#### 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 of when is essentially radial and is multiply-connected domains.

#### 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, if can be written as a finite sum of finite products of Toeplitz operators with -symbols, then is compact if and only if has a Berezin transform which vanishes at the boundary of the disk . 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 symbol . 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 is a bounded multiply-connected domain in the complex plane , whose boundary consists of finitely many simple closed smooth analytic curves where are positively oriented with respect to and if . 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 . 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 Section 2 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

Let be a bounded multiply-connected domain in the complex plane , whose boundary consists of finitely many simple closed smooth analytic curves where are positively oriented with respect to and if . We also assume that is the boundary of the unbounded component of . Let be the bounded component of , and the unbounded component of , respectively, so that .

For we consider the usual -space . The Bergman space , consisting of all holomorphic functions which are -integrable, is a closed subspace of with the inner product given by for . The Bergman projection is the orthogonal projection it is well-known that for any we have where is the Bergman reproducing kernel of . For the Toeplitz operator is defined by where is the standard multiplication operator. A simple calculation shows that We use the symbol to indicate the punctured disk . Let be any one of the domains , .

We call the reproducing kernel of , and we use the symbol to indicate the normalized reproducing kernel; that is, .

For any , the space of bounded operators on , we define , the Berezin transform of , by where .

If , then we indicate with the symbol the Berezin transform of the associated Toeplitz operator , and we have We remind the reader that it is well known that and we have . It is possible, in the case of bounded symbols, to give a characterization of compactness using the Berezin transform (see [5, 6]).

We remind the reader that any bounded multiply-connected domain in the complex plane , whose boundary consists of finitely many simple closed smooth analytic curves , 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 where and for . Here and with if and . Before we state the main result of this paper, we need to give a few definitions.

*Definition 2.1. *Let be a canonical bounded multiply-connected domain. One says that the set of functions is a -partition for iffor every , is a Lipschitz, -function;for every there exists an open set and an such that and the support of are contained in and
for there exists an open set and an such that and the support of are contained in and
for every , , the set is not empty and the function
for any the following equation
holds.

We also need the following.

*Definition 2.2. *A function is said to be essentially radial if there exists a conformally equivalent canonical bounded domain such that, if the map is the conformal mapping from onto , thenfor every and for some , one has
when ,for and for some , one has
when .

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 is essentially radial *via *. Moreover, we stress that in what follows, when we are working with a general multiply-connected domain and we have a conformal equivalence , we always assume that the -partition is given on and transferred to through in the natural way.

*Definition 2.3. *If is an essentially radial function via , for any where is a -partition for then one defines the sequences
as follows: if ,
and if ,

At this point we can state the main result.

Theorem 2.4. *Let be an essentially radial function via and for any where is a -partition for . If the operator is bounded and if for any the sequence satisfies the following
**
then the operator is compact if and only if
*

#### 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 is the disk, , where , then 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 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, if is radial, then where, by definition, Therefore to show that the vanishing of the Berezin transform implies compactness is equivalent, given that is diagonal and to show that implies , 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 disk plays a very important role; for this reason we need the following.

Theorem 3.1. *There exists an isomorphism such that
**
Moreover, for any one has that , and, for any , the Bergman kernels and satisfy the following equation:
*

*Proof. *Suppose that ; this means that is holomorphic on , then we can write down the Laurent expansion of about 0, and we have
This implies that ; therefore we have
The last equation, together with the fact that is square-integrable, implies that if . Then we can conclude that has an holomorphic extension on . We define
in this way: if , then if and
Then and . If , we have just shown that . Clearly is injective and surjective, in fact if , then is an element of and . Then is an isomorphism of onto and . Moreover, observing that implies for any , we conclude that .

Finally, it is easy to verify that for any we have
and this fact implies, by the definition of the Bergman reproducing kernel, that
for any .

In order to better explain our intuition, we remind the reader that we proved that, if is an essentially radial function where is a bounded multiply-connected domain and if we define where where is a -partition for , then the fact that the operator is bounded (compact) is equivalent to fact, that for any , the operators 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. *Let be an essentially radial function, if one defines where and is a -partition for . Then for the bounded operator the following are equivalent:*(1)*the operator is compact;*(2)*for any one has
*

*Proof. *Since , we know that the operator is bounded, and we know that the boundedness (compactness) is equivalent to the fact that for any the operators are bounded (compact). If , we observe that if we consider the following sets and and the maps where and and we use Proposition 1.1 in [8], we can claim that where is an isomorphism of the Hilbert spaces. Therefore is compact if and only if is compact. We also notice that the previous theorem implies that function is an orthonormal basis for , and this, in turn, implies that the compactness of the operator is equivalent to the fact that for the sequence we have where, by definition,
To complete the proof we observe that, since is radial and , then, after a change of variable, we can rewrite the last integral, and hence the formula
must hold for any . For the case the proof is similar.

Now we can prove the following.

Theorem 3.3. *Let be an essentially radial function via , if one defines where and is a -partition for . Then for the bounded operator the following are equivalent:*(1)*the operator is compact;*(2)*for any one has
*

*Proof. *We know that is a regular domain, and therefore if is a conformal mapping from onto then the Bergman kernels of and are related via and the operator is an isometry from onto (see [8, Proposition 1.1]). In particular we have and this implies that . Therefore the operator is bounded if and only if for any the operators are bounded (compact). Hence we can conclude that the operator is bounded (compact) if for any we have
where, by definition, if ,
and if ,

Theorem 3.4. *Let be an essentially radial function, if one defines where and is a -partition for and the operator is bounded (compact) and if for any the sequences are such that
**
is finite, then the operator is compact if and only if
*

*Proof. *We know that the operator under examination is bounded (compact) if and only if for any the operators
are bounded (compact). If , we observe that if we consider the following sets and and the following maps
where and and we use Proposition 1.1 in [8], we can claim that
where is an isomorphism of Hilbert’s spaces. Therefore is compact if and only if is compact. Since and is an orthonormal basis, a simple calculation shows that
therefore our assumption on that
implies (see [9, Theorem 6]) that the compactness of is equivalent to the fact that the Berezin transform vanishes at the boundary. Since the case is immediate, we can conclude, from what we proved so far, that for any we have that compactness is equivalent to the fact that . To complete the proof we set, for any , where is in the -partition for . By definition of -partition, it follows that is , and we can write
Since for any the quantity can be written as
we observe that the function is bounded on the set and vanishes at the boundary.

In fact, to prove this we remind the reader that there exists an isomorphism such that and the Bergman kernels and satisfy the following equation . If we define and , then for all . The well-known fact that the reproducing kernel of the unit disk is given by implies that therefore, by conformal mapping, that the reproducing kernel of is for all . If we define by and using the fact that , we obtain that for all . Since and, for , , then we prove that if .

Hence, for the function , we have
Hence it follows that is bounded on the set and goes to zero as it approaches the boundary. Since and has a finite measure, we can conclude that, by the dominated convergence theorem, we have
Now we observe that Lemma 2 in [4] implies that
goes to zero if and only , and a simple calculation shows that
Hence, as a consequence, we have shown that, if the conditions in the hypothesis hold, then
and this completes the proof since .

Now we can prove the following.

Theorem 3.5. *Let be an essentially radial function via and for any where is a -partition for . If the operator is bounded and if for any the sequence satisfies the following
**
then the operator is compact if and only if
*

*Proof. *We know that is a regular domain, and therefore, if is a conformal mapping from onto then the Bergman kernels of and are related via and the operator is an isometry from onto (see [8, Proposition 1.1]). In particular we have and this implies that . Therefore the operator is bounded (compact) if and only if the operator is bounded (compact). In the previous theorem we proved that the operator in exam is bounded (compact) if and only if for any the operators are bounded (compact). Hence, since the sequences satisfy the stated properties, we can conclude that the operators are compact if and only if for any we have
Therefore it follows that
and, since is a conformal mapping, this implies that

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

Theorem 3.6. *Let be an essentially radial function via and for any where is a -partition for . If the operator is bounded and if for any the sequence satisfies the following
**
then the operator is compact if and only if
*

Finally, we observe that it is also to recover as corollary the following.

Corollary 3.7. *Let be an essentially radial symbol via the conformal equivalence . If one defines where and is a -partition for . Let us assume that is in and that there is a constant such that for **
and for **
Then the operator is compact if and only if
*

The last corollary was also proved, in different way, in [4].