Pointwise mutipliers of Orlicz function spaces and factorization

Abstract

In the paper we find representation of the space of pointwise multipliers between two Orlicz function spaces, which appears to be another Orlicz space and the formula for the Young function generating this space is given. Further, we apply this result to find necessary and sufficient conditions for factorization of Orlicz function spaces.

1 Introduction

Given two Orlicz spaces $L^{\varphi _1},L^{\varphi }$ over the same measure space, the space of pointwise multipliers $M(L^{\varphi _1},L^{\varphi })$ is the space of all functions x, such that $xy\in L^{\varphi }$ for each $y\in L^{\varphi _1}$, equipped with the operator norm. The problem of identifying such spaces was investigated by many authors, starting from Shragin [14], Ando [1], O’Neil [11] and Zabreiko–Rutickii [16], who gave a number of partial answers.

These investigations were continued in number of directions and results were presented in different forms. One of them is the following result from Maligranda–Nakaii paper [8], which states that if for two given Young functions $\varphi ,\varphi _1$ there is a third one $\varphi _2$ satisfying

$$\begin{aligned} \varphi _1^{-1}\varphi _2^{-1}\approx \varphi ^{-1}, \end{aligned}$$

(1.1)

then

$$\begin{aligned} M(L^{\varphi _1},L^{\varphi })=L^{\varphi _2}. \end{aligned}$$

(1.2)

This result, however, neither gives any information when such a function $\varphi _2$ exists, nor says anything how to find it. Further, it was proved in [5, Cor. 6.2] that condition (1.1) is necessary for a wide class of $\varphi ,\varphi _1$ functions satisfying some additional properties, but at the same time Example 7.8 from [5] ensures that in general it is not a case, i.e. there are functions $\varphi ,\varphi _1$ such that no Young function $\varphi _2$ satisfies (1.1), while $M(L^{\varphi _1},L^{\varphi })=L^{\infty }$, which is also Orlicz space generated by the function $\varphi _2$ defined as $\varphi _2(t)=0$ for $0\le t\le 1$ and $\varphi _2(t)=\infty$ for $1< t$. In particular, these functions do not satisfy (1.1), although (1.2) holds.

On the other hand, there is a natural candidate for a function $\varphi _2$ satisfying

$$\begin{aligned} M(L^{\varphi _1},L^{\varphi })=L^{\varphi _2}. \end{aligned}$$

Such a function is the following generalization of Young conjugate function (a kind of generalized Legendre transform considered also in convex analysis, for example in [15]) defined for two Orlicz functions $\varphi ,\varphi _1$ as

$$\begin{aligned} \varphi \ominus \varphi _1(t)=\sup _{s>0}\{\varphi (st)-\varphi _1(s)\}. \end{aligned}$$

The function $\varphi \ominus \varphi _1$ is called to be conjugate to $\varphi _1$ with respect to $\varphi$.

Also in [5] this construction was compared with condition (1.1) and it happens that very often $\varphi _2=\varphi \ominus \varphi _1$ satisfies (1.1) (cf. [5, Thm. 7.9]), but once again Example 7.8 from [5] shows that $\varphi _2=\varphi \ominus \varphi _1$ need not satisfy (1.1). In that example, anyhow, there holds $L^{\infty }=L^{\varphi \ominus \varphi _1}$, so that $M(L^{\varphi _1},L^{\varphi })=L^{\varphi \ominus \varphi _1}$. Therefore, it is natural to expect that in general

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

(1.3)

as was already conjectured in [5]. In fact, such theorem was stated for Orlicz N-functions by Maurey in [10], but his proof depends heavily on the false conjecture, that the construction $\varphi \ominus \varphi _1$ enjoys involution property, i.e. $\varphi \ominus (\varphi \ominus \varphi _1)=\varphi _1$ (see Example 7.12 in [5] for counterexample).

On the other hand, the formula (1.3) was already proved for Orlicz sequence spaces by Djakov and Ramanujan in [4], where they used a slightly modified construction $\varphi \ominus \varphi _1$ (the supremum is taken only over $0<s\le 1$). This modification appeared to be appropriate for sequence case, because then only behaviour of Young functions for small arguments is important, while cannot be used for function spaces. Anyhow, we will borrow some ideas from [4].

In our main Theorem 1 we prove that (1.3) holds in full generality for Orlicz function spaces, as well over finite and infinite nonatomic measure. Then we use this result to find that $\varphi _2=\varphi \ominus \varphi _1$ satisfies (1.1) if and anly if $L^{\varphi _1}$ factorizes $L^{\varphi }$, which completes the discussion from [6].

2 Notation and preliminaries

Let $L^0=L^0(\Omega ,\Sigma ,\mu )$ be the space of all classes of equivalence (with respect to equality $\mu$-a.e.) of $\mu$-measurable, real valued functions on $\Omega$, where $(\Omega ,\Sigma ,\mu )$ is a $\sigma$-finite complete measure space. A Banach space $X\subset L^0$ is called the Banach ideal space if it satisfies the so called ideal property, i.e. $x\in L^0,y\in X$ with $|x|\le |y|$ implies $x\in X$ and $\Vert x\Vert _X\le \Vert y\Vert _X$ (here $|x|\le |y|$ means that $|x(t)|\le |y(t)|$ for $\mu$-a.e. $t\in \Omega$), and it contains a weak unit, i.e. a function $x\in X$ such that $x(t)>0$ for $\mu$-a.e. $t\in \Omega$. When $(\Omega ,\Sigma ,\mu )$ is purely nonatomic measure space, the respective space is called Banach function space (abbreviation B.f.s.), while in case of $\mathbb {N}$ with counting measure we shall speak about Banach sequence space. A Banach ideal space X satisfies the Fatou property when given a 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 $\Vert x\Vert _X\le \sup _n\Vert x_n\Vert _X$.

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

Given two Banach ideal spaces X, Y over the same measure space $(\Omega ,\Sigma ,\mu )$, the space of pointwise multipliers from X to Y is defined as

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

with the natural operator norm

$$\begin{aligned} \Vert y\Vert _{M(X,Y)}=\sup _{\Vert x\Vert _X\le 1}\Vert xy\Vert _Y. \end{aligned}$$

When there is no risk of confusion we will just write $\Vert \cdot \Vert _M$ for the norm of M(X, Y). A space of pointwise multipliers may be trivial, for example for nonatomic measure space $M(L^p,L^q)=\{0\}$ when $1\le p<q$, and therefore it need not be a Banach function space in the sense of above definition. Anyhow, it is a Banach space with the ideal property (see for example [9]). To provide some intuition for multipliers let us recall that $M(L^p,L^q)=L^r$ when $p>q\ge 1$, $1/p+1/r=1/q$ and $M(X,L^1)=X'$, where $X'$ is the Köthe dual of X (see [5, 6, 9] for more examples).

A function $\varphi :[0,\infty )\rightarrow [0,\infty ]$ will be called a Young function if it is convex, non-decreasing and $\varphi (0)=0$. We will need the following parameters

$$\begin{aligned} a_{\varphi }=\sup \{t\ge 0:\varphi (t)=0\} \text { and } b_{\varphi }=\sup \{t\ge 0:\varphi (t)<\infty \}. \end{aligned}$$

A Young fuction $\varphi$ is called Orlicz function when $b_{\varphi }=\infty$. For a Young function $\varphi$ by $\varphi ^{-1}$ we understand the right-continuous inverse defined as $\varphi ^{-1}(v) = \inf \{u\ge 0: \varphi (u)>v\}$ for $v\ge 0$.

Let $\varphi$ be a Young function. The Orlicz space $L^{\varphi }$ is defined as

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

where the modular $I_\varphi$ is given by

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

and the Luxemburg–Nakano norm is defined as

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

We point out here that the function $\varphi \equiv 0$ is excluded from the definition of Young functions, but we allow $\varphi (u)=\infty$ for each $u>0$ and understand that in this case $L^{\varphi }=\{0\}$.

We will often use the following relation between norm and modular. For $x\in L^{\varphi }$

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

(2.1)

(see for example [7]).

Given two Orlicz functions $\varphi ,\varphi _1$, the conjugate function $\varphi \ominus \varphi _1$ of $\varphi _1$ with respect to $\varphi$ is defined by

$$\begin{aligned} \varphi \ominus \varphi _1(u)=\sup _{0\le s}\{\varphi (su)-\varphi _1(s)\}, \end{aligned}$$

for $u\ge 0$. Since we need to deal with Young functions, one may be confused by possibility of appearance of indefinite symbol $\infty -\infty$ in the above definition, when $b_{\varphi },b_{\varphi _1}<\infty$. To avoid such a situation we understand that for Young functions $\varphi ,\varphi _1$ the conjugate function $\varphi \ominus \varphi _1$ is defined as

$$\begin{aligned} \varphi \ominus \varphi _1(u)=\left\{ \begin{array}{ll} {\sup }_{0\le s<\infty }\{\varphi (su)-\varphi _1(s)\}, &{} \text {when }b_{\varphi _1}=\infty ,\\ {\sup }_{0<s<b_{\varphi _1}}\{\varphi (su)-\varphi _1(s)\}, &{} \text {when }b_{\varphi _1}<\infty \text { and }\varphi _1(b_{\varphi _1})=\infty ,\\ {\sup }_{0<s\le b_{\varphi _1}}\{\varphi (su)-\varphi _1(s)\}, &{} \text { when }b_{\varphi _1}<\infty \text { and }\varphi _1(b_{\varphi _1})<\infty . \end{array} \right. \end{aligned}$$

Notice that it is just a natural generalization of conjugate function in a sense of Young, i.e. when $\varphi (u)=u$ we get the classical conjugate function $\varphi _1^*$ to $\varphi _1$. Of course, functions $\varphi ,\varphi _1$ and $\varphi \ominus \varphi _1$ satisfy the generalized Young inequality, i.e.

$$\begin{aligned} \varphi (uv)\le \varphi \ominus \varphi _1(u)+\varphi _1(v) \end{aligned}$$

for each $u,v\ge 0$.

We will also need the following construction.

Definition 1

For two Young functions $\varphi ,\varphi _1$ and $0<a< b_{\varphi _1}$ we define

$$\begin{aligned} \varphi \ominus _a \varphi _1(u)=\sup _{0\le s\le a}\{\varphi (su)-\varphi _1(s)\}, u\ge 0. \end{aligned}$$

Such defined function $\varphi \ominus _a \varphi _1$ enjoys the following elementary properties.

Lemma 2

Let $\varphi ,\varphi _1$ be two Young functions.

• (i) $\varphi \ominus _a \varphi _1$ is Young function for each $0<a< b_{\varphi _1}$.

• (ii) For each $t\ge 0$ there holds

  $$\begin{aligned} \lim \limits _{a\rightarrow b_{\varphi _1}^{-}}\varphi \ominus _a \varphi _1(u)=\varphi \ominus \varphi _1(u). \end{aligned}$$

Remark 3

Notice that dilations of Young functions do not change Orlicz spaces, i.e. when $\varphi$ is a Young function and $\psi$ is defined by $\psi (u)=\varphi (au)$ for some $a>0$, then $L^{\varphi }=L^{\psi }$. It gives a reason to expect that dilating $\varphi ,\varphi _1$ results in dilation of $\varphi \ominus \varphi _1$. In fact, let $\varphi ,\varphi _1$ be Young functions and put $\psi (u)=\varphi (au)$, $\psi _1(u)=\varphi _1(bu)$. Then

$$\begin{aligned} \psi \ominus \psi _1(u)=\sup _{0<s}(\varphi (aus)-\varphi _1(bs))=\sup _{0<s} (\varphi (aus/b)-\varphi _1(s))=\varphi \ominus \varphi _1(au/b). \end{aligned}$$

Moreover, if $b_{\varphi }=b_{\varphi _1}<\infty$, then supremum in the definition of $\varphi \ominus \varphi _1$ is attained for each $u<1$, i.e. for each $u<1$ there is $0<s<b_{\varphi _1}$ such that $\varphi \ominus \varphi _1(u)=\varphi (us)-\varphi _1(s)$. In particular, $b_{\varphi \ominus \varphi _1}=1$.

Remark 4

Let us also recall that a fundamental function $f_{\varphi }$ of an Orlicz space $L^{\varphi }$ is defined for $0\le t \le \mu (\Omega )$ as $f_{\varphi }(t)=\Vert \chi _{A}\Vert _{\varphi }$, where $\mu (A)=t$. Notice that it is well defined, since $\Vert \chi _{A}\Vert _{\varphi }$ does not depend on particular choice of measurable set $A\subset \Omega$ with $\mu (A)=t$ (in general Orlicz spaces belong to the class of the so called rearrangement invariant spaces—see for example [2] for respective definitions). Further, it is well known that $f_{\varphi }$ is given by the formula $f_{\varphi }(t)=\frac{1}{\varphi ^{-1}(1/t)}$, for $0<t<\mu (\Omega )$ and $f_{\varphi }(0)=0$. In particular, the fundamental function of $L^{\varphi }$ is right-continuous at 0 if and only if $b_{\varphi }=\infty$, or equivalently, $b_{\varphi }=\infty$ if and only if for each $\varepsilon >0$ there is $\delta >0$ such that if $A\in \Sigma$, $\mu (A)<\delta$ then $\Vert \chi _{A}\Vert _{\varphi }<\varepsilon$.

3 Multipliers of Orlicz function spaces

Since now on we are interested only in Orlicz function spaces, so that the underlying measure space $(\Omega ,\Sigma ,\mu )$ is understood to be purely nonatomic for all spaces below and to the end of the paper.

Lemma 5

Let $\varphi ,\varphi _1$ be Young functions such that $b_{\varphi }<\infty$ and $b_{\varphi _1}=\infty .$ Then

$$\begin{aligned} M(L^{\varphi _1},L^{\varphi })=\{0\}. \end{aligned}$$

Proof

The proof follows immediately from Proposition 3.2 in [5], since under our assumptions $L^{\varphi _1}\not \subset L^\infty$ but $L^{\varphi }\subset L^\infty$. $\square$

Lemma 6

Let $\varphi ,\varphi _1$ be Young functions and $b_{\varphi }<\infty$. Then

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

Proof

Suppose that $M(L^{\varphi _1},L^{\varphi })\not \subset L^{\infty }$. Then there exists $0\le y\in M(L^{\varphi _1},L^{\varphi })$ such that $\Vert y\Vert _{M}=1$ and for each $n>0$

$$\begin{aligned} \mu (\{t\in \Omega : y(t)\ge n\})>0. \end{aligned}$$

Denote $A_n=\{t\in \Omega : y(t)\ge n\}$ for $n\in \mathbb {N}$. Then $\left||n\chi _{A_n}\right||_{M}\le 1$ and for $A_{n_0}$ chosen in such a way that $\mu (A_{n_0})<\infty$, it follows

$$\begin{aligned} \Vert y\Vert _{M}\ge \Vert n\chi _{A_n}\Vert _{M}\ge \frac{n}{\Vert \chi _{A_{n_0}}\Vert _{\varphi _1}}\Vert \chi _{A_n}\chi _{A_{n_0}}\Vert _{\varphi }=\frac{n}{\Vert \chi _{A_{n_0}}\Vert _{\varphi _1}}\Vert \chi _{A_n}\Vert _{\varphi }\ge \frac{nb_{\varphi }^{-1}}{\Vert \chi _{A_{n_0}}\Vert _{\varphi _1}}, \end{aligned}$$

for each $n>n_0$. This contradiction shows that $M(L^{\varphi _1},L^{\varphi })\subset L^{\infty }$. $\square$

We are in a position to prove the main theorem.

Theorem 1

Let $\varphi ,\varphi _1$ be Young functions. Then

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

Proof

The inclusion

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

(3.1)

is well known (see [1, 5, 8] or [11]) and follows from equivalence of generalized Young inequality and inequality $\varphi _1^{-1}(\varphi \ominus \varphi _1)^{-1}\lesssim \varphi ^{-1}$. For the completeness of presentation we present the proof which employs the generalized Young inequality directly. If $\varphi \ominus \varphi _1(u)=\infty$ for each $u>0$ then $L^{\varphi \ominus \varphi _1}=\{0\}$ and inclusion trivially holds. Suppose $L^{\varphi \ominus \varphi _1}\ne \{0\}$, i.e. $\varphi \ominus \varphi _1(u)<\infty$ for some $u>0$. Let $y\in L^{\varphi \ominus \varphi _1}$ and $x\in L^{\varphi _1}$ be such that

$$\begin{aligned} \left||y\right||_{\varphi \ominus \varphi _1}\le \frac{1}{2} \text { and } \left||x\right||_{\varphi _1}\le \frac{1}{2}. \end{aligned}$$

Then generalized Young inequality gives

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

Consequently $y x\in L^{\varphi }$ and $\left||y x\right||_{\varphi }\le 1$. Therefore, $L^{\varphi \ominus \varphi _1}\subset M(L^{\varphi _1},L^{\varphi })$ and

$$\begin{aligned} \left||y\right||_{M}\le 4\left||y\right||_{\varphi \ominus \varphi _1}. \end{aligned}$$

To prove the second inclusion it is enough to indicate a constant $c>0$ such that for each simple function $y\in M(L^{\varphi _1},L^{\varphi })$ there holds

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

(3.2)

In fact, it follows directly from the Fatou property of both $L^{\varphi \ominus \varphi _1}$ and $M(L^{\varphi _1},L^{\varphi })$ spaces (it is elementary fact that M(X, Y) has the Fatou property when Y has so). Let $0\le y\in M(L^{\varphi _1},L^{\varphi })$ and $0\le y_n\uparrow y$ $\mu$-a.e., where $y_n$ are simple functions. Then, by (3.2),

$$\begin{aligned} \Vert y_n\Vert _{\varphi \ominus \varphi _1}\le c\Vert y_n\Vert _M\rightarrow c\Vert y\Vert _M \end{aligned}$$

and so the Fatou property of $L^{\varphi \ominus \varphi _1}$ implies $y\in L^{\varphi \ominus \varphi _1}$ and $\Vert y\Vert _{\varphi \ominus \varphi _1}\le c\Vert y\Vert _M$.

The proof of (3.2) will be divided into four cases, depending on finiteness of $b_{\varphi }$ and $b_{\varphi _1}$.

Consider firstly the most important case $b_{\varphi }=b_{\varphi _1}=\infty$. Let $0\le y\in M(L^{\varphi _1},L^{\varphi })$ be a simple function of the form $y=\sum _k a_k\chi _{B_k}$ and such that $\left||y\right||_{M}\le \frac{1}{2}$. We will show that for each $a>1$

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

Let $a>1$ be arbitrary. For each $a_k$ there exists $b_k\ge 0$ such that

$$\begin{aligned} \varphi (a_kb_k)=\varphi \ominus _a\varphi _1(a_k)+\varphi _1(b_k). \end{aligned}$$

This is, for $x=\sum _k b_k\chi _{B_k}$, there holds $\varphi (xy)=\varphi \ominus _a\varphi _1(x)+\varphi _1(y)$. Note that from definition of $\varphi \ominus _a\varphi _1$ we have $x(t)\le a$ for each $t\in \Omega$. Further, since $b_{\varphi _1}=\infty$, there exists $t_a>0$ such that $\Vert \chi _{A}\Vert _{\varphi _1}\le \frac{1}{a}$ for each $A\subset \Omega$ with $\mu (A)<t_a$ (see Remark 4). Suppose $\mu (\Omega )=\infty$. Since $(\Omega ,\Sigma ,\mu )$ is $\sigma$-finite and atomless, we can divide $\Omega$ into a sequence of pairwise disjoint sets $(A_n)$ with $\mu (A_n)= t_a$ for each $n\in \mathbb {N}$ and $\Omega =\bigcup A_n$. In the case of $\mu (\Omega )<\infty$ the sequence $(A_n)$ may be chosen finite and such that $\mu (A_n)=\delta \le t_a$ for each $n=1,\ldots ,N$ with $\Omega =\bigcup A_n$.

In any case, for each $A_n$ we have

$$\begin{aligned} \Vert y x\chi _{A_n}\Vert _{\varphi }\le \Vert y\Vert _M\Vert x\chi _{A_n}\Vert _{\varphi _1}\le \frac{a}{2} \Vert \chi _{A_n}\Vert _{\varphi _1}\le \frac{1}{2}, \end{aligned}$$

because $\mu (A_n)\le t_a$ and $x(t)\le a$ for $t\in \Omega$. In consequence, using inequality $\varphi _1(x)\le \varphi (yx)$, we have for each $A_n$

$$\begin{aligned} I_{\varphi _1}(x\chi _{A_n})\le I_{\varphi }(y x\chi _{A_n})\le \Vert y x\chi _{A_n}\Vert _{\varphi }\le \frac{1}{2}. \end{aligned}$$

(3.3)

Define now

$$\begin{aligned} x_n=\sum _{k=1}^nx\chi _{A_k}. \end{aligned}$$

We claim that $I_{\varphi _1}(x_n)\le \frac{1}{2}$ for each n. It will be shown by induction. For $n=1$ it comes from (3.3). Let $n>1$ and suppose

$$\begin{aligned} I_{\varphi _1}(x_{n-1})\le \frac{1}{2}. \end{aligned}$$

It follows

$$\begin{aligned} I_{\varphi _1}(x_{n})=I_{\varphi _1}(x_{n-1})+I_{\varphi _1}(x\chi _{A_n})\le 1, \end{aligned}$$

thus $\left||x_n\right||_{\varphi _1}\le 1$. Moreover, inequality

$$\begin{aligned} \left||y x_n\right||_{\varphi }\le \frac{1}{2}\left||x_n\right||_{\varphi _1}\le \frac{1}{2} \end{aligned}$$

together with $\varphi _1(x)\le \varphi (yx)$ imply

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

It means we proved the claim and can proceed with the proof.

Clearly, $x_n\uparrow x$ $\mu$-a.e., thus from the Fatou property of $L^{\varphi _1}$ we obtain that $x\in L^{\varphi _1}$ and

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

Finally, inequalities $\varphi \ominus _a\varphi _1(y)\le \varphi (yx)$ and $\Vert y x\Vert _{\varphi }\le \frac{1}{2}\Vert x\Vert _{\varphi _1}\le \frac{1}{2}$ give

$$\begin{aligned} I_{\varphi \ominus _a\varphi _1}(y)\le I_{\varphi }(y x)\le \left||y x\right||_{\varphi }\le \frac{1}{2}. \end{aligned}$$

Applying the Fatou Lemma we obtain

$$\begin{aligned} I_{\varphi \ominus \varphi _1}(y)= \int \varphi \ominus \varphi _1(y)d\mu \le \liminf _{a\rightarrow \infty }\int \varphi \ominus _a\varphi _1(y)d\mu \le \frac{1}{2}. \end{aligned}$$

In consequence $y\in L^{\varphi \ominus \varphi _1}$ with $\Vert y\Vert _{\varphi \ominus \varphi _1}\le 1$. This gives also constant for inclusion, i.e.

$$\begin{aligned} \Vert y\Vert _{\varphi \ominus \varphi _1}\le 2\Vert y\Vert _{M}, \end{aligned}$$

when $y\in M(L^{\varphi _1},L^{\varphi })$.

Let us consider the second case, this is $b_{\varphi }=\infty$ and $b_{\varphi _1}<\infty$. Without loss of generality we can assume that $b_{\varphi _1}>1$ (see Remark 3). Let $0\le y\in M(L^{\varphi _1},L^{\varphi })$ be a simple function satisfying $\Vert y\Vert _{M}\le \frac{1}{2b_{\varphi _1}}$. Notice that $b_{\varphi }=\infty$ with $b_{\varphi _1}<\infty$ imply that $b_{\varphi \ominus \varphi _1}=\infty$. Moreover, as before, there exists a simple function x such that $0<x(t)\le b_{\varphi _1}$ for each $t\in \Omega$ and

$$\begin{aligned} \varphi (yx)=\varphi \ominus \varphi _1(y)+\varphi _1(x) \end{aligned}$$

(see Remark 3). As before, we can find $t_0>0$ such that $\mu (A)<t_0$ implies $\Vert \chi _{A}\Vert _{\varphi _1}\le 1$. Selecting the sequence $(A_n)$ like previously, but with $\mu (A_n)\le t_0$ for each $A_n$, we obtain

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

Define further

$$\begin{aligned} x_n=\sum _{k=1}^nx\chi _{A_k}. \end{aligned}$$

Then it may be proved by the same induction as before, that $I_{\varphi _1}(x_n)\le \frac{1}{2}$ for each n. Following respective steps from previous case we get

$$\begin{aligned} \Vert y\Vert _{\varphi \ominus \varphi _1}\le 2b_{\varphi _1}\left||y\right||_{M}. \end{aligned}$$

Let now $b_{\varphi },b_{\varphi _1}<\infty$. We can assume that $b_{\varphi _1}=b_{\varphi }=1$ (see Remark 3). From Lemma 6 it follows that there exists a constant $c\ge 1$ such that for each $y\in M(L^{\varphi _1},L^{\varphi })$ we have

$$\begin{aligned} \Vert y\Vert _{\infty }\le c\Vert y\Vert _{M}. \end{aligned}$$

Let $0\le y\in M(L^{\varphi _1},L^{\varphi })$ be a simple function and $\Vert y\Vert _{M}\le \frac{1}{4c}$. We have $y(t)\le \frac{1}{4}\le \frac{b_{\varphi \ominus \varphi _1}}{2}$ (cf. Remark 3) for almost every $t\in \Omega$, therefore $\varphi \ominus \varphi _1(y(t))<\infty$. Consequently, we can choose a simple function x satisfying

$$\begin{aligned} \varphi (yx)=\varphi \ominus \varphi _1(y)+\varphi _1(x). \end{aligned}$$

Then $x(t)\le b_{\varphi }=1$ for each $t\in \Omega$. Further, we can find $t_0>0$ so that inequality

$$\begin{aligned} \Vert \chi _{A}\Vert _{\varphi _1}\le 2 \end{aligned}$$

is fulfilled for each A with $\mu (A)\le t_0$, just because $\lim _{t\rightarrow 0^+} f_{\varphi }(t)=b_{\varphi }=1$. Choosing a sequence $(A_n)$ as in previous cases we get

$$\begin{aligned} \Vert y x\chi _{A_n}\Vert _{\varphi }\le \frac{1}{4c}\Vert \chi _{A_n}\Vert _{\varphi _1}\le \frac{1}{2}. \end{aligned}$$

Once again we can show by induction that for each $x_n=\sum _{k=1}^nx\chi _{A_k}$ there holds $I_{\varphi _1}(x_n)\le \frac{1}{2}$. Therefore $\Vert x_n\Vert _{\varphi _1}\le 1$ and, by the Fatou property of $L^{\varphi _1}$, $\Vert x\Vert _{\varphi _1}\le 1$. It follows

$$\begin{aligned} \Vert y x\Vert _{\varphi }\le 1 \end{aligned}$$

and by inequality $\varphi \ominus \varphi _1(y)\le \varphi (yx)$ we obtain

$$\begin{aligned} I_{\varphi \ominus \varphi _1}(y)\le I_{\varphi }(y x)\le \Vert y x\Vert _{\varphi }\le 1. \end{aligned}$$

In consequence

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

Finally, there left the trivial case of $b_{\varphi }<\infty$, $b_{\varphi _1}=\infty$ to consider. However, Lemma 5 with the embedding (3.1) give

$$\begin{aligned} L^{\varphi \ominus \varphi _1}=M(L^{\varphi _1},L^{\varphi })=\{0\} \end{aligned}$$

and the proof is finished. $\square$

4 Factorization

Recall that given two B.f.s. X, Y over the same measure space, we say that X factorizes Y when

$$\begin{aligned} X\odot M(X,Y)=Y, \end{aligned}$$

where

$$\begin{aligned} X\odot M(X,Y)=\{z\in L^0:z=xy \mathrm{\ for\ some\ }x\in X, y\in M(X,Y)\}. \end{aligned}$$

The idea of such factorization goes back to Lozanovskii, who proved that each B.f.s. factorizes $L^1$. For more informations on factorization and its importance we send a reader to papers [3, 6] and [13] which are devoted mainly to this subject.

Also in [6] one may find a discussion on factorization of Orlicz spaces (and even more general Calderón–Lozanovskii spaces). Having in hand our representation $M(L^{\varphi _1},L^{\varphi })= L^{\varphi \ominus \varphi _1}$ we are able to complete this discussion by proving sufficient and necessary conditions for factorization in terms of respective Young functions.

We say that equivalence $\varphi _1^{-1}\varphi _2^{-1}\approx \varphi ^{-1}$ holds for all [large] arguments when there are constants $c,C>0$ such that

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

for all $u\ge 0$ [for some $u_0> 0$ and all $u>u_0$].

Theorem 2

Let $\varphi ,\varphi _1$ be two Young functions. Then $L^{\varphi _1}$ factorizes $L^{\varphi }$, i.e.

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

if and only if

• (i) equivalence $\varphi _1^{-1}(\varphi \ominus \varphi _1)^{-1}\approx \varphi ^{-1}$ is satisfied for all arguments when $\mu (\Omega )=\infty$.

• (ii) equivalence $\varphi _1^{-1}(\varphi \ominus \varphi _1)^{-1}\approx \varphi ^{-1}$ is satisfied for large arguments when $\mu (\Omega )<\infty$,

Proof

In the light of Theorem 1

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

Therefore $L^{\varphi _1}$ factorizes $L^{\varphi }$ if and only if $L^{\varphi _1}\odot L^{\varphi \ominus \varphi _1}=L^{\varphi }$. The latter, however, is equivalent with $\varphi _1^{-1}(\varphi \ominus \varphi _1)^{-1}\approx \varphi ^{-1}$ for all, or for large arguments, depending on $\Omega$, as proved in Corollary 6 from [6]. $\square$