# Packing Constant in Orlicz Sequence Spaces Equipped with the p-Amemiya Norm

Authors: Xin He; Jijie Yu; Yunan Cui; Xin Huo
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

The problem of packing spheres in Orlicz sequence spacel
Φ
,
p equipped with the p-Amemiya norm is studied, and a geometric characteristic about the reflexivity of l
Φ
,
p is obtained, which contains the relevant work about l
p
(
p
>
1
) and classical Orlicz spaces l
Φ discussed by Rankin, Burlak, and Cleaver. Moreover the packing constant as well as Kottman constant in this kind of spaces is calculated.


---

## Body

## 1. Introduction and Preliminaries

The packing constant is an important and interesting geometric parameter for studying the geometric structure, isometric embedding, noncompactness, and reflexivity in Banach spaces [1–4]. Let X be a Banach space. We denote by B
(
X
) the unit ball of X and by S
(
X
) the unit sphere of X. The packing constant P
(
X
) of X is the real number such that if r
≤
P
(
X
), then an infinite number of spheres of radius r can be packed in B
(
X
), and if r
>
P
(
X
), only a finite number of spheres can be done. It began in the 1950s studying the packing constant of special sequence spaces. Burlak et al. [1] proved that P
(
l
1
)
=
P
(
l
∞
)
=
1
/
2 and P
(
l
p
)
=
1
/
(
1
+
2
1
-
(
1
/
p
)
) for 1
<
p
<
∞. Rankin found P
(
l
2
) and P
(
l
p
)
(
p
>
1
) in 1955 and 1958, respectively. In 1976, Cleaver discussed Orlicz sequence space l
Φ
∘ equipped with the Orlicz norm under a strong condition, and he found upper and lower bounds of P
(
l
Φ
∘
). In 1983, Ye investigated Orlicz sequence space l
Φ equipped with the Luxemburg norm and obtained a formula for P
(
l
Φ
) [5].In this paper, an analogue for Orlicz sequence spaces equipped with the p-Amemiya norm is illustrated, and some useful definitions and lemmas are presented.Definition 1 (see [1]).
The packing constant of a Banach spaceX is defined by
(1)
P
(
X
)
=
sup
⁡
{
∥
x
i
-
x
j
∥
r
>
0
:
there
exists
{
x
i
}
i
=
1
∞
,
∥
x
i
∥
≤
1
-
r
,
∥
x
i
-
x
j
∥
≥
2
for
i
,
j
∈
N
,
i
≠
j
∥
x
i
-
x
j
∥
}
.
It is obvious that P
(
X
)
=
0, if dim
⁡
⁡
X
<
∞.Lemma 2 (see [2]).
LetX be an infinite-dimensional Banach space. Define
(2)
K
(
X
)
=
sup
⁡
{
inf
⁡
⁡
{
∥
x
n
-
x
m
∥
:
n
≠
m
}
:
{
x
n
}
n
=
1
∞
⊂
S
(
X
)
}
,
which is called the Kottman constant of X. Then
(3)
P
(
X
)
=
K
(
X
)
K
(
X
)
+
2
.It is known that1
≤
K
(
X
)
≤
2. Due to Riesz lemma, it can be summarised that K
(
X
)
≥
1 for any infinite-dimensional Banach space X. Finite-dimensional spaces have Kottman constant equal to zero. Furthermore, Elton and Odell in [6] proved that if X is an infinite-dimensional Banach space, then there exists an ε
>
0 such that K
(
X
)
≥
1
+
ε. Consequently, 1
/
3
≤
P
(
X
)
≤
1
/
2. Hudzik proved that P
(
Y
)
=
1
/
2 and K
(
Y
)
=
2 for every nonreflexive Banach lattice Y [7].Recall that a Banach spaceX is said to be P-convex (see [2]) if P
(
n
,
X
)
<
1
/
2, for some n
∈
N, n
≥
2, where
(4)
P
(
n
,
X
)
=
sup
⁡
⁡
{
∥
x
i
-
x
j
∥
r
>
0
:
there
exist
{
x
i
}
i
=
1
n
,
∥
x
i
∥
≤
1
-
r
,
v
v
v
v
v
v
v
∥
x
i
-
x
j
∥
≥
2
r
for
i
≠
j
∥
x
i
-
x
j
∥
}
.Kottman [2] has proved that any P-convex Banach space is reflexive.The packing problem in Orlicz sequence spaces was investigated in [8–11]. The packing constant for Musielak-Orlicz sequence spaces and Cesaro sequence spaces have been calculated in [12, 13].For any mapΦ
:
R
→
[
0
,
∞
], define
(5)
a
Φ
=
max
⁡
⁡
{
u
≥
0
:
Φ
(
u
)
=
0
}
,
b
Φ
=
max
⁡
⁡
{
u
≥
0
:
Φ
(
u
)
<
∞
}
.
A map Φ is said to be an Orlicz function, if Φ
(
0
)
=
0; Φ is not identically equal to zero; it is even and convex on the interval (
-
b
Φ
,
b
Φ
) and left-continuous at b
Φ.For every Orlicz functionΦ, we define its complementary function Ψ
:
R
→
[
0
,
∞
] by the formula
(6)
Ψ
(
v
)
=
sup
⁡
{
u
|
v
|
-
Φ
(
u
)
:
u
≥
0
}
.
The complementary function Ψ is also an Orlicz function. The convex modular I
Φ is defined on l
0 (the space of all real sequences) by I
Φ
(
x
)
=
∑
i
=
1
∞
Φ
(
x
(
i
)
) for any x
=
(
x
(
i
)
).Definition 3 (see [14–16]).
The Orlicz sequence space is defined as the set(7)
l
Φ
=
{
x
=
(
x
(
i
)
)
:
I
Φ
(
λ
x
)
<
∞
,
for
some
λ
>
0
}
.
The Luxemburg norm and the Orlicz norm are expressed as
(8)
∥
x
∥
Φ
=
inf
⁡
{
λ
>
0
:
I
Φ
(
x
λ
)
≤
1
}
,
∥
x
∥
Φ
∘
=
inf
⁡
k
>
0
1
k
(
1
+
I
Φ
(
k
x
)
)
,
respectively. The Orlicz space equipped with the Luxemburg norm and the Orlicz norm are denoted by l
Φ and l
Φ
∘, respectively.
For any1
≤
p
≤
∞ and u
≥
0, define
(9)
s
p
(
u
)
=
{
(
1
+
u
p
)
1
/
p
,
for
1
≤
p
<
∞
,
max
⁡
⁡
{
1
,
u
}
,
for
p
=
∞
and define s
Φ
,
p
(
x
)
=
s
p
∘
I
Φ
(
x
) for all 1
≤
p
≤
∞. Note that the functions s
p and s
Φ
,
p are convex. Moreover, the function s
p is increasing on R
+, for 1
≤
p
<
∞, but the function s
∞ is increasing on the interval [
1
,
∞
) only.Definition 4 (see [17, 18]).
Let1
≤
p
≤
∞. For any x
=
(
x
(
i
)
), define the p-Amemiya norm by the formula
(10)
∥
x
∥
Φ
,
p
=
inf
⁡
k
>
0
1
k
s
Φ
,
p
(
k
x
)
.
The Orlicz space equipped with the p-Amemiya norm will be denoted by l
Φ
,
p.
It is known that∥
x
∥
Φ
,
1
=
∥
x
∥
Φ
∘ and ∥
x
∥
Φ
,
∞
=
∥
x
∥
Φ. If 1
≤
p
<
∞,  x
≠
0, then
(11)
1
2
∥
x
∥
Φ
∘
≤
∥
x
∥
Φ
≤
∥
x
∥
Φ
,
p
≤
2
1
/
p
∥
x
∥
Φ
<
2
1
/
p
∥
x
∥
Φ
∘
.
(See [17].)
Letp
+ be the right-hand side derivative of Φ on [
0
,
b
Φ
) and put p
+
(
b
Φ
)
=
lim
⁡
u
→
b
Φ
-
p
+
(
u
). Define the function α
p
:
l
Φ
,
p
→
[
-
1
,
∞
] by
(12)
α
p
(
x
)
=
{
I
Φ
p
-
1
(
x
)
I
Ψ
(
p
+
(
|
x
|
)
)
-
1
,
1
≤
p
<
∞
,
-
1
,
p
=
∞
,
I
Φ
(
x
)
≤
1
,
I
Ψ
(
p
+
(
|
x
|
)
)
,
p
=
∞
,
I
Φ
(
x
)
>
1
and the functions k
p
*
:
l
Φ
,
p
→
[
0
,
∞
) and k
p
*
*
:
l
Φ
,
p
→
[
0
,
∞
) by
(13)
k
p
*
(
x
)
=
inf
⁡
⁡
{
k
≥
0
:
α
p
(
k
x
)
≥
0
}
,
(
with
inf
⁡
⁡
ϕ
=
∞
)
,
k
p
*
*
(
x
)
=
inf
⁡
⁡
{
k
≥
0
:
α
p
(
k
x
)
≤
0
}
.
It is obvious that k
p
*
(
x
)
≤
k
p
*
*
(
x
) for every 1
≤
p
≤
∞ and x
∈
l
Φ
,
p.
SetK
p
(
x
)
=
{
0
<
k
<
∞
:
k
p
*
(
x
)
≤
k
≤
k
p
*
*
(
x
)
}.Definition 5 (see [14]).
We say an Orlicz functionΦ satisfies the Δ
2
(
0
)-condition (Φ
∈
Δ
2
(
0
), for short) if there exist constants K
≥
2 and u
0
>
0 such that Φ
(
u
0
)
>
0 and
(14)
Φ
(
2
u
)
≤
K
Φ
(
u
)
for
every
|
u
|
≤
u
0
.
For more details about Orlicz spaces, we refer the reader to [15, 16, 18, 19].Lemma 6 (see [20]).
Assume thatΦ
∈
Δ
2
(
0
), 1
≤
p
<
∞. Then, for any L
>
0 and ε
>
0, there exists δ
>
0 such that for any x
,
y
∈
l
0 there holds the implication
(15)
(
I
Φ
(
x
)
≤
L
)
∧
(
I
Φ
(
y
)
≤
δ
)
⟹
|
I
Φ
p
(
x
+
y
)
-
I
Φ
p
(
x
)
|
<
ε
.


## 2. Main Results

Assume thatΦ
∈
Δ
2
(
0
), 1
≤
p
<
∞. Then, for any x
∈
S
(
l
Φ
,
p
) and k
>
1, there exists a unique d
x
,
k
>
0 such that
(16)
I
Φ
p
(
k
x
d
x
,
k
)
=
k
p
-
1
2
p
.
Set
(17)
d
x
=
inf
⁡
{
d
x
,
k
:
k
>
1
}
,
d
=
sup
⁡
{
d
x
:
x
∈
S
(
l
Φ
,
p
)
}
.
Then d
x
>
1 and 1
<
d
≤
2. Denote
(18)
k
′
=
inf
⁡
⁡
{
k
:
k
∈
K
p
(
x
)
,
∥
x
∥
Φ
,
p
=
1
}
,
k
′′
=
sup
⁡
⁡
{
k
:
k
∈
K
p
(
x
)
,
∥
x
∥
Φ
,
p
=
1
}
.In the sequel, the packing constantl
Φ
,
p is calculated, and the main results of this paper are proposed.Theorem 7.
IfΦ
∈
Δ
2
(
0
), 1
≤
p
<
∞, then K
(
l
Φ
,
p
)
=
d and P
(
l
Φ
,
p
)
=
d
/
(
d
+
2
).Proof.
For anyε
>
0, there exists x
∈
S
(
l
Φ
,
p
) such that d
x
>
d
-
ε, so d
x
,
k
>
d
-
ε for all k
>
1. Define
(19)
x
n
=
∑
i
=
1
∞
‍
x
(
i
)
e
2
n
-
1
(
2
i
-
1
)
,
∀
n
∈
N
.
Then {
x
n
} have pairwise disjoint supports and ∥
x
n
∥
Φ
,
p
=
∥
x
∥
Φ
,
p
=
1  (
n
∈
N
). For all n
≠
m and all k
>
1,
(20)
1
k
(
1
+
I
Φ
p
(
k
x
n
-
x
m
d
-
ε
)
)
1
/
p
=
1
k
(
1
+
2
p
I
Φ
p
(
k
x
d
-
ε
)
)
1
/
p
>
1
k
(
1
+
2
p
I
Φ
p
(
k
x
d
x
,
k
)
)
1
/
p
=
1
k
(
1
+
2
p
·
k
p
-
1
2
p
)
1
/
p
=
1
.
Then ∥
x
n
-
x
m
∥
Φ
,
p
=
inf
⁡
k
>
0
(
1
/
k
)
(
1
+
I
Φ
p
(
k
(
x
n
-
x
m
)
)
)
1
/
p
≥
d
-
ε, so we have K
(
l
Φ
,
p
)
≥
d, since ε is arbitrary.
In the following,K
(
l
Φ
,
p
)
≤
d will be illustrated as an important part of our results.
For any sequence{
x
n
}
⊂
S
(
l
Φ
,
p
), which means that x
n
=
(
x
n
(
i
)
)
i, ∥
x
n
∥
Φ
,
p
=
∥
∑
i
=
1
∞
x
n
(
i
)
e
i
∥
Φ
,
p
=
1, for any n
∈
N, then {
∥
x
n
(
i
)
e
i
∥
Φ
,
p
}
n is bounded for all i
∈
N.
Since{
∥
x
n
(
1
)
e
1
∥
Φ
,
p
}
n is bounded, there exists a subsequence {
x
1
n
}
⊂
{
x
n
} such that {
∥
x
1
n
(
1
)
e
1
∥
Φ
,
p
}
n is convergent, but {
∥
x
1
n
(
2
)
e
2
∥
Φ
,
p
}
n is bounded, so there exists a subsequence {
x
2
n
}
⊂
{
x
1
n
} such that {
∥
x
2
n
(
2
)
e
2
∥
Φ
,
p
}
n is convergent. In a similar way, using the diagonal method, we can find a subsequence {
x
n
n
}
⊂
{
x
n
} such that, for any i
∈
N, {
∥
x
n
n
(
i
)
e
i
∥
Φ
,
p
}
n
≥
i is convergent. Denoting ∥
e
i
∥
Φ
,
p
=
s
i and setting ∥
x
n
n
(
i
)
e
i
∥
Φ
,
p
→
b
i as n
→
∞, then |
x
n
n
(
i
)
|
→
b
i
/
s
i as n
→
∞ for all i
∈
N.
Letx
=
(
b
i
/
s
i
)
i, |
x
n
n
|
=
(
|
x
n
n
(
i
)
|
)
i, and z
n
=
|
x
n
n
|
-
x. Then
(21)
z
n
(
i
)
⟶
0
as
n
⟶
∞
∀
i
∈
N
,
sep
(
z
n
)
=
sep
(
|
x
n
n
|
)
≥
sep
(
x
n
)
.
Since Φ
∈
Δ
2, then x
∈
S
(
l
Φ
,
p
). For any ε
>
0, there exists i
0
∈
N, such that ∥
∑
i
=
i
0
+
1
∞
x
(
i
)
e
i
∥
Φ
,
p
<
ε. Moreover, |
x
n
n
(
i
)
|
→
x
(
i
) as n
→
∞ for i
=
1
,
…
,
i
0. So we have
(22)
∥
z
n
∥
Φ
,
p
=
∥
∑
i
=
1
∞
‍
(
|
x
n
n
(
i
)
|
-
x
(
i
)
)
e
i
∥
Φ
,
p
≤
∥
∑
i
=
1
i
0
‍
(
|
x
n
n
(
i
)
|
-
x
(
i
)
)
e
i
∥
Φ
,
p
+
∥
∑
i
=
i
0
+
1
∞
‍
|
x
n
n
(
i
)
|
e
i
∥
Φ
,
p
+
ε
,
and, consequently, limsup
⁡
n
∥
z
n
∥
Φ
,
p
≤
1
+
ε.
For the aboveε
>
0, since l
Φ
,
p is order continuous, there exists i
1
∈
N such that ∥
∑
i
=
i
1
+
1
∞
z
n
1
(
i
)
e
i
∥
Φ
,
p
<
ε for n
1
=
1. Take n
2
>
n
1 such that ∥
∑
i
=
1
i
1
z
n
2
(
i
)
e
i
∥
Φ
,
p
<
ε. And for n
2, there exists i
2
>
i
1 such that ∥
∑
i
=
i
2
+
1
∞
z
n
2
(
i
)
e
i
∥
Φ
,
p
<
ε. Then
(23)
sep
(
z
n
)
≤
∥
z
n
1
-
z
n
2
∥
Φ
,
p
≤
∥
∑
i
=
1
i
1
‍
z
n
1
(
i
)
e
i
-
∑
i
=
i
1
+
1
i
2
‍
z
n
2
(
i
)
e
i
∥
Φ
,
p
+
∥
∑
i
=
i
1
+
1
∞
‍
z
n
1
(
i
)
e
i
∥
Φ
,
p
+
∥
∑
i
=
1
i
1
‍
z
n
2
(
i
)
e
i
∥
Φ
,
p
+
∥
∑
i
=
i
2
+
1
∞
‍
z
n
2
(
i
)
e
i
∥
Φ
,
p
≤
∥
∑
i
=
1
i
1
‍
z
n
1
(
i
)
e
i
-
∑
i
=
i
1
+
1
i
2
‍
z
n
2
(
i
)
e
i
∥
Φ
,
p
+
3
ε
.
Take n
3
>
n
2 such that ∥
∑
i
=
1
i
2
z
n
3
(
i
)
e
i
∥
Φ
,
p
<
ε, and for n
3, there exists i
3
>
i
2 such that ∥
∑
i
=
i
3
+
1
∞
z
n
3
(
i
)
e
i
∥
Φ
,
p
<
ε. Then
(24)
∥
z
n
1
-
z
n
3
∥
Φ
,
p
≤
∥
∑
i
=
1
i
1
‍
z
n
1
(
i
)
e
i
-
∑
i
=
i
2
+
1
i
3
‍
z
n
3
(
i
)
e
i
∥
Φ
,
p
+
∥
∑
i
=
i
1
+
1
∞
‍
z
n
1
(
i
)
e
i
∥
Φ
,
p
+
∥
∑
i
=
1
i
2
‍
z
n
3
(
i
)
e
i
∥
Φ
,
p
+
∥
∑
i
=
i
3
+
1
∞
‍
z
n
3
(
i
)
e
i
∥
Φ
,
p
≤
∥
∑
i
=
1
i
1
‍
z
n
1
(
i
)
e
i
-
∑
i
=
i
2
+
1
i
3
‍
z
n
3
(
i
)
e
i
∥
Φ
,
p
+
3
ε
,
∥
z
n
2
-
z
n
3
∥
Φ
,
p
≤
∥
∑
i
=
i
1
+
1
i
2
‍
z
n
2
(
i
)
e
i
-
∑
i
=
i
2
+
1
i
3
‍
z
n
3
(
i
)
e
i
∥
Φ
,
p
+
∥
∑
i
=
1
i
1
‍
z
n
2
(
i
)
e
i
∥
Φ
,
p
+
∥
∑
i
=
i
2
+
1
∞
‍
z
n
2
(
i
)
e
i
∥
Φ
,
p
+
∥
∑
i
=
1
i
2
‍
z
n
3
(
i
)
e
i
∥
Φ
,
p
+
∥
∑
i
=
i
3
+
1
∞
‍
z
n
3
(
i
)
e
i
∥
Φ
,
p
≤
∥
∑
i
=
i
1
+
1
i
2
‍
z
n
2
(
i
)
e
i
-
∑
i
=
i
2
+
1
i
3
‍
z
n
3
(
i
)
e
i
∥
Φ
,
p
+
4
ε
.
Analogously, we can find by induction a subsequence {
z
n
k
} of {
z
n
} and {
i
k
}
⊂
N such that n
1
<
n
2
<
⋯
<
n
k
<
⋯, i
1
<
i
2
<
⋯
<
i
k
<
⋯ such that, for any k
∈
N,
(25)
∥
∑
i
=
1
i
k
-
1
‍
z
n
k
(
i
)
e
i
∥
Φ
,
p
<
ε
,
∥
∑
i
=
i
k
+
1
∞
‍
z
n
k
(
i
)
e
i
∥
Φ
,
p
<
ε
,
∥
z
n
1
-
z
n
k
∥
Φ
,
p
≤
∥
∑
i
=
1
i
1
‍
z
n
1
(
i
)
e
i
-
∑
i
=
i
k
-
1
+
1
i
k
‍
z
n
k
(
i
)
e
i
∥
Φ
,
p
+
3
ε
,
∥
z
n
l
-
z
n
k
∥
Φ
,
p
≤
∥
∑
i
=
i
l
-
1
+
1
i
l
‍
z
n
l
(
i
)
e
i
-
∑
i
=
i
k
-
1
+
1
i
k
‍
z
n
k
(
i
)
e
i
∥
Φ
,
p
+
4
ε
,
∀
1
<
l
<
k
.
Since ∥
∑
i
=
1
∞
z
n
(
i
)
e
i
∥
Φ
,
p
≤
1
+
ε, ∥
∑
i
=
i
k
-
1
+
1
i
k
z
n
k
(
i
)
e
i
∥
Φ
,
p
/
(
1
+
ε
)
≤
1 for all k
∈
N. Therefore, for any l
,
k
∈
N,
(26)
∥
z
n
l
-
z
n
k
∥
Φ
,
p
≤
(
1
+
ε
)
×
∥
∑
i
=
i
l
-
1
+
1
i
l
z
n
l
(
i
)
e
i
∥
∑
i
=
i
l
-
1
+
1
i
l
z
n
l
(
i
)
e
i
∥
Φ
,
p
-
∑
i
=
i
k
-
1
+
1
i
k
z
n
k
(
i
)
e
i
∥
∑
i
=
i
k
-
1
+
1
i
k
z
n
k
(
i
)
e
i
∥
Φ
,
p
∥
Φ
,
p
+
4
ε
.
Setting y
k
=
∑
i
=
i
k
-
1
+
1
i
k
z
n
k
(
i
)
e
i
/
∥
∑
i
=
i
k
-
1
+
1
i
k
z
n
k
(
i
)
e
i
∥
Φ
,
p (for all k
∈
N), then
(27)
{
y
m
}
⊂
S
(
l
Φ
,
p
)
,
supp
⁡
(
y
l
)
∩
supp
⁡
(
y
m
)
=
ϕ
.
In this way, we get
(28)
K
(
l
Φ
,
p
)
≤
sep
(
x
n
)
≤
sep
(
z
n
)
≤
sep
(
z
n
i
)
≤
(
1
+
ε
)
∥
y
m
-
y
l
∥
Φ
,
p
+
4
ε
.
For anyε
>
0, by the definition of d, there exists k
m
>
1 such that d
y
m
,
k
m
<
d
+
ε, where d
y
m
,
k
m satisfies the equality I
Φ
p
(
k
m
y
m
/
d
y
m
,
k
m
)
=
(
k
m
p
-
1
)
/
2
p
(
m
∈
N
).
Setting∥
y
m
-
y
l
∥
=
λ
m
l and taking k
m
l
∈
K
p
(
(
y
m
-
y
l
)
/
λ
), we have
(29)
1
=
∥
y
m
-
y
l
λ
m
l
∥
Φ
,
p
=
1
k
m
l
(
1
+
I
Φ
p
(
k
m
l
(
y
m
-
y
l
λ
m
l
)
)
)
1
/
p
=
1
k
m
l
(
1
+
(
I
Φ
(
k
m
l
(
y
m
λ
m
l
)
)
+
I
Φ
(
k
m
l
(
y
l
λ
m
l
)
)
)
p
)
1
/
p
.
Then
(30)
(
k
m
l
p
-
1
)
1
/
p
=
I
Φ
(
k
m
l
(
y
m
λ
m
l
)
)
+
I
Φ
(
k
m
l
(
y
l
λ
m
l
)
)
.
Now we obtain that λ
m
l
≤
max
⁡
⁡
{
d
y
m
,
k
m
l
,
d
y
l
,
k
m
l
}. If not, λ
m
l
>
max
⁡
⁡
{
d
y
m
,
k
m
l
,
d
y
l
,
k
m
l
}, we have
(31)
I
Φ
p
(
k
m
l
(
y
m
λ
m
l
)
)
<
k
m
l
p
-
1
2
p
,
I
Φ
p
(
k
m
l
(
y
l
λ
m
l
)
)
<
k
m
l
p
-
1
2
p
,
whence
(32)
(
I
Φ
(
k
m
l
(
y
m
λ
m
l
)
)
+
I
Φ
(
k
m
l
(
y
l
λ
m
l
)
)
)
p
<
(
(
k
m
l
p
-
1
2
p
)
1
/
p
+
(
k
m
l
p
-
1
2
p
)
1
/
p
)
p
=
k
m
l
p
-
1
.
This is a contradiction. Hence,
(33)
∥
y
m
-
y
l
∥
Φ
,
p
=
λ
m
l
≤
max
⁡
⁡
{
d
y
m
,
k
m
l
,
d
y
l
,
k
m
l
}
≤
d
.
So K
(
l
Φ
,
p
)
≤
(
1
+
ε
)
d
+
4
ε; we get K
(
l
Φ
,
p
)
≤
d due to the arbitrariness of ε.Theorem 8.
IfΦ
∉
Δ
2
(
0
), 1
≤
p
<
∞, then K
(
l
Φ
,
p
)
=
2.Proof.
Denote(34)
l
α
=
{
x
∈
l
Φ
,
p
:
lim
⁡
n
→
∞
∥
(
0
,
…
,
0
,
x
(
n
+
1
)
,
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
x
(
n
+
2
)
,
…
)
∥
Φ
,
p
=
0
lim
⁡
n
→
∞
∥
(
0
,
…
,
0
,
x
(
n
+
1
)
,
}
.
SinceΦ
∉
Δ
2
(
0
), then l
α
≠
l
Φ
,
p; so for ε
>
0, according to Riesz lemma, there exists x
ε
∈
S
(
l
Φ
,
p
) satisfying dist
⁡
(
x
ε
,
l
α
)
>
1
-
ε. Then we have
(35)
∥
(
0
,
…
,
0
,
x
ε
(
n
+
1
)
,
x
ε
(
n
+
2
)
,
…
)
∥
Φ
,
p
>
1
-
ε
.
Since
(36)
lim
⁡
m
→
∞
∥
(
0
,
…
,
0
,
x
ε
(
n
+
1
)
,
…
,
x
ε
(
m
)
,
0
,
…
)
∥
Φ
,
p
>
1
-
ε
,
there exists a subsequence {
n
i
}
⊂
N such that n
1
<
n
2
<
⋯
<
n
k
<
⋯ and
(37)
∥
(
0
,
…
,
0
,
x
ε
(
n
i
+
1
)
,
…
,
x
ε
(
n
i
+
1
)
,
0
,
…
)
∥
Φ
,
p
>
1
-
ε
.
Let
(38)
x
1
=
(
-
x
ε
(
1
)
,
…
,
-
x
ε
(
n
1
)
,
x
ε
(
n
1
+
1
)
,
…
,
x
ε
(
n
2
)
,
x
ε
(
n
2
+
1
)
,
…
)
,
x
2
=
(
x
ε
(
1
)
,
…
,
x
ε
(
n
1
)
,
-
x
ε
(
n
1
+
1
)
,
…
,
-
x
ε
(
n
2
)
,
x
ε
(
n
2
+
1
)
,
…
)
.
Then for any m
,
l
∈
N,
(39)
∥
x
m
-
x
l
∥
Φ
,
p
=
2
∥
(
…
,
0
,
x
ε
(
n
m
-
1
+
1
)
,
…
,
x
ε
(
n
m
)
,
0
,
…
,
vvvvvvv
0
,
x
ε
(
n
l
-
1
+
1
)
,
…
,
x
ε
(
n
l
)
,
0
,
…
)
∥
Φ
,
p
≥
2
∥
(
0
,
…
,
0
,
x
ε
(
n
m
-
1
+
1
)
,
…
,
x
ε
(
n
m
)
,
0
,
…
)
∥
Φ
,
p
≥
2
(
1
-
ε
)
.
Due to the arbitrariness of ε
>
0, we have K
(
l
Φ
,
p
)
=
2.Lemma 9.
IfΦ
∈
Δ
2
(
0
)
∩
∇
2
(
0
), 1
≤
p
<
∞, then
(40)
1
<
k
′
≤
k
′′
<
∞
.Proof.
(1) SinceΦ
∈
Δ
2
(
0
) and the norm convergence and the modular convergence are equivalent, there exists c
>
0 such that
(41)
inf
⁡
∥
x
∥
Φ
,
p
=
1
I
Φ
(
x
)
=
c
>
0
.
For anyx
∈
S
(
l
Φ
,
p
) and k
∈
K
p
(
x
), we have
(42)
1
=
∥
x
∥
Φ
,
p
=
1
k
(
1
+
I
Φ
p
(
k
x
)
)
1
/
p
,
so k
=
(
1
+
I
Φ
p
(
k
x
)
)
1
/
p
≥
1; then
(43)
k
′
=
inf
⁡
∥
x
∥
Φ
,
p
=
1
k
=
inf
⁡
∥
x
∥
Φ
,
p
=
1
(
1
+
I
Φ
p
(
k
x
)
)
1
/
p
≥
inf
⁡
∥
x
∥
Φ
,
p
=
1
(
1
+
I
Φ
p
(
x
)
)
1
/
p
≥
(
1
+
c
p
)
1
/
p
>
1
.
(2) IfΦ
∈
∇
2
(
0
), then there exists α
>
1 such that
(44)
u
p
+
(
u
)
≥
α
Φ
(
u
)
,
(
|
u
|
≤
q
+
(
Ψ
-
1
(
1
c
p
-
1
)
)
)
.
For any x
∈
S
(
l
Φ
,
p
) and k
∈
K
p
(
x
), we have 1
<
k
′
≤
k
≤
k
p
*
*
(
x
); then for any ε
∈
(
0
,
k
′
-
1
), we get
(45)
1
≥
I
Φ
p
-
1
(
(
k
-
ε
)
x
)
I
Ψ
(
p
+
(
|
(
k
-
ε
)
x
|
)
)
≥
I
Φ
p
-
1
(
x
)
I
Ψ
(
p
+
(
|
(
k
-
ε
)
x
|
)
)
≥
c
p
-
1
∑
i
=
1
∞
‍
Ψ
(
p
+
(
|
(
k
-
ε
)
x
(
i
)
|
)
)
≥
c
p
-
1
Ψ
(
p
+
(
|
(
k
-
ε
)
x
(
i
)
|
)
)
,
(
∀
i
=
1,2
,
…
)
;
whence |
(
k
-
ε
)
x
(
i
)
|
≤
q
+
(
Ψ
-
1
(
1
/
c
p
-
1
)
). Moreover, according to the Young inequality
(46)
1
≥
I
Φ
p
-
1
(
(
k
-
ε
)
x
)
I
Ψ
(
p
+
(
|
(
k
-
ε
)
x
|
)
)
≥
c
p
-
1
∑
i
=
1
∞
‍
Ψ
(
p
+
(
|
(
k
-
ε
)
x
(
i
)
|
)
)
≥
c
p
-
1
∑
i
=
1
∞
‍
{
|
(
k
-
ε
)
x
(
i
)
|
p
+
(
|
(
k
-
ε
)
x
(
i
)
|
)
v
v
v
v
v
v
v
v
v
-
Φ
(
(
k
-
ε
)
x
(
i
)
)

|
(
k
-
ε
)
x
(
i
)
|
p
+
(
|
(
k
-
ε
)
x
(
i
)
|
)
}
≥
c
p
-
1
(
α
-
1
)
I
Φ
(
(
k
-
ε
)
x
)
≥
c
p
-
1
(
α
-
1
)
(
k
-
ε
)
I
Φ
(
x
)
≥
c
p
(
α
-
1
)
(
k
-
ε
)
,
since ε
>
0 is arbitrary, we deduce that k
≤
1
/
(
α
-
1
)
c
p
<
∞.Ye et al. [21] have proved that Orlicz function space as well as Orlicz sequence space equipped with the Luxemburg norm is P-convex if and only if it is reflexive; that is, Φ satisfies the suitable Δ
2-condition and ∇
2-condition (i.e., the Δ
2-condition at zero in the sequence case). We will prove now an analogous result for l
Φ
,
p in terms of P
(
l
Φ
,
p
).Theorem 10.
IfΦ
∈
Δ
2
(
0
)
∩
∇
2
(
0
), 1
≤
p
<
∞, then P
(
l
Φ
,
p
)
<
1
/
2.Proof.
IfΦ
∈
Δ
2
(
0
), then due to Lemma 6, for any ε
>
0, there exists δ
>
0 such that
(47)
(
I
Φ
(
x
)
≤
1
)
∧
(
I
Φ
(
y
)
<
δ
k
′′
2
(
2
-
δ
)
)
⟹
I
Φ
p
(
x
+
y
)
≤
I
Φ
p
(
x
)
+
ε
(
(
k
′
)
p
-
1
)
,
where k
′′
=
sup
⁡
⁡
{
k
:
k
∈
K
p
(
x
)
,
∥
x
∥
ϕ
,
p
=
1
}. According to Lemma 9, k
′′
<
∞. Set inf
⁡
∥
x
∥
Φ
,
p
=
1
I
Φ
(
x
)
=
c
>
0. If K
(
l
Φ
,
p
)
=
2, then there exists x
∈
S
(
l
Φ
,
p
) such that d
x
>
2
-
δ, so d
x
,
k
>
2
-
δ for all k
∈
K
p
(
x
).
Since∈
∇
2
(
0
), we can find θ
>
1 such that
(48)
Φ
(
u
2
)
≤
1
2
θ
Φ
(
u
)
,
(
|
u
|
≤
q
+
(
Ψ
-
1
(
1
c
p
-
1
)
)
)
.
Let us notice that
(49)
I
Φ
(
δ
k
2
(
2
-
δ
)
x
)
≤
∥
δ
k
2
(
2
-
δ
)
x
∥
Φ
,
p
=
δ
k
2
(
2
-
δ
)
≤
δ
k
′′
2
(
2
-
δ
)
.
Thus,
(50)
k
p
-
1
2
p
=
I
Φ
p
(
k
x
d
x
,
k
)
<
I
Φ
p
(
k
x
2
-
δ
)
=
I
Φ
p
(
k
x
2
+
δ
k
x
2
(
2
-
δ
)
)
≤
I
Φ
p
(
k
x
2
)
+
ε
(
(
k
′
)
p
-
1
)
≤
1
(
2
θ
)
p
I
Φ
p
(
k
x
)
+
ε
(
k
p
-
1
)
=
1
(
2
θ
)
p
(
k
p
-
1
)
+
ε
(
k
p
-
1
)
=
(
1
(
2
θ
)
p
+
ε
)
(
k
p
-
1
)
;
we have 1
/
2
p
≤
1
/
(
2
θ
)
p
+
ε. Since ε is arbitrary, we obtain θ
<
1; this is a contradiction. Therefore, K
(
l
Φ
,
p
)
<
2 and P
(
l
Φ
,
p
)
<
1
/
2.Corollary 11.
IfX
=
l
p
1
(
1
<
p
1
<
∞
), then
(51)
K
(
l
p
1
)
=
2
1
/
p
1
,
P
(
l
p
1
)
=
1
1
+
2
1
-
(
1
/
p
1
)
.Proof.
For anyx
∈
l
p
1,
(52)
∥
x
∥
Φ
,
p
=
(
p
1
-
1
)
-
1
/
p
q
1
p
1
1
/
p
-
1
/
p
1
∥
x
∥
l
p
1
=
(
p
1
-
1
)
-
1
/
p
q
1
p
1
1
/
p
-
1
/
p
1
Φ
-
1
(
I
Φ
(
x
)
)
,
where Φ
(
u
)
=
|
u
|
p
1
/
p
1 and 1
/
p
+
1
/
q
=
1, 1
/
p
1
+
1
/
q
1
=
1.
In fact, sinceΦ
(
u
)
=
|
u
|
p
1
/
p
1, then Φ
(
∥
x
∥
l
p
1
)
=
I
Φ
(
x
) and Φ
-
1
(
u
)
=
(
p
1
u
)
1
/
p
1. Set
(53)
f
(
k
)
=
1
k
p
(
1
+
I
Φ
p
(
k
x
)
)
=
1
k
p
(
1
+
(
k
p
1
∥
x
∥
l
p
1
p
1
p
1
)
p
)
.
By f
′
(
k
)
=
0, we get k
0
=
(
p
1
-
1
)
-
1
/
p
p
1
p
1
1
/
p
1
1
/
∥
x
∥
l
p
1. Since f
′′
(
k
0
)
<
0, we have
(54)
∥
x
∥
Φ
,
p
=
inf
⁡
k
>
0
1
k
(
1
+
I
Φ
p
(
k
x
)
)
1
/
p
=
(
f
(
k
0
)
)
1
/
p
=
(
p
1
-
1
)
-
1
/
p
q
1
p
1
1
/
p
-
1
/
p
1
∥
x
∥
l
p
1
.
Setα
=
(
p
1
-
1
)
-
1
/
p
q
1
p
1
1
/
p
-
1
/
p
1. From the equation
(55)
k
p
-
1
2
p
=
I
Φ
p
(
k
x
d
x
,
k
)
=
Φ
p
(
1
α
∥
k
x
d
x
,
k
∥
Φ
,
p
)
=
Φ
p
(
k
α
d
x
,
k
)
,
we deduce that d
x
,
k
=
(
k
/
α
)
(
Φ
-
1
(
(
(
k
p
-
1
)
/
2
p
)
1
/
p
)
)
-
1. Therefore,
(56)
K
(
l
p
1
)
=
d
=
sup
⁡
∥
x
∥
Φ
,
p
=
1
inf
⁡
⁡
k
>
1
d
x
,
k
=
1
α
inf
⁡
⁡
k
>
1
{
k
(
Φ
-
1
(
(
k
p
-
1
2
p
)
1
/
p
)
)
-
1
}
=
(
p
1
-
1
)
1
/
p
q
1
p
1
-
1
/
p
2
1
/
p
1
inf
⁡
k
>
1
k
(
k
p
-
1
)
1
/
p
p
1
=
2
1
/
p
1
.
We have K
(
l
p
1
)
=
2
1
/
p
1. So P
(
l
p
1
)
=
1
/
(
1
+
2
)
1
-
(
1
/
p
1
) for 1
<
p
1
<
∞.



---

*Source: 626491-2014-06-10.xml*