On the class of b-weakly compact operators

Abstract

We study the b-weakly compact operators using the Banach-Saks sets. More precisely, we will establish that an operator T from a Banach lattice E into a Banach space Y is b-weakly compact if and only if T carries b-order bounded subsets of E onto Banach-Saks subsets of Y. Next we give a sequential characterization of these operators without requiring the sequences to be disjoint. Also, we describe the relationships between b-weakly compact, and b-L-weakly compact operators.

1 Introduction

The class of b-weakly compact operators were introduced by S. Alpay, B. Altin and C. Tonyali in [3]. Since then, this concept has been studied by many authors; see, for instance, [2, 4, 6]. Recall that an operator T from a Banach lattice E to a Banach space X is said b-weakly compact whenever T carries each b-order bounded subset of E into a relatively weakly compact subset of X. A subset B of E is said b-order bounded if it is order bounded in $E^{''}$(the topological bidual of E). It is not difficult to check that an order bounded subset of E is b-order bounded. However, the unit ball of $c_0$ is b-order bounded but not order bounded. Note that each weakly compact operator T is b-weakly compact, but the converse is not always true. In fact, the identity operator $Id_{L_1[0,1]}:L_1[0,1]\longrightarrow L_1[0,1]$ is b-weakly compact, but not weakly compact (see Example $2.6\; (a)$ in [3]). Some characterization of b-weakly compact operators are given by Alpay et al ([3], Proposition 2.8) and B. Altin ([4], Proposition 1). More precisely, if T is a bounded operator from a Banach lattice E into a Banach space X, the following assertions are equivalent:

• T is b-weakly compact.

• $\lim \limits _{n}\Vert Tx_n\Vert =0$ for every b-order bounded disjoint sequence $(x_n)_{n\in {\mathbb {N}}}$ of E

• $(Tx_n)_{n\in {\mathbb {N}}}$ is norm convergent for every positive increasing sequence $(x_n)_{n\in {\mathbb {N}}}$ of the closed unit ball $B_{E}$ of E.

The main aim of the present paper is studying b-weakly compact operators using the Banach-Saks sets. In Sect. 2 we introduce some basic definitions and facts concerning Banach-Saks and b-order bounded sets. In particular, we prove that the notions of an L-weakly compact and a Banach-Saks set coincide for intervals. In Sect. 3 we present some characterizations of the b-weakly compact operators. Mainly, we prove that an operator T from a Banach lattice E into a Banach space Y is b-weakly compact if and only if T carries b-order bounded subsets of E onto Banach-Saks subsets of Y if and only if $\lim _n \Vert Tx_n\Vert = 0$ for every $b-$order bounded sequence $(x_n)_{n\in {\mathbb {N}}}$ of $E_+$ satisfying that the sequence of arithmetic means $\left( \frac{1}{n}\sum \nolimits _{k=1}^{n}x_k\right) _n$ converge in norm to zero (Theorems 3.3 and 3.6). In Sect. 4, we establish some relationships between the notions of a b-weakly compact, and a b-L-weakly compact operator. In particular, we prove that these two notions coincide for positive operators between two Banach lattices E and F such that F has an order continuous norm.

We refer the reader to [1, 16] for any unexplained terms from the theory of Banach lattices and operators.

2 Banach-Saks and b-order bounded sets

A Banach lattice is a Banach space (E, ||.||) such that E is a Riesz space and its norm satisfies the following property: For each $x,y\in E$ such that $|x|\le |y|,$ we have $||x||\le ||y||.$ Note that the topological dual $E^{\prime },$ endowed with the dual norm and the dual order is a Banach lattice. The maximum, respectively the minimum of the set $\{ x_i, 1 \le i \le n\}$ is denoted by $\vee _{i=1}^{n}x_i,$ respectively $\wedge _{i=1}^{n}x_i$. A net $(u_{\alpha })$ in a Banach lattice is said to be disjoint whenever $|u_{\alpha }| \wedge |u_{\beta }|=0$ holds for $\alpha \ne \beta .$

Recall from [15] that a subset A of a Banach space X is called Banach-Saks if each bounded sequence $(x_n)_{n\in {\mathbb {N}}}$ of A has a subsequence $(y_n)_{n\in {\mathbb {N}}}$ whose arithmetic means converge in norm. That is, there exists $y \in E$ such that:

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\Vert \frac{1}{n}\sum _{k=1}^{n}y_k - y\Vert =0. \end{aligned}$$

In relying upon Proposition 2.3 in [15], a Banach-Saks set is weakly relatively compact. The converse statement is in general not true [7]. We have the following result.

Lemma 2.1

Let $(h_n)_{n\in {\mathbb {N}}}$ be a b-order bounded disjoint sequence of a Banach lattice E. Then $\lim _n\Vert \frac{1}{n}\sum _{k=1}^{n}h_k\Vert =0.$

Proof

Let $(h_n)_{n\in {\mathbb {N}}}$ be a disjoint sequence of E such that $0 \le h_n \le x''$ holds for all $n\in {\mathbb {N}}$ and for some $x'' \in E^{''}.$ Observe that $0\le \vee _{i=1}^{m}h_i= \sum _{i=1}^{m}h_i \le x''$ for all $m \in {\mathbb {N}}$. Therefore, $0 \le \frac{1}{n}\sum _{i=1}^{n}h_i \le \frac{x''}{n}$ for all $n \in {\mathbb {N}},$ which implies that

$$\begin{aligned} \Vert \frac{1}{n}\sum _{i=1}^{n}h_i\Vert \le \frac{\Vert x''\Vert }{n} \rightarrow 0. \end{aligned}$$

$\square$

Recall that a Banach lattice E is said to be order continuous if $\lim _{\alpha } \Vert x_{\alpha }\Vert = 0$ for every decreasing net $(x_{\alpha })_{\alpha }$ in E such that $\inf (x_{\alpha })=0.$ Let E be an order continuous Banach lattice, an element $e \in E$ is said to be a weak unit if for $h \in E,$ $e \wedge h = 0$ implies $h=0.$ The set of all positive vectors of E is denoted by $E_{+}.$ The ideal generated by a vector $x\in E$ is denoted by $E_x$ and is given by

$$\begin{aligned} E_x=\{y\in E; \ \exists \;\uplambda >0 \quad \hbox {with}\quad |y|\le \uplambda |x| \}. \end{aligned}$$

Recall from ([16], Definition 3.6.1) that A bounded subset S of E is said to be L-weakly compact , if $\Vert x_n\Vert \longrightarrow 0$ for every disjoint sequence $(x_n)_n$ in the solid hull of S. The solid hull of S is given by

$$\begin{aligned} \text {Sol}(S)= \{x \in E: |x| \le |a| \hbox { for some } a \in S\}. \end{aligned}$$

The maximal closed ideal in E on which the induced norm is order continuous is denoted by $E^a$. A Grothendieck type characterization of L-weakly compact sets is expressed as follows.

Theorem 2.2

(Proposition 3.6.2 in [16]) Let A be a non-empty bounded subset of E. The following assertions are equivalent:

• (1) A is L-weakly compact.

• (2) For each $\epsilon > 0$ there exists some $u \in (E^a)_+$ such that $A \subset [-u,u] + \epsilon B_E$, where $B_E$ is the closed unit ball of E.

The notions of L-weakly compact and Banach-Saks sets coincide for intervals. The details follow.

Theorem 2.3

Let E be a Banach lattice, and let $b \in E_+$. Then $[-b,b]$ is L-weakly compact if and only if it is Banach-Saks.

Proof

Let $b \in E_+$ be such that $[-b,b]$ is L-weakly compact. Since $[-b,b] \subseteq E^a$ and $E^a$ is a Banach lattice with order continuos norm (where $E^a$ is the maximal closed ideal in E on which the induced norm is order continuous), we conclude from Lemma 2.3 in [12] that $[-b,b]$ is a Banach-Saks in $E^a$. So, $[-b,b]$ is a Banach-Saks set in E.

Conversely, if $[-b,b]$ is Banach-Saks, it follows from Proposition 2.3 in [15], that $[-b,b]$ is weakly compact, so by Corollary 5.54 in [1], $[-b,b]$ is $L-$weakly compact. $\square$

Note that Theorem 2.3 is not true for arbitrary order bounded subsets. Indeed, if $(e_n)$ denotes the sequence of the basic unit vectors of $l_\infty$, then $(e_n)$ is an order bounded Banach-Saks set of $l_{\infty }$ but not L-weakly compact.

A Banach lattice E is said to be a Kantorovich-Banach space (or briefly a KB-space) whenever every increasing norm bounded sequence of $E_+$ is norm convergent ([1], Definition 4.58). For instance, each reflexive Banach lattice is a KB-space ( [1], Theorem 4.70). Also, for $1<p<\infty$ the space $L^{p}[0,1]$ is an example of a KB-space ([5], Proposition 2.1).

Recall from ([1], p. 52) that an ideal I of E is called a $\sigma$-ideal whenever for every sequence $(x_n)_{n\in {\mathbb {N}}}$ of I, if $\sup (x_n) =x$ in E, then $x \in I.$

Theorem 2.4

Let E be a Banach lattice, then the following statements are equivalent:

• (1) E is a $\sigma$-ideal of $E^{\prime \prime }$.

• (2) E is KB-space.

• (3) Every b-order bounded subset A of E has the Banach-Saks property.

• (4) Every b-order bounded subset A of E is relatively weakly compact.

Proof

• $(1) \Longrightarrow (2)$ Let $(x_n)$ be a norm bounded sequence in E satisfying $0 \le x_n \uparrow$. Then $0 \le x_n \uparrow x''$ holds in $E''$ for some $x''$ (see page 232 in [1]). By hypothesis, $x'' \in E.$ Since E is an ideal in $E'',$ it has an order continuous norm (see Theorem 4.9 in [1]). So, by Theorem 2.4.2 iii) of [16], $(x_n)$ is convergent. Thus E is KB-space.

• $(2) \Longrightarrow (3)$ Let A be a b-order bounded subset of E. By Proposition 2.1 in [3], A is order bounded, so there exists $b\in E^{+}$ such that $A\subseteq [-b,b].$ The rest of the proof follows from Theorem 2.4.2 in [16] and Theorem 2.3.

• $(3) \Longrightarrow (4)$ Follows immediately from Proposition 2.3 in [15].

• $(4) \Longrightarrow (1)$ Relying on our hypothesis we have $I:E\rightarrow E$ is b-weakly compact. Hence, we deduce from Proposition 2.10 in [3] that E is a KB-space, which implies that E is a band of $E^{\prime \prime }$ ([1], Theorem 4.60). In particular E is a $\sigma$-ideal of $E^{\prime \prime }.$

$\square$

Notice that the equivalence (2) , (4) of Theorem 2.4 is exactly Proposition 2.10 of [3].

3 Some characterizations of b-weakly compact operators

The main objective of this section is to characterize the b-weakly compact operators. For this, we need to fix some notations and recall some definitions.

Recall from [11] that an operator T between a Banach lattice E and a Banach space Y is said to be order weakly compact if $T([-x,x])$ is relatively weakly compact for every positive element $x \in E.$ Order weakly compact operators are characterized as follows.

Theorem 3.1

([16], Theorem 3.4.6) Suppose that T is a bounded operator from a Banach lattice E into a Banach space Y. Then there exist a Banach lattice G,  a lattice homomorphism $\phi :E\rightarrow G,$ and an operator $R:G\rightarrow Y$ with $T=R\phi$ such that G has order continuous norm if and only if T is order weakly compact.

A Grothendieck type characterization of the Banach-Saks sets is the next.

Lemma 3.2

A subset B of a Banach space X is Banach-Saks if and only if for each $\epsilon > 0$ there exists a Banach-Saks subset S of X such that

$$\begin{aligned} B \subset S + \epsilon B_X. \end{aligned}$$

Proof

If B is Banach-Saks, and $\epsilon > 0,$ then $B \subset B+ \epsilon B_X.$

Conversely, let $(x_n)_{n\in {\mathbb {N}}}$ be a bounded sequence of B and let $\epsilon > 0.$ From our hypothesis, there exists a Banach-Saks set S such that $\{x_n, \ n \in {\mathbb {N}}\} \subset S + \epsilon B_X,$ and hence $x_n= y_n + \epsilon z_n,$ where $(y_n) \subset S$ and $(z_n) \subset B_X.$ Without loss of generality we can assume that the sequence $(\frac{1}{n}\sum _{k=1}^{n}y_k)_n$ converges in norm to some $y \in X.$ Then, there exists $N_{0}\in {\mathbb {N}}$ such that for all $n \ge N_0$ we have

$$\begin{aligned} ||\left( \frac{1}{n}\sum _{k=1}^{n}y_k\right) -y||\le \epsilon . \end{aligned}$$

Let $n \ge N_0,$ then

$$\begin{aligned} ||\left( \frac{1}{n}\sum _{k=1}^{n}x_k\right) -y||&=||\;\left[ \frac{1}{n}\sum _{k=1}^{n}\left( y_k+\epsilon z_k\right) \right] -y\;||\\&\le ||\;\left( \frac{1}{n}\sum _{k=1}^{n}y_k\right) -y\;||+ \frac{\epsilon }{n} \,||\sum _{k=1}^{n}z_k\;||\\&\le ||\;\left( \frac{1}{n}\sum _{k=1}^{n}y_k\right) -y\;||+ \frac{\epsilon }{n} \sum _{k=1}^{n}||z_k||\\&\le \epsilon +\epsilon = 2 \epsilon .\\ \end{aligned}$$

Consequently, the sequence $\left( \frac{1}{n}\sum _{k=1}^{n}x_k\right) _{n\in {\mathbb {N}}}$ converges in norm to y. $\square$

From Proposition 2.8 in [3], T is b-weakly compact if and only if $(T x_n)_n$ is norm convergent to zero for every b-order bounded disjoint sequence $(x_n)_{n\in {\mathbb {N}}}$ of $E_+.$ Our next theorem characterizes the b-weakly compact operators using the Banach-Saks sets.

Theorem 3.3

Let E be a Banach lattice and Y a Banach space. If $T: E \rightarrow Y$ is a bounded operator, then the following assertions are equivalent:

• (1) T is b-weakly compact.

• (2) T carries b-order bounded subsets of E onto Banach-Saks subsets of Y.

Proof

• $(2)\Longrightarrow (1)$ According to ( [15], Proposition 2.3), every Banach-Saks set is relatively weakly-compact. This leads up to the result.

• $(1)\Longrightarrow (2)$ Let B be a b-order bounded subset of E. Since $B^{+}:=\{x^{+}, x\in B\}$ and $B^{-}:=\{x^{-}, x\in B\}$ are both b-order bounded subsets of $E_+$ and $B \subset B^+-B^-,$ it is enough to show that T(A) is a Banach-Saks subset of Y for each b-order bounded subset A of $E_+.$

For this, let A be a b-order bounded subset of $E^{+}.$ If $(w_n)$ is a disjoint sequence in the solid hull of A, then $(w_n)_n$ is also b-order bounded, and therefore $\lim _n\Vert Tw_n\Vert =0$ ([3], Proposition 2.8). Now, let $\epsilon > 0$ be fixed. By Theorem 4.36 in [1], there exists some $u_{\epsilon } \in E_{+}$ such that $\Vert T[(x-u_{\epsilon })^{+}] \Vert < \epsilon , \; \hbox {for all}\; x\in A.$ Using the equality $x=x\wedge u_{\epsilon }+(x-u_{\epsilon })^{+}$, we see that $Tx \in T([-u_{\epsilon },u_{\epsilon }])+\epsilon B_{Y},$ and hence $T(A)\subseteq T([-u_{\epsilon },u_{\epsilon }])+\epsilon B_{Y}.$ According to Lemma 3.2 it remains to show that $T[-u_{\epsilon },u_{\epsilon }]$ is Banach-Saks.

Since T is b-weakly compact, in particular it is order weakly compact, it follows from Theorem 3.1 that there exist an order continuous Banach lattice G,  a lattice homomorphism $\phi :$ E $\longrightarrow$ G and a bounded operator R  : G $\longrightarrow$ Y,  with $T = R \phi .$ Clearly, $\phi [-u_{\epsilon },u_{\epsilon }]$ is an order bounded subset of G. From the order continuity of G,  it follows that $\phi [-u_{\epsilon },u_{\epsilon }]$ is L- weakly compact ([1], Theorem 4.14). Therefore, by Lemma 2.3 in [12], $\phi [-u_{\epsilon },u_{\epsilon }]$ is Banach-Saks. Since R is bounded, it is easy to see that $T[-u_{\epsilon },u_{\epsilon }]$ is likewise Banach-Saks, and the proof is concluded. $\square$

If E is an order continuous Banach lattice which has a weak unit, then there exist a probability space $(\Omega ,\Sigma , \mu )$, an order ideal I of $L_1(\Omega ,\Sigma , \mu ),$ a lattice norm $\Vert$ . $\Vert _I$ on I and an order isometry j from E onto (I,$\Vert$ . $\Vert _I$ ) such that the canonical inclusion from I into $L_1(\Omega ,\Sigma , \mu )$ is continuous with $\Vert f\Vert _1 \le \Vert f\Vert _I$ (see Theorem 1.b.14 in [14]). This implies that j is continuous as an operator from E into $L_1(\Omega ,\Sigma , \mu ).$ Note that a separable subspace X of an order continuous Banach lattice E is included in some closed order ideal Y of E with a weak unit (see Proposition 1.a.9 in [14]). Thus, ${E_X}$ ( the closed ideal generated by X ) has a weak unit. An operator $T: E \rightarrow X$ is $M-$weakly compact if for every bounded disjoint sequence $(w_n)$ we have $\Vert Tw_n\Vert \rightarrow 0$ ( [16]).

At this state of analysis we need this following result.

Theorem 3.4

(Theorem 1.2.8 in [17]) Let $(x_n)_n$ be a normalized sequence of a Banach lattice E with order continuous norm. Then,

• (1) either $(\Vert x_n\Vert _{L_{1}})$ is bounded away from zero,

• (2) or there exist a subsequence $(x_{n_k})$ and a disjoint sequence $(z_k)\subset E$ such that $\Vert z_k-x_{n_k}\Vert \longrightarrow 0.$

To continue our discussion, we need the next Lemma:

Lemma 3.5

Let Y be a Banach space and E be a Banach lattices such that $E'$ has an order continuous norm. For every bounded linear operator $T: E \rightarrow Y$ the following assertions are equivalent.

• (1) T is $M-$weakly compact.

• (2) $\Vert Tx_n\Vert \rightarrow 0$ as $n \rightarrow + \infty$ for every bounded sequence $(x_n)$ of $E_+$ satisfying $x_n \rightarrow 0$ in $\sigma (E,E')$ as $n \rightarrow + \infty .$

Proof

$(1) \Rightarrow (2)$ Suppose that T is $M-$weakly compact. Then there exist a reflexive Banach lattice G,  an $M-$weakly compact lattice homomorphism $\phi :$ E $\longrightarrow$ G and an $M-$weakly compact operator R  : G $\longrightarrow$ Y with $T = R \phi$ (see Exercice 10 page 338 in [1]). Now let $(x_n)$ be a bounded sequence of $E_+$ satisfying $x_n \rightarrow 0$ in $\sigma (E,E')$ as $n \rightarrow + \infty ,$ so that $\phi x_n \rightarrow 0$ in $\sigma (G,G')$ as $n \rightarrow + \infty .$ Since $V :=[\phi x_n]$, the closure of the subspace spanned by the vectors $(\phi x_n)_n,$ is a separable subspace of G,  it follows from Proposition 1.a.9 in [14] that ${E_V}$ is an order ideal with a weak order unit. By applying ([14], $\text { Theorem 1.b.14}),$ we infer that ${E_V}$ can be represented as a dense order ideal of $L_1(\Omega ,\Sigma , \mu )$ for some probability measure $\mu ,$ such that the formal inclusion

$$\begin{aligned} i: {E_V} \hookrightarrow L_1(\Omega ,\Sigma , \mu ) \end{aligned}$$

is continuous. It follows that $(i(\phi x_n))_n$ converges weakly to 0 in $L_1(\Omega ,\Sigma , \mu ).$ Since $L_1(\Omega ,\Sigma , \mu )$ has the positive schur property, $\lim _n\Vert i(\phi x_n)\Vert _1=0$. Now, let $(y_n)$ be an arbitrary subsequence of $(x_n).$ Since $\lim _n\Vert i(\phi y_n)\Vert _1=0$, it follows from Theorem 3.4 that

• (1) either $\Vert \phi y_n\Vert _1 \ge \gamma \Vert \phi y_n\Vert$ for some $\gamma > 0,$

• (2) or there is a subsequence $(z_n)_{n\in {\mathbb {N}}}$ of $(y_n)$ and a disjoint sequence $(w_n)$ in the solid hull of $(\phi z_n)$ such that $\Vert \phi z_n-w_n\Vert \longrightarrow 0.$

Assume first that (1) is satisfied, then $(\Vert \phi y_n\Vert )$ and hence $(\Vert Ty_n\Vert )$ converges to 0. Next, suppose that (2) is satisfied. Since $\Vert \phi z_n-w_n\Vert \rightarrow 0,$ so $\Vert Tz_n-Rw_n\Vert \rightarrow 0.$ On the other hand, since the disjoint sequence $(w_n)$ is bounded and R is $M-$weakly compact, then $\lim \Vert Rw_n\Vert =0,$ which implies $\lim \Vert Tz_n\Vert =0.$ Thus, we have shown that every subsequence of $(Tx_n)$ has a subsequence that is norm convergent to zero. This leads up to $\lim \Vert Tx_n\Vert = 0$, which concludes the proof.

$(2) \Rightarrow (1)$ This assertion follows from Theorem 2.4.14 in [16].

$\square$

Let E be a Banach lattice, $x'' \in E'',$ and let $I_{x''}$ be the principal ideal generated by $x''$ in $E''.$ By Theorem 4.21 in [1] the ideal $Y_{x''}= E \cap I_{x''}$ under the norm $\Vert .\Vert _{\infty }$ defined by

$$\begin{aligned} \Vert x\Vert _{\infty }=\inf \{ \uplambda > 0; \quad |x| \le \uplambda |x''| \}; \quad x \in Y_{x''} , \end{aligned}$$

is an AM-space.

The next result gives a sequential characterization of b-weakly compact operators in the spirit of ([3], Proposition 2.8) without requiring the sequences to be disjoint.

Theorem 3.6

Let E be a Banach lattice and Y a Banach space. If $T: E \rightarrow Y$ is a bounded operator, then the following assertions are equivalent:

• (1) T is b-weakly compact.

• (2) $\Vert Tx_n\Vert \rightarrow 0$ as $n \rightarrow + \infty$ for every b-order bounded sequence $(x_n)$ of $E_+$ satisfying $0\le x_n \le x''$ for all $n \in {\mathbb {N}}$ and $x_n \rightarrow 0$ in $\sigma (Y_{x''},Y'_{x''})$ as $n \rightarrow + \infty$ for some $x'' \in E''.$

Proof

• $(1) \Rightarrow (2)$ Let $(x_n)$ be a bounded sequence of $E_+$ satisfying $0\le x_n \le x''$ for all $n \in {\mathbb {N}}$ and $x_n \rightarrow 0$ in $\sigma (Y_{x''},Y'_{x''})$ as $n \rightarrow + \infty$ for some $x'' \in E''.$ Let $T_{x''}$ be the restriction of the operator T to $Y_{x''}.$ Since T is b-weakly compact, then $T_{x''}$ is weakly compact. Thus, by Theorem 5.62 in [1], $T_{x''}$ is $M-$weakly compact. Since $Y_{x''}'$ has an order continuous norm, it follows from Lemma 3.5 that $\Vert Tx_n\Vert \rightarrow 0.$

• $(2) \Rightarrow (1)$ Let $(w_n)_n$ be a disjoint sequence of E satisfying $0\le w_n \le x''$ for all $n \in {\mathbb {N}}$ for some $x'' \in E''.$ Since $(w_n)$ is an order bounded sequence of $I_{x''}$ (the principal ideal generated by $x''$ in $E''$ under the norm $\Vert .\Vert _{\infty }$), then $w_n \rightarrow 0$ in $\sigma (I_{x''},I'_{x''})$ as $n \rightarrow + \infty$ (see Lemma 2.1), an so $\Vert Tw_n\Vert \rightarrow 0$ as $n \rightarrow + \infty .$ Consequently, by Proposition 2.8 in [3], T is b-weakly compact.

$\square$

Theorem 3.7

Let E be a Banach lattice and Y be a Banach space. If $T: E \rightarrow Y$ is a bounded operator, then the following assertions are equivalent:

• (1) T is b-weakly compact.

• (2) There is no b-order bounded disjoint sequence of unit vectors $(w_n)$ in E such that the restriction of T to the subspace $[w_n]$ is an isomorphism.

Proof

• $(1) \Longrightarrow (2)$ Let $(w_n)_n$ be a b-order bounded disjoint sequence of unit vectors in E. Suppose that $T_{|[w_n]}$ is an isomorphism. Since T is b-weakly compact, it follows from Proposition 2.8 in [3] that $\lim \limits _{n }\Vert Tw_n\Vert =0,$ and so $\lim \limits _{n }\Vert w_n\Vert =0.$ This clearly leads to a contradiction.

• $(2) \Longrightarrow (1)$ Suppose that T is not b-weakly compact. Again by Proposition 2.8 in [3] there is a positive b-order bounded disjoint sequence $(w_n)$ of unit vectors in E such that $\Vert Tw_n\Vert > 1$ for all $n \in {\mathbb {N}}.$ Now, observe that there is some $x'' \in E^{''},$ such that

  $$\begin{aligned} 0 \le \sum \limits _{i=1}^{n}w_i= \vee _{i=1}^nw_i \le x'', \end{aligned}$$

  and therefore $\Vert \sum \limits _{i=1}^{n}w_i\Vert \le \Vert x''\Vert .$ The rest of the proof follows from Proposition 2.3.13 in [16].

$\square$

Recall that an operator T between a Banach lattice E and a Banach space Y is said to be disjointly strictly singular if, there is no disjoint sequence of non null vectors $(x_n)_n$ in E such that the restriction of T to the subspace $[x_n]$ spanned by the vectors $(x_n)_n$ is an isomorphism [13].

Corollary 3.8

Let E be a Banach lattice and X a Banach space. Then every disjointly strictly singular operator $T:E\rightarrow X$ is b-weakly compact.

4 Relationships with b-L-weakly compact operators

The class of b-L-weakly compact operators was introduced by D. Lhaimer et al in their paper [9]. An operator T between two Banach lattices E and F is called b-L-weakly compactif it maps b-order bounded subsets of E into L-weakly compact subsets of F. The notions of b-weakly compact and b-L-weakly compact operators may coincide. The next result provides a condition for this to happen.

Theorem 4.1

Let E and F be Banach lattices such that F has an order continuous norm. If $T: E \rightarrow F$ is a positive operator, then the following assertions are equivalent:

• (1) T is b-weakly compact.

• (2) T carries b-order bounded subsets of E onto Banach-Saks subsets of F.

• (3) T is b-L-weakly compact.

Proof

• $(1)\Leftrightarrow (2):$ See Theorem 3.3.

• $(3)\Rightarrow (1)$ According to ( [16], Proposition 3.6.5), every L-weakly compact subset of a Banach lattice is relatively weakly compact. This yields the result.

It remains to show that (1) $\Longrightarrow$ (3).

For this, let A be a b-order bounded subset of E,  and let $\epsilon$ be given. Arguing as in the proof of Theorem 3.3, we see that

$$\begin{aligned} TA \subseteq T[-u,u] + \epsilon B_F, \end{aligned}$$

for some $u \in E_+.$ Since T is positive, $T[-u,u] \subseteq [-Tu,Tu]$. Consequently,

$$\begin{aligned} TA \subseteq [-Tu,Tu] + \epsilon B_F. \end{aligned}$$

Now taking into account the facts that $T u \in F = F^a,$ we conclude that TA is $L-$weakly compact (by Theorem 2.2). Thus T is $b-L-$weakly compact. $\square$

Next, we provide a Grothendieck type characterization of the L-weakly compact sets.

Lemma 4.2

A subset B of a Banach lattice E is L-weakly compact if and only if for each $\epsilon > 0$ there exist an L-weakly compact subset L of E satisfying

$$\begin{aligned} B \subset L + \epsilon B_E. \end{aligned}$$

Proof

If B is L-weakly compact, then $B \subset B + \epsilon B_X$ for all $\epsilon > 0 .$

Conversely, let B be a subset of Banach lattice E such that for each $\epsilon > 0$, there exists an $L-$weakly compact subset L of E satisfying $B \subseteq L + \epsilon B_E.$ By Theorem 2.2, we have $L \subseteq [-u,u] + \epsilon B_E.$ for some $u \in (E^a)_+.$ Consequently, $B \subseteq [-u,u] + 2\epsilon B_E$, and by applying Theorem 2.2 once more, we conclude that B is $L-$weakly compact. $\square$

Recall from [10] that an operator T from a Banach lattice E into a Banach lattice F is called order L-weakly compact whenever T[0, x] is an L-weakly compact subset of F for each $x \in E_+.$

Theorem 4.3

Let E and F be two Banach lattices. If $T: E \rightarrow F$ is a bounded operator, then the following assertions are equivalent:

• (1) T is b-L-weakly compact.

• (2) T is both order L-weakly compact and b-weakly compact.

Proof

• $(1) \Longrightarrow (2)$ Let T be a b-L-weakly compact operator. According to ([16], Proposition 3.6.5), every L-weakly compact subset of F is relatively weakly compact. Then T is b-weakly compact. On the other hand, since [0, x] is b-order bounded for each $x \in E_+$, it follows that T is order L-weakly compact.

• $(2) \Longrightarrow (1)$ Let A be a b-order bounded set of E,  and let $\epsilon >0.$ Arguing as in the proof of Theorem 3.3 , we see that there exists some $u_{\epsilon } \in E_{+}$ such that

  $$\begin{aligned} T(A) \subset T[-u_{\epsilon },u_{\epsilon }] + \epsilon B_F. \end{aligned}$$

  Since T is order L-weakly compact, $T[-u_{\epsilon },u_{\epsilon }]$ is L-weakly compact subset of F. The rest of the proof follows from Lemma 4.2.

$\square$