# Kadec-Klee Property in Orlicz Function Spaces Equipped with S-Norms

Authors: Jiaqi Dong; Yunan Cui; Marek Wisła
Journal: Journal of Function Spaces
 (2022)

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## Abstract

Using some new techniques, the necessary and sufficient conditions for Kadec-Klee property of Orlicz function spaces equipped with s-norms are presented. An original method that was used in the process of inquiry and the obtained results also systematically complete and broaden the characterization of Kadec-Klee property of Orlicz spaces.


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## Body

## 1. Introduction and Preliminaries

Orlicz spaces, introduced by W. Orlicz in 1932, form a wide class of Banach spaces of measurable functions (in the case of atomless measure) or sequences (in the case of counting measure) (see [1]). On Orlicz spaces of measurable functions, the classical Orlicz and Luxemburg norm can be defined by use of the Amemiya formula: xΦo=infk>01/k1+IΦkx and xΦ=infk>01/kmax1,IΦkx, respectively, where Φ is an Orlicz function and IΦx=∫GΦxtdt. Based on these statements, H. Hudzik and L. Maligrada introduced Orlicz spaces equipped with the p-Amemiya norms, where p∈1,∞ and defined xΦp=infk>01/k1+IΦpkx1/p in 2000 (see [2, 3]). M. Wisła presented a universal and general method of introducing norms (s-norms) in Orlicz spaces in 2019, and the introduction of Orlicz spaces equipped with s-norms covers all the cases mentioned above.


### 1.1. Introduction

In the following, byN we will denote the sets of natural numbers and R and R+ the sets of real and nonnegative real numbers, respectively. By SX and BX, we will denote the unit sphere and the unit ball of the Banach space X, respectively.LetG,Σ,μ be a finite nonatomic measure space, and L0=L0μ be the set of all (μ-equivalence) classes of Σ-measurable real functions defined on G.Definition 1.
A continuous functionΦ:R⟶R+ is called convex if(1)Φu+v2≤Φu+Φv2,for all u,v∈R. If, in addition, the two sides of formula (1) are not equal for all u≠v, then we say Φ is strictly convex.Definition 2.
A functionΦ: R⟶R+ goes by name of an Orlicz function if Φ is nonnegative, even, convex satisfying Φ0=0 and limu⟶∞Φu/u=∞.
Ψv=supuv−Φu:u≥0 for all v∈R is called the complementary function of Φu in the sense of Young. Obviously, Ψ is also an Orlicz function.Definition 3.
IΦ is introduced as a modular of Φ by(2)IΦu=∫GΦutdt,for all u∈L0μ.
Then, the Orlicz spaceLΦ and its subspace EΦ are generated by an Orlicz function and Φ are linear spaces of measurable functions defined by the following formulas:(3)LΦ=u:IΦλu<∞for someλ>0;EΦ=u:IΦλu<∞for anyλ>0.
For eachu∈LΦ, let Luxemburg norm be defined by(4)xΦ=infk>0,IΦxk≤1,and Orlicz norm by(5)xΦo=infk>01k1+IΦkx.Definition 4 (see [4]).
A functions:0,∞⟶1,∞ will be called an outer function, if it is convex and(6)Maxu,1≤su≤u+1forallu≥0.Definition 5 (see [4]).
Lets be an outer function and Φ be an Orlicz function. Then, the function(7)xΦ,s=infk>01ksIΦkx,is a norm that will be called s-norm on the Orlicz space LΦ.
Combing the definitions of various norms, the inequality(8)xΦ≤xΦ,s≤xΦo,holds. Denote(9)LΦ,s=LΦ,⋅Φ,s,EΦ,s=EΦ,⋅Φ,s.Definition 6 (see [4]).
s∗ is an outer function that is conjugate to s in the Hölder sense.
For anyx∈LΦ,s and y∈LΨ,s∗ the following Hölder inequality holds.(10)∫Gxtytdt≤xΦ,syΨ,s∗.Definition 7 (see [4]).
For an outer functions and its right-hand derivative s+′, define(11)ωv=∫0vs′+−1tdt,whenever v∈0,1.
For allu∈0,∞, v∈0,∞, define(12)βsu,v=1−ωs+′u−vs+′u,and for all x∈LΦ,s/0,(13)k∗x=infk>0:βsIΦkx,IΨp+kx≤0,k∗∗x=supk>0:βsIΦkx,IΨp+kx≥0.Lemma 1 (see [4]).
For anyx∈LΦ,s/0, if and only if k∈0,∞∩kx, where kx=k∗x,k∗∗x for short, we have(14)xΦ,s=1ksIΦkx.Definition 8.
We say that an Orlicz functionΦ satisfies condition Δ2 (Φ∈Δ2, for short) if there exists K>0 and u0>0, such that(15)Φ2u≤KΦu,u≥u0.Definition 9.
LetX be a Banach space. If xn∈SX, x∈SX, and xn⟶wx imply xn⟶xn⟶∞, then we say that X has the Kadec-Klee property (see [5–7]).Lemma 2.
For anyf∈LΦ,s∗ has a unique decomposition,(16)f=φ+vφ∈F,v∈EΦ,s∗,where F is the set of all singular functionals on EΦ,s.Lemma 3 (see [8]).
IfΦ is strictly convex, then for any a,b⊂0,1 and D,ε>0, there exists δ>0, such that(17)Φλu+1−λv≤1−δλΦu+1−λΦv,whenever λ∈a,b,u≤D,v≤D,u−v≥ε.Lemma 4 (see [8]).
A subsetA=xn⊂LΦ,s is LΨ,s∗ weakly compact (i.e., there exists a subsequence xni⊂xn and x∈LΦ,s, such that vxni⟶vx for each v∈LΨ,s∗) if(18)limλ⟶0supIΦλxnλ=0.Lemma 5 (see [8]).
For anyxn,x∈LΦ,s, assume that IΦknxn⟶IΦk0x and knxnt⟶μk0xt; if in addition, Φ∈Δ2, then(19)‖xn−x‖Φ,s⟶0.


## 1.1. Introduction

In the following, byN we will denote the sets of natural numbers and R and R+ the sets of real and nonnegative real numbers, respectively. By SX and BX, we will denote the unit sphere and the unit ball of the Banach space X, respectively.LetG,Σ,μ be a finite nonatomic measure space, and L0=L0μ be the set of all (μ-equivalence) classes of Σ-measurable real functions defined on G.Definition 1.
A continuous functionΦ:R⟶R+ is called convex if(1)Φu+v2≤Φu+Φv2,for all u,v∈R. If, in addition, the two sides of formula (1) are not equal for all u≠v, then we say Φ is strictly convex.Definition 2.
A functionΦ: R⟶R+ goes by name of an Orlicz function if Φ is nonnegative, even, convex satisfying Φ0=0 and limu⟶∞Φu/u=∞.
Ψv=supuv−Φu:u≥0 for all v∈R is called the complementary function of Φu in the sense of Young. Obviously, Ψ is also an Orlicz function.Definition 3.
IΦ is introduced as a modular of Φ by(2)IΦu=∫GΦutdt,for all u∈L0μ.
Then, the Orlicz spaceLΦ and its subspace EΦ are generated by an Orlicz function and Φ are linear spaces of measurable functions defined by the following formulas:(3)LΦ=u:IΦλu<∞for someλ>0;EΦ=u:IΦλu<∞for anyλ>0.
For eachu∈LΦ, let Luxemburg norm be defined by(4)xΦ=infk>0,IΦxk≤1,and Orlicz norm by(5)xΦo=infk>01k1+IΦkx.Definition 4 (see [4]).
A functions:0,∞⟶1,∞ will be called an outer function, if it is convex and(6)Maxu,1≤su≤u+1forallu≥0.Definition 5 (see [4]).
Lets be an outer function and Φ be an Orlicz function. Then, the function(7)xΦ,s=infk>01ksIΦkx,is a norm that will be called s-norm on the Orlicz space LΦ.
Combing the definitions of various norms, the inequality(8)xΦ≤xΦ,s≤xΦo,holds. Denote(9)LΦ,s=LΦ,⋅Φ,s,EΦ,s=EΦ,⋅Φ,s.Definition 6 (see [4]).
s∗ is an outer function that is conjugate to s in the Hölder sense.
For anyx∈LΦ,s and y∈LΨ,s∗ the following Hölder inequality holds.(10)∫Gxtytdt≤xΦ,syΨ,s∗.Definition 7 (see [4]).
For an outer functions and its right-hand derivative s+′, define(11)ωv=∫0vs′+−1tdt,whenever v∈0,1.
For allu∈0,∞, v∈0,∞, define(12)βsu,v=1−ωs+′u−vs+′u,and for all x∈LΦ,s/0,(13)k∗x=infk>0:βsIΦkx,IΨp+kx≤0,k∗∗x=supk>0:βsIΦkx,IΨp+kx≥0.Lemma 1 (see [4]).
For anyx∈LΦ,s/0, if and only if k∈0,∞∩kx, where kx=k∗x,k∗∗x for short, we have(14)xΦ,s=1ksIΦkx.Definition 8.
We say that an Orlicz functionΦ satisfies condition Δ2 (Φ∈Δ2, for short) if there exists K>0 and u0>0, such that(15)Φ2u≤KΦu,u≥u0.Definition 9.
LetX be a Banach space. If xn∈SX, x∈SX, and xn⟶wx imply xn⟶xn⟶∞, then we say that X has the Kadec-Klee property (see [5–7]).Lemma 2.
For anyf∈LΦ,s∗ has a unique decomposition,(16)f=φ+vφ∈F,v∈EΦ,s∗,where F is the set of all singular functionals on EΦ,s.Lemma 3 (see [8]).
IfΦ is strictly convex, then for any a,b⊂0,1 and D,ε>0, there exists δ>0, such that(17)Φλu+1−λv≤1−δλΦu+1−λΦv,whenever λ∈a,b,u≤D,v≤D,u−v≥ε.Lemma 4 (see [8]).
A subsetA=xn⊂LΦ,s is LΨ,s∗ weakly compact (i.e., there exists a subsequence xni⊂xn and x∈LΦ,s, such that vxni⟶vx for each v∈LΨ,s∗) if(18)limλ⟶0supIΦλxnλ=0.Lemma 5 (see [8]).
For anyxn,x∈LΦ,s, assume that IΦknxn⟶IΦk0x and knxnt⟶μk0xt; if in addition, Φ∈Δ2, then(19)‖xn−x‖Φ,s⟶0.


## 2. Main Results

Theorem 1.
Orlicz spaceLΦ,s has the Kadec-Klee property if and only if(1)
Φ∈Δ2(2)
Φ is strictly convex on RProof.
Necessity. (i) IfΦ∉Δ2, there exists x∈SLΦ,s/EΦ,s, such that limn⟶∞x−xnΦ,s=θx>0, where θx=infλ>0,IΦx/λ<∞ and IΦx−xn⟶0 as n⟶∞.
For anyn,m>0, put(20)Gn=t∈G:|xt|≤n,Gn,m=t∈G:n≤|xt|≤m.
Usinglimm⟶∞xχGn,mΦ,s, we have(21)limm⟶∞xχGn,mΦ,s≥θx.
Hence, for eachn∈N, there exist mn>n>0 satisfying xχGn,mnΦ,s≥θx/2. Define xn¯=xχGn,mn. Then,(22)xn¯Φ,s≥θx2.
Let us prove thatxn¯⟶w0. For any f=v+φ∈LΦ,s∗, we have φxn¯=0, thanks to xn¯∈EΦ,s and(23)fxn¯=vxn¯=∫Gvtxn¯tdt=∫Gn,mnvtxtdt.
Using∫Gvtxtdt<∞ and limn⟶∞mGn,mn=0, the quality(24)limn⟶∞∫Gn,mnvtxt=0,holds.
Putynt=xt−xn¯t. Then, yn⟶wx, and(25)1≥ynΦ,s≥xχG|GnΦ,s⟶xΦ,s=1,that is ynΦ,s=1. But,(26)yn−xΦ,s=xn¯Φ,s≥θx2,this contract with that LΦ,s has the Kadec-Klee property.
(ii) IfΦ is not strictly convex, there exists x∈SLΦ,s, such that x is not an extreme point, i.e., there exists k0∈k∗x,k∗∗x, such that mt∈G:k0xt∉SΦ>0. Therefore, there exist 0<a<b, such that(27)Φu=Au+B,for some A,B∈R+ when u∈a,b.
Put(28)G0=t∈G:k0xt∈a,b,Gn=t∈G0:k0xt∈a+1n,b−1n.
Then,limn⟶∞mGn=mG0. Without loss of generality, we may assume that xt⩾0 for each t∈G0.
Therefore, there exists an0∈N, such that mGn0>0. Let c=1/mGn0∫Gn0xtdt. Then, a/k0<c<b/k0. Put x0t=cχGn0+xtχG|Gn0 and δ=k0/n0b. Then, we have kxt∈a,b for any k∈k0−δ,k0+δ and t∈Gn0. Hence,(29)IΦkxt=∫Gn0Φkxtdt+∫G|Gn0Φkxtdt=∫Gn0Akxt+Bdt+∫G|Gn0Φkxtdt=AkcmGn0+BmGn0+∫G|Gn0Φkxtdt=ΦkcmGn0+∫G|Gn0Φkxtdt=IΦkx0t,i.e., IΦkx0t=IΦkxt. Then, we have that βsIΦkx0,IΨp+kx0≤0 when βsIΦkx,IΨp+kx≤0 and βsIΦkx0,IΨp+kx0≥0 when βsIΦkx,IΨp+kx≥0. By formula (27), we have(30)xΦ,s=1k0sIΦk0xt=1k0sIΦk0x0t=x0Φ,s.
DivideG0 into two disjoint sets G11 and G12, such that(31)mG11=mG12=12mGn0,and divide G11 and G12 into two disjoint sets G21, G22 and G23, G24, respectively, such that(32)mG21=mG22=mG23=mG24=122mGn0.
By induction as above, we obtain a sequenceGni, such that(33)mGn2i−1=mGn2i=12nmG0,for every n∈N and i=1,2,…,2n. Define(34)xnt=c+εχ∪i=12n−1Gn2i−1+c−εχ∪i=12n−1Gn2i+xtχG|Gn0.
Since, fork0∈k∗x,k∗∗x, the following equalities hold:(35)IΦk0xnt=∫G11Φk0c+εdt+∫G12Φk0c−εdt+∫G|Gn0Φk0xtdt=∫Gn0Φk0cdt+∫G|Gn0Φk0xtdt=IΦk0x0t,whence xΦ,s=x0Φ,s=1/k0sIΦk0x0t=1/k0sIΦk0xnt=xnΦ,s.
Letxn¯t=xnt−xtχG|Gn0; we get that limλ⟶0supIΦλxn¯/λ=0, without loss of generality, using Lemma 4, we may assume that xn¯⟶LΨ,s∗x¯ for some x¯∈LΦ,s; using xn¯∈EΦ,s, we have that xn¯⟶wx¯ holds. Put x˜=x¯+xtχG|Gn0. Then, xn⟶wx˜.
Letyt∈SLΨ,s∗ be a support functional of x0t. Denote d=1/mGn0∫Gn0ytdt and y0t=dχGn0+ytχG|Gn0. We get the following:(36)y0x0=∫Gy0tx0tdt=∫Gn0y0tx0tdt+∫G|Gn0ytxtdt=∫Gn0cddt+∫G|Gn0ytxtdt=c∫Gn0ytdt+∫G|Gn0ytxtdt=yx0=x0Φ,s=1,whence y0Ψ,s∗=y0Ψ,s∗⋅x0Φ,s≥y0x0=yx0=xΦ,s=1.
On the other hand, takek>0, satisfying yΨ,s∗=1/ks∗IΨky. Then,(37)y0Ψ,s∗≤1ks∗IΨky0t=1ks∗∫GΨky0tdt=1ks∗∫Gn0Ψkddt+∫G|Gn0Ψkytdt=1ks∗∫Gn01mGn0∫Gn0Ψkytdtdt+∫G|Gn0Ψkytdt=1ks∗IΨkyt=1.
Hence, we have thaty0Ψ,s∗=1 holds.
Sincexn⟶wx˜, x˜Φ,s≤liminfn⟶∞xnΦ,s=1, and x˜Φ,s≥y0x˜=1, so x˜Φ,s=1.
Furthermore,(38)xn−xmΦ,s=2εχG11Φ,s.
It follows thatxn is not a Cauchy sequence, a contradiction.
Sufficiency. Assume thats is an outer function, and Orlicz function Φ is strictly convex satisfying the Δ2 condition. Take xn,x∈SLΦ,s satisfying xn+xΦ,s⟶2 and xn⟶wx.
For anyx∈LΦ,s, let fk=1/ksIΦkx and Ga=t:xt>a for some a>0 with mGa>0. By the definition of s, we have(39)fk=1ksIΦkx≥IΦkxk≥∫GaΦkxkdt≥∫GaΦkakdt≥ΦkakmGa,and(40)fk=1ksIΦkx≥1k.
Sincelimu⟶∞Φu/u=∞, we obtain limk⟶∞fk=∞ and limk⟶0fk=∞, i.e., kx≠∅. Then, there exist kn∈k∗xn,k∗∗xn, such that(41)xnΦ,s=1knsIΦknxn=1.First, we will prove thatxn⟶x in measure. Suppose that knxn↛μk0x as n⟶∞, there exist ε0>0,δ0>0, such that mt∈G:knxn−k0x≥ε0≥δ0 for any n∈N.(i)
Assume thatkn is bounded, and we have supn>0kn=k¯<∞.Hence,(42)0<k0k0+k¯≤k0k0+kn;knk0+kn≤k¯k0+k¯<1.Usingkn=sIΦknxn, we have(43)k¯≥kn=sIΦknxn≥∫GΦknxntdt≥∫t∈G:knxnt>kΦknxntdt≥Φkmt∈G:knxnt>k.Then, there existsk∈N, such that mt∈G:knxnt>k≤k¯/Φk<δ0/3. Set(44)Gn=t∈G:knxnt≤kk0xt≤kknxnt−k0xt≥ε0.Then,mGn<δ0/3. In virtue of Lemma 3, there exists δ1>0, such that(45)Φk0knk0+knxnt+xt≤1−δ1k0k0+knΦknxnt+knk0+knΦk0xt.Whenevert∈Gn. Combining with the definition of the functions s and Φ as well as with formulas (42) and (45), we obtain(46)2←xn+xΦ,s≤k0+knk0knsIΦk0knk0+knxnt+xt≤k0+knk0kns∫GnΦk0knk0+knxnt+xtdt+∫G|GnΦk0+knk0knxnt+xtdt≤k0+knk0kns1−δ1k0k0+kn∫GnΦknxntdt+knk0+kn∫GnΦk0xtdt+k0k0+kn∫G|GnΦknxntdt+knk0+kn∫G|GnΦk0xtdt≤k0+knk0knsk0k0+kn∫GΦknxntdt+knk0+k0∫GΦk0xtdt−sδ1k0k0+kn∫GnΦknxntdt+knk0+kn∫GnΦk0xtdt≤2−k0+knk0knsδ1k0k0+kn∫GnΦknxntdt+knk0+kn∫GnΦk0xtdt≤2−δ11kn∫GnΦknxntdt+1k0∫GnΦk0xtdt≤2−2δ1k¯∫GnΦknxnt+Φk0xtdt2≤2−2δ1k¯∫GnΦknxnt−k0xt2dt<2−2δ1k¯Φε02δ03,A contradiction. Consequently,knxn⟶μk0x as n⟶∞.(ii)
If the sequencekn is not bounded and without loss of generality, we may assume that limn⟶+∞kn=+∞ and set yn=xn+x/2,ln=knk0/kn+k0; then, yn⟶wx, limn⟶∞ynΦ,s=1, and the following inequalities hold.(47)2limn⟶∞ynΦ,s⩽limn⟶∞1lnsIΦlnyn=limn⟶∞kn+k0knk0sIΦk0kn+k0knxn+knkn+k0k0x⩽limn⟶∞1kn1+IΦknxn+1k01+IΦk0x=2,It is clear that the sequenceln is bounded, and(48)limn⟶∞ln=limn⟶∞knk0kn+k0=limn⟶∞k01+k0/kn=k0.Similar as the proof above, we havelnxn⟶μk0x.Second, we will show thatlimn⟶∞kn=k0. We only need to prove that knxnt converges k0xt weakly star.For anyε>0 and v∈EΨ,s∗, using that v has absolutely continuous norm, there exists δ>0, E⊂G, and mE<δ, such that(49)vχEΨ,s∗<ε3k¯.By Yegolov’s theorem, there existse0⊂G with me0<δ for which knxnt⟶k0xt uniformly t∈G|e0.Hence, there isn0∈N satisfying(50)knxn−k0xχG|e0Φ,s⩽ε3vΨ,s∗,Whenevern≥n0 and t∈G|e0.So,(51)∫Gknxntvtdt−∫Gk0xtvtdt≤∫G|e0knxnt−k0xt‖vtdt+∫e0knxntvtdt+∫e0k0xtvtdt≤knxn−k0xχG|e0Φ,svΨ,s∗+k¯xnΦ,svχe0Ψ,s∗+k¯xΦ,svχe0Ψ,s∗≤ε3+ε3+ε3=ε,i.e.,knxn⟶w∗k0x. By xn⟶wx, we can get that kn⟶k0 holds.Finally, by formula (41), we have s−1kn=IΦknxn and s−1k0=IΦk0x. Hence, we get that IΦknxn⟶IΦk0x holds. In virtue of Lemma 5, we obtain xn−xΦ,s⟶0.



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