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```DEMONSTRATIO MATHEMATICA
Vol. XLVI
No 1
2013
Belmesnaoui Aqzzouz, Khalid Bouras
WEAK AND ALMOST DUNFORD–PETTIS OPERATORS
ON BANACH LATTICES
Abstract. We introduce the class of Banach lattices with the AM-compactness property and we use it to characterize Banach lattices on which each positive weak Dunford–
Pettis operator is almost Dunford–Pettis and conversely.
1. Introduction and notation
Almost Dunford–Pettis operators and weak Dunford–Pettis operators are
two different classes that contain strictly the space of Dunford–Pettis operators. The first one is introduced as a class that defines the positive Schur
property and the second one as that defines Dunford–Pettis property.
Let us recall from [14] that an operator T from a Banach lattice E
into a Banach space F is said to be almost Dunford–Pettis, if the sequence
(kT (xn )k) converges to 0 for every weakly null sequence (xn ) consisting of
pairewise disjoint elements in E.
Also, recall from [2] that an operator T from a Banach space X into
another Banach space Y is called weak Dunford–Pettis, if the sequence
(yn′ (T (xn ))) converges to 0 whenever (xn ) converges weakly to 0 in X and
(yn′ ) converges weakly to 0 in Y . Alternatively, T is weak Dunford–Pettis, if
T maps relatively weakly compact sets of X into Dunford–Pettis sets of Y
(see Theorem 5.99 of [2]).
The common between these two classes of operators is that each one of
them satisfies the domination problem (see [4] (resp. [10])) and any one of
them satisfies the duality problem (see [6] (resp. [5])).
As announced at the beginning, an almost Dunford–Pettis operator is
not necessary weak Dunford–Pettis, and conversely, a weak Dunford–Pettis
operator is not necessary almost Dunford–Pettis. For an example, the iden2000 Mathematics Subject Classification: 46A40, 46B40, 46B42.
Key words and phrases: almost Dunford–Pettis operator, weak Dunford–Pettis operator, order continuous norm, discrete vector lattice.
166
B. Aqzzouz, K. Bouras
tity operator of the Lorentz space ∧(ω, 1) is almost Dunford–Pettis (because
∧(ω, 1) has the positive Schur property), but it is not weak Dunford–Pettis
(because ∧(ω, 1) does not has the Dunford–Pettis property) (see Remark 3
of [13], p. 19) and conversely, the identity operator of the Banach lattice c0
is weak Dunford–Pettis but it is not almost Dunford–Pettis.
However, if the Banach lattice F is discrete and reflexive (for example if
F = l2 ), the class of positive almost Dunford–Pettis operators from a Banach lattice E into F coincides with that of positive weak Dunford–Pettis
operators from E into F .
The main purpose of this work is to characterize Banach lattices on
which all positive weak Dunford–Pettis operators are almost Dunford–Pettis
and conversely. To do this, we introduce a new property that we call the
AM-compactness property of a Banach lattice. Also, we give some important
consequences. More precisely, we will begin by characterizing Banach lattices
on which the class of almost Dunford–Pettis operators coincides with that of
Dunford–Pettis operators. After that, we introduce Banach lattices with the
AM-compactness property, and we use this to characterize Banach lattices on
which, each positive almost Dunford–Pettis operator is weak Dunford–Pettis.
Finally, we study the converse situation.
To state our results, we need to fix some notations and recall some definitions. A vector lattice E is Dedekind σ-complete, if every nonempty
countable subset, that is bounded from above, has a supremum. A nonzero
element x of a vector lattice E is discrete, if the order ideal generated by x
equals the subspace generated by x. The vector lattice E is discrete, if it
admits a complete disjoint system of discrete elements. A Banach lattice is
a Banach space (E, k.k) such that, E is a vector lattice and its norm satisfies the following property: for each x, y ∈ E such that |x| ≤ |y|, we have
kxk ≤ kyk. If E is a Banach lattice, its topological dual E ′ , endowed with the
dual norm, is also a Banach lattice. A norm k.k of a Banach lattice E is order
continuous if for each generalized sequence (xα ) such that xα ↓ 0 in E, the sequence (xα ) converges to 0 for the norm k.k, where the notation xα ↓ 0 means
that the sequence (xα ) is decreasing, its infimum exists and inf(xα ) = 0. We
refer to [2] for unexplained terminology on Banach lattices theory.
2. Dunford–Pettis property of almost Dunford–Pettis operators
We will use the term operator T : E → F between two Banach lattices to mean a bounded linear mapping. It is positive if T (x) ≥ 0 in F
whenever x ≥ 0 in E. It is well known that each positive linear mapping
on a Banach lattice is continuous. If an operator T : E → F between
two Banach lattices is positive, then its adjoint T ′ : F ′ → E ′ is likewise
positive.
Weak and almost Dunford–Pettis operators on Banach lattices
167
We need to recall the following characterization of almost Dunford–Pettis
operators established in ([4], Theorem 2.2).
Theorem 2.1. [4] An operator T from a Banach lattice E into a Banach
space X is almost Dunford–Pettis if and only if kT (xn )k → 0 as n → ∞ for
every weakly null sequence (xn )n ⊂ E + .
An operator T : X → Y between two Banach spaces is called Dunford–
Pettis if the image of any weakly null sequences of X is norm convergent
to 0 in Y . Alternatively, T is Dunford–Pettis if T maps relatively weakly
compact sets of X into relatively compact sets of Y .
Note that every Dunford–Pettis operator from a Banach lattice E into
another F is almost Dunford–Pettis, but the converse is false in general.
In fact, since the Banach lattice L1 ([0, 1]) has the positive Schur property,
its identity operator IdL1 ([0,1]) : L1 ([0, 1]) → L1 ([0, 1]) is almost Dunford–
Pettis. But it is not Dunford–Pettis, because L1 ([0, 1]) does not have the
Schur property.
The following result gives some sufficient conditions under which each
almost Dunford–Pettis operator from E into F is Dunford–Pettis.
Theorem 2.2. Let E and F be two Banach lattices.
(1) If the lattice operations in E are weakly sequentially continuous, then
each almost Dunford–Pettis operator from E into F is Dunford–Pettis.
(2) If F is discrete and its norm is order continuous, then each positive
almost Dunford–Pettis operator from E into F is Dunford–Pettis.
Proof. (1) Let T : E → F be an almost Dunford–Pettis operator. We
have to show that T is Dunford–Pettis, let (xn ) be a weakly null sequence
in E. As the lattice operations in E are weakly sequentially continuous,
−
′
then x+
n → 0 and xn → 0 in σ(E, E ). And it follows from Theorem 2.1 that
−
kT (x+
n )k → 0 and kT (xn )k → 0. Therefore,
− + − kT (xn )k = T (x+
n ) − T (xn ) ≤ T (xn ) + T (xn ) → 0 as n → ∞
holds and the proof is finished.
(2) Let T : E → F be a positive almost Dunford–Pettis operator. We
have to show that T is Dunford–Pettis, let W be a relatively weakly compact
set in E and let A be the solid hull of W in E. By Theorem 4.34 of [2],
zn → 0 weakly for every disjoint sequence (zn ) ⊆ A. And as T is almost
Dunford–Pettis, we obtain kT (zn )k → 0. Thus, by Theorem 4.36 of [2],
for each ε > 0 there exists some u ≥ 0, in the order ideal generated by A,
satisfying kT [(|x| − u)+ ]k < ε for all x ∈ A.
Therefore, from |x| = |x| ∧ u + (|x| − u)+ , we obtain
T (|x|) = T (|x| ∧ u) + T [(|x| − u)+ ],
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B. Aqzzouz, K. Bouras
and then
T (|x|) ∈ [−T (u), T (u)] + ε.BF
for all x ∈ A (where BF is the closed unit ball of F ). By |T (x)| ≤ T (|x|)
(because T is positive), and as [−T (u), T (u)] + ε.BF is solid set in F , we
have
T (x) ∈ [−T (u), T (u)] + ε.BF
for all x ∈ A, hence
T (W ) ⊂ [−T (u), T (u)] + ε.BF .
Finally, as F is discrete with an order continuous norm, it follows from
Theorem 6.1 of [15], that [−T (u), T (u)] is norm compact. So, by Theorem
3.1 of [2], T (W ) is norm relatively compact. This proves that T maps
relatively weakly compact sets of E into relatively compact sets of F . Thus
T is a Dunford–Pettis operator.
Remark. Let E and F be two Banach lattices. Even if each positive
almost Dunford–Pettis operator T from E into F is Dunford–Pettis, the
lattice operations in E are not necessary weakly sequentially continuous and
F is not necessary discrete.
In fact, we consider E = L1 [0, 1] and F = L2 [0, 1], then each operator
from L1 [0, 1] into L2 [0, 1] is Dunford–Pettis, but the lattice operations in
E = L1 [0, 1] fail to be weakly sequentially continuous and F = L2 [0, 1] is
not discrete.
Whenever the Banach lattice F is discrete or its lattice operations are
weakly sequentially continuous in Theorem 2.2, we obtain the following theorem.
Theorem 2.3. Let E and F be two Banach lattices such that F is discrete or its lattice operations are weakly sequentially continuous. Then the
following assertions are equivalent:
(i) Each positive almost Dunford–Pettis operator from E into F is Dunford–Pettis.
(ii) One of the following assertions is valid:
(1) The lattice operations in E are weakly sequentially continuous.
(2) The norm of F is order continuous.
Proof. (1)⇒(i) Follows from Theorem 2.2 (1).
(2)⇒(i) Since the norm of F is order continuous, then Corollary 2.3 of
Chen–Wickstead [8] proves that F is discrete iff the lattice operations in F
are weakly sequentially continuous. So we can assume that F is discrete and
hence the result follows from Theorem 2.2 (2).
Weak and almost Dunford–Pettis operators on Banach lattices
169
(i)⇒(ii) Let T and S be two operators from E into F such that 0 ≤ S ≤ T
and T is Dunford–Pettis. As the class of almost Dunford–Pettis operators
between Banach lattices satisfies the domination problem, the operator S is
almost Dunford–Pettis, and hence by i) S is Dunford–Pettis. Finally, Theorem 2 of [12] proves that the lattice operations in E are weakly sequentially
continuous or the norm on F is order continuous.
As a consequence of Theorem 2.3, we obtain the following characterization.
Corollary 2.4. Let E be a Banach lattice. Then the following assertions
are equivalent:
(i) Each positive almost Dunford–Pettis operator from E into l∞ is Dunford–Pettis.
(ii) The lattice operations in E are weakly sequentially continuous.
Proof. Since the Banach lattice l∞ does not have an order continuous norm,
the result follows from Theorem 2.3.
3. Banach lattices with the AM-compactness property
Let us recall from [2] that a norm bounded subset A of a Banach space X
is said to be Dunford–Pettis whenever every weakly compact operator from
X into 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 i.e. supx∈A |fn (x)| → 0 (see Theorem 5.98 of [2]).
We need to introduce the class of Banach lattices with AM-compactness
property. For this, let us recall that an operator T from a Banach lattice
E into a Banach space X is said to be AM-compact if it carries each order
bounded subset of E onto a relatively compact set of X.
A Banach lattice E is said to have the AM-compactness property if every
weakly compact operator defined on E (and taking their values in a Banach
space X) is AM-compact.
As examples, the Banach lattices c0 , ℓ1 , c and c′ have the AM-compactness property, but ℓ∞ does not have this property.
The following proposition gives a characterization of this property.
Proposition 3.1. Let E be a Banach lattice, the following statements are
equivalent:
(i) E has the AM-compactness property.
(ii) For every weakly null sequence (fn ) of E ′ , we have |fn | → 0 for σ(E ′ , E).
(iii) For each x ∈ E + , [−x, x] is a Dunford–Pettis set.
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B. Aqzzouz, K. Bouras
Proof. (i)⇒(ii) Let (fn ) be a weakly null sequence of E ′ , and consider the
operator S : E → c0 defined by
S(x) = (fn (x))∞
n=0
for each x ∈ E.
Then S is weakly compact (Theorem 5.26 of [2]). But according to our
hypothesis, S([−x, x]) is a norm relatively compact subset of c0 for each
x ∈ E + . From this, it follows that |fn | (x) = supz∈[−x,x] |fn (z)| → 0 for each
x ∈ E + (see Exercise 14 of Section 3.2 of [2]).
(ii)⇒(iii) For each x ∈ E + , supy∈[−x,x] |fn (y)| = |fn | (x) → 0 for every
weakly null sequence (fn ) of X ′ . This shows that [−x, x] is a Dunford–Pettis
set for each x ∈ E + .
(iii)⇒(i) Let S be a weakly compact operator from E into an arbitrary
Banach space X. It follows from iii) that for each x ∈ E + , [−x, x] is a
Dunford–Pettis set, and hence S([−x, x]) is a norm relatively compact subset
of Z. This proves that S is an AM-compact operator.
Another property that we need is the following lemma
Lemma 3.2. Let E be a Banach lattice and let A be an order bounded set
of E. If A is |σ| (E, E ′ )-totally bounded, then A is a Dunford–Pettis set.
Proof. Let (fn ) be a weakly null sequence in E ′ , let x ∈ E + such that
|y| ≤ x for every y ∈ A, and fix ε > 0. By Theorem 4.37 of [2] there exists
some f ∈ (E ′ )+ such that
(|fn | − f )+ (x) <
ε
4
for each n.
Since A is |σ| (E, E ′ )-totally bounded, there exists a finite set {x1 , . . . , xk }
⊂ A such that for each z ∈ A we have f (|z − xi |) < 4ε for at least one
1 ≤ i ≤ k. Since fn → 0 weakly, there exists some N with |fn (xi )| < 4ε for
each i = 1, . . . , k and all n ≥ N .
Now, let z ∈ A. Choose 1 ≤ i ≤ k with f (|z − xi |) < 4ε , and note that
|z − xi | ≤ 2x holds. In particular, for n ≥ N we have
ε
|fn (z)| ≤ |fn (z − xi )| + |fn (xi )| ≤ |fn | (|z − xi |) +
4
ε
+
≤ (|fn | − f ) (|z − xi |) + f (|z − xi |) +
4
ε ε
+
≤ 2(|fn | − f ) (x) + + ≤ ε.
4 4
This implies supz∈A |fn (z)| → 0, and then A is a Dunford–Pettis set.
Recall that if E is a Banach lattice, then
– the lattice operations in E ′ are called 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 ′′ ).
Weak and almost Dunford–Pettis operators on Banach lattices
171
– the lattice operations in E ′ 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).
Also, a Banach space X has the Dunford–Pettis property if every weakly
compact operator defined on X (and taking their values in another Banach
space) is Dunford–Pettis.
Remarks. A Banach lattice E may not have the AM-compactness property if
1. its norm is order continuous. In fact, the Banach lattice L2 [0, 1] has
an order continuous norm but does not have the AM-compactness property,
2. it has the Dunford–Pettis property. In fact, l∞ has the Dunford–Pettis
property, but does not have the AM-compactness property.
The following theorem gives some sufficient conditions under which a
Banach lattice has the AM-compactness property.
Theorem 3.3. Let E be a Banach lattice. Then E has the AM-compactness
property if one of the following assertions is valid:
1)
2)
3)
4)
E has the Dunford–Pettis property and its norm is order continuous.
The topological dual E ′ is discrete.
The lattice operations in E ′ are weakly sequentially continuous.
The lattice operations in E ′ are weak∗ sequentially continuous.
Proof. 1). Since the norm of E is order continuous, [−x, x] is weakly
compact for each x ∈ E + . As E has the Dunford–Pettis property, we obtain
[−x, x] is a Dunford–Pettis set. Finally, Proposition 3.1 proves that E has
the AM-compactness property.
2) Since the topological dual E ′ is discrete and Dedekind complete, and
since the weak absolute topology |σ| (E ′ , E) is Lebesgue, it follows from
Corollary 21.13 of [1] that, for each x′ ∈ (E ′ )+ , [−x′ , x′ ] is |σ| (E ′ , E)-totally
bounded. Also, from Theorem 3.27 of [2], this is equivalent to: for each
x ∈ E + , [−x, x] is |σ| (E, E ′ )-totally bounded (see Exercise 8 of section 3.3
of [2]).
Now, Lemma 3.2 implies that, [−x, x] is a Dunford–Pettis set for each
x ∈ E + . Hence, Proposition 3.1 proves that E has the AM-compactness
property.
3) and 4) are obvious.
Remarks. If a Banach lattice E has the AM-compactness property, then
1. its norm is not necessarily order continuous. In fact, the Banach
lattice c has the AM-compactness property (because its topological dual
c′ is discrete (see assertion 2 of Theorem 3.3)), but its norm is not order
continuous,
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B. Aqzzouz, K. Bouras
2. it does not necessarily have the Dunford–Pettis property. In fact, the
Banach lattice l2 has the AM-compactness property (because its topological
dual (l2 )′ = l2 is discrete), but it does not have the Dunford–Pettis property,
3. its topological dual is not necessarily discrete. In fact, the Banach
lattice E = (l∞ )′ has the AM-compactness property (because (l∞ )′ has the
Dunford–Pettis property and its norm is order continuous (see assertion 1 of
Theorem 3.3)), but its topological dual E ′ = (l∞ )′′ is not discrete,
4. the lattice operations in E ′ are not necessarily weak∗ sequentially
continuous. In fact, the Banach lattice E = (l∞ )′ has the AM-compactness
property, but the lattice operations in E ′ = (l∞ )′′ fail to be weak∗ sequentially continuous.
4. Almost Dunford–Pettis operators
which are weak Dunford–Pettis
Note that every Dunford–Pettis operator from a Banach lattice E into
a Banach space X is weak Dunford–Pettis but the converse is false in general.
In fact, since the Banach lattice c0 has the Dunford–Pettis property, its
identity operator Idc0 : c0 → c0 is weak Dunford–Pettis. But it is not
Dunford–Pettis because c0 does not have the Schur property.
However, if F is reflexive, the class of Dunford–Pettis operators and that
of weak Dunford–Pettis operators are equal.
To give a first result, we need the following Lemma.
Lemma 4.1. Let A be a norm bounded set in a Banach space X. If for every
ε > 0 there exists a Dunford–Pettis set Aε of X such that A ⊆ Aε + εBX ,
then A is a Dunford–Pettis set where BX is the closed unit ball of X.
Proof. Let Y be a Banach space, and let T : X → Y be a weakly compact
operator. We have to show that T (A) is a norm relatively compact set of Y .
Let ε > 0. By hypothesis, there exists a Dunford–Pettis set Aε of X such
that A ⊆ Aε + εBX , and hence T (A) ⊆ T (Aε. ) + ε kT k BY where BY is the
closed unit ball of Y .
As Aε is a Dunford–Pettis set, T (Aε ) is norm relatively compact. So
by Theorem 3.1 of [2], T (A) is norm relatively compact. And thus A is a
Dunford–Pettis set.
Theorem 4.2. Let E and F be two Banach lattices. If F has the AMcompactness property, then each positive almost Dunford–Pettis operator
from E into F is weak Dunford–Pettis.
Proof. Let T : E → F be a positive almost Dunford–Pettis operator. We
have to show that T is weak Dunford–Pettis. Let W be a relatively weakly
compact set in E and let A be the solid hull of W in E. It follows from
Weak and almost Dunford–Pettis operators on Banach lattices
173
the proof of Theorem 2.2, that for each ε > 0 there exists u ≥ 0 in the
order ideal generated by A satisfying T (W ) ⊂ [−T (u), T (u)] + ε.BF , where
BF is the closed unit ball of F . As F has the AM-compactness property,
it follows from Proposition 3.1, that [−T (u), T (u)] is a Dunford–Pettis set.
So by Lemma 4.1, T (W ) is a Dunford–Pettis set. This proves that T maps
relatively weakly compact sets in E into Dunford–Pettis set in F . Thus T
is a weak Dunford–Pettis operator.
The following Corollary gives some sufficient conditions under which each
positive almost Dunford–Pettis operator is weak Dunford–Pettis.
Corollary 4.3. Let E and F be two Banach lattices. Then each positive
almost Dunford–Pettis operator T from E into F is weak Dunford–Pettis if
one of the following assertions is valid:
1)
2)
3)
4)
5)
6)
The lattice operations in E are weakly sequentially continuous.
F is discrete and its norm is order continuous.
The topological dual F ′ is discrete.
The lattice operations in F ′ are weakly sequentially continuous.
The lattice operations in F ′ are weak ∗ sequentially continuous.
E or F has the Dunford–Pettis property.
Proof. 1) Let T : E → F be a positive almost Dunford–Pettis operator.
As the lattice operations in E are weakly sequentially continuous, it follows
from Theorem 2.2 that T : E → F is Dunford–Pettis, and hence it is weak
Dunford–Pettis.
2) Let T : E → F be a positive almost Dunford–Pettis operator. As F is
discrete and its norm is order continuous, it follows from Theorem 2.2 that
T : E → F is Dunford–Pettis, and hence it is weak Dunford–Pettis.
For 3), 4) and 5). It follows from Theorem 3.3 that the Banach lattice
F has the AM-compactness property. Hence Theorem 4.2 ends the proof.
6) In this case each operator T : E → F is weak Dunford–Pettis.
Hence each almost Dunford–Pettis operator from E into F is weak Dunford–
Pettis.
5. Weak Dunford–Pettis operators which
are almost Dunford–Pettis
We will need the following characterizations, which are just Corollary 2.6
and Corollary 2.7 of Dodds and Fremlin [9].
Lemma 5.1. Let E be a Banach lattice and let (xn ) be a sequence of E.
Then the following statements are equivalent:
174
B. Aqzzouz, K. Bouras
(1) kxn k → 0.
(2) |xn | → 0 for σ(E, E ′ ) and fn (xn ) → 0 for every bounded disjoint sequence (fn ) ⊂ (E ′ )+ .
Lemma 5.2. Let E be a Banach lattice and let (fn ) be a sequence of E ′ .
Then the following statements are equivalent:
(1) kfn k → 0.
(2) |fn | → 0 for σ(E ′ , E) and fn (xn ) → 0 for every bounded disjoint sequence (xn ) ⊂ E + .
A Banach lattice E is said to be a KB-space, whenever every increasing
norm bounded sequence of E + is norm convergent. As an example, each
reflexive Banach lattice is a KB-space.
It is clear that each KB-space has an order continuous norm, but a Banach lattice with an order continuous norm is not necessary a KB-space.
In fact, the Banach lattice c0 has an order continuous norm but it is not a
KB-space. However, for each Banach lattice E, its topological dual E ′ is a
KB-space if and only if its norm is order continuous.
A Banach lattice F is called a dual KB-space if F is a KB-space and
F = F0′ for some Banach lattice F0 .
The following theorem gives some sufficient conditions under which each
positive weak Dunford–Pettis operator is almost Dunford–Pettis.
Theorem 5.3. Let E and F be two Banach lattices and let T be a positive
weak Dunford–Pettis operator from E into F . Then T is almost Dunford–
Pettis if one of the following assertions is valid:
(i)
(ii)
(iii)
(iv)
(v)
F ′′ has an order continuous norm.
F is a dual KB-space.
F is a discrete KB-space.
F has the positive Schur property.
E has the positive Schur property.
Proof. (i) Let (xn ) be a sequence in E + such that xn → 0 for σ (E, E ′ ).
By Theorem 2.1 and Lemma 5.1, to prove that T is almost Dunford–Pettis,
it suffices to show that fn (T (xn )) → 0 for each norm bounded disjoint
sequence (fn ) in (F ′ )+ . For that, let (fn ) be a disjoint sequence in (F ′ )+
which is bounded for the norm. Since the norm of the topological bidual F ′′
is order continous, it follows from Corollary 2.9 of Dodds–Fremlin [9] that
fn → 0 for σ (F ′ , F ′′ ). Since T is weak Dunfod–Pettis, then fn (T (xn )) → 0.
(ii) Let (xn ) be a sequence in E + such that xn → 0 for σ (E, E ′ ). Since
(T (xn )) ⊂ F0′ , then by Theorem 2.1 and Lemma 5.2, to prove that T is
almost Dunford–Pettis, it suffices to establish that gn (zn ) → 0 for each
norm bounded disjoint sequence (zn ) in F0+ (where gn = T (xn ) ∈ F0′ ). For
Weak and almost Dunford–Pettis operators on Banach lattices
175
that, let (zn ) be a disjoint sequence in F0+ which is bounded for the norm.
Since the norm of the topological dual F0′ is order continous (because F0′ = F
and F is a KB-space), it follows from Corollary 2.9 of Dodds–Fremlin [9] that
zn → 0 for σ (F0 , F0′ ). Now, as the canonical injection τ : F0 → F0′′ is weakly
continuous, then τ (zn ) → 0 in the topology σ(F0′′ , F0′′′ ) = σ(F ′ , F ′′ ) (note
that F0′′ = F ′ and F0′′′ = F ′′ ). Finally, as T is weak Dunfod–Pettis, we
deduce that τ (zn )(T (xn )) → 0. But, we know that
τ (zn )(T (xn )) = τ (zn )(gn ) = gn (zn ) for each n.
Then gn (zn ) → 0.
(iii) Since each discrete KB-space is a dual (see Exercises 5.4.E2 of [11]),
we have just to use (ii).
(iv) and (v) are obvious.
Remarks. 1) Let F be a Banach lattice such that F is a dual KB-space,
then the norm of F ′′ is not necessary order continuous. In fact, if F = l1 (ln∞ ),
note that l1 (ln∞ ) is a dual KB-space, but the norm of the topological bidual
F ′′ is not order continuous (see Example 2.7 of [3]).
2) A Banach lattice F such that the norm of F ′′ is order continuous, does
not have necessary a predual. In fact, if we take F = L1 [0, 1], note that the
norm of F ′′ is order continuous, but F = L1 [0, 1] does not have a predual.
Recall from [2] that a Banach lattice E is said to be lattice embeddable
into another Banach lattice F , whenever there exists a lattice homomorphism
T : E → F and there exist two positive constants K and M satisfying
K kxk ≤ kT (x)k ≤ M kxk for all x ∈ E.
In this case T is called a lattice embedding from E into F , and T (E) is a
closed sublattice of F which can be identified with E.
The following theorem gives some properties whenever each positive
weak Dunford–Pettis operator from a Banach lattice E into a Dedekind
σ-complete Banach lattice F is almost Dunford–Pettis.
Theorem 5.4. Let E and F be two Banach lattices with F beeing Dedekind
σ-complete. If each positive weak Dunford–Pettis operator from E into F is
almost Dunford–Pettis, then one of the following assertions is valid:
(1) E has the positive Schur property.
(2) F has an order continuous norm.
Proof. Assume by way of contradiction that the norm of F is not order continuous and E does not have the positive Schur property. To finish
the proof, we have to construct a positive weak Dunford–Pettis operator
T : E → F , which is not almost Dunford–Pettis.
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B. Aqzzouz, K. Bouras
Since the norm of F is not order continuous, it results from Theorem
4.51 of [2] that l∞ is lattice embeddable in F . Let i : l∞ → F be a lattice
embedding. Then there exist two positive constants K and M satisfying
K kxk∞ ≤ ki (x)kF ≤ M kxk∞ for all x ∈ l∞ .
On the other hand, since E does not have the positive Schur property,
there exists a disjoint weakly null sequence (xm ) in E + , which is not norm
convergent to 0 (Theorem 3.1 of Chen–Wickstead [7]). Now, Corollary 2.6
of Dodds–Fremlin [9], implies the existence of a bounded disjoint sequence
(fm ) ⊂ (E ′ )+ such that fm (xm ) ≥ ε for every m (ε > 0 fixed).
We consider the operator T = i ◦ P : E → l∞ → F where P is the
operator defined by P : E → l∞ , x 7→ (fn (x))n≥0 . It is clear that T is weak
Dunford–Pettis (because l∞ has the Dunford–Pettis property), but it is not
almost Dunford–Pettis. In fact, (xm ) is a disjoint sequence and xm → 0 for
σ(E, E ′ ) but
kT (xm )k = ki(p (xm ))k ≥ K kp (xm )k∞
= K k(fn (xm ))n≥0 k∞ ≥ Kfm (xm ) > K.ε > 0
for each n, this presentes a contradiction. So, E has the positive Schur
property and this ends the proof of the Theorem.
Remarks. 1) The second necessary condition of Theorem 5.4 is not sufficient i.e. there exist Banach lattices E and F such that the norm of F is
order continuous but a positive weak Dunford–Pettis operator T : E → F
is not necessary almost Dunford–Pettis. In fact, if we take E = F = c0 ,
it is clear that the norm of c0 is order continuous and its identity operator
Idc0 : c0 → c0 is weak Dunford–Pettis (because c0 has the Dunford–Pettis
property), but it is not almost Dunford–Pettis.
2) The assumption ”F is Dedekind σ-complete” is essential in Theorem
5.4. In fact, if we take E = l∞ and F = c, the Banach lattice of all convergent
sequences, it is clear that c is not Dedekind σ-complete and each operator
from l∞ into c is Dunford–Pettis and hence each operator from l∞ into c is
almost Dunford–Pettis. But the conditions 1) and 2) of Theorem 5.4 fail i.e.
l∞ does not have the positive Schur property and the norm of c is not order
continuous.
As a consequence of Theoerm 5.4, we derive the following characterization
of the positive Schur property.
Corollary 5.5. Let E be a Banach lattice. Then the following assertions
are equivalent:
(1) Each positive operator from E into l∞ is almost Dunford–Pettis.
(2) E has the positive Schur property.
Weak and almost Dunford–Pettis operators on Banach lattices
177
Proof. (1)⇒(2) Since the Banach lattice l∞ is Dedekind σ-complete, it
follows from Theorem 5.4, that E has the positive Schur property (because
the norm of l∞ is not order continuous).
For the next result, we need to establish the following characterization
of the positive Schur property.
Theorem 5.6. Let E be a Banach lattice. Then the following assertions
are equivalent:
(i) E has an order continuous norm and each positive operator T from E
into c0 is almost Dunford–Pettis.
(ii) E has the positive Schur property.
Proof. (i)⇒(ii) We establish that E has the positive Schur property. Assume that E does not have this property, then there exists a disjoint weakly
null sequence (xn ) in E + which is not norm convergent to 0 (see Theorem
3.1 of [7]). Now, it follows from Lemma 5.1, the existence of a bounded
disjoint sequence (fn ) ⊂ (E ′ )+ such that fn (xn ) ≥ ε for every n ∈ N (ε > 0
fixed).
On the other hand, as the norm of E is order continuous, we have fn → 0
for the topology σ(E ′ , E) (see Corollary 2.4.3 of [11]). Now, consider the
operator S : E → c0 defined by
S(x) = (fm (x))m≥0 for each x ∈ E.
It is easy to see that S is positive, but it is not almost Dunford–Pettis.
In fact, (xn ) is a disjoint weakly null sequence in E, but
kS(xn )k = k(fm (x))m≥0 k∞ ≥ fn (xn ) ≥ ε
for every n. This presentes a contradiction, and hence E has the positive
Schur property.
(ii)⇒(i) Obvious.
If, instead of assuming F is Dedekind σ-complete in Theorem 5.4, we
assume that E has an order continuous norm, we obtain the following result.
Theorem 5.7. Let E and F be two Banach lattices such that E has
an order continuous norm. If each positive weak Dunford–Pettis operator
T : E → F is almost Dunford–Pettis, then one of the following assertions is
valid:
(i) E has the positive Schur property.
(ii) F is a KB-space.
Proof. It suffices to establish that if F is not a KB-space, then E has the
positive Schur property. Suppose that F is not a KB-space and consider an
arbitrary positive operator T : E → c0 . Since c0 is order embeddable in
178
B. Aqzzouz, K. Bouras
F (see Theorem 4.60 of [2]) and c0 has the Dunford–Pettis property, then
T is weak Dunford–Pettis, and hence it is almost Dunford–Pettis by our
assumption. Finally, Theorem 5.6 finishes the proof.
As a consequence of Theorem 5.7, we obtain the following results.
Corollary 5.8. Let E be a Banach lattice. Then the following assertions
are equivalent:
(i) E has an order continuous norm and each positive operator T from E
into c is almost Dunford–Pettis.
(ii) E has the positive Schur property.
Proof. (i)⇒(ii) Since the Banach lattice c is not a KB-space, the result
follows from Theorem 5.7.
(ii)⇒(i) Obvious.
We end this paper by an open question:
Question. Let E and F be two Banach lattices such that F is a KBspace. Does each positive weak Dunford–Pettis operator T : E → F is
almost Dunford–Pettis ?
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Belmesnaoui Aqzzouz
FACULTÉ DES SCIENCES ECONOMIQUES
UNIVERSITÉ MOHAMMED V-SOUISSI
JURIDIQUES ET SOCIALES, DÉPARTEMENT D’ECONOMIE