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Schatten Class Toeplitz Operators on the Bergman Space
We have shown that if the Toeplitz operator on the Bergman space belongs to the Schatten class , then , where is the Berezin transform of , and is the normalized area measure on the open unit disk . Further, if then and . For certain subclasses of , necessary and sufficient conditions characterizing Schatten class Toeplitz operators are also obtained.
Let be the open unit disc in the complex plane Let and be the Hilbert space of complex-valued, absolutely square-integrable, Lebesgue measurable functions on with the inner product Let denote the Banach space of Lebesgue measurable functions on with and let be the space of bounded analytic functions on . Let be the subspace of consisting of analytic functions. The space is called the Bergman space. Since point evaluation at is a bounded linear functional on the Hilbert space , the Riesz representation theorem implies that there exists a unique function in such that for all in . Let be the function on defined by The function is thus the reproducing kernel for the Bergman space and is called the Bergman kernel. It can be shown that the sequence of functions forms the standard orthonormal basis for and . The Bergman kernel is independent of the choice of orthonormal basis and .
Let . These functions are called the normalized reproducing kernels of ; it is clear that they are unit vectors in . For any , let be the analytic mapping on defined by , . An easy calculation shows that the derivative of at is equal to . It follows that the real Jacobian determinant of at is When , weakly and the normalized reproducing kernels , span . Since is a closed subspace of (see ), there exists an orthogonal projection from onto . For , we define the Toeplitz operator on by , . For , let . The function is called the Berezin transform of . Let be the Hankel operator from into defined by . It is easy to check that (see ). For , define as , where is defined as . The operator is called the little Hankel operator on . Let , the Mobius invariant measure on . Let be the set of all bounded linear operators from the Hilbert space into itself, and let be the set of all compact operators in .
Often it is not easy to verify that a linear operator is bounded, and it is even more difficult to determine its norm. No conditions on the matrix entries have been found which are necessary and sufficient for to be bounded, nor has been determined in the general case. For the more general problem we also need analogues of the notions of operator norm. For more details see [2, 3]. The family of norms that has received much attention during the last decade is the Schatten norm.
A proper two-sided ideal in is said to be a norm ideal if there is a norm on satisfying the following properties:(i) is a Banach space;(ii) for all and for all ;(iii) for a rank one operator.
If is a norm ideal, then the norm is unitarily invariant, in the sense that for all in and unitary in . Each proper ideal of is contained in the ideal of compact operators. Two special families of unitarily invariant norms satisfying conditions (i), (ii), and (iii) are the Schatten -norms defined on the set of compact operators and the Ky Fan norms.
For any nonnegative integer , the th singular value of is defined by Here is the operator norm. Clearly, and
Thus is the distance, with respect to the operator norm, of from the set of operators of rank at most in . The spectral theorem shows that the singular values of the compact operator are the square roots of the eigenvalues of as long as is separable and infinite dimensional. Notice that can be defined for any but clearly, if and only if is compact.
The Schatten Von Neumann class , , consists of all operators such that
If , then is a norm, which makes a Banach space. For , does not satisfy the triangle inequality, it is a quasinorm (i.e., for and , a constant), which makes a quasiBanach space. We will be mainly concerned with the range . The space is also called the trace class and is called the Hilbert-Schmidt class. The linear functional trace is defined on by where is an orthonormal basis in . Moreover, the right-hand side does not depend on the choice of the basis. If , the dual space can be identified with with respect to the pairing , , . Here is the dual exponent. With respect to the same pairing one can identify with and with . We refer the reader to [2, 3] for basic facts about Schatten -classes. The Schatten -classes should be seen as gradations of compactness for an operator. Each Schatten -class is dense in the space of compact operators in the operator norm. For this reason, it is of interest, given a certain class of operators, to ask whether or not there are compact operators not in any Schatten -class. For instance, this was proved by Arazy et al.  and Zhu  for Hankel operators on Bergman space.
Hankel operators are closely related to Toeplitz operators. Many problems about Toeplitz operators can also be formulated in terms Hankel operators and vice versa. The singular values of Hankel operators on the Hardy space play a crucial role in rational approximation. The celebrated results of Adamjan et al.  which give the achievable error in approximating a Hankel operator by another one of smaller rank in terms of the singular values of the Hankel operator is an illustration of this. It may be noted here that the Adamjan, Arov, and Krein theorem has had a considerable influence on the treatment of the problem that arises in engineering applications in the context of model reduction, that is, the problem of finding a simple model to replace a relatively complicated one without too great a loss of accuracy. In view of this it would be nice to have a satisfactory characterization for Schatten class Toeplitz operators on the Bergman space.
In this paper we find necessary and sufficient conditions on that will ensure that the Toeplitz operator belong to , . This will provide some quantitative estimates (size estimates of these operators) in terms of norms. We will also use to denote the full algebra of bounded linear operators from the Bergman space into itself.
For and in , let . These are involutive Mobius transformations on . In fact(1);(2), ;(3) has a unique fixed point in .
Given and any measurable function on , we define a function . Since is the real Jacobian determinant of the mapping (see ), is easily seen to be a unitary operator on and . It is also easy to check that , thus is a self-adjoint unitary operator. If and then . This is because and for , . Let be the Lie group of all automorphisms (biholomorphic mappings) of , and the isotropy subgroup at ; that is, .
2. Compact Operators Whose Real and Imaginary Parts Are Positive
Zhu  had shown that if is a nonnegative function on , , then is in the Schatten class if and only if is in . The following is an easy consequence of it.
Proposition 2.1. Suppose that is such that where and . The Toeplitz operator , if and only if . In this case, if and if .
Proof. Suppose and . Notice that and and since is a Banach space and is closed under adjoints, hence and belong to . From , it follows that . Hence . Now suppose , . This implies . From , it follows that . Hence as is a vector space and (see ). Now let and assume . This implies and therefore by , . Hence and by , . Now suppose , . Then and hence . Since is a vector space, .
Example 2.2. Let , . Then only if , and only if . This can be seen as follows. The matrix of with respect to the standard orthonormal basis of is diagonal, and for . Similarly if . If , .
Proposition 2.3. Let and suppose that is not the zero function. If is compact then is not closed.
Proof. Since is compact, hence contains no closed infinite-dimensional subspace of . If now is closed then is finite dimensional. That is, is of finite rank. This implies by  that . This is a contradiction as is not the zero function.
Recall the following.
Suppose that is a positive operator on a Hilbert space and is unit vector in , then (i) for all ; (ii) for all . For proof see .
Suppose that is analytic. Define the composition operator from into itself as . It is shown in  that is a bounded linear operator on and . Given and any measurable function on , we define the function by , where . The map is a composition operator on .
Proposition 2.4. If then is compact if and only if is compact.
Proof. This follows from the fact that and as .
Proposition 2.5. For , .
Proof. Observe that Similarly, we have . Since and for any and all , we have
Let . The generalized Kantorvich constant is defined by for any real number and it is known that for . We state below the known results on the generalized Kantorvich constant . Let be strictly positive operator satisfying , where . Put . Then the following  inequalities (2.4) and (2.5) hold for every unit vector and are equivalent:
The Kantorvich constant is symmetric with respect to and is an increasing function of for , is a decreasing function of for , and . Further, for or , and for .
Proposition 2.6. Let be strictly positive satisfying , where . The following hold.(i)If and then .(ii)If , then .(iii)Let be such that . If then .
Proof. Suppose and . Then
Hence by (2.4), . That is, . Suppose and . Then . Hence from (2.5), it follows that . Since for , hence .
Now assume . Then if then by (2.5), we have and hence . If , then by (2.4) and (2.5), if and then and .
3. Schatten Class Operators
In this section, we will obtain conditions to describe Schatten class Toeplitz operators on Bergman space . The results of this paper hold for Bergman spaces where is any bounded symmetric domain in . For simplicity, we consider only the case of the open unit disk in .
Let . The space is properly contained in (see ) and if then is bounded on and there is a constant such that .
Theorem 3.1. Suppose and . Then the following hold. If , then . If then and .
Proof. Suppose . Then
That is, . If , then
Now . Thus . That is, and . Suppose . Then by Heinz inequality , it follows that
Hence and therefore . Thus .
Now suppose . Then the change of the order of integration is justified by the positivity of the integrand. Hence . Similarly if then as . By Marcinkiewicz interpolation theorem it follows that if then for . Now suppose , . We will prove . The case is trivial. By interpolation we need only to prove the result for . Suppose and is the standard orthonormal basis for . Now and Thus and .
It is not so difficult to verify the conditions in Theorem 3.1.
Example 3.2. Let . Then . This can be verified as follows: and changing the variable to , this reduces to . Thus .
Example 3.3. Let . Then Thus . Direct computation reveals that and .
Example 3.4. Let . Then by Example 2.2, and
Thus and hence .
It may be noted that the space , contains no nonzero harmonic functions and even no nonzero constants. To see this, for example, for , let This is a nonnegative and nondecreasing function of and . So must tend to as . Thus , and therefore .
But although there is no nonzero harmonic functions in , there are plenty of subharmonic functions. Consider the function . We have verified in Example 3.3 that . Suppose that is real-valued subharmonic and . The subharmonicity of implies that . Hence . Let . It can be verified that and note that since . In other words, is also subharmonic. Proceeding by induction, if we define on then we obtain is subharmonic for all and is a nondecreasing sequence of functions.
Corollary 3.5. If , the space of bounded harmonic functions on and , then if and only if .
Proof. The proof of the corollary follows from the above discussion and the fact  that if and only if .
Corollary 3.6. If is a real-valued bounded subharmonic function on , , then if and only if .
Proof. By Theorem 3.1, if then . Conversely if is real-valued, subharmonic, bounded on and then is also subharmonic and the subharmonicity of implies that . Hence as implies and the result follows from Theorem 3.1.
Corollary 3.7. If then the Toeplitz operator if and only if .
Proof. Notice that if then the following two possibilities hold.() For all , either or .() There exists such that and .
Suppose that () holds and let be such that and . Then this implies and . Hence has an eigenvalue. From , it follows that is a constant and is not compact unless .
Now suppose that () holds, . Then for all , . Let . Note that for all . It follows from Proposition 2.5 and from the previous discussion.
Notice compact implies that is compact. Thus as . Similarly since is compact, as .
Thus as . Hence it follows that as . So as where . Since is compact, as . Thus as . Similarly since is compact we can show that as . Thus as . Hence as . It follows therefore that almost everywhere and hence . Therefore, .
We say majorizes if for all .
Corollary 3.8. If , , and majorizes then .