Kadec-Klee Properties of Calderón-Lozanovskiĭ Function Spaces

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., [3–9]). The complete characterization of Kadec-Klee properties with respect to local 𝐻_𝑙 (global 𝐻_𝑔) convergence in measure for Orlicz function spaces 𝐿_𝜑 has been presented in [10]. The Orlicz spaces 𝐿_𝜑 are special case of Calderón-Lozanovskiĭ spaces 𝐸_𝜑. Thus, it is natural to investigate properties 𝐻_𝑙 and 𝐻_𝑔 in the spaces 𝐸_𝜑 generated by arbitrary Köthe spaces 𝐸. Since 𝐿¹∈𝐻_𝑙 and 𝐸_𝜑=𝐿_𝜑 when 𝐸=𝐿¹, so finding full criteria of 𝐻_𝑙 and 𝐻_𝑔 in Orlicz spaces is simpler than in general Calderón-Lozanovskiĭ spaces. The sufficient conditions of 𝐻_𝑙 in Calderón-Lozanovskiĭ function spaces 𝐸_𝜑 have been proved in [4, 6]. Note that the implication 𝐸_𝜑∈(𝐻_𝑙)⇒𝜑∈Δ^𝐸2 follows from the fact that property 𝐻_𝑙 implies order continuity in Köthe spaces and from results concerning copy of 𝑙^∞ in 𝐸_𝜑 (see [5, 10]). The implication 𝐸_𝜑∈(𝐻_𝑙)⇒𝐸∈(𝐻_𝑙) has not been discussed at all. Moreover, property 𝐻_𝑔 in 𝐸_𝜑 has not been considered early. We present full criteria of properties 𝐻_𝑙 and 𝐻_𝑔 in Calderón-Lozanovskiĭ function spaces 𝐸_𝜑. Furthermore, in the case of property 𝐻_𝑔, we prove some general results concerning Köthe spaces which will be essential in studying this property in spaces 𝐸_𝜑. The crucial difficulty in the proofs concerning property 𝐻_𝑔 in spaces 𝐸_𝜑 is the fact that property 𝐻_𝑔 needs not imply order continuity in general. Note that if 𝐸∈(𝐻_𝑙), then 𝐸∈(𝑂𝐶) (see [10]), so looking for complete characterizations in the spaces 𝐸_𝜑, the case of property 𝐻_𝑙 is easier than 𝐻_𝑔. As an application, we get the criteria of properties 𝐻_𝑙 and 𝐻_𝑔 in Orlicz-Lorentz function spaces (Λ_𝜔)_𝜑. A sufficient condition for property 𝐻_𝑙 of (Λ_𝜔)_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 ℝ, 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 space 𝐸=(𝐸,‖⋅‖_𝐸) is said to be a Köthe space if 𝐸 is a linear subspace of 𝐿⁰ and(i)if 𝑥∈𝐸, 𝑦∈𝐿⁰, and |𝑦|≤|𝑥|  𝜇-a.e., then 𝑦∈𝐸 and ‖𝑦‖_𝐸≤‖𝑥‖_𝐸;(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 (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,2^ℕ,𝑚), then we will say that 𝐸 is a Köthe sequence space.

The set 𝐸₊={𝑥∈𝐸∶𝑥≥0} is called the positive cone of 𝐸. For any subset 𝐴⊂𝐸, define 𝐴₊=𝐴∩𝐸₊.

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 𝑥∼𝑦, 𝑥,𝑦∈𝐿⁰(𝐼) (i.e., they have the same distribution functions 𝑑_𝑥=𝑑_𝑦, where 𝑑_𝑥(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, where𝑥^∗(𝑡)=inf𝜆>0∶𝑑_𝑥(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 𝑥∈𝐸 is said to have order continuous norm if for any sequence (𝑥_𝑚) in 𝐸 such that 0≤𝑥_𝑚≤|𝑥| and 𝑥_𝑚→0  𝜇-a.e. we have ‖𝑥_𝑚‖_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, 𝐸_𝑎 stands for the subspace of order continuous elements of 𝐸. It is known that 𝑥∈𝐸_𝑎 if and only if ‖𝑥𝜒_𝐴𝑛‖_𝐸↓0 for any sequence {𝐴_𝑛} satisfying 𝐴_𝑛↓∅ (i.e., 𝐴_𝑛⊃𝐴_𝑛+1 and 𝜇(⋂^∞𝑛=1𝐴_𝑛)=0). Clearly, 𝜇(⋂^∞𝑛=1𝐴_𝑛)=0 if and only if 𝜒_𝐴𝑛→0  𝜇-a.e. in 𝑇.

Recall that 𝐸 is said to have Kadec-Klee property (𝐸∈(KK) for short) whenever ‖𝑥_𝑛−𝑥‖→0 for any 𝑥 and {𝑥_𝑛} in 𝐸 satisfying 𝑥_𝑛→𝑥 in the weak topology 𝜎(𝐸,𝐸^∗) and ‖𝑥_𝑛‖→‖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 (𝑥→_𝑛𝜇𝑥), by the convergence in measure on every set of finite measure (𝑥→_𝑛𝜇𝑥 locally), or by the uniform convergence (𝑥_𝑛⇉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∈(𝐻_𝑔), 𝐸∈(𝐻_𝑙), and 𝐸∈(𝐻_𝑢)). Clearly, 𝐸∈(𝐻_𝑙)⇒𝐸∈(𝐻_𝑔)⇒𝐸∈(𝐻_𝑢). 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∶ℝ→[0,∞], it is convex, even, vanishing, and continuous at zero, left continuous on (0,∞) and not identically equal to zero. Denote𝑎_𝜑=sup{𝑢≥0∶𝜑(𝑢)=0},𝑏_𝜑=sup{𝑢≥0∶𝜑(𝑢)<∞}.(2.2) We write 𝜑>0 when 𝑎_𝜑=0 and 𝜑<∞ if 𝑏_𝜑=∞. Let 𝜑_𝑟=𝜑𝜒_𝐺𝜑, where𝐺_𝜑=𝑎^𝜑𝑎,𝑏_𝜑if𝜑𝑏_𝜑<∞,_𝜑,𝑏_𝜑otherwise.(2.3) Define on 𝐿⁰ a convex semimodular 𝐼_𝜑 by𝐼_𝜑(𝑥)=‖𝜑∘𝑥‖^𝐸∞iofth𝜑er∘w𝑥is∈e,𝐸,(2.4) where (𝜑∘𝑥)(𝑡)=𝜑(𝑥(𝑡)), 𝑡∈𝑇. By the Calderón-Lozanovskiĭ space 𝐸_𝜑 we mean𝐸_𝜑=𝑥∈𝐿⁰∶𝐼_𝜑(𝑐𝑥)<∞forsome𝑐>0,(2.5) equipped with so called Luxemburg norm defined by‖𝑥‖_𝜑=inf𝜆>0∶𝐼_𝜑𝑥𝜆≤1.(2.6) We generally assume that if 𝑏_𝜑<∞, then 𝑎_𝜑<𝑏_𝜑, because when 0<𝑎_𝜑=𝑏_𝜑, then 𝐸_𝜑=𝐿^∞ and ‖𝑥‖_𝜑=(1/𝑏_𝜑)‖𝑥‖_∞.

If 𝐸=𝐿¹(𝑒=𝑙¹), then 𝐸_𝜑(𝑒_𝜑) is the Orlicz function (sequence) space equipped with the Luxemburg norm. If 𝐸=Λ_𝜔—the Lorentz function space (𝑒=𝜆_𝜔—the Lorentz sequence space), then 𝐸_𝜑(𝑒_𝜑) is the corresponding Orlicz-Lorentz function (sequence) space denoted by (Λ_𝜔)_𝜑((𝜆_𝜔)_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 0≤𝑥_𝑛↑𝑥∈𝐿⁰ with (𝑥_𝑛)^∞𝑛=1 in 𝐸 and sup_𝑛‖𝑥_𝑛‖_𝐸<∞, then 𝑥∈𝐸 and ‖𝑥‖_𝐸=lim_𝑛‖𝑥_𝑛‖_𝐸. Since 𝐸 has the Fatou property, 𝐸_𝜑 has also this property, whence 𝐸_𝜑 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 𝑢⩾0, then we say that 𝜑 satisfies condition Δ₂(ℝ₊). We write for short 𝜑∈Δ₂(0), 𝜑∈Δ₂(∞), and 𝜑∈Δ₂(ℝ₊), respectively.

For a Köthe space 𝐸 and an Orlicz function 𝜑, we say that 𝜑 satisfies condition Δ^𝐸2 (𝜑∈Δ^𝐸2 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 symbol 𝐸↪𝐹 stands for the continuous embedding of the space 𝐸 into the space 𝐹.

Relationships between the modular 𝐼_𝜑 and the norm ‖⋅‖_𝜑 are collected in [8].

3. Results

3.1. Köthe Spaces and Orlicz Functions

Given a Köthe function space 𝐸, we setm_𝐸=sup_{𝐵,𝜒𝐵∈𝐸}sup_{[(𝐴𝑛)⊂𝐵,𝜇(𝐴𝑛)→0]}inf_𝑛‖‖𝜒_𝐴𝑛‖‖_𝐸.(3.1)

Remark 3.1. It is easy to see that if 𝐸 is a symmetric Köthe function space, then m_𝐸=inf_{𝐴∈Σ,𝜇(𝐴)>0}‖‖𝜒_𝐴‖‖_𝐸.(3.2)

Recall that if 𝑥∈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 each 𝑥∈𝐹, there is 𝐴∈Σ, with 𝜇(𝐴)=0, such that supp𝑥⊂𝐴∪supp𝐹,(ii)there is 𝑥∈𝐹 with 𝜇(supp𝐹⧵supp𝑥)=0.

Lemma 3.2. Suppose that 𝐸 is a Köthe function space. Consider the following statements:(i)𝐸∈(𝑂𝐶),(ii)𝐸∈(𝐻_𝑔)₊,(iii)For any 𝑢∈𝐸 and each sequence {𝐴_𝑛} in Σ the implication 𝜇𝐴_𝑛⟶0⟹‖‖𝑢𝜒_𝐴𝑛‖‖⟶0holds.(3.3)(iv)m_𝐸=0, (v)supp𝐸_𝑎=𝑇, (vi)𝐸_𝑎≠{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 supp𝐸_𝑎≠𝑇, then for each 𝑥∈𝐸 with supp𝑥⊂𝑇⧵𝑠upp𝐸_𝑎 we have 𝑥∉𝐸_𝑎. Let 𝑇₀⊂𝑇⧵supp𝐸_𝑎 be such that 0<𝜇(𝑇₀)<∞ and 𝜒_𝑇0∈𝐸. Then 𝑥₁=𝜒_𝑇0∉𝐸_𝑎 and consequently there is a sequence (𝐴_𝑛) of measurable pairwise disjoint sets with ⋃^∞𝑛=1𝐴_𝑛=𝑇₀ and a number 𝛿>0 satisfying ‖𝜒_𝐴𝑛‖≥𝛿 for each 𝑛 (see [19]). Since 𝜇(𝐴_𝑛)→0, we get m_𝐸≥𝛿>0.
(ii) ⇒ (iii). Assume that (iii) does not hold, then we find an element 𝑢∈𝐸, a number 𝛿>0, and a sequence (𝐴_𝑖) with 𝜇(𝐴_𝑖)→0 such that inf_𝑖‖𝑢𝜒_𝐴𝑖‖_𝐸≥𝛿. Without loss of generality, we may assume that ∑^∞𝑖=1𝜇(𝐴_𝑖)<∞. Set 𝑥_𝑘=sup𝑢𝜒_𝐴𝑖∶𝑖≥𝑘∈𝐿⁰.(3.4) Let 𝐴=⋃_𝑖𝐴_𝑖. Clearly, 𝑥_𝑘≤𝑢𝜒_𝐴∈𝐸. Put 𝑦=𝑢𝜒_𝐴 and 𝑦_𝑘=𝑦−𝑥_𝑘. Since {𝑥_𝑘} is decreasing, so {𝑦_𝑘} is increasing. Next, 𝑥_𝑘→0 in measure, because, for each 𝜀>0, 𝜇𝑡∈𝑇∶𝑥_𝑘(𝑡)>𝜀≤𝜇^∞_𝑖=𝑘𝐴_𝑖≤^∞_𝑖=𝑘𝜇𝐴_𝑖⟶0as𝑘⟶∞.(3.5) In particular, by 𝐸∈(𝐹𝑃), ‖𝑦_𝑘‖→‖𝑦‖. Finally, ‖𝑦_𝑘−𝑦‖=‖𝑥_𝑘‖≥‖𝑢𝜒_𝐴𝑘‖≥𝛿. Thus, 𝐸∉(𝐻_𝑔)₊.
Assume that 𝐸 is symmetric. We prove that (vi) ⇒ (iv). If m_𝐸>0, then we find a set 𝐵 with 𝜒_𝐵∈𝐸 and a sequence {𝐴_𝑛}⊂𝐵, 𝜇(𝐴_𝑛)→0 such that inf_𝑛‖𝜒_𝐴𝑛‖_𝐸≥m_𝐸/2. Let 𝑥∈𝐸⧵{0}. We prove that 𝑥∈𝐸⧵𝐸_𝑎. There is a number 𝛿>0 such that the set 𝐷={𝑡∈𝑇∶|𝑥(𝑡)|≥𝛿} has positive measure. We find a sequence {𝐶_𝑘} in 𝐷 of pairwise disjoint sets and a subsequence {𝐴_𝑛𝑘} of {𝐴_𝑛} satisfying 𝜇(𝐶_𝑘)≤(1/2^𝑘)𝜇(𝐷) and 𝜇(𝐶_𝑘)≥𝜇(𝐴_𝑛𝑘). Then, by the symmetry of 𝐸, ‖‖𝑥𝜒_𝐶𝑘‖‖_𝐸≥‖‖𝛿𝜒_𝐶𝑘‖‖_𝐸≥‖‖𝛿𝜒_𝐴𝑛𝑘‖‖_𝐸≥𝛿m^𝐸2,(3.6) for any 𝑘∈ℕ, whence 𝑥∉𝐸_𝑎.
Now we will prove that the implication (vi) ⇒ (iv) is not true in general in nonsymmetric case. Take 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‖_∞.(3.7) Obviously, supp𝐸_𝑎=[0,1] and m_𝐸=1.
Now we will see that implications (i) ⇒ (iii) ⇒ (iv) cannot be reversed in general. If 𝐸=Λ_𝜔—the Lorentz space over (0,∞) with Lebesgue measure and ∫^∞0𝜔<∞, then 𝐸∈(𝐻_𝑔) and 𝐸∉(𝑂𝐶)(see [1, 8]), whence (iii) ⇏ (i) in general.
Finally, let 𝐸=𝐿_𝜑[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 {𝐴_𝑛} with 𝜇(𝐴_𝑛)→0 and ‖𝑢𝜒_𝐴𝑛‖_𝜑=1. On the other hand, for each 𝐵∈Σ with 𝑢=𝜒_𝐵∈𝐸, we claim that 𝜒_𝐵∈𝐸_𝑎, whence m_𝐸=0. Indeed, for any sequence {𝐴_𝑛} satisfying 𝐴_𝑛↓∅ and for each 𝜆>0, we have 𝐼_𝜑𝜆𝑢𝜒_𝐴𝑛=𝜑(𝜆)¹⁰𝜒_𝐴𝑛𝑑𝜇=𝜑(𝜆)𝜇𝐴_𝑛⟶0,(3.8) whence ‖𝑢𝜒_𝐴𝑛‖_𝜑→0. Thus, the implication (iii) ⇒ (iv) cannot be reversed in general.

Recall that if 𝐸∈(𝐻_𝑙)₊, then 𝐸∈(𝑂𝐶) (see [10]) and, as we have seen above, property 𝐻_𝑔 needs not imply 𝑂𝐶 in general. However, from the above result, it follows that if 𝐸∈(𝐻_𝑔)₊, then m_𝐸=0 and consequently supp𝐸_𝑎=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 𝜒_𝐴∉𝐸 only in a nonsymmetric Köthe space. However, if 𝜇(𝐴)<∞ and 𝜒_𝐴∉𝐸, then there is an increasing sequence (𝐴_𝑛) of subsets of 𝐴 satisfying ⋃_𝑛𝐴_𝑛=𝐴 and 𝜒_𝐴𝑛∈𝐸 for each 𝑛.

Proof. Let 𝑥₀ be a weak unit of 𝐸, that is, 𝑥₀(𝑡)>0 for each 𝑡∈𝑇. It is enough to take 𝐴_𝑛={𝑡∈𝐴∶𝑥₀(𝑡)>1/𝑛} for each 𝑛.

Remark 3.4. Let 𝐸 be a Köthe space. If 𝐸∈(𝐻_𝑢)₊, then ‖𝑥𝜒_𝐵𝑘‖→0 for each 𝑥∈𝐸₊, where 𝐵_𝑘={𝑡∈𝑇∶𝑥(𝑡)<1/𝑘}.

Proof. Let 𝑥∈𝐸₊. Clearly, 𝑥𝜒_{𝑇⧵𝐵𝑘}↑𝑥 and sup_𝑘‖𝑥𝜒_{𝑇⧵𝐵𝑘}‖<∞. By 𝐸∈(𝐹𝑃), we get ‖𝑥𝜒_{𝑇⧵𝐵𝑘}‖→‖𝑥‖. Moreover, 𝑥−𝑥𝜒_{𝑇⧵𝐵𝑘}→0 uniformly. Then ‖𝑥𝜒_{𝑇⧵𝐵𝑘}−𝑥‖→0, because 𝐸∈(𝐻_𝑢)₊.
It is easy to see that the above is not true if 𝐸∉(𝐻_𝑢)₊. It is enough to take 𝐸=𝐿^∞𝜔(0,∞)=𝑥∈𝐿⁰∶‖𝑥‖=supess_𝑡≥0𝜔(𝑡)||𝑥(𝑡)||<∞,(3.9) where 𝜔(𝑡)=𝑡 and 𝑥(𝑡)=1/𝑡. Then ‖𝑥𝜒_𝐵𝑘‖=1 for each 𝑘. On the other hand, taking 𝑥_𝑛=𝑥𝜒_[0,𝑛], we get ‖𝑥‖=‖𝑥_𝑛‖=1, 𝑥_𝑛→𝑥 uniformly and ‖𝑥_𝑛−𝑥‖=1, whence 𝐸∉(𝐻_𝑢)₊.

It is known that 𝐸∈(𝐻_𝑙) if and only if 𝐸∈(𝐻_𝑙)₊ whenever 𝐸∈(𝑂𝐶) (see [6]). Since (𝐻_𝑙)₊⇒(𝑂𝐶) (see [10]), so 𝐸∈(𝐻_𝑙) if and only if 𝐸∈(𝐻_𝑙)₊. The same proof as in [6] shows that if 𝐸∈(𝑂𝐶), then 𝐸∈(𝐻_𝑔) if and only if 𝐸∈(𝐻_𝑔)₊. However, the assumption 𝐸∈(𝑂𝐶) is now too strong in this case because there are Köthe spaces satisfying 𝐸∈(𝐻_𝑔) and 𝐸∉(𝑂𝐶). Hence, the following result is more natural and precise.

Lemma 3.5. Suppose 𝐸 is a Köthe space. Then 𝐸∈(𝐻_𝑔) if and only if 𝐸∈(𝐻_𝑔)₊.

Proof. The necessity is obvious. We prove the sufficiency. Let 𝐸∈(𝐻_𝑔)₊. Assume that 𝐸 is a Köthe function space. Let ‖𝑥_𝑛‖→‖𝑥‖ and 𝑥_𝑛→𝑥 in measure. For each 𝜂>0, we get 𝜇𝑡∈𝑇∶||||𝑥_𝑛(𝑡)||−||𝑥(𝑡)||||≥𝜂≤𝜇𝑡∈𝑇∶||𝑥_𝑛(𝑡)−𝑥(𝑡)||≥𝜂⟶0,(3.10) whence |𝑥_𝑛|→|𝑥| in measure. By the assumption 𝐸∈(𝐻_𝑔)₊, we conclude that ‖‖||𝑥_𝑛||−|𝑥|‖‖⟶0.(3.11) Then we find a subsequence (|𝑥_𝑛𝑚|)_𝑚 of (|𝑥_𝑛|)_𝑛, an element 𝑦∈𝐸₊, and a sequence 0≤𝛽_𝑚→0 such that ||𝑥_𝑛𝑚|−|𝑥||≤𝛽_𝑚𝑦 for each 𝑚 (see [12]). Denote (|𝑥_𝑛𝑚|)_𝑚 still by (|𝑥_𝑛|)_𝑛 and (𝛽_𝑚) by (𝛽_𝑛). Thus, we may assume that ||𝑥_𝑛|−|𝑥||≤𝑦 for all 𝑛. Set 𝑢=|𝑥|+𝑦. Let 𝜀>0. Applying Remark 3.4, with respective sets 𝐵_𝑘=𝑡∶𝑢(𝑡)<1𝑘,(3.12) take 𝑘 big enough to satisfy ‖𝑢𝜒_𝐵𝑘‖<𝜀/8. Moreover, ‖|𝑥_𝑛|−|𝑥|‖<𝜀/8 for sufficiently large 𝑛. Setting 𝐶_𝑛=𝑡∈𝑇∶0≤sgn𝑥(𝑡)𝑥_𝑛(𝑡)≤1,𝐷_𝑛=𝑡∈𝑇∶sgn𝑥(𝑡)𝑥_𝑛(𝑡)=−1,(3.13) we get ‖‖𝑥−𝑥_𝑛‖‖≤≤5‖‖𝜀|𝑥|−||𝑥_𝑛||𝜒_𝐶𝑛‖‖+‖‖𝑥−𝑥_𝑛𝜒_{𝐷𝑛∩𝐵𝑘}‖‖+‖‖𝑥−𝑥_𝑛𝜒_{𝐷𝑛⧵𝐵𝑘}‖‖8+‖‖𝑥−𝑥^𝑛𝜒_{𝐷𝑛⧵𝐵𝑘}‖‖.(3.14) Take 0<𝛿<𝜀16𝑘min‖𝑦1‖,‖𝑥1‖.(3.15) Denote 𝐹𝐺^𝑘𝑛=𝑡∈𝐷_𝑛⧵𝐵_𝑘∶||𝑥_𝑛(𝑡)−𝑥(𝑡)||>𝛿_𝑘𝑛,=𝑡∈𝐷_𝑛⧵𝐵_𝑘∶||𝑥_𝑛(𝑡)−𝑥(𝑡)||≤𝛿.(3.16) Note that 𝑥→_𝑛𝜇𝑥 and consequently 𝜇(𝐹^𝑘𝑛)→0 as 𝑛→∞. Then, by Lemma 3.2 (iii), ‖‖𝑥−𝑥_𝑛𝜒_{𝐷𝑛⧵𝐵𝑘}‖‖≤≤‖‖‖‖2𝑢𝑥𝜒−_𝐹𝑘𝑛𝑥_𝑛𝜒_𝐹^𝑘𝑛‖‖+‖‖𝑥−𝑥_𝑛𝜒_𝐺^𝑘𝑛‖‖‖‖+‖‖𝑥−𝑥_𝑛𝜒_𝐺^𝑘𝑛‖‖<𝜀8+‖‖𝑥−𝑥^𝑛𝜒_𝐺^𝑘𝑛‖‖,(3.17) for sufficiently large 𝑛. Finally, divide set 𝐺^𝑘𝑛 into two subsets 𝐻^𝑘𝑛=𝑡∈𝐺^𝑘𝑛∶||𝑥(𝑡)||≥12𝑘,𝐼^𝑘𝑛=𝑡∈𝐺^𝑘𝑛∶𝑦(𝑡)≥12𝑘.(3.18) Clearly, 𝐻^𝑘𝑛∪𝐼^𝑘𝑛=𝐺^𝑘𝑛 because 𝐺^𝑘𝑛⊂𝑇⧵𝐵_𝑘=𝑡∶||𝑥(𝑡)||+𝑦(𝑡)≥1𝑘.(3.19) Then, by (3.15), ‖‖𝑥−𝑥_𝑛𝜒_𝐺^𝑘𝑛‖‖≤≤2𝛿𝑘𝛿‖‖𝜒‖‖|_𝐻𝑘𝑛𝑥‖‖|𝜒+‖‖_𝐻𝑘𝑛𝜒_𝐼𝑘𝑛‖‖≤𝛿‖‖2𝑘|𝑥|𝜒_𝐻𝑘𝑛‖‖+‖‖2𝑘𝑦𝜒_𝐼𝑘𝑛‖‖‖‖+‖‖𝑦𝜒_𝐼𝑘𝑛‖‖<𝜀4.(3.20) Combining (3.14), (3.17), and (3.20), we get ‖𝑥−𝑥_𝑛‖≤𝜀.
If 𝐸 is a Köthe sequence space, the proof is the same, because for a counting measure space (ℕ,2^ℕ,𝑚) for each 𝑢∈𝐸 and any sequence (𝐴_𝑛) in 2^ℕ with 𝑚(𝐴_𝑛)→0, we have ‖𝑢𝜒_𝐴𝑛‖→0.

The straightforward calculation shows that we may equivalently take 𝑥 and {𝑥_𝑛} in 𝑆(𝐸) in the definition of property 𝐻_𝑙 (see also the proof below). However, the analogous problem for property 𝐻_𝑔 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 𝑥_𝑛→𝑥 in measure and ‖𝑥_𝑛‖→‖𝑥‖≠0, then 𝑥_𝑛/‖𝑥_𝑛‖→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 each 𝜀,𝜂>0, we get 𝜇𝑡∶||||𝑥^𝑛(𝑡)‖‖𝑥_𝑛‖‖−‖𝑥𝑥(‖𝑡)||||>𝜀=𝜇𝑡∶||||𝑥^𝑛(𝑡)≤𝜇𝑡∶‖‖𝑥1_𝑛‖‖−‖‖𝑥𝑥(_𝑛𝑡‖‖)+‖‖𝑥𝑥(_𝑛𝑡‖‖)−‖𝑥𝑥(‖𝑡)||||>𝜀‖‖𝑥_𝑛‖‖||𝑥^𝑛(𝑡)−𝑥(𝑡)||+||||1≤𝜇𝑡∶1‖‖𝑥_𝑛‖‖−‖𝑥1‖||||||𝑥(𝑡)||>𝜀‖‖𝑥_𝑛‖‖||𝑥^𝑛(𝑡)−𝑥(𝑡)||>𝜀2+𝜇𝑡∶||||‖‖𝑥1_𝑛‖‖−‖𝑥1‖||||||𝑥(𝑡)||>2𝜀(3.2.1) Since ‖𝑥_𝑛‖→‖𝑥‖, so ‖𝑥_𝑛‖>(1/2)‖𝑥‖ for 𝑛≥𝑁₀, whence 𝜇𝑡∶1‖‖𝑥_𝑛‖‖||𝑥^𝑛(𝑡)−𝑥(𝑡)||>𝜀2≤𝜇𝑡∶||𝑥^𝑛(𝑡)−𝑥(𝑡)||>𝜀‖𝑥‖4<𝜂2,(3.22) for 𝑛≥𝑁₁>𝑁₀. Moreover, applying the symmetry of 𝐸, we get 𝜇𝑡∶||||1‖‖𝑥_𝑛‖‖−‖𝑥1‖||||||𝑥(𝑡)||>2𝜀<2𝜂for𝑛≥𝑁².(3.23) Thus, 𝜇𝑡∶||||𝑥^𝑛(𝑡)‖‖𝑥_𝑛‖‖−‖𝑥𝑥(‖𝑡)||||>𝜀<𝜂,(3.24) for 𝑛≥max{𝑁₁,𝑁₂}.
Considering the second assertion, take 𝐸=𝐿^11/𝑡₃(1,∞)=𝑥∈𝐿⁰((1,∞),Σ,𝑚)∶‖𝑥‖=^∞1||𝑥(𝑡)||1𝑡³𝑑𝑡<∞.(3.25) Set 𝑥(𝑡)=𝑡 and 𝑥_𝑛(𝑡)=𝑡+(1/𝑛), then ‖𝑥‖=1, ‖𝑥_𝑛‖=1+(1/2𝑛), and 𝑥_𝑛→𝑥 in measure. However, 𝜇𝑡∶||||𝑥^𝑛(𝑡)‖‖𝑥_𝑛‖‖−‖𝑥𝑥(‖𝑡)||||>1=𝜇𝑡∶|||2𝑡𝑛−+21|||>1=∞,(3.26) for each 𝑛, whence (𝑥_𝑛/‖𝑥_𝑛‖)↛(𝑥/‖𝑥‖) in measure.

Corollary 3.7. Assume that 𝐸 is a symmetric Köthe function space. Then 𝐸∈(𝐻_𝑔) if and only if ‖𝑥_𝑛−𝑥‖→0 for any 𝑥∈𝑆(𝐸) and (𝑥_𝑛) in 𝑆(𝐸) satisfying 𝑥_𝑛→𝑥 in measure.

Proof. The necessity is obvious. We prove the sufficiency. Let 𝑥_𝑛→𝑥 in measure and ‖𝑥_𝑛‖→‖𝑥‖≠0. Taking 𝑦_𝑛=𝑥_𝑛/‖𝑥_𝑛‖ and 𝑦=𝑥/‖𝑥‖, we get 𝑦_𝑛→𝑦 in measure, by Lemma 3.6. By the assumption, we conclude that ‖𝑦_𝑛−𝑦‖→0. Notice that ||||1‖‖𝑥_𝑛‖‖‖‖𝑥^𝑛−𝑥‖‖−‖𝑥‖||||1≤‖‖‖𝑥‖‖𝑥_𝑛‖‖−‖𝑥1‖_𝑛||||||||‖‖𝑥_𝑛‖‖−‖‖𝑥𝑥_𝑛‖‖−−‖‖𝑥𝑥_𝑛‖‖+‖𝑥𝑥‖‖‖‖=‖‖‖‖‖𝑥𝑥^𝑛_𝑛‖‖−‖𝑥𝑥‖‖‖‖⟶0.(3.27) Since ‖𝑥_𝑛‖→‖𝑥‖≠0, so ‖𝑥_𝑛−𝑥‖→0 as desired.

We will need in the sequel the following.

Lemma 3.8. If 𝜑∈Δ₂(∞), then for any 𝑎>0, there are 𝜎∈(0,1] and 𝑢₁>𝑎_𝜑+𝑎 such that inf_{𝑢0∈(0,𝑎)}inf_𝑢≥𝑢1𝜑𝑢−𝑢₀𝜑(𝑢)−𝜑𝑢₀≥𝜎.(3.28)

Proof. Suppose conversely that there is 𝑎>0 such that for any 𝜎∈(0,1] and each 𝑢₁>𝑎_𝜑+𝑎, there holds inf_{𝑢0∈(0,𝑎)}inf_𝑢≥𝑢1𝜑𝑢−𝑢₀𝜑(𝑢)−𝜑𝑢₀<𝜎.(3.29) Taking 𝜎_𝑛=1/2^𝑛 and 𝑢_𝑛→∞, we get inf_{𝑢0∈(0,𝑎)}inf_{𝑢≥𝑢𝑛}𝜑(𝑢−𝑢₀)/(𝜑(𝑢)−𝜑(𝑢₀))<1/2^𝑛. We may assume that 𝑢_𝑛>2𝑎 for each 𝑛. Hence, for each 𝑛, one may choose 𝑣_𝑛≥𝑢_𝑛 and 𝑢₀∈(0,𝑎) with 𝜑(𝑣_𝑛−𝑢₀)/(𝜑(𝑣_𝑛)−𝜑(𝑢₀))≤1/2^𝑛. Consequently, we get 2^𝑛𝜑12𝑣^𝑛<2^𝑛𝜑𝑣_𝑛−𝑢₀<𝜑𝑣_𝑛−𝜑𝑢₀≤𝜑𝑣_𝑛,(3.30) for each 𝑛. Taking 𝑏_𝑛=(1/2)𝑣_𝑛, one has 2^𝑛𝜑(𝑏_𝑛)<𝜑(2𝑏_𝑛), whence 𝜑∉Δ₂(∞).

Lemma 3.9. (i) Suppose that 𝐸 is a symmetric Köthe space. If 𝑥_𝑛→𝑥 in measure and 𝜑∘𝑥_𝑛, 𝜑∘𝑥∈𝐸 are finitely valued a.e., then 𝜑∘𝑥_𝑛→𝜑∘𝑥 in measure. Furthermore, for a nonsymmetric Köthe space, the above implication is not true in general.
(ii) If 𝑥_𝑛→𝑥 in measure, 𝑥_𝑛,𝑥∈𝐸₊ and 𝜑^𝑟−1∘𝑥_𝑛, 𝜑^𝑟−1∘𝑥 are well-defined functions, then 𝜑^𝑟−1∘𝑥_𝑛→𝜑^𝑟−1∘𝑥 in measure.

Proof. (i) Let 𝑥−→_𝑛𝜇𝑥.
(A) Suppose that 𝑏_𝜑=∞. Set 𝐴_𝑘=𝑡∈𝑇∶||𝑥(𝑡)||>𝑘.(3.31) Then, by symmetry of 𝐸,  𝜇(𝐴_𝑘)→0. Let 𝜂>0 and 𝜀>0. Take 𝑘₀ such that 𝜇(𝐴_𝑘0)<𝜂/3. We have 𝜇≤𝑡∈𝜇𝑇𝑡∶∈||𝜑𝑇⧵𝑥𝐴_𝑛(𝑡_𝑘0)−𝜑(𝑥(𝑡))||≥𝜀∶||𝜑𝑥_𝑛(𝑡)−𝜑(𝑥(𝑡))||≥𝜀+𝜂3.(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 |𝑢−𝑠|<𝛿⇒|𝜑(𝑢)−𝜑(𝑠)|<𝛾 is true for all 𝑢,𝑠∈[−𝑘₀,𝑘₀]. Set 𝛿=𝛿(𝜀,2𝑘₀) and 𝐴^1𝑛=𝑡∈𝑇⧵𝐴_𝑘0∶||𝑥_𝑛(𝑡)||>2𝑘₀,𝐴^2𝑛=𝑡∈𝑇⧵𝐴_𝑘0∶||𝑥_𝑛(𝑡)||≤2𝑘₀.(3.33) Consequently 𝜇≤𝑡∈𝜇𝑇𝐴∶_1𝑛||𝜑𝑥_𝑛(𝑡)−𝜑(𝑥(𝑡))||≥𝜀≤𝜇𝑡∈𝐴+_2𝑛𝜇𝑡∈𝐴^2𝑛∶||𝜑𝑥_𝑛(𝑡)−𝜑(𝑥(𝑡))||≥𝜀+3𝜂∶||𝑥_𝑛(𝑡)−𝑥(𝑡)||≥𝛿+2𝜂3≤𝜂(3.34) for sufficiently large 𝑛, because 𝐴^1𝑛={𝑡∈𝑇∶|𝑥_𝑛(𝑡)−𝑥(𝑡)|≥𝑘₀}. Thus 𝜑∘𝑥→_𝑛𝜇𝜑∘𝑥.
(B) Assume 𝜑(𝑏_𝜑)<∞. The proof is simpler than above because 𝑥_𝑛(𝑡)≤𝑏_𝜑 and 𝑥(𝑡)≤𝑏_𝜑 for 𝜇-a.e. 𝑡∈𝑇 and 𝜑 is uniformly continuous on the interval [−𝑏_𝜑,𝑏_𝜑].
(C) Suppose that 𝑏_𝜑<∞, 𝜑(𝑏_𝜑)=∞. Then 𝑥_𝑛(𝑡)<𝑏_𝜑 and 𝑥(𝑡)<𝑏_𝜑 for 𝜇-a.e. 𝑡∈𝑇. Set 𝐴_𝑘=𝑡∈𝑇∶||𝑥(𝑡)||>𝑏_𝜑−1𝑘.(3.35) Then 𝜇(𝐴_𝑘)→0, because 𝐸 is symmetric. Let 𝜂>0 and 𝜀>0. Take 𝑘₀ such that 𝜇(𝐴_𝑘0)<𝜂/3. We have 𝜇≤𝑡∈𝜇𝑇𝑡∶∈||𝜑𝑇⧵𝑥𝐴_𝑛(𝑡_𝑘0)−𝜑(𝑥(𝑡))||≥𝜀∶||𝜑𝑥_𝑛(𝑡)−𝜑(𝑥(𝑡))||≥𝜀+𝜇𝐴_𝑘0.(3.36) Since 𝜑 is uniformly continuous on the interval [−𝑢₀,𝑢₀] for each 𝑢₀<𝑏_𝜑, so for each 𝑢₀<𝑏_𝜑 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,𝑏_𝜑−1/2𝑘₀) and 𝐴^1𝑛=𝑡∈𝑇⧵𝐴_𝑘0∶||𝑥_𝑛(𝑡)||>𝑏_𝜑−1𝐴_2𝑛2𝑘₀,=𝑡∈𝑇⧵𝐴_𝑘0∶||𝑥_𝑛(𝑡)||≤𝑏_𝜑−12𝑘₀.(3.37) Consequently 𝜇≤𝑡∈𝜇𝑇𝐴∶_1𝑛||𝜑𝑥_𝑛(𝑡)−𝜑(𝑥(𝑡))||≥𝜀≤𝜇𝑡∈𝐴+_2𝑛𝜇𝑡∈𝐴^2𝑛∶||𝜑𝑥_𝑛(𝑡)−𝜑(𝑥(𝑡))||≥𝜀+3𝜂∶||𝑥_𝑛(𝑡)−𝑥(𝑡)||≥𝛿+2𝜂3≤𝜂(3.38) for sufficiently large 𝑛, because 𝐴^1𝑛={𝑡∈𝑇∶|𝑥_𝑛(𝑡)−𝑥(𝑡)|≥1/2𝑘₀}. Thus 𝜑∘𝑥→_𝑛𝜇𝜑∘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 𝜑(𝑢)=𝑢²,𝑥=_𝑛𝑛𝜒_{[𝑛,𝑛+1]},𝑥_𝑘=_𝑛𝑛+1𝑘𝜒^{[𝑛,𝑛+1]}.(3.39) Then 𝑥_𝑘→𝑥 in measure. On the other hand, |𝜑∘𝑥_𝑘−𝜑∘𝑥|=∑_𝑛((1/𝑘²)+(2𝑛/𝑘))𝜒_{[𝑛,𝑛+1]}, whence 𝜑∘𝑥_𝑘↛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 each 𝜀>0, there is 𝛿>0 such that the implication ||𝜑^𝑟−1(𝑢)−𝜑^𝑟−1(𝑣)||>𝜀⟹|𝑢−𝑣|>𝛿(3.40) is true for all 𝑢,𝑣≥0. First, notice that for each 𝜀>0, there is 𝛿>0 such that the implication ||𝜑^𝑟−1(𝑢)−𝜑^𝑟−1(0)||>𝜀⟹|𝑢|>𝛿(3.41) is true for all 𝑢≥0, because 𝜑^𝑟−1 is continuous at zero. Next, defining 𝑔(𝑢)=𝜑^𝑟−1(𝑢)−𝜑^𝑟−1(0), we see that 𝑔 is subadditive on ℝ₊, because 𝑔=𝜓⁻¹, where 𝜓 is an Orlicz function with 𝜓>0. Consequently, 𝜑^𝑟−1(𝑢−𝑣)−𝜑^𝑟−1(0)=𝑔(𝑢−𝑣)≥𝑔(𝑢)−𝑔(𝑣)=𝜑^𝑟−1(𝑢)−𝜑^𝑟−1(𝑣).(3.42) Hence, if |𝜑^𝑟−1(𝑢)−𝜑^𝑟−1(𝑣)|>𝜀, then |𝜑^𝑟−1(𝑢−𝑣)−𝜑^𝑟−1(0)|>𝜀, whence |𝑢−𝑣|>𝛿, by (3.41). Thus, the claim is proved. Assume that 𝑥_𝑛, 𝑥∈𝐸₊ and 𝜑^𝑟−1∘𝑥_𝑛, 𝜑^𝑟−1∘𝑥 are well-defined functions and 𝑥_𝑛→𝑥 in measure, then, applying (3.40), for each 𝜀>0, we get 𝜇𝑡∈𝑇∶||𝜑^𝑟−1∘𝑥_𝑛−𝜑^𝑟−1∘𝑥(𝑡)||>𝜀≤𝜇𝑡∈𝑇∶||𝑥_𝑛−𝑥(𝑡)||>𝛿⟶0.(3.43)

3.2. Calderón-Lozanovskiĭ Function Spaces

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

It is known that 𝐸_𝜑∈(𝑂𝐶) if and only if 𝐸∈(𝑂𝐶) and 𝜑∈Δ^𝐸2 (see Corollary  12 from [9]). Considering some weaker property than 𝑂𝐶, we get the following.

Lemma 3.10. Suppose that 𝑏_𝜑=∞, then 𝐸_𝑎≠{0} if and only if (𝐸_𝜑)_𝑎≠{0}.

Proof. Necessity
Let 𝑧∈𝐸_𝑎⧵{0}. We find a number 𝑚>0 such that 𝜇(𝐴)=𝜇{𝑡∈𝑇∶|𝑧(𝑡)|>1/𝑚}>0. Taking 𝑤=𝜑⁻¹(1)𝜒_𝐴, we get 𝜑∘𝑤∈𝐸_𝑎. Moreover, since 𝑏_𝜑=∞, 𝜑 satisfies the local Δ^𝐸2 condition with respect to the element 𝑤 (see [9]). Consequently, 𝑤∈(𝐸_𝜑)_𝑎, by Theorem  11 from [9].
Sufficiency
We apply Theorem  11 from [9]. Let 𝑥∈(𝐸_𝜑)_𝑎⧵{0}, then 𝜑∘𝑥∈𝐸_𝑎. If 𝜇{𝑡∶|𝑥(𝑡)|>𝑎_𝜑}>0, then 𝜇(𝐴_𝑚)>0 for some 𝑚∈ℕ, where 𝐴_𝑚={𝑡∈𝑇∶|𝑥(𝑡)|≥(1+(1/𝑚))𝑎_𝜑}. Thus, 𝜒_𝐴𝑚∈𝐸_𝑎, and we are done, if 𝜇{𝑡∶|𝑥(𝑡)|≤𝑎_𝜑}=𝜇(𝐴)>0 then 𝜒_𝐴∈𝐸_𝑎⧵{0}, by Theorem  11 from [9].

Theorem 3.11. 𝐸_𝜑∈(𝐻_𝑙) if and only if(a)𝐸∈(𝐻_𝑙), (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 𝐸_𝜑∈(𝐻_𝑙), so 𝐸_𝜑∈(𝑂𝐶) (see [10, Proposition  2.1]), and consequently, 𝐸∈(𝑂𝐶), 𝜑<∞, and 𝜑∈Δ^𝐸2 (see [9, Corollary  12]). Notice that 𝐸∈(𝑂𝐶) implies that 𝐸↪^╱𝐿^∞. Indeed, if 𝐸↪𝐿^∞, then there is a decreasing sequence (𝐴_𝑛) in Σ with 0<𝜇(𝐴_𝑛)→0 and 𝜒_𝐴𝑛∈𝐸, whence 1=‖𝜒_𝐴𝑛‖_𝐿∞≤𝑀‖𝜒_𝐴𝑛‖_𝐸, so 𝐸∉(𝑂𝐶). Thus, (b) is proved. Suppose that 𝐸∉(𝐻_𝑙). Recall that property 𝐻_𝑙 can be equivalently considered only on the positive cone 𝐸₊ [6, Proposition  1]. Consequently, we find 𝑥 and {𝑥_𝑛} in (𝑆(𝐸))₊ with 𝑥→_𝑛𝜇𝑥 locally in measure and ‖𝑥_𝑛−𝑥‖_𝐸≥𝜀. Set 𝑦_𝑛=𝜑^𝑟−1∘𝑥_𝑛 and 𝑦=𝜑^𝑟−1∘𝑥, then 𝐼_𝜑(𝑦)=𝐼_𝜑(𝑦_𝑛)=1, whence 𝑦,𝑦_𝑛∈𝑆(𝐸_𝜑). Moreover, 𝑦→_𝑛𝜇𝑦 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 ‖‖𝑦_𝑛−𝑦‖‖_𝜑≥𝜂,(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 ‖‖𝜑∘𝑦_𝑛−𝜑∘𝑦‖‖_𝐸≥𝜀.(3.45) Denote 𝐴_𝑛=𝑡∈𝑇∶𝜑𝑦_𝑛(𝑡)≥𝜑(𝑦(𝑡)),𝐵_𝑛=𝑇⧵𝐴_𝑛.(3.46) By (3.45), we get max‖‖𝜑∘𝑦_𝑛−𝜑∘𝑦𝜒_𝐴𝑛‖‖_𝐸,‖‖𝜑∘𝑦_𝑛−𝜑∘𝑦𝜒_𝐵𝑛‖‖_𝐸≥𝜀2,(3.47) for each 𝑛. We assume that ‖(𝜑∘𝑦_𝑛−𝜑∘𝑦)𝜒_𝐴𝑛‖_𝐸≥𝜀/2 for infinitely many 𝑛∈ℕ, because otherwise the proof is analogous. We divide the proof into two cases.(I)Suppose that 𝐿^∞↪𝐸 and 𝜑∈Δ₂(∞). Let 𝑐=𝜑^𝑟−1(𝜀/32‖𝜒_𝑇‖_𝐸)>𝑎_𝜑. Denoting𝐴^1𝑛=𝑡∈𝐴_𝑛∶𝑦_𝑛(𝑡)<𝑐,𝐴^2𝑛=𝑡∈𝐴_𝑛∶𝑦_𝑛(𝑡)≥𝑐,(3.48) we get 𝜀2≤‖‖𝜑∘𝑦^𝑛−𝜑∘𝑦𝜒_𝐴𝑛‖‖_𝐸≤‖‖𝜑∘𝑦_𝑛−𝜑∘𝑦𝜒_𝐴^2𝑛‖‖_𝐸+𝜀4.(3.49) Thus, ‖‖𝜑∘𝑦_𝑛−𝜑∘𝑦𝜒_𝐴^2𝑛‖‖_𝐸≥𝜀4.(3.50) Set 𝑐₁=𝑐+𝑎^𝜑2,𝐵^1𝑛=𝑡∈𝐴^2𝑛∶𝑦(𝑡)<𝑐₁,𝐵^2𝑛=𝑡∈𝐴^2𝑛∶𝑦(𝑡)≥𝑐₁.(3.51) The remaining proof of the case (I) we divide into two parts.(1)Suppose that ‖(𝜑∘𝑦_𝑛−𝜑∘𝑦)𝜒_𝐵^1𝑛‖_𝐸≥𝜀/8. By Lemma 3.8, for 𝑎=𝑐₁, there are 𝜎>0 and 𝑢₁>0 such that inf_{𝑢0∈0,𝑐1}inf_𝑢≥𝑢1𝜑𝑢−𝑢₀𝜑(𝑢)−𝜑𝑢₀≥𝜎.(3.52) Set 𝐶^1𝑛=𝑡∈𝐵^1𝑛∶𝑦_𝑛(𝑡)<𝑢₁,𝐶^2𝑛=𝑡∈𝐵^1𝑛∶𝑦_𝑛(𝑡)≥𝑢₁.(3.53)(a)Assume that ‖(𝜑∘𝑦_𝑛−𝜑∘𝑦)𝜒_𝐶^1𝑛‖_𝐸≥𝜀/16. Taking 0<𝛼<(𝑐−𝑐₁)/𝑢₁, we get ‖‖‖𝜑∘𝑦^𝑛−𝑦𝛼𝜒^𝐶1𝑛‖‖‖_𝐸≥‖‖‖𝜑𝑐−𝑐¹𝛼𝜒^𝐶1𝑛‖‖‖_𝐸≥‖‖𝜑𝑢₁𝜒_𝐶^1𝑛‖‖_𝐸≥‖‖𝜑∘𝑦_𝑛−𝜑∘𝑦𝜒_𝐶^1𝑛‖‖_𝐸≥𝜀16.(3.54) Thus, ‖𝑦_𝑛−𝑦‖_𝜑≥𝜂₁=min{1,𝛼𝜀/16}.(b)Let ‖(𝜑∘𝑦_𝑛−𝜑∘𝑦)𝜒_𝐶^2𝑛‖_𝐸≥𝜀/16. Thus, by Lemma 3.8, ‖‖𝜑∘𝑦_𝑛−𝑦𝜒_𝐶^2𝑛‖‖_𝐸≥𝜎2‖‖𝜑∘𝑦^𝑛−𝜑∘𝑦𝜒_𝐶^2𝑛‖‖_𝐸≥𝜎𝜀32,(3.55) and consequently ‖𝑦_𝑛−𝑦‖_𝜑≥𝜂₂=min{1,𝜎𝜀/32} (cf. [8, Lemma  1.1]).(2) Assume that ‖(𝜑∘𝑦_𝑛−𝜑∘𝑦)𝜒_𝐵^1𝑛‖_𝐸<𝜀/8, then, by (3.50), ‖‖𝜑∘𝑦_𝑛−𝜑∘𝑦𝜒_𝐵^2𝑛‖‖_𝐸≥𝜀8.(3.56) Since 𝜑∈Δ₂(∞), 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 𝑢≥𝑐₁ (see [20, Theorem  1.13(4)]). Moreover, we can choose 𝑎>0 satisfying 𝑎1+𝑎𝜑𝑐¹‖‖𝜒_𝑇‖‖_𝐸<𝜀32.(3.58) Let 𝐵^𝑛21=𝑡∈𝐵^2𝑛∶𝑦_𝑛−𝑦(𝑡)<𝑎𝑐¹𝐵1+𝑎_𝑛22,=𝑡∈𝐵^2𝑛∶𝑦_𝑛−𝑦(𝑡)≥𝑎𝑐¹1+𝑎,(3.59) then, by (3.57) and (3.58), using convexity of 𝜑, we get ||𝜑∘𝑦_𝑛−𝜑∘𝑦||𝜒_𝐵^𝑛21=𝜑∘𝑦_𝑛−𝑦+𝑦−𝜑∘𝑦𝜒_𝐵^𝑛21≤𝑎1+𝑎𝜑∘1+𝑎𝑎𝑦^𝑛−𝑦+11+𝑎𝜑∘((1+𝑎)𝑦)−𝜑∘𝑦𝜒^𝐵𝑛21≤𝑎1+𝑎𝜑∘1+𝑎𝑎𝑦^𝑛−𝑦+11+𝑎𝑙𝜑∘𝑦−𝜑∘𝑦𝜒^𝐵𝑛21=𝑎1+𝑎𝜑∘1+𝑎𝑎𝑦^𝑛−𝑦+𝜀/32−𝑎1+𝑎𝜑∘𝑦𝜒^𝐵𝑛21.(3.60) Note that 𝑓(𝑎)=(𝜀/32−𝑎)/(1+𝑎) is a decreasing function of 𝑎>0. Moreover, 𝑎≤𝜀/32 by convexity of 𝜑 and inequality (3.57). Hence, by (3.58), ‖(𝜑∘𝑦_𝑛−𝜑∘𝑦)𝜒_𝐵^𝑛21‖_𝐸<𝜀/16. Then, by (3.56), ‖‖𝜑∘𝑦_𝑛−𝜑∘𝑦𝜒_𝐵^𝑛22‖‖_𝐸≥𝜀16.(3.61) 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 ||𝜑∘𝑦_𝑛−𝜑∘𝑦||𝜒_𝐵^𝑛22=𝜑∘𝑦_𝑛−𝑦+𝑦−𝜑∘𝑦𝜒_𝐵^𝑛22≤𝜑∘𝑦^𝑛−𝑦𝛾+𝑦−𝜑∘𝑦𝜒^𝐵𝑛22≤𝑎1+𝑎𝜑∘1+𝑎𝑎𝑦^𝑛−𝑦𝛾+1+1𝑎𝜑∘((1+𝑎)𝑦)−𝜑∘𝑦𝜒^𝐵𝑛22≤𝑎1+𝑎𝑘²𝜑∘𝑦^𝑛−𝑦𝛾+1+1𝑎𝑙𝜑∘𝑦−𝜑∘𝑦𝜒^𝐵𝑛22=𝑎1+𝑎𝑘²𝜑∘𝑦^𝑛−𝑦𝛾+𝜀/132+−𝑎𝑎𝜑∘𝑦𝜒^𝐵𝑛22.(3.62) Then ‖𝜑∘((𝑦_𝑛−𝑦)/𝛾)‖_𝐸≥𝜀(1+𝑎)/32𝑎𝑘₂, and consequently ‖𝑦_𝑛−𝑦‖_𝜑≥𝜂₃=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 𝐿↪^∞╱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. Theorem 3.19 below presents the full characterization of property 𝐻_𝑔. We divide the whole proof into several lemmas for readers convenience.

Lemma 3.12. If 𝐸_𝜑∈(𝐻_𝑔)₊, then m_𝐸=0.

Proof. If m_𝐸>0, then there are 𝛿>0, a set 𝐵,𝜇(𝐵)>0,𝜒_𝐵∈𝐸, and a sequence (𝐴_𝑛)⊂𝐵 with 𝜇(𝐴_𝑛)→0 and inf_𝑛‖𝜒_𝐴𝑛‖_𝐸≥𝛿. Without loss of generality, we may assume that ∑^∞𝑖=1𝜇(𝐴_𝑖)<∞. Set 𝑧_𝑛=sup𝜒_𝐴𝑖∶𝑖≥𝑛∈𝐿⁰.(3.63) Let 𝐴=⋃_𝑖𝐴_𝑖. Clearly, 𝑧_𝑛≤𝜒_𝐴≤𝜒_𝐵∈𝐸. Put 𝑦=𝜒_𝐴 and 𝑦_𝑛=𝑦−𝑧_𝑛. Choose 𝑎>0 such that 𝑎∈𝑎𝑎_𝜑=𝑏_𝜑if𝜑𝑏_𝜑‖‖𝜒_𝐴‖‖_𝐸<1,𝑎∈𝑎_𝜑,𝑏_𝜑if𝜑𝑏_𝜑<∞,𝜑𝑏_𝜑‖‖𝜒_𝐴‖‖_𝐸≥1,,𝑏_𝜑if𝜑𝑏_𝜑=∞,(3.64) and 𝜑(𝑎)‖𝜒_𝐴‖_𝐸=1 in the last two cases. Take 𝑥=𝑎𝑦 and 𝑥_𝑛=𝑎𝑦_𝑛, then ‖𝑥_𝑛‖_𝜑≤‖𝑥‖_𝜑=1. It is obvious that ‖𝑥_𝑛‖_𝜑=1 when 𝑎=𝑏_𝜑. We show that ‖𝑥_𝑛‖_𝜑→1 in the case 𝑎<𝑏_𝜑. Note that 𝑧_𝑛→0 in measure (see the proof of Lemma 3.2), 𝑧_𝑛↓, and consequently ‖𝑦_𝑛‖_𝐸→‖𝑦‖_𝐸, because 𝐸∈(𝐹𝑃). If ‖𝑥_𝑛‖_𝜑≤1−𝛼 for some 𝛼>0 and infinitely many 𝑛, then we get 1=𝜑(𝑎)‖‖𝜒_𝐴‖‖_𝐸<𝜑𝑎1−𝛼‖‖𝜒^𝐴‖‖_𝐸⟵𝜑𝑎1−𝛼‖‖𝑦^𝑛‖‖_𝐸=𝐼_𝜑𝑥^𝑛1−𝛼≤1,(3.65) which is a contradiction. Moreover, 𝑥_𝑛→𝑥 in measure. Since ‖‖𝑥−𝑥_𝑛‖‖_𝜑≥𝐼_𝜑𝑥−𝑥_𝑛≥𝜑(𝑎)‖‖𝜒_𝐴𝑛‖‖_𝐸≥𝜑(𝑎)𝛿,(3.66) we conclude that 𝐸_𝜑∉(𝐻_𝑔)₊.

Remark 3.13. If 𝐸_𝜑∈(𝐻_𝑔)₊, then 𝐸↪^╱𝐿^∞.

Proof. It is immediate, since if 𝐸↪𝐿^∞, then m_𝐸>0. By Lemma 3.12, we get 𝐸_𝜑∉(𝐻_𝑔)₊.

From Lemmas 3.12 and 3.2, we get immediately the following.

Corollary 3.14. If 𝐸_𝜑∈(𝐻_𝑔)₊, then supp𝐸_𝑎=𝑇.

Lemma 3.15. If 𝐸_𝜑∈(𝐻_𝑔), then 𝜑(𝑏_𝜑)=∞.

Proof. Suppose that 𝐸_𝜑∈(𝐻_𝑔) and 𝜑(𝑏_𝜑)<∞. By Lemma 3.12, we get m_𝐸=0. Let 𝐴∈Σ, 𝜇(𝐴)>0, and 𝜇(𝑇⧵𝐴)>0 be such that 𝜒_𝐴∈𝐸 and 𝜑(𝑏_𝜑)‖𝜒_𝐴‖_𝐸≤1/2. Let (𝐴_𝑛)⊂𝑇⧵𝐴 be such that 0<𝜇(𝐴_𝑛)→0 and 𝜑(𝑏_𝜑)‖𝜒_𝐴𝑛‖_𝐸≤1/2. Taking 𝑥=𝑏_𝜑𝜒_𝐴 and 𝑥_𝑛=𝑏_𝜑𝜒_{𝐴∪𝐴𝑛}, we get ‖𝑥‖_𝜑=‖𝑥_𝑛‖_𝜑=1, 𝑥_𝑛→𝑥 in measure and ‖𝑥_𝑛−𝑥‖_𝜑=1, whence 𝐸_𝜑∉(𝐻_𝑔).

Lemma 3.16. Suppose that (𝑏_𝜑<∞ and 𝜑(𝑏_𝜑)=∞) or (𝑏_𝜑=∞ and 𝜑∉Δ₂(∞)). If 𝐸_𝑎≠{0}, then 𝐸_𝜑∉(𝐻_𝑔).

Proof. It is enough to apply order isometry 𝑃∶𝑙^∞→𝐸_𝜑 constructed in the proof of [5, Theorem  1] and take 𝑧=𝜒_ℕ and 𝑧_𝑛=𝑧−𝑒_𝑛, where 𝑒_𝑛 is the 𝑛th unit vector in 𝑙^∞. If 𝑦=𝑃𝑧 and 𝑦_𝑛=𝑃𝑧_𝑛, then 𝑦_𝑛→𝑦 in measure, because, without loss of generality, one can take in that proof a sequence {𝐴_𝑛} satisfying min{𝜇(𝐴_𝑛),‖𝜒_𝐴𝑛‖_𝐸}≤2^−𝑛. Moreover, ‖𝑦_𝑛‖_𝜑=‖𝑦‖_𝜑=‖𝑦_𝑛−𝑦‖_𝜑=1.

Lemma 3.17. (i) If 𝐿↪^∞╱𝐸, 𝑎_𝜑>0, 𝜑(𝑏_𝜑)=∞, then 𝐸_𝜑∉(𝐻_𝑔).
(ii) If 𝐿↪^∞╱𝐸, supp𝐸_𝑎=𝑇, 𝑎_𝜑=0, 𝑏_𝜑=∞ and 𝜑∉Δ₂(0),
then 𝐸_𝜑∉(𝐻_𝑔).

Proof. (i) Suppose that 𝐿↪^∞╱𝐸, 𝜑(𝑏_𝜑)=∞, and 𝑎_𝜑>0. If 𝑇₁∪𝑇₂=𝑇, 𝑇₁∩𝑇₂=∅, 𝜇(𝑇₁)>0, and 𝜇(𝑇₂)>0, then