Some notes on $b$-weakly compact operators

TL;DR

This paper investigates properties of b-weakly compact operators on Banach lattices, establishing conditions for their modulus, and exploring their relationships with Dunford-Pettis and other operator classes.

Contribution

It introduces new conditions under which the modulus of an operator exists and is b-weakly compact, and characterizes b-weakly compact operators in relation to Dunford-Pettis operators.

Findings

The modulus of a b-weakly compact operator exists under new conditions.

Every Dunford-Pettis operator from a Banach lattice is b-weakly compact.

Order bounded operators into KB-spaces have b-weakly compact moduli.

Abstract

In this paper, we will study some properties of b-weakly compact operators and we will investigate their relationships to some variety of operators on the normed vector lattices. With some new conditions, we show that the modulus of an operator $T$ from Banach lattice $E$ into Dedekind complete Banach lattice $F$ exists and is $b$-weakly operator whenever $T$ is a $b$-weakly compact operator. We show that every Dunford-Pettis operator from a Banach lattice $E$ into a Banach space $X$ is b-weakly compact, and the converse holds whenever $E$ is an $AM$-space or the norm of $E^\prime$ is order continuous and $E$ has the Dunford-Pettis property. We also show that each order bounded operator from a Banach lattice into a $KB$-space admits a $b$-weakly compact modulus.

Equations

$$K(E,X)\subseteq W(E,X)\subseteq W_b(E,X)\subseteq W_o(E,X).$$

$$T\in W_b(E,F)$$

$$x^{\prime }(\sum _{i=1}^k{x}_n)=x^{\prime }(\underset{i=1}{\overset{k}{\bigvee }}{x}_n)\le x^{\prime \prime }(x^{\prime }),\text{for all}k\text{holds.}$$

$$\Vert x_n-x_m\Vert \le \Vert x_n-y\Vert +\Vert y-x_m\Vert \to 0,$$

$$\{y\in G:0\le y\le x_n\}\uparrow x_n,$$

$$\Vert T{x}_n-T{z}_{mn}\Vert \le \Vert T\Vert \Vert x_n-z_{mn}\Vert \le \Vert T\Vert \Vert x_n-y_{mn}\Vert \to 0.$$

$$\Vert T{x}_n-y\Vert \le \Vert T{x}_n-T{z}_{mn}\Vert +\Vert T{z}_{mn}-y\Vert .$$

$$0$$

$$\le |T{y}_{m_k}-z|+|T{y}_{m_{k^{\prime }}}-z|\le z_{\beta }+z_{\beta }\downarrow 0,$$

$$\Vert y^{\prime \prime }{\Vert }_{\infty }=\inf \{\lambda >0:|y^{\prime \prime }|\le \lambda x^{\prime \prime }\},y^{\prime \prime }\in I_{x^{\prime \prime }},$$

Full text

Keywords: Banach lattice, order continuous norm, b-weakly compact operator, Dunford-Pettis operator.

MSC(2010): Primary 46B42; Secondary 47B60.

∗Corresponding author

††copyright: ©0: Iranian Mathematical Society

1. Introduction

An operator $T$ from a Banach lattice $E$ into a Banach space $X$ is said to be b-weakly compact, if it maps each subset of $E$ which is b-order bounded (i.e. order bounded in the topological bidual $E^{\prime\prime}$) into a relatively weakly compact subset of $X$. Note that in [3], the class of b-weakly compact operators is introduced by Alpay-Altin-Tonyali and several interesting characterizations were given in [3, 5, 10, 11]. After that, a series of papers which gave different characterizations of this class of operators were published [3, 4, 5, 6, 7, 8, 9, 10, 11]. The most beautiful property of the class of b-weakly compact operators is that it satisfies the domination property as proved in [3]. But one of shortcomings of this class is that the modulus of a $b$-weakly compact operator need not be $b$-weakly compact. Note that each weakly compact operator is b-weakly compact, but the converse is not true in general. In [9], authors characterized Banach lattices on which each b-weakly compact operator is weakly compact. In [11], authors proved that an operator $T$ from a Banach lattice $E$ into a Banach space $X$ is $b$-weakly compact if and only if $(Tx_{n})$ is norm convergent for every positive increasing sequence $(x_{n})$ of the closed unit ball $B_{E}$ of $E$.

1.1. Some basic definitions

An element $e>0$ in a Riesz space $E$ is said to be an order unit whenever for each $x\in E$ there exists some $\lambda>0$ with $|x|\leq\lambda e$. For example $\ell^{\infty}$ has order unit, but $c_{0}$ has not. A sequence $(x_{n})$ in a vector lattice is said to be disjoint whenever $|x_{n}|\wedge|x_{m}|=0$ holds for $n\neq m$. Let $E$ be a Riesz space. The subset $E^{+}=\{x\in E:x\geqslant 0\}$ is called the positive cone of $E$ and the elements of $E^{+}$ are called the positive elements of $E$. For an operator $T:E\rightarrow F$ between two Riesz spaces we shall say that its modulus $|T|$ exists ( or that $T$ possesses a modulus) whenever $|T|:=T\vee(-T)$ exists in the sense that $|T|$ is the supremum of the set $\{-T,T\}$ in $\mathcal{L}(E,F)$. An operator $T:E\rightarrow F$ between two Riesz spaces is called order bounded if it maps order bounded subsets of $E$ into order bounded subsets of $F$. An operator $T:E\rightarrow F$ between two Riesz spaces is positive if $T(x)\geq 0$ in $F$ whenever $x\geq 0$ in $E$. Note that each positive linear mapping on a Banach lattice is continuous. It is clear that every positive operator is order bounded, but the converse is not true in general. A Banach lattice $E$ has order continuous norm if $\|x_{\alpha}\|\rightarrow 0$ for every decreasing net $(x_{\alpha})_{\alpha}$ with $\inf_{\alpha}x_{\alpha}=0$. If $E$ is a Banach lattice, its topological dual $E^{\prime}$, endowed with the dual norm and dual order is also a Banach lattice. A Banach lattice $E$ is said to be an $AM$-space if for each $x,y\in E$ such that $|x|\wedge|y|=0$, we have $\|x+y\|=max\{\|x\|,\|y\|\}$. A Banach lattice $E$ is said to be $KB$-space whenever each increasing norm bounded sequence of $E^{+}$ is norm convergent. A subset $A$ of a Riesz space $E$ is called b-order bounded in $E$ if it is order bounded in the bidual $E^{\thicksim\thicksim}$. An operator $T:E\rightarrow F$ between two Banach spaces is called a Dunford-Pettis operator whenever $x_{n}\xrightarrow{w}0$ implies $Tx_{n}\xrightarrow{\|\cdot\|}0$. Recall that an operator $T$ from a Banach lattice $E$ into a Banach space $X$ is said to be order weakly compact, if it maps each order bounded subset of $E$ into a relatively weakly compact subset of $X$, i.e., $T[-x,x]$ is relatively weakly compact in $X$ for each $x\in E^{+}$. Assume that $E$ and $F$ are normed lattice. A positive linear operator $T:E\rightarrow F$ is called almost interval preserving if $T[0,x]$ is dense in $[0,Tx]$ for every $x\in E^{+}$. Let $E$ be a vector lattice. A sequence $\{x_{n}\}_{1}^{\infty}\subset E$ is called order convergent to $x$ as $n\to\infty$ if there exists a sequence $\{y_{n}\}^{\infty}_{1}$ such that $y_{n}\downarrow 0$ as $n\rightarrow\infty$ and $|x_{n}-x|\leq y_{n}$ for all $n$. We will write $x_{n}\xrightarrow{o_{1}}x$ when $\{x_{n}\}$ is order convergent to $x$. A sequence $\{x_{n}\}$ in a vector lattice $E$ is strongly order convergent to $x\in E$, denoted by $x_{n}\xrightarrow{o_{2}}x$ whenever there exists a net $\{y_{\beta}\}_{\beta\in\mathcal{B}}$ in $E$ such that $y_{\beta}\downarrow 0$ and that for every $\beta\in\mathcal{B}$, there exists $n_{0}$ such that $|x_{n}-x|\leq y_{\beta}$ for all $n\geq n_{0}$. It is clear that every order convergent sequence is strongly order convergent, but two convergence are different, for information see, [1]. A net $(x_{\alpha})_{\alpha}$ in Banach lattice $E$ is unbounded norm convergent (or, $un$-convergent for short) to $x\in E$ if $|x_{\alpha}-x|\wedge u\xrightarrow{\|\cdot\|}0$ for all $u\in E^{+}$. We denote this convergence by $x_{\alpha}\xrightarrow{un}x$. This convergence has been introduced and studied in [12, 17]. For terminology concerning Banach lattice theory and positive operators, we refer the reader to the excellent books of [2, 14, 16].

2. Main Results

The collections of b-weakly compact operators, order weakly compact operators, weakly compact operators and compact operators will be denoted by $W_{b}(E,X)$, $W_{o}(E,X)$, $W(E,X)$ and $K(E,X)$, respectively, whenever there is not confused. We have the following relationships between these spaces:

[TABLE]

In [3], authors show that the above inclusion may be proper. For example, note that each weakly compact operator is a $b$-weakly operator, but the converse may be false in general. Of course the identity operator $I:L^{1}[0,1]\rightarrow L^{1}[0,1]$ is a $b$-weakly operator, but is not weakly compact. Now let $E$ be a Banach lattice such that the norm of $E^{\prime}$ is order continuous and let $X$ be a Banach space. Then, by [9, Theorem 2.2], it is obvious that each $b$-weakly operator $T:E\rightarrow X$ is weakly compact.

The next example due to Z. L. Chen and A. W. Wickstead in [18] shows that the order bounded $b$-weakly compact operators from a Banach lattice into a Dedekind complete Banach lattice do not form a lattice, i.e., a modulus of a $b$-weakly compact operator need not be $b$-weakly compact.

Example 2.1**.**

Let $E=C[0,1]$, $F=l_{\infty}(F_{n})$ where $F_{n}=(l_{\infty},\|\cdot\|_{n})$ and $\|(\lambda_{k})\|_{n}=\max\{\|(\lambda_{k})\|_{\infty},n\limsup(|\lambda_{k}|)\}$ for all $(\lambda_{k})\in l_{\infty}$. Then for each $n\in\mathbb{N}$, $F_{n}$ is a Dedekind complete $AM$-space, hence so is $F$. Define $T_{n}:E\rightarrow F_{n}$ by $T_{n}(f)=(2^{n}.\int_{I_{n}}f.r_{k}dt)_{k=1}^{\infty}\in F_{n}$ for all $f\in E$, where $r_{n}$ is the $n^{,}$th Rademacher function on $[0,1]$ and $I_{n}=(2^{-n},2^{-n+1})$.

Now define $T:E\rightarrow F$ by $T(f)=(\frac{1}{n}T_{n}(f))_{n=1}^{\infty}$. Then $T$ is a weakly compact operator. So, $T$ is a $b$-weakly compact operator and its modulus $|T|$ exists and $|T|$ is not order weakly compact hence not $b$-weakly compact. So, $W_{b}(E,F)$ is not a Riesz space.

Alpay and Altin in [5] show that for $b$-weakly compact operator $T$ from a Banach lattice $E$ into a Dedekind complete Banach lattice $F$, the modulus of $T$ exist and is $b$-weakly compact operator whenever $F$ is $AM$-space with order unit. Now in the following theorems, we show that $T\in W_{b}(E,F)$ whenever $|T|\in W_{b}(E,F)$ and with some new conditions, we show that the modulus of $T$ exists and is $b$-weakly operator whenever $T$ is a $b$-weakly compact operator.

Theorem 2.2**.**

Let $E$ and $F$ be normed vector lattices. We have the following assertions.

1. (1)

If $T:E\rightarrow F$ is an order bounded operator and $F$ is $KB$-space, then $T$ and $|T|$ are $b$-weakly operator. 2. (2)

If $|T|$ is a $b$-weakly compact operator, then

[TABLE]

Proof.

1. (1)

Since every $KB$-space has order continuous norm, so $F$ is a Dedekind complete Banach lattice. Then, by Theorem 1.18 from [2], $|T|$ exists. Now, let $(x_{n})$ be a positive increasing sequence in $B_{E}$. By Theorem 4.3 of [2], $T^{+}(x_{n})$ is a Positive increasing norm bounded sequence in $F$. Since $F$ is a $KB$-space, $T^{+}(x_{n})$ is norm convergent. Thus, $T^{+}$ is $b$-weakly compact. Similarly, $T^{-}$ is $b$-weakly compact. It follows from the identities $T=T^{+}+T^{-}$ and $|T|=T^{+}+T^{-}$ that $T$ and $|T|$ are $b$-weakly compact operators, so we are done. 2. (2)

Since $0\leq T^{-},T^{+}\leq|T|$, then by using Corollary 2.9 from [3], $T^{-}$ and $T^{+}$ are two $b$-weakly compact operators, so $T$ is a $b$-weakly compact operator.

∎

Theorem 2.3**.**

Let $T$ be an order bounded operator from Banach lattice $E$ into Dedekind complete Banach lattice $F$. If $c_{0}$ dose not embed in $F$, then $T$ and $|T|$ are $b$-weakly compact operators.

Proof.

At first, assume that $T$ is a positive operator. By Theorem 4.3 [2], $T$ is continuous. Thus by Proposition 1 [5], it suffices to show that $(Tx_{n})$ is norm convergent to zero for each $b$-order bounded disjoint sequence $(x_{n})$ in $E^{+}$. Now let $(x_{n})$ be a $b$-order bounded and disjoint sequence in $E^{+}$. It follows that there is a $0\leq x^{\prime\prime}\in E^{\prime\prime}$ such that $0\leq x_{n}\leq x^{\prime\prime}$ for all $n$. Then for each $0\leq x^{\prime}\in E^{\prime}$, we have

[TABLE]

Hence $x^{\prime}(\sum_{i=1}^{\infty}x_{n})<\infty$ for all $0\leq x^{\prime}\in E^{\prime}$. Then for each $0\leq y^{\prime}\in F^{\prime}$ we have $\sum_{i=1}^{k}y^{\prime}(Tx_{n})=\sum_{i=1}^{k}T^{\prime}y^{\prime}(x_{n})<+\infty$. It follows that the sequence $(s_{m}=\sum_{n=1}^{m}Tx_{n})$ is weakly bounded, and by Theorem 2.5.5 [15], it is norm bounded. Since $c_{0}$ dose not embed in $F$, by Theorem 4.60 [2], $F$ is a $KB$-space. Since the sequence $(s_{m})$ is positive, increasing and norm bounded, so it is norm convergent, and so the sequence $(\sum_{n=m}^{k}Tx_{n})$ is norm convergent to zero. It follows that $\lim\|Tx_{n}\|=0$. Therefore, $T$ is a $b$-weakly compact operator. Now, let $T$ be an order bounded operator. By the identities $T=T^{+}-T^{-}$ and $|T|=T^{+}+T^{-}$, it follows that $T$ and $|T|$ are $b$-weakly compact operators. ∎

Proposition 2.4**.**

Let $E$ and $F$ be two Banach lattices. Then we have the following assertions.

1. (1)

If $T$ is a positive and order weakly compact operator from $E$ onto $F$, then $F$ has $\sigma$-order continuous norm. 2. (2)

If a positive $b$-weakly compact operator $T:E\rightarrow F$ between Banach lattices is surjective, then the norm of $F$ is $\sigma$-order continuous.

Proof.

1. (1)

Assume that $\{y_{n}\}_{n}\subseteq F$ with $y_{n}\downarrow 0$. Since $T$ is surjective, there is an element $x_{1}\in E$ such that $Tx_{1}=y_{1}$. It is clear that $\{y_{n}\}_{n}\subseteq[0,y_{1}]\subseteq T([0,x_{1}])$. Since $T([0,x_{1}])$ is relatively weakly compact, there is a subsequence $\{y_{n_{j}}\}_{j}$ of $\{y_{n}\}_{n}$ such that $y_{n_{j}}\xrightarrow{w}y_{0}\in F$. Since $\{y_{n_{j}}\}_{j}$ is a decreasing sequence, by Theorem 3.52 from [2], $y_{n_{j}}\xrightarrow{\|\cdot\|}y_{0}\in F$. It follows from $y_{n}\downarrow 0$ that  $y_{0}=0$. Hence $\|y_{n}\|\rightarrow 0$. Thus $F$ has order continuous norm. 2. (2)

Follows from (1).

∎

Now by [2, Theorem 4.11] we have the following result:

Corollary 2.5**.**

Let $E$ be a normed lattice with order continuous norm. Then the norm completion of $E$ has order continuous norm.

Proof.

We prove that Theorem 4.11 (2) of [2] holds. Let $0\leq x_{n}\uparrow\leq x$ holds in $E$. It follows from [2, Corollary 4.10] that $E$ is Dedekind complete. Put $y=\sup{x_{n}}$. So, $y-x_{n}\downarrow 0$ in $E$. Therefore, $\|y-x_{n}\|\to 0$. We have

[TABLE]

hence $\{x_{n}\}$ is a norm Cauchy sequence. ∎

Proposition 2.6**.**

Let $E$ and $F$ be two Banach lattices such that $E$ has order unit and $F$ has order continuous norm. Then every order bounded operator $T:E\rightarrow F$ is b-weakly compact.

Proof.

Let $A$ be a b-order bounded subset of $E$. Since $E$ has an order unit, $A$ is order bounded in $E$. Therefore, $T(A)$ is order bounded in $F$. Since $F$ has an order continuous norm, $T(A)$ is a relatively weakly compact. Thus $T$ is b-weakly compact operator. ∎

Example 2.7**.**

Every order bounded operator from $\ell^{\infty}$ into $c_{0}$ is b-weakly compact.

Theorem 2.8**.**

Let $E$ and $F$ be two Banach lattices where $E$ has order continuous norm. Let $G$ be an order dense sublattice of $E$ and let $T$ be a positive operator from $E$ into $F$. If $T\in W_{b}(G,F)$, then $T\in W_{b}(E,F)$.

Proof.

Let $\{x_{n}\}$ be a positive increasing sequence in $E$ with $\sup_{n}\|x_{n}\|<\infty$. Since $G$ is order dense in $E$, by Theorem 1.34 from [2], we have

[TABLE]

for each $n$. Let $\{y_{mn}\}_{m=1}^{\infty}\subset G$ with $0\leq y_{mn}\uparrow x_{n}$ for each $n$. Put $z_{mn}=\vee_{i=1}^{n}y_{mi}$ and $0\leq T\in L(E,F)$. It follows that $z_{mn}\uparrow_{m}x_{n}$ and $\sup_{m,n}\|z_{mn}\|\leq\sup_{n}\|x_{n}\|<\infty$. Now, if $T\in W_{b}(G,F)$, then $\{Tz_{mn}\}$ is norm convergent to some point $y\in F$. Now, we have the following inequalities

[TABLE]

Thus by the following inequality proof holds

[TABLE]

∎

Definition 2.9**.**

Let $E$ and $F$ be two vector lattices. $L^{(1)}_{c}(E,F)$ (resp. $L^{(2)}_{c}(E,F)$) is the collection of operators $T\in L_{b}(E,F)$, which $x_{n}\xrightarrow{o_{1}}0$ ($x_{n}\xrightarrow{o_{2}}0$) implies $Tx_{n_{k}}\xrightarrow{o_{1}}0$ (resp. $Tx_{n_{k}}\xrightarrow{o_{2}}0$) whenever $\{x_{n_{k}}\}$ is a subsequence of $\{x_{n}\}$.

In [1], there are some examples which shows that two classifications of operators $L^{(1)}_{c}(E,F)$ and $L^{(2)}_{c}(E,F)$ are different.

Theorem 2.10**.**

Let $E$, $F$ be two Banach lattices such that $E$ has order continuous norm. Then

1. (1)

$W_{b}(E,F)^{+}\subseteq L^{(2)}_{c}(E,F)$. 2. (2)

If $W_{b}(E,F)$ is vector lattice and $F$ Dedekind complete, then $W_{b}(E,F)$ is an order ideal of $L^{(1)}_{c}(E,F)=L^{(2)}_{c}(E,F)$.

Proof.

1. (1)

Let $T$ be a positive $b$-weakly compact operator and let $\{x_{n}\}\subset E$ be a strongly order convergence sequence in $E$. Without lose of generality, we set $0\leq x_{n}\xrightarrow{o_{2}}0$, which follows $\{x_{n}\}$ is norm convergent to zero. Set $\{x_{n_{j}}\}$ as subsequence with $\sum_{k=1}^{+\infty}\|x_{n_{j}}\|<+\infty$. Define $y_{m}=\sum_{j=1}^{m}x_{n_{j}}$. Then $0\leq y_{m}\uparrow$ and $\sup_{m}\|y_{m}\|<\infty$. Since $T$ is $b$-weakly compact operator, $\{Ty_{m}\}$ is norm convergent to some point $z\in F$. Now by [13], page 7, it has a subsequence as $\{Ty_{m_{k}}\}$ which is strongly order convergent to $z\in F$. Thus there is $\{z_{\beta}\}\subset F^{+}$ and that for each $\beta$ there exists $n_{0}$ $|Ty_{m_{k}}-z|\leq z_{\beta}\downarrow 0$ whenever $k\geq n_{0}$. If we set $k\geq k^{\prime}\geq n_{0}$, then we have the following inequalities

[TABLE]

which shows that $T\in L^{(2)}_{c}(E,F)$ and proof immediately follows. 2. (2)

By equality $T=T^{+}-T^{-}$ and Theorem 1.7 from [1], we have $L^{(2)}_{c}(E,F)=L^{(1)}_{c}(E,F)$. Since $W_{b}(E,F)$ is a vector lattice, it follows from part (1) that $W_{b}(E,F)$ is a subspace of $L^{(1)}_{c}(E,F)$. Now proof follows from the fact that $W_{b}(E,F)$ satisfies the domination property.

∎

By using conditions of Theorem 2.10, we can design the following question.

Question. Is $W_{b}(E,F)$ a band in $L^{(1)}_{c}(E,F)=L^{(2)}_{c}(E,F)$?

Proposition 2.11**.**

Let $E$ and $F$ be two Banach lattices such that $F$ is a $KB$-space. Then every bounded operator $T:E\rightarrow F$ is b-weakly compact.

Proof.

By using [5, Proposition 1], it is enough to show that $\{Tx_{n}\}_{n}$ is norm convergent for each b-order bounded increasing sequence $\{x_{n}\}_{n}$ in $E^{+}$. Let $\{x_{n}\}_{n}$ be a b-order bounded increasing sequence in $E^{+}$. Since $F$ is a $KB$-space, by [2, Theorem 4.63], there exists a $KB$-space $G$, a lattice homomorphism $R:E\rightarrow G$ and an operator $S:G\rightarrow F$ such that $T=S\circ R$. Since $R$ is a lattice homomorphism, $R$ is a positive operator and therefore is b-order bounded. Then $R(x_{n})$ is a b-order bounded increasing sequence in $G$. Since $G$ is a $KB$-space, $R(x_{n})$ is norm convergent in $G$. It follows that $S\circ R(x_{n})$ is also norm convergent in $F$. Hence $\{T(x_{n})\}$ is norm convergent in $F$. This completes the proof. ∎

Example 2.12**.**

Let $E$ be a Banach lattice. Then every bounded operator from $E$ into $\ell^{1}$ is b-weakly compact.

In the following proposition, we show that each Dunford-Pettis operator is b-weakly compact, the converse is not always true. In fact, the identity operator of the Banach lattice $\ell^{2}$ is b-weakly compact, but it is not Dunford-Pettis. Recall that if $E$ is a Banach lattice and if $0\leqslant{x^{\prime\prime}}\in{E^{\prime\prime}}$, then the principal ideal $I_{x^{\prime\prime}}$ generated by $x^{\prime\prime}\in E^{\prime\prime}$ under the norm ${\|\cdot\|}_{\infty}$ defined by

[TABLE]

is an $AM$-space with unit $x^{\prime\prime}$, which its closed unit ball coincides with the order interval $[-x^{\prime\prime},x^{\prime\prime}]$.

Lemma 2.13**.**

Let $E$ be a Banach lattice. Then every b-order bounded disjoint sequence in $E$ is weakly convergent to zero.

Proof.

Let $\{x_{n}\}_{n}$ be a disjoint sequence in $E$ such that $\{x_{n}\}_{n}\subseteq{[-x^{\prime\prime},x^{\prime\prime}]}$ for some $x^{\prime\prime}\in E^{\prime\prime}$. Let $Y=I_{x^{\prime\prime}}\cap E$ and equip $Y$ with the order unit norm ${\|\cdot\|}_{\infty}$ generated by $x^{\prime\prime}$. The space $(Y,{\|\cdot\|}_{\infty})$ is an $AM$-space, so, $Y^{\prime}$ is an $AL$-space and hence its norm is order continuous. Now, by [14, Theorem 2.4.14] we see that $x_{n}\xrightarrow{w}0$. ∎

Proposition 2.14**.**

Every Dunford-Pettis operator from a Banach lattice $E$ into a Banach space $X$ is b-weakly compact.

Proof.

Let $T$ be a Dunford-Pettis operator from a Banach lattice $E$ into a Banach space $X$. By [5, Proposition 1], it is enough to show that $\{Tx_{n}\}_{n}$ is norm convergent to zero for each b-order bounded disjoint sequence $\{x_{n}\}_{n}$ in $E^{+}$. Let $\{x_{n}\}_{n}$ be a b-order bounded disjoint sequence in $E^{+}$. As the canonical embedding of $E$ into $E^{\prime\prime}$ is a lattice homomorphism, $\{x_{n}\}_{n}$ is an order bounded disjoint sequence in $E^{\prime\prime}$. Thus by preceding lemma, $\{x_{n}\}_{n}$ is $\sigma(E,E^{\prime})$ convergent to zero in $E$. Now, since $T$ is Dunford-Pettis, $\{Tx_{n}\}_{n}$ is norm convergent to zero. This completes the proof. ∎

As a consequence of [2, Theorem 5.82], [9, Theorem 2.2] and [11, Theorem 2.3], we have the following results.

Corollary 2.15**.**

Let $E$ be a Banach lattice and let $X$ be a Banach space. Then each b-weakly compact operator from $E$ into $X$ is Dunford-Pettis, if one of the following assertions is valid:

1. (1)

$E$* is an $AM$-space.* 2. (2)

The norm of $E^{\prime}$ is order continuous and $E$ has the Dunford-Pettis property (i.e. each weakly compact operator from a Banach space $E$ into another $F$ is Dunford-Pettis).

For the next results we need the following lemma:

Lemma 2.16**.**

1. (1)

If an operator $T$ from a Banach space $X$ into a Banach space $Y$ is compact and $T(X)$ is closed, then $T(X)$ is finite-dimensional. As a consequence, if $T:X\rightarrow Y$ is a surjective compact operator between Banach spaces, then $Y$ is finite-dimensional. 2. (2)

If $T:X\rightarrow Y$ is a weakly compact operator between Banach spaces and $T(X)$ is closed, then $T(X)$ is reflexive. As a consequence, if $T:X\rightarrow Y$ is a surjective weakly compact operator between Banach spaces, then $Y$ is reflexive.

Proof.

1. (1)

Let $T:X\rightarrow Y$ be a compact operator between Banach spaces. Since $\overline{T(X)}=T(X)$, $T(X)$ is a Banach space. If $U$ denotes the open ball of $X$, then $T(U)$ is an open set in $T(X)$. On the other hand, $\overline{T(U)}$ is compact so, $T(X)$ is locally compact and hence $T(X)$ is finite-dimensional. 2. (2)

Let $T:X\rightarrow Y$ be a weakly compact operator between Banach spaces. Since $T(X)$ is closed, $T(X)$ is a Banach space and from equality $T(X)=\bigcup_{n\in\mathbb{N}}nT(B_{X})$, we see that $T(B_{X})$ contains a closed ball of $T(X)$. On the other hand, $\overline{T(B_{X})}$ is weakly compact, so, that closed ball is weakly compact, therefore $T(X)$ is reflexive.

∎

Proposition 2.17**.**

Let $E$ be a Banach lattice and let $X$ be a non-reflexive Banach space. If $T:E\rightarrow X$ is a surjective b-weakly compact operator, then the norm of $E^{\prime}$ is not order continuous.

Proof.

If the norm of $E^{\prime}$ is order continuous then by using [9, Theorem 2.2] and Lemma 2.16, $X$ is reflexive which is a contradiction. ∎

Proposition 2.18**.**

Let $E$, $F$ be two Banach lattices and $F^{\prime}$ has order continuous norm. Suppose that $T:E\rightarrow F$ is an almost interval preserving, injective and b-weakly compact operator which has a closed range. Then $E$ is reflexive.

Proof.

Since $T(E)$ is closed, $T(E)$ is a Banach space, so, $T_{1}:E\rightarrow T(E)$ is a bijective operator between two Banach spaces. Then $T_{1}^{\prime}:T(E)^{\prime}\rightarrow E^{\prime}$ is bijective. Without lose generality we replace $T^{\prime}$ with $T_{1}^{\prime}$. On the other hand, since $T$ is an almost interval preserving, by Theorem 1.4.19 [14], $T^{\prime}$ is a lattice homomorphism, and so by Theorem 2.15 [2], $T^{\prime}$ and ${(T^{\prime})^{-1}}$ are both positive operators. Since $F^{\prime}$ has order continuous norm, by [2, Theorem 4.59], $F^{\prime}$ is a $KB$-space, and so $T^{\prime}$ is b-weakly compact operator and the norm of $E^{\prime}$ is order continuous. Since $T$ is a b-weakly compact, by [9, Theorem 2.2], $T$ is weakly compact. So, $T^{\prime}$ is weakly compact. Now, by Lemma 2.16, $E^{\prime}$ is reflexive and then $E$ is reflexive. This completes the proof. ∎

Recall that a nonzero element $x$ of a Banach lattice $E$ is discrete if the order ideal generated by $x$ is equal to the subspace generated by $x$. The vector lattice $E$ is discrete if it admits a complete disjoint system of discrete elements. For example the Banach lattice $\ell^{2}$ is discrete but $L^{1}[0,1]$ is not.

Proposition 2.19**.**

Let $E$ be a Banach lattice and let $X$ be a Banach space and let $T:E\rightarrow X$ be an injective b-weakly compact operator which its range is closed. If one the following conditions holds, then $E$ is finite dimensional.

1. (1)

$E$* is an $AM$-space with order continuous norm.* 2. (2)

$E$* is an $AM$-space and $E^{\prime}$ is discrete.*

Proof.

According to the proof of Proposition 2.18, $T^{\prime}:X^{\prime}\rightarrow E^{\prime}$ is a surjective operator. By [11, Proposition 2.3], if one of the above conditions holds, then $T$ is a compact operator. Thus, $T^{\prime}$ is compact. Now, by Lemma 2.16, $E^{\prime}$ is finite dimensional. Hence, $E$ is finite dimensional. ∎

Definition 2.20**.**

An operator $T:E\rightarrow F$ between two normed vector lattices is unbounded $b$-weakly compact if $\{Tx_{n}\}$ is $un$-convergent for every positive increasing sequence $\{x_{n}\}_{n}$ in the closed unit ball $B_{E}$ of $E$.

For normed vector lattices $E$ and $F$, the collection of unbounded $b$-weakly compact operators will be denoted by $UW_{b}(E,F)$. If a Banach lattice $F$ has strong unit, by using Theorem 2.3 [17], we have $W_{b}(E,F)=UW_{b}(E,F)$. It is obvious that every $b$-weakly compact operator is unbounded $b$-weakly compact, but the following example shows that the converse does not hold, in general.

Example 2.21**.**

Let $I_{c_{0}}$ be the identity mapping from $c_{0}$ into itself. Then $I_{c_{0}}$ is an unbounded $b$-weakly compact operator. But $I_{c_{0}}$ is not a $b$-weakly compact operator.

The following characterization is obvious.

Proposition 2.22**.**

Let $E$ and $F$ be two normed vector lattices and let $T$ be an operator from $E$ into $F$. Then the following are equivalent:

1. (1)

$T$* is unbounded $b$-weakly compact.* 2. (2)

$\{Tx_{n}\}$* is norm convergent for each $b$-order bounded increasing sequence $\{x_{n}\}$ in $E^{+}$.*

It is easy to see that the class of unbounded $b$-weakly compact operators satisfies the domination property.

Proposition 2.23**.**

Let $E$ and $F$ be two normed vector lattices and let $S$ and $T$ be two operators from $E$ into $F$ with $0\leq S\leq T$. If $T$ is unbounded $b$-weakly compact then $S$ is likewise unbounded $b$-weakly compact.

Theorem 2.24**.**

Let $E$, $F$ be two Banach lattices and let $T\in L(E,F)$. If for an ideal $I$ of $E$ the restriction $T|_{I}:I\rightarrow F$ is a surjective homomorphism which is also a $b$-weakly compact operator, then $T\in UW_{b}(E,F)$.

Proof.

Let $\{x_{n}\}$ be a positive increasing sequence in $E$ with $\sup_{n}\|x_{n}\|<\infty$ and let $x\in I$. Then $x_{n}\wedge x\in I$, $0\leq x_{n}\wedge x\uparrow$ and $\sup_{n}\|x_{n}\wedge x\|\leq\|x_{n}\|<\infty$. Since $T\in W_{b}(I,F)$, $\{T(x_{n}\wedge x)\}$ is convergent for each $x\in I$. As $T$ is homomorphism and surjective, $\{T(x_{n})\wedge y\}$ is convergent for all $y\in F$ and proof follows. ∎

Theorem 2.25**.**

Let $E$ and $F$ be two Banach lattices and let $T:E\rightarrow F$ be a surjective homomorphism. Then by one of the following conditions we have $T\in UW_{b}(E,F)$.

1. (1)

$E$* is a $KB$-space.* 2. (2)

$F$* has order continuous norm.*

Proof.

If $E$ is a $KB$-space, then $T\in W_{b}(E,F)$ and proof follows.

Assume that $F$ has order continuous norm. Let $\{x_{n}\}\subset E^{+}$ be an increasing sequence such that $\sup\|x_{n}\|<\infty$ and let $x\in E^{+}$. Set $y_{n}=x_{n}\wedge x$, which follows that $y_{n}\uparrow\leq x$ and $\sup_{n}\|y_{n}\|\leq\|x\|$. Since $T$ is lattice homomorphism, $T$ is positive, which follows $Ty_{n}\uparrow\leq Tx$. By using Theorem 4.11 [2], $\{Ty_{n}\}$ is norm Cauchy, and so is norm convergence in $F$. On the other hand, $T(x_{n}\wedge x)=Tx_{n}\wedge Tx$. Therefore, $Tx_{n}\wedge y$ is norm convergent for each $y\in F$. ∎