POSITIVE ALMOST DUNFORD-PETTIS OPERATORS AND THEIR DUALITY.

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1 POSITIVE ALMOST DUNFORD-PETTIS OPERATORS AND THEIR DUALITY. BELMESNAOUI AQZZOUZ, AZIZ ELBOUR, AND ANTHONY W. WICKSTEAD Abstract. We study some properties of almost Dunford-Pettis operators and we characterize pairs of Banach lattices for which the adjoint of an almost Dunford-Pettis operator inherits the same property and look at conditions under which an operator is almost Dunford-Pettis whenever its adjoint is. 1. Introduction and notation Recall from [13] that a linear operator from a Banach lattice E into a Banach space F is almost Dunford-Pettis if T (x n ) 0 for every weakly null sequence (x n ) consisting of pairwise disjoint elements in E. Like the Dunford-Pettis operators, [2], there is no automatic duality result for the class of almost Dunford-Pettis operators. In fact, the identity operator Id L 1 [0,1] : L 1 [0, 1] L 1 [0, 1] is almost Dunford-Pettis but its adjoint Id L [0,1] : L [0, 1] L [0, 1] is not almost Dunford- Pettis. Conversely, the identity operator of the Banach lattice c 0, is not almost Dunford-Pettis but its adjoint, which is the identity operator of the Banach lattice l 1, is almost Dunford-Pettis. In this paper we study the duality of positive almost Dunford-Pettis operators. After some preliminaries, we introduce and study a condition on Banach lattices which will recur in later results. We characterize those Banach lattices on which every positive operator into c 0 is almost Dunford-Pettis. After that we establish a necessary and sufficient condition on the pair of Banach lattices E and F which guarantees that if T : E F is almost Dunford-Pettis then so is T : F E. We also provide some (incomplete) results on the converse problem. Our terminology concerning Banach lattices and positive operators is standard and we refer the reader to [1] or [5] for any unexplained terms Mathematics Subject Classification. 46A40, 46B40, 46B42. Key words and phrases. almost Dunford-Pettis operator, order continuous norm, positive Schur property, KB-space. 1

2 2 BELMESNAOUI AQZZOUZ, AZIZ ELBOUR, AND ANTHONY W. WICKSTEAD 2. Preliminaries In this short section we gather, for the convenience of the reader, some definitions and technical results that we will need later. A Banach space E is said to have the Schur property if every weakly convergent sequence to 0 in E is norm convergent to zero. A Banach lattice E has the positive Schur property if each weakly null sequence with positive terms in E converges to zero in norm. It is pointed out on page 16 of [11] that E has the positive Schur property if and only if each weakly null disjoint sequence with positive terms in E converges to zero in norm. The Banach space l 1 has the Schur property, whilst the Banach lattice L 1 [0, 1] has the positive Schur property but not the Schur property. For more information about this notion see [11]. Proposition 2.1. A Banach lattice E does not have the positive Schur property if and only if there exists a disjoint weakly null sequence (x n ) in E + with x n = 1 for all n. Proof. If E does not have the positive Schur property we know that there exists a disjoint weakly null sequence (y n ) of E + such that (y n ) is not norm convergent to 0. By passing to a subsequence if necessarily, we may assume that there exists some α > 0 with y n α for all n. Put x n = y n / y n for all n and it is easy to see that our requirements are satisfied. The converse is easy. In the work that follows we will frequently encounter situations where we have sequences in both a Banach lattice and its dual which interact in some way. We gather here some known results which will be useful to us in such a setting. The first of these is just a particular case of Theorem 2.4 of [3]. Proposition 2.2. If A is a solid subset of a Banach lattice E and (f n ) a sequence in E + such that (1) sup{f n (x) : x A} < for n = 1, 2,..., (2) lim n f n (x) = 0 for each x A, (3) lim n f n (x n ) = 0 for each disjoint sequence (x n ) in A, then lim n sup{f n (x) : x A} = 0. Specializing Theorem 2.6 of [3] gives us: Proposition 2.3. Let E be a Banach lattice. A weak null sequence (x n ) in E + is norm convergent to 0 if and only if f n (x n ) 0 for each disjoint norm bounded sequence (f n ) in E +. And dually, Corollary 2.7 of [3] gives us:

3 POSITIVE ALMOST DUNFORD-PETTIS OPERATORS AND THEIR DUALITY.3 Proposition 2.4. Let E be a Banach lattice. A weak* null sequence (f n ) in E + is norm convergent to 0 if and only if f n (x n ) 0 for each disjoint norm bounded sequence (x n ) in E +. Finally, we note a simple result which is a consequence of Theorem of [14]. Proposition 2.5. Let E be a Banach lattice. If (x n ) is a disjoint sequence in E + with x n = 1 for all n, then there exists a positive disjoint sequence (g n ) in E with g n = g n (x n ) = 1 for all n and g n (x m ) = 0 for n m. 3. A bi-sequential condition One possibility that will occur in several of our results later may be worthy of isolating for possible further study, especially in view of what it reduces to in some special situations. Definition 3.1. We say that a Banach lattice E has the bi-sequence property if for each weak null disjoint sequence (x n ) in E and each weak* null sequence (f n ) in E +, we have f n (x n ) 0. Notice the asymmetry in the conditions on the two sequences. That in E has to be disjoint whilst that in E is assumed positive. In fact the modulus of a weak null disjoint sequence must be weak null so we could have assumed also that (x n ) is positive. In the very important special case that E has order continuous norm, then this reduces to a condition that has been previously studied. Proposition 3.2. A Banach lattice E, which has order continuous norm, has the bi-sequence property if and only if it has the positive Schur property. Proof. Once we recall that if E has the positive Schur property then every positive weakly null sequence converges in norm to 0, it is clear that the bi-sequence property must hold. Suppose that (x n ) is a weak null disjoint sequence in E +. By Proposition 2.3 we need only prove that f n (x n ) 0 for each disjoint norm bounded sequence (f n ) in E +. Since the norm of E is order continuous, it follows from Theorem of [5] that such a sequence (f n ) is weak* null. Hence the bi-sequence property tells us that f n (x n ) 0 and we are done. The bi-sequence property may similarly be simplified if we assume that E has order continuous norm.

4 4 BELMESNAOUI AQZZOUZ, AZIZ ELBOUR, AND ANTHONY W. WICKSTEAD Definition 3.3. A Banach lattice E has the dual positive Schur property if every weak null sequence in E + is norm null. Proposition 3.4. A Banach lattice E, for which E has order continuous norm, has the bi-sequence property if and only if it has the dual positive Schur property. Proof. To see that the dual positive Schur property implies the bisequence property we need only the fact that a weak null sequence in E must be norm bounded. If we assume the bi-sequence property and assume that (f n ) is a weak* null sequence in E + then by Proposition 2.4 it suffices to show that f n (x n ) 0 for each disjoint norm bounded sequence (x n ) in E +. But since the norm of E is order continuous, it follows from Theorem of [5] that (x n ) is weakly null so the bi-sequence condition tells us that f n (x n ) 0 and we are done. The dual positive Schur property should surely be as worthy of study as the positive Schur property itself yet it has not previously occurred in the literature, possibly because its Banach space analogue is of no great interest. Indeed, the Josefson-Nissenzweig Theorem, [4] and [8], tells us that only finite-dimensional Banach spaces have the property that weak* null sequences in their duals are norm null. Clearly any space C(K), where K is a compact Hausdorff space, has the dual positive Schur property, for if f C(K) + and 1 K denotes the constantly one function on K, then f = f(1 K ) so that if (f n ) is weak null then f n = f n (1 K ) 0. In particular, the order direct sum of a Banach lattice with the positive Schur property and one with the dual positive Schur property, such as l 1 c, will have the bi-sequence property. In an effort to stimulate the further study of the bi-sequence property let us pose a rather explicit question, which we have no great expectation of being true. Questions 3.5. Can every Banach lattice with the bi-sequence property be written as the direct sum of a Banach lattice with the positive Schur property and one with the dual positive Schur property? Just as there are Banach lattices with the positive Schur Property other than the classical examples of AL-spaces, so we expect there to be other examples of Banach lattices with the dual positive Schur property. We limit ourselves to looking amongst the rather obvious candidates, AM-spaces, and asking which of these have the dual positive Schur property. It might have been thought that only those with an order unit would have the dual positive Schur property. Such a guess turns out to be only partly correct.

5 POSITIVE ALMOST DUNFORD-PETTIS OPERATORS AND THEIR DUALITY.5 Proposition 3.6. An AM-space E with a topological order unit has the dual positive Schur property if and only if E has a strong order unit. Proof. If E has a strong order unit then, up to an equivalent norm, we are in the C(K) case which we know has the dual positive Schur property and we need only note that the dual positive Schur property is preserved when we change to an equivalent norm. Suppose that e is a topological order unit for E. Let Φ be the set of non-zero extreme points of the positive part of the unit ball in E, all of which have norm 1. If there is α > 0 such that φ(e) α for all φ Φ then e is a strong order unit for E. If not, pick a sequence (φ n ) in Φ with φ n (e) 0. For any x E + and ɛ > 0, let m N be such that x x (me) < ɛ/2, which is possible as e is a topological order unit for E. Now pick n 0 such( that n > ) n 0 implies that φ n (e) ɛ/(2m) which in turn means that φ n x (me) φn (me) = mφ n (e) m ( ɛ/(2m) ) = ɛ/2. We now see that φ n (x) < ɛ if n > n 0, showing that (φ n ) is weak null, even though every φ n has norm equal to 1, so that the dual positive Schur property does not hold. In particular the only separable AM-spaces with the dual positive Schur property are those which are isomorphic to C(M) for M a compact metric space. If we do not make the assumption that the AM-space E has a topological order unit then the conclusion of Proposition 3.6 fails, even if we were to restrict our attention to the next simplest class of AM-spaces which are isomorphic to C 0 (Σ) for some locally compact Hausdorff space Σ. Proposition 3.7. There is a non-compact locally compact Hausdorff space Σ such that C 0 (Σ) has the dual positive Schur property. Proof. Let I be an uncountable set and take E to be the Banach lattice of all bounded functions on I of countable support, with the pointwise order and supremum norm. E is a Dedekind complete AM-space with a norm that is Fatou, and therefore satisfies Nakano s condition ([6] and [7], or see [9]) for E to be isometrically isomorphic to C 0 (Σ) for some locally compact Hausdorff space Σ. It is clear that E does not have a strong order unit, so that the space Σ is not compact. We need only prove that E has the dual positive Schur property. We claim first that if f E + then there is a countable subset C f I such that if x E is disjoint from χ Cf then f(x) = 0. By Zorn s lemma, there is a maximal disjoint subset A of elements x E + with x = 1 and f(x) > 0. Let A n = {x A : f(x) > 1/n}, so that A = n=1 A n. If A is uncountable, then there is m N such that A m is uncountable.

6 6 BELMESNAOUI AQZZOUZ, AZIZ ELBOUR, AND ANTHONY W. WICKSTEAD In particular, A m is infinite so that we can find a sequence (x n ) in A m. As E is an AM-space, its dual is an AL-space so has order continuous norm and we can apply Theorem of [5] to conclude that x n 0 weakly. But f(x n ) > 1/m for all n N, so that f(x n ) f(0), which is a contradiction. Take C f to be the union of the supports of all x A, which is countable as each x has countable support and there are only countably many x. If y E + with y χ Cf then y x for all x A, so if f(y) > 0 we would contradict the maximality of A. Now let (f n ) be a weak null sequence in E +. Let J = n=1 C f n, which is countable so that χ J E. If n N and ɛ > 0 there is x E + with x 1 and f n (x) > f n ɛ. We may write x = x χ J +x χ I\J. As x χ I\J χ J and therefore x χ I\J χ Cfn we have f n (x χ I\J ) = 0 so that f n f n (χ J ) f n (x χ J ) = f n (x) f n ɛ. It follows that f n = f n (χ J ) for all n N. The conclusion is now clear. Questions 3.8. (1) Characterize those AM-spaces which have the dual positive Schur property. (2) Characterize the Banach lattices which have the dual positive Schur property. Note that a closed order ideal in a Banach lattice with the dual positive Schur property must itself have the dual positive Schur property. No similar statement is true for the bi-sequence property, In fact, l has the bi-sequence property but c 0 does not. It is easy to see that if E has the dual positive Schur property then E has the positive Schur property. The converse is false in general as c 0 does not have the dual positive Schur property even though c 0 = l 1 has the full Schur property. 4. Almost Dunford-Pettis operators into c 0. The following characterization, which we will use repeatedly, is just Remark 1 of [13]. Lemma 4.1. An operator T from a Banach lattice E into a Banach space F is almost Dunford-Pettis if and only if T (x n ) 0 for every weakly null disjoint sequence (x n ) in E +. Theorem 4.2. The following assertions about a Banach lattice E are equivalent: (1) Every positive operator T : E c 0 is almost Dunford-Pettis. (2) E has the bi-sequence property.

7 POSITIVE ALMOST DUNFORD-PETTIS OPERATORS AND THEIR DUALITY.7 (3) For each weak null disjoint sequence (x n ) in E + and each weak* null sequence (f n ) in E +, we have lim (sup{f n(x m ) : n N}) = 0. m + Proof. (1) (2) Let (x n ) be a weak null disjoint sequence in E and (f n ) a weak* null sequence in E +. Consider the operator T : E c 0 defined by T (x) = ( f k (x) ). k=1 It is clear that T is positive and does take values in c 0. By our hypothesis T is almost Dunford-Pettis. Thus, since (x n ) is a weakly null disjoint sequence in E, T (x n ) 0. As T (x n ) f n (x n ) for all n, we conclude that f n (x n ) 0 and we are done. (2) (3) Let (x n ) be a weakly null disjoint sequence in E + and (f n ) a weak* null sequence in E +. Let A be the solid hull of the relatively weakly compact subset W = {x n : n N}. It is immediate that sup{f n (x) : x A} is finite for each n = 1, 2,..., as A is norm bounded. We also have f n (x) 0 for each x A, as f n 0 weakly. It follows from Theorem 4.34 of [1] that every disjoint sequence (y n ) in A converges weakly to zero, so it is immediate from (2) that f n (y n ) 0 for every disjoint sequence (y n ) in A. We may thus use Proposition 2.2 to conclude that lim sup{f n(x) : x A} = 0. n If ε > 0 there thus exists n 0 such that sup{f n (x) : x A} ε for all n > n 0. In particular, 0 f n (x m ) ε for all n > n 0 and all m N. On the other hand, since (x m ) is weakly null then there exists m 0 such that 0 f n (x m ) ε for all m > m 0 and n = 1, 2,..., n 0. Thus 0 f n (x m ) ε for all n N and all m > m 0. Hence 0 sup{f n (x m ) : n N} ε for all m > m 0. This implies that lim (sup{f n(x m ) : n N}) = 0. m (3) (1) Every positive operator T : E c 0 is determined by a weak* null sequence (f n ) in E +, i.e. T (x) = ( f k (x) ) for all x E. k=1 Let (x n ) be a weakly null disjoint sequence in E + and note that T (x n ) = ( f k (x n ) ) = sup{f k (x n ) : k N} for all n N. k=1 By (3) we conclude that T (x n ) 0, so that T is almost Dunford- Pettis and we are done.

8 8 BELMESNAOUI AQZZOUZ, AZIZ ELBOUR, AND ANTHONY W. WICKSTEAD If one assumes in Theorem 4.2 that the Banach lattice E has order continuous norm, then we may make use of Proposition 3.2 to obtain the following, slightly improved, result Corollary 4.3. Let E be a Banach lattice with order continuous norm. Then the following assertions are equivalent: (1) Each bounded operator T : E c 0 is almost Dunford-Pettis. (2) Each positive operator T : E c 0 is almost Dunford-Pettis. (3) E has the positive Schur property. Proof. The only extra ingredient here is the fact that all bounded operators are almost Dunford-Pettis. If T : E c 0 is bounded and (x n ) is a weakly null disjoint sequence (x n ) in E + then x n 0 as E has the positive Schur property so that T (x n ) 0 and T is almost Dunford-Pettis. 5. Duality of positive almost Dunford-Pettis operators. We first address the question of when the adjoint of a positive almost Dunford-Pettis operator has to be almost Dunford-Pettis. Theorem 5.1. Let E and F be two Banach lattices. The following conditions are equivalent: (1) If T : E F is a positive almost Dunford-Pettis operator then the adjoint T : F E is almost Dunford-Pettis. (2) At least one of the following assertions is valid: (a) E has order continuous norm. (b) F has the positive Schur property. Proof. Suppose that (2)(a) holds and that T : E F is positive and almost Dunford-Pettis. Let (f n ) be a disjoint sequence in F + such that f n 0 for σ(f, F ). We have to prove that T (f n ) 0. It is clear that 0 T (f n ) 0 for σ(e, E). By Proposition 2.4, it suffices to show that T (f n )(x n ) 0 for each disjoint norm bounded sequence (x n ) in E +. As the norm of E is order continuous, it follows from Corollary of [5] that for such a sequence we have x n 0 for σ(e, E ). Now, since T is almost Dunford-Pettis, T (x n ) 0. As f n 0 for σ(f, F ), (f n ) is norm bounded. Hence T (f n )(x n ) = f n ( T (xn ) ) 0 and we are done. To see that (2)(b) implies (1), we will prove that in fact the adjoint of every operator T : E F is almost Dunford-Pettis. To see this, let (f n ) be a disjoint sequence in F + such that f n 0 for σ(f, F ). Since F has the positive Schur property, f n 0 and hence T (f n ) 0 from which the result follows.

9 POSITIVE ALMOST DUNFORD-PETTIS OPERATORS AND THEIR DUALITY.9 If we assume that (1) holds but (2) fails, then the norm of E is not order continuous and F does not have the positive Schur property. Since the norm of E is not order continuous, it follows from Theorem of [5] that there exists a positive order bounded disjoint sequence (ϕ n ) of E + with ϕ n = 1 for all n. Let 0 ϕ E be such that 0 ϕ n ϕ for all n. Define the operator U : E l 1 by U(x) = (ϕ n (x)) n=1 for x E. Since n=1 ϕ n(x) n=1 ϕ n( x ) ϕ( x ) for each x E, the operator U does indeed take values in l 1 and it is clearly positive. On the other hand, since F does not have the positive Schur property, it follows from Proposition 2.1 that there exists a disjoint weakly null sequence (f n ) in F + with f n = 1 for all n. As f n = sup{f n (y) : y F +, y = 1}, for each n there exists y n F + with y n = 1 and f n (y n ) 1 2. Define a positive operator V : l 1 F by V ( (λ n ) ) = n=1 λ ny n, which series is certainly norm convergent. Let T = V U : E l 1 F so that T (x) = n=1 ϕ n(x)y n for x E. It is clear that T is Dunford-Pettis as if x n 0 weakly in E then V (x n ) 0 weakly in l 1 and therefore, as l 1 has the Schur property, V (x n ) 0 and hence T (x n ) = ( U V (xn ) ) 0. So certainly T is almost Dunford-Pettis. But its adjoint T : F E is not almost Dunford-Pettis. To see this, note that if h F then T (h) = n=1 h(y n)ϕ n. In particular, for every k we have T (f k ) = n=1 f k(y n )ϕ n f k (y k )ϕ k 0 and hence T (f k ) f k (y k )ϕ k = f k (y k ) 1. As (f 2 k) is disjoint and weakly null, T is not almost Dunford-Pettis. Corollary 5.2. The following conditions on a Banach lattice E are equivalent: (1) If T : E E is a positive almost Dunford-Pettis operator then the adjoint T : E E is almost Dunford-Pettis. (2) E has order continuous norm. Now, we study the converse situation. We begin by giving sufficient conditions for which a positive operator, whose adjoint is almost Dunford-Pettis, must itself be almost Dunford-Pettis. Theorem 5.3. Let E and F be Banach lattices. If at least one of the following three conditions holds then each positive operator T : E F, such that T : F E is almost Dunford-Pettis, must itself be almost Dunford-Pettis. (1) E has the positive Schur property. (2) E has the bi-sequence property and F has order continuous norm.

10 10BELMESNAOUI AQZZOUZ, AZIZ ELBOUR, AND ANTHONY W. WICKSTEAD (3) F may be written as an order direct sum F = G H where the bidual G is a KB-space and H has the positive Schur property. Proof. In case (1), every weakly null sequence in E + is norm null, so that every bounded operator T : E F is almost Dunford-Pettis. In case (2), again every positive operator T : E F is almost Dunford-Pettis. In fact, let (x n ) be a weakly null disjoint sequence in E +. It is clear that 0 T (x n ) ( 0 for σ(f, F ). Thus by Proposition 2.3, it suffices to show that f n T (xn ) ) 0 for each disjoint norm bounded sequence (f n ) in F +. Since the norm in F is order continuous, it ( follows from Theorem of [5] that (f n ) is weak* null, and hence T (f n ) ) ( is weak* null in E +. Finally, by (2)(b), we have f n T (xn ) ) = T (f n )(x n ) 0 and we are done. In case (3), consider first operators into G where we know that G is a KB-space and therefore has order continuous norm. Let T : E G be a positive operator such that T : G E is almost Dunford- Pettis. Let (x n ) be a weakly null disjoint sequence in E + and note that 0 T (x n ) 0 for σ(g, G ). Let (f n ) be a disjoint norm bounded sequence in G +. Since the norm of G is order continuous, it follows from Theorem of [5] that (f n ) is a weakly null sequence in G. Now, as the adjoint T : G E ( is almost Dunford-Pettis, we conclude that T (f n ) 0. Hence f n T (xn ) ) = T (f n )(x n ) 0 and we use Proposition 2.3 to conclude that T (x n ) 0 showing that T is almost Dunford-Pettis. If H has the positive Schur property then every positive operator T : E H is almost Dunford-Pettis. For if (x n ) is a weakly null disjoint sequence in E +, note that 0 T (x n ) 0 for σ(h, H ) so that T (x n ) 0 and T is almost Dunford-Pettis. To prove (3) in general, suppose that T : E G H is positive and such that T : G H E is almost Dunford-Pettis. We will let P B denote the band projection onto the band B. It is clear that both T G and T H are almost Dunford-Pettis from which our previous work shows that both P G T and P H T are almost Dunford-Pettis hence so is T itself. If we assume that the Banach lattice F is Dedekind σ-complete, we obtain the following necessary conditions. Theorem 5.4. Let E and F be two Banach lattices such that F is Dedekind σ-complete. If every positive operator T : E F, for which T : F E is almost Dunford-Pettis, must itself be almost Dunford- Pettis then one of the following assertions holds: (1) E has the positive Schur property.

11 POSITIVE ALMOST DUNFORD-PETTIS OPERATORS AND THEIR DUALITY.11 (2) E has the bi-sequence property and F has order continuous norm. (3) F is a KB-space. Proof. Under the stated assumptions, it suffices to establish the following two separate claims. (α) If the norm of F is not order continuous, then E has the positive Schur property. (β) If F is not a KB-space, then f n (x n ) 0 for each weakly null disjoint sequence (x n ) in E and each weak* null sequence (f n ) in E +. Assume that E does not have the positive Schur property and that the norm of F is not order continuous. Since E does not have the positive Schur property, it follows from Proposition 2.1 that there is a disjoint weakly null sequence (x n ) in E + with x n = 1 for all n. Hence, by Proposition 2.5, there exists a positive disjoint sequence (g n ) in E with g n = g n (x n ) = 1 for all n and g n (x m ) = 0 if n m. Consider the operator U : E l defined by U(x) = ( g n (x) ) for each x E n=1 which is clearly a positive operator taking values in l. On the other hand, since the norm of F is not order continuous, it follows from Theorem of [5] that there exists an order bounded disjoint sequence (y n ) in F + which is not norm convergent to zero. We can assume, without loss of generality, that y n = 1 and that there is y F + with 0 y n y for all n. Thus, since F is Dedekind σ- complete, it results from the proof of Theorem of Zaanen, [14], that the positive operator V : l F defined by the order convergent series V ( (λ n ) ) = n=1 λ ny n, for (λ n ) l, is a lattice isomorphism of l into F. Now, we consider the positive operator T = V U : E l F defined by the order convergent series T (x) = n=1 g n(x)y n for x E. Its adjoint T : F l E is almost Dunford-Pettis. In fact, if (f n ) is a weakly null disjoint sequence in F +, then 0 V (f n ) 0 for σ(l, l ). Since l has the positive Schur property, it follows that V (f n ) 0. Hence T (f n ) = ( U V (f n ) ) 0 and T is almost Dunford-Pettis. However, note that (x n ) is a disjoint weakly null sequence in E + and that U(x n ) = e n, where e n is the n th standard basis vector in l. Thus T (x n ) = ( V U(xn ) ) = V (en ) = y n = 1 for each n so that T is not almost Dunford-Pettis. We have now established claim (α). To prove claim (β), let us suppose that F is not a KB-space. Then it follows from, for example, Theorem 4.61 of [1] that c 0 is lattice

12 12BELMESNAOUI AQZZOUZ, AZIZ ELBOUR, AND ANTHONY W. WICKSTEAD embeddable in F i.e. there exists a lattice homomorphism S : c 0 F and two strictly positive constants K and M such that K (α n ) S ( (αn ) ) M (αn ) for all (α n ) c 0. Let (x n ) be a weakly null disjoint sequence in E and (f n ) a weak* null sequence in E +. We need only prove that f n (x n ) 0. It is clear that the operator R : E c 0, defined by R(x) = ( f k (x) ) is positive k=1 and does indeed map E linearly into c 0. Let T = S R : E c 0 F. It is clear that T : F c 0 = l 1 E is Dunford-Pettis and hence almost Dunford-Pettis. By our assumption T is almost Dunford-Pettis. Thus, since (x n ) is a weakly null disjoint sequence in E, we have T (x n ) 0. But, for all n, T (x n ) = S R(x n ) = S ( (f k (x n )) k=1) K ( f k (x n ) ) K f n (x n ) so that f n (x n ) 0. As a consequence of Theorem 5.4, we obtain the following characterization of the positive Schur property which supplements our results from 4. Corollary 5.5. The following conditions on a Banach lattice E are equivalent: (1) Every bounded operator T : E l is almost Dunford-Pettis. (2) Every positive operator T : E l is almost Dunford-Pettis. (3) E has the positive Schur property. Proof. The equivalence of (2) and (3) follows from the theorem as l does not have order continuous norm so is certainly not a KB-space. Clearly (1) (2), whilst (2) (1) is because every bounded operator into l is the difference of two positive operators. The assumption that F is Dedekind σ-complete is essential for Theorem 5.4 to hold. For instance, if we take E = l and F = c, the Banach lattice of all convergent sequences then it is clear that every operator T : l c is weakly compact, see the proof of Proposition 1 in [12], hence is Dunford-Pettis, as l has the Dunford-Pettis property and therefore is almost Dunford-Pettis. Yet none of the three possible conditions listed in Theorem 5.4 holds. How do Theorems 5.3 and 5.4 match up to each other? Apart from the assumption of Dedekind σ-completeness in the latter case, the gap between the two results is that one of the necessary conditions is that F be a KB-space whilst a sufficient condition is that F be the sum G H where G is a KB-space and H has the positive Schur property. k=1

13 POSITIVE ALMOST DUNFORD-PETTIS OPERATORS AND THEIR DUALITY.13 We have no example of a KB-space which cannot be written as such a sum. Now, if in Theorem 5.4, instead of assuming that the Banach lattice F is Dedekind σ-complete, we assume that the Banach lattice E has order continuous norm, we obtain a slightly different set of necessary conditions. Theorem 5.6. Let E and F be Banach lattices such that E has order continuous norm. If each positive operator T : E F, such that T : F E is almost Dunford-Pettis, must itself be almost Dunford- Pettis then one of the following assertions is valid : (1) E has the positive Schur property. (2) F is a KB-space. Proof. Suppose that E does not have the positive Schur property and F is not a KB-space. Since E does not have the positive Schur property, it follows from Proposition 2.1 that there exists a disjoint weakly null sequence (x n ) in E + with x n = 1 for all n. Hence, by Proposition 2.5 there exists a positive disjoint sequence (g n ) in E with g n = g n (x n ) = 1 for all n and g n (x m ) = 0 for n m. As the norm in E is order continuous, it follows from Corollary of [5] that g n 0 for σ(e, E). Hence we may define a positive operator U : E c 0 by U(x) = ( g n (x) ) for x E. On the other hand, since n=1 F is not a KB-space, it follows from Theorem 4.61 of [1] that c 0 is lattice embeddable in F so there exists a lattice embedding V : c 0 F and strictly positive constants K and M such that K (α n ) V ( (αn ) ) M (αn ) for all (α n ) c 0. Now consider the positive operator T = V U : E c 0 F. It is clear that its adjoint T : F c 0 = l 1 E is Dunford-Pettis and hence almost Dunford-Pettis. But T is not almost Dunford-Pettis. To see this, note that (x n ) is a disjoint weakly null sequence in E +, that as above we have U(x n ) = e n and that T (x n ) = V ( U(x n ) ) = V (e n ) K e n = K > 0 for each n. Thus T (x n ) 0. The example given after Theorem 5.4 shows also that the hypothesis of order continuity of the norm may not be omitted from the statement of Theorem 5.6 A consequence of Theorem 5.6 gives us a characterization of KBspaces.

14 14BELMESNAOUI AQZZOUZ, AZIZ ELBOUR, AND ANTHONY W. WICKSTEAD Corollary 5.7. The following conditions on a Banach lattice F are equivalent: (1) Every bounded operator T : c 0 F is compact. (2) Every bounded operator T : c 0 F is almost Dunford-Pettis. (3) Every positive operator T : c 0 F is compact. (4) Every positive operator T : c 0 F is almost Dunford-Pettis. (5) F is a KB-space. Proof. That (1) (2) (4) and that (1) (3) (4) are clear. As c 0 does not have the positive Schur property, (4) (5) follows immediately from Theorem 5.6. If F is a KB-space and T : c 0 F is any bounded operator then Theorem 5.27 of [1] tells us that T is weakly compact. Its adjoint T : F c 0 = l 1 is thus also weakly compact and hence compact, because l 1 has the Schur property, so that T itself is compact. References [1] C. D. Aliprantis and O. Burkinshaw, Positive operators, Springer, Dordrecht, Reprint of the 1985 original. MR [2] B. Aqzzouz, K. Bourass, and A. Elbour, Some generalizations on positive Dunford-Pettis operators, Results in Mathematics (to appear in 2009). [3] P. G. Dodds and D. H. Fremlin, Compact operators in Banach lattices, Israel J. Math. 34 (1979), no. 4, (1980). MR (81g:47037) [4] Bengt Josefson, Weak sequential convergence in the dual of a Banach space does not imply norm convergence, Ark. Mat. 13 (1975), MR (51 #11067) [5] P. Meyer-Nieberg, Banach lattices, Universitext, Springer-Verlag, Berlin, MR (93f:46025) [6] H. Nakano, Über die Charakterisierung des allgemeinen C-Raumes, Proc. Imp. Acad. Tokyo 17 (1941), (German). MR (7,249h) [7], Über normierte teilweisegeordnete Moduln, Proc. Imp. Acad. Tokyo 17 (1941), (German). MR (7,249g) [8] A. Nissenzweig, W sequential convergence, Israel J. Math. 22 (1975), no. 3-4, MR (52 #14939) [9] A. W. Wickstead, An isomorphic version of Nakano s characterization of C 0 (Σ), Positivity 11 (2007), no. 4, MR (2008g:46032) [10] W. Wnuk, Some characterizations of Banach lattices with the Schur property, Rev. Mat. Univ. Complut. Madrid 2 (1989), no. suppl., Congress on Functional Analysis (Madrid, 1988). MR (91f:46033) [11], Banach lattices with properties of the Schur type a survey, Confer. Sem. Mat. Univ. Bari (1993), no. 249, 25. MR (94h:46031) [12], Remarks on J. R. Holub s paper concerning Dunford-Pettis operators, Math. Japon. 38 (1993), no. 6, MR (94i:46028) [13], Banach lattices with the weak Dunford-Pettis property, Atti Sem. Mat. Fis. Univ. Modena 42 (1994), no. 1, MR (95g:46034)

15 POSITIVE ALMOST DUNFORD-PETTIS OPERATORS AND THEIR DUALITY.15 [14] A. C. Zaanen, Riesz spaces. II, North-Holland Mathematical Library, vol. 30, North-Holland Publishing Co., Amsterdam, MR (86b:46001) Université Mohammed V-Souissi, Faculté des Sciences Economiques, Juridiques et Sociales, Département d Economie, B.P. 5295, SalaAljadida, Morocco. address: baqzzouz@hotmail.com Université Ibn Tofail, Faculté des Sciences, Département de Mathématiques, B.P. 133, Kénitra, Morocco. address: azizelbour@hotmail.com Pure Mathematics Research Centre, Queens University Belfast, Northern Ireland. address: A.Wickstead@qub.ac.uk