MONOTONICITY IN ORLICZ-LORENTZ SEQUENCE SPACES EQUIPPED WITH THE ORLICZ NORM∗
2017-01-21WanzhongGONG巩万中DaoxiangZHANG张道祥
Wanzhong GONG(巩万中) Daoxiang ZHANG(张道祥)
Department of Mathematics,Anhui Normal University,Wuhu 241000,China
MONOTONICITY IN ORLICZ-LORENTZ SEQUENCE SPACES EQUIPPED WITH THE ORLICZ NORM∗
Wanzhong GONG(巩万中)†Daoxiang ZHANG(张道祥)
Department of Mathematics,Anhui Normal University,Wuhu 241000,China
E-mail:gongwanzhong@shu.edu.cn;zdxiang1012@163.com
In Orlicz-Lorentz sequence space λ◦ϕ,ωwith the Orlicz norm,uniform monotonicity,points of upper local uniform monotonicity and lower local uniform monotonicity are characterized.Moreover,the monotonicity coefcient in λ◦ϕ,ωare discussed.
Orlicz-Lorentz sequence space;Orlicz norm;point of upper(lower)local uniform monotonicity;uniform monotonicity;monotone coefcient
2010 MR Subject Classifcation46B20
1 Introduction
A Banach lattice X with a lattice norm k·k is said to be strictly monotone(STM for short) [1]if for any x∈X+(positive cone in X)and any y∈X+{0},we have kx+yk>kxk.A point x∈S(X+):=S(X)∩X+is said to be upper monotone[2]if,for any y∈X+{0},kx+yk>1. A point x∈S(X+)is said to be lower monotone[2]if,for any y∈X+{0}and y≤ x, kx−yk<1.An equivalent condition for X being strictly monotone[1]is that any point x∈S(X+)is lower monotone.But lower monotone points and upper monotone points are diferent,see[2].X is called upper locally uniformly monotone(ULUM)[3]if for any ε>0 and x∈S(X+),there exists δ(x,ε)>0 such that y∈X+and kyk≥ε imply kx+yk≥1+δ(x,ε). If for any ε>0 and x∈S(X+),there is δ(x,ε)>0 such that kx−yk≤1−δ(x,ε)whenever y∈X+,kyk≥ε and y≤x,then X is said to be lower locally uniformly monotone(LLUM) [3].We can analogously defne points of lower local uniform monotonicity and points of upper local uniform monotonicity.We say that X is uniformly monotone(UM)[4]if for any ε∈(0,1) there exists δ(ε)∈(0,1)such that kx+yk>1+δ(ε)whenever x∈S(X+),y∈X+and kyk≥ε.For ε∈[0,1],defne ηX(ε)=inf{kx+yk−1:x,y∈X+,kxk=1,kyk≥ε}.We call m(X)=sup{ε∈[0,1]:ηX(ε)=0}the monotone coefcient[5]of X.
It is well known that some rotundity properties of Banach spaces were widely applied in ergodic theory,fxed point theory,probability theory and approximation theory,and in many cases these rotundity properties can be replaced by respective monotonicity properties when werestrict ourselves to a Banach space being a Banach lattice[3].Roughly speaking,monotonicity properties played in Banach lattices similar role as rotundity properties in Banach spaces, and so for monotonicity points and rotundity points.Therefore in recent years monotonicity properties and monotonicity points were widely investigated in Musielak-Orlicz,Orlicz-Lorentz, Orlicz-Sobolev,Calder´on-Lozanovskiˇi spaces[2,3,7,8,19].In addition,some geometric properties concerning with the dual spaces of Orlicz-Lorentz spaces were researched by many mathematicians,where the Orlicz norm play a important role.In this paper we mainly give the criteria for Orlicz-Lorentz sequence spaces λ◦ϕ,ωwith the Orlicz norm being UM,a point in the space being upper locally uniformly monotone and lower locally uniformly monotone.At last we get the monotone coefcients of Orlicz-Lorentz sequence spaces with the Luxemburg norm and the Orlicz norm.
and the non-increasing rearrangement of x,
endowed with the Orlicz norm[9]
or the Banach space λϕ,ωequipped with the Luxemburg norm[10]
Similarly as in the Orlicz space theory[11],denote
Recall that ϕ satisfes δ2-condition if there exist k>0 and u0>0 such that ϕ(2u)≤kϕ(u) for all 02 Lemmas
In recent years,Wang and Ning extended some properties in Orlicz space to Orlicz-Lorentz spaces[9].
Lemma 2.1(see[9]) Let x∈λ◦ϕ,ω,we have
where k∗=inf{h>0:ρψ,ω(p(h|x|))≥1},and k∗∗=sup{h>0:ρψ,ω(p(h|x|))≤1}.
Lemma 2.2(see[9])
By Lemma 1.1 in[23],and similarly as the proof of Lemma 1.40 in[11],we can get
Lemma 2.3Suppose ϕ∈δ2,then for any L>0 and ε>0,there exists δ>0 such that
whenever ρϕ,ω(u)≤L and ρϕ,ω(v)≤δ.
Lemma 2.4Let x∈(λ◦ϕ,ω)+with ρϕ,ω(x)< ∞.Then ρϕ,ω(x−[x]n)→ 0,where [x]n=(x(1),x(2),···,x(n),0···).
Lemma 2.5Suppose x∈(λ◦ϕ,ω)+and δ>0 be such that the set A:={j∈N:δ≤x∗(j)} is nonempty.Then for any i∈A,there exists a constant b=b(x,δ,i)>0 such that
ProofCase 1δ Since we only need to check that Otherwise,x∗(i)=x∗(i+1)=x∗(i+2)=···=x∗(n)=x∗(i)−δ since ω is non-increasing. This contradiction yields that the inequality above holds.It follows the inequalityρϕ,ω(x)> ρϕ,ω(x∗−δei).Set b:=12(ρϕ,ω(x)−ρϕ,ω(x∗−δei)).Then b=b(x,δ,i)>0 satisfes Lemma 2.6Suppose x∈(λ◦ϕ,ω)+and δ>0 satisfy that A:={j∈N:δ≤x∗(j)}is nonempty.Then there exists a constant c=c(x,δ)>0 such that the inequality holds for all h∈A. ProofThere is a bijection σ:N→N(ifµSx<∞)or σ:Sx→N(ifµSx=∞)such that x=x∗◦σ.Clearly h∈H:={σ−1(j):x∗(j)≥δ}andµH<∞.By Lemma 2.5,one get where b′(x,δ,h)>0.Set c=min{b′(x,δ,h):h∈H}.Then c=c(x,δ)>0 satisfes the demand. ? 2.1Monotonicity inλ◦ϕ,ω Theorem 2.7λ◦ϕ,ωis STM. ProofLet us choose arbitrary x∈S?(λ◦ϕ,ω)+?:=S(λ◦ϕ,ω)T(λ◦ϕ,ω)+and y∈λ◦ϕ,ω,y?0, and let k∈K(x)and h∈K(x+y).In the following we will consider two cases. Case Ih 6∈K(x).By Lemma 2.1,we have Case IIh∈K(x).For y?0,there exists i0∈N such that y(i0)>0.We know that there is a bijection σ:N→N(ifµSx<∞)or σ:Sx→N(ifµSx=∞)such that x=x∗◦σ. If i0∈Sx,assuming that σ(i0)=j0,one can get If i06∈Sx,then h0:=µ{i∈N:x(i)≥y(i0)}<∞.Without loss of generality,we may assume that h0≥1.So x∗(h0+1) Hence it follows that kx+yk◦ϕ,ω=1h(1+ρϕ,ω(h(x+y)))>1h(1+ρϕ,ω(hx))=kxk◦ϕ,ω=1,i.e., x is upper monotone,and λ◦ϕ,ωis STM. Theorem 2.8A point x∈S?(λ◦ϕ,ω)+is upper locally uniformly monotone if and only if ϕ∈δ2. ProofSufciencyIf ϕ∈δ2and x is not upper locally uniformly monotone,then there exist{xn}⊂λ◦ϕ,ω,xn≥0 such that kxnk◦ϕ,ω≥ε0∈R+for any n∈N and kx+xnk◦ϕ,ω→1 as n→∞.Now denoting k=k∗x,kn=k∗x+xn,by the defnition we can get kn≤k.Without loss of generality,we can assume that kn→k0as n→∞. If k0 a contradiction!Therefore we may assume that kn→k as n→∞. By Corollary 5.1 in[24],we know that λϕ,ωis ULUM.Hence for the above ε0>0,there exists δ=δ(ε0)>0 such that ρϕ,ω(k(x+xn))> ρϕ,ω (kx)+δ.Set u=kn(x+xn),v= (k−kn)(x+xn).By kn→k and Lemma 2.3,there exist N∈N such that when n>N.Therefore,by the defnition of the Orlicz norm,we have,for n>N, a contradiction which shows that x is upper locally uniformly monotone. If ϕ 6∈δ2,in view of the proof of Theorem 8 in[25],There exists a w0satisfying Sw0= N,ρϕ,ω(w0)<∞and θϕ,ω(w0)=1.For the above subset I0,there exists a w=w|I0such thatFrom Lemma 2.4 and the orthogonally sub-additive convexity of ρϕ,ωwe get where An={n+1,n+2,···}∩I0.Thus kx+ynk→1,a contradiction with x being upper locally uniformly monotone. Theorem 2.9x∈S((λ◦ϕ,ω)+)is lower locally uniformly monotone if and only if θϕ,ω(x)= 0. ProofNecessityIf θϕ,ω(x)=ε>0,denote yn:=x−[x]n.Then θϕ,ω(yn)=θϕ,ω(x). So kynk◦ϕ,ω≥θϕ,ω(yn)=ε0>0.But kx−ynk◦ϕ,ω=k[x]nk◦ϕ,ω→kxk◦ϕ,ω=1,a contradiction with x being lower locally uniformly monotone. SufciencyIf x is not lower locally uniformly monotone,then there exist ε>0 and {xn:0?xn≤x}satisfying kxnk◦ϕ,ω≥2ε and kx−xnk◦ϕ,ω→1. Thus kxnk◦ϕ,ω≤2kxnkϕ,ω≤ε,a contradiction with kxnk◦ϕ,ω≥2ε. So there exist an ε0>0,a subsequence of{xn}still denoted by{xn},and{in:in=i(n)} such that xn(in)≥ε0for any n∈N.Since ρϕ,ω(x)<∞,n0:=µ{i∈Sx:x(i)≥ε0}<∞.In virtue of Lemma 2.6,there is a δ=δ(x,ε0)>0 independent of n,such that ρϕ,ω(kx−kxn)≤ρϕ,ω(kx−kε0ein)≤ρϕ,ω(kx)−kδ. By θϕ,ω(x)=0 implying ρϕ,ω(2x)<∞,we have where k∈[k∗,k∗∗],a contradiction with kx−xnk◦ϕ,ω→1. Corollary 2.10The following conditions are equivalent: 1.ϕ∈δ2; 2.λ◦ϕ,ωis ULUM; 3.λ◦ϕ,ωis LLUM. Theorem 2.11λ◦ϕ,ωis UM if and only if ϕ∈δ2,and ω is regular. Set fi=µ({1,2,···,ei}∩A)and gi=µ({1,2,···,ei}A),then fi+gi=ei.Choosing arbitrarily k0∈K(x),we have k0>1, and Defning and so When i∈N2,combiningfi2p≤gi≤fiwith fi+gi=eiwe get ei≥2gi.So using the regularity and monotonicity of ω,we have For i∈N3,by fi Therefore Which follows that Obviously δ:=ah(ε)satisfes the demand. 2.2Monotone Coefcients in Orlicz-Lorentz Sequence Space In 1999,L¨u,Wang and Wang gave the monotone coefcients in Orlicz function space [5].Here we investigate similarly the monotone coefcients in Orlicz-Lorentz sequence space. Combining Theorem 4.4 in[26]with our Theorem 2.11 we immediately get Theorem 2.12If ϕ∈δ2and ω be regular,then m(λ◦ϕ,ω)=0 and m(λϕ,ω)=0. 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