# Strongly Extreme Points and Middle Point Locally Uniformly Convex in Orlicz Spaces Equipped with s-Norm

Authors: Yunan Cui; Yujia Zhan
Journal: Journal of Function Spaces
 (2019)

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## Abstract

As is well known, the extreme points and strongly extreme points play important roles in Banach spaces. In this paper, the criterion for strongly extreme points in Orlicz spaces equipped with s-norm is given. We complete solved criterion-Orlicz space that generated by Orlicz function. And the sufficient and necessary conditions for middle point locally uniformly convex in Orlicz spaces equipped with s-norm are obtained.


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## Body

## 1. Introduction

The extreme point set plays a crucial role in function analysis, convex analysis, and optimization. In fact, any compact convex set is the convex hull of its extreme point set, and the result is called Krein-Milman theorem. The notion of a dentable subset of a Banach space was introduced by Rieffel [1] in conjunction with a Radon-Nikodym theorem for Banach space-valued measures. Subsequent work by Maynard [2] and by Davis and Phelps [3] has shown that those Banach spaces in which Rieffels Radon-Nikodym theorem is valid are precisely the ones in which every bounded closed convex set is dentable. This is a real breakthrough in studying the nature of Radon-Nikodym as a geometric property. In 1988, Bor-Luh Lin, Pei-Kee Lin, and S. L. Troyanski [4] described the characteristic of denting points and obtained the notion that there is a close relationship between denting points and strongly extreme points. It is easy to see that every denting point of K is a strongly extreme point of K and it is known [Ken Kunen and Haskeil Rosenthal, Martingale proofs of some geometric results in Banach space theory, Pacific J. Math. 100 (1982), 153-175] that every strongly extreme point of K is a weak∗-extreme point of K. Orlicz space is a special kind of Banach space; it was introduced by the famous Polish mathematician W. Orlicz in 1932. The theory of Orlicz space [5, 6] has been greatly developed because of its important theoretical properties and application value. Up to now, the criterion of an element in the unit sphere of Orlicz spaces equipped with the Orlicz norm [5, 7], the Luxemburg norm [5], and p-Amemiya norm [8] has been given. In this paper, we use a new technique to study the strongly extreme point in Orlicz spaces generated by Orlicz function and equipped with a new norm, namely, s-norm. The criterion of strongly extreme points is given, and the sufficient and necessary conditions for middle point locally uniformly convex in Orlicz spaces equipped with s-norm are obtained.


## 2. Preliminaries

Throughout this paper,X will denote a Banach space and X∗ stands for the dual space of X. We denote by (G,Σ,μ) the nonatomic Σ-measure finite space. By B(X) and S(X) we denote the unit ball and the unit sphere of X, respectively. By R and N we will denote the sets of real and natural numbers, respectively.A mappingΦ:R→[0,∞) is said to be an Orlicz function if it is even, continuous, convex, and Φ(0)=0, limu→∞⁡Φ(u)=∞. Moreover, if Φ satisfies limu→0⁡Φ(u)/u=0 and limu→∞⁡Φ(u)/u=∞, Φ is called N-function. Let p+(t) be the right-hand derivative of Φ, where the function Ψ is defined by the formula (1)Ψu=sup⁡uv-Φv:v≥0 and called complementary function to Φ in the sense of Young.We say that an Orlicz functionΦ satisfies △2-condition for large u∈R  (Φ∈△2 for short) if there exist u0>0 and K>2 such that (2)Φ2u≤KΦu whenever u>u0.LetL0 denote the set of all measure real functions on G. For a given Orlicz function Φ we define on L0 a convex function IΦ:L0→[0,∞] (called a pseudomodular; see [6]) by (3)IΦx=∫GΦxtdt. It is clear that IΦ(x)=∫supp(x)Φ(x(t))dt; here supp(x)=t∈G:xt≠0.The Orlicz spaceLΦ generated by an Orlicz function Φ is defined by the formula (4)LΦ=x∈L0:IΦλx<+∞for  someλ>0, and its subspace EΦ is defined by (5)EΦ=x∈L0:IΦλx<+∞for  allλ>0.This space is usually equipped with the Luxemburg norm [5] (6)x=inf⁡k>0:IΦxk≤1, or with the Orlicz norm (Amemiya norm) [5] (7)xΦo=infk>0⁡1k1+IΦkx.A functions:0,∞→1,∞ will be called an outer function, if it is convex and (8)max1,u≤su≤1+ufor  allu≥0.In 2017, M.Wisła introduceds-norm.Definition 1.
Lets be an outer function. Denote s-norm on Orlicz spaces by the formula(9)xΦ,s=infk>0⁡1ksIΦkx.
It is easy to getxΦ,s=x if s(u)=max1,u and xΦ,s=xΦo if s(u)=1+u ([8]). Then we have x≤xΦ,s≤xΦo.In this paper, byLΦ,s we will denote an Orlicz space equipped with the s-norm.Definition 2.
Lets+′(u) be the right-hand derivative of s. For all 0≤v≤1, define (10)ωv=∫0vs+′-1tdtDefinition 3.
Lets be an outer function. For all 0≤u<∞ and 0≤v<∞, (11)βsu,v=1-ωs+′u-vs+′u, the function βs(u,v) is nonincreasing. For any x∈LΦ,s∖0, define ([9]) (12)k∗x=inf⁡k>0:βsIΦkx,IΨp+kx≤0,k∗∗x=inf⁡k>0:βsIΦkx,IΨp+kx≥0.
LetK(x)=[k∗(x),k∗∗(x)]. Then xΦ,s=1/ks(IΦ(kx)) if and only if K(x)≠∅.Definition 4.
A pointx∈S(X) is said to be an extreme point of B(X) if for any y,z∈B(X) with x=y+z/2, then implies y=z.
The set of all extreme points of the unit ballB(X) will be denoted by ExtB(X). X is said to be strictly convex if ExtB(X)=S(X).Definition 5.
A pointx∈S(X) is said to be a strongly extreme point of B(X) if for any xn⊆X,yn⊆X with xn→1, yn→1 and xn+yn/2=x there holds xn-yn→0 as n→∞.
It is obvious that a strongly extreme point is an extreme point.X is said a middle point locally uniformly convex Banach space if and only if each point on S(X) is a strongly extreme point.Definition 6.
Letu0>0. If for every v,w∈R such that v≠w and v+w/2=u0, we have Φ(u0)<1/2Φ(v)+1/2Φ(w), then u0 is called to be a strictly convex point of Φ(u). The set of all strictly convex points of Φ(u) will be denoted by SΦ.
For the results concerning strongly extreme points and convexities in Orlicz spaces which are generated byN-function and equipped with the Orlicz norm, the Luxemburg norm, and p-Amemiya norm, we refer a reader to [10–17].


## 3. Main Theorem

Lemma 7.
(1) Iflimu→∞Φ(u)/u=∞ then K(x)≠∅ for any x∈LΦ,s∖0;
(2) Iflimu→∞Φ(u)/u=A<∞ and K(x)=∅ then xΦ,s=Ax1.Proof.
(1) Supposelimu→∞Φ(u)/u=∞. We have limk→∞IΨ(p+(kx))=∞. Since for any 0≤v≤1, ω(v)∈[0,1], then (13)limk→∞⁡βsIΦkx,IΨp+kx=limk→∞⁡1-ωs+′IΦkx-IΨp+kxs+′IΦkx<0. So k∗(x)<∞, whence K(x)≠∅.
(2) ByK(x)=∅, we have k∗(x)=∞, and then (14)xΦ,s=infk>0⁡1ksIΦkx=limk→∞⁡1ksIΦkx≤limk→∞⁡1k1+IΦkx=limk→∞⁡1k+∫suppxΦkxtkxtxtdt=Ax1, and (15)xΦ,s=limk→∞⁡1ksIΦkx≥limk→∞⁡1kIΦkx=limk→∞⁡∫suppxΦkxtkxtxtdt=Ax1.Therefore xΦ,s=Ax1.Corollary 8.
K ( x ) = ∅ if and only if μ(supp(x))<1-ω(1)/Ψ(A) for any x∈LΦ,s∖0.Proof.
Necessity. We know that IΦ(kx)→∞ as k→∞. By IΦ(kx)≤s(IΦ(kx))≤1+IΦ(kx), we can get 1≤limk→∞s+′(IΦ(kx))≤1. That is, limk→∞s+′(IΦ(kx))=1. By K(x)=∅, we have k∗(x)=∞. Then (16)βsIΦkx,IΨp+kx>0, for all k>0. Since βs(u,v) is nonincreasing, we have (17)limk→∞⁡βsIΦkx,IΨp+kx=limk→∞⁡1-ωs+′IΦkx-IΨp+kxs+′IΦkx=1-ω1-suppxΨA>0, whence μ(supp(x))<1-ω(1)/Ψ(A).
Here we infer thatω(1)<1. If ω(1)=1 we have limk→∞⁡βs(IΦ(kx),IΨ(p+(kx)))=-supp(x)Ψ(A)<0, a contradiction.
Sufficiency. By the definitions of s(u) and ω(v), s+′(u)≤1 and ω(s+′(u))≤ω(1) for any u>0. Therefore for all k>0(18)βsIΦkx,IΨp+kx=1-ωs+′IΦkx-IΨp+kxs+′IΦkx≥1-ω1-IΨp+kx=1-ω1-μsuppxΨA>0, whence k∗(x)=∞, i.e., K(x)=∅.Theorem 9.
Suppose thats(u)>1 when u>0 and Φ is an Orlicz function. A point x0∈S(LΦ,s) is a strongly extreme point if and only if Φ∈△2 and k0x0(t)∈SΦ for k0∈K(x0).Proof.
Necessity. As we know that a strongly extreme point is an extreme point, we only need to prove that x0∈ExtB(LΦ,s) implies k0x0(t)∈SΦ for k0∈K(x0). Firstly, we will prove that if x0∈ExtB(LΦ,s), then K(x0)≠∅. If K(x0)=∅, we will have k∗(x0)=∞ which implies that μ(supp(x0))<1-ω(1)/Ψ(A) holds. There exists a>0 such that μ(t∈G:x0t>a)>0. Put C=t∈G:x0t>a and 0<μ(C)<1-ω(1)/Ψ(A). Divide C into two sets C1 and C2 with C1∩C2=∅ and μ(C1)=μ(C2). Take ε∈(0,a) and define (19)yt=x0t,t∈G∖C1∪C2x0t-ε,t∈C1x0t+ε,t∈C2,zt=x0t,t∈G∖C1∪C2x0t+ε,t∈C1x0t-ε,t∈C2. Then x0=(y+z)/2, y≠z. Moreover supp(y)⊆supp(x0), supp(z)⊆supp(x0). We have (20)yΦ,s=Ay1=A∫Gytdt=A∫C1x0t-εdt+∫C2x0t+εdt+∫G∖C1∪C2x0tdt=A∫Gx0tdt=Ax01=x0Φ,s=1.
Similarly, we can getzΦ,s=1.
Next we will show thatk0x0(t)∈SΦ.
Suppose thatμ(t∈G:k0x0t∉SΦ)>0 for k0∈K(x0). There exists an interval (a,b) such that μ(t∈G:a/k0+ε<x0t<b/k0-ε)>0(ε>0), and Φ is affine on (a,b): Φ(x)=px+q. Divide t∈G:a/k0+ε<x0t<b/k0-ε into two sets E and F with E∩F=∅ and μ(E)=μ(F). Define (21)yt=x0t,t∈G∖E∪Fx0t-ε,t∈Ex0t+ε,t∈F,zt=x0t,t∈G∖E∪Fx0t+ε,t∈Ex0t-ε,t∈F.Then x0=y+z/2, y≠z. Thus (22)IΦk0y=∫E∪FΦk0ytdt+∫G∖E∪FΦk0ytdt=∫Epk0x0t-ε+qdt+∫Fpk0x0t+ε+qdt+∫G∖E∪FΦk0x0tdt=∫E∪Fpk0x0t+qdt+∫G∖E∪FΦk0x0tdt=∫E∪FΦk0x0tdt+∫G∖E∪FΦk0x0tdt=IΦk0x0, whence yΦ,s≤1/k0s(IΦ(k0y))=1/k0s(IΦ(k0x0))=x0Φ,s=1. In the same way, we can prove zΦ,s≤1. This contradicts the fact that x0 is an extreme point of B(LΦ,s).
In order to complete this proof, we need to prove that ifΦ∉△2 there are no strongly extreme points in S(LΦ,s).
SupposeΦ∉△2. Then limu→∞Φ(u)/u=+∞.
In fact, iflimu→∞Φ(u)/u=A<+∞, there exists u0>0 such that A/2u<Φ(u)<3A/2u holds for every u>u0. Then we have Φ(2u)<3/2A(2u)<6A/2u≤6Φ(u); it implies Φ∈△2, a contradiction.
For anyx0∈S(LΦ,s), there exists k0>0 such that (23)1=x0Φ,s=1k0sIΦk0x0. Since x0∈S(LΦ,s), we can find d>0 such that μ(t∈G:x0t≤d)>0. By Φ∉△2, there exist un>0 and un↑∞ such that Φ(2un)>2nΦ(un)(n=1,2,⋯). We may assume that 1/Φ(u1)<μ(t∈G:x0t≤d). Take Gn⊂t∈G:x0t≤d with Gm∩Gn=∅ for any m≠n, satisfying μ(Gn)=1/2nΦ(un)(n=1,2,⋯). Define (24)xnt=x0t,t∈G∖Gnx0t+unk0,t∈Gn,ynt=x0t,t∈G∖Gnx0t-unk0,t∈Gn. Then x0=xn+yn/2, xnt=xn′t+xn′′(t), here xn′(t)=x0χG∖Gn(t)+un/k0χGn(t), xn′′(t)=x0χGn(t).
Notice that(25)xn′′Φ,s=x0χGnΦ,s≤dχGnΦ,s→0n→∞. We have xn′Φ,s≥x0χG∖GnΦ,s≥x0Φ,s-x0χGnΦ,s, that is, lim_n→∞⁡xn′Φ,s≥x0Φ,s=1. And (26)xn′Φ,s=infk>01ksIΦkxn′≤1k0sIΦk0xn′=1k0s∫GΦk0x0χG∖Gnt+unk0χGntdt=1k0s∫G∖GnΦk0x0χG∖Gntdt+∫GnΦunχGntdt≤1k0sIΦk0x0+ΦunμGn=1k0sIΦk0x0+12n. Consequently, lim¯n→∞⁡xn′Φ,s≤x0Φ,s=1. Hence limn→∞⁡xnΦ,s=1. In the same way, we have limn→∞⁡ynΦ,s=1. But IΦ(k0(xn-yn))=∫GnΦk02unt/k0dt=Φ(2un)μ(Gn)≥1(n=1,2,⋯), which implies xn-ynΦ,s=1/k02unχGnΦ,s≥1/k02unχGn≥1/k0, a contradiction.
Sufficiency. Let Φ∈△2, x0∈S(LΦ,s) with k0x0(t)∈SΦ for k0∈K(x0). Take any xn,yn∈LΦ,s with xnΦ,s→1,ynΦ,s→1 and xn+yn=2x0.
Take sequenceskn and hn of positive numbers such that (27)xnΦ,s≥1knsIΦknxn-1n,ynΦ,s≥1hnsIΦhnyn-1n.
Define(28)xn′=xn+x02andyn′=yn+x02, then xn′+yn′=2x0 and lim¯n→∞⁡xn′Φ,s≤1,lim¯n→∞⁡yn′Φ,s≤1.
Now we will prove thatlimn→∞⁡xn′Φ,s=1 and limn→∞⁡yn′Φ,s=1. Otherwise, we can assume that limn→∞⁡xn′Φ,s<1; there exist δ>0 and n0∈N such that, for every n≥n0, (29)xn′Φ,s≤1-δ,yn′Φ,s≤1+δ2.
Then(30)1=x0Φ,s=xn′+yn′2Φ,s≤121-δ+1+δ2<1, a contradiction.
Sincexn′-yn′Φ,s→0 if and only if xn-ynΦ,s→0(n→∞), we will consider the sequences xn′ and yn′, where xn′ and yn′ in place of xn and yn.
Putkn′=2knk0/kn+k0 and hn′=2hnk0/hn+k0. Then kn′ and hn′ are bounded. Since xn′Φ,s→1(n→∞), we have (31)1←xn′Φ,s≤1kn′s(IΦkn′xn′=kn+k02knk0s(IΦknk0kn+k0xn+x0≤121k0sIΦk0x0+1knsIΦknxn≤12x0Φ,s+xnΦ,s+1n→1n→∞, whence it follows that (32)1kn′sIΦkn′xn′→1n→∞.
Analogously,(33)1hn′sIΦhn′yn′→1n→∞.
Putd=supnkn′,hn′<∞. Assume that kn′→k and hn′→h as n→∞. Now we prove k,h≥1. Since (34)1←1kn′sIΦkn′xn′n→∞, then (35)sIΦkn′xn′→kn→∞, and if k<1, consequently, s(IΦ(kn′xn′))<1 as n→∞, a contradiction. Thus k≥1. Similarly, h≥1. Then we have k/k+h,h/k+h∈[1/1+d,d/1+d].
Step 1. We will show that k0=2kh/k+h. In fact (36)1=x0Φ,s=1k0sIΦk0x0≤kn′+hn′2kn′hn′sIΦ2kn′hn′kn′+hn′x0≤kn′+hn′2kn′hn′sIΦkn′hn′kn′+hn′xn′+yn′≤kn′+hn′2kn′hn′sIΦhn′kn′+hn′kn′xn′+IΦkn′kn′+hn′hn′yn′≤121kn′sIΦkn′xn′+1hn′sIΦhn′yn′→1n→∞, whence 2kn′hn′/kn′+hn′→2kh/k+h=k0∈K(x0) as n→∞.
Step 2.We will show that kn′xn′-k0x0→μ0(n→∞).
Firstly, we will prove thatkxn′-hyn′→μ0(n→∞). Otherwise, there exist σ0,ε0>0 such that μ(t∈G:kxn′t-hyn′t≥σ0)≥ε0. Let D=Φ-1(3/ε0) and D1=2kD. Let Gn=t∈G:kxn′t≤D1,hyn′t≤D1,kxn′t-hyn′t≥σ0. It can be easy to calculate that μ(Gn)>ε0/3. In fact, since limn→∞⁡xn′Φ,s=1, xn′ is bounded in norm. Without loss of generality, we may assume that 2≥xn′Φ,s≥xn′; then (37)1≥1Φxn′2>∫t∈G:xn′t/2>DΦxn′t2dt>ΦDμt∈G:xn′t2>D=3ε0μt∈G:xn′t2>D, whence μ(t∈G:xn′t/2>D)<ε0/3. We have μ(t∈G:kxn′t>D1)<ε0/3. Hence (38)μGn≥μt∈G:kxn′t-hyn′t≥σ0-μt∈G:kxn′t>D1-μt∈G:hyn′t>D1>ε0-ε03-ε03=ε03. We know that SΦ is a close set. Let (39)F=x,y:x≤D1,y≤D1,x-y)≥σ0,hk+hx+kk+hy∈SΦ.F is a bounded close set. For every (x,y)∈F, the continuous function is(40)Φh/k+hx+k/k+hyh/k+hΦx+k/k+hΦy<1. Set maximum value equal to 1-δ(δ>0). For every (x,y)∈F, we have (41)Φhk+hx+kk+hy≤1-δhk+hΦx+kk+hΦy. Since k0x0(t)∈SΦ, we have (42)hk+hkxn′t+kk+hhyn′t=2khk+hx0t=k0x0∈SΦ, for t∈G. Therefore, (kxn′(t),hyn′(t))∈F, i.e., for t∈Gn, and (43)Φhk+hkxn′t+kk+hhyn′t≤1-δhk+hΦkxn′t+kk+hΦhyn′t. Hence (44)xn′+yn′Φ,s≤k+hkhsIΦkhk+hxn′+yn′≤k+hkhs∫GΦkhk+hxn′t+yn′tdt≤k+hkhs1-δ∫Gnhk+hΦkxn′t+kk+hΦhyn′tdt+∫G∖Gnhk+hΦkxn′t+kk+hΦhyn′tdt=k+hkhs∫Ghk+hΦkxn′t+kk+hΦhyn′tdt-δ∫Gnhk+hΦkxn′t+kk+hΦhyn′tdt≤1ksIΦkxn′+1hsIΦhyn′-k+hkhsδ∫Gnhk+hΦkxn′t+kk+hΦhyn′tdt-1. Notice that(45)IΦk-kn′xn′≤k-kn′IΦxn′→0n→∞. Since Φ∈△2, there holds (46)IΦkxn′-IΦkn′xn′=IΦkn′xn′+k-kn′xn′-IΦkn′xn′→0n→∞. Thus (47)0≤1ksIΦkxn′-xn′Φ,s≤1ksIΦkxn′-1kn′sIΦkn′xn′+1n→0n→∞. Similarly, we can get 1/hs(IΦ(hyn′))-yn′Φ,s→0(n→∞). Then xn′+yn′Φ,s≤2-k+h/kh(s2δ/1+dΦσ0/2ε0/3-1) as n→∞. By s(u)>1 when u>0, we have limn→∞⁡xn′+yn′Φ,s<2. The contradiction shows that kxn′-hyn′→μ0.
Since s-norm is equivalent with the Luxemburg norm, their weak topology and weak star topology are all equivalent. ThenLΦ,s is w∗ compact. Take xn′′⊂xn′, yn′′⊂yn′ such that xn′′→w∗x′ and yn′′→w∗y′. We can get x′+y′=2x0. By (48)xΦ,s=sup∫Gxtytdt:y∈BLΦ,s∗, where B(LΦ,s∗) represents the unit ball of the dual space of EΦ,s, we have y∈LΨ and (49)xΦ,s≥sup∫Gxtytdt:y∈BEΦ,s∗. For any ε>0, there exists y∈B(LΦ,s∗) such that (50)xΦ,s-ε≤∫Gxtytdt. Put (51)ynt=yt,yt≤n0,yt>n. Then yn(t)∈B(EΦ,s∗) and (52)∫Gxtytdt=limn→∞⁡∫Gxtyntdt. By the definition of “lim”, for any ε>0, there exists n0∈N such that (53)∫Gxtytdt-ε≤∫Gxtyntdt, whenever n≥n0.
Thus(54)xΦ,s-2ε≤∫Gxtyntdt. By arbitrariness of ε and combining with the above proof, we can obtain (55)xΦ,s=sup∫Gxtytdt:y∈BEΦ,s∗. Therefore (56)2=2x0Φ,s≤x′Φ,s+y′Φ,s≤lim_n→∞⁡xn′Φ,s+lim_n→∞⁡yn′Φ,s=2. This shows x′Φ,s=y′Φ,s=1.
As we know2x0=xn′+yn′; then k(2x0-yn′)-hyn′→μ0. It implies that k(2x0-yn′)-hyn′→w∗0. Combining with the proof above yn′→w∗y′ and y′Φ,s=1, we have y′=2k/k+hx0. As a result, 2k=k+h. So k=h. We have xn′-yn′→μ0 as n→∞. Namely, (57)2xn′-x0=xn′-yn′→μ0n→∞. By the proof above, we get 1<k0=k; thus (58)kn′xn′-k0x0→μ0n→∞.
Step 3. We will show that IΦ(kn′xn′)→IΦ(k0x0). In fact (59)sIΦk0x0=k0,sIΦkn′xn′→kn→∞, so s(IΦ(kn′xn′))→s(IΦ(k0x0))(n→∞). By the fact that s(u)>1 and s(u)-1>0, now s(u) is strictly monotonous on [u,+∞). Hence, we have (60)IΦkn′xn′→IΦk0x0n→∞.Corollary 10.
Lets(u)>1 with u>0 and Φ be an Orlicz function. LΦ,s is middle point locally uniformly convex if and only if Φ∈△2 and LΦ,s is strictly convex.



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