# The Banach-Saks Properties in Orlicz-Lorentz Spaces

Authors: Anna Kamińska; Han Ju Lee
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

The Banach-Saks index of an Orlicz-Lorentz spaceΛ
φ
,
w
(
I
) for both function and sequence case, is computed with respect to its Matuszewska-Orlicz indices of φ. It is also shown that an Orlicz-Lorentz function space has weak Banach-Saks (resp., Banach-Saks) property if and only if it is separable (resp., reflexive).


---

## Body

## 1.  Introduction

Let(
X
,
∥
·
∥
) be a real Banach space. A bounded sequence {
x
n
}
⊂
X is called aweak Banach-Saks sequence whenever there exists a subsequence {
y
n
}
⊂
{
x
n
} such that its Cesàro means converge in norm to zero; that is,
(1)
1
n
∑
i
=
1
n
‍
y
i
⟶
0
.
The space X is said to satisfy theweak Banach-Saks property, in short X
∈
(
w
B
S
), if every weakly null sequence {
x
n
} in X is a weak Banach-Saks sequence. Recall that a Banach space X satisfies theBanach-Saks property
(
B
S
) if for every bounded sequence {
x
n
} in X, there is a subsequence {
y
j
} such that its Cesàro means converge; that is, the sequence {
(
1
/
m
)
∑
j
=
1
m
‍
y
j
} is convergent in norm. It is well known that a Banach space has the (
B
S
)-property if and only if it is reflexive and it has the (
w
B
S
)-property [1].Schreier constructed a separable Banach space without the weak Banach-Saks property [2] (which is called a Schreier space) and later it was shown by Baernstein [3] that there is a reflexive separable Banach space without the weak Banach-Saks property. Hence the space ℓ
∞ does not have the weak Banach-Saks property, since every separable space is embeddable in ℓ
∞.Following Johnson [4], given 1
≤
p
≤
∞, a bounded sequence {
x
n
}
⊂
X is called a B
S
p
sequence if there exists a subsequence {
y
n
}
⊂
{
x
n
} such that
(2)
sup
⁡
m
∈
ℕ
m
-
1
/
p
∥
∑
i
=
1
m
‍
y
i
∥
<
∞
.
Here m
1
/
∞
=
1 for all m
∈
ℕ. We then say that X has the p-Banach-Saks property, shortly X
∈
(
B
S
p
), if each weakly null sequence contains a B
S
p subsequence. It is clear that every Banach space has a (
B
S
1
) property. TheBanach-Saks index is the number γ
(
X
)
=
sup
⁡
{
p
:
X
∈
(
B
S
p
)
}. The set Γ
(
X
)
=
{
p
:
X
∈
(
B
S
p
)
} is always an interval [
1
,
γ
(
X
)
) or [
1
,
γ
(
X
)
]. The Bessaga-Pełczyński method [5, Proposition 1
.
a
.
12] shows that Γ
(
ℓ
p
)
=
[
1
,
p
] and the Schur property of ℓ
1 shows that Γ
(
ℓ
1
)
=
[
1
,
∞
]. The Banach-Saks index of Nakano sequence space was computed in [6].The stronger property (S
p) was introduced by Knaust and Odell in [7]. It is said that X has property (
S
p
), 1
<
p
≤
∞, if every weakly null sequence in X has a subsequence {
x
k
} so that there is a constant C
>
0 such that, for all m
∈
ℕ and all real sequences a
=
(
a
n
)
∈
ℓ
p,
(3)
∥
∑
j
=
1
m
‍
a
j
x
j
∥
≤
C
∥
a
∥
p
,
where ∥
a
∥
p is the ℓ
p norm.It is clear that(
S
p
)
⇒
(
B
S
p
)
⇒
(
w
B
S
) for all 1
<
p
≤
∞. The Elton c
0-theorem [8, Theorem III
3.5] states that (
B
S
∞
)
⇔
(
S
∞
) and a Banach space X has (
B
S
∞
) if and only if every normalized weakly null sequence contains a subsequence which is equivalent to the unit vector basis of c
0, which is equivalent to the case in which every subspace of X has the Dunford-Pettis property (for details see [9]). In general, the two properties (
B
S
p
) and (
S
p
) are not equivalent if 1
<
p
<
∞ [10]; however, Rakov [11, Theorem 3] showed that if 1
<
q
<
p
<
∞, then (
B
S
p
) implies (
S
q
). Various weak Banach-Saks properties and their relations were shown in [6] for Musielak-Orlicz sequence spaces.In this paper we compute the Banach-Saks index of Orlicz-Lorentz spaceΛ
φ
,
w
(
I
) and show that Orlicz-Lorentz function space Λ
φ
,
w
(
I
) has the weak Banach-Saks property (resp., Banach-Saks property) if and only if it is separable (resp., reflexive). We start with several preliminary results about Orlicz-Lorentz spaces.


## 2. Preliminaries

LetI
=
ℕ or I
=
(
0
,
a
) with 0
<
a
≤
∞, equipped with the counting measure if I
=
ℕ and the Lebesgue measure if I
=
(
0
,
a
). If A
⊂
I is a measurable set, we denote by |
A
| its measure. By L
0
(
I
) we denote the collection of all real valued measurable functions on I. In the case when I
=
ℕ the elements are sequences x
=
(
x
(
n
)
), and in the other cases they are real valued Lebesgue measurable functions x.For a measurable functionx, the decreasing rearrangement of x will be denoted by x
*, defined as x
*
(
t
)
=
inf
⁡
{
s
>
0
:
d
|
x
|
(
s
)
≤
t
}, t
≥
0, where d
x
(
s
)
=
|
{
t
:
x
(
t
)
≥
s
}
|, s
≥
0. For two measurable functions x and y, the relation x
≺
≺
y means that, for all t
≥
0, we have
(4)
∫
0
t
‍
x
*
(
s
)
d
s
≤
∫
0
t
‍
y
*
(
s
)
d
s
.For a sequencex
=
(
x
(
n
)
), x
* will denote the decreasing rearrangement of x. That is, letting x
~
(
t
)
=
∑
n
=
1
∞
‍
x
(
n
)
χ
[
n
-
1,1
), we have x
*
(
n
)
=
x
~
*
(
n
-
1
) for all n
∈
ℕ. For sequences x and y, x
≺
≺
y means that for all n
∈
ℕ, we have
(5)
∑
k
=
1
n
‍
x
*
(
k
)
≤
∑
k
=
1
n
‍
y
*
(
k
)
.Letφ
:
ℝ
+
→
ℝ
+ be anOrlicz function; that is, φ is convex, strictly increasing, and φ
(
0
)
=
0. Let the weight function w
∈
L
0
(
I
) be nonnegative and nonincreasing. For I
=
(
0
,
a
), 0
<
a
≤
∞, theOrlicz-Lorentz space
Λ
φ
,
w
(
I
)
=
Λ
φ
,
w
(
0
,
a
) is the collection of all functions f
∈
L
0
(
0
,
a
) such that
(6)
∥
f
∥
=
∥
f
∥
φ
,
w
=
inf
⁡
{
ϵ
>
0
:
I
(
f
ϵ
)
≤
1
}
<
∞
,
where I
(
f
)
=
I
φ
,
w
(
f
)
=
∫
I
‍
φ
(
f
*
)
w. Analogously in case when I
=
ℕ, theOrlicz-Lorentz sequence space
d
(
φ
,
w
)
=
Λ
φ
,
w
(
ℕ
) contains all real sequences x
=
(
x
(
n
)
) such that
(7)
∥
x
∥
=
∥
x
∥
φ
,
w
=
inf
⁡
{
ϵ
>
0
:
I
(
x
ϵ
)
≤
1
}
<
∞
,
where I
(
x
)
=
I
φ
,
w
(
x
)
=
∑
n
=
1
∞
‍
φ
(
x
*
(
n
)
)
w
(
n
). It is well known that the space (
Λ
φ
,
w
(
I
)
,
∥
·
∥
) is a rearrangement invariant (r.i.) Banach function space satisfying the Fatou property [12–15]. The space Λ
φ
,
w
0
(
0
,
a
) (resp., d
0
(
φ
,
w
)) is the order continuous part of Λ
φ
,
w
(
0
,
a
) (resp., d
(
φ
,
w
)). In the sequence case d
0
(
φ
,
w
) is the closure of the linear span of the unit vectors (
e
i
) in d
(
φ
,
w
). If 1
≤
p
<
∞ and φ
(
u
)
=
u
p, then d
(
φ
,
w
)
=
d
(
p
,
w
) and Λ
φ
,
w
(
0
,
a
)
=
Λ
p
,
w
(
0
,
a
) [16]. If w
≡
1, then Λ
φ
,
w
(
0
,
a
)
=
L
φ
(
0
,
a
) is an Orlicz function space, and d
(
φ
,
w
)
=
ℓ
φ is an Orlicz sequence space. In this case also Λ
φ
,
w
0
(
0
,
a
)
=
L
φ
0
(
0
,
a
) and d
0
(
φ
,
w
)
=
ℓ
φ
0 are subspaces of order continuous elements in Orlicz spaces.We may assume thatφ
(
1
)
=
1. In fact, the space Λ
ψ
,
w
(
I
), where ψ
(
t
)
=
φ
(
α
t
) with φ
(
α
)
=
1, is isometric to Λ
φ
,
w
(
I
); we have ∥
f
∥
ψ
,
w
=
α
∥
f
∥
φ
,
w.Definition 1.
Given an Orlicz functionφ, one defines the growth conditions connected to itslower andupper Matuszewska-Orlicz indices in three different categories, at zero, at infinity, and on ℝ
+.
We say that, for1
≤
p
<
∞, φ
∈
Δ
0
*
p, φ
∈
Δ
∞
*
p, and φ
∈
Δ
*
p, whenever
(8)
sup
⁡
0
<
a
≤
1
0
<
t
<
1
⁡
φ
(
a
t
)
φ
(
t
)
a
p
<
∞
,
sup
⁡
0
<
a
≤
1
t
>
1
⁡
φ
(
a
t
)
φ
(
t
)
a
p
<
∞
,
sup
⁡
0
<
a
≤
1
t
>
0
⁡
φ
(
a
t
)
φ
(
t
)
a
p
<
∞
,
respectively. Analogously we say that, given 1
≤
q
<
∞, φ
∈
Δ
0
q, φ
∈
Δ
∞
q, and φ
∈
Δ
q, whenever
(9)
inf
⁡
0
<
a
≤
1
0
<
t
<
1
⁡
φ
(
a
t
)
φ
(
t
)
a
q
>
0
,
inf
⁡
0
<
a
≤
1
t
>
1
⁡
φ
(
a
t
)
φ
(
t
)
a
q
>
0
,
inf
⁡
0
<
a
≤
1
t
>
0
⁡
φ
(
a
t
)
φ
(
t
)
a
q
>
0
,
respectively.
The Matuszewska-Orlicz indices, lowerα
φ
0, α
φ
∞, and α
φ and upper β
φ
0, β
φ
∞, and β
φ, are then defined as follows:
(10)
α
φ
0
=
sup
⁡
⁡
{
p
:
φ
∈
Δ
0
*
p
}
,
α
φ
∞
=
sup
⁡
⁡
{
p
:
φ
∈
Δ
∞
*
p
}
,
α
φ
=
sup
⁡
{
p
:
φ
∈
Δ
*
p
}
,
β
φ
0
=
inf
⁡
⁡
{
q
:
φ
∈
Δ
0
q
}
,
β
φ
∞
=
inf
⁡
⁡
{
q
:
φ
∈
Δ
∞
q
}
,
β
φ
=
inf
⁡
{
q
:
φ
∈
Δ
q
}
.
It is clear thatα
φ
=
min
⁡
(
α
φ
0
,
α
φ
∞
) and β
φ
=
max
⁡
(
β
φ
0
,
β
φ
∞
). We say that φ satisfiesconditions
Δ
2
0, Δ
2
∞, and Δ
2, whenever
(11)
sup
⁡
0
<
t
<
1
φ
(
2
t
)
φ
(
t
)
<
∞
,
sup
⁡
t
>
1
φ
(
2
t
)
φ
(
t
)
<
∞
,
sup
⁡
t
>
0
φ
(
2
t
)
φ
(
t
)
<
∞
,
respectively. Let φ
1 and φ
2 be Orlicz functions. We call themequivalent at zero,equivalent at infinity, or justequivalent, if there is D
>
0 such that φ
1
(
D
-
1
t
)
≤
φ
2
(
t
)
≤
φ
1
(
D
t
), for all 0
<
t
<
1, t
>
1, or for all t
>
0, respectively. It is well known and easy to prove that Matuszewska-Orlicz indices of Orlicz functions are preserved by the corresponding equivalence relations (e.g., see [17]); that is, if φ
1 and φ
2 are equivalent (resp., equivalent at 0 and equivalent at ∞), then α
φ
1
=
α
φ
2 (resp., α
φ
1
0
=
α
φ
2
0 and α
φ
1
∞
=
α
φ
2
∞). Similar equalities hold for upper indices. Recall also that any upper index of φ is finite if and only if φ satisfies the corresponding condition Δ
2 [17, 18].
We also have that if the Orlicz functionsφ
1 and φ
2 are equivalent (resp., equivalent at infinity and equivalent at zero), then Λ
φ
1
,
w
(
0
,
∞
)
=
Λ
φ
2
,
w
(
0
,
∞
) (resp., Λ
φ
1
,
w
(
0
,
a
)
=
Λ
φ
2
,
w
(
0
,
a
), a
<
∞; d
(
φ
1
,
w
)
=
d
(
φ
2
,
w
)) with equivalent norms.
The spaceΛ
φ
,
w
(
0
,
∞
) (resp., Λ
φ
,
w
(
0
,
a
) (a
<
∞); d
(
φ
,
w
)) is separable if and only if φ
∈
Δ
2 and W
(
∞
)
=
∞ (resp., φ
∈
Δ
2
∞; φ
∈
Δ
2
0 and W
(
∞
)
=
∞), if a
=
∞ (resp., a
<
∞; I
=
ℕ). The separability is also equivalent to the case in which the space Λ
φ
,
w
(
0
,
∞
) does not contain the isomorphic copy of ℓ
∞ [12, 15, 19].Lemma 2.
Let1
≤
p
,
q
<
∞;(1)
([20, Lemma 6], [21, Lemma 3], and [22]) An Orlicz function φ
∈
Δ
*
p (resp., φ
∈
Δ
∞
*
p; φ
∈
Δ
0
*
p) if and only if there exists an Orlicz function ψ equivalent (resp., equivalent at ∞; equivalent at 0) to φ such that ψ
(
t
1
/
p
) is convex;
(2)
([20, Lemma 5], [21, Lemma 2], and [22]) An Orlicz function φ
∈
Δ
q (resp., φ
∈
Δ
∞
q; φ
∈
Δ
0
q) if and only if there exists an Orlicz function ψ equivalent (resp., equivalent at ∞ and equivalent at 0) to φ such that ψ
(
t
1
/
q
) is concave.We will use the following functions:(12)
W
(
t
)
=
∫
0
t
‍
w
(
s
)
d
s
for
t
∈
(
0
,
a
)
,
or
W
(
m
)
=
∑
i
=
1
m
‍
w
(
i
)
for
m
∈
ℕ
.
Recall that the weight w isregular whenever there exists C
>
0 such that W
(
s
)
≤
C
s
w
(
s
) for all s
∈
I. Since always s
w
(
s
)
≤
W
(
s
), the weight w is regular if the functions W
(
s
) and s
w
(
s
) are equivalent on I. Notice that for any regular weight w on I
=
(
0
,
∞
) or I
=
ℕ it holds that W
(
∞
)
=
∞.A Banach lattice is said to bep
-convex (resp., p-concave) for some 1
≤
p
<
∞ if there is a constant K
>
0 such that
(13)
∥
(
∑
i
=
1
n
‍
|
x
i
|
p
)
1
/
p
∥
≤
K
(
∑
i
=
1
n
‍
∥
x
i
∥
p
)
1
/
p
(
resp
.
(
∑
i
=
1
n
‍
∥
x
i
∥
p
)
1
/
p
≤
K
∥
(
∑
i
=
1
n
‍
|
x
i
|
p
)
1
/
p
∥
)
for every choice of vectors x
1
,
…
,
x
n in X. A Banach lattice is said to satisfyupper
p
-estimate (resp.,lower
p
-estimate) if the definition of p-convexity (resp., p-concavity) holds true for any choice of disjointly supported elements x
1
,
…
,
x
n in X. It is known that, given 1
≤
p
<
∞, if X is p-convex (resp., p-concave), then X is r-convex (resp., r-concave) for 1
≤
r
≤
p (resp., r
≥
p) [23, 24].Theorem 3 (see [25]).
The spaceΛ
φ
,
w
(
0
,
∞
) (resp., Λ
φ
,
w
(
0
,
a
) (a
<
∞); d
(
φ
,
w
)) has finite concavity if and only if φ satisfies the Δ
2 (resp., Δ
2
∞; Δ
2
0) condition and the weight w is regular.Recall that a Banach lattice has the finite concavity if and only if it is of finite cotype [26].Theorem 4 (see [19]).
Let1
≤
p
≤
∞;(1)
[19, Theorem 7.18] For I
=
(
0
,
∞
), the following conditions are equivalent:
(i)
ℓ
p (replaced by c
0 when p
=
∞) is order isomorphic to a sublattice of Λ
φ
,
w
0
(
0
,
∞
),
(ii)
eitherp
∈
[
α
φ
0
,
β
φ
0
]
∪
[
α
φ
∞
,
β
φ
∞
]
=
[
α
φ
,
β
φ
] or, for some c
>
0, ∫
0
∞
‍
φ
(
c
t
-
1
/
p
)
w
(
t
)
d
t
<
∞;
(2)
[19, Theorem 6.13] For I
=
(
0
,
a
)
,
a
<
∞, the following assertions are equivalent:
(i)
ℓ
p (replaced by c
0 when p
=
∞) is order isomorphic to a sublattice of Λ
φ
,
w
0
(
0
,
a
),
(ii)
p
∈
[
α
φ
∞
,
β
φ
∞
],
(iii)
ℓ
p (replaced by c
0 when p
=
∞) is order isomorphic to a sublattice of L
φ
0
(
0
,
a
);
(3)
[19, Theorem 7.8] For I
=
ℕ the following statements are equivalent (where c
0 is meant in place of ℓ
p if p
=
∞):
(i)
ℓ
p is order isomorphic to a sublattice of d
0
(
w
,
φ
),
(ii)
ℓ
p is isomorphic to a subspace of d
0
(
w
,
φ
),
(iii)
p
∈
[
α
φ
0
,
β
φ
0
],
(iv)
ℓ
p is isomorphic to a subspace of h
φ.


## 3. Main Results

Theorem 5.
If1
<
p
<
∞ and φ
∈
Δ
*
p (resp., φ
∈
Δ
∞
*
p and φ
∈
Δ
0
*
p), then Λ
φ
,
w
(
0
,
∞
) (resp., Λ
φ
,
w
(
0
,
a
)
,
a
<
∞; d
(
φ
,
w
)) is p-convex. If Λ
φ
,
w
(
0
,
∞
) (resp., Λ
φ
,
w
(
0
,
a
)
,
a
<
∞; d
(
φ
,
w
)) is p-convex for some 1
<
p
<
∞, then p
≤
α
φ (resp., p
≤
α
φ
∞; p
≤
α
φ
0).Proof.
(1) By Lemma2 we assume that ψ
(
t
)
=
φ
(
t
1
/
p
) is convex. Then we have the equality ∥
f
∥
Λ
ψ
,
w
(
p
)
(
I
)
=
∥
f
∥
φ
,
w, where Λ
ψ
,
w
(
p
)
(
I
) is a p-convexification of Λ
ψ
,
w
(
I
). Indeed,
(14)
∥
f
∥
Λ
ψ
,
w
(
p
)
(
I
)
=
∥
|
f
|
p
∥
ψ
,
w
1
/
p
=
inf
⁡
⁡
{
ϵ
1
/
p
>
0
:
∫
I
‍
ψ
(
(
|
f
|
p
)
*
ϵ
)
≤
1
}
=
inf
⁡
{
ϵ
1
/
p
>
0
:
∫
I
‍
φ
(
f
*
ϵ
1
/
p
)
≤
1
}
=
∥
f
∥
φ
,
w
.
Sinceψ is a convex Orlicz function, the space Λ
ψ
,
w
(
I
) is a Banach space, so its p-convexification Λ
ψ
,
w
(
p
)
(
I
)
=
Λ
φ
,
w
(
I
) is p-convex [26].
We only show the second part in the case ofI
=
(
0
,
∞
). If p
>
α
φ
=
min
⁡
(
α
φ
0
,
α
φ
∞
), then, by Theorem 4(1), there is α
φ
≤
r
<
p such that ℓ
r is an order copy in Λ
φ
,
w
(
0
,
∞
), and thus the space cannot be p-convex.Example 6.
In general, the parallel statement on concavity in Theorem5 does not hold. As shown by Pisier in [26, Example 1
.
f
.
19], the Lorentz space d
(
p
,
w
), where w
n
=
n
1
/
p
-
(
n
-
1
)
1
/
p and 1
<
p
<
2, is not 2-concave even though the function φ
(
t
)
=
t
p satisfies condition Δ
2 and w is regular.Proposition 7.
Let1
<
p
≤
2. If φ
∈
Δ
*
p and φ
∈
Δ
2 if a
=
∞ (resp., φ
∈
Δ
∞
*
p and φ
∈
Δ
2
∞ in the case when a
<
∞) and w is regular, then Λ
φ
,
w
(
0
,
a
), 0
<
a
≤
∞, has type p.Proof.
By Theorem5 the space is p-convex, and thus in view of Theorem 3 and by applying the well-known fact [26, Proposition 1
.
f
.
3], the space has type p.Rakov [27, Theorem 1] showed that if a Banach space has type 1
<
p
≤
2, then it has p-Banach-Saks property. Hence we get the following.Theorem 8.
If a Banach spaceX has type p for some 1
<
p
≤
2, then X
∈
(
B
S
p
). Consequently p
≤
γ
(
X
).Notice that for every separable r.i. spaceE on [
0,1
], it is shown in [28] that γ
(
E
)
≤
2. Because every r.i. function space X on [
0
,
∞
) contains the r.i. space X
~
=
{
f
χ
[
0,1
]
:
f
∈
X
} isometrically, we get the following.Proposition 9 (see [29]).
IfX is a separable r.i. space on (
0
,
a
), 0
<
a
≤
∞, then γ
(
X
)
≤
2.Recall that a Banach spaceX is said to be p-smooth (1
≤
p
≤
2) if its modulus of smoothness ρ
X
(
ϵ
) is majorized by the function ϵ
p. That is, there is a constant A
>
0 such that, for all ϵ
>
0,
(15)
ρ
X
(
ϵ
)
:
=
inf
⁡
{
∥
x
∥
=
∥
y
∥
=
1
}
∥
x
+
ϵ
y
∥
+
∥
x
-
ϵ
y
∥
-
2
2
≤
A
ϵ
p
.Rakov [27, Theorem 1] showed that if X is a p-smooth space, then it has the (
S
p
) property. It is well known [26] that a Banach lattice is isomorphic to a p-smooth Banach space for some 1
<
p
≤
2 if and only if it has the type p and satisfies a lower q-estimate for some 1
≤
q
<
∞. Notice also that the finite concavity is equivalent to the lower q-estimate for some q
<
∞ [26]. Then we obtain immediately the following corollary as a consequence of definition of lower indices and by Propositions 7 and 9 and Theorem 8.Corollary 10.
Let1
<
p
≤
2. If φ
∈
Δ
*
p and φ
∈
Δ
2 if a
=
∞ (resp., φ
∈
Δ
∞
*
p and φ
∈
Δ
2
∞ in the case when a
<
∞) and w is regular, then Λ
φ
,
w
(
0
,
a
), 0
<
a
≤
∞, has the S
p-property.
Consequently, ifφ
∈
Δ
2 when a
=
∞ (resp., φ
∈
Δ
2
∞ in the case when a
<
∞) and w is regular, then
(16)
min
⁡
(
2
,
α
φ
)
≤
γ
(
Λ
φ
,
w
(
0
,
∞
)
)
,
min
⁡
(
2
,
α
φ
∞
)
≤
γ
(
Λ
φ
,
w
(
0
,
a
)
)
,
000000000000000
i
i
i
000
a
<
∞
.The next corollary is a result of the facts thatγ
(
ℓ
∞
)
=
1 and that if φ does not satisfy appropriate condition Δ
2, then Λ
φ
,
w
(
I
) contains an isomorphic copy of ℓ
∞.Corollary 11.
Ifφ
∉
Δ
2 when a
=
∞, φ
∉
Δ
2
∞ in the case when a
<
∞, and φ
∉
Δ
2
0 when I
=
ℕ, then
(17)
γ
(
Λ
φ
,
w
(
0
,
∞
)
)
=
1
,
γ
(
Λ
φ
,
w
(
0
,
a
)
)
=
1
,
0
i
00000000000000
a
<
∞
,
γ
(
d
(
φ
,
w
)
)
=
1
.Corollary 12.
An Orlicz-Lorentz spaceΛ
φ
,
w
0
(
0
,
∞
) (resp., Λ
φ
,
w
0
(
0
,
a
), a
<
∞; d
0
(
w
,
φ
)) does not contain an isomorphic copy of ℓ
1 if and only if α
φ
>
1 (resp., α
φ
∞
>
1; α
φ
0
>
1).Proof.
This hypothesis for sequence spaces is an immediate consequence of Theorem4(3).
Let nowa
<
∞. From [19, Corollary 3.5(2), (3)], the space Λ
φ
,
w
0
(
0
,
a
) contains an isomorphic copy of ℓ
1 if and only if either ℓ
1 is an order isomorphic copy, that is, in view of Theorem 4(2), α
φ
∞
=
1, or the sequence of functions min
⁡
(
t
-
1
,
n
) is bounded in L
φ
(
0
,
a
). However α
φ
∞
>
1 implies that there exists p
>
1 such that φ
(
β
u
)
≥
K
β
p
φ
(
u
) for all u
,
β
≥
1 and some K
>
0. Hence for any constant c
>
0 and large enough n
∈
ℕ,
(18)
∫
0
a
‍
φ
(
c
min
⁡
(
t
-
1
,
n
)
)
d
t
≥
∫
0
1
/
n
‍
φ
(
c
n
)
d
t
≥
∫
0
1
/
n
‍
K
c
p
n
p
φ
(
1
)
d
t
=
K
φ
(
c
)
n
p
-
1
⟶
∞
.
Thus whenever α
φ
∞
>
1 the sequence min
⁡
(
t
-
1
,
n
) is not bounded, and the proof is done in case of finite interval.
Assume thata
=
∞. Again applying Corollary 3.5(2) in [19], we get dichotomy that ℓ
1 is an isomorphic copy in Λ
φ
,
w
0
(
0
,
∞
) if and only if either ℓ
1 is an order isomorphic copy in Λ
φ
,
w
0
(
0
,
∞
) or the sequence min
⁡
(
t
-
1
χ
(
0,1
)
(
t
)
,
n
) is bounded in L
φ
(
0
,
∞
). Assuming now that α
φ
>
1, we can show similarly as above that the second alternative cannot occur. Recall then that, by Theorem 4(1), ℓ
1 is an order isomorphic copy in the space if and only if either α
φ
=
1 or, for some c
>
0, ∫
0
∞
‍
φ
(
c
t
-
1
)
w
(
t
)
d
t
<
∞. We will exclude again the second possibility. In fact if α
φ
>
1, then for some p
>
1 and K
>
0 and for all β
≥
1 and u
≥
0, φ
(
β
u
)
≥
K
β
p
φ
(
u
). Therefore for every c
>
0 and some t
0
<
1 such that w
(
t
0
)
>
0, we have
(19)
∫
0
∞
‍
φ
(
c
t
-
1
)
w
(
t
)
d
t
≥
∫
0
t
0
‍
φ
(
c
t
-
1
)
w
(
t
0
)
d
t
≥
K
φ
(
c
)
w
(
t
0
)
∫
0
t
0
‍
t
-
p
d
t
=
∞
,
and this completes the proof.Corollary 13.
Assume thatΛ
φ
,
w
(
0
,
a
), 0
<
a
≤
∞, does not contain an isomorphic copy of ℓ
1, equivalently that α
φ
>
1 in case a
=
∞ and α
φ
∞
>
1 in case a
<
∞. Then
(20)
γ
(
Λ
φ
,
w
(
0
,
∞
)
)
≤
min
⁡
(
2
,
α
φ
)
,
γ
(
Λ
φ
,
w
(
0
,
a
)
)
≤
min
⁡
⁡
(
2
,
α
φ
∞
)
,
00000000000000000
i
0
a
<
∞
.Proof.
By Proposition9 the Banach-Saks index of any r.i. space over the interval (
0
,
a
) cannot be bigger than 2. On the other hand, recall that Γ
(
ℓ
p
)
=
[
1
,
p
] for 1
<
p
<
∞ and Γ
(
c
0
)
=
Γ
(
ℓ
1
)
=
[
1
,
∞
]. The space ℓ
∞ does not have w
B
S property, so Γ
(
ℓ
∞
)
=
1. Hence Theorem 4 shows that
(21)
γ
(
Λ
φ
,
w
(
0
,
a
)
)
≤
A
∶
=
inf
⁡
{
p
>
1
:
ℓ
p
is
an
isomorphic
copy
in
Λ
φ
,
w
(
0
,
a
)
}
,
where A
≤
α
φ when a
=
∞ and A
≤
α
φ
∞ if a
<
∞.Finally by Corollaries10 and 11 we get the result on Banach-Saks index in the Orlicz-Lorentz spaces over the interval (
0
,
a
), a
≤
∞.Theorem 14.
Assume thatΛ
φ
,
w
(
0
,
a
), 0
<
a
≤
∞, does not contain an isomorphic copy of ℓ
1 and equivalently that α
φ
>
1 in case a
=
∞ and α
φ
∞
>
1 in case a
<
∞. Assume also that φ
∈
Δ
2 when a
=
∞, φ
∈
Δ
2
∞ when a
<
∞, and w is regular. Then
(22)
min
⁡
(
2
,
α
φ
)
=
γ
(
Λ
φ
,
w
(
0
,
∞
)
)
,
min
⁡
⁡
(
2
,
α
φ
∞
)
=
γ
(
Λ
φ
,
w
(
0
,
a
)
)
,
0
i
i
00000000000000000
a
<
∞
.Using the Bessaga-Pełczynski method, we get the following.Theorem 15.
Let1
<
p
<
∞. If an order continuous sequence space X is p-convex, then X has the (
S
p
) property. Consequently p
≤
γ
(
X
).Proof.
SinceX is order continuous, the standard unit vectors {
e
n
} form a basis of X. Suppose that {
x
n
} is a normalized weakly null sequence in X. By the Bessaga-Pełczyński method [5, Proposition 1
.
a
.
12], we may assume that x
n
=
y
n
+
z
n for all n, where {
y
n
} is a block basic sequence of {
e
n
} and ∥
z
n
∥
≤
1
/
2
n for all n. Then for real numbers a
1
,
…
,
a
n with ∑
j
=
1
n
‍
|
a
j
|
p
≤
1, the p-convexity of X and disjointness of the supports of y
n’s show that
(23)
∥
∑
j
=
1
n
‍
a
j
x
j
∥
≤
∥
∑
j
=
1
n
‍
a
j
y
j
∥
+
∑
j
=
1
n
‍
|
a
j
|
∥
z
j
∥
≤
∥
(
∑
j
=
1
n
‍
|
a
j
y
j
|
p
)
1
/
p
∥
+
(
∑
j
=
1
n
‍
|
a
j
|
p
)
1
/
p
(
∑
j
=
1
n
‍
1
2
j
q
)
1
/
q
≤
2
C
(
∑
j
=
1
n
‍
|
a
j
|
p
)
1
/
p
+
(
∑
j
=
1
n
‍
|
a
j
|
p
)
1
/
p
(
∑
j
=
1
n
‍
1
2
j
q
)
1
/
q
≤
(
2
C
+
1
)
(
∑
j
=
1
n
‍
|
a
j
|
p
)
1
/
p
,
where C is the constant from the p-convexity of X and q is the conjugate of p satisfying 1
/
p
+
1
/
q
=
1. The proof is done.As a consequence of the above Theorems15 and 5 about p-convexity and Theorem 4 about ℓ
p-copies, we get the Banach-Saks index in sequence spaces as follows.Theorem 16.
Assume thatd
(
φ
,
w
) does not contain an isomorphic copy of ℓ
1 and equivalently that α
φ
0
>
1. Assume also that φ
∈
Δ
2
0 and W
(
∞
)
=
∞. Then
(24)
γ
(
d
(
φ
,
w
)
)
=
α
φ
0
.Proof.
By Theorems5 and 15, α
φ
0
≤
γ
(
d
(
φ
,
w
)
) and
(25)
γ
(
d
(
φ
,
w
)
)
≤
inf
⁡
{
p
>
1
:
ℓ
p
⊂
d
(
φ
,
w
)
}
.
Since γ
(
ℓ
p
)
=
p (1
<
p
<
∞), Theorem 4 completes the proof.We finish with a result on the weak Banach-Saks property.Theorem 17.
A separable Orlicz-Lorentz spaceΛ
φ
,
w
(
0
,
a
), a
≤
∞, has the weak Banach-Saks property.Proof.
Let{
x
n
}
⊂
Λ
φ
,
w
(
0
,
a
) be weakly null and assume without loss of generality that ∥
x
n
∥
=
1, n
∈
ℕ. Then by [29, Proposition 3.2] there exists a subsequence {
x
n
′
} splitted into three subsequences
(26)
x
n
′
=
y
n
+
z
n
+
d
n
,
such that {
y
n
}
,
{
z
n
}
,
{
d
n
}
⊂
Λ
φ
,
w
(
0
,
a
) are bounded and(a)
y
1
*
=
y
n
*, ∥
y
n
∥
≤
1, and y
n
→
0 weakly;
(b)
z
n
z
m
=
0 for n
≠
m, ∥
z
n
∥
≤
2, z
n
→
0 in measure, and z
n
→
0 weakly;
(c)
∥
d
n
∥
→
0.We need to show that there is a subsequence of {
x
n
′
} whose Cesàro means approach zero in norm. It is obvious that we can do it for the sequence {
d
n
} satisfying (c). We will show this now for the sequence {
y
n
}. By [29, Proposition 4.3] there exists a subsequence {
y
n
(
k
)
}
⊂
{
y
n
}, x
∈
L
1
(
0
,
a
)
+
L
∞
(
0
,
a
), such that
(27)
1
N
∑
k
=
1
N
‍
y
m
(
k
)
⟶
x
in L
1
(
0
,
a
)
+
L
∞
(
0
,
a
) for all further subsequences {
y
m
(
k
)
}
⊂
{
y
n
(
k
)
} (we can assume that y
m
(
1
)
=
y
1). Hence
(28)
1
N
∑
k
=
1
N
‍
y
m
(
k
)
⟶
x
in
measure
on
⁡
(
0
,
a
)
.
(Recall that if x
n
→
0 in any r.i. space X, then x
n
→
0 in measure.) Since y
n
(
k
)
→
0 weakly in Λ
φ
,
w
(
0
,
a
), we have, for any subsequence (
m
(
k
)
) of (
n
(
k
)
),
(29)
1
N
∑
k
=
1
N
‍
y
m
(
k
)
⟶
0
weakly
in
L
1
(
0
,
a
)
+
L
∞
(
0
,
a
)
.
Applying now [29, Lemma 4.4] we get that
(30)
1
N
∑
k
=
1
N
‍
y
m
(
k
)
⟶
0
in
L
1
(
0
,
a
)
+
L
∞
(
0
,
a
)
for all subsequences {
y
m
(
k
)
}
⊂
{
y
n
(
k
)
}. Let now {
y
m
(
k
)
} be an arbitrary subsequence of {
y
n
(
k
)
}. Set
(31)
a
N
=
1
N
∑
k
=
1
N
‍
y
m
(
k
)
.
By y
m
(
k
)
*
=
y
1
* and by subadditivity of the operator t
↦
∫
0
t
‍
f
*, f
∈
L
0
(
I
), we get, for t
≥
0,
(32)
∫
0
t
‍
(
1
N
∑
k
=
1
N
‍
y
m
(
k
)
)
*
≤
1
N
∑
k
=
1
N
‍
∫
0
t
‍
y
m
(
k
)
*
=
∫
0
t
‍
y
1
*
;
that is,
(33)
a
N
≺
≺
y
1
.
Since a
N
→
0 in measure and a
N
≺
≺
y
1 for every N
∈
ℕ, by [29, Proposition 4.2],
(34)
1
N
∑
k
=
1
N
‍
y
m
(
k
)
⟶
0
in Λ
φ
,
w
(
0
,
a
), which finishes the case (a).
In the case (b), by [19, Theorem 5.5], there exists a subsequence {
z
n
(
k
)
}
⊂
{
z
n
} such that any further subsequence {
z
m
(
k
)
} of {
z
n
(
k
)
} is equivalent to the unit vector basis in some Orlicz sequence space ℓ
ψ
0. Thus z
n
(
k
)
→
0 weakly in ℓ
ψ
0, and by w
B
S-property of ℓ
ψ
0 (see, e.g., [27]), there is a subsequence {
z
k
′
}
⊂
{
z
n
(
k
)
} such that (
1
/
n
)
∑
k
=
1
N
‍
z
k
′
→
0 in ℓ
ψ
0 and thus in Λ
φ
,
w
(
0
,
a
). This completes case (b) and proves the theorem.A Banach spaceX is said to have the(resp., weakly) alternative signs Banach-Saks property (ABS) (resp., (wABS)) if, for every bounded (resp., weakly null) sequence (
x
n
)
n
=
1
∞, there is a subsequence (
y
n
) and a sequence (
ϵ
n
) of signs such that the Cesàro averages are norm convergent to zero. That is,
(35)
lim
⁡
n
→
∞
1
n
∥
∑
i
=
1
n
‍
ϵ
i
y
i
∥
=
0
.
It is shown by Rosenthal [30, 31] that if X has the (ABS), then X has the weak Banach-Saks property. In case that X does not contain ℓ
1, then (ABS) is equivalent to the weak Banach-Saks property [31]. It is clear that ℓ
1 does not have the (ABS) and has the weak Banach-Saks property. Hence a Banach space X has the (ABS) if and only if X has the weak Banach-Saks property and does not contain ℓ
1. The notion of (wABS) was introduced in [32] and it was shown that (wABS) is equivalent to (wBS).Corollary 18.
LetI
=
(
0
,
a
) for some 0
<
a
≤
∞. Then the following are equivalent: (1)
Orlicz-Lorentz spaceΛ
φ
,
w
(
I
) has the Banach-Saks property (BS);
(2)
Orlicz-Lorentz spaceΛ
φ
,
w
(
I
) is reflexive;
(3)
Orlicz-Lorentz spaceΛ
φ
,
w
(
I
) does not contain isomorphic copies of c
0 or ℓ
1;
(4)
Orlicz-Lorentz spaceΛ
φ
,
w
(
I
) is separable and does not contain an isomorphic copy of ℓ
1;
(5)
Orlicz-Lorentz spaceΛ
φ
,
w
(
I
) has the weak Banach-Saks property (wBS) and does not contain an isomorphic copy of ℓ
1;
(6)
Orlicz-Lorentz spaceΛ
φ
,
w
(
I
) has the alternative signs Banach-Saks property (ABS).Proof.
Recall that a Banach lattice is reflexive if and only if it does not contain isomorphic copies ofc
0 or ℓ
1 [26, Theorem 1
.
c
.
5]. Hence we get (
2
)
⇔
(
3
). Recall also that a Banach lattice X is a KB space if and only if X does not contain an isomorphic copy of c
0 [33, Theorem 14.12]. That is, if X is a Köthe function space, then it is equivalent to the case in which X is order continuous and satisfies the Fatou property. Since an Orlicz-Lorentz space satisfies the Fatou property and the order continuity is equivalent to the separability of the Orlicz-Lorentz space, (
3
)
⇔
(
4
) follows.
Since a nonseparable Orlicz-Lorentz space containsℓ
∞ and it does not have the weak Banach-Saks property, Theorem 17 shows that (
4
)
⇔
(
5
). The preceding argument of the corollary shows that (
5
)
⇔
(
6
). Since (6) implies both (5) and (2), (6) implies (1). The proof is done.



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*Source: 423198-2014-03-13.xml*