Research Article | Open Access

Belmesnaoui Aqzzouz, Khalid Bouras, Mohammed Moussa, "Duality Property for Positive Weak Dunford-Pettis Operators", *International Journal of Mathematics and Mathematical Sciences*, vol. 2011, Article ID 609287, 12 pages, 2011. https://doi.org/10.1155/2011/609287

# Duality Property for Positive Weak Dunford-Pettis Operators

**Academic Editor:**Yuri Latushkin

#### 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 from a Banach space into another is called Dunford-Pettis if it carries weakly compact subsets of onto compact subsets of . The operator is said to be weak Dunford-Pettis if converges to 0 whenever converges weakly to 0 in and converges weakly to 0 in .

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 has the Schur property, its identity operator is Dunford-Pettis and then weak Dunford-Pettis, but its adjoint , which is the identity operator of the Banach space , is not weak Dunford-Pettis (because the Banach space 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 and are two Banach spaces such that is reflexive, then the class of weak Dunford-Pettis operators from into coincides with that of Dunford-Pettis operators from into , and therefore some results of [5] can be applied here to give some answers to our duality problem.

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

Also, if and are two Banach spaces such that or has the Dunford-Pettis property, then each operator from into is weak Dunford-Pettis, and hence each weak Dunford-Pettis has an adjoint 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 from a Banach lattice into a Banach space is said to be AM-compact if it carries each order-bounded subset of onto a relatively compact set of . In [7], we used this class of operators to introduce Banach lattices which satisfy the AM-compactness property. In fact, a Banach lattice is said to have the AM-compactness property if every weakly compact operator defined on , and taking values in a Banach space , is AM-compact. For an example, the Banach lattice does not have the AM-compactness property, but has the AM-compactness property.

It follows from [7, Proposition 3.1] that a Banach lattice has the AM-compactness property if and only if for every weakly null sequence of , we have for .

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

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

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

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

A Banach space is said to have the Schur property if every sequence in weakly convergent to zero is norm convergent to zero. For an example, the Banach space 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 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 be a Banach lattice. Then has the Schur property if one of the following assertions is valid: *(1)*the norm of is order continuous, has the AM-compactness property and the weak Dunford-Pettis property,*(2)*the norms of and are order continuous and has the Dunford-Pettis property,*(3)*the topological dual is discrete with an order continuous norm and has the weak Dunford-Pettis property.*

*Proof. *(1) Let be a sequence such that in . Since has the AM-compactness property, then in (Proposition 3.1 of [7]).

Now, by Corollary 2.7 of Dodds and Fremlin [8], to show that , it suffices to prove that for every norm-bounded disjoint sequence . To this end, let be a such sequence of . Since the norm of is order continuous, it follows from Corollary 2.9 of Dodds and Fremlin [8] that in . And as has the weak Dunford-Pettis property, we obtain . This proves that has the Schur property.

For (2) and (3), it follows from Theorem 2.1 that has the AM-compactness property. Finally, assertion (1) of the present theorem ends the proof.

*Remarks 2.4. *(1) There exists a Banach lattice which has the AM-compactness property but its topological dual does not have the Schur property. In fact, consider , it has the AM-compactness property but does not have the Schur property.

(2) If the topological dual , of a Banach lattice , has the Schur property, then is discrete, and hence 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 and be two Banach spaces, and let be an operator from into . If the adjoint is weak Dunford-Pettis from into , then is weak Dunford-Pettis.*

*Proof. *Let (resp., ) be a sequence of (resp., of ) such that in (resp., in ). We have to prove that . For this, let be the canonical injection of into its topological bidual . Since is continuous for the topologies and , we obtain for .

Now, as in and the adjoint is weak Dunford-Pettis from into , we deduce that . But we know that
Hence , and this ends the proof.

Let us recall from [5] that a norm-bounded subset of a Banach space is said to be Dunford-Pettis whenever every weakly compact operator from to an arbitrary Banach space carries to a norm relatively compact set of . This is equivalent to saying that is Dunford-Pettis if and only if every weakly null sequence of converges uniformly to zero on the set , that is, (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 and be two Banach lattices. Then each positive weak Dunford-Pettis operator has an adjoint which is Dunford-Pettis (and then weak Dunford-Pettis) if one of the following assertions is valid: *(1)*the norm of is order continuous and has the AM-compactness property,*(2)*the norm of is order continuous and has the AM-compactness property,*(3)*the norms of and are order continuous,*(4)* has the Schur property.*

*Proof. *For (1), (2), and (3), let be a positive weak Dunford-Pettis operator and let be a sequence such that in . In the three cases we have in , in fact, consider the following.(1)As in and has the AM-compactness property, then for .(2)Since in and has the AM-compactness property, then in . Hence, in . Now, from for each , we conclude that in .(3)Since the norm of is order continuous, is weakly compact for each . As is weak Dunford-Pettis, we conclude that is a Dunford-Pettis set, and then for each , . Now, from for each , we obtain for each , and hence in .On the other hand, by Corollary 2.7 of Dodds and Fremlin [8], to prove that , it suffices to show that for every norm-bounded disjoint sequence . To this end, let be a norm-bounded disjoint sequence of . Since the norm of is order continuous, it follows from Corollary 2.9 of Dodds and Fremlin [8] that in . Hence, as is a weak Dunford-Pettis operator, we obtain . And from
we derive that , and hence is Dunford-Pettis.(4)In this case, each operator has an adjoint which is Dunford-Pettis.

*Remarks 3.3. *There exist Banach lattices and and a weakly Dunford-Pettis operator from into such that the adjoint is not Dunford-Pettis in the following situations:(1)if the topological dual has an order continuous norm. In fact, if , we note that has an order continuous norm and its identity operator is weak Dunford-Pettis but its adjoint is not Dunford-Pettis. However, it is weak Dunford-Pettis because has the Dunford-Pettis property,(2)if has the AM-compactness property (resp., has the AM-compactness property, has an order continuous norm). In fact, if , we note that has the AM-compactness property (resp. its norm is order continuous) and its identity operator is weak Dunford-Pettis but its adjoint is not Dunford-Pettis. However, it is weak Dunford-Pettis because has the Dunford-Pettis property.

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

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

*Proof. *For (1), (2), (3), (4), and (5), it follows from Theorem 2.1 that or has the AM-compactness property. Since the norm of is order continuous, Theorem 3.2 implies that each positive weak Dunford-Pettis operator has an adjoint which is Dunford-Pettis (and then weak Dunford-Pettis).

(6) Follows from (3) of Theorem 3.2.

(7) In this case each operator has an adjoint which is weak Dunford-Pettis.

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

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

*Proof. *Assume by way of contradiction that the norm of is not order continuous and does not have the Schur property. We have to construct a positive weak Dunford-Pettis operator such that its adjoint is not Dunford-Pettis.

Since the norm of is not order continuous, it follows from the proof of Theorem 1 of Wickstead [9] the existence of a sublattice of , which is isomorphic to , and a positive projection .

On the other hand, since does not have the Schur property, there exists a weakly null sequence such that for all . Moreover, there exists a sequence with and some such that for all .

Now, we consider the operator , where is the operator defined by
Since has the Dunford-Pettis property, the operator is weak Dunford-Pettis. But its adjoint is not Dunford-Pettis. Indeed, the sequence is weakly null in . And as the operator is surjective, there exist such that , where is the closed unit ball of or . Hence
where is the canonical bases of .

Then for all , and we conclude that is not Dunford-Pettis. This presents a contradiction.

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

Whenever , we obtain the following characterization.

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

*Proof. *(1)*⇒*(2). By Theorem 3.5, the norm of is order continuous. We have just to prove that the norm of is order continuous. Assume that the norm of is not order continuous, and since is Dedekind -complete, then contains a closed sublattice isomorphic to and there is a positive projection . Let be the canonical injection of into . Consider the operator defined by
Since has the Dunford-Pettis property, the positive operator is weak Dunford-Pettis. But its adjoint is not Dunford-Pettis. If not, the adjoint of the composed operator
would be Dunford-Pettis. But is not Dunford-Pettis (because does not have the Schur property). This presents a contradiction, and hence 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 from a Banach lattice into a Banach space is said to be almost Dunford-Pettis if the sequence converges to 0 for every weakly null sequence consisting of pairwise disjoint elements in .

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 is almost Dunford-Pettis but its adjoint, which is the identity of the Banach space , 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 and be two Banach lattices. Then each positive almost Dunford-Pettis operator has an adjoint which is Dunford-Pettis if one of the following assertions is valid: *(1)*the norm of is order continuous and has the AM-compactness property,*(2)*the norm of is order continuous and has the AM-compactness property,*(3)* 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 be a positive almost Dunford-Pettis operator, and let be a sequence such that in . By the uniform boundedness Theorem, there exists some such that for all . In the two cases we have in . In fact, consider the following.(1)As in and has the AM-compactness property, then in .(2)As in , and since has the AM-compactness property, then in . Hence, in and from for each , we conclude that in .Now to prove that , it suffices to show that in every norm-bounded disjoint sequence (Corollary 2.7 of Dodds and Fremlin [8]). To this end, let be a norm-bounded disjoint sequence of .Since the norm of is order continuous, it follows from Corollary 2.9 of Dodds and Fremlin [8] that in . Hence, as is almost Dunford-Pettis operator, we obtain . Now, from
we see that, and hence is Dunford-Pettis.(3)In this case each operator has an adjoint which is Dunford-Pettis.

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

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

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

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

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

Finally, we note that there exists a positive weak Dunford-Pettis (resp., Dunford-Pettis) operator whose adjoint is not almost Dunford-Pettis. In fact, the identity operator of the Banach lattice is weak Dunford-Pettis (resp., Dunford-Pettis) operator but its adjoint, which is the identity of the Banach lattice , 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 and be two Banach lattices. Then the following assertions are equivalent: *(1)*each positive weak Dunford-Pettis (resp., Dunford-Pettis, almost Dunford-Pettis) operator has an adjoint which is almost Dunford-Pettis,*(2)*one of the following assertions is valid:(a) the norm of is order continuous,(b) has the positive Schur property.*

*Proof. *(1)*⇒*(2). Assume by way of contradiction that the norm of is not order continuous and does not have the positive Schur property. We have to construct a positive weak Dunford-Pettis (resp., Dunford-Pettis, almost Dunford-Pettis) operator such that its adjoint is not almost Dunford-Pettis.

Since the norm of is not order continuous, it follows from the proof of Theorem 1 of Wickstead [9] the existence of a sublattice of , which is isomorphic to , and a positive projection .

On the other hand, since does not have the positive Schur property, it follows from Theorem 3.1 of [10] the existence of a disjoint weakly null sequence such that does not converge to zero for the norm. Moreover, there exists a sequence with , and some , a subsequence of such that for all .

Now, we consider the composed operator
where is defined by

Since has the Schur property, the operator is weak Dunford-Pettis (resp. Dunford-Pettis, almost Dunford-Pettis), but its adjoint is not almost Dunford-Pettis. Indeed, is a disjoint weakly null sequence in . And since the operator is surjective, there exist such that where is the closed unit ball of . Hence
where is the canonical bases of .

Then for every , and we conclude that is not almost Dunford-Pettis. This presents a contradiction.

(2), (a)*⇒*(1). Let be a disjoint sequence of such that in . We have to prove that converges to 0 for the norm of . By using Corollary 2.7 of Dodds-Fremlin [8], it suffices to prove that in and for every norm-bounded disjoint sequence . In fact, as is a weakly null sequence with pairwise disjoint terms, it follows from Remark 1 of Wnuk [11] that in , and then for . Now, since for each , then in , and hence in .

On the other hand, since the norm of is order continuous, it follows from Corollary 2.9 of Dodds and Fremlin [8] that in . Hence, as is a weak Dunford-Pettis (resp., Dunford-Pettis, almost Dunford-Pettis) operator, we obtain , and this proves that is almost Dunford-Pettis.

(2), (b)*⇒*(1). Obvious.

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