# Extreme Points and Rotundity in Musielak-Orlicz-Bochner Function Spaces Endowed with Orlicz Norm

Authors: Shaoqiang Shang; Yunan Cui; Yongqiang Fu
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

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.


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## Body

## 1. Introduction

Let(X,∥·∥) be a real Banach space. S(X) and B(X) 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 pointx∈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 ExtB(X)=S(X), then X is said to be rotund. A point x∈S(X) is said to be strongly extreme point if for any {xn}n=1∞,{yn}n=1∞∈X with ∥xn∥→1,∥yn∥→1, and x=(1/2)(xn+yn), there holds ∥xn-yn∥→0(n→∞). If the set of all strongly extreme points of B(X) is equal to S(X), 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 S(X) and strongly extreme points of S(X) is proved. Namely, a Banach space X is midpoint local uniformly rotund if and only if every point of S(X) 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∈S(X) is a strongly extreme point in X, then κ(x) 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 Musielak-Orlicz-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)
forμ-a.e, t∈T, M(t,0)=0, limu→∞M(t,u)=∞ and M(t,u′)<∞ for some u′>0;(2)
forμ-a.e, t∈T, M(t,u) is convex on [0,∞) with respect to u;(3)
for eachu∈[0,∞),M(t,u) is a μ-measurable function of t on T.Letp(t,u) denote the right derivative of M(t,·) at u∈R+ (where if M(t,u)=∞, let p(t,u)=∞) and let q(t,·) be the generalized inverse function of p(t,·) defined on R+ by(1.1)q(t,v)=supu≥0{u≥0:p(t,u)≤v}.
Then N(t,v)=∫0vq(t,s)ds for any v∈R and μ-a.e. t∈T. It is well known that there holds the Young inequality uv≤M(t,u)+N(t,v) for μ-a.e. t∈T. And uv=M(t,u)+N(t,u)⇔u=q(t,v) or v=q(t,u). Let (1.2)e(t)=sup{u>0:M(t,u)=0},E(t)=sup{u>0:M(t,u)<∞}.For fixedt∈T and v≥0, if there exists ϵ∈(0,1) such that (1.3)M(t,v)=12M(t,v+ɛ)+12M(t,v-ɛ)<∞,
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 fixedt∈T, if Kt=Φ, then we call that M(t,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 (1.4)ρM(u)=∫TM(t,∥u(t)∥)dt.
Put(1.5)LM0(X)={u∈XT:ρM(λu)<∞forsomeλ>0}.
Then the Musielak-Orlicz-Bochner function space(1.6)∥u∥0=infk>01k[1+ρM(ku)]
is Banach space. If X=R, LM0(R) is said to be Musielak-Orlicz function space. Set(1.7)K(u)={k>0:1k(1+ρM(ku))=∥u∥0}.
In particular, the set K(u) can be nonempty. To show that, we give a proposition.Proposition 1.1.
Iflimu→∞(M(t,u)/u)=∞μ-a.e.  t∈T,  then K(u)≠ϕ  for any u∈LM0(X).Proof.
For anyu∈LM0(X), there exists a>0 such that μT0>0, where T0={t∈T:∥u(t)∥≥a}. It is easy to see that T0=∪n=1∞Gn, where
(1.8)Gn={t∈T0:M(t,v)v≥3∥u∥0a·μT0,v≥n}.
Noticing that G1⊂G2⊂⋯⊂Gn⊂⋯, then limn→∞μGn=μT0. Hence there exists n1 such that μGn1>(1/2)μT0. This means that if k>n1/a, we have
(1.9)1k[1+∫TM(t,k∥u(t)∥)dt]≥∫Gn1M(t,k∥u(t)∥)kdt≥∫Gn1M(t,ka)kdt=a∫Gn1M(t,ka)kadt≥a∫Gn1M(t,n1)n1dt≥a3∥u∥0a·μT0·μGn1>32∥u∥0.
This implies that if (1/kn)(1+ρM(knu))→∥u∥0(n→∞), then sequence {kn}n=1∞ 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
(1.10)limn→∞1kn[1+ρM(knu)]=limn→∞1kn+limn→∞∫TM(t,∥knu(t)∥)kndt=1k0+∫TM(t,∥k0u(t)∥)k0dt=1k0[1+ρM(k0u)].
If k1≥k2≥⋯kn≥⋯≥k0, by dominated convergence theorem, we have
(1.11)limn→∞1kn[1+ρM(knu)]=limn→∞1kn+limn→∞∫TM(t,∥knu(t)∥)kndt=1k0+∫TM(t,∥k0u(t)∥)k0dt=1k0[1+ρM(k0u)].
Therefore (1/kn)[1+ρM(knu)]→(1/k0)[1+ρM(k0u)](n→∞), namely, (1/k0)[1+ρM(k0u)]=∥u∥0. This implies k0∈K(u).


## 2. Main Results

In order to obtain the main theorems of this paper, we first give some lemmas.Lemma 2.1.
IfK(u)=ϕ, then ∥u∥0=∫TA(t)·∥u(t)∥dt, where A(t)=limu→∞(M(t,u)/u).Proof.
By proof of Proposition1.1, we know that if K(u)=ϕ, then there exists {kn}n=1∞ such that (1/kn)(1+ρM(knu))→∥u∥0 and kn→∞ as n→∞. Without loss of generality, we may assume that k1≤k2≤⋯kn≤⋯. By Levi theorem, we have
(2.1)∥u∥0=limn→∞1kn(1+∫TM(t,kn∥u(t)∥)dt)=limn→∞1kn(1+∫{t∈T:∥u(t)∥≠0}M(t,kn∥u(t)∥)dt)=limn→∞1kn+limn→∞∫{t∈T:∥u(t)∥≠0}M(t,kn∥u(t)∥)kn∥u(t)∥·∥u(t)∥dt=∫{t∈T:∥u(t)∥≠0}limn→∞M(t,kn∥u(t)∥)kn∥u(t)∥·∥u(t)∥dt=∫{t∈T:∥u(t)∥≠0}A(t)·∥u(t)∥dt=∫TA(t)·∥u(t)∥dt.
Hence the conclusion of the lemma is true.Lemma 2.2.
If the setK(u) consists of one element from (0,+∞), then ∥u∥0<∫TA(t)·∥u(t)∥dt.Proof.
Pickk1>k2>0; then we have
(2.2)ρM(k1uχEn)≥∫Tk1∥u(t)χEn(t)∥·p(t,∥k2u(t)χEn(t)∥)dt-ρN(p(k2uχEn)),ρM(k2uχEn)=∫Tk2∥u(t)χEn(t)∥·p(t,∥k2u(t)χEn(t)∥)dt-ρN(p(k2uχEn)),
where
(2.3)En={t∈T:k2∥u(t)∥≤n,p(t,∥k2u(t)∥)≤n}.
It follows that
(2.4)1k1[1+ρM(k1uχEn)]-1k2[1+ρM(k2uχEn)]=k1-k2k1k2(-1+k2k1-k2[ρM(k1uχEn)-ρM(k2uχEn)]-ρM(k2uχEn))≥k1-k2k1k2(-1+k2k1-k2∫T(k1-k2)∥u(t)χEn(t)∥·p(t,∥k2u(t)χEn(t)∥)dt-ρM(k2uχEn)∫T(k1-k2)∥u(t)χEn(t)∥·p(t,∥k2u(t)χEn(t)∥)dt)=k1-k2k1k2(ρN(p(k2uχEn))-1).
Let n→∞; then we obtain
(2.5)1k1[1+ρM(k1uχE)]≥k1-k2k1k2(ρN(p(k2uχE))-1)+1k2[1+ρM(k2uχE)],
where
(2.6)E={t∈T:p(t,∥k2u(t)∥)<∞}.
If p(t,∥k2u(t)∥)=∞, then M(t,k1∥u(t)∥)≥M(t,k2∥u(t)∥)=∞. Hence we have
(2.7)1k1[1+ρM(k1u)]≥k1-k2k1k2(ρN(p(k2u))-1)+1k2[1+ρM(k2u)].
Moreover, there exists k0∈R+ such that ρN(p(ku))≥1, whenever k≥k0. This means that function F(k)=(1/k)[1+ρM(ku)] is nondecreasing, when n is large enough. Pick sequence {kn}n=1∞ such that 0<k1<k2⋯<kn<⋯. By Levi theorem, we have
(2.8)∫TA(t)·∥u(t)∥dt=∫Tlimn→∞M(t,kn∥u(t)∥)kn∥u(t)∥·∥u(t)∥dt=limn→∞∫TM(t,kn∥u(t)∥)kn∥u(t)∥·∥u(t)∥dt=limn→∞1kn∫TM(t,kn∥u(t)∥)dt=limn→∞1kn[1+∫TM(t,kn∥u(t)∥)dt]>1l[1+∫TM(t,l∥u(t)∥)dt]=∥u∥0,
where {l}=K(u). Hence the conclusion of the lemma is true.Lemma 2.3 (see [3]).
LetLM0(R) be rotund, then M(t,u) is strictly convex with respect to u for almost all t∈T.Theorem 2.4.
LetLM0(X) be Musielak-Orlicz-Bochner function spaces, then u∈S(LM0(X)) is an extreme point of B(LM0(X)) if and only if (a)
the setK(u) consists of one element from (0,+∞);(b)
v,w∈XT with u=λv+(1-λ)w and ∥u(t)∥=∥v(t)∥=∥w(t)∥μ-a.e that on T implies v=w, where λ∈(0,1);(c)
μ{t∈T:k∥u(t)∥∈Kt}=0, where k∈K(u).Proof.
Necessity. (a1) Suppose that u is an extreme point of the unit ball B(LM0) and K(u)=ϕ, then ∥u∥0=∫TA(t)·∥u(t)∥dt by Lemma 2.1. Decompose T into T1 and T2 such that ∫T1A(t)·∥u(t)∥dt=∫T2A(t)·∥u(t)∥dt. Pick ɛ∈(0,1). Put
(2.9)u1(t)=(u(t)+ɛu(t))χT1+(u(t)-ɛu(t))χT2,u2(t)=(u(t)-ɛu(t))χT1+(u(t)+ɛu(t))χT2.
Obviously, u=(1/2)(u1+u2) and u1≠u2. Moreover, we have
(2.10)∥u1∥0≤∫TA(t)·∥u1(t)∥dt=∫T1A(t)·∥u(t)+ɛu(t)∥dt+∫T2A(t)·∥u(t)-ɛu(t)∥dt=∫T1A(t)·∥u(t)∥dt+∫T1ɛA(t)·∥u(t)∥dt+∫T2A(t)·∥u(t)∥dt-∫T2ɛA(t)·∥u(t)∥dt=∫T1A(t)·∥u(t)∥dt+∫T2A(t)·∥u(t)∥dt=∫TA(t)·∥u(t)∥dt=1.
Similarly, we have ∥u2∥0≤1. Hence u1,u1∈B(LM0(X)). Therefore u∈S(LM0(X)) is not an extreme point of B(LM0(X)), a contradiction. Hence K(u)≠ϕ. Suppose that ∥u∥0=∫TA(t)·∥u(t)∥dt. Similarly, we get a contradiction.
The necessity of (b) is obvious.
(c) Set(2.11)H1={t∈T:u(t)=0,e(t)>0},H2={t∈T:2M(t,k∥u(t)∥)=M(t,k∥u(t)∥+ɛ)+M(t,k∥u(t)∥-ɛ),∥u(t)∥≠0},
where {k}∈K(u). Suppose that (c) does not hold. Then μH1>0 or μH2>0.
IfμH1>0, then for any x∈S(X), by setting
(2.12)(v(t),w(t))={(12ke(t)x,-12ke(t)x),t∈H1,(u(t),u(t)),t∈T∖H1
we have u≠w and u=(1/2)(v+w). Moreover, we have
(2.13)∥v∥0≤1k(1+ρM(ku))=1k(1+∫H1M(t,k∥12ke(t)∥)dt+∫T∖H1M(t,k∥u(t)∥)dt)=1k(1+∫T∖H1M(t,k∥u(t)∥)dt)≤∥u∥0.
Similarly, we have∥w∥0≤1. Hence v,w∈B(LM0(X)). Therefore u∈S(LM0(X)) is not an extreme point of B(LM0(X)), a contradiction.
IfμH2>0, it is easy to see that H2⊂∪n=1∞{t∈T:∥u(t)∥≠0,M(t,k∥u(t)∥)=(1/2)M(t,(1+1/n)k∥u(t)∥)+(1/2)M(t,(1-1/n)k∥u(t)∥)}, where {k}∈K(u). Then there exists n0∈N such that
(2.14)H={1n0t∈T:u(t)≠0,M(t,k∥u(t)∥)=12M(t,(1+1n0)k∥u(t)∥)+12M(t,(1-1n0)k∥u(t)∥)<∞}
is not a noll set. Decompose H into E and F such that ∫Ep(t,(1/n0)k∥u(t)∥)dt=∫Fp(t,(1/n0)k∥u(t)∥)dt. Define
(2.15)(v(t),w(t))={((1+1n0)u(t),(1-1n0)u(t)),t∈E,((1-1n0)u(t),(1+1n0)u(t)),t∈F,(u(t),u(t)),t∈T∖(E∪F).
Then u≠w and u=(1/2)(v+w). Furthermore, we have
(2.16)∥v∥0≤1k(1+ρM(kv))=1k(1+∫T∖(E∪F)M(t,k∥u(t)∥)dt+∫EM(t,(1+1n0)k∥u(t)∥)dt+∫FM(t,(1-1n0)k∥u(t)∥)dt)=1k(1+∫T∖(E∪F)M(t,k∥u(t)∥)dt+∫EM(t,k∥u(t)∥)dt+∫Ep(t,1n0k∥u(t)∥)dt+∫FM(t,k∥u(t)∥)dt-∫Fp(t,1n0k∥u(t)∥)dt)=1k(1+∫T∖(E∪F)M(t,k∥u(t)∥)dt+∫EM(t,k∥u(t)∥)dt+∫FM(t,k∥u(t)∥)dt)=1k(1+ρM(ku))=∥u∥0=1.
Similarly, we have ∥w∥0≤1. Hence v,w∈B(LM0(X)). Therefore u∈S(LM0(X)) is not an extreme point, a contradiction. Hence (c) is true.
(a2) IfK(u)≠ϕ and u∈S(LM0(X)) is an extreme point, suppose that there exists k1,k2∈K(u) satisfying k1≠k2. Definek=k1k2/(k1+k2),
(2.17)2=∥u∥0+∥u∥0=k1+k2k1k2[1+k2k1+k2ρM(k1u)+k1k1+k2ρM(k2u)]=k1+k2k1k2[1+k2k1+k2∫TM(t,∥k1u(t)∥)dt+k1k1+k2∫TM(t,∥k2u(t)∥)dt]≥k1+k2k1k2[1+∫TM(t,k2k1+k2∥k1u(t)∥+k1k1+k2∥k2u(t)∥)dt]=k1+k2k1k2[1+∫TM(t,∥2k1k2k1+k2u(t)∥)dt]=212k[1+ρM(2ku)]≥2∥u∥0=2.
This implies that
(2.18)∥u∥0=12k[1+ρM(2ku)]
(i.e., 2k∈K(u)),
(2.19)k2k1+k2M(t,k1∥u(t)∥)+k1k1+k2M(t,k2∥u(t)∥)=M(t,2k∥u(t)∥).
Since k1∥u(t)∥≠k2∥u(t)∥ on {t∈T:∥u(t)∥≠0}, then 2k∥u(t)∥∈Kt on {t∈T:∥u(t)∥≠0}, a contradiction. Therefore (a) is true.
Sufficiency. We first prove that for u,u1,u2∈S(LM0(X)) with u=(1/2)(u1+u2) at least one of the sets K(u1) or K(u2) is nonempty. Suppose that K(u1)=ϕ and K(u2)=ϕ. Hence we have
(2.20)1=∥12(u1+u2)∥0<∫TA(t)·∥12(u1(t)+u2(t))∥dt≤12∫TA(t)·∥u1(t)∥dt+12∫TA(t)·∥u2(t)∥dt=12∥u1∥0+12∥u2∥0=1,
a contradiction. This contradiction shows that K(u1)≠ϕ or K(u2)≠ϕ.
Now we will prove thatK(u1)≠ϕ and K(u2)≠ϕ. Otherwise, we can assume without loss of generality that K(u1)≠ϕ and K(u2)=ϕ. Put
(2.21)[u1,u)={(1-λ)u1+λu:0<λ<1},(u,u2]={(1-λ)u+λu2:0<λ<1}.
Next we will prove that K(y)≠ϕ for all y∈[u1,u) and K(y)=ϕ for all y∈(u,u2]. Assume first for the contrary that this is u3∈[u1,u) such that K(u3)=ϕ. Then there exists λ3∈[0,1) such that u3=(1-λ3)u1+λ3u. Since u1=2u-u2, we have
(2.22)u3=(1-λ3)(2u-u2)+λ3u=(2-λ3)u-(1-λ3)u2.
Hence u=(1/(2-λ3))u3+((1-λ3)/(2-λ3))u2. Therefore
(2.23)1=∥u∥0<∫TA(t)·∥12-λ3u3(t)+1-λ32-λ3u2(t)∥dt≤12-λ3∫TA(t)·∥u3(t)∥dt+1-λ32-λ3∫TA(t)·∥u2(t)∥dt=12-λ3∥u3∥0+1-λ32-λ3∥u2∥0=1,
a contradiction.
Assume now for the contrary that this isu4∈(u,u2] such that K(u4)≠ϕ. We can find u5∈(u,u2] such that u=(1/2)(u4+u5) and u4≠u5. Therefore there are k4≥1 and k5≥1 such that(2.24)∥u4∥0=1k4[1+ρM(k4u4)],∥u5∥0=1k5[1+ρM(k5u5)].
By the convexity of the modularρM we have
(2.25)ρM(2k4k5k4+k5u)=ρM(k4k5k4+k5(u4+u5))=ρM(k5k4+k5k4u4+k4k4+k5k5u5)=∫TM(t,∥k5k4+k5k4u4(t)+k4k4+k5k5u5(t)∥)dt≤∫TM(t,k5k4+k5∥k4u4(t)∥+k4k4+k5∥k5u5(t)∥)dt≤k5k4+k5∫TM(t,∥k4u4(t)∥)dt+k4k4+k5∫TM(t,∥k5u5(t)∥)dt=k5k4+k5ρM(k4u4)+k4k4+k5ρM(k5u5).
Hence
(2.26)2=2∥u∥0≤k4+k5k4k5(1+ρM(k4k5k4+k52u))≤k4+k5k4k5(1+k5k4+k5ρM(k4u4)+k4k4+k5ρM(k5u5))≤1k4(1+ρM(k4u4))+1k5(1+ρM(k5u5))=2.
Consequently, all inequalities from the last three lines are equalities in fact. Therefore 2(k4k5/(k4+k5))=k and
(2.27)M(t,∥ku(t)∥)=k5k4+k5M(t,∥k4u4(t)∥)+k4k4+k5M(t,∥k5u5(t)∥)
for μ-a.e  t∈T. By μ{t∈T:k∥u(t)∥∈Kt}=0, it follows that ∥ku(t)∥=∥k4u4(t)∥=∥k5u5(t)∥ for μ-a.e  t∈T. And we have ku(t)=(k5/(k4+k5))k4u4(t)+(k4/(k4+k5))k5u5(t) for μ-a.e  t∈T. By (b), we have ku=k4u4=k5u5. Since u4,u5,u∈S(LM0(X)), we get k4=k5=k, which gives u4=u5=u. This contradicts the inequality u4≠u5. Thus K(y)=ϕ for any y∈(u,u2]. Take un=(1-1/n)u+(1/n)u2 for all n∈N. Then un∈(u,u2] for all n∈N. Hence K(un)=ϕ, and consequently ∥un∥0=∫TA(t)·∥un(t)∥dt for all n∈N. Note that ∥un-u∥0→0(n→∞) and limn→∞∥un(t)∥=∥u(t)∥ for μ-a.e,t∈T. Since K(u)={k}, with 0<k<∞, we have
(2.28)limn→∞∥un∥0=∥u∥0=limn→∞∫TA(t)·∥un(t)∥dt≥∫TA(t)·∥u(t)∥dt>∥u∥0,
a contradiction. Therefore K(u1)≠ϕ and K(u2)≠ϕ. Now repeating the same procedure as above, putting u1 and u2 instead of u4 and u5, respectively, we get
(2.29)k1u1(t)=k2u2(t)=ku(t)
for μ-a.e,t∈T. Hence, by the fact that u1,u2,u∈S(LM0(X)), we have k1=k2=k, and consequently, u1=u2=u. Thus u is an extreme point of B(LM0(X)).Corollary 2.5 (see [3]).
u∈S(LM0(R)) is an extreme point of B(LM0(R)) if and only if (a)
the setK(u) consists of one element from (0,+∞);(b)
μ{t∈T:k|u(t)|∈Kt}=0, where k∈K(u).Finally, we investigate the rotundity ofLM0(X).Theorem 2.6.
LM0(X) is rotund if and only if: (a)
for anyu∈S(LM0(X)), the set K(u) consists of one element from (0,+∞);(b)
X is rotund;(c)
M(t,u) is strictly convex with respect to u for almost all t∈T.Proof.
Sufficiency is obvious by Theorem2.4.
Necessity. (a) is obvious by (a) of Theorem2.4. LM0(R) is isometrically isomorphic to closed subspace of LM0(X), thus LM0(R) is rotund. By Lemma 2.3, (c) is obvious.
If (b) is not true, then there existx,y,z∈S(X) with 2x=y+z and y≠z. Pick h(t)∈S(LM0(X)), then there exists d>0 such that μH>0, where H={t∈T:∥h(t)∥≥d}. Since h(t)∈S(LM0(X)), then there exists k′>0 such that(2.30)∫HM(t,k′∥h(t)∥)dt≤∫TM(t,k′∥h(t)∥)dt<∞.
Set
(2.31)u(t)=d·x·χH(t),v(t)=d·y·χH(t),w(t)=d·z·χH(t).
We have
(2.32)∫TM(t,k′∥u(t)∥)dt=∫HM(t,k′d)dt≤∫HM(t,k∥h(t)∥)dt<∞.
This implies that u(t)∈LM0(X). Similarly, we have v(t),w(t)∈LM0(X). It is easy to see that ∥u∥0=∥v∥0=∥w∥0. Then
(2.33)∥u∥u∥0∥0=∥v∥u∥0∥0=∥w∥u∥0∥0=1,u∥u∥0=12·v∥u∥0+12·w∥u∥0.
However, ∥u(t)∥=∥v(t)∥=∥w(t)∥ for t∈T. By (b) of Theorem 2.4, we have u=w. Hence v/∥u∥0=w/∥u∥0. So u is not an extreme point of B(LM0(X)). Contradicting the rotundity of LM0(X).



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*Source: 914183-2010-09-20.xml*