being in the analytic Besov space B p if p > γ n , and f constant if 0 < p ≤ γ n [1,2,5,11,15,16]. The cut-off point is γ n = 1 if n = 1, and γ n = 2n if n ≥ 2. The equivalence between (a) and (b) was conjectured (at least for p ≥ 1) in 1991 by K. Zhu in [16]. It was previously known that, if p > 2n n+1+α , then (a) is equivalent to where MO α (f ) is a "global" mean oscillation type function. The equivalence between (a) and (c) for p > 2n n+1+α was proved in several steps: K. Zhu [16] proved the case p ≥ 2; J. Xia [12,13] obtained the case max (1, 2n n+1+α ) < p ≤ 2, and the last case 2n n+1+α < p ≤ 1 has been proved recently by J. Isralowitz [4]. It is also well known that condition (c) can not characterize the membership on the Schatten ideals on the missing range 0 < p ≤ 2n/(n + 1 + α), since on this range, condition (c) implies f is a constant (see [19, p.233]). Now we recall the concepts and definitions.
We denote by B n the open unit ball of C n , and let dv be the usual Lebesgue volume measure on B n , normalized so that the volume of B n is one. We fix a real parameter α with α > −1 and write dv α (z) = c α (1 − |z| 2 ) α dv(z), where c α is a positive constant chosen so that v α (B n ) = 1. The weighted Bergman space A 2 α := A 2 α (B n ) is the closed subspace of L 2 α := L 2 (B n , dv α ) consisting of holomorphic functions. It is a Hilbert space with inner product The corresponding norm is denoted by f α . The orthogonal (Bergman) projection P α : L 2 (B n , dv α ) → A 2 α (B n ) is an integral operator given by Given a function f ∈ L 2 (B n , dv α ), the Hankel operator H f with symbol f is where M f denotes the operator of multiplication by f . It is well known that the simultaneous study of the Hankel operators H f and Hf is equivalent to the study of the commutator where H f := H f P α acts now on L 2 α .
Let H and K be separable Hilbert spaces, and let 0 < p < ∞. A compact operator T from H to K is said to belong to the Schatten class S p = S p (H, K) if its sequence of singular numbers belongs to the sequence space ℓ p (the singular numbers are the square roots of the eigenvalues of the positive operator T * T , where T * is the Hilbert adjoint of T ). For p ≥ 1, the class S p is a Banach space with the norm T p = ( n |λ n | p ) 1/p , while for 0 < p < 1 one has [7, Theorem 2.8] the inequality S + T p p ≤ S p p + T p p . Also, if A is a bounded operator on H, B a bounded operator on K, and T is in S p , then BT A is in S p . We refer to [19,Chapter 1] for a brief account on Schatten classes.
For z ∈ B n and r > 0, the Bergman metric ball at z is given by D(z, r) = w ∈ B n : β(z, w) < r , where β(z, w) denotes the hyperbolic distance between z and w induced by the Bergman metric. If f is locally square integrable with respect to the volume measure on B n , the mean oscillation of f at the point z ∈ B n in the Bergman metric is where the averaging function f r is given by It is well known [16,19] that the simultaneous boundedness and compactness of the Hankel operators H f and H f acting on the Bergman space A 2 α can be characterized in terms of the properties of the function MO r (f ). The Hankel operators H f and H f are both bounded if and only if MO r (f ) ∈ L ∞ (B n ); and compact if and only if MO r (f ) ∈ C 0 (B n ). The same characterization holds using a more "global" oscillation function that we introduce next. For any f ∈ L 2 (B n , dv α ) and z ∈ B n , let where B α (g) denotes the Berezin transform of a function g ∈ L 1 (B n , dv α ) defined as where k z are the normalized reproducing kernels of A 2 α , that is, k z = K z / K z α with K z being the reproducing kernel of A 2 α at the point z, given by In order to prove Theorem 1, we must introduce a more general Berezin type transform B α,t f , and a more general "mean oscillation" function MO α,t (f ). For α > −1 and t ≥ 0, let We also denote by h t z to be its normalized function, that is, h t If g ∈ L 1 (B n , dv α ), the Berezin type transform B α,t (g) is defined as It is easy to see that and that one has also the following double integral expression The idea to use the function MO α,t (f ) in the study of Hankel operators has been also suggested by other authors independently (see [4,14] for example). We have the following result.
Theorem 2. Let α > −1, r > 0, f ∈ L 2 (B n , dv α ), and 0 < p < ∞. Then, for each t ≥ 0 such that p > 2n/(n + 1 + α + 2t), we have It is easy to check that, for any z ∈ B n and r > 0, one has From this expression it follows that the behaviour of the local mean oscillation function MO r (f ) is independent of the parameter α. Also, from this and the double integral expression of MO α,t (f ), it is straightforward to see that MO r (f )(z) ≤ C MO α,t (f )(z). From this observation and Theorem 2, we see that Theorem 1 is equivalent to the following one.
Theorem 3. Let α > −1, f ∈ L 2 (B n , dv α ) and 0 < p < ∞. The following are equivalent: From this, it can be seen that the conjecture stated at the end of [4] is also true.
The paper is organized as follows. After some preliminaries given in Section 2, we prove Theorem 2 in Section 3. All the implications in Theorem 1 are proved in Section 4, except the necessity in the case 0 < p < 2. This part is proved in Section 5 (the case 2n/(n + 1 + α) < p < 2), and in Section 6 where we deduce the last case from the previous one in a tricky way.
We are not worried on the computation of the exact values of certain constants when are not depending on the important quantities involved, so that we use C to denote a positive constant like that, whose exact value may change at different occurrences, and sometimes we use the notation A B to indicate 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.

Some known lemmas
We need a well-known result on decomposition of the unit ball B n . A sequence {a k } of points in B n is called a separated sequence (in the Bergman metric) if there exists a positive constant δ > 0 such that β(a i , a j ) > δ for any i = j. By Theorem 2.23 in [18], 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: The sets D(a k , r/4) are mutually disjoint. (iii) Each point z ∈ B n belongs to at most N of the sets D(a k , 4r).
Any sequence {a k } satisfying the above conditions is called an r-lattice in the Bergman metric. Obviously any r-lattice is separated.
We need the following well known integral estimate that has become very useful in this area of analysis (see [18,Theorem 1.12] for example).
We also need the following well known discrete version of the previous lemma.
Lemma B. Let {z k } be a separated sequence in B n , and let n < t < s. Then Lemma B can be deduced from Lemma A after noticing that, if a sequence {z k } is separated, then there is a constant r > 0 such that the Bergman metric balls D(z k , r) are pairwise disjoints.
We also need the following version of Lemma A, with an extra (unbounded) factor β(z, w) in the integrand.
Proof. Pick ε > 0 so that t − c ε > −1 and s − c ε > 0. Since β(z, w) grows logarithmically, we have Here ϕ z denotes the Möbius transformation sending z to 0. It follows from the basic identity The desired result then follows from Lemma A.
The corresponding discrete version is stated below.
We also need the following elementary result.
Proof. It follows from the double integral expression of the mean oscillation that From this, the result is easily deduced.

Proof of Theorem 2
Let {a k } be an (r/3)-lattice on B n . Because r > 0 is arbitrary, due to Lemma 2.3, it is enough to prove Due to the triangle inequality, we see that and because of the symmetry of the terms, in order to establish (3.1) it is enough to show that Here we use the notation Now we need the following technical lemma. For a, z ∈ B n , we have Proof. Let ξ be a point in the lattice with β(z, ξ) ≤ r/3. Since and a similar estimate can be obtained for |f (a) − f r (a)|, it is enough to prove Denote by γ(t), 0 ≤ t ≤ 1, the geodesic in the Bergman metric going from z to a. Let By the triangle inequality, we have For each m, take a point ξ m in the lattice with β(z m , ξ m ) < r/3. It is not difficult to see [19, p.211]). Then This gives Because the Möbius transformation ϕ z sends the geodesic joining z and a to the geodesic joining 0 and ϕ z (a), we have Developing this inequality using the basic identity (2.1) together with its polarized analogue which gives |1 − a, z m | ≤ 2 |1 − a, z |.
Putting these inequality into (3.4), with the help of the estimate |1 − ξ m , a | ≍ |1 − z m , a | (see [18, p.63]), we obtain From here, the result easily follows, since and Hölder's inequality yields Finally, since N (1 + β(a, z)), the inequality completes the proof of the lemma.
Returning to the estimate for A 1 (f, z), putting the inequality of Lemma 3.1 into (3.3), with d = 1 2 (n + 1 + α + 2t) − ε, where ε > 0 is taken so that pd > n, we see that A 1 (f, z) is less than constant times with δ > 0 taken so that α − 2δ > −1 and pd − pδ > n. By Lemma B and Lemma 2.2, we have where g z denotes the holomorphic function on B n given by Now we use the identity . To see this, since K t z (w) = K t w (z), by the reproducing formula . Therefore, (4.1) together with the boundedness of P α+t on L 2 (B n , dv α ) yields Finally, This proves the result with constant C = (1 + P α+t ). Observe that, when t = 0, since P α = 1, one gets C = 1 since in (4.2) one has the term P α (f k z − g z k z ) α , and thus it is not necessary to use again the triangle inequality.
The case t = 0 of Lemma 4.1 appears in [3] and [16], with a proof that seems to be specific of the Hilbert space case. Observe that our proof is flexible enough to work when studying Hankel operators acting on A p α (see [17], where some version of Lemma 4.1 for t = 0 in this setting was proved with a different method).
The following inequality is also satisfied: The following result can be found in [8,Lemma 2].
If we apply this lemma with the positive operator T = H * f H f , then due to (4.3) and Lemma 4.1, we obtain the necessity in Theorem 3 for p ≥ 2 and the sufficiency for p ≤ 2. This together with the inequality MO r (f )(z) MO α,t (f )(z) gives the implication (a) implies (b) in Theorem 1 for p ≥ 2, and if we use Theorem 2 we see that (b) implies (a) for p ≤ 2. Summarizing, the following proposition has been proved.
Next, we consider the Hankel operator H γ f defined by Since for γ > α, the projection P γ is bounded on L 2 α and P γ P α = P α , we have

Hence the result follows.
Proposition 4.4. Let α > −1, f ∈ L 2 α and 2 < p < ∞. If MO r (f ) ∈ L p (B n , dλ n ) then H f and H f are both in S p (A 2 α , L 2 α ). Proof. This follows from Theorem 2 with t = 0 and the well know fact that MO α,0 (f ) ∈ L p (B n , dλ n ) implies the conclusion of the Proposition. However, we will provide a selfcontained proof based on Lemma 3.1. Note that the condition implies that both H α f and H α f are compact (just take a look at Lemma 2.3 which implies MO r (f )(z) → 0 as |z| → 1); and in view of Lemma 4.3, the operators H γ f and H γ f are also compact for all γ > α. Since for γ big enough, say γ = α + 4t with pt > n. By [19,Theorem 1.33], it suffices to prove that n H γ f e n p α ≤ C for any orthonormal set {e n } of A 2 α , with a constant C not depending on the choice of the orthonormal set. Let ε > 0 so that α − ε > −1. By Cauchy-Schwarz and Lemma A, we have Now, Fubini's theorem, Hölder's inequality with exponent p/2 > 1 and e n α = 1 yield dv α+ p 2 (γ−α+ε) (w). Because {e n } is an orthonormal set, we can use the inequality Set Take a lattice {ξ k } and apply Lemma 3.1 with d = t and δ > 0 satisfying pt − pδ > n and α − ε − 2δ > −1, to obtain By Lemma A and Lemma 2.1, we have This, together with Lemma A gives n H γ f e n p α Bn This finishes the proof.
Taking into account Propositions 4.2 and 4.4, in order to complete the proof of Theorem 1 it remains to show that (a) implies (b) for 0 < p < 2. This is done in the next two sections.

5.
Necessity: the case 2n n+1+α < p < 2 This follows immediately from condition (c) and the fact that MO r (f )(z) MO α (f )(z). However, we are going to give a direct proof of the implication (a) ⇒ (b) in that case. There are several reasons for doing that. First, to be self-contained, and also because the proof we give here has some independent interest and works equally well in the full range 2n n+1+α < p < 2, while the proof of (c) given in [12,13] worked only for p > 1, and the proof of (c) given in [4] was specific of the case 2n n+1+α < p ≤ 1. Finally, the proof given here can be of some help when looking for some extensions of Theorem 1 in some contexts where the equivalence with (c) is not available.
The proof follows the lines of [19,Theorem 7.16], a method that seems to have its roots on previous work of S. Semmes [10] and D. Luecking [6], but with much more complicated estimates. If H f is in S p , then H γ f is also in S p for all γ > α, as a consequence of Lemma 4.3. Let γ = α + t, with t > 0 taken big enough so that all the applications of the lemmas appearing in Section 2 are going to be correct (for example, satisfying pt > 10n). Let Λ = {ξ k } be a regular (r/3)-lattice in the Bergman metric. Fix a sufficiently large positive radius R, and partition the lattice {ξ k } into M subsequences so that the Bergman metric between any two points in each subsequence is at least R. Let {a k } be such a subsequence.
Fix an orthonormal basis {e k } for A 2 α (B n ), and define an operator A on A 2 α (B n ) by The boundedness of A follows easily from the fact that {a k } is separated in the Bergman metric. Let We split these operators as T 1 = D 1 + E 1 , and T 2 = D 2 + E 2 , where D 1 and D 2 are the diagonal operators defined by By the triangle inequality, Since A is a bounded operator, we have Since D 1 and D 2 are positive diagonal operators, then In the same way, we also have D 2 Proof. As in the proof of Lemma 2.1, it is enough to show that j:β(a j ,w)>R for R large enough. Taking into account Lemma B, the elementary proof is left to the interested reader.
We return to the proof of (5.4). Since 0 < p ≤ 2, by [19, Proposition 1.29], we have For the other pairs (ℓ, m), we have the same inequality without the ε but, due to the triangle inequality, one has β(ξ ℓ , ξ m ) > R/4. Thus, we assume that we are in this last case. Then We estimate J ℓ,m (f ) with the help of Lemma 3.1. Let d = t/4 and take δ > 0 satisfying p 2 (n + 1 + α − 3δ) > n, and also p(d − δ) > n. By Lemma 3.1 with the same meaning for N p (f, u) and h δ (z, w) as in that lemma, and taking into account that Then, bearing in mind that γ = α + t, h δ (z, w) = 1 + β(z, w) min(1 − |z|, 1 − |w|) −δ , and the triangle inequality for β(z, w), the result follows easily from Lemma A and Lemma 2.1.
Putting the estimate of Lemma 5.2 into (5.7), and taking into account (5.6) and the symmetry of the two terms, it is enough to show where A p (f ) is given by the expression with H δ,p (ξ ℓ , ξ m ) = h δ (ξ ℓ , ξ m ) p/2 N p (f, ξ ℓ ) 1/2 N p (f, ξ m ) 1/2 .
Using the inequality 2AB ≤ A 2 + B 2 , we have and We begin with the estimate for B p (f ). Because we are assuming that β(ξ ℓ , ξ m ) > R/4, by Lemma 5.1, given ε > 0, we can take R big enough so that Therefore, (5.10) In the estimate for C p (f ) is when we use our assumption p 2 (n + 1 + α) > n. Since δ > 0 has been taken so that p 2 (n + 1 + α − 3δ) > n, we can apply Lemma 5.1 to get Then, proceeding as before, we obtain Joining (5.11), (5.10) and (5.9), we have proved that (5.8) holds. The proof is complete.
6. The last case: 0 < p ≤ 2n n+1+α In order to prove this case, we will fix a number β > α satisfying p(n + 1 + β) > 2n. We will show that condition (a) of Theorem 1 implies that both H β f and H β f are in S p (A 2 β , L 2 β ). Then the case already proved will give MO r (f ) ∈ L p (B n , dλ n ).