# Weak Compactness of Almost Limited Operators

Authors: Aziz Elbour; Nabil Machrafi; Mohammed Moussa
Journal: Journal of Function Spaces
 (2014)

---

## Abstract

This paper is devoted to the relationship between almost limited operators and weakly compact operators. We show that ifF is a σ-Dedekind complete Banach lattice, then every almost limited operator T
:
E
→
F is weakly compact if and only if E is reflexive or the norm of F is order continuous. Also, we show that if E is a σ-Dedekind complete Banach lattice, then the square of every positive almost limited operator T
:
E
→
E is weakly compact if and only if the norm of E is order continuous.


---

## Body

## 1. Introduction

Throughout this paperX
,
Y will denote real Banach spaces, and E
,
F will denote real Banach lattices. B
X is the closed unit ball of X and B
E
+
:
=
B
E
∩
E
+ is the positive part of B
E. We will use the term operator T
:
X
→
Y between two Banach spaces to mean a bounded linear mapping. We refer to [1, 2] for unexplained terminology of the Banach lattice theory and positive operators.Let us recall that a norm bounded setA in a Banach space X is calledlimited, if every weak
* null sequence f
n in X
* converges uniformly to zero on A; that is, sup
⁡
x
∈
A
f
n
x
→
0. An operator T
:
X
→
Y is said to belimited whenever T
B
X is a limited set in Y, equivalently, whenever T
*
f
n
→
0 for every weak
* null sequence f
n
⊂
Y
*.Recently, the authors of  [3] considered the disjoint version of limited sets by introducing the class of almost limited sets in Banach lattices. From [3] a norm bounded subset A of a Banach lattice E is said to bealmost limited, if every disjoint weak
* null sequence (
f
n
) in E
* converges uniformly to zero on A.From [4], an operator T
:
X
→
E is calledalmost limited if T
B
X is an almost limited set in E, equivalently, T
*
f
n
→
0 for every disjoint weak
* null sequence f
n
⊂
E
*. Note that an almost limited operator need not be weakly compact. In fact, the identity operator of the Banach lattice l
∞ is almost limited but it is not weakly compact.In this paper, we characterize pairs of Banach latticesE, F for which every almost limited operator T
:
E
→
F is weakly compact. More precisely, we will prove that if F is a σ-Dedekind complete Banach lattice, then every almost limited operator T
:
E
→
F is weakly compact if and only if E is reflexive or the norm of F is order continuous (Theorem 5). Next, we will prove that if E is a σ-Dedekind complete Banach lattice, then the square of every positive almost limited operator T
:
E
→
E is weakly compact if and only if the norm of E is order continuous (Theorem 9). As consequences, we will give some interesting results.


## 2. Main Results

Let us recall that a Banach latticeE is said to have the dual positive Schur property if f
n
→
0 for every weak
* null sequence f
n
⊂
E
*
+, equivalently, f
n
→
0 for every weak
* null sequence f
n
⊂
E
*
+ consisting of pairwise disjoint terms (Proposition 2.3 of [5]). A Banach lattice E has the property (
d
) whenever f
n
∧
f
m
=
0 and f
n
→
w
*
0 in E
* imply f
n
→
w
*
0. It should be noted, by Proposition 1.4 of [5], that every σ-Dedekind complete Banach lattice has the property (
d
) but the converse is not true in general. In fact, the Banach lattice l
∞
/
c
0 has the property (
d
) but it is not σ-Dedekind complete [5, Remark 1.5].Our first result shows that we can restrict sequences appearing in the definition of almost limited operatorT
:
X
→
E to positive disjoint sequences if the Banach lattice E has the property (
d
).Proposition 1.
An operatorT
:
X
→
E from a Banach space X into a Banach lattice E with the property (
d
) is almost limited if and only if T
*
f
n
→
0 for every w
e
a
k
* null sequence f
n in E
* consisting of positive and pairwise disjoint elements.Proof.
The “only if” part is trivial. For the “if” part, letf
n
⊂
E
* be a disjoint weak
* null sequence. As E has the property (
d
), f
n
→
w
*
0. Using the inequalities 0
≤
f
n
+
≤
f
n and 0
≤
f
n
-
≤
f
n, we see that f
n
+ and f
n
- are disjoint weak
* null sequences of E
*. So, from our hypothesis, we see that T
*
f
n
+
→
0 and T
*
f
n
-
→
0. This implies that T
*
f
n
→
0, and hence T is almost limited.The next result follows immediately from Proposition 2.3 of [5] combined with Proposition 1.Corollary 2.
A Banach latticeE with the property (
d
) has the dual positive Schur property if and only if the identity operator on E is almost limited.The following result shows that if a positive almost limited operatorT
:
E
→
F has its range in a Banach lattice with the property (
d
), then every positive operator S
:
E
→
F that it dominates (i.e., 0
≤
S
≤
T) is also almost limited.Proposition 3.
LetE and F be two Banach lattices such that F has the property (
d
). If a positive operator S
:
E
→
F is dominated by an almost limited operator, then S itself is almost limited.Proof.
LetS
,
T
:
E
→
F be two operators such that 0
≤
S
≤
T and T is almost limited. Let (
f
n
) be a disjoint sequence in F
*
+ such that f
n
→
w
*
0. As T is almost limited, T
*
(
f
n
)
→
0. Using the inequalities 0
≤
S
*
(
f
n
)
≤
T
*
(
f
n
), we see that S
*
(
f
n
)
≤
T
*
(
f
n
) for all n, from which we get S
*
(
f
n
)
→
0. Now, by Proposition 1,  S is well almost limited.The next remark will be useful in further considerations.Remark 4.
We have the following.(1)
Consider the scheme of operatorsX
→
R
Y
→
S
F. It is easy to see that if S is an almost limited operator, then S
∘
R is likewise almost limited.
(2)
Consider the scheme of operatorsX
→
R
E
→
S
F.
(a)
IfR is an almost limited operator, then S
∘
R is not necessarily almost limited. In fact, by a result in [6], there exists a nonregular operator S
:
l
∞
→
c
0, which is certainly not compact. So by Proposition 4.3 of [4], S is not almost limited. If R
:
l
∞
→
l
∞ is the identity operator on l
∞, then R is almost limited but S
∘
R
=
S is not almost limited.
(b)
However, ifE has the dual positive Schur property (e.g., E
=
l
∞) and F has the property (
d
) and S is positive, then T
=
S
∘
R is an almost limited operator. In fact, according to Proposition 1, let f
n
⊂
F
* be a positive disjoint weak
* null sequence. Clearly 0
≤
S
*
f
n
→
w
*
0 holds in E
*. Since E has the dual positive Schur property, then S
*
f
n
→
0, and hence T
*
f
n
=
R
*
S
*
f
n
→
0, as desired.Our next major result characterizes pairs of Banach latticesE, F for which every positive almost limited operator T
:
E
→
F is weakly compact.Theorem 5.
LetE and F be two Banach lattices such that F is σ-Dedekind complete. Then the following assertions are equivalent.(1)
Every almost limited operator
T
:
E
→
F
is weakly compact.
(2)
Every positive almost limited operator
T
:
E
→
F
is weakly compact.
(3)
One of the following statements is valid:
(a)
E
is reflexive;
(b)
the norm of
F
is order continuous.Proof.
1
⇒
2 The proof is obvious.
2
⇒
3 Assume by way of contradiction that E is not reflexive and the norm of F is not order continuous. We have to construct a positive almost limited operator T
:
E
→
F which is not weakly compact.
Indeed, since the norm ofF is not order continuous, then by Corollary 2.4.3 of [2] we may assume that l
∞ is a closed sublattice of F. As E is not reflexive, then E
* is not reflexive, and hence the closed unit ball B
E
* of E
* is not weakly compact. So, from B
E
*
⊂
B
E
*
+
-
B
E
*
+, we see that B
E
*
+ is not weakly compact. Then, by the Eberlein-Šmulian theorem one can find a sequence (
f
n
) in B
E
*
+ which does not have any weakly convergent subsequence. Consider the positive operator T
:
E
→
l
∞
⊂
F defined by
(1)
T
x
=
f
n
x
n
=
1
∞
for all x
∈
E. By Remark 4(2b) T is an almost limited operator. But T is not weakly compact. In fact, if T were weakly compact, then T
*
:
l
∞
*
→
E
* would weakly be compact. Note that T
*
λ
n
n
=
1
∞
=
∑
n
=
1
∞
λ
n
f
n for every λ
n
n
=
1
∞
∈
l
1
⊂
l
∞
*. So, if e
n is the usual basis element in l
1, then T
*
e
n
=
f
n so that (
f
n
) would have a weakly convergent subsequence. This contradicts the choice of (
f
n
). Therefore, T is not weakly compact, as desired.
a
⇒
1 In this case, every operator from E into F is weakly compact.
b
⇒
1 By Theorem 4.2 of [4] we see that T is L-weakly compact, and by Theorem 5.61 of [1] T is well weakly compact.By a similar proof as the previous theorem, we obtain the following result.Theorem 6.
LetX be a Banach space and let F be a σ-Dedekind complete Banach lattice. Then the following assertions are equivalent.(1)
Every almost limited operatorT
:
X
→
F is weakly compact.
(2)
One of the following statements is valid:
(a)
X is reflexive;
(b)
the norm ofF is order continuous.As a consequence of Theorem5, we obtain an operator characterization of order continuity of the norm of a σ-Dedekind complete Banach lattice.Corollary 7.
LetE be a σ-Dedekind complete Banach lattice. Then the following statements are equivalent.(1)
Every almost limited operatorT from E into E is weakly compact.
(2)
Every positive almost limited operatorT from E into E is weakly compact.
(3)
The norm ofE is order continuous.Another consequence of Theorem5 is the following result.Corollary 8.
For a Banach latticeE, the following statements are equivalent.(1)
Every positive operatorT
:
E
→
F from E to an arbitrary infinite dimensional AM-space is weakly compact.
(2)
Every positive operatorT
:
E
→
l
∞ is weakly compact.
(3)
E is reflexive.Proof.
1
⇒
2 and 3
⇒
1 are obvious.
2
⇒
3 The proof follows from Theorem 5.The following result characterizes Banach latticeE for which every positive almost limited operator T
:
E
→
E has a weakly compact square.Theorem 9.
LetE be a σ-Dedekind complete Banach lattice. Then the following statements are equivalent.(1)
Every positive almost limited operatorT from E into E is weakly compact.
(2)
For every positive almost limited operatorT from E into E, the operator T
2 is weakly compact.
(3)
The norm ofE is order continuous.Proof.
1
⇒
2 The proof is obvious.
2
⇒
3 Assume by way of contradiction that the norm of E is not order continuous. So, by Theorem 4.14 of [1], there exists a disjoint sequence (
u
n
)
⊂
E
+ satisfying u
n
=
1 and 0
≤
u
n
≤
u for all n and for some u
∈
E
+. We can now proceed analogously to the proof of Proposition 0.5.5 of [7]. Let g
n
∈
E
+
* be of norm one such that g
n
(
u
n
)
=
u
n
=
1 and let P
n be the band projection onto u
n
d
d, where u
n
d
d is the band generated by u
n. If f
n
=
g
n
∘
P
n, then f
n
∧
f
m
=
0 for n
≠
m, sup
⁡
n
f
n
≤
1, and f
n
(
u
m
)
=
δ
n
m. Hence the operator S
:
l
∞
→
E defined by
(2)
S
t
n
n
=
1
∞
=
o
∑
n
=
1
∞
‍
t
n
u
n
is a lattice isomorphism from l
∞ into E, where o
∑
n
=
1
∞
‍
t
n
u
n denotes the order limit of the sequence of the partial sums ∑
n
=
1
m
‍
t
n
u
n for each t
n
n
=
1
∞
∈
l
∞. Also, let R
:
E
→
l
∞ be the positive operator defined by
(3)
R
x
=
f
n
x
n
=
1
∞
.
So, by Remark 4(2b), the positive operator T
=
S
∘
R
:
E
→
F defined by
(4)
T
x
=
o
∑
n
=
1
∞
‍
f
n
x
u
n
is almost limited. But T is not weakly compact. In fact, let x
n
=
∑
k
=
1
n
‍
u
k for each n, and note that 0
≤
x
n
↑
≤
u. Clearly T
(
u
n
)
=
u
n, and hence T
(
x
n
)
=
x
n for all n. If x is a weak limit of a subsequence of x
n, then it is easy to see that x
n
↑
x and x
n
→
w
x must hold. By Theorem 3.52 of [1] we have x
n
-
x
→
0, and hence x
n
+
1
-
x
n
→
0. But this contradicts x
n
+
1
-
x
n
=
u
n
+
1
=
1 for all n. Thus x
n has no weakly convergent subsequence, and hence T is not weakly compact, as desired.
3
⇒
1 The proof follows from Theorem 5.Finally, note that a weakly compact operatorT
:
X
→
F need not be almost limited. In fact, the identity operator of the Banach lattice l
2 is weakly compact but it is not almost limited. However, if F has the positive Schur property, then the two classes coincide. The details follow.Proposition 10.
An operatorT
:
X
→
F from a Banach space X to a Banach lattice F with the positive Schur property is weakly compact if and only if it is almost limited.Proof.
The “if” part follows from Theorem6. For the “only if” part, assume that T
:
X
→
F is weakly compact. It follows from Theorem 3.4 of [8] that T is L-weakly compact, and hence T is almost limited [4, Theorem 4.2].



---

*Source: 263159-2014-12-03.xml*