# Kadec-Klee Properties of Calderón-Lozanovskiĭ Function Spaces

Authors: Paweł Kolwicz
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

We study Kadec-Klee properties with respect to global (local) convergence in measure. First, we present some results concerning Köthe spaces and Orlicz functions.
Next, we shall give full criteria for Kadec-Klee properties with respect to global (local) convergence in measure in Calderón-Lozanovskiĭ function spaces. In particular, we obtain the full characterizations of Kadec-Klee properties in Orlicz-Lorentz function spaces, which have not been presented until now.


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## Body

## 1. Introduction

The Kadec-Klee properties are fundamental notions in the theory of Banach function spaces (see [1, 2]). On the other hand, the Calderón-Lozanovskiĭ spaces are important class of Banach lattices, especially because of their role in the interpolation theory. Geometry of Calderón-Lozanovskiĭ spaces has been intensively developed during the last decades (see, e.g., [3–9]). The complete characterization of Kadec-Klee properties with respect to local Hl (global Hg) convergence in measure for Orlicz function spaces Lφ has been presented in [10]. The Orlicz spaces Lφ are special case of Calderón-Lozanovskiĭ spaces Eφ. Thus, it is natural to investigate properties Hl and Hg in the spaces Eφ generated by arbitrary Köthe spaces E. Since L1∈Hl and Eφ=Lφ when E=L1, so finding full criteria of Hl and Hg in Orlicz spaces is simpler than in general Calderón-Lozanovskiĭ spaces. The sufficient conditions of Hl in Calderón-Lozanovskiĭ function spaces Eφ have been proved in [4, 6]. Note that the implication Eφ∈(Hl)⇒φ∈Δ2E follows from the fact that property Hl implies order continuity in Köthe spaces and from results concerning copy of l∞ in Eφ (see [5, 10]). The implication Eφ∈(Hl)⇒E∈(Hl) has not been discussed at all. Moreover, property Hg in Eφ has not been considered early. We present full criteria of properties Hl and Hg in Calderón-Lozanovskiĭ function spaces Eφ. Furthermore, in the case of property Hg, we prove some general results concerning Köthe spaces which will be essential in studying this property in spaces Eφ. The crucial difficulty in the proofs concerning property Hg in spaces Eφ is the fact that property Hg needs not imply order continuity in general. Note that if E∈(Hl), then E∈(OC) (see [10]), so looking for complete characterizations in the spaces Eφ, the case of property Hl is easier than Hg. As an application, we get the criteria of properties Hl and Hg in Orlicz-Lorentz function spaces (Λω)φ. A sufficient condition for property Hl of (Λω)φ over a finite, atomless measure space has been presented in [6, Theorem  1]. Note also that the case of Kadec-Klee properties with respect to pointwise (uniform) convergence in Calderón-Lozanovskiĭ sequence spaces has been solved in [11].


## 2. Preliminaries

Letℝ, ℝ+, ℕ be the sets of real, nonnegative real, and positive integer numbers, respectively. As usual, S(X) (resp., B(X)) stands for the unit sphere (resp., the closed unit ball) of a real Banach space (X,∥·∥X).Let(T,Σ,μ) be a σ-finite and complete measure space. By L0=L0(T) we mean the set of all μ-equivalence classes of real-valued measurable functions defined on T.A Banach spaceE=(E,∥·∥E) is said to be a Köthe space if E is a linear subspace of L0 and(i)
ifx∈E, y∈L0, and |y|≤|x|  μ-a.e., then y∈E and ∥y∥E≤∥x∥E;(ii)
there exists a functionx in E that is positive on the whole T (see [2, 12]).EveryKöthe space is a Banach lattice under the natural partial order (x≥0 if x(t)≥0 for μ-a.e. t∈T). In particular, if we consider the space E over the nonatomic measure μ, then we will say that E is a Köthe function space.If we replace the measure space (T,Σ,μ) by the counting measure space (ℕ,2ℕ,m), then we will say that E is a Köthe sequence space.The setE+={x∈E:x≥0} is called the positive cone ofE. For any subset A⊂E, define A+=A∩E+.By thesymmetric Köthe function spaceE (symmetric Banach function space) on I, where I=[0,1] or I=[0,∞) with the Lebesgue measure m, we mean a Köthe space E=(E,∥·∥E) with the additional property that for any two equimeasurable functions x~y, x,y∈L0(I) (i.e., they have the same distribution functions dx=dy, where dx(λ)=m({t∈I:|x(t)|>λ), λ≥0) and x∈E we have y∈E and ∥x∥E=∥y∥E. In particular, ∥x∥E=∥x*∥E, where(2.1)x*(t)=inf⁡{λ>0:dx(λ)≤t},t>0.
For basic properties of symmetric spaces and rearrangements, we refer to [13, 14]. In the sequel, considering symmetric Köthe function space E over measure space (T,Σ,μ) we mean T=[0,1] or T=[0,∞) with the Lebesgue measure μ.A pointx∈E is said to have order continuous norm if for any sequence (xm) in E such that 0≤xm≤|x| and xm→0  μ-a.e. we have ∥xm∥E→0. A Köthe space E is called order continuous(E∈(OC)) if every element of E has an order continuous norm (see [2, 12, 15]). As usual, Ea stands for the subspace of order continuous elements of E. It is known that x∈Ea if and only if ∥xχAn∥E↓0 for any sequence {An} satisfying An↓∅ (i.e., An⊃An+1 and μ(⋂n=1∞An)=0). Clearly, μ(⋂n=1∞An)=0 if and only if χAn→0  μ-a.e. in T.Recall thatE is said to have Kadec-Klee property (E∈(KK) for short) whenever ∥xn-x∥→0 for any x and {xn} in E satisfying xn→x in the weak topology σ(E,E*) and ∥xn∥→∥x∥ (see [2]). This property is sometimes called the Radon-Riesz property or property H. It has been considered in many classes of Banach spaces (see [1, 4, 16–18]). If we replace the weak convergence σ(E,E*) by the convergence in measure (xn→μx), by the convergence in measure on every set of finite measure (xn→μx locally), or by the uniform convergence (xn⇉x), then we say that E has the Kadec-Klee property with respect to convergence in measure, local convergence in measure, or uniform convergence, respectively (we will write E∈(Hg), E∈(Hl), and E∈(Hu)). Clearly, E∈(Hl)⇒E∈(Hg)⇒E∈(Hu). Moreover, the converse of any of these implications is not true in general (see Sections 3.1 and 3.2 below).Given a propertyP, we will say that a Köthe space E satisfies property P+(E∈(P)+) if E satisfies property P restricted only to nonnegative elements.In the whole paper,φ denotes an Orlicz function, that is, φ:ℝ→[0,∞], it is convex, even, vanishing, and continuous at zero, left continuous on (0,∞) and not identically equal to zero. Denote(2.2)aφ=sup⁡{u≥0:φ(u)=0},bφ=sup⁡{u≥0:φ(u)<∞}.
We write φ>0 when aφ=0 and φ<∞ if bφ=∞. Let φr=φχGφ, where(2.3)Gφ={[aφ,bφ]ifφ(bφ)<∞,[aφ,bφ)otherwise.
Define on L0 a convex semimodular Iφ by(2.4)Iφ(x)={‖φ∘x‖Eifφ∘x∈E,∞otherwise,
where (φ∘x)(t)=φ(x(t)), t∈T. By the Calderón-Lozanovskiĭ space Eφ we mean(2.5)Eφ={x∈L0:Iφ(cx)<∞for somec>0},
equipped with so called Luxemburg norm defined by(2.6)‖x‖φ=inf⁡{λ>0:Iφ(xλ)≤1}.
We generally assume that if bφ<∞, then aφ<bφ, because when 0<aφ=bφ, then Eφ=L∞ and ∥x∥φ=(1/bφ)∥x∥∞.IfE=L1(e=l1), then Eφ(eφ) is the Orlicz function (sequence) space equipped with the Luxemburg norm. If E=Λω—the Lorentz function space (e=λω—the Lorentz sequence space), then Eφ(eφ) is the corresponding Orlicz-Lorentz function (sequence) space denoted by (Λω)φ((λω)φ) and equipped with the Luxemburg norm (see [5, 8]).We will assume in the whole paper thatE has the Fatou property, that is, if 0≤xn↑x∈L0 with (xn)n=1∞ in E and supn∥xn∥E<∞, then x∈E and ∥x∥E=limn∥xn∥E. Since E has the Fatou property, Eφ has also this property, whence Eφ is a Banach space.We say an Orlicz functionφ satisfies conditionΔ2(0) (resp., Δ2(∞)) if there exist K>0 and u0>0 such that φ(u0)>0  (resp., φ(u0)<∞), and the inequality φ(2u)⩽Kφ(u) holds for all u∈[0,u0] (resp., u∈[u0,∞)). If there exists K>0 such that φ(2u)⩽Kφ(u) for all u⩾0, then we say that φ satisfies conditionΔ2(ℝ+). We write for short φ∈Δ2(0), φ∈Δ2(∞), and φ∈Δ2(ℝ+), respectively.For a Köthe spaceE and an Orlicz function φ, we say that φ satisfies conditionΔ2E (φ∈Δ2E for short) if(1)
φ∈Δ2(0) whenever E↪L∞,(2)
φ∈Δ2(∞) whenever L∞↪E,(3)
φ∈Δ2(ℝ+) whenever neither L∞↪E nor E↪L∞ (see [5]),where the symbolE↪F stands for the continuous embedding of the space E into the space F.Relationships between the modularIφ and the norm ∥·∥φ are collected in [8].


## 3. Results

### 3.1. Köthe Spaces and Orlicz Functions

Given a Köthe function spaceE, we set(3.1)mE=sup⁡(B,χB∈E)sup⁡[(An)⊂B,μ(An)→0]inf⁡n‖χAn‖E.Remark 3.1.
It is easy to see that ifE is a symmetric Köthe function space, then
(3.2)mE=inf⁡A∈Σ,μ(A)>0‖χA‖E.Recall that ifx∈L0, then suppx={t∈T:x(t)≠0}. For a Köthe space E⊂L0(T,Σ,μ), let F⊂E be a subset. By the support of F, we mean the set suppF∈Σ satisfying(i)
for eachx∈F, there is A∈Σ, with μ(A)=0, such that suppx⊂A∪suppF,(ii)
there isx∈F with μ(suppF∖suppx)=0.Lemma 3.2.
Suppose thatE is a Köthe function space. Consider the following statements:(i)
E∈(OC),(ii)
E∈(Hg)+,(iii)
For anyu∈E and each sequence {An} in Σ the implication
(3.3)μ(An)⟶0⟹‖uχAn‖⟶0holds.(iv)
mE=0,(v)
suppEa=T,(vi)
Ea≠{0}.Then (i) ⇒ (iii) ⇒ (iv) ⇒ (v) ⇒ (vi) and (ii) ⇒ (iii). For a symmetric Köthe function space we have additionally (vi) ⇒ (iv). Furthermore, for a nonsymmetric Köthe function space, the implication (vi) ⇒ (iv) is not true in general. Finally, the implications (i) ⇒ (iii) ⇒ (iv) cannot be reversed in general.Proof.
The implications (i)⇒ (iii) ⇒ (iv) and (v) ⇒ (vi) are obvious.
(iv)⇒ (v). If suppEa≠T, then for each x∈E with suppx⊂T∖suppEa we have x∉Ea. Let T0⊂T∖suppEa be such that 0<μ(T0)<∞ and χT0∈E. Then x1=χT0∉Ea and consequently there is a sequence (An) of measurable pairwise disjoint sets with ⋃n=1∞An=T0 and a number δ>0 satisfying ∥χAn∥≥δ for each n (see [19]). Since μ(An)→0, we get mE≥δ>0.
(ii)⇒ (iii). Assume that (iii) does not hold, then we find an element u∈E, a number δ>0, and a sequence (Ai) with μ(Ai)→0 such that infi∥uχAi∥E≥δ. Without loss of generality, we may assume that ∑i=1∞μ(Ai)<∞. Set
(3.4)xk=sup⁡{uχAi:i≥k}∈L0.
Let A=⋃iAi. Clearly, xk≤uχA∈E. Put y=uχA and yk=y-xk. Since {xk} is decreasing, so {yk} is increasing. Next, xk→0 in measure, because, for each ɛ>0,
(3.5)μ{t∈T:xk(t)>ɛ}≤μ{⋃i=k∞Ai}≤∑i=k∞μ(Ai)⟶0ask⟶∞.
In particular, by E∈(FP), ∥yk∥→∥y∥. Finally, ∥yk-y∥=∥xk∥≥∥uχAk∥≥δ. Thus, E∉(Hg)+.
Assume thatE is symmetric. We prove that (vi) ⇒ (iv). If mE>0, then we find a set B with χB∈E and a sequence {An}⊂B, μ(An)→0 such that infn∥χAn∥E≥mE/2. Let x∈E∖{0}. We prove that x∈E∖Ea. There is a number δ>0 such that the set D={t∈T:|x(t)|≥δ} has positive measure. We find a sequence {Ck} in D of pairwise disjoint sets and a subsequence {Ank} of {An} satisfying μ(Ck)≤(1/2k)μ(D) and μ(Ck)≥μ(Ank). Then, by the symmetry of E,
(3.6)‖xχCk‖E≥‖δχCk‖E≥‖δχAnk‖E≥δmE2,
for any k∈ℕ, whence x∉Ea.
Now we will prove that the implication (vi)⇒ (iv) is not true in general in nonsymmetric case. Take
(3.7)E=(L1[0,1]⊕L∞[1,2])1={x∈L0[0,2]:x=u+w,u∈L1[0,1],w∈L∞[1,2],‖x‖=‖u‖1+‖w‖∞}.
Obviously, suppEa=[0,1] and mE=1.
Now we will see that implications (i)⇒ (iii) ⇒ (iv) cannot be reversed in general. If E=Λω—the Lorentz space over (0,∞) with Lebesgue measure and ∫0∞ω<∞, then E∈(Hg) and E∉(OC)(see [1, 8]), whence (iii) ⇏ (i) in general.
Finally, letE=Lφ[0,1]—the Orlicz space with φ<∞ and φ∉Δ2(∞). By the proof of Theorem  1 from [5], we find an element u∈E and a sequence {An} with μ(An)→0 and ∥uχAn∥φ=1. On the other hand, for each B∈Σ with u=χB∈E, we claim that χB∈Ea, whence mE=0. Indeed, for any sequence {An} satisfying An↓∅ and for each λ>0, we have
(3.8)Iφ(λuχAn)=φ(λ)∫01χAndμ=φ(λ)μ(An)⟶0,
whence ∥uχAn∥φ→0. Thus, the implication (iii) ⇒ (iv) cannot be reversed in general.Recall that ifE∈(Hl)+, then E∈(OC) (see [10]) and, as we have seen above, property Hg needs not imply OC in general. However, from the above result, it follows that if E∈(Hg)+, then mE=0 and consequently suppEa=T. Note also that the implication (vi) ⇒ (v) for a symmetric Köthe function space has been proved in [8, Proposition  2.2].Remark 3.3.
It may happen thatμ(A)<∞ and χA∉E only in a nonsymmetric Köthe space. However, if μ(A)<∞ and χA∉E, then there is an increasing sequence (An) of subsets of A satisfying ⋃nAn=A and χAn∈E for each n.Proof.
Letx0 be a weak unit of E, that is, x0(t)>0 for each t∈T. It is enough to take An={t∈A:x0(t)>1/n} for each n.Remark 3.4.
LetE be a Köthe space. If E∈(Hu)+, then ∥xχBk∥→0 for each x∈E+, where Bk={t∈T:x(t)<1/k}.Proof.
Letx∈E+. Clearly, xχT∖Bk↑x and supk∥xχT∖Bk∥<∞. By E∈(FP), we get ∥xχT∖Bk∥→∥x∥. Moreover, x-xχT∖Bk→0 uniformly. Then ∥xχT∖Bk-x∥→0, because E∈(Hu)+.
It is easy to see that the above is not true ifE∉(Hu)+. It is enough to take
(3.9)E=Lω∞(0,∞)={x∈L0:‖x‖=supesst≥0⁡ω(t)|x(t)|<∞},
where ω(t)=t and x(t)=1/t. Then ∥xχBk∥=1 for each k. On the other hand, taking xn=xχ[0,n], we get ∥x∥=∥xn∥=1, xn→x uniformly and ∥xn-x∥=1, whence E∉(Hu)+.It is known thatE∈(Hl) if and only if E∈(Hl)+ whenever E∈(OC) (see [6]). Since (Hl)+⇒(OC) (see [10]), so E∈(Hl) if and only if E∈(Hl)+. The same proof as in [6] shows that if E∈(OC), then E∈(Hg) if and only if E∈(Hg)+. However, the assumption E∈(OC) is now too strong in this case because there are Köthe spaces satisfying E∈(Hg) and E∉(OC). Hence, the following result is more natural and precise.Lemma 3.5.
SupposeE is a Köthe space. Then E∈(Hg) if and only if E∈(Hg)+.Proof.
The necessity is obvious. We prove the sufficiency. LetE∈(Hg)+. Assume that E is a Köthe function space. Let ∥xn∥→∥x∥ and xn→x in measure. For each η>0, we get
(3.10)μ{t∈T:||xn(t)|-|x(t)||≥η}≤μ{t∈T:|xn(t)-x(t)|≥η}⟶0,
whence |xn|→|x| in measure. By the assumption E∈(Hg)+, we conclude that
(3.11)‖|xn|-|x|‖⟶0.
Then we find a subsequence (|xnm|)m of (|xn|)n, an element y∈E+, and a sequence 0≤βm→0 such that ||xnm|-|x||≤βmy for each m (see [12]). Denote (|xnm|)m still by (|xn|)n and (βm) by (βn). Thus, we may assume that ||xn|-|x||≤y for all n. Set u=|x|+y. Let ɛ>0. Applying Remark 3.4, with respective sets
(3.12)Bk={t:u(t)<1k},
take k big enough to satisfy ∥uχBk∥<ɛ/8. Moreover, ∥|xn|-|x|∥<ɛ/8 for sufficiently large n. Setting
(3.13)Cn={t∈T:0≤sgn⁡[x(t)xn(t)]≤1},Dn={t∈T:sgn⁡[x(t)xn(t)]=-1},
we get
(3.14)‖x-xn‖≤‖(|x|-|xn|)χCn‖+‖(x-xn)χDn∩Bk‖+‖(x-xn)χDn∖Bk‖≤5ɛ8+‖(x-xn)χDn∖Bk‖.
Take
(3.15)0<δ<ɛ16kmin⁡{1‖y‖,1‖x‖}.
Denote
(3.16)Fnk={t∈Dn∖Bk:|xn(t)-x(t)|>δ},Gnk={t∈Dn∖Bk:|xn(t)-x(t)|≤δ}.
Note that xn→μx and consequently μ(Fnk)→0 as n→∞. Then, by Lemma 3.2 (iii),
(3.17)‖(x-xn)χDn∖Bk‖≤‖(x-xn)χFnk‖+‖(x-xn)χGnk‖≤‖2uχFnk‖+‖(x-xn)χGnk‖<ɛ8+‖(x-xn)χGnk‖,
for sufficiently large n. Finally, divide set Gnk into two subsets
(3.18)Hnk={t∈Gnk:|x(t)|≥12k},Ink={t∈Gnk:y(t)≥12k}.
Clearly, Hnk∪Ink=Gnk because
(3.19)Gnk⊂T∖Bk={t:|x(t)|+y(t)≥1k}.
Then, by (3.15),
(3.20)‖(x-xn)χGnk‖≤δ(‖χHnk‖+‖χInk‖)≤δ(‖2k|x|χHnk‖+‖2kyχInk‖)≤2kδ(‖|x|χHnk‖+‖yχInk‖)<ɛ4.
Combining (3.14), (3.17), and (3.20), we get ∥x-xn∥≤ɛ.
IfE is a Köthe sequence space, the proof is the same, because for a counting measure space (ℕ,2ℕ,m) for each u∈E and any sequence (An) in 2ℕ with m(An)→0, we have ∥uχAn∥→0.The straightforward calculation shows that we may equivalently takex and {xn} in S(E) in the definition of property Hl (see also the proof below). However, the analogous problem for property Hg is a little more complicated. The crucial point is described in the following lemma.Lemma 3.6.
Suppose thatE is a symmetric Köthe function space. If xn→x in measure and ∥xn∥→∥x∥≠0, then xn/∥xn∥→x/∥x∥ in measure. Moreover, the above implication is not true in general if E is a nonsymmetric Köthe function space.Proof.
Having our assumptions, for eachɛ,η>0, we get
(3.21)μ{t:|xn(t)‖xn‖-x(t)‖x‖|>ɛ}=μ{t:|xn(t)‖xn‖-x(t)‖xn‖+x(t)‖xn‖-x(t)‖x‖|>ɛ}≤μ{t:1‖xn‖|xn(t)-x(t)|+|1‖xn‖-1‖x‖||x(t)|>ɛ}≤μ{t:1‖xn‖|xn(t)-x(t)|>ɛ2}+μ{t:|1‖xn‖-1‖x‖||x(t)|>ɛ2}.
Since ∥xn∥→∥x∥, so ∥xn∥>(1/2)∥x∥ for n≥N0, whence
(3.22)μ{t:1‖xn‖|xn(t)-x(t)|>ɛ2}≤μ{t:|xn(t)-x(t)|>ɛ‖x‖4}<η2,
for n≥N1>N0. Moreover, applying the symmetry of E, we get
(3.23)μ{t:|1‖xn‖-1‖x‖||x(t)|>ɛ2}<η2forn≥N2.
Thus,
(3.24)μ{t:|xn(t)‖xn‖-x(t)‖x‖|>ɛ}<η,
for n≥max{N1,N2}.
Considering the second assertion, take(3.25)E=L1/t31(1,∞)={x∈L0((1,∞),Σ,m):‖x‖=∫1∞|x(t)|1t3dt<∞}.
Set x(t)=t and xn(t)=t+(1/n), then ∥x∥=1, ∥xn∥=1+(1/2n), and xn→x in measure. However,
(3.26)μ{t:|xn(t)‖xn‖-x(t)‖x‖|>1}=μ{t:|t-22n+1|>1}=∞,
for each n, whence (xn/∥xn∥)↛(x/∥x∥) in measure.Corollary 3.7.
Assume thatE is a symmetric Köthe function space. Then E∈(Hg) if and only if ∥xn-x∥→0 for any x∈S(E) and (xn) in S(E) satisfying xn→x in measure.Proof.
The necessity is obvious. We prove the sufficiency. Letxn→x in measure and ∥xn∥→∥x∥≠0. Taking yn=xn/∥xn∥ and y=x/∥x∥, we get yn→y in measure, by Lemma 3.6. By the assumption, we conclude that ∥yn-y∥→0. Notice that
(3.27)|1‖xn‖‖xn-x‖-‖x‖|1‖xn‖-1‖x‖||≤‖xn‖xn‖-x‖xn‖-(-x‖xn‖+x‖x‖)‖=‖xn‖xn‖-x‖x‖‖⟶0.
Since ∥xn∥→∥x∥≠0, so ∥xn-x∥→0 as desired.We will need in the sequel the following.Lemma 3.8.
Ifφ∈Δ2(∞), then for any a>0, there are σ∈(0,1] and u1>aφ+a such that
(3.28)inf⁡u0∈(0,a)inf⁡u≥u1φ(u-u0)φ(u)-φ(u0)≥σ.Proof.
Suppose conversely that there isa>0 such that for any σ∈(0,1] and each u1>aφ+a, there holds
(3.29)inf⁡u0∈(0,a)inf⁡u≥u1φ(u-u0)φ(u)-φ(u0)<σ.
Taking σn=1/2n and un→∞, we get infu0∈(0,a)infu≥unφ(u-u0)/(φ(u)-φ(u0))<1/2n. We may assume that un>2a for each n. Hence, for each n, one may choose vn≥un and u0∈(0,a) with φ(vn-u0)/(φ(vn)-φ(u0))≤1/2n. Consequently, we get
(3.30)2nφ(12vn)<2nφ(vn-u0)<φ(vn)-φ(u0)≤φ(vn),
for each n. Taking bn=(1/2)vn, one has 2nφ(bn)<φ(2bn), whence φ∉Δ2(∞).Lemma 3.9.
(i) Suppose thatE is a symmetric Köthe space. If xn→x in measure and φ∘xn, φ∘x∈E are finitely valued a.e., then φ∘xn→φ∘x in measure. Furthermore, for a nonsymmetric Köthe space, the above implication is not true in general.
(ii) Ifxn→x in measure, xn,x∈E+ and φr-1∘xn, φr-1∘x are well-defined functions, then φr-1∘xn→φr-1∘x in measure.Proof.
(i) Letxn→μx.
(A) Suppose thatbφ=∞. Set
(3.31)Ak={t∈T:|x(t)|>k}.
Then, by symmetry of E,  μ(Ak)→0. Let η>0 and ε>0. Take k0 such that μ(Ak0)<η/3. We have
(3.32)μ{t∈T:|φ(xn(t))-φ(x(t))|≥ε}≤μ{t∈T∖Ak0:|φ(xn(t))-φ(x(t))|≥ε}+η3.
Since φ is uniformly continuous on the interval [-k0,k0], so for each γ>0 there is δ=δ(γ,k0)>0 such that the implication |u-s|<δ⇒|φ(u)-φ(s)|<γ is true for all u,s∈[-k0,k0]. Set δ=δ(ε,2k0) and
(3.33)An1={t∈T∖Ak0:|xn(t)|>2k0},An2={t∈T∖Ak0:|xn(t)|≤2k0}.
Consequently
(3.34)μ{t∈T:|φ(xn(t))-φ(x(t))|≥ε}≤μ(An1)+μ{t∈An2:|φ(xn(t))-φ(x(t))|≥ε}+η3≤μ{t∈An2:|xn(t)-x(t)|≥δ}+2η3≤η
for sufficiently large n, because An1={t∈T:|xn(t)-x(t)|≥k0}. Thus φ∘xn→μφ∘x.
(B) Assumeφ(bφ)<∞. The proof is simpler than above because xn(t)≤bφ and x(t)≤bφ for μ-a.e. t∈T and φ is uniformly continuous on the interval [-bφ,bφ].
(C) Suppose thatbφ<∞, φ(bφ)=∞. Then xn(t)<bφ and x(t)<bφ for μ-a.e. t∈T. Set
(3.35)Ak={t∈T:|x(t)|>bφ-1k}.
Then μ(Ak)→0, because E is symmetric. Let η>0 and ε>0. Take k0 such that μ(Ak0)<η/3. We have
(3.36)μ{t∈T:|φ(xn(t))-φ(x(t))|≥ε}≤μ{t∈T∖Ak0:|φ(xn(t))-φ(x(t))|≥ε}+μ(Ak0).
Since φ is uniformly continuous on the interval [-u0,u0] for each u0<bφ, so for each u0<bφ and γ>0 there is δ=δ(γ,u0)>0 such that the implication |u-s|<δ⇒|φ(u)-φ(s)|<γ is true for all u,s∈[-u0,u0]. Put δ=δ(ε,bφ-1/2k0) and
(3.37)An1={t∈T∖Ak0:|xn(t)|>bφ-12k0},An2={t∈T∖Ak0:|xn(t)|≤bφ-12k0}.
Consequently
(3.38)μ{t∈T:|φ(xn(t))-φ(x(t))|≥ε}≤μ(An1)+μ{t∈An2:|φ(xn(t))-φ(x(t))|≥ε}+η3≤μ{t∈An2:|xn(t)-x(t)|≥δ}+2η3≤η
for sufficiently large n, because An1={t∈T:|xn(t)-x(t)|≥1/2k0}. Thus φ∘xn→μφ∘x.
Now we will show that in nonsymmetric case the above implication is not true in general. Consider the Lebesgue measure space ((0,∞),Σ,m). Take
(3.39)φ(u)=u2,x=∑nnχ[n,n+1],xk=∑n(n+1k)χ[n,n+1].
Then xk→x in measure. On the other hand, |φ∘xk-φ∘x|=∑n((1/k2)+(2n/k))χ[n,n+1], whence φ∘xk↛φ∘x in measure. Clearly, such element φ∘x can be well defined only in some nonsymmetric Köthe space E.
(ii) First, we claim that for eachɛ>0, there is δ>0 such that the implication
(3.40)|φr-1(u)-φr-1(v)|>ɛ⟹|u-v|>δ
is true for all u,v≥0. First, notice that for each ɛ>0, there is δ>0 such that the implication
(3.41)|φr-1(u)-φr-1(0)|>ɛ⟹|u|>δ
is true for all u≥0, because φr-1 is continuous at zero. Next, defining g(u)=φr-1(u)-φr-1(0), we see that g is subadditive on ℝ+, because g=ψ-1, where ψ is an Orlicz function with ψ>0. Consequently,
(3.42)φr-1(u-v)-φr-1(0)=g(u-v)≥g(u)-g(v)=φr-1(u)-φr-1(v).
Hence, if |φr-1(u)-φr-1(v)|>ɛ, then |φr-1(u-v)-φr-1(0)|>ɛ, whence |u-v|>δ, by (3.41). Thus, the claim is proved. Assume that xn, x∈E+ and φr-1∘xn, φr-1∘x are well-defined functions and xn→x in measure, then, applying (3.40), for each ɛ>0, we get
(3.43)μ{t∈T:|(φr-1∘xn-φr-1∘x)(t)|>ɛ}≤μ{t∈T:|(xn-x)(t)|>δ}⟶0.


### 3.2. Calderón-Lozanovskiĭ Function Spaces

We still assume below thatE is a Köthe function space.It is known thatEφ∈(OC) if and only if E∈(OC) and φ∈Δ2E (see Corollary  12 from [9]). Considering some weaker property than OC, we get the following.Lemma 3.10.
Suppose thatbφ=∞, then Ea≠{0} if and only if (Eφ)a≠{0}.Proof.
Necessity
Letz∈Ea∖{0}. We find a number m>0 such that μ(A)=μ{t∈T:|z(t)|>1/m}>0. Taking w=φ-1(1)χA, we get φ∘w∈Ea. Moreover, since bφ=∞, φ satisfies the local Δ2E condition with respect to the element w (see [9]). Consequently, w∈(Eφ)a, by Theorem  11 from [9].
Sufficiency
We apply Theorem  11 from [9]. Let x∈(Eφ)a∖{0}, then φ∘x∈Ea. If μ{t:|x(t)|>aφ}>0, then μ(Am)>0 for some m∈ℕ, where Am={t∈T:|x(t)|≥(1+(1/m))aφ}. Thus, χAm∈Ea, and we are done, if μ{t:|x(t)|≤aφ}=μ(A)>0 then χA∈Ea∖{0}, by Theorem  11 from [9].Theorem 3.11.
Eφ∈(Hl) if and only if(a)
E∈(Hl),(b)
φ∈Δ2(∞) whenever L∞↪E and φ∈Δ2(ℝ+) provided L∞↪E.Proof.
Since the sufficiency has been proved in [4, Theorem  11] (we need also to apply [7, Lemma  2]), we need only to prove the necessity. Since Eφ∈(Hl), so Eφ∈(OC) (see [10, Proposition  2.1]), and consequently, E∈(OC), φ<∞, and φ∈Δ2E (see [9, Corollary  12]). Notice that E∈(OC) implies that E↪L∞. Indeed, if E↪L∞, then there is a decreasing sequence (An) in Σ with 0<μ(An)→0 and χAn∈E, whence 1=∥χAn∥L∞≤M∥χAn∥E, so E∉(OC). Thus, (b) is proved. Suppose that E∉(Hl). Recall that property Hl can be equivalently considered only on the positive cone E+ [6, Proposition  1]. Consequently, we find x and {xn} in (S(E))+ with xn→μx locally in measure and ∥xn-x∥E≥ɛ. Set yn=φr-1∘xn and y=φr-1∘x, then Iφ(y)=Iφ(yn)=1, whence y,yn∈S(Eφ). Moreover, yn→μy locally in measure, because, by σ finiteness of the measure space (T,Σ,μ), the local convergence in measure can be equivalently replaced by convergence μ-a.e. (also Lemma 3.9(ii) can be applied). It is enough to prove that there is η>0 with
(3.44)‖yn-y‖φ≥η,
for infinitely many n. The similar condition has been proved in [8, Theorem  2.11], but under additional assumption that φ>0. Here, the lack of this assumption requires new techniques in comparison with the respective proof in [8, Theorem  2.11]. We have
(3.45)‖φ∘yn-φ∘y‖E≥ɛ.
Denote
(3.46)An={t∈T:φ(yn(t))≥φ(y(t))},Bn=T∖An.
By (3.45), we get
(3.47)max⁡{‖(φ∘yn-φ∘y)χAn‖E,‖(φ∘yn-φ∘y)χBn‖E}≥ɛ2,
for each n. We assume that ∥(φ∘yn-φ∘y)χAn∥E≥ɛ/2 for infinitely many n∈ℕ, because otherwise the proof is analogous. We divide the proof into two cases.(I)
Suppose thatL∞↪E and φ∈Δ2(∞). Let c=φr-1(ɛ/32∥χT∥E)>aφ. Denoting(3.48)An1={t∈An:yn(t)<c},An2={t∈An:yn(t)≥c},
we get
(3.49)ɛ2≤‖(φ∘yn-φ∘y)χAn‖E≤‖(φ∘yn-φ∘y)χAn2‖E+ɛ4.
Thus,
(3.50)‖(φ∘yn-φ∘y)χAn2‖E≥ɛ4.
Set
(3.51)c1=c+aφ2,Bn1={t∈An2:y(t)<c1},Bn2={t∈An2:y(t)≥c1}.
The remaining proof of the case (I) we divide into two parts.(1)
Suppose that∥(φ∘yn-φ∘y)χBn1∥E≥ɛ/8. By Lemma 3.8, for a=c1, there are σ>0 and u1>0 such that
(3.52)inf⁡u0∈(0,c1)inf⁡u≥u1φ(u-u0)φ(u)-φ(u0)≥σ.
Set
(3.53)Cn1={t∈Bn1:yn(t)<u1},Cn2={t∈Bn1:yn(t)≥u1}.
(a)
Assume that∥(φ∘yn-φ∘y)χCn1∥E≥ɛ/16. Taking 0<α<(c-c1)/u1, we get
(3.54)‖(φ∘(yn-yα))χCn1‖E≥‖(φ(c-c1α))χCn1‖E≥‖φ(u1)χCn1‖E≥‖(φ∘yn-φ∘y)χCn1‖E≥ɛ16.
Thus, ∥yn-y∥φ≥η1=min{1,αɛ/16}.(b)
Let∥(φ∘yn-φ∘y)χCn2∥E≥ɛ/16. Thus, by Lemma 3.8,
(3.55)‖φ∘(yn-y)χCn2‖E≥σ2‖(φ∘yn-φ∘y)χCn2‖E≥σɛ32,
and consequently ∥yn-y∥φ≥η2=min{1,σɛ/32} (cf. [8, Lemma  1.1]).(2)
Assume that∥(φ∘yn-φ∘y)χBn1∥E<ɛ/8, then, by (3.50),
(3.56)‖(φ∘yn-φ∘y)χBn2‖E≥ɛ8.
Since φ∈Δ2(∞), for l=1+ɛ/32 and u1=c1, there is a=a(l,u1)∈(0,1) such that
(3.57)φ((1+a)u)≤lφ(u),
for every u≥c1 (see [20, Theorem  1.13(4)]). Moreover, we can choose a>0 satisfying
(3.58)a1+aφ(c1)‖χT‖E<ɛ32.
Let
(3.59)Bn21={t∈Bn2:(yn-y)(t)<ac11+a},Bn22={t∈Bn2:(yn-y)(t)≥ac11+a},
then, by (3.57) and (3.58), using convexity of φ, we get
(3.60)|φ∘yn-φ∘y|χBn21=(φ∘(yn-y+y)-φ∘y)χBn21≤(a1+aφ∘(1+aa(yn-y))+11+aφ∘((1+a)y)-φ∘y)χBn21≤(a1+aφ∘(1+aa(yn-y))+11+alφ∘y-φ∘y)χBn21=(a1+aφ∘(1+aa(yn-y))+(ɛ/32-a1+a)φ∘y)χBn21.
Note that f(a)=(ɛ/32-a)/(1+a) is a decreasing function of a>0. Moreover, a≤ɛ/32 by convexity of φ and inequality (3.57). Hence, by (3.58), ∥(φ∘yn-φ∘y)χBn21∥E<ɛ/16. Then, by (3.56),
(3.61)‖(φ∘yn-φ∘y)χBn22‖E≥ɛ16.
Since φ∈Δ2(∞), for u3=c1 and β=(1+a)/a, there is k2>0 such that φ(βu)≤k2φ(u) for each u≥u3 (see [20]). Taking 0<γ<a/(1+a) and applying (3.57), we get
(3.62)|φ∘yn-φ∘y|χBn22=(φ∘(yn-y+y)-φ∘y)χBn22≤(φ∘(yn-yγ+y)-φ∘y)χBn22≤(a1+aφ∘(1+aa(yn-yγ))+11+aφ∘((1+a)y)-φ∘y)χBn22≤(a1+ak2φ∘(yn-yγ)+11+alφ∘y-φ∘y)χBn22=(a1+ak2φ∘(yn-yγ)+(ɛ/32-a1+a)φ∘y)χBn22.
Then ∥φ∘((yn-y)/γ)∥E≥ɛ(1+a)/32ak2, and consequently ∥yn-y∥φ≥η3=min{1,γɛ(1+a)/32ak2} (see [8, Lemma  1.1]).
Combining cases (1) and (2), we get (3.44) with η=min{η1,η2,η3}.
(II)
Suppose thatL∞↪E. Since φ∈Δ2(ℝ+), so for every l>1, there is a=a(l)∈(0,1) such that φ((1+a)u)≤lφ(u) for every u≥0 (see [20, Theorem  1.13(4)]). Moreover, for every β>0, there is k>0 such that φ(βu)≤kφ(u) for each u≥0. Then the proof is analogous as in case (I) (it is simpler and shorter).Now we will consider the Kadec-Klee property with respect to global convergence in measure. Theorem3.19 below presents the full characterization of property Hg. We divide the whole proof into several lemmas for readers convenience.Lemma 3.12.
IfEφ∈(Hg)+, then mE=0.Proof.
IfmE>0, then there are δ>0, a set B,μ(B)>0,χB∈E, and a sequence (An)⊂B with μ(An)→0 and infn∥χAn∥E≥δ. Without loss of generality, we may assume that ∑i=1∞μ(Ai)<∞. Set
(3.63)zn=sup⁡{χAi:i≥n}∈L0.
Let A=⋃iAi. Clearly, zn≤χA≤χB∈E. Put y=χA and yn=y-zn. Choose a>0 such that
(3.64)a=bφifφ(bφ)‖χA‖E<1,a∈(aφ,bφ]ifφ(bφ)<∞,φ(bφ)‖χA‖E≥1,a∈(aφ,bφ)ifφ(bφ)=∞,
and φ(a)∥χA∥E=1 in the last two cases. Take x=ay and xn=ayn, then ∥xn∥φ≤∥x∥φ=1. It is obvious that ∥xn∥φ=1 when a=bφ. We show that ∥xn∥φ→1 in the case a<bφ. Note that zn→0 in measure (see the proof of Lemma 3.2), zn↓, and consequently ∥yn∥E→∥y∥E, because E∈(FP). If ∥xn∥φ≤1-α for some α>0 and infinitely many n, then we get
(3.65)1=φ(a)‖χA‖E<φ(a1-α)‖χA‖E⟵φ(a1-α)‖yn‖E=Iφ(xn1-α)≤1,
which is a contradiction. Moreover, xn→x in measure. Since
(3.66)‖x-xn‖φ≥Iφ(x-xn)≥φ(a)‖χAn‖E≥φ(a)δ,
we conclude that Eφ∉(Hg)+.Remark 3.13.
IfEφ∈(Hg)+, then E↪L∞.Proof.
It is immediate, since ifE↪L∞, then mE>0. By Lemma 3.12, we get Eφ∉(Hg)+.From Lemmas3.12 and 3.2, we get immediately the following.Corollary 3.14.
IfEφ∈(Hg)+, then suppEa=T.Lemma 3.15.
IfEφ∈(Hg), then φ(bφ)=∞.Proof.
Suppose thatEφ∈(Hg) and φ(bφ)<∞. By Lemma 3.12, we get mE=0. Let A∈Σ, μ(A)>0, and μ(T∖A)>0 be such that χA∈E and φ(bφ)∥χA∥E≤1/2. Let (An)⊂T∖A be such that 0<μ(An)→0 and φ(bφ)∥χAn∥E≤1/2. Taking x=bφχA and xn=bφχA∪An, we get ∥x∥φ=∥xn∥φ=1, xn→x in measure and ∥xn-x∥φ=1, whence Eφ∉(Hg).Lemma 3.16.
Suppose that (bφ<∞ and φ(bφ)=∞) or (bφ=∞ and φ∉Δ2(∞)). If Ea≠{0}, then Eφ∉(Hg).Proof.
It is enough to apply order isometryP:l∞→Eφ constructed in the proof of [5, Theorem  1] and take z=χℕ and zn=z-en, where en is the nth unit vector in l∞. If y=Pz and yn=Pzn, then yn→y in measure, because, without loss of generality, one can take in that proof a sequence {An} satisfying min{μ(An),∥χAn∥E}≤2-n. Moreover, ∥yn∥φ=∥y∥φ=∥yn-y∥φ=1.Lemma 3.17.
(i) IfL∞↪E, aφ>0, φ(bφ)=∞, then Eφ∉(Hg).
(ii) IfL∞↪E, suppEa=T, aφ=0, bφ=∞ and φ∉Δ2(0),
thenEφ∉(Hg).Proof.
(i) Suppose thatL∞↪E, φ(bφ)=∞, and aφ>0. If T1∪T2=T, T1∩T2=∅, μ(T1)>0, and μ(T2)>0, then χT1∉E or χT2∉E. Assume that χT1∉E. Setting x=aφχT1 and xn=x+anχAn, where (An)⊂T2, χAn∈E, μ(An)→0, and φ(an)∥χAn∥E=1, we get ∥x∥φ=∥xn∥φ=1, xn→x in measure and ∥xn-x∥φ=1, whence Eφ∉(Hg).
(ii) The proof is analogous to that of [10, Proposition  2.3]. In fact, it shows even more that in this case, Eφ∉(Hu). By Lemma 3.10, we get (Eφ)a≠{0}. Let x∈(Eφ)a, x≥0, and ∥x∥φ=1.
First, we prove thatφ∘(2x)∈Ea. Since x∈S((Eφ)a), so φ∘x∈Ea and φ satisfies the local Δ2E condition with respect to the element x by Theorem  11 from [9] (notice aφ=0, whence Δ2E(xχT∖A)=Δ2E(x) in Theorem  11 from [9]). Furthermore, by Lemma  9 from [9], we conclude that φ∘(2x)∈Ea, because bφ=∞.
We claim that there is a sequence{An} in Σ satisfying χAn∉E and Iφ(2xχAn)≤2-n for each n. Since the measure space (T,Σ,μ) is σ-finite, there is an increasing sequence {Cn} with 0<μ(Cn)<∞ for any n∈ℕ and ⋃nCn=T. Applying Remark 3.3, without loss of generality, we may assume that χCn∈E for each n (see also [12]). Consequently, ∥φ∘(2x)χT∖Cn∥E→0. Moreover, χT∖Cn∉E, because χT∉E and χCn∈E. Thus, the claim is proved with {An} being the respective subsequence of {T∖Cn}. Since φ∉Δ2(0), there is a sequence {un} in ℝ+ with un→0 and
(3.67)φ(2un)>2nφ(un),
for any n∈ℕ. Notice that if χA∈Ea, then the submeasure νA(B)=∥χB∥E, B∈Σ∩A has the Darboux property in the interval [0,∥χA∥E] (see [21]). We claim that for any n, there is a set Bn⊂An with
(3.68)φ(un)‖χBn‖E=2-n.
Let n∈ℕ. First, by E∈(FP), there is a set Dn⊂An with μ(Dn)<∞, χDn∈E, and φ(un)∥χDn∥E>2·2-n. Next, since suppEa=T and E∈(FP), there is a set En⊂Dn with χEn∈Ea and φ(un)∥χEn∥E>(3/2)·2-n. Finally, there is a set Bn⊂En with φ(un)∥χBn∥E=2-n. Taking
(3.69)xn=x+un2χBn=xχT∖Bn+(x+un2)χBn,
we can complete arguments as in the proof of [10,Proposition  2.3].Lemma 3.18.
Suppose thatE is a symmetric Köthe function space.(i)
IfL∞↪E, E∈(Hg) and φ∈Δ2(∞), then Eφ∈(Hg).(ii)
IfL∞↪E, E∈(Hg) and φ∈Δ2(ℝ+), then Eφ∈(Hg).Proof.
(i) Applying Lemma3.5 and Corollary 3.7, it is enough to take x,xn∈(S(Eφ))+, n∈ℕ, and xn→x in measure. Then ∥φ∘x∥E=∥φ∘xn∥E=1, by φ∈Δ2(∞)=Δ2E. Moreover, φ∘xn→μφ∘x by Lemma 3.9(i). Consequently, ∥φ∘xn-φ∘x∥E→0, because E∈(Hg). Applying superadditivity φ on ℝ+, we get
(3.70)‖φ∘(xn-x)‖E≤‖φ∘xn-φ∘x‖E⟶0.
Take λ>1. We need to show that
(3.71)‖φ∘(λ(xn-x))‖E⟶0.
Let ɛ>0. The remaining part of the proof we divide into two parts.
(A) Suppose thataφ>0. Take δ>0 such that φ((1+δ)aφ)∥χT∥E<ɛ/3. Put
(3.72)An={t∈T:λ|xn(t)-x(t)|>(1+δ)aφ},An1={t∈An:|xn(t)-x(t)|>(1+δ)aφ},An2={t∈An:|xn(t)-x(t)|≤(1+δ)aφ}.
By φ∈Δ2(∞), there is K(λ,(1+δ)aφ)=K>0 such that φ(λu)≤Kφ(u) for each u≥(1+δ)aφ. Consequently, by (3.70),
(3.73)‖φ∘(λ(xn-x))‖E≤‖φ∘(λ(xn-x))χAn‖E+‖φ∘(λ(xn-x))χT∖An‖E≤‖φ∘(λ(xn-x))χAn1‖E+‖φ∘(λ(xn-x))χAn2‖E+φ((1+δ)aφ)‖χT∖An‖E≤K‖φ∘(xn-x)χAn1‖E+φ(λ(1+δ)aφ)‖χAn2‖E+ɛ3<23ɛ+φ(λ(1+δ)aφ)‖χAn2‖E,
for sufficiently large n. Recall that mE=0, because E∈(Hg) (see Lemma 3.2). Moreover, An2⊂An, whence μ(An2)≤μ(An)→0, because xn→x in measure. Consequently, ∥χAn2∥E→0. Thus, ∥φ∘(λ(xn-x))∥E≤ɛ for sufficiently large n.
(B) Ifaφ=0, then we follow as in (A) but with the number u0>0 in place of (1+δ)aφ satisfying φ(λu0)∥χT∥E<ɛ/2. We also set An={t∈T:|xn(t)-x(t)|>u0}.
(ii) The proof is analogous and simpler than in case (i).Theorem 3.19.
Suppose thatE is a symmetric Köthe function space. The Calderón-Lozanovskiĭ function space Eφ has the Kadec-Klee property with respect to convergence in measure if and only if:(a)
E∈(Hg),(b)
φ∈Δ2(∞) whenever L∞↪E and φ∈Δ2(ℝ+) provided L∞↪E.Proof.
The necessity By Lemma 3.12, we conclude that mE=0, whence suppEa=T. We also have E↪L∞ (see Remark 3.13). By Lemmas 3.15 and 3.16, we get bφ=∞ and φ∈Δ2(∞). We consider two cases.(A)
Assume thatL∞↪E. To prove E∈(Hg), we follow analogously as in the proof of condition E∈(Hl) in Theorem 3.11 (case I). We apply also Lemma 3.5, Corollary 3.7, and Lemma 3.9(ii).(B)
Assume thatL∞↪E, then φ∈Δ2(0), by Lemma 3.17. Consequently, φ∈Δ2(ℝ+), and to prove that E∈(Hg), see the proof of Theorem 3.11 (case II).
Thesufficiency is done by Lemma 3.18.Remark 3.20.
PropertyHu does not imply Hg in general. If L∞↪E, then χT∈Eφ, because there is λ>0 with φ(λ)<∞, whence φ∘(λχT)∈E. Consequently, taking xn→x uniformly, we get ∥xn-x∥φ≤ɛ∥χT∥φ for sufficiently large n. Hence, Eφ∈(Hu). However, if φ∉Δ2(∞), then Eφ∉(Hg).Denote by(I,Σ,m) the Lebesgue measure space with I=[0,1] or I=[0,∞). Recall that the function ω:[0,γ)→R+ with γ=m(I) is said to be the weight function, if it is nonnegative, nonincreasing and locally integrable function with the respect to the Lebesgue measure m. Then the Lorentz function space Λω consists of all functions x∈L0(I,Σ,m) such that ∥x∥=∫0γx*(t)ω(t)dt<∞.Applying Theorems3.11 and 3.19 with E=Λω, we get the criteria for Orlicz-Lorentz function spaces. Note that the implication (iii) ⇒ (i) in the case (a) has been proved directly in [6, Theorem  1]. The sufficient condition for case (b)(ii) has been proved in [22, Theorem  6] but under additional assumption.Corollary 3.21.
Let(Λω)φ be the Orlicz-Lorentz function space.(a)
IfI=[0,1], then the following assertions are equivalent:
(i)
(Λω)φ∈(Hl),(ii)
(Λω)φ∈(Hg),(iii)
φ∈Δ2(∞).(b)
Suppose thatI=[0,∞).
(i)
(Λω)φ∈(Hl) if and only if ∫0∞ω=∞ and φ∈Δ2(ℝ+),(ii)
(Λω)φ∈(Hg) if and only if φ∈Δ2(ℝ+).Proof.
Recall thatΛω∈(Hg) for any weight ω [1, Corollary  1.3]. Thus, (a) and (b) (ii) follow immediately. Clearly, each symmetric Köthe space is separable if and only if it is order continuous (see [13]). Consequently, from [1,Theorem  3.3], [8,Lemma  3.2], it follows that Λω∈(Hl) if and only if ∫0∞ω=∞. Thus, (b) (i) follows.Obviously, sinceL1∈(Hl), from Theorems 3.11 and 3.19 with E=L1, we obtain the characterization of Hl and Hg for Orlicz spaces proved directly in [10, Theorem  2.1].


## 3.1. Köthe Spaces and Orlicz Functions

Given a Köthe function spaceE, we set(3.1)mE=sup⁡(B,χB∈E)sup⁡[(An)⊂B,μ(An)→0]inf⁡n‖χAn‖E.Remark 3.1.
It is easy to see that ifE is a symmetric Köthe function space, then
(3.2)mE=inf⁡A∈Σ,μ(A)>0‖χA‖E.Recall that ifx∈L0, then suppx={t∈T:x(t)≠0}. For a Köthe space E⊂L0(T,Σ,μ), let F⊂E be a subset. By the support of F, we mean the set suppF∈Σ satisfying(i)
for eachx∈F, there is A∈Σ, with μ(A)=0, such that suppx⊂A∪suppF,(ii)
there isx∈F with μ(suppF∖suppx)=0.Lemma 3.2.
Suppose thatE is a Köthe function space. Consider the following statements:(i)
E∈(OC),(ii)
E∈(Hg)+,(iii)
For anyu∈E and each sequence {An} in Σ the implication
(3.3)μ(An)⟶0⟹‖uχAn‖⟶0holds.(iv)
mE=0,(v)
suppEa=T,(vi)
Ea≠{0}.Then (i) ⇒ (iii) ⇒ (iv) ⇒ (v) ⇒ (vi) and (ii) ⇒ (iii). For a symmetric Köthe function space we have additionally (vi) ⇒ (iv). Furthermore, for a nonsymmetric Köthe function space, the implication (vi) ⇒ (iv) is not true in general. Finally, the implications (i) ⇒ (iii) ⇒ (iv) cannot be reversed in general.Proof.
The implications (i)⇒ (iii) ⇒ (iv) and (v) ⇒ (vi) are obvious.
(iv)⇒ (v). If suppEa≠T, then for each x∈E with suppx⊂T∖suppEa we have x∉Ea. Let T0⊂T∖suppEa be such that 0<μ(T0)<∞ and χT0∈E. Then x1=χT0∉Ea and consequently there is a sequence (An) of measurable pairwise disjoint sets with ⋃n=1∞An=T0 and a number δ>0 satisfying ∥χAn∥≥δ for each n (see [19]). Since μ(An)→0, we get mE≥δ>0.
(ii)⇒ (iii). Assume that (iii) does not hold, then we find an element u∈E, a number δ>0, and a sequence (Ai) with μ(Ai)→0 such that infi∥uχAi∥E≥δ. Without loss of generality, we may assume that ∑i=1∞μ(Ai)<∞. Set
(3.4)xk=sup⁡{uχAi:i≥k}∈L0.
Let A=⋃iAi. Clearly, xk≤uχA∈E. Put y=uχA and yk=y-xk. Since {xk} is decreasing, so {yk} is increasing. Next, xk→0 in measure, because, for each ɛ>0,
(3.5)μ{t∈T:xk(t)>ɛ}≤μ{⋃i=k∞Ai}≤∑i=k∞μ(Ai)⟶0ask⟶∞.
In particular, by E∈(FP), ∥yk∥→∥y∥. Finally, ∥yk-y∥=∥xk∥≥∥uχAk∥≥δ. Thus, E∉(Hg)+.
Assume thatE is symmetric. We prove that (vi) ⇒ (iv). If mE>0, then we find a set B with χB∈E and a sequence {An}⊂B, μ(An)→0 such that infn∥χAn∥E≥mE/2. Let x∈E∖{0}. We prove that x∈E∖Ea. There is a number δ>0 such that the set D={t∈T:|x(t)|≥δ} has positive measure. We find a sequence {Ck} in D of pairwise disjoint sets and a subsequence {Ank} of {An} satisfying μ(Ck)≤(1/2k)μ(D) and μ(Ck)≥μ(Ank). Then, by the symmetry of E,
(3.6)‖xχCk‖E≥‖δχCk‖E≥‖δχAnk‖E≥δmE2,
for any k∈ℕ, whence x∉Ea.
Now we will prove that the implication (vi)⇒ (iv) is not true in general in nonsymmetric case. Take
(3.7)E=(L1[0,1]⊕L∞[1,2])1={x∈L0[0,2]:x=u+w,u∈L1[0,1],w∈L∞[1,2],‖x‖=‖u‖1+‖w‖∞}.
Obviously, suppEa=[0,1] and mE=1.
Now we will see that implications (i)⇒ (iii) ⇒ (iv) cannot be reversed in general. If E=Λω—the Lorentz space over (0,∞) with Lebesgue measure and ∫0∞ω<∞, then E∈(Hg) and E∉(OC)(see [1, 8]), whence (iii) ⇏ (i) in general.
Finally, letE=Lφ[0,1]—the Orlicz space with φ<∞ and φ∉Δ2(∞). By the proof of Theorem  1 from [5], we find an element u∈E and a sequence {An} with μ(An)→0 and ∥uχAn∥φ=1. On the other hand, for each B∈Σ with u=χB∈E, we claim that χB∈Ea, whence mE=0. Indeed, for any sequence {An} satisfying An↓∅ and for each λ>0, we have
(3.8)Iφ(λuχAn)=φ(λ)∫01χAndμ=φ(λ)μ(An)⟶0,
whence ∥uχAn∥φ→0. Thus, the implication (iii) ⇒ (iv) cannot be reversed in general.Recall that ifE∈(Hl)+, then E∈(OC) (see [10]) and, as we have seen above, property Hg needs not imply OC in general. However, from the above result, it follows that if E∈(Hg)+, then mE=0 and consequently suppEa=T. Note also that the implication (vi) ⇒ (v) for a symmetric Köthe function space has been proved in [8, Proposition  2.2].Remark 3.3.
It may happen thatμ(A)<∞ and χA∉E only in a nonsymmetric Köthe space. However, if μ(A)<∞ and χA∉E, then there is an increasing sequence (An) of subsets of A satisfying ⋃nAn=A and χAn∈E for each n.Proof.
Letx0 be a weak unit of E, that is, x0(t)>0 for each t∈T. It is enough to take An={t∈A:x0(t)>1/n} for each n.Remark 3.4.
LetE be a Köthe space. If E∈(Hu)+, then ∥xχBk∥→0 for each x∈E+, where Bk={t∈T:x(t)<1/k}.Proof.
Letx∈E+. Clearly, xχT∖Bk↑x and supk∥xχT∖Bk∥<∞. By E∈(FP), we get ∥xχT∖Bk∥→∥x∥. Moreover, x-xχT∖Bk→0 uniformly. Then ∥xχT∖Bk-x∥→0, because E∈(Hu)+.
It is easy to see that the above is not true ifE∉(Hu)+. It is enough to take
(3.9)E=Lω∞(0,∞)={x∈L0:‖x‖=supesst≥0⁡ω(t)|x(t)|<∞},
where ω(t)=t and x(t)=1/t. Then ∥xχBk∥=1 for each k. On the other hand, taking xn=xχ[0,n], we get ∥x∥=∥xn∥=1, xn→x uniformly and ∥xn-x∥=1, whence E∉(Hu)+.It is known thatE∈(Hl) if and only if E∈(Hl)+ whenever E∈(OC) (see [6]). Since (Hl)+⇒(OC) (see [10]), so E∈(Hl) if and only if E∈(Hl)+. The same proof as in [6] shows that if E∈(OC), then E∈(Hg) if and only if E∈(Hg)+. However, the assumption E∈(OC) is now too strong in this case because there are Köthe spaces satisfying E∈(Hg) and E∉(OC). Hence, the following result is more natural and precise.Lemma 3.5.
SupposeE is a Köthe space. Then E∈(Hg) if and only if E∈(Hg)+.Proof.
The necessity is obvious. We prove the sufficiency. LetE∈(Hg)+. Assume that E is a Köthe function space. Let ∥xn∥→∥x∥ and xn→x in measure. For each η>0, we get
(3.10)μ{t∈T:||xn(t)|-|x(t)||≥η}≤μ{t∈T:|xn(t)-x(t)|≥η}⟶0,
whence |xn|→|x| in measure. By the assumption E∈(Hg)+, we conclude that
(3.11)‖|xn|-|x|‖⟶0.
Then we find a subsequence (|xnm|)m of (|xn|)n, an element y∈E+, and a sequence 0≤βm→0 such that ||xnm|-|x||≤βmy for each m (see [12]). Denote (|xnm|)m still by (|xn|)n and (βm) by (βn). Thus, we may assume that ||xn|-|x||≤y for all n. Set u=|x|+y. Let ɛ>0. Applying Remark 3.4, with respective sets
(3.12)Bk={t:u(t)<1k},
take k big enough to satisfy ∥uχBk∥<ɛ/8. Moreover, ∥|xn|-|x|∥<ɛ/8 for sufficiently large n. Setting
(3.13)Cn={t∈T:0≤sgn⁡[x(t)xn(t)]≤1},Dn={t∈T:sgn⁡[x(t)xn(t)]=-1},
we get
(3.14)‖x-xn‖≤‖(|x|-|xn|)χCn‖+‖(x-xn)χDn∩Bk‖+‖(x-xn)χDn∖Bk‖≤5ɛ8+‖(x-xn)χDn∖Bk‖.
Take
(3.15)0<δ<ɛ16kmin⁡{1‖y‖,1‖x‖}.
Denote
(3.16)Fnk={t∈Dn∖Bk:|xn(t)-x(t)|>δ},Gnk={t∈Dn∖Bk:|xn(t)-x(t)|≤δ}.
Note that xn→μx and consequently μ(Fnk)→0 as n→∞. Then, by Lemma 3.2 (iii),
(3.17)‖(x-xn)χDn∖Bk‖≤‖(x-xn)χFnk‖+‖(x-xn)χGnk‖≤‖2uχFnk‖+‖(x-xn)χGnk‖<ɛ8+‖(x-xn)χGnk‖,
for sufficiently large n. Finally, divide set Gnk into two subsets
(3.18)Hnk={t∈Gnk:|x(t)|≥12k},Ink={t∈Gnk:y(t)≥12k}.
Clearly, Hnk∪Ink=Gnk because
(3.19)Gnk⊂T∖Bk={t:|x(t)|+y(t)≥1k}.
Then, by (3.15),
(3.20)‖(x-xn)χGnk‖≤δ(‖χHnk‖+‖χInk‖)≤δ(‖2k|x|χHnk‖+‖2kyχInk‖)≤2kδ(‖|x|χHnk‖+‖yχInk‖)<ɛ4.
Combining (3.14), (3.17), and (3.20), we get ∥x-xn∥≤ɛ.
IfE is a Köthe sequence space, the proof is the same, because for a counting measure space (ℕ,2ℕ,m) for each u∈E and any sequence (An) in 2ℕ with m(An)→0, we have ∥uχAn∥→0.The straightforward calculation shows that we may equivalently takex and {xn} in S(E) in the definition of property Hl (see also the proof below). However, the analogous problem for property Hg is a little more complicated. The crucial point is described in the following lemma.Lemma 3.6.
Suppose thatE is a symmetric Köthe function space. If xn→x in measure and ∥xn∥→∥x∥≠0, then xn/∥xn∥→x/∥x∥ in measure. Moreover, the above implication is not true in general if E is a nonsymmetric Köthe function space.Proof.
Having our assumptions, for eachɛ,η>0, we get
(3.21)μ{t:|xn(t)‖xn‖-x(t)‖x‖|>ɛ}=μ{t:|xn(t)‖xn‖-x(t)‖xn‖+x(t)‖xn‖-x(t)‖x‖|>ɛ}≤μ{t:1‖xn‖|xn(t)-x(t)|+|1‖xn‖-1‖x‖||x(t)|>ɛ}≤μ{t:1‖xn‖|xn(t)-x(t)|>ɛ2}+μ{t:|1‖xn‖-1‖x‖||x(t)|>ɛ2}.
Since ∥xn∥→∥x∥, so ∥xn∥>(1/2)∥x∥ for n≥N0, whence
(3.22)μ{t:1‖xn‖|xn(t)-x(t)|>ɛ2}≤μ{t:|xn(t)-x(t)|>ɛ‖x‖4}<η2,
for n≥N1>N0. Moreover, applying the symmetry of E, we get
(3.23)μ{t:|1‖xn‖-1‖x‖||x(t)|>ɛ2}<η2forn≥N2.
Thus,
(3.24)μ{t:|xn(t)‖xn‖-x(t)‖x‖|>ɛ}<η,
for n≥max{N1,N2}.
Considering the second assertion, take(3.25)E=L1/t31(1,∞)={x∈L0((1,∞),Σ,m):‖x‖=∫1∞|x(t)|1t3dt<∞}.
Set x(t)=t and xn(t)=t+(1/n), then ∥x∥=1, ∥xn∥=1+(1/2n), and xn→x in measure. However,
(3.26)μ{t:|xn(t)‖xn‖-x(t)‖x‖|>1}=μ{t:|t-22n+1|>1}=∞,
for each n, whence (xn/∥xn∥)↛(x/∥x∥) in measure.Corollary 3.7.
Assume thatE is a symmetric Köthe function space. Then E∈(Hg) if and only if ∥xn-x∥→0 for any x∈S(E) and (xn) in S(E) satisfying xn→x in measure.Proof.
The necessity is obvious. We prove the sufficiency. Letxn→x in measure and ∥xn∥→∥x∥≠0. Taking yn=xn/∥xn∥ and y=x/∥x∥, we get yn→y in measure, by Lemma 3.6. By the assumption, we conclude that ∥yn-y∥→0. Notice that
(3.27)|1‖xn‖‖xn-x‖-‖x‖|1‖xn‖-1‖x‖||≤‖xn‖xn‖-x‖xn‖-(-x‖xn‖+x‖x‖)‖=‖xn‖xn‖-x‖x‖‖⟶0.
Since ∥xn∥→∥x∥≠0, so ∥xn-x∥→0 as desired.We will need in the sequel the following.Lemma 3.8.
Ifφ∈Δ2(∞), then for any a>0, there are σ∈(0,1] and u1>aφ+a such that
(3.28)inf⁡u0∈(0,a)inf⁡u≥u1φ(u-u0)φ(u)-φ(u0)≥σ.Proof.
Suppose conversely that there isa>0 such that for any σ∈(0,1] and each u1>aφ+a, there holds
(3.29)inf⁡u0∈(0,a)inf⁡u≥u1φ(u-u0)φ(u)-φ(u0)<σ.
Taking σn=1/2n and un→∞, we get infu0∈(0,a)infu≥unφ(u-u0)/(φ(u)-φ(u0))<1/2n. We may assume that un>2a for each n. Hence, for each n, one may choose vn≥un and u0∈(0,a) with φ(vn-u0)/(φ(vn)-φ(u0))≤1/2n. Consequently, we get
(3.30)2nφ(12vn)<2nφ(vn-u0)<φ(vn)-φ(u0)≤φ(vn),
for each n. Taking bn=(1/2)vn, one has 2nφ(bn)<φ(2bn), whence φ∉Δ2(∞).Lemma 3.9.
(i) Suppose thatE is a symmetric Köthe space. If xn→x in measure and φ∘xn, φ∘x∈E are finitely valued a.e., then φ∘xn→φ∘x in measure. Furthermore, for a nonsymmetric Köthe space, the above implication is not true in general.
(ii) Ifxn→x in measure, xn,x∈E+ and φr-1∘xn, φr-1∘x are well-defined functions, then φr-1∘xn→φr-1∘x in measure.Proof.
(i) Letxn→μx.
(A) Suppose thatbφ=∞. Set
(3.31)Ak={t∈T:|x(t)|>k}.
Then, by symmetry of E,  μ(Ak)→0. Let η>0 and ε>0. Take k0 such that μ(Ak0)<η/3. We have
(3.32)μ{t∈T:|φ(xn(t))-φ(x(t))|≥ε}≤μ{t∈T∖Ak0:|φ(xn(t))-φ(x(t))|≥ε}+η3.
Since φ is uniformly continuous on the interval [-k0,k0], so for each γ>0 there is δ=δ(γ,k0)>0 such that the implication |u-s|<δ⇒|φ(u)-φ(s)|<γ is true for all u,s∈[-k0,k0]. Set δ=δ(ε,2k0) and
(3.33)An1={t∈T∖Ak0:|xn(t)|>2k0},An2={t∈T∖Ak0:|xn(t)|≤2k0}.
Consequently
(3.34)μ{t∈T:|φ(xn(t))-φ(x(t))|≥ε}≤μ(An1)+μ{t∈An2:|φ(xn(t))-φ(x(t))|≥ε}+η3≤μ{t∈An2:|xn(t)-x(t)|≥δ}+2η3≤η
for sufficiently large n, because An1={t∈T:|xn(t)-x(t)|≥k0}. Thus φ∘xn→μφ∘x.
(B) Assumeφ(bφ)<∞. The proof is simpler than above because xn(t)≤bφ and x(t)≤bφ for μ-a.e. t∈T and φ is uniformly continuous on the interval [-bφ,bφ].
(C) Suppose thatbφ<∞, φ(bφ)=∞. Then xn(t)<bφ and x(t)<bφ for μ-a.e. t∈T. Set
(3.35)Ak={t∈T:|x(t)|>bφ-1k}.
Then μ(Ak)→0, because E is symmetric. Let η>0 and ε>0. Take k0 such that μ(Ak0)<η/3. We have
(3.36)μ{t∈T:|φ(xn(t))-φ(x(t))|≥ε}≤μ{t∈T∖Ak0:|φ(xn(t))-φ(x(t))|≥ε}+μ(Ak0).
Since φ is uniformly continuous on the interval [-u0,u0] for each u0<bφ, so for each u0<bφ and γ>0 there is δ=δ(γ,u0)>0 such that the implication |u-s|<δ⇒|φ(u)-φ(s)|<γ is true for all u,s∈[-u0,u0]. Put δ=δ(ε,bφ-1/2k0) and
(3.37)An1={t∈T∖Ak0:|xn(t)|>bφ-12k0},An2={t∈T∖Ak0:|xn(t)|≤bφ-12k0}.
Consequently
(3.38)μ{t∈T:|φ(xn(t))-φ(x(t))|≥ε}≤μ(An1)+μ{t∈An2:|φ(xn(t))-φ(x(t))|≥ε}+η3≤μ{t∈An2:|xn(t)-x(t)|≥δ}+2η3≤η
for sufficiently large n, because An1={t∈T:|xn(t)-x(t)|≥1/2k0}. Thus φ∘xn→μφ∘x.
Now we will show that in nonsymmetric case the above implication is not true in general. Consider the Lebesgue measure space ((0,∞),Σ,m). Take
(3.39)φ(u)=u2,x=∑nnχ[n,n+1],xk=∑n(n+1k)χ[n,n+1].
Then xk→x in measure. On the other hand, |φ∘xk-φ∘x|=∑n((1/k2)+(2n/k))χ[n,n+1], whence φ∘xk↛φ∘x in measure. Clearly, such element φ∘x can be well defined only in some nonsymmetric Köthe space E.
(ii) First, we claim that for eachɛ>0, there is δ>0 such that the implication
(3.40)|φr-1(u)-φr-1(v)|>ɛ⟹|u-v|>δ
is true for all u,v≥0. First, notice that for each ɛ>0, there is δ>0 such that the implication
(3.41)|φr-1(u)-φr-1(0)|>ɛ⟹|u|>δ
is true for all u≥0, because φr-1 is continuous at zero. Next, defining g(u)=φr-1(u)-φr-1(0), we see that g is subadditive on ℝ+, because g=ψ-1, where ψ is an Orlicz function with ψ>0. Consequently,
(3.42)φr-1(u-v)-φr-1(0)=g(u-v)≥g(u)-g(v)=φr-1(u)-φr-1(v).
Hence, if |φr-1(u)-φr-1(v)|>ɛ, then |φr-1(u-v)-φr-1(0)|>ɛ, whence |u-v|>δ, by (3.41). Thus, the claim is proved. Assume that xn, x∈E+ and φr-1∘xn, φr-1∘x are well-defined functions and xn→x in measure, then, applying (3.40), for each ɛ>0, we get
(3.43)μ{t∈T:|(φr-1∘xn-φr-1∘x)(t)|>ɛ}≤μ{t∈T:|(xn-x)(t)|>δ}⟶0.


## 3.2. Calderón-Lozanovskiĭ Function Spaces

We still assume below thatE is a Köthe function space.It is known thatEφ∈(OC) if and only if E∈(OC) and φ∈Δ2E (see Corollary  12 from [9]). Considering some weaker property than OC, we get the following.Lemma 3.10.
Suppose thatbφ=∞, then Ea≠{0} if and only if (Eφ)a≠{0}.Proof.
Necessity
Letz∈Ea∖{0}. We find a number m>0 such that μ(A)=μ{t∈T:|z(t)|>1/m}>0. Taking w=φ-1(1)χA, we get φ∘w∈Ea. Moreover, since bφ=∞, φ satisfies the local Δ2E condition with respect to the element w (see [9]). Consequently, w∈(Eφ)a, by Theorem  11 from [9].
Sufficiency
We apply Theorem  11 from [9]. Let x∈(Eφ)a∖{0}, then φ∘x∈Ea. If μ{t:|x(t)|>aφ}>0, then μ(Am)>0 for some m∈ℕ, where Am={t∈T:|x(t)|≥(1+(1/m))aφ}. Thus, χAm∈Ea, and we are done, if μ{t:|x(t)|≤aφ}=μ(A)>0 then χA∈Ea∖{0}, by Theorem  11 from [9].Theorem 3.11.
Eφ∈(Hl) if and only if(a)
E∈(Hl),(b)
φ∈Δ2(∞) whenever L∞↪E and φ∈Δ2(ℝ+) provided L∞↪E.Proof.
Since the sufficiency has been proved in [4, Theorem  11] (we need also to apply [7, Lemma  2]), we need only to prove the necessity. Since Eφ∈(Hl), so Eφ∈(OC) (see [10, Proposition  2.1]), and consequently, E∈(OC), φ<∞, and φ∈Δ2E (see [9, Corollary  12]). Notice that E∈(OC) implies that E↪L∞. Indeed, if E↪L∞, then there is a decreasing sequence (An) in Σ with 0<μ(An)→0 and χAn∈E, whence 1=∥χAn∥L∞≤M∥χAn∥E, so E∉(OC). Thus, (b) is proved. Suppose that E∉(Hl). Recall that property Hl can be equivalently considered only on the positive cone E+ [6, Proposition  1]. Consequently, we find x and {xn} in (S(E))+ with xn→μx locally in measure and ∥xn-x∥E≥ɛ. Set yn=φr-1∘xn and y=φr-1∘x, then Iφ(y)=Iφ(yn)=1, whence y,yn∈S(Eφ). Moreover, yn→μy locally in measure, because, by σ finiteness of the measure space (T,Σ,μ), the local convergence in measure can be equivalently replaced by convergence μ-a.e. (also Lemma 3.9(ii) can be applied). It is enough to prove that there is η>0 with
(3.44)‖yn-y‖φ≥η,
for infinitely many n. The similar condition has been proved in [8, Theorem  2.11], but under additional assumption that φ>0. Here, the lack of this assumption requires new techniques in comparison with the respective proof in [8, Theorem  2.11]. We have
(3.45)‖φ∘yn-φ∘y‖E≥ɛ.
Denote
(3.46)An={t∈T:φ(yn(t))≥φ(y(t))},Bn=T∖An.
By (3.45), we get
(3.47)max⁡{‖(φ∘yn-φ∘y)χAn‖E,‖(φ∘yn-φ∘y)χBn‖E}≥ɛ2,
for each n. We assume that ∥(φ∘yn-φ∘y)χAn∥E≥ɛ/2 for infinitely many n∈ℕ, because otherwise the proof is analogous. We divide the proof into two cases.(I)
Suppose thatL∞↪E and φ∈Δ2(∞). Let c=φr-1(ɛ/32∥χT∥E)>aφ. Denoting(3.48)An1={t∈An:yn(t)<c},An2={t∈An:yn(t)≥c},
we get
(3.49)ɛ2≤‖(φ∘yn-φ∘y)χAn‖E≤‖(φ∘yn-φ∘y)χAn2‖E+ɛ4.
Thus,
(3.50)‖(φ∘yn-φ∘y)χAn2‖E≥ɛ4.
Set
(3.51)c1=c+aφ2,Bn1={t∈An2:y(t)<c1},Bn2={t∈An2:y(t)≥c1}.
The remaining proof of the case (I) we divide into two parts.(1)
Suppose that∥(φ∘yn-φ∘y)χBn1∥E≥ɛ/8. By Lemma 3.8, for a=c1, there are σ>0 and u1>0 such that
(3.52)inf⁡u0∈(0,c1)inf⁡u≥u1φ(u-u0)φ(u)-φ(u0)≥σ.
Set
(3.53)Cn1={t∈Bn1:yn(t)<u1},Cn2={t∈Bn1:yn(t)≥u1}.
(a)
Assume that∥(φ∘yn-φ∘y)χCn1∥E≥ɛ/16. Taking 0<α<(c-c1)/u1, we get
(3.54)‖(φ∘(yn-yα))χCn1‖E≥‖(φ(c-c1α))χCn1‖E≥‖φ(u1)χCn1‖E≥‖(φ∘yn-φ∘y)χCn1‖E≥ɛ16.
Thus, ∥yn-y∥φ≥η1=min{1,αɛ/16}.(b)
Let∥(φ∘yn-φ∘y)χCn2∥E≥ɛ/16. Thus, by Lemma 3.8,
(3.55)‖φ∘(yn-y)χCn2‖E≥σ2‖(φ∘yn-φ∘y)χCn2‖E≥σɛ32,
and consequently ∥yn-y∥φ≥η2=min{1,σɛ/32} (cf. [8, Lemma  1.1]).(2)
Assume that∥(φ∘yn-φ∘y)χBn1∥E<ɛ/8, then, by (3.50),
(3.56)‖(φ∘yn-φ∘y)χBn2‖E≥ɛ8.
Since φ∈Δ2(∞), for l=1+ɛ/32 and u1=c1, there is a=a(l,u1)∈(0,1) such that
(3.57)φ((1+a)u)≤lφ(u),
for every u≥c1 (see [20, Theorem  1.13(4)]). Moreover, we can choose a>0 satisfying
(3.58)a1+aφ(c1)‖χT‖E<ɛ32.
Let
(3.59)Bn21={t∈Bn2:(yn-y)(t)<ac11+a},Bn22={t∈Bn2:(yn-y)(t)≥ac11+a},
then, by (3.57) and (3.58), using convexity of φ, we get
(3.60)|φ∘yn-φ∘y|χBn21=(φ∘(yn-y+y)-φ∘y)χBn21≤(a1+aφ∘(1+aa(yn-y))+11+aφ∘((1+a)y)-φ∘y)χBn21≤(a1+aφ∘(1+aa(yn-y))+11+alφ∘y-φ∘y)χBn21=(a1+aφ∘(1+aa(yn-y))+(ɛ/32-a1+a)φ∘y)χBn21.
Note that f(a)=(ɛ/32-a)/(1+a) is a decreasing function of a>0. Moreover, a≤ɛ/32 by convexity of φ and inequality (3.57). Hence, by (3.58), ∥(φ∘yn-φ∘y)χBn21∥E<ɛ/16. Then, by (3.56),
(3.61)‖(φ∘yn-φ∘y)χBn22‖E≥ɛ16.
Since φ∈Δ2(∞), for u3=c1 and β=(1+a)/a, there is k2>0 such that φ(βu)≤k2φ(u) for each u≥u3 (see [20]). Taking 0<γ<a/(1+a) and applying (3.57), we get
(3.62)|φ∘yn-φ∘y|χBn22=(φ∘(yn-y+y)-φ∘y)χBn22≤(φ∘(yn-yγ+y)-φ∘y)χBn22≤(a1+aφ∘(1+aa(yn-yγ))+11+aφ∘((1+a)y)-φ∘y)χBn22≤(a1+ak2φ∘(yn-yγ)+11+alφ∘y-φ∘y)χBn22=(a1+ak2φ∘(yn-yγ)+(ɛ/32-a1+a)φ∘y)χBn22.
Then ∥φ∘((yn-y)/γ)∥E≥ɛ(1+a)/32ak2, and consequently ∥yn-y∥φ≥η3=min{1,γɛ(1+a)/32ak2} (see [8, Lemma  1.1]).
Combining cases (1) and (2), we get (3.44) with η=min{η1,η2,η3}.
(II)
Suppose thatL∞↪E. Since φ∈Δ2(ℝ+), so for every l>1, there is a=a(l)∈(0,1) such that φ((1+a)u)≤lφ(u) for every u≥0 (see [20, Theorem  1.13(4)]). Moreover, for every β>0, there is k>0 such that φ(βu)≤kφ(u) for each u≥0. Then the proof is analogous as in case (I) (it is simpler and shorter).Now we will consider the Kadec-Klee property with respect to global convergence in measure. Theorem3.19 below presents the full characterization of property Hg. We divide the whole proof into several lemmas for readers convenience.Lemma 3.12.
IfEφ∈(Hg)+, then mE=0.Proof.
IfmE>0, then there are δ>0, a set B,μ(B)>0,χB∈E, and a sequence (An)⊂B with μ(An)→0 and infn∥χAn∥E≥δ. Without loss of generality, we may assume that ∑i=1∞μ(Ai)<∞. Set
(3.63)zn=sup⁡{χAi:i≥n}∈L0.
Let A=⋃iAi. Clearly, zn≤χA≤χB∈E. Put y=χA and yn=y-zn. Choose a>0 such that
(3.64)a=bφifφ(bφ)‖χA‖E<1,a∈(aφ,bφ]ifφ(bφ)<∞,φ(bφ)‖χA‖E≥1,a∈(aφ,bφ)ifφ(bφ)=∞,
and φ(a)∥χA∥E=1 in the last two cases. Take x=ay and xn=ayn, then ∥xn∥φ≤∥x∥φ=1. It is obvious that ∥xn∥φ=1 when a=bφ. We show that ∥xn∥φ→1 in the case a<bφ. Note that zn→0 in measure (see the proof of Lemma 3.2), zn↓, and consequently ∥yn∥E→∥y∥E, because E∈(FP). If ∥xn∥φ≤1-α for some α>0 and infinitely many n, then we get
(3.65)1=φ(a)‖χA‖E<φ(a1-α)‖χA‖E⟵φ(a1-α)‖yn‖E=Iφ(xn1-α)≤1,
which is a contradiction. Moreover, xn→x in measure. Since
(3.66)‖x-xn‖φ≥Iφ(x-xn)≥φ(a)‖χAn‖E≥φ(a)δ,
we conclude that Eφ∉(Hg)+.Remark 3.13.
IfEφ∈(Hg)+, then E↪L∞.Proof.
It is immediate, since ifE↪L∞, then mE>0. By Lemma 3.12, we get Eφ∉(Hg)+.From Lemmas3.12 and 3.2, we get immediately the following.Corollary 3.14.
IfEφ∈(Hg)+, then suppEa=T.Lemma 3.15.
IfEφ∈(Hg), then φ(bφ)=∞.Proof.
Suppose thatEφ∈(Hg) and φ(bφ)<∞. By Lemma 3.12, we get mE=0. Let A∈Σ, μ(A)>0, and μ(T∖A)>0 be such that χA∈E and φ(bφ)∥χA∥E≤1/2. Let (An)⊂T∖A be such that 0<μ(An)→0 and φ(bφ)∥χAn∥E≤1/2. Taking x=bφχA and xn=bφχA∪An, we get ∥x∥φ=∥xn∥φ=1, xn→x in measure and ∥xn-x∥φ=1, whence Eφ∉(Hg).Lemma 3.16.
Suppose that (bφ<∞ and φ(bφ)=∞) or (bφ=∞ and φ∉Δ2(∞)). If Ea≠{0}, then Eφ∉(Hg).Proof.
It is enough to apply order isometryP:l∞→Eφ constructed in the proof of [5, Theorem  1] and take z=χℕ and zn=z-en, where en is the nth unit vector in l∞. If y=Pz and yn=Pzn, then yn→y in measure, because, without loss of generality, one can take in that proof a sequence {An} satisfying min{μ(An),∥χAn∥E}≤2-n. Moreover, ∥yn∥φ=∥y∥φ=∥yn-y∥φ=1.Lemma 3.17.
(i) IfL∞↪E, aφ>0, φ(bφ)=∞, then Eφ∉(Hg).
(ii) IfL∞↪E, suppEa=T, aφ=0, bφ=∞ and φ∉Δ2(0),
thenEφ∉(Hg).Proof.
(i) Suppose thatL∞↪E, φ(bφ)=∞, and aφ>0. If T1∪T2=T, T1∩T2=∅, μ(T1)>0, and μ(T2)>0, then χT1∉E or χT2∉E. Assume that χT1∉E. Setting x=aφχT1 and xn=x+anχAn, where (An)⊂T2, χAn∈E, μ(An)→0, and φ(an)∥χAn∥E=1, we get ∥x∥φ=∥xn∥φ=1, xn→x in measure and ∥xn-x∥φ=1, whence Eφ∉(Hg).
(ii) The proof is analogous to that of [10, Proposition  2.3]. In fact, it shows even more that in this case, Eφ∉(Hu). By Lemma 3.10, we get (Eφ)a≠{0}. Let x∈(Eφ)a, x≥0, and ∥x∥φ=1.
First, we prove thatφ∘(2x)∈Ea. Since x∈S((Eφ)a), so φ∘x∈Ea and φ satisfies the local Δ2E condition with respect to the element x by Theorem  11 from [9] (notice aφ=0, whence Δ2E(xχT∖A)=Δ2E(x) in Theorem  11 from [9]). Furthermore, by Lemma  9 from [9], we conclude that φ∘(2x)∈Ea, because bφ=∞.
We claim that there is a sequence{An} in Σ satisfying χAn∉E and Iφ(2xχAn)≤2-n for each n. Since the measure space (T,Σ,μ) is σ-finite, there is an increasing sequence {Cn} with 0<μ(Cn)<∞ for any n∈ℕ and ⋃nCn=T. Applying Remark 3.3, without loss of generality, we may assume that χCn∈E for each n (see also [12]). Consequently, ∥φ∘(2x)χT∖Cn∥E→0. Moreover, χT∖Cn∉E, because χT∉E and χCn∈E. Thus, the claim is proved with {An} being the respective subsequence of {T∖Cn}. Since φ∉Δ2(0), there is a sequence {un} in ℝ+ with un→0 and
(3.67)φ(2un)>2nφ(un),
for any n∈ℕ. Notice that if χA∈Ea, then the submeasure νA(B)=∥χB∥E, B∈Σ∩A has the Darboux property in the interval [0,∥χA∥E] (see [21]). We claim that for any n, there is a set Bn⊂An with
(3.68)φ(un)‖χBn‖E=2-n.
Let n∈ℕ. First, by E∈(FP), there is a set Dn⊂An with μ(Dn)<∞, χDn∈E, and φ(un)∥χDn∥E>2·2-n. Next, since suppEa=T and E∈(FP), there is a set En⊂Dn with χEn∈Ea and φ(un)∥χEn∥E>(3/2)·2-n. Finally, there is a set Bn⊂En with φ(un)∥χBn∥E=2-n. Taking
(3.69)xn=x+un2χBn=xχT∖Bn+(x+un2)χBn,
we can complete arguments as in the proof of [10,Proposition  2.3].Lemma 3.18.
Suppose thatE is a symmetric Köthe function space.(i)
IfL∞↪E, E∈(Hg) and φ∈Δ2(∞), then Eφ∈(Hg).(ii)
IfL∞↪E, E∈(Hg) and φ∈Δ2(ℝ+), then Eφ∈(Hg).Proof.
(i) Applying Lemma3.5 and Corollary 3.7, it is enough to take x,xn∈(S(Eφ))+, n∈ℕ, and xn→x in measure. Then ∥φ∘x∥E=∥φ∘xn∥E=1, by φ∈Δ2(∞)=Δ2E. Moreover, φ∘xn→μφ∘x by Lemma 3.9(i). Consequently, ∥φ∘xn-φ∘x∥E→0, because E∈(Hg). Applying superadditivity φ on ℝ+, we get
(3.70)‖φ∘(xn-x)‖E≤‖φ∘xn-φ∘x‖E⟶0.
Take λ>1. We need to show that
(3.71)‖φ∘(λ(xn-x))‖E⟶0.
Let ɛ>0. The remaining part of the proof we divide into two parts.
(A) Suppose thataφ>0. Take δ>0 such that φ((1+δ)aφ)∥χT∥E<ɛ/3. Put
(3.72)An={t∈T:λ|xn(t)-x(t)|>(1+δ)aφ},An1={t∈An:|xn(t)-x(t)|>(1+δ)aφ},An2={t∈An:|xn(t)-x(t)|≤(1+δ)aφ}.
By φ∈Δ2(∞), there is K(λ,(1+δ)aφ)=K>0 such that φ(λu)≤Kφ(u) for each u≥(1+δ)aφ. Consequently, by (3.70),
(3.73)‖φ∘(λ(xn-x))‖E≤‖φ∘(λ(xn-x))χAn‖E+‖φ∘(λ(xn-x))χT∖An‖E≤‖φ∘(λ(xn-x))χAn1‖E+‖φ∘(λ(xn-x))χAn2‖E+φ((1+δ)aφ)‖χT∖An‖E≤K‖φ∘(xn-x)χAn1‖E+φ(λ(1+δ)aφ)‖χAn2‖E+ɛ3<23ɛ+φ(λ(1+δ)aφ)‖χAn2‖E,
for sufficiently large n. Recall that mE=0, because E∈(Hg) (see Lemma 3.2). Moreover, An2⊂An, whence μ(An2)≤μ(An)→0, because xn→x in measure. Consequently, ∥χAn2∥E→0. Thus, ∥φ∘(λ(xn-x))∥E≤ɛ for sufficiently large n.
(B) Ifaφ=0, then we follow as in (A) but with the number u0>0 in place of (1+δ)aφ satisfying φ(λu0)∥χT∥E<ɛ/2. We also set An={t∈T:|xn(t)-x(t)|>u0}.
(ii) The proof is analogous and simpler than in case (i).Theorem 3.19.
Suppose thatE is a symmetric Köthe function space. The Calderón-Lozanovskiĭ function space Eφ has the Kadec-Klee property with respect to convergence in measure if and only if:(a)
E∈(Hg),(b)
φ∈Δ2(∞) whenever L∞↪E and φ∈Δ2(ℝ+) provided L∞↪E.Proof.
The necessity By Lemma 3.12, we conclude that mE=0, whence suppEa=T. We also have E↪L∞ (see Remark 3.13). By Lemmas 3.15 and 3.16, we get bφ=∞ and φ∈Δ2(∞). We consider two cases.(A)
Assume thatL∞↪E. To prove E∈(Hg), we follow analogously as in the proof of condition E∈(Hl) in Theorem 3.11 (case I). We apply also Lemma 3.5, Corollary 3.7, and Lemma 3.9(ii).(B)
Assume thatL∞↪E, then φ∈Δ2(0), by Lemma 3.17. Consequently, φ∈Δ2(ℝ+), and to prove that E∈(Hg), see the proof of Theorem 3.11 (case II).
Thesufficiency is done by Lemma 3.18.Remark 3.20.
PropertyHu does not imply Hg in general. If L∞↪E, then χT∈Eφ, because there is λ>0 with φ(λ)<∞, whence φ∘(λχT)∈E. Consequently, taking xn→x uniformly, we get ∥xn-x∥φ≤ɛ∥χT∥φ for sufficiently large n. Hence, Eφ∈(Hu). However, if φ∉Δ2(∞), then Eφ∉(Hg).Denote by(I,Σ,m) the Lebesgue measure space with I=[0,1] or I=[0,∞). Recall that the function ω:[0,γ)→R+ with γ=m(I) is said to be the weight function, if it is nonnegative, nonincreasing and locally integrable function with the respect to the Lebesgue measure m. Then the Lorentz function space Λω consists of all functions x∈L0(I,Σ,m) such that ∥x∥=∫0γx*(t)ω(t)dt<∞.Applying Theorems3.11 and 3.19 with E=Λω, we get the criteria for Orlicz-Lorentz function spaces. Note that the implication (iii) ⇒ (i) in the case (a) has been proved directly in [6, Theorem  1]. The sufficient condition for case (b)(ii) has been proved in [22, Theorem  6] but under additional assumption.Corollary 3.21.
Let(Λω)φ be the Orlicz-Lorentz function space.(a)
IfI=[0,1], then the following assertions are equivalent:
(i)
(Λω)φ∈(Hl),(ii)
(Λω)φ∈(Hg),(iii)
φ∈Δ2(∞).(b)
Suppose thatI=[0,∞).
(i)
(Λω)φ∈(Hl) if and only if ∫0∞ω=∞ and φ∈Δ2(ℝ+),(ii)
(Λω)φ∈(Hg) if and only if φ∈Δ2(ℝ+).Proof.
Recall thatΛω∈(Hg) for any weight ω [1, Corollary  1.3]. Thus, (a) and (b) (ii) follow immediately. Clearly, each symmetric Köthe space is separable if and only if it is order continuous (see [13]). Consequently, from [1,Theorem  3.3], [8,Lemma  3.2], it follows that Λω∈(Hl) if and only if ∫0∞ω=∞. Thus, (b) (i) follows.Obviously, sinceL1∈(Hl), from Theorems 3.11 and 3.19 with E=L1, we obtain the characterization of Hl and Hg for Orlicz spaces proved directly in [10, Theorem  2.1].


## 4. Open Questions

(1) We have seen that Lemma3.9(i) is crucial in the proof of sufficiency in the main Theorem 3.19. Moreover, Lemma 3.9(i) is not true for a nonsymmetric Köthe space. Consequently, natural questions are as follows(a)
are conditions presented in Theorem3.19 sufficient also in a nonsymmetric case (different proof)? or, if the answer is no,(b)
how the necessary and sufficient conditions forHg are in general?Looking for full criteria in nonsymmetric case, we are not allowed to apply Corollary3.7 in the proof of sufficiency of Theorem 3.19.Note also that the symmetry ofE is applied in the proof of necessity of Theorem 3.19, by using Corollary 3.7. However, it seems that these parts go also if we drop the symmetry of E (after minor changes).(2) We say thatE has property Hgs provided for any x∈S(E) and (xn) in S(E) with xn→x in measure we have ∥xn-x∥→0. Clearly, by Corollary 3.7, E∈(Hgs) if and only if E∈(Hg) whenever E is a symmetric Köthe function space. Are properties Hgs and Hg different in nonsymmetric case?



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*Source: 314068-2012-01-19.xml*