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

Abstract

We study the problem of the boundedness and compactness of 𝑇_𝜙 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 𝑇_𝜙.

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 𝛾_𝑗(𝑗=1,2,…,𝑛) where 𝛾_𝑗 are positively oriented with respect to Ω and 𝛾_𝑗∩𝛾_𝑖=∅ 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 Ω_𝑗(𝑗=2,…,𝑛) the unbounded component of ℂ⧵𝛾_𝑗, respectively, so that Ω=∩^𝑛𝑗=1Ω_𝑗.

For 𝑑𝜈=(1/𝜋)𝑑𝑥𝑑𝑦, we consider the usual 𝐿²-space 𝐿²(Ω)=𝐿²(Ω,𝑑𝜈). The Bergman space 𝐿^2𝑎(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,𝑔∈𝐿²(Ω,𝑑𝜈). The Bergman projection is the orthogonal projection 𝑃∶𝐿²(Ω,𝑑𝜈)⟶𝐿^2𝑎(Ω,𝑑𝜈).(1.2) It is well-known that for any 𝑓∈𝐿²(Ω,𝑑𝜈), we have 𝑃𝑓(𝑤)=_Ω𝑓(𝑧)𝐾^Ω(𝑧,𝑤)𝑑𝜈(𝑧),(1.3) where 𝐾^Ω is the Bergman reproducing kernel of Ω. For 𝜑∈𝐿^∞(Ω,𝑑𝜈), the Toeplitz operator 𝑇_𝜑∶𝐿^2𝑎(Ω,𝑑𝜈)→𝐿^2𝑎(Ω,𝑑𝜈) is defined by 𝑇_𝜑=𝑃𝑀_𝜑, where 𝑀_𝜑 is the standard multiplication operator. A simple calculation shows that 𝑇_𝜑𝑓(𝑧)=_Ω𝜑(𝑤)𝑓(𝑤)𝐾^Ω(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 𝜑∈𝐿²(Ω,𝑑𝜈) for the Toeplitz operators 𝑇_𝜑 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 𝐿^2𝑎(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=1Ω_𝑗, where Ω₁ is the bounded component of ℂ⧵𝛾₁, and Ω_𝑗(𝑗=2,…,𝑛) is the unbounded component of ℂ⧵𝛾_𝑗. 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,Ω_𝑗(𝑗=2,…,𝑛).

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

For any 𝐴∈ℬ(𝐿^2𝑎(Γ,𝑑𝜈)), we define 𝐴, the Berezin transform of 𝐴, by 𝐴(𝑤)=⟨𝐴𝑘^Γ𝑤,𝑘^Γ𝑤⟩=_Γ𝐴𝑘^Γ𝑤(𝑧)𝑘^Γ𝑤(𝑧)𝑑𝜈(𝑧),(2.1) where 𝑘^Γ𝑤(⋅)=𝐾^Γ(⋅,𝑤)𝐾^Γ(𝑤,𝑤)^−1/2.

If 𝜑∈𝐿^∞(Γ), then we indicate with the symbol 𝜑 the Berezin transform of the associated Toeplitz operator 𝑇_𝜑, and we have 𝜑(𝑤)=_Γ𝜑(𝑧)||𝑘^Γ𝑤(𝑧)||²𝑑𝜈(𝑧).(2.2) We remind the reader that it is well known that 𝐴∈𝒞^∞𝑏(Γ), and we have ‖𝐴‖_∞≤‖𝐴‖_ℬ(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 ℂ, whose boundary 𝜕Ω consists of finitely many simple closed smooth analytic curves 𝛾_𝑗(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=1𝐷_𝑖 where 𝐷₁={𝑧∈ℂ∶|𝑧|<1} and 𝐷_𝑗={𝑧∈ℂ∶|𝑧−𝑎_𝑗|>𝑟_𝑗} for 𝑗=2,…,𝑛. Here 𝑎_𝑗∈𝐷₁ and 0<𝑟_𝑗<1 with |𝑎_𝑗−𝑎_𝑘|>𝑟_𝑗+𝑟_𝑘 if 𝑗≠𝑘 and 1−|𝑎_𝑗|>𝑟_𝑗. Before we state the main results of this paper we need to give a few definitions.

Definition 2.1. Let Ω=∩^𝑛𝑖=1Ω_𝑖 be a canonical bounded multiply-connected domain. We say that the set of 𝑛+1 functions 𝔓={𝑝₀,𝑝₁,…,𝑝_𝑛} is a 𝜕-partition for Ω if(1)for every 𝑗=0,1,…,𝑛,𝑝_𝑗∶Ω→[0,1] is a Lipschitz, 𝐶^∞-function,(2)for every 𝑗=2,…,𝑛, there exists an open set 𝑊_𝑗⊂Ω and an 𝜖_𝑗>0 such that 𝑈_𝜖𝑗={𝜁∈Ω∶𝑟_𝑗<|𝜁−𝑎_𝑗|<𝑟_𝑗+𝜖_𝑗}, and the support of 𝑝_𝑗 is contained in 𝑊_𝑗 and 𝑝_𝑗(𝜁)=1,∀𝜁∈𝑈_𝜖𝑗,(2.3)(3)for 𝑗=1, there exists an open set 𝑊₁⊂Ω and an 𝜖₁>0 such that 𝑈_𝜖1={𝜁∈Ω∶1−𝜖₁<|𝜁|<1} and the support of 𝑝₁ is contained in 𝑊₁ and 𝑝₁(𝜁)=1,∀𝜁∈𝑈_𝜖1,(2.4)(4)for every 𝑗,𝑘=1,…,𝑛,𝑊_𝑗∩𝑊_𝑘=∅, the set Ω⧵(⋃^𝑛𝑗=1𝑊_𝑗) is not empty and the function 𝑝₀(𝜁)=1,∀𝜁∈^𝑛_𝑗=1𝑊_𝑗^𝑐𝑝∩Ω,₀(𝜁)=0,∀𝜁∈𝑈_𝜖𝑘,𝑘=1,…,𝑛,(2.5)(5) for any 𝜁∈Ω, the following equation: ^𝑛_𝑘=0𝑝_𝑘(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 𝜑∶Ω=∩^𝑛𝑖=1Ω_𝑖→ℂ is said to be essentially radial if there exists a conformally equivalent canonical bounded domain 𝐷=∩^𝑛𝑖=1𝐷_𝑖, 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 𝜖_𝑘>0, we have 𝜑∘Θ⁻¹(𝑧)=𝜑∘Θ⁻¹||𝑧−𝑎_𝑘||,(2.7) when 𝑧∈𝑈_𝜖𝑘={𝜁∈Ω∶𝑟_𝑘<|𝜁−𝑎_𝑘|<𝑟_𝑘+𝜖_𝑘},(2)for 𝑘=1 and for some 𝜖₁>0, we have 𝜑∘Θ⁻¹(𝑧)=𝜑∘Θ⁻¹(|𝑧|),(2.8) when 𝑧∈𝑈_𝜖1={𝜁∈Ω∶1−𝜖₁<|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 Θ∶∩^𝑛ℓ=1Ω_ℓ→∩^𝑛ℓ=1𝐷_ℓ.

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 Θ∶∩^𝑛ℓ=1Ω_ℓ→∩^𝑛ℓ=1𝐷_ℓ, we always assume that the 𝜕-partition is given on ∩^𝑛ℓ=1𝐷_ℓ and transferred to ∩^𝑛ℓ=1Ω_ℓ 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 Θ∶∩^𝑛ℓ=1Ω_ℓ→∩^𝑛ℓ=1𝐷_ℓ, if one defines 𝜑_𝑗=𝜑⋅𝑝_𝑗, where 𝑗=1,…,𝑛 and 𝔓 is a 𝜕-partition for Ω, then the following are equivalent: (1)the operator 𝑇_𝜑∶𝐿^2𝑎(Ω,𝑑𝜈)⟶𝐿^2𝑎(Ω,𝑑𝜈)(2.9) is bounded (compact).(2)for any 𝑗=1,…,𝑛 the sequences 𝛾_𝜑𝑗={𝛾_𝜑𝑗(𝑚)}_𝑚∈ℕare in ℓ_∞(ℤ₊)(𝑐₀(ℤ₊)) where, by definition, if 𝑗=2,…,𝑛, 𝛾_𝜑𝑗(𝑚)=𝑟_𝑗^∞𝑟_𝑗𝜑_𝑗∘Θ⁻¹𝑟^{𝑗(2𝑚+1)/2(𝑚+1)}𝑠^1/2(𝑚+1)+𝑎_𝑗1𝑠²𝑑𝑠,∀𝑚∈ℤ₊,(2.10) and if 𝑗=1𝛾_𝜑1(𝑚)=¹⁰𝜑₁∘Θ⁻¹𝑠^1/2(𝑚+1)𝑑𝑠,∀𝑚∈ℤ₊.(2.11)

3. The Structure of 𝐿^2𝑎(Ω) and Some Estimates about the Bergman Kernel

From now on, we will assume that Ω=∩^𝑛𝑗=1Ω_𝑗 where Ω₁={𝑧∈ℂ∶|𝑧|<1} and Ω_𝑗={𝑧∈ℂ∶|𝑧−𝑎_𝑗|>𝑟_𝑗} for 𝑗=2,…,𝑛. Here, 𝑎_𝑗∈Ω₁ and 0<𝑟_𝑗<1 with |𝑎_𝑗−𝑎_𝑘|>𝑟_𝑗+𝑟_𝑘 if 𝑗≠𝑘 and 1−|𝑎_𝑗|>𝑟_𝑗. We will indicate with the symbol Δ_0,1 the punctured disk Ω₁⧵{0}.

With the symbols 𝐾^Ω𝑗(𝑧,𝑤),𝐾^Ω(𝑧,𝑤),𝐾^Δ(𝑧,𝑤), we denote the Bergman kernel on Ω_𝑗,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 Δ_0,1 and the disk Ω₁, we have 𝐿^𝑝𝑎(Δ_0,1)=𝐿^𝑝𝑎(Ω₁), if 𝑝≥2, and, for any (𝑧,𝑤)∈Δ²,𝐾^Δ(𝑧,𝑤)=𝐾^Ω1(𝑧,𝑤) (see [7, 8]). This fact has an important and simple consequence. In fact, if we consider Δ_𝑎,𝑟={𝑧∈ℂ∶0<|𝑧−𝑎|<𝑟} and 𝑂_𝑎,𝑟={𝑧∈ℂ∶|𝑧−𝑎|>𝑟}, we can conclude that 𝐾^{𝑂𝑎,𝑟}(𝑧,𝑤)=𝑟²𝑟²−(𝑧−𝑎)⋅(𝑤−𝑎)²,∀(𝑧,𝑤)∈𝑂_𝑎,𝑟×𝑂_𝑎,𝑟.(3.1)

To see this, we use the well-known fact that the reproducing kernel of the unit disk is given by (1−𝑧𝑤)⁻², therefore we have 𝐾^Δ0,1(𝑧,𝑤)=11−𝑧⋅𝑤²,∀(𝑧,𝑤)∈Δ_0,1×Δ_0,1.(3.2) This implies, by conformal mapping, that the reproducing kernel of Δ_𝑎,𝑟 is 𝐾^Δ𝑎,𝑟(𝑧,𝑤)=𝑟²𝑟²−(𝑧−𝑎)⋅(𝑤−𝑎)²,∀(𝑧,𝑤)∈Δ_𝑎,𝑟×Δ_𝑎,𝑟.(3.3) Now, we define 𝜑∶Δ_𝑎,𝑟→𝑂_𝑎,𝑟 by 𝜑(𝑧)=(𝑧−𝑎)⁻¹𝑟²+𝑎,(3.4) and we use the well-known fact that the Bergman kernels of Δ_𝑎,𝑟 and 𝜓(Δ_𝑎,𝑟)=𝑂_𝑎,𝑟 are related via 𝐾^{𝑂𝑎,𝑟}(𝜑(𝑧),𝜑(𝑤))𝜑^(𝑧)𝜑^(𝑤)=𝐾^Δ𝑎,𝑟(𝑧,𝑤)(3.5) to obtain that 𝐾^{𝑂𝑎,𝑟}(𝑧,𝑤)=𝑟²𝑟²−(𝑧−𝑎)⋅(𝑤−𝑎)²,∀(𝑧,𝑤)∈𝑂_𝑎,𝑟×𝑂_𝑎,𝑟.(3.6) Since Ω₁=𝑂_0,1 and, for 𝑗=2,…,𝑛,𝑂_{𝑎𝑗,𝑟𝑗}=Ω_𝑗, then the last equations implies that 𝐾^Ω1(𝑧,𝑤)=11−𝑧⋅𝑤²𝐾_Ω𝑗,(𝑧,𝑤)=𝑟^2𝑗𝑟^2𝑗−𝑧−𝑎_𝑗⋅𝑤−𝑎_𝑗²(3.7) if 𝑗=2,…,𝑛.

We also note that if we define 𝐸^Ω(𝑧,𝑤)=𝐾^Ω(𝑧,𝑤)−^𝑛_𝑗=1𝐾^Ω𝑗(𝑧,𝑤),(3.8) we can prove the following.

Lemma 3.1. (1) 𝐸^Ω is conjugate symmetric about z and w. For each 𝑤∈Ω,𝐸^Ω(⋅,𝑤) is conjugate analytic on Ω and 𝐸^Ω∈𝐶^∞(Ω×Ω).
(2) There are neighborhoods 𝑈_𝑗 of 𝜕Ω_𝑗(𝑗=1,…,𝑛) and a constant 𝐶>0 such that 𝑈_𝑗∩𝑈_𝑘 is empty if 𝑗≠𝑘 and ||𝐾^Ω(𝑧,𝑤)−𝐾^Ω𝑗(𝑧,𝑤)||<𝐶,(3.9) for 𝑧∈Ω and 𝑤∈𝑈_𝑗.
(3) 𝐸^Ω∈𝐿^∞(Ω×Ω).

Proof. (1) Since the Bergman kernels 𝐾^Ω and 𝐾^Ω𝑗 have these properties (see [9]), by the definition of 𝐸^Ω, we get (1).
(2) The proof is given in [7, 8].
(3) Using the fact that 𝐾^Ω1(𝑧,𝑤)=11−𝑧⋅𝑤²𝐾_Ω𝑗,(𝑧,𝑤)=𝑟^2𝑗𝑟^2𝑗−𝑧−𝑎_𝑗⋅𝑤−𝑎_𝑗²,(3.10) for 𝑗=2,...,𝑛 and (1) and (2), we get (3).

We observe that we can choose 𝑅_𝑗>𝑟_𝑗 for 𝑗=2,…,𝑛 and 𝑅₁<1 such that 𝐺_𝑗={𝑧∶𝑟_𝑗<|𝑧−𝑎_𝑗|<𝑅_𝑗}(𝑗=2,…,𝑛) and 𝐺₁={𝑧∶𝑅₁<|𝑧|<1}, then we have 𝐺_𝑗⊂𝑈_𝑗, where 𝑈_𝑗 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)∈𝐺_𝑖×Ω∪Ω×𝐺_𝑖, one has ||𝐾||𝐾^Ω(𝑧,𝑤)||<𝐷||𝐾^Ω𝑗(𝑧,𝑤)_Ω𝑗||,(𝑧,𝑤)||<||𝐾^Ω(𝑧,𝑤)||+ℳ,(3.11)(2)for any 𝑧∈Ω, one has 𝐾^Ω𝑗(𝑧,𝑧)<𝐾^Ω(𝑧,𝑧).

Proof. By the explicit formula of the Bergman kernels 𝐾^Ω𝑖, there are constants 𝐶_𝑖 and 𝑀_𝑖 such that ||𝐾^Ω𝑖(𝑧,𝑤)||≥𝐶_𝑖,(3.12) for (𝑧,𝑤)∈(𝐺_𝑖×Ω)∪(Ω×𝐺_𝑖) and ||𝐾^Ω𝑖(𝑧,𝑤)||≤𝑀_𝑖(3.13) if (𝑧,𝑤)∉𝐺_𝑖×𝐺_𝑖 for 𝑖=1,2,…,𝑛. From the last Lemma, it follows that ||𝐾^Ω(𝑧,𝑤)||≤||𝐾^Ω𝑖(𝑧,w)||+𝐶≤1+𝐶𝐶_𝑖||𝐾^Ω𝑖||𝐾(𝑧,𝑤)_Ω𝑖||,(𝑧,𝑤)||≤<||||𝐾𝐾^Ω(𝑧,𝑤)||+||𝐸^Ω(𝑧,𝑤)||+_𝑗≠𝑖||𝐾^Ω𝑗(𝑧,𝑤)_Ω||(𝑧,𝑤)||+‖𝐸^Ω‖_∞+_𝑖≠𝑗𝑀_𝑗,(3.14) whenever (𝑧,𝑤)∈(𝐺_𝑖×Ω)∪(Ω×𝐺_𝑖). If we call 𝒟 the biggest number among {1+𝐶/𝐶_𝑗} and we let ℳ=‖𝐸^Ω‖_∞+∑^𝑛𝑗=1𝑀_𝑗, 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=1Ω_𝑗. This also suggests that we should look for a representation of the elements of 𝐿^2𝑎(Ω) that reflects this fact. For this reason, we give the following.

Definition 3.3. Given Ω=∩^𝑛𝑗=1Ω_𝑗 with Ω₁={𝑧∈ℂ∶|𝑧|<1} and Ω_𝑗={𝑧∈ℂ∶|𝑧−𝑎_𝑗|>𝑟_𝑗}, for any 𝑓∈𝐿^2𝑎(Ω), we define 𝑛+1 functions 𝑃₀𝑓,𝑃₁𝑓,𝑃₂𝑓,…,𝑃_𝑛𝑓 as follows: if 𝑧∈Ω, then we set, for 𝑗=1, 𝑃₁𝑓(𝑧)=12𝜋𝑖⋅𝛾̂₁𝑓(𝜁)𝜁−𝑧𝑑𝜁,(3.15) for 𝑗=2,3,…,𝑛, 𝑃_𝑗𝑓=12𝜋𝑖⋅𝛾̂_𝑗𝑓(𝜁)𝜁−𝑧𝑑𝜁−21𝜋𝑖⋅𝛾̂_𝑗𝑓(𝜁)𝑑𝜁,(3.16) and for 𝑗=0, 𝑃₀𝑓=^𝑛_𝑗=221𝜋𝑖⋅𝛾̂_𝑗𝑓(𝜁)𝑑𝜁1𝑧−𝑎_𝑗,(3.17) where 𝛾̂_𝑗(𝑗=1,…,𝑛) are the circles which center at 𝑎_𝑗(𝑎₁=0) and lie in 𝐺_𝑗 (see Lemma 3.2), respectively, so that 𝑧 is exterior to 𝛾̂_𝑗(𝑗=2,…,𝑛) and interior to 𝛾̂₁.

It is important that the reader notices that the Cauchy theorem implies that our definition is independent from how we choose 𝛾̂₁,…,𝛾̂_𝑛. Moreover, it is important to notice that the domains of the functions 𝑃₂𝑓,…,𝑃_𝑛𝑓 are actually the sets Ω₂,…,Ω_𝑛. In the next Lemma, we give more information about this representation.

Lemma 3.4. For 𝑓∈𝐿^2𝑎(Ω), one can write it uniquely as 𝑓(𝑧)=^𝑛_𝑗=1𝑃_𝑗𝑓(𝑧)+𝑃₀𝑓(𝑧),(3.18) with 𝑃_𝑗𝑓∈𝐿^2𝑎(Ω_𝑗),𝑃₀𝑓∈𝐿^2𝑎(Ω)∩𝐶^∞(Ω),𝑃_𝑘(𝑃_𝑗𝑓)=0 if 𝑗≠𝑘, and moreover, there exists a constant 𝑀₁ such that, for 𝑗=0,1,…,𝑛, one has ‖‖𝑃_𝑗𝑓‖‖_Ω≤‖‖𝑃_𝑗𝑓‖‖_Ω𝑗≤𝑀₁‖𝑓‖_Ω.(3.19) In particular, if 𝑓∈𝐿^2𝑎(Ω_𝑖), then 𝑃_𝑖𝑓=𝑓 and ‖𝑓‖_Ω𝑖≤𝑀₁‖𝑓‖_Ω,(3.20) for 𝑖=1,…,𝑛.

Proof. Let 𝑓 be any function analytic on Ω. For any 𝑧∈Ω, let 𝛾_𝑖(𝑖=1,…,𝑛) be the circles which center at 𝑎_𝑖(𝑎₁=0) and lie in 𝐺_𝑖, respectively, so that 𝑧 is exterior to 𝛾_𝑖(𝑖=2,…,𝑛) and interior to 𝛾₁. Using Cauchy's Formula, we can write 𝑓(𝑧)=^𝑛_𝑗=112𝜋𝑖⋅_𝛾𝑗𝑓(𝜁)𝜁−𝑧𝑑𝜁.(3.21) Let 𝑓_𝑗(𝑧)=12𝜋𝑖⋅_𝛾𝑗𝑓(𝜁)𝜁−𝑧𝑑𝜁.(3.22) By Cauchy's Formula, the value 𝑓_𝑗(𝑧) does not depend on the choice of 𝛾_𝑗 if 1≤𝑗≤𝑛 and 𝑓(𝑧)=∑^𝑛𝑗𝑓_𝑗(𝑧). Of course, each 𝑓_𝑗 is well defined for all 𝑧∈Ω_𝑗 and analytic in Ω_𝑗. In addition, if 𝑗≠1, we have that 𝑓_𝑗(𝑧)→0 as |𝑧|→∞. Writing the Laurent expansion at 𝑎_𝑗 of 𝑓_𝑗, we have 𝑓₁(𝑧)=^∞_𝑘=0𝛼_1,𝑘𝑧^𝑘,(3.23) and, for 𝑗≠1, 𝑓_𝑗(𝑧)=^−∞_𝑘₌₋₁𝛼_𝑗,𝑘𝑧−𝑎_𝑗^𝑘,(3.24) and these series converge to 𝑓_𝑗 uniformly and absolutely on any compact subset of Ω_𝑗, respectively. We remark that the coefficients are given by the following formula: 𝛼_𝑗,𝑘=12𝜋𝑖_𝛾𝑗𝑓(𝜁)𝜁−𝑎_𝑗^𝑘+1𝑑𝜁,(3.25) where 𝑘≥0 if 𝑗=1 and 𝑘≤−1 if 𝑗≠1 and 𝛾_𝑗⊂𝐺_𝑗,1≤𝑗≤𝑛. Moreover, if 𝑓 is holomorphic in some Ω_𝑗 and 𝑓(𝑧)→0 as |𝑧|→∞ when 𝑖≠1, then 𝛼_𝑗𝑘=0 for all 𝑗≠𝑖 by Cauchy's theorem and, therefore, 𝑓_𝑗=0.
Now, we define 𝑃₁𝑓=𝑓₁ and 𝑃_𝑗𝑓(𝑧)=^−∞_𝑘₌₋₂𝛼_𝑗𝑘𝑧−𝑎_𝑗^𝑘,(3.26) for 𝑗=2,3,…,𝑛 and 𝑃₀𝑓(𝑧)=^𝑛_𝑗=2𝛼_𝑗,−1𝑧−𝑎_𝑗⁻¹,(3.27) then 𝑓(𝑧)=∑^𝑛𝑖=0𝑃_𝑖𝑓(𝑧) for all 𝑧∈Ω and 𝑃_𝑘(𝑃_𝑗𝑓)=0 if 0≠𝑘≠𝑗≠0 as we have proved above.
We claim that 𝑓∈𝐿^2𝑎(Ω) implies that 𝑃_𝑖𝑓∈𝐿^2𝑎(Ω_𝑗) for 𝑗=1,2,…,𝑛, respectively. Indeed, since each annulus 𝐺_𝑗 is contained in Ω,𝑓∈𝐿^2𝑎(Ω) implies that 𝑓 is an element of 𝐿^2𝑎(𝐺_𝑖) for all 𝑖=1,2,…,𝑛.
For any fixed 𝑖, note that 𝑃_𝑗𝑓(0≠𝑗≠𝑖) and 𝑃₀𝑓−𝛼_𝑗,−1⋅(𝑧−𝑎_𝑗)⁻¹ are analytic on 𝐺_𝑖∪(ℂ/Ω_𝑖) and lim_|𝑧|→∞𝑃_𝑗𝑓(𝑧)=0 for 𝑗≠1. Expanding them as Laurent series, it follows that:(1)If 𝑖=1, then 𝑃_𝑗𝑓=∑^+∞𝑘=1𝛽_𝑗𝑘/𝑧^𝑘 for 𝑗≠1,(2)If 𝑖≠1, then 𝑃_𝑗𝑓(𝑧)=^+∞_𝑘=0𝛽_𝑗𝑘𝑧−𝑎_𝑖^𝑘,(3.28) for 0≠𝑗≠𝑖 and 𝑃₀𝑓(𝑧)=^+∞_𝑘=0𝛽_0𝑘𝑧−𝑎_𝑖^𝑘+𝛼^𝑖,−1𝑧−𝑎_𝑖.(3.29) It is obvious that, in any case, these series converge uniformly and absolutely on 𝐺_𝑖. Observing that each 𝐺_𝑖 is an annulus at 𝑎_𝑖, we have, by direct computation, that ⟨𝑓,𝑓⟩_𝐺𝑖≥⟨𝑃_𝑖𝑓,𝑃_𝑖𝑓⟩_𝐺𝑖+||𝛼_𝑖,−1||²ln𝑅_𝑖−ln𝑟_𝑖(3.30) if 𝑖≠1 and ⟨𝑓,𝑓⟩_𝐺1≥⟨𝑃₁𝑓,𝑃₁𝑓⟩_𝐺1.(3.31) Therefore, for any 𝑖=1,…,𝑛, there exists a constant 𝑀^ such that ‖‖𝑃_𝑖𝑓||𝛼‖‖_𝐺𝑖≤‖𝑓‖_𝐺𝑖≤‖𝑓‖_Ω,(∗)_𝑖,−1||≤𝑀^⋅‖𝑓‖_Ω.(∗∗) From the definition of 𝑃_𝑗𝑓, we derive ‖‖𝑃₁𝑓‖‖^2𝐺₁=^+∞₀||𝛼_1𝑘||²1−𝑅^12𝑘+2‖‖𝑃𝑘+1_𝑖,𝑓‖‖^2𝐺_𝑖=^−∞_𝑘₌₋₂|𝛼|^2𝑖𝑘𝑟^𝑖2𝑘+2−𝑅^𝑖2𝑘+2𝑘+1,(3.32) for 𝑖=2,…,𝑛. The convergence of these series is guaranteed by the conditions ∗ and ∗∗. Since 𝑅₁<1 and 𝑟_𝑖<𝑅_𝑖, it follows that 𝑃_𝑖𝑓∈𝐿^2𝑎(Ω_𝑖) and ‖‖𝑃₁𝑓‖‖^2Ω₁=^+∞₀||𝛼_1𝑘||²‖‖𝑃𝑘+1_𝑖,𝑓‖‖^2Ω_𝑖=^−∞_𝑘₌₋₂||𝛼_1𝑘||²𝑟^𝑖2𝑘+2𝑘+1,(3.33) for 𝑖=2,…,𝑛. Comparing the expression of ‖𝑃_𝑖𝑓‖_Ω𝑖 with the expression of ‖𝑃_𝑖𝑓‖_𝐺𝑖, it follows that ‖𝑃_𝑖𝑓‖_Ω𝑖<𝑀⋅‖𝑃_𝑖𝑓‖_𝐺𝑖 for some constant 𝑀 for 𝑖=1,…,𝑛. Hence, ‖𝑃_𝑖𝑓‖_Ω𝑖<𝑀⋅‖𝑃_𝑖𝑓‖_Ω. Moreover, if we define 𝑀^=Max{‖(𝑧−𝑎_𝑖)⁻¹‖_Ω}, from the inequalities ‖𝑃_𝑖𝑓‖_𝐺𝑖≤‖𝑓‖_𝐺𝑖≤‖𝑓‖_Ω and |𝛼_𝑖,−1|≤𝑀^⋅‖𝑓‖_Ω and the definition of 𝑃₀, it follows that ‖𝑃₀𝑓‖_Ω≤𝑛⋅𝑀^⋅𝑀^⋅‖𝑓‖_Ω.
If 𝑓∈𝐿^2𝑎(Ω_𝑖) for some 𝑖∈{1,2,…,𝑛}, note that lim𝑓(𝑧)=0 as |𝑧|→∞ for 𝑖≠1, then 𝑓(𝑧)=𝑃_𝑖𝑓(𝑧)+𝛼_𝑖,−1(𝑧−𝑎_𝑖)⁻¹ if 𝑖≠1 and 𝑃₁𝑓=𝑓 if 𝑖=1. For 𝑖≠1, since 𝑓∈𝐿^2𝑎(Ω_𝑖)⊂𝐿^2𝑎(Ω) implies that 𝑃_𝑖𝑓∈𝐿^2𝑎(Ω_𝑖), then 𝛼_𝑖,−1⋅(𝑧−𝑎_𝑖)⁻¹∈𝐿^2𝑎(Ω_𝑖). We must have 𝛼_𝑖,−1=0 and, consequently, 𝑃₀𝑓=0. Hence, in any case, 𝑓∈𝐿^2𝑎(Ω_𝑖) implies 𝑓=𝑃_𝑖𝑓 and 𝑃_𝑗𝑓=0 if 𝑖≠𝑗, and this remark completes our proof.

Lemma 3.5. If {𝑓_𝑛} is a bounded sequence in 𝐿^2𝑎(Ω) and 𝑓_𝑛→0 weakly in 𝐿^2𝑎(Ω), then 𝑃_𝑗𝑓_𝑛→0 weakly on 𝐿^2𝑎(Ω_𝑗) for 𝑗=1,…,𝑛 and 𝑃₀𝑓_𝑛→0 uniformly on Ω.

Proof. By the previous Lemma, we know that the linear transformations {𝑃_𝑗} are bounded operators, then 𝑓_𝑛→0 weakly in 𝐿^2𝑎(Ω) implies that 𝑃_𝑗𝑓_𝑛→0 weakly on 𝐿^2𝑎(Ω_𝑗) for 𝑗=1,…,𝑛. For the same reason, 𝑃₀𝑓_𝑛→0 weakly in 𝐿^2𝑎(Ω) and then 𝑃₀𝑓_𝑛(𝜁)→0 for any 𝜁∈Ω. Since 𝑃₀𝑓_𝑚=^𝑛_𝑖=2𝛼_𝑖,−1(𝑚)𝜁−𝑎_𝑖,(3.34) by the estimates given in the last lemma, we have that |𝛼_𝑖,−1(𝑚)|<𝑀‖𝑓_𝑚‖_Ω. The boundedness of {‖𝑓_𝑚‖_Ω} implies that the family of continuous functions {𝑃₀𝑓_𝑚} is uniformly bounded and equicontinuous on Ω, then, by Arzela-Ascoli's Theorem, we have that 𝑃₀𝑓_𝑚→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 𝑇_𝜑 to be bounded or compact.

Before we state the next Theorem, we remind the reader that 𝐾^Ω(𝜁,𝑧)=𝐸^Ω(𝜁,𝑧)+^𝑛_ℓ=1𝐾^Ωℓ(𝜁,𝑧),(4.1) where 𝐸^Ω∈𝐿^∞(Ω×Ω) and, for all ℓ=1,…,𝑛, we have 𝐾^Ωℓ(𝜁,𝑧)=𝐾^Ωℓ(𝜁,𝑧),∀𝜁,𝑧∈Ω×Ω,(4.2) where 𝐾^Ωℓ is the reproducing kernel of Ω_ℓ. If we use the symbol 𝐾^Ω0 to indicate 𝐸^Ω, we can write 𝐾^Ω(𝜁,𝑧)=^𝑛_ℓ=0𝐾^Ωℓ(𝜁,𝑧).(4.3) We also remind the reader that if 𝐼∶𝐿^2𝑎(Ω)→𝐿^2𝑎(Ω) is the identity operator, then 𝐼=^𝑛_ℓ=0𝑃_ℓ,(4.4) where 𝑃_ℓ∶𝐿^2𝑎(Ω)→𝐿^2𝑎(Ω) is a bounded operator for all ℓ=0,1,…,𝑛 with 𝑃_ℓ𝑓∈𝐿^2𝑎(Ω_ℓ) if ℓ=1,…,𝑛 and 𝑃₀𝑓∈𝒞^∞(Ω) and 𝑃_𝑘𝑃_ℓ=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 𝐾^Ω.

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

Lemma 4.1. Let 𝜑∈𝐿²(𝐷) be an essentially radial function where 𝐷=∩^𝑛𝑗=1𝐷_𝑗 with 𝐷₁={𝑧∈ℂ∶|𝑧|<1} and 𝐷_𝑗={𝑧∈ℂ∶|𝑧−𝑎_𝑗|>𝑟_𝑗} for 𝑗=2,…,𝑛. If one defines 𝜑_𝑗=𝜑⋅𝑝_𝑗 where 𝑗=1,…,𝑛 and 𝔓={𝑝₀,𝑝₁,…,𝑝_𝑛} is a 𝜕-partition for 𝐷,then the following are equivalent: (1)the operator 𝑇_𝜑∶𝐿^2𝑎(𝐷,𝑑𝜈)⟶𝐿^2𝑎(𝐷,𝑑𝜈)(4.5) is bounded (compact);(2)for any 𝑗=1,…,𝑛, the operators 𝑇_𝜑𝑗∶𝐿^2𝑎𝐷_𝑗,𝑑𝜈⟶𝐿^2𝑎𝐷_𝑗,𝑑𝜈(4.6) are bounded (compact).

Proof. Let {𝑝₀,𝑝₁,…,𝑝_𝑛} be a partition of the unit on 𝐷=∩^𝑛𝑗=1𝐷_𝑗, which is a canonical domain. Now, we notice that for all 𝑓∈𝐿²(𝐷) and for all 𝑤∈𝐷, we have the following: 𝑇_𝜑𝑓(𝑤)==_𝐷𝜑(𝑧)𝑓(𝑧)𝐾^𝐷(𝑧,𝑤)𝑑𝑣(𝑧)_𝑛_𝑗=0_𝐷=𝜑(𝑧)𝑓(𝑧)𝐾^𝐷𝑗(𝑧,𝑤)𝑑𝑣(𝑧)_𝑛_𝑛𝑗=0_𝑘=0_𝐷=𝜑(𝑧)𝑝_𝑘(𝑧)𝑓(𝑧)𝐾^𝐷𝑗(𝑧,𝑤)𝑑𝑣(𝑧)_𝑛_𝑛𝑗=0_𝑘=0𝑇_𝑗𝑘𝑓(𝑤),(4.7) where, by definition, we have 𝑇_𝑗𝑘𝑓(𝑤)=_𝐷𝜑(𝑧)𝑝_𝑘(𝑧)𝐾^𝐷𝑗(𝑧,𝑤)𝑓(𝑧)𝑑𝑣(𝑤)𝑑𝑣(𝑧).(4.8)

Claim 1. The operator 𝑇_𝑗0 is Hilbert-Schmidt for any 𝑗=0,1,…,𝑛.

Proof. We observe that, by definition, we have 𝑇_𝑗0𝑓(𝑤)=_𝐷𝜑(𝑧)𝑝₀(𝑧)𝐾^𝐷𝑗(𝑧,𝑤)𝑓(𝑧)𝑑𝑣(𝑧),(4.9) therefore, if we define ℐ₁=_𝐷||𝜑(𝑧)𝑝₀(𝑧)𝐾^𝐷𝑗(𝑧,𝑤)||²𝑑𝑣(𝑧)𝑑𝑣(𝑤),(4.10) we have ℐ₁=_𝐷||𝜑(𝑧)𝑝₀(𝑧)||²_𝐷≤||𝐾^𝐷𝑗(𝑧,𝑤)||²𝑑𝑣(𝑤)𝑑𝑣(𝑧)_𝐷≤||𝜑M(a𝑧x)𝑝_{0𝑧∈supp(𝑝0)}(𝑧)||²||𝐾^𝐷𝑗(𝑧,𝑧)||𝑑𝑣(𝑧)≤Max_{𝑧∈supp(𝑝0)}||𝑝₀(𝑧)||²𝐾^𝐷𝑗(𝑧,𝑧)_𝐷||𝜑_𝑗(𝑧)||²𝑑𝑣(𝑧)||𝑝₀(𝑧)||²𝐾^𝐷𝑗(𝑧,𝑧)⋅‖𝜑‖^2𝐷,2<∞.(4.11) This implies that for any 𝑡=0,1,…,𝑛, 𝑇_𝑡0 is Hilbert-Schmidt. Therefore, the operator ^𝑛_𝑡=0𝑇_𝑡0(4.12) is Hilbert-Schmidt, and this completes the proof of the claim.

Claim 2. The operator 𝑇_0𝑘 is Hilbert-Schmidt for any 𝑘=0,1,…,𝑛.

Proof. We observe that, by definition, we have 𝑇_0𝑘𝑓(𝑤)=_𝐷𝜑(𝑧)𝑝_𝑘(𝑧)𝐾^𝐷0(𝑧,𝑤)𝑓(𝑧)𝑑𝑣(𝑧),(4.13) therefore, if we define ℐ₂=_𝐷||𝜑(𝑧)𝑝_𝑘(𝑧)𝐾^𝐷0(𝑧,𝑤)||²𝑑𝑣(𝑧)𝑑𝑣(𝑤),(4.14) we have ℐ₂=≤_𝐷||𝜑M(a𝑧x)𝑝_{0(𝑧,𝑤)∈𝐷×𝐷}(𝑧)||²||𝐾^𝐷0(𝑧,𝑤)||²𝑑𝑣(𝑤)𝑑𝑣(𝑧)||𝐾^𝐷0(𝑧,𝑤)||²⋅𝑣(𝐷)⋅_𝐷≤Max_{(𝑧,𝑤)∈𝐷×𝐷}||𝜑(𝑧)𝑝₀(𝑧)||²𝑑𝑣(𝑧)||𝐾^𝐷0(𝑧,𝑤)||²⋅𝑣(𝐷)⋅‖𝜑‖^2𝐷,2<∞.(4.15) This implies that for any 𝑡=0,1,…,𝑛, 𝑇_0𝑡 is Hilbert-Schmidt. Therefore, the following ^𝑛_𝑡=0𝑇_0𝑡(4.16) is Hilbert-Schmidt, and this completes the proof of the claim.

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

Proof. We observe that 𝑇_𝑗𝑘𝑓(𝑤)=_𝐷𝜑(𝑧)𝑝_𝑘(𝑧)𝐾^𝐷𝑗(𝑧,𝑤)𝑓(𝑧)𝑑𝑣(𝑤)𝑑𝑣(𝑧).(4.17) To start, we give the following: 𝒩_𝑗𝑖(𝑧,𝑤)=^def𝜑_𝑗(𝑧)⋅𝐾^𝐷𝑖(𝑧,𝑤).(4.18) We will show that Fubini theorem and the properties of the 𝜕-partition imply that _𝐷||𝒩_𝑗𝑖(𝑧,𝑤)||²𝑑𝑣(𝑤)𝑑𝑣(𝑧)<∞.(4.19) In fact, we have _𝐷||𝒩_𝑗𝑖(𝑧,𝑤)||²=_𝐷_𝐷=||𝒩_𝑗𝑖(𝑧,𝑤)||²𝑑𝑣(𝑤)𝑑𝑣(𝑧)_𝐷=||𝜑_𝑗(𝑧)||²||𝐾^𝐷𝑖(𝑧,𝑤)||²𝑑𝑣(𝑤)𝑑𝑣(𝑧)_𝐷||𝜑_𝑗(𝑧)||²_𝐷=||𝐾^𝐷𝑖(𝑧,𝑤)||²𝑑𝑣(𝑤)𝑑𝑣(𝑧)_𝐷=||𝜑_𝑗(𝑧)||²𝐾^𝐷𝑖(𝑧,𝑧)𝑑𝑣(𝑧)_𝐷≤||𝜑M(𝑧a)x||²||𝑝_{𝑗𝑧∈supp(𝑝𝑗)}(𝑧)||²𝐾^𝐷𝑖(𝑧,𝑧)𝑑𝑣(𝑧)||𝑝_𝑗(𝑧)||²𝐾^𝐷𝑖(𝑧,𝑧)⋅‖𝜑‖^2𝐷,2<∞.(4.20)
Therefore, we can write that 𝑇_𝜑=𝒦+^𝑛_ℓ=1𝑇_𝜑ℓℓ,(4.21) where 𝒦 is a compact operator.
We also observe that Lemma 3.4 implies that 𝑇_𝜑ℓℓ=∑^𝑛𝑗=0𝑇_𝜑ℓℓ𝑃_𝑗, and we prove that the operator 𝑇_𝜑ℓℓ𝑃_𝑗 is compact if 𝑗≠ℓ and 𝑗,ℓ=1,…,𝑛.

Proof. In order to simplify the notation, we define the operator 𝑅_𝑗,ℓ=𝑇_𝜑ℓℓ𝑃_𝑗=𝐾^𝐷ℓ𝑀_𝜑𝑝ℓ𝑃_𝑗. To prove our statement, it is enough to prove that if we take a bounded sequence {𝑓_𝑛} in 𝐿²(𝐷) such that 𝑓_𝑛→0 weakly, then we can prove that ‖𝑅_𝑗,ℓ𝑓_𝑛‖₂→0. We know that the continuity of 𝑃_ℓ implies that 𝑃_𝑗𝑓_𝑘→0 weakly on 𝐻²(𝐷_𝑙), and {‖𝑃_𝑗𝑓_𝑘‖_𝐷ℓ} is bounded by Lemma 3.5. Since it is a sequence of holomorphic functions, we know that {𝑃_𝑗𝑓_𝑘} is uniformly bounded on any compact subset of 𝐷_ℓ. Therefore, the sequence {𝑃_𝑗𝑓_𝑘} is a normal family of functions. Since 𝑃_𝑗𝑓_𝑘(𝜁)→0 for any 𝜁∈𝐷_𝑗, then 𝑃_𝑗𝑓_𝑘 converges uniformly on any compact subset of 𝐷_𝑗 and consequently on 𝐹=supp(𝑝_ℓ). To complete the proof, we remind the reader that if we define the operators 𝑄_ℓ∶𝐿²(𝐷)→𝐿²(𝐷), for ℓ=1,2,…,𝑛, in this way 𝑄_ℓ𝑓(𝑧)=_𝐷𝑓(𝜁)||𝐾^𝐷ℓ(𝜁,𝑧)||𝑑𝑣(𝜁).(4.22) It is possible to prove, with the help of Schur's test (see [11] ), that 𝑄_ℓ is a bounded operator (see [5]). Now, we observe that ||𝑅_𝑗,ℓ𝑓_𝑘(𝜁)||≤Sup||𝑃_𝑗𝑓_𝑘(𝜁)||∶𝜁∈𝐹⋅||𝑄_𝑗||𝒳_𝐹𝜑𝑝_𝑠||(𝜁)||,(4.23) then, by using the fact that 𝑄_ℓ is bounded, we have ‖‖𝑅_𝑗,ℓ𝑓_𝑘‖‖_𝐷≤Sup||𝑃_𝑗𝑓_𝑘(𝜁)||∶𝜁∈𝐹⋅𝑀⋅‖‖𝜑₁𝑝_𝑠‖‖_𝐷,2⟶0,(4.24) and this completes the proof of our claim. Notice also that using the same strategy, we can prove that each 𝑇_𝜑ℓℓ𝑃₀ is compact.

Therefore, we have 𝑇_𝜑=𝒦+^𝑛_ℓ=1𝑇_𝜑ℓℓ=𝒦+𝐾₁+^𝑛_ℓ=1𝑇_𝜑ℓℓ𝑃_ℓ,(4.25) where 𝒦,𝐾₁ are compact operators. Since 𝑃^2𝑡=𝑃_𝑡,𝑃_𝑡𝑃_𝑠=0 and if 𝑗≠ℓ, then 𝑇_𝜑 is bounded (compact) if and only if the operators 𝑇_𝜑ℓℓ𝑃_ℓ are bounded (compact) operators.

Since 𝑃_ℓ𝐿^2𝑎(𝐷)=𝐿^2𝑎(𝐷_ℓ), then it follows that the operator 𝑇_𝜑ℓℓ𝑃_ℓ is bounded (compact) if and only if 𝑇_𝜑ℓℓis 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 𝐷=∩^𝑛𝑗=1𝐷_𝑗 with 𝐷₁={𝑧∈ℂ∶|𝑧|<1} and 𝐷_𝑗={𝑧∈ℂ∶|𝑧−a_𝑗|>𝑟_𝑗} for 𝑗=2,…,𝑛. If one defines 𝜑_𝑗=𝜑⋅𝑝_𝑗 where 𝑗=1,…,𝑛 and 𝔓={𝑝₀,𝑝₁,…,𝑝_𝑛} is a 𝜕-partition for 𝐷then the following are equivalent: (1) the operator 𝑇_𝜑∶𝐿^2𝑎(𝐷,𝑑𝜈)⟶𝐿^2𝑎(𝐷,𝑑𝜈)(4.26) is bounded (compact).(2) for any 𝑗=1,…,𝑛, the sequences 𝛾_𝜑𝑗={𝛾_𝜑𝑗(𝑚)}_𝑚∈ℕare in ℓ_∞(ℤ₊)(𝑐₀(ℤ₊)) where, by definition, if 𝑗=2,…,𝑛𝛾_𝜑𝑗(𝑚)=𝑟_𝑗^∞𝑟_𝑗𝜑_𝑗𝑟^{𝑗(2𝑚+1)/2(𝑚+1)}𝑠^1/2(𝑚+1)+𝑎_𝑗1𝑠²𝑑𝑠∀𝑚∈ℤ₊,(4.27) and for 𝑗=1, 𝛾_𝜑1(𝑚)=¹⁰𝜑₁𝑠^1/2(𝑚+1)𝑑𝑠,∀𝑚∈ℤ₊.(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 𝑇_𝜑𝑗∶𝐿²𝐷_𝑗,𝑑𝜈⟶𝐿^2𝑎𝐷_𝑗,𝑑𝜈(4.29) are bounded (compact). If 𝑗=2,…,𝑛, we observe that if we consider the following sets Δ_0,1={𝑧∈ℂ∶0<|𝑧−𝑎|<1} and Δ_{𝑎𝑗,𝑟𝑗}={𝑧∈ℂ∶0<|𝑧−𝑎_𝑗|<𝑟_𝑗} and the following maps Δ_𝛼0,1−−−→Δ−−_{𝑎𝑗,𝑟𝑗}^𝛽−→𝐷_𝑗,(4.30) where 𝛼(𝑧)=𝑎_𝑗+𝑟_𝑗𝑧 and 𝛽(𝑤)=(𝑤−𝑎_𝑗)⁻¹𝑟^2𝑗+𝑎_𝑗 and we use Proposition 1.1 in [8], we can claim that 𝑇_𝜑𝑗=𝑉^{−1𝛽∘𝛼}𝑇_{𝜑𝑗∘𝛽∘𝛼}𝑉_𝛽∘𝛼,(4.31) where 𝑉_𝛽∘𝛼∶𝐿²(Δ_0,1)→𝐿²(𝐷_𝑗) is an isomorphism of Hilbert spaces. Therefore, 𝑇_𝜑𝑗 is bounded (compact) if and only if 𝑇_{𝜑𝑗∘𝛽∘𝛼} is bounded (compact). We also know that this, in turn, is equivalent to the fact that the sequence 𝛾_𝜑𝑗=𝛾_𝜑𝑗(𝑚)_𝑚∈ℕ(4.32) is in ℓ_∞(ℤ₊)(𝑐₀(ℤ₊)), where 𝛾_𝜑𝑗(𝑚)=¹⁰𝜑_𝑗∘𝛽∘𝛼𝑟^1/2(𝑚+1)𝑑𝑟,∀𝑚∈ℤ₊.(4.33) To complete the proof, we observe that since 𝜑_𝑗 is radial and 𝛽∘𝛼(𝑟)=𝑟⁻¹𝑟_𝑗+𝑎_𝑗 then, after a change of variable, we can rewrite the last integral, and therefore the formula 𝛾_𝜑𝑗(𝑚)=𝑟_𝑗^∞𝑟_𝑗𝜑_𝑗𝑟^{𝑗(2𝑚+1)/2(𝑚+1)}𝑠^1/2(𝑚+1)+𝑎_𝑗1𝑠²𝑑𝑠,∀𝑚∈ℤ₊(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 𝜑_𝑗=𝜑⋅𝑝_𝑗 where 𝑗=1,…,𝑛 and 𝔓 is a 𝜕-partition for Ω, then the following conditions are equivalent: (1)the operator 𝑇_𝜑∶𝐿^2𝑎(Ω,𝑑𝜈)⟶𝐿^2𝑎(Ω,𝑑𝜈)(4.35) is bounded (compact);(2)for any 𝑗=1,…,𝑛, the sequences 𝛾_𝜑𝑗={𝛾_𝜑𝑗(𝑚)}_𝑚∈ℕare in ℓ_∞(ℤ₊)(𝑐₀(ℤ₊)) where, by definition, if 𝑗=2,…,𝑛𝛾_𝜑𝑗(𝑚)=𝑟_𝑗^∞𝑟_𝑗𝜑_𝑗∘Θ⁻¹𝑟^{𝑗(2𝑚+1)/2(𝑚+1)}𝑠^1/2(𝑚+1)+𝑎_𝑗1𝑠²𝑑𝑠,∀𝑚∈ℤ₊,(4.36) and for 𝑗=1𝛾_𝜑1(𝑚)=¹⁰𝜑₁∘Θ⁻¹𝑠^1/2(𝑚+1)𝑑𝑠,∀𝑚∈ℤ₊.(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 𝐾^𝐷(Θ(𝑧),Θ(𝑤))Θ^(𝑧)Θ^(𝑤)=𝐾^Ω(𝑧,𝑤), and the operator 𝑉_Θ𝑓=Θ^⋅𝑓∘Θ is an isometry from 𝐿²(𝐷) onto𝐿²(Ω) (see Proposition 1.1 in [8]). In particular, we have 𝑉_Θ𝑃^𝐷=𝑃^Ω𝑉_Θ and this implies that 𝑉_Θ𝑇_𝜑=𝑇_𝜑∘Θ⁻¹𝑉_Θ. Therefore, the operator 𝑇_𝜑 is bounded (compact) if and only if the operator 𝑇_𝜑∘Θ⁻¹∶𝐿²(𝐷,𝑑𝜈)→𝐿^2𝑎(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 𝑇_{𝜑𝑗∘Θ}⁻¹∶𝐿^2𝑎𝐷_𝑗,𝑑𝜈⟶𝐿^2𝑎𝐷_𝑗,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 𝛾_𝜑𝑗={𝛾_𝜑𝑗(𝑚)}_𝑚∈ℕ are in ℓ_∞(ℤ₊)(𝑐₀(ℤ₊)) where, by definition, if 𝑗=2,…,𝑛, we have 𝛾_𝜑𝑗(𝑚)=𝑟_𝑗^∞𝑟_𝑗𝜑_𝑗∘Θ⁻¹𝑟^{𝑗(2𝑚+1)/2(𝑚+1)}𝑠^1/2(𝑚+1)+𝑎_𝑗1𝑠²𝑑𝑠,∀𝑚∈ℤ₊,(4.39) and for 𝑗=1, 𝛾_𝜑1(𝑚)=¹⁰𝜑₁∘Θ⁻¹𝑠^1/2(𝑚+1)𝑑𝑠,∀𝑚∈ℤ₊,(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 𝐵_𝜑𝑗(𝑠)=𝑟_𝑗^𝑠𝑟_𝑗𝜑_𝑗∘Θ⁻¹𝑟^𝑗1/2𝑥^1/2+𝑎_𝑗1𝑥²𝑑𝑥,(4.41) and for 𝑗=1, we set 𝐵_𝜑1(𝑠)=^1𝑠𝜑₁∘Θ⁻¹𝑥^1/2𝑑𝑥.(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 𝜑_𝑗=𝜑⋅𝑝_𝑗 where 𝑗=1,…,𝑛 and 𝔓 is a 𝜕-partition for Ω, then for the operator 𝑇_𝜑∶𝐿^2𝑎(Ω,𝑑𝜈)→𝐿^2𝑎(Ω,𝑑𝜈) the following hold true: (1)if for any 𝑗=1,…,𝑛|||𝐵_𝜑𝑗(s)|||=𝑂𝑟_𝑗−𝑠as𝑠⟶𝑟_𝑗,(4.43) then 𝑇_𝜑 is bounded;(2)if for any 𝑗=1,…,𝑛|||𝐵_𝜑𝑗(𝑠)|||=𝑜𝑟_𝑗−𝑠as𝑠⟶𝑟_𝑗,(4.44) then 𝑇_𝜑 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 𝛾_𝜑𝑗={𝛾_𝜑𝑗(𝑚)}_𝑚∈ℕ are in ℓ_∞(ℤ₊)(𝑐₀(ℤ₊)) where, by definition, if 𝑗=2,…,𝑛, 𝛾_𝜑𝑗(𝑚)=𝑟_𝑗^∞𝑟_𝑗𝜑_𝑗∘Θ⁻¹𝑟^{𝑗(2𝑚+1)/2(𝑚+1)}𝑠^1/2(𝑚+1)+𝑎_𝑗1𝑠²𝑑𝑠∀𝑚∈ℤ₊,(4.45) and for 𝑗=1𝛾_𝜑𝑗(𝑚)=¹⁰𝜑₁∘Θ⁻¹𝑠^1/2(𝑚+1)𝑑𝑠∀𝑚∈ℤ₊,(4.46) and, in virtue of [1]'s main result, it is true that 𝛾_𝜑𝑗={𝛾_𝜑𝑗(𝑚)}_𝑚∈ℕ are in ℓ_∞(ℤ₊) if for any 𝑗=1,…,𝑛, |||𝐵_𝜑𝑗(𝑠)|||=𝑂𝑟_𝑗−𝑠as𝑠⟶𝑟_𝑗,(4.47) and 𝛾_𝜑𝑗={𝛾_𝜑𝑗(𝑚)}_𝑚∈ℕ are in 𝑐₀(ℤ₊)) if for any 𝑗=1,…,𝑛|||𝐵_𝜑𝑗(𝑠)|||=𝑜𝑟_𝑗−𝑠as𝑠⟶𝑟_𝑗.(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 𝜑_𝑗=𝜑⋅𝑝_𝑗 where 𝑗=1,…,𝑛 and 𝔓 is a 𝜕-partition for Ω and if 𝜑≥0 a.e. in Ω, then for the operator 𝑇_𝜑∶𝐿^2𝑎(Ω,𝑑𝜈)→𝐿^2𝑎(Ω,𝑑𝜈), the following hold true: (1)𝑇_𝜑 is bounded if and only if |||𝐵_𝜑𝑗(𝑠)|||=𝑂𝑟_𝑗−𝑠as𝑠⟶𝑟_𝑗,(4.49) for any 𝑗=1,…,𝑛,(2)𝑇_𝜑 is compact if and only if |||𝐵_𝜑𝑗(𝑠)|||=𝑜𝑟_𝑗−𝑠as𝑠⟶𝑟_𝑗,(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 𝑇_𝜑∶𝐿^2𝑎(Ω,𝑑𝜈)→𝐿^2𝑎(Ω,𝑑𝜈) has an essentially radial positive symbol 𝜑≥0 such that for some ℓ=1,…,𝑛, the following lim_𝛿→0inf_{dist(𝑧,𝜕Ωℓ)<𝛿}𝜑(𝑧)=∞(4.51) holds, then the operator 𝑇_𝜑 is unbounded. Moreover, if 𝑇_𝜑 is bounded and the symbol is an unbounded essentially radial function, then it must be true that around any 𝜕Ω_ℓ, the symbol has an oscillating behavior.

In order to present an application, we consider a family of examples. Let us consider the case where Ω=∩^𝑛𝑗=1Ω_𝑗 with Ω₁={𝑧∈ℂ∶|𝑧|<1} and Ω_𝑗={𝑧∈ℂ∶|𝑧−𝑎_𝑗|>𝑟_𝑗} for 𝑗=2,…,𝑛. Let 𝜑∈𝐿²(Ω) be a function that can be written in the following way: 𝜑=^𝑛_ℓ=1𝜑(ℓ),(4.52) where, for any ℓ=1,2,…,𝑛,𝜑(ℓ) is radial, that is, 𝜑(ℓ)=𝜑(ℓ)(|𝑧−𝑎_ℓ|) and satisfies inf_{||𝑧−𝑎ℓ||>𝑟ℓ+𝜖ℓ}𝜑(ℓ)||𝑧−𝑎_ℓ||=𝑚_ℓ>0,sup_{||𝑧−𝑎ℓ||>𝑟ℓ+𝜖ℓ}𝜑(ℓ)||𝑧−𝑎_ℓ||=𝑀_ℓ<∞(4.53) if ℓ=2,…,𝑛 and inf_{|𝑧|<1−𝜖1}𝜑(1)(|𝑧|)=𝑚₁>0,sup_{||𝑧−𝑎ℓ||<1−𝜖1}𝜑(1)(|𝑧|)=𝑀₁<∞.(4.54) if ℓ=1. As a consequence of our results, we can conclude that (1)𝑇_𝜑 is bounded if there exists a constant 𝒞₁ such that for any 𝑗=2,…,𝑛, limsup_{𝑠→𝑟ℓ}||||𝑟_ℓ𝑠−𝑟_ℓ^𝑠𝑟_ℓ𝜑(ℓ)𝑟^ℓ1/2𝑥^1/2+𝑎_ℓ1𝑥²limsup_𝑠→1𝑑𝑥||||<𝒞₁,||||11−𝑠^1𝑠𝜑(ℓ)𝑥^1/2𝑑𝑥||||<𝒞₁,(4.55) for any 𝑗=1,(2)𝑇_𝜑 is compact if for any 𝑗=2,…,𝑛lim_{𝑠→𝑟ℓ}𝑟_ℓ𝑠−𝑟_ℓ^𝑠𝑟_ℓ𝜑(ℓ)𝑟^ℓ1/2𝑥^1/2+𝑎_ℓ1𝑥²lim_𝑠→1𝑑𝑥=0,11−𝑠^1𝑠𝜑(ℓ)𝑥^1/2𝑑𝑥=0,(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 𝑇_𝜑 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 𝐷=∩^𝑛𝑖=1𝐷_𝑖 where 𝐷₁={𝑧∈ℂ∶|𝑧|<1} and 𝐷_𝑗={z∈ℂ∶|𝑧−𝑎_𝑗|>𝑟_𝑗} for 𝑗=2,…,𝑛 where 𝑎_𝑗∈𝐷₁ and 0<𝑟_𝑗<1 with |𝑎_𝑗−𝑎_𝑘|>𝑟_𝑗+𝑟_𝑘 if 𝑗≠𝑘 and 1−|𝑎_𝑗|>𝑟_𝑗. If we denote with the symbol Ψ∶Ω⟶𝐷(4.57) the conformal equivalence between Ω and 𝐷, then we can define, on Ω, the map 𝜑_𝐮,𝐯=^𝑛_ℓ=1𝜑_{𝑢ℓ,𝑣ℓ}(ℓ),(4.58) where, for any ℓ, we have 𝜑_{𝑢ℓ,𝑣ℓ}(ℓ)(𝑧)=−1−𝑢_ℓ1−distΨ(𝑧),𝜕𝐷_ℓ²^−𝑢ℓsin1−distΨ(𝑧),𝜕𝐷_ℓ²^−𝑣ℓ+𝑣_ℓ1−distΨ(𝑧),𝜕𝐷^ℓ²^{−𝑣ℓ−𝑢ℓ}cos1−distΨ(𝑧),𝜕𝐷_ℓ²^−𝑣ℓ,(4.59) where 𝑏_ℓ,𝑎_ℓ∈(0,∞). It is very easy to see that if we denote with 𝒬_ℓ=𝑢_ℓ,𝑣_ℓ∈(0,∞)²∣𝑣_ℓ+𝑢_ℓ<1,𝑢_ℓ<𝑣_ℓ,(4.60) then on the set of parameters 𝒬₁×𝒬₂⋯×𝒬_𝑛, the operator 𝑇_{𝜑𝐚,𝐛} 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 𝑇_𝜙 on 𝐿^2𝑎, we define 𝑇_𝜙, the Berezin transform of 𝑇_𝜙, by 𝜙(𝑤)=⟨𝑇_𝜙𝑘_𝑤,𝑘_𝑤⟩=_Ω𝜙(𝑧)||𝑘_𝑤(𝑧)||²𝑑𝜈(𝑧),(4.61) where 𝑘_𝑤(⋅)=𝐾(⋅,𝑤)𝐾(𝑤,𝑤)^−1/2. It is quite simple to show that if an operator 𝐴∈ℬ(𝐿^2𝑎(Γ,𝑑𝜈)) is compact, then 𝐴, 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, 𝑆=∑^𝑚𝑗∏^𝑚𝑗_𝑘𝑇_{𝜑𝑖,𝑘}, where 𝜑_𝑖,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 𝑇_𝜑 with the help of the sequences 𝜸_𝜑𝑗={𝛾_𝜑𝑗(𝑚)}. 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 𝑇_𝜑(𝑧)=(1−|𝑧|²)²∑(𝑛+1)⟨𝑇_𝜑𝑒_𝑛,𝑒_𝑛⟩|𝑧|^2𝑛. Therefore to show that the vanishing of the Berezin Transform implies compactness is equivalent, given that 𝑇_𝜑 is diagonal, to show that lim_|𝑧|→1(1−|𝑧|²)²∑(𝑛+1)⟨𝑇_𝜑𝑒_𝑛,𝑒_𝑛⟩|𝑧|^2𝑛=0 implies lim_𝑛→∞⟨𝑇_𝜑𝑒_𝑛,𝑒_𝑛⟩=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 𝐷=∩^𝑛𝑗=1𝐷_𝑗 with 𝐷₁={𝑧∈ℂ∶|𝑧|<1} and 𝐷_𝑗={𝑧∈ℂ∶|𝑧−𝑎_𝑗|>𝑟_𝑗} for 𝑗=2,…,𝑛. If one defines 𝜑_𝑗=𝜑⋅𝑝_𝑗 where 𝑗=1,…,𝑛 and 𝔓={𝑝₀,𝑝₁,…,𝑝_𝑛} is a 𝜕-partition for 𝐷. Let us assume that 𝛾_𝜙𝑗={𝛾_𝜙𝑗(𝑚)}_𝑚∈ℕis in ℓ_∞(ℤ₊) and that there is a constant 𝒞₂ such that for 𝑗=2,…,𝑛, sup_{𝜏∈[𝑎𝑗+𝑟𝑗,∞)}|||||𝜑_𝑗(𝜏)−𝜏−𝑎^𝑗𝜏−𝑟_𝑗−𝑎_𝑗^𝜏𝑎_{𝑗+𝑟𝑗}𝜑_𝑗(𝑦)𝑟^𝑗𝑦−𝑎_𝑗²𝑑𝑦|||||<𝒞²,(4.62) and for 𝑗=1sup_𝜏∈[0,1]||||𝜑₁(𝜏)−11−𝜏^1𝜏𝜑₁(𝑠)𝑑𝑠||||<𝒞₂,(4.63) then the operator 𝑇_𝜑∶𝐿^2𝑎(𝐷,𝑑𝜈)→𝐿^2𝑎(𝐷,𝑑𝜈) is compact if and only if lim_{𝑤→𝜕𝐷}𝑇_𝜑(𝑤)=0.(4.64)

Proof. We know that the operator 𝑇_𝜑∶𝐿^2𝑎(𝐷,𝑑𝜈)→𝐿^2𝑎(𝐷,𝑑𝜈) is bounded if and only if for any 𝑗=1,…,𝑛 the operators 𝑇_𝜑𝑗∶𝐿^2𝑎(𝐷_𝑗,𝑑𝜈)→𝐿^2𝑎(𝐷_𝑗,𝑑𝜈) are bounded. Since we assume that 𝛾_𝜙𝑗={𝛾_𝜙𝑗(𝑚)}_𝑚∈ℕ is in ℓ_∞(ℤ₊), then we can conclude that 𝑇_𝜑 is bounded. As we have done before if we fix 𝑗=2,…,𝑛, by using Δ_𝛼0,1−→Δ−→_{𝑎𝑗,𝑟𝑗}^𝛽𝐷_𝑗 with Δ_0,1={𝑧∈ℂ∶0<|𝑧−𝑎|<1} and Δ_{𝑎𝑗,𝑟𝑗}={𝑧∈ℂ∶0<|𝑧−𝑎_𝑗|<𝑟_𝑗}, and the maps Δ_𝛼0,1−→Δ−→_{𝑎𝑗,𝑟𝑗}^𝛽𝐷_𝑗 where 𝛼(𝑧)=𝑎_𝑗+𝑟_𝑗𝑧 and 𝛽(𝑤)=(𝑤−𝑎_𝑗)⁻¹𝑟^2𝑗+𝑎_𝑗, we can claim that 𝑇_𝜑𝑗=𝑉^{−1𝛽∘𝛼}𝑇_{𝜑𝑗∘𝛽∘𝛼}𝑉_𝛽∘𝛼 where 𝑉_𝛽∘𝛼∶𝐿²(Δ_0,1)→𝐿²(𝐷_𝑗) is an isomorphism of Hilbert spaces. Therefore, 𝑇_𝜑𝑗 is compact if and only if 𝑇_{𝜑𝑗∘𝛽∘𝛼} is compact. We also know that this, in turn, is equivalent to the vanishing of the Berezin transform if the function (𝜑∘𝛽∘𝛼)_Ber,𝑗(𝑠)=𝜑_𝑗∘𝛽∘𝛼(𝑠)−11−𝑠^1𝑠𝜑_𝑗∘𝛽∘𝛼(𝑡)𝑑𝑡(4.65) is bounded. Since 𝜑_𝑗 is radial and 𝛽∘𝛼(𝑡)=𝑡⁻¹𝑟_𝑗+𝑎_𝑗, then, after a change of variable under the sign of integral, we can rewrite the last integral, and therefore we obtain the formula (𝜑∘𝛽∘𝛼)_Ber,𝑗(𝑠)=𝜑_𝑗∘𝛽∘𝛼(𝑠)−11−𝑠^{𝑎𝑗+𝑟𝑗}_{(𝑟𝑗/𝑠)+𝑎𝑗}𝜑_𝑗(𝑦)−𝑟^𝑗=𝜑_𝑗𝑦−𝑎_𝑗²𝑑𝑦∘𝛽∘𝛼(𝑠)−11−𝑠^{(𝑟𝑗/𝑠)+𝑎𝑗}_{𝑎𝑗+𝑟𝑗}𝜑_𝑗(𝑦)𝑟^𝑗=𝜑_𝑗𝑦−𝑎_𝑗²𝑑𝑦∘𝛽∘𝛼(𝑠)−11−𝑠^{(𝑟𝑗/𝑠)+𝑎𝑗}_{𝑎𝑗+𝑟𝑗}𝜑_𝑗(𝑦)𝑟^𝑗𝑦−𝑎_𝑗²𝑑𝑦.(4.66) Moreover, if we define 𝜏=𝑠⁻¹𝑟_𝑗+𝑎_𝑗, we can write 𝜑_Ber,𝑗(𝜏)=𝜑_𝑗(𝜏)−𝜏−𝑎^𝑗𝜏−𝑟_𝑗−𝑎_𝑗^𝜏𝑎_{𝑗+𝑟𝑗}𝜑_𝑗(𝑦)𝑟^𝑗𝑦−𝑎_𝑗²𝑑𝑦.(4.67) Therefore, if we assume that this function is bounded, we can conclude (see [2]) that from lim_𝑤→Δ0,1𝑇_{𝜑𝑗∘𝛽∘𝛼}(𝑤)=0, it follows that 𝑇_{𝜑𝑗∘𝛽∘𝛼} is compact for 𝑗=2,…,𝑛. Therefore, we can infer that 𝑇_𝜑𝑗 is compact. We also observe that lim_𝑤→Δ0,1𝑇_{𝜑𝑗∘𝛽∘𝛼}(𝑤)=0(4.68) if and only if lim_{𝑤→𝜕𝐷𝑗}𝑇_𝜑𝑗(𝑤)=0. To prove this fact, we observe that, by definition, we have 𝑇_{𝜑𝑗∘𝛽∘𝛼}(𝑧)=𝑇_{𝜑𝑗∘𝛽∘𝛼}𝑘^Δ0,1_𝑧,𝑘^Δ0,1_𝑧=_Δ0,1𝑇_{𝜑𝑗∘𝛽∘𝛼}𝑘^Δ0,1_𝑧(𝑤)𝑘^Δ0,1_𝑧(𝑤)𝑑𝑤,(4.69) where 𝑘^Δ0,1_𝑧(⋅)=𝐾^Δ0,1_𝑧(⋅,𝑧)𝐾^Δ0,1_𝑧(𝑧,𝑧)^−1/2.(4.70) Let us take (𝛽∘𝛼)⁻¹∶𝐷_𝑗→Δ_0,1. Since (𝐽_𝐑(𝛽∘𝛼)⁻¹)(𝑤) is |((𝛽∘𝛼)⁻¹)^(𝑤)|² and there exists 𝜁∈𝐷_𝑗 such that (𝛽∘𝛼)(𝑧)=𝜁, we obtain 𝑇_{𝜑𝑗∘𝛽∘𝛼}(𝑧)=_𝐷𝑗𝐴𝑉^{(𝛽∘𝛼)−1}𝑘^Δ0,1_{(𝛽∘𝛼)−1(𝜍)}𝑘^Δ0,1_{(𝛽∘𝛼)−1(𝜍)}(𝛽∘𝛼)⁻¹(𝛽∘𝛼)⁻¹^𝑑𝑤,(4.71) where 𝐴=(𝑉_{(𝛽∘𝛼)}⁻¹𝑇_{𝜑𝑗∘𝛽∘𝛼}𝑉_{(𝛽∘𝛼)}). Given the relationship between 𝑘^Δ0,1_{(𝛽∘𝛼)−1(𝜁)}(𝛽∘𝛼)⁻¹(𝑤)(4.72) and 𝑘^𝐷𝑗_𝜁(𝑤), we obtain 𝑇_{𝜑𝑗∘𝛽∘𝛼}(𝑧)=_𝐷𝑗𝑇^𝜑_𝑗𝑘^𝐷𝑗_𝜁(𝑤)𝑘^𝐷𝑗_𝜁(𝑤)𝑑𝑤.(4.73) Therefore, it follows that 𝑇_{𝜑𝑗∘𝛽∘𝛼}(𝑧)=𝑇_𝜑𝑗((𝛽∘𝛼)(𝑧)). 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 lim_{𝑤→𝜕𝐷}𝑇_𝜑(𝑤)=0(4.74) then 𝑇_𝜑 is a compact operator. To complete the proof, we observe that the compactness of 𝑇_𝜑∶𝐿^2𝑎(𝐷,𝑑𝜈)→𝐿^2𝑎(𝐷,𝑑𝜈) implies the vanishing of the Berezin transform since 𝑘_𝑤 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 𝜑_𝑗=𝜑⋅𝑝_𝑗 where 𝑗=1,…,𝑛 and 𝔓 is a 𝜕-partition for Ω. Let us assume that 𝛾_𝜙𝑗={𝛾_𝜙𝑗(𝑚)}_𝑚∈ℕis in ℓ_∞(ℤ₊) and that there is a constant 𝐶₃ such that for 𝑗=2,…,𝑛, sup_{𝜏∈[𝑎𝑗+𝑟𝑗,∞)}|||||𝜑_𝑗∘Θ(𝜏)−𝜏−𝑎^𝑗𝜏−𝑟_𝑗−𝑎_𝑗^𝜏𝑎_{𝑗+𝑟𝑗}𝜑_𝑗∘Θ(y)𝑟^𝑗𝑦−𝑎_𝑗²𝑑𝑦|||||<𝒞³,(4.75) and for 𝑗=1sup_𝜏∈[0,1]||||𝜑₁∘Θ(𝜏)−11−𝜏^1𝜏𝜑₁∘Θ(𝑠)𝑑𝑠||||<𝒞₃,(4.76) then the operator 𝑇_𝜑∶𝐿^2𝑎(Ω,𝑑𝜈)→𝐿^2𝑎(Ω,𝑑𝜈) is compact if and only if lim_{𝑤→𝜕Ω}𝑇_𝜑(𝑤)=0.(4.77)