International Journal of Mathematics and Mathematical Sciences
Volume 2011, Issue 1 164843
Research Article
Open Access

Toeplitz Operators on the Bergman Space of Planar Domains with Essentially Radial Symbols

Roberto C. Raimondo, 

Corresponding Author

Roberto C. Raimondo

Division of Mathematics, Faculty of Statistics, University of Milano-Bicocca, Via Bicocca degli Arcimboldi 8, 20126 Milan, Italy unimib.it

Department of Economics, University of Melbourne, Parkville, VIC 3010, Australia unimelb.edu.au

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First published: 22 August 2011
https://doi.org/10.1155/2011/164843
Citations: 1
Academic Editor: B. N. Mandal

Abstract

We study the problem of the boundedness and compactness of Tϕ when ϕ ∈ L2(Ω) and Ω is a planar domain. We find a necessary and sufficient condition while imposing a condition that generalizes the notion of radial symbol on the disk. We also analyze the relationship between the boundary behavior of the Berezin transform and the compactness of Tϕ.

1. Introduction

Let Ω be a bounded multiply-connected domain in the complex plane ℂ, whose boundary ∂Ω consists of finitely many simple closed smooth analytic curves γj  (j = 1,2, …, n) where γj are positively oriented with respect to Ω and γj∩γi = ∅ if i ≠ j. We also assume that γ1 is the boundary of the unbounded component of ℂ∖Ω. Let Ω1 be the bounded component of ℂ∖γ1, and Ωj  (j = 2, …, n) the unbounded component of ℂ∖γj, respectively, so that .

For dν = (1/π)dxdy, we consider the usual L2-space L2(Ω) = L2(Ω, dν). The Bergman space , consisting of all holomorphic functions which are L2-integrable, is a closed subspace of L2(Ω, dν) with the inner product given by
(1.1)
for f, g ∈ L2(Ω, dν). The Bergman projection is the orthogonal projection
(1.2)
It is well-known that for any f ∈ L2(Ω, dν), we have
(1.3)
where KΩ is the Bergman reproducing kernel of Ω. For φ ∈ L∞(Ω, dν), the Toeplitz operator is defined by Tφ = PMφ, where Mφ is the standard multiplication operator. A simple calculation shows that
(1.4)
For square-integrable symbols, the Toeplitz operator is densely defined but is not necessarily bounded; therefore, the problem of finding necessary and sufficient conditions on the function φ ∈ L2(Ω, dν) for the Toeplitz operators Tφ to be bounded or compact is a natural one, and it has been studied by many authors. Several important results have been established when the symbol has special geometric properties. In fact, in the context of radial symbols on the disk, many papers have been written with quite surprising results (see [1] of Grudsky and Vasilevski, [2] of Zorboska, and [3] of Korenblum and Zhu) showing that operators with unbounded radial symbols can have a very rich structure. In fact, in the case of a continuous symbol, the compactness of the Toeplitz operators depends only on the behavior of the symbol on the boundary of the disk and this is similar to what happens in the Hardy space case, even though in the case of Bergman space, the Toeplitz operator with continuous radial symbol is a compact perturbation of a scalar operator and in the Hardy space case a Toeplitz operator with radial symbol is just a scalar operator. In the case of unbounded radial symbols, a pivotal role is played by the fact that in the Bergman space setting, contrary to the Hardy space setting, there is an additional direction that Grudsky and Vasileski term as inside the domain direction: symbols that are nice with respect to the circular direction may have very complicated behavior in the radial direction. Of course, in the context of arbitrary planar domains, it is not possible to use the notion of radial symbol. We go around this difficulty by making two simple observations. To start, it is necessary to notice that the structure of the Bergman kernel suggests that there is in any planar domain an internal region that we can neglect when we are interested in boundedness and compactness of Toeplitz operators with square integrable symbols, therefore the inside the domain direction counts up to a certain point. The second observation consists in exploiting the geometry of the domain and conformal equivalence in order to partially recover the notion of radial symbol. For these reasons, we study the problem for planar domains when the Toeplitz operator symbols have an almost-radial behavior and, for this class, we give a necessary and sufficient condition for boundedness and compactness. We also address the problem of the characterization of compactness by using the Berezin transform. In fact, under a growth condition for the almost-radial symbol, we show that the Berezin transform vanishes to the boundary if and only if the operator is compact.

The paper is organized as follows. In Section 2, we describe the setting where we work, give the relevant definitions, and state our main result. In Section 3, we collect results about the Bergman kernel for a planar domain and the structure of . In Section 4, we prove the main result and study several important consequences.

2. Preliminaries

Let Ω be the bounded multiply-connected domain given at the beginning of Section 1, that is, , where Ω1 is the bounded component of ℂ∖γ1, and Ωj  (j = 2, …, n) is the unbounded component of ℂ∖γj. We use the symbol Δ to indicate the punctured disk {z ∈ ℂ∣0 < |z| < 1}. Let Γ be any one of the domains Ω, Δ, Ωj  (j = 2, …, n).

We call KΓ(z, w) the reproducing kernel of Γ and we use the symbol kΓ(z, w) to indicate the normalized reproducing kernel, that is, kΓ(z, w) = KΓ(z, w)/KΓ(w, w) 1/2.

For any , we define , the Berezin transform of A, by
(2.1)
where .
If φ ∈ L∞(Γ), then we indicate with the symbol the Berezin transform of the associated Toeplitz operator Tφ, and we have
(2.2)
We remind the reader that it is well known that , and we have . It is possible, in the case of bounded symbols, to give a characterization of compactness using the Berezin transform (see [4, 5]).

We remind the reader that any Ω bounded multiply-connected domain in the complex plane ℂ, whose boundary ∂Ω consists of finitely many simple closed smooth analytic curves γj  (j = 1,2, …, n), is conformally equivalent to a canonical bounded multiply-connected domain whose boundary consists of finitely many circles (see [6]). This means that it is possible to find a conformally equivalent domain where D1 = {z ∈ ℂ : |z| < 1} and Dj = {z ∈ ℂ:|z − aj | > rj} for j = 2, …, n. Here aj ∈ D1 and 0 < rj < 1 with |aj − ak| > rj + rk if j ≠ k and 1 − |aj| > rj. Before we state the main results of this paper we need to give a few definitions.

Definition 2.1. Let be a canonical bounded multiply-connected domain. We say that the set of n + 1 functions 𝔓 = {p0, p1, …, pn} is a ∂-partition for Ω if

  • (1)

    for every j = 0,1, …, n,   pj : Ω → [0,1] is a Lipschitz, C∞-function,

  • (2)

    for every j = 2, …, n, there exists an open set Wj⊂Ω and an ϵj > 0 such that , and the support of pj is contained in Wj and

    (2.3)

  • (3)

    for j = 1, there exists an open set W1⊂Ω and an ϵ1 > 0 such that and the support of p1 is contained in W1 and

(2.4)
  • (4)

    for every j, k = 1, …, n,   Wj∩Wk = ∅, the set is not empty and the function

(2.5)
  • (5)

    for any ζ ∈ Ω, the following equation:

    (2.6)
    holds.

We need to point out two facts about the definition above: (i) that near each connected component of the boundary there is only one function which is different from zero (note that this implies that the function must be equal to 1), and (ii) far away from the boundary only the function p0 is different from zero.

Definition 2.2. A function is said to be essentially radial if there exists a conformally equivalent canonical bounded domain , such that if the map Θ : Ω → D is the conformal mapping from Ω onto D, then

  • (1)

    for every k = 2, …, n and for some ϵk > 0, we have

    (2.7)
    when z∈,

  • (2)

    for k = 1 and for some ϵ1 > 0, we have

    (2.8)
    when z∈.

The reader should note that in the case where it is necessary to stress the use of a specific conformal equivalence, we will say that the map φ is essentially radial via .

Before we proceed, the reader should notice that the definition, in the case of the disk, just says that, when we are near to the boundary, the values depend only on the distance from the center of the disk, so the function is essentially radial. In the general case, to formalize the fact that the values depend essentially on the distance from the boundary, we can simplify our analysis if we use the fact that this type of domain is conformally equivalent to a canonical bounded multiply-connected domain whose boundary consists of finitely many circles. For this type of domain the idea of essentially radial symbol is quite natural. For this reason, we use this simple geometric intuition to give the general definition.

Before we state the main result, we stress that in what follows, when we are working with a general multiply-connected domain and we have a conformal equivalence , we always assume that the ∂-partition is given on and transferred to through Θ in the natural way.

At this point, we can state the main result.

Theorem 2.3. Let φ ∈ L2(Ω) be an essentially radial function via , if one defines φj = φ · pj, where j = 1, …, n and 𝔓 is a ∂-partition for Ω, then the following are equivalent:

  • (1)

    the operator

    (2.9)
    is bounded (compact).

  • (2)

    for any j = 1, …, n the sequences are in ℓ∞(ℤ+)(c0(ℤ+)) where, by definition, if j = 2, …, n,

    (2.10)
    and if j = 1
    (2.11)

3. The Structure of and Some Estimates about the Bergman Kernel

From now on, we will assume that where Ω1 = {z ∈ ℂ:|z | < 1} and Ωj = {z ∈ ℂ:|z − aj | > rj} for j = 2, …, n. Here, aj ∈ Ω1 and 0 < rj < 1 with |aj − ak | > rj + rk if j ≠ k and 1−|aj | > rj. We will indicate with the symbol Δ0,1 the punctured disk Ω1∖{0}.

With the symbols , we denote the Bergman kernel on Ωj,   Ω, and Δ, respectively.

In order to gain more information about the kernel of a planar domain, it is important to remind the reader that for the the punctured disk Δ0,1 and the disk Ω1, we have , if p ≥ 2, and, for any (see [7, 8]). This fact has an important and simple consequence. In fact, if we consider Δa,r = {z ∈ ℂ : 0<|z − a | < r} and Oa,r = {z ∈ ℂ : |z − a| > r}, we can conclude that
(3.1)
To see this, we use the well-known fact that the reproducing kernel of the unit disk is given by , therefore we have
(3.2)
This implies, by conformal mapping, that the reproducing kernel of Δa,r is
(3.3)
Now, we define φ : Δa,r → Oa,r by
(3.4)
and we use the well-known fact that the Bergman kernels of Δa,r and ψ(Δa,r) = Oa,r are related via
(3.5)
to obtain that
(3.6)
Since Ω1 = O0,1 and, for , then the last equations implies that
(3.7)
if j = 2, …, n.
We also note that if we define
(3.8)
we can prove the following.

Lemma 3.1. (1) EΩ is conjugate symmetric about z and w. For each w ∈ Ω,   EΩ(·, w) is conjugate analytic on Ω and .

(2) There are neighborhoods Uj of ∂Ωj  (j = 1, …, n) and a constant C > 0 such that Uj∩Uk is empty if j ≠ k and

(3.9)
for z ∈ Ω and w ∈ Uj.

(3) EΩ ∈ L∞(Ω × Ω).

Proof. (1) Since the Bergman kernels KΩ and have these properties (see [9]), by the definition of EΩ, we get (1).

(2) The proof is given in [7, 8].

(3) Using the fact that

(3.10)
for j = 2, …, n and (1) and (2), we get (3).

We observe that we can choose Rj > rj for j = 2, …, n and R1 < 1 such that Gj = {z : rj<|z − aj | < Rj}  (j = 2, …, n) and G1 = {z : R1<|z | < 1}, then we have , where Uj is the same as in Lemma 3.1. We also have the following.

Lemma 3.2. There are constants 𝒟 > 0 and ℳ > 0 such that

  • (1)

    for any (z, w) ∈ Gi × Ω ∪ Ω × Gi, one has

    (3.11)

  • (2)

    for any z ∈ Ω, one has .

Proof. By the explicit formula of the Bergman kernels , there are constants Ci and Mi such that

(3.12)
for (z, w)∈(Gi × Ω)∪(Ω × Gi) and
(3.13)
if (z, w) ∉ Gi × Gi for i = 1,2, …, n. From the last Lemma, it follows that
(3.14)
whenever (z, w)∈(Gi × Ω)∪(Ω × Gi). If we call 𝒟 the biggest number among {1 + C/Cj} and we let , then we get the first claimed estimate. The proof of (2) can be found in [8, 10].

It is clear from what we wrote so far that we put a strong emphasis on the fact that the domain under analysis Ω is actually the intersection of other domains, that is, . This also suggests that we should look for a representation of the elements of that reflects this fact. For this reason, we give the following.

Definition 3.3. Given with Ω1 = {z ∈ ℂ:|z | < 1} and Ωj = {z ∈ ℂ:|z − aj | > rj}, for any , we define n + 1 functions P0f, P1f, P2f, …, Pnf as follows: if z ∈ Ω, then we set, for j = 1,

(3.15)
for j = 2,3, …, n,
(3.16)
and for j = 0,
(3.17)
where are the circles which center at aj  (a1 = 0) and lie in Gj (see Lemma 3.2), respectively, so that z is exterior to and interior to .

It is important that the reader notices that the Cauchy theorem implies that our definition is independent from how we choose . Moreover, it is important to notice that the domains of the functions P2f, …, Pnf are actually the sets Ω2, …, Ωn. In the next Lemma, we give more information about this representation.

Lemma 3.4. For , one can write it uniquely as

(3.18)
with if j ≠ k, and moreover, there exists a constant M1 such that, for j = 0,1, …, n, one has
(3.19)
In particular, if , then Pif = f and
(3.20)
for i = 1, …, n.

Proof. Let f be any function analytic on Ω. For any z ∈ Ω, let γi  (i = 1, …, n) be the circles which center at ai  (a1 = 0) and lie in Gi, respectively, so that z is exterior to γi  (i = 2, …, n) and interior to γ1. Using Cauchy′s Formula, we can write

(3.21)
Let
(3.22)
By Cauchy′s Formula, the value fj(z) does not depend on the choice of γj if 1 ≤ j ≤ n and . Of course, each fj is well defined for all z ∈ Ωj and analytic in Ωj. In addition, if j ≠ 1, we have that fj(z) → 0 as |z | → ∞. Writing the Laurent expansion at aj of fj, we have
(3.23)
and, for j ≠ 1,
(3.24)
and these series converge to fj uniformly and absolutely on any compact subset of Ωj, respectively. We remark that the coefficients are given by the following formula:
(3.25)
where k ≥ 0 if j = 1 and k ≤ −1 if j ≠ 1 and γj ⊂ Gj,  1 ≤ j ≤ n. Moreover, if f is holomorphic in some Ωj and f(z) → 0 as |z | → ∞ when i ≠ 1, then αjk = 0 for all j ≠ i by Cauchy′s theorem and, therefore, fj = 0.

Now, we define P1f = f1 and

(3.26)
for j = 2,3, …, n and
(3.27)
then for all z ∈ Ω and Pk(Pjf) = 0 if 0 ≠ k ≠ j ≠ 0 as we have proved above.

We claim that implies that for j = 1,2, …, n, respectively. Indeed, since each annulus Gj is contained in implies that f is an element of for all i = 1,2, …, n.

For any fixed i, note that Pjf  (0 ≠ j ≠ i) and P0f − αj,−1 · (z − aj) −1 are analytic on and lim |z|→∞Pjf(z) = 0 for j ≠ 1. Expanding them as Laurent series, it follows that:

  • (1)

    If i = 1, then for j ≠ 1,

  • (2)

    If i ≠ 1, then

    (3.28)
    for 0 ≠ j ≠ i and
    (3.29)

It is obvious that, in any case, these series converge uniformly and absolutely on . Observing that each Gi is an annulus at ai, we have, by direct computation, that
(3.30)
if i ≠ 1 and
(3.31)
Therefore, for any i = 1, …, n, there exists a constant M′ such that
( * )
( * * )
From the definition of Pjf, we derive
(3.32)
for i = 2, …, n. The convergence of these series is guaranteed by the conditions * and **. Since R1 < 1 and ri < Ri, it follows that and
(3.33)
for i = 2, …, n. Comparing the expression of with the expression of , it follows that for some constant M for i = 1, …, n. Hence, . Moreover, if we define M′′ = Max {∥(z − ai) −1∥Ω}, from the inequalities and |αi,−1 | ≤ M′ · ∥f∥Ω and the definition of P0, it follows that ∥P0f∥Ω ≤ n · M′ · M′′ · ∥f∥Ω.

If for some i ∈ {1,2, …, n}, note that lim f(z) = 0 as |z | → ∞ for i ≠ 1, then f(z) = Pif(z) + αi,−1(z − ai) −1 if i ≠ 1 and P1f = f if i = 1. For i ≠ 1, since implies that , then . We must have αi,−1 = 0 and, consequently, P0f = 0. Hence, in any case, implies f = Pif and Pjf = 0 if i ≠ j, and this remark completes our proof.

Lemma 3.5. If {fn} is a bounded sequence in and fn → 0 weakly in , then Pjfn → 0 weakly on for j = 1, …, n and P0fn → 0 uniformly on Ω.

Proof. By the previous Lemma, we know that the linear transformations {Pj} are bounded operators, then fn → 0 weakly in implies that Pjfn → 0 weakly on for j = 1, …, n. For the same reason, P0fn → 0 weakly in and then P0fn(ζ) → 0 for any ζ ∈ Ω. Since

(3.34)
by the estimates given in the last lemma, we have that |αi,−1(m)| < M∥fm∥Ω. The boundedness of {∥fm∥Ω} implies that the family of continuous functions {P0fm} is uniformly bounded and equicontinuous on , then, by Arzela-Ascoli′s Theorem, we have that P0fm → 0 uniformly on Ω.

4. Canonical Multiply-Connected Domains and Essentially Radial Symbols

In this section, we investigate, with the help of the results established in the previous section, necessary and sufficient conditions on the essentially radial function φ ∈ L2(Ω, dν) for the Toeplitz operator Tφ to be bounded or compact.

Before we state the next Theorem, we remind the reader that
(4.1)
where EΩ ∈ L∞(Ω × Ω) and, for all ℓ = 1, …, n, we have
(4.2)
where is the reproducing kernel of Ωℓ. If we use the symbol to indicate EΩ, we can write
(4.3)
We also remind the reader that if is the identity operator, then
(4.4)
where is a bounded operator for all ℓ = 0,1, …, n with if ℓ = 1, …, n and and PkPℓ = 0 if k ≠ ℓ (see Lemma 3.4).

In order to make our notation a little simpler, when we use a kernel operator we will denote it by the name of its kernel function. For example, the Bergman projection will be denoted by the symbol KΩ.

We are now in a position to prove the following result.

Lemma 4.1. Let φ ∈ L2(D) be an essentially radial function where with D1 = {z ∈ ℂ:|z | < 1} and Dj = {z ∈ ℂ:|z − aj | > rj} for j = 2, …, n. If one defines φj = φ · pj where j = 1, …, n and 𝔓 = {p0, p1, …, pn} is a ∂-partition for D,   then the following are equivalent:

  • (1)

    the operator

    (4.5)
    is bounded (compact);

  • (2)

    for any j = 1, …, n, the operators

    (4.6)
    are bounded (compact).

Proof. Let {p0, p1, …, pn} be a partition of the unit on , which is a canonical domain. Now, we notice that for all f ∈ L2(D) and for all w ∈ D, we have the following:

(4.7)
where, by definition, we have
(4.8)

Claim 1. The operator Tj0 is Hilbert-Schmidt for any j = 0,1, …, n.

Proof. We observe that, by definition, we have

(4.9)
therefore, if we define
(4.10)
we have
(4.11)
This implies that for any t = 0,1, …, n, Tt0 is Hilbert-Schmidt. Therefore, the operator
(4.12)
is Hilbert-Schmidt, and this completes the proof of the claim.

Claim 2. The operator T0k is Hilbert-Schmidt for any k = 0,1, …, n.

Proof. We observe that, by definition, we have

(4.13)
therefore, if we define
(4.14)
we have
(4.15)
This implies that for any t = 0,1, …, n, T0t is Hilbert-Schmidt. Therefore, the following
(4.16)
is Hilbert-Schmidt, and this completes the proof of the claim.

Claim 3. The operator Tij is Hilbert-Schmidt if i ≠ j ≠ 0 and j, i = 1, …, n.

Proof. We observe that

(4.17)
To start, we give the following:
(4.18)
We will show that Fubini theorem and the properties of the ∂-partition imply that
(4.19)
In fact, we have
(4.20)

Therefore, we can write that

(4.21)
where 𝒦 is a compact operator.

We also observe that Lemma 3.4 implies that , and we prove that the operator is compact if j ≠ ℓ and j,   ℓ = 1, …, n.

Proof. In order to simplify the notation, we define the operator . To prove our statement, it is enough to prove that if we take a bounded sequence {fn} in L2(D) such that fn → 0 weakly, then we can prove that ∥Rj,ℓfn∥2 → 0. We know that the continuity of Pℓ implies that Pjfk → 0 weakly on H2(Dl), and is bounded by Lemma 3.5. Since it is a sequence of holomorphic functions, we know that {Pjfk} is uniformly bounded on any compact subset of Dℓ. Therefore, the sequence {Pjfk} is a normal family of functions. Since Pjfk(ζ) → 0 for any ζ ∈ Dj, then Pjfk converges uniformly on any compact subset of Dj and consequently on F = supp (pℓ). To complete the proof, we remind the reader that if we define the operators Qℓ : L2(D) → L2(D), for ℓ = 1,2, …, n, in this way

(4.22)
It is possible to prove, with the help of Schur′s test (see [11] ), that Qℓ is a bounded operator (see [5]). Now, we observe that
(4.23)
then, by using the fact that Qℓ is bounded, we have
(4.24)
and this completes the proof of our claim. Notice also that using the same strategy, we can prove that each is compact.

Therefore, we have
(4.25)
where 𝒦, K1 are compact operators. Since and if j ≠ ℓ, then Tφ is bounded (compact) if and only if the operators are bounded (compact) operators.

Since , then it follows that the operator is bounded (compact) if and only if is bounded (compact).

We are finally, with the help of [1]′s main result, in a position to prove the main result of this paper.

Theorem 4.2. Let φ ∈ L2(D) be an essentially radial function where with D1 = {z ∈ ℂ:|z | < 1} and Dj = {z ∈ ℂ:|z − aj | > rj} for j = 2, …, n. If one defines φj = φ · pj where j = 1, …, n and 𝔓 = {p0, p1, …, pn} is a ∂-partition for D  then the following are equivalent:

  • (1)

    the operator

    (4.26)
    is bounded (compact).

  • (2)

    for any j = 1, …, n, the sequences are in ℓ∞(ℤ+)(c0(ℤ+)) where, by definition, if j = 2, …, n

    (4.27)
    and for j = 1,
    (4.28)

Proof. In the previous theorem, we proved that the operator under examination is bounded (compact) if and only if for any j = 1, …, n the operators

(4.29)
are bounded (compact). If j = 2, …, n, we observe that if we consider the following sets Δ0,1 = {z ∈ ℂ : 0<|z − a | < 1} and and the following maps
(4.30)
where α(z) = aj + rjz and and we use Proposition 1.1 in [8], we can claim that
(4.31)
where Vβ∘α : L2(Δ0,1) → L2(Dj) is an isomorphism of Hilbert spaces. Therefore, is bounded (compact) if and only if is bounded (compact). We also know that this, in turn, is equivalent to the fact that the sequence
(4.32)
is in ℓ∞(ℤ+)(c0(ℤ+)), where
(4.33)
To complete the proof, we observe that since φj is radial and β∘α(r) = r−1rj + aj then, after a change of variable, we can rewrite the last integral, and therefore the formula
(4.34)
must hold for any j = 2, …, n. The case j = 1 is immediate.

Now, we can prove the following.

Theorem 4.3. Let φ ∈ L2(Ω) be an essentially radial function via the conformal equivalence Θ : Ω → D, define φj = φ · pj where j = 1, …, n and 𝔓 is a ∂-partition for Ω, then the following conditions are equivalent:

  • (1)

    the operator

    (4.35)
    is bounded (compact);

  • (2)

    for any j = 1, …, n, the sequences are in ℓ∞(ℤ+)(c0(ℤ+)) where, by definition, if j = 2, …, n

    (4.36)
    and for j = 1
    (4.37)

Proof. We know that Ω is a regular domain, and therefore if Θ is a conformal mapping from Ω onto D then the Bergman kernels of Ω and Θ(Ω) = D, are related via , and the operator VΘf = Θ′ · f∘Θ is an isometry from L2(D) ontoL2(Ω) (see Proposition 1.1 in [8]). In particular, we have VΘPD = PΩVΘ and this implies that . Therefore, the operator Tφ is bounded (compact) if and only if the operator is bounded (compact). In the previous theorem we proved that the operator in exam is bounded (compact) if and only if for any j = 1, …, n the operators

(4.38)
are bounded (compact). Hence, we can conclude that the operator is bounded (compact) if and only if for any j = 1, …, n the sequences are in ℓ∞(ℤ+)(c0(ℤ+)) where, by definition, if j = 2, …, n, we have
(4.39)
and for j = 1,
(4.40)
and this completes the proof.

We now introduce a set of functions that will allow us to further explore the structure of Toeplitz operators with radial-like symbols. For j = 2, …, n, we define
(4.41)
and for j = 1, we set
(4.42)

We obtain the following useful theorem.

Theorem 4.4. Let φ ∈ L2(Ω) be an essentially radial function via the conformal equivalence Θ : Ω → D. If one defines φj = φ · pj where j = 1, …, n and 𝔓 is a ∂-partition for Ω, then for the operator the following hold true:

  • (1)

    if for any j = 1, …, n

    (4.43)
    then Tφ is bounded;

  • (2)

    if for any j = 1, …, n

    (4.44)
    then Tφ is compact.

Proof. To prove the first, we observe that our main theorem implies that the boundedness (compactness) of the operator is equivalent to the fact that for any j = 1, …, n the sequences are in ℓ∞(ℤ+)(c0(ℤ+)) where, by definition, if j = 2, …, n,

(4.45)
and for j = 1
(4.46)
and, in virtue of [1]′s main result, it is true that are in ℓ∞(ℤ+) if for any j = 1, …, n,
(4.47)
and are in c0(ℤ+)) if for any j = 1, …, n
(4.48)

It is also useful to observe that in the case of a positive symbol, we can prove that the condition above is necessary and sufficient. In fact (see [1]), we have the following.

Theorem 4.5. Let φ ∈ L2(Ω) be an essentially radial function via the conformal equivalence Θ : Ω → D. If we define φj = φ · pj where j = 1, …, n and 𝔓 is a ∂-partition for Ω and if φ ≥ 0 a.e. in Ω, then for the operator , the following hold true:

  • (1)

    Tφ is bounded if and only if

    (4.49)
    for any j = 1, …, n,

  • (2)

    Tφ is compact if and only if

    (4.50)
    for any j = 1, …, n.

Proof. The proof is an immediate consequence of Theorem 3.5 in [1] and the theorem above.

There are a few useful observations that we can make at this point. If the Toeplitz operator has an essentially radial positive symbol φ ≥ 0 such that for some ℓ = 1, …, n, the following
(4.51)
holds, then the operator Tφ is unbounded. Moreover, if Tφ is bounded and the symbol is an unbounded essentially radial function, then it must be true that around any ∂Ωℓ, the symbol has an oscillating behavior.
In order to present an application, we consider a family of examples. Let us consider the case where with Ω1 = {z ∈ ℂ:|z | < 1} and Ωj = {z ∈ ℂ:|z − aj | > rj} for j = 2, …, n. Let φ ∈ L2(Ω) be a function that can be written in the following way:
(4.52)
where, for any ℓ = 1,2, …, n,   φ(ℓ) is radial, that is, φ(ℓ) = φ(ℓ)(|z − aℓ|) and satisfies
(4.53)
if ℓ = 2, …, n and
(4.54)
if ℓ = 1. As a consequence of our results, we can conclude that
  • (1)

    Tφ is bounded if there exists a constant 𝒞1 such that for any j = 2, …, n,

    (4.55)
    for any j = 1,

  • (2)

    Tφ is compact if for any j = 2, …, n

    (4.56)
    for j = 1.

It is also possible to show that the sufficient conditions may fail, but the operator is still bounded or even compact. In fact, we can show that given any planar bounded multiply-connected domain Ω, whose boundary ∂Ω consists of finitely many simple closed smooth analytic curves, there exist unbounded functions φ ∈ L2(Ω) such that Tφ is compact even when the sufficient conditions are not satisfied. To prove this claim, we observe that for the domain Ω there exists a conformally equivalent domain where D1 = {z ∈ ℂ:|z | < 1} and Dj = {z ∈ ℂ:|z − aj | > rj} for j = 2, …, n where aj ∈ D1 and 0 < rj < 1 with |aj − ak | > rj + rk if j ≠ k and 1−|aj | > rj. If we denote with the symbol
(4.57)
the conformal equivalence between Ω and D, then we can define, on Ω, the map
(4.58)
where, for any ℓ, we have
(4.59)
where bℓ, aℓ ∈ (0, ∞). It is very easy to see that if we denote with
(4.60)
then on the set of parameters 𝒬1 × 𝒬2 ⋯ ×𝒬n, the operator is bounded and compact.
In the last part of this paper, we concentrate on the relationship between compact operators and the Berezin transform. We remind the reader that given a Toeplitz operator for any Tϕ on , we define , the Berezin transform of Tϕ, by
(4.61)
where kw(·) = K(·, w)K(w, w) −1/2. It is quite simple to show that if an operator is compact, then , the Berezin transform of A, must vanish at the boundary. However, it is possible to show (see [12]) that there are bounded operators which are not compact but whose Berezin transforms vanish at the boundary. In a beautiful paper, Axler and Zheng have proved (see [4]) that if D is the disk, , where φi,k ∈ L∞(D), then S is compact if and only if its Berezin transform vanishes at the boundary of the disk. Their fundamental result has been extended in several directions, in particular when Ω is a general smoothly bounded multiply-connected planar domain [5].

So far, we have characterized the boundedness and compactness of the operator Tφ with the help of the sequences . However, we did not so far try to characterize the compactness in terms of the Berezin transform. In the next theorem, under a certain condition, we will show that the Berezin transform characterization of compactness still holds in this context.

Before we state and prove the next result, we would like to say a few words about the intuition behind it. In the case of the disk, it is possible to show that when the operator is radial then its Berezin transform has a very special form. In fact, if φ : 𝔻 → ℂ is radial then . Therefore to show that the vanishing of the Berezin Transform implies compactness is equivalent, given that Tφ is diagonal, to show that lim |z|→1(1 − |z|2) 2∑(n + 1)〈Tφen, en〉|z|2n = 0 implies lim n→∞〈Tφen, en〉 = 0. Korenblum and Zhu realized this in the their seminal paper [3] and, along this line, more was discovered by Zorboska (see [2]) and Grudsky and Vasilevski (see [1]). In the case of a multiply-connected domain, it is not possible to write things so neatly; however, we can exploit our estimates near the boundary to use similar arguments. This is the content of what we prove in the following.

Theorem 4.6. Let φ ∈ L2(D) be an essentially radial function where with D1 = {z ∈ ℂ:|z | < 1} and Dj = {z ∈ ℂ:|z − aj | > rj} for j = 2, …, n. If one defines φj = φ · pj where j = 1, …, n and 𝔓 = {p0, p1, …, pn} is a ∂-partition for D. Let us assume that is in ℓ∞(ℤ+) and that there is a constant 𝒞2 such that for j = 2, …, n,

(4.62)
and for j = 1
(4.63)
then the operator is compact if and only if
(4.64)

Proof. We know that the operator is bounded if and only if for any j = 1, …, n the operators are bounded. Since we assume that is in ℓ∞(ℤ+), then we can conclude that Tφ is bounded. As we have done before if we fix j = 2, …, n, by using with Δ0,1 = {z ∈ ℂ : 0<|z − a | < 1} and , and the maps where α(z) = aj + rjz and , we can claim that where Vβ∘α : L2(Δ0,1) → L2(Dj) is an isomorphism of Hilbert spaces. Therefore, is compact if and only if is compact. We also know that this, in turn, is equivalent to the vanishing of the Berezin transform if the function

(4.65)
is bounded. Since φj is radial and β∘α(t) = t−1rj + aj, then, after a change of variable under the sign of integral, we can rewrite the last integral, and therefore we obtain the formula
(4.66)
Moreover, if we define τ = s−1rj + aj, we can write
(4.67)
Therefore, if we assume that this function is bounded, we can conclude (see [2]) that from , it follows that is compact for j = 2, …, n. Therefore, we can infer that is compact. We also observe that
(4.68)
if and only if . To prove this fact, we observe that, by definition, we have
(4.69)
where
(4.70)
Let us take (β∘α)−1 : Dj → Δ0,1. Since (JR(β∘α)−1)(w) is |((β∘α)−1) ′(w)|2 and there exists ζ ∈ Dj such that (β∘α)(z) = ζ, we obtain
(4.71)
where . Given the relationship between
(4.72)
and , we obtain
(4.73)
Therefore, it follows that . The case j = 1 is immediate.

Hence, we observe that from what we have proved so far, we can infer with the help of Lemmas 3.1 and 3.2 in Section 2 that, under the stated condition, if

(4.74)
then Tφ is a compact operator. To complete the proof, we observe that the compactness of implies the vanishing of the Berezin transform since kw converges weakly and uniformly to zero as w → ∂D.

Finally, we also observe that as a simple consequence, we obtain the following.

Corollary 4.7. Let φ ∈ L2(Ω) be an essentially radial function via the conformal equivalence Θ : Ω → D. If one defines φj = φ · pj where j = 1, …, n and 𝔓 is a ∂-partition for Ω. Let us assume that is in ℓ∞(ℤ+) and that there is a constant C3 such that for j = 2, …, n,

(4.75)
and for j = 1
(4.76)
then the operator is compact if and only if
(4.77)

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