Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2010, Article ID 914183, 13 pages
doi:10.1155/2010/914183
Research Article
Extreme Points and Rotundity in
Musielak-Orlicz-Bochner Function Spaces
Endowed with Orlicz 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 17 June 2010; Accepted 27 August 2010
Academic Editor: Paul Eloe
Copyright q 2010 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 extreme point and rotundity of Musielak-Orlicz-Bochner function spaces equipped
with Orlicz norm are given. Although criteria for extreme point of Musielak-Orlicz function spaces
equipped with the Orlicz norm were known, we can easily deduce them from our main results.
1. Introduction
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.
A point x ∈ A is said to be extreme point of A if 2x y z and y, z ∈ A imply y z.
The set of all extreme points of A is denoted by ExtA. If ExtBX SX, then X is said to be
∞
rotund. A point x ∈ SX is said to be strongly extreme point if for any {xn }∞
n1 , {yn }n1 ∈ X
with xn → 1, yn → 1, and x 1/2xn yn , there holds xn − yn → 0 n → ∞. If
the set of all strongly extreme points of BX is equal to SX, then X is said to be midpoint
local uniform rotund.
The notion of extreme point plays an important role in some branches of mathematics.
For example, the Krein-Milman theorem, Choquet integral representation theorem, Rainwater theorem on convergence in weak topology, Bessaga-Pelczynski theorem, and Elton test
unconditional convergence are strongly connected with this notion. In 1, using the principle
of locally reflexivity, a remarkable theorem describing connections between extreme points of
SX and strongly extreme points of SX is proved. Namely, a Banach space X is midpoint
2
Abstract and Applied Analysis
local uniformly rotund if and only if every point of SX is an extreme point in X ∗∗ . Another
proof of this theorem based on Goldstein’s theorem is given in 2. Analyzing the proof of
this fact one can easily see its local version, namely, if x ∈ SX is a strongly extreme point in
X, then κx is an extreme point in X ∗∗ , where κ is the mapping of canonical embedding of X
into X ∗∗ .
The criteria for extreme point and rotundity in the classical Musielak-Orlicz function
spaces have been given in 3 already. However, because of the complication of MusielakOrlicz-Bochner function spaces equipped with Orlicz norm, at present, the criteria for
extreme point and rotundity have not been discussed yet. The aim of this paper is to
give criteria for extreme point and rotundity of Musielak-Orlicz-Bochner function spaces
equipped with Orlicz norm. By the result of this paper, it is easy to see that the result of
3 is true.
Let T, , μ be nonatomic measurable 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 pt, u denote the right derivative of Mt, · at u ∈ R where if Mt, u ∞, let pt, u ∞ and let qt, · be the generalized inverse function of pt, · defined on R by
qt, v sup u ≥ 0 : pt, u ≤ v .
u≥0
1.1
v
Then Nt, v 0 qt, sds for any v ∈ R and μ-a.e. t ∈ T . It is well known that there holds
the Young inequality uv ≤ Mt, u Nt, v for μ-a.e. t ∈ T . And uv Mt, u Nt, u ⇔
u qt, v or v qt, u. Let
et sup{u > 0 : Mt, u 0},
Et sup{u > 0 : Mt, u < ∞}.
1.2
For fixed t ∈ T and v ≥ 0, if there exists ∈ 0, 1 such that
Mt, v 1
1
Mt, v ε Mt, v − ε < ∞,
2
2
1.3
then we call v a nonstrictly convex points of M with respects to t. The set of all nonstrictly
convex point of M with respect to t is denoted by Kt .
For fixed t ∈ T , if Kt Φ, then we call that Mt, u is strictly convex with respect u
for t.
Moreover, for a given Banach space X, · , we denote by XT the set of all strongly
μ-measurable functions from T to X, and for each u ∈ XT , define the modular of u by
ρM u Mt, utdt.
T
1.4
Abstract and Applied Analysis
3
Put
L0M X u ∈ XT : ρM λu < ∞ for some λ > 0 .
1.5
Then the Musielak-Orlicz-Bochner function space
1
1 ρM ku
k>0 k
1.6
u0 inf
is Banach space. If X R, L0M R is said to be Musielak-Orlicz function space. Set
Ku 1
k>0:
1 ρM ku u0 .
k
1.7
In particular, the set Ku can be nonempty. To show that, we give a proposition.
φ for any u ∈ L0M X.
Proposition 1.1. If limu → ∞ Mt, u/u ∞ μ-a.e. t ∈ T , then Ku /
Proof. For any u ∈ L0M X, there exists a > 0 such that μT0 > 0, where T0 {t ∈ T : ut ≥ a}.
It is easy to see that T0 ∪∞
n1 Gn , where
Gn 3u0
Mt, v
≥
t ∈ T0 :
, v≥n .
v
a · μT0
1.8
Noticing that G1 ⊂ G2 ⊂ · · · ⊂ Gn ⊂ · · · , then limn → ∞ μGn μT0 . Hence there exists n1 such
that μGn1 > 1/2μT0 . This means that if k > n1 /a, we have
Mt, kut
1
1
dt ≥
Mt, kutdt ≥
k
k
T
Gn
Gn
1
a
Gn1
Mt, ka
dt ≥ a
ka
1
Mt, ka
dt
k
0
Gn1
Mt, n1 3u
3
dt ≥ a
· μGn1 > u0 .
n1
a · μT0
2
1.9
This implies that if 1/kn 1 ρM kn u → u0 n → ∞, then sequence {kn }∞
n1 is bounded.
Without loss of generality, we may assume that kn → k0 . Without loss of generality, we may
assume that k1 ≤ k2 ≤ · · · kn ≤ · · · ≤ k0 or k1 ≥ k2 ≥ · · · kn ≥ · · · ≥ k0 . If k1 ≤ k2 ≤ · · · kn ≤ · · · ≤
k0 , by Levi theorem, we have
1
1
Mt, kn ut
lim
dt
lim
1 ρM kn u lim
n → ∞ kn
n → ∞ kn
n→∞ T
kn
Mt, k0 ut
1
dt
k0
k0
T
1
1 ρM k0 u .
k0
1.10
4
Abstract and Applied Analysis
If k1 ≥ k2 ≥ · · · kn ≥ · · · ≥ k0 , by dominated convergence theorem, we have
1
1
lim
1 ρM kn u lim
n → ∞ kn
n → ∞ kn
n→∞
lim
1
k0
T
T
Mt, kn ut
dt
kn
Mt, k0 ut
dt
k0
1.11
1
1 ρM k0 u .
k0
Therefore 1/kn 1 ρM kn u → 1/k0 1 ρM k0 u n → ∞, namely, 1/k0 1 ρM k0 u u0 . This implies k0 ∈ Ku.
2. Main Results
In order to obtain the main theorems of this paper, we first give some lemmas.
Lemma 2.1. If Ku φ, then u0 T At · utdt , where At limu → ∞ Mt, u/u.
Proof. By proof of Proposition 1.1, we know that if Ku φ, then there exists {kn }∞
n1 such
that 1/kn 1 ρM kn u → u0 and kn → ∞ as n → ∞. Without loss of generality, we
may assume that k1 ≤ k2 ≤ · · · kn ≤ · · · . By Levi theorem, we have
1
Mt, kn utdt
1
u lim
n → ∞ kn
T
0
1
lim
n → ∞ kn
1
0}
{t∈T :ut /
1
lim
n → ∞ kn
n→∞
lim
{t∈T :ut / 0}
Mt, kn utdt
{t∈T :ut / 0}
Mt, kn ut
lim
· utdt
n→∞
kn ut
{t∈T :ut / 0}
At · utdt
At · utdt.
Mt, kn ut
· utdt
kn ut
T
Hence the conclusion of the lemma is true.
2.1
Abstract and Applied Analysis
5
Lemma 2.2. If the set Ku consists of one element from 0, ∞, then u0 <
T
At · utdt .
Proof. Pick k1 > k2 > 0; then we have
ρM k1 uχEn ≥
ρM k2 uχEn T
T
k1 utχEn t · p t, k2 utχEn t dt − ρN p k2 uχEn ,
k2 utχEn t · p t, k2 utχEn t dt − ρN p k2 uχEn ,
2.2
where
En t ∈ T : k2 ut ≤ n, pt, k2 ut ≤ n .
2.3
It follows that
1 1
1 ρM k1 uχEn −
1 ρM k2 uχEn
k1
k2
k1 − k2
k2 −1 ρM k1 uχEn − ρM k2 uχEn − ρM k2 uχEn
k1 k2
k1 − k2
k2
k1 − k2
−1 ≥
k1 − k2 utχEn t · p t, k2 utχEn t dt
k1 k2
k1 − k2 T
2.4
−ρM k2 uχEn
k1 − k2 ρN p k2 uχEn − 1 .
k1 k2
Let n → ∞; then we obtain
k1 − k2 1
1
1 ρM k1 uχE ≥
ρN p k2 uχE − 1 1 ρM k2 uχE ,
k1
k1 k2
k2
2.5
E t ∈ T : pt, k2 ut < ∞ .
2.6
where
6
Abstract and Applied Analysis
If pt, k2 ut ∞, then Mt, k1 ut ≥ Mt, k2 ut ∞. Hence we have
k 1 − k2 1
1
1 ρM k1 u ≥
ρN pk2 u − 1 1 ρM k2 u .
k1
k1 k2
k2
2.7
Moreover, there exists k0 ∈ R such that ρN pku ≥ 1, whenever k ≥ k0 . This means that
function Fk 1/k1 ρM ku is nondecreasing, when n is large enough. Pick sequence
{kn }∞
n1 such that 0 < k1 < k2 · · · < kn < · · · . By Levi theorem, we have
At · utdt T
lim
T n→∞
lim
n→∞
T
1
n → ∞ kn
Mt, kn ut
· utdt
kn ut
Mt, kn ut
· utdt
kn ut
lim
Mt, kn utdt
2.8
T
1
1 Mt, kn utdt
lim
n → ∞ kn
T
>
1
1
Mt, lutdt u0 ,
l
T
where {l} Ku. Hence the conclusion of the lemma is true.
Lemma 2.3 see 3. Let L0M R be rotund, then Mt, u is strictly convex with respect to u for
almost all t ∈ T .
Theorem 2.4. Let L0M X be Musielak-Orlicz-Bochner function spaces, then u ∈ SL0M X is an
extreme point of BL0M X if and only if
a the set Ku consists of one element from 0, ∞;
b v, w ∈ XT with u λv 1 − λw and ut vt wtμ-a.e that on T implies
v w, where λ ∈ 0, 1;
c μ{t ∈ T : kut ∈ Kt } 0, where k ∈ Ku.
Proof. Necessity. a1 Suppose that u is an extreme point of the unit ball BL0M and Ku φ,
then u0 T At · utdt by Lemma 2.1. Decompose T into T1 and T2 such that T1 At ·
utdt T2 At · utdt . Pick ε ∈ 0, 1. Put
u1 t ut εutχT1 ut − εutχT2 ,
u2 t ut − εutχT1 ut εutχT2 .
2.9
Abstract and Applied Analysis
7
u2 . Moreover, we have
Obviously, u 1/2u1 u2 and u1 /
At · u1 tdt
u1 0 ≤
T
At · ut εutdt T1
At · utdt T1
−
εAt · utdt T1
At · ut dt
T2
2.10
εAt · utdt
T2
At · utdt At · ut − εutdt
T2
T1
At · utdt
T2
At · utdt
T
1.
Similarly, we have u2 0 ≤ 1. Hence u1 , u1 ∈ BL0M X. Therefore u ∈ SL0M X is not an
φ. Suppose that u0 T At ·
extreme point of BL0M X, a contradiction. Hence Ku /
utdt . Similarly, we get a contradiction.
The necessity of b is obvious.
c Set
H1 {t ∈ T : ut 0, et > 0},
H2 {t ∈ T : 2Mt, kut Mt, kut ε Mt, kut − ε,
ut / 0},
where {k} ∈ Ku. Suppose that c does not hold. Then μH1 > 0 or μH2 > 0.
If μH1 > 0, then for any x ∈ SX, by setting
⎧
1
1
⎪
⎨
etx, − etx , t ∈ H1 ,
2k
2k
vt, wt ⎪
⎩
t ∈ T \ H1
ut, ut,
2.11
2.12
we have u /
w and u 1/2v w. Moreover, we have
1
1 ρM ku
k
1
1
M t, k
Mt, kutdt
1
2k et dt k
H1
T \H1
1
1
Mt, kutdt
k
T \H1
v0 ≤
≤ u0 .
2.13
8
Abstract and Applied Analysis
Similarly, we have w0 ≤ 1. Hence v, w ∈ BL0M X. Therefore u ∈ SL0M X is not
an extreme point of BL0M X, a contradiction.
0, Mt, kut If μH2 > 0, it is easy to see that H2 ⊂ ∪∞
n1 {t ∈ T : ut /
1/2Mt, 1 1/nkut 1/2Mt, 1 − 1/nkut}, where {k} ∈ Ku. Then there
exists n0 ∈ N such that
H
t ∈ T : ut / 0,
1
1
1
1
Mt, kut M t, 1 kut M t, 1 −
kut < ∞
2
n0
2
n0
2.14
is not a noll set. Decompose H into E and F such that
pt, 1/n0 kutdt. Define
F
⎧
1
⎪
⎪
ut,
1−
1
⎪
⎪
n0
⎪
⎪
⎪
⎪
⎪
⎪
⎨
1
vt, wt 1
−
ut,
1
⎪
⎪
n0
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
ut, ut,
E
pt, 1/n0 kutdt
1
ut , t ∈ E,
n0
1
ut , t ∈ F,
n0
2.15
t ∈ T \ E ∪ F.
Then u / w and u 1/2v w. Furthermore, we have
1
1 ρM kv
k
1
1
1
Mt, kutdt M t, 1 kut dt
k
n0
T \E∪F
E
1
M t, 1 −
kut dt
n0
F
1
1
Mt, kutdt Mt, kutdt
k
T \E∪F
E
2.16
1
1
p t, kut dt Mt, kutdt −
p t, kut dt
n
n
0
0
E
F
F
1
1
Mt, kutdt Mt, kutdt Mt, kutdt
k
T \E∪F
E
F
v0 ≤
1
1 ρM ku
k
u0 1.
Abstract and Applied Analysis
9
Similarly, we have w0 ≤ 1. Hence v, w ∈ BL0M X. Therefore u ∈ SL0M X is not an
extreme point, a contradiction. Hence c is true.
a2 If Ku /
φ and u ∈ SL0M X is an extreme point, suppose that there exists
k1 , k2 ∈ Ku satisfying k1 / k2 . Define k k1 k2 /k1 k2 ,
2 u0 u0
k1 k 2
k1
k2
ρM k1 u ρM k2 u
1
k1 k2
k1 k2
k1 k2
k1
k2
k1 k2
Mt, k1 utdt Mt, k2 utdt
1
k1 k2
k1 k2 T
k1 k2 T
k1 k2
k2
k1
≥
M t,
1
k1 ut k2 ut dt
k1 k2
k1 k2
k1 k2
T
2.17
2k1 k2
k1 k2
dt
M t, ut
1
k1 k2
k1 k2
T
2
1 1 ρM 2ku
2k
≥ 2u0
2.
This implies that
u0 1 1 ρM 2ku
2k
2.18
i.e., 2k ∈ Ku,
k2
k1
Mt, k1 ut Mt, k2 ut Mt, 2kut.
k1 k2
k1 k2
2.19
k2 ut on {t ∈ T : ut /
0}, then 2kut ∈ Kt on {t ∈ T : ut /
0}, a
Since k1 ut /
contradiction. Therefore a is true.
10
Abstract and Applied Analysis
Sufficiency. We first prove that for u, u1 , u2 ∈ SL0M X with u 1/2u1 u2 at least
one of the sets Ku1 or Ku2 is nonempty. Suppose that Ku1 φ and Ku2 φ. Hence
we have
0
1
1 u1 u2 2
<
T
1
≤
2
1
dt
At · u
t
t
u
1
2
2
1
At · u1 tdt 2
T
At · u2 tdt
2.20
T
1
1
u1 0 u2 0
2
2
1,
a contradiction. This contradiction shows that Ku1 /
φ or Ku2 /
φ.
φ and Ku2 /
φ. Otherwise, we can assume without
Now we will prove that Ku1 /
φ and Ku2 φ. Put
loss of generality that Ku1 /
u1 , u {1 − λu1 λu : 0 < λ < 1},
u, u2 {1 − λu λu2 : 0 < λ < 1}.
2.21
Next we will prove that Ky /
φ for all y ∈ u1 , u and Ky φ for all y ∈ u, u2 . Assume
first for the contrary that this is u3 ∈ u1 , u such that Ku3 φ. Then there exists λ3 ∈ 0, 1
such that u3 1 − λ3 u1 λ3 u. Since u1 2u − u2 , we have
u3 1 − λ3 2u − u2 λ3 u 2 − λ3 u − 1 − λ3 u2 .
2.22
Hence u 1/2 − λ3 u3 1 − λ3 /2 − λ3 u2 . Therefore
0
1 u <
T
≤
1
2 − λ3
At · u3 tdt T
1 − λ3
1
u3 0 u2 0
2 − λ3
2 − λ3
1,
a contradiction.
1
1 − λ3
At · u3 t u2 t
dt
2 − λ3
2 − λ3
1 − λ3
2 − λ3
At · u2 tdt
T
2.23
Abstract and Applied Analysis
11
φ. We can find
Assume now for the contrary that this is u4 ∈ u, u2 such that Ku4 /
u5 . Therefore there are k4 ≥ 1 and k5 ≥ 1
u5 ∈ u, u2 such that u 1/2u4 u5 and u4 /
such that
u4 0 1
1 ρM k4 u4 ,
k4
u5 0 1
1 ρM k5 u5 .
k5
2.24
By the convexity of the modular ρM we have
ρM
2k4 k5
u
k4 k5
ρM
k 4 k5
u4 u5 k4 k5
k4
k5
k 4 u4 k 5 u5
k4 k5
k4 k5
k5
k4
dt
M t, k
u
k
u
t
t
4
4
5
5
k k
k4 k5
4
5
T
k5
k4
≤
M t,
k4 u4 t k5 u5 t dt
k4 k5
k4 k5
T
k4
k5
Mt, k4 u4 tdt Mt, k5 u5 tdt
≤
k4 k5 T
k4 k5 T
ρM
2.25
k4
k5
ρM k4 u4 ρM k5 u5 .
k4 k5
k4 k5
Hence
2 2u0
k 4 k5
k 4 k5
2u
1 ρM
k4 k5
k4 k5
k4 k 5
k5
k4
≤
ρM k4 u4 ρM k5 u5 1
k4 k5
k4 k5
k4 k5
≤
≤
2.26
1
1
1 ρM k4 u4 1 ρM k5 u5 k4
k5
2.
Consequently, all inequalities from the last three lines are equalities in fact. Therefore
2k4 k5 /k4 k5 k and
Mt, kut k5
k4
Mt, k4 u4 t Mt, k5 u5 t
k4 k5
k4 k5
2.27
for μ-a.e t ∈ T . By μ{t ∈ T : kut ∈ Kt } 0, it follows that kut k4 u4 t k5 u5 t
for μ-a.e t ∈ T . And we have kut k5 /k4 k5 k4 u4 tk4 /k4 k5 k5 u5 t for μ-a.e t ∈ T .
By b, we have ku k4 u4 k5 u5 . Since u4 , u5 , u ∈ SL0M X, we get k4 k5 k, which gives
12
Abstract and Applied Analysis
u5 . Thus Ky φ for any y ∈ u, u2 . Take
u4 u5 u. This contradicts the inequality u4 /
for
all
n
∈
N.
Then
u
∈ u, u2 for all n ∈ N. Hence Kun φ, and
un 1 − 1/nu 1/nu
2
n
consequently un 0 T At · un tdt for all n ∈ N. Note that un − u0 → 0 n → ∞
and limn → ∞ un t ut for μ-a.e, t ∈ T . Since Ku {k}, with 0 < k < ∞, we have
lim un 0 u0 lim
n→∞
n→∞
At · utdt > u0 ,
At · un tdt ≥
T
2.28
T
a contradiction. Therefore Ku1 /
φ and Ku2 /
φ. Now repeating the same procedure as
above, putting u1 and u2 instead of u4 and u5 , respectively, we get
k1 u1 t k2 u2 t kut
2.29
for μ-a.e, t ∈ T . Hence, by the fact that u1 , u2 , u ∈ SL0M X, we have k1 k2 k, and
consequently, u1 u2 u. Thus u is an extreme point of BL0M X.
Corollary 2.5 see 3. u ∈ SL0M R is an extreme point of BL0M R if and only if
a the set Ku consists of one element from 0, ∞;
b μ{t ∈ T : k|ut| ∈ Kt } 0, where k ∈ Ku.
Finally, we investigate the rotundity of L0M X.
Theorem 2.6. L0M X is rotund if and only if:
a for any u ∈ SL0M X, the set Ku consists of one element from 0, ∞;
b X is rotund;
c Mt, u is strictly convex with respect to u for almost all t ∈ T .
Proof. Sufficiency is obvious by Theorem 2.4.
Necessity. a is obvious by a of Theorem 2.4. L0M R is isometrically isomorphic to
closed subspace of L0M X, thus L0M R is rotund. By Lemma 2.3, c is obvious.
If b is not true, then there exist x, y, z ∈ SX with 2x y z and y /
z. Pick ht ∈
SL0M X, then there exists d > 0 such that μH > 0, where H {t ∈ T : ht ≥ d}. Since
ht ∈ SL0M X, then there exists k > 0 such that
M t, k ht dt ≤
H
M t, k ht dt < ∞.
2.30
T
Set
ut d · x · χH t,
vt d · y · χH t,
wt d · z · χH t.
2.31
Mt, khtdt < ∞.
2.32
We have
M t, k ut dt T
M t, k d dt ≤
H
H
Abstract and Applied Analysis
13
This implies that ut ∈ L0M X. Similarly, we have vt, wt ∈ L0M X. It is easy to see that
u0 v0 w0 . Then
u 0 v 0 w 0
1,
u0 u0 u0 u
0
u
v
1
1 w
·
·
.
2 u0 2 u0
2.33
However, ut vt wt for t ∈ T . By b of Theorem 2.4, we have u w. Hence
v/u0 w/u0 . So u is not an extreme point of BL0M X. Contradicting the rotundity of
L0M X.
Acknowledgment
The authors would like to thank the anonymous referees for some suggestions to improve
the paper. This work was supported by Heilong Jiang Natural Science Fund A200902.
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