1 Introduction and Preliminaries

An operator T from a Banach lattice E to a Banach space X is said to be b-weakly compact, if the image of every b-order bounded subset of E (that is, order bounded in the topological bidual \(E^{\prime \prime }\) of E) under T is relatively weakly compact. The authors in [8] proved that an operator T from a Banach lattice E into a Banach space X is b-weakly compact if and only if \(\{Tx_n\}_n\) is norm convergent for every positive increasing sequence \(\{x_n\}_n\) of the closed unit ball \(B_E\) of E. The class of all b-weakly compact operators between E and X will be denoted by \(W_b(E,X)\). The class of b-weakly compact operators was firstly introduced by Alpay et al. [3]. One of the interesting properties of the class of b-weakly compact operators is that it satisfies the domination property. Some more investigations on \(W_b(E,X)\) were done by [3,4,5, 7, 8].

In this paper, we continue the investigation on \(W_b(E,X)\). In the first section, we provide some prerequisites. The second section is devoted to the main results. We mainly focus on the inclusion relationship between \(W_b(E,X)\) with some known class of operators. We also study those Banach lattices for which the modulus of an order bounded operator is b-weakly compact.

1.1 Some Basic Definitions

Let E be a vector lattice. An element \(e>0\) in E is said to be an order unit whenever for each \(x\in E\) there exists a \(\lambda >0\) with \(|x| \le \lambda e\). A sequence \((x_n)\) in a vector lattice is said to be disjoint whenever \(|x_n| \wedge |x_m| =0\) holds for \(n\ne m\). A vector lattice is called Dedekind complete whenever every nonempty bounded above subset has a supremum. For an operator \(T:E\rightarrow F\) between two vector lattices, we shall say that its modulus |T| exists whenever \(|T|:=T\vee (-T)\) exists in the sense that |T| is the supremum of the set \(\{-T,T\}\). An operator \(T:E\rightarrow F\) between two vector lattices is called order bounded if it maps order bounded subsets of E into order bounded subsets of F. An operator \(T: E\rightarrow F\) is said to be positive if \(T(x)\ge 0\) in F whenever \(x\ge 0\) in E. Note that each positive linear operator on a Banach lattice is continuous and order bounded. A Banach lattice E has order continuous norm if \(\Vert x_\alpha \Vert \rightarrow 0\) for every decreasing net \((x_\alpha )\) 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 \(\Vert x\vee y\Vert = max (\Vert x\Vert , \Vert y\Vert )\). A Banach lattice E is said to be a KB-space whenever each increasing norm bounded sequence of \(E^+\) is norm convergent. It is known that every reflexive Banach lattice is a KB-spaces. Moreover, each KB-space has order continuous norm. 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 \(x\in E^+\).

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 \((x_n)\subset E\) is called order convergent to x if there exists a sequence \((y_n)\) such that \(y_n \downarrow 0\) and for some \(n_0\), \(|x_n -x| \le y_n\) for all \(n\ge n_0\). 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 \((y_\beta )\) in E such that \(y_\beta \downarrow 0\) and that for every \(\beta \), there exists a \(n_0\) such that \(|x_n -x| \le y_\beta \) for every \(n\ge n_0\). It is clear that every order convergent sequence is strongly order convergent.

For more information concerning the Banach lattice and the related topics, we refer the reader to [2, 10].

2 Results

We commence with the following result showing that if an operator is b-weakly compact on some order dense sublattice, then it will be b-weakly compact on the whole space.

Theorem 2.1

Let E and F be two Banach lattices such that the norm of E is order continuous and let \(T:E\rightarrow F\) be a positive operator. Then, for an order dense sublattice G of E,  if \(T\in W_b(G,F)\) then \(T\in W_b(E,F)\).

Proof

Let \((x_n)\) be a bounded positive increasing sequence in E. Since G is order dense in E, from [2, Theorem 1.34] we have \(\{y\in G:~0\le y\le x_n\}\uparrow x_n,\) for every n. Let \((y_{mn})_m\subset G\) with \(0\le y_{mn}\uparrow x_n\) for every n and set \(z_{mn}=\vee _{i=1}^n y_{mi}\). It follows that \(z_{mn}\uparrow _m x_n\) and \(\sup _{m,n}\Vert z_{mn}\Vert \le \sup _n\Vert x_n\Vert <\infty \). Now, if \(T\in W_b(G,F)\), then \((Tz_{mn})\) is norm convergent to some \(y\in F\). Then, from

$$\begin{aligned} \Vert Tx_n-Tz_{mn}\Vert \le \Vert T\Vert \Vert x_n-z_{mn}\Vert \le \Vert T\Vert \Vert x_n-y_{mn}\Vert \rightarrow 0, \end{aligned}$$

we get

$$\begin{aligned} \Vert Tx_n-y\Vert \le \Vert Tx_n-Tz_{mn}\Vert +\Vert Tz_{mn}-y\Vert ; \end{aligned}$$

and this completes the proof. \(\square \)

Definition 2.2

Let E and F be two vector lattices. We define \(L^{(1)}_c(E,F)\) (resp. \(L^{(2)}_c(E,F)\)) as the collection of all order bounded operators T for which \(x_n\xrightarrow {o_1}0\) (resp. \(x_n\xrightarrow {o_2}0\)) implies \(Tx_{n_k} \xrightarrow {o_1}0\) (resp. \(Tx_{n_k} \xrightarrow {o_2}0\)) for some subsequence \((x_{n_k})\) of \((x_{n})\).

It should be noted that \(L^{(1)}_c(E,F)=L^{(2)}_c(E,F)\) when F is Dedekind complete, see for example [1], in which there are also examples showing that these two collections can be different.

Theorem 2.3

Let E and F be two Banach lattices such that the norm of E is order continuous. Then

  1. (1)

    \(W_b(E,F)^+ \subseteq L^{(2)}_c(E,F)\).

  2. (2)

    If \(W_b(E,F)\) is a vector lattice and F is 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 \(T\in W_b(E,F)^+\) and let \((x_n)\subset E\) be a strongly order convergent sequence. Without lose of generality, we assume that \(0\le x_n\xrightarrow {o_2} 0\), which follows that \((x_n)\) is norm convergent to zero. Set \((x_{n_j})\) as a subsequence with \(\sum _{j=1}^{\infty }\Vert x_{n_j} \Vert <\infty \). Define \(y_m=\sum _{j=1}^m x_{n_j}\). Then, \((y_m)\) is bounded and \(0\le y_m\uparrow \). Since T is a b-weakly compact operator, \((Ty_m)\) is norm convergent to some point \(z\in F\). Using [9, Lemma 3.11], \((Ty_m)\) has a subsequence \((Ty_{m_k})\) strongly order convergent to \(z\in F\). Thus, there exists a net \((z_\beta )\subset F^+\) with the property that for each \(\beta \) there exists some \(n_0\) such that if \(k\ge n_0\), then \(\vert Ty_{m_k}-z\vert \le z_\beta \downarrow 0\). Consequently

    $$\begin{aligned} 0\le Tx_{n_{m_k}}&\le \vert Ty_{m_k}-Ty_{m_{k'}}\vert \\&\le \vert Ty_{m_k}-z\vert +\vert Ty_{m_{k'}}-z\vert \\&\le z_\beta +z_\beta \downarrow 0, \end{aligned}$$

    for every \(k\ge k' \ge n_0\), which confirms that \(T\in L^{(2)}_c(E,F)\) as required.

  2. (2)

    By [2, Corollary 4.10], E is Dedekind complete, so by Dedekind completeness of F, \( L^{(2)}_c(E,F)=L^{(1)}_c(E,F)\). Furthermore, 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.

\(\square \)

We need the following elementary lemma in some of the forthcoming results in this section.

Lemma 2.4

Let E and X be Banach spaces and let \(T:E\rightarrow X\) be a bounded linear operator with closed range. Then

  1. (1)

    If T is compact, then T is of finite rank.

  2. (2)

    If T is weakly compact, then T(E) is reflexive.

Proof

  1. (1)

    If T is compact, then T(U) is relatively norm compact, where U is the open unit ball in E. On the other hand, by the open mapping theorem, T(U) is open. It follows that the Banach subspace T(E) of X is locally compact, so it must be of finite dimensional, as claimed.

  2. (2)

    If T is weakly compact, then T(B) is relatively weakly compact, where B denotes the closed unit ball in E. This fact together with the equality \(T(E)=\bigcup _{n\in {\mathbb {N}}}nT(B)\) implies that the unit closed ball in T(E) is weakly compact. Consequently, T(E) must be reflexive.

\(\square \)

Proposition 2.5

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 [7, Theorem 2.2] T must be weakly compact and Lemma 2.4 implies that X is reflexive which is a contradiction.

Proposition 2.6

Let E and F be two Banach lattices such that the norm of \(F^\prime \) is order continuous. If \(T:E \rightarrow F\) is an injective almost interval preserving and b-weakly compact operator with closed range, then E is reflexive.

Proof

Since T has closed range, we may assume, without loss of generality, that T is onto. Thus, \(T:E \rightarrow F\) is a bijection between two Banach spaces and it follows that \(T^\prime :F^\prime \rightarrow E^\prime \) is also a bijection. On the other hand, since T is almost interval preserving, by [10, Theorem 1.4.19], \(T^\prime \) is a lattice homomorphism, so by [2, Theorem 2.15], both \(T^\prime \) and \({(T')^{-1}}\) are positive operators. Since the norm of \(F^\prime \) is order continuous, by [2, Theorem 4.59], \(F^\prime \) is a KB-space, so \(T^\prime \) is b-weakly compact and the norm of \(E^\prime \) is also order continuous. Since T is b-weakly compact, by [7, Theorem 2.2], T is weakly compact. It follows that \(T^\prime \) is also weakly compact. Now Lemma 2.4 implies that \(E^\prime \) must be reflexive, so E is reflexive, as claimed.

Proposition 2.7

Let E be a Banach lattice, X be a Banach space and let \(T:E\rightarrow X\) be an injective b-weakly compact operator with closed range. Then, E is finite-dimensional when either of the following conditions hold.

  1. (1)

    E is an AM-space with order continuous norm.

  2. (2)

    E is an AM-space and \(E^\prime \) is discrete.

Proof

Similar to the proof of Proposition 2.6, we may assume that T is onto, so \(T^\prime :X^\prime \rightarrow E^\prime \) is a also onto. Then by [8, Proposition 2.3], T is a compact operator under either of the conditions (1) and (2). Thus, \(T^\prime \) is compact. Now, by Lemma 2.4, \(E^\prime \) is finite dimensional. Hence, E is finite dimensional.

Theorem 2.8

For two Banach lattices E and F, if E has order unit and the norm of F is order continuous, then every order bounded operator \(T:E\rightarrow F\) is b-weakly compact operator.

Proof

Let \(T:E\rightarrow F\) be a bounded operator and take a bounded increasing sequence \((x_{n})\) in E. Let \(e \in E^{+}\) be an order unit for E. For each \(x \in E\), the norm \(\Vert x\Vert _{\infty }=\inf \lbrace \lambda >0 : \vert x \vert \le \lambda e\rbrace \) on E is equivalent to the original norm, so it follows that \(\mathop {\text {sup}}_{n} \Vert x_{n}\Vert _{\infty }< \infty \). For each \(n\in {\mathbb {N}} \), there exists a \(\lambda _n>0\) such that \(\lambda _{n} \le \Vert x_{n}\Vert _{\infty }+1\) and \(\vert x_{n}\vert \le \lambda _{n}e\). Then, we get

$$\begin{aligned} 0<\lambda =\text {sup} \lambda _{n} \le \mathop {\text {sup}}_{n} \Vert x_{n}\Vert _{\infty }+1 < \infty . \end{aligned}$$

In particular, \(( x_{n}) \subseteq [- \lambda e , \lambda e]\). Now since F has order continuous norm, it is Dedekind complete, and so \(T^+\) exists. It follows that \(( T^+ x_{n}) \subseteq T^+[-\lambda e , \lambda e]\subseteq [- T^+\lambda e , T^+\lambda e]\). By [2, Theorem 4.9], \([- T^+\lambda e , T^+\lambda e]\) is weakly compact, and so there is a subsequence \(( T^{+}x_{n_{j}})\) of \((T^{+}x_{n})\) which is weakly convergent to some point \(z \in F\). Since \((T^{+}x_{n})\) is an increasing sequence, \((T^{+}x_{n})\) is norm convergent to z, so \(T^{+}\in W_b(E,F)\). A similar argument reveals that \(T^{-}\in W_b(E,F)\). We thus conclude that \(T=T^{+}-T^{-} \in W_b(E,F)\). \(\square \)