# Monotonicities in Orlicz Spaces Equipped with Mazur-OrliczF-Norm

Authors: Xinran Bai; Yunan Cui; Joanna Kończak
Journal: Journal of Function Spaces
 (2020)

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## Abstract

Some basic properties in Orlicz spaces and Orlicz sequence spaces that are generated by monotone function equipped with the Mazur-OrliczF-norm are studied in this paper. We give some relationships between the modulus and the Mazur-Orlicz F-norm. We obtain an interesting result that the norm of an element in line segments is formed by two elements on the unit sphere less than or equal to 1 if and only if that the monotone function is a convex function. The criterion that Orlicz spaces and Orlicz sequence spaces that are generated by monotone function equipped with the Mazur-Orlicz F-norm are strictly monotone or lower locally uniform monotone is presented.


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## Body

## 1. Introduction

Geometry of Banach space has important applications in the control theory, fixed-point theory, ergodic theory, probability theory, and vector analytic function theory. Recall that the monotonicity properties are restrictions of appropriate rotundity properties to the set of couples of comparable elements in the positive cone of a Banach lattice (see [1–5]). Consequently, then in many cases, good convex properties can be replaced successfully by respective monotonicity properties (see [6–8]). It is well known that monotonicity properties (strict and uniform monotonicity) play an analogous role in the best dominated approximation problems in Banach lattices as the respective rotundity properties (strict and uniform rotundity) in the best approximation problems in Banach spaces (see [9, 10]). Moreover, monotonicity properties are applicable in the ergodic theory, since they provide a tool for estimating a norm (see [11–13]). In this paper, based on the paper in 2019 by Cui et al., after removing the continuous condition of generating function (see [6]), we found some conclusions completely different from those in the Banach spaces we studied in the past, which provided us with a new understanding and enlightenment for studying geometric properties of seminormed linear spaces in the future.


## 2. General Auxiliary and New Results

Definition 1.
LetG,Σ,μbe aσ-finite measure space andL0μbe the space of all (equivalence) classes ofΣ-measurable real-valued funcions defined onG. A functionΦ:−∞,∞⟶0,∞is called an Orlicz function ifΦu>0for allu≠0limu→0+0Φu=0 , and evenΦis the nondecreasing function andlimu→∞Φu=∞ .Any Orlicz functionΦdetermines a functionalIΦ:L0μ⟶0,∞defined by the formulaIΦf=∫GΦ∣ft∣dμtand called the modular. The order ideal(1)LΦμ=f∈L0μ:IΦrf<∞forsomer>0
inL0μ is called an Orlicz space.
Let us define a lattice idealEΦμ in LΦμas
(2)EΦμ=f∈L0μ:IΦrf<∞forallr>0.
The spaceLΦμ is linear with respect to the following lattice F-norm, called the Mazur-Orlicz F-norm (see [6]):
(3)∥f∥F=infλ>0:IΦf/λ≤λ.
Note that ifΦ is a convex function, then LΦμ, endowed with the equivalent lattice norm (called the Luxemburg norm) fΦ=infλ>0:IΦf/λ≤1, becomes a Banach lattice. If Φ is right continuous and X is a Banach space, increasing function on 0,+∞ such that Φ0=0. Define
(4)LΦμ,X=f∈L0μ,X:∫GΦ∥fx∥ρdμx<∞forsomeρ>0.
The properties ofΦ yield that if for f∈LΦμ,X, we put
(5)fΦ=infρ>0:∫GΦ∥fx∥ρdμx<ρ,then ·Φ satisfies the axioms of a Δ-norm in LΦμ,X (see [14]). Moreover, ·Φ is equivalent to an F-norm under which LΦμ,X is complete, whence LΦμ,X,·Φ is an F-space.
In the case of the counting measure space, the spaceLΦμ=lΦ is always nontrivial and
(6)lΦ=x∈l0:∀λ>0∃nλ∈ℕ∑n=nλ∞Φλxn<∞≕hΦ.Remark 2.
IfΦis not right continuous at zero, thenlimu→0+0Φu=a>0. If we takexnt=1/nχGt, thenxntis convergent to 0 uniformly, butIΦxn=∫GΦ1/nχGtdt=Φ1/nmG≥amGfor alln∈N. This means that the caselimu→0+0Φu>0. We do not even have the fact that the uniform convergence of the sequencexnimplies its modular convergence.Lemma 3.
LΦμis a linear space.Proof.
Obviously, ifΦ is a monotone increasing function, then for all u,v∈R and α∈0,1 we have
(7)Φαu+1−αv≤Φu+Φv.
Take anyf,g∈LΦμ, by the definition of the LΦμ; there exists λ1,λ2>0 such that
(8)∫GΦλ1fxdx<∞,∫GΦλ2fxdx<∞.
We chooseλ0=minλ1,λ2, that is,
(9)∫GΦλ0fxdx<∞,∫GΦλ0fxdx<∞.
Assume now thata≠0 or b≠0, and we conclude that
(10)∫GΦλ0a+bafx+bgxdx≤∫GΦλ0a+bafx+bgxdx≤∫GΦλ0fxdx+∫GΦλ0gxdx<∞.
Hence,LΦμ is a linear space.Definition 4 (see [6]).
AnF-normed Orlicz spaceLΦμ,·Fis said to be strictly monotone (we writeLΦμ∈SM) if for anyx,y∈LΦμsuch that0≤y≤x, we haveyF<xFwhenevery≠x(or equivalentlyx−yE<xEwhennevery≠0).Definition 5 (see [6]).
AnF-normed Orlicz spaceLΦμ,·Fis said to be lower locally uniformly monotone (we writeLΦμ∈LLUM) if for anyx∈LΦμandxnn=1∞inLΦμsuch that0≤xn≤xfor alln∈ℕandxnF⟶xFasn⟶∞, the conditionx−xnF⟶0asn⟶∞holds.Definition 6.
We say thatΦsatisfiesΔ2-condition (Φ∈Δ2for short) whenever there are constantsK>0andu0>0 such that (see [15])
(11)Φ2u≤KΦuu≥u0.Definition 7.
We say thatΦsatisfiesΔl-condition (Φ∈Δlfor short) whenever there are constantsK1>0andu1>0such that(12)Φlu≤K1Φuu≥u1.Lemma 8.
Φ∈Δ2if and only ifΦ∈Δlfor alll>1.Proof.
Sufficiency: given thatl>1, choose an integer n0 such that ln0>2. Then, by (12), we have
(13)Φ2u≤Φln0u<K1n0Φuu≥u1.
Necessity: now choose an integern1 such that 2n1>l. Then, by (11) we get
(14)Φlu≤Φ2n1u≤Kn1Φuu≥u0which ends the proof.Definition 9.
We say thatΦsatisfiesδ2-condition (Φ∈δ2for short) whenever there are constantsk1>0andu2>0such that(15)Φ2u≤k1Φu∣u∣≤u2.Theorem 10.
IfΦis an Orlicz function such thatΦ∈Δ2, then for anyε>0, there existsK>0such that(16)IΦ2f≤KIΦf+ε,for all f∈LΦμ.Proof.
ByΦ∈Δ2, there exists u1>0 and K1>0 such that Φ2u≤K1Φu (∀u≥u1). Let us assume that 0<u0<u1 and for any u∈u0,u1 and Φu0>0, we get
(17)Φ2uΦu≤Φ2u1Φu0.
We chooseK=maxK1,Φ2u1/Φu0. We can hence easily get that Φ2u≤KΦu for any u≥u0. By limu→0+0Φu=0, for any ε>0, there exists δ>0 such that Φ2u0mG<ε, whenever u0<δ. Defining Gn=x∈G:∣fx∣≤u0, we obtain
(18)IΦ2f=∫GnΦ2fxdx+∫G\GnΦ2fxdx≤Φ2u0mG+K∫GΦfxdx≤KIΦf+ε.Theorem 11.
For anyf∈LΦμ,IΦf/fF=fFif and only ifΦis right continuous andΦ∈Δ2.Proof.
Sufficiency: for anyf∈LΦμ\0 the function Fλ=IΦλf/∥f∥F is right continuous on the interval 0,∞. By the definition of fF, we have F1≤fF and for any λ>1, we have Fλ>fF. Using limλ→1+0Φλfx⟶Φfx and the Lebesgue dominated convergence theorem, we obtain
(19)fF≥IΦf∥f∥F=limλ→1+∫GΦλftfFdt,which finishes the proof of the implication IΦf/fF=fF.
Necessity: assume thatΦ does not satisfy the Δ2-condition. This yields that there exists a strictly increasing sequence unn=1∞ of position numbers such that Φu1mG≥1 and
(20)Φ1+1nun>2nΦunn=1,2,3,⋯.
Next, choose a sequenceGnn=1∞ of pairwise disjoint sets in G such that mGn=1/2nΦunn=1,2,3,⋯. Defining x=∑n=1∞unχGn, we have
(21)IΦx=∑n=1∞∫GnΦxtdt=∑n=1∞ΦunmGn=∑n=1∞12n=1.
On the other hand, we have that for anyλ∈0,1, there exists n1∈N such that 1+1/n<λ for any n≥n1. Hence,
(22)IΦλx=∫GΦλxtdt=∑n=1∞ΦλunmGn≥∑n=1∞Φ1+1nunmGn≥∑n=1∞2nΦunmGn=∑n=1∞1=∞.
Consequently, this shows thatΦ∈Δ2 is a necessary condition.
Next, we will prove the necessary of the right-continuous function ofΦ. If Φ is not right continuous on R, then there exists a point u0∈0,+∞ such that Φu0<limu→u0+0Φu=b; that is, for any λ<1, we get Φu0/λ≥b. Take any G0⊂G with Φu0mG0<1. Let us take c>0 large enough. Since limu→∞Φu=+∞, we have
(23)Φu0mG0+ΦcmGG0>1.
SinceG is nonatomic, we can find G2⊂G\G0 such that
(24)Φu0mG0+ΦcmG1=1−b−Φu02mG0.
Let us define(25)x=u0χG0+cχG1
Then,IΦx=Φu0mG0+ΦcmG1<1, whence we obtain that x∈LΦμ and ∥x∥F≤1. On the other hand, for any λ>1, we have
(26)IΦxλ=Φu0λmG0+ΦcλmG1>bmG0+ΦcmG1=bmG0+1−b−Φu02mG0−Φu0mG0=1+b−Φu02mG0>1,which gives that ∥x∥F≥1 yields ∥x∥F=1, a contradiction, which finishes the proof of necessity.For convenience, from now on, we writeSLΦμ=f∈LΦμ:∥f∥F=1.which contradicts the assumption yF=xF. Therefore, yF<xF.Theorem 12.
Assume thatx,y∈SLΦμandα∈0,1. Then,∥αx+1−αy∥F≤1if and only ifΦis a convex function onR.Proof.
Sufficiency: for anyx,y∈SLΦμ, we have IΦx≤1 and IΦy≤1. Assume that z=αx+1−αy. Then, IΦz≤αIΦx+1−αIΦy≤1. Hence, ∥αx+1−αy∥F≤1. This finishes the proof.
Necessity: ifΦ is not a convex function, then there exists u>v, such that Φu+v/2>1/2Φu+Φv. Let us choose G1,G2⊂G such that mG1=mG2 and ΦumG1+ΦvmG2≤1. Take any c>0 such that
(27)ΦumG1+ΦvmG2+ΦcmG3=1.
Define(28)x=uχG1+vχG2+cχG3,y=vχG1+uχG2+cχG3,z=x+y2.
We obtain(29)IΦz=Φu+v2mG1+Φu+v2mG2+ΦcmG3>12Φu+ΦvmG1+12Φv+ΦumG2+ΦcmG3=12IΦx+IΦy=1.
Consequently,∥x∥F>1 is a contradiction.Example 13.
Put(30)Φu=u,∣u∣≤1,1,1≤u<2,2,2≤u≤3,3u,u>3.
Then,Φ satisfies the Δ2-condition and Φ is right continuous. Define
(31)xt=1,x∈0,1,0,x∈1,2,yt=3,x∈0,12,0,x∈12,2.
Then,y≠x and IΦx=Φ1m0,1=1 and IΦy=Φ3m0,1/2=1. Moreover, ∥y∥F=∥x∥F=1. Now let us define
(32)zt=13x+23y=73,x∈0,12,13,x∈12,1,0,x∈1,2.
Then,IΦz=Φ7/3m0,1/2+Φ1/3m1/2,1=2·1/2+1/3·1/2=1+1/6>1. This means that zF>1.Remark 13.
This example shows that the norm of an element in a line segmentx,yconsisting of two elementsxandyin a unit sphere may be greater than 1. This is essentially different from Banach spaceX. For Banach spaceX ,for anyx,y∈SXand for anyλ∈0,1 ,we get thatx,y⊂BX=x∈X:∥x∥F≤1.Using Theorem 12, ifΦis a convex function, we conclude that if∥x∥F≤1, yF≤1, then∥λx+1−λy∥F≤1.Theorem 14.
The Orlicz spaceLΦμequipped with the Mazur-OrliczF-norm·Fis strictly monotone if and only ifΦis strictly increasing onR+andΦsatisfies theΔ2-condition andΦis right continuous.Proof.
Sufficiency: let0≤y≤x, y≠x, then there exists e0⊂G and me0>0 such that 0≤yt<xtt∈e0. Assuming for the contrary that yF=xF, we have by Theorem 11,
(33)yF=∫G\e0ΦytyFdt+∫e0ΦytyFdt<∫G\e0ΦxtyFdt+∫e0ΦxtyFdt=∫G\e0ΦxtxFdt+∫e0ΦxtxFdt=xF,Necessity: ifΦ is not right continuous on R, then there exists a point u0∈0,+∞ such that Φu0<limu→u0+0Φu=b; that is,for any λ<1, we get Φu0/λ≥b. Take any G0⊂G with Φu0mG0<1. Let us take c>0 large enough. Since limu→∞Φu=+∞, we obtain
(34)Φu0mG0+ΦcmGG0>1.SinceG is nonatomic, we can find G1⊂G\G0∪G1 such that
(35)Φu0mG0+ΦcmG1=1−b−Φu02mG0.Moreover, takeG2⊂G\G0∪G1, where d is a fixed positive number such that
(36)Φu0mG0+ΦcmG1+ΦdmG2=1.Let us define(37)x=u0χG0+cχG1,y=u0χG0+cχG1+dχG2.Then,IΦx=Φu0mG0+ΦcmG1<1, whence we obtain that x∈LΦμ and xF≤1. On the other hand, we have that for any λ>1,
(38)IΦxλ=Φu0λmG0+ΦcλmG1>bmG0+ΦcmG1=bmG0+1−b−Φu02mG0−Φu0mG0=1+b−Φu02mG0>1,which gives that ∥x∥F≥1 yields ∥x∥F=1. By (36), we obtain ∥x∥F=∥y∥F=1. Since 0≤x≤y and x≠y, this means that the space LΦμ,·F is not strictly monotone.We will prove the necessity of theΔ2-condition. Assume first that Φ does not satisfy the Δ2-condition. This yields that there exists a strictly increasing sequence unn=1∞ of positive numbers such that Φu1mG≥1 and Φ1+1/nun>2nΦun. Let us choose first G1⊂G such that Φu1mG1=1/2. We have Φu2mGG1≥Φu1mG1≥1/2. Since G is nonatomic, we can find G2⊂G\G1 such that Φu2mG2=1/4. Continuing this procedure by induction, we can find a sequence Gnn=1∞ of measurable sets such that Gn⊂G\∪k=1n−1Gn and ΦunmGn=2−n, for any n∈N, n≥2. Let us define
(39)y=∑n=2∞unχGn,x=∑n=1∞unχGn.Then,0≤y≤x, y≠x and IΦx=∑n=1∞ΦunmGn=∑n=1∞2−n=1, whence we obtain that x,y∈LΦμ and ∥x∥F≤1. On the other hand, we have that for any λ∈0,1, there exists m∈N, m≥2, such that 1/λ>1+1/n for any n≥m, whence
(40)IΦyλ≥∑n=m∞ΦunλmGn≥∑n=m∞Φ1+1nunmGn>∑n=m∞2nΦunmGn=∑n=m∞1=∞,which gives that yF≥λ and by the arbitrariness of λ∈0,1, we get yF≥1. The inequalities 0≤y≤x, xF≤1, and yF≥1 yield yF=xF. This equality shows that space LΦμ,·F is not strictly monotone.Now, we will prove the necessity of strict monotonicity ofΦ on R+. Assume for the contrary that Φ is constant on some interval a,b⊂R+, where 0<a<b<∞, so we have Φa>0. Take A1∈Σ such that 0<ΦamA1<1, and find a positive number c such that 0<ΦamA1+ΦcmG\A1>1; using the measure space G,Σ,μ which is nonatomic, there exists A2⊂GA1 such that
(41)ΦamA1+ΦcmA2=1.Define(42)xt=aχA1+cχA2,yt=bχA1+cχA2.We obviously have0≤x≤y, y≠x, IΦy=IΦx=1 which yields xF=yF=1; this means that ∥·∥F is not strictly monotone.Theorem 15.
LΦμis lower locally uniformly monotone if and only ifΦis strictly monotone and a right continuous function andΦsatisfies theΔ2-condition.Proof.
In order to prove the sufficiency of the theorem, assume that0≤xn≤x∈LΦμ\0 and xnF⟶xnF as n⟶∞. We need to prove that xn−xF⟶0 as n⟶∞.First, we will prove thatxn⟶x in measure. Assuming that xn↛x in measure, without loss of generality, then there exists ε0>0, δ0>0 such that mt∈G:xt−xnt≥ε0≥δ0. Define
(43)Gn=t∈G:xt−xnt≥ε0and znt=xtχG\Gn+xt−ε0χGn for each n∈N. It is clear that 0≤xnt≤znt≤xt hold. Using condition Φ∈Δ2, we know that ∫GΦxtdt<∞ and then, there exists D>0 such that
(44)mt∈G:∣xt∣>D<δ03.Therefore, by conditions (43) and (44), we get that
(45)mt∈G:xt≤D,znt≤D,xt−znt≥ε0≥δ03.Next, we will prove that there existsδ1>0 such that
(46)Φu+ε0≥Φu+δ10≤u≤D.It is obvious thatΦu+ε0>Φu, since Φ is strictly increasing. If (46) does not hold, then ∃un∈0,D, such that
(47)Φun+ε0≤Φun+1n.By the compactness theorem, assume thatun⟶u0. If un↘u0, since Φ is right continuous, we get Φu0+ε0≤Φu0, a contradiction. If un↗u0, since limun→u0Φun≤Φu0, we have
(48)limn→∞Φun=b≤Φu0.Moreover,un+ε0↗u0+ε0; we have known that there exists n1∈N,n>n1 such that Φun+ε0>Φu0, which gives limn→∞Φun+ε0>Φu0. This shows that
(49)Φu0<limn→∞Φun+ε0≤limn→∞Φun=b≤Φu0,a contradiction.PutGnD=t∈G:xnt/∥xn∥F≥D for each n∈N. From condition (44) there exists a D>0 such that mGnD≤δn/2. Put Gn1=Gn\GnD for each n∈N. Then,
(50)mGn1≥mGn−mGnD≥δ02.Therefore,(51)xt∥xn∥F−xnt∥xn∥F≥ε0∥xn∥F≥ε0∥x∥Ft∈Gn1.Without loss of generality, we may assume that∥x∥F/2≤∥xn∥F for each n∈N. Put ε1=2ε0/∥x∥F. Applying condition (46), there exists a δ1≥0 such that Φu+ε1≥Φu+δ1 when u∈0,D.Using0≤xnt≤xt and ∥xn∥F↗∥x∥F as n⟶∞, we have Φxt/∥xn∥F⟶Φxt/∥xn∥F thanks to Φ that is right continuous. Then, there exists a constant α>0 such that Φxt/∥xn∥F≤Φαxt for all n∈N and μ−a.e.t∈Ω. Since Φ∈Δ2, we have ∫GΦαxtdt<∞. By the Lebesgue dominated convergence theorem, we obtain
(52)∫GΦxt∥xn∥Fdt⟶∫GΦxt∥x∥Fdt=xF.Consequently,(53)∥x∥F⟵∫GΦxt∥xn∥Fdt≥∫Gn1Φxnt+ε0∥xn∥Fdt+∫G\Gn1Φxt∥xn∥Fdt≥∫Gn1Φxnt∥xn∥F+ε0∥x∥Fdt+∫G\Gn1Φxt∥xn∥Fdt≥∫Gn1Φxnt∥xn∥Fdt+∫Gn1δ1dt+∫G\Gn1Φxt∥xn∥Fdt=∫Gn1Φxnt∥xn∥Fdt+δ0δ12+∫G\Gn1Φxt∥xn∥Fdt≥∫Gn1Φxnt∥xn∥Fdt+δ0δ12+∫G\Gn1Φxnt∥xn∥Fdt=∫GΦxnt∥xn∥Fdt+δ0δ12=∥xn∥F+δ0δ12,which contradicts with the assumption limn→∞∥xn∥F=∥x∥F. Therefore, xn is convergent to x in measure.Hence,xn/∥xn∥F⟶μx/∥x∥F. By the fact that there is absolute continuity of integral, for any ε>0, there exists δ2>0 such that
(54)∫E0Φxt∥x∥Fdt<ε,for every measureable set E⊂G with mE<δ2.Consequently, this implieslimn→∞∫E0Φxnt/∥xn∥Fdt=∫E0Φxt/∥x∥Fdt. By the Egorov theorem, for the above δ2>0, there exists E0⊂G such that mE0<δ2 and
(55)xnt∥xn∥F⟶xt∥x∥Funiformlyint∈G\E0.Sincelimu→0+0Φu=0, we get
(56)limn→∞∫G\E0Φxt∥x∥F−xnt∥xn∥Fdt=0.Consequently,(57)∫GΦxt∥x∥F−xnt∥xn∥Fdt=∫G\E0Φxt∥x∥F−xnt∥xn∥Fdt+∫E0Φxt∥x∥F−xnt∥xn∥Fdt≤∫G\E0Φxt∥x∥F−xnt∥xn∥Fdt+∫E0Φxt∥x∥Fdt≤ε+ε=2ε.By the arbitrariness ofε, we obtain that ∥x−xn∥E⟶0 as n⟶∞ holds.In virtue of Theorem14, we have a necessity that is obvious.Theorem 16.
For anyx∈lΦ\0, IΦx/∥x∥F=∥x∥Fif and only if(i)
Φ∈δ2(ii)
Φ is right continuousProof.
Necessity:(i)
IfΦ∉δ2, then there exists an element x∈lΦ such that IΦx<1 and IΦx/λ=∞ for any 0<λ<1. Hence, ∥x∥F=1 and IΦx/∥x∥F<∥x∥F(ii)
If there exists a pointu0∈0,+∞ such that Φu0<limu→u0+0Φu=b, that is, for any 0<λ<1, we get Φu0/λ≥bPutx=bu0,0,⋯. Then, IΦx/b=Φu0<b and for any 0<λ<1, we have IΦx/λb=Φbu0/λb≥b. Hence, ∥x∥F=b and IΦx/∥x∥F<∥x∥F.The sufficiency of the proof is similar as Theorem11.Theorem 17.
The following statements are equivalent:(1)
The Orlicz sequence spacelΦ equipped with the Mazur-Orlicz F-norm is lower locally uniformly monotone(2)
The Orlicz sequence spacelΦ equipped with the Mazur-Orlicz F-norm is strictly monotone(3)
(a)Φ is strictly increasing on R+
(b)Φ∈δ2
(c)Φ is right continuousProof.
The implication (1)⟹(2) is clear.
(2)⟹(3) The proof is similar as the proof of Theorem 15. We only give the proof that Φ is right continuous, if there exists a point u0∈0,+∞ such that Φu0<limu→u0+0Φu=b. That is, for any 0<λ<1, we get Φu0/λ≥b. Put x=bu0,0,⋯. Then, IΦx/∥x∥F<∥x∥F, Since limu→0+0Φu=0, there exists a c>0 such that Φbu0/b+Φc/b≤∥x∥F=b. Put y=bu0,c,0,⋯. Then, b=∥x∥F≤∥y∥F≤b, 0≤x≤y and x≠y.
(3)⟹(1) The idea is the same as the proof of Theorem 15, and the proof is simpler than Theorem 15. We only need to replace convergence by the measure of convergence by coordinate.



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