Volume 2011, Article ID 875649, 15 pagesdoi:10.1155/2011/875649 Research Article Nonsquareness and Locally Uniform Nonsquareness in Orlicz-Bochner Function Spaces Endowed with Luxemburg
Trang 1Volume 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
1 Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
2 Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, China
Received 5 July 2010; Accepted 12 February 2011
Academic Editor: Nikolaos Papageorgiou
Copyrightq 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 knownsee 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 in3,4 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
LetX, · 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 δ x > 0 such that min{x y/2, x − y/2} < 1 − δ x , where y ∈ SX.
Trang 2Let 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 lim u → 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 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
Put
L M
u t ∈ X T:
G
M λutdt < ∞ for some λ > 0
The linear set L Mendowed with the Luxemburg norm
u infλ > 0 : ρ M u
λ
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
M 2u ≤ KMu u ≥ u0. 1.4 First let us recall a known result that will be used in the further part of the paper
ρ M u n −→ 0 ⇐⇒ u n −→ 0, ρ M u n −→ 1 ⇐⇒ u n −→ 1n −→ ∞. 1.5
2 Main Results
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
Trang 3Case 1 If x < y, then
x y ≤ x x
y · y
1− x y · y
< x x y − x
x y
2.2
or
x − y < x y . 2.3
Case 2 If x ≥ y, then
x y ≤ y
x · x y
x
y − 1 · y
< y y x − y
x y
2.4
or
x − y < x y . 2.5 This impliesx 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 ∈ SL M and δ > 0 such that ρ M 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{r n}∞n1and 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
, 2n M
1 1
2n
r n
μE n 2−n δ.
2.6
Therefore, if we define v ∞
n1 r n χ E n , then for any l > 1, we have
ρ M lv ∞
n1
ρ M
lr n χ E n
≥ ∞
nm
ρ M
11
n
r n χ E n
>
∞
nm
2n ρ M
1 1
2n
r n χ E n
≥ ∞
nm
2n M
1 1
2n
r n
μE n ∞
nm
2n· 2−n δ ∞,
ρ M v ∞
n1
M r n μE n <
∞
n1
M r n cμE n <
∞
n1
M
1 1
2n
r n
μE n δ.
2.7
Trang 4This yields
v 1, ρ M u ± v ≤ ρ M u ∞
n1
M r n cμE n 1 − δ δ 1. 2.8
Hence,u ± v ≤ 1 But u v u − v ≤ 2u 2, and we deduce that u v u − v 1.
Moreover, we have1/2u v 1/2u − v 1 and 1/2u v − 1/2u − v} 1,
a contradiction with nonsquareness of L M
Ifb is not true, then there exist x, y ∈ SX such that x y 1/2x y
1/2x − y Pick α > 0 such thatT Mαdt 1 Put
u t α · x · χ T t, v t α · y · χ T t. 2.9 Then we have
ρ M u
T
M αxdt
T
M αdt 1,
ρ M v
T
M αy dt
T
M αdt 1.
2.10
It is easy to see u, v ∈ SL M We know that
u t vt
2 α · x y
2 · χ T t, u t − vt
2 α · x − y
2 · χ T t. 2.11 Hence, we have
ρ M u v
2
T
M
α · x y2 dt
T
M αdt 1,
ρ M
u − v
2
T
M
α · x − y2 dt
T
M αdt 1.
2.12
It is easy to see1/2u v, 1/2u − v ∈ SL M, a contradiction!
Sufficiency Suppose that there exists u, v ∈ SL M such that
u v
12u v
12u − v
1. 2.13
We will derive a contradiction for each of the following two cases
Trang 5Case 1 μ{t ∈ T : ut / 0} ∩ {t ∈ T : vt / 0} 0 Let G {t ∈ T : ut / 0} Hence, we
have
1
2ρ M u 1
2ρ M v 1
2
G
M utdt 1
2
T\G
M vtdt
1 2
G
M ut vtdt 1
2
T\G
M ut vtdt
>
G
M
1
2ut vt
dt
T\G
M
1
2ut vt
dt
T
M
1
2ut vt
dt
ρ M
1
2u v
.
2.14
Since M ∈ Δ2, we have ρ M u ρ M v 1 Hence, ρ M u v/2 < 1 This implies u
v/2 < 1, a contradiction!
Case 2 μ{t ∈ T : ut / 0} ∩ {t ∈ T : vt / 0} > 0 ByLemma 2.2, without loss of
generality, we may assume that there exists T1⊂ {t ∈ T : ut / 0} ∩ {t ∈ T : vt / 0} such
thatut vt > ut vt, t ∈ T1and μT1> 0 Therefore,
1
2ρ M u1
2ρ M v 1
2
T
M utdt 1
2
T
M vtdt
T
1
2M ut 1
2M vtdt
≥
T1
M
1
2ut 1
2vt
dt
T\T1
M
1
2ut1
2vt
dt
>
T1
M
1
2ut vt
dt
T\T1
M
1
2ut vt
dt
ρ M u v
2
.
2.15
Since M ∈ Δ2, we have ρ M u ρ M v 1 Hence, ρ M u v/2 < 1 This implies
u v/2 < 1, a contradiction!
a M ∈ Δ2;
b X is locally uniformly nonsquare.
Trang 6In 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
y / 0 x y 2≤ y ≤ r1
> 0 2.16
b If x n → x, then lim n → ∞ δx n δx, where
δ x inf
y / 0 x y 2≤ y ≤ r1
. 2.17
Proof a Since X is locally uniformly nonsquare, we have η x > 0 and η λx λη x , where λ > 0
and
η x inf
In fact, since X is locally uniformly nonsquare, we have
η x inf
y x y
x · inf y
2− min
x x
y
y ,
x x −
y
y
:x y > 0> 0,
η λx inf
y λx y
λ · inf
y
x
λ1y − min x 1λ y
, x −1λ y
:x
1λ y
> 0
λ · inf
y x y
λ · η x
2.19
Trang 7Case 1 If x ≥ y, then
x y ≤ 1− y
x x y
y
x x
≤
1− y
x x
y
y
x x
≤ x − y y y − η y/xx
≤ x y − η r2/xx
2.20
or
x − y ≤ x y − η r2/xx 2.21
Case 2 If x < y, then
x y ≤ x
y · y x
1− x y · y
≤ x x − η x y − x
x y − η x
≤ x y − η x
2.22
or
x − y ≤ x y − η x 2.23 Therefore, we get, the following inequality
inf
2/xx , η x
> 0 2.24
holds
b1 Suppose that lim supn → ∞ 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
y0∈ X such that
x y0 0 , x − y0 < δ x a
8, r1≤ y0 ≤ r2. 2.25
We will derive a contradiction for each of the following two cases
Trang 8Case 1 x y0 x − y0 Since x n → x n → ∞, there exists n0such thatx n0− x < a/8.
Therefore,
x n0 y0 − x n0 y0 ≤ x x n0− x y0 − x n0 y0
≤ x x n0− x y0 − x y0 − x n0− x
x y0 − x y0 2x n0− x
≤ δx a8 2x n0− x
< δ x a
8 2 ·a 8
δx 3
8a.
2.26
This implies δx n0 ≤ δx 3/8a, a contradiction!
Case 2 x − y0 / x y0 Without loss of generality, we may assume x − y0 > x y0 r, where r > 0 Since x n → x n → ∞, there exists n0such thatx n0−x < min{1/8a, 1/8r}.
Therefore, we have
x n
0− y0 x − y0 x n0− x
≥ x − y0 − x n0− x
≥ x − y0 − 1
8r,
x n
0 y0 x y0 x n0− x
≤ x y0 x n0− x
≤ x y0 18r.
2.27
This implies
x n0− y0 ≥ x − y0 − 18r ≥ x y0 r − 18r ≥ x y0 18r ≥ x n0 y0 . 2.28 Similarly, we have
x n0 y0 − x n0 y0 ≤ δx 38a. 2.29 Therefore, we have
x n0 y0 n
0 y0 , x n
0− y0 ≤ δx 3
8a. 2.30
This implies δx n ≤ δx 3/8a, a contradiction! Hence, lim sup n → ∞ δx n ≤ δx.
Trang 9b2 Suppose that lim infn → ∞ δx n < δx, where x n → x n → ∞ Then there exist
b > 0 and subsequence {n} of {n}, such that δx − δx n ≥ b Since x n → x n → ∞, then there exist n0 ∈ N such that x n0− x n < 1/8b, whenever n ≥ n0 By definition of δx n0,
there exist y0∈ X such that
x n0 y0 n
0 y0 , x n
0− y0 < δ x n
0 b
8, r2≤ y0 ≤ r1. 2.31
Therefore, we have
x n y0 n y0 , x n − y0
x n0− x n0 x n y0 n
0− x n0 x n y0 , x n
0− x n0 x n − y0
≤ x n0 1
8b y0 n
0 y0 , x n
0− y0 1
8b
x n0 y0 n
0 y0 , x n
0− y0 1
4b
< δ x n0 1
8b 1
4b
< δ x n0 3
8b
2.32
whenever n ≥ n0 Since x n → x n → ∞, there exists n1 > n0 such that|ηx − ηx n1| <
1/8b, where
η x x y0 0 , x − y0 . 2.33
Hence, we have
η x n1 > ηx −1
8b ≥ δ x −1
8b ≥ δ x n0 b −1
8b δ x n0 7
8b. 2.34
This implies
x n1 y0 n
1 y0 , x n
1− y0 ≥ δx n0 7
8b, 2.35
which contradict2.32 Hence, lim infn → ∞ δx n ≥ δx.
Combingb1 with b2, we get limn → ∞ δx n δx This completes the proof.
Trang 10Proof of Theorem 2.3 Necessity By Theorem 2.1, M ∈ Δ2 If b is not true, then there exist
x ∈ SX, {y n}∞
n1 ⊂ SX such that 1/2x y n → 1 and 1/2x − y n → 1 as n → ∞ Pick α > 0 such that
T Mαdt 1 Put
u t α · x · χ T t, v n t α · y n · χ T t. 2.36 Then we have
ρ M u
T
M αxdt
T
M αdt 1,
ρ M v n
T
M αy n dt
T
M αdt 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 · χ T t, u t − v n t
2 α · x − y n
2 · χ T t. 2.38
Moreover, we have Mα·xy n /2 ≤ Mα, Mα·x−y 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
M αdt 1,
lim
n → ∞
T
M
α ·
x − y n
2
dt
T
lim
n → ∞ M
α ·
x − y n
2
dt
T
M αdt 1.
2.39
It is easy to see ρ M 1/2uv n → 1, ρ M 1/2u−v n → 1 as n → ∞ ByLemma 1.1, we have1/2u v n → 1 and 1/2u − v n → 1 as n → ∞, a contradiction with locally uniform nonsquareness of L M
Sufficiency Suppose that there exist u ∈ SL M , {v n}∞n1 ⊂ SL M such that 1/2u
v n → 1, 1/2u − v n → 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 μG n > ε0, where G n {t ∈ T : v n t ≥ σ0} Put
H n
t ∈ T : σ0≤ v n t ≤ M−1
4
ε0
We have
1
T
M v n tdt ≥
G \H M v n tdt ≥ 4
ε0 · μG n \ H n . 2.41
Trang 11This implies μG n \ H n ≤ 1/4ε0 Hence, μH n > 1/2ε0 We define a function
η t inf
y / 0
ut y 0≤ y ≤ M−1
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
h n t → ut μ-a.e on T0, where h nis simple function Hence,
η n t inf
y / 0
h n t y n ty , h n t−y : σ0≤ y ≤M−1
4
ε0
2.43
is μ-measurable ByLemma 2.4, we have η n t → ηt μ-a.e on T0 Then ηt is μ-measurable.
Using
T ⊃
∞
i1
t ∈ T0: 1
i 1 < η t ≤ 1
i
we get that there exists η0> 0 such that μH < 1/8ε0, where
H t ∈ T0: ηt ≤ 2η0
Let E n H n \ H, E1
n H n ∩ {t ∈ T : ut / 0} \ H, E2
n H n ∩ {t ∈ T : ut 0} \ H It
is easy to see E n E1
n ∪ E2
n , E1
n ∩ E2
n φ and μE n ≥ 3/8ε0 If t ∈ E1
n, byLemma 2.4, we have
ut v n t − min{ut v n t, ut − v n t} ≥ ηt ≥ 2η0. 2.46
Without loss of generality, we may assume that there exists F1
n ⊂ E1
nsuch that
μF n1≥ 1
2μE1n , ut v n t − ut v n t ≥ 2η0t ∈ F n1. 2.47
Moreover, for any u ≥ v > 0, we have
1
2M u − M
1
2u
−
1
2M v − M
1
2v
1 2
u
0
p tdt −
1/2u
0
p tdt − 1
2
v
0
p tdt
1/2v
0
p tdt
1 2
u
0
p tdt −1
2
v
0
p tdt
−
1/2u
0
p tdt −
1/2v
0
p tdt
1 2
u
v
p tdt −
1/2u
1/2v p tdt
≥
1/2u1/2v
v
p tdt −
1/2u
1/2v p tdt ≥ 0.
2.48
Trang 12Hence, if t ∈ E2
n, then
1
2M v n t 1
2M v n t ≥ 1
2M σ0 − M σ0
2
Let F n F1
n ∪ E2
n Then μF n ≥ 1/8ε0 Therefore,
1
2ρ M u 1
2ρ M v n − ρ M u v n
2
1
2
T
M utdt 1
2
T
M v n tdt −
T
M
ut v
n t
2
dt
T
1
2M ut 1
2M v n t − M
ut v
n t
2
dt
≥
F1
1
2M ut 1
2M v n t − M
ut v
n t
2
dt
E2
1
2M ut 1
2M v n t − M
ut v
n t
2
dt
≥
F1
M
1
2ut 1
2v n t
− M
ut v
n t
2
dt
E2
1
2M ut 1
2M v n t − M
ut v
n t
2
dt
≥
F1
M
ut v
n t
2 η0
− M
ut v
n t
2
dt
E2
1
2M v n t − M
v
n t
2
dt
≥
F1M
η0
dt
E2
1
2M σ0 − M σ0
2
dt
≥
F n
min
M
η0
,1
2M σ0 − M σ0
2
dt
min
M
η0
,1
2M σ0 − M σ0
2
· μF n
≥ min
M
η0
,1
2M σ0 − M σ0
2
·1
8ε0.
2.50
By Lemma 1.1, we have ρ M u ρ M v n 1, ρ M u v n /2 → 1 as n → ∞ This is in
contradiction with1/2ρ M u 1/2ρ M v n − ρ M u v n /2 ≥ min{Mη0, 1/2Mσ0 −
Mσ0/2} · 1/8ε0
... Δ2;b X is locally uniformly nonsquare.
Trang 6In order to prove the theorem,... following two cases
Trang 8Case x y0 x − y0 Since... derive a contradiction for each of the following two cases
Trang 5Case μ{t ∈ T : ut / 0} ∩ {t