# The Intersection of Upper and Lower Semi-Browder Spectrum of Upper-Triangular Operator Matrices

Authors: Shifang Zhang; Huaijie Zhong; Long Long
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

WhenA∈B(H) and B∈B(K) are given, we denote by MC the operator acting on the infinite-dimensional separable Hilbert space H⊕K of the form MC=(AC0B). In this paper, it is proved that there exists some operator C∈B(K,H) such that MC is upper semi-Browder if and only if there exists some left invertible operator C∈B(K,H) such that MC is upper semi-Browder. Moreover, a necessary and sufficient condition for MC to be upper semi-Browder for some C∈G(K,H) is given, where G(K,H) denotes the subset of all of the invertible operators of B(K,H).


---

## Body

## 1. Introduction

It is well known that ifH is a Hilbert space, T is a bounded linear operator defined on H, and H1 is an invariant closed subspace of T, then T can be represented in the following form:
(1)T=(**0*):H1⊕H1⊥⟶H1⊕H1⊥,
which motivated the interest in 2×2 upper-triangular operator matrices. For recent investigations on this subject, see references [1–23].Throughout this paper, letH and K be separable infinite-dimensional complex Hilbert spaces, and let B(H,K) be the set of all bounded linear operators from H into K; when H=K, we write B(H,H) as B(H). For A∈B(H), B∈B(K), and C∈B(K,H), we have MC=(AC0B)∈B(H⊕K). For T∈B(H,K), let R(T) and N(T) denote the range and the kernel of T, respectively, and denote that α(T)=dimN(T) and β(T)=dimK/R(T). If T∈B(H), the ascent asc(T) of T is defined to be the smallest nonnegative integer k which satisfies and N(Tk)=N(Tk+1). If such k does not exist, then the ascent of T is defined as infinity. Similarly, the descent des(T) of T is defined as the smallest nonnegative integer k for which R(Tk)=R(Tk+1) holds. If such k does not exist, then des(T) is defined as infinity, too. If the ascent and the descent of T are finite, then they are equal (see [6]). For T∈B(H), if R(T) is closed and α(T)<∞, then T is said to be an upper semi-Fredholm operator; if β(T)<∞, which implies that R(T) is closed, then T is said to be a lower semi-Fredholm operator. If T∈B(H) is either upper or lower semi-Fredholm operator, then T is said to be a semi-Fredholm operator. If both α(T)<∞ and β(T)<∞, then T is said to be a Fredholm operator. For a semi-Fredholm operator T, its index ind(T) is defined by ind(T)=α(T)-β(T).For a semi-Fredholm operatorT∈B(H), its shift Samuel multiplicity s_mul(T) and backward shift Samuel multiplicity b.s._mul(T) are defined, respectively, by the following (see [24]):
(2)s_mul(T)=limk→∞β(Tk)k,b.s._mul(T)=limk→∞α(Tk)k.
Moreover, it has been proved that s_mul(T), b.s._mul(T)∈{0,1,2,…,∞} and that ind(T)=b.s._mul(T)-s_mul(T). These two invariants refine the Fredholm index and can be regarded as the stabilized dimensions of the kernel and the cokernel (see [24]).In this paper, the sets of invertible operators and left invertible operators fromH into K are denoted by G(H,K) and Gl(H,K), respectively; the sets of all Fredholm operators, upper semi-Fredholm operators, and lower semi-Fredholm operators from H into K are denoted by Φ(H,K), Φ+(H,K), and Φ-(H,K), respectively; the sets of all Browder operators, upper semi-Browder operators, and lower semi-Browder operators, on H are defined, respectively, by the following:
(3)Φb(H):={T∈Φ(H):asc(T)=des(T)<∞},Φab(H):={T∈Φ+(H):asc(T)<∞},Φsb(H):={T∈Φ-(H):des(T)<∞}.
Moreover, for T∈B(H), we introduce its corresponding spectra as follows.The spectrum is given asσ(T)={λ∈ℂ:T-λI∉G(H)}.The left spectrum is given asσl(T)={λ∈ℂ:T-λI∉Gl(H)}.The essential spectrum is defined asσe(T)={λ∈ℂ:T-λI∉Φ(H)}.The upper semi-Fredholm spectrum is defined asσSF+(T)={λ∈ℂ:T-λI∉Φ+(X)}.The lower semi-Fredholm spectrum is presented asσSF-(T)={λ∈ℂ:T-λI∉Φ-(X)}.The Browder spectrum is presented asσb(T)={λ∈ℂ:T-λI∉Φb(H)}.The upper semi-Browder spectrum is defined asσab(T)={λ∈ℂ:T-λI∉Φab(X)}.The lower semi-Browder spectrum is presented asσsb(T)={λ∈ℂ:T-λI∉Φsb(X)}.Using the Samuel multiplicities, Zhang and Wu (see [20]) gave a necessary and sufficient condition for which MC∈Φab(H⊕K) for some C∈B(K,H) and characterized the set of ∩C∈B(K,H)σab(MC). In this paper, our main goal is to characterize the intersection of ∩C∈Gl(K,H)σab(MC) and ∩C∈G(K,H)σab(MC). This paper is organized as follows. In Section 2, we give a necessary and sufficient condition for which MC∈Φab(H⊕K) for some C∈Gl(K,H) and get
(4)⋂C∈Gl(K,H)‍σab(MC)=⋂C∈B(K,H)‍σab(MC).
In Section 3, we give a necessary and sufficient condition for which MC∈Φab(H⊕K) for some C∈G(K,H) and get
(5)⋂C∈G(K,H)‍σab(MC)=⋂C∈Φ(K,H)‍σab(MC)=(⋂C∈B(K,H)‍σab(MC))∪{λ∈ℂ:B-λiscompact}.For the sake of convenience, we now present some lemmas which will be used in the sequel.Lemma 1 (see [20, 24]).
An operatorT∈B(H) is semi-Fredholm if and only if  T can be decomposed into the following form with respect to some orthogonal decomposition H=H1⊕H2⊕H3:(6)T=(T1T12T130T2T2300T3):H1⊕H2⊕H3⟶H1⊕H2⊕H3,
where dim(H3)<∞, T1 is a right invertible operator, T3 is a finite nilpotent operator, T2 is a left invertible operator, and min{ind(T1),-ind(T2)}<∞. Moreover, ind(T1)=α(T1)=b.s._mul(T), ind(T2)=-β(T2)=-s_mul(T), and ind(T)=α(T1)-β(T2).Lemma 2 (see [18]).
LetA∈B(H), B∈B(K), and C∈B(K,H).(1)
IfA∈Φb(H), then B∈Φab(K) if MC∈Φab(H⊕K) for some C∈B(K,H).(2)
IfMC∈Φab(H⊕K) for some C∈B(K,H), then A∈Φab(H).(3)
IfA∈Φab(H) and B∈Φab(K), then MC∈Φab(H⊕K) for any C∈B(K,H).(4)
IfB∈Φb(K), then A∈Φab(H) if MC∈Φab(H⊕K) for some C∈B(K,H);A∈Φsb(H) if MC∈Φsb(H⊕K) for some C∈B(K,H).(5)
IfMC∈Φb(H⊕K) for some C∈B(K,H), then A∈Φab(H) and B∈Φsb(K).(6)
If two ofA, B, and MC are Browder, then so is the third.Lemma 3 (see [20]).
LetT∈B(H). Then, T is upper semi-Browder if T can be decomposed into the following form with respect to some orthogonal decomposition H=H1⊕H2:(7)T=(T1T120T2),
where dim(H1)<∞, T1 is nilpotent, T2 is left invertible, and β(T2)=s_mul(T)=-ind(T).Lemma 4 (see [20]).
LetT∈B(H). Then, T is lower semi-Browder if T can be decomposed into the following form with respect to some orthogonal decomposition H=H1⊕H2:
(8)T=(T1T120T2),
where dim(H2)<∞, T1 is right invertible, T2 is nilpotent, and α(T1)=b.s._mul(T)=ind(T).Lemma 5 (see [20]).
For any givenA∈B(H) and B∈B(K), MC∈Φab(H⊕K) for some C∈B(K,H) if A∈Φab(H) and
(9)s_mul(A)=∞ifB∉Φ+(K),b.s._mul(B)≤s_mul(A)ifB∈Φ+(K).Lemma 6 (see [9]).
For any givenA∈B(H) and B∈B(K), MC is left invertible for some C∈B(K,H) if A is left invertible and
(10)a(B)≤β(A)ifR(B)isclosed,β(A)=∞ifR(B)isnotclosed.Lemma 7 (see [25]).
LetV be a linear subspace of H. Then, the following statements are equivalent.(1)
Any bounded operatorA∈B(H) with R(A)⊆V is compact.(2)
V contains no closed infinite-dimensional subspace.


## 2.⋂C∈B(K,H)‍σab(MC) and ⋂C∈Gl(K,H)‍σab(MC)

In [1, 20], the authors have proved that
(11)⋂C∈B(K,H)‍σb(MC)=σab(A)∪σsb(B)∪{Bλ∈ℂ:α(A-λ)=σab(A)∪σsb(B)iii∪ppp+α(B-λ)≠β(A-λ)=σab(A)∪σsb(B)iii∪ppp+β(B-λ)}.
They, moreover, proved that
(12)⋂C∈B(K,H)‍σab(MC)=σab(A)∪{λ∈ℂ:λ∈σSF+(B),=σab(A)∪iiippps_mul(A-λ)<∞}s.miiiul∪{λ∈Φ(A)∩Φ+(B):b.s._mul(B-λ)ppppppp>s_mul(A-λ)}.Comparing the above two kinds of spectra with the upper semi-Weyl spectrum and Weyl spectrum, one may expect that the following equality holds:(13)⋂C∈B(K,H)‍σab(MC)=σab(A)∪{λ∈ℂ:λ∈σSF+(B),β(A-λ)<∞}=σab(A)∪.∪{βλ∈ℂ:α(A-λ)+α(B-λ)=σab(A)∪.pp.>β(A-λ)+β(B-λ)}.
However, it is not that case, as the following example shows.Example 8.
LetA be the unilateral shift on ℓ2, that is,
(14)V:ℓ2⟶ℓ2,{z1,z2,…}↦{0,z1,z2,…},
and let the operators A and B be defined by
(15)A=V,B=((V*)200V5):ℓ2⊕ℓ2⟶ℓ2⊕ℓ2.
Then, we have b.s._mul(B)=2>s_mul(A)=1, while α(A)+α(B)=2<β(A)+β(B)=6. Moreover, 0∈⋂C∈B(K,H)‍σab(MC), while 0∉σab(A)∪{λ∈ℂ:λ∈σSF+(B) and β(A-λ)<∞}∪{λ∈ℂ:α(A-λ) +α(B-λ)>β(A-λ)+β(B-λ)}. Thus, (13) does not hold.In spite of the above counter example, we have the following.Proposition 9.
For any givenA∈B(H) and B∈B(K), one has
(16)⋂C∈B(K,H)‍σab(MC)⊇σab(A)⋂C∉B(K,H)‍∪{λ∈ℂ:λ∈σSF+(B),β(A-λ)<∞}⋂C∉B(K,H)‍∪{Bλ∈ℂ:α(A-λ)+α(B-λ)⋂CB(Kiooooo,H)‍>β(A-λ)+β(B-λ)}.Proof.
From the proof of Theorem  2.3 in [20], we know that when A∈Φab(H), s_mul(A)<∞ if and only if β(A)<∞. Combining this fact with Corollary 2.5 of [20], it is easy to see that
(17)⋂C∈B(K,H)‍σab(MC)⊇σab(A)∪{λ∈ℂ:λ∈σSF+(B),⋂C∉B(K,H)‍σab(MC)iii⊇σab(A)∪β(A-λ)<∞}.
Noting that β(B-λ)<∞ implies that R(B-λ) is closed, it follows from corollary 2.5 of [2] that
(18)⋂C∈B(K,H)‍σab(MC)⊇⋂C∈B(K,H)‍σaw(MC)⋂C∉B(K,H)‍σab(MC)⊇{βλ∈ℂ:α(A-λ)+α(B-λ)⋂C∉B(K,H)‍σab(MC)iiiiiii>β(A-λ)+β(B-λ)},
where σaw(MC)={λ∈ℂ:MC-λ is not uppersemi-Fredholm operator with index less than or equal to 0}.Now, we are ready to present the main result of this section.Theorem 10.
For any givenA∈B(H) and B∈B(K), one has
(19)⋂C∈Gl(K,H)‍σab(MC)=⋂C∈B(K,H)‍σab(MC).Proof.
Since⋂C∈Gl(K,H)‍σab(MC)⊇⋂C∈B(K,H)‍σab(MC) is obvious, it is sufficient to prove that if MC∈Φab(H⊕K), then there exists some left invertible operator Q∈B(K,H) such that MQ∈Φab(H⊕K).
Suppose thatMC∈Φab(H⊕K). It follows from Lemma 5 that A∈Φab(H) and
(20)s_mul(A)=∞ifB∉Φ+(K),b.s._mul(B)≤s_mul(A)ifB∈Φ+(K).
There are two cases to consider.
Case  1. Assume that A∈Φab(H), s_mul(A)=∞, and B∉Φ+(K). Then, it follows from Lemma 3 that A can be decomposed into the following form:
(21)A=(A1A120A2):H1⊕H2⟶H1⊕H2,
where dim(H1)<∞, A1 is nilpotent, A2 is a left invertible operator, and β(A2)=s_mul(A)=∞. So, we can let
(22)Q=(00V):K⟶H1⊕R(A2)⊕(H2⊖R(A2)),
where V∈B(K,(H2⊖R(A2))) is unitary. Obviously, Q is left invertible. Now, MQ can be rewritten as
(23)MQ=(A1A1200A2000V00B):H1⊕H2⊕K⟶H1⊕R(A2)⊕(H2⊖R(A2))⊕K.
Since A2 is left invertible and V is invertible, then there exist unique A2′ and V′ such that A2′A2=IH2 and V′V=IK, and
(24)(A2′000V′0)(A200V0B)=IH2⊕IK.
This implies that (A200V0B) is left invertible. And, hence, Lemma 2 leads to MQ∈Φab(H⊕K).
Case  2. Assume that A∈Φab(H), b.s._mul(B)≤s_mul(A), and B∈Φ+(K). Then, it follows from Lemma 3 that A can be decomposed into the following form:
(25)A=(A1A120A2):H1⊕H2⟶H1⊕H2,
where dim(H1)<∞, A1 is nilpotent, A2 is a left invertible operator, and β(A2)=s_mul(A). By the assumption that B∈Φ+(K) and Lemma 1, we know that B can be decomposed into the following form with respect to some orthogonal decomposition K=K1⊕K2⊕K3:(26)B=(B1**0B2*00B3),
where dim(K3)<∞, B1 is a right invertible operator, B2 is a left invertible operator, B3 is a finite nilpotent operator, and the parts marked by * can be any operators. Moreover, ∞>α(B1)=b.s._mul(B). Thus, β(A2)≥α(B1), and then there exists some left invertible C1∈B(N(B1),H2⊖R(A2)). Noting that dim((K1⊖N(B1))⊕K2⊕K3)=dim(H1⊕R(A2))=∞, we can let C2∈G((K1⊖N(B1))⊕K2⊕K3,H1⊕R(A)). Consider
(27)Q=(C100C2):N(B1)⊕[K1⊖N(B1)⊕K2⊕K3]⟶(H2⊖R(A2))⊕[H1⊕R(A2)].
Obviously,Q is left invertible, and MQ can be rewritten as
(28)MQ=(A1A12C110C12C130A21C210C22C23000C10000B110**0000B2*00000B3):H1⊕H2⊕(K1⊖N(B1)⊕N(B1))piiii.p..⊕K2⊕K3⟶H1⊕R(A2)⊕(H2⊖dR(A2))⊕K1oioiii⊕K2⊕K3,
where A21 and B11 are invertible and C1 and B2 are left invertible. Similar to the proof of Case 1, through direct calculation we can show that (A21C21000C10B110) is left invertible. Also since dim(H1)<∞ and dim(K3)<∞, we have A1∈Φb(H1) and B3∈Φb(K3). Thus, it follows from Lemma 2 that MC∈Φab(H⊕K).By duality, we have the following.Theorem 11.
For any givenA∈B(H) and B∈B(K), one has
(29)⋂C∈Gr(K,H)‍σsb(MC)=⋂C∈B(K,H)‍σsb(MC).


## 3.⋂C∈G(K,H)‍σab(MC) and ⋂C∈Φ(K,H)‍σab(MC)

In this section, we give the characterization of invertible and Fredholm perturbations of upper semi-Browder spectra of2×2 upper-triangular matrices. We begin with some lemmas.Lemma 12 (see [19]).
For a given pair(A,B)∈B(H)×B(K), if either A or B is a compact operator, then, for each C∈Φ(K,H), MC is not a semi-Fredholm operator.
In particular, ifB is not compact, then MC is not semi-Browder for any invertible operator C.Lemma 13.
The following statements are equivalent.(i)
B is not compact.(ii)
For each givenA∈Φab(H), if β(A)=∞, then there exists an operator C∈G(K,H) such that MC is an upper semi-Browder operator.(iii)
For each givenA∈Φab(H), if β(A)=∞, then there exists an operator C∈Φ(K,H) such that MC is an upper semi-Browder operator.Proof.
Obviously, we only need to prove the implications (i)⇒ (ii) and (iii) ⇒ (i).(iii)⇒ (i). If B is compact, then it follows from Lemma 12 that MC is not a semi-Fredholm operator for each C∈Φ(K,H), which contradicts with (iii). Thus, B is not compact.(i)⇒ (ii). Suppose that B is not compact. Then, we consider the following two cases.
Case  1. Assume that R(B) is closed. It follows from Lemma 3 that A can be decomposed into the following form with respect to some orthogonal decomposition H=H1⊕H2:(30)A=(A1A120A2):H1⊕H2⟶H1⊕H2,
where dim(H1)<∞, A1 is nilpotent, and A2 is a left invertible operator. Noting that β(A)=∞, we have β(A2)=∞. Since the assumption that B is not compact, we have that dimN(B)⊥=∞. Also since β(A2)=∞, let R(A2)⊥=H3⊕H4 with dim(H3)=dimN(B) and dim(H4)=∞. Define an operator C:K→H by
(31)C=(C100C2):N(B)⊕N(B)⊥⟶H3⊕(H1⊕R(A2)⊕H4),
where C1∈B(N(B),H3) and C2∈B(N(B)⊥,H1⊕R(A2)⊕H4) are invertible operators. Obviously, C∈B(K,H) is invertible. Next, we claim that MC is an upper semi-Browder operator. To see this, MC can be rewritten as
(32)MC=(A1A12C1100A22C210000C100C41000B10):H1⊕H2⊕N(B)⊥⊕N(B)⟶H1⊕R(A2)⊕H3⊕H4⊕K,
where A22∈B(H2,R(A2)) is invertible and B1∈B(N(B)⊥,R(B)) is left invertible. By Lemma 2 and the fact that A1∈Φb(H1), it is sufficient to prove that
(33)M1=:(A22C21000C10C4100B10)
is semi-Browder. For this, we only need to show that M1 is left invertible. In fact, since A22 is invertible and B1 and C1 are left invertible, we can set A22′, B1′, and C1′ such that
(34)A22′A22=IH2,B1′B1=IN(B)⊥,C1′C1=IN(B).
Direct calculation shows that
(35)(A22′00-A22′C21B1′000B1′0C1′00)(A22C21000C10C4100B10)=(IH2000IN(B)⊥000IN(B)),
which implies that M1 is left invertible. Noting that A1∈Φ(H1), by Lemma 2 we have that MC is upper semi-Browder.
Case  2. Assume that R(B) is not closed. If B is not compact, then by Lemma 7, R(B) contains a closed infinite-dimensional subspace. Without loss of generality, suppose that K1~ is a closed subspace of R(B) with dimK1~=∞ and dimK1~⊥=∞. Let K1={x∈N(B)⊥:Bx∈K1~}. Thus, K1 is a closed subspace of N(B)⊥, and dim(K1)=∞. Denote K2=N(B)⊥⊖K1. Without loss of generality, we may assume that dim(K2)=∞ (otherwise, suppose that {en}n=1∞ is an orthonormal basis of K1. Denote K1′=span{en:n=2,4,6,…} and K1′~={Bx:x∈K1′}, then K1 and K1~ can be replaced by K1′ and K1′~, resp.). Since β(A2)=∞, let R(A2)⊥=H3⊕H4 with dim(H3)=dimN(B) and dimH4=∞. Define an operator C:K→H by
(36)C=(C1000C2000C3):K1⊕K2⊕N(B)⟶(H1⊕R(A2))⊕H4⊕H3,
where C1, C2, and C3 are unitary operators. Obviously, C is invertible. MC can be rewritten as
(37)MC=(A1A12C11000A22C21000000C3000C2000B11B120000B220):H1⊕H2⊕K1⊕K2⊕N(B)⟶H1⊕R(A2)⊕H3⊕H4⊕K1~⊕K1~⊥,
where A22 and B11 are invertible and C1=(C11C21).
Next, we prove thatMC∈Φab(H⊕K). Noting that dim(H1)<∞, then, by Lemma 2, it is sufficient to prove that
(38)M1=:(A22C2100000C300C200B11B12000B220)
is left invertible. For this, let A22′, B11′, C1′, and C2′ be operators satisfying
(39)A22′A22=IH2,B11′B11=IK1,C2′C2=IK2,C3′C3=IN(B).
Direct calculation shows that
(40)(A22′0A22′C21B11′B12C2′-A22′C21B11′000B11′B12C2′B11′000C2′000C3′00)×(A22C2100000C300C200B11B12000B220)=(IH20000IK10000IK20000IN(B)),
which implies that M1 is left invertible.
Combining Case 1 with Case 2, the lemma is proved.Similarly, we have the following.Lemma 14.
The following statements are equivalent:(i)
A is not compact.(ii)
For each givenB∈Φsb(H), if α(B)=∞, then there exists an operator C∈G(K,H) such that MC is a lower semi-Browder operator.(iii)
For each givenB∈Φsb(H), if α(B)=∞, then there exists an operator C∈Φ(K,H) such that MC is a lower semi-Browder operator.
One is now ready to prove the main result of this section.Theorem 15.
For a given pair(A,B)∈B(H)×B(K), one has
(41)⋂C∈G(K,H)‍σab(MC)=⋂C∈Φ(K,H)‍σab(MC)=(⋂C∈B(K,H)‍σab(MC))∪{λ∈ℂ:B-λiscompact}.Proof.
According to Lemma12, it is clear that
(42)⋂C∈G(K,H)‍σab(MC)⊇⋂C∈Φ(K,H)‍σab(MC)⊇(⋂C∈B(K,H)‍σab(MC))∪{λ∈ℂ:B-λiscompact}.
For the conversion, without loss of generality, suppose that
(43)0∉(⋂C∈B(K,H)σab(MC))∪{λ∈ℂ:B-λiscompact}.
Then,B is not compact, and there exists some C∈B(K,H) such that MC∈Φab(H⊕K), and, hence, A∈Φab(H).
Case  1.β(A)=∞. It follows from Lemma 13 that there exists some C∈G(K,H) such that MC is an upper semi-Browder operator. This implies that λ∉⋂C∈G(K,H)‍σab(MC). In this case, we have proved. Consider that
(44)⋂C∈G(K,H)‍σab(MC)⊆(⋂C∈B(K,H)‍σab(MC))∪{λ∈ℂ:B-λiscompact}.
Case  2. Consider β(A)<∞. This implies that A∈Φ(H), and, thus, B∈Φ+(K) since MC∈Φab(H⊕K). It follows from Lemma 5 that b.s._mul(B)≤s_mul(A). Moreover, using Lemmas 1 and 3, we have
(45)A=(A1A120A2):H1⊕H2↦H1⊕H2,B=(B1**0B2*00B3):K1⊕K2⊕K3↦K1⊕K2⊕K3,
where dim(H1)<∞, A1 is nilpotent, A2 is a left invertible operator, dim(K3)<∞, B1 is a right invertible operator, B2 is a left invertible operator, B3 is a finite nilpotent operator, and the parts marked by * can be any operators. Moreover, β(A2)=s_mul(A), and α(B1)=b.s._mul(B). Hence, α(B1)≤β(A2). Now, put H2=R(A2)⊕H3⊕H4, where dim(H3)=α(B1)<∞. Noting that dimH1⊕R(A2)⊕H4=N(B1)⊥⊕K2⊕K3=∞, there exist unitaries C33∈B(N(B1),H3) and C′∈B(H1⊕R(A2)⊕H4,N(B1)⊥⊕K2⊕K3). Let C=(C3300C′). Obviously, C=(C3300C′)∈G(H⊕K).
Consider that operator(46)MC=(AC0B):H⊕K→H⊕K=(A1A12C110C13C140A2C210C23C24000C330000C410C43C4400B110**0000B2*00000B3):H1⊕H2⊕N(B1)⊥⊕N(B1)⊕K2⊕K3⟶H1⊕R(A2)⊕H3⊕H4⊕K1⊕K2⊕K3,
where
(47)C′=(C11C13C14C21C23C24C41C43C44):H1⊕H2⊕N(B1)⊥⟶H1⊕R(A2)⊕H4,B1=(B110).
We claim that MC∈Φab(H⊕K). In fact, since A1 and B3 are Browder operators, then, by Lemma 2, it is sufficient to show that
(48)M=:(A2C210C2300C3300C410C430B110*000B2):H2⊕N(B1)⊥⊕N(B1)⊕K2⟶R(A2)⊕H3⊕H4⊕K1⊕K2
is upper semi-Browder. Observe that A2 and B2 are left invertible; C33 and B11 are invertible. Direct calculation shows that M is injective. Since A∈Φ(H) and B∈Φ+(K), we have MC∈Φab(H⊕K), and, hence, M is an upper semi-Fredholm operator. Thus, M is left invertible. Combining this with Lemma 2 yields MC∈Φab(H⊕K), which means that λ∉⋂C∈G(K,H)‍σab(MC). Thus,
(49)⋂C∈G(K,H)‍σab(MC)⊆⋂C∈B(K,H)‍σab(MC)∪{λ∈ℂ:B-λiscompact}.
Combining Case 1 with Case 2 leads to(50)⋂C∈G(K,H)‍σab(MC)=⋂C∈ϕ(K,H)‍σab(MC)=(⋂C∈B(K,H)‍σab(MC))∪{λ∈ℂ:B-λiscompact}.
This completes the proof.By duality, we haveTheorem 16.
For a given pair(A,B)∈B(H)×B(K), one has
(51)⋂C∈G(K,H)‍σsb(MC)=⋂C∈Φ(K,H)‍σsb(MC)=(⋂C∈B(K,H)‍σsb(MC))∪{λ∈ℂ:A-λiscompact}.



---

*Source: 373147-2013-09-10.xml*