# Uniformly Nonsquare in Orlicz Space Equipped with the Mazur-OrliczF-Norm

Authors: Yunan Cui; Tongyu Wang
Journal: Journal of Function Spaces
 (2021)

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## Abstract

The definition of uniformly nonsquareness in Banach spaces is extended toF-normed spaces. Most of the results from this paper concern (uniformly) nonsquareness in the sense of James or in the sense of Schäffer in Orlicz spaces equipped with the Mazur-Orlicz F-norm. It is well known that uniform nonsquareness in the sense of Schäffer and in the sense of James are equivalent in Banach spaces. In this paper, we found that uniform nonsquareness in the sense of James and in the sense of Schäffer are not equivalent for F-normed spaces. Criteria for Orlicz spaces equipped with the Mazur-Orlicz F-norm to be nonsquare and uniformly nonsquare in the sense of James or in the sense of Schäffer are given.


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## Body

## 1. Introduction and Preliminaries

As well known, Orlicz space is a generalization of classical Lebesgue space. The theory of Orlicz space has important applications in control theory, fixed point theory, ergodic theory, probability theory, and theory of vector analytic function and has been intensively developed during the last decades. In 2018, Cui et al. discussed the monotonicity of Orlicz space that generated by the monotone continuous function equipped with Mazur-OrliczF-norm (see [1, 2]). In 2020, Bai et al. given criteria that Orlicz spaces that generated by the monotone function equipped with Mazur-Orlicz F-norm have strictly monotonicity and upper locally uniform monotonicity, and they get the conclusion that λx+1−λyF≤1 for each x,y∈SLΦμ and λ∈0,1 if and only if Φ is convex function on R. So, in order to studying geometric properties of Orlicz spaces equipped with the Mazur-Orlicz F-norm, we need to assume that Φ is convex see [3].InspiredB-convex spaces, in 1964, the definition of uniformly nonsquare in normed linear space was introduced by James (see [4]). In 1976, the concept of uniformly nonsquare in normed linear space was introduced by Schäffer (see [5]). A lot of results concerning with uniformly nonsquare in Banach space are known. Among the great number of papers concerning this topic, we list here a little [4–10]. Particularly, whether or not uniformly nonsquare Banach space has fixed point of nonexpansive mapping has been discussed as an open problem. Until 2005, Garca-Falset et al. solved the open problem and obtained that uniformly nonsquare Banach space has fixed point property (see [11]).The aim of this paper is to give criteria that Orlicz spaces equipped with theMazur OrliczF-norm are nonsquare and uniformly nonsquare in the sense of James or in the sense of Schäffer.


### 1.1. Introduction

Denoted byℕ and ℝ the sets of natural and real numbers, respectively. Let ℝ+≔0,+∞. Given any real linear space X, the functional ⋅:X→ℝ+ is called an F-norm if the following conditions are satisfied:
(i)
x=0 if and only if x=0(ii)
−x=x for all x∈X(iii)
x+y≤x+y for all x,y∈X(iv)
λnxn−λx→0 whenever xn−x→0 and λn→λ for any x∈X, xnn=1∞ in X and λ∈ℝ and λnn=1∞ in ℝWe say thatX=X,⋅ is an F-normed space if it is complete with respect to the F-norm topology.Definition 1.
LetG,Σ,μ be a finite measure space and L0 be the space of all (equivalence) classes of Σ-measurable real-valued functions defined on G. A function Φ: −∞,+∞→0,∞ is called an Orlicz function if Φu>0 for all u≠0, even, convex, and limu→∞Φu=∞. Any Orlicz function Φ determines a mapping IΦ:L0→0,+∞ defined by the formula IΦf=∫GΦftdμ called the modular. The order ideal LΦ=f∈L0:IΦrf<∞ for some r>0 in L0 is called an Orlicz space.The spaceLΦ is an F-normed space with respect to the following lattice F-norm, called the Mazur-Orlicz F-norm [12]:
(1)fF=infλ>0:IΦf/λ≤λ.Definition 2.
We say thatΦ satisfies Δ2-condition (Φ∈Δ2, for short) there are constants K>0 and u0>0 such that
(2)Φ2u≤KΦu,whenever ∣u∣≥u0.Definition 3.
LetΦ be an Orlicz function, p be the right derivative of Φ, and q be the right-inverse function of p. Then, we call
(3)Ψv=∫0∣v∣qsds,the complementary function of Φ.Definition 4.
We sayΦ satisfies ∇2-condition (Φ∈∇2, for short) there exist u0>0 and δ1>0 such that
(4)Φ2u≥2+δ1Φu,whenever ∣u∣≥u0. It is well known that Φ∈∇2 if and only if Ψ∈Δ2.Lemma 5.
(see [13]).
Φ∈Δ2 if and only if there exist constants α>0 and u0>0 such that
(5)upuΦu<α,whenever ∣u∣≥u0.The characteristic of convexity ofX is defined by ε0X=supε∈0,2:δXε=0, where δXε=inf1−1/2x+y:x≤1,y≤1,x−y≥ε is the Clarkson modulus of convexity of X. A Banach space X is said to be uniformly nonsquare whenever ε0X<2. Since X satisfies the definition of a finite tree, we can get that X is superreflexive (see [4]). Now, we extend the definition of nonsquare and uniformly nonsquare to F-normed space.Definition 6.
AnF-normed space X is said to be nonsquare in the sense of James if for any r>0 it is verified that
(6)minx+yF,x−yF<2r,whenever xF=r,yF=r.Definition 7.
AnF-normed space X is said to be nonsquare in the sense of Schäffer if for any r>0 it is verified that
(7)minx+yF,x−yF>r,whenever xF=r,yF=r.Definition 8.
AnF-normed space X is said to be uniformly nonsquare in the sense of James if for any r>0 there exists δ>0 such that
(8)maxminx+yF,x−yF,xF=r,yF=r<2r−δ,whenever xF=r,yF=r.Definition 9.
AnF-normed space X is said to be uniformly nonsquare in the sense of Schäffer if for any r>0 there exists δ>0 such that
(9)maxminx+yF,x−yF,xF=r,yF=r>r+δ,whenever xF=r,yF=r.


## 1.1. Introduction

Denoted byℕ and ℝ the sets of natural and real numbers, respectively. Let ℝ+≔0,+∞. Given any real linear space X, the functional ⋅:X→ℝ+ is called an F-norm if the following conditions are satisfied:
(i)
x=0 if and only if x=0(ii)
−x=x for all x∈X(iii)
x+y≤x+y for all x,y∈X(iv)
λnxn−λx→0 whenever xn−x→0 and λn→λ for any x∈X, xnn=1∞ in X and λ∈ℝ and λnn=1∞ in ℝWe say thatX=X,⋅ is an F-normed space if it is complete with respect to the F-norm topology.Definition 1.
LetG,Σ,μ be a finite measure space and L0 be the space of all (equivalence) classes of Σ-measurable real-valued functions defined on G. A function Φ: −∞,+∞→0,∞ is called an Orlicz function if Φu>0 for all u≠0, even, convex, and limu→∞Φu=∞. Any Orlicz function Φ determines a mapping IΦ:L0→0,+∞ defined by the formula IΦf=∫GΦftdμ called the modular. The order ideal LΦ=f∈L0:IΦrf<∞ for some r>0 in L0 is called an Orlicz space.The spaceLΦ is an F-normed space with respect to the following lattice F-norm, called the Mazur-Orlicz F-norm [12]:
(1)fF=infλ>0:IΦf/λ≤λ.Definition 2.
We say thatΦ satisfies Δ2-condition (Φ∈Δ2, for short) there are constants K>0 and u0>0 such that
(2)Φ2u≤KΦu,whenever ∣u∣≥u0.Definition 3.
LetΦ be an Orlicz function, p be the right derivative of Φ, and q be the right-inverse function of p. Then, we call
(3)Ψv=∫0∣v∣qsds,the complementary function of Φ.Definition 4.
We sayΦ satisfies ∇2-condition (Φ∈∇2, for short) there exist u0>0 and δ1>0 such that
(4)Φ2u≥2+δ1Φu,whenever ∣u∣≥u0. It is well known that Φ∈∇2 if and only if Ψ∈Δ2.Lemma 5.
(see [13]).
Φ∈Δ2 if and only if there exist constants α>0 and u0>0 such that
(5)upuΦu<α,whenever ∣u∣≥u0.The characteristic of convexity ofX is defined by ε0X=supε∈0,2:δXε=0, where δXε=inf1−1/2x+y:x≤1,y≤1,x−y≥ε is the Clarkson modulus of convexity of X. A Banach space X is said to be uniformly nonsquare whenever ε0X<2. Since X satisfies the definition of a finite tree, we can get that X is superreflexive (see [4]). Now, we extend the definition of nonsquare and uniformly nonsquare to F-normed space.Definition 6.
AnF-normed space X is said to be nonsquare in the sense of James if for any r>0 it is verified that
(6)minx+yF,x−yF<2r,whenever xF=r,yF=r.Definition 7.
AnF-normed space X is said to be nonsquare in the sense of Schäffer if for any r>0 it is verified that
(7)minx+yF,x−yF>r,whenever xF=r,yF=r.Definition 8.
AnF-normed space X is said to be uniformly nonsquare in the sense of James if for any r>0 there exists δ>0 such that
(8)maxminx+yF,x−yF,xF=r,yF=r<2r−δ,whenever xF=r,yF=r.Definition 9.
AnF-normed space X is said to be uniformly nonsquare in the sense of Schäffer if for any r>0 there exists δ>0 such that
(9)maxminx+yF,x−yF,xF=r,yF=r>r+δ,whenever xF=r,yF=r.


## 2. Main Results

Lemma 10.
Suppose thatΦ∈Δ2. If there exists ε>0 such that IΦx/r≥r+ε for anyr>0, then there exists δ>0 which satisfy xF>r+δ.Proof.
If there exists a sequencexn⊂LΦ such that IΦxn/r≥r+ε and xnF→r, then
(10)IΦxnr→r,thanks to Φ∈Δ2, a contradiction.Theorem 11.
LΦ is nonsquare in the sense of Schäffer if and only if Φ∈Δ2.Proof.
Necessity. Assume thatΦ∉Δ2. We show that there exists a strictly increasing sequence numbers un such that for each n∈N, limn→∞un=+∞ and
(11)Φ1+1nun>2nΦun.
We takeG0,G1⊂G and G1∩G0=∅ which satisfy μG0=μG1>0. There exists a sequence Gn0n=1∞ of pairwise disjoint sets in G0, for some n1∈N, such that μG10≥1/Φun1⋅r/22, μGn0=r/2⋅1/2nΦun+n1, n=1,2⋯. Let x=r∑n=1∞un+n1χGn0. Then, suppx⊂G0, and we can get the following
(12)IΦxr=∫G0Φ∑n=1∞un+n1tχGn0dt=∑n=1∞∫Gn0Φun+n1tdt=∑n=1∞Φun+n1μGn0=∑n=1∞Φun+n112nΦun+n1⋅r2=r2∑n=1∞12n=r2.
On the other hand, for anyλ∈0,1, there exists n0∈N such that
(13)1λ≥1+1n+n1,whenever n≥n0.
Combing (11) and (13), the following inequalities are found to be true:
(14)IΦxλr≥∑n=n0∞Φun+n1λμGn0≥∑n=n0∞Φ1+1n+n1un+n1μGn0>∑n=n0∞2n+n1Φun+n1μGn0=∑n=n0∞2n+n112n⋅r2=r2⋅2n1∑n=n0∞1=+∞.
Similar to the selection method of sequenceGn0n=1∞, the selection of sequence Gn1n=1∞ of pairwise disjoint sets in G1 satisfies μGn1=r/2⋅1/2n+n1Φun+n1, n=1,2⋯. Supposing y=r∑n=1∞un+n1χGn1, we verity that IΦy/r=r/2 and IΦy/λr=+∞. According to the definition of F-norm, we can obtain that xF=yF=r. Furthermore, by suppx∩suppy=∅, we deduce that
(15)IΦx+yr=IΦx−yr=IΦxr+IΦyr=r,which implies that x+yF=x−yF=r. Namely
(16)maxx+yF,x−yF=r,contradiction with maxx+yF,x−yF>r, the proof of necessity is completed.Sufficiency. Assume on the contrary that LΦ is not nonsquare in the sense of Schäffer, then LΦ has elements x,y satisfying xF=r,yF=r and x+yF=r, x−yF=r for some r>0. Using Φ∈Δ2, we conclude that IΦx+y/r=IΦx−y/r=r.Due to the convexity ofΦ, we have the following inequality
(17)αt=12Φxt+ytr+Φxt−ytr≥maxΦxtr,Φytr,and then(18)IΦαt=12∫GΦxt+ytrdt+∫GΦxt−ytrdt=12IΦx+yr+IΦx−yr=12r+r=r.Observing that the Orlicz functionΦ has the property Φ∣x∣−∣y∣≤∣Φx−Φy∣, the following inequality can be obtained
(19)IΦ∣x∣−∣y∣r=∫GΦ∣xt∣−∣yt∣rdt≤∫GΦxtrdt−∫GΦytrdt≤∫Gαt−Φxtrdt+∫Gαt−Φytrdt=0.Hence∣x∣=∣y∣ holds.SinceΦu=∫0uptdt, the following formula can be obtained
(20)Φ2u=∫02uptdt=∫0uptdt+∫u2uptdt>2∫0uptdt=2Φu,whenever u>0.From the above analysis,(21)∫GΦxt+ytrdt+∫GΦxt−ytrdt≥∫GΦ∣xt∣+∣yt∣rdt=∫GΦ2∣xt∣rdt>2∫GΦ∣xt∣rdt=2r.Therefore, we obtain that(22)IΦx+y/r+IΦx−y/r2>r.It means thatmaxIΦx+y/r,IΦx−y/r>r, hence maxx+yF,x−yF>r, a contradiction. The proof is complete.Theorem 12.
The following conditions are equivalent:(i)
LΦ is uniformly nonsquare in the sense of James(ii)
LΦ is nonsquare in the sense of James(iii)
Φ∈Δ2Proof.
It is obvious thati⇒ii. We only need to prove ii⇒iii and iii⇒i.
ii⇒iii. Assume that Φ does not satisfy Δ2-condition. Since the idea is similar to the proof of Theorem 11, we can findG0,G1⊂G, μG0>0, μG1>0, and G1∩G0=∅. Construct x,y∈LΦ such that suppx⊂G0 and suppy⊂G1. Then, IΦx/r=r/2 and IΦy/r=r/2 and IΦx/λr=+∞ and IΦy/λr=+∞ for any λ∈0,1. For convenience, denote u=x+y and v=x−y which shows that uF=vF=r.
Otherwise, we letu+v=2x and u−v=2y; then, IΦu+v/2r=r/2, and IΦu−v/2r=r/2. By the arbitrariness of λ∈0,1, we obtain that IΦu+v/2λr=+∞ and IΦu−v/2λr=+∞.
Then, by the definition of Mazur-OrliczF-norm, we conclude that
(23)u+vF=u−vF=2r.
It shows thatLΦ is not a nonsquare in the sense of James. Namely, we have Φ∈Δ2.
We prove the implicationiii⇒i. Obviously, by the convexity of Φ, we have
(24)IΦx+y2r≤12IΦxr+IΦyr=r.
For the sake of simplicity, we setz=x+y. Next, we want to prove that the inequality
(25)supzF:IΦz2r≤r<2r,holds.
SinceIΦz/2r≤r<2r, we have zF≤2r,which implies that supzF:IΦz/2r≤r≤2r holds.
Without loss of generality, we may assumesupzF:IΦz/2r≤r=2r. Then, there exists zn⊂LΦ which satisfies IΦzn/2r≤r and znF→2r as n→∞.
SinceΦ∈Δ2, for a fixed δ=1/4 find 0<ε<1 such that
(26)Φ1+εu<1+δΦu,for every u≥u0, where u0=1/2Φ−1r/4μG. Namely,
(27)Φ1+εu<54Φu.
SinceznF→2r as n→∞, we can obtain limn→∞2r/znF=1. Then, there exists n0∈N such that
(28)2rznF≤1+ε,whenever n≥n0.
Combining conditions (27) and (28) and IΦzn/2r≤r, we derive a contradiction:
(29)2r=IΦznznF=IΦ2rznF⋅zn2r≤∫GΦ1+εznt2rdt=∫Gu0Φ1+εznt2rdt+∫G∣Gu0Φ1+εznt2rdt≤Φ2u0μG+54∫GΦznt2rdt=3r2<2r,where Gu0=t∈G:znt/2r≤u0.
We get thatsupx+yF:IΦx+y/2r≤r<2r. Hence, we can obtained that
(30)minx+yF,x−yF:xF=yF=r≤supx+yF:IΦx+y2r≤r<2r,,which shows that LΦ is uniformly nonsquare in the sense of James.Theorem 13.
LΦ is uniformly nonsquare in the sense of Schäffer if and only if Φ∈Δ2 and Φ∈∇2.Proof.
Since the necessity ofΦ∈Δ2 is similar to the one given in Theorem 11, we only state the results without the proof. We only need to prove the necessity of ∇2-condition.
Suppose that∇2-condition is invalid. There exists a strictly increasing sequence ξnn=1∞ which satisfy limn→∞ξn=+∞ and
(31)Φξn≥1−1/n2Φ2ξn.
Without loss of generality, we may takeGn⊂G with
(32)ΦξnμGn=rn=1,2⋯.
Letxn=rξnχGn, it means that IΦxn/r=r. Then, xnF=r.
DividedGn into disjoint two subsets Gn′ and Gn″ with μGn′=μGn″=1/2μGn.
Supposeyn=rξnχGn′−ξnχGn″, we also have IΦyn/r=r and ynF=r.
Consequently, by the convexity ofΦ, we deduce that
(33)IΦxn+ynr=IΦ2ξnχGn′=Φ2ξn12μGn≥2Φξn12μGn=ΦξnμGn=r,then the following inequality is found to be true
(34)xn+ynF≥r.
To prove the reverse inequalities, notice that by (31) and (32), we can obtain the following inequality
(35)IΦxn+ynr=IΦ2ξnχGn′≤21−1/nΦξnμGn′=11−1/nΦξnμGn=r⋅11−1/n.
That isIΦ1−1/nxn+yn/1−1/nr≤r/1−1/n, it follows that
(36)nn−1xn+ynF≤r1−1/n.
Next, we will prove thatlimn→∞1/n−1xn+ynF=0. Otherwise, we assume that limn→∞¯1/n−1xn+ynF>0. Without loss of generality, we may suppose there exists ε0>0, which satisfies the following inequality
(37)limn→∞1n−1xn+ynF>ε0.
Taken0∈N large enough for which r/n−1ε0<1 whenever n≥n0. By the convexity of Φ and (37), this lead to a contradiction:
(38)ε0<limn→∞IΦ1/n−1xn+ynε0=limn→∞IΦrn−1ε0xn+ynr≤limn→∞rn−1ε0IΦxn+ynr≤limn→∞rn−1ε0⋅r1−1/n=limn→∞r2nn−12ε0→0.
By inequality (36), we have the following results
(39)xn+ynF=nn−1xn+yn−1n−1xn+ynF≤nn−1xn+ynF−1n−1xn+ynF≤r1−1/n−xn+ynn−1F.
It is easily to see that(40)limn→∞¯xn+ynF≤r.
From (34) and (40), it follows that limn→∞xn+ynF=r.
Using the method of the proof oflimn→∞xn+ynF=r, we can obtain that limn→∞xn−ynF=r. This implies that LΦ is not uniformly nonsquare in the sense of Schäffer, that is Φ∈∇2.
Sufficiency. Assume for the contrary that LΦ is not uniformly nonsquare in the sense of Schäffer. We can find subsequences xn,yn⊂LΦ satisfying xnF=r and ynF=r and xn+ynF→r and xn−ynF→r for some r>0.
Without loss of generality, we may takeu1>0 such that
(41)Φu1mG≤r2.
SinceΦ∈∇2, there exists δ1>0 such that
(42)Φ2u≥2+δ1Φu,whenever u≥u1.
Put(43)αnt=12Φxnt+yntr+Φxnt−yntrn=1,2⋯.
The following equation can be obtained by integrating both sides of the above formula(44)∫Gαntdt=12∫GΦxnt+yntrdt+∫GΦxnt−yntrdt→r.
SinceΦ∈Δ2, there exists K≥2 and u0>0 such that
(45)Φ2u≤KΦu+1,whenever u≥u0.
Hence, we have the following results(46)IΦxn+ynr=IΦ2xn+2yn2r≤12IΦ2xnr+IΦ2ynr≤12KIΦxnr+1+KIΦynr+1=12KxnF+1+KynF+1=Kr+1.
Put(47)L=Kr+1,ε=δ1r6.
UsingΦ∈Δ2, there exists δ>0 such that
(48)IΦx+y−IΦx<ε,whenever IΦx≤L and IΦy≤δ.
Obviously,(49)IΦ∣xn∣−∣yn∣r≤∫GΦxntr−Φyntrdt≤∫Gαnt−Φxntrdt+∫Gαnt−Φyntrdt→0.
We can choosen0∈N such that
(50)IΦ∣xn∣−∣yn∣r≤δ,whenever n≥n0.
Therefore, we have(51)IΦ∣xn∣+∣yn∣r+∣xn∣−∣yn∣r−IΦ∣xn∣+∣yn∣r<ε,this shows that
(52)IΦ2xnr−IΦ∣xn∣+∣yn∣r<ε.
DefineGu1=t∈G:xnt/r≥u1, then we derive that
(53)∫GΦxnt+yntrdt+∫GΦxnt−yntrdt≥∫GΦ∣xnt∣+∣ynt∣rdt≥∫GΦ2xntrdt−ε=∫Gu1Φ2xntrdt+∫G∣Gu1Φ2xntrdt−ε≥2+δ1∫Gu1Φxntrdt+2∫G∣Gu1Φxntrdt−ε=2∫GΦxntrdt+δ1∫Gu1Φxntrdt−ε=2∫GΦxntrdt+δ1∫GΦxntrdt−∫G∣Gu1Φxntrdt−ε≥2r+rδ12−ε≥2r+13rδ1.
It follows thatmaxIΦxn+yn/r,IΦxn−yn/r≥r+δ1r/6 holds. Thanks to Lemma 10, there exists δ′>0 such thatmaxxn+ynF,xn−ynF≥r+δ′, which contradicts with the assumption xn+ynF→r and xn−ynF→r. So LΦ is uniformly nonsquare in the sense of Schäffer. The proof is complete.Example 14.
Put(54)Φu=e∣u∣−∣u∣−1.
The derivative of the functionΦu is pt=et−1. Obviously,
(55)limu→∞u⋅puΦu=limu→∞ueu−1eu−u−1=+∞.
So, from the Lemma 1.3, we immediately haveΦu does not satisfy the Δ2-condition.
On the other hand, we can get thatqs=lns+1. Thus
(56)Ψv=∫0vqsds=∫0vlns+1ds=1+vln1+v−vv≥0.
Then(57)limv→∞v⋅qvΨv=limv→∞v⋅lnv+11+vln1+v−v=1+limv→∞vv+1ln1+v=1.
Using Lemma 1.3 in [6], we have Ψv satisfy the Δ2-condition, i.e., Φ∈∇2. Then, LΨ is uniformly nonsquare in the sense of James and is not uniformly nonsquare in the sense of Schäffer. The proof is complete.



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