# Duality Property for Positive Weak Dunford-Pettis Operators

Authors: Belmesnaoui Aqzzouz; Khalid Bouras; Mohammed Moussa
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## 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.


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## Body

## 1. Introduction and Notation

Let us recall that an operatorT 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 yn′(T(xn)) converges to 0 whenever (xn) converges weakly to 0 in E and (yn′) 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(ln2) has the Schur property, its identity operator Idl1(ln2) is Dunford-Pettis and then weak Dunford-Pettis, but its adjoint Idl∞(ln2), which is the identity operator of the Banach space l∞(ln2), is not weak Dunford-Pettis (because the Banach space l∞(ln2) 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, ifE 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, ifE 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, ifE 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 operatorT 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 (fn) of E′, we have |fn|→0 for σ(E′,E).On the other hand, ifE is a Banach lattice, then(1)
the lattice operations in the topological dualE′ are called sequentially continuous if the sequence (|fn|) converges to 0 in σ(E′,E′′) whenever the sequence (fn) converges to 0 in σ(E′,E′′);(2)
the lattice operations inE′ are called weak* sequentially continuous if the sequence (|fn|) converges to 0 in the weak* topology σ(E′,E) whenever the sequence (fn) 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(xn)∥) converges to 0 for every weakly null sequence (xn) 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]).
LetE be a Banach lattice. Then E has the AM-compactness property if one of the following assertions is valid:(1)
the norm ofE is order continuous and E has the Dunford-Pettis property,(2)
the topological dualE′ is discrete,(3)
the lattice operations inE′ are weakly sequentially continuous,(4)
the lattice operations inE′ are weak* sequentially continuous.Remarks 2.2.
There exists a Banach latticeE such that(1)
the norm ofE′ 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 ofE′ 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 ofE 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 ofE 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 dualE′ 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 dualE′ 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 spaceE 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 spacec0 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.
LetE be a Banach lattice. Then  E′ has the Schur property if one of the following assertions is valid: (1)
the norm ofE′ is order continuous,  E has the AM-compactness property and the weak Dunford-Pettis property,(2)
the norms ofE and  E′ are order continuous and  E has the Dunford-Pettis property,(3)
the topological dualE′ is discrete with an order continuous norm and  E has the weak Dunford-Pettis property.Proof.
(1) Let(fn)⊂E′ be a sequence such that fn→0 in σ(E′,E′′). Since E has the AM-compactness property, then |fn|→0 in σ(E′,E) (Proposition 3.1 of [7]).
Now, by Corollary 2.7 of Dodds and Fremlin [8], to show that ∥fn∥→0, it suffices to prove that fn(xn)→0 for every norm-bounded disjoint sequence (xn)⊂E+. To this end, let (xn) 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 xn→0 in σ(E,E′). And as E has the weak Dunford-Pettis property, we obtain fn(xn)→0. This proves that E′ has the Schur property.
For (2) and (3), it follows from Theorem2.1 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 latticeF 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 dualF′, 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.
LetE 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(xn) (resp., (yn′)) be a sequence of E (resp., of F′) such that xn→0 in σ(E,E′) (resp., yn′→0 in σ(F′,F′′)). We have to prove that yn′(T(xn))→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(xn)→0 for σ(E′′,E′′′).
Now, asyn′→0 in σ(F′,F′′) and the adjoint T′ is weak Dunford-Pettis from F′ into E′, we deduce that τ(x)(T′(yn′))→0. But we know that
(3.1)τ(xn)(T′(yn′))=T′(yn′)(xn)=yn′(T(xn))for eachn.
Hence yn′(T(xn))→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 (fn) of X′ converges uniformly to zero on the set A, that is, supx∈A|fn(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.
LetE 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 ofE′ is order continuous and  E has the AM-compactness property,(2)
the norm ofE′ is order continuous and  F has the AM-compactness property,(3)
the norms ofE and  E′ are order continuous,(4)
F′ has the Schur property.Proof.
For (1), (2), and (3), letT:E→F be a positive weak Dunford-Pettis operator and let (fn)⊂F′ be a sequence such that fn→0 in σ(F′,F′′). In the three cases we have |T′(fn)|→0 in σ(E′,E), in fact, consider the following.(1)
AsT′(fn)→0 in σ(E′,E′′) and E has the AM-compactness property, then |T′(fn)|→0 for σ(E′,E).(2)
Sincefn→0 in σ(F′,F′′) and F has the AM-compactness property, then |fn|→0 in σ(F′,F). Hence, T′(|fn|)→0 in σ(E′,E). Now, from |T′(fn)|≤T′(|fn|) for each n, we conclude that |T′(fn)|→0 in σ(E′,E).(3)
Since the norm ofE 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+, supy∈T([-x,x])|fn(y)|→0. Now, from supy∈T([-x,x])|fn(y)|=|T′(fn)|(x) for each n, we obtain |T′(fn)|(x)→0 for each x∈E+, and hence |T′(fn)|→0 in σ(E′,E).On the other hand, by Corollary 2.7 of Dodds and Fremlin [8], to prove that ∥T′(fn)∥→0, it suffices to show that [T′(fn)](xn)→0 for every norm-bounded disjoint sequence (xn)⊂E+. To this end, let (xn) 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 xn→0 in σ(E,E′). Hence, as T is a weak Dunford-Pettis operator, we obtain fn(T(xn))→0. And from
(3.2)[T′(fn)](xn)=fn(T(xn))for eachn,
we derive that [T′(fn)](xn)→0, and hence T′ is Dunford-Pettis.(4)
In this case, each operatorT:E→F has an adjoint T′:F′→E′ which is Dunford-Pettis.Remarks 3.3.
There exist Banach latticesE 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 dualE′ 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 Idl∞: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)
ifE 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 Idl1:l1→l1 is weak Dunford-Pettis but its adjoint Idl∞:l∞→l∞ is not Dunford-Pettis. However, it is weak Dunford-Pettis because l∞ has the Dunford-Pettis property.As a consequence of Theorems2.1 and 3.2, we obtain the following.Corollary 3.4.
LetE 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 dualE′ is discrete with an order continuous norm,(2)
the norm ofE′ is order continuous and  F′ is discrete,(3)
the norm ofE′ is order continuous and the lattice operations in  F′ are weakly sequentially continuous,(4)
the norm ofE′ is order continuous and the lattice operations in  F′ are  weak* sequentially continuous,(5)
the norms ofE′ and  F are order continuous and  F has the Dunford-Pettis property,(6)
the norms ofE 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 Theorem2.1 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 Theorem3.2.
(7) In this case each operatorT:E→F has an adjoint T′:F′→E′ which is weak Dunford-Pettis.For the converse of Theorem3.2, we have the following.Theorem 3.5.
LetE 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 ofE′ is order continuous,(2)
F′ has the Schur property.Proof.
Assume by way of contradiction that the norm ofE′ 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 ofE′ 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, sinceF′ does not have the Schur property, there exists a weakly null sequence (fn)⊂F′ such that ∥fn∥=1 for all n. Moreover, there exists a sequence (yn)⊂F+ with ∥yn∥≤1 and some ε0>0 such that |fn(yn)|≥ε0 for all n.
Now, we consider the operatorT=S∘P:E→l1→F, where S is the operator defined by
(3.3)S:l1→F,(λn)⟼∑nλnyn.
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 (fn) is weakly null in F′. And as the operator P:E→l1 is surjective, there exist δ>0 such that δ·Bl1⊂P(BE), where BH is the closed unit ball of H=E or l1. Hence
(3.4)‖T′(fn)‖=sup⁡x∈BE|T′(fn)(x)|=sup⁡x∈BE|fn(T(x))|=sup⁡x∈BE|fn∘S(p(x))|≥δ⋅sup⁡(λi)i∈Bl1|fn∘S((λi))|≥δ⋅|fn∘S((en))|≥δ⋅|fn(yn)|>δ.ε0,
where (ei)i=1∞ is the canonical bases of l1.
Then∥T′(fn)∥>δ·ε0 for all n, and we conclude that T′ is not Dunford-Pettis. This presents a contradiction.Remarks 3.6.
LetE 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 ofE 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.WheneverE=F, we obtain the following characterization.Theorem 3.7.
LetE be a Dedekind  σ-complete Banach lattice. Then the following assertions are equivalent: (1)
each positive weak Dunford-Pettis operatorT from  E into  E has an adjoint which is Dunford-Pettis,(2)
the norms ofE 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
(3.5)T=i∘P:E⟶l∞⟶E.
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
(3.6)P∘T∘i:l∞⟶E⟶E⟶l∞
would be Dunford-Pettis. But (P∘T∘i)′=(Idl∞)′=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 operatorT from a Banach lattice E into a Banach space F is said to be almost Dunford-Pettis if the sequence (∥T(xn)∥) converges to 0 for every weakly null sequence (xn) 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 spacel1 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.
LetE 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 ofE′ is order continuous and  E has the AM-compactness property,(2)
the norm ofE′ 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 Theorem3.2. In fact, let T:E→F be a positive almost Dunford-Pettis operator, and let (fn)⊂F be a sequence such that fn→0 in σ(F′,F′′). By the uniform boundedness Theorem, there exists some α>0 such that ∥fn∥≤α for all n. In the two cases we have |T′(fn)|→0 in σ(E′,E). In fact, consider the following.(1)
AsT′(fn)→0 in σ(E′,E′′) and E has the AM-compactness property, then |T′(fn)|→0 in σ(E′,E).(2)
Asfn→0 in σ(F′,F′′), and since F has the AM-compactness property, then |fn|→0 in σ(F′,F). Hence, T′(|fn|)→0 in σ(E′,E) and from |T′(fn)|≤T′(|fn|) for each n, we conclude that |T′(fn)|→0 in σ(E′,E).Now to prove that∥T′(fn)∥E′→0, it suffices to show that [T′(fn)](xn)→0 in every norm-bounded disjoint sequence (xn)⊂E+ (Corollary 2.7 of Dodds and Fremlin [8]). To this end, let (xn) be a norm-bounded disjoint sequence of E+.Since the norm ofE′ is order continuous, it follows from Corollary 2.9 of Dodds and Fremlin [8] that xn→0 in σ(E,E′). Hence, as T is almost Dunford-Pettis operator, we obtain ∥T(xn)∥F→0. Now, from
(4.1)|[T′(fn)](xn)|=|fn(T(xn))|≤α⋅‖T(xn)‖Ffor eachn,
we see that[T′(fn)](xn)→0, and hence T′ is Dunford-Pettis.(3)
In this case each operatorT:E→F has an adjoint T′:F′→E′ which is Dunford-Pettis.Remarks 4.2.
LetE 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 dualE′ 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)
IfE 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 Idl1:l1→l1 is almost Dunford-Pettis but the adjoint Idl∞:l∞→l∞ is not Dunford-Pettis.(3)
IfF 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 Idl1:l1→l1 is almost Dunford-Pettis, but the adjoint Idl∞:l∞→l∞ is not Dunford-Pettis.For the converse of Theorem4.1, we obtain the following.Theorem 4.3.
LetE 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 ofE′ is order continuou,(2)
F′ has the Schur property.Proof.
The proof is the same as that of Theorem3.5 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.
LetE 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) operatorT: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.
LetE and  F be two Banach lattices. Then the following assertions are equivalent: (1)
each positive weak Dunford-Pettis (resp., Dunford-Pettis, almost Dunford-Pettis) operatorT:E→F has an adjoint  T′:F′→E′ which is almost Dunford-Pettis,(2)
one of the following assertions is valid:
(a)
the norm ofE′ 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 ofE′ 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, sinceF′ does not have the positive Schur property, it follows from Theorem 3.1 of [10] the existence of a disjoint weakly null sequence (fn)⊂(F′)+ such that (fn) does not converge to zero for the norm. Moreover, there exists a sequence (yn)⊂F+ with ∥yn∥≤1, and some ε>0, a subsequence (gn) of (fn) such that gn(yn)≥ε for all n.
Now, we consider the composed operator(4.2)T=S∘P:E⟶l1⟶F,
where S is defined by
(4.3)S:l1→F,(λn)⟼∑nλnyn.
Sincel1 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, (gn) is a disjoint weakly null sequence in F′. And since the operator P:E→l1 is surjective, there exist δ>0 such that δ·Bl1⊂P(BE) where BH is the closed unit ball of H=E,l1. Hence
(4.4)‖T′(gn)‖=sup⁡x∈BE|T′(gn)(x)|=sup⁡x∈BE|gn(T(x))|=sup⁡x∈BE|gn∘S(p(x))|≥δ⋅sup⁡(λi)i∈Bl1|gn∘S((λi))|≥δ⋅|gn∘S((en))|≥δ⋅|gn(yn)|>δ.ε0,
where (ei)i=1∞ is the canonical bases of l1.
Then∥T′(gn)∥>δ·ε0 for every n, and we conclude that T′ is not almost Dunford-Pettis. This presents a contradiction.
(2), (a)⇒(1). Let (fn) be a disjoint sequence of F′ such that fn→0 in σ(F′,F′′). We have to prove that (T′(fn)) converges to 0 for the norm of E′. By using Corollary 2.7 of Dodds-Fremlin [8], it suffices to prove that |T′(fn)|→0 in σ(E′,E) and [T′(fn)](xn)→0 for every norm-bounded disjoint sequence (xn)⊂E+. In fact, as (fn) is a weakly null sequence with pairwise disjoint terms, it follows from Remark 1 of Wnuk [11] that |fn|→0 in σ(F′,F′′), and then T′(|fn|)→0 for σ(E′,E′′). Now, since |T′(fn)|≤T′(|fn|) for each n, then |T′(fn)|→0 in σ(E′,E′′), and hence |T′(fn)|→0 in σ(E′,E).
On the other hand, since the norm ofE′ is order continuous, it follows from Corollary 2.9 of Dodds and Fremlin [8] that xn→0 in σ(E,E′). Hence, as T is a weak Dunford-Pettis (resp., Dunford-Pettis, almost Dunford-Pettis) operator, we obtain [T′(fn)](xn)=fn(T(xn))→0, and this proves that T′ is almost Dunford-Pettis.
(2), (b)⇒(1). Obvious.



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*Source: 609287-2011-08-16.xml*