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

1Département d’Economie, Faculté des Sciences Economiques, Juridiques et Sociales, Université Mohammed V-Souissi, BP 5295, Sala Al Jadida, Morocco

2Département 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 spacel¹l²_nhas the Schur property, its identity operator Id_l1l²nis Dunford-Pettis and then weak Dunford-Pettis, but its adjoint Id_l∞l²n, which is the identity operator of the Banach spacel^∞l²_n, is not weak Dunford-Pettisbecause the Banach spacel^∞l_n²does not have the Dunford-Pettis propertysee4, page 22. However, each operator is weak Dunford-Pettis if its adjoint is.

On the other hand, ifEandFare two Banach spaces such thatF is reflexive, then the class of weak Dunford-Pettis operators fromEintoFcoincides with that of Dunford-Pettis

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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 latticeL²0,1does not have the AM-compactness property, butl¹has 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, 2the topological dualEis discrete,

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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 ofEl^∞, 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 ofE l²andE l², are order continuous butl²does 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 andl²does 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 spacel¹has 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 assertions is valid:

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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 thatf_n → 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 thatf_nxn → 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 x_n → 0 in σE, E. And as E has the weak Dunford-Pettis property, we obtainf_nxn → 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 Tis weak Dunford-Pettis fromFintoE, thenTis 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

τxn T

y_n T

y_n

xn y_nTxn 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 zero on the setA, that is, sup_x∈A|fnx| → 0see Theorem 5.98 of5.

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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.

2Sincef_n → 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 sup_y∈T−x,x|fny|

|Tfn|x for each n, we obtain |Tfn|x → 0 for each x ∈ E, and hence

|Tfn| → 0 inσE, E.

On the other hand, by Corollary 2.7 of Dodds and Fremlin 8, to prove that Tfn → 0, it suﬃces to show thatTfnxn → 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 x_n → 0 in σE, E. Hence, as T is a weak Dunford-Pettis operator, we obtainf_nTxn → 0. And from

T fn

xn fnTxn for eachn, 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 Dunford-Pettis. However, it is weak Dunford-Pettis becausel^∞has the Dunford- Pettis property,

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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 Id_l¹ :l¹ → l¹is 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 Theorems2.1and3.2, 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, 2Fhas the Schur property.

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 withyn ≤1 and someε₀>0 such that|fnyn| ≥ε₀for alln.

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Now, we consider the operatorT S◦P : E → l¹ → F, whereSis the operator defined by

S:l¹ → F, λn−→

n

λ_ny_n. 3.3

Sincel¹has the Dunford-Pettis property, the operatorTis weak Dunford-Pettis. But its adjoint T:F → Eis not Dunford-Pettis. Indeed, the sequencefnis weakly null inF. And as the operatorP :E → l¹is surjective, there existδ >0 such thatδ·B_l1 ⊂PBE, whereB_His the closed unit ball ofHEorl¹. Hence

T

fnsup

x∈BE

T fn

xsup

x∈BE

fnTxsup

x∈BE

fn◦S

px

≥δ· sup

λii∈B_l1

fn◦Sλi≥δ·fn◦Sen≥δ·fn

yn> 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 Ewhich 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^∞intol¹which is Dunford-Pettisbecausel¹has the Schur property, butFl^∞ 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 operatorT fromcintol^∞ has an adjointTfroml^∞ 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,

2the norms ofEandEare order continuous.

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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 henceEhas 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, 3Fhas the Schur 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 thatf_n → 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| → 0 inσE, E.

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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 Fremlin8thatx_n → 0 inσE, E. Hence, asT is almost Dunford-Pettis operator, we obtainTxn_F → 0. Now, from

T f_n

x_nf_nTx_n≤α· Tx_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 propertyand thenT 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 l¹has 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 l¹has 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 T :E → Fhas an adjointT:F → Ewhich is Dunford-Pettis, then one of the following assertions is valid:

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1the norm ofEis order continuou, 2Fhas the Schur property.

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 Ewhich 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^∞intol¹which is Dunford-Pettisbecausel¹has the Schur property, butFl^∞ 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 latticel¹is 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 operators and almost Dunford-Pettis operators.

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 yn ≤1, and someε >0, a subsequencegnoffnsuch thatg_nyn≥εfor alln.

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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

Sincel¹has the Schur property, the operatorT is weak Dunford-Pettisresp. 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·B_l¹ ⊂ PBEwhereBHis the closed unit ball of HE, l¹. Hence

T

gnsup

x∈BE

T gn

xsup

x∈BE

gnTxsup

x∈BE

gn◦S

px

≥δ· sup

λii∈Bl1

g_n◦Sλi≥δ·g_n◦Sen≥δ·g_n

y_n> 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|Tfn| → 0 inσE, E.

On the other hand, since the norm ofEis order continuous, it follows from Corollary 2.9 of Dodds and Fremlin 8 that x_n → 0 in σE, E. Hence, as T is a weak Dunford- Pettis resp., Dunford-Pettis, almost Dunford-Pettis operator, we obtain Tfnxn

f_nTxn → 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 Spaces, vol. 2 of Contemporary Mathematics, pp. 15–60, Chapel Hill, NC, USA, 1980.

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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 Fisico dell’Universit`a di Modena, vol. 42, no. 1, pp. 227–236, 1994.

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