Property (Bv) and Tensor Product

Abstract

In this article, a-Browder’s classical theorem is considered through the property $\left(Bv\right)$, and we show that if two operators are norm equivalent, then property $\left(Bv\right)$ holds for one if and only if it holds for the other. The necessary conditions for the difference of the spectrum and essential approximate point spectrum of the tensor product of two operators coincide with the product of differences between the spectrum and the essential approximate point spectrum of its two factors are investigated. We also discuss the necessary conditions for the tensor product of two operators to verify the property $\left(Bv\right)$ and simultaneously give the equivalence between their various spectra with their Browder spectrum, likewise with the Drazin spectrum.

1. Preliminaries and Introduction

Let $\mathfrak{L}\left(\mathfrak{X}\right)$ be the Banach algebra of all bounded linear operators on a complex Banach space $\mathfrak{X}.$ We put $\mathfrak{I}\in \mathfrak{L}\left(\mathfrak{X}\right)$ the identity operator, and the resolvent of an operator $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ is defined as $\rho \left(\mathfrak{T}\right)=\mathbb{C}\backslash \sigma \left(\mathfrak{T}\right),$ where $\sigma \left(\mathfrak{T}\right)$ denotes spectrum of an operator $\mathfrak{T}$ which is defined as

$$\sigma \left(\mathfrak{T}\right)=\{\lambda \in \mathbb{C}:\lambda \mathfrak{I}-\mathfrak{T}\mathrm{is} \mathrm{not} \mathrm{invertible}\}.$$

The other classical spectra of $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ such as the approximate point spectrum $A_{\sigma }\left(\mathfrak{T}\right)$, surjective spectrum $S_{\sigma }\left(\mathfrak{T}\right),$ essential spectrum or Fredholm spectrum $F_{\sigma }\left(\mathfrak{T}\right)$, Weyl spectrum $W_{\sigma }\left(\mathfrak{T}\right)$, Browder spectrum $B_{\sigma }\left(\mathfrak{T}\right)$, Drazin spectrum $D_{\sigma }\left(\mathfrak{T}\right)$, upper semi-Fredholm spectrum $F_{\sigma }^{+}\left(\mathfrak{T}\right)$, lower semi-Fredholm spectrum $F_{\sigma }^{-}\left(\mathfrak{T}\right)$, upper semi-Browder spectrum $B_{\sigma }^{+}\left(\mathfrak{T}\right),$ essential approximate point or upper semi-Weyl spectrum $E{A}_{\sigma }\left(\mathfrak{T}\right)$ and left Drazin spectrum $D_{\sigma }^l\left(\mathfrak{T}\right)$ are defined analogously to $\sigma \left(\mathfrak{T}\right)$. In addition, we denote by $p\left(\mathfrak{T}\right)$ and $q\left(\mathfrak{T}\right)$ the ascent and descent of $\mathfrak{T}$, respectively. See [1] for more details.

Some spectra mentioned above have been generalized by Berkani et al. ([2,3,4]). So, we have B-Weyl spectrum $B{W}_{\sigma }\left(\mathfrak{T}\right)$, upper semi B-Weyl spectrum $B{W}_{\sigma }^{+}\left(\mathfrak{T}\right)$, B-Fredholm spectrum $B{F}_{\sigma }\left(\mathfrak{T}\right)$ and upper semi B-Fredholm spectrum $B{F}_{\sigma }^{+}\left(\mathfrak{T}\right)$. Recall that the upper semi B-Browder spectrum coincides with the left Drazin invertible spectrum, similarly the B-Browder spectrum coincides with Drazin spectrum, see [5] or [6].

There are several studies which give the relationship between some of the above mentioned spectra, but they do not give conditions for the spectra to coincide in a non-trivial way. Furthermore, an important breakthrough in operator theory is due to the fact that the various spectral properties are defined in a special way as a function of spectra, such as the classical Weyl’s theorem. We mention some of the notations and those properties to be used in this article.

For $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right),$ if $\mathfrak{A}\subseteq \sigma \left(\mathfrak{T}\right),$ then ${\mathfrak{A}}^{\prime }$ denotes the complement of $\mathfrak{A}$ in $\sigma \left(\mathfrak{T}\right).$ i.e., ${\mathfrak{A}}^{\prime }=\sigma \left(\mathfrak{T}\right)\backslash \mathfrak{A}$, and we denote by $iso\mathfrak{B},$ the set of isolated points in the set $\mathfrak{B}\subseteq \mathbb{C}.$

• ${\mathrm{\Pi }}_a^0\left(\mathfrak{T}\right):=A_{\sigma }\left(\mathfrak{T}\right)\backslash B_{\sigma }^{+}\left(\mathfrak{T}\right)$.
• ${\mathrm{\Pi }}_a\left(\mathfrak{T}\right):=A_{\sigma }\left(\mathfrak{T}\right)\backslash D_{\sigma }^l\left(\mathfrak{T}\right)$.
• ${\pi }_{00}\left(\mathfrak{T}\right):=\{\lambda \in iso\sigma \left(\mathfrak{T}\right):0<\alpha (\lambda \mathfrak{I}-\mathfrak{T})<\infty \}.$

Next, we recall the definition of some spectral properties.

Definition 1.

An operator $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ is said to satisfy:

• Browder’s theorem [7] , if $\sigma \left(\mathfrak{T}\right)\backslash W_{\sigma }\left(\mathfrak{T}\right)=B_{\sigma }^{\prime }\left(\mathfrak{T}\right).$
• a-Browder’s theorem [8] , say $\mathfrak{T}\in \left(aBt\right)$, if $A_{\sigma }\left(\mathfrak{T}\right)\backslash E{A}_{\sigma }\left(\mathfrak{T}\right)={\mathrm{\Pi }}_a^0\left(\mathfrak{T}\right).$
• Weyl’s theorem [9] , if $\sigma \left(\mathfrak{T}\right)\backslash W_{\sigma }\left(\mathfrak{T}\right)={\pi }_{00}\left(\mathfrak{T}\right)$.
• Property $\left(gaz\right)$ [10] , if $B{W}_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right):={\mathrm{\Pi }}_a\left(\mathfrak{T}\right)$.
• Property $\left(Bv\right)$ [11] , say $\mathfrak{T}\in \left(J_{bv}\right)$ if $E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right)=B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).$

Recall that for every bounded operator $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right),$$B_{\sigma }^{\prime }\left(\mathfrak{T}\right)\subseteq iso\sigma \left(\mathfrak{T}\right)$ is true, see ([12], Proposition 50.2), and an operator $\mathfrak{T}$ is said to be a-polaroid if every point $\lambda \in iso{A}_{\sigma }\left(\mathfrak{T}\right)$ is a pole of the resolvent of $\mathfrak{T}$, i.e., $\lambda \in D_{\sigma }^{\prime }\left(\mathfrak{T}\right)$. In addition, it is denoted by $\alpha \left(\mathfrak{T}\right):=\dim \ker \mathfrak{T}$, $\beta \left(\mathfrak{T}\right):=\mathrm{codim}\mathfrak{T}\left(\mathfrak{X}\right)=\dim \mathfrak{X}/\mathfrak{T}\left(\mathfrak{X}\right)$, the index of $\mathfrak{T}$ is defined by $\mathrm{ind}\mathfrak{T}:=\alpha \left(\mathfrak{T}\right)-\beta \left(\mathfrak{T}\right)$. So, $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ is an upper semi-Weyl operator if $\mathrm{ind}\mathfrak{T}\le 0$ and $\mathfrak{T}\left(\mathfrak{X}\right)$ is closed. See [1] for more details.

Remark 1.

By ([8], Theorem 3.3) for $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$, we have that

$$\mathfrak{T}\in \left(J_{bv}\right)\iff \mathfrak{T}\in \left(a{B}_t\right).$$

Note that by Theorems 6, 7 and 2 of [13], we have the next theorem.

Theorem 1.

Let $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$. If $int\left(E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right)\right)=Ø$, then

1. 

$\mathfrak{T}\in \left(J_{bv}\right)$.

2. 

$\sigma \left(\mathfrak{T}\right)=A_{\sigma }\left(\mathfrak{T}\right)$.

The analysis of the spectrum of an operator has seen some development thanks to a powerful property called the single-valued extension property, introduced by Dunford and Schwarz in [14]. In fact, this property plays an important role in the Laursen and Neumann [15] and Aiena [1] books. Finch defines it in a local version in [16].

Definition 2.

([16]). An operator $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ is said to have the single valued extension property at ${\lambda }_0\in C$ (abbreviated $\mathcal{SVEP}$ at ${\lambda }_0$) if for every open disc $D_{{\lambda }_0}\subseteq C$ centered at ${\lambda }_0$, the only analytic function $f:D_{{\lambda }_0}⟶\mathfrak{X}$ which satisfies the equation

$$(\lambda \mathfrak{I}-\mathfrak{T})f\left(\lambda \right)=0forall\lambda \in D_{{\lambda }_0}$$

is f $\equiv 0.$

The operator $\mathfrak{T}$ is said to have $\mathcal{SVEP}$ if it has $\mathcal{SVEP}$ at every point $\lambda \in C.$ Evidently, every $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ has $\mathcal{SVEP}$ at each point of the resolvent set $\rho \left(\mathfrak{T}\right).$ From ([1], Theorem 3.8), we have for $\lambda \in C$ that:

$$\begin{array}{ccc}\hfill p(\lambda \mathfrak{I}-\mathfrak{T})<\infty & \Rightarrow & \mathfrak{T}has\mathcal{SVEP}at\lambda .\hfill \end{array}$$

(1)

$$\begin{array}{ccc}\hfill \lambda \in iso{A}_{\sigma }\left(\mathfrak{T}\right)& \Rightarrow & \mathfrak{T}has\mathcal{SVEP}at\lambda .\hfill \end{array}$$

(2)

On the other hand, given Banach spaces $\mathfrak{X}$ and $\mathfrak{Y},$ let $\mathfrak{X}\otimes \mathfrak{Y}$ be the tensor product of $\mathfrak{X}$ and $\mathfrak{Y}$ as defined in [17]. The tensor product between two operators is defined as follows.

Definition 3.

([17]). The tensor product of $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ and $\mathfrak{S}\in \mathfrak{L}\left(\mathfrak{Y}\right)$ on $\mathfrak{X}\otimes \mathfrak{Y}$ is the operator given by

$$(\mathfrak{T}\otimes \mathfrak{S})(\sum _i{x}_i\otimes y_i)=\sum _{\mathfrak{i}}\mathfrak{T}{x}_i\otimes \mathfrak{S}{y}_i.$$

The study of the tensor product between two operators has influence on spectral theory. The tensor product of two operators and its stability with respect to Weyl’s and Browder’s theorem were initiated by Kubrusly and Duggal in [18]. Then, Duggal [19], Rashid [20] and Rashid and Prasad [21] continued these studies with Weyl and Browder type theorems. In addition, recently in [22,23] we can see a strong study linked to the tensor product of two operators.

The different spectra of the tensor product of two operators $\mathfrak{T}$ and $\mathfrak{S}$ are established as follows in [17,24] as follows:

• $\sigma (\mathfrak{T}\otimes \mathfrak{S})=\sigma \left(\mathfrak{T}\right).\sigma \left(\mathfrak{S}\right)$.
• $A_{\sigma }(\mathfrak{T}\otimes \mathfrak{S})=A_{\sigma }\left(\mathfrak{T}\right).A_{\sigma }\left(\mathfrak{S}\right)$.
• $F_{\sigma }(\mathfrak{T}\otimes \mathfrak{S})=\sigma \left(\mathfrak{T}\right).F_{\sigma }\left(\mathfrak{S}\right)\cup F_{\sigma }\left(\mathfrak{T}\right).\sigma \left(\mathfrak{S}\right)$.
• $F_{\sigma }^{+}(\mathfrak{T}\otimes \mathfrak{S})=A_{\sigma }\left(\mathfrak{T}\right).F_{\sigma }^{+}\left(\mathfrak{S}\right)\cup F_{\sigma }^{+}\left(\mathfrak{T}\right).A_{\sigma }\left(\mathfrak{S}\right)$.
• $E{A}_{\sigma }(\mathfrak{T}\otimes \mathfrak{S})\subseteq A_{\sigma }\left(\mathfrak{T}\right).E{A}_{\sigma }\left(\mathfrak{S}\right)\cup E{A}_{\sigma }\left(\mathfrak{T}\right).A_{\sigma }\left(\mathfrak{S}\right)$.
• $B_{\sigma }^{+}(\mathfrak{T}\otimes \mathfrak{S})=A_{\sigma }\left(\mathfrak{T}\right).B_{\sigma }^{+}\left(\mathfrak{S}\right)\cup B_{\sigma }^{+}\left(\mathfrak{T}\right).A_{\sigma }\left(\mathfrak{S}\right)$.

Recall that the essential approximate point spectrum of $\mathfrak{T}\otimes \mathfrak{S},$ for $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ and $\mathfrak{S}\in \mathfrak{L}\left(\mathfrak{Y}\right)$ verifies the spectral identity if

$$E{A}_{\sigma }(\mathfrak{T}\otimes \mathfrak{S})=A_{\sigma }\left(\mathfrak{T}\right).E{A}_{\sigma }\left(\mathfrak{S}\right)\cup E{A}_{\sigma }\left(\mathfrak{T}\right).A_{\sigma }\left(\mathfrak{S}\right).$$

In this paper, we consider the property $\left(Bv\right)$ or equivalently a-Browder’s theorem. This has been studied through the methods of the local spectral theory, through localized $\mathcal{SVEP}$, under a proper closed subspace of $\mathfrak{X}$ and also under some topological conditions and others. So, it has a lot of influence on the development of the spectral theory because the class of operators satisfying the property $\left(Bv\right)$ is stronger than the class of operators satisfying other properties, such as those seen in [10,22,25].

In view of the many existing studies on a-Browder’s theorem, it is important to continue studying the property $\left(Bv\right)$. Thus, for further development of operator theory, in this paper, combining local spectral theory with the properties of the tensor product of two operators, we give new results on the property $\left(Bv\right)$. Namely:

In Section 2, taking into account that new partitions of the spectrum of an operator are always investigated, we show that the spectrum of an operator satisfying property $\left(Bv\right)$ can be partitioned into an essential approximate spectrum and a surjective spectrum. We also show that the spectrum of an a-polaroid operator satisfying property $\left(Bv\right)$ coincides with its approximate point spectrum, and the essential spectrum coincides with its upper semi-Fredholm spectrum.

In Section 3, since the question “are there infinite operators satisfying a-Browder’s theorem?” has not been answered, we prove that $\mathfrak{T}$ verifies property $\left(Bv\right)$ if and only if every nonzero scalar multiple of $\mathfrak{T}$ verifies property $\left(Bv\right).$ In particular, the same is true for the two operators that are norm equivalent. Thus, it is observed that there exist infinite $\mathfrak{T}$ in $\mathfrak{L}\left(\mathfrak{X}\right)$ verifying a-Browder’s theorem.

In Section 4, in view of the fact that the relationships between the different spectra are always under research, assuming that the essential approximate point spectrum verifies the spectral identity for $\mathfrak{T}\otimes \mathfrak{S}$ with $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ and $\mathfrak{S}\in \mathfrak{L}\left(\mathfrak{Y}\right),$ the necessary conditions for the equality $E{A}_{\sigma }^{\prime }(\mathfrak{T}\otimes \mathfrak{S})=E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right).E{A}_{\sigma }^{\prime }\left(\mathfrak{S}\right)$ to hold and further, for the tensor product of two operators satisfying the property $\left(Bv\right)$, the Fredholm spectrum, the Weyl spectrum, the Browder spectrum, the upper Fredholm spectrum, the upper Weyl spectrum and the upper Browder spectrum to coincide with each other and the same for Berkani type spectra are investigated.

In Section 5, we list some conclusions.

2. On Spectra and the Property (Bv)

In this section, we give partitions of the spectrum of $\mathfrak{T}$ for operators that satisfy the property $\left(Bv\right).$ The approximate point spectrum equals the spectrum, and the upper semi-Fredholm spectrum equals the essential spectrum if $\mathfrak{T}$ is a-polaroid.

Aiena [1] has shown that the spectrum of $\mathfrak{T}$ is the union of $\mathrm{\Xi }\left(\mathfrak{T}\right)$ and surjective spectrum, i.e., $\sigma \left(\mathfrak{T}\right)=\mathrm{\Xi }\left(\mathfrak{T}\right)\cup S_{\sigma }\left(\mathfrak{T}\right)=A_{\sigma }\left(\mathfrak{T}\right)\cup S_{\sigma }\left(\mathfrak{T}\right),$ where

$$\mathrm{\Xi }\left(\mathfrak{T}\right)=\{\lambda \in \mathbb{C}:\mathfrak{T}does{n}^{\prime }thave\mathcal{SVEP}at\lambda \}.$$

Thus, we obtain the following Theorem.

Theorem 2.

If $\mathfrak{T}\in \left(J_{bv}\right),$ then $\sigma \left(\mathfrak{T}\right)=E{A}_{\sigma }\left(\mathfrak{T}\right)\cup S_{\sigma }\left(\mathfrak{T}\right)$.

Proof. 

Let $\lambda \in \sigma \left(\mathfrak{T}\right).$ If $\lambda \notin E{A}_{\sigma }\left(\mathfrak{T}\right),$ then $\lambda \notin B_{\sigma }^{+}\left(\mathfrak{T}\right).$ This implies that $\mathfrak{T}$ has $\mathcal{SVEP}$ at $\lambda$, i.e., $\lambda \notin \mathrm{\Xi }\left(\mathfrak{T}\right).$ Hence, $\lambda \in S_{\sigma }\left(\mathfrak{T}\right)$ and thus $\sigma \left(\mathfrak{T}\right)=E{A}_{\sigma }\left(\mathfrak{T}\right)\cup S_{\sigma }\left(\mathfrak{T}\right).$ □

By Theorems 1 and 2, it results that if the interior of $E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right)$ is empty, then the approximate point spectrum coincides with the union of the upper semi Browder spectrum and the surjective spectrum.

Corollary 1.

Let $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$. If $int\left(E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right)\right)=Ø$, then $A_{\sigma }\left(\mathfrak{T}\right)=B_{\sigma }^{+}\left(\mathfrak{T}\right)\cup S_{\sigma }\left(\mathfrak{T}\right)$.

Next, we find a condition through Riesz operators for an operator $\mathfrak{T}$ to satisfy property $\left(Bv\right)$.

Theorem 3.

Let $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right).$ If $\mathfrak{K}\mathfrak{T}=\mathfrak{T}\mathfrak{K}$, for some Riesz operator $\mathfrak{K}$ such that $A_{\sigma }(\mathfrak{T}+\mathfrak{R})=E{A}_{\sigma }\left(\mathfrak{T}\right)$, then $\mathfrak{T}\in \left(J_{bv}\right)$.

Proof. 

It is known that, see [1],

$$B_{\sigma }^{+}\left(\mathfrak{T}\right)=\bigcap _{\mathfrak{K}\in \mathcal{R}\left(\mathfrak{X}\right),\mathfrak{K}\mathfrak{T}=\mathfrak{T}\mathfrak{K}}{A}_{\sigma }(\mathfrak{T}+\mathfrak{K}).$$

By hypothesis we have that $A_{\sigma }(\mathfrak{T}+\mathfrak{K})=E{A}_{\sigma }\left(\mathfrak{T}\right)$, whereby if $\lambda \notin E{A}_{\sigma }\left(\mathfrak{T}\right),$ then $\lambda \notin B_{\sigma }^{+}\left(\mathfrak{T}\right)$. Hence, $E{A}_{\sigma }\left(\mathfrak{T}\right)=B_{\sigma }^{+}\left(\mathfrak{T}\right)$, i.e., $\mathfrak{T}$ verifies a-Browder’s theorem, or equivalently $\mathfrak{T}\in \left(J_{bv}\right)$. □

For an a-polaroid operator that satisfies property $\left(Bv\right)$, we obtain that the approximate point spectrum becomes the spectrum, and the upper semi-Fredholm spectrum becomes the essential spectrum. Indeed,

Theorem 4.

Let $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ be an a-polaroid operator. If $\mathfrak{T}\in \left(J_{bv}\right),$ then

1. 

$S_{\sigma }\left(\mathfrak{T}\right)\subseteq A_{\sigma }\left(\mathfrak{T}\right)$.

2. 

$F_{\sigma }^{-}\left(\mathfrak{T}\right)\subseteq F_{\sigma }^{+}\left(\mathfrak{T}\right).$

Proof. 

1. By ([26], Theorem 3.1), we have $B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right)=B_{\sigma }^{\prime }\left(\mathfrak{T}\right)$, or equivalently $B_{\sigma }^{+}\left(\mathfrak{T}\right)=B_{\sigma }\left(\mathfrak{T}\right).$ So, as $E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right)=B_{\sigma }^{\prime }\left(\mathfrak{T}\right)$ and $B_{\sigma }^{\prime }\left(\mathfrak{T}\right)\subseteq iso\sigma \left(\mathfrak{T}\right)$, we deduce that $int\left(E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right)\right)=Ø$. By Theorem 1, we have that $\sigma \left(\mathfrak{T}\right)=A_{\sigma }\left(\mathfrak{T}\right).$ Hence, $S_{\sigma }\left(\mathfrak{T}\right)\subseteq A_{\sigma }\left(\mathfrak{T}\right)$.

2. As in part 1., we have that $B_{\sigma }\left(\mathfrak{T}\right)=B_{\sigma }^{+}\left(\mathfrak{T}\right)$. Since $\mathfrak{T}$ verifies property $\left(Bv\right)$, it results that $E{A}_{\sigma }\left(\mathfrak{T}\right)=B_{\sigma }^{+}\left(\mathfrak{T}\right)=B_{\sigma }\left(\mathfrak{T}\right).$ On the other hand, let $\lambda \notin F_{\sigma }^{+}\left(\mathfrak{T}\right)$.

If $\lambda \notin E{A}_{\sigma }\left(\mathfrak{T}\right)$, then $\lambda \notin B_{\sigma }\left(\mathfrak{T}\right)$ and hence $\lambda \notin F_{\sigma }\left(\mathfrak{T}\right).$

If $\lambda \in E{A}_{\sigma }\left(\mathfrak{T}\right)$, then $\mathrm{ind}(\lambda \mathfrak{I}-\mathfrak{T})>0.$ Therefore, $\lambda \mathfrak{I}-\mathfrak{T}$ is both upper semi-Fredholm and lower semi-Fredholm; hence $\lambda \notin F_{\sigma }\left(\mathfrak{T}\right).$

So, in both cases $F_{\sigma }\left(\mathfrak{T}\right)=F_{\sigma }^{+}\left(\mathfrak{T}\right)$ and hence $F_{\sigma }^{-}\left(\mathfrak{T}\right)\subseteq F_{\sigma }^{+}\left(\mathfrak{T}\right).$ □

3. Property (Bv) and Norm Equivalent Operators

For $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$, we define $\mathrm{\Gamma }\left(\mathfrak{T}\right):=E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right)\backslash B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right)$. Clearly,

$$\mathfrak{T}\in \left(J_{bv}\right) \mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if} \mathrm{\Gamma }\left(\mathfrak{T}\right)=Ø.$$

We recall that an operator $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ is Drazin invertible if there exists an operator $\mathfrak{S}\in \mathfrak{L}\left(\mathfrak{X}\right)$ (called the Drazin inverse of $\mathfrak{T}$) and an integer $n\ge 0$ such that

$$\mathfrak{T}\mathfrak{S}=\mathfrak{S}\mathfrak{T},\mathfrak{S}\mathfrak{T}\mathfrak{S}=\mathfrak{S},{\mathfrak{T}}^n\mathfrak{S}\mathfrak{T}={\mathfrak{T}}^n.$$

Of conclusions in [27], we deduce that for $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ is Drazin invertible with Drazin inverse $\mathfrak{S}$, it results that:

$$\mathrm{\Gamma }\left(\mathfrak{T}\right)=Ø \mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if} \mathrm{\Gamma }\left(\mathfrak{S}\right)=Ø.$$

In [28], the concept of equivalent norm operator was introduced and studied under linear operators with closed range. Two operators $\mathfrak{S}$ and $\mathfrak{T}$ in $\mathfrak{L}(\mathfrak{X},\mathfrak{Y})$ are said to be norm equivalent if there exist two positive real numbers $k_1$ and $k_2$ such that

$$k_1\left|\right|\mathfrak{S}x\left|\right|\le \left|\right|\mathfrak{T}x\left|\right|\le k_2\left|\right|\mathfrak{S}x\left|\right|,\mathrm{for}\mathrm{all}x\in \mathfrak{X}.$$

In particular, the condition that $k_1\left|f\left(\mathfrak{T}x\right)\right|\le \left|f\left(\mathfrak{S}x\right)\right|\le k_2\left|f\left(\mathfrak{T}x\right)\right|$ for all $x\in \mathfrak{X}$ and $f\in {\mathfrak{Y}}^{*}$ with $k_1>0,k_2>0$ was studied in detail. This condition implies equivalent norm.

In this section, we show that if $\mathfrak{T},\mathfrak{S}\in \mathfrak{L}(\mathfrak{X},\mathfrak{Y})$ are norm equivalent, then

$$\mathrm{\Gamma }\left(\mathfrak{T}\right)=Ø \mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if} \mathrm{\Gamma }\left(\mathfrak{S}\right)=Ø.$$

In order to test the latter, in the following theorem we first show that a nonzero scalar multiple of an operator $\mathfrak{T}$ that verifies the property $\left(Bv\right),$ also verifies the property $\left(Bv\right).$

Theorem 5.

Let $0\ne \mu \in \mathbb{C}$ and $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$. Then,

1. 

$\mu E{A}_{\sigma }\left(\mathfrak{T}\right)=E{A}_{\sigma }\left(\mu \mathfrak{T}\right).$

2. 

$\mu B_{\sigma }^{+}\left(\mathfrak{T}\right)=B_{\sigma }^{+}\left(\mu \mathfrak{T}\right).$

3. 

$\mathrm{\Gamma }\left(\mathfrak{T}\right)=Ø$ if and only if $\mathrm{\Gamma }\left(\mu \mathfrak{T}\right)=Ø$.

Proof. 

1. Let $\lambda \in \mathbb{C}$, observe that:

(a)

${\displaystyle \alpha (\lambda \mathfrak{I}-\mu \mathfrak{T})=\alpha \left(\mu \left(\frac{\lambda }{\mu }\mathfrak{I}-\mathfrak{T}\right)\right)=\alpha \left(\frac{\lambda }{\mu }\mathfrak{I}-\mathfrak{T}\right)}$.

(b)

${\displaystyle (\lambda \mathfrak{I}-\mu \mathfrak{T})\left(\mathfrak{X}\right)=\left(\mu \left(\frac{\lambda }{\mu }\mathfrak{I}-\mathfrak{T}\right)\right)\left(\mathfrak{X}\right)=\mu \left(\frac{\lambda }{\mu }\mathfrak{I}-\mathfrak{T}\right)\left(\mathfrak{X}\right)}$.

By part (b), we have that ${\displaystyle \beta (\lambda \mathfrak{I}-\mu \mathfrak{T})=\beta \left(\frac{\lambda }{\mu }\mathfrak{I}-\mathfrak{T}\right)}$ and by part (a) it turns out $\mathrm{ind}(\frac{\lambda }{\mu }\mathfrak{I}-\mathfrak{T})=\mathrm{ind}(\lambda \mathfrak{I}-\mu \mathfrak{T})$.

Thus, if $\lambda \notin E{A}_{\sigma }(\mu \mathfrak{T})$, then $\mathrm{ind}(\lambda \mathfrak{I}-\mu \mathfrak{T})<0$ and $(\lambda \mathfrak{I}-\mu \mathfrak{T})\left(\mathfrak{X}\right)$ is closed, whereby $\mathrm{ind}(\frac{\lambda }{\mu }\mathfrak{I}-\mathfrak{T})<0$ and by part (b) we obtain that $(\frac{\lambda }{\mu }\mathfrak{I}-\mathfrak{T})\left(\mathfrak{X}\right)$ is closed. Hence, ${\displaystyle \frac{\lambda }{\mu }\notin E{A}_{\sigma }\left(\mathfrak{T}\right)}$ and so $\lambda \notin \mu E{A}_{\sigma }\left(\mathfrak{T}\right)$, we deduce that

$$\mu E{A}_{\sigma }\left(\mathfrak{T}\right)\subset E{A}_{\sigma }\left(\mu \mathfrak{T}\right).$$

Similarly, it results that $E{A}_{\sigma }\left(\mu \mathfrak{T}\right)\subset \mu E{A}_{\sigma }\left(\mathfrak{T}\right)$. Therefore, $E{A}_{\sigma }\left(\mu \mathfrak{T}\right)=\mu E{A}_{\sigma }\left(\mathfrak{T}\right)$.

2.

In similar way to 1., we have that

$${\displaystyle \alpha {(\lambda \mathfrak{I}-\mu \mathfrak{T})}^p=\alpha \left({\mu }^p{\left(\frac{\lambda }{\mu }\mathfrak{I}-\mathfrak{T}\right)}^p\right)=\alpha {\left(\frac{\lambda }{\mu }\mathfrak{I}-\mathfrak{T}\right)}^p.}$$

Therefore, $p(\lambda \mathfrak{I}-\mu \mathfrak{T})=p\left({\displaystyle \frac{\lambda }{\mu }}\mathfrak{I}-\mathfrak{T}\right)$ and so $\mu B_{\sigma }^{+}\left(\mathfrak{T}\right)=B_{\sigma }^{+}\left(\mu \mathfrak{T}\right).$

3.

If $\mathrm{\Gamma }\left(\mathfrak{T}\right)=Ø$ then $\mathfrak{T}\in \left(J_{bv}\right)$, so $E{A}_{\sigma }\left(\mathfrak{T}\right)=B_{\sigma }^{+}\left(\mathfrak{T}\right).$ By parts 1 and 2, We obtain that $E{A}_{\sigma }\left(\mu \mathfrak{T}\right)=B_{\sigma }^{+}\left(\mu \mathfrak{T}\right)$ equivalently $\mu \mathfrak{T}\in \left(J_{bv}\right).$ Hence $\mathrm{\Gamma }\left(\mu \mathfrak{T}\right)=Ø$. Similarly $\mathrm{\Gamma }\left(\mu \mathfrak{T}\right)=Ø$ implies $\mathrm{\Gamma }\left(\mathfrak{T}\right)=Ø$. □

Now, we can show that if $\mathfrak{T}$ and $\mathfrak{S}$ are norm equivalent, then $\mathfrak{T}\in \left(J_{bv}\right)$ iff $\mathfrak{S}\in \left(J_{bv}\right)$.

Theorem 6.

Let $\mathfrak{T},\mathfrak{S}\in \mathfrak{L}(\mathfrak{X},\mathfrak{Y})$ such that $k_1\left|f\left(\mathfrak{T}x\right)\right|\le \left|f\left(\mathfrak{S}x\right)\right|\le k_2\left|f\left(\mathfrak{T}x\right)\right|$ for all $x\in \mathfrak{X}$ and $f\in {\mathfrak{X}}^{*}$ with $k_1,k_2$ two positive real numbers. Then, the following holds:

1. 

$\mathrm{\Gamma }\left(\mathfrak{T}\right)=Ø$ if and only if $\mathrm{\Gamma }\left(\mathfrak{S}\right)=Ø$.

2. 

$\mathrm{\Gamma }\left(\mathfrak{T}\right)=Ø$⇒$\sigma \left(\mathfrak{S}\right)=E{A}_{\sigma }\left(\mathfrak{S}\right)\cup S_{\sigma }\left(\mathfrak{S}\right)$.

Proof. 

1. By ([28], Theorem 2.7), for some $0\ne \mu \in \mathbb{C}$ result that $\mathfrak{S}=\mu \mathfrak{T}$. Then, the result follows by Theorem 5 part 3.

2. Since $\mathrm{\Gamma }\left(\mathfrak{T}\right)=Ø$ implies $\mathrm{\Gamma }\left(\mathfrak{S}\right)=Ø$, so by Theorem 2, it turns out that $\sigma \left(\mathfrak{S}\right)=E{A}_{\sigma }\left(\mathfrak{S}\right)\cup S_{\sigma }\left(\mathfrak{S}\right).$ □

By Theorem 3, we obtain the following corollary.

Corollary 2.

Let $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right).$ If $\mathfrak{R}\mathfrak{T}=\mathfrak{T}\mathfrak{R}$, for some Riesz operator $\mathfrak{R}$ such that $A_{\sigma }(\mathfrak{T}+\mathfrak{R})=E{A}_{\sigma }\left(\mathfrak{T}\right)$, then $\mu \mathfrak{T}\in \left(J_{bv}\right)$, for each $\mu \ne 0$.

Remark 2.

It is already known that there are several classes of operators that satisfy a-Browder’s theorem. Then, by the results of the present section, we have that: there exist infinite $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ such that $\mathrm{\Gamma }\left(\mathfrak{T}\right)=Ø$, or the same $\mathfrak{T}\in \left(J_{bv}\right)$, or verify a-Browder’s theorem.

4. Property (Bv) for Tensor Product

Throughout this section, we assume that $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right)$ and $\mathfrak{S}\in \mathfrak{L}\left(\mathfrak{Y}\right)$. Thus, we focus on obtaining the conditions that $E{A}_{\sigma }^{\prime }(\mathfrak{T}\otimes \mathfrak{S})=E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right).E{A}_{\sigma }^{\prime }\left(\mathfrak{S}\right)$. This will make it possible to see the equality between the different spectra of $\mathfrak{T}\otimes \mathfrak{S}$. In addition, with an example of one of our main results, we show an important stability result with respect to the property $\left(Bv\right)$ under tensor product. Thus, we can test the stability of the various spectra for the tensor product.

Note that the set $B_{\sigma }^{{+}^{\prime }}(\mathfrak{T}\otimes \mathfrak{S})$ is not equal to $B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)$ in general. However, we do have an inclusion.

Theorem 7.

$B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)\subseteq B_{\sigma }^{{+}^{\prime }}(\mathfrak{T}\otimes \mathfrak{S})$.

Proof. 

Let the factorization of $\lambda$ be $0\ne \lambda =\mu \nu$ such that $\mu \in B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right)=\sigma \left(\mathfrak{T}\right)\backslash B_{\sigma }^{+}\left(\mathfrak{T}\right)$ and $\nu \in B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)=\sigma \left(\mathfrak{S}\right)\backslash B_{\sigma }^{+}\left(\mathfrak{S}\right).$ This implies that $\lambda \notin \sigma \left(\mathfrak{T}\right).B_{\sigma }^{+}\left(\mathfrak{S}\right)\cup B_{\sigma }^{+}\left(\mathfrak{T}\right).\sigma \left(\mathfrak{S}\right).$ Hence it follows from the tensor product of upper semi-Browder spectrum identity that $\lambda \in \sigma (\mathfrak{T}\otimes \mathfrak{S})\backslash B_{\sigma }^{+}(\mathfrak{T}\otimes \mathfrak{S})=B_{\sigma }^{{+}^{\prime }}(\mathfrak{T}\otimes \mathfrak{S}).$ Therefore $B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)\subseteq B_{\sigma }^{{+}^{\prime }}(\mathfrak{T}\otimes \mathfrak{S}).$ □

Next, we show that if the spectrum coincides with the approximate point spectrum for $\mathfrak{T}$ and $\mathfrak{S},$ then the equality holds in the above Theorem.

Theorem 8.

If $\sigma \left(\mathfrak{T}\right)=A_{\sigma }\left(\mathfrak{T}\right)$ and $\sigma \left(\mathfrak{S}\right)=A_{\sigma }\left(\mathfrak{S}\right)$, then $B_{\sigma }^{{+}^{\prime }}(\mathfrak{T}\otimes \mathfrak{S})=B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)$.

Proof. 

By Theorem 7, it is sufficient to prove $B_{\sigma }^{{+}^{\prime }}(\mathfrak{T}\otimes \mathfrak{S})\subseteq B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)$. Let $\lambda \in B_{\sigma }^{{+}^{\prime }}(\mathfrak{T}\otimes \mathfrak{S})=\sigma (\mathfrak{T}\otimes \mathfrak{S})\backslash B_{\sigma }^{+}(\mathfrak{T}\otimes \mathfrak{S}).$ Then $\lambda \notin A_{\sigma }\left(\mathfrak{T}\right).B_{\sigma }^{+}\left(\mathfrak{S}\right)\cup B_{\sigma }^{+}\left(\mathfrak{T}\right).A_{\sigma }\left(\mathfrak{S}\right),$ thus for any factorization $0\ne \lambda =\mu \nu \in \sigma (\mathfrak{T}\otimes \mathfrak{S}),$ it results that $\mu \nu \notin A_{\sigma }\left(\mathfrak{T}\right).B_{\sigma }^{+}\left(\mathfrak{S}\right)$ and $\mu \nu \notin A_{\sigma }\left(\mathfrak{S}\right).B_{\sigma }^{+}\left(\mathfrak{T}\right).$ As $\mu \in A_{\sigma }\left(\mathfrak{T}\right)$ and $\nu \in A_{\sigma }\left(\mathfrak{S}\right)$ we have that $\nu \notin B_{\sigma }^{+}\left(\mathfrak{S}\right)$ and $\mu \notin B_{\sigma }^{+}\left(\mathfrak{T}\right).$ This implies that $\mu \in B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right)$ and $\nu \in B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right).$ That is $\lambda =\mu \nu \in B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right).$ Therefore, $B_{\sigma }^{{+}^{\prime }}(\mathfrak{T}\otimes \mathfrak{S})\subseteq B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)$ and so $B_{\sigma }^{{+}^{\prime }}(\mathfrak{T}\otimes \mathfrak{S})=B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)$. □

Hereafter in the remaining part of this section, we assume that the essential approximate point spectrum verifies the spectral identity for $\mathfrak{T}\otimes \mathfrak{S}$.

In the following theorem, we discuss the necessary conditions for $\mathfrak{T},\mathfrak{S}\in \left(J_{bv}\right),$ so that the equality $B_{\sigma }^{{+}^{\prime }}(\mathfrak{T}\otimes \mathfrak{S})=B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)$ becomes $E{A}_{\sigma }^{\prime }(\mathfrak{T}\otimes \mathfrak{S})=E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right).E{A}_{\sigma }^{\prime }\left(\mathfrak{S}\right)$.

Theorem 9.

If $\mathfrak{T},\mathfrak{S}\in \left(J_{bv}\right)$ such that $\sigma \left(\mathfrak{T}\right)=A_{\sigma }\left(\mathfrak{T}\right)$ and $\sigma \left(\mathfrak{S}\right)=A_{\sigma }\left(\mathfrak{S}\right),$ then $E{A}_{\sigma }^{\prime }(\mathfrak{T}\otimes \mathfrak{S})=E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right).E{A}_{\sigma }^{\prime }\left(\mathfrak{S}\right)$.

Proof. 

Assume $\mathfrak{T},\mathfrak{S}\in \left(J_{bv}\right)$, so $\mathfrak{T}$ and $\mathfrak{S}$ verify a-Browder’s theorem, whereby $\mathfrak{T}\otimes \mathfrak{S}$ verifies a-Browder’s theorem, consequently $\mathfrak{T}\otimes \mathfrak{S}\in \left(J_{bv}\right)$. Thus:

$$B_{\sigma }^{{+}^{\prime }}(\mathfrak{T}\otimes \mathfrak{S})=E{A}_{\sigma }^{\prime }(\mathfrak{T}\otimes \mathfrak{S}),B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right)=E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right),B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)=E{A}_{\sigma }^{\prime }\left(\mathfrak{S}\right).$$

By Theorem 8 it results that $B_{\sigma }^{{+}^{\prime }}(\mathfrak{T}\otimes \mathfrak{S})=B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).B_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)$. Therefore, $E{A}_{\sigma }^{\prime }(\mathfrak{T}\otimes \mathfrak{S})=E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right).E{A}_{\sigma }^{\prime }\left(\mathfrak{S}\right)$. □

Theorem 4 allows us to obtain the same result for a-polaroid operators in the class of operators verifying property $\left(Bv\right).$

Theorem 10.

If $\mathfrak{T},\mathfrak{S}\in \left(J_{bv}\right)$ are a-polaroid, then $E{A}_{\sigma }^{\prime }(\mathfrak{T}\otimes \mathfrak{S})=E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right).E{A}_{\sigma }^{\prime }\left(\mathfrak{S}\right)$.

The following theorem establishes three equalities, whereby various spectral properties are satisfied for the tensor product $\mathfrak{T}\otimes \mathfrak{S}$, for example the property $\left(gaz\right)$.

Theorem 11.

If $\mathfrak{T},\mathfrak{S}\in \left(J_{bv}\right)$ are two a-polaroid operators, then:

1. 

$\sigma (\mathfrak{T}\otimes \mathfrak{S})=A_{\sigma }(\mathfrak{T}\otimes \mathfrak{S})$.

2. 

$F_{\sigma }^{+}(\mathfrak{T}\otimes \mathfrak{S})=F_{\sigma }(\mathfrak{T}\otimes \mathfrak{S})$.

3. 

$B{W}_{\sigma }^{+}(\mathfrak{T}\otimes \mathfrak{S})=D_{\sigma }(\mathfrak{T}\otimes \mathfrak{S})$.

Proof. 

1. It is obtained by applying part (1) of Theorem 4.

2. Let $\lambda \notin F_{\sigma }^{+}(\mathfrak{T}\otimes \mathfrak{S})=A_{\sigma }\left(\mathfrak{T}\right).F_{\sigma }^{+}\left(\mathfrak{S}\right)\cup F_{\sigma }^{+}\left(\mathfrak{T}\right).A_{\sigma }\left(\mathfrak{S}\right).$ By Theorem 4, we have $\lambda \notin \sigma \left(\mathfrak{T}\right).F_{\sigma }\left(\mathfrak{S}\right)\cup F_{\sigma }\left(\mathfrak{T}\right).\sigma \left(\mathfrak{S}\right)=F_{\sigma }(\mathfrak{T}\otimes \mathfrak{S}).$ Hence, $F_{\sigma }(\mathfrak{T}\otimes \mathfrak{S})\subseteq F_{\sigma }^{+}(\mathfrak{T}\otimes \mathfrak{S}).$ Since the reverse inclusion is always true, $F_{\sigma }^{+}(\mathfrak{T}\otimes \mathfrak{S})=F_{\sigma }(\mathfrak{T}\otimes \mathfrak{S}).$

3. By hypothesis $\mathfrak{T},\mathfrak{S}\in \left(J_{bv}\right)$, so $\mathfrak{T}$ and $\mathfrak{S}$ verify a-Browder’s Theorem; by Theorem 4, we have that $\sigma \left(\mathfrak{T}\right)=A_{\sigma }\left(\mathfrak{T}\right)$ and $\sigma \left(\mathfrak{S}\right)=A_{\sigma }\left(\mathfrak{S}\right)$, thus by ([25], Theorem 3.2), we obtain that $\mathfrak{T}$ and $\mathfrak{S}$ verify property $\left(gaz\right)$. We have assumed that

$$E{A}_{\sigma }(\mathfrak{T}\otimes \mathfrak{S})=A_{\sigma }\left(\mathfrak{T}\right).E{A}_{\sigma }\left(\mathfrak{S}\right)\cup E{A}_{\sigma }\left(\mathfrak{T}\right).A_{\sigma }\left(\mathfrak{S}\right).$$

Therefore, by ([23], Corollary 1) we obtain that $B{W}_{\sigma }^{+}(\mathfrak{T}\otimes \mathfrak{S})=D_{\sigma }(\mathfrak{T}\otimes \mathfrak{S})$. □

The following Theorem is the main result of this section which establishes the equality between the different spectra of $\mathfrak{T}\otimes \mathfrak{S}$.

Theorem 12.

If ${\mathfrak{S}}_0:=\mathfrak{T}\otimes \mathfrak{S}$ and $\mathrm{int}(F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right))=\mathrm{int}(F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right))=Ø$, then:

1. 

$\mathrm{\Gamma }\left({\mathfrak{S}}_0\right)=Ø$.

2. 

$F_{\sigma }^{+}\left({\mathfrak{S}}_0\right)=F_{\sigma }\left({\mathfrak{S}}_0\right)=E{A}_{\sigma }\left({\mathfrak{S}}_0\right)=W_{\sigma }\left({\mathfrak{S}}_0\right)=B_{\sigma }^{+}\left({\mathfrak{S}}_0\right)=B_{\sigma }\left({\mathfrak{S}}_0\right)$. And $B{F}_{\sigma }^{+}\left({\mathfrak{S}}_0\right)=B{F}_{\sigma }\left({\mathfrak{S}}_0\right)=B{W}_{\sigma }^{+}\left({\mathfrak{S}}_0\right)=B{W}_{\sigma }\left({\mathfrak{S}}_0\right)=D_{\sigma }^l\left({\mathfrak{S}}_0\right)=D_{\sigma }\left({\mathfrak{S}}_0\right).$

Proof. 

1. By hypothesis, it results that $\mathrm{int}\left(E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right)\right)=\mathrm{int}\left(E{A}_{\sigma }^{\prime }\left(\mathfrak{S}\right)\right)=Ø$, so by Theorem 1, we obtain that $\mathfrak{T},\mathfrak{S}\in \left(J_{bv}\right)$, $\sigma \left(\mathfrak{T}\right)=A_{\sigma }\left(\mathfrak{T}\right)$ and $\sigma \left(\mathfrak{S}\right)=A_{\sigma }\left(\mathfrak{S}\right).$ Hence, by Theorem 9 we obtain that $E{A}_{\sigma }^{\prime }(\mathfrak{T}\otimes \mathfrak{S})=E{A}_{\sigma }^{\prime }\left(\mathfrak{T}\right).E{A}_{\sigma }^{\prime }\left(\mathfrak{S}\right)$ and so $\mathrm{int}\left(E{A}_{\sigma }^{\prime }\left({\mathfrak{S}}_0\right)\right)=Ø$; by Theorem 1 we have that ${\mathfrak{S}}_0\in \left(J_{bv}\right)$, or $\mathrm{\Gamma }\left({\mathfrak{S}}_0\right)=Ø$.

2. Let $\lambda \in F_{\sigma }^{{+}^{\prime }}\left({\mathfrak{S}}_0\right)=\sigma \left({\mathfrak{S}}_0\right)\backslash F_{\sigma }^{+}\left({\mathfrak{S}}_0\right).$ So $\lambda \notin A_{\sigma }\left(\mathfrak{T}\right).F_{\sigma }^{+}\left(\mathfrak{S}\right)\cup F_{\sigma }^{+}\left(\mathfrak{T}\right).A_{\sigma }\left(\mathfrak{S}\right)$ and if $0\ne \lambda =\mu \nu \in \sigma \left({\mathfrak{S}}_0\right)$ it results that $\mu \nu \notin A_{\sigma }\left(\mathfrak{T}\right).F_{\sigma }^{+}\left(\mathfrak{S}\right)$ and $\mu \nu \notin F_{\sigma }^{+}\left(\mathfrak{T}\right).A_{\sigma }\left(\mathfrak{S}\right).$ Since $\mu \in \sigma \left(\mathfrak{T}\right)=A_{\sigma }\left(\mathfrak{T}\right)$ and $\nu \in \sigma \left(\mathfrak{S}\right)=A_{\sigma }\left(\mathfrak{S}\right)$, we have that $\nu \notin F_{\sigma }^{+}\left(\mathfrak{S}\right)$ and $\mu \notin F_{\sigma }^{+}\left(\mathfrak{T}\right)$. This implies that $\mu \in F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right)$ and $\nu \in F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right).$ Hence, $\lambda =\mu \nu \in F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)$. Therefore, $F_{\sigma }^{{+}^{\prime }}\left({\mathfrak{S}}_0\right)\subseteq F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right).F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right).$ By hypothesis, we conclude that $\mathrm{int}(F_{\sigma }^{{+}^{\prime }}\left({\mathfrak{S}}_0\right))=Ø$. Hence, $\mathrm{int}\left(E{A}_{\sigma }^{\prime }\left({\mathfrak{S}}_0\right)\right)=Ø$ and the results follow by ([13], Section 5). □

Corollary 3.

If $\mathrm{int}(F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right))=\mathrm{int}(F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right))=Ø$ and ${\mathfrak{S}}_0:=\mathfrak{T}\otimes \mathfrak{S}$, then $A_{\sigma }\left({\mathfrak{S}}_0\right)=F_{\sigma }^{+}\left({\mathfrak{S}}_0\right)\cup S_{\sigma }\left({\mathfrak{S}}_0\right)$.

Proof. 

As in the proof of part 2.-3. of Theorem 12, we obtain that $\mathrm{int}\left(E{A}_{\sigma }^{\prime }\left({\mathfrak{S}}_0\right)\right)=Ø$, also $F_{\sigma }^{+}\left({\mathfrak{S}}_0\right)=B_{\sigma }^{+}\left({\mathfrak{S}}_0\right).$ By using Corollary 1, we obtain that $A_{\sigma }\left({\mathfrak{S}}_0\right)=F_{\sigma }^{+}\left({\mathfrak{S}}_0\right)\cup S_{\sigma }\left({\mathfrak{S}}_0\right)$. □

Example 1.

If ${\mathfrak{T}}_1\in \mathfrak{L}\left(\mathfrak{X}\right)$ and ${\mathfrak{T}}_2\in \mathfrak{L}\left(\mathfrak{X}\right)$ are two left m-invertible contractions such that $\sigma \left({\mathfrak{T}}_i\right)\subseteq \mathrm{\Gamma }$, $i=1,2,$ then $\lambda \in \sigma \left({\mathfrak{T}}_i\right)$ is a pole of ${\mathfrak{T}}_i$ if and only if $\left(\lambda \mathfrak{I}-{\mathfrak{T}}_i\right)\left(\mathfrak{X}\right)$ is closed (see [29], for definition and details). On the other hand, for $i=1,2$, it results that $\forall \lambda \in F_{\sigma }^{{+}^{\prime }}\left({\mathfrak{T}}_i\right),$$\left(\lambda \mathfrak{I}-{\mathfrak{T}}_i\right)\left(\mathfrak{X}\right)$ is closed; hence $E{A}_{\sigma }^{\prime }\left({\mathfrak{T}}_i\right)\subseteq F_{\sigma }^{{+}^{\prime }}\left({\mathfrak{T}}_i\right)\subseteq D_{\sigma }^{\prime }\left({\mathfrak{T}}_i\right)\subseteq iso\sigma \left({\mathfrak{T}}_i\right)$, whereby $\mathrm{int}\left(F_{\sigma }^{{+}^{\prime }}\left({\mathfrak{T}}_i\right)\right)=Ø.$ Hence, Theorem 12 applies to ${\mathfrak{S}}_{00}:={\mathfrak{T}}_1\otimes {\mathfrak{T}}_2.$

Example 2.

If ${\mathfrak{Q}}_1\in \mathfrak{L}\left(\mathfrak{X}\right)$ and ${\mathfrak{Q}}_2\in \mathfrak{L}\left(\mathfrak{Y}\right)$ are two quasi-nilpotent operators commuting with $\mathfrak{T}$ and $\mathfrak{S}$, respectively. Then, by Corollaries $3.24$ and $3.18$, of [30], we have that $\sigma \left(\mathfrak{T}\right)=\sigma (\mathfrak{T}+{\mathfrak{Q}}_1)$, $F_{\sigma }^{+}\left(\mathfrak{T}\right)=F_{\sigma }^{+}(\mathfrak{T}+{\mathfrak{Q}}_1)$, $\sigma \left(\mathfrak{S}\right)=\sigma (\mathfrak{S}+{\mathfrak{Q}}_2)$, $F_{\sigma }^{+}\left(\mathfrak{S}\right)=F_{\sigma }^{+}(\mathfrak{S}+{\mathfrak{Q}}_2)$. Hence, $F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right)=F_{\sigma }^{{+}^{\prime }}(\mathfrak{T}+{\mathfrak{Q}}_1)$ and $F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right)=F_{\sigma }^{{+}^{\prime }}(\mathfrak{S}+{\mathfrak{Q}}_2)$. Thus, if $\mathrm{int}(F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{T}\right))=\mathrm{int}(F_{\sigma }^{{+}^{\prime }}\left(\mathfrak{S}\right))=Ø$, then Theorem 12 applies to ${\mathfrak{S}}_{000}:=(\mathfrak{T}+{\mathfrak{Q}}_1)\otimes (\mathfrak{S}+{\mathfrak{Q}}_2).$

5. Conclusions

• In general, spectrum of an operator can be written in the form of union of approximate point spectrum and surjective spectrum. We have shown that the spectrum of an operator verifying property $\left(Bv\right)$ coincides with the union of essential approximate point spectrum and surjective spectrum. See Theorem 2.
• For an a-polaroid operator $\mathfrak{T}\in \mathfrak{L}\left(\mathfrak{X}\right),$ the spectrum and essential spectrum coincide with the approximate point spectrum and upper semi-Fredholm spectrum respectively, if $\mathfrak{T}$ verifies property $\left(Bv\right).$ See Theorem 4.
• We have proved that any nonzero multiple of an operator satisfying property $\left(Bv\right)$ also satisfies property $\left(Bv\right)$, and particularly if two operators are norm equivalent, then property $\left(Bv\right)$ holds for one if and only if it holds for the other. See Theorems 5 and 6. In addition, there exist infinite operators which verifies the Browder theorem, see Remark 2.
• Finally, we obtain that different spectra of the two-factor tensor product coincide when the interior of the difference between the spectrum and the upper semifredholm spectrum of each factor is empty. See Theorem 12.