Pointwise multipliers of Musielak–Orlicz spaces and factorization

Abstract

We prove that the space of pointwise multipliers between two distinct Musielak–Orlicz spaces is another Musielak–Orlicz space and the function defining it is given by an appropriately generalized Legendre transform. In particular, we obtain characterization of pointwise multipliers between Nakano spaces. We also discuss factorization problem for Musielak–Orlicz spaces and exhibit some differences between Orlicz and Musielak–Orlicz cases.

1 Introduction

Given two function spaces X and Y (over the same measure space), the space of pointwise multipliers M(X, Y) is the space of all functions f such that $fg\in Y$ for each $g\in X$. M(X, Y) may be regarded as a generalized Köthe dual space (cf. [3, 17]) and a basic question is to identify M(X, Y) for a given spaces X and Y. Many authors have investigated this problem for Orlicz spaces and many characterizations (mainly partial) have been given—see for example Shragin [23], Ando [1], O’Neil [22], Zabreiko–Rutickii [24], Maurey [18], Maligranda–Persson [17] and Maligranda–Nakai [16]. In 2000 Djakov and Ramanujan settled the problem for Orlicz sequence spaces and, recently, in [13] the authors established an analogous characterization for Orlicz function spaces. In both cases, the space of pointwise multipliers $M(L^{\varphi _1},L^{\varphi })$ between Orlicz spaces is proved to be just another Orlicz space, i.e.

$$\begin{aligned} M(L^{\varphi _1},L^{\varphi })=L^{\varphi \ominus \varphi _1}, \end{aligned}$$

(1.1)

where the function $\varphi \ominus \varphi _1$ is generalized Young conjugate (generalized Legendre transform) of $\varphi _1$ with respect to $\varphi$. Observe that the above characterization generalizes, in the evident way, the classical Köthe duality formula for Orlicz spaces, this is

$$\begin{aligned} (L^{\varphi _1})'=M(L^{\varphi _1},L^{1})=L^{\varphi _1^*}, \end{aligned}$$

(1.2)

where $\varphi _1^*$ is the Young conjugate of $\varphi _1$ (i.e. $\varphi _1^*=id\ominus \varphi _1$). Let us also mention here, that the identification as in (1.1) seems to be the most desirable, since the function $\varphi \ominus \varphi _1$ is given in an explicit and constructive way, in contrast to theorems from [16] and [11], which have rather existential character (cf. [1, 18, 22,23,24]).

In the paper we focus on the multipliers of Musielak–Orlicz spaces. Such investigations have been already initiated by Nakai [20] (cf. [21]). Under a number of assumptions on functions $\varphi , \varphi _1$ he generalized results of [16] to the Musielak–Orlicz setting. Since this method is not constructive (see discussion in [13]), we are not going to employ it. Instead of that we will use ideas of [5] and [13] to prove that the representation (1.1) holds also in the Musielak–Orlicz case, for an arbitrary $\sigma$—finite measure space and without any additional assumptions on Musielak–Orlicz functions $\varphi , \varphi _1$.

The paper is organized as follows. In Sect. 2 we give necessary definitions on Banach ideal space and Musielak–Orlicz spaces. We also define the function $\varphi \ominus \varphi _1$ (Young conjugate of $\varphi _1$ with respect to $\varphi$) for Musieak–Orlicz functions $\varphi , \varphi _1$.

The next section contains a number of technical lemmas concerning Musielak–Orlicz spaces and multipliers. Consequently, we are ready to prove the representation theorem in the third section. Finally, the last section is devoted to discussion on factorization and differences between Orlicz and Musielak–Orlicz cases. In particular, we give an example showing that inequality $\varphi _1^{-1}(\varphi \ominus \varphi _1)^{-1}\succ \varphi ^{-1}$ is not necessary condition for factorization of Musielak–Orlicz spaces, unlike in the Orlicz spaces case (cf. [13, Theorem 2]).

2 Notation and preliminaries

Trough the paper we will assume that $(\Omega ,\Sigma ,\mu )$ is a $\sigma$-finite, complete measure space. For a given set $A\in \Sigma$ we will denote the non-atomic part and purely atomic part of A by $A^c$ and $A^a$, respectively. Notice, that there may be at most countably many atoms in $\Omega$, since the measure space is $\sigma$-finite. To simplify the notion, we will assume that all atoms in $\Omega$, if there are any, are singletons.

Let $L^0=L^0(\Omega ,\Sigma ,\mu )$ be the space of classes of equivalence of $\mu$-measurable, real valuable and $\mu$-a.e. finite functions. A Banach space $X\subset L^0$ is called the Banach ideal space if it satisfies the ideal property, i.e. $x\in L^0,y\in X$ and $|x|\le |y|$ implies $x\in X$ and $\left\Vert x\right\Vert _X\le \left\Vert y\right\Vert _X$ ($|x|\le |y|$ means that $|x(t)|\le |y(t)|$ for $\mu$-a.e. $t\in \Omega$).

For $x\in L^0$ we define its support as $\hbox {supp}(x):=\{t\in \Omega :x(t)\ne 0\}.$ A support $\hbox {supp}X$ of a Banach ideal space X is defined as a measurable subset of $\Omega$ such that:

(i) for each $x\in X$ there is $A\in \Sigma$ with $\mu (A)=0$ such that $\hbox {supp}(x)\subset \hbox {supp}X\cup A$,

(ii) there is $x\in X$ such that $\mu (\hbox {supp}X- \hbox {supp}(x))=0$.

Notice that according to the above definition $\hbox {supp}X$ is not unique, thus we rather write a support, than the support of X.

For any measurable $F\subset \Omega$ and a Banach ideal space X we define

$$\begin{aligned} X[F]:=\{x\in X: \mu (\hbox {supp}(x){\setminus } F)=0\} \mathrm{\ with\ the\ norm\ }\left\Vert x\right\Vert _{X[F]}=\Vert x\chi _F\Vert _{X}. \end{aligned}$$

Given a Banach ideal space X on $\Omega$ and a positive measurable weight function v, the weighted space X(v) is defined as

$$\begin{aligned} X(v):=\{x\in L^0:xv\in X\} \mathrm{\ with\ the\ norm\ }\Vert x\Vert _{X(v)}=\Vert xv\Vert _X. \end{aligned}$$

Writing $X=Y$ for two Banach ideal spaces X, Y we mean that they are equal as set, but norms are just equivalent. Recall also that for Banach ideal spaces X, Y the inclusion $X\subset Y$ is always continuous, i.e. there is $c>0$ such that $\left\Vert x\right\Vert _Y\le c\left\Vert x\right\Vert _X$ for each $x\in X$.

A Banach ideal space X satisfies the Fatou property ($X\in (FP)$ for short) if for each sequence $(x_n)\subset X$ satisfying $x_n\uparrow x$ $\mu$-a.e. and $\sup _n\Vert x_n\Vert _X<\infty$, there holds $x\in X$ and $\left\Vert x\right\Vert _X\le \sup _n\left\Vert x_n\right\Vert _X$.

Given two Banach ideal spaces X, Y over the same measure space $(\Omega ,\Sigma ,\mu )$ we define their pointwise product space

$$\begin{aligned} X\odot Y=\{x\cdot y\in L^0: x\in X, y\in Y\}, \end{aligned}$$

with a quasi–norm

$$\begin{aligned} \left\Vert z\right\Vert _{X\odot Y}=\inf \{\left\Vert x\right\Vert _X\left\Vert y\right\Vert _Y:z=xy\}. \end{aligned}$$

If additionally $\hbox {supp}X =\Omega$, then the space of pointwise multipliers from X to Y is defined as

$$\begin{aligned} M(X,Y)=\{y\in L^0: xy\in Y\quad \hbox {for all }x\in X\}, \end{aligned}$$

with the natural operator norm

$$\begin{aligned} \left\Vert y\right\Vert _{M}=\sup _{\left\Vert x\right\Vert _X\le 1}\left\Vert xy\right\Vert _Y. \end{aligned}$$

When there is no risk of confusion we will just write $\Vert \cdot \Vert _M$ for the norm of M(X, Y).

If Banach ideal spaces X and Y have the Fatou property then both spaces M(X, Y) and $X\odot Y$ have the Fatou property [11, 12, 17].

We will need the following easy observation concerning the space of pointwise multipliers. Let $A,B\subset \Omega$ be measurable sets such that $A\cup B=\Omega$. Given a Banach ideal space X over $\Omega$, we can decompose it as

$$\begin{aligned} X=X[A]\oplus X[B], \end{aligned}$$

with the (equivalent) norm given by $\Vert x\Vert _{X[A]\oplus X[B]}=\Vert x\chi _A\Vert _{X[A]}+\Vert x\chi _B\Vert _{X[B]}$. It is easy to see that the space of pointwise multipliers respects such a “decomposition”, i.e. M(X, Y) may be written as follows

$$\begin{aligned} M(X,Y)=M(X[A]\oplus X[B],Y[A]\oplus Y[B])=M(X[A],Y[A])\oplus M(X[B],Y[B]). \end{aligned}$$

(2.1)

In another words, determining the space of pointwise multipliers between two Banach ideal spaces, we may determine it on A and on B separately.

A function $\varphi :[0,\infty )\rightarrow [0,\infty ]$ will be called the Young function if it satisfies $\varphi (0)=0$, $\lim \limits _{u\rightarrow \infty }\varphi (u)=\infty$ and is convex on $[0,b_{\varphi })$ (or on $[0,b_{\varphi }]$ when $\varphi (b_{\varphi })<\infty$), where

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

We point out here that we allow $\varphi (u)=\infty$ for each $u>0$. In such a case the corresponding Orlicz space $L^{\varphi }$ contains only the zero function.

Given a measure space $(\Omega ,\Sigma ,\mu )$, a function $\varphi :\Omega \times [0,\infty )\rightarrow [0,\infty ]$ is called the Musielak–Orlicz function if the following conditions hold:

• 1. $\varphi (t,\cdot )$ is a Young function for $\mu$-a.e. $t\in \Omega$,

• 2. $\varphi (\cdot ,u)$ is a measurable function for each $u\in [0,\infty )$.

Let $\varphi$ be a Musielak–Orlicz function. We define the convex modular $I_\varphi$ on $L^0$ as

$$\begin{aligned} I_{\varphi }(x)=\int _{\Omega }\varphi (t,|x(t)|)d\mu (t). \end{aligned}$$

The Musielak–Orlicz space $L^{\varphi }$ is defined as

$$\begin{aligned} L^{\varphi }=\{x\in L^0:I_{\varphi }(\lambda x)<\infty \mathrm{\ for\ some}\ \lambda >0\} \end{aligned}$$

and is equipped with the Luxemburg–Nakano norm

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

It is known that Musielak–Orlicz spaces have the Fatou property. Moreover, it follows immediately from the definition, that $\hbox {supp}L^{\varphi }=\{t\in \Omega : b_{\varphi }(t)>0\}$ (up to a set of measure zero). In case, $\varphi$ does not depend on t, the Musielak–Orlicz space $L^{\varphi }$ is just the Orlicz space.

For a given Musielak–Orlicz function $\varphi$ we define two useful (functions) parameters

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

It is known, that both $a_{\varphi }$ and $b_{\varphi }$ are measurable [4, Proposition 5.1].

Furthermore, we define the right-continuous inverse at point $t\in \Omega$

$$\begin{aligned} \varphi ^{-1}(t,u):=\inf \{v\ge 0:\varphi (t,v)> u\}. \end{aligned}$$

Properties of $\varphi ^{-1}$ for a Young function $\varphi$ have been collected in [11, Lemma 3.1].

The following basic relation between the norm and the modular will be used frequently through the paper

$$\begin{aligned} \Vert x\Vert _{\varphi }\le 1 \Rightarrow I_{\varphi }(x)\le \left\Vert x\right\Vert _{\varphi }, \end{aligned}$$

(2.2)

for $x\in L^{\varphi }$ (see [15, Theorem 1.1]). More information on Musielak–Orlicz and Orlicz spaces can be found for example in [7,8,9, 19].

3 Auxiliary results

Recall that our goal is to describe the space of pointwise multipliers $M(L^{\varphi _1},L^{\varphi })$ between two Musielak–Orlicz spaces and thus we will operate on two Musielak–Orlicz functions $\varphi , \varphi _1$, both defined over the same measure space $\Omega$. The result will be given in terms of the third Musielak–Orlicz function $\varphi \ominus \varphi _1$—the Young conjugate of $\varphi _1$ with respect to $\varphi$. In order to define it we need to introduce the following decomposition of the continuous part of the domain $\Omega$ depending on behaviour of both $\varphi , \varphi _1$. Let $\varphi , \varphi _1$ be two Musielak–Orlicz functions. We define the following sets:

$$\begin{aligned} \Omega _{0,0}&:=\{t\in \Omega ^c:b_{\varphi _1}(t)=b_{\varphi }(t)=\infty \},\\ \Omega _{0,\infty }&:=\{t\in \Omega ^c:b_{\varphi _1}(t)=\infty ,\ b_{\varphi }(t)<\infty \},\\ \Omega _{\infty ,0}&:=\{t\in \Omega ^c:0<b_{\varphi _1}(t)<\infty ,\ b_{\varphi }(t)=\infty \},\\ \Omega _{\infty ,\infty }&:=\{t\in \Omega ^c:0<b_{\varphi _1}(t)<\infty ,\ b_{\varphi }(t)<\infty \},\\ \Omega _{\infty }&:=\Omega _{\infty ,\infty }\cup \Omega _{\infty ,0}. \end{aligned}$$

Given two Musielak–Orlicz functions $\varphi , \varphi _1$ over the same measure space $\Omega$, the Young conjugate of $\varphi _1$ with respect to $\varphi$ is defined as

$$\begin{aligned}&\varphi \ominus \varphi _1(t,u) \\&\quad := {\left\{ \begin{array}{ll} \sup \{ \varphi (t,su)-\varphi _1(t,s): 0\le s < b_{\varphi _1}(t)\} &{}\text{ if } t\in \Omega ^c, \\ \sup \{\varphi (t,su)-\varphi _1(t,s): 0\le s \le \min \{1/\varphi ^{-1}\left( t,\frac{1}{\mu (\{t\})}\right) ,\frac{b_{\varphi _1}(t)}{2}\}\} &{}\text{ if } t\in \Omega ^a. \\ \end{array}\right. }\end{aligned}$$

Observe firstly that such defined function $\varphi \ominus \varphi _1$ satisfies $b_{\varphi \ominus \varphi _1}(t)>0$ for each $t\in \Omega ^a$. Moreover, it is easy to see, that for $t\in \Omega _{0,\infty }$

$$\begin{aligned}&\varphi \ominus \varphi _1(t,u)= {\left\{ \begin{array}{ll} 0 &{}\text{ if } u=0,\\ \infty &{}\text{ if } u>0. \end{array}\right. }\end{aligned}$$

In consequence,

$$\begin{aligned} \hbox {supp}(L^{\varphi \ominus \varphi _1})=(\Omega _{0,0}\cup \Omega _{\infty }\cup \Omega ^a)\cap \hbox {supp}(L^{\varphi }) \end{aligned}$$

(3.1)

It may be instructive to realize what is $\varphi \ominus \varphi _1$, when $\varphi ,\varphi _1$ are Nakano functions.

Example 1

Let $p,q:\Omega \rightarrow [1,\infty )$ be two measurable functions and define $\varphi (t,u)=\frac{1}{q(t)}u^{q(t)}$, $\varphi _1(t,u)=\frac{1}{p(t)}u^{p(t)}$ for $t\in \Omega , u\ge 0$. Assume that $q(t)\le p(t)$ for a.e. $t\in \Omega$. One can easily calculate that

$$\begin{aligned} \varphi \ominus \varphi _1(t,u)=\frac{1}{r(t)}u^{r(t)} \end{aligned}$$

where $\frac{1}{p(t)}+\frac{1}{r(t)}=\frac{1}{q(t)}$ for a.e. $t\in \Omega$.

Finally, we shall see that $\varphi \ominus \varphi _1$ satisfies assumptions of Musielak–Orlicz functions.

Lemma 2

Given two Musielak–Orlicz functions $\varphi , \varphi _1$ over the same measure space $\Omega$, the function $\varphi \ominus \varphi _1$ is also the Musielak–Orlicz function on $\Omega$.

Proof

It is already known that $\varphi \ominus \varphi _1(t,\cdot )$ is a Young function for a.e. $t\in \Omega$ (see [11, 13]). Thus we need only to prove measurability of $\varphi \ominus \varphi _1(\cdot ,u)$ for each $u\ge 0$. Each component of $\Omega$ may be considered separately. We will explain only situation on $\Omega _{\infty }$, since the remaining cases are either simpler, or evident. Firstly we observe that for each $u\ge 0$ and $t\in \Omega _{\infty }$

$$\begin{aligned} \varphi \ominus \varphi _1(t,u)= & {} \sup \{ \varphi (t,su)-\varphi _1(t,s): 0\le s< b_{\varphi _1}(t)\}\\= & {} \sup \{ \varphi (t,ub_{\varphi _1}(t)v)-\varphi _1(t,b_{\varphi _1}(t)v): 0\le v < 1\}\\= & {} \sup \{ \varphi (t,ub_{\varphi _1}(t)v)-\varphi _1(t,b_{\varphi _1}(t)v): v\in \mathbb {Q}\cap [0,1)\}. \end{aligned}$$

However, functions $\varphi (\cdot ,ub_{\varphi _1}(\cdot )v)$ and $\varphi _1(\cdot ,b_{\varphi _1}(\cdot )v)$ are measurable by properties of Musielak–Orlicz functions. In consequence, $\varphi \ominus \varphi _1(\cdot ,u)$ is measurable, as the supremum of countable collection of measurable functions.

In the proof of the main theorem, we are going to imitate inductive argument used in [5] and in [13]. In order to do it we need a kind of decomposition of the measure space $\Omega$. The following two lemmas provide it.

Lemma 3

Let $(\Omega ,\Sigma ,\mu )$ be non-atomic and let $\varphi$ be a Musielak–Orlicz function such that $b_{\varphi }(t)=\infty$ for $\mu$-a.e. $t\in \Omega$. For each $a>0$ there exists a sequence of pairwise disjoint measurable sets $(A_n)$ such that $\bigcup \limits _{n\in \mathbb {N}}A_n=\Omega$ and

$$\begin{aligned}\left\Vert \chi _{A_n}\right\Vert _{\varphi }\le \frac{1}{a}\end{aligned}$$

for every $n\in \mathbb {N}$.

Proof

Fix $a>0$. Define the sets

$$\begin{aligned}B_n:=\{t\in \Omega :n-1\le \varphi (t,a)< n\}\end{aligned}$$

for $n\in \mathbb {N}$. Evidently, each $B_n$ is measurable, since the function $\varphi (\cdot ,a)$ is measurable. Moreover $\bigcup \nolimits _{n\in \mathbb {N}}B_n=\Omega$ and $(B_n)$ is a sequence of pairwise disjoint sets. Since we operate on a non-atomic measure space, each $B_n$ may be divided further into a sequence (finite or not) of pairwise disjoint sets $(C^n_j)_{j\in I_n}$ such that $\bigcup \nolimits _{j\in I_n}C^n_j=B_n$ and $\mu (C^n_j)\le \frac{1}{n}$ for each $j\in I_n$. In consequence, we have for $n\in \mathbb {N}$ and $j\in I_n$

$$\begin{aligned} I_{\varphi }\left( a\chi _{C^n_j}\right)&=\int _{C^n_j}\varphi (t,a)d\mu (t)\\&\le \mu (C^n_j)\sup \limits _{t\in C^n_j} \varphi (t,a) \le 1. \end{aligned}$$

It follows that

$$\begin{aligned}\left\Vert \chi _{C^n_j}\right\Vert _{\varphi }\le \frac{1}{a},\end{aligned}$$

for every $n\in \mathbb {N}$ and $j\in I_n$. Finally, we get the desired sequence $(A_n)$ just by rearranging the (doubly indexed) sequence $(C^n_j)$.

Lemma 4

Let $(\Omega ,\Sigma ,\mu )$ be non-atomic and let $\varphi$ be a Musielak–Orlicz function such that $0<b_{\varphi }(t)<\infty$ for $\mu$-a.e. $t\in \Omega$. There exists a sequence of pairwise disjoint measurable sets $(A_n)$ such that $\bigcup \limits _{n\in \mathbb {N}}A_n=\Omega$ and for each $n\in \mathbb {N}$

$$\begin{aligned} \left\Vert \chi _{A_n}\right\Vert _{\varphi }\le \frac{2}{\mathop {\hbox {ess sup}}\limits _{t\in A_n}\{b_{\varphi }(t)\}}. \end{aligned}$$

Proof

For each $k\in \mathbb {Z}$ define

$$\begin{aligned}B_k:=\{t\in \Omega :2^{k-1}<b_{\varphi }(t)\le 2^{k}\}.\end{aligned}$$

Evidently, sets $B_k$ are measurable, since $b_{\varphi }$ is a measurable function. Next, for each $k\in \mathbb {Z}$ and $n\in \mathbb {N}$ we define

$$\begin{aligned} B_{k,n}:=\{t\in B_k:n-1\le \varphi (t,2^{k-1})< n\}. \end{aligned}$$

Then the doubly indexed sequence $(B_{k,n})$ consists of pairwise disjoint measurable sets such that $\bigcup \limits _{n\in \mathbb {N},k\in \mathbb {Z}}B_{k,n}=\Omega$. Denote

$$\begin{aligned}I:=\{(k,n)\in \mathbb {Z}^2:B_{k,n}\ne \emptyset \}.\end{aligned}$$

For each $(k,n)\in I$ we can further decompose $B_{k,n}$ into a (finite or not) sequence $(C^{k,n}_j)_{j\in I_{k,n}}$ of pairwise disjoint measurable sets in such a way that $\bigcup \limits _{j\in I_{k,n}}C^{k,n}_j=B_{k,n}$ and $\mu (C^{k,n}_j)\le \frac{1}{n}$ for each $j\in I_{k,n}$. Finally, for every $(k,n)\in I$ and $j\in I_{k,n}$ we have

$$\begin{aligned} I_{\varphi }\left( 2^{k-1}\chi _{C^{k,n}_j}\right)= & {} \int _{C^{k,n}_j}\varphi (t,2^{k-1})d\mu (t)\\\le & {} \mu (C^{k,n}_j)\sup \limits _{t\in C^{k,n}_j} \varphi (t,2^{k-1}) \le 1. \end{aligned}$$

In consequence,

$$\begin{aligned} \left\Vert \chi _{C^{k,n}_j}\right\Vert _{\varphi }\le \frac{1}{2^{k-1}}\le \frac{2}{\mathop {\hbox {ess sup}}\limits _{t\in B_{k,n}}\{b_{\varphi }(t)\}}\le \frac{2}{\mathop {\hbox {ess sup}}\limits _{t\in C^{k,n}_j}\{b_{\varphi }(t)\}}. \end{aligned}$$

Similarly as before, the desired sequence is obtained after rearranging the (triple indexed) sequence $(C^{k,n}_j)$.

Fact 5

If a Musielak–Orlicz function $\varphi$ is such that $b_{\varphi }(t)<\infty$ for a.e. $t\in \Omega$, then

$$\begin{aligned} L^{\varphi }\subset L^{\infty }(1/b_{\varphi }). \end{aligned}$$

Proof

Let $0\le y\notin L^{\infty }(1/b_{\varphi })$. For each $n\in \mathbb {N}$ we define sets

$$\begin{aligned} A_n=\{t\in \Omega : n\le \frac{y(t)}{b_{\varphi }(t)}\}. \end{aligned}$$

Then there is $N\in \mathbb {N}$ such that $\mu (A_n)>0$ for $n\ge N$. Fix $a>0$ and choose $n\ge N$ satisfying $an>2$. We can see that

$$\begin{aligned} nb_{\varphi }\chi _{A_n}\le y. \end{aligned}$$

In consequence,

$$\begin{aligned} I_{\varphi }\left( ay\right) \ge I_{\varphi }\left( anb_{\varphi }\chi _{A_n}\right) \ge I_{\varphi }\left( 2b_{\varphi }\chi _{A_n}\right) =\infty . \end{aligned}$$

Since $a>0$ was arbitrary we conclude that $y\notin L^{\varphi }.$

Lemma 6

Let $(\Omega ,\Sigma ,\mu )$ be non-atomic and let $\varphi , \varphi _1$ be two Musielak–Orlicz functions such that $0<b_{\varphi _1}(t)<\infty$ and $0< b_{\varphi }(t)<\infty$ for $\mu$-a.e. $t\in \Omega$. Then

$$\begin{aligned} M(L^{\varphi _1},L^{\varphi })\subset L^{\infty }(b_{\varphi _1}/b_{\varphi }). \end{aligned}$$

Proof

Let $0\le y\notin L^{\infty }(v)$, where $v(t):=\frac{b_{\varphi _1}(t)}{b_{\varphi }(t)}$. For each $n\in \mathbb {N}$ we define

$$\begin{aligned} A_n=\{t\in \Omega :n\le y(t)v(t)< n+1\}. \end{aligned}$$

Then there exist infinitely many $n\in \mathbb {N}$ for which $\mu (A_n)>0$. Denote the set of such n’s by I. Next, since $\Omega$ is non-atomic, for each $n\in I$ there is $B_n\subset A_n$ such that $\mu (B_n)>0$ and

$$\begin{aligned} \int _{B_n}\varphi _1\left( t,\frac{b_{\varphi _1}(t)}{2}\right) d\mu (t)\le \frac{1}{2^n}. \end{aligned}$$

We define

$$\begin{aligned} f(t):=\sum \limits _{n=1}^{\infty }\frac{b_{\varphi _1}(t)}{2}\chi _{B_n}. \end{aligned}$$

Then

$$\begin{aligned}I_{\varphi _1}\left( f\right) =\int \varphi _1(t,f(t))d\mu (t)=\sum \limits _{n=1}^{\infty }\int _{B_n}\varphi _1\left( t,\frac{b_{\varphi _1}(t)}{2}\right) d\mu (t)\le 1.\end{aligned}$$

It means that $f\in L^{\varphi _1}$ and $\left\Vert f\right\Vert _{\varphi _1}\le 1$. However,

$$\begin{aligned} y(t)f(t)\ge \frac{1}{2}y(t)b_{\varphi _1}(t)\ge \frac{n}{2}b_{\varphi }(t)\quad \text { for a.e. } t\in B_n, \end{aligned}$$

which implies that $yf\notin L^{\varphi }$, since $L^{\varphi }\subset L^{\infty }(\frac{1}{b_{\varphi }})$ by Fact 5. Consequently, $y\notin M(L^{\varphi _1},L^{\varphi })$ and the proof is finished.

Lemma 7

Suppose $(\Omega ,\Sigma ,\mu )$ is non-atomic and let $\varphi , \varphi _1$ be Musielak–Orlicz functions such that $\hbox {supp}L^{\varphi _1}=\Omega$. Then

$$\begin{aligned} \mu (\hbox {supp}M(L^{\varphi _1},L^{\varphi }){\setminus }\Omega _{0,0}\cup \Omega _{\infty })=0. \end{aligned}$$

Proof

We need only to show that $\mu (\Omega _{0,\infty }\cap \hbox {supp}M(L^{\varphi _1},L^{\varphi }))=0$. Suppose, for a contrary, there exists $A\subset \Omega _{0,\infty }$ such that $\mu (A)>0$ and $\chi _A\in M(L^{\varphi _1},L^{\varphi })$. Let $C\subset A$ be chosen in such a way that $\mu (C)>0$ and $\inf \limits _{t\in C}b_{\varphi }(t)=\delta >0$. From Lemma 3 it follows that for each $n\in \mathbb {N}$ there exists $A_n\subset C$ such that $\mu (A_n)>0$ and

$$\begin{aligned} \left\Vert \chi _{A_n}\right\Vert _{\varphi _1}\le \frac{1}{n}. \end{aligned}$$

Moreover, by Fact 5, we know that $L^{\varphi }[\Omega _{0,\infty }]\subset L^{\infty }(\frac{1}{b_{\varphi }})[\Omega _{0,\infty }]$ with some inclusion constant $c>0$. It means

$$\begin{aligned} \left\Vert \chi _{A_n}\right\Vert _{\varphi }&\ge c^{-1}\left\Vert \chi _{A_n}\right\Vert _{L^{\infty }(\frac{1}{b_{\varphi }})}\\&\ge c^{-1}\sup \limits _{t\in A_n}\frac{1}{b_{\varphi }(t)}\\&=\frac{1}{c\inf \limits _{t\in A_n}b_{\varphi }(t)}\\&\ge \frac{1}{c\inf \limits _{t\in C}b_{\varphi }(t)}=\frac{1}{c\delta }. \end{aligned}$$

Finally, for each $n\in \mathbb {N}$ define $x_n:=n\chi _{A_n}$. Then $\left\Vert x_n\right\Vert _{L^{\varphi _1}}\le 1$ and it follows

$$\begin{aligned} \left\Vert \chi _A\right\Vert _{M}\ge \left\Vert x_n\chi _A\right\Vert _{\varphi }=\left\Vert n\chi _{A_n}\right\Vert _{\varphi }\ge \frac{n}{c\delta }, \end{aligned}$$

for each $n\in \mathbb {N}$. In consequence, $\chi _A\not \in M(L^{\varphi _1},L^{\varphi })$ which contradicts our assumption.

Of course, the supremum in definition of function $\varphi \ominus \varphi _1$ need not be attained. To avoid such a situation, we introduce a truncated version of $\varphi \ominus \varphi _1$ (cf. [13, Definition 1]). Namely, for $a>0$ we define the function $\varphi \ominus _a \varphi _1$ in the following way

$$\begin{aligned}&\varphi \ominus _a\varphi _1(t,u) \\&\quad := {\left\{ \begin{array}{ll} \sup \{ \varphi (t,su)-\varphi _1(t,s): 0\le s\le a \} &{}\text{ if } t\in \Omega _{0,0}\cup \Omega _{0,\infty } \\ \sup \{\varphi (t,su)-\varphi _1(t,s): 0\le s \le \frac{a}{a+1}b_{\varphi _1}(t)\} &{}\text{ if } t\in \Omega _{\infty } \\ \sup \{\varphi (t,su)-\varphi _1(t,s): 0\le s \le \min \{1/\varphi ^{-1}\left( t,\frac{1}{\mu (\{t\})}\right) ,\frac{b_{\varphi _1}(t)}{2}\}\} &{}\text{ if } t\in \Omega ^a \\ \end{array}\right. }\end{aligned}$$

Using exactly the same reasoning as in the proof of Lemma 2 one can see that $\varphi \ominus _a\varphi _1$ is the Musielak–Orlicz function over $\Omega$. Furthermore, it is easy to see that

$$\begin{aligned} b_{\varphi \ominus _a\varphi _1}(t)=\frac{(a+1)b_{\varphi }(t)}{ab_{\varphi _1}(t)} \end{aligned}$$

(3.2)

for $t\in \Omega _{\infty }$.

Lemma 8

Let $(\Omega ,\Sigma ,\mu )$ be non-atomic and let $\varphi , \varphi _1$ be Musielak–Orlicz functions such that $\hbox {supp}L^{\varphi _1}=\Omega$. If $A\subset \hbox {supp}L^{\varphi }{\setminus }\Omega _{\infty ,0}$ is a set of positive measure and numbers $a>1,u>0$ satisfy $\varphi \ominus _a\varphi _1\left( t,\frac{3}{2}u\right) <\infty$ for a.e. $t\in A$, then the function $x:A\rightarrow \mathbb {R}_+$, defined by

$$\begin{aligned} x(t):=\max \left\{ 0\le v\le \min \left\{ a, \frac{a}{a+1}b_{\varphi _1}(t)\right\} :\varphi _1(t,v)+\varphi \ominus _a\varphi _1(t,u)=\varphi (t,uv)\right\} , \end{aligned}$$

is measurable.

Proof

Without loss of generality we may assume that $\varphi _1(t,\cdot ),\varphi (t,\cdot )$ are Young functions for each $t\in A$. Fix $u>0$ and $a>1$ satisfying

$$\begin{aligned} \varphi \ominus _a\varphi _1\left( t,\frac{3}{2}u\right) <\infty \quad \hbox {for a.e.} t\in A \end{aligned}$$

(3.3)

and let x be like in the statement. Let $(r_k)$ be a dense sequence in [0, a]. For each $k,n\in \mathbb {N}$ define

$$\begin{aligned} B^n_k:=\left\{ t\in A: r_k\le \frac{a}{a+1}b_{\varphi _1}(t),\ \varphi _1(t,r_k)+\varphi \ominus _a\varphi _1(t,u)-\varphi (t,ur_k)<1/n \right\} \end{aligned}$$

and

$$\begin{aligned} q^n_k:=r_k\chi _{B^n_k}. \end{aligned}$$

Just notice that by the definition of $\varphi \ominus _a\varphi _1$

$$\begin{aligned}0\le \varphi _1(t,v)+\varphi \ominus _a\varphi _1(t,w)-\varphi (t,wv)\end{aligned}$$

for a.e. $t\in \Omega$ and $w,v\ge 0$. Therefore,

$$\begin{aligned}\varphi (t,ur_k)<\infty ,\end{aligned}$$

because for every $k\in \mathbb {N}$ we have $\varphi _1(t,r_k)<\infty$ and $\varphi \ominus _a\varphi _1(t,u)<\infty$. Of course, functions $q^n_k$ are measurable, since sets $B^n_k$ are measurable. We will show that

$$\begin{aligned} x=\limsup _{k,n\rightarrow \infty } q^n_k. \end{aligned}$$

(3.4)

Firstly we will explain the inequality $\limsup _{k,n\rightarrow \infty } q^n_k\le x$. Suppose, for a contradiction, that for some $t_0\in A$ and some $\delta >0$ there holds

$$\begin{aligned} \limsup _{k,n\rightarrow \infty } q^n_k(t_0)>x(t_0)+\delta . \end{aligned}$$

This implies that there is a (singly-indexed) sequence $(q^{n_i}_{k_i})$ such that $\min \{a,\frac{a}{a+1}b_{\varphi _1}(t_0)\}\ge q^{n_i}_{k_i}(t_0)>x(t_0)+\delta$, $n_i,k_i\rightarrow \infty$ and

$$\begin{aligned} \varphi _1(t_0,u_{q^{n_i}_{k_i}}(t_0))+\varphi \ominus _a\varphi _1(t_0,u)-\varphi (t_0,u_{q^{n_i}_{k_i}}(t_0))<1/n_i \end{aligned}$$

(3.5)

for each $i=1,2,3,\ldots$. On the other hand, there is a subsequence $(q_{j}):=(q^{n_{i_j}}_{k_{i_j}})$ of $(q^{n_i}_{k_i})$ and $q_0>x(t_0)$ such that $q_j(t_0)\rightarrow q_0$. However, by (3.5) and continuity of respective functions, we get

$$\begin{aligned} \varphi _1(t_0,q_{0})+\varphi \ominus _a\varphi _1(t_0,u)-\varphi (t_0,uq_{0})=0, \end{aligned}$$

which contradicts maximality of $x(t_0)$ and proves inequality $\limsup _{k,n\rightarrow \infty } q^n_k\le x$.

To see the opposite inequality fix $t\in A$ and denote

$$\begin{aligned} C_n:= \left\{ 0\le v\le \min \left\{ a,\frac{a}{a+1}b_{\varphi _1}(t)\right\} :\varphi _1(t,v)+\varphi \ominus _a\varphi _1(t,u)-\varphi (t,uv)<1/n\right\} . \end{aligned}$$

We see that sets $C_n$ are open and non-empty, since $x(t)\in C_n$ for each n. Therefore, one can select a sequence $(r_{n_i})$ such that $r_{n_i}\in C_{i}$ and $r_{n_i}\rightarrow x(t)$. Then $t\in B^i_{n_i}$ for each $i=1,2,3,\ldots$ and, consequently,

$$\begin{aligned} x(t)\le \limsup _{k,n\rightarrow \infty } q^n_k(t), \end{aligned}$$

which finally proves measurability of x.

4 Pointwise multipliers

Theorem 9

Let $\varphi , \varphi _1$ be Musielak–Orlicz functions over a measure space $(\Omega ,\Sigma ,\mu )$ and assume that $\hbox {supp}L^{\varphi _1}=\Omega$. Then

$$\begin{aligned}M(L^{\varphi _1},L^{\varphi })=L^{\varphi \ominus \varphi _1}.\end{aligned}$$

Proof

Without loss of generality we can assume that $\hbox {supp}(L^{\varphi })=\Omega$, since

$$\begin{aligned} M(L^{\varphi _1},L^{\varphi })[\Omega {\setminus }\hbox {supp}(L^{\varphi })]=\{0\}=L^{\varphi \ominus \varphi _1}[\Omega {\setminus }\hbox {supp}(L^{\varphi })], \end{aligned}$$

where the second equality follows from (3.1). The proof of inclusion

$$\begin{aligned} L^{\varphi \ominus \varphi _1}\subset M(L^{\varphi _1},L^{\varphi }) \end{aligned}$$

is the same as in the case of Orlicz spaces and we omit it (see for example [13, Lemma 6]).

We only need to prove the remaining inclusion

$$\begin{aligned} M(L^{\varphi _1},L^{\varphi })\subset L^{\varphi \ominus \varphi _1} \end{aligned}$$

Let $0\le y\in M(L^{\varphi _1},L^{\varphi })$ be a simple function such that $\left\Vert y\right\Vert _{M}\le \frac{1}{4c}$, where $c\ge 1$ is the constant of inclusion

$$\begin{aligned} M(L^{\varphi _1},L^{\varphi })[\Omega _{\infty ,\infty }]\subset L^{\infty }(b_{\varphi _1}/b_{\varphi })[\Omega _{\infty ,\infty }] \end{aligned}$$

(cf. Lemma 6). We will show that

$$\begin{aligned} I_{\varphi \ominus _a\varphi _1}(y)\le 1 \end{aligned}$$

(4.1)

for every $a>1$.

To prove this inequality, for each $a>1$ we will construct a function x(t) on $\Omega$ and a family of pairwise disjoint sets $(A_n)$ satisfying:

• (i) $\varphi \ominus _a\varphi _1(t,y(t))=\varphi (t,x(t)y(t))-\varphi _1(t,x(t))$ for a.e $t\in \Omega$,

• (ii) $\left\Vert xy\chi _{A_n}\right\Vert _{\varphi }\le \frac{1}{2}$ for each $n\in \mathbb {N}$,

• (iii) $\hbox {supp}(M(L^{\varphi _1},L^{\varphi }))\subset \bigcup \limits _{n\in \mathbb {N}}A_n$,

• (iv) $x\in L^{\varphi _1}$ and $\left\Vert x\right\Vert _{\varphi _1}\le 1.$

Let $a>1$. Since y is a simple function we can write it in the form

$$\begin{aligned} y=\sum \limits _{k=0}^n b_k\chi _{B_k}+\sum \limits _{k=0}^m d_k\chi _{\{\omega _k\}}, \end{aligned}$$

where for every k we have $b_k,d_k>0$, $B_k\subset \Omega _{\infty }\cup \Omega _{0,0}$, $\mu (B_k)<\infty$ and $\omega _k$’s are atoms. In order to construct the desired function x, we will apply Lemma 8 for each $b_k$ and $B_k$. First of all we need to show that assumptions of Lemma 8 are fulfilled, i.e. for each $0\le k\le n$ we have $\varphi \ominus _a\varphi _1\left( t,\frac{3}{2}b_k\right) <\infty$ for a.e. $t\in B_k$. Let $0\le k\le n$. Then for a.e. $t\in B_k$ we have

$$\begin{aligned} b_k=y(t)\le \frac{b_{\varphi \ominus \varphi _1}(t)}{4}\le \frac{b_{\varphi \ominus _a\varphi _1}(t)}{2}, \end{aligned}$$

since, by Lemma 6,

$$\begin{aligned} \left\Vert yb^{-1}_{\varphi \ominus \varphi _1}\chi _{\Omega _{\infty ,\infty }}\right\Vert _{\infty }\le c\left\Vert y\chi _{\Omega _{\infty ,\infty }}\right\Vert _{M}\le \frac{1}{4} \end{aligned}$$

and

$$\begin{aligned} b_{\varphi \ominus _a\varphi _1}(t)=\infty \end{aligned}$$

for a.e. $t\in \Omega _{0,0}\cup \Omega _{\infty ,0}$. Consequently $\varphi \ominus _a\varphi _1(t,\frac{3}{2}b_k)<\infty$ for a.e. $t\in B_k$. Thus using Lemma 8 for the set $B_k$ and the number $b_k$ we obtain measurable function $x_k(t)$ on $B_k$ such that

$$\begin{aligned} \varphi \ominus _a\varphi _1(t,y(t))=\varphi (t,x_k(t)y(t))-\varphi _1(t,x_k(t)) \end{aligned}$$

and $0\le x_k(t)\le \min \{a, \frac{a}{a+1}b_{\varphi _1}(t)\}$ for a.e. $t\in B_k$.

Now we will consider the atomic part. For every $0\le k\le m$ let $c_k>0$ satisfy

$$\begin{aligned} 0\le c_k\le \min \left\{ 1/\varphi ^{-1}\left( \omega _k,\frac{1}{\mu (\{\omega _k\})}\right) ,\frac{b_{\varphi _1}(\omega _k)}{2}\right\} \end{aligned}$$

and

$$\begin{aligned} \varphi \ominus _a\varphi _1(\omega _k,y(\omega _k))=\varphi (\omega _k,c_ky(\omega _k))-\varphi _1(\omega _k,c_k). \end{aligned}$$

Such numbers exist, since the supremum in definition of $\varphi \ominus _a\varphi _1$ is taken over a compact set.

The function satisfying (i) is defined as

$$\begin{aligned} x(t) := {\left\{ \begin{array}{ll} x_k(t) &{}\text{ if } t\in B_k,\ 0\le k\le n, \\ c_k &{}\text{ if } t=\omega _k,\ 0\le k\le m, \\ 0 &{}\text{ if } t\notin \hbox {supp}(y). \\ \end{array}\right. } \end{aligned}$$

In the next step we will determine sets $(A_n)$ satisfying (ii) and (iii).

We start with $\Omega _{\infty }$. By Lemma 4 there exists a sequence of pairwise disjoint measurable sets $(A^1_n)$ such that $\bigcup \limits _{n\in \mathbb {N}}A^1_n=\Omega _{\infty }$ and

$$\begin{aligned} \left\Vert \chi _{A^1_n}\right\Vert _{\varphi _1}\le \frac{2}{\sup \limits _{t\in A^1_n}\{b_{\varphi _1}(t)\}}, \end{aligned}$$

for every $n\in \mathbb {N}$. Since $0\le x(t)< b_{\varphi _1}(t)$, we have

$$\begin{aligned} \left\Vert xy\chi _{A^1_n}\right\Vert _{\varphi }\le \sup _{t\in A^1_n}\{b_{\varphi _1}(t)\}\left\Vert y\right\Vert _M\left\Vert \chi _{A^1_n}\right\Vert _{\varphi _1}\le \frac{1}{2}, \end{aligned}$$

(4.2)

and therefore sets $(A^1_n)$ satisfy (ii).

Secondly, by Lemma 3, there exists a sequence $(A^2_n)$ of pairwise disjoint measurable sets such that $\bigcup \limits _{n\in \mathbb {N}}A^2_n=\Omega _{0,0}$ and

$$\begin{aligned} \left\Vert \chi _{A^2_n}\right\Vert _{\varphi _1}\le \frac{1}{a}. \end{aligned}$$

Moreover, we have

$$\begin{aligned} \left\Vert xy\chi _{A^2_n}\right\Vert _{\varphi }\le \frac{a}{2}\left\Vert \chi _{A^2_n}\right\Vert _{\varphi _1}\le \frac{1}{2}, \end{aligned}$$

(4.3)

because $x(t)\le a$.

Considering the atomic part, let’s observe that for each $\omega \in \Omega ^a$

$$\begin{aligned} \left\Vert xy\chi _{\{\omega \}}\right\Vert _{\varphi }\le \frac{1}{2\varphi ^{-1}(\omega ,\frac{1}{\mu (\{\omega \})})}\left\Vert \chi _{\{\omega \}}\right\Vert _{\varphi _1}=\frac{1}{2}, \end{aligned}$$

(4.4)

where the last equality follows by $\left\Vert \chi _{\{\omega \}}\right\Vert _{\varphi _1}=\varphi ^{-1}(\omega ,\frac{1}{\mu (\{\omega \})})$. Therefore, we can take atoms as desired sets.

Finally, it is enough to renumerate the sequences $(A^1_n),(A^2_n),(\{\omega \})_{\omega \in \hbox {supp}(M(L^{\varphi _1},L^{\varphi }))^a}$ into one sequence $(A_n)$. By Lemma 7,

$$\begin{aligned} \hbox {supp}(M(L^{\varphi _1},L^{\varphi }))\subset \bigcup _{n\in \mathbb {N}}A_n, \end{aligned}$$

thus the construction of desired sets $(A_n)$ is finished.

It just left to show that (iv) is fulfilled, i.e.

$$\begin{aligned} \left\Vert x\right\Vert _{\varphi _1}\le 1. \end{aligned}$$

In order to prove it, we define functions $x_n:=\sum \limits _{k=1}^n x\chi _{A_k}$ and we will inductively show that

$$\begin{aligned} I_{\varphi _1}\left( x_n\right) \le \frac{1}{2}. \end{aligned}$$

Since $x_n\uparrow x$ a.e., from the Fatou property, it will follow that $x\in L^{\varphi _1}$ and

$$\begin{aligned} \left\Vert x\right\Vert _{\varphi _1}\le \sup \limits _n\left\Vert x_n\right\Vert _{\varphi _1}\le 1. \end{aligned}$$

Firstly we need to show that for every $k\in \mathbb {N}$ there holds

$$\begin{aligned} \left\Vert x\chi _{A_k}\right\Vert _{\varphi _1}\le \frac{1}{2}. \end{aligned}$$

(4.5)

From the equality

$$\begin{aligned} \varphi \ominus _a\varphi _1(t,y(t))=\varphi (t,x(t)y(t))-\varphi _1(t,x(t)) \end{aligned}$$

we obtain two inequalities

$$\begin{aligned} \varphi _1(t,x(t))\le & {} \varphi (t,x(t)y(t)) \quad \hbox {for a.e. } t\in \Omega ,\end{aligned}$$

(4.6)

$$\begin{aligned} \varphi \ominus _a\varphi _1(t,y(t))\le & {} \varphi (t,x(t)y(t)) \hbox { for a.e.}\,t\in \Omega . \end{aligned}$$

(4.7)

From (4.6) and by inequality $\left\Vert xy\chi _{A_k}\right\Vert _{\varphi }\le \frac{1}{2}$ we have

$$\begin{aligned} I_{\varphi _1}\left( x\chi _{A_k}\right) =\int _{A_k}\varphi _1(t,x(t))d\mu (t)\le \int _{A_k}\varphi (t,y(t)x(t))d\mu (t)=I_{\varphi }\left( yx\chi _{A_k}\right) \le \frac{1}{2} \end{aligned}$$

(4.8)

for every $k\in \mathbb {N}$, where the last inequality follows from (2.2).

In particular, $I_{\varphi _1}\left( x_1\right) \le \frac{1}{2}$, and we can proceed with the induction. Let $n\ge 1$ and suppose that

$$\begin{aligned} I_{\varphi _1}\left( x_n\right) \le \frac{1}{2}. \end{aligned}$$

We have

$$\begin{aligned} I_{\varphi _1}\left( x_{n+1}\right) =I_{\varphi _1}\left( x_n\right) +I_{\varphi _1}\left( x\chi _{A_{n+1}}\right) \le 1 \end{aligned}$$

and thus $\left\Vert x_{n+1}\right\Vert _{\varphi _1}\le 1.$

Similarly, as in inequality (4.8), we obtain

$$\begin{aligned} I_{\varphi _1}\left( x_{n+1}\right) \le I_{\varphi }\left( yx_{n+1}\right) \le \frac{1}{2}, \end{aligned}$$

by $\left\Vert yx_{n+1}\right\Vert _{\varphi }\le \frac{1}{2}\left\Vert x_{n+1}\right\Vert _{\varphi _1}\le \frac{1}{2}$. It means that (4.5) is proved and (iv) follows.

Finally, we are ready to show that $I_{\varphi \ominus _a\varphi _1}(y)\le 1$. We have

$$\begin{aligned} \left\Vert yx\right\Vert _{\varphi }\le \left\Vert y\right\Vert _M\left\Vert x\right\Vert _{\varphi _1}\le \frac{1}{2} \end{aligned}$$

and from inequality (4.7) we obtain

$$\begin{aligned} I_{\varphi \ominus _a\varphi _1}\left( y\right) =\int \varphi \ominus _a\varphi _1(t,y(t))d\mu (t)\le \int \varphi (t,y(t)x(t))d\mu (t)= I_{\varphi }\left( yx\right) \le 1. \end{aligned}$$

Clearly, $\varphi \ominus _a\varphi _1(t,y(t))\uparrow \varphi \ominus \varphi _1(t,y(t))$ for a.e. $t\in \Omega$ when $a\uparrow \infty$. Applying the Fatou lemma we have

$$\begin{aligned} I_{\varphi \ominus \varphi _1}\left( y\right) =\int \limits _{\Omega }\varphi \ominus \varphi _1(t,y(t))d\mu (t) \le \liminf \limits _{a\rightarrow \infty }\int \limits _{\Omega }\varphi \ominus _a\varphi _1(t,y(t))d\mu (t)\le 1, \end{aligned}$$

which proves the inequality (4.1). It means that $y\in L^{\varphi \ominus \varphi _1}$ and

$$\begin{aligned} \left\Vert y\right\Vert _{\varphi \ominus \varphi _1}\le 1. \end{aligned}$$

Concluding, if $0\le y\in M(L^{\varphi _1},L^{\varphi })$ is a simple function, then $y\in L^{\varphi \ominus \varphi _1}$ and

$$\begin{aligned} \left\Vert y\right\Vert _{\varphi \ominus \varphi _1}\le 4c\left\Vert y\right\Vert _M. \end{aligned}$$

Thus the theorem is proved for positive simple functions. We will once again use the Fatou property to complete the argument for an arbitrary function.

Let now $y\in M(L^{\varphi _1},L^{\varphi })$ be arbitrary. There exists a sequence of simple functions $(y_n)$ such that $0\le y_n\uparrow \left|y\right|$ a.e. on $\Omega$. Since $M(L^{\varphi _1},L^{\varphi })$ is a Banach ideal space, $\left\Vert y_n\right\Vert _M\le \left\Vert y\right\Vert _M$ for every $n\in \mathbb {N}$. From the Fatou property of $L^{\varphi \ominus \varphi _1}$ we have $y\in L^{\varphi \ominus \varphi _1}$ and

$$\begin{aligned} \left\Vert y\right\Vert _{\varphi \ominus \varphi _1}\le \sup \limits _{n\in \mathbb {N}}\left\Vert y_n\right\Vert _{\varphi \ominus \varphi _1} \le 4c\left\Vert y_ n\right\Vert _M\le 4c\left\Vert y\right\Vert _M, \end{aligned}$$

which finishes the proof.

In the special case of variable exponent spaces we have the following corollary. It has been recently proved in [10] using elementary methods. Recall that the variable exponent space (or Nakano space) is defined as $L^{p(\cdot )}:=L^{\varphi }$, where $\varphi (t,u)=u^{p(t)}$, for a measurable function $p:\Omega \rightarrow [1,\infty )$.

Corollary 10

Let $(\Omega ,\Sigma ,\mu )$ be non-atomic and let $p,q:\Omega \rightarrow [1,\infty )$ be two measurable functions satisfying $q(t)\le p(t)$ for $\mu$-a.e. $t\in \Omega$. Then

$$\begin{aligned} M(L^{p(\cdot )},L^{q(\cdot )})=L^{r(\cdot )}, \end{aligned}$$

where $\frac{1}{p(t)}+\frac{1}{r(t)}=\frac{1}{q(t)}$ for $\mu$-a.e. $t\in \Omega$.

Proof

First of all, observe that each Nakano space $L^{p(\cdot )}$ may be equivalently defined by the Musielak–Orlicz function $\varphi _p(t,u)=\frac{1}{p(t)}u^{p(t)}$. In fact, we see that for $\varphi (t,u)=u^{p(t)}$ there holds

$$\begin{aligned} \varphi \left( t,\frac{u}{2}\right) =\left( \frac{u}{2}\right) ^{p(t)}\le \frac{1}{p(t)}u^{p(t)}=\varphi _p(t,u)\le \varphi (t,u), \end{aligned}$$

for each $t\in \Omega$ and $u>0$, which means that $L^{p(\cdot )}=L^{\varphi _p}$. Now the proof follows directly from Example 1 and the above theorem.

5 Pointwise products

If $\varphi , \varphi _1, \varphi _2$ are Musielak–Orlicz functions we write $\varphi _1^{-1}\varphi _2^{-1}\prec \varphi ^{-1}$ if there exists a constant $C>0$ such that

$$\begin{aligned} C\varphi _1^{-1}(t,u)\varphi _2^{-1}(t,u)\le \varphi ^{-1}(t,u) \end{aligned}$$

for a.e. $t\in \Omega$ and each $u\ge 0$. Similarly, we write $\varphi _1^{-1}\varphi _2^{-1}\succ \varphi ^{-1}$ if there exists a constant $C>0$ such that for a.e. $t\in \Omega$ and each $u\ge 0$

$$\begin{aligned} C\varphi _1^{-1}(t,u)\varphi _2^{-1}(t,u)\ge \varphi ^{-1}(t,u). \end{aligned}$$

Moreover, $\varphi _1^{-1}\varphi _2^{-1}\approx \varphi ^{-1}$ means that $\varphi _1^{-1}\varphi _2^{-1}\prec \varphi ^{-1}$ and $\varphi _1^{-1}\varphi _2^{-1}\succ \varphi ^{-1}$.

Recall the classical Lozanovskii factorization theorem (see [14, Theorem 6], cf. [6]) which says that each Banach ideal space E factorizes $L^1$, this is

$$\begin{aligned} E\odot M(E,L^1)=L^1. \end{aligned}$$

Generalizing this idea, for a couple of Banach ideal spaces E, F we say that E factorizes F if

$$\begin{aligned} E\odot M(E,F)=F, \end{aligned}$$

(see [12, Section 6] for a discussion of the general factorization problem). Recently the authors proved in [13, Theorem 2] that for a pair of Young functions $\varphi , \varphi _1$, the function space $L^{\varphi _1}$ may be factorized by $L^{\varphi }$ if and only if

$$\begin{aligned} \varphi _1^{-1}(\varphi \ominus \varphi _1)^{-1}\approx \varphi ^{-1}. \end{aligned}$$

(5.1)

That result is based on Theorem 5 in [12], which states that in the case of non-atomic and finite measure space, given three Young functions $\varphi ,\varphi _0, \varphi _1$, there holds

$$\begin{aligned} L^{\varphi }\odot L^{\varphi _1}=L^{\varphi _0} \end{aligned}$$

if and only if

$$\begin{aligned} \varphi _1^{-1}\varphi _0^{-1}\approx \varphi ^{-1}. \end{aligned}$$

In this section we will show that, in the case of Musielak–Orlicz spaces, the condition (5.1) is sufficient, but not necessary to have the factorization

$$\begin{aligned} L^{\varphi }\odot L^{\varphi _1}=L^{\varphi _0}. \end{aligned}$$

An immediate consequence of Theorem 9 is the following inclusion.

Lemma 11

Let $\varphi , \varphi _1$ be Musielak–Orlicz functions over $(\Omega ,\Sigma ,\mu )$. If $\hbox {supp}(L^{\varphi _1})=\Omega$ then

$$\begin{aligned} L^{\varphi }\subset L^{\varphi \ominus \varphi _1}\odot L^{\varphi _1} \end{aligned}$$

Proof

Let $x\in L^{\varphi \ominus \varphi _1}$ and $y\in L^{\varphi _1}$. Then, since $M(L^{\varphi _1},L^{\varphi })=L^{\varphi \ominus \varphi _1}$, we see that

$$\begin{aligned} xy\in L^{\varphi } \end{aligned}$$

and

$$\begin{aligned} \left\Vert xy\right\Vert _{\varphi }\le \left\Vert x\right\Vert _{M}\left\Vert y\right\Vert _{\varphi _1}\le c\left\Vert x\right\Vert _{\varphi \ominus \varphi _1}\left\Vert y\right\Vert _{\varphi _1}. \end{aligned}$$

Lemma 12

Let $\varphi ,\varphi _0, \varphi _1$ be Musielak–Orlicz functions over $(\Omega ,\Sigma ,\mu )$. Assume that $\varphi _1^{-1}\varphi _0^{-1}\succ \varphi ^{-1}$ and $\hbox {supp}L^{\varphi _1}=\Omega$. Then

$$\begin{aligned} L^{\varphi }\subset L^{\varphi _0}\odot L^{\varphi _1}. \end{aligned}$$

Proof

Denote by $c\ge 1$ the constant of inclusion

$$\begin{aligned} L^{\varphi }[\Omega _{\infty ,\infty }]\subset L^{\infty }(b_{\varphi }^{-1})[\Omega _{\infty ,\infty }]. \end{aligned}$$

Let $0\le z\in L^{\varphi }$ be such that $\left\Vert z\right\Vert _{\varphi }=\frac{2}{3c}$. Put $y(t):=\varphi (t,z(t))$. We have $y(t)<\infty$ a.e., since $z(t)\le \frac{2}{3}b_{\varphi }(t)$. For $i=0,1$, define

$$\begin{aligned}z_i(t) := {\left\{ \begin{array}{ll} \varphi _i^{-1}(t,y(t))\sqrt{\frac{z(t)}{\varphi _0^{-1}(t,y(t))\varphi _1^{-1}(t,y(t))}} &{}\text{ if } t\in \hbox {supp}(z)\\ 0 &{}\text{ if } t\notin \hbox {supp}(z) \end{array}\right. }\end{aligned}$$

Note that $z=z_0z_1$. We will show that $z_i\in L^{\varphi _i}$ for $i=0,1$. Let $D>0$ be such that

$$\begin{aligned} D\varphi _1^{-1}(t,u)\varphi _0^{-1}(t,u)\ge \varphi ^{-1}(t,u). \end{aligned}$$

We claim that

$$\begin{aligned} \varphi _i\left( t,\frac{z_i(t)}{\sqrt{D}}\right) \le y(t). \end{aligned}$$

(5.2)

If $y(t)=0$ then

$$\begin{aligned} z_i(t)=a_{\varphi }(t)\sqrt{\frac{z(t)}{a_{\varphi _0}(t)a_{\varphi _1}(t)}} \le a_{\varphi }(t)\sqrt{\frac{a_{\varphi }(t)}{a_{\varphi _0}(t)a_{\varphi _1}(t)}} \le a_{\varphi }(t)\sqrt{D}\end{aligned}$$

thus

$$\begin{aligned}\varphi _i\left( t,\frac{z_i(t)}{\sqrt{D}}\right) =0. \end{aligned}$$

If $y(t)>0$ then

$$\begin{aligned} z_i(t)&=\varphi _i^{-1}(t,y(t))\sqrt{\frac{z(t)}{\varphi _0^{-1}(t,y(t))\varphi _1^{-1}(t,y(t))}} \\&\le \varphi _i^{-1}(t,y(t))\sqrt{\frac{Dz(t)}{\varphi ^{-1}(t,y(t))}} \\&= \varphi _i^{-1}(t,y(t))\sqrt{\frac{Dz(t)}{z(t)}}=\varphi _i^{-1}(t,y(t))\sqrt{D}. \end{aligned}$$

Therefore,

$$\begin{aligned} \varphi _i\left( t,\frac{z_i(t)}{\sqrt{D}}\right) \le \varphi _i(t,\varphi _i^{-1}(t,y(t)))=y(t) \end{aligned}$$

and the claim is proved. Integrating both sides in (5.2) we obtain

$$\begin{aligned} I_{\varphi _i}\left( \frac{z_i}{\sqrt{D}}\right) \le I_{\varphi }\left( z\right) \le 1, \end{aligned}$$

for $i=0,1$. It follows, that

$$\begin{aligned} \left\Vert z_i\right\Vert _{\varphi _i}\le \sqrt{D}\le \sqrt{2Dc\left\Vert z\right\Vert _{\varphi }}. \end{aligned}$$

This means that $z\in L^{\varphi _0}\odot L^{\varphi _1}$ and

$$\begin{aligned} \left\Vert z\right\Vert _{L^{\varphi _0}\odot L^{\varphi _1}}\le 2Dc\left\Vert z\right\Vert _{\varphi }. \end{aligned}$$

Recall that for Musielak–Orlicz functions $\varphi , \varphi _1$, the generalized Young inequality implies that

$$\begin{aligned} \varphi _1^{-1}(\varphi \ominus \varphi _1)^{-1}\prec \varphi ^{-1} \end{aligned}$$

(see for example [11]).

Corollary 13

Let $\varphi , \varphi _1$ be Musielak–Orlicz functions over a measure space $(\Omega ,\Sigma ,\mu )$. If $\varphi _1^{-1}(\varphi \ominus \varphi _1)^{-1}\approx \varphi ^{-1}$ then $L^{\varphi _1}$ factorizes $L^{\varphi }$.

We finish the paper providing an example, which shows that the opposite implication does not hold. In particular, Theorem 2 in [13] cannot be directly generalized to Musielak–Orlicz spaces.

Example 14

Consider $\Omega =[0,1/2)$ with the Lebesgue measure. Define the following Musielak–Orlicz functions

$$\begin{aligned}&\varphi (t,u)=\max \{u-t,0\},\\&\varphi _1(t,u)=u, \end{aligned}$$

for $t\in \Omega$ and $u\ge 0$. Then $L^{\varphi }=L^{\varphi _1}=L^1$. Moreover, we have

$$\begin{aligned} \varphi \ominus \varphi _1(t,u)= {\left\{ \begin{array}{ll} 0 &{}\text{ if } 0\le u\le 1\\ \infty &{}\text{ if } u>1, \end{array}\right. } \end{aligned}$$

thus $L^{\varphi \ominus \varphi _1}=L^{\infty }$. In consequence, the factorization

$$\begin{aligned} L^{\varphi _1}\odot L^{\varphi \ominus \varphi _1}=L^{\varphi } \end{aligned}$$

holds. On the other hand an easy computations show that

$$\begin{aligned} (\varphi \ominus \varphi _1)^{-1}(t,u)=1,\ \varphi ^{-1}(t,u)=u+t,\ \varphi _1^{-1}(t,u)=u. \end{aligned}$$

We have $\varphi _1^{-1}(t,u)(\varphi \ominus \varphi _1)^{-1}(t,u)=u$, thus there is no constant D such that

$$\begin{aligned} D\varphi _1^{-1}(t,u)(\varphi \ominus \varphi _1)^{-1}(t,u)\ge \varphi ^{-1}(t,u) \end{aligned}$$

for every $t\in \Omega$ and $u\ge 0$ (take for example $u=0$ and $t>0$). Hence

$$\begin{aligned} \varphi _1^{-1}(\varphi \ominus \varphi _1)^{-1}\nsucc \varphi ^{-1}. \end{aligned}$$