Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2011, Article ID 361525, 16 pages
doi:10.1155/2011/361525
Research Article
Nonsquareness in Musielak-Orlicz-Bochner
Function Spaces
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 20 October 2010; Revised 3 January 2011; Accepted 14 February 2011
Academic Editor: Paul Eloe
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.
The criteria for nonsquareness in the classical Orlicz function spaces have been given already.
However, because of the complication of Musielak-Orlicz-Bochner function spaces, at present the
criteria for nonsquareness have not been discussed yet. In the paper, the criteria for nonsquareness
of Musielak-Orlicz-Bochner function spaces are given. As a corollary, the criteria for nonsquareness
of Musielak-Orlicz function spaces are given.
1. Introduction
A lot of nonsquareness concepts in Banach spaces are known. Nonsquareness are the
important notion in geometry of Banach space. One of reasons is that the property is strongly
related to the fixed point property see 1. The criteria for nonsquareness in the classical
Orlicz function spaces have been given in 2 already. However, because of the complication
of Musielak-Orlicz-Bochner function spaces, at present the criteria for nonsquareness have
not been discussed yet. The aim of this paper is to give criteria nonsquareness of MusielakOrlicz-Bochner function spaces. As a corollary, the criteria for nonsquareness of MusielakOrlicz function spaces are given. The topic of this paper is related to the topic of 3–8.
Let X, · be a real Banach space. SX and BX denote the unit sphere and
unit ball, respectively. By X ∗ , denote the dual space of X. Let N, R, and R denote the set
natural number, reals, and nonnegative reals, respectively. Let us recall some geometrical
notions concerning nonsquareness. A Banach space X is said to be nonsquare space if for
any x, y ∈ SX we have min{1/2x y, 1/2x − y} < 1. A Banach space X is
said to be uniformly nonsquare space if for any x, y ∈ SX, there exists δ > 0 such that
min{1/2x y, 1/2x − y} < 1 − δ.
2
Abstract and Applied Analysis
Let T, , μ be nonatomic measure space. Suppose that a function M : T × 0, ∞ →
0, ∞ satisfies the following conditions:
1 for μ-a.e, t ∈ T, Mt, 0 0, limu → ∞ Mt, u ∞ and Mt, u < ∞ for some u > 0,
2 for μ-a.e, t ∈ T, Mt, u is convex on 0, ∞ with respect to u,
3 for each u ∈ 0, ∞, Mt, u is a μ-measurable function of t on T.
Let et sup{u > 0 : Mt, u 0}. It is well known that et is μ-measurable see 2.
Moreover, for a given Banach space X, · , we denote by XT , the set of all strongly
μ-measurable function from T to X, and for each u ∈ XT , define the modular of u by
ρM u Mt, utdt.
1.1
T
Put
LM X u ∈ XT : ρM λu < ∞ for some λ > 0 .
1.2
It is well known that Musielak-Orlicz-Bochner function space LM X is Banach spaces
equipped with the Luxemburg norm
u
≤1
u inf λ > 0 : ρM
λ
1.3
or Orlicz’s norm
u0 inf
1
k>0 k
1 ρM ku .
1.4
In particular, LM R and L0M R are said to be Musielak-Orlicz function space. Set
supp u {t ∈ T : ut /
0},
Ku k>0:
1
1 ρM ku u0 .
k
1.5
In particular, the set Ku can be nonempty. To show that, we give a proposition.
Proposition 1.1 see 9. If limu → ∞ Mt, u/u ∞ μ-a.e. t ∈ T, then Ku /
φ for any u ∈
L0M X.
We define a function
1
1
σt sup u ≥ 0 : M t, u Mt, u .
2
2
1.6
Function σt will be used in the further part of the paper. Moreover, σt is μ-measurable. To
show that, we give a proposition.
Abstract and Applied Analysis
3
Proposition 1.2. Function σt is μ-measurable.
Proof. Pick a dense set {ri }∞
i1 in 0, ∞ and set
Bk 1
1
t ∈ T : M t, rk Mt, rk ,
2
2
qk t rk χBk t k ∈ N.
1.7
It is easy to see that for all k ∈ N, σt ≥ qk t μ-a.e on T. Hence, supk≥1 qk t ≤ σt. For
μ-a.e t ∈ T, arbitrarily choose ε ∈ 0, σt. Then, there exists rk ∈ σt − ε, σt such that
Mt, 1/2rk 1/2Mt, rk , that is, qk t ≥ rk > σt − ε. Since ε is arbitrary, we find
supk≥1 qk t ≥ σt. Thus, supk≥1 qk t σt.
It is easy to prove the following proposition.
Proposition 1.3. For any α ∈ 0, 1, if ut ≤ σt, then Mt, αut αMt, ut.
Proposition 1.4. For any α ∈ 0, 1, if ut < σt < vt, then Mt, αut 1 − αvt <
αMt, ut 1 − αMt, vt.
Proposition 1.5. For any α ∈ 0, 1, if σt < vt, then Mt, αvt < αMt, vt.
Definition 1.6 see 2. We say that Mt, u satisfies condition ΔM
∈ Δ if there exist
K ≥ 1 and a measureable nonnegative function δt on T such that T Mt, δtdt < ∞ and
Mt, 2u ≤ KMt, u for almost all t ∈ T and all u ≥ δt.
First, we give some results that will used in the further part of the paper.
Lemma 1.7 see 2. Suppose M ∈ Δ. Then ρM u 1 ⇔ u 1.
Lemma 1.8 see 9. Let L0M X be Musielak-Orlicz-Bochner function spaces, then, if Ku φ,
one has u0 T At · utdt, where At limu → ∞ Mt, u/u.
2. Main Results
Theorem 2.1. LM X is nonsquare space if and only if
a M ∈ Δ,
b for any u, v ∈ SLM X, one has μ{t ∈ supp u ∩ supp v : ut vt > 2et} > 0
or μ{t ∈ T : ut > σt} ∪ {t ∈ T : vt > σt} > 0,
c X is nonsquare space.
In order to prove the theorem, we give a lemma.
Lemma 2.2. Let X be nonsquare space, then for any x, y /
0, one has
x y − min x y, x − y > 0.
2.1
4
Abstract and Applied Analysis
Proof. For any x, y /
0, without loss of generality, we may assume x ≤ y. Since X is
nonsquare space, we have
x y > min x x x x y x y
, x −
y x y
.
y 2.2
Therefore, by 2.2, we obtain
x
x
x y ≤ x · y 1 − · y
y y < x x y − x
x y
2.3
or
x − y ≤ x −
x
x
· y 1 − · y y y < x x y − x
x y.
2.4
This implies x y − min{x y, x − y} > 0. This completes the proof.
Proof of Theorem 2.1. Necessity. a If LM X is nonsquare space, then LM R is nonsquare
space, because LM R is isometrically embedded into LM X. Since LM R is nonsquare
space, then M ∈ Δ which follows from the theorem proved in more general case see 10, 11.
Namely, if M ∈
/ Δ, then LM R contains isometric copy of l∞ .
If b is not true, then there exist u, v ∈ SLM X such that μ{t ∈ supp u ∩ supp v :
ut vt > 2et} 0 and μ{t ∈ T : ut > σt} ∪ {t ∈ T : vt > σt} 0. Let
G supp u ∩ supp v. We have
1
1
ρM u ρM v
2
2
1
1
Mt, utdt Mt, vtdt
2 T
2 T
1
1
Mt, ut Mt, vtdt
2
2
T
1
1
1
1
≥
M t, ut vt dt M t, ut vt dt
2
2
2
2
G
T \G
Abstract and Applied Analysis
1
1
≥
M t, ut vt dt M t, ut vt dt
2
2
G
T \G
u v
ρM
.
2
5
2.5
By μ{t ∈ supp u ∩ supp v : ut vt > 2et} 0 and μ{t ∈ T : ut > σt} ∪ {t ∈
T : vt > σt} 0, we obtain that two inequalities of 2.5 are equations. This implies
1/2ρM u 1/2ρM v ρM u v/2. By Lemma 1.7, we have ρM u v/2 1. Thus,
1/2u v 1. Similarly, we have 1/2u − v 1, a contradiction!
c Pick ht ∈ SLM X, then there exists d > 0 such that μE > 0, where E {t ∈ T :
ht ≥ d}. Put h1 t d ·x0 ·χE t, where x0 ∈ SX. It is easy to see that h1 t ∈ LM X\ {0}.
Hence, there exists k > 0 such that k · h1 t ∈ SLM X. By Lemma 1.7, we have
1
Mt, k · h1 tdt T
Mt, k · d · x0 dt.
2.6
E
Let α k · d. Then, E Mt, αdt 1. The necessity of c follows from the fact that X is
isometrically embedded into LM X. Namely, defining the operator I : X → LM X by
Ix α · x · χE t,
x ∈ X.
2.7
Hence, for any x ∈ X \ {0}, we have
ρM
Ix
x
Ix αx
dt M t, M
t,
Mt, αdt 1.
dt
x x
E
E
E
2.8
By Lemma 1.7, we have Ix/xLM X 1, hence IxLM X x.
Sufficiency. The proof requires the consideration of two cases separately.
Case 1. μ{t ∈ T : ut > σt} ∪ {t ∈ T : vt > σt} > 0. Without loss of generality, we
may assume μ{t ∈ T : ut > σt} > 0. Let F {t ∈ T : ut > σt}. Put
F1 {t ∈ F : ut vt > ut vt},
F2 {t ∈ F : ut vt > ut − vt},
F3 {t ∈ F : vt 0}.
2.9
6
Abstract and Applied Analysis
Since X is nonsquare space, we have μF1 ∪ F3 > 0 or μF2 ∪ F3 > 0 by Lemma 2.2. Without
loss of generality, we may assume μF1 ∪ F3 > 0. Moreover, we have
1
1
1
ρM u ρM v − ρM
u v
2
2
2
1
1
1
Mt, utdt Mt, vtdt − M t, ut vt dt
2 T
2 T
2
T
1
1
1
Mt, ut Mt, vt − M t, ut vt dt
2
2
T 2
1
1
1
Mt, ut Mt, vt − M t, ut vt dt
≥
2
2
F1 ∪F3 2
1
1
1
M t, ut vt − M t, ut vt dt
≥
2
2
2
F1 ∪F3
2.10
≥ 0.
Let E {t ∈ F1 ∪ F3 : vt ≥ σt}, E1 {t ∈ E : vt σt 0}. Then, μF1 ∪ F3 \ E \
E1 > 0 or μE \ E1 > 0. By F {t ∈ T : ut > σt} ⊃ F1 ∪ F3 ⊃ F1 ∪ F3 \ E \ E1 , we
obtain
1
1
1
Mt, ut Mt, vt − M t, ut vt dt
2
2
F1 ∪F3 2
1
1
1
M t, ut vt − M t, ut vt dt,
>
2
2
2
F1 ∪F3
2.11
whenever μF1 ∪ F3 \ E \ E1 > 0. By F {t ∈ T : ut > σt} ⊃ F1 ∪ F3 ⊃ E \ E1 , we
obtain
1
1
1
M t, ut vt − M t, ut vt dt > 0,
2
2
2
F1 ∪F3
2.12
whenever μE \ E1 > 0. This means that one of three inequalities of 2.10 is strict inequality.
By ρM u ρM v 1, we have ρM 1/2u v < 1. By Lemma 1.7, we have 1/2u v
< 1.
Case 2. μ{t ∈ T : ut > σt} ∪ {t ∈ T : vt > σt} 0. By b, we have μ{t ∈ supp u ∩
supp v : ut vt > 2et} > 0. Let G {t ∈ supp u ∩ supp v : ut vt > 2et}.
Put
G1 {t ∈ G : ut vt > ut vt},
G2 {t ∈ G : ut vt > ut − vt}.
2.13
Abstract and Applied Analysis
7
Since X is nonsquare space, we have μG1 > 0 or μG2 > 0 by Lemma 2.2. Without loss of
generality, we may assume μG1 > 0. Hence,
1
1
1
M t, ut vt dt >
M t, ut vt dt.
2
2
2
G1
G1
2.14
Therefore, by 2.14, we have
1
1
ρM u ρM v
2
2
1
1
Mt, ut Mt, vtdt
2
2
T
≥
>
1
1
1
1
M t, ut vt dt M t, ut vt dt
2
2
2
2
G1
T \G1
2.15
1
1
M t, ut vt dt M t, ut vt dt
2
2
G1
T \G1
ρM
u v
2
.
By ρM u ρM v 1, we have ρM uv/2 < 1. By Lemma 1.7, we have 1/2uv < 1.
This completes the proof.
Corollary 2.3. LM R is nonsquare space if and only if
a M ∈ Δ,
b for any u, v ∈ SLM R, one has μ{t ∈ supp u ∩ supp v : |ut| |vt| > 2et} > 0 or
μ{t ∈ T : |ut| > σt} ∪ {t ∈ T : |vt| > σt} > 0.
Theorem 2.4. Let et 0 μ-a.e on T. Then, LM X is nonsquare space if and only if
a M ∈ Δ,
b ρM σ < 2,
c X is nonsquare space.
Proof. Necessity. By Theorem 2.1, a and c are obvious. Suppose that ρM σ ≥ 2. Then, there
exists E ∈ Σ such that μE > 0, ρM σ · χT \E 1 and ρM σ · χD 1, where D ⊂ T \ E. Set
ut x · σt · χT \E t,
vt x · σt · χD t,
2.16
where x ∈ SX. It is easy to see that u v 1, μ{t ∈ supp u ∩ supp v : ut vt >
2et} 0 and μ{t ∈ T : ut > σt} ∪ {t ∈ T : vt > σt} 0. Contradicting
Theorem 2.1.
8
Abstract and Applied Analysis
Sufficiency. We only need to prove that for any u, v ∈ SLM X, if μsupp u ∩ supp v 0, then μ{t ∈ T : ut > σt} ∪ {t ∈ T : vt > σt} > 0. Suppose that there exist
u, v ∈ SLM X such that μsupp u ∩ supp v 0, μ{t ∈ T : ut > σt} ∪ {t ∈ T : vt >
σt} 0. By μsupp u ∩ supp v 0, we have ρM u ρM v ρM u v ≤ ρM σ < 2. This
implies ρM u < 1 or ρM v < 1. Hence, u < 1 or v < 1, a contradiction! This completes
the proof.
Theorem 2.5. L0M X is nonsquare space if and only if
a for any u ∈ L0M X \ {0}, one has Ku /
φ,
b at least one of the conditions
b1 kl/k l ∈
/ Ku v ∩ Ku − v,
b2 μ{t ∈ T : kut − σt > 0} ∪ {t ∈ T : lvt − σt > 0} > 0,
b3 μ{t ∈ supp u ∩ supp v : kl/k lut vt > et} > 0
is true, where k ∈ Ku, l ∈ Kv and u, v ∈ SL0M X,
c X is nonsquare space.
Proof. Necessity. a Suppose that there exists u ∈ L0M X \ {0} such that Ku φ, then
0
T At · utdt
by Lemma 1.8. Decompose T into disjoint sets T1 and T2 such that
u
At
·
utdt
At
· utdt. Put
T1
T2
u1 t 2utχT1 ,
2.17
u2 t 2utχT2 .
Obviously, u 1/2u1 u2 . Pick sequence {kn }∞
n1 ⊂ R such that kn → ∞ as n → ∞. Let
T0 supp u1 . By Levi theorem, we have
1
1 Mt, kn u1 tdt
n → ∞ kn
T
Mt, kn u1 t
lim
u1 tdt
n→∞ T
kn u1 t
0
Mt, kn u1 t
lim
u1 tdt
n
→
∞
kn u1 t
T0
Atu1 tdt
u1 0 ≤ lim
2.18
T
At2utdt.
T1
Therefore,
u1 0 ≤
At · 2utdt
T1
At · utdt T1
At · utdt T1
At · utdt −
T2
At · utdt
T2
Abstract and Applied Analysis
At · utdt At · utdt
T1
9
T2
At · utdt u0 .
T
2.19
Similarly, we have u2 0 ≤ u0 . By u 1/2u1 u2 , we obtain u0 u1 0 u2 0 1/2u1 u2 0 . By u1 t − u2 t 2utχT1 −2utχT2 , we have u0 1/2u1 − u2 0 .
Therefore,
0 0
1
1
0
u
−
u
u1 0 u2 0 u
u
2 2 u .
2 1
2 1
2.20
This implies
u1 0 u2 0 min u1 u2 0 , u1 − u2 0 ,
2.21
a contradiction!
If b is not true, then there exist u, v ∈ SL0M X such that kl/k l ∈ Ku v ∩
Ku − v, μ{t ∈ T : kut − σt > 0} ∪ {t ∈ T : lvt − σt > 0} 0 and μ{t ∈
supp u ∩ supp v : kl/k lut vt > et} 0, where k ∈ Ku, l ∈ Kv. Let
E T \ supp u ∩ supp v. It is easy to see that if t ∈ E, then ut · vt 0 on E. This
implies
kl
kl
M t,
ut vt
kl
kl
kl
M t,
ut vt
kl
t ∈ E.
2.22
Therefore, by 2.22, we have
kl
kl
kl
M t,
M t,
ut vt dt ut vt dt.
kl
kl
kl
E
E
2.23
By μ{t ∈ supp u ∩ supp v : kl/k lut vt > et} 0, we have
kl
kl
M t,
ut vt dt
kl
kl
T \E
kl
M t,
vt
dt.
ut
kl
T \E
0
2.24
By 2.23 and 2.24, we have
kl
kl
kl
M t,
M t,
ut vt dt ut vt dt.
kl
kl
kl
T
T
2.25
10
Abstract and Applied Analysis
By μ{t ∈ T : kut − σt > 0} ∪ {t ∈ T : lvt − σt > 0} 0, we have
T
k
l
Mt, kut Mt, lvt dt
kl
kl
kl
kl
M t,
ut vt dt.
kl
kl
T
2.26
Therefore, by 2.25 and 2.26, we have
1
1
1 ρM ku 1 ρM lv
k
l
kl
l
k
1
ρM ku ρM lv
kl
kl
kl
u0 v0 l
k
kl
1
Mt, kutdt Mt, lvtdt
kl
kl T
kl T
kl
k
l
Mt, kut Mt, lvt dt
1
kl
kl
T kl
kl
kl
kl
1 M t,
ut vt dt
kl
kl
kl
T
kl
kl
1 M t,
ut vt dt
kl
kl
T
kl
kl
1 ρM
v
u
kl
kl
2.27
u v0 .
Similarly, we have u0 v0 u − v0 . This implies u0 v0 u v0 u − v0 , a
contradiction!
c Pick ht ∈ SL0M X, then there exists d > 0 such that μE > 0, where E {t ∈ T :
ht ≥ d}. Put h1 t d ·x0 ·χE t, where x0 ∈ SX. It is easy to see that h1 t ∈ L0M X\ {0}.
Hence, there exists l > 0 such that l · h1 t ∈ SL0M X. The necessity of c follows from the
fact that X is isometrically embedded into L0M X. Namely, defining the operator I : X →
L0M X by
Ix ld · x · χE t,
x ∈ X.
2.28
Abstract and Applied Analysis
11
It is easy to see that Ix0 ∈ SL0M X. Hence, for any x ∈ X \ {0}, we have
1
1 ρM k · Ix
k>0 k
Ix0 inf
1
1
Mt, k · ldxdt
k>0 k
E
inf
1
1
Mt, k · xldx0 dt
k>0 k
E
inf
1
inf 1 ρM k · xIx0 k>0 k
2.29
x · Ix0 0
x · Ix0 0
x.
Sufficiency. Suppose that there exists u, v ∈ SL0M X such that u0 v0 1/2u v0 1/2u − v0 1. Let k ∈ Ku, l ∈ Kv. We will derive a contradiction
for each of the following two cases.
Case 1. μ{t ∈ T : ut /
0} ∪ {t ∈ T : vt /
0} \ {t ∈ T : σt > 0} 0. By Lemma 2.2,
we have ut vt > min{ut vt, ut − vt}t ∈ supp u ∩ supp v. Put
T1 t ∈ supp u ∩ supp v : ut vt > ut vt ,
T2 t ∈ supp u ∩ supp v : ut vt > ut − vt .
Moreover, we have
1
1
1 ρM ku 1 ρM lv
k
l
l
k
kl
1
Mt, kutdt Mt, lvtdt
kl
kl T
kl T
u0 v0 kl
k
l
1
Mt, kut Mt, lvt dt
kl
kl
T kl
kl
kl
kl
≥
1 M t,
ut vt dt
kl
kl
kl
T
2.30
12
Abstract and Applied Analysis
kl
kl
1
M t,
≥
ut vt dt
kl
k
l
T1
kl
kl
1
M t,
ut vt dt
kl
kl
T \T1
kl
kl
1 ρM
u v
kl
kl
≥ u v0 .
2.31
By u0 v0 u v0 , three inequalities of 2.31 are equation. This implies
k
kl
l
Mt, kut Mt, lvt dt
1
kl
kl
T kl
kl
kl
kl
1 M t,
ut vt dt .
kl
kl
kl
T
2.32
Next, we will prove μ{t ∈ T : kut − σt > 0} ∪ {t ∈ T : lvt − σt > 0} 0.
Suppose that μ{t ∈ T : kut − σt > 0} ∪ {t ∈ T : lvt − σt > 0} > 0. Without loss
of generality, we may assume μ{t ∈ T : kut − σt > 0} > 0. Therefore, by 2.32, we have
lvt − σt ≥ 0 μ-a.e. on {t ∈ T : kut − σt > 0}. Hence, μ{t ∈ T : kut − σt >
0} ∩ {t ∈ T : lvt − σt ≥ 0} > 0. Since three inequalities of 2.31 are equation, we deduce
kl
kl
kl
M t,
M t,
ut vt dt ut vt dt.
kl
kl
kl
T
T
2.33
Moreover, it is easy to see
kl
kl
ut vt > σt ≥ et
kl
kl
2.34
on {t ∈ T : kut − σt > 0} ∩ {t ∈ T : lvt − σt ≥ 0}. Therefore, by 2.33 and 2.34, we
have
kl
kl
kl
ut vt ut vt
kl
kl
kl
2.35
μ-a.e. on {t ∈ T : kut − σt > 0} ∩ {t ∈ T : lvt − σt ≥ 0}. Since X is nonsquare space,
we have
kl
kl
kl
kl
ut vt > min
ut vt,
ut − vt
kl
kl
kl
kl
2.36
Abstract and Applied Analysis
13
on {t ∈ T : kut − σt > 0} ∩ {t ∈ T : lvt − σt ≥ 0}. Thus,
kl
kl
kl
ut vt >
ut − vt
kl
kl
kl
2.37
μ-a.e. on {t ∈ T : kut − σt > 0} ∩ {t ∈ T : lvt − σt ≥ 0}. Therefore, by 2.34 and
2.37, we have
1
1
1 ρM ku 1 ρM lv
k
l
kl
kl
kl
≥
1 M t,
ut vt dt
kl
kl
kl
T
u0 v0 >
kl
kl
1 M t,
ut − vt dt
kl
kl
T
2.38
kl
kl
1 ρM
u − v
kl
kl
≥ u − v0 .
This implies u0 v0 > u − v0 , a contradiction! Hence, μ{t ∈ T : kut − σt > 0}
∪ {t ∈ T : lvt − σt > 0} 0. Since three inequalities of 2.31 are equation, we deduce
kl/k l ∈ Ku v and μ{t ∈ T1 : kl/k lut vt > et} 0. By b, we have
/ Ku−v,
kl/kl ∈
/ Ku−v or μ{t ∈ T2 : kl/klutvt > et} > 0. If kl/kl ∈
then
1
1
1 ρM ku 1 ρM lv
k
l
kl
kl
kl
1 M t,
≥
ut vt dt
kl
kl
kl
T
u0 v0 kl
kl
1 M t,
≥
ut − vt dt
kl
kl
T
kl
kl
1 ρM
u − v
kl
kl
> u − v0 .
2.39
14
Abstract and Applied Analysis
This implies u0 v0 > u − v0 , a contradiction! If μ{t ∈ T2 : kl/k lut vt >
et} > 0, then
1
1
1 ρM ku 1 ρM lv
k
l
kl
kl
kl
1 M t,
≥
ut vt dt
kl
kl
kl
T
kl
kl
1
M t,
>
ut − vt dt
kl
kl
T2
kl
kl
M t,
1
ut − t dt
kl
kl
T \T2
kl
kl
1 ρM
u − v
kl
kl
u0 v0 ≥ u − v0 .
2.40
This implies u0 v0 > u − v0 , a contradiction!
Case 2. μ{t ∈ T : ut /
0} ∪ {t ∈ T : vt /
0} \ {t ∈ T : σt > 0} > 0. Let E {t ∈ T :
ut 0} \ {t ∈ T : σt > 0}. Put
/ 0} ∪ {t ∈ T : vt /
E1 {t ∈ E : ut · vt 0},
E2 {t ∈ E : ut · vt > 0}.
2.41
Then, μE1 > 0 or μE2 > 0. If μE1 > 0, then
E1
k
l
Mt, kutdt Mt, lvt dt
kl
kl
kl
kl
>
M t,
ut vt dt.
kl
kl
E1
Therefore, by 2.42, we have
1
1
1 ρM ku 1 ρM lv
k
l
k
kl
l
1
Mt, kut Mt, lvt dt
kl
kl
T kl
u0 v0 2.42
Abstract and Applied Analysis
kl
kl
kl
1
M t,
>
ut vt dt
kl
kl
kl
E1
15
kl
kl
kl
1
M t,
ut vt dt
kl
kl
kl
T \E1
≥
kl
kl
1 M t,
ut vt dt
kl
kl
T
kl
kl
1 ρM
u v
kl
kl
≥ u v0 .
2.43
This implies u0 v0 > u v0 , a contradiction! If μE2 > 0, then μE21 > 0 or μE22 > 0 by
Lemma 2.2, where
E21 {t ∈ E2 : ut vt > ut vt},
E22 {t ∈ E2 : ut vt > ut − vt}.
Without loss of generality, we may assume μE21 > 0. Hence,
kl
kl
kl
M t,
M t,
ut vt dt >
ut vt dt.
kl
kl
kl
E21
E21
2.44
2.45
Therefore, by 2.45, we have
1
1
1 ρM ku 1 ρM lv
k
l
kl
kl
kl
1 M t,
≥
ut vt dt
kl
kl
kl
T
kl
kl
1
>
M t,
ut vt dt
kl
kl
E21
u0 v0 kl
kl
1
M t,
ut vt dt
kl
kl
T \E21
kl
kl
1 ρM
u v
kl
kl
≥ u v0 .
This implies u0 v0 > u v0 , a contradiction! This completes the proof.
2.46
16
Abstract and Applied Analysis
Corollary 2.6. L0M R is nonsquare space if and only if
φ,
a for any u ∈ L0M R, one has Ku /
b at least one of the conditions
b1 kl/k l ∈
/ Ku v ∩ Ku − v,
b2 μ{t ∈ T : k|ut| − σt > 0} ∪ {t ∈ T : l|vt| − σt > 0} > 0,
b3 μ{t ∈ supp u ∩ supp v : kl/k l|ut| |vt| > et} > 0
is true, where k ∈ Ku, l ∈ Kv and u, v ∈ SL0M R.
Acknowledgments
The authors would like to thank the anonymous referee for the suggestions to improve the
manuscript. This work was supported by Academic Leaders Fund of Harbin University of
Science and Technology and supported by China Natural Science Fund 11061022.
References
1 J. Garcı́a-Falset, E. Llorens-Fuster, and E. M. Mazcuñan-Navarro, “Uniformly nonsquare Banach
spaces have the fixed point property for nonexpansive mappings,” Journal of Functional Analysis, vol.
233, no. 2, pp. 494–514, 2006.
2 S. T. Chen, “Geometry of Orlicz spaces,” Dissertationes Math, vol. 356, p. 204, 1996.
3 C. Castaing and R. Pluciennik, “Property H in Köthe-Bochner spaces,” Indagationes Mathematicae,
vol. 7, no. 4, pp. 447–459, 1996.
4 H. Hudzik and M. Mastyło, “Strongly extreme points in Köthe-Bochner spaces,” The Rocky Mountain
Journal of Mathematics, vol. 23, no. 3, pp. 899–909, 1993.
5 H. Hudzik and B. Wang, “Approximate compactness in Orlicz spaces,” Journal of Approximation
Theory, vol. 95, no. 1, pp. 82–89, 1998.
6 H. Hudzik and A. Narloch, “Relationships between monotonicity and complex rotundity properties
with some consequences,” Mathematica Scandinavica, vol. 96, no. 2, pp. 289–306, 2005.
7 H. Hudzik and M. Wisła, “On extreme points of Orlicz spaces with Orlicz norm,” Collectanea
Mathematica, vol. 23, pp. 899–909, 1993.
8 J. Cerdà, H. Hudzik, and M. Mastyło, “Geometric properties of Köthe-Bochner spaces,” Mathematical
Proceedings of the Cambridge Philosophical Society, vol. 120, no. 3, pp. 521–533, 1996.
9 S. Shang, Y. Cui, and Y. Fu, “Extreme points and rotundity in Musielak-Orlicz-Bochner function
spaces endowed with Orlicz norm,” Abstract and Applied Analysis, vol. 2010, Article ID 914183, 13
pages, 2010.
10 H. Hudzik, “Banach lattices with order isometric copies of l∞ ,” Indagationes Mathematicae, vol. 9, no.
4, pp. 521–527, 1998.
11 P. Foralewski and H. Hudzik, “Some basic properties of generalized Calderon-Lozanovskiı̆ spaces,”
Collectanea Mathematica, vol. 48, no. 4–6, pp. 523–538, 1997.
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