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 Fu1
1
2
Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, China
Correspondence should be addressed to Shaoqiang Shang, sqshang@163.com
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
strongly related to the fixed point property see 2. The criteria for nonsquareness and
locally uniform nonsquareness in the classical Orlicz function spaces have been given in 3, 4
already. However, because of the complicated structure of Orlicz-Bochner function spaces
equipped with the Luxemburg norm, the criteria for nonsquareness and locally uniform
nonsquareness of them have not been found yet. The aim of this paper is to give criteria
for nonsquareness and locally uniform nonsquareness of Orlicz-Bochner function spaces
equipped with Luxemburg norm.
Let X, · be a real Banach space. SX and BX denote the unit sphere and unit
ball, respectively. Let us recall some geometrical notions concerning nonsquareness. A Banach
space X is said to be nonsquare if for any x, y ∈ SX we have min{xy/2, x−y/2} <
1. A Banach space X is said to be uniformly nonsquare if there exists δ > 0 such that for any
x, y ∈ SX, min{x y/2, x − y/2} < 1 − δ. A Banach space X is said to be locally
uniformly nonsquare if for any x ∈ SX, there exists δx > 0 such that min{x y/2, x −
y/2} < 1 − δx , where y ∈ SX.
2
Journal of Inequalities and Applications
Let R be set of real numbers. A function M : R → R is called an N-function if M is
convex, even, M0 0, Mu > 0 u /
0 and limu → 0 Mu/u 0, and limu → 0 Mu/u
∞.
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 XT the set of all strongly μmeasurable function from T to X, and for each u ∈ XT , we define the modular of u by
ρM u Mutdt.
1.1
G
Put
LM
ut ∈ XT :
Mλutdt < ∞ for some λ > 0 .
1.2
G
The linear set LM endowed with the Luxemburg norm
u
≤1
u inf λ > 0 : ρM
λ
1.3
is a Banach space. We say that an Orlicz function M satisfies condition Δ2 M ∈ Δ2 if there
exist K > 2 and u0 ≥ 0 such that
M2u ≤ KMu
u ≥ u0 .
1.4
First let us recall a known result that will be used in the further part of the paper.
Lemma 1.1 see 3. Suppose M ∈ Δ2 . Then
ρM un −→ 0 ⇐⇒ un −→ 0,
ρM un −→ 1 ⇐⇒ un −→ 1n −→ ∞.
1.5
2. Main Results
Theorem 2.1. LM is nonsquare if and only if
a M ∈ Δ2 ;
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, x − y > 0.
2.1
Journal of Inequalities and Applications
3
Proof.
Case 1. If x < y, then
x y ≤ x x
x
· y 1 − · y
y y
< x x y − x
x y
2.2
or
x − y < x y.
2.3
Case 2. If x ≥ y, then
y x
x y ≤ · x y − 1 · y
x
y
2.4
< y y x − y
x y
or
x − y < x y.
2.5
This implies x y − min{x y, x − y} > 0. This completes the proof.
Proof of Theorem 2.1. a Necessity. Suppose that M ∈
/ Δ2 , then there exist u ∈ SLM and δ > 0
such that ρM u 1 − δ < 1. Pick c > 0 such that E {t ∈ T : ut ≤ c} is not a null set. Since
∞
M∈
/ Δ2 , there exist sequence {rn }∞
n1 and disjont subsets {En }n1 of E such that
rn > 2nc,
M
1
1
1
1
rn > 2n M
rn ,
n
2n
Therefore, if we define v ρM lv ≥
∞
1
1
rn μEn 2−n δ.
2n
2.6
then for any l > 1, we have
∞
∞
1
1
ρM lrn χEn ≥
ρM
2n ρM
1
1
rn χEn >
rn χEn
n
2n
nm
nm
n1
∞
∞
2n M
1
nm
ρM v n1 rn χEn ,
2n M
∞
n1
Mrn μEn <
∞
1
rn μEn 2n · 2−n δ ∞,
2n
nm
∞
n1
Mrn cμEn <
∞
1
M
1
rn μEn δ.
2n
n1
2.7
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Journal of Inequalities and Applications
This yields
v 1,
ρM u ± v ≤ ρM u ∞
Mrn cμEn 1 − δ δ 1.
2.8
n1
Hence, u ± v ≤ 1. But u v u − v ≤ 2u 2, and we deduce that u v u − v 1.
Moreover, we have 1/2u v 1/2u − v 1 and 1/2u v − 1/2u − v} 1,
a contradiction with nonsquareness of LM .
If b is not true, then there exist
x, y ∈ SX such that x y 1/2x y 1/2x − y. Pick α > 0 such that T Mαdt 1. Put
ut α · x · χT t,
vt α · y · χT t.
2.9
Then we have
Mαxdt ρM u T
ρM v Mαdt 1,
T
M αy dt T
2.10
Mαdt 1.
T
It is easy to see u, v ∈ SLM . We know that
xy
ut vt
α·
· χT t,
2
2
x−y
ut − vt
α·
· χT t.
2
2
2.11
Hence, we have
x y
dt M α ·
Mαdt 1,
ρM
2
2 T
T
x − y
u−v
dt ρM
M α ·
Mαdt 1.
2
2 T
T
u v
2.12
It is easy to see 1/2u v, 1/2u − v ∈ SLM , a contradiction!
Sufficiency. Suppose that there exists u, v ∈ SLM such that
1
1
u v u v u − v
1.
2
2
We will derive a contradiction for each of the following two cases.
2.13
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
1
1
ρM u ρM v 2
2
2
1
2
Mutdt G
1
2
T \G
Mvtdt
1
Mut vtdt 2
G
T \G
Mut vtdt
1
1
>
M
M
ut vt dt ut vt dt
2
2
G
T \G
2.14
1
M
ut vt dt
2
T
ρM
1
u v .
2
Since M ∈ Δ2 , we have ρM u ρM v 1. Hence, ρM u v/2 < 1. This implies 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 T1 ⊂ {t ∈ T : ut / 0} ∩ {t ∈ T : vt / 0} such
that ut vt > ut vt, t ∈ T1 and μT1 > 0. Therefore,
1
1
1
ρM u ρM v 2
2
2
T
Mutdt T
1
1
1
1
M
ut vt dt
ut vt dt
2
2
2
2
T \T1
1
1
M
ut vt dt ut vt dt
2
2
T \T1
T1
M
T1
ρM
Mvtdt
T
M
>
1
1
Mut Mvtdt
2
2
≥
1
2
2.15
u v
2
.
Since M ∈ Δ2 , we have ρM u ρM v 1. Hence, ρM u v/2 < 1. This implies
u v/2 < 1, a contradiction!
Theorem 2.3. LM is locally uniformly nonsquare if and only if
a M ∈ Δ2 ;
b X is locally uniformly nonsquare.
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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, r1 ≥ r2 > 0, we have
inf x y − min x y, x − y : x ∈ X, r2 ≤ y ≤ r1 > 0
y/
0
2.16
b If xn → x, then limn → ∞ δxn δx, where
δx inf x y − min x y, x − y : x ∈ X, r2 ≤ y ≤ r1 .
y/
0
2.17
Proof. a Since X is locally uniformly nonsquare, we have ηx > 0 and ηλx ληx , where λ > 0
and
ηx inf x y − min x y, x − y : x y > 0 .
y
2.18
In fact, since X is locally uniformly nonsquare, we have
ηx inf x y − min x y, x − y : x y > 0
y
x
x
y y x · inf 2 − min , − : x y > 0 > 0,
y
x y x y ηλx inf λx y − min λx y, λx − y : λx y > 0
y
1 1 1 1 λ · inf x y − min x y, x − y : x y > 0
y
λ
λ
λ
λ
λ · inf x y − min x y, x − y : x y > 0
y
λ · ηx .
2.19
Journal of Inequalities and Applications
7
Case 1. If x ≥ y, then
x y ≤ 1−
≤ 1−
y y xy
x
x
x y y x y x
x
x 2.20
≤ x − y y y − ηy/xx
≤ x y − ηr2 /xx
or
x − y ≤ x y − ηr /xx .
2
2.21
x
x
x y ≤ · y x 1 − · y
y
y ≤ x x − ηx y − x
x y − ηx
≤ x y − ηx
2.22
x − y ≤ x y − ηx .
2.23
Case 2. If x < y, then
or
Therefore, we get, the following inequality
inf x y − min x y, x − y : x ∈ X ≥ min ηr2 /xx , ηx > 0
y
2.24
holds.
b1 Suppose that lim supn → ∞ δxn > δx, where xn → x n → ∞. Then there exist
a > 0 and subsequence {n} of {n}, such that δxn − δx ≥ a. By definition of δx, there exist
y0 ∈ X such that
a
x y0 − min x y0 , x − y0 < δx ,
8
r1 ≤ y0 ≤ r2 .
We will derive a contradiction for each of the following two cases.
2.25
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Journal of Inequalities and Applications
Case 1. x y0 x − y0 . Since xn → x n → ∞, there exists n0 such that xn0 − x < a/8.
Therefore,
xn0 y0 − xn0 y0 ≤ x xn0 − x y0 − xn0 y0 ≤ x xn0 − x y0 − x y0 − xn0 − x
x y0 − x y0 2xn0 − x
a
2xn0 − x
8
a
a
< δx 2 ·
8
8
≤ δx 2.26
3
δx a.
8
This implies δxn0 ≤ δx 3/8a, a contradiction!
x y0 . Without loss of generality, we may assume x − y0 > x y0 r,
Case 2. x − y0 /
where r > 0. Since xn → x n → ∞, there exists n0 such that xn0 −x < min{1/8a, 1/8r}.
Therefore, we have
xn − y0 x − y0 xn − x
0
0
≥ x − y0 − xn0 − x
1
≥ x − y0 − r,
8
xn y0 x y0 xn − x
0
0
≤ x y0 xn0 − x
2.27
1
≤ x y0 r.
8
This implies
xn − y0 ≥ x − y0 − 1 r ≥ x y0 r − 1 r ≥ x y0 1 r ≥ xn y0 .
0
0
8
8
8
2.28
Similarly, we have
3
xn0 y0 − xn0 y0 ≤ δx a.
8
2.29
3
xn0 y0 − min xn0 y0 , xn0 − y0 ≤ δx a.
8
2.30
Therefore, we have
This implies δxn0 ≤ δx 3/8a, a contradiction! Hence, lim supn → ∞ δxn ≤ δx.
Journal of Inequalities and Applications
9
b2 Suppose that lim infn → ∞ δxn < δx, where xn → x n → ∞. Then there exist
b > 0 and subsequence {n} of {n}, such that δx − δxn ≥ b. Since xn → x n → ∞, then
there exist n0 ∈ N such that xn0 − xn < 1/8b, whenever n ≥ n0 . By definition of δxn0 ,
there exist y0 ∈ X such that
b
xn0 y0 − min xn0 y0 , xn0 − y0 < δxn0 ,
8
r2 ≤ y0 ≤ r1 .
2.31
Therefore, we have
xn y0 − min xn y0 , xn − y0 xn0 − xn0 xn y0 − min xn0 − xn0 xn y0 , xn0 − xn0 xn − y0 1
1
≤ xn0 b y0 − min xn0 y0 , xn0 − y0 b
8
8
1
xn0 y0 − min xn0 y0 , xn0 − y0 b
4
2.32
1
1
< δxn0 b b
8
4
3
< δxn0 b
8
whenever n ≥ n0 . Since xn → x n → ∞, there exists n1 > n0 such that |ηx − ηxn1 | <
1/8b, where
ηx x y0 − min x y0 , x − y0 .
2.33
1
1
1
7
ηxn1 > ηx − b ≥ δx − b ≥ δxn0 b − b δxn0 b.
8
8
8
8
2.34
7
xn1 y0 − min xn1 y0 , xn1 − y0 ≥ δxn0 b,
8
2.35
Hence, we have
This implies
which contradict 2.32. Hence, lim infn → ∞ δxn ≥ δx.
Combing b1 with b2, we get limn → ∞ δxn δx. This completes the proof.
10
Journal of Inequalities and Applications
Proof of Theorem 2.3. Necessity. By Theorem 2.1, M ∈ Δ2 . If b is not true, then there exist
such that 1/2x yn → 1 and 1/2x − yn → 1 as n → ∞.
x ∈ SX, {yn }∞
n1 ⊂ SX
Pick α > 0 such that T Mαdt 1. Put
ut α · x · χT t, vn t α · yn · χT t.
2.36
Then we have
Mαxdt ρM u T
M αyn dt ρM vn Mαdt 1,
T
T
2.37
Mαdt 1.
T
It is easy to see u, vn ∈ SLM . We know that
x yn
ut vn t
α·
· χT t,
2
2
x − yn
ut − vn t
α·
· χT t.
2
2
2.38
Moreover, we have Mα·xyn /2 ≤ Mα, Mα·x−yn /2 ≤ Mα. By the dominated
convergence theorem, we have
x yn x yn dt dt M α·
lim M α · Mαdt 1,
2 2 T
T n→∞
T
x − yn x − yn dt dt M α·
lim M α · Mαdt 1.
2 2 T
T n→∞
T
lim
n→∞
lim
n→∞
2.39
It is easy to see ρM 1/2u vn → 1, ρM 1/2u − vn → 1 as n → ∞. By Lemma 1.1, we
have 1/2u vn → 1 and 1/2u − vn → 1 as n → ∞, a contradiction with locally
uniform nonsquareness of LM .
Sufficiency. Suppose that there exist u ∈ SLM , {vn }∞
n1 ⊂ SLM such that 1/2u vn → 1, 1/2u − vn → 1 as n → ∞. We will derive a contradiction for each of the
following two cases.
Case 1. There exist ε0 > 0, σ0 > 0 such that μGn > ε0 , where Gn {t ∈ T : vn t ≥ σ0 }. Put
Hn t ∈ T : σ0 ≤ vn t ≤ M
−1
4
ε0
.
2.40
We have
1
Mvn tdt ≥
T
Gn \Hn
Mvn tdt ≥
4
· μGn \ Hn .
ε0
2.41
Journal of Inequalities and Applications
11
This implies μGn \ Hn ≤ 1/4ε0 . Hence, μHn > 1/2ε0 . We define a function
ηt inf ut y − min ut y, ut − y : σ0 ≤ y ≤ M−1
y
/0
4
ε0
2.42
on T0 , where T0 {t ∈ T : ut /
0}. By Lemma 2.4, we have ηt > 0 μ-a.e on T0 . Let
hn t → ut μ-a.e on T0 , where hn is simple function. Hence,
4
ηn t inf hn t y −min hn ty, hn t−y : σ0 ≤ y ≤ M−1
y/
0
ε0
2.43
is μ-measurable. By Lemma 2.4, we have ηn t → ηt μ-a.e on T0 . Then ηt is μ-measurable.
Using
∞ t ∈ T0 :
T⊃
i1
1
1
< ηt ≤
,
i1
i
2.44
we get that there exists η0 > 0 such that μH < 1/8ε0 , where
H t ∈ T0 : ηt ≤ 2η0 .
2.45
Let En Hn \ H, En1 Hn ∩ {t ∈ T : ut / 0} \ H, En2 Hn ∩ {t ∈ T : ut 0} \ H. It
is easy to see En En1 ∪ En2 , En1 ∩ En2 φ and μEn ≥ 3/8ε0 . If t ∈ En1 , by Lemma 2.4, we have
ut vn t − min{ut vn t, ut − vn t} ≥ ηt ≥ 2η0 .
2.46
Without loss of generality, we may assume that there exists Fn1 ⊂ En1 such that
1 1
μE ,
2 n
μFn1 ≥
ut vn t − ut vn t ≥ 2η0 t ∈ Fn1 .
2.47
Moreover, for any u ≥ v > 0, we have
1
1
1
1
Mu − M
u − Mv − M
v
2
2
2
2
1/2u
1/2v
1 u
1 v
ptdt −
ptdt −
ptdt ptdt
2 0
2 0
0
0
1/2v
u
1/2u
1
1 v
ptdt −
ptdt −
ptdt −
ptdt
2 0
2 0
0
0
≥
1
2
u
ptdt −
v
1/2u1/2v
v
1/2u
1/2v
ptdt
ptdt −
1/2u
1/2v
ptdt ≥ 0.
2.48
12
Journal of Inequalities and Applications
Hence, if t ∈ En2 , then
σ 1
1
1
0
Mvn t Mvn t ≥ Mσ0 − M
> 0.
2
2
2
2
2.49
Let Fn Fn1 ∪ En2 . Then μFn ≥ 1/8ε0 . Therefore,
u v 1
1
n
ρM u ρM vn − ρM
2
2
2
1
1
ut vn t
dt
Mutdt Mvn tdt − M
2 T
2 T
2
T
1
1
ut vn t
Mut Mvn t − M
dt
2
2
T 2
1
1
ut vn t
Mut Mvn t − M
dt
≥
2
2
Fn1 2
1
1
ut vn t
Mut Mvn t − M
dt
2
2
En2 2
1
1
ut vn t
dt
≥
M
ut vn t − M
2
2
2
Fn1
1
1
ut vn t
Mut Mvn t − M
dt
2
2
En2 2
ut vn t
ut vn t
η0 − M
dt
≥
M
2
2
Fn1
1
vn t
Mvn t − M
dt
2
En2 2
σ 1
0
Mσ0 − M
dt
M η0 dt ≥
2
Fn1
En2 2
σ 1
0
≥
dt
min M η0 , Mσ0 − M
2
2
Fn
σ 1
0
min M η0 , Mσ0 − M
· μFn
2
2
σ 1
1
0
· ε0 .
≥ min M η0 , Mσ0 − M
2
2
8
2.50
By Lemma 1.1, we have ρM u ρM vn 1, ρM u vn /2 → 1 as n → ∞. This is in
contradiction with 1/2ρM u 1/2ρM vn − ρM u vn /2 ≥ min{Mη0 , 1/2Mσ0 −
Mσ0 /2} · 1/8ε0 .
Journal of Inequalities and Applications
13
Case 2. For any ε > 0, σ > 0, there exists N such that μ{t ∈ T : vn t ≥ σ} < ε whenever
n > N. By the Riesz theorem, without loss of generality, we may assume that vn t → 0 μ−
a.e on T . Using
∞ 0} ⊃
{t ∈ T : ut /
n1
1
1
< ut ≤
,
t∈T :
n1
n
2.51
we get that there exist r > 0 such that μT1 < 1/8μT0 , where
T1 {t ∈ T : 0 < ut < d},
T0 {t ∈ T : ut / 0}.
2.52
Since M is N-function, we can choose 0 < h < d such that 1/2Md 1/2Mh − Md h/2 > 0. Since vn t → 0μ− a.e on T , by the Egorov theorem, there exists N1 such that
vn t < h, t ∈ F whenever n > N1 , where F ⊂ T , μT \ F < 1/8μT0 . Next, we will prove
that if u1 ≥ u2 ≥ v2 ≥ v1 > 0, then
u v 1
u v 1
1
1
1
1
2
2
Mu1 Mv1 − M
≥ Mu2 Mv2 − M
.
2
2
2
2
2
2
2.53
In fact, we have
u v 1
u v 1
1
1
1
1
1
2
Mu1 Mv1 − M
− Mu1 Mv2 − M
2
2
2
2
2
2
u v 1
u v 1
1
1
1
2
Mv1 − M
− Mv2 − M
2
2
2
2
1
2
v1
ptdt −
u1 v1 /2
0
0
v
u1 v2 /2
1 2
ptdt −
ptdt −
ptdt
2 0
0
u1 v1 /2
u1 v2 /2
v1
1 v2
1
ptdt −
ptdt ptdt −
ptdt
2 0
2 0
0
0
1
−
2
≥
v2
ptdt v1
u1 v2 /2
u1 v1 /2
u1 v2 /2
u1 v1 /2
ptdt −
ptdt
v2
v1 v2 /2
ptdt ≥ 0.
2.54
14
Journal of Inequalities and Applications
Moreover, we have
u v 1
u v 1
1
2
2
2
Mu1 − M
− Mu2 − M
2
2
2
2
u2 v2 /2
u1
u1 v2 /2
1
1 u2
ptdt −
ptdt −
ptdt −
ptdt
2 0
2 0
0
0
1
2
≥
u1
ptdt −
u1 v2 /2
u2
u1 u2 /2
u2 v2 /2
ptdt −
ptdt
u1 v2 /2
u2
2.55
u2 v2 /2
ptdt ≥ 0.
By 2.54 and 2.55, we have
u v 1
u v 1
1
1
1
1
2
2
Mu1 Mv1 − M
≥ Mu2 Mv2 − M
.
2
2
2
2
2
2
2.56
This shows that if t ∈ T2 T0 \ T1 ∪ T \ F, then
Mut Mvn t − 2M
ut vn t
2
≥ Md Mh − 2M
dh
.
2
2.57
It is easy to see μT2 ≥ 1/4μT0 . Therefore,
u v 1
1
n
ρM u ρM vn − ρM
2
2
2
1
1
ut vn t
dt
Mutdt Mvn tdt − M
2 T
2 T
2
T
1
1
ut vn t
Mut Mvn t − M
dt
2
2
T 2
1
1
ut vn t
Mut Mvn t − M
≥
dt
2
2
T 2
1
1
ut vn t
Mut Mvn t − M
dt
≥
2
2
T2 2
1
dh
1
Md Md − M
dt
≥
2
2
T2 2
1
dh
1
1
Md Md − M
≥
· μT0 ,
2
2
2
4
2.58
for n large enough. By Lemma 1.1, we have ρM u ρM vn 1, ρM uvn /2 → 1 as n → ∞,
which contradicts 1/2ρM u 1/2ρM vn − ρM u vn /2 ≥ 1/2Md 1/2Md −
Md h/2 · 1/4μT0 , for n large enough. This completes the proof.
Journal of Inequalities and Applications
15
Corollary 2.5. The following statements are equivalent:
a LM is locally uniformly nonsquare if and only if LM is nonsquare;
b X is locally uniformly nonsquare.
Acknowledgments
The authors would like to thank the anonymous referee for some suggestions to improve
the manuscript. This work was supported by China Natural Science Fund under Grant no.
11061022.
References
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2, pp. 494–514, 2006.
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Harbin, China, 1986.