Right and Left Weyl Operator Matrices in a Banach Space Setting

Abstract

Let be Banach spaces. The operator matrix of the form acting between and is investigated. By using row and column operators, equivalent conditions are obtained for to be left Weyl, right Weyl, and Weyl for some , respectively. Based on these results, some sufficient conditions are also presented. As applications, some discussions on Hamiltonian operators are given in the context of Hilbert spaces.

1. Introduction

Throughout this paper, and are always reserved to denote some Banach spaces. If is a bounded linear operator from to , we write , and if , we write instead of . For , the range and the kernel of are denoted by and , respectively. Write and . The sets of all left and right Fredholm operators are, respectively, defined as

They are natural extensions of Hilbert space cases since every closed subspace is complemented in the whole space. The set of Fredholm operators is defined as

An operator is said to be semi-Fredholm if , for which the index is defined as . The left Weyl, right Weyl, and Weyl operators are collected inrespectively.

An operator is said to be left (resp. right) invertible if there exists an operator such that (resp. ). If is both left and right invertible, then is invertible. As it is well known, the sets of all left invertible, right invertible, and invertible operators are as follows:

Again, we have the abbreviations of the above classes of operators like . Now let . As usual, the spectrum , left spectrum , right spectrum , Fredholm spectrum , left Fredholm spectrum , right Fredholm spectrum , Weyl spectrum , left Weyl spectrum , and right Weyl spectrum will be the collections of the numbers such that does not belong to the corresponding classes of operators, i.e., , , , , , , , , and , respectively.

We say that is relatively regular or simply regular if there exists some such that . In this case, is called an inner generalized inverse of . Obviously, the classes of left or right invertible, invertible, semi-Fredholm, and Fredholm operators are all regular. If is a closed subspace in Banach space , then is said to be topologically complemented or simply complemented if there exists another closed subspace of such that and ; for such complementary subspaces and of , we write . It is well known that is regular if and only if and are complemented subspaces of . Denote by and the complementary subspaces with and , respectively.

For given and , definewhere is an unknown element. The spectrum and its various subdivisions of are considered in many papers such as [1–7] and the references therein. Although most of these papers worked in the context of Hilbert spaces, some results on the invertibility and Fredholm theory were established in Banach spaces [5–7]. In this note, we investigate the left and right Weyl spectra of upper triangular operator matrices in a Banach space setting. Our main tools are the regularity of an operator and its equivalent form, which are closely related to some appropriate space decompositions.

2. Preliminaries

This section is devoted to collecting some basic results. Although most of them are well-known standard results on Fredholm operators, we list them here for convenience of later proofs.

Lemma 1 (see [8]). Let be a semi-Fredholm operator and be a compact operator. Then, is semi-Fredholm and .
The following lemma is obvious, so its proof will be omitted.

Lemma 2. Assume that and have bounded inverses defined on and, respectively. Then, and (and hence ) have the same closedness as, where.

Lemma 3 (see [9]). For, we haveNote that the last two inclusions are strict in general, and due to the closed range theorem, equality holds here precisely when is closed.
The sets of all upper and lower semi-Fredholm operators are defined byrespectively. It is obvious that , , and .

Lemma 4 (see [8]). Let with closed range. Then, , , and

Lemma 5 (see [8]). Let and.(i)If is right Fredholm, then is right Fredholm.(ii)If is left Fredholm, then is left Fredholm.(iii)If is Fredholm, then is right Fredholm and is left Fredholm.

Lemma 6 (see [8]). Let and. If both and are upper semi-Fredholm or both are lower semi-Fredholm, then.

3. Main Results and Proofs

In this section, we present the main results of this paper and their proofs. First, we establish the right Weylness of .

Theorem 1. Let and be given operators. Then, there exists such that is right Weyl if and only if is right Fredholm, and one of the following statements is fulfilled:(i), and there exists such that the row operator is a right Fredholm operator with .(ii) is right Weyl.

Proof. Let be right Fredholm. If assertion (ii) holds, then taking gives the conclusion. We now assume that assertion (i) is true. There are two possible cases depending on the dimension of .(i)Case 1: dim . In this case, is obviously a right Weyl operator. By Lemma 1, we see that is a right Weyl operator for any .(ii)Case 2: dim. In this case, is complemented in and is complemented in , since is right Fredholm. Then, and have the decompositionsConsequently, can be written asDefine operator byAs an operator from to , has the following matrix form:It is not difficult to see that is complemented in and is complemented in , and thatThus, . This shows that is right Weyl.
Conversely, assume that there exists such that is right Weyl. Obviously, is right Fredholm. In view of the space decompositions in (10), as an operator from to , can be represented asEvidently, andThen,If , then is of finite rank, and hence, from (17) and Lemmas 1 and 2, it follows that is right Weyl, i.e., assertion (ii) is valid. We now assume . Since is right Weyl, from (17), we equivalently have that is right Weyl. Thus, is exactly an operator such that the row operator is right Fredholm, and , i.e., ind. Taking yields assertion (i).
From Theorem 1, the following perturbation result of right Weyl spectrum is obvious.

Corollary 1. Let and be given operators. Then,whereIf having certain special properties for the given diagonal entries, one can further analyze the right Weylness of on the basis of Theorem 1.

Corollary 2. Let and be given operators. If is right Fredholm, then there exists such that is right Weyl if and only if is right Weyl.

Proof. Assume that there exists such that is right Weyl. From the assumption and the right Fredholmness of , it follows that and , and hence if , as required in assertion (i) of Theorem 1, we still conclude that is right Weyl, i.e., assertion (ii) holds. By virtue of Theorem 1, the desired result is obvious.

Corollary 3. Let and be given operators. If is Weyl, then there exists such that is right Weyl if and only if is right Weyl.

Proof. From the proof of Corollary 2, it is sufficient to notice that is right Weyl if and only if is right Weyl when is Weyl.
Dually, we have the following left Weyl description.

Theorem 2. Let and be given operators. Then, there exists such that is left Weyl if and only if is left Fredholm, and one of the following statements is fulfilled:(i), and there exists such that the column operator is a left Fredholm operator with.(ii) is left Weyl.

Proof. Let be left Fredholm. Similar to the proof of Theorem 2, it suffices to consider the case that assertion (i) is true under the condition dim. Obviously, is complemented in and is complemented in , and henceThen,Define operator byAt this point, has the following matrix form:Evidently, is complemented in and is complemented in . In addition, and which impliesThus, is left Weyl.
Conversely, assume that there exists such that is left Weyl. It is clear that is left Fredholm, and as an operator from to , can be written asThen, with the aid of the operators andwe getIf , then is of finite rank, which implies that is left Weyl, i.e., assertion (ii) is valid. We now assume that . Since is left Weyl, it follows from (27) that is left Weyl. Thus, is a desired operator such that the column operator is left Fredholm, and , i.e., . Taking yields assertion (i).

Remark 1. Theorem 2 is dual to Theorem 1 to some extent. However, we cannot directly resort to conjugate relationship, since Lemma 4 is not valid for left and right Fredholm operators. Actually, implies, but the converse is not true in general (see, e.g., [10] for more details).

Corollary 4. Let and be given operators. Then,where

Corollary 5. Let and be given operators. If is left Fredholm, then there exists such that is left Weyl if and only if is left Weyl.

Corollary 6. Let and be given operators. If is Weyl, then there exists such that is left Weyl if and only if is left Weyl.
In [[6], Theorem 3.6], the author described the Weylness of on a Banach space. Using the row operator and the column operator in Theorems 1 and 2, we will give a different statement.

Theorem 3. Let and be given operators. Then, there exists such that is Weyl if and only if is left Fredholm, is right Fredholm, and one of the following statements is fulfilled:(i), and there exists and such that is right Fredholm with, is left Fredholm with, and.(ii) is Weyl.

Proof. Assume that is left Fredholm and is right Fredholm. If assertion (ii) is valid, then taking yields the desired sufficiency. We now assume that assertion (i) holds.
Obviously, , , and hence decompositions (10) and (20) are still satisfied. Then, the operators and in assertion (i) can be written aswhere . Since is right Fredholm, , and is clearly complemented in . Similarly, the left Fredholmness of implies that and is complemented in . Note that the assumptions and in (i) imply thatrespectively. Hence,Takingwe claim that is Weyl.
Indeed, as an operator from to ,where and . Then,are invertible operators such thatFrom Lemmas 5 and 6, we infer that has the same semi-Fredholmness and index as the operator matrix on the right hand side of (36). Thus, is clearly closed, and , and hence by (32). This shows that is Weyl, as claimed.
Conversely, assume that there exists such that is Weyl. Obviously, is left Fredholm, and is a right Fredholm. Then, decompositions (10) and (20) still hold true, and as an operator from to , has a new block representation:where and . Then,andIf either or holds, then is of finite rank, which together with (39) implies that is Weyl, i.e., assertion (ii) is valid. We now assume that . From (39), we equivalently have that is Weyl, which implies that . Taking gives a right Fredholm row operator with , while taking yields a left Fredholm column operator with . Also, . This proves assertion (i). More generally, and also serve as the desired operators, since and are invertible.
As a direct consequence of Theorem 3, we have the following.

Corollary 7. Let and be given operators. Then,where is the collection of all for which there is not any pair of operators such that is right Fredholm with , is left Fredholm with , and. Note that,, and are defined as in Corollaries 1 and 4.

Corollary 8. Let and be given operators. If is right Fredholm or is left Fredholm, then there exists such that is Weyl if and only if is Weyl.

proof. Assume that there exists such that is Weyl. Then, we always see that the relations do not hold, whenever is right Fredholm or is left Fredholm. From Theorem 3, it follows that is Weyl. The rest of the proof is trivial.

Corollary 9. Let and be given operators. If (resp.) is Weyl, then there exists such that is Weyl if and only if (resp.) is Weyl.

proof. By symmetry of the result, we only prove the case when is Weyl. The sufficiency is obvious. For necessity, let be Weyl for some . Since is Weyl, . Then, assertion (i) of Theorem 3 cannot be satisfied. Hence, is Weyl, which is equivalent to the fact that is Weyl.
Based on the embedded relationship of certain spaces, the sufficient conditions under which is left and right Weyl in a Banach space setting are given, respectively.

Definition 1. (see 6, Definition 4.2]). For two Banach spaces and , we say that can be embedded in and write if there exists a left invertible operator . Note that if and only if there exists a right invertible operator .

Theorem 4. Let and be given operators such that. Then, there exists such that is right Weyl if the following statements are fulfilled:(i) is right Fredholm.(ii) is complemented in.(iii)There exists such that is right Fredholm.

proof. Assume that assertions (i), (ii), and (iii) hold. If , then is right Fredholm if and only if is right Fredholm, and hence which together with the right Fredholmness of and the relation implies that , i.e., . This means that is right Weyl, and taking gives the theorem.
Now let . If , again, , and is the desired operator. It remains to consider the case of . In this case, the decompositions in (10) and the second equality of (20) are still valid, and we further haveObviously, is right invertible and write for some right inverse of . Sincethe operator is right Fredholm. Hence,with and . Since , from , it follows thatIt is clear that . This together with (42) implies thatTherefore, operator is the desired one satisfying assertion (i) of Theorem 1, and hence we get the conclusion that there exists such that is right Weyl.

Theorem 5. Let and be given operators such that. Then, there exists such that is left Weyl if the following statements are fulfilled:(i) is left Fredholm.(ii) is complemented in .(iii)There exists such that is left Fredholm.

proof. Assume that assertions (i), (ii), and (iii) hold. If , then is left Fredholm if and only if is left Fredholm, and hence which together with the left Fredholmness of and the relation implies that , i.e., . This means that is left Weyl, and taking proves the theorem.
Now let . If , again, , and is the desired operator. It remains to consider the case of . In this case, the decompositions in the first equality of (17) and (20) are still valid, and we further haveObviously, is left invertible and write for some left inverse of . Sincethe operator is left Fredholm. Hence,with and . Since , from , it follows thatIt is clear that . This together with (47) implies thatTherefore, taking operator above will give assertion (i) of Theorem 2, and hence this completes the proof.

Remark 2. In particular, Theorems 4 and 5 hold if and are regular, respectively.

4. Some Extensions and Applications on a Hilbert Space

In the Hilbert space setting, we have the following results as corollaries of Theorems 1, 2, and 3. In what follows, and are always complex separable Hilbert spaces; denote by the orthogonal sum of and , and write and , as usual, for an operator between Hilbert spaces.

Corollary 10. Let and be given operators. Then, there exists such that is right Weyl if and only if is right Fredholm, and one of the following holds:(i).(ii) is right Weyl.

proof. Comparing Theorem 1 with this corollary, we have to show that, under the condition , there naturally exists an operator such that the row operator is right Fredholm with ind. In fact, assume first that . If , then every unitary operators from to can be chosen as the desired . If, however, , then any right invertible operator meets our choice. We now let . At this point, decomposing into the orthogonal sum of two infinite dimensional closed subspaces will reduce to the case of .
The following is a dual result of Corollary 10, and it gets proved directly by Corollary 10.

Corollary 11. Let and be given operators. Then, there exists such that is left Weyl if and only if is left Fredholm, and one of the following holds:(i).(ii) is left Weyl.

Corollary 12. Let and be given operators. Then, there exists such that is Weyl if and only if is left Fredholm, is right Fredholm, and one of the following holds:(i).(ii) is Weyl.

proof. Assume that is left Fredholm, is right Fredholm, and . To complete the proof, it suffices to show that there exists and such that is right Fredholm with , is left Fredholm with , and .
In fact, since is left Fredholm and is right Fredholm, are complemented, are complemented, and one has the following decompositions:Since , we havewhere and are infinite dimensional closed subspaces of and , respectively. Pick an arbitrary bounded operator from into such that . Take