Schatten class Toeplitz operators acting on large weighted Bergman spaces

A full description of the membership in the Schatten ideal $S_ p(A^2_{\omega})$ for $0<p<\infty$ of the Toeplitz operator acting on large weighted Bergman spaces is obtained.


INTRODUCTION
Let H(D) denote the space of all analytic functions on D, where D is the open unit disk in the complex plane C. A weight is a positive function ω ∈ L 1 (D, dA), with dA(z) = dxdy π being the normalized area measure on D. For 0 < p < ∞, the weighted Bergman space A p ω is the space of all functions f ∈ H(D) such that We are going to study Toeplitz operators acting on these weighted Bergman spaces, for a certain class E of weights. The prototype is the exponential type weight (1.1) ω α (z) = exp −1 (1 − |z| 2 ) α , α > 0, but he class E also contains non-radial weights. For the weights ω in the class E, the point evaluations L z are bounded linear functionals on A p ω for each z ∈ D. In particular, the space A 2 ω is a reproducing kernel Hilbert space: for each z ∈ D, there are functions K z ∈ A 2 ω with L z = K z A 2 ω such that L z f = f (z) = f, K z ω , where is the natural inner product in L 2 (D, ωdA). The function K z is called the reproducing kernel for the Bergman space A 2 ω and has the property that K z (ξ) = K ξ (z). The study of the properties of the Bergman spaces with exponential type weights has attracted a lot of attention in recent years [1,4,6,7,8,12,13], and new techniques different from the ones used on standard Bergman spaces are required. Notice that, in [1], the space A p ω is denoted by A p (ω p/2 ). Let ω ∈ E. The Toeplitz operator T ω µ with symbol µ is given by We suppose that µ is a finite positive Borel measure on D that satisfies the condition D |K z (ξ)| 2 ω(ξ) dµ(ξ) < ∞. (1.2) Then, the Toeplitz operator T µ is well-defined on a dense subset of A p ω , 1 ≤ p < ∞. In fact, by [1,Corollary 6.4], the set E of finite linear combinations of reproducing kernels is dense in A p ω , and it follows from the condition (1.2) and the Cauchy-Schwarz inequality that T µ (f ) is well defined for any f ∈ E. Theorem 1.1. Let ω ∈ E and 1 ≤ p ≤ q < ∞. Then T µ : A p ω → A q ω is bounded if and only if for each δ > 0 small enough, one has Moreover, Here µ δ is the averaging function defined as We refer to Section 2 for the definition of the function τ (z) associated to the weight ω, and D(δτ (z)) denotes an euclidian disk centered at z and radius δτ (z).

Theorem 1.2.
Let ω ∈ E, 1 ≤ q < p < ∞ and let µ be a finite positive Borel measure on D. The following conditions are equivalent: (i) The Toeplitz operator T µ : A p ω → A q ω is bounded. (ii) For each sufficiently small δ > 0, the function µ δ belongs to L .
The corresponding description on compactness is also obtained. Also (see Theorem 6.6) we prove that, for 0 < p < ∞, the Toeplitz operator is in Schatten class S p (A 2 ω ) if and only if the averaging function µ δ is in L p (D, dλ τ ), where dλ τ = τ (z) −2 dA(z). This completes the characterization obtained by Lin and Rochberg in [9], where the necessity of the previous condition when 0 < p < 1 was left open.
Throughout this work, the letter C will denote an absolute constant whose value may change at different occurrences. We also use the notation a b to indicate that there is a constant C > 0 with a ≤ Cb, and the notation a ≍ b means that a b and b a.

PRELIMINARIES AND BASIC PROPERTIES
A positive function τ on D is said to be in the class L if satisfies the following two properties: (A) There is a constant c 1 such that τ (z) ≤ c 1 (1 − |z|) for all z ∈ D; (B) There is a constant c 2 such that |τ (z) − τ (ζ)| ≤ c 2 |z − ζ| for all z, ζ ∈ D. We also use the notation where c 1 and c 2 are the constants appearing in the previous definition. For a ∈ D and δ > 0, we use D(δτ (a)) to denote the euclidian disc centered at a and radius δτ (a). It is easy to see from conditions (A) and (B) (see [12,Lemma 2.1]) that if τ ∈ L and z ∈ D(δτ (a)), then for sufficiently small δ > 0, that is, for δ ∈ (0, m τ ). This fact will be used many times in this work.
for all f ∈ H(D) and all δ > 0 sufficiently small.
A consequence of the above result is that the Bergman space A p ω is a Banach space when 1 ≤ p < ∞ and a complete metric space when 0 < p < 1. The following lemma on coverings is due to Oleinik, see [11].
The multiplicity N in the previous Lemma is independent of δ, and it is easy to see that one can take, for example, N = 256. Any sequence satisfying the conditions in Lemma B will be called a (δ, τ )-lattice.

Definition 2.2.
A weight ω is in the class E if ω ∈ L * and its associated function τ satisfies the condition The class E contains the exponential type weights given by (1.1), but also includes non-radial weights (see [1] for an example).

2.1.
Reproducing kernels estimates. The next result (see [3,8,12] for (a), and [1] for part (b)) provides useful estimates involving reproducing kernels. Theorem A. Let K z be the reproducing kernel of A 2 ω where ω is a weight in the class E. Then (a) For each z ∈ D, one has For weights in the class E, and points close to the diagonal, one has the following well-known estimate (see [9,Lemma 3.6] for example) for all δ ∈ (0, m τ ) sufficiently small.
We also need the following result appearing in [2, Lemma 2.2].

Lemma C.
Let ω ∈ E, and K z be be the reproducing kernel for A 2 (ω). For 0 < p < ∞ and α ∈ R, there exists a constant C > 0 such that As a direct application of Lemma C and (2.3), one gets the following estimate for the p-norm of the reproducing kernels. Let ω ∈ E and 0 < p < ∞. Then, for each z ∈ D, one has

CARLESON TYPE MEASURES
Let 0 < p, q < ∞. We say that µ is a q-Carleson measure for A p ω if there exists a finite positive constant C such that Thus, µ is q-Carleson for A p ω when the inclusion I µ : A p ω → L q (D, µ) is bounded. Next results were essentially proved in [12]. Since the conditions on the weights are slightly different we give a sketch of the proofs.
Proof. Suppose first that I µ : A p ω → L q (D, dµ) is bounded. If δ ∈ (0, m τ ) is sufficiently small then, due to (2.3), part (a) in Theorem A and the fact that τ (z) ≍ τ (a) for z ∈ D(δτ (a)), we get
Thus, taking into account the estimate for the norm in A p ω of the reproducing kernel K a given in (2.4) together with the boundedness of I µ , we obtain Conversely, suppose that (3.1) holds, and let f ∈ A p ω . Let {z j } be a (δ, τ )-lattice. Then, by the properties of the lattice given in Lemma B, and Lemma A, we have Since q/p ≥ 1, using the finite multiplicity of the covering, we obtain For each a ∈ D, we use the notation k p,a for the normalized reproducing kernels in A p ω , that is, k p,a = K p,a / K p,a A p ω . Lemma 3.2. Let ω ∈ E and 0 < p < ∞. Then k p,a converges to zero uniformly on compact subsets of D as |a| → 1 − .

Theorem 3.3.
Let ω ∈ E and µ be a finite positive Borel measure on D. Let 0 < p ≤ q < ∞. Then is compact if and only if for each sufficiently small δ > 0 we have Proof. One only needs to follow the proof given in [12] with the help of Lemma 3.2 and minor modifications. The details are left to the interested reader.
For the case 0 < q < p < ∞, we need the following lemma.

Lemma 3.4.
Let ω ∈ E, 0 < p < ∞ and, for δ ∈ (0, m τ ), let {z k } be a (δ, τ )-lattice on D. The function given by Proof. We left as an exercise for the reader to check that the partial sums defining F converges uniformly on compact subsets of D showing that F defines an analytic function on D. For 0 < p ≤ 1, using Lemma C we get For the case p > 1, let By Hölder's inequality we have On the other hand, using Lemma A, the lattice properties and Lemma C we have Therefore, applying Lemma C again we obtain

Theorem 3.5.
Let ω ∈ E and let µ be a finite positive Borel measure on D. Let 0 < q < p < ∞. The following conditions are equivalent:

Moreover, one has
.
Proof. It is obvious that (a) implies (b). To prove (b) ⇒ (c), for an arbitrary sequence λ = {λ k } ∈ ℓ p , and t ∈ (0, 1), consider the function where r k (t) is a sequence of Rademacher functions see Appendix A of [5] and From here the proof follows the same lines as in [12]. Finally, the proof of (c) ⇒ (a) can be done exactly as in [12].

BOUNDEDNESS
In this section we describe the boundedness of Toeplitz operators on large Bergman spaces. For δ ∈ (0, m τ ), we consider the averaging function µ δ defined on D by .
. This together with the norm estimate given in (2.2) and the fact that τ (z) ≍ τ (a) for z ∈ D(δτ (a)) gives Therefore, by Lemma A and the estimate of the norm of the reproducing kernels given in (2.4), we obtain Conversely, suppose that (1.3) holds. We first prove that Indeed, by Lemma A, we have If q > 1. by Hölder's inequality, we obtain Therefore, by Fubini's theorem and Lemma C, we obtain If q = p = 1, we arrive at this point directly after the use of Fubini's theorem. Consider the measure ν given by Since (1.3) holds, it follows from Theorem 3.1 that the identity I ν : This finishes the proof.
Before going to the proof of Theorem 1.2 we need the following result.
On the other hand, by Fubini's theorem and τ (z) ≍ τ (s), for s ∈ D(δτ (z)), we have Combining this with (4.5), we get The proof is complete.

Proof of Theorem 1.2. (i) ⇒ (ii)
For an arbitrary sequence λ = {λ k } ∈ ℓ p , consider the function where r k (t) is a sequence of Rademacher functions and {z k } is a (δ, τ )-lattice on D. By Lemma 3.4, In other words, we have Integrating with respect to t from 0 to 1, applying Fubini's theorem and invoking Khinchine's inequality (see [10] for example), we obtain Let χ k denote the characteristic function of the set D(3δτ (z k )). If 0 < q < 2, since the covering {D(3δτ (z k ))} of D has finite multiplicity N, Hölder's inequality gives For q ≥ 2, we have the trivial inequality All together, we obtain Hence, by Lemma A, we arrive at the inequality As in (4.1), for δ ∈ (0, m τ ) small enough, we have That is, Bearing in mind (4.6), this gives Then, by the duality between ℓ p/q and ℓ p p−q we conclude that This is the discrete version of our condition. To obtain the continuous version, simply note that .
This finishes the proof of this implication.
To deal with the case 1 < q < ∞, let {z j } be a (δ, τ )-lattice on D. By Lemma A, for ξ ∈ D(δτ (z j )), we have Then Since q > 1, we can use Hölder's inequality to get Since the covering has finite multiplicity, it follows that Therefore, By Fubini's Theorem, Thus, by Lemma 4.1, we finally obtain The proof is complete.

COMPACTNESS
We begin with a useful criteria for compactness of Toeplitz operators acting on large weighted Bergman spaces. The proof, usually omitted for these type of results, is standard and can be done following the lines of [15]. In order to offer no doubt on the validity of that, we offer the proof here. Before doing that we need a preliminary result. Proof. Let ε > 0, and fix 0 < r < 1. Since {f n } converges to zero uniformly on compact subsets, there exists a natural number n 0 such that sup |ξ|≤R |f n (ξ)| < ε, for all n ≥ n 0 , with r + (1−r) 2 ≤ R < 1.

Now we are going to prove that sup
|z|≤r |T µ f n (z)| → 0. By Theorem A for M ≥ 1 and for all z, ξ ∈ D
Proposition 5.2. Let 1 ≤ p, q < ∞ and ω ∈ E. Then T µ : A p ω → A q ω is compact, if and only if, for any bounded sequence {f n } in A p ω converging to zero uniformly on compact subsets of D, one has lim n T µ f n A q ω = 0. Proof. Suppose first that lim n T µ f n A q ω = 0 for any bounded sequence {f n } in A p ω converging to zero uniformly on compact subsets of D. We want to prove that T µ is compact. Let {f n } be a bounded sequence in A p ω . By Lemma A, {f n } is uniformly bounded on compact subsets of D. Therefore, by Montel's Theorem, there is a subsequence f n k such that f n k → f uniformly on compact subsets of D, for some f ∈ H(D). Using Fatou's lemma, it is easy to see that f must be in A p ω . By our assumption, lim Conversely, assume that T µ is compact and let {f n } be a bounded sequence in A p ω such that f n → 0 uniformly on compact subsets of D. If the conclusion is false, then there exists an ε > 0 and a subsequence {f n j } such that (5.1) T µ f n j A q ω ≥ ε, for all j = 1, 2, 3, ... Since T µ is compact, we can find a further subsequence {f n j k } k and f ∈ A q ω with T µ f n j k − f A q ω → 0, as k → ∞. By Lemma A, for any z ∈ D Hence, T µ f n j k − f → 0 uniformly on compact subsets of D. Moreover, since f n j k → 0 uniformly on compact subsets of D, from Lemma 5.1 we get f ≡ 0. Hence T µ f n j k A q ω → 0 as k → ∞ which contradicts (5.1). The proof is complete.
With the help of the previous proposition, we can characterize the compactness of the Toeplitz operator T µ , result that is stated below. Theorem 5.3. Let 1 ≤ p ≤ q < ∞ and ω ∈ E. Then T µ : A p ω → A q ω is compact if and only if for each δ ∈ (0, m τ ) sufficiently small, one has Proof. Let k p,z be the normalized reproducing kernels in A p (ω p/2 ). From (4.2) in the proof of Theorem 1.1, and the estimate for the p-norm of K z in (2.4), we have By Lemma 3.2, k p,z converges to zero uniformly on compact subsets of D as |z| → 1 − . Thus, if T µ is compact, from Proposition 5.2 we obtain (5.2).
Conversely, suppose (5.2) holds, and let {f n } be a bounded sequence in A p ω converging to zero uniformly on compact subsets of D. According to Proposition 5.2, we must show that T µ f n A q ω → 0. By (4.4) in the proof of Theorem 1.1, we have . By Theorem 5.3, our assumption (5.2) implies that I ν : dν) is compact, which implies that f n L q (D,ν) tends to zero. Hence T µ f n A q ω → 0 proving that T µ is compact. The proof is complete.

MEMBERSHIP IN SCHATTEN CLASSES
Let H be a separable Hilbert space, and 0 < p < ∞. The Schatten class S p = S p (H) consists of those compact operators T on H for which its sequence of singular numbers {λ n } 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 the inequality S + T p p ≤ S p p + T p p . Also, one has T ∈ S p if and only if T * T ∈ S p/2 . We refer to [16, Chapter 1] for a brief account on Schatten classes.
In this section, we are going to describe those positive Borel measures µ for which the Toeplitz operator T µ belongs to the the Schatten ideal S p (A 2 ω ), for ω ∈ E. We recall the reader that we are still assuming that (1.2) holds. In order to obtain such a characterization, we need to introduce first some concepts.
We define the ω-Berezin transform B ω µ of the measure µ as where k z are the normalized reproducing kernels in A 2 ω . We also recall that, for δ ∈ (0, m τ ), the averaging function µ δ is given by We also consider the measure λ τ given by Proposition 6.1. Let 0 < p < ∞, and ω ∈ E. The following conditions are equivalent : . This is essentially proved in the proof of Theorem 3.1, where the same computations give Since B ω µ is in L p (D, dλ τ ), this gives (b).
If p > 1, we use Lemma A in order to obtain Thus, by Hölder's inequality,

This gives
Putting this estimate in the previous inequality, we finally get finishing the proof for the case p > 1. Now, if 0 < p ≤ 1, proceeding as before using Lemma A, we have Since 0 < p ≤ 1, this gives Integrating this inequality against the measure dλ τ (z), and using the norm estimate K z is less than constant times n µ δ (z n ) p τ (z n ) 2(p−1) D(3δτ (zn)) D |K ξ (z)| 2p ω(z) p τ (z) 2(p−1) dA(z) ω(ξ) p dA(ξ).

By Lemma C, we have
This, together with the previous estimate, yield The proof is complete.
Next Lemma is the analogue to our setting of a well known result for standard Bergman spaces.

Lemma 6.2.
Let ω ∈ E, and T be a positive operator on A 2 ω . Let T be the Berezin transform of the operator T defined by T (z) = T k z , k z ω , z ∈ D.
Proof. Let p > 0. The positive operator T is in S p if and only if T p is in the trace class S 1 . Fix an orthonormal basis {e k } of A 2 ω . Since T p is positive, it belongs to the trace class if and only if k T p e k , e k ω < ∞. Let S = √ T p . Then Now, by Fubini's theorem and Parseval's identity, we have Hence, both (a) and (b) are consequences of the inequalities (see [16,Proposition 1.31]) This finishes the proof of the lemma.
Proof. If B ω µ is in L p (D, dλ τ ), then it is easy to see that T µ is bounded on A 2 ω (just use the discrete version in Proposition 6.1 to see that the condition in Theorem 1.1 holds). Therefore, the result is a consequence of Lemma 6.2 since T µ (z) = B ω µ(z). Now we are almost ready for the characterization of Schatten class Toeplitz operators, but we need first some technical lemmas on properties of lattices. We use the notation , z, ζ ∈ D. Proof. Let K be the number of points of the lattice contained in D m (ζ). Due to the Lipschitz condition (B), we have Then As done before, we also have τ (z j ) ≤ C m τ (ζ), if z j ∈ D m (ζ). From this we easily see that for some constant c. Since the sets {D( δ 4 τ (z j ))} are pairwise disjoints, we have Therefore, we get Next, we use the result just proved to decompose any (δ, τ )-lattice into a finite number of "big" separated subsequences. Lemma 6.5. Let τ ∈ L and δ ∈ (0, m τ ). Let m be a positive integer. Any (δ, τ )-lattice {z j } on D can be partitioned into M subsequences such that, if a j and a k are different points in the same subsequence, then d τ (a j , a k ) ≥ 2 m δ.
Proof. Let K be the number given by Lemma 6.4. From the lattice {z j } extract a maximal (2 m δ)subsequence, that is, we select one point ξ 1 in our lattice, and then we continue selecting points ξ n of the lattice so that d τ (ξ n , ξ) ≥ 2 m δ for all previous selected point ξ. We stop once the subsequence is maximal, that is, when all the remaining points x of the lattice satisfy d τ (x, ξ x ) < 2 m δ for some ξ x in the subsequence. With the remaining points of the lattice we extract another maximal (2 m δ)subsequence, and we repeat the process until we get M = K + 1 maximal (2 m δ)-subsequences. If no point of the lattice is left, we are done. On the other hand, if a point ζ in the lattice is left, this means that there are M = K + 1 distinct points x ζ (at least one for each subsequence) in the lattice with d τ (ζ, x ζ ) < 2 m δ, in contradiction with the choice of K from Lemma 6.4. The proof is complete. Now we are ready for the main result of this Section, that characterizes the membership in the Schatten ideals of the Toeplitz operator acting on A 2 ω . For p ≥ 1, the equivalence of T µ being in S p and condition (c) was obtained in [9], where the sufficiency of (c) for 0 < p < 1 was also established. They left open the necessity for 0 < p < 1, a problem that is solved here. Theorem 6.6. Let ω ∈ E and 0 < p < ∞. The following conditions are equivalent: Proof. By Proposition 6.1, the statements (b), (c) and (d) are equivalent. Also, according to Proposition 6.3, it remains to prove that (d) implies (a) for p > 1, and that (a) implies (c) when 0 < p < 1.
Putting this into (6.1) and taking into account that K ζ By [16,Theorem 1.27] this proves that T µ is in S p with T µ Sp µ δ L p (D,dλτ ) .
Next, let 0 < p < 1, and suppose that T µ ∈ S p (A 2 ω ). We will prove that (c) holds. The method for this proof has his roots in previous work of S. Semmes [14] and D. Luecking [10]. Let {z n } be a (δ, τ )-lattice on D with δ ∈ (0, 1 4 m τ ). We want to show that { µ δ (z n )} is in ℓ p . To this end, we fix a large positive integer m ≥ 2 and apply Lemma 6.5 to partition the lattice {z n } into M subsequences such that any two distinct points a j and a k in the same subsequence satisfy d τ (a j , a k ) ≥ 2 m δ. Let {a n } be such a subsequence and consider the measure where χ n denotes the characteristic function of D(δτ (a n )). Since m ≥ 2, the disks D(δτ (a n )) are pairwise disjoints. Since T µ is in S p and 0 ≤ ν ≤ µ, then 0 ≤ T ν ≤ T µ which implies that T ν is also in S p . Moreover, T ν Sp ≤ T µ Sp . Fix an orthonormal basis {e n } for A 2 ω and define an operator B on A 2 ω by B n λ n e n = n λ n k an , where k an are the normalized reproducing kernels of A 2 ω . By Lemma 3.4, the operator B is bounded.
We split the operator T as T = D + E, where D is the diagonal operator on A 2 ω defined by T e n , e n ω f, e n ω e n , f ∈ A 2 ω , and E = T − D. By the triangle inequality, Since D is positive diagonal operator, and B ω ν(a n ) ν δ (a n ) = µ δ (a n ), we have B ω ν(a n ) p ≥ C 1 n µ δ (a n ) p . If n = k, then d τ (a n , a k ) ≥ 2 m δ. Thus, for ξ ∈ D(δτ (a j )), is not difficult to see that either d τ (ξ, a n ) ≥ 2 m−2 δ or d τ (ξ, a k ) ≥ 2 m−2 δ.
With this notation and taking into account (6.4), we have (6.5) E p Sp ≤ n.k:k =n I nk (µ) p .