Quasinormed spaces generated by a quasimodular

Abstract

In this paper, we introduce the notion of a quasimodular and we prove that the respective Minkowski functional of the unit quasimodular ball becomes a quasinorm. In this way, we refer to and complete the well-known theory related to the notions of a modular and a convex modular that lead to the F-norm and to the norm, respectively. We use the obtained results to consider the basic properties of quasinormed Calderón–Lozanovskiĭ spaces $E_{\varphi}$, where the lower Matuszewska–Orlicz index $\alpha _{\varphi}$ plays the key role. Our studies are conducted in a full possible generality.

1 Introduction

The subject of quasinormed or F-normed spaces (which are a natural generalization of normed spaces) has been the object of research by many authors for a long time (see [1, 4, 10–16, 22, 25, 27]). Recall also that the so-called Δ-norm is a generalization of both a quasinorm and an F-norm (see [12]). From a different point of view, the theory of modular spaces has also been developed (see [21, 26]).

We will try to link these threads. Namely, we will introduce a new notion of the functional ρ that we will call a quasimodular. This is an essential generalization of the concept of a convex semimodular. Furthermore, this quasimodular ρ is defined in such a way that the respective Minkowski functional of the unit quasimodular ball gives a quasinorm. We also check if the basic properties from the classical theory of modular spaces are still true for quasimodular spaces.

Then, we apply our general concept to a special class of quasimodular spaces, that is, the quasinormed Calderón–Lozanovskiĭ spaces $E_{\varphi}$ generated by a quasimodular $\rho _{\varphi}^{E}$. We focus mainly on the basic relations between a positive lower Matuszewska–Orlicz index $\alpha _{\varphi}^{E}$ and the fact that $\rho _{\varphi}^{E}$ $(\|\cdot \|_{\varphi})$ is a quasimodular (a quasinorm), respectively. We study carefully these relations in both directions. We will finish the article by characterizing the quasinormed Orlicz–Lorentz (Orlicz) spaces as spaces generated by a suitable quasimodular. We admit nonconvex, nondecreasing, and degenerated Orlicz functions, which give the full generality of studies. Some geometric properties of these spaces have been studied in [14] and, in the particular case for $E=L^{1}$, in [16], but the authors consider them directly with a quasinorm (not as spaces generated by a “modular”).

2 Preliminaries

Definition 2.1

Given a real vector space X the functional $x\mapsto \Vert x\Vert$ is called a quasinorm if the following three conditions are satisfied:

• (i) $\Vert x\Vert =0$ if and only if $x=0$;

• (ii) $\Vert ax\Vert =|a|\Vert x\Vert$ for any $x\in X$ and $a\in \mathbb{R}$;

• (iii) there exists $C=C_{X}\geq 1$ such that $\Vert x+y\Vert \leq C(\Vert x\Vert +\Vert y\Vert )$ for all $x,y\in X$.

For $0< p\leq 1$, the functional $x\mapsto \Vert x\Vert _{1}$ is called a p-norm if it satisfies the first two conditions of the quasinorm and the condition $\Vert x+y\Vert _{1}^{p}\leq \Vert x\Vert _{1}^{p}+\Vert y\Vert _{1}^{p}$ for any $x,y\in X$. Clearly, each p-norm is a quasinorm. By the Aoki–Rolewicz theorem (cf. [12, Theorem 1.3 on page 7], [25, page 86]), given a quasinorm $\Vert \cdot \Vert$, if $0< p\leq 1$ is such that $C=2^{1/p-1}$, then there exists a p-norm $\Vert \cdot \Vert _{1}$ which is equivalent to $\Vert \cdot \Vert$, that is

$$ \Vert x \Vert _{1}\leq \Vert x \Vert \leq 2C \Vert x \Vert _{1} $$

(2.1)

for all $x\in X$. The quasinorm $\Vert \cdot \Vert$ induces a metric topology on X: in fact a metric can be defined by $d(x,y)=\Vert x-y\Vert _{1}^{p}$. We say that $X=(X,\Vert \cdot \Vert )$ is a quasi-Banach space if it is complete for this metric. Let us note that a lot of important information and results on quasi-Banach spaces can be found in [11], see also [12].

Recall that a quasi-Banach lattice E is called order continuous ($E\in ({\mathrm{OC}})$) if for each sequence $x_{n}\downarrow 0$, that is $x_{n}\geq x_{n+1}$ and $\inf_{n}x_{n}=0$, we have $\Vert x_{n}\Vert _{E}\rightarrow 0$ (see [17, 23, 28]).

3 Quasimodular spaces

In this section, we introduce the notions of a quasimodular and a quasimodular space.

Definition 3.1

Let X be a real linear space. We say that a function $\rho :X\rightarrow {}[ 0,\infty ]$ is a quasimodular whenever for all $x,y\in X$ the following conditions are satisfied:

• (i) $\rho ( 0 ) =0$ and the condition $\rho (\lambda x )\leq 1$ for all $\lambda >0$ implies that $x=0$;

• (ii) $\rho ( -x ) =\rho ( x )$;

• (iii) $\rho (\lambda x)$ is a nondecreasing function of λ, where $\lambda \geq 0$;

• (iv) There is $M\geq 1$ such that

  $$ \rho (\alpha x+\beta y )\leq M \bigl[\rho (x )+ \rho (y ) \bigr] $$

  provided $\alpha ,\beta \geq 0$ and $\alpha +\beta =1$;

• (v) There is a constant $p>0$ such that for all $\varepsilon >0$ and all $A>0$ there exists $K=K(\varepsilon ,A)\geq 1$ such that

  $$ \rho (ax )\leq Ka^{p}\rho (x )+\varepsilon $$

  for any $0< a\leq 1$ whenever $\rho (x)\leq A$.

Remark 3.2

$(i)$ The above definition, in particular the condition $(v)$, has been introduced in such a way as to cover the largest possible class of mappings ρ and to provide a quasinorm (see Theorem 3.4). As we will show in Theorem 4.10, condition $(v)$ can be simplified in some particular cases (more precisely, some particular modulars satisfy condition $(v)$ in a simpler or stronger form).

$(ii)$ Obviously, a convex modular (more precisely a convex semimodular) defined in [26] is in particular a quasimodular. As we will show in Example 4.15$(ii)$ and $(iii)$, the concepts of quasimodular and modular (more precisely a semimodular, which induces an F-norm) defined in [26] are incomparable.

If ρ is a quasimodular on X, then

$$ X_{\rho}:= \Bigl\{ x\in X\colon \lim_{\lambda \rightarrow 0}\rho ( \lambda x)=0 \Bigr\} $$

is called a quasimodular space. It is easy to show that $X_{\rho}$ is a linear subspace of X. We also obtain the following:

Lemma 3.3

For any quasimodular ρ, we have

$$ X_{\rho}= \bigl\{ x\in X\colon \rho (\lambda x)< \infty \textit{ for some }\lambda >0 \bigr\} . $$

Proof

Note that in order to prove this lemma, it is enough to show that if $\rho (\lambda _{0}x)<\infty$ for some $\lambda _{0}>0$, then $\lim_{\lambda \rightarrow 0}\rho (\lambda x)=0$. Let $x\in X$ and $\rho (\lambda _{0}x)<\infty$ for some $\lambda _{0}>0$. We prove that for any $\varepsilon \in (0,\rho (\lambda _{0}x))$ there exists $\delta =\delta (\varepsilon )>0$ such that $\rho (\lambda x)<\varepsilon$, whenever $\lambda <\delta$. Let $\varepsilon \in (0,\rho (\lambda _{0}x))$ be fixed and take $K=K(\frac{\varepsilon}{2},\rho (\lambda _{0}x))$ from condition $(v)$ of Definition 3.1. Then, for $\lambda <\delta$, where $\delta =\lambda _{0}(\frac{\varepsilon}{2K\rho (\lambda _{0}x)})^{1/p}$, we obtain

$$ \rho (\lambda x )=\rho \biggl( \frac{\lambda}{\lambda _{0}}\lambda _{0}x \biggr)\leq K \biggl( \frac{\lambda}{\lambda _{0}} \biggr)^{p}\rho (\lambda _{0}x )+\varepsilon /2< \varepsilon . $$

 □

Theorem 3.4

Let ρ be a quasimodular on X. Then, the functional

$$ \Vert x \Vert _{\rho }=\inf \bigl\{ \lambda >0:\rho (x/\lambda )\leq 1 \bigr\} $$

is a quasinorm on $X_{\rho}$.

Proof

The condition $\Vert 0 \Vert _{\rho}=0$ is obvious. Suppose $x\neq 0$. Then, by condition $(i)$, there exists $\lambda >0$ such that $\rho (\lambda x )>1$ and, by condition $(iii)$, $\Vert x \Vert _{\rho }\geq 1/\lambda$.

For any $x\in X_{\rho}$ and all $\alpha \in \mathbb{R}$, exactly the same way as in [26], we obtain

$$ \Vert \alpha x \Vert _{\rho}=\inf \biggl\{ \lambda >0:\rho \biggl( \frac{\alpha x}{\lambda} \biggr)\leq 1 \biggr\} = \vert \alpha \vert \inf \biggl\{ \lambda / \vert \alpha \vert >0:\rho \biggl(\frac{x}{\lambda /\alpha} \biggr) \leq 1 \biggr\} = \vert \alpha \vert \Vert x \Vert _{\rho}. $$

Finally, we prove the quasitriangle inequality. Let $0<\varepsilon <\frac{1}{2M}$ be fixed and take $K=K(\varepsilon ,1)$ (where constants M and K arise from conditions $(iv)$ and $(v)$ of Definition 3.1). Defining

$$ C= \biggl(\frac{K}{\frac{1}{2M}-\varepsilon} \biggr)^{1/p}, $$

for all $x,y\in X_{\rho }$ and any $\delta >0$, we obtain

$$\begin{aligned} & \rho \biggl( \frac{x+y}{C ( \Vert x \Vert _{\rho }+ \Vert y \Vert _{\rho }+\delta ) } \biggr) \\ &\quad =\rho \biggl( \frac{ \Vert x \Vert _{\rho}+\delta /2}{ \Vert x \Vert _{\rho }+ \Vert y \Vert _{\rho}+\delta} \frac{x}{C( \Vert x \Vert _{\rho}+\delta /2)}+ \frac{ \Vert y \Vert _{\rho}+\delta /2}{ \Vert x \Vert _{\rho}+ \Vert y \Vert _{\rho }+\delta} \frac{y}{C( \Vert y \Vert _{\rho}+\delta /2)} \biggr) \\ &\quad \leq M \biggl[\rho \biggl( \frac{x}{C( \Vert x \Vert _{\rho}+\delta /2)} \biggr) + \rho \biggl( \frac{y}{C( \Vert y \Vert _{\rho}+\delta /2)} \biggr) \biggr] \\ &\quad \leq M \biggl[\frac{K}{C^{p}}\cdot \rho \biggl( \frac{x}{( \Vert x \Vert _{\rho}+\delta /2)} \biggr)+ \varepsilon +\frac{K}{C^{p}}\cdot \rho \biggl( \frac{y}{( \Vert y \Vert _{\rho}+\delta /2)} \biggr)+ \varepsilon \biggr] \\ &\quad \leq 2M \biggl(\frac{K}{C^{p}}+\varepsilon \biggr)=1, \end{aligned}$$

whence

$$ \Vert x+y \Vert _{\rho }\leq C \bigl( \Vert x \Vert _{\rho }+ \Vert y \Vert _{\rho }+\delta \bigr). $$

By the arbitrariness of δ we have $\Vert x+y \Vert _{\rho }\leq C( \Vert x \Vert _{\rho }+ \Vert y \Vert _{\rho})$. □

By the definition of quasinorm, we obtain immediately the following:

Lemma 3.5

Let ρ be a quasimodular on X. Then, for any $x\in X_{\rho}$ the following statements hold:

$(i)$ If $\rho (x)\leq 1$, then $\|x\|_{\rho}\leq 1$;

$(ii)$ If ρ is left continuous $(\lim_{\lambda \rightarrow 1-}\rho (\lambda x)=\rho (x) \textit{ for all }x\in X_{\rho})$, then $\rho (x)\leq 1$ whenever $\|x\|_{\rho}\leq 1$;

$(iii)$ If $\|x\|_{\rho}<1$, then $\rho (x)\leq 1$;

$(iv)$ If ρ is right continuous $(\lim_{\lambda \rightarrow 1+}\rho (\lambda x)=\rho (x) \textit{ for all }x\in X_{\rho})$, then $\|x\|_{\rho}<1$ whenever $\rho (x)<1$.

Remark 3.6

As we will show in Example 4.15$(iv)$ the implication, if $\|x\|_{\rho}<1$ then $\rho (x)<1$, is not always true. Recall that this implication is true for any modular as well as for any convex modular (see [26]).

Lemma 3.7

For each sequence $( x_{n} )$ in $X_{\rho }$ we have $\lim_{n\rightarrow \infty} \Vert x_{n} \Vert _{\rho }=0$ if and only if $\lim_{n\rightarrow \infty}\rho (\lambda x_{n} )=0$ for all $\lambda >0$.

Proof

The implication, if $\lim_{n\rightarrow \infty}\rho (\lambda x_{n} )=0$ for all $\lambda >0$ then $\lim_{n\rightarrow \infty} \Vert x_{n} \Vert _{\rho }=0$ is obvious. Let now $\|x_{n}\|_{\rho}\rightarrow 0$. Fix $\lambda >0$ and $\varepsilon \in (0,1)$ and let $K=K(\varepsilon /2,1)$ be the constant from condition $(v)$ of Definition 3.1. Then, there exists $n_{\lambda ,\varepsilon}$ such that $\|\lambda x_{n}\|_{\rho}\leq (\varepsilon /4K)^{1/p}$ for all $n\geq n_{\lambda ,\varepsilon}$. Hence, for each $a\in ((\varepsilon /4K)^{1/p},(\varepsilon /2K)^{1/p})$ we obtain

$$ \rho (\lambda x_{n})=\rho \biggl(a\frac{\lambda x_{n}}{a} \biggr) \leq Ka^{p}\rho \biggl(\frac{\lambda x_{n}}{a} \biggr)+ \frac{\varepsilon}{2}\leq \varepsilon . $$

By the arbitrariness of ε, for any $\lambda >0$ we obtain $\lim_{n\rightarrow \infty}\rho (\lambda x_{n})=0$. □

4 Quasinormed Calderón–Lozanowskiĭ spaces

A triple $(T,\Sigma ,\mu )$ stands for a positive, complete, and σ-finite measure space and $L^{0}=L^{0}(T,\Sigma ,\mu )$ denotes the space of all (equivalence classes of) Σ-measurable functions $x:T\rightarrow \mathbb{R}$. For every $x\in L^{0}$ we denote $\operatorname{supp}x=\{ t\in T:x(t)\neq 0\}$. Moreover, for any $x,y\in L^{0}$, we write $x\leq y$, if $x(t)\leq y(t)$ almost everywhere with respect to the measure μ on the set T.

A quasinormed lattice [quasi-Banach lattice] $E=(E,\leq ,\Vert \cdot \Vert _{E})$ is called a quasinormed ideal space [quasi-Banach ideal space (or a quasi-Köthe space)] if it is a linear subspace of $L^{0}$ satisfying the following conditions:

• (i) If $x\in L^{0}$, $y\in E$ and $|x|\leq |y|$ μ-a.e., then $x\in E$ and $\Vert x\Vert _{E}\leq \Vert y\Vert _{E}$;

• (ii) There exists $x\in E$ that is strictly positive on the whole T.

By $E_{+}$ we denote the positive cone of E, that is, $E_{+}={\{x\in E:x\geq 0\}}$. Let $C_{E}$ be the constant from the quasitriangle inequality for E. In turn, by $E(w)$ we denote the weighted quasinormed ideal space, that is,

$$ E(w)=\bigl\{ x\in L^{0}\colon xw\in E\bigr\} $$

with the quasinorm $\|x\|_{E(w)}=\|xw\|_{E}$, where $w:T\rightarrow (0,\infty )$ is a measurable weight function.

We say that a quasinormed ideal space E has the Fatou property, if for all $x\in L^{0}$ and any $( x_{n} ) _{n=1}^{\infty }$ in $E_{+}$ such that $x_{n}\uparrow |x|$ μ-a.e and $\sup_{n\in {N}}\Vert x_{n}\Vert _{E}<\infty$, we obtain $x\in E$ and $\lim_{n}\Vert x_{n}\Vert _{E}=\Vert x\Vert _{E}$. It is well known that E has the Fatou property if and only if for each $x\in L^{0}$ and all $( x_{n} ) _{n=1}^{\infty }$ in E such that $x_{n}\rightarrow x$ μ-a.e and $\liminf_{n\in {N}}\Vert x_{n}\Vert _{E}<\infty$, we have $x\in E$ and $\Vert x\Vert _{E}\leq \liminf_{n}\Vert x_{n}\Vert _{E}$ (cf. [2, Lemma 1.5 on page 4]).

Lemma 4.1

Let E be a quasinormed ideal space.

• (i) If $\lim_{n\rightarrow \infty}\|x-x_{n}\|_{E}=0$, where $x\in E$ and $(x_{n})_{n=1}^{\infty}$ is a sequence in E, then $x_{n}\rightarrow x$ locally in measure.

• (ii) For any Cauchy sequence $(x_{n})_{n=1}^{\infty}$ in E there exists $x\in L^{0}$ such that $x_{n}\rightarrow x$ locally in measure.

Proof

This lemma can be proved analogously as Theorem 1 on page 96 in [17]. Indeed, assuming in (2) on page 96 that $\|x_{n}-x\|_{E}<\frac{\varepsilon}{2^{n}C_{E}^{n}}$, we obtain $\|\chi _{B_{n}}\|_{E}<\frac{1}{2^{n}C_{E}^{n}}$ (see (5) on page 96) and, in consequence:

$$ \Vert \chi _{D_{n}} \Vert _{E}\leq \Vert \chi _{C_{ms_{m}}} \Vert _{E}\leq \Biggl\Vert \sum _{k=m+1}^{m+s_{m}}\chi _{B_{k}} \Biggr\Vert _{E}\leq \sum_{k=m+1}^{m+s_{m}}C_{E}^{(k-m)} \Vert \chi _{B_{k}} \Vert _{E}< \sum _{k=m+1}^{m+s_{m}} \frac{C_{E}^{(k-m)}}{2^{k}C_{E}^{k}}< \frac{1}{2^{m}}, $$

(see page 97, line 5). □

Lemma 4.2

[14, Lemma 2.1] A quasinormed ideal space E with the Fatou property is complete.

Proof

We recall a short proof of this lemma for the sake of completeness. By the Aoki–Rolewicz theorem, it is enough to show that for any Cauchy sequence $(x_{n})_{n=1}^{\infty}$ in E there exists $x\in E$ such that $\lim_{n\rightarrow \infty}\|x-x_{n}\|_{E}=0$. If $(x_{n})_{n=1}^{\infty}$ is a Cauchy sequence in E, then by Lemma 4.1$(ii)$, $x_{n}\rightarrow x$ locally in measure for some $x\in L^{0}$. Without loss of generality (passing to subsequences and applying the double extract convergence theorem, if necessary), we can assume $x_{n}\rightarrow x$ μ-a.e. Hence, by the Fatou property, we obtain $x\in E$ and $\|x-x_{n}\|_{E}\leq \liminf_{m\rightarrow \infty}\|x_{m}-x_{n}\|_{E}$ for each $n\in \mathbb{N}$, which completes the proof. □

The following basic fact, very well known for Banach ideal spaces (see [17, Lemma 2, p. 97]), is also true for quasi-Banach ideal spaces.

Lemma 4.3

Let $( E, \Vert \cdot \Vert _{E} )$ be a quasi-Banach ideal space. If $\Vert x_{n} \Vert _{E}\rightarrow 0$, then there exists a subsequence $( x_{n_{k}} ) _{k=1}^{\infty }$, an element $y\in E_{+}$ and a sequence $\varepsilon _{k}\downarrow 0$ such that $\vert x_{n_{k}} \vert \leq \varepsilon _{k}\cdot y$ for each k.

Proof

If $\Vert x_{n} \Vert _{E}\rightarrow 0$, then we can find a subsequence $( x_{n_{k}} ) _{k=1}^{\infty }$ such that $\Vert x_{n_{k}} \Vert <\frac{1}{C^{k}_{E}2^{k}}$ for any $k\in \mathbb{N}$. Consequently,

$$ \sum_{k=1}^{\infty}C_{E}^{k} \Vert k\cdot x_{n_{k}} \Vert \leq \sum_{k=1}^{\infty}C^{k}_{E} \frac{k}{C^{k}_{E}2^{k}}= \sum_{k=1}^{\infty } \frac{k}{2^{k}}< \infty . $$

By Theorem 1.1 from [25], $y:=\sum_{k=1}^{\infty} \vert k\cdot x_{n_{k}} \vert \in E_{+}$. Thus, $\vert x_{n_{k}} \vert \leq \frac{y}{k}$ for all k. Taking $\varepsilon _{k}=1/k$ we complete the proof. □

From now on, we will assume that E is a quasi-Banach ideal space with the Fatou property. During our studies we will consider three natural classes of E:

$( 1 )$ neither $L_{\infty}\subset E$ nor $E\subset L_{\infty}$;

$( 2 )$ $L_{\infty}\subset E$;

$( 3 )$ $E\subset L_{\infty}$.

Let $T=[0,\gamma )$, $0<\gamma \leq \infty$, and μ be the Lebesgue measure. Also, the space $E=L_{p}$, $0< p<\infty$, belongs to the class $(1)$ if $\gamma =\infty$ and the class $(2)$ otherwise. Moreover, the space $L_{1}\cap L_{\infty}$ belongs to the class $(3)$ whenever $\gamma =\infty$. Let now $T=\mathbb{N}$ and $\mu =m$ be a counting measure. Then, the space $l_{p}$, $p\in (0,\infty )$, belongs to class $(3)$, the weighted sequence space $l_{1}(w)$, $w=(w(n))_{n=1}^{\infty}$ and $\sum_{n=1}^{\infty}w(n)<\infty$, belongs to class $(2)$ and the Cesàro sequence space $ces_{p}$, $1< p<\infty$ (see [19] for the respective definition), belongs to class $(1)$.

Remark 4.4

Let $E\subset L_{\infty }$. Then, by the closed-graph theorem that is still true for quasi-Banach spaces (see [12, Theorem. 1.6]), the inclusion is continuous, whence there exists a constant $D_{E}>0$ such that

$$ \Vert x \Vert _{L_{\infty}}\leq D_{E} \Vert x \Vert _{E} $$

(4.1)

for each $x\in E$. Defining

$$ a_{E}=\inf \bigl\{ \Vert \chi _{A} \Vert _{E}:\chi _{A} \in E,\mu (A )>0 \bigr\} , $$

(4.2)

by (4.1), we have $a_{E}\geq 1/D_{E}>0$.

Definition 4.5

A function $\varphi :[0,\infty )\rightarrow [0,\infty ]$ is called an Orlicz function if φ is nondecreasing, vanishing, and right continuous at 0, continuous on $(0,b_{\varphi})$, where

$$ b_{\varphi }=\sup \bigl\{ u\geq 0:\varphi ( u )< \infty \bigr\} $$

and left continuous at $b_{\varphi}$. In the whole paper, excluding Remark 4.14 and Example 4.15$(iii)$, we will assume that $\lim_{u\rightarrow \infty}\varphi (u)=\infty$.

Let

$$ a_{\varphi}=\sup \bigl\{ u\geq 0:\varphi (u ) =0 \bigr\} . $$

By $\varphi ^{-1}$ we denote the generalized inverse of the function φ defined by

$$ \varphi ^{-1}(v)=\inf \bigl\{ u\geq 0\colon \varphi (u)>v\bigr\} \quad \text{for }v\in [0, \infty )\text{ and }\varphi ^{-1}( \infty )=\lim_{v\rightarrow \infty} \varphi ^{-1}(v) $$

(see [20]).

The following lemma is an easy exercise (see [20, Lemma 3.1], for a little different convex case).

Lemma 4.6

For any Orlicz function φ we have:

• (i) Let $u\in [0,b_{\varphi})$. Then, $\varphi ^{-1}(\varphi (u))>u$ if φ is constant on the interval $[u,u+\delta )$ for some $\delta >0$ and $\varphi ^{-1}(\varphi (u))=u$ otherwise.

• (ii) If $b_{\varphi}<\infty$, then $\varphi ^{-1}(\varphi (b_{\varphi}))=b_{\varphi}$ and $\varphi ^{-1}(\varphi (u))=b_{\varphi}< u$ for $b_{\varphi}< u<\infty$.

• (iii) If either $b_{\varphi}=\infty$ or $b_{\varphi}<\infty$ with $\varphi (b_{\varphi} )=\infty$, then $\varphi (\varphi ^{-1}(u))=u$ for any $u\in [0,\infty )$.

• (iv) If $b_{\varphi}<\infty$ and $\varphi (b_{\varphi} )<\infty$, then $\varphi (\varphi ^{-1}(u))=u$ for $u\in [0,\varphi (b_{\varphi})]$ and $\varphi (\varphi ^{-1}(u))=\varphi (b_{\varphi})< u$ for $u>\varphi (b_{\varphi})$.

From $(i)$ and $(ii)$, in particular, we obtain:

• (v) $\varphi ^{-1}(\varphi (u))=u$ for $a_{\varphi}\leq u< b_{\varphi}$ if either $b_{\varphi}=\infty$ or $b_{\varphi}<\infty$ with $\varphi (b_{\varphi} )=\infty$ and φ is strictly increasing on $[a_{\varphi },b_{\varphi})$.

• (vi) $\varphi ^{-1}(\varphi (u))=u$ for $a_{\varphi}\leq u\leq b_{\varphi}$ if $b_{\varphi}<\infty$ and $\varphi (b_{\varphi} )<\infty$ and φ is strictly increasing on $[a_{\varphi},b_{\varphi}]$.

Finally, note that from $(iii)$ and $(iv)$ and $(i)$ and $(ii)$ we obtain

• (vii) $\varphi (\varphi ^{-1}(u))\leq u$ for all $u\in [0,\infty )$ and $u\leq \varphi ^{-1}(\varphi (u))$ if $\varphi (u)<\infty$.

Recall that for any Orlicz function φ the lower Matuszewska–Orlicz index $\alpha _{\varphi}$ for all arguments is defined by the formula

$$\begin{aligned} \alpha ^{a}_{\varphi} =&\sup \bigl\{ p\in \mathbb{R}:\text{there exists }K \geq 1\text{ such that }\varphi (au)\leq Ka^{p} \varphi (u) \\ &\text{for any }u\geq 0\text{ and }0< a\leq 1\bigr\} . \end{aligned}$$

Analogously, the lower Matuszewska–Orlicz indices for large and for small arguments are defined as

$$\begin{aligned} \alpha ^{\infty}_{\varphi} =&\sup \bigl\{ p\in \mathbb{R}: \text{there exist }K\geq 1\text{ and }u_{0}>0\text{ such that } \varphi (u_{0})< \infty \text{ and } \\ &\varphi (au)\leq Ka^{p}\varphi (u)\text{ for any }u\geq u_{0} \text{ and }0< a\leq 1\bigr\} \end{aligned}$$

and

$$\begin{aligned} \alpha ^{0}_{\varphi} =&\sup \bigl\{ p\in \mathbb{R}:\text{there exist }K \geq 1\text{ and }u_{0}>0\text{ such that } \varphi (au)\leq Ka^{p} \varphi (u) \\ &\text{for any }0\leq u\leq u_{0}\text{ and }0< a \leq 1\bigr\} , \end{aligned}$$

respectively.

Remark 4.7

$( i )$ If $0< a_{\varphi }< b_{\varphi }$, then $\alpha _{\varphi}^{0}=\infty$ and we may extend the key inequality in the definition of $\alpha _{\varphi }^{0}$ to any $u_{0}>a_{\varphi}$ such that $\varphi (u_{0})<\infty$. Indeed, let $p>0$. For every $0\leq u\leq a_{\varphi }$ we have $\varphi (au )=0=Ka^{p}\varphi (u )$ for each $K>0$ and $0< a\leq 1$. Take any $u_{0}>a_{\varphi}$ such that $\varphi (u_{0})<\infty$. If $a_{\varphi }< u\leq u_{0}$ and $0< a<\frac{a_{\varphi }}{u_{0}}$, then we have

$$ \varphi ( au ) \leq \varphi ( au_{0} ) \leq \varphi ( a_{\varphi } ) =0\leq Ka^{p}\varphi ( u ) $$

for every $K>0$. Moreover,

$$ \sup_{a_{\varphi }< u\leq u_{0}}\sup_{\frac{a_{\varphi }}{u_{0}} \leq a\leq 1} \frac{\varphi ( au ) }{a^{p}\varphi ( u ) } \leq \sup_{ \frac{a_{\varphi }}{u_{0}}\leq a\leq 1}\frac{1}{a^{p}}= \biggl( \frac{u_{0}}{a_{\varphi }} \biggr) ^{p}. $$

Thus, for $K= ( \frac{u_{0}}{a_{\varphi }} ) ^{p}$ we have

$$ \varphi ( au ) \leq Ka^{p}\varphi ( u ) $$

for all $0< a\leq 1$ and $0\leq u\leq u_{0}$.

$( ii )$ If $a_{\varphi }=0$ and $\alpha _{\varphi }^{0}>0$ then we may extend the key inequality in the definition of $\alpha _{\varphi }^{0}$ to any $u_{1}$ such that $\varphi ( u_{1} ) <\infty$. Indeed, suppose there is $p>0$, $u_{0}>0$ and $K>0$ such that

$$ \varphi ( au ) \leq Ka^{p}\varphi ( u ) $$

(4.3)

for all $0< a\leq 1$ and $0\leq u\leq u_{0}$. Take $u_{1}>u_{0}$ satisfying $\varphi ( u_{1} ) <\infty$. Then,

$$ \sup_{u_{0}< u\leq u_{1}}\sup_{\frac{u_{0}}{u_{1}}\leq a\leq 1} \frac{\varphi ( au ) }{a^{p}\varphi ( u ) } \leq \sup_{\frac{u_{0}}{u_{1}}\leq a\leq 1}\frac{1}{a^{p}}= \biggl( \frac{u_{1}}{u_{0}} \biggr) ^{p}. $$

Set $K_{1}= ( \frac{u_{1}}{u_{0}} ) ^{p}$. Now, we claim that

$$ K_{2}:=\sup_{u_{0}< u\leq u_{1}}\sup_{0< a< \frac{u_{0}}{u_{1}}} \frac{\varphi (au ) }{a^{p}\varphi (u )}< \infty . $$

Otherwise, for each $n\in \mathbb{N}$ we can find $u_{0}< u_{n}\leq u_{1}$ and $0< a_{n}<\frac{u_{0}}{u_{1}}$ such that

$$ \varphi ( a_{n}u_{n} ) >na_{n}^{p} \varphi ( u_{n} ). $$

Denote $b_{n}:=\frac{a_{n}u_{n}}{u_{0}}$. Then, $b_{n}<1$ and

$$ \varphi ( b_{n}u_{0} ) =\varphi ( a_{n}u_{n} )>na_{n}^{p} \varphi ( u_{n} ) =nb_{n}^{p} \biggl( \frac{u_{0}}{u_{n}} \biggr) ^{p}\varphi ( u_{n} )\geq nb_{n}^{p} \biggl( \frac{u_{0}}{u_{1}} \biggr)^{p}\varphi (u_{0} ). $$

On the other hand, by inequality (4.3),

$$ \varphi ( b_{n}u_{0} ) \leq Kb_{n}^{p} \varphi ( u_{0} ), $$

which gives a contradiction and proves the claim. Finally, setting $K_{3}=\max \{ K,K_{1},K_{2} \}$ we conclude that

$$ \varphi ( au ) \leq K_{3}a^{p}\varphi ( u ) $$

for all $0< a\leq 1$ and $0\leq u\leq u_{1}$.

$(iii )$ If $\alpha _{\varphi}^{\infty }>0$ then we may extend the key inequality in the definition of $\alpha _{\varphi }^{\infty }$ to any $u_{1}>a_{\varphi }$. Indeed, suppose there is $p>0$, $u_{0}>0$, and $K>0$ such that $\varphi (u_{0})<\infty$ and

$$ \varphi ( au ) \leq Ka^{p}\varphi ( u ) $$

for all $0< a\leq 1$ and $u\geq u_{0}$. Take $u_{1}< u_{0}$ satisfying $\varphi (u_{1} )>0$. Then,

$$ \sup_{u_{1}\leq u< u_{0}}\sup_{0< a\leq 1} \frac{\varphi ( au ) }{a^{p}\varphi ( u ) } \leq \sup_{0< a\leq 1} \frac{\varphi (au_{0} ) }{a^{p}\varphi (u_{1} ) } \leq \sup_{0< a\leq 1} \frac{Ka^{p}\varphi (u_{0} ) }{a^{p}\varphi ( u_{1} ) }= \frac{K\varphi (u_{0} )}{\varphi (u_{1} )}. $$

Taking $K_{1}=\max \{K, \frac{K\varphi (u_{0} )}{\varphi ( u_{1} ) } \} = \frac{K\varphi ( u_{0} ) }{\varphi ( u_{1} ) }$ we obtain that

$$ \varphi ( au ) \leq K_{1}a^{p}\varphi ( u ) $$

for all $0< a\leq 1$ and $u\geq u_{1}$.

$(iv )$ From the above consideration, we conclude immediately that if $\alpha _{\varphi }^{\infty }>0$ and $\alpha _{\varphi }^{0}>0$ then $\alpha _{\varphi }^{a}>0$.

Example 4.8

Taking $\varphi _{1} (u )=\ln (1+u )$, for $u\geq 0$, we easily obtain

$$ \lim_{u\rightarrow \infty} \frac{\varphi _{1} (au )}{\varphi _{1} (u )}=1 $$

for any $a\in (0,1)$, whence $\alpha ^{\infty}_{\varphi _{1}}=0$. Analogously, defining $\varphi _{2}(0)=0$ and

$$ \varphi _{2}(u)=\frac{1}{\ln (1+\frac{1}{u} )} \quad \text{for } u>0, $$

we obtain

$$ \lim_{u\rightarrow 0^{+}} \frac{\varphi _{2} (au )}{\varphi _{2} (u )}=1 $$

for any $a\in (0,1)$ and, in consequence, $\alpha ^{0}_{\varphi _{2}}=0$.

For any pair E and φ we define the lower Matuszewska–Orlicz index $\alpha ^{E}_{\varphi}$, by the formula

$$ \alpha ^{E}_{\varphi}:=\textstyle\begin{cases} \alpha ^{a}_{\varphi}, & \text{when neither }L_{\infty}\subset E \text{ nor }E\subset L_{\infty }, \\ \alpha ^{\infty}_{\varphi}, & \text{when }L_{\infty}\subset E, \\ \alpha ^{0}_{\varphi}, & \text{when }E\subset L_{\infty}. \end{cases} $$

Given a quasi-Banach ideal space E and an Orlicz function φ, we define on $L^{0}$ a functional $\rho _{\varphi }^{E}$, by

$$ \rho _{\varphi}^{E}(x):=\textstyle\begin{cases} \Vert \varphi ( \vert x \vert ) \Vert _{E} & \text{if }\varphi ( \vert x \vert ) \in E,\text{ } \vspace{1mm} \\ \infty & \text{otherwise.} \end{cases} $$

It is well known that if φ is a convex Orlicz function and E is a Banach ideal space then $\rho _{\varphi}^{E}$ is a convex modular. The result below concerns the more general case (the index $\alpha _{\varphi}^{E}$ plays a role of substitution of convexity of φ).

Theorem 4.9

Let E be a quasi-Banach ideal space and φ be an Orlicz function. If $\alpha ^{E}_{\varphi}>0$, then $\rho _{\varphi }^{E}$ is quasimodular (see Definition 3.1).

Proof

Obviously, $\rho _{\varphi}^{E}(0)=0$ and $\rho _{\varphi}^{E}(-x)=\rho _{\varphi}^{E}(x)$ for any $x\in L^{0}$. Let $x\neq 0$. Then, there exist $A\in \Sigma$ with $\mu (A)>0$ and $n\in \mathbb{N}$ such that $\frac{1}{n}\chi _{A}\leq |x|$. If $\varphi (|x|)\notin E$, then $\rho _{\varphi }^{E}(x)=\infty >1$, while, if $\varphi (|x|)\in E$, by $\varphi (\frac{1}{n})\chi _{A}\leq \varphi (|x|)$, we obtain $\chi _{A}\in E$. Since $\lim_{u\rightarrow \infty}\varphi (u)=\infty$, we can find $\lambda _{A}>0$ such that $\varphi (\lambda _{A}/n)>1/\|\chi _{A}\|_{E}$ and, in consequence, $\rho _{\varphi }^{E}(\lambda _{A}x)=\|\varphi (\lambda _{A}|x|)\|_{E} \geq \|\varphi (\lambda _{A}/n)\chi _{A}\|_{E}>1$.

For every $x\in L^{0}$ and all $0\leq \lambda _{1}\leq \lambda _{2}$ we have $\varphi (\lambda _{1}|x(t)|)\leq \varphi (\lambda _{2}|x(t)|)$ for μ-a.e. $t\in T$, whence $\rho _{\varphi}^{E}(\lambda _{1}x)\leq \rho _{\varphi}^{E}(\lambda _{2}x)$.

Let now $x,y\in L^{0}$ and $\alpha ,\beta \geq 0$, $\alpha +\beta =1$. Then,

$$ \varphi \bigl( \bigl\vert \alpha x(t)+\beta y(t) \bigr\vert \bigr)\leq \varphi \bigl(\max \bigl( \bigl\vert x(t) \bigr\vert , \bigl\vert y(t) \bigr\vert \bigr)\bigr) \leq \varphi \bigl( \bigl\vert x(t) \bigr\vert \bigr)+ \varphi \bigl( \bigl\vert y(t) \bigr\vert \bigr) $$

for μ-a.e. $t\in T$. Hence, if $\varphi (|x|)\in E$ and $\varphi (|y|)\in E$, we obtain

$$\begin{aligned} \rho _{\varphi}^{E} (\alpha x+\beta y ) =& \bigl\Vert \varphi \bigl( \vert \alpha x+\beta y \vert \bigr) \bigr\Vert _{E}\leq \bigl\Vert \varphi \bigl( \vert x \vert \bigr) + \varphi \bigl( \vert y \vert \bigr) \bigr\Vert _{E} \\ \leq & C_{E} \bigl( \bigl\Vert \varphi \bigl( \vert x \vert \bigr) \bigr\Vert _{E}+ \bigl\Vert \varphi \bigl( \vert y \vert \bigr) \bigr\Vert _{E} \bigr)=C_{E} \bigl(\rho _{ \varphi}^{E} (x )+\rho _{\varphi}^{E} (y ) \bigr). \end{aligned}$$

Obviously, the inequality $\rho _{\varphi}^{E} (\alpha x+\beta y )\leq C_{E} ( \rho _{\varphi}^{E} (x )+\rho _{\varphi}^{E} (y ) )$ holds true, when $\varphi (|x|)\notin E$ or $\varphi (|y|)\notin E$.

Finally, we will prove that $\rho _{\varphi}^{E}$ satisfies condition $(v)$. Without loss of generality, we can suppose that $0<\rho _{\varphi}^{E}(x)<\infty$ (then, in particular, we have $a_{\varphi}< b_{\varphi}$). We will consider three cases.

If $L_{\infty}\subset E$, then for any $\varepsilon >0$ there exists $u_{1}\in (a_{\varphi},b_{\varphi})$ such that $C_{E}\cdot \varphi (u_{1})\|\chi _{T}\|_{E}<\varepsilon$. Since $\alpha ^{E}_{\varphi}=\alpha ^{\infty}_{\varphi}>0$, by Remark 4.7, we obtain that there exist numbers $p>0$ and $K=K(\varepsilon )\geq 1$ such that $\varphi (au)\leq Ka^{p}\varphi (u)$ for all $u\geq u_{1}$ and $0< a\leq 1$. Defining $B=\{t\in T\colon |x(t)|\geq u_{1}\}$, for any $a\in (0,1]$ we obtain

$$\begin{aligned} \rho _{\varphi}^{E}(ax) =& \bigl\Vert \varphi \bigl(a \vert x \vert \bigr) \bigr\Vert _{E}\leq C_{E} \bigl( \bigl\Vert \varphi \bigl(a \vert x \vert \bigr) \chi _{B} \bigr\Vert _{E}+ \bigl\Vert \varphi \bigl(a \vert x \vert \bigr)\chi _{T\backslash B} \bigr\Vert _{E} \bigr) \\ \leq & C_{E}Ka^{p} \bigl\Vert \varphi \bigl( \vert x \vert \bigr) \bigr\Vert _{E}+C_{E}\varphi (u_{1}) \Vert \chi _{T} \Vert _{E}\leq C_{E}Ka^{p}\rho _{\varphi}^{E}(x)+ \varepsilon . \end{aligned}$$

(4.4)

In the case when neither $L_{\infty}\subset E$ nor $E\subset L_{\infty}$, analogously as above, we obtain

$$ \rho _{\varphi}^{E}(ax)= \bigl\Vert \varphi \bigl(a \vert x \vert \bigr) \bigr\Vert _{E}\leq Ka^{p} \bigl\Vert \varphi \bigl( \vert x \vert \bigr) \bigr\Vert _{E}=Ka^{p} \rho _{\varphi}^{E}(x). $$

(4.5)

Let now $E\subset L_{\infty}$, take $A>0$ and assume that $\rho _{\varphi}^{E}(x)=\|\varphi (|x|)\|_{E}\leq A$. By the closed-graph theorem, there exists a constant $D_{E}>0$ such that $\|\varphi (|x|)\|_{L^{\infty}}\leq D_{E}\|\varphi (|x|)\|_{E}\leq AD_{E}$. Thus, $\varphi (|x(t)|)\leq \min (AD_{E},\varphi (b_{\varphi}))$ for μ-a.e. $t\in T$, whence, by Lemma 4.6, $|x(t)|\leq u_{2}$ for the same t, where $u_{2}=\varphi ^{-1}(\min (AD_{E},\varphi (b_{\varphi})))$. Simultaneously, by $\alpha ^{E}_{\varphi}=\alpha ^{0}_{\varphi}>0$ and Remark 4.7, we obtain that there exist $p>0$ and $K=K(A)\geq 1$ such that $\varphi (au)\leq Ka^{p}\varphi (u)$ for any $u\leq u_{2}$ and $0< a\leq 1$. Therefore, $\rho _{\varphi}^{E}(ax)\leq Ka^{p}\rho _{\varphi}^{E}(x)$. □

Now, we will show that, depending on the embeddings between E and $L_{\infty}$, the condition $\alpha _{\varphi}^{E}>0$ is even equivalent to a certain condition that is close to the point $(v)$ of Definition 3.1.

Theorem 4.10

$(i)$ Assume that neither $L_{\infty}\subset E$ nor $E\subset L_{\infty}$. Then, $\alpha ^{a}_{\varphi}>0$ if and only if there exist constants $p>0$ and $K\geq 1$ such that for all $x\in L^{0}$ and $0< a\leq 1$ we have $\rho _{\varphi}^{E} (ax )\leq Ka^{p}\rho _{\varphi}^{E} (x )$.

$(ii)$ Let $L_{\infty}\subset E$. Then, $\alpha ^{\infty}_{\varphi}>0$ if and only if there is a constant $p>0$ such that for all $v_{0}>0$ there exists $K=K(v_{0})\geq 1$ such that for each $x\in L^{0}$ satisfying $|x(t)|\geq v_{0}$ for μ-a.e. $t \in T$ and any $0< a\leq 1$ we have $\rho _{\varphi}^{E} (ax )\leq Ka^{p}\rho _{\varphi}^{E} (x )$.

$(iii)$ Let $E\subset L_{\infty}$. Then, $\alpha ^{0}_{\varphi}>0$ if and only if there is a constant $p>0$ such that for every $A>0$ there exists $K=K(A)\geq 1$ such that for any $x\in L^{0}$ satisfying $\rho _{\varphi}^{E}(x)\leq A$ and every $0< a\leq 1$ we have $\rho _{\varphi}^{E} (ax )\leq Ka^{p}\rho _{\varphi}^{E} (x )$.

Proof

The necessity of statements $(i)$ and $(iii)$ follows from the proof of Theorem 4.9. Now, we will show the sufficiency of $(iii)$. Let $D\in \Sigma$ be such that $\mu (D)>0$ and $\chi _{D}\in E$. Take $u_{0}>0$ satisfying $0<\varphi (u_{0})<\infty$ and define $A:=\varphi (u_{0})\|\chi _{D}\|_{E}$. Then, for any $u\leq u_{0}$ and any $a\in (0,1]$ we obtain

$$\begin{aligned} \varphi (au) \Vert \chi _{D} \Vert _{E} =& \bigl\Vert \varphi (au)\chi _{D} \bigr\Vert _{E}=\rho _{ \varphi}^{E} (au\chi _{D} ) \\ \leq & K(A)a^{p}\rho _{\varphi}^{E} (u\chi _{D} )=K(A)a^{p} \bigl\Vert \varphi (u)\chi _{D} \bigr\Vert _{E}=K(A)a^{p}\varphi (u) \Vert \chi _{D} \Vert _{E}. \end{aligned}$$

Hence, by the arbitrariness of u and a, we obtain $\alpha ^{0}_{\varphi}>0$. Analogously, we can prove the sufficiency of the condition $\alpha ^{a}_{\varphi}>0$ in $(i)$.

Now, we will prove statement $(ii)$, that is, let $L_{\infty}\subset E$. First, let us note that, by the proof of Theorem 4.9, we obtain the implication: if $\alpha ^{\infty}_{\varphi}>0$, then there is a constant $p>0$ such that for each $\varepsilon >0$ there exists $K=K(\varepsilon )\geq 1$ such that $\rho _{\varphi}^{E} (ax )\leq Ka^{p}\rho _{\varphi}^{E} ( x )+\varepsilon$ for any $x\in L^{0}$ and $0< a\leq 1$.

Let $\alpha ^{\infty}_{\varphi}>0$ and take $p\in (0,\alpha ^{\infty}_{\varphi})$, $v_{0}>0$ and $x\in L^{0}$ such that $\rho _{\varphi}^{E}(x)<\infty$ and $|x(t)|\geq v_{0}$ for μ-a.e. $t\in T$. By Remark 4.7 (especially $(iii)$ and $(iv)$), there exists $K=K(v_{0})$ such that

$$ \varphi \bigl(a \bigl\vert x(t) \bigr\vert \bigr)\leq Ka^{p} \varphi \bigl( \bigl\vert x(t) \bigr\vert \bigr) $$

for all $a\in (0,1]$ and μ-a.e. $t\in T$, whence $\rho _{\varphi}^{E} (ax )\leq Ka^{p}\rho _{\varphi}^{E} (x )$.

Finally, we will prove the opposite implication. Take $u_{0}>0$ satisfying $0<\varphi (u_{0})<\infty$ and define $v_{0}=u_{0}$. Then, for any $u\geq u_{0}$ and any $a\in (0,1]$ we have

$$ \varphi (au) \Vert \chi _{T} \Vert _{E}=\rho _{\varphi}^{E} (au\chi _{T} )\leq K(v_{0})a^{p}\rho _{\varphi}^{E} (u \chi _{T} )=K(v_{0})a^{p} \varphi (u) \Vert \chi _{T} \Vert _{E} $$

and, in consequence, $\alpha ^{\infty}_{\varphi}>0$. □

Definition 4.11

Let a quasi-Banach ideal space E and an Orlicz function φ be such that $\alpha _{\varphi}^{E}>0$. Then, the Calderón–Lozanovskiĭ space $E_{\varphi}$ is defined by

$$ E_{\varphi}=\Bigl\{ x\in L^{0}:\lim_{\lambda \rightarrow 0} \rho _{\varphi}^{E}( \lambda x)=0\Bigr\} . $$

By Theorem 4.9 and Lemma 3.3, $E_{\varphi}$ is a quasimodular space and

$$ E_{\varphi}=\bigl\{ x\in L^{0}:\rho _{\varphi }^{E}( \lambda x)< \infty \text{ for some }\lambda >0\bigr\} . $$

Moreover, by Theorem 3.4, the functional

$$ \Vert x \Vert _{\varphi }=\inf \bigl\{ \lambda >0:\rho _{\varphi}^{E} (x/\lambda )\leq 1 \bigr\} , $$

is a quasinorm, called a Luxemburg–Nakano quasinorm. It is easy to show that $E_{\varphi}=(E_{\varphi},\leq ,\|\cdot \|_{\varphi})$ is a quasinormed ideal space.

Remark 4.12

It is known that there is a connection between the Banach ideal space $E_{\varphi}$ (where φ is a convex Orlicz function and E is a Banach ideal space) and the normed Calderón–Lozanovskii interpolation construction $\psi ( E,L_{\infty} )$ (where ψ is a homogeneous, concave function on $\mathbb{R}_{+}^{2}$) – see [24, Example 2, p. 178]. However, there is a similar relation if φ is a nonconvex Orlicz function, E is a quasi-Banach ideal space, and ψ is positively homogeneous, nondecreasing with respect to each variable. On the other hand, such an approach leads to quasi-Banach lattices $\psi (E_{0},E_{1} )$ that have application in interpolation theory (see [27]).

We also obtain the following:

Lemma 4.13

For any quasi-Banach ideal space E and any Orlicz function φ the following assertions hold:

• (i) For each $x\in E_{\varphi}$ the function $f_{x} (\alpha ):=\rho _{\varphi}^{E}(\alpha x)$, for $\alpha >0$, is nondecreasing and left continuous.

• (ii) For any $x\in E_{\varphi }$ we have $\|x\|_{\varphi}\leq 1$ if and only if $\rho _{\varphi}^{E}(x)\leq 1$.

• (iii) The quasinormed ideal space $E_{\varphi}$ has the Fatou property and, in consequence, $E_{\varphi}$ is complete.

Proof

The assertion $(i)$ follows immediately from the properties of φ and E (recall that E has the Fatou property). Next, by $(i)$ and Lemma 3.5, we obtain $(ii)$. Proceeding analogously as in [6, Theorem 12] (see also [14, Lemma 2.2$(ii)$]), we obtain that $E_{\varphi}$ has the Fatou property. Hence, by Lemma 4.2, $E_{\varphi}$ is complete. □

Remark 4.14

Now, we will show the naturalness of the assumption $\lim_{u\rightarrow \infty}\varphi (u)=\infty$ in the definition of Orlicz function (see Definition 4.5). Obviously, if $\alpha ^{a}_{\varphi}>0$ or $\alpha ^{\infty}_{\varphi}>0$, then $\lim_{u\rightarrow \infty}\varphi (u)=\infty$. Simultaneously, for $\varphi (u)=\min (u^{2},1)$, we have $\alpha ^{0}_{\varphi}>0$ and $\lim_{u\rightarrow \infty}\varphi (u)=1$. Let $E\subset L_{\infty}$, $\lim_{u\rightarrow \infty}\varphi (u)<\infty$ and $\alpha ^{E}_{\varphi}=\alpha ^{0}_{\varphi}>0$. Then, condition (i) of Definition 3.1 holds whenever $\lim_{u\rightarrow \infty}\varphi (u)>1/a_{E}$, where $a_{E}$ is defined by formula (4.2), and, in consequence, $\rho _{\varphi}^{E}$ is quasimodular. Moreover, defining the new Orlicz function ψ, by $\psi (u)=\varphi (u)$ for $u\in [0,\varphi ^{-1}(1/a_{E})]$ and $\psi (u)=u-(\varphi ^{-1}(1/a_{E})-1/a_{E})$ for $u>\varphi ^{-1}(1/a_{E})$, we obtain $\lim_{u\rightarrow \infty}\psi (u)=\infty$ and $(E_{\varphi},\|\cdot \|_{\varphi})\equiv (E_{\psi},\|\cdot \|_{\psi})$.

Now, we will show (among others) that the notion of modular and quasimodular are incomparable.

Example 4.15

Assume $T=[0,\infty )$ and μ is the Lebesgue measure.

$(i)$ If $E=L_{1}$ and $\varphi (u)=u^{p}$ for $u\geq 0$, $p>0$, then $\rho _{\varphi}^{E}$ is a quasimodular (a convex modular for $p>1$) as well as a modular (see [26]). We have

$$ \Vert x \Vert _{\varphi}= \Vert x \Vert _{p}= \biggl( \int _{0}^{\infty} \bigl\vert x(t) \bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}}. $$

Simultaneously, the F-norm is given by

$$ \big\vert \Vert x \Vert \big\vert _{\varphi}:=\inf \bigl\{ \lambda >0\colon \rho _{\varphi}^{E} (x/\lambda )\leq \lambda \bigr\} = \biggl( \int _{0}^{ \infty} \bigl\vert x(t) \bigr\vert ^{p}\,dt \biggr)^{\frac{1}{1+p}} $$

(see [26]). Note also that $|\Vert x_{n}\Vert |_{\varphi}\rightarrow 0$ if and only if $\|x\|_{\varphi}\rightarrow 0$, but there do not exist constants $A,B>0$ such that $A \|x\|_{\varphi} \leq |\Vert x\Vert |_{\varphi} \leq B \|x\|_{\varphi}$ for all $x \in L_{\varphi}$.

$(ii)$ Let $E=L_{(1/4)}$ and $\varphi (u)=u^{2}$ for $u\geq 0$. Obviously, $\rho _{\varphi}^{E}$ is a quasimodular. Simultaneously, for $x=\chi _{[0,1)}$ and $y=\chi _{[1,2)}$ we obtain

$$ \rho _{\varphi}^{E} \biggl(\frac{1}{2}x+ \frac{1}{2}y \biggr)=4>2=\rho _{ \varphi}^{E}(x)+\rho _{\varphi}^{E}(y). $$

Thus, $\rho _{\varphi}^{E}$ is not a modular.

$(iii)$ If $E=L_{1}$ and $\varphi (u)=\arctan (u)$ for $u\geq 0$ (see Definition 4.5), then $\rho _{\varphi}^{E}$ is a modular (see [26]). Now, we will show that $\rho _{\varphi}^{E}$ is not a quasimodular, more precisely, $\rho _{\varphi}^{E}$ does not satisfy condition $(v)$ of Definition 3.1. Let $p>0$ and take $A=\pi /2$ and $\varepsilon =\pi /8$. Then, for $x_{n}=n\chi _{[0,1)}$ and $a_{n}=1/n$, we obtain $\rho _{\varphi}^{E}(x_{n})\leq \pi /2$, $\rho _{\varphi}^{E}(a_{n}x_{n})=\pi /4$, and $\lim_{n\rightarrow \infty}(a_{n})^{p}=0$. In consequence, for any $K\geq 1$ there exists $n\in \mathbb{N}$ such that $\rho _{\varphi}^{E}(a_{n}x_{n})>Ka_{n}^{p}\rho _{\varphi}^{E}(x_{n})+ \pi /8$.

$(iv)$ Let $E=L_{1}$, $\varphi (u)=u$ for $u\in [0,1]$ and $\varphi (u)=\max (1,u-1)$ for $u>1$. For $x=\chi _{[0,1)}$ we have $\rho _{\varphi}^{E}(x)=\rho _{\varphi}^{E}(2x)=1$. Simultaneously, $\rho _{\varphi}^{E}(x/\lambda )>1$ for $\lambda <1/2$, so $\|x\|_{\varphi}=\frac{1}{2}$ (see Remark 3.6).

Recall the notion of uniform monotonicity (see, for example, [3, 9, 22]) that plays an important role in the theory of Banach lattices. A quasi-Banach lattice $(E, \Vert \cdot \Vert _{E} )$ is said to be uniformly monotone provided for each $\varepsilon >0$ there exists $\delta =\delta (\varepsilon )>0$ such that for all $x,y\in E_{+}$ with $\Vert x \Vert _{E}=1$ we have $\Vert x+y \Vert _{E}\geq 1+\delta$ whenever $\Vert y \Vert _{E}\geq \varepsilon$.

Lemma 4.16

If E is uniformly monotone, then for each $\varepsilon _{1}>0$ and $A>0$ there exists $\delta _{1}=\delta _{1} (\varepsilon _{1},A ) =\delta (\frac{\varepsilon _{1}}{A} )>0$ (here the function $\delta (\cdot )$ comes from the definition of the uniform monotonicity) such that for all $x,y\in E_{+}$ we have $\Vert x+y \Vert _{E}\geq \Vert x \Vert _{E} (1+\delta _{1} )$ whenever $\Vert y \Vert _{E}\geq \varepsilon _{1}$ and $\Vert x \Vert _{E}\leq A$.

Proof

The proof can be found in [9] (the same for the quasinorm). However, we will need the precise form of the function $\delta _{1} (\varepsilon _{1},A )$, so we present the proof for the reader’s convenience.

Let $\varepsilon _{1}>0$ and $A>0$. Take $x,y\in E_{+}$ such that $\Vert y \Vert _{E}\geq \varepsilon _{1}$ and $\Vert x \Vert _{E}\leq A$. Denote

$$ \widetilde{x}=\frac{x}{ \Vert x \Vert _{E}}\quad \text{and} \quad \widetilde{y}= \frac{y}{ \Vert x \Vert _{E}}. $$

Then, $\Vert \widetilde{x} \Vert _{E}=1$ and $\Vert \widetilde{y} \Vert _{E}\geq \frac{\varepsilon _{1}}{A}$. By the uniform monotonicity of E, there exists $\delta _{1}=\delta _{1} (\varepsilon _{1},A )=\delta (\frac{\varepsilon _{1}}{A} )>0$ such that $\Vert \widetilde{x}+\widetilde{y} \Vert _{E}\geq 1+ \delta _{1}$, whence

$$ \Vert x+y \Vert _{E}\geq \Vert x \Vert _{E} (1+ \delta _{1} ). $$

 □

Recall that the assumption $\alpha _{\varphi }^{E}>0$ is important to show that $\rho _{\varphi }^{E}$ is a quasimodular (see Theorem 4.9 and also Theorem 4.10) and, consequently, that $\Vert \cdot \Vert _{\varphi}$ is a quasinorm. However, if we know nothing about the index $\alpha _{\varphi }^{E}$ then we may still define a functional

$$ \rho _{\varphi }^{E}(x):=\textstyle\begin{cases} \Vert \varphi ( \vert x \vert ) \Vert _{E} & \text{if }\varphi ( \vert x \vert )\in E, \\ \infty & \text{otherwise} \end{cases} $$

and the set

$$ E_{\varphi}=\Bigl\{ x\in L^{0}:\lim_{\lambda \rightarrow 0} \rho _{\varphi}^{E}( \lambda x)=0\Bigr\} . $$

Then, it is easy to see that $E_{\varphi }$ is a linear space and we may consider the functional

$$ \Vert x \Vert _{\varphi}=\inf \bigl\{ \lambda >0:\rho _{\varphi}^{E} (x/\lambda )\leq 1 \bigr\} ,\quad \text{for }x \in E_{\varphi}, $$

which satisfies the conditions $(i )$ and $( ii )$ of the quasinorm definition. We are going to show that, under some natural assumptions, the condition $\alpha _{\varphi }^{E}>0$ can be even necessary, that is to say, if the functional $\Vert \cdot \Vert _{\varphi}$ is a quasinorm, then $\alpha _{\varphi}^{E}>0$. Since the result below is only some illustration of how natural is the assumption that $\alpha _{\varphi}^{E}>0$, we will limit ourselves only to one case of the ideal space E.

Theorem 4.17

Suppose φ is a finitely valued, strictly increasing Orlicz function. Let $(E, \Vert \cdot \Vert _{E} )$ be a uniformly monotone, p-normed ideal space over nonatomic measure space $(T,\Sigma ,\mu )$ for some $0< p\leq 1$. Assume that neither $L_{\infty }\subset E$ nor $E\subset L_{\infty}$. If $(E_{\varphi},\Vert \cdot \Vert _{\varphi} )$ is a quasinormed space, then $\alpha _{\varphi}^{a}>0$.

Proof

Denote by $C\geq 1$ the constant from the quasitriangle inequality for $(E_{\varphi},\Vert \cdot \Vert _{\varphi} )$. Let $\delta _{0}=\delta (1/2 )$ be the constant from the definition of the uniform monotonicity of E. Fix $s>0$. Recall also that if E is uniformly monotone, then E is order continuous (see [22, Proposition 2.4]). Thus, by the proof of Theorem 2.4 in [14], the function ν, defined by $\nu (A)=\|\chi _{A}\|_{E}$ for all $A\in \Sigma$, $\chi _{A}\in E$, is the submeasure in the sense of [5, Definition 1], whence by [5, Theorem 10], ν has the Darboux property. In consequence, we can find a set $A\in \Sigma$, $\chi _{A}\in E$ satisfying

$$ \varphi (s )= \frac{1}{ \Vert \chi _{A} \Vert _{E} (1+\delta _{0} )}. $$

Take a set $B\in \Sigma$ of positive measure such that $\chi _{B}\in E$, $A\cap B=\emptyset$ and $\Vert \chi _{B} \Vert _{E}=\frac{1}{2} \Vert \chi _{A} \Vert _{E}$. Applying Lemma 4.16 we conclude that

$$ \Vert \chi _{A\cup B} \Vert _{E}\geq \Vert \chi _{A} \Vert _{E} (1+\delta _{0} ). $$

(4.6)

It is well known that

$$ \Vert \chi _{A} \Vert _{\varphi}= \frac{1}{\varphi ^{-1} (\frac{1}{ \Vert \chi _{A} \Vert _{E}} )}, $$

where $\varphi ^{-1}$ is the general right-inverse to φ. Indeed, $\Vert \varphi (\frac{\chi _{A}}{\lambda} ) \Vert _{E}\leq 1$ if and only if $\frac{1}{\lambda}\leq \varphi ^{-1} ( \frac{1}{ \Vert \chi _{A} \Vert _{E}} )$. In consequence,

$$ C\geq \frac{ \Vert \chi _{A}+\chi _{B} \Vert _{\varphi}}{ \Vert \chi _{A} \Vert _{\varphi}+ \Vert \chi _{B} \Vert _{\varphi}} \geq \frac{ \Vert \chi _{A}+\chi _{B} \Vert _{\varphi}}{2 \Vert \chi _{A} \Vert _{\varphi}}= \frac{ \Vert \chi _{A\cup B} \Vert _{\varphi}}{2 \Vert \chi _{A} \Vert _{\varphi}}= \frac{\varphi ^{-1} (\frac{1}{ \Vert \chi _{A} \Vert _{E}} )}{2\varphi ^{-1} (\frac{1}{ \Vert \chi _{A\cup B} \Vert _{E}} )}. $$

Moreover, also applying (4.6), we obtain

$$ 2C\varphi ^{-1} \biggl( \frac{1}{ \Vert \chi _{A} \Vert _{E} (1+\delta _{0} )} \biggr)\geq 2C\varphi ^{-1} \biggl( \frac{1}{ \Vert \chi _{A\cup B} \Vert _{E}} \biggr)\geq \varphi ^{-1} \biggl(\frac{1}{ \Vert \chi _{A} \Vert _{E}} \biggr) $$

and consequently we obtain

$$ \varphi \biggl[2C\varphi ^{-1} \biggl( \frac{1}{ \Vert \chi _{A} \Vert _{E} (1+\delta _{0} )} \biggr) \biggr]\geq \varphi \biggl[\varphi ^{-1} \biggl( \frac{1}{ \Vert \chi _{A} \Vert _{E}} \biggr) \biggr]= \frac{1}{ \Vert \chi _{A} \Vert _{E}}. $$

Taking $C_{1}=2C$, we have

$$ (1+\delta _{0} )\varphi (s )\leq \varphi \bigl[2C\varphi ^{-1} \bigl(\varphi (s ) \bigr) \bigr]= \varphi [ 2Cs ]=\varphi [C_{1}s ] $$

for each $s>0$. For every $a\geq 1$ there is $m\in \mathbb{N}$ such that $C_{1}^{m-1}\leq a< C_{1}^{m}$. Fix $p>0$ satisfying $p=\frac{\ln (1+\delta _{0} )}{\ln C_{1}}$. Then, $(1+\delta _{0} )^{m-1}= (C_{1}^{m-1} )^{p}$ and

$$ \varphi (as )\geq \varphi \bigl(C_{1}^{m-1}s \bigr) \geq (1+\delta _{0} )^{m-1}\varphi (s )= \bigl(C_{1}^{m-1} \bigr)^{p}\varphi ( s )\geq \biggl( \frac{a}{C_{1}} \biggr)^{p}\varphi (s )=a^{p}C_{1}^{-p} \varphi (s ) $$

for each $s>0$. Setting $u:=as$ and $b:=1/a$ we conclude that

$$ \varphi (bu )\leq b^{p}C_{1}^{p}\varphi (u ) $$

for any $u>0$ and each $b\in (0,1]$. This means that $\alpha _{\varphi}^{a}>0$. □

5 The quasinormed Orlicz–Lorentz spaces and Orlicz spaces

Take $I=[0,1]$ or $I=[0,\infty )$ with the Lebesgue measure μ. Let $\omega :I\rightarrow \mathbb{R_{+}}$ be a measurable function with $\int _{0}^{t}\omega (s)\,ds<\infty$ for each $t \in I$. We assume that there is a constant $C>0$ such that $\int _{0}^{2t}\omega (s)\,ds\leq C\int _{0}^{t}\omega (s)\,ds$ for each $t\in \frac{1}{2}I$, which implies that the space

$$ \Lambda _{1,\omega}= \biggl\{ f\in L^{0}\colon \lVert f \rVert _{\omega}= \int _{I} f^{\ast}(s)\omega (s)\,ds< \infty \biggr\} , $$

where $f^{\ast}$ is the nonincreasing rearrangement of f – see [2, 23], is the quasinormed ideal space with the Fatou property (see [13]) and it is called the Lorentz funtion space $\Lambda _{1,\omega}$. The Lorentz sequence space $\lambda _{1,\omega}$ over the counting measure space $(\mathbb{N},2^{\mathbb{N}},m )$ we define analogously (see [15]).

Note that,

$(1)$ neither $L_{\infty }\subset \Lambda _{1,\omega}$ nor $\Lambda _{1,\omega}\subset L_{\infty }$, whenever $I=[0,\infty )$ and $\int _{0}^{\infty}\omega (s)\,ds=\infty$;

$(2)$ $L_{\infty}\subset \Lambda _{1,\omega}$,whenever $I=[0,1]$ or ($I=[0, \infty )$ and $\int _{0}^{\infty}\omega (s)\,ds<\infty$);

$(3)$ $\lambda _{1,\omega}\subset l_{\infty}$ (furthermore, $\lambda _{1,\omega}=l_{\infty}$ provided $\sum_{i=1}^{\infty}\omega (i)<\infty$).

If E is a Lorentz function (sequence) space $\Lambda _{1,w}$ ($\lambda _{1,w}$), then Calderón–Lozanovskiĭ space $E_{\varphi }$ is the corresponding Orlicz–Lorentz function (sequence) space $\Lambda _{\varphi ,w}$ ($\lambda _{\varphi ,w}$). If $E=L_{1}$ ($E=l_{1}$), then the space $E_{\varphi }$ becomes the Orlicz function (sequence) space $L_{\varphi}$ ($l_{\varphi}$) (cf. [16]). On the other hand, if $\varphi (u)=u^{p}$, $1\leq p<\infty$ [$0< p<1$] then $E_{\varphi }$ is the p-convexification [concavification] $E^{(p)}$ of E with the quasinorm $\Vert x\Vert _{E^{(p)}}=\Vert |x|^{p}\Vert _{E}^{1/p}$. If $\varphi (u)=0$ for $0\leq u\leq 1$ and $\varphi (u)=\infty$ for $u>1$, then $E_{\varphi}=L_{\infty}$ ($E_{\varphi}=l_{\infty}$) with equality of the norms.

It is well known that Orlicz–Lorentz spaces $\Lambda _{\varphi ,\omega}$ (in particular, the Lorentz spaces $\Lambda _{p,\omega}$ or the Orlicz spaces $L_{\varphi}$) have been studied directly by many authors (see, for example, [13–16] and the references therein).

Applying Theorems 3.4 and 4.9 with $E=\Lambda _{1,w}$ or $E=\lambda _{1,w}$ we conclude immediately:

Corollary 5.1

$(i )$ Let $E=\Lambda _{1,w} (I,\Sigma ,\mu )$ be such that $\mu (I)<\infty$ or $(\mu (I)=\infty \textit{ and }\int _{0}^{\infty}\omega (s)\,ds<\infty )$. If $\alpha _{\varphi}^{\infty}>0$, then the functional $\rho _{\varphi}^{\Lambda _{1,w}}(\cdot )$ is a quasimodular and the functional $\Vert \cdot \Vert _{\varphi ,w}$ is a quasinorm (called a Luxemburg–Nakano quasinorm) in $\Lambda _{\varphi ,w}$.

$(ii )$ Let $E=\Lambda _{1,w} (I,\Sigma ,\mu )$ be such that $\mu (I)= \infty$ and $\int _{0}^{\infty}\omega (s)\,ds=\infty$. If $\alpha _{\varphi}^{a}>0$, then the functional $\rho _{\varphi}^{\Lambda _{1,w}}(\cdot )$ is a quasimodular and the functional $\Vert \cdot \Vert _{\varphi ,w}$ is a quasinorm in $\Lambda _{\varphi ,w}$.

$(iii )$ Let $E=\lambda _{1,w}$ and $\sum_{i=1}^{\infty}\omega (i)=\infty$. If $\alpha _{\varphi}^{0}>0$, then the functional $\rho _{\varphi}^{\lambda _{1,w}}(\cdot )$ is a quasimodular and the functional $\Vert \cdot \Vert _{\varphi ,w}$ is a quasinorm in $\lambda _{\varphi ,w}$.

Remark 5.2

The second conclusion in statement $(iii )$ has been proved directly in [15, Proposition 1.3], but the authors did not consider the quasimodular, only the quasinorm.

Applying the above corollary with $\omega \equiv 1$ and Theorem 4.17 with $E=L^{1}$ we obtain:

Corollary 5.3

$(i )$ Let $E=L_{1} (I,\Sigma ,\mu )$ with a finite measure μ. If $\alpha _{\varphi}^{\infty}>0$, then the functional $\rho _{\varphi}^{L_{1}}(\cdot )$ is a quasimodular and the functional $\Vert \cdot \Vert _{\varphi}$ is a quasinorm (called a Luxemburg–Nakano quasinorm) in $L_{\varphi}$.

$(ii )$ Let $E=L_{1} (I,\Sigma ,\mu )$ with an infinite measure μ. If $\alpha _{\varphi}^{a}>0$, then the functional $\rho _{\varphi}^{L_{1}}(\cdot )$ is a quasimodular and the functional $\Vert \cdot \Vert _{\varphi}$ is a quasinorm in $L_{\varphi}$.

$(iii )$ Let $E=L_{1} (I,\Sigma ,\mu )$ with an infinite measure μ. Assume that the function φ is finitely valued and strictly increasing. Then, the functional $\Vert \cdot \Vert _{\varphi}$ is a quasinorm in $L_{\varphi}$ if and only if $\alpha _{\varphi}^{a}>0$.

$(iv )$ If $E=l_{1}$ and $\alpha _{\varphi}^{0}>0$, then the functional $\rho _{\varphi}^{l_{1}}(\cdot )$ is a quasimodular and the functional $\Vert \cdot \Vert _{\varphi}$ is a quasinorm in $l_{\varphi}$.

Remark 5.4

The statement $(iii )$ has been proved directly in [16, Theorem 1.8].

6 Further research and open ends

6.1 Further research

It is well known that the relations between the modular and the norm play a crucial role in the metric geometry of normed Orlicz spaces (normed Calderón–Lozanovskiĭ spaces). The following basic results have many applications:

Lemma 6.1

Let E be a Banach ideal space and φ be a convex Orlicz function. Then,

$( i )$ if $\rho _{\varphi }^{E}(x_{n})\rightarrow 1$, then $\Vert x_{n}\Vert _{\varphi }\rightarrow 1$ for all sequences $( x_{n} )$ in $E_{\varphi }$. In particular,

$( ii )$ if $\rho _{\varphi }^{E}(x)=1$, then $\Vert x\Vert _{\varphi }=1$ for every $x\in E_{\varphi }$.

$( iii )$ if $\Vert x_{n}\Vert _{\varphi }\rightarrow 0$, then $\rho _{\varphi }^{E}(x_{n})\rightarrow 0$ for all sequences $( x_{n} )$ in $E_{\varphi }$.

Suppose additionally that $\varphi \in \Delta _{2}^{E}$. Then,

$( iv )$ if $\Vert x_{n}\Vert _{\varphi }\rightarrow 1$, then $\rho _{\varphi }^{E}(x_{n})\rightarrow 1$ for all sequences $( x_{n} )$ in $E_{\varphi }$. In particular,

$( v )$ if $\Vert x\Vert _{\varphi }=1$, then $\rho _{\varphi }^{E}(x)=1$ for each $x\in E_{\varphi }$.

Assume additionally that $\varphi \in \Delta _{2}^{E}$ and $a_{\varphi }=0$. Then,

$( vi )$ if $\rho _{\varphi }^{E}(x_{n})\rightarrow 0$, then $\Vert x_{n}\Vert _{\varphi }\rightarrow 0$ for all sequences $( x_{n} )$ in $E_{\varphi }$.

The known proofs of properties $( i )$–$( vi )$ cannot be applied for the nonconvex Orlicz function φ. In [7] and [8] we presented new proofs of the above lemma for the quasimodular and the quasinorm. We have shown that for properties $(i )$ and $( ii )$ we need additionally the condition $\Delta _{\varepsilon }^{E}$, which is a substitute for convexity. Moreover, to show the properties $( iv )$ and $( v )$ the so-called condition $\Delta _{2-str}^{E}$ is required (the condition $\Delta _{2-str}^{E}$ is essentially stronger than $\Delta _{2}^{E}$, in general). Finally, the condition $(iii)$ comes from Lemma 3.7 and the condition $( vi )$ is true in the same form as above (see [8]). Next, applying also some new techniques, we described order isomorphic and order linearly isometric copies of $l^{\infty }$ in the quasinormed Calderón–Lozanovskiĭ spaces $E_{\varphi }$ (a number of theorems describe these copies in the natural language of suitable properties of the quasinormed ideal space E and the nondecreasing Orlicz function φ – see [7]). In [8] we characterized the monotonicity properties of quasinormed Calderón–Lozanovskiĭ spaces $E_{\varphi }$, that is, we described the strict monotonicity, the uniform monotonicity, and the respective orthogonal counterparts of the quasinormed Calderón–Lozanovskiĭ spaces $E_{\varphi }$.

6.2 Open ends

The theory of modular spaces has been widely investigated in [26], see also [24]. Next, the modular spaces equipped with the additional measure structure (called modular function spaces) has been studied in [21]. Furthermore, the authors of [18] considered the modular function spaces from the geometry and the fixed-point theory point of view. All the monographs deal with the modulars (convex modulars) that lead to the F-normed (normed) spaces, respectively. It seems natural to study some aspects of the theory developed in the monographs [18] and [21] in the context of quasimodulars.