A Note on the Order Lozanovsky Spectrum for Positive Operators

Abstract

In this note we consider the order Lozanovsky spectrum of a positive operator on a complex Banach lattice and show that it generally contains the order Weyl spectrum, settling a weaker form of a question raised in Alekhno (Positivity 13(1):3–20, 2009).

1 Introduction

During the past two decades, Fredholm theory in the context of ordered structures has seen some development [2, 3, 8,9,10]. Motivated by a result of Lozanovsky [1, p.199], Alekhno introduced the Lozanovsky spectrum of an arbitrary positive operator in [2] to further study the interplay between Fredholm theory and ordering in the setting of positive operators on Banach lattices. In particular, the Lozanovsky spectrum was useful in completely solving a domination problem involving band irreducible operators [2, Theorem 8].

Since the non-emptiness of the Lozanovksy spectrum is not immediately clear, Alekhno asked in [3, Section 5] whether the Lozanovsky spectrum of a positive operator T defined on an arbitrary complex Banach lattice always contains the (non-empty) Weyl spectrum of T. To date, this remains an open question. It is the purpose of this note to settle the relevant question for the order-counterparts of the aforementioned two spectra in the context of regular operators on complex Banach lattices; that is, we prove that the order Lozanovsky spectrum of any positive operator T acting on an arbitrary complex Banach lattice always contains the order Weyl spectrum of T. As a consequence, we recover Alekhno’s result in [3] which gives an affirmative answer to the aforementioned question for the classical Banach lattices.

The outline of this paper is as follows. After recording some definitions and basic facts in Sect. 2 and stating a few auxiliary results in Sect. 3, we accumulate in Sect. 4 some domination results involving compact and r-compact operators. In particular, we have that, if a positive operator T acting on an arbitrary complex Banach lattice dominates (a not necessarily positive operator) S and T is r-compact, then $S^3$ is also r-compact. This crucial result is then used in Sect. 5 to prove that the order Weyl spectrum of a positive operator T on any complex Banach lattice is in general contained in the order Lozanovsky spectrum of T. We conclude this paper with some consequences of the latter result in Sect. 6.

2 Preliminaries

Throughout this paper, E will denote a complex Banach lattice (that is, $E = E_{{\mathbb {R}}} + iE_{{\mathbb {R}}},$ where $E_{{\mathbb {R}}}$ is a real Banach lattice) and

$$\begin{aligned}E_{+}:= \{x:= x_1 + ix_2 \in E: x = |x| = \sup _{\theta \in [0, 2\pi ]}[x_1 \cos \theta + x_2 \sin \theta ]\}\end{aligned}$$

its positive cone. The algebras of all bounded linear operators on E and all compact operators on E will be indicated by ${\mathcal {L}}(E)$ and ${\mathcal {K}}(E)$, respectively. An operator $T \in {\mathcal {L}}(E)$ (and we write $T:= \text{ Re } T + i \text{ Im } T$, where the real and imaginary parts Re T and Im T are bounded linear operators on $E_{{\mathbb {R}}}$) is said to be positive (written $T \ge 0$) if $T = \text{ Re } T$ and $TE_{+} \subseteq E_{+}$. The cone of all positive bounded linear operators on E will be denoted by K.

We call $T \in {\mathcal {L}}(E)$ regular if both its real and imaginary parts are regular on $E_{{\mathbb {R}}},$ that is, Re T and Im T are differences between two positive operators on $E_{{\mathbb {R}}}$. The algebra of all regular operators on E will be denoted by ${\mathcal {L}}^r(E).$ Let ${\mathcal {F}}(E)$ indicate the ideal of all finite-rank operators on E. Then $T \in {\mathcal {L}}^r(E)$ is said to be r-compact if it can be approximated in the r-norm

$$\begin{aligned}||T||_r:= \inf \{||S||: S \ge 0 \text{ and } |Tx| \le S|x| \text{ for } \text{ all } x\in E\}\end{aligned}$$

by operators in ${\mathcal {F}}(E)$ (cf. [5, Definition 1.2]). We use the notation ${\mathcal {K}}^r(E)$ to denote the ideal of all r-compact operators on E. Obviously, ${\mathcal {F}}(E) \subseteq {\mathcal {K}}^r(E) \subseteq {\mathcal {K}}(E)$.

The spectrum (respectively, essential spectrum) of $T \in {\mathcal {L}}(E)$ and the order spectrum (respectively, order essential spectrum) of $T \in {\mathcal {L}}^r(E)$ will be denoted by

$$\begin{aligned}{} & {} \sigma (T):= \{\lambda \in {\mathbb {C}}: T - \lambda \text{ is } \text{ not } \text{ invertible } \text{ in } {\mathcal {L}}(E)\}\\{} & {} \quad (\text{ respectively, } \sigma _{e}(T):= \{ \lambda \in {\mathbb {C}}: (T - \lambda ) + {\mathcal {K}}(E) \text{ is } \text{ not } \text{ invertible } \text{ in } {\mathcal {L}}(E)/ {\mathcal {K}}(E) \}) \end{aligned}$$

and

$$\begin{aligned}{} & {} \sigma _o(T):= \{\lambda \in {\mathbb {C}}: T - \lambda \text{ is } \text{ not } \text{ invertible } \text{ in } {\mathcal {L}}^r(E)\}\\{} & {} \quad (\text{ respectively, } \sigma _{oe}(T):= \{ \lambda \in {\mathbb {C}}: (T - \lambda ) + {\mathcal {K}}^r(E) \text{ is } \text{ not } \text{ invertible } \text{ in } {\mathcal {L}}^r(E)/ {\mathcal {K}}^r(E) \}), \end{aligned}$$

respectively. Indeed, the (essential) spectrum is always contained in the order (essential) spectrum, that is, $\sigma _e(T) \subseteq \sigma _{oe}(T)$ and $\sigma (T) \subseteq \sigma _o(T).$ From [7, Examples 5.1 and (A1)] we have that these inclusions are proper in general. The spectral radius of $T \in {\mathcal {L}}(E)$ and the order spectral radius of $T \in {\mathcal {L}}^r(E)$ will be denoted, respectively, by $r(T):= \sup \nolimits _{\lambda \in \sigma (T)}{|\lambda |}\ \ \text {and}\ \ r_o(T):= \sup \nolimits _{\lambda \in \sigma _o(T)}{|\lambda |}.$ We remind the reader of the identity $r(T) = r_o(T)$ for all positive operators T on E (cf. [13, p.79]).

Now for a bounded linear operator T on E, we recall that the Weyl spectrum of T is the set

$$\begin{aligned}\sigma _{w}(T):= \{\lambda \in {\mathbb {C}}: T - \lambda \text{ is } \text{ not } \text{ a } \text{ Weyl } \text{ operator } \text{ on } E\},\end{aligned}$$

where an operator on E is said to be Weyl if it can be written as a sum of an invertible operator and a compact operator on E. Consequently,

$$\begin{aligned} \sigma _{w}(T) = \bigcap _{S \in {\mathcal {K}}(E)} \sigma (T + S) = \bigcap _{S \in K \cap {\mathcal {K}}(E)} \sigma (T - S), \end{aligned}$$

where the second identity follows from [3, Theorem 16]. If, in addition, T is positive, then Alekhno in [2, p.384] defined the Lozanovsky spectrum of T as the set

$$\begin{aligned} \sigma _{l}(T):= \hspace{-0.9cm}\bigcap _{\begin{array}{c} 0 \le Q \le T \\ Q \le R \hspace{0.1cm} \mathrm { \small {for \hspace{0.1cm} some} } \hspace{0.1cm} R \in {\mathcal {K}}(E) \end{array}} \hspace{-0.9cm} \sigma (T - Q). \end{aligned}$$

We remark here that in [10], the first-named author and Mouton attempted to understand the Lozanovsky spectrum from a general point of view by introducing the aforementioned spectrum for an arbitrary positive element of a general ordered Banach algebra (OBA), and applying the results achieved in the more general setting to the OBAs $({\mathcal {L}}(E), K)$ and $({\mathcal {L}}^r(E), K).$

In an analogous way, we can look at the order Weyl spectrum of $T \in {\mathcal {L}}^r(E)$, which is defined by

$$\begin{aligned}\sigma _{ow}(T):= \{\lambda \in {\mathbb {C}}: T - \lambda \text{ is } \text{ not } \text{ an } \text{ order } \text{ Weyl } \text{ operator } \text{ on } E\},\end{aligned}$$

where an operator on E is order Weyl if it can be written as a sum of an invertible operator in ${\mathcal {L}}^r(E)$ and a r-compact operator on E. From [7, Definition (4.1)] and [8, Example 3.1.2], we have that

$$\begin{aligned} \sigma _{ow}(T) = \bigcap _{S \in {\mathcal {K}}^r(E)} \sigma _o(T + S) = \bigcap _{S \in K \cap {\mathcal {K}}^r(E)} \sigma _o(T - S). \end{aligned}$$

For a positive T, the order Lozanovsky spectrum of T is now simply defined by

$$\begin{aligned}\sigma _{ol}(T):= \hspace{-0.9cm} \bigcap _{\begin{array}{c} 0 \le Q \le T \\ Q \le R \hspace{0.1cm} \mathrm { \small {for \hspace{0.1cm} some} } \hspace{0.1cm} R \in {\mathcal {K}}^r(E) \end{array}} \hspace{-0.9cm} \sigma _o(T - Q). \end{aligned}$$

(Here we replaced the spectrum by the order spectrum, and ${\mathcal {K}}(E)$ by ${\mathcal {K}}^r(E)$.) Some properties of the order Lozanovsky spectrum of a positive operator can be found in [10]. Evidently, and in view of [3, Theorem 24(a)] and [7, Theorem 4.2(a)],

where

$$\begin{aligned} \sigma _{oel}(T):= \bigcap \limits _{\begin{array}{c} 0 \le Q \le T \\ Q \le R \in {\mathcal {K}}(E) \end{array}} \hspace{-0.4cm} \sigma _o(T - Q) \end{aligned}$$

is the Lozanovsky’s order essential spectrum of T, which should not be confused with the term “order Lozanovsky spectrum of T”. As it is clear from the inclusion-scheme above, the Weyl spectrum is known to be contained in a set generally larger than the Lozanovsky spectrum. By adapting the ideas in the proof of the latter result, we are able to establish the inclusion $\sigma _{ow}(T) \subseteq \sigma _{ol}(T)$ for an arbitrary positive operator T on a general complex Banach lattice (see Theorem 5.1). Finally, for any unexplained terminology and notations concerning the theory of positive operators on Banach lattices, we refer the reader to [1, 12].

3 Auxiliary Results

In this section we will record a few useful properties of positive operators acting on complex Banach lattices, and begin by recalling the notion of ‘domination’ between operators. Specifically, for $0 \le T \in {\mathcal {L}}(E)$, we say that T dominates $S \in {\mathcal {L}}(E)$ (or S is dominated by T) if $|Sx| \le T|x|$ for all $x \in E$.

Lemma 3.1

Suppose that T is a positive operator on a complex Banach lattice E that is dominated by a positive r-compact (respectively, compact) operator, say S. If R is a regular operator on $E_{{\mathbb {R}}},$ then there exists a positive r-compact (respectively, compact) operator on E that dominates RT.

Proof

For $x \in E$, there exist positive operators $R_1$ and $R_2$ such that

$$\begin{aligned} |(RT)x|= & {} |((R_1 - R_2)T)x| \\= & {} |(R_1T)x - (R_2T)x| \\\le & {} |(R_1T)x| + |(R_2T)x| \\\le & {} (R_1T)|x| + (R_2T)|x| \\\le & {} (R_1S)|x| + (R_2S)|x|, \end{aligned}$$

where the second last inequality follows from the facts that $R_1T$ and $R_2T$ are positive operators on E. Hence, if $S \ge 0$ is r-compact (respectively, compact), then RT is dominated by the positive r-compact (respectively, compact) operator $(R_1 + R_2)S$. $\square$

Corollary 3.2

If T is a positive operator on a complex Banach lattice E that is dominated by a positive r-compact (respectively, compact) operator S, then RT is dominated by a positive r-compact (respectively, compact) operator for every regular operator R on E.

Proof

From Lemma 3.1 we have the existence of positive r-compact (respectively, compact) operators $S_1$ and $S_2$ on E such that $S_1$ dominates $(\text{ Re } R)T$ and $S_2$ dominates $(\text{ Im } R)T$. Let $x \in E$. Using the properties of the modulus, it now follows that

$$\begin{aligned} |(RT)x|= & {} |((\text{ Re } R)T)x + i((\text{ Im } R)T)x| \\\le & {} |((\text{ Re } R)T)x| + |((\text{ Im } R)T)x| \\\le & {} S_1|x| + S_2|x| \\= & {} (S_1 + S_2)|x|. \end{aligned}$$

We conclude the result from the fact that $S_1 + S_2$ is a positive r-compact (respectively, compact) operator on E. $\square$

The following proposition is an extension of [6, Proposition 2(i)]. We also emphasize here that a direct consequence of this result—Lemma 3.4—will be useful in the proof of Example 5.2.

Proposition 3.3

Consider the Banach lattice $E:= \bigoplus _{i=1}^n \ell ^2$, where $\ell ^2$ is the Banach lattice of all square-summable sequences. If $B, C \in {\mathcal {L}}^r(E)$ is such that $|B| \le C$, then $C \in {\mathcal {K}}(E)$ implies that $B \in {\mathcal {K}}^r(E)$.

Proof

Assume that $B,C\in {\mathcal {L}}^r(E)$. By the construction of the space E (as a direct sum of $\ell ^2$ spaces), we can write $B=B_1\oplus \cdots \oplus B_n$ and $C=C_1\oplus \cdots \oplus C_n$, where $B_1,\ldots ,B_n,C_1,\ldots ,C_n\in {\mathcal {L}}^r(\ell ^2)$. Now suppose that $|B| \le C$ and $C \in {\mathcal {K}}(E)$. Then this translates to $|B_i|\le C_i$ and $C_i\in {\mathcal {K}}(\ell ^2)$ for all $i \in \{1, \ldots , n\}$, which (in view of [6, Proposition 2(i)]) shows that $B_i\in {\mathcal {K}}^r(\ell ^2)$ for all $i\in \{1, \ldots , n\}$. By construction we have that $B\in {\mathcal {K}}^r(E)$. $\square$

In order to establish our next result, some notions and facts need to be recalled. Let T be a positive operator on a complex Banach lattice E. In [2], Alekhno also introduced and examined the lower Weyl spectrum $\omega ^{-}(T)$ of T, which we recall is defined by

$$\begin{aligned}\omega ^{-}(T):= \bigcap \limits _{\begin{array}{c} 0 \le Q \le T \\ Q \in {\mathcal {K}}(E) \end{array}} \sigma (T - Q).\end{aligned}$$

Indeed, this set is generally larger than the Lozanovksy spectrum of T and is non-empty since it always contains the Weyl spectrum of T. We also consider the order lower Weyl spectrum $\omega _{o}^{-}(T)$ of T, given by

$$\begin{aligned} \omega _{o}^{-}(T):= \bigcap \limits _{\begin{array}{c} 0 \le Q \le T \\ Q \in {\mathcal {K}}^r(E) \end{array}} \sigma _o(T - Q). \end{aligned}$$

Note that from [10, Proposition 3.2], when considering the ordered Banach algebra $({\mathcal {L}}^r(E), K)$ and canonical homomorphism $\pi _r: {\mathcal {L}}^r(E) \rightarrow {\mathcal {L}}^r(E)/ {\mathcal {K}}^r(E)$, we have that the order lower Weyl spectrum is a non-empty compact subset of the complex plane. Moreover, it contains the lower Weyl spectrum, as well as the order Weyl and order Lozanovksy spectra. We show next that, for positive operators on direct sums of $\ell ^2$, the (order) Lozanovsky spectrum coincides with the (order) lower Weyl spectrum.

Lemma 3.4

If T is a positive operator on $E:= \bigoplus _{i=1}^n \ell ^2$, then the following identities hold:

• (a) $\sigma _{l}(T) = \omega ^{-}(T)$

• (b) $\sigma _{ol}(T) = \omega ^{-}_o(T)$

Proof

We start by proving the non-trivial inclusion in (a), so let $\lambda \notin \sigma _{l}(T)$. Then there exists a positive operator $Q^*$ on E such that $Q^* \le S$ for some $S \in {\mathcal {K}}(E),$ $Q^* \le T$ and $\lambda \notin \sigma (T - Q^*).$ Since $S \in {\mathcal {K}}(E),$ we have from Proposition 3.3 that $Q = |Q| \in {\mathcal {K}}^r(E) \subseteq {\mathcal {K}}(E),$ and hence $\lambda \notin \omega _{o}^{-}(T),$ establishing the identity in (a). Now using the inclusion ${\mathcal {K}}^r(E) \subseteq {\mathcal {K}}(E)$ again, a similar argument as in the proof of (a) can be applied to prove the statement in (b). $\square$

In the following section we discuss the domination properties of compact and r-compact operators on general complex Banach lattices.

4 Domination Theorems

The first result we point out is the classical domination theorem for compact operators, which states that the cube of any positive operator dominated by a compact operator is also compact.

Theorem 4.1

[12, Corollary 3.7.15(i)] Let E be a complex Banach lattice and $S, T \in {\mathcal {L}}(E)$ be such that $0 \le S \le T$. If $T \in {\mathcal {K}}(E)$, then $S^3 \in {\mathcal {K}}(E)$.

We recall that a positive operator $T \in {\mathcal {L}}(E)$ dominates an operator $S \in {\mathcal {L}}(E)$ (not necessarily positive) if and only if $\pm S \le T$. If, in addition, S is positive, then $0 \le S \le T$ obviously implies that $\pm S \le T.$

There is a more general form of Theorem 4.1 that was utilised in [2, Theorem 7] to prove that the spectral radius r(T) of a positive operator T belongs to the Lozanovsky spectrum of T whenever it is an element of the essential spectrum of T. Since this result is important to our discussion, we state it next.

Theorem 4.2

[4, Remark 1, p.295] Let E be a complex Banach lattice and $S, T \in {\mathcal {L}}(E)$. If $T \ge 0$ dominates S, then $T \in {\mathcal {K}}(E)$ implies that $S^3 \in {\mathcal {K}}(E)$.

We remark here that Theorem 4.2 also played a crucial role in proving that the Weyl spectrum of a positive operator T on an arbitrary complex Banach lattice is a subset of the Lozanovsky’s order essential spectrum of T [7, Theorem 4.2(a)].

Next we point out an analogue of Theorem 4.1 for r-compact operators that was established by Martinez and Mazon in [11] for Dedekind complete Banach lattices. Using the following result by Wickstead—a type of Dodds–Fremlin theorem for r-compact operators—we show in Theorem 4.4 that the condition in [11, Lemma 2.7] that the Banach lattice is Dedekind complete can be omitted.

Theorem 4.3

[14, Theorem 4.1] Let E be a complex Banach lattice such that both E and $E^*$ have order continuous norms. If  $0 \le S \le T$ and $T \in {\mathcal {K}}^r(E)$, then $S \in {\mathcal {K}}^r(E)$.

Theorem 4.4

Let E be a complex Banach lattice and $S, T \in {\mathcal {L}}^r(E)$ be such that $0 \le S \le T$. If $T \in {\mathcal {K}}^r(E)$, then $S^3 \in {\mathcal {K}}^r(E).$

Proof

The proof of this result is similar to the proof of [12, Corollary 3.7.14(i)]; we need only replace the Dodds–Fremlin theorem [12, Theorem 3.17.13] by Theorem 4.3. $\square$

In order to prove that the order Weyl spectrum is generally contained in the order Lozanovsky spectrum (where we come across an operator—not necessarily positive—that is dominated by a positive r-compact operator), we need a stronger form of Theorem 4.4, that is, an r-compact version of Theorem 4.2. This is given next:

Theorem 4.5

Let E be a complex Banach lattice and $S, T \in {\mathcal {L}}^r(E)$. If $T \ge 0$ dominates S, then $T \in {\mathcal {K}}^r(E)$ implies that $S^3 \in {\mathcal {K}}^r(E)$.

Proof

Let $S, T \in {\mathcal {L}}^r(E)$ be such that $\pm S \le T$, that is $-T \le S \le T$, which is equivalent to $0 \le S + T \le 2T$. If $T \in {\mathcal {K}}^r(E)$, then from Theorem 4.4 we have that $(S+T)^3 \in {\mathcal {K}}^r(E)$, and therefore $S^3 \in {\mathcal {K}}^r(E)$ since ${\mathcal {K}}^r(E)$ is an ideal. $\square$

5 The Order Lozanovsky Spectrum Contains the Order Weyl Spectrum

As a result of Theorem 4.5—an extension of the classical domination theorem for r-compact operators—we can now prove that the order Weyl spectrum is in general contained in the order Lozanovsky spectrum.

Theorem 5.1

Let T be a positive operator on an arbitrary complex Banach lattice E. Then $\sigma _{ow}(T) \subseteq \sigma _{ol}(T)$.

Proof

Suppose that $\lambda \notin \sigma _{ol}(T)$. Then there exists a positive operator $Q^*$ on E such that $Q^* \le S$ for some $S \in {\mathcal {K}}^r(E),$ $Q^* \le T$ and $\lambda \notin \sigma _o(T - Q^*),$ that is, $R:= \lambda I - (T - Q^*)$ is invertible in ${\mathcal {L}}^r(E)$. By Corollary 3.2, $R^{-1}Q^*$ is dominated by a positive r-compact operator, and therefore $(R^{-1} Q^*)^3 \in {\mathcal {K}}^r(E)$ in view of Theorem 4.5. Using the spectral mapping theorem for the order essential spectrum, we have that

$$\begin{aligned} (\sigma _{oe}(R^{-1} Q^*))^3 = \sigma _{oe}((R^{-1} Q^*)^3) = \{0\}, \end{aligned}$$

and hence (by using the spectral mapping theorem again) $\sigma _{oe}(I - R^{-1} Q^*) = \{1\}.$ From [7, Theorem 4.6] it then follows that $\omega _o(I - R^{-1} Q^*) = \{1\}$. Now since $0 \notin \omega _o(I - R^{-1} Q^*),$ the operator $I -R^{-1} Q^*$ is order Weyl, and hence $\lambda I -T = R(I - R^{-1}Q^*)$ is an order Weyl operator on E. This gives $\lambda \notin \sigma _{ow}(T),$ which completes the proof. $\square$

It should be emphasize here that Theorem 5.1 could not have been obtained from the results in [10], established in the general setting of ordered Banach algebras. This is because of the commutativity assumption in [10, Proposition 5.6] and the requirement that the cone $\pi _r(K)$, where $\pi _r: {\mathcal {L}}^r(E) \rightarrow {\mathcal {L}}^r(E)/ {\mathcal {K}}^r(E)$ is the canonical homomorphism on ${\mathcal {L}}^r(E),$ be proper in [10, Corollary 5.11].

We illustrate next, using Alekhno’s example [3, Example 19], that the inclusion in Theorem 5.1 is proper in general.

Example 5.2

Consider the Banach lattice $E:= \ell ^2 \oplus \ell ^2 \oplus \ell ^2$ and $T: E \rightarrow E$ defined by $T = T_1 \oplus T_2 \oplus T_3$, where $T_1$ is the forward shift operator on $\ell ^2$, $T_2$ the backward shift operator on $\ell ^2$ and $T_3 = \frac{1}{2}T_1$. Then

$$\begin{aligned} \sigma _{ow}(T) \subsetneq \sigma _{ol}(T). \end{aligned}$$

Proof

Indeed, from [3, Example 19] we have that the (order) Weyl spectrum $\sigma _w(T)$ of T is given by

$$\begin{aligned}\left\{ \lambda \in {\mathbb {C}}: |\lambda | \le \frac{1}{2}\right\} \cup \{\lambda \in {\mathbb {C}}: |\lambda | = 1\},\end{aligned}$$

and the lower Weyl spectrum $\omega ^{-}(T)$ of T by the unit disk $\{ \lambda \in {\mathbb {C}}: |\lambda | \le 1\},$ which coincides with $\sigma (T)$ and $\sigma _o(T)$. Since

$$\begin{aligned}\sigma _o(T) = \omega ^{-}(T) \subseteq \omega _{o}^{-}(T) \subseteq \sigma _o(T),\end{aligned}$$

it follows that $\omega _{o}^{-}(T) = \{ \lambda \in {\mathbb {C}}: |\lambda | \le 1 \}$, and hence $\sigma _{ol}(T) = \{ \lambda \in {\mathbb {C}}: |\lambda | \le 1 \}$ by Lemma 3.4(b). This establishes the strict inclusion. $\square$

6 Consequences

This section is concerned with some consequences of Theorem 5.1. In [10], Benjamin and Mouton concluded that the spectral radius r(T) of a positive operator T is an element of the order Lozanovsky spectrum of T whenever it lies inside the order Weyl spectrum of T. Since we now know that the order Weyl spectrum is entirely contained in the order Lozanovsky spectrum, [10, Corollary 5.5] given next is simply a special case of Theorem 5.1.

Corollary 6.1

[10, Corollary 5.5] If T is a positive operator on an arbitrary complex Banach lattice E, then $r(T) \in \sigma _{ow}(T)$ implies that $r(T) \in \sigma _{ol}(T).$

In the following corollary we give the complete inclusion-scheme for the order-counterparts in ${\mathcal {L}}^r(E).$

Corollary 6.2

The (generally strict) inclusions

$$\begin{aligned}\sigma _{oe}(T) \subseteq \sigma _{ow}(T) \subseteq \sigma _{ol}(T) \subseteq \omega _o^{-}(T) \subseteq \sigma _o(T)\end{aligned}$$

hold for an arbitrary positive operator T on a general complex Banach lattice E.

The authors do not know what the general relationship between the order Weyl spectrum and the Lozanovsky’s order essential spectrum (defined on p.3 and p.4, respectively) is.

By a complex AL-space (respectively, AM-space) we mean the complexification of a real AL-space (respectively, AM-space). Moreover, it is known that, if E is either an AL-space or a Dedekind complete AM-space with unit, then ${\mathcal {L}}^r(E) = {\mathcal {L}}(E)$ and ${\mathcal {K}}^r(E) = {\mathcal {K}}(E)$ (see, for instance, [1, Theorem 3.9] and [9, Lemma 2.1]). For the classical Banach lattices, the inclusion-scheme on p.4 now reduces to:

Corollary 6.3

If E is either an AL-space or a Dedekind complete complete AM-space with unit, then

$$\begin{aligned} \sigma _w(T) = \sigma _{ow}(T) \subseteq \sigma _{ol}(T) = \sigma _{oel}(T) = \sigma _{l}(T) \end{aligned}$$

for all positive operators T on E. In particular, $\sigma _{l}(T) \ne \emptyset .$

Remark 6.4

It should be noted that Corollary 6.3 is not a new result. In fact, Alekhno in [3, Theorem 23] established the inclusion $\sigma _w(T) \subseteq \sigma _{l}(T)$ for a positive operator T with an almost d-empty pure spectrum. From [3, Example 22(a) &(b)] and the remark above it, we have that every positive operator on either an AL-space or an AM-space has almost d-empty pure spectrum.

As an immediate consequence of Corollary 6.2 we have:

Corollary 6.5

For any positive operator T acting on an arbitrary complex Banach lattice, the order Lozanovsky spectrum $\sigma _{ol}(T)$ of T is a non-empty compact subset of the complex plane.

Note that, to date, it is not known whether the Lozanovsky spectrum $\sigma _{l}(T)$ of a positive operator T on an arbitrary complex Banach lattice E is a non-empty set. However, in Proposition 6.7, we will describe further conditions (on either the Banach lattice or the positive operator) under which the Lozanovsky spectrum is non-empty.

The following result deals with the spectral radius of a positive operator. In a way it can be viewed as a type of Corollary 6.2, where an arbitrary complex Banach lattice is now considered and the inclusion and identities in Corollary 6.2 are replaced by (forward and reverse) implications which involve the spectral radius. Recall the identity $r(T) = r_o(T)$ for all $0 \le T \in {\mathcal {L}}(E).$

Proposition 6.6

Let T be a positive operator on an arbitrary complex Banach lattice. Consider the following statements:

• (a) $r(T) \in \sigma _{ow}(T)$

• (b) $r(T) \in \sigma _{ol}(T)$

• (c) $r(T) \in \sigma _{l}(T)$

• (d) $r(T) \in \sigma _{oel}(T)$

Then the following implications hold: (a) $\Rightarrow$ (b) $\Leftarrow$ (c) $\Leftrightarrow$ (d).

Moreover, if $\sigma _o(T)$ is totally disconnected, then (a) $\Rightarrow$ (c), and if $\sigma _o(T)$ (or just $\sigma (T)$ in view of [13, Corollary, p.84]) has no isolated points, then (b) $\Leftrightarrow$ (c).

Proof

The forward implications (a) $\Rightarrow$ (b) and (d) $\Rightarrow$ (c) follow from Theorem 5.1 and [3, Theorem 24(b)], respectively, while (c) $\Rightarrow$ (d) and (d) $\Rightarrow$ (b) are clear from the inclusion-scheme on p.4.

For the second statement, first assume that $\sigma _{o}(T)$ is totally disconnected. By [7, Corollary 6.3], we have that $\sigma _e(T) = \sigma _{oe}(T)$, and hence the (unique) unbounded components of the complements of the (order) essential and (order) Weyl spectra coincide in view of [7, Theorems 4.6 and 7.1]. The condition $r(T) \in \sigma _{ow}(T)$ is therefore equivalent to $r(T) \in \sigma _{w}(T)$, which in turn implies that $r(T) \in \sigma _{oel}(T)$ (recall the inclusion-scheme on p.4). The result then follows from the equivalence of statements (c) and (d).

Now let us assume that $\sigma _o(T)$ has no isolated points. We show next that $r(T) \in \sigma _{oel}(T)$ if and only if $r(T) \in \sigma _{o}(T),$ from which we conclude that (b) $\Leftrightarrow$ (c) as the inclusions $\sigma _{oel}(T) \subseteq \sigma _{ol}(T) \subseteq \sigma _{o}(T)$ hold in general and (c) $\Leftrightarrow$ (d). Now since $\sigma (T)$ has no isolated points, the set $\sigma (T) \backslash \sigma _e(T)$ simply consists of the bounded components of the complement of $\sigma _e(T)$. From the condition $r(T) = r_o(T) \in \sigma _o(T)$ (which may in fact be replaced by $r(T) \in \sigma (T)$ as T is positive), together with the inclusion-scheme on p.4, it then follows that $r(T) \in \sigma _e(T) \subseteq \sigma _w(T) \subseteq \sigma _{oel}(T)$. This completes the proof. $\square$

The authors do not know whether the implication (a) $\Rightarrow$ (c) holds if the assumption that the order spectrum of T is totally disconnected is dropped. If this is true, then it is worth noting that such result presents a stronger version of Corollary 6.1. Further note that, when there is implication $r(T) \in \sigma _{ol}(T) \Rightarrow r(T) \in \sigma _{oel}(T)$, then the Lozanovsky spectrum of T will be a non-empty set whenever the unbounded components of the complements of $\sigma _{ol}(T)$ and $\sigma _o(T)$ coincide. We now conclude the paper with a summary of the conditions under which the Lozanovsky spectrum is a non-empty set.

Proposition 6.7

If T is a positive operator on an arbitrary complex Banach lattice E, then each of the following conditions ensures that the Lozanovsky spectrum of T is a non-empty subset of the complex plane:

• (a) E is either an AL-space or an AM-space,

• (b) E is the direct sum of $\ell ^2$,

• (c) the spectrum of T has no isolated points.

Proof

(a) follows from [3, Theorem 23], while (b) is clear from Lemma 3.4(a). For the proof of (c), assume that $\sigma (T)$ has no isolated points. In view of the proof of Proposition 6.6 and [3, Theorem 24(b)] (see also Proposition 6.6), we have that $r(T) \in \sigma _l(T)$ since $r(T) \in \sigma _o(T).$ $\square$

It not clear whether the condition ‘the order spectrum of T is totally disconnected’ generally implies the non-emptiness of the Lozanovsky spectrum of T.