#### Abstract

We study the problem of the boundedness and compactness of when and 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

#### 1. Introduction

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
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 for the Toeplitz operators 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 . In Section 4, we prove the main result and study several important consequences.

#### 2. Preliminaries

Let be the bounded multiply-connected domain given at the beginning of Section 1, that is, , where is the bounded component of , and is the unbounded component of . 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 , 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 [4, 5]).

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 [6]). 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 results of this paper we need to give a few definitions.

*Definition 2.1. *Let be a canonical bounded multiply-connected domain. We say that the set of functions is a -partition for if(1)for every is a Lipschitz, -function,(2)for every , there exists an open set and an such that , and the support of is contained in and
(3)for , there exists an open set and an such that and the support of is contained in and (4)for every , the set is not empty and the function (5) for any , the following equation:
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 is different from zero.

*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 , then(1)for every and for some , we have
when ,(2)for and for some , we have
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 *.

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 , we always assume that the -partition is given on and transferred to through in the natural way.

At this point, we can state the main result.

Theorem 2.3. *Let be an essentially radial function via , if one defines , where and is a -partition for , then the following are equivalent: *(1)*the operator
is bounded (compact).*(2)*for any the sequences are in where, by definition, if ,
and if *

#### 3. The Structure of and Some Estimates about the Bergman Kernel

From now on, we will assume that where and for . Here, and with if and . We will indicate with the symbol the punctured disk .

With the symbols , we denote the Bergman kernel on , and , 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 and the disk , we have , if , and, for any (see [7, 8]). This fact has an important and simple consequence. In fact, if we consider and , we can conclude that

To see this, we use the well-known fact that the reproducing kernel of the unit disk is given by , therefore we have This implies, by conformal mapping, that the reproducing kernel of is Now, we define by and we use the well-known fact that the Bergman kernels of and are related via to obtain that Since and, for , then the last equations implies that if .

We also note that if we define we can prove the following.

Lemma 3.1. *(1) is conjugate symmetric about z and w. For each is conjugate analytic on and . **(2) There are neighborhoods of and a constant such that is empty if and
**
for and . ** (3) .*

* Proof. *(1) Since the Bergman kernels and have these properties (see [9]), by the definition of , we get (1).

(2) The proof is given in [7, 8].

(3) Using the fact that
for and (1) and (2), we get (3).

We observe that we can choose for and such that and , then we have , where is the same as in Lemma 3.1. We also have the following.

Lemma 3.2. *There are constants and such that *(1)*for any , one has
*(2)*for any , one has .*

*Proof. *By the explicit formula of the Bergman kernels , there are constants and such that
for and
if for . From the last Lemma, it follows that
whenever . If we call the biggest number among and we let , 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 is actually the intersection of other domains, that is, . This also suggests that we should look for a representation of the elements of that reflects this fact. For this reason, we give the following.

*Definition 3.3. *Given with and , for any , we define functions as follows: if , then we set, for ,
for ,
and for ,
where are the circles which center at and lie in (see Lemma 3.2), respectively, so that is exterior to and interior to .

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

Lemma 3.4. *For , one can write it uniquely as
**
with if , and moreover, there exists a constant such that, for , one has
**
In particular, if , then and
**
for .*

* Proof. *Let be any function analytic on . For any , let be the circles which center at and lie in , respectively, so that is exterior to ) and interior to . Using Cauchy's Formula, we can write
Let
By Cauchy's Formula, the value does not depend on the choice of if and . Of course, each is well defined for all and analytic in . In addition, if , we have that as . Writing the Laurent expansion at of , we have
and, for ,
and these series converge to uniformly and absolutely on any compact subset of , respectively. We remark that the coefficients are given by the following formula:
where if and if and ,. Moreover, if is holomorphic in some and as when , then for all by Cauchy's theorem and, therefore, .

Now, we define and
for and
then for all and if as we have proved above.

We claim that implies that for , respectively. Indeed, since each annulus is contained in implies that is an element of for all .

For any fixed , note that and are analytic on and for . Expanding them as Laurent series, it follows that:(1)If , then for ,(2)If , then
for and
It is obvious that, in any case, these series converge uniformly and absolutely on . Observing that each is an annulus at , we have, by direct computation, that
if and
Therefore, for any , there exists a constant such that
From the definition of , we derive
for . The convergence of these series is guaranteed by the conditions and . Since and , it follows that and
for . Comparing the expression of with the expression of , it follows that for some constant for . Hence, . Moreover, if we define , from the inequalities and and the definition of , it follows that .

If for some , note that as for , then if and if . For , since implies that , then . We must have and, consequently, . Hence, in any case, implies and if , and this remark completes our proof.

Lemma 3.5. *If is a bounded sequence in and weakly in , then weakly on for and uniformly on .*

* Proof. *By the previous Lemma, we know that the linear transformations are bounded operators, then weakly in implies that weakly on for . For the same reason, weakly in and then for any . Since
by the estimates given in the last lemma, we have that . The boundedness of implies that the family of continuous functions is uniformly bounded and equicontinuous on , then, by Arzela-Ascoli's Theorem, we have that uniformly on .

#### 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 for the Toeplitz operator to be bounded or compact.

Before we state the next Theorem, we remind the reader that where and, for all , we have where is the reproducing kernel of . If we use the symbol to indicate , we can write We also remind the reader that if is the identity operator, then where is a bounded operator for all with if and and if (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 .

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

Lemma 4.1. *Let be an essentially radial function where with and for . If one defines where and is a -partition for then the following are equivalent: *(1)*the operator
is bounded (compact);*(2)*for any , the operators
are bounded (compact).*

*Proof. *Let be a partition of the unit on , which is a canonical domain. Now, we notice that for all and for all , we have the following:
where, by definition, we have

*Claim 1. *The operator is Hilbert-Schmidt for any .

*Proof. *We observe that, by definition, we have
therefore, if we define
we have
This implies that for any , is Hilbert-Schmidt. Therefore, the operator
is Hilbert-Schmidt, and this completes the proof of the claim.

*Claim 2. *The operator is Hilbert-Schmidt for any .

*Proof. *We observe that, by definition, we have
therefore, if we define
we have
This implies that for any , is Hilbert-Schmidt. Therefore, the following
is Hilbert-Schmidt, and this completes the proof of the claim.

*Claim 3. *The operator is Hilbert-Schmidt if and .

*Proof. *We observe that
To start, we give the following:
We will show that Fubini theorem and the properties of the -partition imply that
In fact, we have

Therefore, we can write that
where is a compact operator.

We also observe that Lemma 3.4 implies that , and we prove that the operator is compact if and .

*Proof. *In order to simplify the notation, we define the operator . To prove our statement, it is enough to prove that if we take a bounded sequence in such that weakly, then we can prove that . We know that the continuity of implies that weakly on , and is bounded by Lemma 3.5. Since it is a sequence of holomorphic functions, we know that is uniformly bounded on any compact subset of . Therefore, the sequence is a normal family of functions. Since for any , then converges uniformly on any compact subset of and consequently on . To complete the proof, we remind the reader that if we define the operators , for , in this way
It is possible to prove, with the help of Schur's test (see [11] ), that is a bounded operator (see [5]). Now, we observe that
then, by using the fact that is bounded, we have
and this completes the proof of our claim. Notice also that using the same strategy, we can prove that each is compact.

Therefore, we have where are compact operators. Since and if , then is bounded (compact) if and only if the operators are bounded (compact) operators.

Since , then it follows that the operator is bounded (compact) if and only if is 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 be an essentially radial function where with and for . If one defines where and is a -partition for then the following are equivalent: *(1) *the operator
is bounded (compact).*(2) *for any , the sequences are in where, by definition, if
and for ,
*

* Proof. *In the previous theorem, we proved 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 spaces. Therefore, is bounded (compact) if and only if is bounded (compact). We also know that this, in turn, is equivalent to the fact that the sequence
is in , where
To complete the proof, we observe that since is radial and then, after a change of variable, we can rewrite the last integral, and therefore the formula
must hold for any . The case is immediate.

Now, we can prove the following.

Theorem 4.3. *Let be an essentially radial function via the conformal equivalence , define where and is a -partition for , then the following conditions are equivalent: *(1)*the operator
is bounded (compact);*(2)*for any , the sequences are in where, by definition, if
and for *

*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 Proposition 1.1 in [8]). 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, we can conclude that the operator is bounded (compact) if and only if for any the sequences are in where, by definition, if , we have
and for ,
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 , we define and for , we set

We obtain the following useful theorem.

Theorem 4.4. *Let be an essentially radial function via the conformal equivalence . If one defines where and is a -partition for , then for the operator the following hold true: *(1)*if for any
then is bounded;*(2)*if for any
then 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 the sequences are in where, by definition, if ,
and for
and, in virtue of [1]'s main result, it is true that are in if for any ,
and are in if for any

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 be an essentially radial function via the conformal equivalence . If we define where and is a -partition for and if a.e. in , then for the operator , the following hold true: *(1)* is bounded if and only if
for any ,*(2)* is compact if and only if
for any .*

* 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 has an essentially radial positive symbol such that for some , the following holds, then the operator is unbounded. Moreover, if is bounded and the symbol is an unbounded essentially radial function, then it must be true that around any , the symbol has an oscillating behavior.

In order to present an application, we consider a family of examples. Let us consider the case where with and for . Let be a function that can be written in the following way:
where, for any is radial, that is, and satisfies
if and
if . As a consequence of our results, we can conclude that (1) is bounded if there exists a constant such that for any ,
for any ,(2) is compact if for any *for *.

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 , whose boundary consists of finitely many simple closed smooth analytic curves, there exist unbounded functions such that is compact even when the sufficient conditions are not satisfied. To prove this claim, we observe that for the domain there exists a conformally equivalent domain where and for where and with if and . If we denote with the symbol the conformal equivalence between and , then we can define, on , the map where, for any , we have where . It is very easy to see that if we denote with