# The Modulus of Nearly Uniform Smoothness in Orlicz Sequence Spaces

Authors: Shaoyong Zhang; Meiling Zhang; Yujia Zhan
Journal: Journal of Function Spaces
 (2019)

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## Abstract

It is well known that the modulus of nearly uniform smoothness related with the fixed point property is important in Banach spaces. In this paper, we prove that the modulus of nearly uniform smoothness in Köthe sequence spaces without absolutely continuous norm isΓX(t)=t. Meanwhile, the formula of the modulus of nearly uniform smoothness in Orlicz sequence spaces equipped with the Luxemburg norm is given. As a corollary, we get a criterion for nearly uniform smoothness of Orlicz sequence spaces equipped with the Luxemburg norm. Finally, the equivalent conditions of R(a,l(Φ))<1+a and RW(a,l(Φ))<1+a are given.


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## Body

## 1. Introduction

In last century, the fixed point property has been studied by many scholars. A Banach spaceX is said to have the fixed point property (FPP, for short) if every nonexpansive mapping T:C→C(1)Tx-Ty≤x-y,∀x,y∈C acting on a nonempty bounded closed and convex subset C of X has a fixed point. A natural generalization of FPP is the weak fixed point property (WFPP, for short). A Banach space X is said to have the WFPP, whenever it satisfies the above condition from the definition of FPP with “weakly compact” in place of “bounded closed”. In 1965 Kirk [1] proved that any reflexive Banach space with normal structure has the FPP and he also asserts that a Banach space with weak normal structure has WFPP. In this connection, see also the paper [2, 3]. In 1989 Prus [4] introduced a property of a Banach space X, called nearly uniform smoothness and he got that a Banach space is nearly uniform convex if and only if its dual space X∗ is nearly uniformly smooth. In paper [5], it is proved that weak nearly uniformly smooth Banach spaces have the FPP, which strengthens the result by S. Prus [6]. In [7], Domínguez-Benavides proved an important result on the existence of fixed points for nonexpansive mappings. In order to do this, he defined, for a Banach space X and a nonnegative real number a, the parameter(2)Ra,X≔supliminfn→∞xn+x, where the supremum is taken over all x∈X with x≤a and all weakly null sequences (xn) in B(X) such that D[(xn)]≔limsupn→∞,m→∞xn-xm≤1. He also defined the coefficient(3)MX≔sup1+aRa,X:a≥0. The result that a Banach space X has the WFPP, whenever M(X)>1 was obtained in [7] using an embedding into l∞(X)/c0(X).In this paper we prove thatΓX(t)=t in Ko¨the sequence spaces without absolutely continuous norm. Furthermore, the formula of the modulus of nearly uniform smoothness in Orlicz sequence spaces equipped with the Luxemburg norm is given. We find the criteria for nearly uniform smoothness of Orlicz sequence spaces equipped with the Luxemburg. Finally, we prove that the condition of R(a,l(Φ))<1+a is equivalent to Φ∈δ2. RW(a,l(Φ))=Γl(Φ)(a) and RW(a,l(Φ))<1+a if and only if exist δ>0 and u0>0 such that Φ(u/1+a)≤1-δ/1+aΦ(u) or Φ(u/a)≥1-δ/aΦ(u) when |u|≤u0.


## 2. Preliminaries

Throughout this paper,X is a Banach space which is assumed not to have the Schur property, in which X has a weakly convergent sequence that is not norm convergent. By S(X) and B(X) we denote, respectively, the unit sphere and the unit ball of X, and by l0 we denote the set of all real sequences.Inspiring on the underlying ideas in the definition of nearly uniformly smooth spaces given by Prus [4] as well as on the modulus of uniform smoothness, in [8] Domínguez-Benavides defined the modulus of nearly uniform smoothness.Definition 1.
The modulus of nearly uniform smoothness is defined by(4)ΓXt≔supinfn>1x1+txn+x1-txn2-1,where the supremum is taken over all basic sequences(xn) in B(X).Definition 2.
For each0<a≤1, we define (5)RWa,X≔supliminfn→∞xn+x∧liminfn→∞xn-x:xn∈BX,xn→ω0,x≤a.In [9], they have the important result that if a Banach space X is reflexive and RW(a,X)<1+a for some a>0 then Γ′X(0)<1 that the Banach space X has the FPP.They use the above result to give a direct proof that the uniformly nonsequare Banach space has the FPP.In paper [8] it is shown that a Banach space X is nearly uniformly smooth if and only if X is reflexive and (6)Γ′X0≔limt→0+ΓXtt=0 and (7)ΓXt≔supinfn>1x1+txn+x1-txn2-1, where the supremum is taken over all weak null sequences (xn) in B(X).The following theorem was proved in [10].Theorem 3.
IfX is a reflexive Banach space and Γ′X(0)<1/2, X has the FPP.Definition 4.
A Banach spaceX is called a Ko¨the sequence space if for every x∈l0 and y∈X satisfying |x(i)|≤|y(i)| for all i∈N, we have x∈X and x≤y and if there is a x=(x(i))⊂X with x(i)>0 for all i∈N(see [11, 12]). Let (8)Xa≔x∈X:limn→∞0,0,…0,xn+1,xn+2,…=0.A Köthe sequence spaceX is said to be norm absolutely continuous if Xa=X.Definition 5.
A Köthe sequence spaceX is said to have the semi-Fatou property if for every sequence {xn}⊂X and x∈X satisfying |xn(i)|↑|x(i)| for all i∈N there holds xn→x.Definition 6.
Φ : R → R + is called an Orlicz function if it has the following properties:
(1)Φ is even, continuous, convex and Φ(0)=0;
(2)Φ(u)>0 for all u≠0.Definition 7.
For every Orlicz functionΦ we define the complementary function Ψ:R→[0,∞] by the formula (9)Ψu≔supu>0uv-Φu for every v∈R.Definition 8.
We say thatΦ satisfies the δ2-condition (Φ∈δ2 for short) if there exist k>0,u0>0 such that the inequality (10)Φ2u≤kΦu whenever |u|≤u0.We say thatΦ satisfies the δ2¯-condition (Φ∈δ2¯ for short) if its complementary function Ψ satisfies the δ2-condition.The Orlicz sequence spacel(Φ) is defined to be the set {x∈l0:IΦ(λx)=∑i=1∞‍Φ(λx(i))<∞ for some λ>0} equipped with the Luxemburg norm (11)xΦ≔infk>0:IΦxk≤1. To simplify notations, we put l(Φ)=(lΦ,·(Φ)).The basic informations on Orlicz spaces can be found in [13–15].


## 3. Main Results

Theorem 9.
LetX be a Ko¨the sequence space with the Semi-Fatou property. If the norm of X is not absolutely continuous, ΓX(t)=t.Proof.
Suppose that the norm ofX is not absolutely continuous. Then there exist ε0>0 and x0∈S(X) such that (12)limn→∞∑i=n+1∞x0iei=ε0,where ei=(0,0,…,1ith,0,…).
Take anyε>0. By limn→∞∑i=n+1∞‍x0(i)ei=ε0, there exists n1∈N such that (13)∑i=n1∞x0iei≤1+εε0. Notice that (14)limm→∞∑i=n1+1mx0iei=∑i=n1+1∞x0iei, so there exists n2>n1 such that (15)1-εε0≤∑i=n1+1n2x0iei≤1+εε0. In this way, we get a sequence {ni} of natural numbers such that (16)1-εε0≤∑j=ni+1ni+1x0iei≤1+εε0,i=1,2,…. Put xi=∑j=ni+1ni+1‍x0(j)ej and vk=∑j=nk+1∞‍x0(j)ej. Then
(a)xi→ε0 as i→∞;
(b)xi→w0 as i→∞. It is well known that, for any Köthe space X, we have (17)X∗=X′⊕S, where S is the space of all singular functionals over X. X′={y∈l0:theseries∑i=1∞‍x(i)y(i) converges for all x∈X}. This means that every f∈X∗ is uniquely represented in the form (18)f=Ty+φ, where φ∈S and for y∈X′ the function Ty is defined by (19)Tyx=∑i=1∞xiyi for all x∈X.
Consider that the series∑i=1∞‍x(i)y(i) converges. We have (20)limn→∞∑j=1∞xnjyj=limi→∞∑j=ni+1ni+1xijyj=0.
(c) Putzi=vi/vi and wk=xk-xk+1/xk-xk+1 for all i,k∈N. Then for k>i the following holds: (21)12zi+twn+zi-twn=12vivi+txk-xk+1xk-xk+1+vivi-txk-xk+1xk-xk+1≥12xkvi+txkxk-xk+1+xk+1vi+txk+1xk-xk+1≥1-εε01vi+t1xk-xk+1≥1-ε1+ε1+t.
By the arbitrariness ofε, we get ΓX(t)≥t. It is clear that ΓX(t)≤t. Therefore ΓX(t)=t.Theorem9 also can be derived from the result of Hudzik and Mastyło [16].Corollary 10.
IfΦ∉δ2, ΓX(t)=t for any t>0.Proof.
SinceΦ∉δ2, we have that the norm of l(Φ) is not absolutely continuous. Hence ΓX(t)=t for any t>0.Theorem 11.
LetX be a reflexive Ko¨the sequence space and the norm of X be absolutely continuous. Then (22)ΓXt≔supinfn>1x1+txn-1, where the supremum is taken over all basic sequences (xn) in S(X).Proof.
It is clear that if{xn} is a basic sequence in B(X), (xn/xn) is also a basic sequence in S(X). We only prove that liminfn→∞(x1+txn-x1-txn)=0. For any ε>0, by the norm of X which is absolutely continuous, there is a i0∈N such that (23)∑i=i0+1∞x1iei<ε, and since {xn} is a basic sequence, we have that the sequence {xn} converge to 0 in coordinate, and there exists a n1∈N such that (24)∑i=1i0xniei<ε whenever n>n1. So (25)x1+txn-x1-txn≤∑i=1i0x1iei+t∑i=i0+1∞xniei-∑i=1i0x1iei-t∑i=i0+1∞xniei+2ε1+t. Noting that X is a Köthe space, we have (26)∑i=1i0x1iei+t∑i=i0+1∞xniei=∑i=1i0x1iei-t∑i=i0+1∞xniei. Therefore (27)x1+txn-x1-txn≤2ε1+t whenever n>n1, that is, liminfn→∞(x1+txn-x1-txn)=0.In order to get the criterion that an Orlicz sequence spacel(Φ) is nearly uniformly smooth, we may assume that Φ∈δ2¯ and Ψ∈δ2 and that the space l(Φ) is reflexive. Put N(x)={i∈N:x(i)≠0}. For any x∈S(l(Φ)) with finite N(x) and t>0, we define cx,n(t) as follows: (28)cx,nt≔supcx,y,t>0:IΦxcx,y,t+IΦtycx,y,t=1fory∈SlΦwithn≤minNy<∞. It is clear that {cx,n(t)}n=1∞ is decreasing. Hence cx(t)=limn→∞cx,n(t).Theorem 12.
Supposed thatΦ∈δ2 and Ψ∈δ2. For an Orlicz sequence space l(Φ), there holds (29)ΓlΦt≔supcxt:x∈SlΦwith  finiteNx-1.Proof.
Let(30)dΦ≔supcxt:x∈SlΦwith  finiteNx. Then for any ε∈(0,dΦ), there exists x(Φ)=1 with finite N(x) such that cx>dΦ-ε. By the definition of cx(t), there exists n1∈N such that (31)supcx,y,t>0:IΦxcx,y+IΦtycx,y=1fory∈SlΦandminNy≥n1>dΦ-ε whenever n≥n1. By the definition of cx(t), there exists y1∈S(l(Φ)) with minN(y1)>n1 such that cx,y1,t>dΦ-ε, that is to say, (32)IΦxdΦ-ε+IΦty1dΦ-ε>1. Hence there exists n2>n1 such that (33)IΦxdΦ-ε+∑i=n1+1n2Φity1idΦ-ε>1. Since n2>n1, we also have (34)supcx,y>0:IΦxcx,y+IΦtycx,y=1andminNy≥n2>dΦ-ε, and there exists y2∈l(Φ) with minN(y2)>n1 such that cx,y2>dΦ-ε, that is to say, (35)IΦxdΦ-ε+IΦty2dΦ-ε>1. Hence there also exists y2∈S(l(Φ)) with minN(y2)>maxN(y1) such that (36)IΦxdΦ-ε+∑i=n2+1n3Φity2idΦ-ε>1. In such a way, there exist a sequence {yk}k=1∞⊂S(l(Φ)) and a sequence of natural numbers n1<n2<n3<⋯ such that (37)IΦxdΦ-ε+∑i=nk+1nk+1ΦtykidΦ-ε>1 for all k∈N. It is clear that yk→w0.
For anyk>i0, we have (38)IΦx+tykdΦ-ε=∑i=nk+1nk+1ΦitykidΦ-ε+IΦxidΦ-ε>1, that is to say, (39)x+tykΦ>dΦ-ε. Therefore Γl(Φ)(t)≥dΦ-ε-1 and by the arbitrariness of ε, we have Γl(Φ)(t)≥dΦ-1.
Now we prove thatΓl(Φ)(t)≤dΦ-1. By the definition of dΦ, we always have (40)limn→∞supcx,y,t>0:IΦxcx,y,t+IΦtycx,y,t=1fory∈SlΦandminNy≥n≤dΦ for any x∈S(lΦ) with finite N(x).
Taking any weak null sequence{xn}⊂S(l(Φ)) and ε>0, by Φ∈δ2 there exists i0>0 such that ∑i=i0+1∞‍x1(i)ei(Φ)<ε. Since xn→w0, there is a n0∈N such that ∑i=1i0‍xn(i)ei(Φ)<ε when n≥no. Hence (41)x1+txnΦ≤∑i=1i0x1i+txniei+∑i=i0+1∞x1i+txnieiΦ≤∑i=1i0x1iei+∑i=i0+1∞txnieiΦ+1+tε. We next estimate the ∑i=1i0‍x1(i)ei+∑i=i0+1∞‍txn(i)ei(Φ), when n>n0. Put z1=∑i=1i0‍x1(i)ei/∑i=1i0‍x1(i)ei(Φ) and zn=∑i=i0+1∞‍xn(i)ei/∑i=i0+1∞‍xn(i)ei(Φ) for n>n0. There exists n1≥n0 such that if IΦz1/cz1,zn,t+IΦtzn/cz1,zn,t=1, then cz1,zn,t≤dΦ+ε for n≥n1. (42)IΦ∑i=1i0z1iei+∑i=i0+1∞tznieidΦ+ε=IΦz1dΦ+ε+IΦtzndΦ+ε≤IΦz1cz1,zn,t+IΦtzncz1,zn,t=1, that is, (43)∑i=1i0z1iei+∑i=i0+1∞tznieiΦ≤dΦ+εwhen n≥n0.
Whenn≥n1. It follows that (44)∑i=1i0xiei-∑i=i0+1∞xnieiΦ≤∑i=1i0x1ieiΦ∑i=1i0z1iei+∑i=i0+1∞xnieiΦ∑i=i0+1∞tznieiΦ≤∑i=1i0z1iei+∑i=i0+1∞tznieiΦ≤dΦ+ε. By the arbitrariness of ε, we get the inequality Γl(Φ)(t)≤dΦ-1. Summing up, we obtain the equality Γl(Φ)(t)=dΦ-1.Lemma 13.
LetΦ∈δ¯2. Then for any ε>0 and u0>0, there exists δ>0 such that (45)Φtu≤tεΦu whenever |u|≤Φ-1(1) and t∈(0,δ).Proof.
ByΦ∈δ¯2, for any u0>0, there exists θ∈(0,1) such that (46)Φu2≤θ2Φu whenever |u|≤Φ-1(1).
For anyε>0, there is a n∈N such that θn<ε. Put δ=1/2n. Then (47)Φδu=Φu2n≤θn2nΦu≤δεΦu whenever |u|≤Φ-1(1). Hence (48)Φtu=Φtδδu≤tδΦδu≤tδδεΦu=tεΦu whenever t∈(0,δ) and |u|≤Φ-1(1).Corollary 14.
An Orlicz sequence spacel(Φ) is nearly uniformly smooth if and only if Φ∈δ2 and Φ∈δ¯2.Proof.
SinceΦ∈δ2 and Φ∈δ¯2 imply that the space l(Φ) is reflexive, we only need to prove the sufficiency. For any x∈S(lΦ) with finite N(x) that is max{i:i∈N(x)}<n, and y∈S(lΦ) with minN(y)≥n. For any ε>0, by Lemma 13, there exists δ>0 such that Φ(tu)≤tεΦ(u) when |u|≤Φ-1(1) and t∈(0,δ). Hence if cx,y,t≥1 and t∈(0,δ)(49)1=IΦxcx,y,t+IΦtycx,y,t≤1cx,y,tIΦx+tεcx,y,tIΦy=1cx,y,t1+tε. This shows that cx,y,t≤1+tε, that is, Γl(Φ)≤tε when t∈(0,δ). Therefore Γ′l(Φ)(0)=0.Corollary 15.
Letlp(1<p<∞) be a Lebesgue sequence space. Then Γlp(t)=(1+tp)1/p-1.Proof.
It is well known that the Lebesgue sequence spacelp(1<p<∞) is an Orlicz sequence space generated by Φ(u)=|u|p. For any x,y∈S(lp), we now consider the following:(50)IΦxcx,y,t+IΦtycx,y,t=1. That is, (51)∑i=1∞xicx,y,tp+∑i=1∞tyicx,y,tp=1. Hence (52)1cx,y,tp∑i=1∞xip+tpcx,y,tp∑i=1∞yip=1+tpcx,y,tp=1, that is to say, cx,y,t=(1+tp)1/p. So Γl(p)(t)=(1+tp)1/p-1 for t>0.It is also well known that ifΦ is an Orlicz function, limu→0Φ(u)/u=a. If a>0, l(Φ) has Schur property [17].Theorem 16.
Suppose thatΦ∈δ2 and a∈(0,1). RW(a,l(Φ)))<1+a if and only if exist δ>0 and u0>0 such that Φ(u/1+a)≤1-δ/1+aΦ(u) or Φ(u/a)≥1+δ/aΦ(u) when |u|≤u0.Proof.
Necessity. If there exists a sequence un↘0 such that (53)Φun1+a≥1-1/n1+aΦun and (54)Φuna≥1+1/naΦun for each n∈N, take ln∈N such that (55)lnΦun1+a≤1<ln+1Φun1+a. Putting xn=un,un,…,un︷ln,0,…, then IΦxn/1+a→1. So xn(Φ)→1+a as n→∞. (56)xnΦ≤IΦxn=lnΦun≤ln1+a1-1/nΦun1+a≤1+a1-1/n. Therefore (57)limn→∞IΦxn=1+a. Take kn∈N such that (58)knΦun≤1<kn+1Φun. It is clear that kn<ln for n large enough. Let y1(n)=un,un,…,un︷ln-kn,0,…, (59)y2n=0,0,…,0,︷lnun,un,…,un︷kn,0,…,…,⋯ymn=0,0,…,0,︷ln+m-2knun,un,…,un︷kn,0,…,…. Then IΦ(ym(n))→1 and IΦy1(n)=IΦ(xn)-IΦ(ym(n))→a as n→∞.
Since(60)IΦy1na=ln-knΦuna≥1+1/naln-knΦun=1+1/naIΦy1n→1, that is, y1(n)(Φ)→a.
By(61)limλ→0supIΦλymnλ:m=1,2,…=limλ→0IΦλxnλ=0, we have ym(n)→w0 as m→∞.
Therefore(62)RWa,lΦ≥supliminfm→∞y1n+ymnΦ:n=1,2,…=1+a, that is to say RW(a,l(Φ))=1+a.
Sufficiency. Suppose that there exist δ>0 and u0>0 such that Φu/1+a≤1-δ/1+aΦ(u) or Φu/a≥1+δ/aΦ(u) when |u|≤u0. That means there exist δ>0 and u0>0 such that Φ(u)≤1-δ/1+aΦ((1+a)u) or Φ(u)≤a/1+δΦu/a when |u|≤u0. It is well known that if Φ is an Orlicz function, limu→0Φ(u)/u=b. If b>0, l(Φ) has Schur property. That is to say, the norm convergence of sequence in l(Φ) coincides with the weak convergence. In order to discuss the coefficient RW(a,l(Φ)) we may assume b=0. Put f(u)=(1+a)Φ(u)/Φ((1+a)u). By b=0, we have f(u)<1. Since the function f is continuous in [u0,Φ-1(1)], there exists a u1∈[u0,Φ-1(1)] such that f(u)≤f(u1) for each u∈[u0,Φ-1(1)]. Suppose f(u1)=1. Then (1+a)Φ(u1)/Φ((1+a)u1)=1 and Φ(u) is affine function in [0,v1], where v1=(1+a)u1. This contradicts with b=0. Therefore (1+a)Φ(u)/Φ((1+a)u)=f(u1). This implies Φ(u)≤f(u1)/1+aΦ((1+a)u) when u∈[u0,Φ-1(1)]. It is shown that there exist δ>0 and u0>0 such that Φ(u)≤1-δ/1+aΦ((1+a)u) or Φ(u)≤a/1+δΦu/a when |u|≤Φ-1(1). Using Φ∈δ2 there exists a β>0 such that IΦ(x)≥β when x(Φ)=a.
Put(63)Ba=u∈0,Φ-11:Φua≥1+δaΦu and (64)B1+a=u∈0,Φ-11:Φu1+a≥1-δ1+aΦu. Let Na={i:x(i)∈Ba} and N1+a={i:x(i)∈B1+a}.
We consider the following two cases.
Case I (∑i∈Na‍Φ(x(i))≥β/2). In this case, from (65)1≥IΦxa=∑i∈NaΦxia+∑i∈N1+aΦxia≥1+δa∑i∈NaΦxi+1a∑i∈N1+aΦxi=1aIΦx+δβ2a, we get (66)IΦx≤a-δβ2a. For any y∈B(lΦ) with supp(y)={i∈N:y(i)≠0}∩supp(x)=∅, the following inequality holds: (67)IΦx+y=IΦx+IΦy≤a-δβ2a+1. Therefore x+y(Φ)≤IΦ(x+y)≤a-δβ/2a+1, that is to say RW(a,l(Φ))≤a-δβ/2a+1<1+a.
Case II (∑i∈N1+a‍Φ(x(i))≥β/2). In this case, for any y∈B(lΦ) with supp(y)={i∈N:y(i)≠0}∩supp(x)=∅, the following inequality holds: (68)IΦx+y1+a=IΦx1+a+IΦy1+a≤∑i∈NaΦxi1+a+∑i∈N1+aΦxi1+a+11+aIΦy≤11+a∑i∈NaΦxi+1-δ1+a∑i∈N1+aΦxi+11+a=11+aIΦx+11+a-δβ21+a. From IΦ(x)≤x(Φ)=a≤1, we have (69)IΦx+y1+a≤1-δβ21+a, that is to say (70)x+yΦ≤2+a-δβ2. Therefore RW(a,l(Φ))≤2+a-δβ/2<1+a.Theorem 17.
Fora∈(0,1), R(a,l(Φ))<1+a if and only if Φ∈δ2.Proof.
It is easy to get the following formula:(71)Ra,X=supliminfn→∞xn+xΦ:xn→w0,xn,x⊂BX,xn-xmΦ≤1. Take any weak null sequence {xn}⊂B(l(Φ)) with xn-xm(Φ)≤1. Hence IΦ(xn-xm)≤1. Using xn→w0, for any ε∈(0,1/2) and subsequence {xni} of {xn} such that IΦ(xni)≤1/2+ε. By Φ∈δ2, there exists δ∈(0,1) such that xni(Φ)≤δ. Therefore for any x∈lΦ,x(Φ)=a. It holds (72)xni+xΦ≤xniΦ+xΦ≤δ+a<1+a.



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