Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 30, Number 1, February 2025
Page(s) 60 - 68
DOI https://doi.org/10.1051/wujns/2025301060
Published online 12 March 2025

© Wuhan University 2025

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 Weyl type theorem of bounded linear operator can well reflect the structure characteristics and distribution of operator's spectral[1-5]. Therefore, the Weyl type theorem is an important topic in spectral theory. In recent years, the research scope of Weyl type theorems has been extended from general operators to operator functions, operator matrices, etc. Numerous significant results have been obtained[6-9]. Property (WE) and property (R) are the latest variations of Weyl type theorems, which have attracted the attention and research of operator theorists[10-12]. In this paper, by decomposing and constructing the operator spectrum and using the topological uniform scaling property, we give a new method for bounded linear operators and operator functions to satisfy the property (WE) and property (R). Furthermore, the perturbation of property (WE) and property (R) is characterized, and the necessary and sufficient conditions for the operator function to have both the property (WE) and property (R) are studied.

Throughout this paper, H denotes a complex separable infinite dimensional Hilbert space. Let B(H) be the algebra of all bounded linear operators on H. For an operator TB(H) we shall denote by n(T) the dimension of the kernel N(T), and by d(T) the codimension of the range R(T). We call TB(H) is an upper semi-Fredholm operator if n(T)< and R(T) is closed. If TB(H) is an upper semi-Fredholm operator and n(T)=0, we call T a bounded below operator. If d(T)<, T is a lower semi-Fredholm operator. An operator TB(H) is said to be Fredholm if R(T) is closed and both n(T) and d(T) are finite. If TB(H) is an upper (or a lower) semi-Fredholm operator, the index of T, ind(T), is defined to be ind(T)=n(T)-d(T). The ascent of T, asc(T), is the least non-negative integer n such that N(Tn)=N(Tn+1) and the descent, des(T), is the least non-negative integer n such that R(Tn)=R(Tn+1). We call T a Drazin invertible operator if asc(T)=des(T)<. The operator T is Weyl if it is Fredholm of index zero, and T is said to be Browder if it is Fredholm "of finite ascent and descent". Let σ(T) be the spectrum of T and σa(T) be the approximate point spectrum of T. We write σw(T), σb(T), σe(T), σSF(T), σea(T) and σab(T) for the Weyl spectrum of T, the Browder spectrum of T, the essential spectrum of T, the semi-Fredholm spectrum of T, the essential approximate point spectrum of T and the Browder essential approximate point spectrum of T. Let ρ(T)=C\σ(T), ρa(T)=C\σa(T), ρw(T)=C\σw(T), ρb(T)=C\σb(T), ρe(T)=C\σe(T), ρab(T)=C\σab(T) (C denotes the set of complex numbers). TB(H) is called an isoloid operator if isoσ(T)σp(T) where σp(T)={λC:n(T-λI)>0}. For a set XC, we write isoX, accX and X for the set of isolated points, accumulation points and boundary points set of X. We denote by σ0(T) the set of all normal eigenvalues of T, thus σ0(T)=σ(T)\σb(T). TB(H) satisfies property (WE) if σ(T)\σw(T)=E(T), where E(T)={λisoσ(T):n(T-λI)>0}[10]. TB(H) satisfies property (R) if σa(T)\σab(T)=π00(T), where π00(T)={λisoσ(T):0<n(T-λI)<+}[10].

In this paper, we mainly study property (WE) and property (R) for bounded linear operators and its functions. Some meaningful conclusions are obtained.

1 Judgement of Property (WE) and Property (R) for Bounded Linear Operators

Although both property (WE) and property (R) are variations of Weyl's theorem based on their definitions, there is no necessary connection between them.

Remark 1   (i) TB(H) satisfies property (WE)T satisfies property (R).

Let A,BB(l2) be defined by A(x1,x2,x3,)=(0,x1,x2,), B(x1,x2,x3,)=(0,x2,x3,). Suppose that T=(A00B). Then σ(T)={λC:|λ|1}=σw(T), E(T)=π00(T)=, σa(T)={λC:|λ|=1}{0}, σab(T)={λC:|λ|=1}. T satisfies property (WE), but property (R) does not hold for T.

(ii) TB(H) satisfies property (R)T satisfies property (WE).

Let A,BB(l2) be defined by A(x1,x2,x3,)=(0,x1,x2,), B(x1,x2,x3,)=(x2,x3,), T=(A00B). We have that σ(T)={λC:|λ|1}σw(T), σa(T)=σab(T)={λC:|λ|1}. π00(T)=E(T)=. So T satisfies property (R) and T does not have property (WE).

(iii) TB(H) has property (R) and property (WE)σw(T)=σab(T)[ρa(T)σ(T)] and π00(T)=E(T)=σ0(T).

Topological uniform descent is an important property of operators, which is widely used in spectral theory. If TB(H), then for each nonnegative integer n, T induces a linear transformation from the vector space R(Tn)/R(Tn+1) to R(Tn+1)/R(Tn+2). We denote kn(T) the dimension of the null space of the induced map and let k(T)=n=0kn(T). The following definition was introduced by Grabiner[13]. Let TB(H), if there is a nonnegative integer d for which kn(T)=0 for nd and R(Tn) is closed in the operator range topology of R(Td) for nd, then we say that T has topological uniform descent.

It can be shown that if T is upper semi-Fredholm, then T has topological uniform descent. Let ρτ(T)={λC:T-λI has topological uniform descent}, στ(T)=C\ρτ(T). Grabiner discovered many properties of topological uniform descent. We will use the following property (Ref. [13], Corollary 4.9): Suppose that TB(H), λσ(T), if T-λI has topological uniform descent, then λ is a pole of T. Next, we will discuss property (WE) and property (R) by using the topological uniform descent.

Lemma 1   Let TB(H), the following statements are equivalent:

(1) T satisfies property (WE);

(2) σb(T)=[στ(T)accσ(T)]accσw(T){λσ(T):n(T-λI)=0};

(3) σb(T)=[στ(T)accσa(T)]accσea(T){λσ(T):n(T-λI)=0}acc{λC:n(T-λI)<d(T-λI)}.

Proof   (1)(2). Suppose that T satisfies property (WE). If λ0[στ(T)accσ(T)]accσw(T){λσ(T):n(T-λI)=0}, then there exists a deleted neighborhood B°(λ0) centered on λ0 such that for any λB°(λ0), T-λI is a Weyl operator. Since T has property (WE), we know that T-λI is a Browder operator. Then we get that λ0ρ(T)σ(T). We assume that λ0σ(T). If λ0ρτ(T), we get that T-λ0I is Drazin invertible. According to n(T-λ0I)>0, we have λ0E(T). Since T satisfies property (WE), we can get that T-λ0I is a Browder operator. If λ0accσ(T), we can also get that λ0E(T) and T-λ0I is Browder operators. The inclusion "" is obviously true.

(2)(1). Since σ(T)\σw(T)][στ(T)accσa(T)]= and [σ(T)\σw(T)][accσw(T){λσ(T):n(T-λI)=0}]=, we know that σ(T)\σw(T)E(T). Similarly, E(T)σ(T)\σw(T). Hence T satisfies property (WE).

The fact that accσea(T)acc{λC:n(T-λI)<d(T-λI)}=accσw(T) implies that (2)(3).

The closeness of operator range is very important in spectral theory. According to the closeness of operator range, the following spectral set is defined: σc(T)={λC:R(T-λI) is not closed}.

Lemma 2   Let TB(H), the following statements are equivalent:

(1) T satisfies property (R);

(2) σb(T)=[στ(T)accσ(T)]accσab(T){λσ(T):n(T-λI)=0}{λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T

d ( T - λ I ) } σ c ( T ) ] ;

(3) σb(T)=[στ(T)accσa(T)]accσab(T){λσ(T):n(T-λI)=0}{λC:n(T-λI)=}[acc{λC:n(T-λI)<-λ

I ) } σ c ( T ) ]

Proof   (1)(2). Suppose λ0[στ(T)accσ(T)]accσab(T){λσ(T):n(T-λI)=0}{λC:n(T-λI)=}[acc{λ

C : n ( T - λ I ) < d ( T - λ I ) } σ c ( T ) ] . Without loss of generality, we assume that λ0σ(T). Then we have 0<n(T-λ0I)<. There exists a deleted neighborhood B°(λ0;ε) centered on λ0 such that for any λB°(λ0;ε), λρab(T). If λ0accσ(T), then λ0π00(T). Since T satisfies property (R), we have that T-λ0I is a Browder operator. Suppose that λ0ρτ(T), if λ0ρc(T), then T-λ0I is an upper semi-Fredholm operator. By λB°(λ0;ε), λρab(T) and λ0isoσa(T), we know that λ0σa(T)\σab(T). Because T has property (R), we can get that λ0ρb(T). If λ0acc{λC:n(T-λI)<d(T-λI)}, then exists a deleted neighborhood B°(λ0)B°(λ0;ε) centered on λ0 such that for any λB°(λ0), λρb(T). Therefore λ0σ(T). Since λ0ρτ(T) and n(T-λ0I)<, we can also get that λ0ρb(T).

(2)(1). Suppose that λ0σa(T)\σab(T). It follows that λ0[στ(T)accσ(T)]accσab(T){λσ(T):n(T-λI)=0}{λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. So T-λ0I is a Browder operator. Then we have σa(T)\σab(T)π00(T). Similarly, we can get the inclusion "". Hence T satisfies property (R).

If (2) holds, then T has property (R). This implies that {λC:0<n(T-λI)<}ρc(T)acc[ρa(T)σ(T)]=, where ρc(T)=C\σc(T). It follows that accσ(T)=accσa(T){acc[ρa(T)σ(T)]σc(T)}{acc[ρa(T)σ(T)]ρc(T){λ

σ ( T ) : n ( T - λ I ) = 0 } } { a c c [ ρ a ( T ) σ ( T ) ] ρ c ( T ) { λ C : n ( T - λ I ) = } } . Hence [στ(T)accσ(T)]{[στ(T)accσa(T)]

{ λ σ ( T ) : n ( T - λ I ) = 0 } { λ C : n ( T - λ I ) = } [ a c c { λ C : n ( T - λ I ) < d ( T - λ I ) } σ c ( T ) ] } . Then we have that σb(T)[στ

( T ) a c c σ ( T ) ] a c c σ a b ( T ) { λ σ ( T ) : n ( T - λ I ) = 0 } { λ C : n ( T - λ I ) = } [ a c c { λ C : n ( T - λ I ) < d ( T - λ I ) } σ c ( T ) ]

The inclusion "" is obviously true. This implies that (3) holds. For the converse, if (3) holds, by accσa(T)accσ(T) we know that (2) holds. Therefore (2)(3).

Based on Lemma 1 and Lemma 2, the following results demonstrate that the two properties can be valid at the same time:

Theorem 1   T B ( H ) satisfies property (R) and property (WE) if and only if σb(T)=[στ(T)accσa(T)][accσab(T)σw(T)]{λσ(T):n(T-λI)=0}{λaccσ(T):n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)].

Proof   "". Suppose T has property (R). Using Lemma 2, we have that σb(T)=[στ(T)accσa(T)]accσab(T){λσ(T):n(T-λI)=0}{λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. Since T has property (WE), we know that ρw(T)=ρb(T) and E(T)=σ0(T). Hence accσab(T)ρw(T)=, {λisoσ(T):n(T-λI)=}=. Thus accσab(T)=[accσab(T)σw(T)][accσab(T)ρw(T)]=accσab(T)σw(T). {λC:n(T-λI)=}={λaccσ(T):n(T-λI)=}{λisoσ(T):n(T-λI)=}={λaccσ(T):n(T-λI)=}. Therefore σb(T)=[στ(T)accσa(T)][accσab(T)σw(T)]{λσ(T):n(T-λI)=0}{λaccσ(T):n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)].

"". By the condition, we get that σb(T)[στ(T)accσa(T)]accσab(T){λσ(T):n(T-λI)=0}{λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. The inclusion "" is obviously true. From Lemma 2, we know that T satisfies property (R).

If λ0σ(T)\σw(T), then λ0[στ(T)accσa(T)][accσab(T)σw(T)]{λσ(T):n(T-λI)=0}{λaccσ(T):n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. Thus T-λ0I is a Browder operator, σ(T)\σw(T)E(T). The converse is similar. Therefore T satisfies property (WE).

Remark 2   (i) In Theorem 1, suppose TB(H) satisfies property (R) and property (WE), then each part of the decomposition of σb(T) cannot be deleted.

(a) Let TB(l2) be defined by T(x1,x2,x3,)=(0,x22,x33,). Hence T has property (R) and property (WE). But σb(T)[accσab(T)σw(T)]{λσ(T):n(T-λI)=0}{λaccσ(T):n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. Thus [στ(T)accσa(T)] cannot be deleted.

(b) Let TB(l2) be defined by T(x1,x2,x3,)=(x2,x3,x4,). Then T has property (R) and property (WE). But σb(T)[στ(T)accσa(T)]{λσ(T):n(T-λI)=0}{λaccσ(T):n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. Thus [accσab(T)σw(T)] cannot be deleted.

(c) Let TB(l2) be defined by T(x1,x2,x3,)=(0,x1,x2,x3,). We know that T has property (R) and property (WE). But σb(T)[στ(T)accσa(T)][accσab(T)σw(T)]{λaccσ(T):n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. Hence {λσ(T):n(T-λI)=0} cannot be deleted.

(d) Let A,BB(l2) be defined by A(x1,x2,)=(0,x1,x2,), B(x1,x2,x3,)=(x1,0,x3,0,). And suppose T=(A00B). Then T satisfies property (R) and property (WE). But σb(T)[στ(T)accσa(T)][accσab(T)σw(T)]{λσ(T):n(T-λI)=0}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. Therefore {λaccσ(T):n(T-λI)=} cannot be deleted.

(e) Let A,BB(l2) be defined by A(x1,x2,)=(0,x1,x2,), B(x1,x2,x3,)=(0,0,x22,x33,). And suppose T=(A00B). Thus T satisfies property (R) and property (WE). But σb(T)[στ(T)accσa(T)][accσab(T)σw(T)]{λσ(T):n(T-λI)=0}{λaccσ(T):n(T-λI)=}. It follows that [acc{λC:n(T-λI)<d(T-λI)}σc(T)] cannot be deleted.

(ii) The conditions in Theorem 1 can be transformed as follows: accσa(T) can be replaced by accσ(T), accσab(T)σw(T) can be replaced by accσab(T)accσw(T).

From Lemma 1 and Lemma 2, we can get the following Corollary.

Corollary 1   Let TB(H). Then:

(1) T satisfies property (WE)E(T)ρτ(T)ρb(T)accσw(T)E(T)ρτ(T)ρb(T)accσea(T)acc{λC:n(T-λI)<d(T-λI)};

(2) T satisfies property (R)π00(T)ρτ(T)ρb(T)accσa(T){λC:n(T-λI)=0}{λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]π00(T)ρτ(T)ρb(T)accσab(T){λC:n(T-λI)=0}{λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-[λI)}σc(T)]].

By Theorem 1 and Corollary 1, we can get the following Corollary.

Corollary 2   Let TB(H). Then T satisfies property (R) and property (WE)E(T)ρτ(T)ρb(T)[accσab(T)

σ w ( T ) ] { λ C : n ( T - λ I ) = 0 } { λ C : n ( T - λ I ) = } [ a c c { λ C : n ( T - λ I ) < d ( T - λ I ) } σ c ( T ) ] .

Proof   Using Theorem 1, T satisfies property (R) and property (WE)σb(T)=[στ(T)accσa(T)][accσab(T)σw

( T ) ] { λ σ ( T ) : n ( T - λ I ) = 0 } { λ a c c σ ( T ) : n ( T - λ I ) = } [ a c c { λ C : n ( T - λ I ) < d ( T - λ I ) } σ c ( T ) ] .

"". E(T)ρτ(T) is obvious. If λ0ρτ(T) and λ0[accσab(T)σw(T)]{λσ(T):n(T-λI)=0}{λaccσ(T):n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)], by Theorem 1 we know that λ0ρb(T).

"". Suppose λ0σ(T) and λ0[στ(T)accσa(T)][accσab(T)σw(T)]{λσ(T):n(T-λI)=0}{λaccσ(T):n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. If λ0ρτ(T), we have λ0ρb(T) by the condition. Suppose λ0accσa(T), if λ0ρc(T), we can get that λ0ρτ(T), then T-λ0I is a Browder operator. If λ0acc{λC:n(T-λI)<d(T-λI)}, using λ0accσa(T) we have λ0isoσ(T). It also follows that λ0ρb(T). From Theorem 1, we can get that T satisfies property (R) and property (WE).

2 The Perturbation of Property (WE) and Property (R)

Suppose TB(H) satisfies property (R) or property (WE) and KB(H) is a compact operator or even a finite rank operator, but we cannot deduce that T+K satisfies property (R) or property (WE). For example, let T,KB(l2) be defined by T(x1,x2,x3,)=(0,x1,x22,x33,), K(x1,x2,x3,)=(0,x1,0,0,). It is obviously that T satisfies property (R) and property (WE) and K is a finite rank operator. However, calculations indicate that T+K does not satisfy either property (R) or property (WE).

The operator F is called a power finite rank operator, if there is a positive integer n such that dimR(Fn)<. A power finite rank operator is a Riesz operator, therefore, when F is a power finite rank operator, operator T is commutative with F, then it has the following properties: σ1(T+F)=σ1(T), where σ1{σw,σb,σe,σSF,σea} and isoσ(T+F)isoσ(T)ρ(T)[14].

Due to the complexity of general compact perturbations, this section focuses on the power finite rank perturbation of property (WE) and property (R).

Lemma 3   Let TB(H) and F is a power finite rank operator with FT=TF. The following statements are equivalent:

(1) T+F satisfies property (WE) and is an isoloid operator;

(2) σb(T)=[στ(T)accσ(T)]accσw(T);

(3) σb(T)=[στ(T)accσa(T)]accσea(T)acc{λC:n(T-λI)<d(T-λI)}.

Proof   (1)(2). Since T+F satisfies property (WE) and σw(T+F)=σw(T), σb(T+F)=σb(T), we have that σw(T)=σb(T). If λ0[στ(T)accσ(T)]accσw(T), when λ0στ(T), by the proof of Lemma 1, we know that T-λ0I is Drazin invertible. Thus λ0isoσ(T+F)ρ(T+F). If n(T+F-λ0I)=, then λ0E(T+F). Since T+F has property (WE), we get that T+F-λ0I is a Browder operator. It is a contradiction. Therefore n(T+F-λ0I)<, hence n(T-λ0I)<. By T-λ0I is Drazin invertible we have that λ0σb(T). Suppose that λ0accσ(T). Without loss of generality, we assume that λ0isoσ(T). Then λ0isoσ(T+F). The fact that T+F is isoloid implies that λ0E(T+F). Because T+F has the property (WE), T+F-λ0I is a Browder operator. We also have that λ0σb(T).

(2)(1). Suppose that λ0σ(T+F)\σw(T+F). Then we have that T-λ0I is a Weyl operator. So λ0[στ(T)accσ(T)]accσw(T). By condition (2), we can get that T-λ0I is a Browder operator. Thus T+F-λ0I is a Browder operator. Then λ0E(T+F). Contrarily, if λ0E(T+F), then λ0ρ(T)isoσ(T). Therefore λ0[στ(T)accσ(T)]accσw(T). Hence T-λ0I and T+F-λ0I are Browder operators.

If there exists λ0isoσ(T+F) such that n(T+F-λ0I)=0, then λ0isoσ(T)ρ(T). By condition (2), we have that T-λ0I is a Browder operator. It follows that T+F-λ0I is a Browder operator. The fact that n(T+F-λ0I)=0 implies that T+F-λ0I is invertible, which is a contradiction. Therefore T+F is an isoloid operator.

(1)(3). Suppose λ0[στ(T)accσa(T)]accσea(T)acc{λC:n(T-λI)<d(T-λI)}. If λ0στ(T), from the proof of (1) (2) we can get that λ0σb(T). If λ0accσa(T), we assume that λ0σ(T) without loss of generality. Since λ0accσea(T)acc{λC:n(T-λI)<d(T-λI)}, we have that λ0isoσ(T). Hence λ0isoσ(T+F). Since T+F is isoloid and satisfies property (WE), we have that T+F-λ0I is a Browder operator. Thus λ0σb(T).

(3)(1). The proof for (3)(1) is the same as the proof for (2)(1).

Lemma 4   Let TB(H) and F is a power finite rank operator with FT=TF. The following statements are equivalent:

(1) T+F is an isoloid operator and satisfies property (R), σa(T+F)=σ(T+F);

(2) σb(T)=[στ(T)accσ(T)]accσab(T){λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)];

(3) σb(T)=[στ(T)accσa(T)]accσab(T){λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)].

Proof   (1)(2). Suppose λ0[στ(T)accσ(T)]accσab(T){λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. We assume that λ0σ(T). If λ0ρτ(T), when λ0acc{λC:n(T-λI)<d(T-λI)}, we can get that λ0σ(T). Then T-λ0I is a Browder operator. When λ0σc(T), we have that T-λ0I is a Weyl operator. Hence λ0ρw(T+F). The fact that T+F satisfies property (R) implies that T+F-λ0I is a Browder operator. Hence λ0ρb(T). If λ0accσ(T), then λ0isoσ(T+F)ρ(T+F). We can assume that λ0isoσ(T+F). Since T+F is an isoloid operator, we have that λ0π00(T+F). Since T+F satisfies property (R), we can get that T+F-λ0I is a Browder operator. Therefore λ0ρb(T).

(2)(1). Let's first prove that T+F is an isoloid operator. If there exists λ0isoσ(T+F) such that n(T+F-λ0I)=0. We assume that λ0σ(T) without loss of generality. Thus n(T-λ0I)<, λ0isoσ(T). Therefore λ0[στ(T)accσ(T)]accσab(T){λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. It follows that T-λ0I is a Browder operator. Then T+F-λ0I is a Browder operator. Hence T+F-λ0I is invertible, which is a contradiction. Thus T+F is an isoloid operator.

Next we will prove σa(T+F)=σ(T+F). If λ0ρa(T+F), then λ0[στ(T)accσ(T)]accσab(T){λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. It implies that both T-λ0I and T+F-λ0I are Browder operators. Hence λ0σ(T+F).

If λ0σa(T+F)\σab(T+F), then λ0[στ(T)accσ(T)]accσab(T){λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. Hence T-λ0I and T+F-λ0I are Browder operators. Thus λ0π00(T+F). Conversely, suppose that λ0π00(T+F), then λ0isoσ(T)ρ(T). Then we also have that λ0[στ(T)accσ(T)]accσab(T){λC:n(T-λI)=}

[ a c c { λ C : n ( T - λ I ) < d ( T - λ I ) } σ c ( T ) ] . Thus T-λ0I and T+F-λ0I are Browder operators. Hence T+F satisfies property (R).

The proof for (1)(3) is the same as the proof for (1)(2).

Then we obtain the finite rank perturbation with two properties held at the same time.

Theorem 2   Let TB(H) and F is a power finite rank operator with FT=TF. The following statements are equivalent:

(1) T+F is an isoloid operator and satisfies property (R) and property (WE), σa(T+F)=σ(T+F);

(2) σb(T)=[στ(T)accσa(T)][accσab(T)σw(T)]{λaccσ(T):n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}

σ c ( T ) ]

Proof   (1)(2). Suppose λ0[στ(T)accσa(T)][accσab(T)σw(T)]{λaccσ(T):n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. If λ0accσab(T), by Lemma 4 we have that λ0σb(T). If λ0σw(T), by Lemma 3 we also get that λ0σb(T).

(2)(1). The statement (2) implies that σw(T)=σb(T). Suppose λ0[στ(T)accσ(T)]accσw(T), then λ0[στ(T)accσa(T)][accσab(T)σw(T)][acc{λC:n(T-λI)<d(T-λI)}σc(T)]. If λ0accσ(T), then λ0{λacc

σ ( T ) : n ( T - λ I ) = } . If λ0ρτ(T), we have that λ0ρ(T)σ(T). Thus T-λ0I is Drazin invertible. Therefore λ0{λaccσ(T):n(T-λI)=}. Hence λ0[στ(T)accσa(T)][accσab(T)σw(T)]{λaccσ(T):n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. Then T-λ0I is a Browder operator. Thus σb(T)=[στ(T)accσ(T)]accσw(T). With Lemma 3 we get that T+F is isoloid and satisfies property (WE).

By condition (2) we have that σb(T)=[στ(T)accσa(T)]accσab(T){λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)]. Then according to Lemma 4 we have that T+F satisfies property (R) and σa(T+F)=σ(T+F). Therefore (1) holds.

3 Property (WE) and Property (R) for Operator Functions

Let H(T) be the class of all complex-valued functions which are analytic on a neighborhood of σ(T) and are not constant on any component of σ(T). For the function's properties, we discuss property (WE) first.

Lemma 5   Let TB(H). For any fH(T), f(T) satisfies property (WE) if and only if λ, μρe(T), ind(T-λI)ind(T-μI)0 and one of the following conditions holds:

(1) σ(T)=[στ(T)accσ(T)]accσw(T){λσ(T):n(T-λI)=0};

(2) σb(T)=[στ(T)accσ(T)]accσw(T).

Proof   "". Suppose (1) holds. Then σ0(T)=, σ(T)=σb(T)=σw(T), thus for any fH(T), σ(f(T))=f(σ(T))=f(σw(T))=σw(f(T)), σ(f(T))\σw(f(T))=. Since E(f(T))f(E(T)), we have that E(f(T))=. So f(T) satisfies property (WE). If (2) holds, then isoσ(T)σ0(T), σw(T)=σb(T). Hence σw(f(T))=f(σw(T))=f(σb(T))=σb(f(T)). It follows that σ(f(T))\σw(f(T))σ0(f(T))E(f(T)). For the converse, let μ0E(f(T)), and let f(T)-μ0I=(T-λ1I)n1(T-λ2I)n2(T-λtI)ntg(T), where λiλj(i,j=1,2,,t), and g(T) is inverse. We assume that λiσ(T), then λiisoσ(T), thus λiσ0(T). It follows that f(T)-μ0I is a Browder operator.

"". If there exists λ1,λ2ρe(T) such that ind(T-λ1I)=n>0, ind(T-λ2I)=-m<0, suppose that f1(T)=(T-λ1I)m(T-λ2I)n, then 0σ(f1(T))\σw(f1(T)). By f1(T) satisfies property (WE), we have that f1(T) is a Browder operator. Hence both T-λ1I and T-λ2I are Browder operators. It is a contradiction. Thus λ, μρe(T), ind(T-λI)ind(T-μI)0.

If σ0(T)=, then the fact that T satisfies property (WE) implies that σ(T)=σw(T)=σb(T), E(T)=. According to Lemma 1, we can get (1) holds. When σ0(T), we can assert that {λisoσ(T):n(T-λI)=0}=. If not, suppose that λ3{λisoσ(T):n(T-λI)=0}, λ4σ0(T). Let f2(T)=(T-λ3I)(T-λ4I), then 0E(f2(T)). Since f2(T) satisfies property (WE), then f2(T) is a Browder operator, so is T-λ3I, which is a contradiction. Therefore {λisoσ(T):n(T-λI)=0}=. By Lemma 1 and its proof, we get that (2) holds.

Similarly, we can obtain the following conclusion. Let TB(H), for any fH(T), f(T) satisfies property (WE) if and only if λ, μρe(T), ind(T-λI)ind(T-μI)0 and one of the following conditions holds:

(1) σ(T)=[στ(T)accσa(T)]accσea(T){λσ(T):n(T-λI)=0}acc{λC:n(T-λI)<d(T-λI)};

(2) σb(T)=[στ(T)accσa(T)]accσea(T)acc{λC:n(T-λI)<d(T-λI)}.

Corollary 3   Let TB(H), for any fH(T), f(T) satisfies property (WE) if and only if λ,μρe(T), ind(T-λI)ind(T-μI)0 and the following conditions holds:

(1) T satisfies property (WE);

(2) If σ0(T), then σb(T)=[στ(T)accσ(T)]accσw(T).

For property (R) of operator functions, we have the following conclusions.

Lemma 6   Let TB(H), for any fH(T), f(T) satisfies property (R) if and only if one of the following conditions holds:

(1) σ(T)=[στ(T)accσ(T)]accσab(T){λσ(T):n(T-λI)=0}{λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)];

(2) σb(T)=[στ(T)accσ(T)]accσab(T){λC:n(T-λI)=}.

Proof   "". If σ0(T)=, then σ(T)=σb(T)=σw(T). By Lemma 2, we can get that (1) holds.

If σ0(T), we can assert that σ(T)=σa(T). If not, suppose that λ1ρa(T)σ(T), λ2σ0(T), let f(T)=(T-λ1I)(T-λ2I), then 0σa(f(T))\σab(f(T)). Since f(T) satisfies property (R), we can get that f(T) is a Browder operator. Then T-λ1I is a Browder operator, which is a contradiction. Therefore σ(T)=σa(T). Similarly, we have {λisoσ(T):n(T-λI)=0}= in this situation. Suppose that λ0[στ(T)accσ(T)]accσab(T){λC:n(T-λI)=}. If λ0στ(T), by T satisfies property (R) and σ(T)=σa(T) we have that λ0σ(T). The fact that n(T-λ0I)< implies that T-λ0I is a Browder operator. If λ0accσ(T), we can assume that λ0σ(T). Since {λisoσ(T):n(T-λI)=0}=, we have λ0π00(T). By T has property (R), we can get that T-λ0I is a Browder operator. Therefore (2) holds.

"". If condition (1) holds, then σ0(T)=, π00(T)=, σa(T)=σab(T). Then for any fH(T), σa(f(T))=f(σa(T))=f(σab(T))=σab(f(T)). Hence σa(f(T))\σab(f(T))=. Since π00(f(T))f(π00(T)), we have that π00(f(T))=. Therefore f(T) satisfies property (R).

If conditions (2) holds, then σab(T)=σb(T), σa(T)=σ(T), {λisoσ(T):n(T-λI)=0}=. Hence σa(f(T))\σab(f(T))σ0(f(T))π00(f(T)). Suppose that μ0π00(f(T)). Let f(T)-μ0I=(T-λ1I)n1(T-λ2I)n2(T-λtI)ntg(T), where λiλj, g(T) is invertible. We can assume that λiσ(T), then λiπ00(T). Since T has property (R), we get λiρb(T). So f(T)-μ0I is a Browder operator. Therefore f(T) satisfies property (R).

By the proof of Lemma 6, we can get that:

Corollary 4   Let TB(H). For any fH(T), f(T) satisfies property (R) if and only if the following conditions holds:

(1) T satisfies property (R);

(2) If σ0(T), then σb(T)=[στ(T)accσ(T)]accσab(T){λC:n(T-λI)=}.

Then we obtain the condition that the operator function satisfies both two properties.

Theorem 3   Let TB(H). For any fH(T), f(T) satisfies property (R) and property (WE) if and only if λ, μρe(T), ind(T-λI)ind(T-μI)0 and:

(1) T satisfies property (R) and property (WE);

(2) If σ0(T), then σb(T)=[στ(T)accσ(T)][accσab(T)σw(T)].

Proof   "". According to Corollary 3 and Corollary 4, we know that we only need to prove (2). Suppose that λ0[στ(T)accσ(T)][accσab(T)σw(T)]. If λ0σw(T), by Corollary 3 we know that T-λ0I is a Browder operator. Suppose λ0accσab(T). When n(T-λ0I)=0, by Corollary 4 we can get that λ0ρb(T). When n(T-λ0I)>0, the fact that σ(T)=σa(T) implies λ0isoσ(T). Thus λ0E(T). We also get that T-λ0I is a Browder operator. The converse is obviously true. Hence (2) holds.

The sufficiency is easy to get from Corollary 3 and Corollary 4.

Corollary 5   Let TB(H). For any fH(T), f(T) satisfies property (R) and property (WE) if and only if λ,μρe(T), ind(T-λI)ind(T-μI)0 and one of the following conditions holds:

(1) σ(T)=[στ(T)accσ(T)]accσab(T){λσ(T):n(T-λI)=0}{λC:n(T-λI)=}[acc{λC:n(T-λI)<d(T-λI)}σc(T)];

(2) σb(T)=[στ(T)accσ(T)]accσab(T){λC:n(T-λI)=}.

Proof   "". If σ0(T)=, then σ(T)=σw(T)=σb(T). Since {λC:n(T-λI)=}={λaccσ(T):n(T-λI)=}, by Theorem 1 and Remark 2 we can get that (1) holds. If σ0(T), the condition (2) hold.

"". According to Theorem 1 and Theorem 3, we know that for any fH(T), f(T) satisfies property (R) and property (WE).

Example 1 Let A,BB(l2) be defined by: A(x1,x2,)=(x2,x3,), B(x1,x2,x3,)=(x1,0,0,0,). Suppose that T=(A00I+B). Then: (1) T satisfies property (R) and property (WE); (2) For any λρe(T), ind(T-λI)0; (3) σ0(T), and σb(T)=[στ(T)accσ(T)][accσab(T)σw(T)].

Therefore by Theorem 3, we can get that for any fH(T), f(T) satisfies property (R) and property (WE).

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