Some results on almost L-weakly and almost M-weakly compact operators

TL;DR

This paper characterizes when semi-compact operators are almost L-weakly or M-weakly compact, linking these properties to the order continuity of the involved Banach space norms and exploring their relation to Dunford-Pettis operators.

Contribution

It provides necessary and sufficient conditions for semi-compact operators to be almost L-weakly or M-weakly compact, based on the order continuity of Banach space norms.

Findings

Semi-compact operators are almost L-weakly compact iff the target space's norm is order continuous.

Positive semi-compact operators are almost M-weakly compact iff the dual space's norm is order continuous.

Relationships between almost L-weakly compact and Dunford-Pettis operators are examined.

Abstract

In this paper, we present some necessary and sufficient conditions for semi-compact operators being almost L-weakly compact (resp. almost M-weakly compact) and the converse. Mainly, we prove that if $X$ is a nonzero Banach space, then every semi-compact operator $T: X\rightarrow E$ is almost L-weakly compact if and only if the norm of $E$ is order continuous. And every positive semi-compact operator $T:E\rightarrow F$ is almost M-weakly compact if and only if the norm of $E'$ is order continuous. Moreover, we investigate the relationships between almost L-weakly compact operators and Dunford-Pettis (resp. almost Dunford-Pettis) operators.

Equations

$$T(x)=\varphi (x)\cdot y$$

$$T({\lambda }_n)=(\sum _{n=1}^{\infty }{\lambda }_n)\cdot e$$

$$T(x)=\varphi (x)\cdot y.$$

$$T^{\prime }(f)=f(y)\cdot \varphi$$

$$S_1(x)={\left(\frac{{\varphi }_n(x)}{\varphi (u_n)}\right)}_{n=1}^{\infty }$$

$$\sum_{n=1}^{\infty}\bigl{\lvert}\frac{\phi_{n}(x)}{\phi(u_{n})}\bigr{\rvert}\leq\frac{1}{\varepsilon}\sum_{n=1}^{\infty}\phi_{n}(\lvert x\rvert)\leq\frac{1}{\varepsilon}\phi(\lvert x\rvert)$$

$$S_2({\lambda }_n)=\sum _{n=1}^{\infty }{\lambda }_n{u}_n$$

$$T(x)=\sum _{n=1}^{\infty }\frac{{\varphi }_n(x)}{\varphi (u_n)}{u}_n$$

$$\lvert f_{n}(T(x_{n}))\rvert\leq\lVert f_{n}\rVert\bigl{\lVert}T(x_{n})\bigr{\rVert}\leq\bigl{\lVert}Tx_{n}\bigr{\rVert},$$

$$\lVert u_{n}\rVert\leq\bigl{\lVert}(\lvert u_{n}\rvert-u)^{+}\bigr{\rVert}+\bigl{\lVert}\lvert u_{n}\rvert\wedge u\bigr{\rVert}\leq 2\varepsilon$$

$$S_1(x)={\left(\frac{{\varphi }_n(x)}{\varphi (u_n)}\right)}_{n=1}^{\infty }$$

$$S_3({\lambda }_n)=\sum _{n=1}^{\infty }{\lambda }_n{y}_n$$

$$T(x)=\sum _{n=1}^{\infty }\frac{{\varphi }_n(x)}{\varphi (u_n)}{y}_n$$

$$\lvert f_{n}(T(x_{n}))\rvert\leq\lVert f_{n}\rVert\bigl{\lVert}T(x_{n})\bigr{\rVert}\leq\bigl{\lVert}Tx_{n}\bigr{\rVert},$$

$$T(x)=\varphi (x)\cdot y$$

Full text

1. introduction

Throughout this paper, $X$ and $Y$ will denote real Banach spaces, $E$ and $F$ will denote real Banach lattices. $B_{X}$ (resp. $B_{E}$) is the closed unit of Banach space $X$ (resp. Banach lattice $E$) and $Sol(A)$ denotes the solid hull of a subset $A$ of a Banach lattice.

Recall that a continuous operator $T:X\rightarrow E$ from a Banach space to a Banach lattice is said semi-compact if and only if for each $\varepsilon>0$ there exists some $u\in E_{+}$ such that $T(B_{X})\subset[-u,u]+\varepsilon B_{E}$. In recent years, K. Bouras et al. [4] introduced two classes of operators of almost L-weakly and almost M-weakly compact. Recall that an operator $T$ from a Banach space $X$ into a Banach lattice $F$ is called almost L-weakly compact if $T$ carries relatively weakly compact subsets of $X$ onto L-weakly compact subsets of $F$. An operator $T$ from a Banach lattice $E$ into a Banach space $Y$ is called almost M-weakly compact if for every disjoint sequence $(x_{n})$ in $B_{E}$ and every weakly convergent sequence $(f_{n})$ of $Y^{\prime}$, we have $f_{n}(T(x_{n}))\rightarrow 0$.

They proved in [4] that an operator $T$ from a Banach space $X$ into a Banach lattice $F$ is almost L-weakly compact if and only if $f_{n}(T(x_{n}))\rightarrow 0$ for every weakly convergent sequence $(x_{n})$ of $X$ and every disjoint sequence $(f_{n})$ of $B_{F^{\prime}}$ ([4, Theorem 2.2]). After that, A. Elbour et al. [6] gave a useful characterization of almost L-weakly compact operator. An operator $T$ from a Banach space $X$ into a Banach lattice $F$ is almost L-weakly compact if and only if $T(X)\subset F^{a}$ and $f_{n}(T(x_{n}))\rightarrow 0$ for every weakly null sequence $(x_{n})$ of $X$ and every disjoint sequence $(f_{n})$ of $B_{F^{\prime}}$ ([6, Proposition 1]).

Recall that a norm $\lVert\cdot\rVert$ of a Banach lattice $E$ is order continuous if for each net $(x_{\alpha})$ in $E$ with $x_{\alpha}\downarrow 0$, one has $\lVert x_{\alpha}\rVert\downarrow 0$. It is easy to see that if $E$ has an order continuous norm, $E=E^{a}$. A Banach lattice is said to have weakly sequentially continuous lattice operations whenever $x_{n}\xrightarrow{w}0$ implies $\lvert x_{n}\rvert\xrightarrow{w}0$. Every $AM$-space has this property. A Banach space is said to have the ** Schur property** whenever every weakly null sequence is norm null, i.e., whenever $x_{n}\xrightarrow{w}0$ implies $\lVert x_{n}\rVert\rightarrow 0$. A Banach space is said to have the ** positive Schur property** whenever every disjoint weakly null sequence is norm null. In [4], it was proved that the identity operator $Id_{E}$ is almost L-weakly compact if and only if $E$ has the positive Schur property ([4, Proposition 2.2]). And the identity operator $Id_{E}$ is almost M-weakly compact if and only if $E^{\prime}$ has the positive Schur property ([4, Corollary 2.1]).

Following from these conclusions, it is easy to see that there exist operators which are semi-compact but not almost L-weakly compact or almost M-weakly compact. And there also exist operators which are almost L-weakly compact (resp. almost M-weakly compact) but not semi-compact.

In this paper, we establish some necessary and sufficient conditions for semi-compact operator being almost L-weakly compact (resp. almost M-weakly compact) and the converse. More precisely, we prove that every semi-compact operator $T$ from a nonzero Banach space $X$ to a Banach lattice $E$ is almost L-weakly compact if and only if the norm of $E$ is order continuous (Theorem 2.1). And every positive semi-compact operator from a Banach lattice $E$ into a nonzero Banach lattice $F$ is almost M-weakly compact if and only if the norm of $E^{\prime}$ is order continuous (Theorem 2.2). We also investigate the conditions under which each almost L-weakly compact operator is semi-compact (Theorems 2.3, 2.5). Moreover, we show each positive almost Dunford-Pettis operator $T:E\rightarrow F$ is almost L-weakly compact if and only if $F$ has an order continuous norm (Proposition 2.8).

All operators in this paper are assumed to be continuous. We refer to  [1, 7] for all unexplained terminology and standard facts on vector and Banach lattices. All vector lattices in this paper are assumed to be Archimedean.

2. main results

There exist operators which are semi-compact but not almost L-weakly compact. For example, the identity operator $Id_{c}:c\rightarrow c$ is semi-compact since $c$ is an AM-space with unit. But it is not almost L-weakly compact since $c$ doesn’t have the positive Schur property.

The following Theorem gives a necessary and sufficient condition under which every semi-compact operator is almost L-weakly compact.

Theorem 2.1**.**

Let $X$ be a nonzero Banach space and $E$ be a Banach lattice. Then the following statements are equivalent:

• (1)

Every semi-compact operator $T:X\rightarrow E$ is almost L-weakly compact;

• (2)

The norm of $E$ is order continuous.

Proof.

$(2)\Rightarrow(1)$ If the norm of $E$ is order continuous, then by Corollary 3.6.14 of [7], semi-compact operator $T:X\rightarrow E$ is L-weakly compact. It is obvious that every L-weakly compact operator is almost L-weakly compact. Hence $T$ is almost L-weakly compact.

$(1)\Rightarrow(2)$ Assume by way of contradiction that the norm of $E$ is not order continuous, we need to construct an operator which is semi-compact but not almost L-weakly compact.

Since the norm of $E$ is not order continuous, by Theorem 4.14 of [1], there exists a vector $y\in E_{+}$ and a disjoint sequence $(y_{n})\subset[-y,y]$ such that $\lVert y_{n}\rVert\nrightarrow 0$. On the other hand, as $X$ is nonzero, we may fix $u\in X$ and pick a $\phi\in X^{\prime}$ such that $\phi(u)=\lVert u\rVert=1$ holds.

Now, we consider operator $T:X\rightarrow E$ defined by

[TABLE]

for each $x\in X$. Obviously, $T$ is semi-compact as it is compact (its rank is one). But it is not an almost L-weakly compact operator. If not, as the singleton $\{u\}$ is a weakly compact subset of $X$, and $T(u)=\phi(u)\cdot y=y$, the singleton $\{y\}$ is an L-weakly compact subset of $E$. Since disjoint sequence $(y_{n})\subset sol(\{y\})$, we have $\lVert y_{n}\rVert\rightarrow 0$, which is a contradiction. ∎

There exist operators which are semi-compact but not almost M-weakly compact. For example, the operator $T:\ell_{1}\rightarrow\ell_{\infty}$ defined by

[TABLE]

for each $(\lambda_{n})\in\ell_{1}$, where $e=(1,1,...)$ is the constant sequence with value 1 [1, p. 322 ]. Obviously, $T$ is semi-compact as it is compact (its rank is one). But based on the argument in [6, p. 3], we know that $T$ is not an almost M-weakly compact operator.

The following Theorem gives a necessary and sufficient condition under which every semi-compact operator is almost M-weakly compact.

Theorem 2.2**.**

Let $E$ and $F$ be two nonzero Banach lattices. Then the following statements are equivalent:

• (1)

Every positive semi-compact operator $T:E\rightarrow F$ is almost M-weakly compact;

• (2)

The norm of $E^{\prime}$ is order continuous.

Proof.

$(2)\Rightarrow(1)$ Since positive operator $T:E\rightarrow F$ is semi-compact, following from Corollary 3.3 of [2], $T^{\prime}:F^{\prime}\rightarrow E^{\prime}$ is an almost Dunford-Pettis operator. As the norm of $E^{\prime}$ is order continuous, by Proposition 2.4 of [4], $T^{\prime}$ is an almost L-weakly compact operator. Following from Theorem 2.5(1) of [4], $T$ is almost M-weakly compact.

$(1)\Rightarrow(2)$ Assume by way of contradiction that the norm of $E^{\prime}$ is not order continuous, we need to construct a positive operator which is semi-compact but not almost M-weakly compact.

Since the norm of $E^{\prime}$ is not order continuous, by Theorem 4.14 of [1], there exists a vector $\phi\in E^{\prime}_{+}$ and a disjoint sequence $(\phi_{n})\subset[-\phi,\phi]$ such that $\lVert\phi_{n}\rVert\nrightarrow 0$. On the other hand, as $F$ is nonzero, we may fix $y\in F_{+}$ and pick a vector $g\in(F^{\prime})_{+}$ such that $g(y)=\lVert y\rVert=1$ holds.

Now, we consider operator $T:E\rightarrow F$ defined by

[TABLE]

for each $x\in E$. Obviously, $T$ is positive and semi-compact operator as it is compact (its rank is one). But it is not an almost M-weakly compact operator. In fact, by Theorem 2.5(1) of [4], we only need to show that its adjoint $T:F^{\prime}\rightarrow E^{\prime}$ defined by

[TABLE]

for any $f\in F^{\prime}$ is almost L-weakly compact. If not, as the singleton $\{g\}$ is a weakly compact subset of $X^{\prime}$, and $T^{\prime}(g)=g(y)\cdot\phi=\phi$, the singleton $\{\phi\}$ is an L-weakly compact subset of $E^{\prime}$. Since disjoint sequence $(\phi_{n})\subset sol(\{\phi\})$, we have $\lVert\phi_{n}\rVert\rightarrow 0$, which is a contradiction. ∎

There also exist operators which are almost L-weakly compact but not semi-compact. For instance, the identity operator $Id_{\ell_{1}}:\ell_{1}\rightarrow\ell_{1}$ is almost L-weakly compact since $\ell_{1}$ has the positive Schur property. But it is not semi-compact. If not, as $\ell_{1}$ is discrete with order continuous norm, $Id_{\ell_{1}}$ is compact, which is impossible.

Next, denote $T:E\rightarrow F$ as a continuous operator , we investigate the conditions under which each almost L-weakly compact operator $T$ is semi-compact.

Based on Theorem 4 of [6], we know that if $E^{\prime}$ has an order continuous norm, then each positive almost L-weakly compact operator $T$ is M-weakly compact, hence semi-compact. Now, we claim that if $E$ is reflexive then each almost L-weakly compact operator $T$ is semi-compact. In fact, if $E$ is reflexive, then $B_{E}$ is a relatively weakly compact subset of $E$. As $T$ is almost L-weakly compact, $T(B_{E})$ is an L-weakly compact subset of $F$. By Proposition 3.6.2 of [7], for every $\varepsilon>0$, there exists a vector $u\in F_{+}^{a}\subset F_{+}$ such that $T(B_{E})\subset[-u,u]+\varepsilon B_{F}$. So $T$ is semi-compact.

The following Theorem gives the conditions under which each positive almost L-weakly compact operator $T$ from $E$ to $E$ is semi-compact.

Theorem 2.3**.**

Let $E$ be a Banach lattice with an order continuous norm. Then the following assertions are equivalent:

• (1)

Each positive almost L-weakly compact operator $T$ from $E$ to $E$ is semi-compact.

• (2)

The norm of $E^{\prime}$ is order continuous.

Proof.

$(2)\Rightarrow(1)$ Follows from Theorem 4 of [6].

$(1)\Rightarrow(2)$ Assume by way of contradiction that the norm of $E^{\prime}$ is not order continuous. To finish the proof, we need to construct a positive almost L-weakly compact operator $T:E\rightarrow E$ which is not semi-compact.

Since the norm of $E^{\prime}$ is not order continuous, it follows from Theorem 116.1 of [8] that there exists a norm bounded disjoint sequence $(u_{n})$ of positive elements in $E$ which does not weakly convergence to zero. Without loss of generality, we may assume that $\lVert u_{n}\rVert\leq 1$ for any $n$. And there exist $\varepsilon>0$ and $0\leq\phi\in E^{\prime}$ such that $\phi(u_{n})>\varepsilon$ for all $n$. Then by Theorem 116.3 of [8], we know that the components $\phi_{n}$ of $\phi$ in the carriers $C_{u_{n}}$ form an order bounded disjoint sequence in $(E^{\prime})_{+}$ such that

$\phi_{n}(u_{n})=\phi(u_{n})$ for all $n$ and  $\phi_{n}(u_{m})=0$ if  $n\neq m.\qquad(*)$

Define the positive operator $S_{1}:E\rightarrow\ell_{1}$ as follows:

[TABLE]

for all $x\in E$. Since

[TABLE]

holds for all $x\in E$, the operator $S_{1}$ is well defined and it is also easy to see that $S_{1}$ is a positive operator.

Now define the operator $S_{2}:\ell_{1}\rightarrow E$ as follows:

[TABLE]

for all $(\lambda_{n})\in\ell_{1}$. As $\sum_{n=1}^{\infty}\lVert\lambda_{n}u_{n}\rVert\leq\sum_{n=1}^{\infty}\lvert\lambda_{n}\rvert<\infty$, $S_{2}$ is well defined and is positive.

Next, we consider the composed operator $T=S_{2}\circ S_{1}:E\rightarrow\ell_{1}\rightarrow E$ defined by

[TABLE]

for all $x\in E$. Now we claim $T$ is an almost L-weakly compact operator. Since $E$ has an order continuous norm, $E=E^{a}$. It suffices to show that $T$ satisfies the condition (b) of Proposition 1 of [6]. Let $x_{n}\xrightarrow{w}0$ in $E$ and $(f_{n})$ be a disjoint sequence in $B_{E^{\prime}}$. It is obvious that $S_{1}(x_{n})$ is a weakly null sequence in $\ell_{1}$. As $\ell_{1}$ has the Schur property, \bigl{\lVert}S_{1}x_{n}\bigr{\rVert}\rightarrow 0. Hence \lVert T(x_{n})\rVert=\bigl{\lVert}S_{2}(S_{1}(x_{n}))\bigr{\rVert}\rightarrow 0. Now following from the inequality

[TABLE]

we obtain that $T$ is an almost L-weakly compact operator.

But $T$ is not a semi-compact operator. In fact, note that $\lVert u_{n}\rVert\leq 1$ and $T(u_{n})=u_{n}$ for all $n$ following from $(*)$. So, if $T$ is semi-compact, then $T(B_{E})$ is almost order bounded in $E$. Hence, $(u_{n})\subset T(B_{E})$ is also almost order bounded. So, $u_{n}\rightarrow 0$ in $\sigma(E,E^{\prime})$, which is a contradiction. ∎

To investigate the necessary and sufficient conditions under which each almost L-weakly compact operator $T:E\rightarrow F$ is semi-compact, we first give the following useful Lemma.

Lemma 2.4**.**

Let $E$ be a Banach lattice with an order continuous norm. If $(u_{n})$ is a norm bounded disjoint sequence of $E$ such that the set $\{u_{n}\}$ is almost order bounded in $E$, then $(u_{n})$ converges to zero in norm.

Proof.

Since $A:=\{u_{n}:n\in\mathcal{N}\}$ is almost order bounded, there exists some $u\in E_{+}$ such that

\bigl{\lVert}(\lvert u_{n}\rvert-u)^{+}\bigr{\rVert}\leq\varepsilon for all  $n$.

On the other hand, since $(\lvert u_{n}\rvert\wedge u)$ is an order bounded disjoint sequence in $E$ and the norm of $E$ is order continuous, following from Theorem 4.14 of [1], $(\lvert u_{n}\rvert\wedge u)$ converges to zero in norm . Hence, there exists some $n_{0}$ such that

\bigl{\lVert}\lvert u_{n}\rvert\wedge u\bigr{\rVert}\leq\varepsilon  for  all  $n\geq n_{0}$.

Now, following from the equality $\lvert u_{n}\rvert=(\lvert u_{n}\rvert-u)^{+}+\lvert u_{n}\rvert\wedge u$, we obtain that

[TABLE]

holds for all $n\geq n_{0}$. So, $u_{n}\rightarrow 0$ in norm. ∎

Theorem 2.5**.**

Let $E$ and $F$ be two Banach lattices such that the norm of $F$ is order continuous. Then the following assertions are equivalent:

• (1)

Each positive almost L-weakly compact operator $T:E\rightarrow F$ is semi-compact;

• (2)

One of the following conditions is valid:

• (a)

The norm of $E^{\prime}$ is order continuous;

• (b)

$E$* or $F$ is finite dimensional.*

Proof.

$(1)\Rightarrow(2)$ Assume $E$ and $F$ are both infinite dimensional. We have to show that the norm of $E^{\prime}$ is order continuous. If not, to finish the proof, we need to construct a positive almost L-weakly compact operator $T:E\rightarrow F$ which is not semi-compact.

Since the norm of $E^{\prime}$ is not order continuous, similarly with the proof of Theorem 2.3, we define the positive operator $S_{1}:E\rightarrow\ell_{1}$ as follows:

[TABLE]

for all $x\in E$. And the operator $S_{1}$ is well defined.

On the other hand, since $F$ is infinite dimensional, by Lemma 2.3 of [3], there exists a disjoint sequence $(y_{n})\subset(B_{F})_{+}$ such that $\lVert y_{n}\rVert=1$. Now define the operator $S_{3}:\ell_{1}\rightarrow F$ as follows:

[TABLE]

for all $(\lambda_{n})\in\ell_{1}$. As $\sum_{n=1}^{\infty}\lVert\lambda_{n}y_{n}\rVert\leq\sum_{n=1}^{\infty}\lvert\lambda_{n}\rvert<\infty$, $S_{3}$ is well defined and is positive.

Next, we consider the composed operator $T=S_{3}\circ S_{1}:E\rightarrow\ell_{1}\rightarrow F$ defined by

[TABLE]

for all $x\in E$. Now we claim $T$ is an almost L-weakly compact operator. Since $F$ has an order continuous norm, $F=F^{a}$. It suffices to show that $T$ satisfies the condition (b) of Proposition 1 of [6]. Let $x_{n}\xrightarrow{w}0$ in $E$ and $(f_{n})$ be a disjoint sequence in $B_{F^{\prime}}$. It is obvious that $S_{1}(x_{n})$ is a weakly null sequence in $\ell_{1}$. As $\ell_{1}$ has the Schur property, \bigl{\lVert}S_{1}x_{n}\bigr{\rVert}\rightarrow 0. Hence \lVert T(x_{n})\rVert=\bigl{\lVert}S_{3}(S_{1}(x_{n}))\bigr{\rVert}\rightarrow 0. Now following from the inequality

[TABLE]

we obtain that $T$ is an almost L-weakly compact operator.

But $T$ is not a semi-compact operator. In fact, note that $\lVert u_{n}\rVert\leq 1$ and $T(u_{n})=y_{n}$ for all $n$ following from $(*)$. So, if $T$ is semi-compact, then $T(B_{E})$ is almost order bounded in $F$. Hence, $(y_{n})\subset T(B_{E})$ is almost order bounded. By Lemma 2.4, $y_{n}\rightarrow 0$ in norm, which is a contradiction.

$(2a)\Rightarrow(1)$ Follows from Theorem 4 of [6].

$(2b)\Rightarrow(1)$ Let $T:E\rightarrow F$ be a positive operator. If $E$ is finite dimensional, $T$ is M-weakly compact. Also, if $F$ is finite dimensional, $T$ is L-weakly compact. Hence, $T$ is semi-compact. ∎

There exist operators which are almost M-weakly compact but not semi-compact. For instance, the identity operator $Id_{c_{0}}:c_{0}\rightarrow c_{0}$ is almost M-weakly compact since $(c_{0})^{\prime}=\ell_{1}$ has the Positive Schur property. But it is not semi-compact. If not, as $c_{0}$ is discrete with order continuous norm, $Id_{c_{0}}$ is compact, which is impossible. Following by Corollary 5 of [6], we have the following assertion.

Theorem 2.6**.**

Let $E$ and $F$ be two Banach lattices. And let $T:E\rightarrow F$ be an order bounded almost M-weakly compact operator. If $F^{\prime\prime}$ has order continuous norm, then $T$ is semi-compact.

By Theorem 1 of [6], we know that every Dunford-Pettis operator $T:X\rightarrow E$ is almost L-weakly compact if and only if $E$ has an order continuous norm. Next, we investigate the conditions under which each almost L-weakly compact operator is Dunford-Pettis.

There exist operators which are almost L-weakly compact but not Dunford-Pettis. For example, the identity operator $Id_{L_{1}[0,1]}:L_{1}[0,1]\rightarrow L_{1}[0,1]$ is almost L-weakly compact as $L_{1}[0,1]$ has the positive Schur property. But it is not Dunford-Pettis as $L_{1}[0,1]$ does not have the Schur property. However, we have the following result.

Proposition 2.7**.**

Let $X$ be a Banach space and $E$ be a Banach lattice. If $E$ has weakly sequentially continuous lattice operations, then each almost L-weakly compact operator $T:X\rightarrow E$ is Dunford-Pettis.

Proof.

Let $(x_{n})$ be a weakly null sequence, we have to show that $T(x_{n})\rightarrow 0$ in norm. Based on Corollary 2.6 of [5], it suffices to show $\lvert T(x_{n})\rvert\xrightarrow{w}0$ in $E$ and $f_{n}(T(x_{n}))\rightarrow 0$ for each positive disjoint sequence $(f_{n})$ in $B_{E^{\prime}}$.

As $x_{n}\xrightarrow{w}0$, then $T(x_{n})\xrightarrow{w}0$. Since the lattice operations of $E$ are weakly sequentially continuous, $\lvert T(x_{n})\rvert\xrightarrow{w}0$ in $E$. On the other hand, as $x_{n}\xrightarrow{w}0$ in $X$ and $T$ is an almost L-weakly compact operator, for each positive disjoint sequence $(f_{n})$ in $B_{E^{\prime}}$, $f_{n}(T(x_{n}))\rightarrow 0$. Hence, we get that $T$ is Dunford-Pettis. ∎

At last, we give a conclusion about the relationships between almost L-weakly compact operators and almost Dunford-Pettis operators. K. Bouras et al. show that if $F$ has an order continuous norm, each positive almost Dunford-Pettis operator $T:E\rightarrow F$ is almost L-weakly compact ([4, Proposition 2.4]). We show that the condition of “$F$ has an order continuous norm” is also necessary.

Proposition 2.8**.**

Let $E$ and $F$ be two nonzero Banach lattices. Then the following statements are equivalent:

• (1)

Each positive almost Dunford-Pettis operator $T:E\rightarrow F$ is almost L-weakly compact.

• (2)

The norm of $F$ is order continuous.

Proof.

$(2)\Rightarrow(1)$ Follows from Proposition 2.4 of [4].

$(1)\Rightarrow(2)$ Assume by way of contradiction that the norm of $F$ is not order continuous. we need to construct a positive operator which is almost Dunford-Pettis but not almost L-weakly compact.

Similarly with the proof of Theorem 2.1. Since the norm of $F$ is not order continuous, by Theorem 4.14 of [1], there exists a vector $y\in F_{+}$ and a disjoint sequence $(y_{n})\subset[-y,y]$ such that $\lVert y_{n}\rVert\nrightarrow 0$. On the other hand, as $E$ is nonzero, we may fix $u\in E_{+}$ and pick a $\phi\in(E^{\prime})_{+}$ such that $\phi(u)=\lVert u\rVert=1$ holds.

Now, we consider operator $T:E\rightarrow F$ defined by

[TABLE]

for each $x\in E$. Obviously, $T$ is a positive operator and is compact (its rank is one). Hence, it is almost Dunford-Pettis. But it is not an almost L-weakly compact operator. If not, as the singleton $\{u\}$ is a weakly compact subset of $E$, and $T(u)=\phi(u)\cdot y=y$, the singleton $\{y\}$ is an L-weakly compact subset of $F$. Since disjoint sequence $(y_{n})\subset sol(\{y\})$, we have $\lVert y_{n}\rVert\rightarrow 0$, which is a contradiction.

∎