# Toeplitz Operators with Essentially Radial Symbols

Authors: Roberto C. Raimondo
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

For Topelitz operators with radial symbols on the disk, there are important results that characterize boundedness, compactness, and its relation to the Berezin transform. The notion of essentially radial symbol is a natural extension, in the context of multiply-connected domains, of the notion of radial symbol on the disk. In this paper we analyze the relationship between the boundary behavior of the Berezin transform and the compactness ofTϕ when ϕ∈L2(Ω) is essentially radial and Ω is multiply-connected domains.


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## Body

## 1. Introduction

Toeplitz operators are object of intense study. Many papers have been dedicated to the study of these concrete class of operators generating many interesting results. A very important tool to study the behavior of these operators is the Berezin transform. This tool is particularly relevant with its connections with quantum mechanics, especially in the case of the Toeplitz operators on the Segal-Bargmann space. In this case, they arises naturally as anti-Wick quantization operators, and there is a natural equivalence between Toeplitz operators and a generalization of pseudodifferential operators, the so-called Weyl’s quantization.In a fundamental paper, Axler and Zheng proved that, ifS∈ℬ(L2(D)) can be written as a finite sum of finite products of Toeplitz operators with L∞-symbols, then S is compact if and only if S has a Berezin transform which vanishes at the boundary of the disk D. As they expected, this result has been extended into several directions, and it has been proved even for operators which are not of the Toeplitz type. Therefore it has been an important open problem to characterize the class of operators for which the compactness is equivalent to the vanishing of the Berezin transform. Since there are operators which are not compact but have a Berezin transform which vanishes at the boundary, it is now clear that the two notions are not equivalent. Moreover, it is possible to show that in the context of Toeplitz operators there are examples of unbounded symbols whose corresponding operators are bounded and even compact.Recently, many papers have been written in the case when the operator has an unbounded radial symbolφ∈L2(D). Of course, for a square-integrable symbol, the Toeplitz operator is densely defined but is not necessarily bounded. However, it is possible (see [1] of Grudsky and Vasilevski, [2] of Zorboska, and [3] of Korenblum and Zhu) to show that operators with unbounded radial symbols can have a very rich structure. Moreover, there is a very neat and elegant way to characterize boundedness and compactness. The reason being that the operators with radial symbols on the disk are diagonal operators. In this context the relation between compactness and the Berezin transform has been studied in depth, and interesting results have been established.In a previous paper (see [4]), the author showed that it is possible to extend the notion of radial symbol when Ω is 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. The key ingredient for this extension is to observe two facts. The first fact is 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 the Toeplitz operators with square integrable symbols. 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 this class of essentially radial symbols, the compactness and boundedness have been studied and necessary and sufficient conditions established. In this paper we carry forward our analysis by investigating the relationship between the compactness and the vanishing of the Berezin transform. It is important to observe that in the case of the disk the analysis uses the fact that the Berezin transform can be easily written in a simple way since we can write explicitly an orthonormal basis, namely the collection of functions {k+1zk}k=0∞. In the case of a planar domain, this is not possible because it is very hard to construct explicitly an orthonormal basis for the Bergman space. However, it is possible to reach interesting results that fully extend what it is known in the case of the disk.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 prove the main result and we study several important consequences.


## 2. Preliminaries

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(2.1)〈f,g〉=∫Ωf(z)g(z)¯dν(z),
for f,g∈L2(Ω,dν). The Bergman projection is the orthogonal projection(2.2)P:L2(Ω,dν)⟶La2(Ω,dν);
it is well-known that for any f∈L2(Ω,dν) we have(2.3)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(2.4)Tφf(z)=∫Ωφ(w)f(w)KΩ(w,z)dν(w).
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)), the space of bounded operators on La2(Γ,dν), we define Ã, the Berezin transform of A, by(2.5)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(2.6)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 [5, 6]).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 [7]). 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 result of this paper, we need to give a few definitions.Definition 2.1.
LetΩ=⋂i=1nΩi be a canonical bounded multiply-connected domain. One says 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 are contained in Wj and
(2.7)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 are contained in W1 and
(2.8)p1(ζ)=1∀ζ∈Uϵ1;(4)
for everyj,k=1,…,n, Wj∩Wk=∅, the set Ω∖(⋃j=1nWj) is not empty and the function
(2.9)p0(ζ)=1∀ζ∈(⋃j=1nWj)c∩Ω,p0(ζ)=0∀ζ∈Uϵk,k=1,…,n,(5)
for anyζ∈Ω the following equation
(2.10)∑k=0npk(ζ)=1holds.We also need the following.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, one has
(2.11)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, one has
(2.12)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. Moreover, we stress that in what follows, when we are working with a general multiply-connected domain and we have a conformal equivalence Θ:⋂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.Definition 2.3.
Ifφ∈L2(Ω) is an essentially radial function via Θ:⋂i=1nΩi→⋂i=1nDi, φj=φ·pj for any j=1,…,n where 𝔓={p0,p1,…,pn} is a ∂-partition for Ω then one defines the n sequences
(2.13)aφ1={aφ1(k)}k∈Z+,aφ2={aφ2(k)}k∈Z+,…,aφn={aφn(k)}k∈Z+
as follows: if j=2,…,n,
(2.14)aφj(k)=rj∫rj∞φj∘Θ-1(rjs+aj)(k+1)rj2k+1s2k+11s2ds∀k∈Z+,
and if j=1,
(2.15)aφ1(k)=∫01φ1∘Θ-1(s)(k+1)s2k+1ds∀k∈Z+.At this point we can state the main result.Theorem 2.4.
Letφ∈L2(Ω) be an essentially radial function via Θ:⋂ℓ=1nΩℓ→⋂ℓ=1nDℓ and φj=φ·pj for any j=1,…,n where 𝔓={p0,p1,…,pn} is a ∂-partition for Ω. If the operator Tφ:La2(Ω,dν)→La2(Ω,dν) is bounded and if for any j=1,…,n the sequence aφj={aφj(k)}k∈ℤ+ satisfies the following
(2.16)sup⁡k∈Z+{|(k+1)aφj(k)-kaφj(k-1)|}<∞,
then the operator Tφ:La2(Ω,dν)→La2(Ω,dν) is compact if and only if
(2.17)lim⁡w→∂ΩTφ̃(w)=0.


## 3. Canonical Multiply-Connected Domains and Essentially Radial Symbols

We concentrate on the relationship between compact Toeplitz operators and the Berezin transform. As we said in the introduction, Axler and Zheng have proved (see [5]) that if D is the disk, S=∑im∏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 [6]. In this section we 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.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(3.1)Tφ̃(z)=(1-|z|2)2∑(n+1)〈Tφen,en〉|z|2n,
where, by definition,(3.2)en(z)=n+1zn∀n∈Z+.
Therefore to show that the vanishing of the Berezin transform implies compactness is equivalent, given that Tφ is diagonal and 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 fact in 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. In fact, for an essentially radial function, the values depend essentially on the distance from the boundary. Moreover, we can simplify our analysis if we use the fact that every multiply-connected domain is conformally equivalent to a canonical bounded multiply-connected domain whose boundary consists of finitely many circles. It is important to stress that in the case of essentially radial symbol it is possible to exploit what has been done in the case of the disk, but the operator is not a diagonal operator, and the Berezin transform is not particularly simple to write in an explicit way.In what follows the punctured diskΔ={z∈ℂ∣0<|z|<1} plays a very important role; for this reason we need the following.Theorem 3.1.
There exists an isomorphismI:L2(Δ)→L2(Ω1) such that
(3.3)I(La2(Δ))=La2(Ω1).
Moreover, for any p≥2 one has that Lap(Δ)=Lap(Ω1), and, for any (z,w)∈Δ2, the Bergman kernels KΔ and KΩ1 satisfy the following equation:
(3.4)KΔ(z,w)=KΩ1(z,w).Proof.
Suppose thatf∈La2(Δ); this means that f is holomorphic on Δ, then we can write down the Laurent expansion of f about 0, and we have
(3.5)f(z)=∑n=-∞∞anzn.
This implies that |f(z)|2=∑n,m=-∞∞anam¯znzm¯; therefore we have
(3.6)∫Δ|f(z)|2dν(z)=∫Δ∑n,m=-∞∞anam¯znzm¯dν(z)=∫02π∫01∑n,m=-∞∞anam¯rn+m+1ei(n-m)θdrdθ=∑n,m=-∞∞anam¯∫02πei(n-m)θdθ∫01rn+m+1dr=2π∑n=-∞∞|an|2∫01r2n+1dr=2π(∑n≠-1∞|an|2[r2n+22n+2]01+|a-1|2∫011rdr).
The last equation, together with the fact that f is square-integrable, implies that an=0 if n≤-1. Then we can conclude that f has an holomorphic extension on Ω1. We define
(3.7)I:L2(Δ)⟶L2(Ω1)
in this way: if g∈L2(Δ), then ℐg(z)=g(z) if z≠0 and
(3.8)Ig(0)=∫Δg(z)dν(z).
Then ℐg∈L2(Ω1) and ∥ℐg∥Ω1=∥g∥Δ. If f∈La2(Δ), we have just shown that ℐf∈La2(Ω1). Clearly ℐ is injective and surjective, in fact if G∈L2(Ω1), then g=G|Δ is an element of L2(Δ) and ℐ(g)=G. Then ℐ is an isomorphism of L2(Δ) onto L2(Ω1) and ℐ(La2(Δ))=La2(Ω1). Moreover, observing that p>2 implies ∥f∥Δ,2≤∥f∥Δ,p for any f∈Lp(Δ), we conclude that Lap(Δ)=Lap(Ω1).
Finally, it is easy to verify that for anyf,g∈La2(Δ) we have
(3.9)〈f,g〉Δ=〈If,Ig〉Ω1,
and this fact implies, by the definition of the Bergman reproducing kernel, that
(3.10)KΔ(z,w)=KΩ1(z,w),
for any (z,w)∈Δ2.In order to better explain our intuition, we remind the reader that we proved that, ifφ∈L2(D) is an essentially radial function where Ω is a bounded multiply-connected domain and if we define φj=φ·pj where j=1,…,n where 𝔓={p0,p1,…,pn} is a ∂-partition for Ω, then the fact that the operator Tφ:La2(Ω,dν)→La2(Ω,dν) is bounded (compact) is equivalent to fact, that for any j=1,…,n, the operators Tφj:La2(Ωj,dν)→La2(Ωj,dν) are bounded (compact) (see [4]).We start our investigation by focusing our attention on the case of bounded symbols. In fact, we prove the following.Theorem 3.2.
Letφ∈L∞(D) be an essentially radial function, if one defines φj=φ·pj where j=1,…,n and 𝔓 is a ∂-partition for D. Then for the bounded operator Tφ the following are equivalent:(1)
the operatorTφ:La2(D,dν)→La2(D,dν) is compact;(2)
for anyj=1,…,n one has
(3.11)lim⁡k→∞aφj(k)=0.Proof.
Sinceφ∈L∞(D), we know that the operator Tφ:La2(Ω,dν)→La2(Ω,dν) is bounded, and we know that the boundedness (compactness) is equivalent to the fact that for any j=1,…,n the operators 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 maps 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 Tφj=Vβ∘α-1Tφj∘β∘αVβ∘α where Vβ∘α:L2(Δ0,1)→L2(Dj) is an isomorphism of the Hilbert spaces. Therefore Tφj is compact if and only if Tφj∘β∘α is compact. We also notice that the previous theorem implies that function {k+1zk} is an orthonormal basis for L2(Δ0,1), and this, in turn, implies that the compactness of the operator Tφj∘β∘α is equivalent to the fact that for the sequence aφj={aφj(k)}k∈ℕ we have limk→∞aφj(k)=0 where, by definition,
(3.12)aφj(k)=∫Δ0,1φj∘β∘α(z)(k+1)zkz¯kdz∀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 hence the formula
(3.13)aφj(k)=rj∫rj∞φj(rjs+aj)(k+1)rj2k+1s2k+11s2ds∀m∈Z+
must hold for any j=2,…,n. For the case j=1 the proof is similar.Now we can prove the following.Theorem 3.3.
Letφ∈L∞(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 for the bounded operator Tφ the following are equivalent:(1)
the operatorTφ:La2(Ω,dν)→La2(Ω,dν) is compact;(2)
for anyj=1,…,n one has
(3.14)lim⁡k→∞aφj(k)=0.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) onto L2(Ω) (see [8, Proposition  1.1]). In particular we have VΘPD=PΩVΘ and this implies that VΘTφ=Tφ∘Θ-1VΘ. Therefore the operator Tφ is bounded if and only if for any j=1,…,n the operators Tφj∘Θ-1:La2(Dj,dν)→La2(Dj,dν) are bounded (compact). Hence we can conclude that the operator is bounded (compact) if for any j=1,…,n we have
(3.15)lim⁡k→∞aφj(k)=0,
where, by definition, if j=2,…,n,
(3.16)aφj(k)=rj∫rj∞φj∘Θ-1(rjs+aj)(k+1)rj2k+1s2k+11s2ds∀k∈Z+,
and if j=1,
(3.17)aφ1(k)=∫01φ1∘Θ-1(s)(k+1)s2k+1ds∀k∈Z+.Theorem 3.4.
Letφ∈L2(D) be an essentially radial function, if one defines φj=φ·pj where j=1,…,n and 𝔓 is a ∂-partition for Ω and the operator Tφ:La2(D,dν)→La2(D,dν) is bounded (compact) and if for any j=1,…,n the sequences aφj={aφj(k)}k∈ℕ are such that
(3.18)sup⁡k∈N{|(k+1)aφj(k)-kaφj(k-1)|}
is finite, then the operator Tφ:La2(D,dν)→La2(D,dν) is compact if and only if
(3.19)lim⁡w→∂DTφ̃(w)=0.Proof.
We know that the operator under examination is bounded (compact) if and only if for anyj=1,…,n the operators
(3.20)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
(3.21)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
(3.22)Tφj=Vβ∘α-1Tφj∘β∘αVβ∘α,
where Vβ∘α:L2(Δ0,1)→L2(Dj) is an isomorphism of Hilbert’s spaces. Therefore Tφj is compact if and only if Tφj∘β∘α is compact. Since Tφj∘β∘α:La2(Δ0,1)→La2(Δ0,1) and {k+1zk}k=0∞ is an orthonormal basis, a simple calculation shows that
(3.23)aφj(k)=〈Tφj∘β∘αk+1zk,k+1zk〉;
therefore our assumption on aφj={aφj(k)}k∈ℕ that
(3.24)sup⁡k∈Z+{|(k+1)aφj(k)-kaφj(k-1)|}<∞
implies (see [9, Theorem  6]) that the compactness of Tφj∘β∘α is equivalent to the fact that the Berezin transform vanishes at the boundary. Since the case j=1 is immediate, we can conclude, from what we proved so far, that for any j=1,2,…,n we have that compactness is equivalent to the fact that limz→∂Dℓφℓ̃Dℓ(z)=0. To complete the proof we set, for any j=1,…,n, 𝒮j={w∈Ω∣pj(w)=1} where {p1,…,pn} is in the ∂-partition for Ω. By definition of ∂-partition, it follows that 𝒮j∩𝒮i=∅ is j≠i, and we can write(3.25)φ̃(z)=〈TφkzD,kzD〉=∫Dφ(w)|kzD(w)|2dw=∫Sjφl(w)|kzD(w)|2dw+∫Ω∩Sjcφl(w)|kzD(w)|2dw.
Since for any ℓ=1,…,n the quantity ∫𝒮ℓφℓ(w)|kzD(w)|2dw can be written as
(3.26)∫Sl(φl(w)(KzDl)2‖KzDl‖22)‖KzDl‖22‖KzD‖22dw+∫Slφl(w)(|kzD(w)|2-(KzDl)2‖KzD‖22)dw,
we observe that the function θℓ(z,w)=(|kzD(w)|2-(KzDℓ)2/∥KzD∥22) is bounded on the set 𝒮ℓ and vanishes at the boundary.
In fact, to prove this we remind the reader that there exists an isomorphismI:L2(Δ0,1)→L2(Ω1) such that I(La2(Δ0,1))=La2(Ω1) and the Bergman kernels KΔ and KΩ1 satisfy the following equation KΔ(z,w)=KΩ1(z,w). If we define Δa,r={z∈ℂ:0<|z-a|<r} and Oa,r={z∈ℂ:|z-a|>r}, then KOa,r(z,w)=r2/(r2-(z-a)·(w-a)¯)2 for all (z,w)∈Oa,r×Oa,r. The well-known fact that the reproducing kernel of the unit disk is given by (1-zw¯)-2 implies that KΔ0,1(z,w)=1/(1-z·w¯)2for all(z,w)∈Δ0,1×Δ0,1 therefore, by conformal mapping, that the reproducing kernel of Δa,r is KΔa,r(z,w)=r2/(r2-(z-a)·(w-a)¯)2 for all (z,w)∈Δa,r×Δa,r. If we define ϕ:Δa,r→Oa,r by ϕ(z)=(z-a)-1r2+a and using the fact that KOa,r(ϕ(z),ϕ(w))ϕ′(z)ϕ′(w)¯=KΔa,r(z,w), we obtain that KOa,r(z,w)=r2/(r2-(z-a)·(w-a)¯)2 for all (z,w)∈Oa,r×Oa,r. Since Ω1=O0,1 and, for j=2,…,n, Oaj,rj=Ωj, then we prove that KDj(z,w)=rj2/(rj2-(z-aj)·(w-aj)¯)2 if j=2,…,n.
Hence, for the functionθℓ(z,w), we have
(3.27)|θl(ζ,z)|=|(∑m=0nKmD(ζ,z)‖KzD‖22)2-(KzDl)2‖KzD‖22|=|(KzDl)2‖KzD‖22(1+∑m≠lnKmD(ζ,z)KzDl(ζ,z))2-(KzDl)2‖KzD‖22|=|(KzDl)2‖KzD‖22(1+∑m≠lm>0nrm2/(rm2-(z-am)⋅(ζ-am)¯)2rl2/(rl2-(z-al)⋅(ζ-al)¯)2+K0D(ζ,z)KzDl(ζ,z))2-(KzDl)2‖KzD‖22|.
Hence it follows that θℓ(z,w) is bounded on the set 𝒮ℓ and goes to zero as it approaches the boundary. Since φ∈L2(D) and D has a finite measure, we can conclude that, by the dominated convergence theorem, we have
(3.28)lim⁡z→∂Dl∫Slφl(w)(|kzD(w)|2-(KzDl)2‖KzD‖22)dw=0.
Now we observe that Lemma  2 in [4] implies that
(3.29)lim⁡z→∂Dl∫D∩Sl(φl(w)(KzDl‖KzDl‖2)2)(‖KzDl‖2‖KzD‖2)2dw
goes to zero if and only limz→∂Dℓφℓ̃Dℓ(z)=0, and a simple calculation shows that
(3.30)∫Ω∩Sjcφl(w)|kzD(w)|2dw=0.
Hence, as a consequence, we have shown that, if the conditions in the hypothesis hold, then
(3.31)lim⁡z→∂Dlφ̃D(z)=0,
and this completes the proof since ∂D=⋃1n∂Dℓ.Now we can prove the following.Theorem 3.5.
Letφ∈L2(Ω) be an essentially radial function via Θ:⋂ℓ=1nΩℓ→⋂ℓ=1nDℓ and φj=φ·pj for any j=1,…,n where 𝔓={p0,p1,…,pn} is a ∂-partition for Ω. If the operator Tφ:La2(Ω,dν)→La2(Ω,dν) is bounded and if for any j=1,…,n the sequence aφj={aφj(k)}k∈ℤ+ satisfies the following
(3.32)sup⁡k∈Z+{|(k+1)aφj(k)-kaφj(k-1)|}<∞,
then the operator Tφ:La2(Ω,dν)→La2(Ω,dν) is compact if and only if
(3.33)lim⁡w→∂ΩTφ̃(w)=0.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) onto L2(Ω) (see [8, Proposition  1.1]). 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 Tφj∘Θ-1:La2(Dj,dν)→La2(Dj,dν) are bounded (compact). Hence, since the sequences aφj={aφj(m)}m∈ℕ satisfy the stated properties, we can conclude that the operators Tφj∘Θ-1:La2(Dj,dν)→La2(Dj,dν) are compact if and only if for any j=1,…,n we have
(3.34)lim⁡z→∂Djφj∘Θ-1̃Dj(z)=0.
Therefore it follows that
(3.35)lim⁡z→∂Dφ∘Θ-1̃D(z)=0,
and, since Θ is a conformal mapping, this implies that
(3.36)lim⁡z→∂Ωφ̃Ω(z)=0.Finally, we also observe that as a simple consequence we obtain the following.Theorem 3.6.
Letφ∈L2(Ω) be an essentially radial function via Θ:⋂ℓ=1nΩℓ→⋂ℓ=1nDℓ and φj=φ·pj for any j=1,…,n where 𝔓={p0,p1,…,pn} is a ∂-partition for Ω. If the operator Tφ:La2(Ω,dν)→La2(Ω,dν) is bounded and if for any j=1,…,n the sequence aφj={aφj(k)}k∈ℤ+ satisfies the following
(3.37)sup⁡k∈Z+{|k(aφj(k)-aφj(k-1))|}<∞,
then the operator Tφ:La2(Ω,dν)→La2(Ω,dν) is compact if and only if
(3.38)lim⁡w→∂ΩTφ̃(w)=0.Finally, we observe that it is also to recover as corollary the following.Corollary 3.7.
Letφ∈L2(Ω) be an essentially radial symbol 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∈ℕ is in ℓ∞(ℤ+) and that there is a constant C3 such that for j=2,…,n(3.39)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(3.40)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
(3.41)lim⁡w→∂ΩTφ̃(w)=0.The last corollary was also proved, in different way, in [4].



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*Source: 492690-2012-01-17.xml*