International Journal of Mathematics and Mathematical Sciences

International Journal of Mathematics and Mathematical Sciences / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 164843 | https://doi.org/10.1155/2011/164843

Roberto C. Raimondo, "Toeplitz Operators on the Bergman Space of Planar Domains with Essentially Radial Symbols", International Journal of Mathematics and Mathematical Sciences, vol. 2011, Article ID 164843, 26 pages, 2011. https://doi.org/10.1155/2011/164843

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

Academic Editor: B. N. Mandal
Received25 Feb 2011
Revised05 Jun 2011
Accepted06 Jun 2011
Published22 Aug 2011

Abstract

We study the problem of the boundedness and compactness of 𝑇𝜙 when 𝜙𝐿2(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 , 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 𝛾𝑗𝛾𝑖= if 𝑖𝑗. We also assume that 𝛾1 is the boundary of the unbounded component of Ω. Let Ω1 be the bounded component of 𝛾1, and Ω𝑗(𝑗=2,,𝑛) the unbounded component of 𝛾𝑗, respectively, so that Ω=𝑛𝑗=1Ω𝑗.

For 𝑑𝜈=(1/𝜋)𝑑𝑥𝑑𝑦, we consider the usual 𝐿2-space 𝐿2(Ω)=𝐿2(Ω,𝑑𝜈). The Bergman space 𝐿2𝑎(Ω,𝑑𝜈), consisting of all holomorphic functions which are 𝐿2-integrable, is a closed subspace of 𝐿2(Ω,𝑑𝜈) with the inner product given by 𝑓,𝑔=Ω𝑓(𝑧)𝑔(𝑧)𝑑𝜈(𝑧)(1.1) for 𝑓,𝑔𝐿2(Ω,𝑑𝜈). The Bergman projection is the orthogonal projection 𝑃𝐿2(Ω,𝑑𝜈)𝐿2𝑎(Ω,𝑑𝜈).(1.2) It is well-known that for any 𝑓𝐿2(Ω,𝑑𝜈), we have 𝑃𝑓(𝑤)=Ω𝑓(𝑧)𝐾Ω(𝑧,𝑤)𝑑𝜈(𝑧),(1.3) where 𝐾Ω is the Bergman reproducing kernel of Ω. For 𝜑𝐿(Ω,𝑑𝜈), the Toeplitz operator 𝑇𝜑𝐿2𝑎(Ω,𝑑𝜈)𝐿2𝑎(Ω,𝑑𝜈) is defined by 𝑇𝜑=𝑃𝑀𝜑, where 𝑀𝜑 is the standard multiplication operator. A simple calculation shows that 𝑇𝜑𝑓(𝑧)=Ω𝜑(𝑤)𝑓(𝑤)𝐾Ω(𝑤,𝑧)𝑑𝜈(𝑤).(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 𝜑𝐿2(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 𝐿2𝑎(Ω,𝑑𝜈). 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, Ω=𝑛𝑗=1Ω𝑗, where Ω1 is the bounded component of 𝛾1, and Ω𝑗(𝑗=2,,𝑛) is the unbounded component of 𝛾𝑗. We use the symbol Δ to indicate the punctured disk {𝑧0<|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 𝐴(𝐿2𝑎(Γ,𝑑𝜈)), we define 𝐴, the Berezin transform of 𝐴, by 𝐴(𝑤)=𝐴𝑘Γ𝑤,𝑘Γ𝑤=Γ𝐴𝑘Γ𝑤(𝑧)𝑘Γ𝑤(𝑧)𝑑𝜈(𝑧),(2.1) where 𝑘Γ𝑤()=𝐾Γ(,𝑤)𝐾Γ(𝑤,𝑤)1/2.

If 𝜑𝐿(Γ), then we indicate with the symbol 𝜑 the Berezin transform of the associated Toeplitz operator 𝑇𝜑, and we have 𝜑(𝑤)=Γ||𝑘𝜑(𝑧)Γ𝑤(||𝑧)2𝑑𝜈(𝑧).(2.2) We remind the reader that it is well known that 𝐴𝒞𝑏(Γ), and we have 𝐴𝐴(𝐿2(Ω)). 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 , 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 𝐷=𝑛𝑖=1𝐷𝑖 where 𝐷1={𝑧|𝑧|<1} and 𝐷𝑗={𝑧|𝑧𝑎𝑗|>𝑟𝑗} for 𝑗=2,,𝑛. Here 𝑎𝑗𝐷1 and 0<𝑟𝑗<1 with |𝑎𝑗𝑎𝑘|>𝑟𝑗+𝑟𝑘 if 𝑗𝑘 and 1|𝑎𝑗|>𝑟𝑗. Before we state the main results of this paper we need to give a few definitions.

Definition 2.1. Let Ω=𝑛𝑖=1Ω𝑖 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,𝑝𝑗Ω[0,1] is a Lipschitz, 𝐶-function,(2)for every 𝑗=2,,𝑛, there exists an open set 𝑊𝑗Ω and an 𝜖𝑗>0 such that 𝑈𝜖𝑗={𝜁Ω𝑟𝑗<|𝜁𝑎𝑗|<𝑟𝑗+𝜖𝑗}, and the support of 𝑝𝑗 is contained in 𝑊𝑗 and 𝑝𝑗(𝜁)=1,𝜁𝑈𝜖𝑗,(2.3)(3)for 𝑗=1, there exists an open set 𝑊1Ω and an 𝜖1>0 such that 𝑈𝜖1={𝜁Ω1𝜖1<|𝜁|<1} and the support of 𝑝1 is contained in 𝑊1 and 𝑝1(𝜁)=1,𝜁𝑈𝜖1,(2.4)(4)for every 𝑗,𝑘=1,,𝑛,𝑊𝑗𝑊𝑘=, the set Ω(𝑛𝑗=1𝑊𝑗) is not empty and the function 𝑝0(𝜁)=1,𝜁𝑛𝑗=1𝑊𝑗𝑐𝑝Ω,0(𝜁)=0,𝜁𝑈𝜖𝑘,𝑘=1,,𝑛,(2.5)(5) for any 𝜁Ω, the following equation: 𝑛𝑘=0𝑝𝑘(𝜁)=1.(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 𝜑Ω=𝑛𝑖=1Ω𝑖 is said to be essentially radial if there exists a conformally equivalent canonical bounded domain 𝐷=𝑛𝑖=1𝐷𝑖, such that if the map ΘΩ𝐷 is the conformal mapping from Ω onto 𝐷, then(1)for every 𝑘=2,,𝑛 and for some 𝜖𝑘>0, we have 𝜑Θ1(𝑧)=𝜑Θ1||𝑧𝑎𝑘||,(2.7) when 𝑧𝑈𝜖𝑘={𝜁Ω𝑟𝑘<|𝜁𝑎𝑘|<𝑟𝑘+𝜖𝑘},(2)for 𝑘=1 and for some 𝜖1>0, we have 𝜑Θ1(𝑧)=𝜑Θ1(|𝑧|),(2.8) when 𝑧𝑈𝜖1={𝜁Ω1𝜖1<|𝜁|<1}.

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 Θ𝑛=1Ω𝑛=1𝐷.

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 Θ𝑛=1Ω𝑛=1𝐷, we always assume that the 𝜕-partition is given on 𝑛=1𝐷 and transferred to 𝑛=1Ω through Θ in the natural way.

At this point, we can state the main result.

Theorem 2.3. Let 𝜑𝐿2(Ω) be an essentially radial function via Θ𝑛=1Ω𝑛=1𝐷, if one defines 𝜑𝑗=𝜑𝑝𝑗, where 𝑗=1,,𝑛 and 𝔓 is a 𝜕-partition for Ω, then the following are equivalent: (1)the operator 𝑇𝜑𝐿2𝑎(Ω,𝑑𝜈)𝐿2𝑎(Ω,𝑑𝜈)(2.9) is bounded (compact).(2)for any 𝑗=1,,𝑛 the sequences 𝛾𝜑𝑗={𝛾𝜑𝑗(𝑚)}𝑚are in (+)(𝑐0(+)) where, by definition, if 𝑗=2,,𝑛, 𝛾𝜑𝑗(𝑚)=𝑟𝑗𝑟𝑗𝜑𝑗Θ1𝑟𝑗(2𝑚+1)/2(𝑚+1)𝑠1/2(𝑚+1)+𝑎𝑗1𝑠2𝑑𝑠,𝑚+,(2.10) and if 𝑗=1𝛾𝜑1(𝑚)=10𝜑1Θ1𝑠1/2(𝑚+1)𝑑𝑠,𝑚+.(2.11)

3. The Structure of 𝐿2𝑎(Ω) and Some Estimates about the Bergman Kernel

From now on, we will assume that Ω=𝑛𝑗=1Ω𝑗 where Ω1={𝑧|𝑧|<1} and Ω𝑗={𝑧|𝑧𝑎𝑗|>𝑟𝑗} for 𝑗=2,,𝑛. Here, 𝑎𝑗Ω1 and 0<𝑟𝑗<1 with |𝑎𝑗𝑎𝑘|>𝑟𝑗+𝑟𝑘 if 𝑗𝑘 and 1|𝑎𝑗|>𝑟𝑗. We will indicate with the symbol Δ0,1 the punctured disk Ω1{0}.

With the symbols 𝐾Ω𝑗(𝑧,𝑤),𝐾Ω(𝑧,𝑤),𝐾Δ(𝑧,𝑤), 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 𝐿𝑝𝑎(Δ0,1)=𝐿𝑝𝑎(Ω1), if 𝑝2, and, for any (𝑧,𝑤)Δ2,𝐾Δ(𝑧,𝑤)=𝐾Ω1(𝑧,𝑤) (see [7, 8]). This fact has an important and simple consequence. In fact, if we consider Δ𝑎,𝑟={𝑧0<|𝑧𝑎|<𝑟} and 𝑂𝑎,𝑟={𝑧|𝑧𝑎|>𝑟}, we can conclude that 𝐾𝑂𝑎,𝑟𝑟(𝑧,𝑤)=2𝑟2(𝑧𝑎)(𝑤𝑎)2,(𝑧,𝑤)𝑂𝑎,𝑟×𝑂𝑎,𝑟.(3.1)

To see this, we use the well-known fact that the reproducing kernel of the unit disk is given by (1𝑧𝑤)2, therefore we have 𝐾Δ0,11(𝑧,𝑤)=1𝑧𝑤2,(𝑧,𝑤)Δ0,1×Δ0,1.(3.2) This implies, by conformal mapping, that the reproducing kernel of Δ𝑎,𝑟 is 𝐾Δ𝑎,𝑟𝑟(𝑧,𝑤)=2𝑟2(𝑧𝑎)(𝑤𝑎)2,(𝑧,𝑤)Δ𝑎,𝑟×Δ𝑎,𝑟.(3.3) Now, we define 𝜑Δ𝑎,𝑟𝑂𝑎,𝑟 by 𝜑(𝑧)=(𝑧𝑎)1𝑟2+𝑎,(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 𝐾𝑂𝑎,𝑟(𝜑(𝑧),𝜑(𝑤))𝜑(𝑧)𝜑(𝑤)=𝐾Δ𝑎,𝑟(𝑧,𝑤)(3.5) to obtain that 𝐾𝑂𝑎,𝑟𝑟(𝑧,𝑤)=2𝑟2(𝑧𝑎)(𝑤𝑎)2,(𝑧,𝑤)𝑂𝑎,𝑟×𝑂𝑎,𝑟.(3.6) Since Ω1=𝑂0,1 and, for 𝑗=2,,𝑛,𝑂𝑎𝑗,𝑟𝑗=Ω𝑗, then the last equations implies that 𝐾Ω11(𝑧,𝑤)=1𝑧𝑤2,𝐾Ω𝑗𝑟(𝑧,𝑤)=2𝑗𝑟2𝑗𝑧𝑎𝑗𝑤𝑎𝑗2(3.7) if 𝑗=2,,𝑛.

We also note that if we define 𝐸Ω(𝑧,𝑤)=𝐾Ω(𝑧,𝑤)𝑛𝑗=1𝐾Ω𝑗(𝑧,𝑤),(3.8) we can prove the following.

Lemma 3.1. (1) 𝐸Ω is conjugate symmetric about z and w. For each 𝑤Ω,𝐸Ω(,𝑤) is conjugate analytic on Ω and 𝐸Ω𝐶(Ω×Ω).
(2) There are neighborhoods 𝑈𝑗 of 𝜕Ω𝑗(𝑗=1,,𝑛) and a constant 𝐶>0 such that 𝑈𝑗𝑈𝑘 is empty if 𝑗𝑘 and ||𝐾Ω(𝑧,𝑤)𝐾Ω𝑗||(𝑧,𝑤)<𝐶,(3.9) for 𝑧Ω and 𝑤𝑈𝑗.
(3) 𝐸Ω𝐿(Ω×Ω).

Proof. (1) Since the Bergman kernels 𝐾Ω and 𝐾Ω𝑗 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 𝐾Ω11(𝑧,𝑤)=1𝑧𝑤2,𝐾Ω𝑗𝑟(𝑧,𝑤)=2𝑗𝑟2𝑗𝑧𝑎𝑗𝑤𝑎𝑗2,(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={𝑧𝑟𝑗<|𝑧𝑎𝑗|<𝑅𝑗}(𝑗=2,,𝑛) and 𝐺1={𝑧𝑅1<|𝑧|<1}, then we have 𝐺𝑗𝑈𝑗, where 𝑈𝑗 is the same as in Lemma 3.1. We also have the following.

Lemma 3.2. There are constants 𝒟>0 and >0 such that (1)for any (𝑧,𝑤)𝐺𝑖×ΩΩ×𝐺𝑖, one has ||𝐾Ω||||𝐾(𝑧,𝑤)<𝐷Ω𝑗||,||𝐾(𝑧,𝑤)Ω𝑗||<||𝐾(𝑧,𝑤)Ω||(𝑧,𝑤)+,(3.11)(2)for any 𝑧Ω, one has 𝐾Ω𝑗(𝑧,𝑧)<𝐾Ω(𝑧,𝑧).

Proof. By the explicit formula of the Bergman kernels 𝐾Ω𝑖, there are constants 𝐶𝑖 and 𝑀𝑖 such that ||𝐾Ω𝑖||(𝑧,𝑤)𝐶𝑖,(3.12) for (𝑧,𝑤)(𝐺𝑖×Ω)(Ω×𝐺𝑖) and ||𝐾Ω𝑖||(𝑧,𝑤)𝑀𝑖(3.13) if (𝑧,𝑤)𝐺𝑖×𝐺𝑖 for 𝑖=1,2,,𝑛. From the last Lemma, it follows that ||𝐾Ω||||𝐾(𝑧,𝑤)Ω𝑖||𝐶(𝑧,w)+𝐶1+𝐶𝑖||𝐾Ω𝑖||,||𝐾(𝑧,𝑤)Ω𝑖||||𝐾(𝑧,𝑤)Ω||+||𝐸(𝑧,𝑤)Ω||+(𝑧,𝑤)𝑗𝑖||𝐾Ω𝑗||<||𝐾(𝑧,𝑤)Ω||(𝑧,𝑤)+𝐸Ω+𝑖𝑗𝑀𝑗,(3.14) whenever (𝑧,𝑤)(𝐺𝑖×Ω)(Ω×𝐺𝑖). If we call 𝒟 the biggest number among {1+𝐶/𝐶𝑗} and we let =𝐸Ω+𝑛𝑗=1𝑀𝑗, 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, Ω=𝑛𝑗=1Ω𝑗. This also suggests that we should look for a representation of the elements of 𝐿2𝑎(Ω) that reflects this fact. For this reason, we give the following.

Definition 3.3. Given Ω=𝑛𝑗=1Ω𝑗 with Ω1={𝑧|𝑧|<1} and Ω𝑗={𝑧|𝑧𝑎𝑗|>𝑟𝑗}, for any 𝑓𝐿2𝑎(Ω), we define 𝑛+1 functions 𝑃0𝑓,𝑃1𝑓,𝑃2𝑓,,𝑃𝑛𝑓 as follows: if 𝑧Ω, then we set, for 𝑗=1, 𝑃11𝑓(𝑧)=2𝜋𝑖̂𝛾1𝑓(𝜁)𝜁𝑧𝑑𝜁,(3.15) for 𝑗=2,3,,𝑛, 𝑃𝑗1𝑓=2𝜋𝑖̂𝛾𝑗𝑓(𝜁)1𝜁𝑧𝑑𝜁2𝜋𝑖̂𝛾𝑗𝑓(𝜁)𝑑𝜁,(3.16) and for 𝑗=0, 𝑃0𝑓=𝑛𝑗=212𝜋𝑖̂𝛾𝑗1𝑓(𝜁)𝑑𝜁𝑧𝑎𝑗,(3.17) where ̂𝛾𝑗(𝑗=1,,𝑛) are the circles which center at 𝑎𝑗(𝑎1=0) and lie in 𝐺𝑗 (see Lemma 3.2), respectively, so that 𝑧 is exterior to ̂𝛾𝑗(𝑗=2,,𝑛) and interior to ̂𝛾1.

It is important that the reader notices that the Cauchy theorem implies that our definition is independent from how we choose ̂𝛾1,,̂𝛾𝑛. 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 𝑓𝐿2𝑎(Ω), one can write it uniquely as 𝑓(𝑧)=𝑛𝑗=1𝑃𝑗𝑓𝑃(𝑧)+0𝑓(𝑧),(3.18) with 𝑃𝑗𝑓𝐿2𝑎(Ω𝑗),𝑃0𝑓𝐿2𝑎(Ω)𝐶(Ω),𝑃𝑘(𝑃𝑗𝑓)=0 if 𝑗𝑘, and moreover, there exists a constant 𝑀1 such that, for 𝑗=0,1,,𝑛, one has 𝑃𝑗𝑓Ω𝑃𝑗𝑓Ω𝑗𝑀1𝑓Ω.(3.19) In particular, if 𝑓𝐿2𝑎(Ω𝑖), then 𝑃𝑖𝑓=𝑓 and 𝑓Ω𝑖𝑀1𝑓Ω,(3.20) for 𝑖=1,,𝑛.

Proof. Let 𝑓 be any function analytic on Ω. For any 𝑧Ω, 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 𝑓(𝑧)=𝑛𝑗=112𝜋𝑖𝛾𝑗𝑓(𝜁)𝜁𝑧𝑑𝜁.(3.21) Let 𝑓𝑗(1𝑧)=2𝜋𝑖𝛾𝑗𝑓(𝜁)𝜁𝑧𝑑𝜁.(3.22) By Cauchy's Formula, the value 𝑓𝑗(𝑧) does not depend on the choice of 𝛾𝑗 if 1𝑗𝑛 and 𝑓(𝑧)=𝑛𝑗𝑓𝑗(𝑧). Of course, each 𝑓𝑗 is well defined for all 𝑧Ω𝑗 and analytic in Ω𝑗. In addition, if 𝑗1, we have that 𝑓𝑗(𝑧)0 as |𝑧|. Writing the Laurent expansion at 𝑎𝑗 of 𝑓𝑗, we have 𝑓1(𝑧)=𝑘=0𝛼1,𝑘𝑧𝑘,(3.23) and, for 𝑗1, 𝑓𝑗(𝑧)=𝑘=1𝛼𝑗,𝑘𝑧𝑎𝑗𝑘,(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: 𝛼𝑗,𝑘=12𝜋𝑖𝛾𝑗𝑓(𝜁)𝜁𝑎𝑗𝑘+1𝑑𝜁,(3.25) where 𝑘0 if 𝑗=1 and 𝑘1 if 𝑗1 and 𝛾𝑗𝐺𝑗,1𝑗𝑛. Moreover, if 𝑓 is holomorphic in some Ω𝑗 and 𝑓(𝑧)0 as |𝑧| when 𝑖1, then 𝛼𝑗𝑘=0 for all 𝑗𝑖 by Cauchy's theorem and, therefore, 𝑓𝑗=0.
Now, we define 𝑃1𝑓=𝑓1 and 𝑃𝑗𝑓(𝑧)=𝑘=2𝛼𝑗𝑘𝑧𝑎𝑗𝑘,(3.26) for 𝑗=2,3,,𝑛 and 𝑃0𝑓(𝑧)=𝑛𝑗=2𝛼𝑗,1𝑧𝑎𝑗1,(3.27) then 𝑓(𝑧)=𝑛𝑖=0𝑃𝑖𝑓(𝑧) for all 𝑧Ω and 𝑃𝑘(𝑃𝑗𝑓)=0 if 0𝑘𝑗0 as we have proved above.
We claim that 𝑓𝐿2𝑎(Ω) implies that 𝑃𝑖𝑓𝐿2𝑎(Ω𝑗) for 𝑗=1,2,,𝑛, respectively. Indeed, since each annulus 𝐺𝑗 is contained in Ω,𝑓𝐿2𝑎(Ω) implies that 𝑓 is an element of 𝐿2𝑎(𝐺𝑖) for all 𝑖=1,2,,𝑛.
For any fixed 𝑖, note that 𝑃𝑗𝑓(0𝑗𝑖) and 𝑃0𝑓𝛼𝑗,1(𝑧𝑎𝑗)1 are analytic on 𝐺𝑖(/Ω𝑖) and lim|𝑧|𝑃𝑗𝑓(𝑧)=0 for 𝑗1. Expanding them as Laurent series, it follows that:(1)If 𝑖=1, then 𝑃𝑗𝑓=+𝑘=1𝛽𝑗𝑘/𝑧𝑘 for 𝑗1,(2)If 𝑖1, then 𝑃𝑗𝑓(𝑧)=+𝑘=0𝛽𝑗𝑘𝑧𝑎𝑖𝑘,(3.28) for 0𝑗𝑖 and 𝑃0𝑓(𝑧)=+𝑘=0𝛽0𝑘𝑧𝑎𝑖𝑘+𝛼𝑖,1𝑧𝑎𝑖.(3.29) It is obvious that, in any case, these series converge uniformly and absolutely on 𝐺𝑖. Observing that each 𝐺𝑖 is an annulus at 𝑎𝑖, we have, by direct computation, that 𝑓,𝑓𝐺𝑖𝑃𝑖𝑓,𝑃𝑖𝑓𝐺𝑖+||𝛼𝑖,1||2ln𝑅𝑖ln𝑟𝑖(3.30) if 𝑖1 and 𝑓,𝑓𝐺1𝑃1𝑓,𝑃1𝑓𝐺1.(3.31) Therefore, for any 𝑖=1,,𝑛, there exists a constant 𝑀 such that 𝑃𝑖𝑓𝐺𝑖𝑓𝐺𝑖𝑓Ω,||𝛼()𝑖,1||𝑀𝑓Ω.() From the definition of 𝑃𝑗𝑓, we derive 𝑃1𝑓2𝐺1=+0||𝛼1𝑘||21𝑅12𝑘+2,𝑃𝑘+1𝑖𝑓2𝐺𝑖=𝑘=2|𝛼|2𝑖𝑘𝑟𝑖2𝑘+2𝑅𝑖2𝑘+2,𝑘+1(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 𝑃𝑖𝑓𝐿2𝑎(Ω𝑖) and 𝑃1𝑓2Ω1=+0||𝛼1𝑘||2,𝑃𝑘+1𝑖𝑓2Ω𝑖=𝑘=2||𝛼1𝑘||2𝑟𝑖2𝑘+2,𝑘+1(3.33) for 𝑖=2,,𝑛. Comparing the expression of 𝑃𝑖𝑓Ω𝑖 with the expression of 𝑃𝑖𝑓𝐺𝑖, it follows that 𝑃𝑖𝑓Ω𝑖<𝑀𝑃𝑖𝑓𝐺𝑖 for some constant 𝑀 for 𝑖=1,,𝑛. Hence, 𝑃𝑖𝑓Ω𝑖<𝑀𝑃𝑖𝑓Ω. Moreover, if we define 𝑀=Max{(𝑧𝑎𝑖)1Ω}, from the inequalities 𝑃𝑖𝑓𝐺𝑖𝑓𝐺𝑖𝑓Ω and |𝛼𝑖,1|𝑀𝑓Ω and the definition of 𝑃0, it follows that 𝑃0𝑓Ω𝑛𝑀𝑀𝑓Ω.
If 𝑓𝐿2𝑎(Ω𝑖) for some 𝑖{1,2,,𝑛}, note that lim𝑓(𝑧)=0 as |𝑧| for 𝑖1, then 𝑓(𝑧)=𝑃𝑖𝑓(𝑧)+𝛼𝑖,1(𝑧𝑎𝑖)1 if 𝑖1 and 𝑃1𝑓=𝑓 if 𝑖=1. For 𝑖1, since 𝑓𝐿2𝑎(Ω𝑖)𝐿2𝑎(Ω) implies that 𝑃𝑖𝑓𝐿2𝑎(Ω𝑖), then 𝛼𝑖,1(𝑧𝑎𝑖)1𝐿2𝑎(Ω𝑖). We must have 𝛼𝑖,1=0 and, consequently, 𝑃0𝑓=0. Hence, in any case, 𝑓𝐿2𝑎(Ω𝑖) implies 𝑓=𝑃𝑖𝑓 and 𝑃𝑗𝑓=0 if 𝑖𝑗, and this remark completes our proof.

Lemma 3.5. If {𝑓𝑛} is a bounded sequence in 𝐿2𝑎(Ω) and 𝑓𝑛0 weakly in 𝐿2𝑎(Ω), then 𝑃𝑗𝑓𝑛0 weakly on 𝐿2𝑎(Ω𝑗) for 𝑗=1,,𝑛 and 𝑃0𝑓𝑛0 uniformly on Ω.

Proof. By the previous Lemma, we know that the linear transformations {𝑃𝑗} are bounded operators, then 𝑓𝑛0 weakly in 𝐿2𝑎(Ω) implies that 𝑃𝑗𝑓𝑛0 weakly on 𝐿2𝑎(Ω𝑗) for 𝑗=1,,𝑛. For the same reason, 𝑃0𝑓𝑛0 weakly in 𝐿2𝑎(Ω) and then 𝑃0𝑓𝑛(𝜁)0 for any 𝜁Ω. Since 𝑃0𝑓𝑚=𝑛𝑖=2𝛼𝑖,1(𝑚)𝜁𝑎𝑖,(3.34) by the estimates given in the last lemma, we have that |𝛼𝑖,1(𝑚)|<𝑀𝑓𝑚Ω. The boundedness of {𝑓𝑚Ω} implies that the family of continuous functions {𝑃0𝑓𝑚} is uniformly bounded and equicontinuous on Ω, then, by Arzela-Ascoli's Theorem, we have that 𝑃0𝑓𝑚0 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 𝜑𝐿2(Ω,𝑑𝜈) for the Toeplitz operator 𝑇𝜑 to be bounded or compact.

Before we state the next Theorem, we remind the reader that 𝐾Ω(𝜁,𝑧)=𝐸Ω(𝜁,𝑧)+𝑛=1𝐾Ω(𝜁,𝑧),(4.1) where 𝐸Ω𝐿(Ω×Ω) and, for all =1,,𝑛, we have 𝐾Ω(𝜁,𝑧)=𝐾Ω(𝜁,𝑧),𝜁,𝑧Ω×Ω,(4.2) where 𝐾Ω is the reproducing kernel of Ω. If we use the symbol 𝐾Ω0 to indicate 𝐸Ω, we can write 𝐾Ω(𝜁,𝑧)=𝑛=0𝐾Ω(𝜁,𝑧).(4.3) We also remind the reader that if 𝐼𝐿2𝑎(Ω)𝐿2𝑎(Ω) is the identity operator, then 𝐼=𝑛=0𝑃,(4.4) where 𝑃𝐿2𝑎(Ω)𝐿2𝑎(Ω) is a bounded operator for all =0,1,,𝑛 with 𝑃𝑓𝐿2𝑎(Ω) if =1,,𝑛 and 𝑃0𝑓𝒞(Ω) and 𝑃𝑘𝑃=0 if 𝑘 (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 𝜑𝐿2(𝐷) be an essentially radial function where 𝐷=𝑛𝑗=1𝐷𝑗 with 𝐷1={𝑧|𝑧|<1} and 𝐷𝑗={𝑧|𝑧𝑎𝑗|>𝑟𝑗} for 𝑗=2,,𝑛. If one defines 𝜑𝑗=𝜑𝑝𝑗 where 𝑗=1,,𝑛 and 𝔓={𝑝0,𝑝1,,𝑝𝑛} is a 𝜕-partition for 𝐷,then the following are equivalent: (1)the operator 𝑇𝜑𝐿2𝑎(𝐷,𝑑𝜈)𝐿2𝑎(𝐷,𝑑𝜈)(4.5) is bounded (compact);(2)for any 𝑗=1,,𝑛, the operators 𝑇𝜑𝑗𝐿2𝑎𝐷𝑗,𝑑𝜈𝐿2𝑎𝐷𝑗,𝑑𝜈(4.6) are bounded (compact).

Proof. Let {𝑝0,𝑝1,,𝑝𝑛} be a partition of the unit on 𝐷=𝑛𝑗=1𝐷𝑗, which is a canonical domain. Now, we notice that for all 𝑓𝐿2(𝐷) and for all 𝑤𝐷, we have the following: 𝑇𝜑𝑓(𝑤)=𝐷𝜑(𝑧)𝑓(𝑧)𝐾𝐷(=𝑧,𝑤)𝑑𝑣(𝑧)𝑛𝑗=0𝐷𝜑(𝑧)𝑓(𝑧)𝐾𝐷𝑗(=𝑧,𝑤)𝑑𝑣(𝑧)𝑛𝑛𝑗=0𝑘=0𝐷𝜑(𝑧)𝑝𝑘(𝑧)𝑓(𝑧)𝐾𝐷𝑗=(𝑧,𝑤)𝑑𝑣(𝑧)𝑛𝑛𝑗=0𝑘=0𝑇𝑗𝑘𝑓(𝑤),(4.7) where, by definition, we have 𝑇𝑗𝑘𝑓(𝑤)=𝐷𝜑(𝑧)𝑝𝑘(𝑧)𝐾𝐷𝑗(𝑧,𝑤)𝑓(𝑧)𝑑𝑣(𝑤)𝑑𝑣(𝑧).(4.8)

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

Proof. We observe that, by definition, we have 𝑇𝑗0𝑓(𝑤)=𝐷𝜑(𝑧)𝑝0(𝑧)𝐾𝐷𝑗(𝑧,𝑤)𝑓(𝑧)𝑑𝑣(𝑧),(4.9) therefore, if we define 1=𝐷||𝜑(𝑧)𝑝0(𝑧)𝐾𝐷𝑗(||𝑧,𝑤)2𝑑𝑣(𝑧)𝑑𝑣(𝑤),(4.10) we have 1=𝐷||𝜑(𝑧)𝑝0||(𝑧)2𝐷||𝐾𝐷𝑗||(𝑧,𝑤)2𝑑𝑣(𝑤)𝑑𝑣(𝑧)𝐷||𝜑(𝑧)𝑝0||(𝑧)2||𝐾𝐷𝑗||(𝑧,𝑧)𝑑𝑣(𝑧)Max𝑧supp(𝑝0)||𝑝0||(𝑧)2𝐾𝐷𝑗(𝑧,𝑧)𝐷||𝜑𝑗||(𝑧)2𝑑𝑣(𝑧)Max𝑧supp(𝑝0)||𝑝0||(𝑧)2𝐾𝐷𝑗(𝑧,𝑧)𝜑2𝐷,2<.(4.11) This implies that for any 𝑡=0,1,,𝑛, 𝑇𝑡0 is Hilbert-Schmidt. Therefore, the operator 𝑛𝑡=0𝑇𝑡0(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 𝑇0𝑘𝑓(𝑤)=𝐷𝜑(𝑧)𝑝𝑘(𝑧)𝐾𝐷0(𝑧,𝑤)𝑓(𝑧)𝑑𝑣(𝑧),(4.13) therefore, if we define 2=𝐷||𝜑(𝑧)𝑝𝑘(𝑧)𝐾𝐷0(||𝑧,𝑤)2𝑑𝑣(𝑧)𝑑𝑣(𝑤),(4.14) we have 2=𝐷||𝜑(𝑧)𝑝0||(𝑧)2||𝐾𝐷0(||𝑧,𝑤)2𝑑𝑣(𝑤)𝑑𝑣(𝑧)Max(𝑧,𝑤)𝐷×𝐷||𝐾𝐷0||(𝑧,𝑤)2𝑣(𝐷)𝐷||𝜑(𝑧)𝑝0||(𝑧)2𝑑𝑣(𝑧)Max(𝑧,𝑤)𝐷×𝐷||𝐾𝐷0||(𝑧,𝑤)2𝑣(𝐷)𝜑2𝐷,2<.(4.15) This implies that for any 𝑡=0,1,,𝑛, 𝑇0𝑡 is Hilbert-Schmidt. Therefore, the following 𝑛𝑡=0𝑇0𝑡(4.16) is Hilbert-Schmidt, and this completes the proof of the claim.

Claim 3. The operator 𝑇𝑖𝑗 is Hilbert-Schmidt if 𝑖𝑗0 and 𝑗,𝑖=1,,𝑛.

Proof. We observe that 𝑇𝑗𝑘𝑓(𝑤)=𝐷𝜑(𝑧)𝑝𝑘(𝑧)𝐾𝐷𝑗(𝑧,𝑤)𝑓(𝑧)𝑑𝑣(𝑤)𝑑𝑣(𝑧).(4.17) To start, we give the following: 𝒩𝑗𝑖(𝑧,𝑤)def=𝜑𝑗(𝑧)𝐾𝐷𝑖(𝑧,𝑤).(4.18) We will show that Fubini theorem and the properties of the 𝜕-partition imply that 𝐷||𝒩𝑗𝑖||(𝑧,𝑤)2𝑑𝑣(𝑤)𝑑𝑣(𝑧)<.(4.19) In fact, we have 𝐷||𝒩𝑗𝑖||(𝑧,𝑤)2=𝐷𝐷||𝒩𝑗𝑖||(𝑧,𝑤)2=𝑑𝑣(𝑤)𝑑𝑣(𝑧)𝐷||𝜑𝑗||(𝑧)2||𝐾𝐷𝑖||(𝑧,𝑤)2=𝑑𝑣(𝑤)𝑑𝑣(𝑧)𝐷||𝜑𝑗||(𝑧)2𝐷||𝐾𝐷𝑖||(𝑧,𝑤)2=𝑑𝑣(𝑤)𝑑𝑣(𝑧)𝐷||𝜑𝑗||(𝑧)2𝐾𝐷𝑖=(𝑧,𝑧)𝑑𝑣(𝑧)𝐷||||𝜑(𝑧)2||𝑝𝑗||(𝑧)2𝐾𝐷𝑖(𝑧,𝑧)𝑑𝑣(𝑧)Max𝑧supp(𝑝𝑗)||𝑝𝑗||(𝑧)2𝐾𝐷𝑖(𝑧,𝑧)𝜑2𝐷,2<.(4.20)
Therefore, we can write that 𝑇𝜑=𝒦+𝑛=1𝑇𝜑,(4.21) where 𝒦 is a compact operator.
We also observe that Lemma 3.4 implies that 𝑇𝜑=𝑛𝑗=0𝑇𝜑𝑃𝑗, and we prove that the operator 𝑇𝜑𝑃𝑗 is compact if 𝑗 and 𝑗,=1,,𝑛.

Proof. In order to simplify the notation, we define the operator 𝑅𝑗,=𝑇𝜑𝑃𝑗=𝐾𝐷𝑀𝜑𝑝𝑃𝑗. To prove our statement, it is enough to prove that if we take a bounded sequence {𝑓𝑛} in 𝐿2(𝐷) such that 𝑓𝑛0 weakly, then we can prove that 𝑅𝑗,𝑓𝑛20. We know that the continuity of 𝑃 implies that 𝑃𝑗𝑓𝑘0 weakly on 𝐻2(𝐷𝑙), and {𝑃𝑗𝑓𝑘𝐷} 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 𝐷. Therefore, the sequence {𝑃𝑗𝑓𝑘} is a normal family of functions. Since 𝑃𝑗𝑓𝑘(𝜁)0 for any 𝜁𝐷𝑗, then 𝑃𝑗𝑓𝑘 converges uniformly on any compact subset of 𝐷𝑗 and consequently on 𝐹=supp(𝑝). To complete the proof, we remind the reader that if we define the operators 𝑄𝐿2(𝐷)𝐿2(𝐷), for =1,2,,𝑛, in this way 𝑄𝑓(𝑧)=𝐷||𝐾𝑓(𝜁)𝐷(||𝜁,𝑧)𝑑𝑣(𝜁).(4.22) It is possible to prove, with the help of Schur's test (see [11] ), that 𝑄 is a bounded operator (see [5]). Now, we observe that ||𝑅𝑗,𝑓𝑘||||𝑃(𝜁)Sup𝑗𝑓𝑘||||𝑄(𝜁)𝜁𝐹𝑗||𝒳𝐹𝜑𝑝𝑠||||,(𝜁)(4.23) then, by using the fact that 𝑄 is bounded, we have 𝑅𝑗,𝑓𝑘𝐷||𝑃Sup𝑗𝑓𝑘||𝜑(𝜁)𝜁𝐹𝑀1𝑝𝑠𝐷,20,(4.24) and this completes the proof of our claim. Notice also that using the same strategy, we can prove that each 𝑇𝜑𝑃0 is compact.

Therefore, we have 𝑇𝜑=𝒦+𝑛=1𝑇𝜑=𝒦+𝐾1+𝑛=1𝑇𝜑𝑃,(4.25) where 𝒦,𝐾1 are compact operators. Since 𝑃2𝑡=𝑃𝑡,𝑃𝑡𝑃𝑠=0 and if 𝑗, then 𝑇𝜑 is bounded (compact) if and only if the operators 𝑇𝜑𝑃 are bounded (compact) operators.

Since 𝑃𝐿2𝑎(𝐷)=𝐿2𝑎(𝐷), then it follows that the operator 𝑇𝜑𝑃 is bounded (compact) if and only if 𝑇𝜑is 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 𝜑𝐿2(𝐷) be an essentially radial function where 𝐷=𝑛𝑗=1𝐷𝑗 with 𝐷1={𝑧|𝑧|<1} and 𝐷𝑗={𝑧|𝑧a𝑗|>𝑟𝑗} for 𝑗=2,,𝑛. If one defines 𝜑𝑗=𝜑𝑝𝑗 where 𝑗=1,,𝑛 and 𝔓={𝑝0,𝑝1,,𝑝𝑛} is a 𝜕-partition for 𝐷then the following are equivalent: (1) the operator 𝑇𝜑𝐿2𝑎(𝐷,𝑑𝜈)𝐿2𝑎(𝐷,𝑑𝜈)(4.26) is bounded (compact).(2) for any 𝑗=1,,𝑛, the sequences 𝛾𝜑𝑗={𝛾𝜑𝑗(𝑚)}𝑚are in (+)(𝑐0(+)) where, by definition, if 𝑗=2,,𝑛𝛾𝜑𝑗(𝑚)=𝑟𝑗𝑟𝑗𝜑𝑗𝑟𝑗(2𝑚+1)/2(𝑚+1)𝑠1/2(𝑚+1)+𝑎𝑗1𝑠2𝑑𝑠𝑚+,(4.27) and for 𝑗=1, 𝛾𝜑1(𝑚)=10𝜑1𝑠1/2(𝑚+1)𝑑𝑠,𝑚+.(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 𝑇𝜑𝑗𝐿2𝐷𝑗,𝑑𝜈𝐿2𝑎𝐷𝑗,𝑑𝜈(4.29) are bounded (compact). If 𝑗=2,,𝑛, we observe that if we consider the following sets Δ0,1={𝑧0<|𝑧𝑎|<1} and Δ𝑎𝑗,𝑟𝑗={𝑧0<|𝑧𝑎𝑗|<𝑟𝑗} and the following maps Δ𝛼0,1Δ𝑎𝑗,𝑟𝑗𝛽𝐷𝑗,(4.30) where 𝛼(𝑧)=𝑎𝑗+𝑟𝑗𝑧 and 𝛽(𝑤)=(𝑤𝑎𝑗)1𝑟2𝑗+𝑎𝑗 and we use Proposition 1.1 in [8], we can claim that 𝑇𝜑𝑗=𝑉1𝛽𝛼𝑇𝜑𝑗𝛽𝛼𝑉𝛽𝛼,(4.31) where 𝑉𝛽𝛼𝐿2(Δ0,1)𝐿2(𝐷𝑗) is an isomorphism of Hilbert spaces. Therefore, 𝑇𝜑𝑗 is bounded (compact) if and only if 𝑇𝜑𝑗𝛽𝛼 is bounded (compact). We also know that this, in turn, is equivalent to the fact that the sequence 𝛾𝜑𝑗=𝛾𝜑𝑗(𝑚)𝑚(4.32) is in (+)(𝑐0(+)), where 𝛾𝜑𝑗(𝑚)=10𝜑𝑗𝑟𝛽𝛼1/2(𝑚+1)𝑑𝑟,𝑚+.(4.33) To complete the proof, we observe that since 𝜑𝑗 is radial and 𝛽𝛼(𝑟)=𝑟1𝑟𝑗+𝑎𝑗 then, after a change of variable, we can rewrite the last integral, and therefore the formula 𝛾𝜑𝑗(𝑚)=𝑟𝑗𝑟𝑗𝜑𝑗𝑟𝑗(2𝑚+1)/2(𝑚+1)𝑠1/2(𝑚+1)+𝑎𝑗1𝑠2𝑑𝑠,𝑚+(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 𝜑𝐿2(Ω) be an essentially radial function via the conformal equivalence ΘΩ𝐷, define 𝜑𝑗=𝜑𝑝𝑗 where 𝑗=1,,𝑛 and 𝔓 is a 𝜕-partition for Ω, then the following conditions are equivalent: (1)the operator 𝑇𝜑𝐿2𝑎(Ω,𝑑𝜈)𝐿2𝑎(Ω,𝑑𝜈)(4.35) is bounded (compact);(2)for any 𝑗=1,,𝑛, the sequences 𝛾𝜑𝑗={𝛾𝜑𝑗(𝑚)}𝑚are in (+)(𝑐0(+)) where, by definition, if 𝑗=2,,𝑛𝛾𝜑𝑗(𝑚)=𝑟𝑗𝑟𝑗𝜑𝑗Θ1𝑟𝑗(2𝑚+1)/2(𝑚+1)𝑠1/2(𝑚+1)+𝑎𝑗1𝑠2𝑑𝑠,𝑚+,(4.36) and for 𝑗=1𝛾𝜑1(𝑚)=10𝜑1Θ1𝑠1/2(𝑚+1)𝑑𝑠,𝑚+.(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 𝐾𝐷(Θ(𝑧),Θ(𝑤))Θ(𝑧)Θ(𝑤)=𝐾Ω(𝑧,𝑤), and the operator 𝑉Θ𝑓=Θ𝑓Θ is an isometry from 𝐿2(𝐷) onto𝐿2(Ω) (see Proposition 1.1 in [8]). In particular, we have 𝑉Θ𝑃𝐷=𝑃Ω𝑉Θ and this implies that 𝑉Θ𝑇𝜑=𝑇𝜑Θ1𝑉Θ. Therefore, the operator 𝑇𝜑 is bounded (compact) if and only if the operator 𝑇𝜑Θ1𝐿2(𝐷,𝑑𝜈)𝐿2𝑎(𝐷,𝑑𝜈) 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 𝑇𝜑𝑗Θ1𝐿2𝑎𝐷𝑗,𝑑𝜈𝐿2𝑎𝐷𝑗,𝑑𝜈(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 𝛾𝜑𝑗={𝛾𝜑𝑗(𝑚)}𝑚 are in (+)(𝑐0(+)) where, by definition, if 𝑗=2,,𝑛, we have 𝛾𝜑𝑗(𝑚)=𝑟𝑗𝑟𝑗𝜑𝑗Θ1𝑟𝑗(2𝑚+1)/2(𝑚+1)𝑠1/2(𝑚+1)+𝑎𝑗1𝑠2𝑑𝑠,𝑚+,(4.39) and for 𝑗=1, 𝛾𝜑1(𝑚)=10𝜑1Θ1𝑠1/2(𝑚+1)𝑑𝑠,𝑚+,(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 𝐵𝜑𝑗(𝑠)=𝑟𝑗𝑠𝑟𝑗𝜑𝑗Θ1𝑟𝑗1/2𝑥1/2+𝑎𝑗1𝑥2𝑑𝑥,(4.41) and for 𝑗=1, we set 𝐵𝜑1(𝑠)=1𝑠𝜑1Θ1𝑥1/2𝑑𝑥.(4.42)

We obtain the following useful theorem.

Theorem 4.4. Let 𝜑𝐿2(Ω) be an essentially radial function via the conformal equivalence ΘΩ𝐷. If one defines 𝜑𝑗=𝜑𝑝𝑗 where 𝑗=1,,𝑛 and 𝔓 is a 𝜕-partition for Ω, then for the operator 𝑇𝜑𝐿2𝑎(Ω,𝑑𝜈)𝐿2𝑎(Ω,𝑑𝜈) the following hold true: (1)if for any 𝑗=1,,𝑛|||𝐵𝜑𝑗|||𝑟(s)=𝑂𝑗𝑠as𝑠𝑟𝑗,(4.43) then 𝑇𝜑 is bounded;(2)if for any 𝑗=1,,𝑛|||𝐵𝜑𝑗|||𝑟(𝑠)=𝑜𝑗𝑠as𝑠𝑟𝑗,(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 𝛾𝜑𝑗={𝛾𝜑𝑗(𝑚)}𝑚 are in (+)(𝑐0(+)) where, by definition, if 𝑗=2,,𝑛, 𝛾𝜑𝑗(𝑚)=𝑟𝑗𝑟𝑗𝜑𝑗Θ1𝑟𝑗(2𝑚+1)/2(𝑚+1)𝑠1/2(𝑚+1)+𝑎𝑗1𝑠2𝑑𝑠𝑚+,(4.45) and for 𝑗=1𝛾𝜑𝑗(𝑚)=10𝜑1Θ1𝑠1/2(𝑚+1)𝑑𝑠𝑚+,(4.46) and, in virtue of [1]'s main result, it is true that 𝛾𝜑𝑗={𝛾𝜑𝑗(𝑚)}𝑚 are in (+) if for any 𝑗=1,,𝑛, |||𝐵𝜑𝑗|||𝑟(𝑠)=𝑂𝑗𝑠as𝑠𝑟𝑗,(4.47) and 𝛾𝜑𝑗={𝛾𝜑𝑗(𝑚)}𝑚 are in 𝑐0(+)) if for any 𝑗=1,,𝑛|||𝐵𝜑𝑗|||𝑟(𝑠)=𝑜𝑗𝑠as𝑠𝑟𝑗.(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 𝜑𝐿2(Ω) be an essentially radial function via the conformal equivalence ΘΩ𝐷. If we define 𝜑𝑗=𝜑𝑝𝑗 where 𝑗=1,,𝑛 and 𝔓 is a 𝜕-partition for Ω and if 𝜑0 a.e. in Ω, then for the operator 𝑇𝜑𝐿2𝑎(Ω,𝑑𝜈)𝐿2𝑎(Ω,𝑑𝜈), the following hold true: (1)𝑇𝜑 is bounded if and only if |||𝐵𝜑𝑗|||𝑟(𝑠)=𝑂𝑗𝑠as𝑠𝑟𝑗,(4.49) for any 𝑗=1,,𝑛,(2)𝑇𝜑 is compact if and only if |||𝐵𝜑𝑗|||𝑟(𝑠)=𝑜𝑗𝑠as𝑠𝑟𝑗,(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 𝑇𝜑𝐿2𝑎(Ω,𝑑𝜈)𝐿2𝑎(Ω,𝑑𝜈) has an essentially radial positive symbol