Duality Property for Positive Weak Dunford-Pettis Operators

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.


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 l 1 l 2 n has the Schur property, its identity operator Id l 1 l 2 n is Dunford-Pettis and then weak Dunford-Pettis, but its adjoint Id l ∞ l 2 n , which is the identity operator of the Banach space l ∞ l 2 n , is not weak Dunford-Pettis because the Banach space l ∞ l 2 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 F is 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.

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 AMcompactness 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 L 2 0, 1 does not have the AM-compactness property, but l 1 has the AM-compactness property.
It follows from 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.property.In fact, consider E l 1 ⊕ l ∞ , the norms of E l 1 ⊕ l ∞ and E l ∞ ⊕ l ∞ , are not order continuous but l 1 ⊕ 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 l 2 , the topological dual E l 2 , is discrete with an order continuous norm and l 2 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 l 1 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 c 0 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.
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 l 1 , 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 .

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 τ x n T y n T y n x n y n T x n for each n.

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 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.
2 Since f n → 0 in σ F , F and F has the AM-compactness property, then 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 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 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, International Journal of Mathematics and Mathematical Sciences 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 l 1 , we note that l 1 has the AM-compactness property resp.its norm is order continuous and its identity operator Id l 1 : l 1 → l 1 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.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 l 1 , and a positive projection P : E → l 1 .
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 ε 0 > 0 such that |f n y n | ≥ ε 0 for all n.Now, we consider the operator T S • P : E → l 1 → F, where S is the operator defined by

3.3
Since l 1 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 → l 1 is surjective, there exist δ > 0 such that δ • B l 1 ⊂ P B E , where B H is the closed unit ball of H E or l 1 .Hence where e i ∞ i 1 is the canonical bases of l 1 .Then T f n > δ • ε 0 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 c 0 and F l ∞ , we observe that each operator T from c 0 into l ∞ has an adjoint T from l ∞ into l 1 which is Dunford-Pettis because l 1 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.

International Journal of Mathematics and Mathematical Sciences
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 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 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.

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.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.
2 As f n → 0 in σ F , F , and since F has the AM-compactness property, then 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 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 S 1 , S 2 : l ∞ → l ∞ such that 0 ≤ S 1 ≤ S 2 , S 2 is compact, and S 1 is not compact.Now, as l ∞ has an order continuous norm, Theorem 5.31 of Aliprantis and Burkinshaw 5 implies that S 1 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 S 1 R • Q.We note that Q is not compact because S 1 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 l 1 , we note that E l 1 has the AM-compactness property and its identity operator Id l 1 : 3 If F has the AM-compactness property.In fact, if we take E F l 1 , we observe that F l 1 has the AM-compactness property and its identity operator Id l 1 : For the converse of Theorem 4.1, we obtain the following.and F l ∞ , we observe that each operator T from c 0 into l ∞ has an adjoint T from l ∞ into l 1 which is Dunford-Pettis because l 1 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 l 1 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.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 l 1 , and a positive projection P : E → l 1 .
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.Since l 1 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 → l 1 is surjective, there exist δ > 0 such that δ • B l where e i ∞ i 1 is the canonical bases of l 1 .Then T g n > δ•ε 0 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  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.

4 F
x∈A |f n x | → 0 see Theorem 5.98 of 5 .International Journal of Mathematics and Mathematical Sciences 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, has the Schur property.

Corollary 3 . 4 .Theorem 3 . 5 . 2 F
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 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 AMcompactness 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.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, has the Schur property.

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.

Theorem 4 . 1 . 3 F
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, has the Schur property.

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.

International
Journal of Mathematics and Mathematical Sciences 11 Now, we consider the composed operator T S • P : E −→ l 1 −→ F, 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 .
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: International Journal of Mathematics and Mathematical Sciences 3 3 the lattice operations in E are weakly sequentially continuous, 4 the lattice operations in E are weak * sequentially continuous.and E are order continuous, but E does not have the Dunford-Pettis property.In fact, consider E l 2 , the norms of E l 2 and E l 2 , are order continuous but l 2 does not have the Dunford-Pettis property; 6 the norms of E and E are not order continuous, but E has the Dunford-Pettis 1 the norm of E is order continuous and E has the Dunford-Pettis property, 2 the topological dual E is discrete, 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 l 1 , 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.fact, if we take E c 0 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 1⊂ P B E where B H is the closed unit ball of H E, l 1 .Hence n • S p x ≥ δ • sup