Duality Property for Positive Weak Dunford Pettis Operators

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Volume 2011, Article ID 609287,12pages doi:10.1155/2011/609287

Research Article

Duality Property for Positive Weak

Dunford-Pettis Operators

Belmesnaoui Aqzzouz,

1

_{Khalid Bouras,}

2

and Mohammed Moussa

2

1_{D´epartement d’Economie, Facult´e des Sciences Economiques, Juridiques et Sociales,}

Universit´e Mohammed V-Souissi, BP 5295, Sala Al Jadida, Morocco

2_{Département de Mathématiques, Faculté des Sciences, Université Ibn Tofail, BP 133, Kénitra, Morocco}

Correspondence should be addressed to Belmesnaoui Aqzzouz,[email protected]

Received 15 December 2010; Accepted 5 May 2011

Academic Editor: Yuri Latushkin

Copyrightq2011 Belmesnaoui Aqzzouz et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 operatorTfrom a Banach spaceEinto anotherFis called Dunford-Pettis if it carries weakly compact subsets ofEonto compact subsets ofF. The operatorTis said to be weak Dunford-Pettis ify_nTxnconverges to 0 wheneverxnconverges weakly to 0 in

Eandy_nconverges weakly to 0 inF.

The class of weak Dunford-Pettis operators was used by Aliprantis and Burkinshaw

1and Kalton and Saab2when they studied the domination property of Dunford-Pettis operators. As this latter class3, 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 spacel1_l2

nhas the Schur property, its identity

operator Id_l1_l2_nis Dunford-Pettis and then weak Dunford-Pettis, but its adjoint Id_l∞_l2_n, which

is the identity operator of the Banach spacel∞l2

n, is not weak Dunford-Pettisbecause the

Banach spacel∞l2

ndoes not have the Dunford-Pettis propertysee4, page 22. However,

each operator is weak Dunford-Pettis if its adjoint is.

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operators fromEintoF, and therefore some results of5can be applied here to give some answers to our duality problem.

Morever, ifEandFare both reflexive, then the class of weak Dunford-Pettis operators fromEintoFcoincides with that of compact operators fromEintoF, and hence ifT: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 fromF intoE is weak Dunford-Pettis, and hence each weak Dunford-PettisT :E → Fhas an adjointT:F → Ewhich is one.

As we have already done for Dunford-Pettis operators3and almost Dunford-Pettis operators6, 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 to5for unexplained terminologies on Banach lattice theory and positive operators.

2. Some Preliminaries

Let us recall that an operatorT from a Banach latticeEinto a Banach spaceX is said to be AM-compact if it carries each order-bounded subset ofE onto a relatively compact set of X. In7, we used this class of operators to introduce Banach lattices which satisfy the AM-compactness property. In fact, a Banach latticeEis said to have the AM-compactness property if every weakly compact operator defined onE, and taking values in a Banach spaceX, is AM-compact. For an example, the Banach latticeL2₀_,₁_{does not have the AM-compactness}

property, butl1_{has the AM-compactness property.}

It follows from7, Proposition 3.1 that a Banach latticeEhas the AM-compactness property if and only if for every weakly null sequence fn of E, we have |fn| → 0 for

σE_{, E}_.

On the other hand, ifEis a Banach lattice, then

1the lattice operations in the topological dualEare called sequentially continuous if the sequence |fn| converges to 0 in σE, E whenever the sequence fn

converges to 0 inσE, E;

2the lattice operations inEare called weak∗sequentially continuous if the sequence

|fn| converges to 0 in the weak∗ topology σE, E whenever the sequencefn

converges to 0 inσE, E.

A Banach spaceresp., Banach latticeEhas the Dunford-Pettisresp., weak Dunford-Pettisproperty if every weakly compact operatorT defined on Eand taking values in a Banach spaceFis Dunford-Pettisresp., almost Dunford-Pettis, i.e., the sequenceTxn

converges to 0 for every weakly null sequencexnconsisting of pairwise disjoint elements

inE.

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

Theorem 2.1see7. LetEbe a Banach lattice. ThenEhas the AM-compactness property if one

of the following assertions is valid:

1the norm ofEis order continuous andEhas the Dunford-Pettis property,

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3the lattice operations inEare weakly sequentially continuous,

4the lattice operations inEare weak∗sequentially continuous.

Remarks 2.2. There exists a Banach latticeEsuch that

1the norm of E is order continuous but E does not have the AM-compactness property nor the weak Dunford-Pettis property. In fact, considerE L2₀_,₁_{, the}

norm of E L20,1, is order continuous but L20,1 does not have the AM-compactness property nor the weak Dunford-Pettis property;

2the norm ofEis not order continuous, butEhas the AM-compactness property or the weak Dunford-Pettis property. In fact, considerEl1_{, the norm of}_E_l∞_{, is not}

order continuous butl1has the AM-compactness property and the weak Dunford-Pettis property;

3Ehas 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;

4Ehas 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;

5the norms of E and E are order continuous, butE does not have the Dunford-Pettis property. In fact, considerE l2_{, the norms of}_E _l2_and_E _l2_{, are order}

continuous butl2_{does not have the Dunford-Pettis property;}

6the norms of E and E are not order continuous, but E has the Dunford-Pettis property. In fact, considerEl1⊕l∞, the norms ofEl1⊕l∞andEl∞⊕l∞, are not order continuous butl1_⊕_l∞_{has the Dunford-Pettis property;}

7the topological dualEis discrete with an order continuous norm, andEdoes not have the weak Dunford-Pettis property. In fact, considerEl2_{, the topological dual}

E _l2_{, is discrete with an order continuous norm and}_l2_{does not have the weak}

Dunford-Pettis property;

8the topological dualEis not discrete and its norm is not order continuous, but it has the weak Dunford-Pettis property. In fact, considerE l∞, the topological dualE l∞, is not discrete and its norm is not order continuous but it has the weak Dunford-Pettis property.

A Banach spaceEis said to have the Schur property if every sequence in Eweakly convergent to zero is norm convergent to zero. For an example, the Banach spacel1_{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 spacec0has the weak Dunford-Pettis propertybecause 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. ThenE has the Schur property if one of the following

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1the norm of E is order continuous,E has the AM-compactness property and the weak Dunford-Pettis property,

2the norms ofEandEare order continuous andEhas the Dunford-Pettis property,

3the topological dual E is discrete with an order continuous norm and E has the weak Dunford-Pettis property.

Proof. 1Let fn ⊂ Ebe a sequence such thatfn → 0 inσE, E. SinceEhas the

AM-compactness property, then|fn| → 0 inσE, E Proposition 3.1 of7.

Now, by Corollary 2.7 of Dodds and Fremlin8, to show thatfn → 0, it suﬃces

to prove thatfnxn → 0 for every norm-bounded disjoint sequencexn⊂E. To this end,

letxnbe a such sequence ofE. Since the norm ofEis order continuous, it follows from

Corollary 2.9 of Dodds and Fremlin 8that xn → 0 in σE, E. And as E has the weak

Dunford-Pettis property, we obtainfnxn → 0. This proves thatEhas the Schur property.

For2and3, it follows fromTheorem 2.1thatEhas the AM-compactness property. Finally, assertion1of the present theorem ends the proof.

Remarks 2.4. 1There exists a Banach latticeFwhich has the AM-compactness property but its topological dualFdoes not have the Schur property. In fact, considerF l1_{, it has the}

AM-compactness property butFl∞does not have the Schur property.

2If the topological dualF, of a Banach latticeF, has the Schur property, thenFis discrete, and henceFhas the AM-compact propertyseeTheorem 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. LetEandFbe two Banach spaces, and letTbe an operator fromEintoF. If the adjoint

T_{is weak Dunford-Pettis from}_F_into_E_{, then}_T_{is weak Dunford-Pettis.}

Proof. Letxn resp.,ynbe a sequence ofE resp., ofFsuch thatxn → 0 in σE, E

resp.,y_n → 0 inσF, F. We have to prove thaty_nTxn → 0. For this, letτ :E → E

be the canonical injection ofE into its topological bidualE. Sinceτ is continuous for the topologiesσE, EandσE, E, we obtainτxn → 0 forσE, E.

Now, asy_n → 0 inσF, Fand the adjointTis weak Dunford-Pettis fromFintoE, we deduce thatτxTy_n → 0. But we know that

τxnTynTynxn ynTxn for eachn. 3.1

Hencey_nTxn → 0, and this ends the proof.

Let us recall from5that a norm-bounded subsetAof a Banach space X is said to be Dunford-Pettis whenever every weakly compact operator fromX to an arbitrary Banach spaceY carriesAto a norm relatively compact set ofY. This is equivalent to saying thatAis Dunford-Pettis if and only if every weakly null sequencefnofXconverges uniformly to

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

Theorem 3.2. LetEandFbe two Banach lattices. Then each positive weak Dunford-Pettis operator

T :E → F has an adjointT:F → Ewhich is Dunford-Pettis (and then weak Dunford-Pettis) if one of the following assertions is valid:

1the norm ofEis order continuous andEhas the AM-compactness property,

2the norm ofEis order continuous andFhas the AM-compactness property,

3the norms ofEandEare order continuous,

4Fhas the Schur property.

Proof. For1,2, and3, letT :E → Fbe a positive weak Dunford-Pettis operator and let

fn⊂Fbe a sequence such thatfn → 0 inσF, F. In the three cases we have|Tfn| → 0

inσE, E, in fact, consider the following.

1AsTfn → 0 inσE, EandEhas the AM-compactness property, then|Tfn| →

0 forσE, E.

2Sincefn → 0 inσF, FandFhas the AM-compactness property, then|fn| → 0

inσF, F. Hence,T|fn| → 0 inσE, E. Now, from|Tfn| ≤T|fn|for eachn,

we conclude that|Tfn| → 0 inσE, E.

3Since the norm ofEis order continuous,−x, xis weakly compact for eachx∈E. AsT is weak Dunford-Pettis, we conclude thatT−x, xis a Dunford-Pettis set, and then for eachx∈E, sup_y_∈_T₋_x,x|fny| → 0. Now, from supy∈T−x,x|fny|

|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 su}_ﬃ_{ces to show that}_T_f_n_x_n _→ _{0 for every norm-bounded} disjoint sequence xn ⊂ E. To this end, let xn be a norm-bounded disjoint

sequence ofE. Since the norm ofEis order continuous, it follows from Corollary 2.9 of Dodds and Fremlin 8 that xn → 0 in σE, E. Hence, as T is a weak

Dunford-Pettis operator, we obtainfnTxn → 0. And from

T_f_n_x_n _f_n_T_x_n _{for each}_n, _3.2

we derive thatTfnxn → 0, and henceTis Dunford-Pettis.

4In this case, each operator T : E → F has an adjoint T : F → E which is Dunford-Pettis.

Remarks 3.3. There exist Banach latticesEandFand a weakly Dunford-Pettis operatorTfrom EintoFsuch that the adjointTis not Dunford-Pettis in the following situations:

1if 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 Idl∞ :l∞ → l∞is weak Dunford-Pettis but its adjoint Id_l∞ :l∞ → l∞is not

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2ifEhas the AM-compactness propertyresp.,Fhas the AM-compactness property, E has an order continuous norm. In fact, ifE F l1_{, we note that} _l1 _{has the}

AM-compactness property resp. its norm is order continuous and its identity operator Idl1 :l1 → l1is 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 Theorems2.1and3.2, we obtain the following.

Corollary 3.4. LetEandFbe two Banach lattices. Then each positive weak Dunford-Pettis operator

T :E → Fhas an adjointT:F → Ewhich is weak Dunford-Pettis if one of the following assertions is valid:

1the topological dualEis discrete with an order continuous norm,

2the norm ofEis order continuous andFis discrete,

3the norm ofEis order continuous and the lattice operations inFare weakly sequentially continuous,

4the norm ofEis order continuous and the lattice operations inFare weak∗sequentially continuous,

5the norms ofEandFare order continuous andFhas the Dunford-Pettis property,

6the norms ofEandEare order continuous,

7EorFhas the Dunford-Pettis property.

Proof. For 1, 2, 3, 4, and 5, it follows from Theorem 2.1 that Eor F has the AM-compactness property. Since the norm ofE is order continuous, Theorem 3.2 implies that each positive weak Dunford-Pettis operatorT :E → Fhas an adjointT :F → Ewhich is Dunford-Pettisand then weak Dunford-Pettis.

6Follows from3ofTheorem 3.2.

7In this case each operatorT :E → Fhas an adjointT : F → Ewhich is weak Dunford-Pettis.

For the converse ofTheorem 3.2, we have the following.

Theorem 3.5. Let Eand F be two Banach lattices. If each positive weak Dunford-Pettis operator

T :E → Fhas an adjointT:F → Ewhich is Dunford-Pettis, then one of the following assertions is valid:

1the norm ofEis order continuous,

Proof. Assume by way of contradiction that the norm ofEis not order continuous andFdoes not have the Schur property. We have to construct a positive weak Dunford-Pettis operator T :E → Fsuch that its adjointT:F → Eis not Dunford-Pettis.

Since the norm ofEis not order continuous, it follows from the proof of Theorem 1 of Wickstead9the existence of a sublatticeHofE, which is isomorphic tol1_{, and a positive}

projectionP:E → l1_.

On the other hand, sinceFdoes not have the Schur property, there exists a weakly null sequencefn⊂ Fsuch thatfn 1 for alln. Moreover, there exists a sequenceyn⊂ F

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Now, we consider the operatorT S◦P : E → l1 _→ _F_{, where}_S_{is the operator}

defined by

S:l1 _→ _F, _λ

n−→

n

λnyn. 3.3

Sincel1_{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} operatorP :E → l1_{is surjective, there exist}_{δ >}_{0 such that}_δ_·_B

l1 ⊂PB_E, whereB_His the

closed unit ball ofHEorl1_{. Hence}

T_f_n_sup

x∈BE

T_f_n_x_sup

x∈BE

fnTxsup

x∈BE

fn◦Spx

≥δ· sup

λii∈B_l1

fn◦Sλi≥δ·fn◦Sen≥δ·fnyn> δ.ε0,

3.4

whereei∞i1is the canonical bases ofl1.

ThenTfn > δ·ε0 for alln, and we conclude thatTis not Dunford-Pettis. This

presents a contradiction.

Remarks 3.6. LetEandFbe two Banach lattices such thatFdoes not have the Schur property. If each positive weak Dunford-Pettis operatorT fromEintoFhas an adjointTfromFinto E_{which is Dunford-Pettis, then}

1Fdoes not necessarily have the AM-compactness property. In fact, if we takeEc0

andFl∞, we observe that each operatorT fromc0intol∞has an adjointTfrom l∞_into_l1_{which is Dunford-Pettis}_because_l1_{has the Schur property}_{, but}_F_l∞ does not have the AM-compactness property,

2the norm of Eis not necessarily order continuous. In fact, if we take E cand F l∞_{, we note that each operator}_T _from_c_into_l∞ _{has an adjoint}_T_from_l∞ intocwhich is Dunford-Pettisbecausechas the Schur property, but the norm of Ecis not order continuous,

3E 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 froml∞intol∞has an adjointTfroml∞intol∞which is Dunford-Pettis

see assertion 2 of Theorem 3.2, butE l∞ does not have the AM-compactness property.

WheneverEF, we obtain the following characterization.

Theorem 3.7. Let E be a Dedekindσ-complete Banach lattice. Then the following assertions are

equivalent:

1each positive weak Dunford-Pettis operator T from E into E has an adjoint which is Dunford-Pettis,

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Proof. 1⇒2. ByTheorem 3.5, the norm ofE is order continuous. We have just to prove that the norm ofEis order continuous. Assume that the norm ofEis not order continuous, and sinceEis Dedekindσ-complete, thenEcontains a closed sublattice isomorphic tol∞and there is a positive projectionP :E → l∞. Leti:l∞ → Ebe the canonical injection ofl∞into E. Consider the operator defined by

T i◦P :E−→l∞−→E. 3.5

Sincel∞has the Dunford-Pettis property, the positive operatorTis weak Dunford-Pettis. But its adjointT:E → Eis 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 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 from3ofTheorem 3.2.

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

In6, 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 latticeEinto a Banach spaceFis said to be almost Dunford-Pettis if the sequenceTxnconverges to 0 for every weakly null

sequencexnconsisting of pairwise disjoint elements inE.

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 spacel∞, 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. LetEandFbe two Banach lattices. Then each positive almost Dunford-Pettis operator

T :E → Fhas an adjointT:F → Ewhich is Dunford-Pettis if one of the following assertions is valid:

1the norm ofEis order continuous andEhas the AM-compactness property,

2the norm ofEis order continuous andFhas the AM-compactness property,

Proof. Note that for1and2, the proof is the same as1and2ofTheorem 3.2. In fact, let T :E → Fbe a positive almost Dunford-Pettis operator, and letfn⊂Fbe a sequence such

thatfn → 0 inσF, F. By the uniform boundedness Theorem, there exists someα >0 such

thatfn ≤αfor alln. In the two cases we have|Tfn| → 0 inσE, E. In fact, consider the

following.

1AsTfn → 0 inσE, EandEhas the AM-compactness property, then|Tfn| →

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2Asfn → 0 inσF, F, and sinceFhas the AM-compactness property, then|fn| →

0 inσF, F. Hence,T|fn| → 0 inσE, Eand from|Tfn| ≤T|fn|for eachn,

we conclude that|Tfn| → 0 inσE, E.

Now to prove thatTfn_E → 0, it suﬃces to show thatTfnxn → 0 in every

norm-bounded disjoint sequencexn ⊂ E Corollary 2.7 of Dodds and Fremlin

8. To this end, letxnbe a norm-bounded disjoint sequence ofE.

Since the norm ofEis order continuous, it follows from Corollary 2.9 of Dodds and Fremlin8thatxn → 0 inσE, E. Hence, asT is almost Dunford-Pettis operator,

we obtainTxnF → 0. Now, from

T_f_n_x_n_f_n_T_x_n_≤_α_·_T_x_n

F for eachn, 4.1

we see thatTfnxn → 0, and henceTis Dunford-Pettis.

3In this case each operator T : E → F has an adjoint T : F → E which is Dunford-Pettis.

Remarks 4.2. LetEandFbe two Banach lattices, and letTbe an operator fromEintoF. Then the adjointTis not necessarily Dunford-Pettis wheneverT is almost Dunford-Pettis in the following situations.

1If the topological dualEhas an order continuous norm. In fact, since the norm ofl∞ is not order continuous and the Banach latticel∞is not discrete, it follows from Theorem 1 of Wickstead9the existence of two positive operatorsS1, S2:l∞ → l∞

such that 0 ≤ S1 ≤ S2, S2 is compact, andS1 is not compact. Now, asl∞ has

an order continuous norm, Theorem 5.31 of Aliprantis and Burkinshaw5implies thatS1is weakly compact. So, by Theorem 5.44 of Aliprantis and Burkinshaw5,

there exist a reflexive Banach latticeG, lattice homomorphismQ :l∞ → G, and a positive operatorR:G → l∞such thatS1 R◦Q. We note thatQis not compact becauseS1is not one.

On the other hand, if we takeE l∞,F G, andT Q, thenT : l∞ → Gis a weakly compact operatorbecauseGis reflexive, and henceT is Dunford-Pettis

l∞_{has the Dunford-Pettis property}_{and then}_T _{is almost Dunford-Pettis. But its} adjointT :G → l∞is not Dunford-Pettisif not, sinceGis reflexive,Twould be compact and soT is compact, which is a contradiction. However, the norm of E _l∞_{is order continuous.}

2IfEhas the AM-compactness property. In fact, if we takeEF l1_{, we note that}

E l1_{has the AM-compactness property and its identity operator Id}

l1 :l1 → l1is

almost Dunford-Pettis but the adjoint Idl∞ :l∞ → l∞is not Dunford-Pettis.

3IfF has the AM-compactness property. In fact, if we takeE F l1, we observe thatF l1_{has the AM-compactness property and its identity operator Id}

l1 :l1 → l1

is almost Dunford-Pettis, but the adjoint Idl∞:l∞ → l∞is not Dunford-Pettis.

For the converse ofTheorem 4.1, we obtain the following.

Theorem 4.3. LetEand F be two Banach lattices. If each positive almost Dunford-Pettis operator

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1the norm ofEis order continuou,

Proof. The proof is the same as that ofTheorem 3.5if we observe that the operatorT in the proof ofTheorem 3.5is almost Dunford-PettisbecauseT admits a factorization through the Banach latticel1_{, which has the Schur property}_.

Remarks 4.4. LetEandFbe two Banach lattices such thatFdoes not have the Schur property. If each positive almost Dunford-Pettis operatorTfromEintoFhas an adjointTfromFinto E_{which is Dunford-Pettis, then}

1E 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 froml∞intol∞has an adjointTfroml∞intol∞which is Dunford-Pettis

see assertion 2 of Theorem 4.1, butE l∞ does not have the AM-compactness property,

2Fdoes not necessarily have the AM-compactness property. In fact, if we takeEc0

andFl∞, we observe that each operatorT fromc0intol∞has an adjointTfrom 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-Pettisresp., Dunford-Pettis operator T : E → F whose adjoint T : F → E is not almost Dunford-Pettis. In fact, the identity operator of the Banach latticel1_{is weak Dunford-Pettis} _{resp., Dunford-Pettis}

operator but its adjoint, which is the identity of the Banach latticel∞, is not almost Dunford-Pettis.

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

Theorem 4.5. LetEandFbe two Banach lattices. Then the following assertions are equivalent:

1each positive weak Dunford-Pettis (resp., Dunford-Pettis, almost Dunford-Pettis) operator

T :E → Fhas an adjointT:F → Ewhich is almost Dunford-Pettis,

2one of the following assertions is valid:

athe norm ofEis order continuous,

bFhas the positive Schur property.

Proof. 1⇒2. Assume by way of contradiction that the norm ofEis not order continuous and F does not have the positive Schur property. We have to construct a positive weak Dunford-Pettisresp., Dunford-Pettis, almost Dunford-PettisoperatorT :E → Fsuch that its adjointT:F → Eis not almost Dunford-Pettis.

Since the norm ofEis not order continuous, it follows from the proof of Theorem 1 of Wickstead9the existence of a sublatticeHofE, which is isomorphic tol1_{, and a positive}

projectionP:E → l1_.

On the other hand, sinceFdoes not have the positive Schur property, it follows from Theorem 3.1 of10the existence of a disjoint weakly null sequence fn ⊂ F such that

fndoes not converge to zero for the norm. Moreover, there exists a sequenceyn⊂Fwith

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Now, we consider the composed operator

T S◦P:E−→l1−→F, 4.2

whereSis defined by

S:l1 → F, λn−→

n

λnyn. 4.3

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, gnis a disjoint weakly null sequence in F. And since the operator P : E → l1

is surjective, there existδ > 0 such thatδ·Bl1 ⊂ PB_EwhereB_His the closed unit ball of

HE, l1_{. Hence}

_T_g_n_sup

x∈BE

_T_g_n_x_sup

x∈BE

_g_n_T_x_sup

x∈BE

_g_n_◦_S

px

≥δ· sup

λii∈Bl1

gn◦Sλi≥δ·gn◦Sen≥δ·gnyn> δ.ε0,

4.4

whereei∞i1is the canonical bases ofl1.

ThenTgn> δ·ε0for everyn, and we conclude thatTis not almost Dunford-Pettis.

This presents a contradiction.

2,a⇒1. Letfnbe a disjoint sequence ofFsuch that fn → 0 inσF, F. We

have to prove thatTfnconverges to 0 for the norm ofE. By using Corollary 2.7 of

Dodds-Fremlin8, it suﬃces to prove that|Tfn| → 0 inσE, EandTfnxn → 0 for every

norm-bounded disjoint sequencexn ⊂ E. In fact, asfnis a weakly null sequence with

pairwise disjoint terms, it follows from Remark 1 of Wnuk11that|fn| → 0 inσF, F, and

thenT|fn| → 0 forσE, E. Now, since|Tfn| ≤T|fn|for eachn, then|Tfn| → 0 in

σE_{, E}_{, and hence}_|_T_f_n_{| →} _{0 in}_σ_E_{, E}_.

On the other hand, since the norm ofEis order continuous, it follows from Corollary 2.9 of Dodds and Fremlin 8 that xn → 0 in σE, E. Hence, as T is a weak

Dunford-Pettis resp., Dunford-Pettis, almost Dunford-Pettis operator, we obtain Tfnxn

fnTxn → 0, and this proves thatTis almost Dunford-Pettis.

2,b⇒1. Obvious.

References

1 C. D. Aliprantis and O. Burkinshaw, “Dunford-Pettis operators on Banach lattices,” Transactions of the

American Mathematical Society, vol. 274, no. 1, pp. 227–238, 1982.

2 N. J. Kalton and P. Saab, “Ideal properties of regular operators between Banach lattices,” Illinois

Journal of Mathematics, vol. 29, no. 3, pp. 382–400, 1985.

3 B. Aqzzouz, K. Bouras, and A. Elbour, “Some generalizations on positive Dunford-Pettis operators,”

Results in Mathematics, vol. 54, no. 3-4, pp. 207–218, 2009.

4 J. Diestel, “A survey of results related to the Dunford-Pettis property. Integration, topology and geometry in linear,” in Proceedings of the Conference on Integration, Topology, and Geometry in Linear

(12)

5 C. D. Aliprantis and O. Burkinshaw, Positive Operators, Springer, Dordrecht, The Netherlands, 2006, Reprint of the 1985 original.

6 B. Aqzzouz, A. Elbour, and A. W. Wickstead, “Positive almost Dunford-Pettis operators and their duality,” Positivity, vol. 15, no. 2, pp. 185–197, 2011.

7 B. Aqzzouz and K. Bouras, “Weak and almost Dunford-Pettisoperators on Banach lattices,” preprint.

8 P. G. Dodds and D. H. Fremlin, “Compact operators in Banach lattices,” Israel Journal of Mathematics, vol. 34, no. 4, pp. 287–320, 1979.

9 A. W. Wickstead, “Converses for the Dodds-Fremlin and Kalton-Saab theorems,” Mathematical

Proceedings of the Cambridge Philosophical Society, vol. 120, no. 1, pp. 175–179, 1996.

10 Z. L. Chen and A. W. Wickstead, “L-weakly and M-weakly compact operators,” Indagationes

Mathe-maticae, vol. 10, no. 3, pp. 321–336, 1999.

11 W. Wnuk, “Banach lattices with the weak Dunford-Pettis property,” Atti del Seminario Matematico e

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