Antinormal Weighted Composition Operators

Abstract

Let , where is set of all positive integers and is the counting measure whose -algebra is the power set of . In this paper, we obtain necessary and sufficient conditions for a weighted composition operator to be antinormal on the Hilbert space . We also determine a class of antinormal weighted composition operators on Hardy space .

1. Introduction

A deep and interesting problem in operator theory is to determine the distance of an operator from a given class of bounded linear operators on a Hilbert space. In particular, the distance between an operator and the set of Hermitian, positive, compact, and unitary operators has been investigated in [1–4]. In 1974, Holmes [5] discovered that there are operators for which their largest possible distance from the set of normal operators can be achieved. He named such operators as antinormal operators and showed that no compact operator is antinormal. Thus, this extremality of operators is a consequence of infinite dimensional phenomenon of the underlying space. Subsequently, this class has been extensively studied by several authors in [6–9]. For the first time, the same problem has been studied in context of composition operators on the Hilbert space in [10] by Tripathi and Lal. In this paper, we investigate antinormality of weighted composition operators on and .

Notation and Terminology. In this paper, and denote the set of all positive integers and the set of all complex numbers, respectively. Also, let be defined as Suppose is a separable complex Hilbert space and denotes the algebra of all bounded linear operators on . Further, for , let and , respectively, denote the null space and the range space of .

We now record certain definitions and results which are useful in our context.

Definition 1. A function is said to be bounded away from zero if there exists a positive real number, say , such that for all .

Definition 2 (see [11]). An operator is said to be Fredholm operator if dimension of and the dimension of the quotient space are both finite.

We also note here that, equivalently, is Fredholm if both and are finite dimensional.

Definition 3. Essential spectrum of an operator is defined as .
Since every invertible operator is Fredholm operator, therefore ; see [12].

Definition 4. Minimum modulus of an operator is defined as .

Definition 5. Essential minimum modulus of an operator is defined as , where .

Definition 6. An operator is said to be antinormal if , where is the class of all normal operators in .

Remark 7. An operator is antinormal if and only if its adjoint is antinormal.

Definition 8. For an operator in , index of is defined as

Remark 9. Observe that .

The following results will be used in the later part of the paper.

Theorem 10 (see [7]). Let . (i)If index, then .(ii)If index, then .

Remark 11. If index, then is not antinormal.

Theorem 12 (see [7]). Let with index. Then, following conditions are equivalent:(i) is antinormal.(ii).(iii), where is the class of all unitary operators in .(iv) for some , isometry and positive compact contraction .

2. Antinormal Weighted Composition Operators on

The Hilbert space is the space of all square summable sequences of complex numbers. Let be self-map and denote the inverse image of under . Let denote the cardinality of . For , a weighted composition transformation induced by and is defined as If is bounded, then it is called weighted composition operator induced by and . The following result gives a necessary and sufficient condition for boundedness of .

Theorem 13. Weighted composition operator on is bounded if and only if . In this case, .

Proof. Let be an element in . Then,Thus, implies that is bounded and Conversely, if is bounded, then we have Therefore, hence the proof.

In the following result, we determined weighted composition operators on which are not antinormal.

Theorem 14. Let be bounded away from zero and be a bijective self-map on . Then, is not antinormal.

Proof. Since is bijective and is bounded away from zero, is invertible on . Hence, index. Consequently, is not antinormal by Remark 11.

The following remark shows that the condition that “ is bounded away from zero” is not essential for the validity of the above theorem.

Remark 15. Let be such that as . Then, it is well known that is compact ([13], p. 89). Hence, is not antinormal.

We now prove the following lemma.

Lemma 16. Suppose is injective but not surjective and for each . Then, is not injective but is injective.

Proof. Since is not surjective, there exists such that is empty. Therefore, Consequently, is not injective.
Now, for , we have Now, a simple computation using the above expression and injectivity of shows that whenever , then . Since , therefore . Hence, is injective.

The following result gives a necessary and sufficient condition for antinormality of when is injective but not surjective.

Theorem 17. Suppose for each and is injective but not surjective. Then, is antinormal if and only if the following conditions hold:(a)For each , , except for finitely many .(b) for infinitely many .

Proof. Since is injective but not surjective and for each , hence by above lemma index. Now, for and for given complex number , we have From the above equation and condition (a) it follows that is finite whenever . Hence, is Fredholm for each . Thus, for each . Again by condition (b) for infinitely many . Hence, is infinite. Therefore, . Thus, . Hence, is antinormal by Theorem 12. Now, since adjoint of an antinormal operator is antinormal, therefore is antinormal. Conversely, if either of the conditions fail, then we claim that is not antinormal. For, if condition (a) fails, there exists an with such that . Therefore, . Hence, is not antinormal. Now, suppose condition (b) fails. Then, by (11), . Therefore, . Hence, is not antinormal.

We now prove the following lemma.

Lemma 18. Suppose is surjective but not injective and for each . Then, is injective but is not injective.

Proof. Let be such that . Therefore, . This implies as . Now, subjectivity implies . Further, Since is not injective, there exist with such that . Now, define and . It is easy to see that and but . Therefore, is not injective.

Using above lemma we prove the following result.

Theorem 19. Suppose for each and is surjective but not injective. Then, is antinormal if and only if the following conditions hold:(a)For each , , except for finitely many ,(b) for infinitely many .

Proof. Since is surjective but not injective and for each , hence by above lemma index. For and , we have Therefore, is finite for each by condition (a). Hence, for each . Also, for infinitely many , so . Thus, . Consequently, is antinormal. Conversely, if either of the conditions is not true, then we claim that is not antinormal. If condition (a) fails, then there exists an with , such that . Therefore, and so is not antinormal. Now, suppose condition (b) fails. Then, by (12), . Therefore, . Thus, is not antinormal in either case.

Theorem 20. Suppose and is neither injective nor surjective.(i)If index, then is not antinormal.(ii)If index, then is antinormal if and only if the following conditions hold:(a)for each , , except for finitely many ,(b)if for infinitely many .

Proof. (i) It follows by Remark 11.
(ii) Proof is same as in case of Theorem 19.

We now give examples, one in each case, satisfying Theorems 17 and 19.

Example 21. Let and Clearly, is injective but not surjective and for . Hence, by Theorem 17, is antinormal.

Example 22. Let Clearly, is surjective but not injective and for . Hence, antinormality of follows from Theorem 19.

3. Antinormal Weighted Composition Operators on Hardy Space

In this section, we continue our line of investigation of antinormal operators. Let and . Hardy space [14] is defined to be the space of all analytic functions on the unit disk with the property The inner product on is defined as follows: Let be analytic self-map. For , weighted composition transformation on is defined by for all and for all . The fact that is bounded linear operator on follows from the well-known Hardy-Littlewood subordination, Theorem [15]. In 2011, Gunatillake [16] has shown that the operator on is invertible if and only if is both bounded and bounded away from zero on the unit disc and is an automorphism of the unit disc.

Following result is an easy consequence of preceding statement.

Theorem 23. If is bounded away from zero and is automorphism of the unit disc, then is not an antinormal composition operator.

Proof. Since is bijective, so index. Hence, is not antinormal by Remark 11.

One of the difficult issues is to find a useful description of the adjoint of for arbitrary . In 2011, Matache [17] showed that if then for each . We have used his idea in the following theorem.

Theorem 24. If self-map for and a.e. on , then is antinormal.

Proof. Let . Then, Therefore . This implies that is invertible for . Hence, is Fredholm for . Thus, whenever . Also is infinite for . Hence, . Therefore, . Since and a.e. on , . Further, is injective, and is not surjective as is not an automorphism. Consequently, index. Hence, the antinormality of follows from Theorem 12.

Following corollary is an immediate consequence of the preceding theorem.

Corollary 25. If self-map , , then is antinormal.

Competing Interests

The authors declare that they have no competing interests.