Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2011, Article ID 875649, 15 pages
doi:10.1155/2011/875649
Research Article
Nonsquareness and Locally Uniform
Nonsquareness in Orlicz-Bochner Function
Spaces Endowed with Luxemburg Norm
Shaoqiang Shang,
1
Yunan Cui,
2
and Yongqiang Fu
1
1
Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
2
Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, China
Correspondence should be addressed to Shaoqiang Shang, [email protected]
Received 5 July 2010; Accepted 12 February 2011
Academic Editor: Nikolaos Papageorgiou
Copyright q 2011 Shaoqiang Shang et al. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
Criteria for nonsquareness and locally uniform nonsquareness of Orlicz-Bochner function spaces
equipped with Luxemburg norm are given. We also prove that, in Orlicz-Bochner function
spaces generated by locally uniform nonsquare Banach space, nonsquareness and locally uniform
nonsquareness are equivalent.
1. Introduction
A lot of nonsquareness concepts in Banach spaces are known see 1. Nonsquareness are
important notions in geometry of Banach space. One of reasons is that these properties are
∞.
Let T, Σ,μ be a nonatomic measurable space. p denotes right derivative of M.
Moreover, for a given Banach space X, ·, we denote by X
T
the set of all strongly μ-
measurable function from T to X, and for each u ∈ X
T
, we define the modular of u by
ρ
M
u
G
M
u
t
dt.
1.1
Put
L
M
inf
λ>0:ρ
M
u
λ
≤ 1
1.3
is a Banach space. We say that an Orlicz function M satisfies condition Δ
2
M ∈ Δ
2
if there
exist K>2andu
0
≥ 0 such that
M
2u
≤ KM
u
u ≥ u
0
. 1.5
2. Main Results
Theorem 2.1. L
M
is nonsquare if and only if
a M ∈ Δ
2
;
b X is nonsquare.
In order to prove the theorem, we give a lemma.
Lemma 2.2. If X is nonsquare, then for any x, y
/
0, we have
x
y
− min
x y
· y
1 −
x
y
·
y
<
x
<
x
y
. 2.3
Case 2. If x≥y, then
x y
≤
y
x
y
x
−
y
x
y
2.4
or
n1
and disjont subsets {E
n
}
∞
n1
of E such that
r
n
> 2nc, M
1
1
n
r
n
> 2
n
M
1
1
2n
r
n
, 2
∞
n1
ρ
M
lr
n
χ
E
n
≥
∞
nm
ρ
M
1
1
n
r
n
χ
E
n
2n
r
n
μE
n
∞
nm
2
n
· 2
−n
δ ∞,
ρ
M
v
∞
n1
M
r
n
2.7
4 Journal of Inequalities and Applications
This yields
v
1,ρ
M
u ± v
≤ ρ
M
u
∞
n1
M
r
n
c
μE
n
1 − δ δ 1.
2.8
ρ
M
u
T
M
αx
dt
T
M
α
dt 1,
ρ
M
v
T
x y
2
· χ
T
t
,
u
t
− v
t
2
α ·
x − y
2
· χ
T
t
.
2.11
Hence, we have
ρ
M
M
u − v
2
T
M
α ·
x − y
2
dt
T
M
α
dt 1.
1
2
u − v
1. 2.13
We will derive a contradiction for each of the following two cases.
Journal of Inequalities and Applications 5
Case 1. μ{t ∈ T : ut
/
0}∩{t ∈ T : vt
/
0}0. Let G {t ∈ T : ut
/
0}. Hence, we
have
1
2
ρ
M
u
v
t
dt
1
2
G
M
u
t
v
t
dt
1
2
T\G
dt
T\G
M
1
2
u
t
v
t
dt
T
M
1
2
u
v/2 < 1, a contradiction!
Case 2. μ{t ∈ T : ut
/
0}∩{t ∈ T : vt
/
0} > 0. By Lemma 2.2, without loss of
generality, we may assume that there exists T
1
⊂{t ∈ T : ut
/
0}∩{t ∈ T : vt
/
0} such
that ut vt > utvt, t ∈ T
1
and μT
1
> 0. Therefore,
1
2
ρ
M
u
1
2
ρ
M
dt
T
1
2
M
u
t
1
2
M
v
t
dt
≥
2
u
t
1
2
v
t
dt
>
T
1
M
1
2
u
t
u v
2
.
2.15
Since M ∈ Δ
2
, we have ρ
M
uρ
M
v1. Hence, ρ
M
u v/2 < 1. This implies
u v/2 < 1, a contradiction!
Theorem 2.3. L
M
is locally uniformly nonsquare if and only if
a M ∈ Δ
2
;
b X is locally uniformly nonsquare.
6 Journal of Inequalities and Applications
In order to prove the theorem, we give a lemma.
Lemma 2.4. If X is locally uniformly nonsquare, then
a For any x
/
0, r
1
≥ r
: x ∈ X, r
2
≤
y
≤ r
1
> 0
2.16
b If x
n
→ x,thenlim
n →∞
δx
n
δx,where
δ
x
inf
y
/
0
≤ r
1
.
2.17
Proof. a Since X is locally uniformly nonsquare, we have η
x
> 0andη
λx
λη
x
, where λ>0
and
η
x
inf
y
x
y
− min
x
y
− min
x y
,
x − y
:
x
,
x
x
−
y
y
:
λx y
,
λx − y
:
λx
y
> 0
λ · inf
y
x
1
λ
y
:
x
1
λ
y
> 0
λ · inf
y
> 0
λ · η
x
.
2.19
Journal of Inequalities and Applications 7
Case 1. If x≥y, then
x y
≤
1 −
y
x
x
x
y
y
x
x
y
− η
r
2
/xx
2.20
or
x − y
≤
x
y
− η
r
2
/xx
x
y
·
y
≤
x
x
− η
x
y
or
x − y
≤
x
y
− η
x
. 2.23
Therefore, we get, the following inequality
inf
y
x
y
b1 Suppose that lim sup
n →∞
δx
n
>δx, where x
n
→ x n →∞. Then there exist
a>0 and subsequence {n} of {n}, such that δx
n
− δx ≥ a. By definition of δx, there exist
y
0
∈ X such that
x
y
0
− min
x y
0
8 Journal of Inequalities and Applications
Case 1. x y
0
x − y
0
. Since x
n
→ x n →∞, there exists n
0
such that x
n
0
− x <a/8.
Therefore,
x
n
0
y
0
−
x
y
0
≤
x
x
n
0
− x
y
0
−
x y
0
0
− x
≤ δ
x
a
8
2
x
n
0
− x
<δ
x
a
8
2 ·
a
8
δ
x
x
n
0
− y
0
x − y
0
x
n
0
− x
≥
x − y
0
−
x
n
0
− x
≤
x y
0
x
n
0
− x
≤
x y
0
1
8
r.
8
r ≥
x y
0
1
8
r ≥
x
n
0
y
0
.
2.28
Similarly, we have
x
n
0
y
0
− min
x
n
0
y
0
,
x
n
0
− y
0
there exist n
0
∈ N such that x
n
0
− x
n
< 1/8b, whenever n ≥ n
0
. By definition of δx
n
0
,
there exist y
0
∈ X such that
x
n
0
y
0
− min
2
≤
y
0
≤ r
1
.
2.31
Therefore, we have
x
n
y
0
− min
x
n
− min
x
n
0
− x
n
0
x
n
y
0
,
x
n
0
− x
n
0
x
n
− y
,
x
n
0
− y
0
1
8
b
x
n
0
y
0
− min
1
8
b
1
4
b
<δ
x
n
0
3
8
b
2.32
whenever n ≥ n
0
. Since x
n
→ x n →∞, there exists n
1
>n
0
such that |ηx − ηx
n
1
. 2.33
Hence, we have
η
x
n
1
>η
x
−
1
8
b ≥ δ
x
−
1
8
b ≥ δ
x
n
0
x
n
1
y
0
,
x
n
1
− y
0
≥ δ
x
n
0
7
8
b,
t
α · x · χ
T
t
,v
n
t
α · y
n
· χ
T
t
. 2.36
Then we have
ρ
M
u
T
dt
T
M
α
dt 1.
2.37
It is easy to see u, v
n
∈ SL
M
. We know that
u
t
v
n
t
2
α ·
x y
n
2
n
/2 ≤ Mα. By the dominated
convergence theorem, we have
lim
n →∞
T
M
α ·
x y
n
2
dt
T
lim
n →∞
M
α ·
x − y
n
2
dt
T
lim
n →∞
M
α ·
x − y
n
2
dt
∞
n1
⊂ SL
M
such that 1/2u
v
n
→1, 1/2u − v
n
→1asn →∞. We will derive a contradiction for each of the
following two cases.
Case 1. There exist ε
0
> 0, σ
0
> 0 such that μG
n
>ε
0
, where G
n
{t ∈ T : v
n
t≥σ
0
}.Put
H
n
dt ≥
G
n
\H
n
M
v
n
t
dt ≥
4
ε
0
· μ
G
n
\ H
n
.
y
− min
u
t
y
,
u
t
− y
: σ
0
≤
t
inf
y
/
0
h
n
t
y
−min
h
n
t
2.43
is μ-measurable. By Lemma 2.4, we have η
n
t → ηt μ-a.e on T
0
. Then ηt is μ-measurable.
Using
T ⊃
∞
i1
t ∈ T
0
:
1
i 1
<η
t
≤
1
i
,
2.44
we get that there exists η
0
> 0 such that μH < 1/8ε
∩{t ∈ T : ut 0} \ H.It
is easy to see E
n
E
1
n
∪ E
2
n
, E
1
n
∩ E
2
n
φ and μE
n
≥ 3/8ε
0
.Ift ∈ E
1
n
,byLemma 2.4, we have
u
t
}
≥ η
t
≥ 2η
0
. 2.46
Without loss of generality, we may assume that there exists F
1
n
⊂ E
1
n
such that
μF
1
n
≥
1
2
μE
1
n
,
u
Moreover, for any u ≥ v>0, we have
1
2
M
u
− M
1
2
u
−
1
2
M
v
− M
1
2
v
1
2
t
dt
1
2
u
0
p
t
dt −
1
2
v
0
p
t
dt
−
p
t
dt
≥
1/2u1/2v
v
p
t
dt −
1/2u
1/2v
p
t
dt ≥ 0.
2.48
12 Journal of Inequalities and Applications
Hence, if t ∈ E
2
n
, then
1
2
− M
σ
0
2
> 0.
2.49
Let F
n
F
1
n
∪ E
2
n
. Then μF
n
≥ 1/8ε
0
. Therefore,
1
2
ρ
M
u
1
2
T
M
v
n
t
dt −
T
M
u
t
v
n
t
− M
u
t
v
n
t
2
dt
≥
F
1
n
1
2
M
u
2
dt
E
2
n
1
2
M
u
t
1
2
M
v
n
t
u
t
1
2
v
n
t
− M
u
t
v
n
t
2
− M
u
t
v
n
t
2
dt
≥
F
1
n
M
u
t
2
n
1
2
M
v
n
t
− M
v
n
t
2
dt
≥
F
F
n
min
M
η
0
,
1
2
M
σ
0
− M
σ
0
2
dt
min
M
η
σ
0
− M
σ
0
2
·
1
8
ε
0
.
2.50
By Lemma 1.1, we have ρ
M
uρ
M
v
n
1, ρ
M
u v
n
/2 → 1asn →∞.Thisisin
contradiction with 1/2ρ
u
t
/
0
}
⊃
∞
n1
t ∈ T :
1
n 1
<
u
t
≤
1
n
,
2.51
we get that there exist r>0 such that μT
}
. 2.52
Since M is N-function, we can choose 0 <h<dsuch that 1/2Md1/2Mh − Md
h/2 > 0. Since v
n
t → 0μ− a.e on T, by the Egorov theorem, there exists N
1
such that
v
n
t <h,t∈ F whenever n>N
1
, where F ⊂ T, μT \ F < 1/8μT
0
. Next, we will prove
that if u
1
≥ u
2
≥ v
2
≥ v
1
> 0, then
1
2
M
u
1
v
2
− M
u
2
v
2
2
.
2.53
In fact, we have
1
2
M
u
1
1
2
M
v
1
− M
2
1
2
M
v
1
− M
u
1
v
1
2
−
1
2
M
v
2
− M
dt −
1
2
v
2
0
p
t
dt −
u
1
v
2
/2
0
p
t
dt
1
2
p
t
dt −
u
1
v
1
/2
0
p
t
dt
−
1
2
v
2
v
1
p
t
t
dt −
v
2
v
1
v
2
/2
p
t
dt ≥ 0.
2.54
14 Journal of Inequalities and Applications
Moreover, we have
1
2
M
u
1
− M
u
0
p
t
dt −
1
2
u
2
0
p
t
dt
−
u
1
v
2
/2
0
p
t
1
v
2
/2
u
2
v
2
/2
p
t
dt
≥
u
1
u
2
/2
u
2
p
t
dt −
u
− M
u
1
v
1
2
≥
1
2
M
u
2
1
2
M
v
2
− M
u
2
v
− 2M
u
t
v
n
t
2
≥ M
d
M
h
− 2M
d h
2
1
2
T
M
u
t
dt
1
2
T
M
v
n
t
t
1
2
M
v
n
t
− M
u
t
v
n
t
− M
u
t
v
n
t
2
dt
≥
T
2
1
2
M
v
n
t
2
dt
≥
T
2
1
2
M
d
1
2
M
d
− M
d h
for n large enough. By Lemma 1.1, we have ρ
M
uρ
M
v
n
1, ρ
M
uv
n
/2 → 1asn →∞,
which contradicts 1/2ρ
M
u1/2ρ
M
v
n
− ρ
M
u v
n
/2 ≥ 1/2Md1/2Md −
Md h/2 · 1/4μT
0
,forn large enough. This completes the proof.
Journal of Inequalities and Applications 15
Corollary 2.5. The following statements are equivalent:
a L
M
is locally uniformly nonsquare if and only if L
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