Journal of Function Spaces

Journal of Function Spaces / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 314068 | 21 pages | https://doi.org/10.1155/2012/314068

Kadec-Klee Properties of Calderón-Lozanovskiĭ Function Spaces

Academic Editor: Henryk Hudzik
Received31 Jan 2011
Accepted02 Feb 2011
Published19 Jan 2012

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.

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., [39]). 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 𝐿1𝐻𝑙 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)𝜑Δ𝐸2 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

Let , +, 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 space 𝐸=(𝐸,𝐸) is said to be a Köthe space if 𝐸 is a linear subspace of 𝐿0 and(i)if 𝑥𝐸, 𝑦𝐿0, and |𝑦||𝑥|  𝜇-a.e., then 𝑦𝐸 and 𝑦𝐸𝑥𝐸;(ii)there exists a function 𝑥 in 𝐸 that is positive on the whole 𝑇 (see [2, 12]).

Every Köthe space is a Banach lattice under the natural partial order (𝑥0 if 𝑥(𝑡)0 for 𝜇-a.e. 𝑡𝑇). 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 (,2,𝑚), then we will say that 𝐸 is a Köthe sequence space.

The set 𝐸+={𝑥𝐸𝑥0} is called the positive cone of 𝐸. For any subset 𝐴𝐸, define 𝐴+=𝐴𝐸+.

By the symmetric Köthe function space 𝐸 (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 𝑥𝑦, 𝑥,𝑦𝐿0(𝐼) (i.e., they have the same distribution functions 𝑑𝑥=𝑑𝑦, where 𝑑𝑥(𝜆)=𝑚({𝑡𝐼|𝑥(𝑡)|>𝜆), 𝜆0) and 𝑥𝐸 we have 𝑦𝐸 and 𝑥𝐸=𝑦𝐸. In particular, 𝑥𝐸=𝑥𝐸, where𝑥(𝑡)=inf𝜆>0𝑑𝑥(𝜆)𝑡,𝑡>0.(2.1) 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 point 𝑥𝐸 is said to have order continuous norm if for any sequence (𝑥𝑚) in 𝐸 such that 0𝑥𝑚|𝑥| and 𝑥𝑚0  𝜇-a.e. we have 𝑥𝑚𝐸0. 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 𝑥𝐸𝑎 if and only if 𝑥𝜒𝐴𝑛𝐸0 for any sequence {𝐴𝑛} satisfying 𝐴𝑛 (i.e., 𝐴𝑛𝐴𝑛+1 and 𝜇(𝑛=1𝐴𝑛)=0). Clearly, 𝜇(𝑛=1𝐴𝑛)=0 if and only if 𝜒𝐴𝑛0  𝜇-a.e. in 𝑇.

Recall that 𝐸 is said to have Kadec-Klee property (𝐸(KK) for short) whenever 𝑥𝑛𝑥0 for any 𝑥 and {𝑥𝑛} in 𝐸 satisfying 𝑥𝑛𝑥 in the weak topology 𝜎(𝐸,𝐸) and 𝑥𝑛𝑥 (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, 1618]). If we replace the weak convergence 𝜎(𝐸,𝐸) by the convergence in measure (𝑥𝑛𝜇𝑥), by the convergence in measure on every set of finite measure (𝑥𝑛𝜇𝑥 locally), or by the uniform convergence (𝑥𝑛𝑥), 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 property 𝑃, 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[0,], it is convex, even, vanishing, and continuous at zero, left continuous on (0,) and not identically equal to zero. Denote𝑎𝜑=sup{𝑢0𝜑(𝑢)=0},𝑏𝜑=sup{𝑢0𝜑(𝑢)<}.(2.2) We write 𝜑>0 when 𝑎𝜑=0 and 𝜑< if 𝑏𝜑=. Let 𝜑𝑟=𝜑𝜒𝐺𝜑, where𝐺𝜑=𝑎𝜑,𝑏𝜑𝑏if𝜑𝜑𝑎<,𝜑,𝑏𝜑otherwise.(2.3) Define on 𝐿0 a convex semimodular 𝐼𝜑 by𝐼𝜑(𝑥)=𝜑𝑥𝐸if𝜑𝑥𝐸,otherwise,(2.4) where (𝜑𝑥)(𝑡)=𝜑(𝑥(𝑡)), 𝑡𝑇. By the Calderón-Lozanovskiĭ space 𝐸𝜑 we mean𝐸𝜑=𝑥𝐿0𝐼𝜑(𝑐𝑥)<forsome𝑐>0,(2.5) equipped with so called Luxemburg norm defined by𝑥𝜑=inf𝜆>0𝐼𝜑𝑥𝜆.1(2.6) We generally assume that if 𝑏𝜑<, then 𝑎𝜑<𝑏𝜑, because when 0<𝑎𝜑=𝑏𝜑, then 𝐸𝜑=𝐿 and 𝑥𝜑=(1/𝑏𝜑)𝑥.

If 𝐸=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 that 𝐸 has the Fatou property, that is, if 0𝑥𝑛𝑥𝐿0 with (𝑥𝑛)𝑛=1 in 𝐸 and sup𝑛𝑥𝑛𝐸<, then 𝑥𝐸 and 𝑥𝐸=lim𝑛𝑥𝑛𝐸. Since 𝐸 has the Fatou property, 𝐸𝜑 has also this property, whence 𝐸𝜑 is a Banach space.

We say an Orlicz function 𝜑 satisfies condition Δ2(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 𝑢0, then we say that 𝜑 satisfies condition Δ2(+). We write for short 𝜑Δ2(0), 𝜑Δ2(), and 𝜑Δ2(+), respectively.

For a Köthe space 𝐸 and an Orlicz function 𝜑, we say that 𝜑 satisfies condition Δ𝐸2 (𝜑Δ𝐸2 for short) if(1)𝜑Δ2(0) whenever 𝐸𝐿,(2)𝜑Δ2() whenever 𝐿𝐸,(3)𝜑Δ2(+) whenever neither 𝐿𝐸 nor 𝐸𝐿 (see [5]),

where the symbol 𝐸𝐹 stands for the continuous embedding of the space 𝐸 into the space 𝐹.

Relationships between the modular 𝐼𝜑 and the norm 𝜑 are collected in [8].

3. Results

3.1. Köthe Spaces and Orlicz Functions

Given a Köthe function space 𝐸, we setm𝐸=sup𝐵,𝜒𝐵𝐸sup[(𝐴𝑛)𝐵,𝜇(𝐴𝑛)0]inf𝑛𝜒𝐴𝑛𝐸.(3.1)

Remark 3.1. It is easy to see that if 𝐸 is a symmetric Köthe function space, then m𝐸=inf𝐴Σ,𝜇(𝐴)>0𝜒𝐴𝐸.(3.2)

Recall that if 𝑥𝐿0, then supp𝑥={𝑡𝑇𝑥(𝑡)0}. For a Köthe space 𝐸𝐿0(𝑇,Σ,𝜇), let 𝐹𝐸 be a subset. By the support of 𝐹, we mean the set supp𝐹Σ satisfying(i)for each 𝑥𝐹, there is 𝐴Σ, with 𝜇(𝐴)=0, such that supp𝑥𝐴supp𝐹,(ii)there is 𝑥𝐹 with 𝜇(supp𝐹supp𝑥)=0.

Lemma 3.2. Suppose that 𝐸 is a Köthe function space. Consider the following statements:(i)𝐸(𝑂𝐶),(ii)𝐸(𝐻𝑔)+,(iii)For any 𝑢𝐸 and each sequence {𝐴𝑛} in Σ the implication 𝜇𝐴𝑛0𝑢𝜒𝐴𝑛0holds.(3.3)(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 supp𝐸𝑎𝑇, then for each 𝑥𝐸 with supp𝑥𝑇𝑠upp𝐸𝑎 we have 𝑥𝐸𝑎. Let 𝑇0𝑇supp𝐸𝑎 be such that 0<𝜇(𝑇0)< and 𝜒𝑇0𝐸. Then 𝑥1=𝜒𝑇0𝐸𝑎 and consequently there is a sequence (𝐴𝑛) of measurable pairwise disjoint sets with 𝑛=1𝐴𝑛=𝑇0 and a number 𝛿>0 satisfying 𝜒𝐴𝑛𝛿 for each 𝑛 (see [19]). Since 𝜇(𝐴𝑛)0, we get m𝐸𝛿>0.
(ii)  (iii). Assume that (iii) does not hold, then we find an element 𝑢𝐸, a number 𝛿>0, and a sequence (𝐴𝑖) with 𝜇(𝐴𝑖)0 such that inf𝑖𝑢𝜒𝐴𝑖𝐸𝛿. Without loss of generality, we may assume that 𝑖=1𝜇(𝐴𝑖)<. Set 𝑥𝑘=sup𝑢𝜒𝐴𝑖𝑖𝑘𝐿0.(3.4) Let 𝐴=𝑖𝐴𝑖. Clearly, 𝑥𝑘𝑢𝜒𝐴𝐸. Put 𝑦=𝑢𝜒𝐴 and 𝑦𝑘=𝑦𝑥𝑘. Since {𝑥𝑘} is decreasing, so {𝑦𝑘} is increasing. Next, 𝑥𝑘0 in measure, because, for each 𝜀>0, 𝜇𝑡𝑇𝑥𝑘(𝑡)>𝜀𝜇𝑖=𝑘𝐴𝑖𝑖=𝑘𝜇𝐴𝑖0as𝑘.(3.5) In particular, by 𝐸(𝐹𝑃), 𝑦𝑘𝑦. Finally, 𝑦𝑘𝑦=𝑥𝑘𝑢𝜒𝐴𝑘𝛿. Thus, 𝐸(𝐻𝑔)+.
Assume that 𝐸 is symmetric. We prove that (vi)  (iv). If m𝐸>0, then we find a set 𝐵 with 𝜒𝐵𝐸 and a sequence {𝐴𝑛}𝐵, 𝜇(𝐴𝑛)0 such that inf𝑛𝜒𝐴𝑛𝐸m𝐸/2. Let 𝑥𝐸{0}. We prove that 𝑥𝐸𝐸𝑎. There is a number 𝛿>0 such that the set 𝐷={𝑡𝑇|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, 𝑥𝜒𝐶𝑘𝐸𝛿𝜒𝐶𝑘𝐸𝛿𝜒𝐴𝑛𝑘𝐸𝛿m𝐸2,(3.6) for any 𝑘, whence 𝑥𝐸𝑎.
Now we will prove that the implication (vi)  (iv) is not true in general in nonsymmetric case. Take 𝐿𝐸=1[]0,1𝐿[]1,21=𝑥𝐿0[]0,2𝑥=𝑢+𝑤,𝑢𝐿1[]0,1,𝑤𝐿[]1,2,𝑥=𝑢1+𝑤.(3.7) 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 0𝜔<, then 𝐸(𝐻𝑔) and 𝐸(𝑂𝐶)(see [1, 8]), whence (iii) (i) in general.
Finally, let 𝐸=𝐿𝜑[0,1]—the Orlicz space with 𝜑< and 𝜑Δ2(). By the proof of Theorem  1 from [5], we find an element 𝑢𝐸 and a sequence {𝐴𝑛} with 𝜇(𝐴𝑛)0 and 𝑢𝜒𝐴𝑛𝜑=1. On the other hand, for each 𝐵Σ with 𝑢=𝜒𝐵𝐸, we claim that 𝜒𝐵𝐸𝑎, whence m𝐸=0. Indeed, for any sequence {𝐴𝑛} satisfying 𝐴𝑛 and for each 𝜆>0, we have 𝐼𝜑𝜆𝑢𝜒𝐴𝑛=𝜑(𝜆)10𝜒𝐴𝑛𝐴𝑑𝜇=𝜑(𝜆)𝜇𝑛0,(3.8) whence 𝑢𝜒𝐴𝑛𝜑0. Thus, the implication (iii)  (iv) cannot be reversed in general.

Recall that if 𝐸(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 that 𝜇(𝐴)< and 𝜒𝐴𝐸 only in a nonsymmetric Köthe space. However, if 𝜇(𝐴)< and 𝜒𝐴𝐸, then there is an increasing sequence (𝐴𝑛) of subsets of 𝐴 satisfying 𝑛𝐴𝑛=𝐴 and 𝜒𝐴𝑛𝐸 for each 𝑛.

Proof. Let 𝑥0 be a weak unit of 𝐸, that is, 𝑥0(𝑡)>0 for each 𝑡𝑇. It is enough to take 𝐴𝑛={𝑡𝐴𝑥0(𝑡)>1/𝑛} for each 𝑛.

Remark 3.4. Let 𝐸 be a Köthe space. If 𝐸(𝐻𝑢)+, then 𝑥𝜒𝐵𝑘0 for each 𝑥𝐸+, where 𝐵𝑘={𝑡𝑇𝑥(𝑡)<1/𝑘}.

Proof. Let 𝑥𝐸+. Clearly, 𝑥𝜒𝑇𝐵𝑘𝑥 and sup𝑘𝑥𝜒𝑇𝐵𝑘<. By 𝐸(𝐹𝑃), we get 𝑥𝜒𝑇𝐵𝑘𝑥. Moreover, 𝑥𝑥𝜒𝑇𝐵𝑘0 uniformly. Then 𝑥𝜒𝑇𝐵𝑘𝑥0, because 𝐸(𝐻𝑢)+.
It is easy to see that the above is not true if 𝐸(𝐻𝑢)+. It is enough to take 𝐸=𝐿𝜔(0,)=𝑥𝐿0𝑥=supess𝑡0||||,𝜔(𝑡)𝑥(𝑡)<(3.9) 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, 𝑥𝑛𝑥 uniformly and 𝑥𝑛𝑥=1, whence 𝐸(𝐻𝑢)+.

It is known that 𝐸(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. Suppose 𝐸 is a Köthe space. Then 𝐸(𝐻𝑔) if and only if 𝐸(𝐻𝑔)+.

Proof. The necessity is obvious. We prove the sufficiency. Let 𝐸(𝐻𝑔)+. Assume that 𝐸 is a Köthe function space. Let 𝑥𝑛𝑥 and 𝑥𝑛𝑥 in measure. For each 𝜂>0, we get 𝜇||||𝑥𝑡𝑇𝑛||||𝑥||||||𝑥(𝑡)(𝑡)𝜂𝜇𝑡𝑇𝑛||(𝑡)𝑥(𝑡)𝜂0,(3.10) whence |𝑥𝑛||𝑥| in measure. By the assumption 𝐸(𝐻𝑔)+, we conclude that ||𝑥𝑛|||𝑥|0.(3.11) Then we find a subsequence (|𝑥𝑛𝑚|)𝑚 of (|𝑥𝑛|)𝑛, an element 𝑦𝐸+, and a sequence 0𝛽𝑚0 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 𝜀>0. Applying Remark 3.4, with respective sets 𝐵𝑘=1𝑡𝑢(𝑡)<𝑘,(3.12) take 𝑘 big enough to satisfy 𝑢𝜒𝐵𝑘<𝜀/8. Moreover, |𝑥𝑛||𝑥|<𝜀/8 for sufficiently large 𝑛. Setting 𝐶𝑛=𝑥𝑡𝑇0sgn(𝑡)𝑥𝑛(𝑡)1,𝐷𝑛=𝑥𝑡𝑇sgn(𝑡)𝑥𝑛(𝑡)=1,(3.13) we get 𝑥𝑥𝑛||𝑥|𝑥|𝑛||𝜒𝐶𝑛+𝑥𝑥𝑛𝜒𝐷𝑛𝐵𝑘+𝑥𝑥𝑛𝜒𝐷𝑛𝐵𝑘5𝜀8+𝑥𝑥𝑛𝜒𝐷𝑛𝐵𝑘.(3.14) Take 𝜀0<𝛿<116𝑘min,1𝑦𝑥.(3.15) Denote 𝐹𝑘𝑛=𝑡𝐷𝑛𝐵𝑘||𝑥𝑛||,𝐺(𝑡)𝑥(𝑡)>𝛿𝑘𝑛=𝑡𝐷𝑛𝐵𝑘||𝑥𝑛||.(𝑡)𝑥(𝑡)𝛿(3.16) Note that 𝑥𝑛𝜇𝑥 and consequently 𝜇(𝐹𝑘𝑛)0 as 𝑛. Then, by Lemma 3.2 (iii), 𝑥𝑥𝑛𝜒𝐷𝑛𝐵𝑘𝑥𝑥𝑛𝜒𝐹𝑘𝑛+𝑥𝑥𝑛𝜒𝐺𝑘𝑛2𝑢𝜒𝐹𝑘𝑛+𝑥𝑥𝑛𝜒𝐺𝑘𝑛<𝜀8+𝑥𝑥𝑛𝜒𝐺𝑘𝑛,(3.17) for sufficiently large 𝑛. Finally, divide set 𝐺𝑘𝑛 into two subsets 𝐻𝑘𝑛=𝑡𝐺𝑘𝑛||||1𝑥(𝑡)2𝑘,𝐼𝑘𝑛=𝑡𝐺𝑘𝑛1𝑦(𝑡).2𝑘(3.18) Clearly, 𝐻𝑘𝑛𝐼𝑘𝑛=𝐺𝑘𝑛 because 𝐺𝑘𝑛𝑇𝐵𝑘=||||1𝑡𝑥(𝑡)+𝑦(𝑡)𝑘.(3.19) Then, by (3.15), 𝑥𝑥𝑛𝜒𝐺𝑘𝑛𝜒𝛿𝐻𝑘𝑛+𝜒𝐼𝑘𝑛𝛿2𝑘|𝑥|𝜒𝐻𝑘𝑛+2𝑘𝑦𝜒𝐼𝑘𝑛2𝑘𝛿|𝑥|𝜒𝐻𝑘𝑛+𝑦𝜒𝐼𝑘𝑛<𝜀4.(3.20) Combining (3.14), (3.17), and (3.20), we get 𝑥𝑥𝑛𝜀.
If 𝐸 is a Köthe sequence space, the proof is the same, because for a counting measure space (,2,𝑚) for each 𝑢𝐸 and any sequence (𝐴𝑛) in 2 with 𝑚(𝐴𝑛)0, we have 𝑢𝜒𝐴𝑛0.

The straightforward calculation shows that we may equivalently take 𝑥 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 that 𝐸 is a symmetric Köthe function space. If 𝑥𝑛𝑥 in measure and 𝑥𝑛𝑥0, then 𝑥𝑛/𝑥𝑛𝑥/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 each 𝜀,𝜂>0, we get 𝜇||||𝑥𝑡𝑛(𝑡)𝑥𝑛𝑥(𝑡)||||||||𝑥𝑥>𝜀=𝜇𝑡𝑛(𝑡)𝑥𝑛𝑥(𝑡)𝑥𝑛+𝑥(𝑡)𝑥𝑛𝑥(𝑡)||||1𝑥>𝜀𝜇𝑡𝑥𝑛||𝑥𝑛||+||||1(𝑡)𝑥(𝑡)𝑥𝑛1||||||||1𝑥𝑥(𝑡)>𝜀𝜇𝑡𝑥𝑛||𝑥𝑛||>𝜀(𝑡)𝑥(𝑡)2||||1+𝜇𝑡𝑥𝑛1||||||𝑥||>𝜀𝑥(𝑡)2.(3.21) Since 𝑥𝑛𝑥, so 𝑥𝑛>(1/2)𝑥 for 𝑛𝑁0, whence 𝜇1𝑡𝑥𝑛||𝑥𝑛||>𝜀(𝑡)𝑥(𝑡)2||𝑥𝜇𝑡𝑛||>(𝑡)𝑥(𝑡)𝜀𝑥4<𝜂2,(3.22) for 𝑛𝑁1>𝑁0. Moreover, applying the symmetry of 𝐸, we get 𝜇||||1𝑡𝑥𝑛1||||||||>𝜀𝑥𝑥(𝑡)2<𝜂2for𝑛𝑁2.(3.23) Thus, 𝜇||||𝑥𝑡𝑛(𝑡)𝑥𝑛𝑥(𝑡)||||𝑥>𝜀<𝜂,(3.24) for 𝑛max{𝑁1,𝑁2}.
Considering the second assertion, take 𝐸=𝐿11/𝑡3(1,)=𝑥𝐿0((1,),Σ,𝑚)𝑥=1||||1𝑥(𝑡)𝑡3𝑑𝑡<.(3.25) Set 𝑥(𝑡)=𝑡 and 𝑥𝑛(𝑡)=𝑡+(1/𝑛), then 𝑥=1, 𝑥𝑛=1+(1/2𝑛), and 𝑥𝑛𝑥 in measure. However, 𝜇||||𝑥𝑡𝑛(𝑡)𝑥𝑛𝑥(𝑡)|||||||𝑥>1=𝜇𝑡𝑡2|||2𝑛+1>1=,(3.26) for each 𝑛, whence (𝑥𝑛/𝑥𝑛)(𝑥/𝑥) in measure.

Corollary 3.7. Assume that 𝐸 is a symmetric Köthe function space. Then 𝐸(𝐻𝑔) if and only if 𝑥𝑛𝑥0 for any 𝑥𝑆(𝐸) and (𝑥𝑛) in 𝑆(𝐸) satisfying 𝑥𝑛𝑥 in measure.

Proof. The necessity is obvious. We prove the sufficiency. Let 𝑥𝑛𝑥 in measure and 𝑥𝑛𝑥0. Taking 𝑦𝑛=𝑥𝑛/𝑥𝑛 and 𝑦=𝑥/𝑥, we get 𝑦𝑛𝑦 in measure, by Lemma 3.6. By the assumption, we conclude that 𝑦𝑛𝑦0. Notice that ||||1𝑥𝑛𝑥𝑛||||1𝑥𝑥𝑥𝑛1||||||||𝑥𝑥𝑛𝑥𝑛𝑥𝑥𝑛𝑥𝑥𝑛+𝑥=𝑥𝑥𝑛𝑥𝑛𝑥𝑥0.(3.27) Since 𝑥𝑛𝑥0, so 𝑥𝑛𝑥0 as desired.

We will need in the sequel the following.

Lemma 3.8. If 𝜑Δ2(), then for any 𝑎>0, there are 𝜎(0,1] and 𝑢1>𝑎𝜑+𝑎 such that inf𝑢0(0,𝑎)inf𝑢𝑢1𝜑𝑢𝑢0𝑢𝜑(𝑢)𝜑0𝜎.(3.28)

Proof. Suppose conversely that there is 𝑎>0 such that for any 𝜎(0,1] and each 𝑢1>𝑎𝜑+𝑎, there holds inf𝑢0(0,𝑎)inf𝑢𝑢1𝜑𝑢𝑢0𝑢𝜑(𝑢)𝜑0<𝜎.(3.29) Taking 𝜎𝑛=1/2𝑛 and 𝑢𝑛, we get inf𝑢0(0,𝑎)inf𝑢𝑢𝑛𝜑(𝑢𝑢0)/(𝜑(𝑢)𝜑(𝑢0))<1/2𝑛. We may assume that 𝑢𝑛>2𝑎 for each 𝑛. Hence, for each 𝑛, one may choose 𝑣𝑛𝑢𝑛 and 𝑢0(0,𝑎) with 𝜑(𝑣𝑛𝑢0)/(𝜑(𝑣𝑛)𝜑(𝑢0))1/2𝑛. Consequently, we get 2𝑛𝜑12𝑣𝑛<2𝑛𝜑𝑣𝑛𝑢0𝑣<𝜑𝑛𝑢𝜑0𝑣𝜑𝑛,(3.30) for each 𝑛. Taking 𝑏𝑛=(1/2)𝑣𝑛, one has 2𝑛𝜑(𝑏𝑛)<𝜑(2𝑏𝑛), whence 𝜑Δ2().

Lemma 3.9. (i) Suppose that 𝐸 is a symmetric Köthe space. If 𝑥𝑛𝑥 in measure and 𝜑𝑥𝑛, 𝜑𝑥𝐸 are finitely valued a.e., then 𝜑𝑥𝑛𝜑𝑥 in measure. Furthermore, for a nonsymmetric Köthe space, the above implication is not true in general.
(ii) If 𝑥𝑛𝑥 in measure, 𝑥𝑛,𝑥𝐸+ and 𝜑𝑟1𝑥𝑛, 𝜑𝑟1𝑥 are well-defined functions, then 𝜑𝑟1𝑥𝑛𝜑𝑟1𝑥 in measure.

Proof. (i) Let 𝑥𝑛𝜇𝑥.
(A) Suppose that 𝑏𝜑=. Set 𝐴𝑘=||𝑥||𝑡𝑇(𝑡)>𝑘.(3.31) Then, by symmetry of 𝐸,  𝜇(𝐴𝑘)0. Let 𝜂>0 and 𝜀>0. Take 𝑘0 such that 𝜇(𝐴𝑘0)<𝜂/3. We have 𝜇||𝜑𝑥𝑡𝑇𝑛||(𝑡)𝜑(𝑥(𝑡))𝜀𝜇𝑡𝑇𝐴𝑘0||𝜑𝑥𝑛(||+𝜂𝑡)𝜑(𝑥(𝑡))𝜀3.(3.32) 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 |𝑢𝑠|<𝛿|𝜑(𝑢)𝜑(𝑠)|<𝛾 is true for all 𝑢,𝑠[𝑘0,𝑘0]. Set 𝛿=𝛿(𝜀,2𝑘0) and 𝐴1𝑛=𝑡𝑇𝐴𝑘0||𝑥𝑛||(𝑡)>2𝑘0,𝐴2𝑛=𝑡𝑇𝐴𝑘0||𝑥𝑛||(𝑡)2𝑘0.(3.33) Consequently 𝜇||𝜑𝑥𝑡𝑇𝑛||𝐴(𝑡)𝜑(𝑥(𝑡))𝜀𝜇1𝑛+𝜇𝑡𝐴2𝑛||𝜑𝑥𝑛(||+𝜂𝑡)𝜑(𝑥(𝑡))𝜀3𝜇𝑡𝐴2𝑛||𝑥𝑛||+(𝑡)𝑥(𝑡)𝛿2𝜂3𝜂(3.34) for sufficiently large 𝑛, because 𝐴1𝑛={𝑡𝑇|𝑥𝑛(𝑡)𝑥(𝑡)|𝑘0}. Thus 𝜑𝑥𝑛𝜇𝜑𝑥.
(B) Assume 𝜑(𝑏𝜑)<. The proof is simpler than above because 𝑥𝑛(𝑡)𝑏𝜑 and 𝑥(𝑡)𝑏𝜑 for 𝜇-a.e. 𝑡𝑇 and 𝜑 is uniformly continuous on the interval [𝑏𝜑,𝑏𝜑].
(C) Suppose that 𝑏𝜑<, 𝜑(𝑏𝜑)=. Then 𝑥𝑛(𝑡)<𝑏𝜑 and 𝑥(𝑡)<𝑏𝜑 for 𝜇-a.e. 𝑡𝑇. Set 𝐴𝑘=||||𝑡𝑇𝑥(𝑡)>𝑏𝜑1𝑘.(3.35) Then 𝜇(𝐴𝑘)0, because 𝐸 is symmetric. Let 𝜂>0 and 𝜀>0. Take 𝑘0 such that 𝜇(𝐴𝑘0)<𝜂/3. We have 𝜇||𝜑𝑥𝑡𝑇𝑛||(𝑡)𝜑(𝑥(𝑡))𝜀𝜇𝑡𝑇𝐴𝑘0||𝜑𝑥𝑛(||𝐴𝑡)𝜑(𝑥(𝑡))𝜀+𝜇𝑘0.(3.36) 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 |𝑢𝑠|<𝛿|𝜑(𝑢)𝜑(𝑠)|<𝛾 is true for all 𝑢,𝑠[𝑢0,𝑢0]. Put 𝛿=𝛿(𝜀,𝑏𝜑1/2𝑘0) and 𝐴1𝑛=𝑡𝑇𝐴𝑘0||𝑥𝑛||(𝑡)>𝑏𝜑12𝑘0,𝐴2𝑛=𝑡𝑇𝐴𝑘0||𝑥𝑛||(𝑡)𝑏𝜑12𝑘0.(3.37) Consequently 𝜇||𝜑𝑥𝑡𝑇𝑛||𝐴(𝑡)𝜑(𝑥(𝑡))𝜀𝜇1𝑛+𝜇𝑡𝐴2𝑛||𝜑𝑥𝑛(||+𝜂𝑡)𝜑(𝑥(𝑡))𝜀3𝜇𝑡𝐴2𝑛||𝑥𝑛||+(𝑡)𝑥(𝑡)𝛿2𝜂3𝜂(3.38) for sufficiently large 𝑛, because 𝐴1𝑛={𝑡𝑇|𝑥𝑛(𝑡)𝑥(𝑡)|1/2𝑘0}. Thus 𝜑𝑥𝑛𝜇𝜑𝑥.
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 𝜑(𝑢)=𝑢2,𝑥=𝑛𝑛𝜒[𝑛,𝑛+1],𝑥𝑘=𝑛1𝑛+𝑘𝜒[𝑛,𝑛+1].(3.39) Then 𝑥𝑘𝑥 in measure. On the other hand, |𝜑𝑥𝑘𝜑𝑥|=𝑛((1/𝑘2)+(2𝑛/𝑘))𝜒[𝑛,𝑛+1], whence 𝜑𝑥𝑘𝜑𝑥 in measure. Clearly, such element 𝜑𝑥 can be well defined only in some nonsymmetric Köthe space 𝐸.
(ii) First, we claim that for each 𝜀>0, there is 𝛿>0 such that the implication ||𝜑𝑟1(𝑢)𝜑𝑟1||(𝑣)>𝜀|𝑢𝑣|>𝛿(3.40) is true for all 𝑢,𝑣0. First, notice that for each 𝜀>0, there is 𝛿>0 such that the implication ||𝜑𝑟1(𝑢)𝜑𝑟1||(0)>𝜀|𝑢|>𝛿(3.41) is true for all 𝑢0, because 𝜑𝑟1 is continuous at zero. Next, defining 𝑔(𝑢)=𝜑𝑟1(𝑢)𝜑𝑟1(0), we see that 𝑔 is subadditive on +, because 𝑔=𝜓1, where 𝜓 is an Orlicz function with 𝜓>0. Consequently, 𝜑𝑟1(𝑢𝑣)𝜑𝑟1(0)=𝑔(𝑢𝑣)𝑔(𝑢)𝑔(𝑣)=𝜑𝑟1(𝑢)𝜑𝑟1(𝑣).(3.42) Hence, if |𝜑𝑟1(𝑢)𝜑𝑟1(𝑣)|>𝜀, then |𝜑𝑟1(𝑢𝑣)𝜑𝑟1(0)|>𝜀, whence |𝑢𝑣|>𝛿, by (3.41). Thus, the claim is proved. Assume that 𝑥𝑛, 𝑥𝐸+ and 𝜑𝑟1𝑥𝑛, 𝜑𝑟1𝑥 are well-defined functions and 𝑥𝑛𝑥 in measure, then, applying (3.40), for each 𝜀>0, we get 𝜇||𝜑𝑡𝑇𝑟1𝑥𝑛𝜑𝑟1||||𝑥𝑥(𝑡)>𝜀𝜇𝑡𝑇𝑛||𝑥(𝑡)>𝛿0.(3.43)

3.2. Calderón-Lozanovskiĭ Function Spaces

We still assume below that 𝐸 is a Köthe function space.

It is known that 𝐸𝜑(𝑂𝐶) if and only if 𝐸(𝑂𝐶) and 𝜑Δ𝐸2 (see Corollary  12 from [9]). Considering some weaker property than 𝑂𝐶, we get the following.

Lemma 3.10. Suppose that 𝑏𝜑=, then 𝐸𝑎{0} if and only if (𝐸𝜑)𝑎{0}.

Proof. Necessity
Let 𝑧𝐸𝑎{0}. We find a number 𝑚>0 such that 𝜇(𝐴)=𝜇{𝑡𝑇|𝑧(𝑡)|>1/𝑚}>0. Taking 𝑤=𝜑1(1)𝜒𝐴, we get 𝜑𝑤𝐸𝑎. Moreover, since 𝑏𝜑=, 𝜑 satisfies the local Δ𝐸2 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 𝜑𝑥𝐸𝑎. If 𝜇{𝑡|𝑥(𝑡)|>𝑎𝜑}>0, then 𝜇(𝐴𝑚)>0 for some 𝑚, where 𝐴𝑚={𝑡𝑇|𝑥(𝑡)|(1+(1/𝑚))𝑎𝜑}. Thus, 𝜒𝐴𝑚𝐸𝑎, and we are done, if 𝜇{𝑡|𝑥(𝑡)|𝑎𝜑}=𝜇(𝐴)>0 then 𝜒𝐴𝐸𝑎{0}, by Theorem  11 from [9].

Theorem 3.11. 𝐸𝜑(𝐻𝑙) if and only if(a)𝐸(𝐻𝑙), (b)𝜑Δ2() whenever 𝐿𝐸 and 𝜑Δ2(+) provided 𝐿𝐸.

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 𝜑Δ𝐸2 (see [9, Corollary  12]). Notice that 𝐸(𝑂𝐶) implies that 𝐸𝐿. Indeed, if 𝐸𝐿, then there is a decreasing sequence (𝐴𝑛) in Σ with 0<𝜇(𝐴𝑛)0 and 𝜒𝐴𝑛𝐸, whence 1=𝜒𝐴𝑛𝐿𝑀𝜒𝐴𝑛𝐸, 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 𝑥𝑛𝜇𝑥 locally in measure and 𝑥𝑛𝑥𝐸𝜀. Set 𝑦𝑛=𝜑𝑟1𝑥𝑛 and 𝑦=𝜑𝑟1𝑥, then 𝐼𝜑(𝑦)=𝐼𝜑(𝑦𝑛)=1, whence 𝑦,𝑦𝑛𝑆(𝐸𝜑). Moreover, 𝑦𝑛𝜇𝑦 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 𝑦𝑛𝑦𝜑𝜂,(3.44) 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 𝜑𝑦𝑛𝜑𝑦𝐸𝜀.(3.45) Denote 𝐴𝑛=𝑦𝑡𝑇𝜑𝑛(𝑡)𝜑(𝑦(𝑡)),𝐵𝑛=𝑇𝐴𝑛.(3.46) By (3.45), we get max𝜑𝑦𝑛𝜒𝜑𝑦𝐴𝑛𝐸,𝜑𝑦𝑛𝜒𝜑𝑦𝐵𝑛𝐸𝜀2,(3.47) for each 𝑛. We assume that (𝜑𝑦𝑛𝜑𝑦)𝜒𝐴𝑛𝐸𝜀/2 for infinitely many 𝑛, because otherwise the proof is analogous. We divide the proof into two cases.(I)Suppose that 𝐿𝐸 and 𝜑Δ2(). Let 𝑐=𝜑𝑟1(𝜀/32𝜒𝑇𝐸)>𝑎𝜑. Denoting𝐴1𝑛=𝑡𝐴𝑛𝑦𝑛(𝑡)<𝑐,𝐴2𝑛=𝑡𝐴𝑛𝑦𝑛(𝑡)𝑐,(3.48) we get 𝜀2𝜑𝑦𝑛𝜒𝜑𝑦𝐴𝑛𝐸𝜑𝑦𝑛𝜒𝜑𝑦𝐴2𝑛𝐸+𝜀4.(3.49) Thus, 𝜑𝑦𝑛𝜒𝜑𝑦𝐴2𝑛𝐸𝜀4.(3.50) Set 𝑐1=𝑐+𝑎𝜑2,𝐵1𝑛=𝑡𝐴2𝑛𝑦(𝑡)<𝑐1,𝐵2𝑛=𝑡𝐴2𝑛𝑦(𝑡)𝑐1.(3.51) The remaining proof of the case (I) we divide into two parts.(1)Suppose that (𝜑𝑦𝑛𝜑𝑦)𝜒𝐵1𝑛𝐸𝜀/8. By Lemma 3.8, for 𝑎=𝑐1, there are 𝜎>0 and 𝑢1>0 such that inf𝑢00,𝑐1inf𝑢𝑢1𝜑𝑢𝑢0𝑢𝜑(𝑢)𝜑0𝜎.(3.52) Set 𝐶1𝑛=𝑡𝐵1𝑛𝑦𝑛(𝑡)<𝑢1,𝐶2𝑛=𝑡𝐵1𝑛𝑦𝑛(𝑡)𝑢1.(3.53)(a)Assume that (𝜑𝑦𝑛𝜑𝑦)𝜒𝐶1𝑛𝐸𝜀/16. Taking 0<𝛼<(𝑐𝑐1)/𝑢1, we get 𝑦𝜑𝑛𝑦𝛼𝜒𝐶1𝑛𝐸𝜑𝑐𝑐1𝛼𝜒𝐶1𝑛𝐸𝜑𝑢1𝜒𝐶1𝑛𝐸𝜑𝑦𝑛𝜒𝜑𝑦𝐶1𝑛𝐸𝜀.16(3.54) Thus, 𝑦𝑛𝑦𝜑𝜂1=min{1,𝛼𝜀/16}.(b)Let (𝜑𝑦𝑛𝜑𝑦)𝜒𝐶2𝑛𝐸𝜀/16. Thus, by Lemma 3.8, 𝑦𝜑𝑛𝜒𝑦𝐶2𝑛𝐸𝜎2𝜑𝑦𝑛𝜒𝜑𝑦𝐶2𝑛𝐸𝜎𝜀,32(3.55) and consequently 𝑦𝑛𝑦𝜑𝜂2=min{1,𝜎𝜀/32} (cf. [8, Lemma  1.1]).(2) Assume that (𝜑𝑦𝑛𝜑𝑦)𝜒𝐵1𝑛𝐸<𝜀/8, then, by (3.50), 𝜑𝑦𝑛𝜒𝜑𝑦𝐵2𝑛𝐸𝜀8.(3.56) Since 𝜑Δ2(), for 𝑙=1+𝜀/32 and 𝑢1=𝑐1, there is 𝑎=𝑎(𝑙,𝑢1)(0,1) such that 𝜑((1+𝑎)𝑢)𝑙𝜑(𝑢),(3.57) for every 𝑢𝑐1 (see [20, Theorem  1.13(4)]). Moreover, we can choose 𝑎>0 satisfying 𝑎𝜑𝑐1+𝑎1𝜒𝑇𝐸<𝜀.32(3.58) Let 𝐵𝑛21=𝑡𝐵2𝑛𝑦𝑛𝑦(𝑡)<𝑎𝑐1,𝐵1+𝑎𝑛22=𝑡𝐵2𝑛𝑦𝑛(𝑦𝑡)𝑎𝑐1,1+𝑎(3.59) then, by (3.57) and (3.58), using convexity of 𝜑, we get ||𝜑𝑦𝑛||𝜒𝜑𝑦𝐵𝑛21=𝑦𝜑𝑛𝜒𝑦+𝑦𝜑𝑦𝐵𝑛21𝑎1+𝑎𝜑1+𝑎𝑎𝑦𝑛+1𝑦𝜒1+𝑎𝜑((1+𝑎)𝑦)𝜑𝑦𝐵𝑛21𝑎1+𝑎𝜑1+𝑎𝑎𝑦𝑛+1𝑦𝜒1+𝑎𝑙𝜑𝑦𝜑𝑦𝐵𝑛21=𝑎1+𝑎𝜑1+𝑎𝑎𝑦𝑛+𝑦𝜀/32𝑎𝜒1+𝑎𝜑𝑦𝐵𝑛21.(3.60) Note that 𝑓(𝑎)=(𝜀/32𝑎)/(1+𝑎) is a decreasing function of 𝑎>0. Moreover, 𝑎𝜀/32 by convexity of 𝜑 and inequality (3.57). Hence, by (3.58), (𝜑𝑦𝑛𝜑𝑦)𝜒𝐵𝑛21𝐸<𝜀/16. Then, by (3.56), 𝜑𝑦𝑛𝜒𝜑𝑦𝐵𝑛22𝐸𝜀.16(3.61) Since 𝜑Δ2(), for 𝑢3=𝑐1 and 𝛽=(1+𝑎)/𝑎, there is 𝑘2>0 such that 𝜑(𝛽𝑢)𝑘2𝜑(𝑢) for each 𝑢𝑢3 (see [20]). Taking 0<𝛾<𝑎/(1+𝑎) and applying (3.57), we get ||𝜑𝑦𝑛||𝜒𝜑𝑦𝐵𝑛22=𝑦𝜑𝑛𝜒𝑦+𝑦𝜑𝑦𝐵𝑛22𝑦𝜑𝑛𝑦𝛾𝜒+𝑦𝜑𝑦𝐵𝑛22𝑎1+𝑎𝜑1+𝑎𝑎𝑦𝑛𝑦𝛾+1𝜒1+𝑎𝜑((1+𝑎)𝑦)𝜑𝑦𝐵𝑛22𝑎𝑘1+𝑎2𝑦𝜑𝑛𝑦𝛾+1𝜒1+𝑎𝑙𝜑𝑦𝜑𝑦𝐵𝑛22=𝑎𝑘1+𝑎2𝑦𝜑𝑛𝑦𝛾+𝜀/32𝑎𝜒1+𝑎𝜑𝑦𝐵𝑛22.(3.62) Then 𝜑((𝑦𝑛𝑦)/𝛾)𝐸𝜀(1+𝑎)/32𝑎𝑘2, and consequently 𝑦𝑛𝑦𝜑𝜂3=min{1,𝛾𝜀(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 that 𝐿𝐸. Since 𝜑Δ2(+), so for every 𝑙>1, there is 𝑎=𝑎(𝑙)(0,1) such that 𝜑((1+𝑎)𝑢)𝑙𝜑(𝑢) for every 𝑢0 (see [20, Theorem  1.13(4)]). Moreover, for every 𝛽>0, there is 𝑘>0 such that 𝜑(𝛽𝑢)𝑘𝜑(𝑢) for each 𝑢0. 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. Theorem 3.19 below presents the full characterization of property 𝐻𝑔. We divide the whole proof into several lemmas for readers convenience.

Lemma 3.12. If 𝐸𝜑(𝐻𝑔)+, then m𝐸=0.

Proof. If m𝐸>0, then there are 𝛿>0, a set 𝐵,𝜇(𝐵)>0,𝜒𝐵𝐸, and a sequence (𝐴𝑛)𝐵 with 𝜇(𝐴𝑛)0 and inf𝑛𝜒𝐴𝑛𝐸𝛿. Without loss of generality, we may assume that 𝑖=1𝜇(𝐴𝑖)<. Set 𝑧𝑛𝜒=sup𝐴𝑖𝑖𝑛𝐿0.(3.63) Let 𝐴=𝑖𝐴𝑖. Clearly, 𝑧𝑛𝜒𝐴𝜒𝐵𝐸. Put 𝑦=𝜒𝐴 and 𝑦𝑛=𝑦𝑧𝑛. Choose 𝑎>0 such that