Orlicz–Lorentz function spaces equipped with the Orlicz norm

Abstract

We investigate Orlicz–Lorentz function spaces equipped with the Orlicz norm generated by any Orlicz function and any non-increasing weight function. As far as we know, this is the first time such a general research is conducted. First we show some basic properties of the Orlicz norm, including its equality to the Amemiya norm, the problem of attainability of infimum in the definition of the Amemiya norm and a formula for the norm of a characteristic function. Then we find criteria for the order continuity and strict monotonicity and we study the problem of existence of order linearly isometric copies of $l^{\infty }$.

1 Introduction

Orlicz–Lorentz spaces, a natural generalization of Orlicz spaces as well as Lorentz spaces, have been intensely studied for the last 30 years. Attention has focused primarily on the Luxemburg norm (see e.g. [5, 8,9,10,11, 15, 16]). In recent years some papers concerning Orlicz–Lorentz spaces equipped with the Orlicz norm have been published (see [12, 26, 27] and also [5]). However, in the case of these papers, as well as in the case of classical Orlicz spaces equipped with the Orlicz norm, it was assumed that the Orlicz function $\varphi$ is an N-function, that means $\varphi (u)\in (0, \infty )$ for $u\in (0, \infty )$, $\lim \nolimits _{u\rightarrow 0}\frac{\varphi (u)}{u}=0$ and $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi (u)}{u}=\infty$ (see [3, 17, 21, 22, 25]). In the case of Orlicz spaces, there was an exception to that rule in the paper [13], where the authors proved the equality of the Orlicz and Amemiya norms in Orlicz spaces defined for any Orlicz function. In this context it is worth mentioning the paper [4], where the problem of attaining infimum in the definition of the Orlicz–Amemiya norm was considered, provided that $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi (u)}{u}=B<\infty$ and $\lim \nolimits _{u\rightarrow \infty }(Bu-\varphi (u))=\infty$ (that means the function $\varphi$ does not have an oblique asymptote at infinity).

Herein we begin a study of Orlicz–Lorentz function spaces equipped with the Orlicz norm generated by any Orlicz function and any non-increasing weight function.

This paper is organized as follows. In Sect. 2 we recall some basic definitions. In Sect. 3, after proving the equality of the Orlicz and Amemiya norms, we show some basic properties of these spaces, among others the problem of attaining infimum in the definition of the Amemiya norm and a formula for the norm of a characteristic function. According to our knowledge, the above-mentioned problems have not been considered so far by any other authors in such a general way, not even in Orlicz spaces. Hence, we hope that the presented results will provide an impetus toward further investigation of not only Orlicz–Lorentz spaces but also Orlicz spaces equipped with the Orlicz norm in full generality. Finally, in Sects. 4 and 5 we give criteria for the order continuity and strict monotonicity of the considered spaces and we study inclusion of respective copies of $l^{\infty }$.

Analogous results for the case of sequence spaces can be found in the paper [7] (see also [6]); however, in the case of function spaces some different techniques had to be implemented in many components of the proofs.

2 Preliminaries

Let $L^{0}=L^{0}([0,\gamma ))$ be the space of all (equivalence classes of) Lebesgue measurable real-valued functions defined on the interval $[0,\gamma )$, where $0<\gamma \le \infty$. For any $x,y\in L^{0}$, we write $x\le y$, if $x(t)\le y(t)$ almost everywhere with respect to the Lebesgue measure m on the interval $[0,\gamma )$.

Given any $x\in L^{0}$ we define its distribution function $\mu _{x}:[0,+\infty )\rightarrow [0,\gamma ]$ by

$$\begin{aligned} \mu _{x}(\lambda )=m\{t\in [0,\gamma ):|x(t)|>\lambda \} \end{aligned}$$

(see [1, 18] and [19]) and the non-increasing rearrangement $x^{*}:[0,\gamma )\rightarrow [0,\infty ]$ of x as

$$\begin{aligned} x^{*}(t)=\inf \{\lambda \ge 0:\mu _{x}(\lambda )\le t\} \end{aligned}$$

(under the convention $\inf \emptyset =\infty$). We say that two functions $x,y\in L^{0}$ are equimeasurable if $\mu _{x}(\lambda )=\mu _{y}(\lambda )$ for all $\lambda \ge 0$. Then we obviously have $x^{*}=y^{*}$.

In the whole paper $\varphi$ denotes an Orlicz function (see [3, 17, 21, 22, 25]), that is, $\varphi :\left[ -\infty , \infty \right] \rightarrow \left[ 0, \infty \right]$ (our definition is extended from ${\mathbb {R}}$ into ${\mathbb {R}}^{e}=\left[ -\infty , \infty \right]$ by assuming that $\varphi \left( -\infty \right) =\varphi \left( \infty \right) =\infty$) and $\varphi$ is convex, even, vanishing and continuous at zero, left continuous on $\left( 0, \infty \right)$ and not identically equal to zero on $\left( -\infty , \infty \right)$. Denote

$$\begin{aligned} a_{\varphi }&=\sup \{ u\ge 0: \varphi \left( u\right) =0\},\\ b_{\varphi }&=\sup \{ u\ge 0: \varphi \left( u\right) <\infty \}. \end{aligned}$$

Note that the left continuity of $\varphi$ on $\left( 0, \infty \right)$ is equivalent to the fact that $\lim \nolimits _{u\rightarrow \left( b_{\varphi }\right) ^{-}} \varphi \left( u\right) =\varphi \left( b_{\varphi }\right)$.

Recall that an Orlicz function $\varphi$ satisfies the condition $\Delta _{2}$ for all $u\in {\mathbb {R}}$ ($\varphi \in \Delta _{2}({\mathbb {R}})$ for short) if there exists a constant $K>0$ such that the inequality

$$\begin{aligned} \varphi (2u)\le K\varphi (u) \end{aligned}$$

(1)

holds for any $u\in {\mathbb {R}}$ (then we have $a_{\varphi }=0$ and $b_{\varphi }=\infty$). Analogously, we say that an Orlicz function $\varphi$ satisfies the condition $\Delta _{2}$ at infinity [at zero] ($\varphi \in \Delta _{2}({\infty })$ [$\varphi \in \Delta _{2}({0})$] for short) if there exist constants $K,u_{0}\in (0,\infty )$ such that $\varphi (u_{0})<\infty$ [$\varphi (u_{0})>0$] and inequality (1) holds for any $u\ge u_{0}$ [$u\le u_{0}$] (then we have $b_{\varphi }=\infty$ [$a_{\varphi }=0$]).

For any Orlicz function $\varphi$ we define its complementary function in the sense of Young by the formula

$$\begin{aligned} \psi (u)=\sup _{v>0}\{|u|v-\varphi (v)\} \end{aligned}$$

for all $u\in {\mathbb {R}}$. It is easy to show that $\psi$ is also an Orlicz function and that $\varphi$ and $\psi$ satisfy the Young inequality

$$\begin{aligned} uv\le \varphi \left( u\right) +\psi \left( v\right) \quad \text {for any} \quad u, v\ge 0. \end{aligned}$$

(2)

In some proofs in this paper the case when the inequality (2) becomes an equality will be important. Let us define a subdifferential $\partial \varphi \left( u\right)$ of $\varphi$ at $u\ge 0$ as follows:

$$\begin{aligned} \partial \varphi \left( u\right) =\{ v\ge 0: \varphi \left( u\right) +\psi \left( v\right) =uv\}. \end{aligned}$$

(3)

By l and p we denote the left and right derivative of $\varphi$ respectively. Then we have

• (i) if $u\in \left[ 0, b_{\varphi }\right)$, then $\partial \varphi \left( u\right) =\left[ l\left( u\right) , p\left( u\right) \right]$,

• (ii) if $u=b_{\varphi }$ and $l\left( b_{\varphi }\right) <\infty$, then $\partial \varphi \left( b_{\varphi }\right) =\left[ l\left( b_{\varphi }\right) , \infty \right)$,

• (iii) if either ($u=b_{\varphi }$ and $l\left( u\right) =\infty$) or $u>b_{\varphi }$, then $\partial \varphi \left( u\right) =\emptyset$.

Let $\omega :[0,\gamma )\rightarrow {\mathbb {R}}_{+}$ be a non-increasing and locally integrable function, called a weight function. We will assume that $\omega$ is a non-zero function.

Given any Orlicz function $\varphi$ and any weight function $\omega$, we define on $L^{0}$ the convex semimodular $I_{\varphi ,\omega }:L^{0}\rightarrow {\mathbb {R}}_{+}^{e}=[0,\infty ]$ by the formula

$$\begin{aligned} I_{\varphi ,\omega }\left( x\right) =\int _{0}^{\gamma }\varphi (x^{*}(t)) \omega (t)dt=\sup _{\sigma }\int _{0}^{\gamma }\varphi (x(\sigma (t)))\omega (t)dt, \end{aligned}$$

(4)

where $\sigma :[0,\gamma )\rightarrow [0,\gamma )$ denotes a measure preserving transformation (see [1]) and the supremum is extended over all measure preserving transformations from $[0,\gamma )$ into itself (see Remark 2.1). The modular space

$$\begin{aligned} \Lambda _{\varphi ,\omega }([0,\gamma ))=\{x\in L^{0}:I_{\varphi ,\omega } \left( \lambda x\right) <\infty \quad \textrm{for}\quad \textrm{some}\quad \lambda >0\}, \end{aligned}$$

generated by $I_{\varphi ,\omega }$, is called the Orlicz–Lorentz function space. In the case when $\gamma <\infty$, we have $L^{\infty }\left( \left[ 0, \gamma \right) \right) \varsubsetneq \Lambda _{\varphi ,\omega }([0,\gamma ))$ whenever $b_{\varphi }=\infty$ and $\Lambda _{\varphi ,\omega }([0,\gamma ))=L^{\infty }\left( \left[ 0, \gamma \right) \right)$ otherwise (see [8, Corollary 1.7(i)]). Let now $\gamma =\infty$. If $b_{\varphi }=\infty$ and ($a_{\varphi }=0$ and $\int _{0}^{\infty }\omega (t)dt=\infty$), then neither of the two inclusions $\Lambda _{\varphi ,\omega }([0,\infty ))\subset L^{\infty }\left( \left[ 0,\infty \right) \right)$ and $L^{\infty }\left( \left[ 0,\infty \right) \right) \subset \Lambda _{\varphi ,\omega }([0,\infty ))$ holds. In this case in particular we have $\lim \nolimits _{t\rightarrow \infty }x^{*}(t)=0$ for any $x\in \Lambda _{\varphi ,\omega }([0,\infty ))$ (equivalently $\mu _{x}\left( \lambda \right) <\infty$ for any $\lambda >0$). If $b_{\varphi }<\infty$, we have $\Lambda _{\varphi ,\omega }([0,\infty ))\subset L^{\infty }\left( \left[ 0,\infty \right) \right)$, whereas if $a_{\varphi }>0$ or $\int _{0}^{\infty }\omega (t)dt<\infty$, then $L^{\infty }\left( \left[ 0,\infty \right) \right) \subset \Lambda _{\varphi ,\omega }([0,\infty ))$ (see [8, Corollary 1.7(ii)]).

The subspace $\Lambda _{\varphi , \omega }^{\theta }\left( \left[ 0, \gamma \right) \right)$ of the space $\Lambda _{\varphi ,\omega }([0,\gamma ))$ is defined by the formula

$$\begin{aligned} \Lambda _{\varphi , \omega }^{\theta }\left( \left[ 0, \gamma \right) \right) =\{ x\in L_{0}: I_{\varphi , \omega }\left( \lambda x\right) <\infty \text { for any } \lambda >0 \}. \end{aligned}$$

It is known that $\Lambda _{\varphi , \omega }^{\theta }\left( \left[ 0, \gamma \right) \right)$ is non-trivial if and only if $b_{\varphi }=\infty$ (see [8, Theorem 3.1(i)]). If $b_{\varphi }=\infty$ and ($\gamma <\infty$ or $\gamma =\infty$ and $\int _{0}^{\infty }\omega (t)dt<\infty$), then $L^{\infty }\left( \left[ 0, \gamma \right) \right) \subset \Lambda _{\varphi , \omega }^{\theta }\left( \left[ 0, \gamma \right) \right)$. Moreover, in this case $\Lambda _{\varphi , \omega }^{\theta }\left( \left[ 0, \gamma \right) \right) =\Lambda _{\varphi ,\omega }([0,\gamma ))$ if and only if $\varphi \in \Delta _{2}\left( \infty \right)$ (see [8, Theorem 2.4]). On the other hand, if $b_{\varphi }=\infty$, $\gamma =\infty$ and $\int _{0}^{\infty }\omega (t)dt=\infty$, then the equality $\Lambda _{\varphi , \omega }^{\theta }\left( \left[ 0, \infty \right) \right) =\Lambda _{\varphi , \omega }\left( \left[ 0, \infty \right) \right)$ holds if and only if $\varphi \in \Delta _{2}\left( {\mathbb {R}}\right)$.

As is known, the space $\Lambda _{\varphi ,\omega }([0,\gamma ))$ equipped with the Luxemburg norm defined by the formula

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }=\inf \left\{ \lambda >0: I_{\varphi , \omega }\left( \frac{x}{\lambda }\right) \le 1\right\} \end{aligned}$$

(5)

is a Banach symmetric space with the Fatou property (for basic properties of symmetric spaces we refer to [1, 18, 19]); recall that a Banach lattice $(E, \Vert \cdot \Vert _{E})$ has the Fatou property if for any $x\in L^{0}$ and $(x_{n})_{n=1}^{\infty }$ in $E_{+}$ (the positive cone of E) such that $x_{n}\nearrow \vert x\vert$ m-a.e. and $\sup _{n}\Vert x_{n}\Vert <\infty$, we have $x\in E$ and $\Vert x\Vert =\lim \nolimits _{n\rightarrow \infty }\Vert x_{n}\Vert$ (see [19]).

Similarly as in the case of Orlicz spaces, we can define in $\Lambda _{\varphi ,\omega }([0,\gamma ))$ two additional norms, namely the Orlicz norm

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{O} =\sup \left\{ \int _{0}^{\gamma }x^{*}(t)y^{*}(t) \omega (t)dt: I_{\psi , \omega }\left( y\right) \le 1\right\} \end{aligned}$$

(6)

and the Amemiya norm

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{A}=\inf _{k>0}\frac{1}{k} \left\{ 1+I_{\varphi , \omega }\left( kx\right) \right\} . \end{aligned}$$

(7)

Remark 2.1

Let $\varphi$ be any Orlicz function and $\omega$ any non-negative, locally integrable and non-trivial weight function (not necessarily non-increasing). If $a_{\varphi }=b_{\varphi }$, then $\Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right) =L^{\infty }\left( \left[ 0, \gamma \right) \right)$ and $\Vert x\Vert _{\varphi , \omega }=\Vert x\Vert _{\varphi , \omega }^{O}=\Vert x\Vert _{\varphi , \omega }^{A}=\Vert x\Vert _{L^{\infty }}/b_{\varphi }$ for any $x\in \Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right)$. If $a_{\varphi }<b_{\varphi }$, then the following conditions are equivalent: the weight function $\omega$ is non-increasing, the second equality in (4) holds true and the semimodular $I_{\varphi , \omega }$ is convex (see [8, Theorem 1.2]). The convexity of the semimodular $I_{\varphi , \omega }$ is in turn an essential condition to prove that the functionals defined by (5) and (7) are norms. The assumption about the non-increasing weight function is also important while proving that the functional defined by (6) is a norm.

3 Some basic results

For any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$ we define the function

$$\begin{aligned} f(k)=f_{x}(k)=\frac{1}{k}\left( 1+I_{\varphi ,\omega }(kx)\right) \end{aligned}$$

(8)

for any $k\in (0,\infty )$, and the constant

$$\begin{aligned} \lambda _{\infty }=\lambda _{\infty }(x)=\sup \left\{ \lambda >0 : I_{\varphi ,\omega }(\lambda x)<\infty \right\} . \end{aligned}$$

(9)

The function f is continuous on the interval $(0,\lambda _{\infty })$ and left continuous at $\lambda _{\infty }$ if $\lambda _{\infty }<\infty$.

Lemma 3.1

The following statements hold true:

• (i) If $k<\lambda _{\infty }=\lambda _{\infty }(x)$, then $I_{\psi ,\omega }\left( p\left( kx^{*}\right) \right) <\infty$.

• (ii) Let $\Vert x\Vert _{\varphi ,\omega }^{O}\le 1$. Additionally, in the case when $x^*(0)=a_{\varphi }>0$ and $p(a_{\varphi })>0$, we assume $x^*(t)<a_{\varphi }$ for every $t\in (0,\gamma )$. Then $I_{\psi ,\omega }\left( p\left( x^{*}\right) \right) \le 1$.

• (iii) If $\Vert x\Vert _{\varphi ,\omega }^{O}\le 1$, then $I_{\varphi , \omega }\left( x\right) \le \Vert x\Vert _{\varphi ,\omega }^{O}$.

• (iv) For any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$ we have $\Vert x\Vert _{\varphi , \omega }\le \Vert x\Vert _{\varphi ,\omega }^{O}\le 2\Vert x\Vert _{\varphi ,\omega }$.

• (v) For any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$ we have $\Vert x\Vert _{\varphi , \omega }\le \Vert x\Vert _{\varphi ,\omega }^{A}\le 2\Vert x\Vert _{\varphi , \omega }$.

Proof

(i) First let us notice that if $b_{\varphi }<\infty$, then by the assumption $k<\lambda _{\infty }$ we obtain $kx^{*}\left( 0\right) <b_{\varphi }$ and consequently $p\left( kx^{*}\left( 0\right) \right) <\infty$. Simultaneously, by $b_{\varphi }<\infty$, we have $b_{\psi }=\infty$, whence $\psi \left( p\left( kx^{*}\left( 0\right) \right) \right) <\infty$.

Assume to the contrary that

$$\begin{aligned} I_{\psi ,\omega }\left( p\left( kx^{*}\right) \right) =\infty . \end{aligned}$$

(10)

If there exists $t_{0}\in \left( 0, \gamma \right)$ such that $\int _{0}^{t_{0}}\psi \left( p\left( kx^{*}(t)\right) \right) \omega (t)dt=\infty$, then $\psi \left( p\left( kx^{*} \left( 0\right) \right) \right) =\infty$ and by the previous remark, we conclude that $b_{\varphi }=\infty$ and $x^{*}\left( 0\right) =\infty$. Without loss of generality we may assume that $x^{*}(t)>x^{*}(t_{0})$ for $t<t_{0}$. Let us define $A=\{t\in \left[ 0, \gamma \right) : \vert x(t)\vert >x^{*} \left( t_{0}\right) \}$ and $y=x\chi _{A}$. We have $y^{*}(t)=x^{*}(t)$ for any $t\in \left[ 0,t_{0}\right)$, whence $\int _{0}^{t_{0}} \psi \left( p\left( ky^{*}(t)\right) \right) \omega (t)dt=\infty$. Define

$$\begin{aligned} y_{n}(t)=\left\{ \begin{array}{ll} y(t) &{}\quad \text{ if }\ |y(t)|\le n, \\ n{{\,\textrm{sgn}\,}}(y(t)) &{}\quad \text{ otherwise }, \end{array}\right. \end{aligned}$$

(11)

for any $t\in A$ and $n\ge n_{0}$, where $n_{0}$ is the smallest integer greater than $x^{*}\left( t_{0}\right)$. Since $\vert y_{n}\vert \nearrow \vert y\vert$, we have $y^{*}_{n}\nearrow y^{*}$, whence $I_{\varphi , \omega }\left( ly_{n}\right) \nearrow I_{\varphi , \omega }\left( ly\right)$ for any $l\in \left( 0, \lambda _{\infty }\right)$. Moreover, by the definition of the sequence $(y_{n})_{n=n_{0}}^{\infty }$, we conclude that there exists a non-increasing and converging to zero sequence $(t_{n})_{n=n_{0}}^{\infty }$ (more precisely $t_{n}=m\{t\in A: |y(t)|>n\}$ for any $n\ge n_{0}$) such that $y^{*}_{n}(t)=y^{*}(t)$ for any $n\ge n_{0}$ and any $t\in [t_{n},t_{0})$. Thus $p(ky^{*}_{n}(t))=p(ky^{*}(t))$ for any $n\ge n_{0}$ and any $t\in [t_{n},t_{0})$, whence $I_{\psi ,\omega }\left( p\left( ky_{n}^{*}\right) \right) \nearrow I_{\psi ,\omega }\left( p\left( ky^{*}\right) \right)$. In particular, there exists $n_{1}\ge n_{0}$ such that $I_{\psi , \omega }\left( p\left( ky_{n}^{*}\right) \right) >1$ for any $n\ge n_{1}$. For any fixed $h\in \left( k, \lambda _{\infty }\right)$ and any $n\ge n_{1}$ we have

$$\begin{aligned} I_{\varphi , \omega }\left( hy_{n}\right)&\ge \int \limits _{0}^{t_{0}}hy_{n}^{*}(t)p \left( ky_{n}^{*}(t)\right) \omega (t)dt -I_{\psi , \omega } \left( p\left( ky_{n}^{*}\right) \right) ,\nonumber \\ I_{\varphi , \omega }\left( ky_{n}\right)&=\int \limits _{0}^{t_{0}}ky_{n}^{*}(t)p \left( ky_{n}^{*}(t)\right) \omega (t)dt -I_{\psi , \omega }\left( p\left( ky_{n}^{*}\right) \right) . \end{aligned}$$

(12)

Therefore, for any $n\ge n_{1}$ we obtain

$$\begin{aligned}&\frac{1}{h}\left( 1+I_{\varphi , \omega }\left( hy_{n}\right) \right) -\frac{1}{k} \left( 1+I_{\varphi , \omega }\left( ky_{n}\right) \right) \nonumber \\&\quad =\frac{h-k}{h\cdot k}\left( -1+\frac{k}{h-k} \left( I_{\varphi , \omega }\left( hy_{n}\right) -I_{\varphi , \omega }\left( ky_{n}\right) \right) -I_{\varphi , \omega }\left( ky_{n}\right) \right) \nonumber \\&\quad \ge \frac{h-k}{h\cdot k}\left( -1+\frac{k}{h-k} \int \limits _{0}^{t_{0}}\left( h-k\right) y_{n}^{*} (t)p \left( ky_{n}^{*}(t)\right) \omega (t)dt-I_{\varphi , \omega }\left( ky_{n}\right) \right) \nonumber \\&\quad =\frac{h-k}{h\cdot k}\left( I_{\psi , \omega } \left( p\left( ky_{n}^{*}\right) \right) -1\right) >0, \end{aligned}$$

(13)

whence for the same n we get

$$\begin{aligned} I_{\psi , \omega }\left( p\left( ky_{n}^{*}\right) \right) \le \frac{k}{h-k}\left( 1+I_{\varphi , \omega }\left( hy\right) \right) +1<\infty , \end{aligned}$$

which gives a contradiction with the fact that $I_{\psi , \omega }\left( p\left( ky_n^{*}\right) \right) \nearrow I_{\psi , \omega }\left( p\left( ky^{*}\right) \right) =\infty$.

Suppose now that $\int _{0}^{t_{0}}\psi \left( p \left( kx^{*}(t)\right) \right) \omega (t)dt<\infty$ for any $t_{0}\in \left[ 0, \gamma \right)$. By (10) we have $\gamma =\infty$, $\int _{0}^{\infty }\omega (t)dt=\infty$ and $\lim \nolimits _{t\rightarrow \infty }kx^{*}(t)\ge a_{\varphi }$. Using again the assumption $k<\lambda _{\infty }$, we obtain $\lim \nolimits _{t\rightarrow \infty }kx^{*}(t)=a_{\varphi }=0$. Define $B_{n}=\{t\in [0,\infty ): |x(t)|\ge 1/n\}$ and $x_{n}=x\chi _{B_{n}}$ for $n\in {\mathbb {N}}$. We have $m(B_{n})<\infty$, $x_{n}\nearrow x$, whence $x_{n}^{*}\nearrow x^{*}$ and $I_{\varphi ,\omega }\left( lx_{n}\right) \nearrow I_{\varphi ,\omega }\left( lx\right)$ for any $l\in \left( 0, \lambda _{\infty }\right)$. By the definition of $x_{n}$, we have $x^{*}_{n}(t)=x^{*}(t)$ for any $n\in {\mathbb {N}}$ and any $t\in [0,m(B_{n}))$. Hence $p(kx^{*}_{n}(t))=p(kx^{*}(t))$ for any $n\in {\mathbb {N}}$ and any $t\in [0,m(B_{n}))$ and, in consequence, $I_{\psi ,\omega }\left( p\left( kx_{n}^{*} \right) \right) \nearrow I_{\psi ,\omega }\left( p \left( kx^{*}\right) \right)$. Therefore, there exists $n_{2}\in {\mathbb {N}}$ such that $I_{\psi ,\omega } \left( p\left( kx_{n}^{*}\right) \right) >1$ for any $n\ge n_{2}$. Take any fixed $h\in \left( k, \lambda _{\infty }\right)$. Proceeding in the same way as in (12) and (13) for any $n\ge n_{2}$, we get

$$\begin{aligned} I_{\psi , \omega }\left( p\left( kx_{n}^{*}\right) \right) \le \frac{k}{h-k}\left( 1+I_{\varphi , \omega }\left( hx\right) \right) +1<\infty , \end{aligned}$$

which gives a contradiction again.

(ii) If $x=0$, then by equalities $\psi \left( p\left( 0\right) \right) =\psi \left( a_{\psi }\right) =0$ we have $I_{\psi , \omega }\left( p(x^{*})\right) =0$. Assume now that $x\ne 0$ and $\Vert x\Vert _{\varphi ,\omega }^{O}\le 1$.

First let $0<x^{*}\left( 0\right) \le a_{\varphi }$. Then, by the condition $x^*(t)<a_{\varphi }$ for every $t\in (0,\gamma )$ in the case when $x^*(0)=a_{\varphi }>0$ and $p(a_{\varphi })>0$, we get $I_{\psi ,\omega }\left( p(x^{*})\right) =0$.

Suppose now that $x^{*}\left( 0\right) >a_{\varphi }\ge 0$ and assume to the contrary that $I_{\psi , \omega }\left( p(x^{*})\right) >1$. If $I_{\psi ,\omega }\left( p(x^{*})\right) <\infty$ then, using the convexity of the modular, we get

$$\begin{aligned} I_{\psi , \omega }\left( \frac{p(x^{*})}{I_{\psi , \omega } \left( p(x^{*})\right) }\right) \le \frac{1}{I_{\psi , \omega } \left( p(x^{*})\right) }\cdot I_{\psi , \omega }\left( p(x^{*})\right) =1. \end{aligned}$$

(14)

Therefore,

$$\begin{aligned} 1&\ge \Vert x\Vert _{\varphi ,\omega }^{O} \ge \int \limits _{0}^{\gamma }x^{*}(t) \frac{p\left( x^{*}(t)\right) }{I_{\psi ,\omega } \left( p(x^{*})\right) }\omega (t)dt\nonumber \\&=\frac{1}{I_{\psi ,\omega }\left( p(x^{*})\right) } \int \limits _{0}^{\gamma }x^{*}(t)p\left( x^{*}(t)\right) \omega (t)dt\nonumber \\&=\frac{1}{I_{\psi ,\omega }\left( p(x^{*})\right) } \left( I_{\varphi , \omega }\left( x\right) +I_{\psi ,\omega } \left( p(x^{*})\right) \right) >1, \end{aligned}$$

(15)

which gives a contradiction, leading us to the conclusion that $I_{\psi ,\omega }\left( p(x^{*})\right) \le 1$.

Let now $I_{\psi ,\omega }\left( p\left( x^{*}\right) \right) =\infty$. First we will show that if $b_{\varphi }<\infty$, then $x^{*}(t)<b_{\varphi }$ for any $t\in (0,\gamma )$. Assume for the contrary that there exists $t_{0}\in (0,\gamma )$ such that $x^{*}(t_{0})\ge b_{\varphi }$. Without loss of generality we can assume that $\omega (t_{0})>0$. We will consider two cases.

Let us assume first that $a_{\varphi }=b_{\varphi }$. By the assumption $x^{*}(0)>a_{\varphi }$ and by the right continuity of the non-increasing rearrangement, we conclude that there exists $t_{1}\in \left( 0, t_{0}\right]$ such that $x^{*}(t_{1})>b_{\varphi }$. At the same time we have $\psi (u)=b_{\varphi }\cdot u$, hence, we can find $u_{0}$ such that

$$\begin{aligned} \int _{0}^{t_{0}}\psi (u_{0})\omega (t)dt=\int _{0}^{t_{0}}b_{\varphi } \cdot u_{0}\cdot \omega (t)dt=1. \end{aligned}$$

Defining $y=u_{0}\chi _{[0, t_{0})}$, we get $y^{*}=y$, $I_{\psi , \omega }(y)=1$ and

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{O}\ge \int _{0}^{t_{0}}x^{*}(t) y^{*}(t)\omega (t)dt>\int _{0}^{t_{0}}b_{\varphi }\cdot u_{0} \cdot \omega (t)dt=1, \end{aligned}$$

which gives a contradiction with the assumption $\Vert x\Vert _{\varphi , \omega }^{O}\le 1$.

Suppose now that $a_{\varphi }<b_{\varphi }$. We have $\psi (u)<b_{\varphi }\cdot u$ for any $u\in {\mathbb {R}}$. Since $b_{\psi }=\infty$, there exists $u_{0}$ such that

$$\begin{aligned} \int _{0}^{t_{0}}\psi (u_{0})\omega (t)dt=1. \end{aligned}$$

Defining $y=u_{0}\chi _{[0, t_{0})}$ again, we have $y^{*}=y$ and $I_{\psi , \omega }(y)=1$. Simultaneously

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{O}\ge \int _{0}^{t_{0}} x^{*}(t)y^{*}(t)\omega (t)dt\ge \int _{0}^{t_{0}}b_{\varphi } \cdot u_{0}\cdot \omega (t)dt>\int _{0}^{t_{0}}\psi (u_{0})\omega (t)dt=1, \end{aligned}$$

which again leads to a contradiction. In conclusion, if $b_{\varphi }<\infty$, then $x^{*}(0)\le b_{\varphi }$ and $x^{*}(t)<b_{\varphi }$ for any $t\in (0,\gamma )$.

Assume now that there exists $t_{0}\in \left( 0,\gamma \right)$ such that $\int _{0}^{t_{0}}\psi \left( p\left( x^{*}(t)\right) \right) \omega (t)dt=\infty$, then $x^{*}\left( 0\right) =b_{\varphi } \le \infty$. Without loss of generality we may assume that $x^{*}(t_{0})<x^{*}(t)<x^{*}(0)$ for any $t\in (0,t_{0})$. Defining $A=\{ t\in \left[ 0, \gamma \right) : \vert x(t)\vert >x^{*}\left( t_{0}\right) \}$ and $y=x\chi _{A}$, analogously as in (i), we have $y^{*}(t)=x^{*}(t)$ for any $t\in \left[ 0,t_{0}\right)$ and $\int _{0}^{t_{0}}\psi \left( p\left( y^{*}(t)\right) \right) \omega (t)dt=\infty$.

Let first $b_{\varphi }=\infty$. Then for the sequence $(y_{n})_{n=n_{0}}^{\infty }$, defined by the formula (11), we get $|y_{n}|\le |y|\le |x|$ for $n\ge n_{0}$, whence $\Vert y_{n}\Vert _{\varphi ,\omega }^{O}\le \Vert x\Vert _{\varphi ,\omega }^{O}\le 1$ for the same n. Simultaneously, $I_{\psi ,\omega }\left( p \left( y_{n}^{*}\right) \right) \nearrow I_{\psi ,\omega } \left( p\left( y^{*}\right) \right)$. So, there exists $n_{1}\ge n_{0}$ such that $1<I_{\psi ,\omega }\left( p\left( y_{n}^{*} \right) \right) <\infty$ for $n\ge n_{1}$. Proceeding analogously as in formulas (14) and (15), we get $\Vert y_{n}\Vert _{\varphi ,\omega }^{O}>1$ for $n\ge n_{1}$, which gives a contradiction.

Suppose now that $b_{\varphi }<\infty$ and define

$$\begin{aligned} z_{n}(t)=\left\{ \begin{array}{ll} y(t) &{} \quad \text{ if }\ |y(t)|\le x^{*}(0)-\frac{1}{n}, \\ \left( x^{*}(0)-\frac{1}{n}\right) {{\,\textrm{sgn}\,}}(y(t)) &{}\quad \text{ otherwise }, \end{array}\right. \end{aligned}$$

for any $t\in A$ and $n\ge n_{2}$, where $n_{2}$ is the smallest natural number such that $x^{*}(0)-\frac{1}{n_{2}}>x^{*} \left( t_{0}\right)$. Obviously $|z_{n}|\le |y|\le |x|$ for $n\ge n_{2}$, whence $\Vert z_{n}\Vert _{\varphi ,\omega }^{O}\le \Vert x\Vert _{\varphi , \omega }^{O}\le 1$ for the same n. Moreover, defining $t_{n}:=m\{t\in A: |y(t)|>x^{*}(0)-\frac{1}{n}\}$ for $n\ge n_{2}$, we get $\lim \nolimits _{n\rightarrow \infty }t_{n}=0$ and $z^{*}_{n}(t)=y^{*}(t)$ for any $n\ge n_{2}$ and any $t\in [t_{n},t_{0})$. In consequence, $I_{\psi ,\omega }\left( p\left( z_{n}^{*}\right) \right) \nearrow I_{\psi ,\omega }\left( p\left( y^{*}\right) \right)$, whence $1<I_{\psi ,\omega }\left( p\left( z_{n}^{*}\right) \right) <\infty$ starting from a certain $n_{3}\ge n_{2}$. Using again (14) and (15), we obtain $\Vert z_{n}\Vert _{\varphi ,\omega }^{O}>1$ for $n\ge n_{3}$, which again leads to a contradiction.

Finally assume that $\int _{0}^{t_{0}}\psi \left( p\left( x^{*}(t) \right) \right) \omega (t)dt<\infty$ for any $t_{0}\in \left[ 0,\gamma \right)$. We have $\gamma =\infty$, $\int _{0}^{\infty }\omega (t)dt=\infty$ and $\lim \nolimits _{t\rightarrow \infty }x^{*}(t)\ge a_{\varphi }$. Defining $w_{n}=x^{*}\chi _{[0,n)}$ for $n\in {\mathbb {N}}$, we have $0\le w_{n}=w_{n}^{*}\le x^{*}$ and $\Vert w_{n}\Vert _{\varphi ,\omega }^{O}\le \Vert x^{*}\Vert _{\varphi ,\omega }^{O}\le 1$ for the same n. Simultaneously, $w_{n}(t)=w_{n}^{*}(t)=x^{*}(t)$ for $n\in {\mathbb {N}}$ and $t\in [0,n)$ and, in consequence, $I_{\psi ,\omega }\left( p\left( w_{n}^{*}\right) \right) \nearrow I_{\psi ,\omega }\left( p\left( x^{*}\right) \right)$. Therefore, we can find $n_{4}\in {\mathbb {N}}$ such that $1<I_{\psi ,\omega }\left( p\left( w_{n}^{*}\right) \right) <\infty$ for $n\ge n_{4}$. Hence, by (14) and (15), we have $\Vert w_{n}\Vert _{\varphi ,\omega }^{O}>1$ for $n\ge n_{4}$, which gives again a contradiction.

(iii) Let $\Vert x\Vert _{\varphi , \omega }^{O}\le 1$. If $0\le x^{*}(0)\le a_{\varphi }$, then we immediately get $I_{\varphi , \omega }(x)=0\le \Vert x\Vert _{\varphi , \omega }^{O}$. Suppose now that $x^{*}(0)>a_{\varphi }\ge 0$. Then by the proof of (ii), we have $I_{\psi , \omega }\left( p\left( x^{*}\right) \right) \le 1$, whence by the equality in the Young inequality (see (3)), we get

$$\begin{aligned} I_{\varphi , \omega }(x)\le I_{\varphi , \omega }(x) +I_{\psi , \omega }\left( p(x^{*})\right) =\int _{0}^{\gamma } x^{*}(t)p(x^{*}(t))\omega (t)dt\le \Vert x\Vert _{\varphi , \omega }^{O}. \end{aligned}$$

(16)

(iv) By (iii) the inequality $I_{\varphi , \omega }\left( \frac{x}{\Vert x\Vert _{\varphi , \omega }^{O}}\right) \le \left\| \frac{x}{\Vert x\Vert _{\varphi , \omega }^{O}}\right\| _{\varphi ,\omega }^{O}=1$ holds true, whence we immediately get $\Vert x\Vert _{\varphi , \omega }\le \Vert x\Vert _{\varphi , \omega }^{O}$. On the other hand, by the Young inequality and left-hand continuity of the modular, we have

$$\begin{aligned} \left\| \frac{x}{\Vert x\Vert _{\varphi , \omega }} \right\| _{\varphi , \omega }^{O}\le I_{\varphi , \omega } \left( \frac{x}{\Vert x\Vert _{\varphi , \omega }}\right) +1\le 2, \end{aligned}$$

whence we eventually get $\Vert x\Vert _{\varphi , \omega }^{O}\le 2\Vert x\Vert _{\varphi , \omega }$.

(v) Finally, by $\Vert x\Vert _{\varphi , \omega }=\inf \nolimits _{k>0}\{\max \left( \frac{1}{k}, \frac{1}{k}I_{\varphi , \omega }(kx)\right) \}$, we obtain $\Vert x\Vert _{\varphi , \omega }\le \Vert x\Vert _{\varphi ,\omega }^{A}\le 2\Vert x\Vert _{\varphi ,\omega }$. $\square$

Remark 3.1

By Lemma 3.1(iv), we conclude that the space $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right) , \Vert \cdot \Vert _{\varphi , \omega }^{O}\right)$ is complete, therefore, it is a Banach lattice as well as a Banach symmetric space. By Beppo Levi’s theorem, it also has the Fatou property. Analogous conclusions for the space $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right) , \Vert \cdot \Vert _{\varphi , \omega }^{A}\right)$ follow from Lemma 3.1(v).

Remark 3.2

In the example below we will show that in the case when $\lambda _{\infty }=\lambda _{\infty }\left( x\right) <\infty$ the value of $I_{\psi , \omega }\left( p\left( \lambda _{\infty }(x)\cdot x^{*}\right) \right)$ is not necessarily finite. We will also prove that if $x^*(0)=x^*(t)=a_{\varphi }>0$ for some $t\in (0,\gamma )$ and $p\left( a_{\varphi }\right) >0$, then the implication $\Vert x\Vert _{\varphi , \omega }^{O}\le 1\Rightarrow I_{\psi , \omega }\left( p(x^{*})\right) \le 1$ does not need to be true.

Example 3.1

Define

$$\begin{aligned}{} & {} \varphi _{1}\left( u\right) =0\quad \textrm{for}\quad u\in \left[ 0, 1\right) \quad \textrm{and}\quad \varphi _{1} \left( u\right) =\left( u-1\right) ^2\quad \textrm{for}\quad u\ge 1, \\{} & {} \varphi _{2}\left( u\right) =0\quad \textrm{for}\quad u\in \left[ 0,1\right) \quad \textrm{and}\quad \varphi _{2} \left( u\right) =u^2-1\quad \textrm{for}\quad u\ge 1 \end{aligned}$$

and $\omega (t)=1$ for any $t\in [0,\gamma )$. We have $a_{\varphi _{1}}=a_{\varphi _{2}}=1$,

$$\begin{aligned} p_{1}\left( u\right) =0\quad \textrm{for}\quad u\in \left[ 0, 1\right) \quad \textrm{and}\quad p_{1}\left( u\right) =2\left( u-1\right) \quad \textrm{for}\quad u\ge 1, \end{aligned}$$

and

$$\begin{aligned} p_{2}\left( u\right) =0\quad \textrm{for}\quad u\in \left[ 0, 1\right) \quad \textrm{and}\quad p_{2}\left( u\right) =2u \quad \textrm{for}\quad u\ge 1. \end{aligned}$$

If $\gamma =\infty$ and $x=\chi _{[0,\infty )}$, then $\lambda _{\infty }=\lambda _{\infty }(x)=1$ for both functions $\varphi _{1}$ and $\varphi _{2}$ and $\Vert x\Vert _{\varphi _{1}, \omega }^{O}=\Vert x\Vert _{\varphi _{2}, \omega }^{O}=1$. Simultaneously, $I_{\psi _{1}, \omega }\left( p_{1}(x^{*})\right) =0$ and $I_{\psi _{2}, \omega }\left( p_{2}(x^{*})\right) =\infty$.

Let now $\gamma \ge 1$, $y=\chi _{[0,1/2)}$ and $z=\chi _{[0,1)}$. We have $\Vert y\Vert _{\varphi _{2}, \omega }^{O}=\Vert z\Vert _{\varphi _{2}, \omega }^{O}=1$ (obviously $\lambda _{\infty }(y)=\lambda _{\infty }(z)=\infty$), $I_{\psi _{2},\omega }\left( p_{2}\left( y^{*}\right) \right) =1$ and $I_{\psi _{2}, \omega }\left( p_{2}\left( z^{*}\right) \right) =2$.

In the first theorem we will show that the Orlicz and Amemiya norms are equal for any Orlicz function $\varphi$ and any weight function $\omega$. Let us recall that Hudzik and Maligranda proved the above-mentioned equality in Orlicz spaces generated by any Orlicz function in [13]. In that paper the reader will also find many important facts concerning all three norms in Orlicz spaces, including the history of research on relations between them.

Theorem 3.1

The Orlicz and Amemiya norms in an Orlicz–Lorentz space $\Lambda _{\varphi ,\omega }([0,\gamma ))$ are equal, that means the equality $\Vert x\Vert _{\varphi , \omega }^{O}=\Vert x\Vert _{\varphi , \omega }^{A}$ holds for any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$.

Proof

It is easy to show that the Orlicz norm is not bigger than Amemiya norm for any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$. In fact, by the Young inequality (2), for any $y\in \Lambda _{\psi , \omega }\left( \left[ 0, \gamma \right) \right)$, $I_{\psi , \omega }\left( y\right) \le 1$, and for any $k>0$ we have

$$\begin{aligned} \int \limits _{0}^{\gamma }x^{*}(t)y^{*}(t)\omega (t)dt&=\frac{1}{k} \int \limits _{0}^{\gamma }k x^{*}(t)y^{*}(t)\omega (t)dt\\&\le \frac{1}{k}\int \limits _{0}^{\gamma }(\varphi (kx^{*}(t)) +\psi (y^{*}(t)))\omega (t)dt\le \frac{1}{k}(1+I_{\varphi , \omega }(kx)). \end{aligned}$$

Hence, $\Vert x\Vert _{\varphi , \omega }^{O}\le \frac{1}{k}\left( 1+I_{\varphi , \omega }\left( kx\right) \right)$ for any $k>0$ and it follows that $\Vert x\Vert _{\varphi , \omega }^{O}\le \Vert x\Vert _{\varphi , \omega }^{A}$.

The converse inequality will be proved for the non-negative simple functions with finite supports first. Set

$$\begin{aligned} x=\sum \limits _{i=1}^{n}a_{i}\chi _{E_{i}}, \end{aligned}$$

where $E_{i}$ are pairwise disjoint subsets of $\left[ 0, \gamma \right)$ with $m\left( E_{i}\right) <\infty$ and $\infty>a_{1}>a_{2}>\ldots>a_{n}>0$. Defining $e_{0}=0$ and $e_{i}=\sum _{j=1}^{i}m\left( E_{i}\right)$ for $i=1, 2, \ldots , n$, we have

$$\begin{aligned} x^{*}=\sum \limits _{i=1}^{n}a_{i}\chi _{\left[ e_{i-1}, e_{i}\right) }. \end{aligned}$$

We will consider three cases.

Case 1. Assume that $b_{\varphi }=\infty$ and $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi \left( u\right) }{u}=\lim \nolimits _{u\rightarrow \infty }p\left( u\right) =\infty$. First, we will show that there exist $k_{1}>0$ and a non-increasing function $y_{1}$ such that $y_{1}(t)\in \left[ l\left( k_{1}x^{*}(t)\right) , p\left( k_{1}x^{*}(t)\right) \right]$ for any $t\in \left[ 0,m\left( {{\,\textrm{supp}\,}}x\right) \right)$ and $I_{\psi ,\omega }\left( y_{1}\right) =1$. Obviously $I_{\psi ,\omega }\left( p(0\cdot x^{*})\right) =0$ and $I_{\psi ,\omega }(p(kx^{*}))$ is a non-decreasing function of $k\ge 0$. Since $I_{\varphi ,\omega }(kx)<\infty$ for any $k\ge 0$, by Lemma 3.1(i), we have $I_{\psi ,\omega }\left( p(kx^{*})\right) <\infty$ for the same k. Simultaneously, by $\lim \nolimits _{u\rightarrow \infty }p\left( u\right) =\infty$, we have $\lim \nolimits _{k\rightarrow \infty }I_{\psi ,\omega }\left( p(kx^{*}) \right) =\infty$. Therefore, denoting

$$\begin{aligned} k_{1}=\sup \{k>0: I_{\psi ,\omega }\left( p(kx^{*}\right) \le 1\}, \end{aligned}$$

we obtain $k_{1}\in (0,\infty )$. Since p is right continuous, we have $1\le I_{\psi ,\omega }\left( p(k_{1}x^{*})\right) <\infty$. At the same time $I_{\psi ,\omega }\left( l(kx^{*})\right) \le I_{\psi ,\omega }\left( p(kx^{*})\right) \le 1$ for any $k\in (0, k_{1})$, whence by the left continuity of l, we get $I_{\psi ,\omega }\left( l(k_{1}x^{*})\right) \le 1$. If $I_{\psi ,\omega }\left( p(k_{1}x^{*})\right) =1$, then $y_{1}:=p(k_{1}x^{*})$. Analogously, if $I_{\psi ,\omega }\left( p(k_{1}x^{*})\right) >1=I_{\psi ,\omega }\left( l(k_{1}x^{*})\right)$, then $y_{1}:=l(k_{1}x^{*})$. Finally, assume $I_{\psi ,\omega }\left( p(k_{1}x^{*})\right)>1 >I_{\psi ,\omega }\left( l(k_{1}x^{*})\right)$. Let $i_{1}$ $(i_{1}=1,\ldots ,n)$ be the smallest number such that

$$\begin{aligned} \sum \limits _{i=1}^{i_{1}}\int _{e_{i-1}}^{e_{i}} \psi (p(k_{1}a_{i}))\omega (t)dt+\sum \limits _{i=i_{1}+1}^{n} \int _{e_{i-1}}^{e_{i}}\psi (l(k_{1}a_{i}))\omega (t)dt\ge 1 \end{aligned}$$

Then we find $u\in [l(k_{1}a_{i_{1}}),p(k_{1}a_{i_{1}})]$ such that

$$\begin{aligned} \sum \limits _{i=1}^{i_{1}-1}\int _{e_{i-1}}^{e_{i}}\psi (p(k_{1}a_{i})) \omega (t)dt+\int _{e_{i_{1}-1}}^{e_{i_{1}}}\psi (u)\omega (t)dt +\sum \limits _{i=i_{1}+1}^{n}\int _{e_{i-1}}^{e_{i}} \psi (l(k_{1}a_{i}))\omega (t)dt=1, \end{aligned}$$

and define

$$\begin{aligned} y_{1}:=\sum \limits _{i=1}^{i_{1}-1}p(k_{1}a_{i}) \chi _{\left[ e_{i-1},e_{i}\right) }+u\chi _{[e_{i_{1}-1}, e_{i_{1}})}+\sum \limits _{i=i_{1}+1}^{n}l(k_{1}a_{i}) \chi _{\left[ e_{i-1},e_{i}\right) }. \end{aligned}$$

Hence

$$\begin{aligned} \Vert x\Vert _{\varphi ,\omega }^{O}&\ge \int \limits _{0}^{\gamma } x^{*}(t)y_{1}^{*}(t)\omega (t)dt=\frac{1}{k_{1}} \int \limits _{0}^{\gamma }k_{1} x^{*}(t)y_{1}^{*}(t)\omega (t)dt\nonumber \\&=\frac{1}{k_{1}}\int \limits _{0}^{\gamma }(\varphi (k_{1}x^{*}(t)) +\psi (y_{1}^{*}(t)))\omega (t)dt=\frac{1}{k_{1}}(1+I_{\varphi , \omega }(k_{1}x)) \ge \Vert x\Vert _{\varphi ,\omega }^{A}. \end{aligned}$$

(17)

Case 2. Let now $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi \left( u\right) }{u}=\lim \nolimits _{u\rightarrow \infty }p\left( u\right) =B<\infty$. Then we have $b_{\varphi }=\infty$ and $b_{\psi }=B$.

Case 2.1. Assume that $\psi \left( B\right) \int _{0}^{m \left( {{\,\textrm{supp}\,}}x\right) }\omega (t)dt>1$ (notice that this inequality holds in particular if $\psi \left( B\right) =\infty$; then $p\left( u\right) <\lim \nolimits _{n\rightarrow \infty }p\left( u\right) =B$ for any $u\ge 0$ and in consequence $\psi \left( p\left( u\right) \right) <\infty$ for the same u). Since $a_{\psi }<b_{\psi }=B$, there exists $\delta >0$ such that $\psi \left( B-\delta \right) \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\omega (t)dt>1$. Let $u_{\delta }>0$ be such that $p\left( u\right) \ge B-\delta$ for any $u\ge u_{\delta }$. Since for $k_{x}>\frac{u_{\delta }}{a_{n}}$ we have $k_{x}x^{*}(t)\ge k_{x} a_{n}>u_{\delta }$ for $t\in \left[ 0, m\left( {{\,\textrm{supp}\,}}x\right) \right)$, we get

$$\begin{aligned} I_{\psi , \omega }\left( p\left( k_{x}x\right) \right) =\int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\psi \left( p \left( k_{x}x^{*}(t)\right) \right) \omega (t)dt\ge \psi \left( B-\delta \right) \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\omega (t)dt>1. \end{aligned}$$

Simultaneously

$$\begin{aligned} I_{\psi , \omega }\left( p\left( k_{x}x\right) \right) \le \psi \left( p\left( k_{x}a_{1}\right) \right) \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\omega (t)dt<\infty . \end{aligned}$$

Therefore, we can find $k_{2}\in \left( 0, k_{x}\right]$ and a non-increasing function $y_{2}$ such that $y_{2}(t)\in \left[ l\left( k_{2}x^{*}(t)\right) , p\left( k_{2}x^{*}(t)\right) \right]$ for any $t\in \left[ 0, m\left( {{\,\textrm{supp}\,}}x\right) \right)$ and $I_{\psi , \omega }\left( y_{2}\right) =1$. Proceeding analogously as in (17), we get $\Vert x\Vert _{\varphi , \omega }^{O}\ge \Vert x\Vert _{\varphi , \omega }^{A}$ again.

Case 2.2. Suppose now that $\psi \left( B\right) \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\omega (t)dt\le 1$ (this inequality holds in particular if $\psi \left( B\right) =0$, that means $a_{\psi }=b_{\psi }=B$). We have $\psi \left( B\right) <\infty$ and $\psi \left( u\right) =\infty$ for any $u>B$. Hence, for any $y\in \Lambda _{\psi , \omega }\left( \left[ 0, \gamma \right) \right)$ such that $I_{\psi , \omega }\left( y\right) \le 1$ we have $y^{*}\left( 0\right) \le B$ and this gives

$$\begin{aligned} \int _{0}^{\gamma }x^{*}(t)y^{*}(t)\omega (t)dt\le B \int _{0}^{\gamma }x^{*}(t)\omega (t)dt. \end{aligned}$$

(18)

Simultaneously, for $z:=B\chi _{\left[ 0, m\left( {{\,\textrm{supp}\,}}x\right) \right) }$ we have

$$\begin{aligned} I_{\psi , \omega }\left( z\right) =\int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) } \psi \left( B\right) \omega (t)dt\le 1 \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{\gamma }x^{*}(t)z^{*}(t)\omega (t)dt=B\int _{0}^{\gamma } x^{*}(t)\omega (t)dt. \end{aligned}$$

(19)

By (18) and (19), we get $\Vert x\Vert _{\varphi , \omega }^{O}=B\int _{0}^{\gamma }x^{*}(t)\omega (t)dt$. Since

$$\begin{aligned}&\lim _{k\rightarrow \infty }\frac{1}{k}I_{\varphi , \omega } \left( kx\right) =\lim _{k\rightarrow \infty }\frac{1}{k} \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\varphi \left( kx^{*}(t) \right) \omega (t)dt \nonumber \\&\quad =\lim _{k\rightarrow \infty }\frac{1}{k}\sum _{i=1}^{n} \int _{e_{i-1}}^{e_{i}}\varphi \left( ka_{i}\right) \omega (t)dt =\sum _{i=1}^{n}a_{i}\lim _{k\rightarrow \infty }\frac{\varphi \left( ka_{i}\right) }{ka_{i}}\int _{e_{i-1}}^{e_{i}}\omega (t)dt \nonumber \\&\quad =B\sum _{i=1}^{n}\int _{e_{i-1}}^{e_{i}}a_{i}\cdot \omega (t)dt =B\int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }x^{*}(t)\omega (t)dt=\Vert x \Vert _{\varphi , \omega }^{O}, \end{aligned}$$

(20)

we obtain eventually

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{A}\le \lim _{k\rightarrow \infty } \frac{1}{k}\left( 1+I_{\varphi , \omega }\left( kx\right) \right) =\Vert x\Vert _{\varphi , \omega }^{O}. \end{aligned}$$

Case 3. Finally we assume that $b_{\varphi }<\infty$. We will consider two subcases.

Case 3.1. If $l\left( b_{\varphi }\right) =\infty$, then $\lim \nolimits _{u\rightarrow \left( b_{\varphi }\right) ^{-}}p \left( u\right) =\infty$. Therefore, we will find $k_{x}=\left( 0, \frac{b_{\varphi }}{a_{1}}\right)$ such that $\psi \left( p \left( k_{x}a_{1}\right) \right) \int _{0}^{e_{1}}\omega (t)dt>1$. Moreover,

$$\begin{aligned} I_{\psi , \omega }(p(k_{x}x))\le \psi (p(k_{x}a_{1})) \int _{0}^{m({{\,\textrm{supp}\,}}x)}\omega (t)dt<\infty . \end{aligned}$$

Hence, there exist $k_{3}>0$ and a non-increasing function $y_{3}$ such that $y_{3}(t)\in \left[ l\left( k_{3}x^{*}(t)\right) , p\left( k_{3}x^{*}(t)\right) \right]$ for any $t\in \left[ 0, m\left( {{\,\textrm{supp}\,}}x\right) \right)$ and $I_{\psi , \omega }\left( y_{3}\right) =1$. Proceeding in the same way as in (17), we get $\Vert x\Vert _{\varphi , \omega }^{O}\ge \Vert x\Vert _{\varphi , \omega }^{A}$.

Case 3.2. Let now $l\left( b_{\varphi }\right) <\infty$ and $k_{x}=\frac{b_{\varphi }}{a_{1}}$. If $I_{\psi , \omega }\left( l\left( k_{x}x\right) \right) \ge 1$, we proceed analogously as in Case 3.1. Assume now that $I_{\psi , \omega }\left( l\left( k_{x}x\right) \right) <1$. Since $\psi$ assumes finite values on the whole ${\mathbb {R}}$, we will find $\beta >l\left( b_{\varphi }\right)$ such that

$$\begin{aligned} \psi \left( \beta \right) \int _{0}^{e_{1}}\omega (t)dt +\int _{e_{1}}^{m\left( {{\,\textrm{supp}\,}}x\right) }\psi \left( l\left( k_{x}x^{*}(t)\right) \right) \omega (t)dt=1. \end{aligned}$$

Defining

$$\begin{aligned} z=\beta \chi _{\left[ 0, e_{1}\right) }+\sum _{i=2}^{n}l \left( k_{x}a_{i}\right) \chi _{\left[ e_{i-1}, e_{i}\right) }, \end{aligned}$$

we get $z^{*}=z$, $I_{\psi , \omega }\left( z\right) =1$ and

$$\begin{aligned} k_{x}x^{*}(t)z^{*}(t)=\varphi \left( k_{x}x^{*}(t)\right) +\psi \left( z^{*}(t)\right) \end{aligned}$$

for any $t\in \left[ 0, m\left( {{\,\textrm{supp}\,}}x\right) \right)$. Hence,

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{O}&\ge \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) } x^{*}(t)z^{*}(t)\omega (t)dt=\frac{1}{k_{x}} \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }k_{x}x^{*}(t)z^{*}(t)\omega (t)dt \\&=\frac{1}{k_{x}}\int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\left( \varphi \left( k_{x}x^{*}(t)\right) +\psi \left( z^{*}(t)\right) \right) \omega (t)dt=\frac{1}{k_{x}}\left( 1+I_{\varphi , \omega } \left( k_{x}x\right) \right) \ge \Vert x\Vert _{\varphi , \omega }^{A}. \end{aligned}$$

Finally we take any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$. Since $\Vert x\Vert _{\varphi , \omega }^{O}=\Vert \vert x\vert \Vert _{\varphi , \omega }^{O}$ and $\Vert x\Vert _{\varphi , \omega }^{A}=\Vert \vert x\vert \Vert _{\varphi , \omega }^{A}$, we can assume without loss of generality that x is non-negative. Then there exists a sequence $\left( x_{m}\right)$ of non-negative simple functions such that $x_{m}\subset \left[ 0,\min \left( \gamma , m\right) \right)$ for $m\in {\mathbb {N}}$ and $x_{m}\nearrow x$ m-a.e. By the previous part of the proof and the Fatou property for both norms (see Remark 3.1), we get

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{O}=\lim _{m\rightarrow \infty } \Vert x_{m}\Vert _{\varphi , \omega }^{O}\ge \lim _{m\rightarrow \infty } \Vert x_{m}\Vert _{\varphi , \omega }^{A}=\Vert x\Vert _{\varphi , \omega }^{A} \end{aligned}$$

and this finishes the proof. $\square$

Remark 3.3

In the paper [13] the reader will find many examples of Orlicz functions suitable for any case considered in the Theorem above.

For our further consideration it will be important to answer the question if the infimum in the formula (7) is attained for any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))\setminus \{\theta \}$; equivalently, if there exists $k=k\left( x\right) >0$ such that

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{O}=\Vert x\Vert _{\varphi ,\omega }^{A} =\frac{1}{k}\{ 1+I_{\varphi , \omega }\left( kx\right) \}. \end{aligned}$$

In order to answer this question, for any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$ we define two constants

$$\begin{aligned} k^{*}&=k^{*}\left( x\right) =\inf \{ k>0: I_{\psi , \omega } \left( p\left( kx^{*}\right) \right) \ge 1\}, \\ k^{**}&=k^{**}\left( x\right) =\sup \{ k>0: I_{\psi , \omega } \left( p\left( kx^{*}\right) \right) \le 1\}. \end{aligned}$$

First we present two lemmas which short proofs will be shown for the sake of completeness.

Lemma 3.2

For any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))\setminus \{\theta \}$ we have

$$\begin{aligned} 0<k^{*}\le k^{**}\le \lambda _{\infty }\le \infty , \end{aligned}$$

where $\lambda _{\infty }$ is defined by the formula (9).

Proof

We will start with the proof of the inequality $0<k^{*}$. Defining $k_{1}=\min \left( 1, \frac{1}{\Vert x\Vert _{\varphi , \omega }^{O}}\right)$, we have $\Vert k_{1}x\Vert _{\varphi , \omega }^{O}\le 1$. If $0<k_{1}\cdot x^{*}\left( 0\right) \le a_{\varphi }$, then for any $k\in (0,k_{1})$ we have $I_{\psi , \omega }\left( p\left( kx^{*}\right) \right) =0$, whence we conclude that $k^{*}\ge k_{1}$. Suppose now that $k_{1}\cdot x^{*}\left( 0\right) >a_{\varphi }\ge 0$. By Lemma 3.1(ii), we get $I_{\psi ,\omega }\left( p\left( k_{1}x^{*}\right) \right) \le 1$. If $I_{\psi , \omega }\left( p\left( k_{1}x^{*}\right) \right) <1$, then we have immediately $k_{1}\le k^{*}$. Assume now that $I_{\psi , \omega }\left( p\left( k_{1}x^{*}\right) \right) =1$. Then there exists $t_{0}>0$ such that $\omega (t_{0})>0$ and $\psi \left( p\left( k_{1}x^{*}\left( t_{0}\right) \right) \right) >\psi \left( p\left( 0\right) \right) =\psi \left( a_{\psi }\right) =0$ and in consequence $p\left( k_{1}x^{*}\left( \frac{t_{0}}{2}\right) \right) \ge p\left( k_{1}x^{*}\left( t_{0}\right) \right) >p \left( 0\right) =a_{\psi }$. Since p is right continuous, there exists $k_{2}\in \left( 0, k_{1}\right)$ such that $p\left( k_{2}x^{*}\left( \frac{t_{0}}{2}\right) \right) \le \frac{1}{2}\left( p\left( 0\right) +p\left( k_{1}x^{*} \left( t_{0}\right) \right) \right)$. Hence,

$$\begin{aligned} \psi \left( p\left( k_{2}x^{*}(t)\right) \right) \le \psi \left( p\left( k_{2}x^{*}\left( \frac{t_{0}}{2}\right) \right) \right) <\psi \left( p\left( k_{1}x^{*}\left( t_{0}\right) \right) \right) \le \psi \left( p\left( k_{1}x^{*}(t)\right) \right) \end{aligned}$$

for any $t\in \left[ \frac{t_{0}}{2}, t_{0}\right]$. This gives $I_{\psi ,\omega }\left( p\left( k_{2}x^{*}\right) \right) <1$, whence $k^{*}\ge k_{2}$.

Assume for the contrary, that $k^{**}<k^{*}$. Let $k_{0}\in (k^{**},k^{*})$ be arbitrary. Recall that $I_{\psi ,\omega }\left( p\left( kx^{*}\right) \right)$ is a non-decreasing function of $k\ge 0$. Therefore, by $k^{**}<k_{0}$, we get $I_{\psi ,\omega }\left( p\left( k_{0}x^{*}\right) \right) >1$. Simultaneously, by $k_{0}<k^{*}$, we have $I_{\psi ,\omega }\left( p\left( k_{0}x^{*}\right) \right) <1$, which gives a contradiction. Finally, we obtain $k^{*}\le k^{**}$.

Now we will show the inequality $k^{**}\le \lambda _{\infty }$. Take any fixed $k<k^{**}$. Since $I_{\psi , \omega }\left( p\left( kx^{*}\right) \right) \le 1$, we have

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{O}&\ge \int _{0}^{\gamma }x^{*}(t)p \left( kx^{*}(t)\right) \omega (t)dt=\frac{1}{k}\int _{0}^{\gamma }kx^{*}(t)p \left( kx^{*}(t)\right) \omega (t)dt \\&=\frac{1}{k}\left( I_{\varphi , \omega }\left( kx\right) +I_{\psi , \omega } \left( p\left( kx^{*}\right) \right) \right) , \end{aligned}$$

whence in particular we get $I_{\varphi , \omega }\left( kx\right) <\infty$. By the arbitrariness of $k<k^{**}$, we receive the respective inequality. The inequality $\lambda _{\infty }\le \infty$ is obvious. $\square$

Lemma 3.3

Let $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$. Then the function $f\left( k\right) =f_{x}\left( k\right)$ (see formula (8)) is strictly decreasing on the interval $\left( 0, k^{*}\right)$ and strictly increasing on the interval $\left( k^{**}, \lambda _{\infty }\right)$ whenever $k^{**}<\lambda _{\infty }$. Moreover, if $k^{*}<k^{**}$, then the function $f\left( k\right)$ is constant on the interval $\left[ k^{*}, k^{**}\right]$ and $f\left( k\right) =\Vert x\Vert _{\varphi , \omega }^{O}$ for any $k\in \left[ k^{*}, k^{**}\right]$ (if $k^{**}=\infty$, then we write $\left[ k^{*}, k^{**}\right)$ in place of $\left[ k^{*}, k^{**}\right]$ respectively).

Proof

As we noticed before, the function $f_{x}$ is continuous on the interval $\left( 0, \lambda _{\infty }\right)$ and left-continuous at $\lambda _{\infty }$ whenever $\lambda _{\infty }<\infty$. Assume that $0<k_{1}<k_{2}<k^{*}$. Since $k^{*}\le \lambda _{\infty }$, we have $I_{\varphi , \omega }\left( k_{1}x\right) <\infty$ and $I_{\varphi , \omega }\left( k_{2}x\right) <\infty$. From the definition of $k^{*}$ it follows directly that $I_{\psi , \omega }\left( p\left( k_{2} x^{*}\right) \right) <1$. Moreover,

$$\begin{aligned} I_{\varphi , \omega }\left( k_{1}x\right)&\ge \int \limits _{0}^{\gamma }k_{1}x^{*}(t) p\left( k_{2}x^{*}(t)\right) \omega (t)dt -I_{\psi , \omega }\left( p\left( k_{2}x^{*}\right) \right) ,\\ I_{\varphi , \omega }\left( k_{2}x\right)&=\int \limits _{0}^{\gamma }k_{2}x^{*}(t)p \left( k_{2}x^{*}(t)\right) \omega (t)dt -I_{\psi , \omega }\left( p\left( k_{2}x^{*}\right) \right) . \end{aligned}$$

Thus,

$$\begin{aligned} f(k_{1})-f(k_{2})&=\frac{1}{k_{1}}\left( 1+I_{\varphi , \omega }\left( k_{1}x\right) \right) -\frac{1}{k_{2}} \left( 1+I_{\varphi , \omega }\left( k_{2}x\right) \right) \\&=\frac{k_{2}-k_{1}}{k_{1}\cdot k_{2}}\left( 1+\frac{k_{2}}{k_{2}-k_{1}} \left( I_{\varphi , \omega }\left( k_{1}x\right) -I_{\varphi , \omega }\left( k_{2}x\right) \right) +I_{\varphi , \omega }\left( k_{2}x\right) \right) \\&\ge \frac{k_{2}-k_{1}}{k_{1}\cdot k_{2}} \left( 1+\frac{k_{2}}{k_{2}-k_{1}}\int \limits _{0}^{\gamma } \left( k_{1}-k_{2}\right) x^{*}(t)p\left( k_{2}x^{*}(t)\right) \omega (t)dt+I_{\varphi , \omega }\left( k_{2}x\right) \right) \\&=\frac{k_{2}-k_{1}}{k_{1}\cdot k_{2}}\left( 1-I_{\psi , \omega } \left( p\left( k_{2}x^{*}\right) \right) \right) >0. \end{aligned}$$

By the arbitrariness of $k_{1}$ and $k_{2}$, we conclude that the function $f\left( k\right)$ is strictly decreasing on the interval $\left( 0, k^{*}\right)$.

Suppose now that $k^{**}<\lambda _{\infty }$. Let $k^{**}<k_{2}<k_{1}<\lambda _{\infty }$. We have $I_{\varphi , \omega }\left( k_{2}x\right) <\infty$ and $I_{\varphi , \omega }\left( k_{1}x\right) <\infty$. Moreover, by Lemma 3.1(i) and definition of $k^{**}$, we have $1<I_{\varphi , \omega }\left( p\left( k_{2}x^{*}\right) \right) <\infty$. Proceeding analogously as above (cf. also (13)), we get

$$\begin{aligned} f\left( k_{1}\right) -f\left( k_{2}\right) \ge \frac{k_{1}-k_{2}}{k_{1} \cdot k_{2}}\left( I_{\psi , \omega }\left( p\left( k_{2}x^{*}\right) \right) -1\right) >0, \end{aligned}$$

whence, by the arbitrariness of $k_{1}$ and $k_{2}$ we infer that the function $f\left( k\right)$ is strictly increasing on the interval $\left( k^{**}, \lambda _{\infty }\right)$.

Finally suppose that $k^{*}<k^{**}$. For any $k\in \left( k^{*}, k^{**}\right)$ we have $I_{\psi , \omega }\left( p\left( kx^{*}\right) \right) =1$, whence

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{O}\le f\left( k\right) =\frac{1}{k}\left( I_{\psi , \omega }\left( p\left( kx^{*}\right) \right) +I_{\varphi , \omega }\left( kx\right) \right) =\int \limits _{0}^{\gamma }x^{*}(t)p(kx^{*} (t))\omega (t)dt\le \Vert x\Vert _{\varphi , \omega }^{O}. \end{aligned}$$

By the continuity (left-continuity at $k^{**}$ if $k^{**}=\lambda _{\infty }<\infty$) of the function f we get $f\left( k^{*}\right) =\Vert x\Vert _{\varphi , \omega }^{O}$ and $f\left( k^{**}\right) =\Vert x\Vert _{\varphi , \omega }^{O}$ whenever $k^{**}<\infty$. $\square$

For any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$ we define the set $K\left( x\right)$ by the formula

$$\begin{aligned} K\left( x\right) ={\left\{ \begin{array}{ll} \left[ k^{*}, k^{**}\right] &{}\quad \text {if}\ k^{**}<\infty , \\ \left[ k^{*}, \infty \right) &{} \quad \text {if}\ k^{*}<k^{**}=\infty , \\ \emptyset &{}\quad \text {if}\ k^{*}=\infty . \end{array}\right. } \end{aligned}$$

Theorem 3.2

Let $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$. If $k^{*}<\infty$, then

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{O}=\frac{1}{k} \left( 1+I_{\varphi , \omega }\left( kx\right) \right) \end{aligned}$$

for any $k\in K\left( x\right)$, while in the case when $k^{*}=\infty$ we get

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{O}=\lim _{k\rightarrow \infty } \frac{1}{k}\left( 1+I_{\varphi , \omega }\left( kx\right) \right) . \end{aligned}$$

The proof of this theorem can be obtained immediately from Lemma 3.3.

Remark 3.4

Let us notice that if $b_{\varphi }=\infty$ and $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi \left( u\right) }{u} =\lim \nolimits _{u\rightarrow \infty }p\left( u\right) =\infty$, then for any $x\in \Lambda _{\varphi ,\omega }([0,\gamma )){\setminus }\{\theta \}$ we have $k^{**}<\infty$. In fact, if $\lambda _{\infty }=\infty$, then there exists $t_{x}\in \left( 0, m\left( {{\,\textrm{supp}\,}}x\right) \right)$ for which we can find $k_{x}>0$ such that

$$\begin{aligned} \int _{0}^{t_{x}}\psi \left( p\left( k_{x}x^{*}\left( t_{x}\right) \right) \right) \omega (t)dt>1 \end{aligned}$$

and $k^{**}\le k_{x}$. In the case when $b_{\varphi }<\infty$, for any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))\setminus \{\theta \}$ we have $x^{*}\left( 0\right) \in \left( 0, \infty \right)$ and in consequence we obtain $k^{**}\le \lambda _{\infty }\le \frac{b_{\varphi }}{x^{*}\left( 0\right) }$.

Assume now that $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi \left( u\right) }{u}=\lim \nolimits _{u\rightarrow \infty }p\left( u\right) =B<\infty$ and $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))\setminus \{\theta \}$. If $\psi \left( B\right) \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\omega (t)dt>1$, then $k^{**}<\infty$. In fact, let $t_{x}\in \left( 0, m\left( {{\,\textrm{supp}\,}}x\right) \right)$ be such that $\psi \left( B\right) \int _{0}^{t_{x}}\omega (t)dt>1$. Since $\lim \nolimits _{u\rightarrow \infty }p\left( u\right) =B$, there exists $u_{x}$ such that $\psi \left( p\left( u_{x}\right) \right) )\int _{0}^{t_{x}}\omega (t)dt>1$. Defining $k_{x}=\frac{u_{x}}{x^{*}\left( t_{x}\right) }$, we have $\psi \left( p\left( k_{x}x^{*}\left( t_{x}\right) \right) \right) )\int _{0}^{t_{x}}\omega (t)dt>1$, whence $k^{**}\le k_{x}$.

In the case when $\psi \left( B\right) \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\omega (t)dt<1$ or ($\psi \left( B\right) \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\omega (t)dt=1$ and $p\left( u\right) <B$ for any $u>0$), we get $k^{*}=\infty$, whence $K\left( x\right) =\emptyset$.

Finally suppose that $\psi \left( B\right) \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\omega (t)dt=1$ and $p\left( u\right) =B$, starting with some $u_{0}>0$. Define $\gamma _{0}=\sup \{ t\ge 0: \omega (t)>0\}$. If $\gamma _{0}<m\left( {{\,\textrm{supp}\,}}x\right)$, then assuming $k_{x}=\frac{u_{0}}{x^{*}\left( \gamma _{0}\right) }$, we have $\psi \left( p\left( k_{x}x^{*}\left( \gamma _{0}\right) \right) \right) \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\omega (t)dt=1$ and in consequence $k^{*}\le \frac{u_{0}}{x^{*}\left( \gamma _{0}\right) }<k^{**}=\infty$. Let now $m\left( {{\,\textrm{supp}\,}}x\right) \le \gamma _{0}$. If $\lim \nolimits _{t\rightarrow m\left( {{\,\textrm{supp}\,}}x\right) ^{-}}x^{*}(t)>0$, then proceeding analogously as above, we obtain $k^{*}<k^{**}=\infty$. Otherwise we get $K\left( x\right) =\emptyset$.

To sum up, if $\psi \left( B\right) \int _{0}^{\gamma }\omega (t)dt<1$ or ($\psi \left( B\right) \int _{0}^{\gamma }\omega (t)dt=1$ and $p\left( u\right) <B$ for any $u>0$), then for any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$ we have $k^{*}=\infty$ or equivalently $K\left( x\right) =\emptyset$. In particular, this is the case when $\psi \left( B\right) =0$. On the other hand, if $\psi \left( B\right) =\infty$, then for any $x\in \Lambda _{\varphi ,\omega }([0,\gamma )){\setminus }\{\theta \}$ we have $k^{**}<\infty$. Let us notice that the equality $\psi \left( B\right) =\infty$ is equivalent to the condition $\lim \nolimits _{u\rightarrow \infty }\left( Bu-\varphi \left( u\right) \right) =\infty$ (and this means that the graph of the function $\varphi$ has no oblique asymptote at infinity). Chen, Cui and Hudzik in the paper [4] were the first to notice the importance of the problem whether the finite limit $\lim \nolimits _{u\rightarrow \infty }\left( Bu-\varphi \left( u\right) \right)$ exists or not.

Below are some examples of functions for which the above cases apply.

Example 3.2

(i):

Let $\varphi \left( u\right) =au$ for $u\ge 0$, $a>0$. Then $p\left( u\right) =B=a$ for $u\ge 0$ and $\psi \left( B\right) =0$. Hence, for any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$ we have $K\left( x\right) =0$.

(ii):

In the case when $\varphi \left( u\right) =u-1+e^{-u}$ for $u\ge 0$, we get $p\left( u\right) =1-e^{-u}<1$ for $u\ge 0$, $B=1$ and $\psi \left( B\right) =1$. Thus, if $\int _{0}^{\gamma }\omega (t)dt\le 1$, then we have $K\left( x\right) =\emptyset$ for any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$. If $\int _{0}^{\gamma }\omega (t)dt>1$, then we find $x, y\in \Lambda _{\varphi ,\omega }([0,\gamma )){\setminus }\{\theta \}$ such that $K\left( x\right) =\emptyset$ and $k^{**}\left( y\right) <\infty$.

(iii):

Now we assume that $\varphi \left( u\right) =u^2$ for $u\in \left[ 0, 1\right]$ and $\varphi \left( u\right) =2u-1$ for $u>1$. We have $p\left( u\right) =2$ for $u\ge 1$, $B=2$ and $\psi \left( B\right) =1$. If $\int _{0}^{\gamma }\omega (t)dt<1$, then $K\left( x\right) =\emptyset$ for any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))$. On the other hand, if $\int _{0}^{\gamma }\omega (t)dt\ge 1$, then there exist $x, y\in \Lambda _{\varphi ,\omega }([0,\gamma )){\setminus }\{\theta \}$ such that $K\left( x\right) =\emptyset$ and $K\left( y\right) \ne \emptyset$. In particular, if $\gamma =\infty$, $\omega (t)=1$ and $x_{t}=\chi _{[0,t)}$ for any $t>0$, then $K\left( x_{t}\right) =\emptyset$ for $t<1$, $K\left( x_{t}\right) =[1,\infty )$ for $t=1$ and $K\left( x_{t}\right) =\{1/\sqrt{t}\}$ whenever $t>1$.

(iv):

Lastly, we assume that $\varphi \left( u\right) =u+1-\left( u+1\right) ^{\frac{1}{2}}$ (see [4, Remark 2]). Then $p\left( u\right) =1-\frac{1}{2} \left( u+1\right) ^{-\frac{1}{2}}$, $B=1$ and $\psi \left( B\right) =\lim \nolimits _{u\rightarrow \infty }\left( u-\varphi \left( u\right) \right) =\lim \nolimits _{u\rightarrow \infty }\left( \left( u+1 \right) ^{\frac{1}{2}}-1\right) =\infty$. Hence, we have $k^{**}<\infty$ for any $x\in \Lambda _{\varphi ,\omega }([0,\gamma ))\setminus \{\theta \}$.

Lemma 3.4

Let $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi \left( u\right) }{u}=B<\infty$. Then for any $x\in \Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right)$ such that $\psi \left( B\right) \int _{0}^{m\left( {{\,\textrm{supp}\,}}x\right) }\omega (t)dt\le 1$ (equivalently $k^{**}=\infty )$ we have

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{O}=B\int _{0}^{\gamma }x^{*}(t)\omega (t)dt. \end{aligned}$$

(21)

Moreover, if $\gamma =\infty$ and $\psi \left( B\right) \int _{0}^{\infty }\omega (t)dt>1$, then $m\left( {{\,\textrm{supp}\,}}x\right) <\infty$.

Proof

By the assumptions we have $\psi \left( B\right) <\infty$ and $\psi \left( u\right) =\infty$ for any $u>B$. Therefore, for any $y\in \Lambda _{\psi , \omega }\left( \left[ 0, \gamma \right) \right)$ satisfying the condition $I_{\psi , \omega }\left( y\right) \le 1$, we have $y(t)\le B$ for m-a.e. $t\in \left[ 0,\gamma \right)$. Hence, we obtain

$$\begin{aligned} \int _{0}^{\gamma }x^{*}(t)y^{*}(t)\omega (t)dt\le B\int _{0}^{\gamma }x^{*}(t)\omega (t)dt. \end{aligned}$$

Simultaneously, for $z:=B\chi _{\left[ 0, m\left( {{\,\textrm{supp}\,}}x\right) \right) }$ we get $I_{\psi ,\omega }\left( z\right) \le 1$ and

$$\begin{aligned} \int _{0}^{\gamma }x^{*}(t)z^{*}(t)\omega (t)dt =B\int _{0}^{\gamma }x^{*}(t)\omega (t)dt. \end{aligned}$$

This yields (21). Now assume additionally that $\gamma =\infty$ and $\psi \left( B\right) \int _{0}^{\infty }\omega (t)dt>1$. Then $a_{\psi }<b_{\psi }=B$ and $0<\psi \left( B\right) <\infty$. Moreover, there exists $t_{0}\in \left[ 0, \infty \right)$ such that $\psi \left( B\right) \int _{0}^{t_{0}}\omega (t)dt>1$, whence we get $m\left( {{\,\textrm{supp}\,}}x\right) <t_{0}$. $\square$

Lemma 3.5

Let $A\subset \left[ 0,\gamma \right)$ be a Lebesgue measurable set such that $\int _{0}^{m(A)}\omega (t)dt<\infty$. If $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi \left( u\right) }{u}=B<\infty$ and $\psi \left( B\right) \int _{0}^{m(A)}\omega (t)dt\le 1$, then

$$\begin{aligned} \Vert \chi _{A}\Vert _{\varphi , \omega }^{O}=B \int _{0}^{m\left( A\right) }\omega (t)dt. \end{aligned}$$

(22)

Otherwise we have

$$\begin{aligned} \Vert \chi _{A}\Vert _{\varphi , \omega }^{O}=\psi ^{-1} \left( \frac{1}{\int _{0}^{m\left( A\right) }\omega (t)dt}\right) \int _{0}^{m\left( A\right) }\omega (t)dt, \end{aligned}$$

(23)

where $\psi ^{-1}$ is the generalized inverse of the function $\psi$ defined by $\psi ^{-1}(v)=\inf \{u\ge 0:\psi (u)>v\}$ for $v\in [0,\infty )$ and $\psi ^{-1}(\infty )=\lim _{v\rightarrow \infty }\psi ^{-1}(v)$ (see [23]).

Proof

Take any Lebesgue measurable set $A\subset \left[ 0,\gamma \right)$ such that $\int _{0}^{m(A)}\omega (t)dt<\infty$ (if $\gamma =\infty$ and $\int \nolimits _{0}^{\infty }\omega (t)d(t)=\infty$, then $m(A)<\infty$). We will consider three cases.

Case 1. First suppose that $b_{\varphi }=\infty$ and $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi \left( u\right) }{u}=\infty$. We have $k^{**}=k^{**}\left( \chi _{A}\right) <\infty$ (see Remark 3.4). If $k^{*}<k^{**}$, then for any $k_{0}\in \left( k^{*}, k^{**}\right)$ we have

$$\begin{aligned} I_{\psi , \omega }\left( p\left( k_{0}\chi _{A}\right) \right) =\psi \left( p\left( k_{0}\right) \right) \int _{0}^{m\left( A\right) } \omega (t)dt=1, \end{aligned}$$

(24)

whence

$$\begin{aligned} p\left( k_{0}\right) =\psi ^{-1}\left( \frac{1}{\int _{0}^{m\left( A\right) } \omega (t)dt}\right) . \end{aligned}$$

(25)

By the equality in the Young’s inequality (see (3)) and equalities (24) and (25), we obtain

$$\begin{aligned} \Vert \chi _{A}\Vert _{\varphi , \omega }^{O}&=\frac{1}{k_{0}} \left\{ 1+I_{\varphi , \omega }\left( k_{0}\chi _{A}\right) \right\} =\frac{1}{k_{0}}\left\{ 1+\varphi \left( k_{0}\right) \int _{0}^{m\left( A\right) }\omega (t)dt\right\} \\&=\frac{1}{k_{0}}\left\{ 1+\left( k_{0}p\left( k_{0}\right) -\psi \left( p\left( k_{0}\right) \right) \right) \int _{0}^{m\left( A\right) }\omega (t)dt\right\} \\&=\frac{1}{k_{0}}\left( 1+k_{0}p\left( k_{0}\right) \int _{0}^{m\left( A\right) }\omega (t)dt-1\right) =p\left( k_{0}\right) \int _{0}^{m\left( A\right) }\omega (t)dt \\&=\psi ^{-1}\left( \frac{1}{\int _{0}^{m\left( A\right) } \omega (t)dt}\right) \int _{0}^{m\left( A\right) }\omega (t)dt. \end{aligned}$$

If $k^{*}=k^{**}$ and $I_{\psi , \omega }\left( p\left( k^{**}\chi _{A}\right) \right) =1$, we proceed analogously as above, replacing $k_{0}$ with $k^{**}$. In the case when $k^{*}=k^{**}$ and $I_{\psi , \omega }\left( p\left( k^{**}\chi _{A}\right) \right) >1$, we get $I_{\psi , \omega }\left( l\left( k^{**}\chi _{A}\right) \right) \le 1$. Thus, there exists $v\in \partial \varphi \left( k^{**}\right) =\left[ l\left( k^{**}\right) , p\left( k^{**}\right) \right]$ (see (3)) such that $\psi \left( v\right) \int _{0}^{m\left( A\right) }\omega (t)dt=1$. Similarly as above, we obtain (23).

Case 2. Assume that $b_{\varphi }<\infty$. Then $k^{*}\le k^{**}\le b_{\varphi }$ (see Remark 3.4). If $k^{*}<b_{\varphi }$, then proceeding in the same way as in Case 1, we get again (23). Suppose now that $k^{*}=k^{**}= b_{\varphi }$. Then we have $I_{\psi , \omega }\left( l\left( k^{**}\chi _{A}\right) \right) \le 1$ and it follows that there exists $w\in \partial \varphi \left( k^{**}\right) =\left[ l\left( b_{\varphi }\right) ,\infty \right)$ (see (3)) such that $\psi \left( w\right) \int _{0}^{m\left( A\right) }\omega (t)dt=1$. Eventually we obtain (23) again.

Case 3. Finally we assume that the equality $\lim \nolimits _{n\rightarrow \infty }\frac{\varphi \left( u\right) }{u}=B<\infty$ holds. If $\psi \left( B\right) \int _{0}^{m\left( A\right) }\omega (t)dt> 1$, then $k^{**}<\infty$ and repeating the proof of Case 1, we get (23). On the other hand, if $\psi \left( B\right) \int _{0}^{m\left( A\right) }\omega (t)dt\le 1$, it follows from Lemma 3.4 that the equality (22) holds. $\square$

Remark 3.5

Let us notice that if $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi \left( u\right) }{u}=B<\infty$ and $\psi \left( B\right) \int _{0}^{m\left( A\right) }\omega (t)dt=1$, then we have

$$\begin{aligned} \Vert \chi _{A}\Vert _{\varphi ,\omega }^{O}=B \int _{0}^{m\left( A\right) }\omega (t)dt=\psi ^{-1} \left( \frac{1}{\int _{0}^{m\left( A\right) }\omega (t)dt}\right) \int _{0}^{m\left( A\right) }\omega (t)dt, \end{aligned}$$

whereas if $\psi \left( B\right) \int _{0}^{m\left( A\right) } \omega (t)dt<1$, then, in virtue of the equality $B=b_{\psi }$, we conclude that the function $\psi$ does not attain the value $\frac{1}{\int _{0}^{m\left( A\right) }\omega (t)dt}$.

4 Order continuity and separability

Let $E\subset L^{0}$ be a Banach ideal space. An element $x\in E$ is said to be order continuous if for any sequence $\left( x_{n}\right)$ in $E_{+}$ (the positive cone of E) with $x_{n}\le \vert x\vert$ and $x_{n}\rightarrow 0$ m-a.e. we have $\Vert x_{n}\Vert _{E}\rightarrow 0$. The subspace $E_{a}$ of all order continuous elements in E is an order ideal in E. The space E is called order continuous if $E_{a}=E$ (see [19]). Since the Lebesgue measure space $([0,\gamma ),\Sigma ,m)$ is separable, the Banach ideal space E is separable if and only if it is order continuous.

Since both the Luxemburg and the Orlicz norms are equivalent (see Lemma 3.1(iv)), the subspace of order continuous elements for both of these norms is the same (as a subset of the elements of the space $\Lambda _{\varphi ,\omega }([0,\gamma ))$). We will denote it as $(\Lambda _{\varphi ,\omega }([0,\gamma )))_a$. The subspace $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right) \right) _{a}$ is non-trivial if and only if $b_{\varphi }=\infty$ (see [8, Theorem 3.1(i)]). Let $b_{\varphi }=\infty$. Then $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right) \right) _{a} =\Lambda ^{\theta }_{\varphi , \omega }\left( \left[ 0, \gamma \right) \right)$ if $\gamma <\infty$ or $\gamma =\infty$ and $\int _{0}^{\infty }\omega (t)dt=\infty$. In the case when $\gamma =\infty$ and $\int _{0}^{\infty }\omega (t)dt<\infty$ we have $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right) \right) _{a}\varsubsetneq \Lambda ^{\theta }_{\varphi , \omega }\left( \left[ 0, \gamma \right) \right)$ (see [8, Theorem 3.1(ii)–(iv)]).

By [15, Theorem 2.4] and the equivalence of the Luxemburg and Orlicz norms, we obtain the following

Theorem 4.1

If $\gamma <\infty$, then the space $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right) , \Vert \cdot \Vert _{\varphi , \omega }^{O}\right)$ is order continuous (equivalently: separable) if and only if $\varphi \in \Delta _{2}\left( \infty \right)$. In the case when $\gamma =\infty$ the space $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right) , \Vert \cdot \Vert _{\varphi , \omega }^{O}\right)$ is order continuous (equivalently: separable) if and only if $\int _{0}^{\infty }\omega (t)dt=\infty$ and $\varphi \in \Delta _{2}\left( {\mathbb {R}}\right)$.

5 Copies of $l^{\infty }$ and strict monotonicity

It is well known that Banach ideal space $\left( E,\Vert \cdot \Vert _{E}\right)$ is not order continuous if and only if it contains an order isomorphic copy of $l^{\infty }$ (see [20]). Moreover, if $\left( E, \Vert \cdot \Vert _{E}\right)$ is an Orlicz–Lorentz space equipped with the Luxemburg norm, then we can obtain a stronger result, namely $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right) , \Vert \cdot \Vert _{\varphi , \omega }\right)$ is not order continuous if and only if it contains an order linearly isometric copy of $l^{\infty }$ (see [15, Theorem 2.4]). The last statement does not hold if we consider the Orlicz norm instead of the Luxemburg norm. Furthermore, the problem of existence of an order linearly isometric copy of $l^{\infty }$ is in the case of the Orlicz norm connected to the strict monotonicity of this space (especially for $\gamma =\infty$). Recall that Banach lattice E is strictly monotone if $\Vert y\Vert _{E}<\Vert x\Vert _{E}$ whenever $x, y\in E_{+}$ (the positive cone of E) and $y\lneqq x$ (see [2]). Analogous results for Orlicz spaces equipped with the Orlicz norm were obtained by Chen, Cui and Hudzik in [4].

Theorem 5.1

If $\gamma =\infty$, then the following conditions are equivalent:

• (i) $\int _{0}^{\infty }\omega (t)dt=\infty$ and $a_{\varphi }=0$.

• (ii) The space $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \infty \right) \right) , \Vert \cdot \Vert _{\varphi , \omega }^{O}\right)$ is strictly monotone.

• (iii) The space $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \infty \right) \right) , \Vert \cdot \Vert _{\varphi , \omega }^{O}\right)$ does not contain an order linearly isometric copy of $l^{\infty }$.

Proof

$(i)\Rightarrow (ii)$. Let $x, y\in \Lambda _{\varphi , \omega }\left( \left[ 0, \infty \right) \right)$ be such that $x\ge 0$ and $y\gneqq 0$. We have $x\lneqq x+y$ and in consequence $x^{*}\le \left( x+y\right) ^{*}$. Moreover, in virtue of Lemma 3.2 in [15], there exists $t_{0}\in \left[ 0, \infty \right)$ such that $x^{*}\left( t_{0}\right) <\left( x+y\right) ^{*}\left( t_{0}\right)$. By the right continuity of $\left( x+y\right) ^{*}$, we conclude that there exists $t_{1}>t_{0}$ such that $x^{*}\left( t_{1}\right) \le x^{*}\left( t_{0}\right) <\left( x+y\right) ^{*}\left( t_{1}\right)$.

If $K\left( x+y\right) \ne \emptyset$, then for any $k_{0}\in K\left( x+y\right)$ we have $I_{\varphi ,\omega }\left( k_{0}x\right) <I_{\varphi , \omega }\left( k_{0}\left( x+y\right) \right)$ and consequently

$$\begin{aligned} \Vert x+y \Vert _{\varphi , \omega }^{O}&=\frac{1}{k_{0}} \{1+I_{\varphi ,\omega }\left( k_{0}\left( x+y\right) \right) \}>\frac{1}{k_{0}}\{ 1+I_{\varphi ,\omega }\left( k_{0}x\right) \}\\&\ge \inf _{k>0}\frac{1}{k}\{ 1+I_{\varphi , \omega }\left( kx\right) \} =\Vert x\Vert _{\varphi , \omega }^{O}. \end{aligned}$$

Let now $K\left( x+y\right) =\emptyset$. By Remark 3.4, we have $\lim \nolimits _{n\rightarrow \infty }\frac{\varphi \left( u\right) }{u} =B<\infty$ and ($\psi \left( B\right)$ $\int _{0}^{m\left( {{\,\textrm{supp}\,}}\left( x+y\right) \right) } \omega (t)dt<1$ or $\psi \left( B\right) \int _{0}^{m\left( {{\,\textrm{supp}\,}}\left( x+y\right) \right) }\omega (t)dt=1$ and $p\left( u\right) <B$ for any $u>0$). Therefore, $K\left( x\right) =\emptyset$ and by Lemma 3.4

$$\begin{aligned} \Vert x\Vert _{\varphi , \omega }^{O}=B\int _{0}^{\infty }x^{*}(t) \omega (t)dt<B\int _{0}^{\infty }(x+y)^{*}(t)\omega (t)dt=\Vert x +y\Vert _{\varphi ,\omega }^{O}. \end{aligned}$$

(26)

By the arbitrariness of x and y, we conclude that the space $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \infty \right) \right) , \Vert \cdot \Vert _{\varphi , \omega }^{O}\right)$ is strictly monotone.

The implication $(ii)\Rightarrow (iii)$ is obvious. Now we will prove the implication $(iii)\Rightarrow (i)$. Assume that $\int _{0}^{\infty }\omega (t)dt<\infty$ and define x by the formula $x(t)=1$ for $t\in \left[ 0, \infty \right)$. Clearly $x\in \Lambda _{\varphi , \omega }\left( \left[ 0, \infty \right) \right)$; moreover, the norm of $y=\frac{x}{\Vert x\Vert _{\varphi , \omega }^{O}}$ is equal to 1. Let $\left( T_{n}\right) _{n=1}^{\infty }$ be a sequence of pairwise disjoint subsets of $\left[ 0,\infty \right)$ such that $m(T_{n})=\infty$ for any $n\in {\mathbb {N}}$ and $\bigcup _{n=1}^{\infty }T_{n}=\left[ 0,\infty \right)$. Define

$$\begin{aligned} y_{n}=y\chi _{T_{n}} \text { for any } n\in {\mathbb {N}}. \end{aligned}$$

Since $y_{n}^{*}=y^{*}$ for any $n\in {\mathbb {N}}$, we get $\Vert y_{n}\Vert _{\varphi , \omega }^{O}=1$ for the same n. Thus, the operator defined by the formula

$$\begin{aligned} P\left( z\right) =\sum _{n=1}^{\infty }z_{n}y_{n} \text { for any } z=\left( z_{n}\right) _{n=1}^{\infty }\in l^{\infty }, \end{aligned}$$

which is linear and positive, is an order isometry of $l^{\infty }$ onto the closed subspace $P\left( l^{\infty }\right)$ of $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \infty \right) \right) , \Vert \cdot \Vert _{\varphi , \omega }^{O}\right)$.

Let now $\int _{0}^{\infty }\omega (t)dt=\infty$ and $a_{\varphi }>0$. Then for y defined by the formula $y(t)=a_{\varphi }$ for any $t\in \left[ 0, \infty \right)$, we have $1+I_{\varphi , \omega }\left( y\right) =1$, $\frac{1}{k}\{ 1+I_{\varphi , \omega }\left( ky\right) \}=\frac{1}{k}>1$ for any $k\in \left( 0, 1\right)$ and $\frac{1}{k}\{ 1+I_{\varphi , \omega }\left( ky\right) \}=\infty$ for any $k>1$, whence $\Vert y\Vert _{\varphi , \omega }^{O}=1$. Defining a sequence $\left( y_{n}\right) _{n=1}^{\infty }$ and an operator P analogously as above, we infer once again that the space $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \infty \right) \right) , \Vert \cdot \Vert _{\varphi , \omega }^{O}\right)$ contains an order linearly isometric copy of $l^{\infty }$. $\square$

Theorem 5.2

Let $\gamma <\infty$. Then the space $\left( \Lambda _{\varphi ,\omega }\left( \left[ 0, \gamma \right) \right) , \Vert \cdot \Vert _{\varphi , \omega }^{O}\right)$ is strictly monotone if and only if $\omega (t)>0$ for any $t\in \left[ 0,\gamma \right)$ and the following implication holds: if there exists $t\in \left[ 0,\gamma \right)$ such that $K\left( x_{t}\right) \ne \emptyset$, where $x_{t}=\chi _{\left[ 0,t\right) }$, then $a_{\varphi }=0$.

Proof

Sufficiency. If $a_{\varphi }=0$, the proof is obtained in the similar way as the proof of the implication $(i)\Rightarrow (ii)$ in Theorem 5.1. Suppose now that $a_{\varphi }>0$; hence, by assumption, we have $K\left( x_{t}\right) =\emptyset$ for any $t\in \left[ 0, \gamma \right)$. By virtue of Remark 3.4, we get $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi \left( u\right) }{u}=B<\infty$ and $\psi \left( B\right) \int _{0}^{\gamma }\omega (t)dt\le 1$. Therefore, by Lemma 3.4, we have

$$\begin{aligned} \Vert z\Vert _{\varphi , \omega }^{O}=B\int _{0}^{\gamma }z^{*}(t)\omega (t)dt \end{aligned}$$

for any $z\in \Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right)$. Take any $x, y\in \Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right)$ such that $x\ge 0$ and $y\gneqq 0$. We have $x\lneqq x+y$, whence $x^{*}\le \left( x+y\right) ^{*}$. Furthermore, by Lemma 3.2 in [15] and by the right continuity of the rearrangement function, there exist $0\le t_{0}<t_{1}\le \gamma$ such that $x^{*}\left( t_1\right) \le x^{*}\left( t_{0}\right) <(x+y)^{*} \left( t_{1}\right) \le (x+y)^{*}\left( t_{0}\right)$. Proceeding analogously as in (26), we get $\Vert x\Vert _{\varphi , \omega }^{O}<\Vert x+y\Vert _{\varphi , \omega }^{O}$, which establishes the sufficiency.

Necessity. Suppose that $\omega (t_{0})=0$ for some $t_{0}\in \left( 0, \gamma \right)$. Without loss of generality we can assume that $\omega (t)>0$ for any $t<t_{0}$. Then for $x=\chi _{\left[ 0,t_{0}\right) }$ and $y=\chi _{\left[ t_{0},\gamma \right) }$ we have $I_{\varphi , \omega }\left( kx\right) =I_{\varphi , \omega }\left( k\left( x+y\right) \right)$ for any $k>0$, whence $\Vert x\Vert _{\varphi , \omega }^{O}=\Vert x+y\Vert _{\varphi , \omega }^{O}$.

Now suppose that $\omega (t)>0$ for any $t\in \left[ 0,\gamma \right)$, $a_{\varphi }>0$ and $K\left( x_{t_{0}}\right) \ne \emptyset$ for some $t_{0}\in \left[ 0, \gamma \right)$. Let $k_{0}\in K\left( x_{t_{0}}\right)$ and $y:=\min \left( 1, \frac{a_{\varphi }}{k_{0}}\right) \chi _{\left[ t_{0}, \gamma \right) }$. We have

$$\begin{aligned} \Vert x_{t_{0}}\Vert _{\varphi , \omega }^{O}=\frac{1}{k_{0}} \{1+I_{\varphi , \omega }\left( k_{0}x_{t_{0}}\right) \} =\frac{1}{k_{0}}\{ 1+I_{\varphi , \omega }\left( k_{0} \left( x_{t_{0}}+y\right) \right) \}\ge \Vert x_{t_{0}} +y\Vert _{\varphi , \omega }^{O}, \end{aligned}$$

whence eventually we get $\Vert x_{t_{0}}\Vert _{\varphi ,\omega }^{O} =\Vert x_{t_{0}}+y\Vert _{\varphi , \omega }^{O}$. $\square$

Remark 5.1

(i):

Let $\gamma =\infty$. By Theorems 4.1 and 5.1 we come to the conclusion that if the space $\left( \Lambda _{\varphi , \omega }\left( \left[ 0, \infty \right) \right) , \Vert \cdot \Vert _{\varphi , \omega }^{O}\right)$ is order continuous, then it is strictly monotone as well. The converse implication does not hold as will be shown in Example 5.1. In the case when $\gamma <\infty$, these properties are not comparable (see Example 5.1).

(ii):

Let us notice that if the space $\Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right)$ is strictly monotone in the case of Luxemburg norm $\Vert \cdot \Vert _{\varphi , \omega }$, then it is also strictly monotone in the case of Orlicz norm $\Vert \cdot \Vert _{\varphi , \omega }^{O}$ (see [10, Corollary 4.4] and Theorems 5.1 and 5.2). As we will show in Example 5.1, there exist Orlicz–Lorentz spaces that are strictly monotone when equipped with the Orlicz norm and do not have that property when equipped with the Luxemburg norm.

Example 5.1

(i):

Since the function $\varphi (u)=e^{u}-1$ for $u\ge 0$ does not satisfy the condition $\Delta _{2}$ at infinity, the space $\Lambda _{\varphi , \omega }\left( \left[ 0, \gamma \right) \right)$, $\gamma \le \infty$ generated by $\varphi$ is not order continuous for both norms (that means for the Orlicz and Luxemburg norms). It is also not strictly monotone in the case of Luxembug norm. However, it is strictly monotone in the case of Orlicz norm, whenever $\omega (t)>0$ for $t\in \left[ 0, \gamma \right)$ if $\gamma <\infty$ or $\int _{0}^{\infty }\omega (t)dt=\infty$ if $\gamma =\infty$.

(ii):

Assume now that $\gamma <\infty$ and $\varphi (u)=\max (0,u-1)$ for $u\ge 0$. Since $\varphi \in \Delta _{2}(\infty )$, the space generated by $\varphi$ is order continuous for both norms. It is not strictly monotone in the case of Luxemburg norm. However, it is strictly monotone in the case of Orlicz norm if and only if $\omega (t)>0$ for $t<\gamma$ and $\int _{0}^{\gamma }\omega (t)dt\le 1$ (we have $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi (u)}{u}=1$ and $\psi (1)=1$).

Theorem 5.3

The space $(\Lambda _{\varphi , \omega }([0, \gamma )), \Vert \cdot \Vert _{\varphi , \omega }^{O})$, where $\gamma <\infty$, contains an order linearly isometric copy of $l^{\infty }$ if and only if $a_{\varphi }=b_{\varphi }$.

Proof

Sufficiency. Suppose $a_{\varphi }=b_{\varphi }$. Then $(\Lambda _{\varphi , \omega }([0, \gamma )), \Vert \cdot \Vert _{\varphi , \omega }^{O})=(L^{\infty }([0, \gamma )), \frac{1}{b_{\varphi }}\Vert \cdot \Vert _{L^{\infty }})$ and as a result the space $(\Lambda _{\varphi , \omega }([0, \gamma )), \Vert \cdot \Vert _{\varphi , \omega }^{O})$ contains an order linearly isometric copy of of $l^{\infty }$.

Necessity. Assume now that $0\le a_{\varphi }<b_{\varphi }\le \infty$ and there exists an order linearly isometry P of $l^{\infty }$ onto a closed subspace $P(l^{\infty })$ of $(\Lambda _{\varphi , \omega }([0,\gamma )),\Vert \cdot \Vert ^{O}_{\varphi ,\omega })$. Then $(\Lambda _{\varphi ,\omega }([0,\gamma )),\Vert \cdot \Vert ^{O}_{\varphi , \omega })$ is not order continuous, whence, by Theorem 4.1, the function $\varphi$ does not satisfy the condition $\Delta _{2}(\infty )$. Therefore, $b_{\varphi }<\infty$ or $b_{\varphi }=\infty$ and $\lim \nolimits _{u\rightarrow \infty }\frac{\varphi (u)}{u}=\infty$ and in consequence for any $x\in \Lambda _{\varphi , \omega }([0, \gamma )){\setminus }\{ 0\}$ the set K(x) is nonempty (see Remark 3.4).

Let $x_{n}=P(e_{n})$, where $e_{n}=(\underbrace{0, \ldots , 0}_{n-1}, 1, 0, \ldots )$. We have $\Vert x\Vert _{\varphi , \omega }^{O}=1$, $\Vert \sum _{n=1}^{\infty }x_{n}\Vert _{\varphi , \omega }^{O}=1$ and ${{\,\textrm{supp}\,}}x_{n}\wedge {{\,\textrm{supp}\,}}x_{m}=\emptyset$ for $n\ne m$. Since $\gamma <\infty$, we get $\lim \nolimits _{n\rightarrow \infty } m({{\,\textrm{supp}\,}}x_{n})=0$.

Suppose first that $a_{\varphi }=0$. By Theorem 5.2, there exists $t_{0}<\gamma$ such that $\omega (t)>0$ for $t<t_{0}$ and $\omega (s)=0$ for $s>t_{0}$. Let $n_{0}$ be such that $m({{\,\textrm{supp}\,}}x_{n})<t_{0}$ for any $n\ge n_{0}$ and let $n_{1}>n_{0}$ be fixed. Then for any $k\in K(x_{n_{0}}+x_{n_{1}})$ we have

$$\begin{aligned} 1&=\Vert x_{n_{0}}+x_{n_{1}}\Vert _{\varphi , \omega }^{O} =\frac{1}{k}\left( 1+I_{\varphi , \omega } \left( k\left( x_{n_{0}}+x_{n_{1}}\right) \right) \right) \\&>\frac{1}{k}\left( 1+I_{\varphi , \omega }\left( kx_{n_{0}} \right) \right) \ge \Vert x_{n_{0}}\Vert _{\varphi , \omega }^{O}=1, \end{aligned}$$

which leads to a contradiction.

Assume now that $0<a_{\varphi }<b_{\varphi }$. If $t_{0}=\sup \{ t: \omega (t)>0\}<\gamma$, then analogously as above we choose $n_{0}$ such that $m({{\,\textrm{supp}\,}}x_{n})<t_{0}$ for any $n\ge n_{0}$. Otherwise, that means if $t_{0}=\gamma$, we will assume $n_{0}=1$. First suppose that there exist $n_{1}$ and $n_{2}$ such that $n_{1}\ne n_{2}$, $\min (n_{1}, n_{2})\ge n_{0}$, $m(\{ t: x_{n_{1}}(t)>a_{\varphi }\})>0$ and $m(\{ t: x_{n_{2}}(t)>a_{\varphi }\})>0$. Since $\Vert x_{n_{1}}+x_{n_{2}}\Vert _{\varphi , \omega }^{O}=1$, there exists $k>1$ such that $k\in K(x_{n_{1}}+x_{n_{2}})$. Hence,

$$\begin{aligned}&\Vert x_{n_{1}}+x_{n_{2}}\Vert _{\varphi , \omega }^{O}=\frac{1}{k} \left( 1+I_{\varphi ,\omega }\left( k\left( x_{n_{1}}+x_{n_{2}}\right) \right) \right) \\&\quad >\frac{1}{k}\left( 1+I_{\varphi ,\omega }\left( kx_{n_{1}}\right) \right) \ge \Vert x_{n_{1}}\Vert _{\varphi , \omega }^{O}=1, \end{aligned}$$

which gives a contradiction. If such $n_{1}$ and $n_{2}$ do not exist, then starting with some $n_{3}\ge n_{0}$, we have $m(\{ t: x_{n}(t)>a_{\varphi }\})=0$ for any $n\ge n_{3}$. Take any fixed $k>1$. Since $\lim \nolimits _{n\rightarrow \infty }m({{\,\textrm{supp}\,}}x_{n})=0$ and for any $n\ge n_{3}$ we have $kx_{n}(t)\le ka_{\varphi }$ for m-a.e. $t\in [0, \gamma )$, we obtain $\lim \nolimits _{n\rightarrow \infty }I_{\varphi , \omega }(kx_{n})=0$. Hence, starting with some $n_{4}\ge n_{3}$, we have constantly

$$\begin{aligned} 1=\Vert x_{n}\Vert _{\varphi , \omega }\le \frac{1}{k} \left( 1+I_{\varphi , \omega }\left( kx_{n}\right) \right) <1, \end{aligned}$$

which again leads to a contradiction. $\square$

Recall that if a Banach space contains a subspace isomorphic to $l^{\infty }$, then it also contains subspaces arbitrarily nearly isometric to $l^{\infty }$ (see [24]). In the case of Orlicz–Lorentz spaces we have the following

Theorem 5.4

Let $b_{\varphi }=\infty$. If $\varphi$ does not satisfy the condition $\Delta _{2}(\infty )$ or $(\varphi$ does not satisfy the condition $\Delta _{2}(0)$ whenever $\gamma =\infty$, $\int _{0}^{\infty }\omega (t)dt=\infty$ and $a_{\varphi }=0)$, then for any $\varepsilon >0$ there exists an operator $P_{\varepsilon }: l^{\infty }\rightarrow \Lambda _{\varphi , \omega }([0, \gamma ))$, which is an order linear $(1+\varepsilon )$-isometry in the case of the Luxemburg norm as well as in the case of the Orlicz norm. Furthermore, $P_{\varepsilon }(c_{0})\hookrightarrow (\Lambda _{\varphi , \omega }([0, \gamma )))_{a}$.

Proof

As we noticed in the previous section, the condition $b_{\varphi }=\infty$ is equivalent to the fact that the subspace $(\Lambda _{\varphi , \omega }([0, \gamma )))_{a}$ is non-trivial. Take any $\varepsilon >0$. Without loss of generality we can assume that $\varepsilon <1$. First suppose that $\varphi \notin \Delta _{2}(\infty )$. Then there exists an increasing sequence $(u_{n})_{n=1}^{\infty }$ such that $\lim \nolimits _{n\rightarrow \infty }u_{n}=\infty$,

$$\begin{aligned} \varphi (u_{n})\int _{0}^{\alpha /2^{n}}\omega (t)dt \ge \frac{\varepsilon }{2^{n}}, \end{aligned}$$

(27)

where $\alpha =\min (1, \gamma )$ and

$$\begin{aligned} \varphi \left( \left( 1+\frac{\varepsilon }{2}\right) u_{n}\right) >\frac{2^{n+1}}{\varepsilon }\varphi (u_{n}). \end{aligned}$$

(28)

Let $(t_{n})_{n=1}^{\infty }$ be such that

$$\begin{aligned} \varphi (u_{n})\int _{0}^{t_{n}}\omega (t)dt=\frac{\varepsilon }{2^{n}} \end{aligned}$$

(29)

for any $n\in {\mathbb {N}}$, by (27) we have $t_{n}\le \frac{\alpha }{2^{n}}$ for the same n. Let $s_{0}=0$, $s_{n}=\sum _{i=1}^{n}t_{i}$ and $x_{n}=u_{n}\chi _{\left[ s_{n-1}, s_{n}\right) }$ for any $n\in {\mathbb {N}}$. Now we define an operator $P_{\varepsilon }$ by the formula

$$\begin{aligned} P_{\varepsilon }(z)=\sum _{n=1}^{\infty }z_{n}x_{n} \text { for any } z=(z_{n})_{n=1}^{\infty }\in l^{\infty }. \end{aligned}$$

It is obvious that $P_{\varepsilon }$ is linear and positive. Moreover, by (29) and by the orthogonal subadditivity of the modular, for any $z\in l^{\infty }$ we have

$$\begin{aligned} I_{\varphi , \omega }\left( \frac{P_{\varepsilon }(z)}{\Vert z \Vert _{\infty }}\right) \le \sum _{n=1}^{\infty } \varphi (u_{n})\int _{0}^{t_{n}}\omega (t)dt=\varepsilon , \end{aligned}$$

whence $P_{\varepsilon }(z)\in \Lambda _{\varphi ,\omega }\left( \left[ 0, \gamma \right) \right)$. Furthermore,

$$\begin{aligned} \left\| \frac{P_{\varepsilon }(z)}{\Vert z\Vert _{\infty }} \right\| ^{O}_{\varphi , \omega }\le 1 +I_{\varphi , \omega } \left( \frac{P_{\varepsilon }(z)}{\Vert z\Vert _{\infty }}\right) \le 1+\varepsilon , \end{aligned}$$

whence

$$\begin{aligned} \Vert P_{\varepsilon }(z) \Vert _{\varphi , \omega } \le \Vert P_{\varepsilon }(z) \Vert ^{O}_{\varphi , \omega } \le (1+\varepsilon )\Vert z\Vert _{\infty } \end{aligned}$$

for any $z\in l^{\infty }$. On the other hand, for any $z=\left( z_{n}\right) _{n=1}^{\infty }\in l^{\infty }$ there exists n(z) such that $\frac{z_{n(z)}}{(1-\varepsilon )\Vert z\Vert _{\infty }}\ge \frac{1}{1-\varepsilon /2}$. Therefore, and by (28) and (29), we get

$$\begin{aligned} I_{\varphi , \omega }\left( \frac{P_{\varepsilon }(z)}{(1-\varepsilon ) \Vert z\Vert _{\infty }}\right)&\ge \varphi \left( \frac{1}{1-\varepsilon /2} u_{n(z)}\right) \int _{0}^{t_{n(z)}}\omega (t)dt\\&\ge \varphi \left( \left( 1+\frac{\varepsilon }{2}\right) u_{n(z)}\right) \int _{0}^{t_{n(z)}}\omega (t)dt\\&>\frac{2^{n(z)+1}}{\varepsilon }\varphi (u_{n(z)})\int _{0}^{t_{n(z)}}\omega (t)dt=2 \end{aligned}$$

for any $z=(z_{n})_{n=1}^{\infty }\in l^{\infty }\setminus \{ 0\}$, which yields

$$\begin{aligned} \Vert P_{\varepsilon }(z)\Vert _{\varphi , \omega }^{O} \ge \Vert P_{\varepsilon }(z)\Vert _{\varphi , \omega } \ge (1-\varepsilon )\Vert z\Vert _{\infty } \end{aligned}$$

for any $z\in l^{\infty }\setminus \{ 0\}$.

Now let us take any $z\in c_{0}$. Then for any $\lambda >0$ we can find $n_{\lambda }\in {\mathbb {N}}$ such that $\lambda \vert z_{n}\vert \le 1$ for any $n\ge n_{\lambda }$. Therefore, by (29), we obtain

$$\begin{aligned} I_{\varphi , \omega }\left( \lambda P_{\varepsilon }(z)\right)&\le \sum _{i=1}^{\infty }\varphi (\lambda z_{n}u_{n})\int _{0}^{t_{n}} \omega (t)dt\\&\le \sum _{n=1}^{n_{\lambda }-1}\varphi (\lambda z_{n}u_{n}) \int _{0}^{t_{n}}\omega (t)dt+\sum _{n_{\lambda }}^{\infty }\varphi (u_{n}) \int _{0}^{t_{n}}\omega (t)dt\\&\le \sum _{n=1}^{n_{\lambda }-1}\varphi (\lambda z_{n}u_{n}) \int _{0}^{t_{n}}\omega (t)dt+\varepsilon \sum _{n_{\lambda }}^{\infty } 2^{-n}<\infty , \end{aligned}$$

whence it follows that $P_{\varepsilon }(z)\in \Lambda _{\varphi , \omega }^{\theta }([0, \gamma ))$ and thereby $P_{\varepsilon }(z)\in (\Lambda _{\varphi , \omega }([0, \gamma )))_{a}$ for any $z\in c_{0}$ (in the case when $\gamma =\infty$ and $\int _{0}^{\infty }\omega (t)dt<\infty$ the order continuity of $P_{\varepsilon }(z)$ follows from the fact that $m({{\,\textrm{supp}\,}}(P_{\varepsilon }(z))\le 1<\infty$ – see proof of Theorem 3.1(iii) in [8]).

Suppose now that $\gamma =\infty$, $\int _{0}^{\infty }\omega (t)dt=\infty$, $a_{\varphi }=0$ and $\varphi$ does not satisfy the condition $\Delta _{2}(0)$. Then there exists a decreasing sequence $(u_{n})_{n=1}^{\infty }$ such that $\lim \nolimits _{n\rightarrow \infty }u_{n}=0$ and

$$\begin{aligned} \varphi \left( \left( 1+\frac{\varepsilon }{2}\right) u_{n}\right) >\frac{2^{n+1}}{\varepsilon }\varphi (u_{n}). \end{aligned}$$

Since $\int _{0}^{\infty }\omega (t)dt=\infty$, for any $n\in {\mathbb {N}}$ we can find $t_{n}$ such that

$$\begin{aligned} \varphi (u_{n})\int _{0}^{t_{n}}\omega (t)dt=\frac{\varepsilon }{2^{n}}. \end{aligned}$$

Defining $s_{0}=0$, $s_{n}=\sum _{i=1}^{n}t_{i}$, $x_{n}=u_{n}\chi _{\left[ s_{n-1}, s_{n}\right) }$ for any $n\in {\mathbb {N}}$ and an operator $P_{\varepsilon }(z)=\sum _{n=1}^{\infty }z_{n}x_{n}$ for any $z\in l^{\infty }$ in the same way as above, we can show that $P_{\varepsilon }$ is order linear $(1+\varepsilon )$-isometry and $P_{\varepsilon }(z)\in \left( \Lambda _{\varphi , \omega }\left( \left[ 0, \infty \right) \right) \right) _{a}$ for any $z\in c_{0}$. $\square$

Remark 5.2

(i):

The last theorem is a generalization of results obtained by Hudzik and Mastyło in [14].

(ii):

Let us notice that in the case of isometries defined in the proof of implication $(iii)\Rightarrow (i)$ in Theorem 5.1 we have $P(z)\notin \left( \Lambda _{\varphi , \omega }\left( \left[ 0, \infty \right) \right) \right) _{a}$ for any $z\in c_{0}{\setminus }\{ 0\}$. The same situation takes place for an isometry defined in the case of an Orlicz–Lorentz space equipped with the Luxemburg norm (see [15, Theorem 2.4]).