Duality Property for Positive Weak Dunford-Pettis Operators

Abstract

We prove that an operator is weak Dunford-Pettis if its adjoint is one but the converse is false in general, and we give some necessary and sufficient conditions under which each positive weak Dunford-Pettis operator has an adjoint which is weak Dunford-Pettis.

1. Introduction and Notation

Let us recall that an operator 𝑇 from a Banach space 𝐸 into another 𝐹 is called Dunford-Pettis if it carries weakly compact subsets of 𝐸 onto compact subsets of 𝐹. The operator 𝑇 is said to be weak Dunford-Pettis if 𝑦^′_𝑛(𝑇(𝑥_𝑛)) converges to 0 whenever (𝑥_𝑛) converges weakly to 0 in 𝐸 and (𝑦^′_𝑛) converges weakly to 0 in 𝐹.

The class of weak Dunford-Pettis operators was used by Aliprantis and Burkinshaw [1] and Kalton and Saab [2] when they studied the domination property of Dunford-Pettis operators. As this latter class [3], weak Dunford-Pettis operators do not satisfy the duality property. In fact, there exist weak Dunford-Pettis operators whose adjoints are not weak Dunford-Pettis. For example, as the Banach space 𝑙¹(𝑙²_𝑛) has the Schur property, its identity operator Id_𝑙¹_(𝑙²_𝑛) is Dunford-Pettis and then weak Dunford-Pettis, but its adjoint Id_𝑙^∞_(𝑙²_𝑛), which is the identity operator of the Banach space 𝑙^∞(𝑙²_𝑛), is not weak Dunford-Pettis (because the Banach space 𝑙^∞(𝑙²_𝑛) does not have the Dunford-Pettis property (see [4], page 22)). However, each operator is weak Dunford-Pettis if its adjoint is.

On the other hand, if 𝐸 and 𝐹 are two Banach spaces such that 𝐹is reflexive, then the class of weak Dunford-Pettis operators from 𝐸 into 𝐹 coincides with that of Dunford-Pettis operators from 𝐸 into 𝐹, and therefore some results of [5] can be applied here to give some answers to our duality problem.

Morever, if 𝐸 and 𝐹 are both reflexive, then the class of weak Dunford-Pettis operators from 𝐸 into 𝐹 coincides with that of compact operators from 𝐸 into 𝐹, and hence if 𝑇∶𝐸→𝐹 is an operator such that 𝑇 is weak Dunford-Pettis, then its adjoint 𝑇^′∶𝐹^′→𝐸^′ is weak Dunford-Pettis.

Also, if 𝐸 and 𝐹 are two Banach spaces such that 𝐸^′ or 𝐹^′ has the Dunford-Pettis property, then each operator from 𝐹^′ into 𝐸^′ is weak Dunford-Pettis, and hence each weak Dunford-Pettis 𝑇∶𝐸→𝐹 has an adjoint 𝑇^′∶𝐹^′→𝐸^′ which is one.

As we have already done for Dunford-Pettis operators [3] and almost Dunford-Pettis operators [6], one of the aims of this paper is to characterize Banach lattices for which each weak Dunford-Pettis operator has an adjoint which is weak Dunford-Pettis.

We refer the reader to [5] for unexplained terminologies on Banach lattice theory and positive operators.

2. Some Preliminaries

Let us recall that an operator 𝑇 from a Banach lattice 𝐸 into a Banach space 𝑋 is said to be AM-compact if it carries each order-bounded subset of 𝐸 onto a relatively compact set of 𝑋. In [7], we used this class of operators to introduce Banach lattices which satisfy the AM-compactness property. In fact, a Banach lattice 𝐸 is said to have the AM-compactness property if every weakly compact operator defined on 𝐸, and taking values in a Banach space 𝑋, is AM-compact. For an example, the Banach lattice 𝐿²[0,1] does not have the AM-compactness property, but 𝑙¹ has the AM-compactness property.

It follows from [7, Proposition 3.1] that a Banach lattice 𝐸 has the AM-compactness property if and only if for every weakly null sequence (𝑓_𝑛) of 𝐸^′, we have |𝑓_𝑛|→0 for 𝜎(𝐸^′,𝐸).

On the other hand, if 𝐸 is a Banach lattice, then(1) the lattice operations in the topological dual 𝐸^′ are called sequentially continuous if the sequence (|𝑓_𝑛|) converges to 0 in 𝜎(𝐸^′,𝐸^′′) whenever the sequence (𝑓_𝑛) converges to 0 in 𝜎(𝐸^′,𝐸^′′);(2) the lattice operations in 𝐸^′ are called weak* sequentially continuous if the sequence (|𝑓_𝑛|) converges to 0 in the weak^∗ topology 𝜎(𝐸^′,𝐸) whenever the sequence (𝑓_𝑛) converges to 0 in 𝜎(𝐸^′,𝐸).

A Banach space (resp., Banach lattice) 𝐸 has the Dunford-Pettis (resp., weak Dunford-Pettis) property if every weakly compact operator 𝑇 defined on 𝐸 (and taking values in a Banach space 𝐹) is Dunford-Pettis (resp., almost Dunford-Pettis, i.e., the sequence (‖𝑇(𝑥_𝑛)‖) converges to 0 for every weakly null sequence (𝑥_𝑛) consisting of pairwise disjoint elements in 𝐸).

We need to recall, from [7], the following sufficient conditions for which a Banach lattice has the AM-compactness property.

Theorem 2.1 (see [7]). Let 𝐸 be a Banach lattice. Then 𝐸 has the AM-compactness property if one of the following assertions is valid:(1)the norm of 𝐸 is order continuous and 𝐸 has the Dunford-Pettis property,(2)the topological dual 𝐸^′ is discrete,(3)the lattice operations in 𝐸^′ are weakly sequentially continuous,(4)the lattice operations in 𝐸^′ are weak* sequentially continuous.

Remarks 2.2. There exists a Banach lattice 𝐸 such that(1)the norm of 𝐸^′ is order continuous but 𝐸 does not have the AM-compactness property nor the weak Dunford-Pettis property. In fact, consider 𝐸=𝐿²[0,1], the norm of 𝐸^′=𝐿²[0,1], is order continuous but 𝐿²[0,1] does not have the AM-compactness property nor the weak Dunford-Pettis property;(2)the norm of 𝐸^′ is not order continuous, but 𝐸 has the AM-compactness property or the weak Dunford-Pettis property. In fact, consider 𝐸=𝑙¹, the norm of 𝐸^′=𝑙^∞, is not order continuous but 𝑙¹ has the AM-compactness property and the weak Dunford-Pettis property; (3)𝐸 has the AM-compact property but not the weak Dunford-Pettis property. In fact, consider 𝐸=𝑙², it has the AM-compactness property but not the weak Dunford-Pettis property;(4)𝐸 has the weak Dunford-Pettis property but not the AM-compactness property. In fact, consider 𝐸=𝑙^∞, it has the weak Dunford-Pettis property but not the AM-compactness property;(5)the norms of 𝐸 and 𝐸^′ are order continuous, but 𝐸 does not have the Dunford-Pettis property. In fact, consider 𝐸=𝑙², the norms of 𝐸=𝑙² and 𝐸^′=𝑙², are order continuous but 𝑙² does not have the Dunford-Pettis property;(6)the norms of 𝐸 and 𝐸^′ are not order continuous, but 𝐸 has the Dunford-Pettis property. In fact, consider 𝐸=𝑙¹⊕𝑙^∞, the norms of 𝐸=𝑙¹⊕𝑙^∞ and 𝐸^′=𝑙^∞⊕(𝑙^∞)^′, are not order continuous but 𝑙¹⊕𝑙^∞ has the Dunford-Pettis property;(7)the topological dual 𝐸^′ is discrete with an order continuous norm, and 𝐸 does not have the weak Dunford-Pettis property. In fact, consider 𝐸=𝑙², the topological dual 𝐸^′=𝑙², is discrete with an order continuous norm and 𝑙² does not have the weak Dunford-Pettis property;(8)the topological dual 𝐸^′ is not discrete and its norm is not order continuous, but it has the weak Dunford-Pettis property. In fact, consider 𝐸=(𝑙^∞)^′, the topological dual 𝐸^′=(𝑙^∞)^′′, is not discrete and its norm is not order continuous but it has the weak Dunford-Pettis property.

A Banach space 𝐸 is said to have the Schur property if every sequence in 𝐸 weakly convergent to zero is norm convergent to zero. For an example, the Banach space 𝑙¹ has the Schur property.

Note that the Schur property implies the Dunford-Pettis property, and hence the weak Dunford-Pettis property, but the weak Dunford-Pettis property does not imply the Schur property. In fact, the Banach space 𝑐₀ has the weak Dunford-Pettis property (because it has the Dunford-Pettis property), but it does not have the Schur property.

The following result gives some sufficient conditions for which the topological dual, of a Banach lattice, has the Schur property.

Theorem 2.3. Let 𝐸 be a Banach lattice. Then 𝐸^′ has the Schur property if one of the following assertions is valid: (1)the norm of 𝐸^′ is order continuous, 𝐸 has the AM-compactness property and the weak Dunford-Pettis property,(2)the norms of 𝐸 and 𝐸^′ are order continuous and 𝐸 has the Dunford-Pettis property,(3)the topological dual 𝐸^′ is discrete with an order continuous norm and 𝐸 has the weak Dunford-Pettis property.

Proof. (1) Let (𝑓_𝑛)⊂𝐸^′ be a sequence such that 𝑓_𝑛→0 in 𝜎(𝐸^′,𝐸^′′). Since 𝐸 has the AM-compactness property, then |𝑓_𝑛|→0 in 𝜎(𝐸^′,𝐸) (Proposition 3.1 of [7]).
Now, by Corollary 2.7 of Dodds and Fremlin [8], to show that ‖𝑓_𝑛‖→0, it suffices to prove that 𝑓_𝑛(𝑥_𝑛)→0 for every norm-bounded disjoint sequence (𝑥_𝑛)⊂𝐸⁺. To this end, let (𝑥_𝑛) be a such sequence of 𝐸⁺. Since the norm of 𝐸^′ is order continuous, it follows from Corollary 2.9 of Dodds and Fremlin [8] that 𝑥_𝑛→0 in 𝜎(𝐸,𝐸^′). And as 𝐸 has the weak Dunford-Pettis property, we obtain 𝑓_𝑛(𝑥_𝑛)→0. This proves that 𝐸^′ has the Schur property.
For (2) and (3), it follows from Theorem 2.1 that 𝐸 has the AM-compactness property. Finally, assertion (1) of the present theorem ends the proof.

Remarks 2.4. (1) There exists a Banach lattice 𝐹 which has the AM-compactness property but its topological dual 𝐹^′ does not have the Schur property. In fact, consider 𝐹=𝑙¹, it has the AM-compactness property but 𝐹^′=𝑙^∞ does not have the Schur property.
(2) If the topological dual 𝐹^′, of a Banach lattice 𝐹, has the Schur property, then 𝐹^′ is discrete, and hence 𝐹 has the AM-compact property (see Theorem 2.1).

3. Duality Property for Weak Dunford-Pettis Operators

Now, we study the duality property of weak Dunford-Pettis operators. Our first result proves that each operator is weak Dunford-Pettis whenever its adjoint is one.

Theorem 3.1. Let 𝐸 and 𝐹 be two Banach spaces, and let 𝑇 be an operator from 𝐸 into 𝐹. If the adjoint 𝑇^′ is weak Dunford-Pettis from 𝐹^′ into 𝐸^′, then 𝑇 is weak Dunford-Pettis.

Proof. Let (𝑥_𝑛) (resp., (𝑦^′_𝑛)) be a sequence of 𝐸 (resp., of 𝐹^′) such that 𝑥_𝑛→0 in 𝜎(𝐸,𝐸^′) (resp., 𝑦^′_𝑛→0 in 𝜎(𝐹^′,𝐹^′′)). We have to prove that 𝑦^′_𝑛(𝑇(𝑥_𝑛))→0. For this, let 𝜏∶𝐸→𝐸^′′ be the canonical injection of 𝐸 into its topological bidual 𝐸^′′. Since 𝜏 is continuous for the topologies 𝜎(𝐸,𝐸^′) and 𝜎(𝐸^′′,𝐸^′′′), we obtain 𝜏(𝑥_𝑛)→0 for 𝜎(𝐸^′′,𝐸^′′′).
Now, as 𝑦^′_𝑛→0 in 𝜎(𝐹^′,𝐹^′′) and the adjoint 𝑇^′ is weak Dunford-Pettis from 𝐹^′ into 𝐸^′, we deduce that 𝜏(𝑥)(𝑇^′(𝑦^′_𝑛))→0. But we know that 𝜏𝑥_𝑛𝑇^′𝑦^′_𝑛=𝑇^′𝑦^′_𝑛𝑥_𝑛=𝑦^′_𝑛𝑇𝑥_𝑛foreach𝑛. (3.1) Hence 𝑦^′_𝑛(𝑇(𝑥_𝑛))→0, and this ends the proof.

Let us recall from [5] that a norm-bounded subset 𝐴 of a Banach space 𝑋 is said to be Dunford-Pettis whenever every weakly compact operator from 𝑋 to an arbitrary Banach space 𝑌 carries 𝐴 to a norm relatively compact set of 𝑌. This is equivalent to saying that 𝐴 is Dunford-Pettis if and only if every weakly null sequence (𝑓_𝑛) of 𝑋^′ converges uniformly to zero on the set 𝐴, that is, sup_𝑥∈𝐀|𝑓_𝑛(𝑥)|→0 (see Theorem 5.98 of [5]).

Now, we give some sufficient conditions for which each positive weak Dunford-Pettis operator has an adjoint which is Dunford-Pettis.

Theorem 3.2. Let 𝐸 and 𝐹 be two Banach lattices. Then each positive weak Dunford-Pettis operator 𝑇∶𝐸→𝐹 has an adjoint 𝑇^′∶𝐹^′→𝐸^′ which is Dunford-Pettis (and then weak Dunford-Pettis) if one of the following assertions is valid: (1)the norm of 𝐸^′ is order continuous and 𝐸 has the AM-compactness property,(2)the norm of 𝐸^′ is order continuous and 𝐹 has the AM-compactness property,(3)the norms of 𝐸 and 𝐸^′ are order continuous,(4)𝐹^′ has the Schur property.

Proof. For (1), (2), and (3), let 𝑇∶𝐸→𝐹 be a positive weak Dunford-Pettis operator and let (𝑓_𝑛)⊂𝐹^′ be a sequence such that 𝑓_𝑛→0 in 𝜎(𝐹^′,𝐹^′′). In the three cases we have |𝑇^′(𝑓_𝑛)|→0 in 𝜎(𝐸^′,𝐸), in fact, consider the following.(1)As 𝑇^′(𝑓_𝑛)→0 in 𝜎(𝐸^′,𝐸^′′) and 𝐸 has the AM-compactness property, then |𝑇^′(𝑓_𝑛)|→0 for 𝜎(𝐸^′,𝐸).(2)Since 𝑓_𝑛→0 in 𝜎(𝐹^′,𝐹^′′) and 𝐹 has the AM-compactness property, then |𝑓_𝑛|→0 in 𝜎(𝐹^′,𝐹). Hence, 𝑇^′(|𝑓_𝑛|)→0 in 𝜎(𝐸^′,𝐸). Now, from |𝑇^′(𝑓_𝑛)|≤𝑇^′(|𝑓_𝑛|) for each 𝑛, we conclude that |𝑇^′(𝑓_𝑛)|→0 in 𝜎(𝐸^′,𝐸).(3)Since the norm of 𝐸 is order continuous, [−𝑥,𝑥] is weakly compact for each 𝑥∈𝐸⁺. As 𝑇 is weak Dunford-Pettis, we conclude that 𝑇([−𝑥,𝑥]) is a Dunford-Pettis set, and then for each 𝑥∈𝐸⁺, sup_{𝑦∈𝑇([−𝑥,𝑥])}|𝑓_𝑛(𝑦)|→0. Now, from sup_{𝑦∈𝑇([−𝑥,𝑥])}|𝑓_𝑛(𝑦)|=|𝑇^′(𝑓_𝑛)|(𝑥) for each 𝑛, we obtain |𝑇^′(𝑓_𝑛)|(𝑥)→0 for each 𝑥∈𝐸⁺, and hence |𝑇^′(𝑓_𝑛)|→0 in 𝜎(𝐸^′,𝐸).On the other hand, by Corollary 2.7 of Dodds and Fremlin [8], to prove that ‖𝑇^′(𝑓_𝑛)‖→0, it suffices to show that [𝑇^′(𝑓_𝑛)](𝑥_𝑛)→0 for every norm-bounded disjoint sequence (𝑥_𝑛)⊂𝐸⁺. To this end, let (𝑥_𝑛) be a norm-bounded disjoint sequence of 𝐸⁺. Since the norm of 𝐸^′ is order continuous, it follows from Corollary 2.9 of Dodds and Fremlin [8] that 𝑥_𝑛→0 in 𝜎(𝐸,𝐸^′). Hence, as 𝑇 is a weak Dunford-Pettis operator, we obtain 𝑓_𝑛(𝑇(𝑥_𝑛))→0. And from 𝑇^′𝑓_𝑛𝑥_𝑛=𝑓_𝑛𝑇𝑥_𝑛foreach𝑛, (3.2) we derive that [𝑇^′(𝑓_𝑛)](𝑥_𝑛)→0, and hence 𝑇^′ is Dunford-Pettis.(4)In this case, each operator 𝑇∶𝐸→𝐹 has an adjoint 𝑇^′∶𝐹^′→𝐸^′ which is Dunford-Pettis.

Remarks 3.3. There exist Banach lattices 𝐸 and 𝐹 and a weakly Dunford-Pettis operator 𝑇 from 𝐸 into 𝐹 such that the adjoint 𝑇^′ is not Dunford-Pettis in the following situations:(1)if the topological dual 𝐸^′ has an order continuous norm. In fact, if 𝐸=𝐹=𝑙^∞, we note that 𝐸^′=(𝑙^∞)^′ has an order continuous norm and its identity operator Id_𝑙^∞∶𝑙^∞→𝑙^∞ is weak Dunford-Pettis but its adjoint Id_(𝑙∞)′∶(𝑙^∞)^′→(𝑙^∞)^′ is not Dunford-Pettis. However, it is weak Dunford-Pettis because (𝑙^∞)^′ has the Dunford-Pettis property,(2)if 𝐸 has the AM-compactness property (resp., 𝐹 has the AM-compactness property, 𝐸 has an order continuous norm). In fact, if 𝐸=𝐹=𝑙¹, we note that 𝑙¹ has the AM-compactness property (resp. its norm is order continuous) and its identity operator Id_𝑙¹∶𝑙¹→𝑙¹ is weak Dunford-Pettis but its adjoint Id_𝑙^∞∶𝑙^∞→𝑙^∞ is not Dunford-Pettis. However, it is weak Dunford-Pettis because 𝑙^∞ has the Dunford-Pettis property.

As a consequence of Theorems 2.1 and 3.2, we obtain the following.

Corollary 3.4. Let 𝐸 and 𝐹 be two Banach lattices. Then each positive weak Dunford-Pettis operator 𝑇∶𝐸→𝐹 has an adjoint 𝑇^′∶𝐹^′→𝐸^′ which is weak Dunford-Pettis if one of the following assertions is valid:(1)the topological dual 𝐸^′ is discrete with an order continuous norm,(2)the norm of 𝐸^′ is order continuous and 𝐹^′ is discrete,(3)the norm of 𝐸^′ is order continuous and the lattice operations in 𝐹^′ are weakly sequentially continuous,(4)the norm of 𝐸^′ is order continuous and the lattice operations in 𝐹^′ are weak^∗ sequentially continuous,(5)the norms of 𝐸^′ and 𝐹 are order continuous and 𝐹 has the Dunford-Pettis property,(6)the norms of 𝐸 and 𝐸^′ are order continuous,(7)𝐸^′ or 𝐹^′has the Dunford-Pettis property.

Proof. For (1), (2), (3), (4), and (5), it follows from Theorem 2.1 that 𝐸 or 𝐹 has the AM-compactness property. Since the norm of 𝐸^′ is order continuous, Theorem 3.2 implies that each positive weak Dunford-Pettis operator 𝑇∶𝐸→𝐹 has an adjoint 𝑇^′∶𝐹^′→𝐸^′ which is Dunford-Pettis (and then weak Dunford-Pettis).
(6) Follows from (3) of Theorem 3.2.
(7) In this case each operator 𝑇∶𝐸→𝐹 has an adjoint 𝑇^′∶𝐹^′→𝐸^′ which is weak Dunford-Pettis.

For the converse of Theorem 3.2, we have the following.

Theorem 3.5. Let 𝐸 and 𝐹 be two Banach lattices. If each positive weak Dunford-Pettis operator 𝑇∶𝐸→𝐹 has an adjoint 𝑇^′∶𝐹^′→𝐸^′ which is Dunford-Pettis, then one of the following assertions is valid: (1)the norm of 𝐸^′ is order continuous, (2)𝐹^′ has the Schur property.

Proof. Assume by way of contradiction that the norm of 𝐸^′ is not order continuous and 𝐹^′ does not have the Schur property. We have to construct a positive weak Dunford-Pettis operator 𝑇∶𝐸→𝐹 such that its adjoint 𝑇^′∶𝐹^′→𝐸^′ is not Dunford-Pettis.
Since the norm of 𝐸^′ is not order continuous, it follows from the proof of Theorem 1 of Wickstead [9] the existence of a sublattice 𝐻 of 𝐸, which is isomorphic to 𝑙¹, and a positive projection 𝑃∶𝐸→𝑙¹.
On the other hand, since 𝐹^′ does not have the Schur property, there exists a weakly null sequence (𝑓_𝑛)⊂𝐹^′ such that ‖𝑓_𝑛‖=1 for all 𝑛. Moreover, there exists a sequence (𝑦_𝑛)⊂𝐹⁺ with ‖𝑦_𝑛‖≤1 and some 𝜀₀>0 such that |𝑓_𝑛(𝑦_𝑛)|≥𝜀₀ for all 𝑛.
Now, we consider the operator 𝑇=𝑆∘𝑃∶𝐸→𝑙¹→𝐹, where 𝑆 is the operator defined by 𝑆∶𝑙¹→𝐹,𝜆_𝑛⟼_𝑛𝜆_𝑛𝑦_𝑛. (3.3) Since 𝑙¹ has the Dunford-Pettis property, the operator 𝑇 is weak Dunford-Pettis. But its adjoint 𝑇^′∶𝐹^′→𝐸^′ is not Dunford-Pettis. Indeed, the sequence (𝑓_𝑛) is weakly null in 𝐹^′. And as the operator 𝑃∶𝐸→𝑙¹ is surjective, there exist 𝛿>0 such that 𝛿⋅𝐵_𝑙¹⊂𝑃(𝐵_𝐸), where 𝐵_𝐻 is the closed unit ball of 𝐻=𝐸 or 𝑙¹. Hence 𝑇^′𝑓_𝑛=sup_{𝑥∈𝐵𝐸}𝑇^′𝑓_𝑛(𝑥)=sup_{𝑥∈𝐵𝐸}𝑓_𝑛(𝑇(𝑥))=sup_{𝑥∈𝐵𝐸}𝑓_𝑛∘𝑆𝑝(𝑥)≥𝛿⋅sup_{𝜆𝑖𝑖∈𝐵𝑙1}𝑓_𝑛∘𝑆𝜆_𝑖≥𝛿⋅𝑓_𝑛∘𝑆𝑒_𝑛≥𝛿⋅𝑓_𝑛𝑦_𝑛>𝛿.𝜀₀, (3.4) where (𝑒_𝑖)^∞_𝑖=1 is the canonical bases of 𝑙¹.
Then ‖𝑇^′(𝑓_𝑛)‖>𝛿⋅𝜀₀ for all 𝑛, and we conclude that 𝑇^′ is not Dunford-Pettis. This presents a contradiction.

Remarks 3.6. Let 𝐸 and 𝐹 be two Banach lattices such that 𝐹^′ does not have the Schur property. If each positive weak Dunford-Pettis operator 𝑇 from 𝐸 into 𝐹 has an adjoint 𝑇^′ from 𝐹^′ into 𝐸^′ which is Dunford-Pettis, then(1)𝐹 does not necessarily have the AM-compactness property. In fact, if we take 𝐸=𝑐₀ and 𝐹=𝑙^∞, we observe that each operator 𝑇 from 𝑐₀ into 𝑙^∞ has an adjoint 𝑇^′ from (𝑙^∞)^′ into 𝑙¹ which is Dunford-Pettis (because 𝑙¹ has the Schur property), but 𝐹=𝑙^∞ does not have the AM-compactness property,(2)the norm of 𝐸 is not necessarily order continuous. In fact, if we take 𝐸=𝑐 and 𝐹=𝑙^∞, we note that each operator 𝑇 from 𝑐 into 𝑙^∞ has an adjoint 𝑇^′ from (𝑙^∞)^′ into 𝑐^′ which is Dunford-Pettis (because 𝑐^′ has the Schur property), but the norm of 𝐸=𝑐 is not order continuous,(3)𝐸 does not necessarily have the AM-compactness property. In fact, if we take 𝐸=𝑙^∞ and 𝐹=(𝑙^∞)^′, we note that each positive weak Dunford-Pettis operator 𝑇 from 𝑙^∞ into (𝑙^∞)^′ has an adjoint 𝑇^′ from (𝑙^∞)^′′ into (𝑙^∞)^′ which is Dunford-Pettis (see assertion 2 of Theorem 3.2), but 𝐸=𝑙^∞ does not have the AM-compactness property.

Whenever 𝐸=𝐹, we obtain the following characterization.

Theorem 3.7. Let 𝐸 be a Dedekind 𝜎-complete Banach lattice. Then the following assertions are equivalent: (1)each positive weak Dunford-Pettis operator 𝑇 from 𝐸 into 𝐸 has an adjoint which is Dunford-Pettis,(2)the norms of 𝐸 and 𝐸^′are order continuous.

Proof. (1)⇒(2).  By Theorem 3.5, the norm of 𝐸^′ is order continuous. We have just to prove that the norm of 𝐸 is order continuous. Assume that the norm of 𝐸 is not order continuous, and since 𝐸 is Dedekind 𝜎-complete, then 𝐸 contains a closed sublattice isomorphic to 𝑙^∞ and there is a positive projection 𝑃∶𝐸→𝑙^∞. Let 𝑖∶𝑙^∞→𝐸 be the canonical injection of 𝑙^∞ into 𝐸. Consider the operator defined by 𝑇=𝑖∘𝑃∶𝐸⟶𝑙^∞⟶𝐸. (3.5) Since 𝑙^∞ has the Dunford-Pettis property, the positive operator 𝑇 is weak Dunford-Pettis. But its adjoint 𝑇^′∶𝐸^′→𝐸^′ is not Dunford-Pettis. If not, the adjoint of the composed operator 𝑃∘𝑇∘𝑖∶𝑙^∞⟶𝐸⟶𝐸⟶𝑙^∞ (3.6) would be Dunford-Pettis. But (𝑃∘𝑇∘𝑖)^′=(Id_𝑙^∞)^′=Id_(𝑙∞)′ is not Dunford-Pettis (because (𝑙^∞)^′ does not have the Schur property). This presents a contradiction, and hence 𝐸 has an order continuous norm.
(2)⇒(1). It follows from (3) of Theorem 3.2.

4. Complements on the Duality of Almost Dunford-Pettis Operators

In [6], we studied the duality for almost Dunford-Pettis operators. In this section we use the AM-compactness property to give some new results.

Let us recall that an operator 𝑇 from a Banach lattice 𝐸 into a Banach space 𝐹 is said to be almost Dunford-Pettis if the sequence (‖𝑇(𝑥_𝑛)‖) converges to 0 for every weakly null sequence (𝑥_𝑛) consisting of pairwise disjoint elements in 𝐸.

Note that the adjoint of a positive almost Dunford-Pettis operator is not necessarily Dunford-Pettis. In fact, the identity operator of the Banach space 𝑙¹ is almost Dunford-Pettis but its adjoint, which is the identity of the Banach space 𝑙^∞, is not Dunford-Pettis.

The following result gives some sufficient conditions for which each positive almost Dunford-Pettis operator has an adjoint which is Dunford-Pettis.

Theorem 4.1. Let 𝐸 and 𝐹 be two Banach lattices. Then each positive almost Dunford-Pettis operator 𝑇∶𝐸→𝐹 has an adjoint 𝑇^′∶𝐹^′→𝐸^′ which is Dunford-Pettis if one of the following assertions is valid: (1)the norm of 𝐸^′ is order continuous and 𝐸 has the AM-compactness property,(2)the norm of 𝐸^′ is order continuous and 𝐹 has the AM-compactness property,(3)𝐹^′ has the Schur property.

Proof. Note that for (1) and (2), the proof is the same as (1) and (2) of Theorem 3.2. In fact, let 𝑇∶𝐸→𝐹 be a positive almost Dunford-Pettis operator, and let (𝑓_𝑛)⊂𝐹 be a sequence such that 𝑓_𝑛→0 in 𝜎(𝐹^′,𝐹^′′). By the uniform boundedness Theorem, there exists some 𝛼>0 such that ‖𝑓_𝑛‖≤𝛼 for all 𝑛. In the two cases we have |𝑇^′(𝑓_𝑛)|→0 in 𝜎(𝐸^′,𝐸). In fact, consider the following.(1)As 𝑇^′(𝑓_𝑛)→0 in 𝜎(𝐸^′,𝐸^′′) and 𝐸 has the AM-compactness property, then |𝑇^′(𝑓_𝑛)|→0 in 𝜎(𝐸^′,𝐸).(2)As 𝑓_𝑛→0 in 𝜎(𝐹^′,𝐹^′′), and since 𝐹 has the AM-compactness property, then |𝑓_𝑛|→0 in 𝜎(𝐹^′,𝐹). Hence, 𝑇^′(|𝑓_𝑛|)→0 in 𝜎(𝐸^′,𝐸) and from |𝑇^′(𝑓_𝑛)|≤𝑇^′(|𝑓_𝑛|) for each 𝑛, we conclude that |𝑇^′(𝑓_𝑛)|→0 in 𝜎(𝐸^′,𝐸).Now to prove that ‖𝑇^′(𝑓_𝑛)‖_𝐸′→0, it suffices to show that [𝑇^′(𝑓_𝑛)](𝑥_𝑛)→0 in every norm-bounded disjoint sequence (𝑥_𝑛)⊂𝐸⁺ (Corollary 2.7 of Dodds and Fremlin [8]). To this end, let (𝑥_𝑛) be a norm-bounded disjoint sequence of 𝐸⁺.Since the norm of 𝐸^′ is order continuous, it follows from Corollary 2.9 of Dodds and Fremlin [8] that 𝑥_𝑛→0 in 𝜎(𝐸,𝐸^′). Hence, as 𝑇 is almost Dunford-Pettis operator, we obtain ‖𝑇(𝑥_𝑛)‖_𝐹→0. Now, from 𝑇^′𝑓_𝑛𝑥_𝑛=𝑓_𝑛𝑇𝑥_𝑛≤𝛼⋅𝑇𝑥_𝑛𝐹foreach𝑛, (4.1) we see that[𝑇^′(𝑓_𝑛)](𝑥_𝑛)→0, and hence 𝑇^′ is Dunford-Pettis.(3)In this case each operator 𝑇∶𝐸→𝐹 has an adjoint 𝑇^′∶𝐹^′→𝐸^′ which is Dunford-Pettis.

Remarks 4.2. Let 𝐸 and 𝐹 be two Banach lattices, and let 𝑇 be an operator from 𝐸 into 𝐹. Then the adjoint 𝑇^′ is not necessarily Dunford-Pettis whenever 𝑇 is almost Dunford-Pettis in the following situations.(1)If the topological dual 𝐸^′ has an order continuous norm. In fact, since the norm of 𝑙^∞ is not order continuous and the Banach lattice (𝑙^∞)^′ is not discrete, it follows from Theorem 1 of Wickstead [9] the existence of two positive operators 𝑆₁,𝑆₂∶𝑙^∞→𝑙^∞ such that 0≤𝑆₁≤𝑆₂, 𝑆₂ is compact, and 𝑆₁ is not compact. Now, as (𝑙^∞)^′ has an order continuous norm, Theorem 5.31 of Aliprantis and Burkinshaw [5] implies that 𝑆₁ is weakly compact. So, by Theorem 5.44 of Aliprantis and Burkinshaw [5], there exist a reflexive Banach lattice 𝐺, lattice homomorphism 𝑄∶𝑙^∞→𝐺, and a positive operator 𝑅∶𝐺→𝑙^∞ such that 𝑆₁=𝑅∘𝑄. We note that 𝑄 is not compact (because 𝑆₁ is not one). On the other hand, if we take 𝐸=𝑙^∞, 𝐹=𝐺, and 𝑇=𝑄, then 𝑇∶𝑙^∞→𝐺 is a weakly compact operator (because 𝐺 is reflexive), and hence 𝑇 is Dunford-Pettis (𝑙^∞ has the Dunford-Pettis property) and then 𝑇 is almost Dunford-Pettis. But its adjoint 𝑇^′∶𝐺^′→(𝑙^∞)^′ is not Dunford-Pettis (if not, since 𝐺^′ is reflexive, 𝑇^′ would be compact and so 𝑇 is compact, which is a contradiction). However, the norm of 𝐸^′=(𝑙^∞)^′ is order continuous. (2)If 𝐸 has the AM-compactness property. In fact, if we take 𝐸=𝐹=𝑙¹, we note that 𝐸=𝑙¹ has the AM-compactness property and its identity operator Id_𝑙¹∶𝑙¹→𝑙¹ is almost Dunford-Pettis but the adjoint Id_𝑙^∞∶𝑙^∞→𝑙^∞ is not Dunford-Pettis. (3)If 𝐹 has the AM-compactness property. In fact, if we take 𝐸=𝐹=𝑙¹, we observe that 𝐹=𝑙¹ has the AM-compactness property and its identity operator Id_𝑙¹∶𝑙¹→𝑙¹ is almost Dunford-Pettis, but the adjoint Id_𝑙^∞∶𝑙^∞→𝑙^∞ is not Dunford-Pettis.

For the converse of Theorem 4.1, we obtain the following.

Theorem 4.3. Let 𝐸 and 𝐹 be two Banach lattices. If each positive almost Dunford-Pettis operator 𝑇∶𝐸→𝐹 has an adjoint 𝑇^′∶𝐹^′→𝐸^′ which is Dunford-Pettis, then one of the following assertions is valid: (1)the norm of 𝐸^′ is order continuou,(2)𝐹^′ has the Schur property.

Proof. The proof is the same as that of Theorem 3.5 if we observe that the operator 𝑇 in the proof of Theorem 3.5 is almost Dunford-Pettis (because 𝑇 admits a factorization through the Banach lattice 𝑙¹, which has the Schur property).

Remarks 4.4. Let 𝐸 and 𝐹 be two Banach lattices such that 𝐹^′ does not have the Schur property. If each positive almost Dunford-Pettis operator 𝑇 from 𝐸 into 𝐹 has an adjoint 𝑇^′ from 𝐹^′ into 𝐸^′ which is Dunford-Pettis, then(1)𝐸 does not necessarily have the AM-compactness property. In fact, if we take 𝐸=𝑙^∞ and 𝐹=(𝑙^∞)^′, we note that each positive almost Dunford-Pettis operator 𝑇 from 𝑙^∞ into (𝑙^∞)^′ has an adjoint 𝑇^′ from (𝑙^∞)^′′ into (𝑙^∞)^′ which is Dunford-Pettis (see assertion 2 of Theorem 4.1), but 𝐸=𝑙^∞ does not have the AM-compactness property,(2)𝐹 does not necessarily have the AM-compactness property. In fact, if we take 𝐸=𝑐₀ and 𝐹=𝑙^∞, we observe that each operator 𝑇 from 𝑐₀ into 𝑙^∞ has an adjoint 𝑇^′ from (𝑙^∞)^′ into 𝑙¹ which is Dunford-Pettis (because 𝑙¹ has the Schur property), but 𝐹=𝑙^∞ does not have the AM-compactness property.

Finally, we note that there exists a positive weak Dunford-Pettis (resp., Dunford-Pettis) operator 𝑇∶𝐸→𝐹 whose adjoint 𝑇^′∶𝐹^′→𝐸^′ is not almost Dunford-Pettis. In fact, the identity operator of the Banach lattice 𝑙¹ is weak Dunford-Pettis (resp., Dunford-Pettis) operator but its adjoint, which is the identity of the Banach lattice 𝑙^∞, is not almost Dunford-Pettis.

Now, we give a characterization on the duality between weak Dunford-Pettis operators and almost Dunford-Pettis operators.

Theorem 4.5. Let 𝐸 and 𝐹 be two Banach lattices. Then the following assertions are equivalent: (1)each positive weak Dunford-Pettis (resp., Dunford-Pettis, almost Dunford-Pettis) operator 𝑇∶𝐸→𝐹 has an adjoint 𝑇^′∶𝐹^′→𝐸^′ which is almost Dunford-Pettis,(2)one of the following assertions is valid:(a)the norm of 𝐸^′ is order continuous,(b)𝐹^′ has the positive Schur property.

Proof. (1)⇒(2).  Assume by way of contradiction that the norm of 𝐸^′ is not order continuous and 𝐹^′ does not have the positive Schur property. We have to construct a positive weak Dunford-Pettis (resp., Dunford-Pettis, almost Dunford-Pettis) operator 𝑇∶𝐸→𝐹 such that its adjoint 𝑇^′∶𝐹^′→𝐸^′ is not almost Dunford-Pettis.
Since the norm of 𝐸^′ is not order continuous, it follows from the proof of Theorem 1 of Wickstead [9] the existence of a sublattice 𝐻 of 𝐸, which is isomorphic to 𝑙¹, and a positive projection 𝑃∶𝐸→𝑙¹.
On the other hand, since 𝐹^′ does not have the positive Schur property, it follows from Theorem 3.1 of [10] the existence of a disjoint weakly null sequence (𝑓_𝑛)⊂(𝐹^′)⁺ such that (𝑓_𝑛) does not converge to zero for the norm. Moreover, there exists a sequence (𝑦_𝑛)⊂𝐹⁺ with ‖𝑦_𝑛‖≤1, and some 𝜀>0, a subsequence (𝑔_𝑛) of (𝑓_𝑛) such that 𝑔_𝑛(𝑦_𝑛)≥𝜀 for all 𝑛.
Now, we consider the composed operator 𝑇=𝑆∘𝑃∶𝐸⟶𝑙¹⟶𝐹, (4.2) where 𝑆 is defined by 𝑆∶𝑙¹→𝐹,𝜆_𝑛⟼_𝑛𝜆_𝑛𝑦_𝑛. (4.3)
Since 𝑙¹ has the Schur property, the operator 𝑇 is weak Dunford-Pettis (resp. Dunford-Pettis, almost Dunford-Pettis), but its adjoint 𝑇^′∶𝐹^′→𝐸^′ is not almost Dunford-Pettis. Indeed, (𝑔_𝑛) is a disjoint weakly null sequence in 𝐹^′. And since the operator 𝑃∶𝐸→𝑙¹ is surjective, there exist 𝛿>0 such that 𝛿⋅𝐵_𝑙¹⊂𝑃(𝐵_𝐸) where 𝐵_𝐻 is the closed unit ball of 𝐻=𝐸,𝑙¹. Hence 𝑇^′𝑔_𝑛=sup_{𝑥∈𝐵𝐸}𝑇^′𝑔_𝑛(𝑥)=sup_{𝑥∈𝐵𝐸}𝑔_𝑛(𝑇(𝑥))=sup_{𝑥∈𝐵𝐸}𝑔_𝑛∘𝑆𝑝(𝑥)≥𝛿⋅sup_{𝜆𝑖𝑖∈𝐵𝑙1}𝑔_𝑛∘𝑆𝜆_𝑖≥𝛿⋅𝑔_𝑛∘𝑆𝑒_𝑛≥𝛿⋅𝑔_𝑛𝑦_𝑛>𝛿.𝜀₀, (4.4) where (𝑒_𝑖)^∞_𝑖=1 is the canonical bases of 𝑙¹.
Then ‖𝑇^′(𝑔_𝑛)‖>𝛿⋅𝜀₀ for every 𝑛, and we conclude that 𝑇^′ is not almost Dunford-Pettis. This presents a contradiction.
(2), (a)⇒(1). Let (𝑓_𝑛) be a disjoint sequence of 𝐹^′ such that 𝑓_𝑛→0 in 𝜎(𝐹^′,𝐹^′′). We have to prove that (𝑇^′(𝑓_𝑛)) converges to 0 for the norm of 𝐸^′. By using Corollary 2.7 of Dodds-Fremlin [8], it suffices to prove that |𝑇^′(𝑓_𝑛)|→0 in 𝜎(𝐸^′,𝐸) and [𝑇^′(𝑓_𝑛)](𝑥_𝑛)→0 for every norm-bounded disjoint sequence (𝑥_𝑛)⊂𝐸⁺. In fact, as (𝑓_𝑛) is a weakly null sequence with pairwise disjoint terms, it follows from Remark 1 of Wnuk [11] that |𝑓_𝑛|→0 in 𝜎(𝐹^′,𝐹^′′), and then 𝑇^′(|𝑓_𝑛|)→0 for 𝜎(𝐸^′,𝐸^′′). Now, since |𝑇^′(𝑓_𝑛)|≤𝑇^′(|𝑓_𝑛|) for each 𝑛, then |𝑇^′(𝑓_𝑛)|→0 in 𝜎(𝐸^′,𝐸^′′), and hence |𝑇^′(𝑓_𝑛)|→0 in 𝜎(𝐸^′,𝐸).
On the other hand, since the norm of 𝐸^′ is order continuous, it follows from Corollary 2.9 of Dodds and Fremlin [8] that 𝑥_𝑛→0 in 𝜎(𝐸,𝐸^′). Hence, as 𝑇 is a weak Dunford-Pettis (resp., Dunford-Pettis, almost Dunford-Pettis) operator, we obtain [𝑇^′(𝑓_𝑛)](𝑥_𝑛)=𝑓_𝑛(𝑇(𝑥_𝑛))→0, and this proves that 𝑇^′ is almost Dunford-Pettis.
(2), (b)⇒(1). Obvious.