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

Authors: Roberto C. Raimondo
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

We study the problem of the boundedness and compactness ofTϕ 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.


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## Body

## 1. Introduction

LetΩ 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 Ω=∩j=1nΩj.Fordν=(1/π)dxdy, we consider the usual L2-space L2(Ω)=L2(Ω,dν). The Bergman space La2(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 (1.1)〈f,g〉=∫Ωf(z)g(z)¯dν(z)
for f,g∈L2(Ω,dν). The Bergman projection is the orthogonal projection (1.2)P:L2(Ω,dν)⟶La2(Ω,dν).
It is well-known that for any f∈L2(Ω,dν), we have (1.3)Pf(w)=∫Ωf(z)KΩ(z,w)dν(z),
where KΩ is the Bergman reproducing kernel of Ω. For φ∈L∞(Ω,dν), the Toeplitz operator Tφ:La2(Ω,dν)→La2(Ω,dν) is defined by Tφ=PMφ, where Mφ is the standard multiplication operator. A simple calculation shows that (1.4)Tφf(z)=∫Ωφ(w)f(w)KΩ(w,z)dν(w).
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 Section2, 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 La2(Ω,dν). In Section 4, we prove the main result and   study several important consequences.


## 2. Preliminaries

LetΩ be the bounded multiply-connected domain given at the beginning of Section 1, that is, Ω=∩j=1nΩj, 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 callKΓ(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 anyA∈ℬ(La2(Γ,dν)), we define Ã, the Berezin transform of A, by (2.1)Ã(w)=〈AkwΓ,kwΓ〉=∫ΓAkwΓ(z)kwΓ(z)¯dν(z),
where kwΓ(·)=KΓ(·,w)KΓ(w,w)-1/2.Ifφ∈L∞(Γ), then we indicate with the symbol φ̃ the Berezin transform of the associated Toeplitz operator Tφ, and we have (2.2)φ̃(w)=∫Γφ(z)|kwΓ(z)|2dν(z).
We remind the reader that it is well known that Ã∈𝒞b∞(Γ), and we have ∥Ã∥∞≤∥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 anyΩ 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 D=∩i=1nDi 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.
LetΩ=∩i=1nΩi 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 everyj=0,1,…,n,pj:Ω→[0,1] is a Lipschitz, C∞-function,(2)
for everyj=2,…,n, there exists an open set Wj⊂Ω and an ϵj>0 such that Uϵj={ζ∈Ω:rj<|ζ-aj|<rj+ϵj}, and the support of pj  is contained in Wj and
(2.3)pj(ζ)=1,∀ζ∈Uϵj,(3)
forj=1, there exists an open set W1⊂Ω and an ϵ1>0 such that Uϵ1={ζ∈Ω:1-ϵ1<|ζ|<1} and the support of p1 is contained in W1 and(2.4)p1(ζ)=1,∀ζ∈Uϵ1,(4)
for everyj,k=1,…,n,Wj∩Wk=∅, the set Ω∖(⋃j=1nWj) is not empty and the function(2.5)p0(ζ)=1,∀ζ∈(⋃j=1nWj)c∩Ω,p0(ζ)=0,∀ζ∈Uϵk,k=1,…,n,(5)
for anyζ∈Ω, the following equation:
(2.6)∑k=0npk(ζ)=1.
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 functionp0 is different from zero.Definition 2.2.
A functionφ:Ω=∩i=1nΩi→ℂ is said to be essentially radial if there exists a conformally equivalent canonical bounded domain D=∩i=1nDi, such that if the map Θ:Ω→D is the conformal mapping from Ω onto D, then(1)
for everyk=2,…,n and for some ϵk>0, we have
(2.7)φ∘Θ-1(z)=φ∘Θ-1(|z-ak|),
when z∈Uϵk={ζ∈Ω:rk<|ζ-ak|<rk+ϵk},(2)
fork=1 and for some ϵ1>0, we have
(2.8)φ∘Θ-1(z)=φ∘Θ-1(|z|),
when z∈Uϵ1={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 mapφ is essentially radial viaΘ:∩ℓ=1nΩℓ→∩ℓ=1nDℓ.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 equivalenceΘ:∩ℓ=1nΩℓ→∩ℓ=1nDℓ, we always assume that the ∂-partition is given on ∩ℓ=1nDℓ and transferred to ∩ℓ=1nΩℓ through Θ in the natural way.At this point, we can state the main result.Theorem 2.3.
Letφ∈L2(Ω) be an essentially radial function via Θ:∩ℓ=1nΩℓ→∩ℓ=1nDℓ, 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)Tφ:La2(Ω,dν)⟶La2(Ω,dν)
is bounded (compact).(2)
for anyj=1,…,n  the sequences  γφj={γφj(m)}m∈ℕare in  ℓ∞(ℤ+)(c0(ℤ+)) where, by definition, if  j=2,…,n,
(2.10)γφj(m)=rj∫rj∞φj∘Θ-1(rj(2m+1)/2(m+1)s1/2(m+1)+aj)1s2ds,∀m∈Z+,
and if j=1(2.11)γφ1(m)=∫01φ1∘Θ-1(s1/2(m+1))ds,∀m∈Z+.


## 3. The Structure ofLa2(Ω) and Some Estimates about the Bergman Kernel

From now on, we will assume thatΩ=∩j=1nΩj 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 symbolsKΩj(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 diskΔ0,1 and the disk Ω1, we have Lap(Δ0,1)=Lap(Ω1), if p≥2, 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={z∈ℂ:0<|z-a|<r} and Oa,r={z∈ℂ:|z-a|>r}, we can conclude that (3.1)KOa,r(z,w)=r2(r2-(z-a)⋅(w-a)¯)2,∀(z,w)∈Oa,r×Oa,r.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 (3.2)KΔ0,1(z,w)=1(1-z⋅w¯)2,∀(z,w)∈Δ0,1×Δ0,1.
This implies, by conformal mapping, that the reproducing kernel of Δa,r is (3.3)KΔa,r(z,w)=r2(r2-(z-a)⋅(w-a)¯)2,∀(z,w)∈Δa,r×Δa,r.
Now, we define φ:Δa,r→Oa,r by (3.4)φ(z)=(z-a)-1r2+a,
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)KOa,r(φ(z),φ(w))φ′(z)φ′(w)¯=KΔa,r(z,w)
to obtain that (3.6)KOa,r(z,w)=r2(r2-(z-a)⋅(w-a)¯)2,∀(z,w)∈Oa,r×Oa,r.
Since Ω1=O0,1 and, for j=2,…,n,Oaj,rj=Ωj, then the last equations implies that (3.7)KΩ1(z,w)=1(1-z⋅w¯)2,KΩj(z,w)=rj2(rj2-(z-aj)⋅(w-aj)¯)2
if j=2,…,n.We also note that if we define(3.8)EΩ(z,w)=KΩ(z,w)-∑j=1nKΩj(z,w),
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 EΩ∈C∞(Ω¯×Ω).
(2) There are neighborhoodsUj of ∂Ωj(j=1,…,n) and a constant C>0 such that Uj∩Uk is empty if j≠k and
(3.9)|KΩ(z,w)-KΩj(z,w)|<C,
for z∈Ω and w∈Uj.
(3)EΩ∈L∞(Ω×Ω).Proof.
(1) Since the Bergman kernelsKΩ 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(3.10)KΩ1(z,w)=1(1-z⋅w¯)2,KΩj(z,w)=rj2(rj2-(z-aj)⋅(w-aj)¯)2,
for j=2,...,n and (1) and (2), we get (3).We observe that we can chooseRj>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 Gj¯⊂Uj, where Uj is the same as in Lemma 3.1. We also have the following.Lemma 3.2.
There are constants𝒟>0 and ℳ>0 such that(1)
for any(z,w)∈Gi×Ω∪Ω×Gi, one has
(3.11)|KΩ(z,w)|<D|KΩj(z,w)|,|KΩj(z,w)|<|KΩ(z,w)|+M,(2)
for anyz∈Ω, one has KΩj(z,z)<KΩ(z,z).Proof.
By the explicit formula of the Bergman kernelsKΩi, there are constants Ci and Mi such that
(3.12)|KΩi(z,w)|≥Ci,
for (z,w)∈(Gi×Ω)∪(Ω×Gi) and
(3.13)|KΩi(z,w)|≤Mi
if (z,w)∉Gi×Gi for i=1,2,…,n. From the last Lemma, it follows that
(3.14)|KΩ(z,w)|≤|KΩi(z,w)|+C≤(1+CCi)|KΩi(z,w)|,|KΩi(z,w)|≤|KΩ(z,w)|+|EΩ(z,w)|+∑j≠i|KΩj(z,w)|<|KΩ(z,w)|+∥EΩ∥∞+∑i≠jMj,
whenever (z,w)∈(Gi×Ω)∪(Ω×Gi). If we call 𝒟 the biggest number among {1+C/Cj} and we let ℳ=∥EΩ∥∞+∑j=1nMj, 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 analysisΩ is actually the intersection of other domains, that is, Ω=∩j=1nΩj. This also suggests that we should look for a representation of the elements of La2(Ω) that reflects this fact. For this reason, we give the following.Definition 3.3.
GivenΩ=∩j=1nΩj with Ω1={z∈ℂ:|z|<1} and Ωj={z∈ℂ:|z-aj|>rj}, for any f∈La2(Ω), we define n+1 functions P0f,P1f,P2f,…,Pnf as follows: if z∈Ω, then we set, for j=1,
(3.15)P1f(z)=12πi⋅∫γ̂1f(ζ)ζ-zdζ,
for j=2,3,…,n,
(3.16)Pjf=12πi⋅∫γ̂jf(ζ)ζ-zdζ-12πi⋅∫γ̂jf(ζ)dζ,
and for j=0,
(3.17)P0f=∑j=2n(12πi⋅∫γ̂jf(ζ)dζ)1z-aj,
where γ̂j(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 γ̂j(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 chooseγ̂1,…,γ̂n. 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.
Forf∈La2(Ω), one can write it uniquely as
(3.18)f(z)=∑j=1n(Pjf)(z)+(P0f)(z),
with Pjf∈La2(Ωj),P0f∈La2(Ω)∩C∞(Ω¯),Pk(Pjf)=0 if j≠k, and moreover, there exists a constant M1 such that, for j=0,1,…,n, one has
(3.19)‖Pjf‖Ω≤‖Pjf‖Ωj≤M1‖f‖Ω.
In particular, if f∈La2(Ωi), then Pif=f and
(3.20)‖f‖Ωi≤M1‖f‖Ω,
for i=1,…,n.Proof.
Letf 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)f(z)=∑j=1n12πi⋅∫γjf(ζ)ζ-zdζ.
Let
(3.22)fj(z)=12πi⋅∫γjf(ζ)ζ-zdζ.
By Cauchy's Formula, the value fj(z) does not depend on the choice of γj if 1≤j≤n and f(z)=∑jnfj(z). 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)f1(z)=∑k=0∞α1,kzk,
and, for j≠1,
(3.24)fj(z)=∑k=-1-∞αj,k(z-aj)k,
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)αj,k=12πi∫γjf(ζ)(ζ-aj)k+1dζ,
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 defineP1f=f1 and
(3.26)Pjf(z)=∑k=-2-∞αjk(z-aj)k,
for j=2,3,…,n and
(3.27)P0f(z)=∑j=2nαj,-1(z-aj)-1,
then f(z)=∑i=0nPif(z) for all z∈Ω and Pk(Pjf)=0 if 0≠k≠j≠0 as we have proved above.
We claim thatf∈La2(Ω) implies that Pif∈La2(Ωj) for j=1,2,…,n, respectively. Indeed, since each annulus Gj is contained in Ω,f∈La2(Ω) implies that f is an element of La2(Gi) for all i=1,2,…,n.
For any fixedi, note that Pjf(0≠j≠i) and P0f-αj,-1·(z-aj)-1 are analytic on G¯i∪(ℂ/Ωi) and lim|z|→∞Pjf(z)=0 for j≠1. Expanding them as Laurent series, it follows that:(1)
Ifi=1, then Pjf=∑k=1+∞βjk/zk for j≠1,(2)
Ifi≠1, then
(3.28)Pjf(z)=∑k=0+∞βjk(z-ai)k,
for 0≠j≠i and
(3.29)P0f(z)=∑k=0+∞β0k(z-ai)k+αi,-1z-ai. 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
(3.30)〈f,f〉Gi≥〈Pif,Pif〉Gi+|αi,-1|2(ln⁡Ri-ln⁡ri)
if i≠1 and
(3.31)〈f,f〉G1≥〈P1f,P1f〉G1.
Therefore, for any i=1,…,n, there exists a constant M′ such that
(*)‖Pif‖Gi≤‖f‖Gi≤‖f‖Ω,(**)|αi,-1|≤M′⋅‖f‖Ω.
From the definition of Pjf, we derive
(3.32)‖P1f‖G12=∑0+∞|α1k|2(1-R12k+2)k+1,‖Pif‖Gi2=∑k=-2-∞|α|ik2(ri2k+2-Ri2k+2)k+1,
for i=2,…,n. The convergence of these series is guaranteed by the conditions * and **. Since R1<1 and ri<Ri, it follows that Pif∈La2(Ωi) and
(3.33)‖P1f‖Ω12=∑0+∞|α1k|2k+1,‖Pif‖Ωi2=∑k=-2-∞|α1k|2ri2k+2k+1,
for i=2,…,n. Comparing the expression of ∥Pif∥Ωi with the expression of ∥Pif∥Gi, it follows that ∥Pif∥Ωi<M·∥Pif∥Gi for some constant M for i=1,…,n. Hence, ∥Pif∥Ωi<M·∥Pif∥Ω. Moreover, if we define M′′=Max{∥(z-ai)-1∥Ω}, from the inequalities ∥Pif∥Gi≤∥f∥Gi≤∥f∥Ω and |αi,-1|≤M′·∥f∥Ω and the definition of P0, it follows that ∥P0f∥Ω≤n·M′·M′′·∥f∥Ω.
Iff∈La2(Ωi) for some i∈{1,2,…,n}, note that limf(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 f∈La2(Ωi)⊂La2(Ω) implies that Pif∈La2(Ωi), then αi,-1·(z-ai)-1∈La2(Ωi). We must have αi,-1=0 and, consequently, P0f=0. Hence, in any case, f∈La2(Ωi) 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 La2(Ω) and fn→0 weakly in La2(Ω), then Pjfn→0 weakly on La2(Ωj) 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 La2(Ω) implies that Pjfn→0 weakly on La2(Ωj) for j=1,…,n. For the same reason, P0fn→0 weakly in La2(Ω) and then P0fn(ζ)→0 for any ζ∈Ω. Since
(3.34)P0fm=∑i=2nαi,-1(m)(ζ-ai),
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 Ω¯, 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 functionφ∈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)KΩ(ζ,z)=EΩ(ζ,z)+∑l=1nKlΩ(ζ,z),
where EΩ∈L∞(Ω×Ω) and, for all ℓ=1,…,n, we have (4.2)KlΩ(ζ,z)=KΩl(ζ,z),∀ζ,z∈Ω×Ω,
where KΩℓ is the reproducing kernel of Ωℓ. If we use the symbol K0Ω to indicate EΩ, we can write (4.3)KΩ(ζ,z)=∑l=0nKlΩ(ζ,z).
We also remind the reader that if I:La2(Ω)→La2(Ω) is the identity operator, then (4.4)I=∑l=0nPl,
where Pℓ:La2(Ω)→La2(Ω) is a bounded operator for all ℓ=0,1,…,n with Pℓf∈La2(Ωℓ) if ℓ=1,…,n and P0f∈𝒞∞(Ω¯) 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 symbolKΩ.We are now in a position to prove the following result.Lemma 4.1.
Letφ∈L2(D) be an essentially radial function where D=∩j=1nDj 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)Tφ:La2(D,dν)⟶La2(D,dν)
is bounded (compact);(2)
for anyj=1,…,n,  the operators
(4.6)Tφj:La2(Dj,dν)⟶La2(Dj,dν)
are bounded (compact).Proof.
Let{p0,p1,…,pn} be a partition of the unit on D=∩j=1nDj, 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)Tφf(w)=∫Dφ(z)f(z)KD(z,w)dv(z)=∑j=0n∫Dφ(z)f(z)KDj(z,w)dv(z)=∑j=0n∑k=0n∫Dφ(z)pk(z)f(z)KDj(z,w)dv(z)=∑j=0n∑k=0nTjkf(w),
where, by definition, we have
(4.8)Tjkf(w)=∫Dφ(z)pk(z)KDj(z,w)f(z)dv(w)dv(z).Claim 1.
The operatorTj0 is Hilbert-Schmidt for any j=0,1,…,n.Proof.
We observe that, by definition, we have(4.9)Tj0f(w)=∫Dφ(z)p0(z)KDj(z,w)f(z)dv(z),
therefore, if we define
(4.10)I1=∬D|φ(z)p0(z)KDj(z,w)|2dv(z)dv(w),
we have
(4.11)I1=∫D|φ(z)p0(z)|2(∫D|KDj(z,w)|2dv(w))dv(z)≤∫D|φ(z)p0(z)|2|KDj(z,z)|dv(z)≤(Max⁡z∈supp⁡(p0)|p0(z)|2KDj(z,z))∫D|φj(z)|2dv(z)≤(Max⁡z∈supp⁡(p0)|p0(z)|2KDj(z,z))⋅‖φ‖D,22<∞.
This implies that for any t=0,1,…,n, Tt0 is Hilbert-Schmidt. Therefore, the operator
(4.12)∑t=0nTt0
is Hilbert-Schmidt, and this completes the proof of the claim.Claim 2.
The operatorT0k is Hilbert-Schmidt for any k=0,1,…,n.Proof.
We observe that, by definition, we have(4.13)T0kf(w)=∫Dφ(z)pk(z)KD0(z,w)f(z)dv(z),
therefore, if we define
(4.14)I2=∬D|φ(z)pk(z)KD0(z,w)|2dv(z)dv(w),
we have
(4.15)I2=∬D|φ(z)p0(z)|2|KD0(z,w)|2dv(w)dv(z)≤(Max⁡(z,w)∈D×D|KD0(z,w)|2)⋅v(D)⋅∫D|φ(z)p0(z)|2dv(z)≤(Max⁡(z,w)∈D×D|KD0(z,w)|2)⋅v(D)⋅‖φ‖D,22<∞.
This implies that for any t=0,1,…,n, T0t is Hilbert-Schmidt. Therefore, the following
(4.16)∑t=0nT0t
is Hilbert-Schmidt, and this completes the proof of the claim.Claim 3.
The operatorTij is Hilbert-Schmidt if i≠j≠0 and j,i=1,…,n.Proof.
We observe that(4.17)Tjkf(w)=∫Dφ(z)pk(z)KDj(z,w)f(z)dv(w)dv(z).
To start, we give the following:
(4.18)Nji(z,w)≝φj(z)⋅KDi(z,w).
We will show that Fubini theorem and the properties of the ∂-partition imply that
(4.19)∬D|Nji(z,w)|2dv(w)dv(z)<∞.
In fact, we have
(4.20)∬D|Nji(z,w)|2=∫D(∫D|Nji(z,w)|2dv(w))dv(z)=∬D|φj(z)|2|KDi(z,w)|2dv(w)dv(z)=∫D|φj(z)|2(∫D|KDi(z,w)|2dv(w))dv(z)=∫D|φj(z)|2KDi(z,z)dv(z)=∫D|φ(z)|2|pj(z)|2KDi(z,z)dv(z)≤(Max⁡z∈supp⁡(pj)|pj(z)|2KDi(z,z))⋅∥φ∥D,22<∞.
Therefore, we can write that(4.21)Tφ=K+∑l=1nTφll,
where 𝒦 is a compact operator.
We also observe that   Lemma3.4 implies that Tφℓℓ=∑j=0nTφℓℓPj, and we prove that the operator TφℓℓPj is compact if j≠ℓ and j,ℓ=1,…,n.Proof.
In order to simplify the notation, we define the operatorRj,ℓ=TφℓℓPj=KℓDMφpℓPj. 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 {∥Pjfk∥Dℓ} 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)Qlf(z)=∫Df(ζ)|KlD(ζ,z)|dv(ζ).
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)|Rj,lfk(ζ)|≤Sup⁡{|Pjfk(ζ)|:ζ∈F}⋅|Qj(|XFφps|)(ζ)|,
then, by using the fact that Qℓ is bounded, we have
(4.24)‖Rj,lfk‖D≤Sup⁡{|Pjfk(ζ)|:ζ∈F}⋅M⋅‖φ1ps‖D,2⟶0,
and this completes the proof of our claim. Notice also that using the same strategy, we can prove that  each TφℓℓP0 is compact.Therefore, we have(4.25)Tφ=K+∑l=1nTφll=K+K1+∑l=1nTφllPl,
where 𝒦,K1 are compact operators. Since Pt2=Pt,PtPs=0 and if j≠ℓ, then Tφ is bounded (compact) if and only if the operators TφℓℓPℓ are bounded (compact) operators.SincePℓLa2(D)=La2(Dℓ), then it follows that the operator TφℓℓPℓ is bounded (compact) if and only if Tφℓℓis 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.
Letφ∈L2(D) be an essentially radial function where D=∩j=1nDj 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  Dthen the following are equivalent: (1)
the operator(4.26)Tφ:La2(D,dν)⟶La2(D,dν)
is bounded (compact).(2)
for anyj=1,…,n,  the sequences  γφj={γφj(m)}m∈ℕare in  ℓ∞(ℤ+)(c0(ℤ+)) where, by definition, if  j=2,…,n(4.27)γφj(m)=rj∫rj∞φj(rj(2m+1)/2(m+1)s1/2(m+1)+aj)1s2ds∀m∈Z+,
and for  j=1,
(4.28)γφ1(m)=∫01φ1(s1/2(m+1))ds,∀m∈Z+.Proof.
In the previous theorem, we proved that the operator under examination is bounded (compact) if and only if for anyj=1,…,n  the operators
(4.29)Tφj:L2(Dj,dν)⟶La2(Dj,dν)
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 Δaj,rj={z∈ℂ:0<|z-aj|<rj} and the following maps
(4.30)Δ0,1→αΔaj,rj→βDj,
where α(z)=aj+rjz and β(w)=(w-aj)-1rj2+aj and we use Proposition 1.1 in [8], we can claim that
(4.31)Tφj=Vβ∘α-1Tφj∘β∘αVβ∘α,
where Vβ∘α:L2(Δ0,1)→L2(Dj) is an isomorphism of Hilbert spaces. Therefore, Tφj is bounded (compact) if and only if Tφj∘β∘α is bounded (compact). We also know that this, in turn, is equivalent to the fact that the sequence
(4.32)γφj={γφj(m)}m∈N
is in ℓ∞(ℤ+)(c0(ℤ+)), where
(4.33)γφj(m)=∫01φj∘β∘α(r1/2(m+1))dr,∀m∈Z+.
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)γφj(m)=rj∫rj∞φj(rj(2m+1)/2(m+1)s1/2(m+1)+aj)1s2ds,∀m∈Z+
must hold for any j=2,…,n. The case  j=1 is immediate.Now, we can prove the following.Theorem 4.3.
Letφ∈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)Tφ:La2(Ω,dν)⟶La2(Ω,dν)
is bounded (compact);(2)
for anyj=1,…,n,  the sequences  γφj={γφj(m)}m∈ℕare in  ℓ∞(ℤ+)(c0(ℤ+)) where, by definition, if  j=2,…,n(4.36)γφj(m)=rj∫rj∞φj∘Θ-1(rj(2m+1)/2(m+1)s1/2(m+1)+aj)1s2ds,∀m∈Z+,
and for  j=1(4.37)γφ1(m)=∫01φ1∘Θ-1(s1/2(m+1))ds,∀m∈Z+.Proof.
We know thatΩ 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=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 VΘTφ=Tφ∘Θ-1VΘ. Therefore, the operator Tφ is bounded (compact) if and only if the operator Tφ∘Θ-1:L2(D,dν)→La2(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
(4.38)Tφj∘Θ-1:La2(Dj,dν)⟶La2(Dj,dν)
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∈ℕ are in ℓ∞(ℤ+)(c0(ℤ+)) where, by definition, if j=2,…,n, we have
(4.39)γφj(m)=rj∫rj∞φj∘Θ-1(rj(2m+1)/2(m+1)s1/2(m+1)+aj)1s2ds,∀m∈Z+,
and for  j=1,
(4.40)γφ1(m)=∫01φ1∘Θ-1(s1/2(m+1))ds,∀m∈Z+,
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. Forj=2,…,n, we define  (4.41)Bφj(s)=rj∫rjsφj∘Θ-1(rj1/2x1/2+aj)1x2dx,
and for j=1, we set (4.42)Bφ1(s)=∫s1φ1∘Θ-1(x1/2)dx.We obtain the following useful theorem.Theorem 4.4.
Letφ∈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 Tφ:La2(Ω,dν)→La2(Ω,dν) the following hold true:(1)
if for anyj=1,…,n(4.43)|Bφj(s)|=O(rj-s)ass⟶rj,
then Tφ is bounded;(2)
if for anyj=1,…,n(4.44)|Bφj(s)|=o(rj-s)ass⟶rj,
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 anyj=1,…,n  the sequences γφj={γφj(m)}m∈ℕ are in ℓ∞(ℤ+)(c0(ℤ+)) where, by definition, if j=2,…,n,
(4.45)γφj(m)=rj∫rj∞φj∘Θ-1(rj(2m+1)/2(m+1)s1/2(m+1)+aj)1s2ds∀m∈Z+,
and for  j=1(4.46)γφj(m)=∫01φ1∘Θ-1(s1/2(m+1))ds∀m∈Z+,
and, in virtue of [1]'s main result, it is true that γφj={γφj(m)}m∈ℕ are in ℓ∞(ℤ+) if for any j=1,…,n,
(4.47)|Bφj(s)|=O(rj-s)ass⟶rj,
and γφj={γφj(m)}m∈ℕ are in c0(ℤ+)) if for any j=1,…,n(4.48)|Bφj(s)|=o(rj-s)ass⟶rj.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.
Letφ∈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 Tφ:La2(Ω,dν)→La2(Ω,dν), the following hold true:(1)
Tφ is bounded if and only if
(4.49)|Bφj(s)|=O(rj-s)ass⟶rj,
for any  j=1,…,n,(2)
Tφ is compact if and only if
(4.50)|Bφj(s)|=o(rj-s)ass⟶rj,
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 operatorTφ:La2(Ω,dν)→La2(Ω,dν) has an essentially radial positive symbol φ≥0 such that   for some ℓ=1,…,n, the following (4.51)lim⁡δ→0(inf⁡dist⁡(z,∂Ωl)<δφ(z))=∞
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 whereΩ=∩j=1nΩj 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)φ=∏l=1nφ(l),
where, for any ℓ=1,2,…,n,φ(ℓ) is radial, that is, φ(ℓ)=φ(ℓ)(|z-aℓ|) and satisfies (4.53)inf⁡|z-al|>rl+ϵlφ(l)(|z-al|)=ml>0,sup⁡|z-al|>rl+ϵlφ(l)(|z-al|)=Ml<∞
if ℓ=2,…,n and (4.54)inf⁡|z|<1-ϵ1φ(1)(|z|)=m1>0,sup⁡|z-al|<1-ϵ1φ(1)(|z|)=M1<∞.
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)limsup⁡s→rl|rls-rl∫rlsφ(l)(rl1/2x1/2+al)1x2dx|<C1,limsup⁡s→1|11-s∫s1φ(l)(x1/2)dx|<C1,
for any j=1,(2)
Tφ is compact if for any j=2,…,n(4.56)lim⁡s→rlrls-rl∫rlsφ(l)(rl1/2x1/2+al)1x2dx=0,lim⁡s→111-s∫s1φ(l)(x1/2)dx=0,forj=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 domainΩ, 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 D=∩i=1nDi 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)Ψ:Ω⟶D
the conformal equivalence between Ω and D, then we can define, on Ω, the map (4.58)φu,v=∏l=1nφul,vl(l),
where, for any ℓ, we have (4.59)φul,vl(l)(z)=-(1-ul)(1-dist⁡(Ψ(z),∂Dl)2)-ulsin(1-dist⁡(Ψ(z),∂Dl)2)-vl+vl(1-dist⁡(Ψ(z),∂Dl)2)-vl-ulcos⁡(1-dist⁡(Ψ(z),∂Dl)2)-vl,
where bℓ,aℓ∈(0,∞). It is very easy to see that if we denote with (4.60)Ql={(ul,vl)∈(0,∞)2∣vl+ul<1,ul<vl},
then on the set of parameters 𝒬1×𝒬2⋯×𝒬n, the operator Tφa,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 anyTϕ on La2, we define Tϕ̃, the Berezin transform of Tϕ, by (4.61)ϕ̃(w)=〈Tϕkw,kw〉=∫Ωϕ(z)|kw(z)|2dν(z),
where kw(·)=K(·,w)K(w,w)-1/2. It is quite simple to show that if an operator A∈ℬ(La2(Γ,dν)) is compact, then Ã, 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=∑jm∏kmjTφi,k, 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 operatorTφ with the help of the sequences γφj={γφj(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, ifφ:𝔻→ℂ is radial then Tφ̃(z)=(1-|z|2)2∑(n+1)〈Tφen,en〉|z|2n. 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 limn→∞〈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.
Letφ∈L2(D) be an essentially radial function where D=∩j=1nDj 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 γϕj={γϕj(m)}m∈ℕis in  ℓ∞(ℤ+)  and that there is a constant 𝒞2 such that for j=2,…,n,
(4.62)sup⁡τ∈[aj+rj,∞)|φj(τ)-τ-ajτ-rj-aj∫aj+rjτφj(y)(rj(y-aj)2)dy|<C2,
and for j=1(4.63)sup⁡τ∈[0,1]|φ1(τ)-11-τ∫τ1φ1(s)ds|<C2,
then the operator Tφ:La2(D,dν)→La2(D,dν) is compact if and only if
(4.64)lim⁡w→∂DTφ̃(w)=0.Proof.
We know that the operatorTφ:La2(D,dν)→La2(D,dν) is bounded if and only if for any j=1,…,n  the operators Tφj:La2(Dj,dν)→La2(Dj,dν) are bounded. Since we assume that γϕj={γϕj(m)}m∈ℕ 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 Δ0,1→αΔaj,rj→βDj with Δ0,1={z∈ℂ:0<|z-a|<1} and Δaj,rj={z∈ℂ:0<|z-aj|<rj}, and the maps Δ0,1→αΔaj,rj→βDj where α(z)=aj+rjz and β(w)=(w-aj)-1rj2+aj, we can claim that Tφj=Vβ∘α-1Tφj∘β∘αVβ∘α where Vβ∘α:L2(Δ0,1)→L2(Dj) is an isomorphism of Hilbert spaces. Therefore, Tφj is compact if and only if Tφj∘β∘α is compact. We also know that this, in turn, is equivalent to the vanishing of the Berezin transform if the function
(4.65)(φ∘β∘α)Ber,j(s)=φj∘β∘α(s)-11-s∫s1φj∘β∘α(t)dt
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)(φ∘β∘α)Ber,j(s)=φj∘β∘α(s)-11-s∫(rj/s)+ajaj+rjφj(y)(-rj(y-aj)2)dy=φj∘β∘α(s)-11-s∫aj+rj(rj/s)+ajφj(y)(rj(y-aj)2)dy=φj∘β∘α(s)-11-s∫aj+rj(rj/s)+ajφj(y)(rj(y-aj)2)dy.
Moreover, if we define τ=s-1rj+aj, we can write
(4.67)φBer,j(τ)=φj(τ)-τ-ajτ-rj-aj∫aj+rjτφj(y)(rj(y-aj)2)dy.
Therefore, if we assume that this function is bounded, we can conclude (see [2]) that from limw→Δ0,1Tφj∘β∘α̃(w)=0, it follows that Tφj∘β∘α̃ is compact for j=2,…,n. Therefore, we can infer that Tφj̃ is compact. We also observe that
(4.68)lim⁡w→Δ0,1Tφj∘β∘α̃(w)=0
if and only if limw→∂DjTφj̃(w)=0. To prove this fact, we observe that, by definition, we have
(4.69)Tφj∘β∘α̃(z)=〈Tφj∘β∘αkzΔ0,1,kzΔ0,1〉=∫Δ0,1Tφj∘β∘αkzΔ0,1(w)kzΔ0,1(w)¯dw,
where
(4.70)kzΔ0,1(⋅)=KzΔ0,1(⋅,z)KzΔ0,1(z,z)-1/2.
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)Tφj∘β∘α̃(z)=∫Dj(A(V(β∘α)-1k(β∘α)-1(ς)Δ0,1))k(β∘α)-1(ς)Δ0,1((β∘α)-1)((β∘α)-1)′¯dw,
where A=(V(β∘α)-1Tφj∘β∘αV(β∘α)). Given the relationship between
(4.72)k(β∘α)-1(ζ)Δ0,1((β∘α)-1(w))
and kζDj(w), we obtain
(4.73)Tφj∘β∘α̃(z)=∫Dj(TφjkζDj)(w)kζDj(w)¯dw.
Therefore, it follows that Tφj∘β∘α̃(z)=Tφj̃((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 Lemmas3.1 and  3.2 in Section 2 that, under the stated condition, if
(4.74)lim⁡w→∂DTφ̃(w)=0
then Tφ is a compact operator. To complete the proof, we observe that the compactness of Tφ:La2(D,dν)→La2(D,dν) 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.
Letφ∈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 γϕj={γϕj(m)}m∈ℕis in  ℓ∞(ℤ+)  and that there is a constant C3 such that for j=2,…,n,
(4.75)sup⁡τ∈[aj+rj,∞)|φj∘Θ(τ)-τ-ajτ-rj-aj∫aj+rjτφj∘Θ(y)(rj(y-aj)2)dy|<C3,
and for j=1(4.76)sup⁡τ∈[0,1]|φ1∘Θ(τ)-11-τ∫τ1φ1∘Θ(s)ds|<C3,
then the operator Tφ:La2(Ω,dν)→La2(Ω,dν) is compact if and only if
(4.77)lim⁡w→∂ΩTφ̃(w)=0.



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*Source: 164843-2011-08-22.xml*