Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 28, Number 5, October 2023
Page(s) 392 - 398
DOI https://doi.org/10.1051/wujns/2023285392
Published online 10 November 2023

© Wuhan University 2023

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

The research of Weyl type theorem is an important subject in spectral theory. In 1909, Weyl discovered Weyl's theorem when he studied the spectrum of the self-adjoint operator[1]. Then Harte and Lee defined Browder's theorem[2]. Rakočević gave two other variations of Weyl's theorem: a-Weyl's theorem and a-Browder's theorem[3,4]. These generalizations are called Weyl type theorems by scholars. The study on Weyl type theorems can well reflect the structural characteristics of spectrums. Hence the research of Weyl type theorem has attracted much attention and got many good results in resent years[5-7]. In this paper, we mainly study a-Browder's theorem and a-Weyl's theorem for bounded linear operators and operator functions by means of the property of the topological uniform descent.

In this paper, denotes a complex separable infinite dimensional Hilbert space. Let be the algebra of all bounded linear operators on . For an operator we shall denote by the dimension of the kernel , and by the codimension of the range . We call is an upper semi-Fredholm operator if is closed and . We say that is a lower semi-Fredholm operator when . An operator is said to be Fredholm if is closed and both and are finite. If is an upper (or a lower) semi-Fredholm operator, the index of , , is defined to be . The ascent of , , is the least non-negative integer such that and the descent, , is the least non-negative integer such that . The operator is Weyl if it is Fredholm of index zero, and is said to be Browder if it is Fredholm "of finite ascent and descent". We call a Drazin invertible operator if . Let be the spectrum of . The approximate point spectrum of is denoted by . The Weyl spectrum , the upper semi-Fredholm spectrum , the Browder spectrum , the Drazin spectrum , are defined by is not Weyl( denotes the set of complex numbers ), is not upper semi-Fredholm, is not Browder, is not Drazin invertible. Let , , , , , . For a set , we write () for the set of isolated (accumulation) points of , and we denote () for interior (boundary) points set of . If , then is a pole of . is called an a-isoloid operator if , where .

satisfies a-Browder's theorem if

where is not upper semi-Fredholm or and is not upper semi-Fredholm or . Let and . The a-Weyl's theorem holds for if and only if

where we write . It can be shown that a-Weyl's theorem a-Browder's theorem, but the converse is not true. Let be defined by . Then and . So satisfies a-Browder's theorem, but a-Weyl's theorem does not hold for .

If satisfies , then is called a Sapher operator[8,9]. The Sapher spectrum is is not Sapher operator. Goldberg defined is not closed[10]. is called a Kato operator if is closed and . Therefore, the Kato spectrum is .

Let , for each nonnegative integer , induces a linear transformation from the vector space to . We denote by the dimension of the null space of the induced map and put . If there is a nonnegative integer for which for and is closed in the operator range topology of for , then we say that has topological uniform descent[11]. If is upper semi-Fredholm, then has topological uniform descent. Let has topological uniform descent, and . We will use the following property which is discovered by Grabiner (Ref.[11], Corollary 4.9):

Lemma 1   Let , . If has topological uniform descent, then .

On the basis of analyzing distribution of various spectrums of bounded linear operators, the sufficient and necessary conditions holding a-Browder's theorem and a-Weyl's theorem are established by means of the property of the topological uniform descent. In addition, the new judgements of a-Browder's theorem and a-Weyl's theorem for operator function are discussed.

1 Judgement of A-Browder's Theorem and A-Weyl's Theorem for Bounded Linear Operator

First, we describe a-Browder's theorem by the relation between topological uniform descent and .

Theorem 1   satisfies a-Browder's theorem if and only if .

Proof   "". Suppose

Then there exists a deleted neighborhood centered on such that for any , . Moreover, for any deleted neighborhood , there exists such that . Let , then we will get that is Browder operator since satisfies a-Browder's theorem and . It follows that . Since , , we know that according to Lemma 1.

"". It's clear that

Suppose . According to perturbation theorem of semi-Fredholm operator, there exists such that if . Then . It follows that . If and , then is Browder operator. Therefore, a-Browder's theorem holds for .

Remark 1   (i) In Theorem 1, suppose satisfies a-Browder's theorem, then each part of the decomposition of cannot be deleted.

(a) Let be defined by . Then , satisfies a-Browder's theorem.

But . Thus cannot be deleted.

(b) Let be defined by . We can get that , a-Browder's theorem holds for . However, ,

which means cannot be deleted.

(c) Let be defined by . Then . satisfies a-Browder's theorem. But . Therefore cannot be deleted.

(d) Let be defined by . We have , which implies a-Browder's theorem holds for . Since , cannot be deleted.

(ii) Since , satisfies a-Browder's theorem if and only if .

(iii) By Theorem 1, if , then satisfies a-Browder's theorem. But the converse is not true. Let be defined by . We can get that a-Browder's theorem holds for , but .

(iv) satisfies a-Browder's theorem and .

In fact, yields .

For the converse, since , by Theorem 1 we get .

(v) satisfies a-Browder's theorem and , where is denumerable.

"". It is clear that . Then we can get that satisfies a-Browder's theorem by Theorem 1. Since , we have that . Since , it follows that is denumerable.

"". Suppose that . If , then by a-Browder's theorem holds for . If , then there exists a neighborhood centered on such that . Since is denumerable, for any , there exists such that is Weyl operator. It follows that since satisfies a-Browder's theorem. We can also get .

By Theorem 1, the following results can be obtained.

Corollary 1   Let . The following statements are equivalent:

(1) satisfies a-Browder's theorem;

(2) ;

(3) ;

(4) ;

(5) .

Proof   (1)(2). Using Theorem 1, we have when satisfies a-Browder's theorem. Since , it follows that (2) holds.

(2)(3). Let and is Kato operator. Then (Ref. [12], Lemma 3.4). Therefore . By (2) we know that (3) holds.

(3)(4). Suppose . Then there exists a deleted neighborhood centered on such that for any , . Since , for any , there exists such that . Let , it follows that

Thus we can get is Browder operator by (3). From the proof of Theorem 1, we have .

(4)(5). Let . Then there exists a deleted neighborhood centered on such that for any , and . Since , we know that for any , there exists

It follows that is Browder operator. This implies that (5) holds.

(5)(1). It is clear that

If , then , therefore . If and , then is Browder operator. We can conclude that satisfies a-Browder's theorem.

Corollary 2   Let . The following statements are equivalent:

(1) satisfies a-Browder's theorem;

(2) ;

(3)

Proof   (1)(2). We only need to prove

.Suppose . Then there exists a deleted neighborhood centered on such that for any , . If , there exists such that for any , . Let . It follows that for any , is Weyl operator. From the a-Browder's theorem holds for and the proof of Theorem 1, we can get that is Browder operator. If , then there exists a deleted neighborhood centered on such that for any , since satisfies a-Browder's theorem. Therefore . Moreover, by and , we know that .

(2)(3). It is clear that . Since , and , we can get that . Also, . Then we have

Hence (2)(3) is true.

(3)(1). We know that , and . If , then . If and , then is Browder operator. Thus satisfies a-Browder's theorem.

In the following, we will discuss the a-Weyl's theorem for .

Theorem 2   Let . The following statements are equivalent:

(1) satisfies a-Weyl's theorem;

(2)

;

(3) .

Proof   (1)(2). Since satisfies a-Weyl's theorem, we know that , . From Corollary 2, we have . Moreover, . Since . We get .

Also, . Hence . Then we know that (2) holds.

(2)(3). Since and , it follows that .

(3)(1). , , . If and , then is Browder operator. Moreover, . Therefore . Similarly, we can prove that . So the a-Weyl's theorem holds for .

Remark 2   (i) By Theorem 2, we can get that if , then satisfies a-Weyl's theorem. But the converse is not true. Let be defined by . Then satisfies a-Weyl's theorem, but .

(ii) satisfies a-Weyl's theorem, and .

In fact, suppose that . Then , so . Similarly, since and , we know that .

For the converse, let . If , then is invertible. If , we can get that is Browder operator by and the a-Weyl's theorem holds for . If , by we can get is Browder operator.

2 Judgement of A-Browder's Theorem and A-Weyl's Theorem for Operator Function

In the following, we will research the a-Browder's theorem and a-Weyl's theorem for operator function by means of the property of the topological uniform descent.

Theorem 3   Let . Then for any polynomial , satisfies a-Browder's theorem if and only if

(1) satisfies a-Browder's theorem;

(2) If , then .

Proof   Suppose satisfies a-Browder's theorem for any polynomial . We only need to prove (2) holds. We can assert that if , then for any , . In fact, suppose there exists , where is finite or . Let , . We can see that is finite. If , let . Moreover, let if . Then . It follows that . It is in contradiction to the fact that . Then we will prove that . If and , then for any deleted neighborhood centered on , there exists such that . Since for any , , we know that is Weyl operator. By satisfing a-Browder's theorem, we can get is Browder operator. Thus . From and , we conclude that is Browder operator.

For the converse, if , then for any , . If , we can get that since but . Let and , where and , . It follows that is upper semi-Fredholm operator and for all . From satisfing a-Browder's theorem, we have that . Hence , satisfies a-Browder's theorem.

By Theorem 3 and the proof procedure, we can get the following results.

Corollary 3   Let . Then for any polynomial , satisfies a-Browder's theorem if and only if

(1) satisfies a-Browder's theorem;

(2) If , then .

Remark 3   (i) Suppose , then satisfies a-Browder's theorem and . This implies that for any polynomial , satisfies a-Browder's theorem. However, the converse is not true. Let . The a-Browder's theorem holds for , but .

for any polynomial , satisfies a-Browder's theorem, .

In fact, suppose . This implies that . So .

For the converse, let and . Then for any deleted neighborhood centered on , there exists such that . Thus is Browder operator. So . By and , we know that is Browder operator.

(ii) for any polynomial , satisfies a-Browder's theorem, .

In the following, we will establish sufficient and necessary conditions for operator functions holding a-Weyl's theorem.

Theorem 4   Let . Then is a-isoloid and for any polynomial , satisfies a-Weyl's theorem if and only if:

(1)

;

(2) If , then .

Proof   "". First, we will prove that is a-isoloid. If there exists and , then . If , then . It is a contradiction. If , then is Browder operator, so . It is a contradiction too. Thus is a-isoloid. From (1) we know that satisfies a-Browder's theorem. By Theorem 3, we have that for any polynomial , satisfies a-Browder's theorem. Suppose , let , where and , . Because is a-isoloid, we get that . From (1) and the proof of Theorem 3, we have . Then , hence . Thus for any polynomial , satisfies a-Weyl's theorem.

"". By Theorem 3 we know (2) holds. Suppose . If , then . In fact, if , from is a-isoloid we can get that . Hence . Since , it follows that is Browder operator. If , from the proof of Theorem 2, we get (1) holds.

Remark 4   (i) By Theorem 2, we can get that if , then for any polynomial , satisfies a-Weyl's theorem. But from Remark 2(ii), we know that the converse is not true. However,

for any polynomial , satisfies a-Weyl's theorem, and .

(ii) Let . If , then for any polynomial , satisfies a-Weyl's theorem. In fact, if , let , where and , . Then . Since , we have . By we know that is Browder operator. Thus is Browder operator. By Remark 3, we can get that satisfies a-Weyl's theorem for any polynomial . But the converse is not true. Let . Then for any polynomial , satisfies a-Weyl's theorem. However, .

if and only if for any polynomial , satisfies a-Weyl's theorem, is a-isoloid and .

In fact, if , we only need to prove is a-isoloid. If and . Then . It follows that is Browder operator. So is invertible. It is in contradiction to the fact that .

For the converse, we only need to prove that . Since is a-isoloid, this implies that . By a-Weyl's theorem holds for , we get that .

Corollary 4   Let . Then is a-isoloid and for any polynomial , satisfies a-Weyl's theorem if and only if:

(1) ;

(2) , .

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