Basic properties of multiplication and composition operators between distinct Orlicz spaces

Abstract

First, we present some simple (and easily verifiable) necessary conditions and sufficient conditions for boundedness of the multiplication operator $M_u$ and composition operator $C_T$ acting from Orlicz space $L^{\Phi _1}(\Omega )$ into Orlicz space $L^{\Phi _2}(\Omega )$ over arbitrary complete, $\sigma$-finite measure space $(\Omega ,\Sigma ,\mu )$. Next, we investigate the problem of conditions on the generating Young functions, the function u, and/or the function $h=d(\mu \circ T^{-1})/d\mu$, under which the operators $M_u$ and $C_T$ are of closed range or finite rank. Finally, we give necessary and sufficient conditions for boundedness of the operators $M_u$ and $C_T$ in terms of techniques developed within the theory of Musielak–Orlicz spaces.

1 Introduction

Given the functions f and T ($f:\Omega \rightarrow {\mathbb {R}}$, $T:\Omega \rightarrow \Omega$), one way to produce, under certain conditions, a new function is to compose them, i.e. to evaluate f at points T(t). The resulting function is denoted by $f\circ T$, and the operator $f\mapsto f\circ T$, as f runs through a linear function space, is a linear operator $C_T$ called the composition operator induced by a mapping T. Another way to produce a new function out of given functions u and f (each from $\Omega$ into ${\mathbb {R}}$) is to multiply them, whenever it makes sense, and this gives rise to the operator called the multiplication operator $M_u$ induced by a function u. A combination of the two methods leads to yet another operator, named the weighted composition operator, defined by $M_{u}C_{T}f=u (f\circ T)$. Composition and multiplication operators received much attention over the past several decades, especially in spaces of measurable functions such as $L^p$-spaces, Bergman spaces, and, to a lesser degree, Orlicz spaces. They also play an important role in the study of operators on Hilbert spaces. Consequently, by now there exists a vast literature on the properties of these transformations in various function spaces; we refer the interested reader to the beautiful books [5, 35, 36], and, for more recent studies, to [9,10,11, 14, 23, 25], for instance.

As said, composition and multiplication operators between classical $L^p$-spaces have been the subject-matter of intensive and extensive study and they feature prominently in operator theory, operator algebras, dynamical systems, and endomorphisms of Banach algebras. Composition operators were used by Banach [2] in his study of isometries between function spaces. In this paper, we aim at a generalization to Orlicz spaces of some results of the paper [38] concerning the classical Lebesgue spaces. Apparently, the paper [24] by Kumar in 1997 was the first attempt at the study of composition operators between Orlicz spaces. In 2004, Cui, Hudzik, Kumar, and Maligranda in [6] considered the composition operator between Orlicz spaces induced by a non-singular measurable transformation and studied its boundedness and compactness. Some other results on the boundedness and compactness of the composition operator between Orlicz spaces were published in [6,7,8, 12, 13, 16, 21, 25, 33].

Our concern in this paper is to state and prove some necessary conditions, sufficient conditions, and some simultaneously necessary and sufficient conditions for the composition and multiplication operator between distinct Orlicz spaces to be bounded or to have closed-range or finite rank. Our results generalize and improve on some recent results to be found in the literature.

2 Preliminaries and basic lemmas

In this section, for the convenience of the reader, we gather some essential facts on Orlicz spaces and prove two basic lemmas for later use. For more detail on Orlicz spaces, see [22, 34].

A function $\Phi :{\mathbb {R}}\rightarrow [0,\infty ]$ is called a Young function if $\Phi$ is convex, even, and $\Phi (0)=0$; we will also assume that $\Phi$ is neither identically zero nor identically infinite on $(0,\infty )$. The fact that $\Phi (0)=0$, along with the convexity of $\Phi$, implies that $\lim \nolimits _{x\rightarrow 0^+}\Phi (x)=0$; while $\Phi \ne 0$, again along with the convexity of $\Phi$, implies that $\lim \nolimits _{x\rightarrow \infty }\Phi (x)=\infty$. We set $a_{\Phi }:=\sup \{x\ge 0:\Phi (x)=0\}$ and $b_{\Phi }:=\sup \{x>0:\Phi (x)<\infty \}.$ Then it can be checked that $\Phi$ is continuous and nondecreasing on $[0,b_{\Phi })$ and strictly increasing on $[a_{\Phi },b_{\Phi })$ if $a_\Phi < b_\Phi$. We also assume the left-continuity of the function $\Phi$ at $b_\Phi$, i.e. $\lim _{x\rightarrow b_\Phi ^-} \Phi (x)=\Phi (b_\Phi )\in [0,\infty ]$.

To each Young function $\Phi$ is associated another convex function $\Psi :{\mathbb {R}}\rightarrow [0,\infty )$ with similar properties, defined by

$$\begin{aligned} \Psi (y)=\sup \{x|y|-\Phi (x):x\ge 0\} \quad (y\in {\mathbb {R}}). \end{aligned}$$

The function $\Psi$ is called the function complementary to $\Phi$ in the sense of Young. Any pair of complementary functions $(\Phi ,\Psi )$ satisfies Young’s inequality $xy\le \Phi (x)+\Psi (y)\,\, (x,y\in R)$.

The generalized inverse of the Young function $\Phi$ is defined by

$$\begin{aligned} \Phi ^{-1}(y)=\inf \{ x\ge 0: \Phi (x)> y\} \quad (y\in [0,\infty )). \end{aligned}$$

Notice that if $x\ge 0$, then $\Phi \big (\Phi ^{-1}(x)\big )\le x$, and if $\Phi (x)<\infty$, we also have $x\le \Phi ^{-1}\big (\Phi (x)\big )$. There are equalities in both cases when $\Phi$ is a Young function vanishing only at zero and taking only finite values. Also, if $(\Phi ,\Psi )$ is a pair of complementary Young functions, then

$$\begin{aligned} x<\Phi ^{-1}(x)\Psi ^{-1}(x)\le 2x \end{aligned}$$

(2.1)

for all $x\ge 0$ (Proposition 2.1.1(ii) in [34]).

By an N-function we mean a Young function vanishing only at zero, taking only finite values, and such that $\lim \nolimits _{x\rightarrow \infty }\Phi (x)/x=\infty$ and $\lim \nolimits _{x\rightarrow 0^+}\Phi (x)/x=0$. Note that then $a_\Phi =0,$ $b_\Phi =\infty$, and, as we said above, $\Phi$ is continuous and strictly increasing on $[0,\infty )$. Moreover, a function complementary to an N-function is again an N-function.

A Young function $\Phi$ is said to satisfy the $\Delta _{2}$-condition at $\infty$ if $\Phi (2x)\!\le \! K\Phi (x) \; ( x\!\ge \! x_{0})$ for some constants $K>0$ and $x_0\!>\!0$ with$\Phi (x_0)\!<\!\infty$. A Young function $\Phi$ satisfies the $\Delta _{2}$-condition globally if $\Phi (2x)\le K\Phi (x) \; ( x\ge 0)$ for some $K>0$.

A Young function $\Phi$ is said to satisfy the $\Delta '$-condition (respectively, the $\nabla '$-condition) at $\infty$, if there exist $c>0$ (respectively, $b>0$) and $x_0>0$ with $\Phi (x_0)<\infty$ such that

$$\begin{aligned} \Phi (xy)\le & {} c\,\Phi (x)\Phi (y) \quad (x,y\ge x_{0})\\ (\text{ respectively, } \Phi (bxy)\ge & {} \Phi (x)\Phi (y) \quad ( x,y\ge x_{0})). \end{aligned}$$

If $x_{0}=0$, these conditions are said to hold globally. Notice that if $\Phi \in \Delta '$, then $\Phi \in \Delta _{2}$ (both at $\infty$ and globally).

Let $\Phi _{1}, \Phi _{2}$ be Young functions. Then $\Phi _1$ is called stronger than $\Phi _2$ at $\infty$, which is denoted by $\Phi _1\mathrel {\overset{{\ell }}{\succ }}\Phi _2$ [or $\Phi _2\mathrel {\overset{{\ell }}{\prec }}\Phi _1$], if

$$\begin{aligned} \Phi _2(x)\le \Phi _1(ax)\quad (x\ge x_0) \end{aligned}$$

for some $a\ge 0$ and $x_0>0$ with $\Phi _2(x_0)<\infty$; if $x_0=0$, this condition is said to hold globally and is then denoted by $\Phi _1\mathrel {\overset{{a}}{\succ }}\Phi _2$ [or $\Phi _2\mathrel {\overset{{a}}{\prec }}\Phi _1$]. We record the following observation for later use.

Lemma 2.1

If $\Phi _1, \Phi _2, \Phi _3$ are Young functions vanishing only at zero, taking only finite values, and such that

$$\begin{aligned} \Phi _1(xy)\le \Phi _2(x)+\Phi _3(y)\quad (x,y\ge 0), \end{aligned}$$

then $\Phi _2\mathrel {\overset{{\ell }}{\nprec }}\Phi _1$, and hence also $\Phi _2\mathrel {\overset{{a}}{\nprec }}\Phi _1$.

Proof

The inequality in the assumptions of the lemma implies that $\Phi _1(\Phi _2^{-1}(x)\Phi _3^{-1}(x))\le 2x$ for all $x\ge 0$, and this, together with subadditivity of $\Phi _1^{-1},$ leads to

$$\begin{aligned} \Phi _2^{-1}(x)\Phi _3^{-1}(x)\le \Phi _1^{-1}(2x)\le 2\Phi _1^{-1}(x) \quad (x\ge 0). \end{aligned}$$

(2.2)

Now suppose, towards a contradiction, that there are $a>0$ and $x_0>0$ such that

$$\begin{aligned} \Phi _2(x)<\Phi _1(a x) \quad (x\ge x_0). \end{aligned}$$

Then, since $\Phi _1^{-1}$ is increasing on $[0,\infty )$, we have $\Phi _1^{-1}(\Phi _2(x))<\Phi _1^{-1}(\Phi _1(ax))= a x$, and so, replacing x with $\Phi _2(x)$ in (2.2), we get

$$\begin{aligned} x\,\Phi _3^{-1}(\Phi _2(x))=\Phi _2^{-1}(\Phi _2(x))\Phi _3^{-1}(\Phi _2(x))\le 2\Phi _1^{-1}(\Phi _2(x)) <2ax \quad (x\ge x_0). \end{aligned}$$

This means, since $\Phi _3$ is increasing on $[0,\infty )$, that $\Phi _1(x)< \Phi _3(2a)=\text{ const }.$ for all $x\ge x_0$, a contradiction, because $\Phi _2(x)\rightarrow \infty$ as $x\rightarrow \infty$. $\square$

Let $(\Omega , \Sigma , \mu )$ be a complete $\sigma$-finite measure space and let $L^0(\Omega )$ be the linear space of equivalence classes of $\Sigma$-measurable real-valued functions on $\Omega$, that is, we identify functions equal $\mu$-almost everywhere on $\Omega$. The support S of a measurable function f is defined by $S(f):=\{t\in \Omega : f(t)\ne 0\}$. For a Young function $\Phi$, the space

$$\begin{aligned} L^{\Phi }(\Omega )=\left\{ f\in L^0(\Omega ):\exists k>0, I_{\Phi }(kf)<\infty \right\} \end{aligned}$$

is a Banach space if it is equipped with the norm

$$\begin{aligned} \Vert f\Vert _{\Phi }=\inf \left\{ k>0:I_{\Phi }(f/k)\le 1\right\} , \end{aligned}$$

where $I_\Phi (f)=\int _{\Omega }\Phi (f(t))d\mu (t)$. The couple $(L^{\Phi }(\Omega ), \Vert \cdot \Vert _{\Phi })$ is called the Orlicz space generated by a Young function $\Phi$. For any $f\in L^0(\Omega )$, we have $I_\Phi (f)\le 1$ if and only if $\Vert f\Vert _{\Phi }\le 1$. In particular, if $I_\Phi (f)\le a\,\,(a>0)$, then $\Vert f\Vert _{\Phi }\le \max (a,1)$. If $\Phi \in \Delta _2$ and $f\in L^\Phi (\Omega )$, then $I_\Phi (kf)<\infty$ for all $k>0$.

Let $\Phi (x)=|x|^p/p$ with $1<p<\infty$; $\Phi$ is then a Young function and $\Psi (x)=|x|^{p'}{/}p'$, with $1{/}p+1{/}p'=1$, is the Young function complementary to $\Phi$. Thus, with this function $\Phi$ we retrieve the classical Lebesgue space $L^p(\Omega )$, i.e. $L^\Phi (\Omega )=L^p(\Omega )$.

Recall that an atom of the measure space $(\Omega ,\Sigma ,\mu )$ is a set ${\mathfrak {A}}\in \Sigma$ with $\mu ({\mathfrak {A}})>0$ such that if $F\in \Sigma$ and $F\subset {\mathfrak {A}}$, then either $\mu (F)=0$ or $\mu (F)=\mu ({\mathfrak {A}})$. A measure space $(\Omega ,\Sigma ,\mu )$ with no atoms is called a non-atomic measure space [29].

Remark 2.2

By saying that $\Phi \in \Delta _2$ we will always mean that the Young function $\Phi$ satisfies the $\Delta _{2}$-condition globally, since we place no restrictions on the underlying measure space $(\Omega ,\Sigma ,\mu )$, which can be finite or infinite, non-atomic, purely atomic, or mixed.

The following lemma is a key tool in some of our investigations.

Lemma 2.3

Let $\Phi _1, \Phi _2$ be Young functions such that $\Phi _2\mathrel {\overset{{\ell }}{\nprec }}\Phi _1$. If E is a non-atomic $\Sigma$-measurable set with positive measure, then there exists $f\in L^{\Phi _1}(\Omega )$ such that $f_{|_E}\notin L^{\Phi _2}(E)$.

Proof

It is well-known that if $L^{\Phi _{1}}(\Omega , \Sigma ,\mu )\subset L^{\Phi _{2}}(\Omega , \Sigma ,\mu )$ for a non-atomic measure space $(\Omega , \Sigma ,\mu )$, then $\Phi _2\mathrel {\overset{{\ell }}{\prec }}\Phi _1$ (see, for instance, [26] Theorem 3.4, and [28, 30, 31] for some extensions). Therefore, if $\Phi _2\mathrel {\overset{{\ell }}{\nprec }}\Phi _1$ and E is a $\Sigma$-measurable subset of $\Omega$ with $\mu (E)>0$, then we can find a function $g\in L^{\Phi _1}(E)$ such that $g\notin L^{\Phi _2}(E)$. Defining $f(t)=g(t)$ for $t\in E$, and $f(t)=0$ for $t \notin E$, we have the desired function. $\square$

Remark 2.4

We will use Lemmas 2.1 and 2.3 in conjunction since the thesis of the former is the assumption of the latter.

Given a measurable function $u\in L_+^0(\Omega )$ the mapping $f \mapsto uf$ is a linear transformation on $L^0(\Omega )$ called the multiplication operator induced by u and denoted $M_u$.

Let $T:\Omega \rightarrow \Omega$ be a measurable transformation. The transformation T is said to be non-singular if $\mu (T^{-1}(A))=0$ for all $A\in \Sigma \cap T(\Omega )$ with $\mu (A)=0$. The condition ensures that the measure $\mu \circ T^{-1}$ defined by $\mu \circ T^{-1}(A)=\mu (T^{-1}(A))$ for all $A\in \Sigma \cap T(\Omega )$ is absolutely continuous with respect to the measure $\mu$ (this fact is usually denoted by $\mu \circ T^{-1}\ll \mu$). Then the Radon–Nikodym theorem guarantees the existence of a non-negative locally integrable function $h:T(\Omega )\rightarrow {\mathbb {R}}_+$, called the Radon–Nikodym derivative of $\mu \circ T^{-1}$ with respect to $\mu$, such that

$$\begin{aligned} \mu \circ T^{-1}(A)=\int _{A}h\, d\mu \quad (A\in \Sigma \cap T(\Omega )). \end{aligned}$$

We do not necessarily assume that the transformation T is surjective or injective.

The non-singular measurable transformation T induces a linear operator $C_{T}$ on $L^0(\Omega )$, called the composition operator, defined by

$$\begin{aligned} C_{T}f(t)=f(T(t))\quad (t\in \Omega ,\; f\in L^0(\Omega )). \end{aligned}$$

Finally, we recall two well-known facts from measure theory; for proofs see, for example, Section 11.9 in [29]:

• (i) If $(\Omega , \Sigma , \mu )$ be a $\sigma$-finite measure space, then for every measurable real-valued function f on $\Omega$ and every atom ${\mathfrak {A}}$, there is a real number, denoted by $f({\mathfrak {A}})$, such that $f=f({\mathfrak {A}})\,\, \mu$-a.e. on ${\mathfrak {A}}$.

• (ii) If $(\Omega , \Sigma , \mu )$ is a $\sigma$-finite measure space that fails to be non-atomic, there is a non-empty countable set of pairwise disjoint atoms $\{{\mathfrak {A}}_n\}_{n\in {\mathbb {N}}}$ with the property that ${\mathfrak {B}}:=\Omega {\setminus } \bigcup _{n\in {\mathbb {N}}}{\mathfrak {A}}_n$ contains no atoms. We assume that both the atoms and their counterimages under T have strictly positive measure. The latter fact means that the Radon–Nikodym derivative h is strictly positive on all atoms from $T(\Omega )$.

We keep the above notation throughout the paper.

3 Boundedness of the multiplication operator

In this section we state various necessary conditions and sufficient conditions under which the multiplication operator $M_u$ between distinct Orlicz spaces is bounded.

Recall the well-known theorem due to O’Neil [32] on the multiplication operator between Orlicz spaces (see also [1] and Theorem 5.4.1 in [33]): suppose that $\Phi _1$, $\Phi _2$, $\Phi _3$ are Young functions such that either (a) there exist $c>0$ and $x_0>0$ with$\Phi _2(x_0)<\infty$ such that $\Phi _2(cxy)\le \Phi _1(x)+\Phi _3(y)$ for all $x,y\ge x_0$ in the case when $\mu (\Omega )<\infty$ or (b) there exists $c>0$ such that $\Phi _2(cxy)\le \Phi _1(x)+\Phi _3(y)$ for all $x,y>0$ in the case when $\mu (\Omega )=\infty$. Let $\Omega$ contain a non-atomic set of positive measure. Then, whenever $u\in L^{\Phi _3}(\Omega )$, the multiplication operator $M_u$ is bounded from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$.

Before going any further, let us consider some related examples.

Example 3.1

• (i) Let $\Phi$ and $\Psi$ be complementary Young functions. Since $\frac{|x|^p}{p}\le \frac{1}{p}(\Phi (x)+\Psi (x^{p-1}))$ and $\frac{|x|^p}{p}\le \frac{1}{p}(\Phi (x^{p-1})+\Psi (x))$, for $x\ge 0$ and $p>2$, by O’Neil’s theorem we get that if $u^{\frac{1}{p-1}}\in L^{\Psi }(\Omega )$ then the operator $M_u$ is bounded from $L^{\Phi }(\Omega )$ into $L^p(\Omega )$, and if $v^{\frac{1}{p-1}}\in L^{\Phi }(\Omega )$ then the operator $M_v$ is bounded from $L^{\Psi }(\Omega )$ into $L^p(\Omega )$.

• (ii) Let $\Omega =[a,b]$, $a,b>1$, $p>1$, and let $\mu$ be Lebesgue measure. If we take $\Phi _1(x)=e^{|x|^p}-|x|^p-1$, $\Phi _2(x)=\frac{|x|^p}{p}$, and $\Phi _3(x)=(1+|x|^p)\log (1+|x|^p)-|x|^p$, then, since $\phi _1(x)=\Phi _1(|x|^{1/p})$ and $\phi _3(x)=\Phi _3(|x|^{1/p})$ are complementary Young functions, Young’s inequality yields $\Phi _2(xy)\le \Phi _1(x)+\Phi _3(y)$ for all $x,y\ge 0$. Now, if $u(t)=\root p \of {t^p-1}$, it is clear that $I_{\Phi _3}(u)<\infty$, i.e. $u\in L^{\Phi _3}(\Omega )$. O’Neil’s theorem implies that $M_u$ is a bounded operator from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$.

• (iii) Let $A=(0,a]$, $B=\{\ln t: t\in {{\mathbb {N}}} ,\ t>a\}$, for $a>0$, and let $\Omega =A\cup B$. Take $\Phi _1(x)=e^{|x|}-|x|-1$, $\Phi _2(x)=(1+|x|)\ln (1+|x|)-|x|$, and for every Lebesgue measurable $C\subset \Omega$ define $\mu (C)=\mu _1(C\cap A)+\mu _2(C\cap B)$, where $\mu _1$ is Lebesgue measure and $\mu _2(\{\ln t\})=1/t^3$ for $\ln t\in B$. If we set $u(t)=1/t^2$, then $M_u$ is not bounded from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$. Indeed, for $f(t)=t$ we have

  $$\begin{aligned} I_{\Phi _1}(f)= & {} \int _\Omega (\Phi _1\circ f)\,d\mu =\int _\Omega \left( e^t-t-1\right) \,d\mu \\= & {} \int _A \left( e^t-t-1\right) \,d\mu +\int _B\left( e^t-t-1\right) \,d\mu <\infty , \end{aligned}$$

  where we used the following calculation

  $$\begin{aligned} \int _B\left( e^t-t-1\right) \,d\mu&=\sum \limits _{n>a}(e^{\ln n}-\ln n-1)\frac{1}{n^3}<\sum \limits _{n>a}\frac{1}{n^2}<\infty . \end{aligned}$$

  Hence $f\in L^{\Phi _1}(\Omega )$. But for any $\lambda >0$ we have

  $$\begin{aligned} I_{\Phi _2}(\lambda M_u f)= & {} \int _\Omega (\Phi _2\circ \lambda M_u f)d\mu =\int _\Omega \Phi _2\left( \frac{\lambda }{t}\right) d\mu \\= & {} \int _\Omega \left( \left( 1+\frac{\lambda }{t}\right) \ln \left( 1+\frac{\lambda }{t}\right) -\frac{\lambda }{t}\right) d\mu \\= & {} \int _A\left( \left( 1+\frac{\lambda }{t}\right) \ln \left( 1+\frac{\lambda }{t}\right) -\frac{\lambda }{t}\right) d\mu \\&+\int _B\left( \left( 1+\frac{\lambda }{t}\right) \ln \left( 1+\frac{\lambda }{t}\right) -\frac{\lambda }{t}\right) d\mu \\> & {} \int _0^a\left( \left( 1+\frac{\lambda }{t}\right) \ln \left( 1+\frac{\lambda }{t}\right) -\frac{\lambda }{t}\right) dt\\ {}= & {} \lambda \int _{\lambda /a}^\infty \left( (1+s)\ln (1+s)-s\right) \frac{1}{s^2}ds\\= & {} \lambda \int _{\lambda /a}^\infty \left( \frac{\ln (1+s)}{s^2}+\frac{\ln (1+s)-1}{s}\right) ds\\ {}> & {} \lambda \int _{\lambda /a}^\infty \left( \frac{\ln (1+s)-1}{s}\right) ds =\infty , \end{aligned}$$

  and so $M_u f \notin L^{\Phi _2}(\Omega )$. The operator $M_u$ does even act from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$ and, by O’Neil’s theorem, we conclude that if a Young function $\Phi _3$ satisfies $\Phi _2(xy)\le \Phi _1(x)+\Phi _3(y)$ for all $x,y\ge 0$, then u does not belong to $L^{\Phi _3}(\Omega )$.

Maligranda and Persson in [27] proved a theorem which can be stated in the following way: assume that $\Phi _1$, $\Phi _2$, $\Phi _3$ are Young functions with values in [0,$\infty$) such that $\Phi _2(xy)\le \Phi _1(x)+\Phi _3(y)$ for all $x,y\ge 0$ and $\Phi _2^{-1}(x)\le \Phi _1^{-1}(x)\Phi _3^{-1}(x)$ for all $x\ge 0$, and assume further that either (A) $\Omega$ is a non-atomic measure space or (B) $\Phi _2(x)=x^r$ and $x^{-r}\Phi _1(x)$ is nondecreasing. Then $M_u$ is a bounded multiplication operator from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$ if and only if $u\in L^{\Phi _3}(\Omega )$.

These results have been generalized and extended to the more general setting of Calderón–Lozanovskiĭ spaces in two recent papers [19, 20] by Kolwicz et al.

We give another necessary condition on the function u so that it induces a bounded multiplication operator $M_u$ between distinct Orlicz spaces; it will be used in the proof of Theorem 4.11.

Proposition 3.2

Let $\Phi _1$ and $\Phi _2$ be Young functions vanishing only at zero, taking only finite values, and such that $\Phi _1$ is an N-function, $\Phi _1\in \Delta '$, $\Phi _2\in \Delta _2$, and $\Phi _3:=\Psi _2\circ \Psi _1^{-1}$ is a Young function. If $u\in L_+^0(\Omega )$ and $M_u$ is a bounded multiplication operator from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$, then $u\in L^{\Psi _3\circ \Psi _1}(\Omega )$.

Proof

Let $u\in L_+^0(\Omega )$ and let $M_u$ be a bounded multiplication operator from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$ induced by the function u. Since $\Phi _1\in \Delta _2$ (as a consequence of the fact that $\Phi _1\in \Delta '$) and $\Phi _2\in \Delta _2$, the dual spaces of $L^{\Phi _1}(\Omega )$ and $L^{\Phi _2}(\Omega )$ are equal to $L^{\Psi _1}(\Omega )$ and $L^{\Psi _2}(\Omega )$, respectively. Hence the adjoint operator $M^{*}_u$ to the bounded operator $M_u$ is bounded and acts from $L^{\Psi _2}(\Omega )$ into $L^{\Psi _1}(\Omega )$.

It is easy to check that $\Phi _1\in \Delta '$ implies that $\Psi _1\in \nabla '$, i.e. there is $b>0$ such that $\Psi _1(bxy)\ge \Psi _1(x)\Psi _1(y)$ for all $x,y\ge 0$.

Let f be any function from $L^{\Phi _3}(\Omega )$. By the definition of $\Phi _3$, we have that the function $\Psi _1^{-1}\circ |f|$ is contained in the space $L^{\Psi _2}(\Omega )$, and so $M^{*}_{u}(\Psi _1^{-1}\circ |f|)$ is contained in the space $L^{\Psi _1}(\Omega )$. Therefore, there is $\lambda >0$ such that $I_{\Psi _1}( \lambda M^{*}_{u}(\Psi _1^{-1}\circ |f|))<\infty$. We have

$$\begin{aligned} \bigg |\int _\Omega (\Psi _1\circ \frac{\lambda }{b}u)f\,d\mu \bigg |\le & {} \int _\Omega (\Psi _1\circ \frac{\lambda }{b}u)|f|\,d\mu \,{\le } \int _\Omega (\Psi _1\circ \frac{\lambda }{b}u)(\Psi _1\circ \Psi _1^{-1}\circ |f|)d\mu \\\le & {} \int _\Omega \Psi _1 \big (b\frac{\lambda }{b}u\Psi _1^{-1}\circ |f|\big )d\mu \,{=} \int _\Omega \Psi _1\big (\lambda M^{*}_{u}(\Psi _1^{-1}\circ |f|)\big )d\mu <\infty . \end{aligned}$$

This means that $\Psi _1\circ \frac{\lambda }{b}u$ belongs to the Köthe dual of $L^{\Phi _3}(\Omega )$, which is equal to $L^{\Psi _3}(\Omega )$. Consequently, there is $\alpha \in (0,1)$ such that $I_{\Psi _3}(\alpha \,\Psi _1\circ \frac{\lambda }{b}u)<\infty$, whence $I_{\Psi _3 \circ \Psi _1}(\alpha \frac{\lambda }{b}u)= I_{\Psi _3}(\Psi _1 \circ \alpha \, \frac{\lambda }{b}u)\le I_{\Psi _3}(\alpha \, \Psi _1\circ \frac{\lambda }{b}u)<\infty$. Hence $u\in L^{\Psi _3\circ \Psi _1}(\Omega )$. $\square$

In the case of a non-atomic complete and $\sigma$-finite measure space, some necessary and sufficient conditions for the boundedness of $M_u$ and $C_T$ from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$ were established in [3]. It turns out that these are equivalent to the conditions for the inclusion of the Orlicz space $L^{\Phi _1}(\Omega )$ into the Musielak–Orlicz space $L^{\Phi _2,u}(\Omega )$ and of the Orlicz space $L^{\Phi _1}(T(\Omega ))$ into the Musielak–Orlicz space $L^{\Phi _{2}}_h (T(\Omega ))$, where $h=d(\mu \circ T^{-1})/ d\mu$, respectively. The spaces $L^{\Phi _2,u}(\Omega )$ and $L^{\Phi _{2}}_h (T(\Omega ))$ are generated over the measure spaces $(\Omega , \Sigma , \mu )$ and $(T(\Omega ),\Sigma \cap T(\Omega ), \mu _{|\Sigma \cap T(\Omega )})$ by the Musielak–Orlicz functions $\Phi _2(x u(t))$ ($t\in \Omega$ and $x\in {\mathbb {R}}$) and $\Phi _2(x)h(t)$ ($t\in T(\Omega )$ and $x\in {\mathbb {R}}$), respectively, $t\in \Omega$ and $x\in {\mathbb {R}}$.

We shall now prove a theorem from which it follows that, if the measure space $(\Omega ,\Sigma ,\mu )$ is non-atomic, for the above inclusions to hold it is necessary that $\Phi _2\mathrel {\overset{{\ell }}{\prec }}\Phi _1$. This might come as a surprise since it means that the simple condition $\Phi _2\mathrel {\overset{{\ell }}{\prec }}\Phi _1$, which does not involve the Radon–Nikodym derivative h, is necessary for the inclusion $L^{\Phi _1}(T(\Omega ))\subset L^{\Phi _2}_h (T(\Omega ))$, which does.

Theorem 3.3

Let $\Phi _1, \Phi _2$ be Young functions such that $\Phi _2\mathrel {\overset{{\ell }}{\nprec }}\Phi _1$ and let $(\Omega ,\Sigma ,\mu )$ be a non-atomic measure space. Then there are no non-zero bounded operators $C_TM_u$ from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$.

Proof

Suppose, to the contrary, that $C_TM_{u}$ is a non-zero bounded linear operator from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$ and let

$$\begin{aligned} E_n=\left\{ t\in \Omega :u(t)>\frac{1}{n}\right\} \bigcap \left\{ t\in T(\Omega ):h(t)>\frac{1}{n}\right\} \end{aligned}$$

For any $n\in {\mathbb {N}}$. Then $\{E_n\}_{n\in {\mathbb {N}}}$ is an increasing sequence of $\Sigma$-measurable sets. Since $C_TM_{u}$ is non-zero, $\mu (E_m)>0$ for some $m\in {\mathbb {N}}$, whence also $\mu (E_n)>0$ for all $n\ge m$. We assume without loss of generality that $\mu (E_n)>0$ for all $n\in {\mathbb {N}}$. Let $F\subset E:=\bigcup _n E_n$ and $0<\mu (F)<\infty$. The assumption that $\Phi _2\mathrel {\overset{{\ell }}{\nprec }}\Phi _1$ implies that an increasing sequence of positive numbers $\{y_n\}$ can be found such that $\Phi _2(y_n)>\Phi _1(2^nn^3y_n)$. Since the measure space $(\Omega ,\Sigma ,\mu )$ is non-atomic, we can choose a sequence $\{F_n\}$ of pairwise disjoint measurable subsets of F such that $F_n\subset E_n$ and $\mu (F_n)=\frac{\Phi _1(y_1)\mu (F)}{2^n\Phi _1(n^3y_n)}$ For any $n \in {\mathbb {N}}$. This is possible because $\nu$ is non-atomic and $\Phi _1(n^3y_n)\ge \Phi _1(y_n)\ge \Phi _1(y_1)$, and so $\mu (F_n)\le \frac{\mu (F)}{2^n}$, and $\sum \nolimits ^\infty _{n=1}\frac{\mu (F)}{2^n}=\mu (F)$.

Define the function $f:=\sum \nolimits ^\infty _{n=1}b_n\chi _{F_n}$, where $b_n:=n^2y_n$, and take arbitrary $\alpha >0$. Then for a natural number $n_0>\alpha$ we have

$$\begin{aligned} I_{\Phi _1}(\alpha f)= & {} \int _\Omega \Phi _1(\alpha f)d\mu =\displaystyle \sum ^{\infty }_{n=1}\int _\Omega \Phi _1(\alpha \, b_n)\chi _{F_n} d\mu \\= & {} \sum ^{n_0}_{n=1}\Phi _1(\alpha \,b_n)\mu (F_n)+\sum \limits _{n> n_0}\Phi _1(\alpha \,b_n)\mu (F_n)\\= & {} \sum ^{n_0}_{n=1}\Phi _1(\alpha \,b_n)\mu (F_n)+\mu (F)\sum _{n> n_0}\frac{\Phi _1(\alpha \,b_n)\Phi _1( y_1)}{2^n\Phi _1(n^3 y_n)}\\\le & {} \sum ^{n_0}_{n=1}\Phi _1(\alpha \,b_n)\mu (F_n)+\mu (F)\sum _{n> n_0}\frac{\Phi _1(n^3y_n)\Phi _1( y_1)}{2^n\Phi _1(n^3 y_n)}<\infty . \end{aligned}$$

This implies that $f\in L^{\Phi _1}(\Omega )$. But for $m_0>0$ such that $\frac{1}{m_0}<\alpha$, we obtain

$$\begin{aligned} I_{\Phi _2}(\alpha C_TM_{u}f)= & {} \int _\Omega \Phi _2(\alpha C_TM_{u}f)d\mu =\int _{T(\Omega )} h(t)\Phi _2(\alpha uf)d\mu \\\ge & {} \sum \limits _{n\ge m_0}\int _{F_n}h(t)\Phi _2(\alpha \,b_n u)d\mu \ge \sum \limits _{n\ge m_0}\int _{F_n}\frac{1}{n}\Phi _2( b_n/n^2)d\mu \\\ge & {} \sum \limits _{n\ge m_0}\int _{F_n}\frac{1}{n}\Phi _2(y_n)d\mu \ge \sum _{n\ge m_0}\frac{1}{n}\Phi _1(2^nn^3 y_n)\mu (F_n)\\\ge & {} \sum _{n\ge m_0}\frac{1}{n}2^n\Phi _1(n^3 y_n)\mu (F_n)\ge \mu (F)\sum _{n\ge m_0}\frac{1}{n}\Phi _1(y_1)=\infty , \end{aligned}$$

which, by the arbitratiness of $\alpha >0$, contradicts the boundedness of $C_TM_u$. (In fact, we even proved that $C_T M_u$ does not act from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$). $\square$

We provide some necessary conditions for the boundedness of the multiplication operator $M_u$ from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$ under the assumption that $\Phi _1(xy)\le \Phi _2(x)+\Phi _3(y)$ for some Young function $\Phi _3$ and for all $x,y\ge 0$.

Theorem 3.4

Let $\Phi _1,\Phi _2, \Phi _3$ be Young functions vanishing only at zero, taking only finite values, and such that $\Phi _1(xy)\le \Phi _2(x)+\Phi _3(y)$ for all $x,y\ge 0$. If $u\in L_+^0(\Omega )$ induces a bounded multiplication operator $M_u:L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$, then

• (i) $u(t)=0$ for $\mu$-a.e. $t\in {\mathfrak {B}}$, the non-atomic part of $\Omega$;

• (ii) $\sup \nolimits _{n\in {\mathbb {N}}}u({\mathfrak {A}}_n)\,\Phi _3^{-1}(\frac{1}{\mu ({\mathfrak {A}}_n)})<\infty$.

Proof

Let the assumptions about $\Phi _i$(i=1,2,3) be satisfied and $M_u$ be bounded. First we prove (i). If $\mu \{t\in {\mathfrak {B}}:u(t)\ne 0\}>0$, then there exists a constant $\delta >0$ such that the set $E=\{t\in {\mathfrak {B}}:u(t)>\delta \}$ has positive measure. Since E is non-atomic, $\mu (E)>0$, and $\Phi _1(xy)\le \Phi _2(x)+\Phi _3(y)$ for all $x,y\ge 0$, by Lemmas 2.1 and 2.3 we have that there exists $f\in L^{\Phi _1}(\Omega )$ such that $f_{|_E}\notin L^{\Phi _2}(E)$, and so

$$\begin{aligned} \infty =\int _E\Phi _2\left( \frac{\delta f(t)}{\Vert M_uf\Vert _{\Phi _2}}\right) d\mu \le \int _\Omega \Phi _2\left( \frac{u(t)f(t)}{\Vert M_uf\Vert _{\Phi _2}}\right) d\mu \le 1, \end{aligned}$$

which is a contradiction. Thus (i) holds.

Now we prove (ii). We may assume that the function u is not identically zero. For each $n\in {\mathbb {N}}$, put $f_n=\Phi _1^{-1}(\frac{1}{\mu ({\mathfrak {A}}_n)})\chi _{{\mathfrak {A}}_n}$. It is clear that $f_n\in L^{\Phi _1}(\Omega )$ and $I_{\Phi _1}(f_n)=1$, whence $\Vert f_n\Vert _{\Phi _1}=1$. Since the operator $M_u$ is bounded, we have

$$\begin{aligned} 1\ge \int _\Omega \Phi _2\left( \frac{u(t) f_n(t)}{\Vert M_uf_n\Vert _{\Phi _2}}\right) d\mu= & {} \int _{{\mathfrak {A}}_n}\Phi _2\left( \frac{u(t)\Phi _1^{-1}(\frac{1}{\mu ({\mathfrak {A}}_n)})}{\Vert M_uf_n\Vert _{\Phi _2}}\right) d\mu \\= & {} \Phi _2\left( \frac{u({\mathfrak {A}}_n)\Phi _1^{-1}(\frac{1}{\mu ({\mathfrak {A}}_n)})}{\Vert M_uf_n\Vert _{\Phi _2}}\right) \mu ({\mathfrak {A}}_n). \end{aligned}$$

Therefore

$$\begin{aligned} \frac{u({\mathfrak {A}}_n)\Phi _1^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) }{\Phi _2^{-1}(1/\mu ({\mathfrak {A}}_n))}\le \Vert M_uf_n\Vert _{\Phi _2}. \end{aligned}$$

(3.1)

Put $x=1/\mu ({\mathfrak {A}}_n)$ in the inequality $\Phi _2^{-1}(x)\Phi _3^{-1}(x)\le 2\Phi _1^{-1}(x)$ derived in lines 2–3 of the proof of Lemma 2.1 and use inequality (3.1) to obtain

$$\begin{aligned} u({\mathfrak {A}}_n)\,\Phi _3^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right)&\le u({\mathfrak {A}}_n)\frac{2\Phi _1^{-1}(1/\mu ({\mathfrak {A}}_n))}{\Phi _2^{-1}(1/\mu ({\mathfrak {A}}_n))}\le 2\Vert M_uf_n\Vert _{\Phi _2}\le 2\Vert M_u\Vert <\infty . \end{aligned}$$

This completes the proof. $\square$

We have a straightforward consequence.

Corollary 3.5

Under the assumptions of Theorem 3.4, if $(\Omega , \Sigma , \mu )$ is a non-atomic measure space, then the multiplication operator $M_u$ is bounded from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$ if and only if $M_u=0$.

Now we present some sufficient conditions for the continuity of the operator $M_u$ from one Orlicz space into another.

Theorem 3.6

Let $\Phi _1, \Phi _2$ be Young functions vanishing only at zero, taking only finite values, and such that $\Phi _1,\Phi _2\in \Delta '$ and $\Phi _2\circ \Phi ^{-1}_1$ is a Young function. Then $u\in L_+^0(\Omega )$ induces a bounded multiplication operator $M_u:L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$ if

• (i) $u(t)=0$ for $\mu$-a.e. $t\in {\mathfrak {B}}$;

• (ii) $\sup \nolimits _{n\in {\mathbb {N}}}\Phi _2[\frac{u({\mathfrak {A}}_n)}{\Phi _1^{-1}(\mu ({\mathfrak {A}}_n))}]\mu ({\mathfrak {A}}_n)<\infty$.

Proof

Suppose that (i) and (ii) hold and set $M:=\sup \nolimits _{n\in {\mathbb {N}}}\Phi _2[\frac{u({\mathfrak {A}}_n)}{\Phi _1^{-1}(\mu ({\mathfrak {A}}_n))}]\mu ({\mathfrak {A}}_n)$. Then for each $f\in L^{\Phi _1}(\Omega )$ we have

$$\begin{aligned} I_{\Phi _2}(M_uf)= & {} \int _\Omega \Phi _2(M_uf)d\mu =\int _{\mathfrak {B}}\Phi _2(u(t)\,f(t))d\mu +\int _{\bigcup _n {\mathfrak {A}}_n}\Phi _2(u(t)\,f(t))d\mu \\= & {} \sum \limits _{n\in {\mathbb {N}}}\int _{{\mathfrak {A}}_n}\Phi _2(u(t)\,f(t))d\mu =\sum \limits _{n\in {\mathbb {N}}}\Phi _2(u({\mathfrak {A}}_n)\,f({\mathfrak {A}}_n))\mu ({\mathfrak {A}}_n)\\= & {} \sum \limits _{n\in {\mathbb {N}}}\Phi _2\left[ \Phi _1^{-1}(\mu ({\mathfrak {A}}_n))\,\Phi _1^{-1}\circ \Phi _1(f({\mathfrak {A}}_n))\frac{u({\mathfrak {A}}_n)}{\Phi _1^{-1}(\mu ({\mathfrak {A}}_n))}\right] \mu ({\mathfrak {A}}_n)\\\le & {} \sum \limits _{n\in {\mathbb {N}}}\Phi _2\left[ \Phi _1^{-1}\Big [c_1\mu ({\mathfrak {A}}_n)\,\Phi _1(f({\mathfrak {A}}_n))\Big ]\,\frac{u({\mathfrak {A}}_n)}{\Phi _1^{-1}(\mu ({\mathfrak {A}}_n))}\right] \mu ({\mathfrak {A}}_n)\\\le & {} c_2\,\sum \limits _{n\in {\mathbb {N}}}\Phi _2\circ \Phi _1^{-1}\Big [c_1\mu ({\mathfrak {A}}_n)\,\Phi _1(f({\mathfrak {A}}_n))\Big ]\,\Phi _2\left[ \frac{u({\mathfrak {A}}_n)}{\Phi _1^{-1}(\mu ({\mathfrak {A}}_n))}\right] \mu ({\mathfrak {A}}_n)\\\le & {} c_2\,\sum \limits _{n\in {\mathbb {N}}}\Phi _2\circ \Phi _1^{-1}\Big [c_1\mu ({\mathfrak {A}}_n)\,\Phi _1(f({\mathfrak {A}}_n))\Big ]\,M\\\le & {} c_2 M\,\Phi _2\circ \Phi _1^{-1}\left[ c_1\sum \limits _{n\in {\mathbb {N}}}\mu ({\mathfrak {A}}_n)\,\Phi _1(f({\mathfrak {A}}_n))\right] , \end{aligned}$$

where $c_1,c_2$ are positive constants related to conditions $\Phi _1\in \Delta '$ and $\Phi _2\in \Delta '$, respectively, and the last inequality follows from the superadditivity of the convex function $\Phi _2\circ \Phi _1^{-1}$ on the interval $[0,\infty )$. Now let $\Vert f\Vert _{\Phi _1}\le 1$. We get

$$\begin{aligned} I_{\Phi _2}(M_uf) \le c_2M\,\Phi _2\circ \Phi _1^{-1}(c_1)<\infty . \end{aligned}$$

This implies that $\Vert M_u f\Vert _{\Phi _2}\le \max (c_2M\,\Phi _2\circ \Phi _1^{-1}(c_1),1)$, and so $M_u$ is bounded. $\square$

Remark 3.7

The following results on the classical Lebesgue spaces, presented in [38] (see also [18]), are an immediate consequence of Theorem 3.6.

• 1. Taking $\Phi _1(x)=|x|^p/p$ and $\Phi _2(x)=|x|^q/q$, where $1<p<q<\infty$, by Theorems 3.4 and 3.6, we obtain that the multiplication operator $M_{u}$ induced by a function $u\in L_+^0(\Omega )$ is bounded from $L^{p}(\Omega )$ into $L^{q}(\Omega )$ if and only if the following conditions hold:

  • (i) $u(t)=0$ for $\mu$-almost all $t\in {\mathfrak {B}}$;

  • (ii) $\sup _{n\in {\mathbb {N}}}\frac{u({\mathfrak {A}}_n)^r}{\mu ({\mathfrak {A}}_n)}<\infty$, where $q^{-1}+r^{-1}=p^{-1}$.

• 2. Similarly, taking $\Phi _1(x)=|x|^p/p$ and $\Phi _2(x)=|x|^q/q$, where $1<q<p<\infty$, by O’Niel’s theorem (see the beginning of Sect. 3) combined with Proposition 3.2 we obtain that the multiplication operator $M_{u}$ induced by a function $u\in L_+^0(\Omega )$ is bounded from $L^{p}(\Omega )$ into $L^{q}(\Omega )$ if and only if $u\in L^r(\Omega )$, where $p^{-1}+r^{-1}=q^{-1}$.

4 Boundedness of the composition operator

In this section, we give necessary conditions and sufficient conditions under which the composition operator $C_T$ acts continuously between distinct Orlicz spaces.

Theorem 4.1

Let $T:\Omega \rightarrow \Omega$ be a non-singular measurable transformation and let $\Phi _1, \Phi _2, \Phi _3$ be Young functions vanishing only at zero, taking only finite values, and such that $\Phi _1(xy)\le \Phi _2(x)+\Phi _3(y)$ for all $x,y\ge 0$. If T induces a bounded composition operator $C_T:L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$, then $h(t) = 0$ for $\mu$-almost all $t \in T(B)$.

Proof

Suppose, towards a contradiction, that $\mu (\{t\in T({\mathfrak {B}}): h(t)\ne 0\})>0$. Then there is a constant $\delta >0$ such that the set $E=\{t\in T({\mathfrak {B}}):h(t)>\delta \}$ has positive measure. Since $\mu (E)>0$, E is non-atomic, and $\Phi _1(xy)\le \Phi _2(x)+\Phi _3(y)$ for all $x,y\ge 0$, by Lemmas 2.1 and 2.3 we have that there exists $f\in L^{\Phi _1}(\Omega )$ such that $f_{|_E}\notin L^{\Phi _2}(E)$. Since $E\subset T(\mathfrak {B})$, we obtain that $C_T f\ne 0$ (otherwise we would have $f_{|_E}\in L^{\Phi _2}(E)$). Together with the fact that $C_T$ is bounded, the above gives

$$\begin{aligned} \infty =\int _E\delta \Phi _2\left( \frac{f(t)}{\Vert C_Tf\Vert _{\Phi _2}}\right) d\mu\le & {} \int _Eh(t)\Phi _2\left( \frac{ f(t)}{\Vert C_Tf\Vert _{\Phi _2}}\right) d\mu \\\le & {} \int _\Omega \Phi _2\left( \frac{C_Tf(t)}{\Vert C_Tf\Vert _{\Phi _2}}\right) d\mu \le 1, \end{aligned}$$

which is a contradiction finishing the proof. $\square$

Remark 4.2

Although the function $\Phi _3$ does not feature in the thesis of Theorem 4.1, it was actually used in the proof via Lemma 2.1.

Theorem 4.3

Let $T:\Omega \rightarrow \Omega$ be a non-singular measurable transformation, and let $\Phi _1, \Phi _2$ be Young functions vanishing only at zero, taking only finite values, and such that $\Phi _1,\Phi _2\in \Delta '$ and $\Phi _2\circ \Phi ^{-1}_1$ is a Young function. Then T induces a bounded composition operator $C_T:L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$ if

• (i) $h(t)=0$ for $\mu$-almost all $t\in T({\mathfrak {B}})$;

• (ii) $\sup \nolimits _{n\in {\mathbb {N}},\, {\mathfrak {A}}_n\subset T(\Omega )}\Phi _2[\frac{1}{\Phi _1^{-1}(\mu ({\mathfrak {A}}_n))}]h({\mathfrak {A}}_n)\mu ({\mathfrak {A}}_n)<\infty$.

Proof

Assume that (i) and (ii) hold and put

$$M:=\sup \nolimits _{n\in {\mathbb {N}},\,{\mathfrak {A}}_n\subset T(\Omega )}\Phi _2[\frac{1}{\Phi _1^{-1}(\mu ({\mathfrak {A}}_n))}]h({\mathfrak {A}}_n)\mu ({\mathfrak {A}}_n).$$

Then for each $f\in L^{\Phi _1}(\Omega )$ we have

$$\begin{aligned} I_{\Phi _2}(C_Tf)= & {} \int _\Omega \Phi _2(C_Tf)d\mu =\int _{T(\mathfrak {B})}h(t)\Phi _2(f(t))d\mu +\int _{\bigcup _n T({\mathfrak {A}}_n)}h(t)\Phi _2(f(t))d\mu \\= & {} \sum \limits _{{\mathfrak {A}}_n\subset T(\Omega )}\int _{{\mathfrak {A}}_n}h(t)\Phi _2(f(t))d\mu =\sum \limits _{{\mathfrak {A}}_n\subset T(\Omega )}h({\mathfrak {A}}_n)\Phi _2(f({\mathfrak {A}}_n))\mu ({\mathfrak {A}}_n)\\= & {} \sum \limits _{{\mathfrak {A}}_n\subset T(\Omega )}h({\mathfrak {A}}_n)\Phi _2\left[ \Phi _1^{-1}(\mu ({\mathfrak {A}}_n))\,\Phi _1^{-1}\circ \Phi _1(f({\mathfrak {A}}_n))\frac{1}{\Phi _1^{-1}(\mu ({\mathfrak {A}}_n))}\right] \mu ({\mathfrak {A}}_n)\\\le & {} \sum \limits _{{\mathfrak {A}}_n\subset T(\Omega )}h({\mathfrak {A}}_n)\Phi _2\left[ \Phi _1^{-1}\Big [c_1\mu ({\mathfrak {A}}_n)\,\Phi _1(f({\mathfrak {A}}_n))\Big ]\frac{1}{\Phi _1^{-1}(\mu ({\mathfrak {A}}_n))}\right] \mu ({\mathfrak {A}}_n)\\\le & {} c_2\,\sum \limits _{{\mathfrak {A}}_n\subset T(\Omega )}h({\mathfrak {A}}_n)\Phi _2\circ \Phi _1^{-1}\Big [c_1\mu ({\mathfrak {A}}_n)\,\Phi _1(f({\mathfrak {A}}_n))\Big ]\,\Phi _2\left[ \frac{1}{\Phi _1^{-1}(\mu ({\mathfrak {A}}_n))}\right] \mu ({\mathfrak {A}}_n)\\\le & {} c_2\,\sum \limits _{{\mathfrak {A}}_n\subset T(\Omega )}\Phi _2\circ \Phi _1^{-1}\Big [c_1\mu ({\mathfrak {A}}_n)\,\Phi _1(f({\mathfrak {A}}_n))\Big ]\,M\\\le & {} c_2M\,\Phi _2\circ \Phi _1^{-1}\left[ c_1\sum \limits _{n\in {\mathbb {N}}}\mu ({\mathfrak {A}}_n)\,\Phi _1(f({\mathfrak {A}}_n))\right] , \end{aligned}$$

where $c_1,c_2$ are positive constants related to conditions $\Phi _1\in \Delta '$ and $\Phi _2\in \Delta '$, respectively, and the last inequality follows from the superadditivity of the convex function $\Phi _2\circ \Phi _1^{-1}$ on the interval $[0,\infty )$. Now let $\Vert f\Vert _{\Phi _1}\le 1$. We obtain

$$\begin{aligned} I_{\Phi _2}(C_Tf)\le c_2\,M\,\Phi _2\circ \Phi _1^{-1}(c_1)<\infty . \end{aligned}$$

Hence $\Vert C_Tf\Vert _{\Phi _2}\le \max (c_2M\,\Phi _2\circ \Phi _1^{-1}(c_1),1)$, and so $C_T$ is bounded. $\square$

Theorem 4.4

Let $T:\Omega \rightarrow \Omega$ be a non-singular measurable transformation, and let $\Phi _1, \Phi _2, \Phi _3$ be Young functions vanishing only at zero, taking only finite values, and such that $\Phi _1(xy)\le \Phi _2(x)+\Phi _3(y)$ for all $x,y\ge 0$. Consider the following statements:

• (i) T induces a bounded composition operator $C_T:L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$.

• (ii) $\mu ({\mathfrak {B}})=0$ and there is a constant $M>0$ such that $\Phi _1^{-1}(\frac{1}{\mu ({\mathfrak {A}}_n)})\le M\Phi _2^{-1}(\frac{1}{h({\mathfrak {A}}_n)\mu ({\mathfrak {A}}_n)})$ for all $n\in {\mathbb {N}}$ such that ${\mathfrak {A}}_n\subset T(\Omega )$.

• (iii) $h(t)=0$ for $\mu$-a.e. $t\in T({\mathfrak {B}})$ and

  $$\begin{aligned} \sup \limits _{n\in {\mathbb {N}},\, {\mathfrak {A}}_n\subset T(\Omega )} \frac{\Phi _2^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) \Phi _3^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) }{\Phi _2^{-1}\left( \frac{1}{h({\mathfrak {A}}_n)\mu ({\mathfrak {A}}_n)}\right) }<\infty . \end{aligned}$$

• (iv) $h(t)=0$ for $\mu$-a.e. $t\in T({\mathfrak {B}})$ and

  $$\begin{aligned} \sup \limits _{n\in {\mathbb {N}},\,{\mathfrak {A}}_n\subset T(\Omega )} \Phi _2^{-1}(h({\mathfrak {A}}_n))\,\Phi _3^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) <\infty . \end{aligned}$$

  Then (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii). Moreover, if $\Phi _2\in \Delta '$, then (ii) $\Rightarrow$ (iv).

Proof

To prove that (i) $\Rightarrow$ (ii), assume that $C_T$ is a bounded operator from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$. It implies that there is a constant $M>0$ such that for every $n\in {\mathbb {N}}$,

$$\begin{aligned} \Vert C_T\chi _{{\mathfrak {A}}_n}\Vert _{\Phi _2} \le M \Vert \chi _{{\mathfrak {A}}_n}\Vert _{\Phi _1} , \end{aligned}$$

where $\chi _{{\mathfrak {A}}_n}$ is the characteristic function of the atom ${\mathfrak {A}}_n\subset T(\Omega )$. Calculating the norm of the characteristic function $\chi _{{\mathfrak {A}}_n}$, we obtain

$$\begin{aligned} \frac{1}{\Phi _2^{-1}\left( \frac{1}{\mu \circ T^{-1}({\mathfrak {A}}_n)}\right) }\le \frac{M}{\Phi _1^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) }, \end{aligned}$$

which yields the desired inequality.

By Theorem 4.1, we have $h(t)=0$ for $\mu$-a.e. $t\in T({\mathfrak {B}})$, hence $\mu (\mathfrak {B})=\mu \circ T^{-1}({T(\mathfrak {B}}))=\int _{T(\mathfrak {B})}h(t)d\mu (t)=0$.

Next we show that (ii) $\Rightarrow$ (iii). For each ${\mathfrak {A}}_n\subset T(\Omega )$ let

$$\begin{aligned} \Phi _1^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right)&\le M\Phi _2^{-1}\left( \frac{1}{h({\mathfrak {A}}_n)\mu ({\mathfrak {A}}_n)}\right) . \end{aligned}$$

(4.1)

Since $\Phi _1(xy)\le \Phi _2(x)+\Phi _3(y)$ for all $x,y\ge 0$, we have $\Phi _3^{-1}(x)\Phi _2^{-1}(x)\le 2 \Phi _1^{-1}(x)$ for all $x\ge 0$. Plugging $x=\frac{1}{\mu ({\mathfrak {A}}_n)}$ in this inequality and using (4.1), we have

$$\begin{aligned} \Phi _3^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) \Phi _2^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right)\le & {} 2 \Phi _1^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) \nonumber \\\le & {} 2M\Phi _2^{-1}\left( \frac{1}{h({\mathfrak {A}}_n)\mu ({\mathfrak {A}}_n)}\right) . \end{aligned}$$

(4.2)

Hence

$$\begin{aligned} \sup \limits _{n\in {\mathbb {N}},\, {\mathfrak {A}}_n\subset T(\Omega )} \frac{\Phi _2^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) \Phi _3^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) }{\Phi _2^{-1}\left( \frac{1}{h({\mathfrak {A}}_n)\mu ({\mathfrak {A}}_n)}\right) }\le 2M<\infty . \end{aligned}$$

Moreover, since $\mu \circ T^{-1}(T(\mathfrak {B}))=\mu (\mathfrak {B})=0$, we have $h(t)=0$ for $\mu$-a.e. $t\in T({\mathfrak {B}})$.

Finally, we prove implication (ii) $\Rightarrow$ (iv) under the assumption that $\Phi _2\in \Delta '$, i.e. that there is a constant $c>0$ such that $\Phi _2^{-1}(x)\Phi _2^{-1}(y)\le \Phi _2^{-1}(cxy)\; (x,y\ge 0)$. We may assume that $c\ge 1$. Plugging $x=1/h({\mathfrak {A}}_n)\mu ({\mathfrak {A}}_n)$ and $y=h({\mathfrak {A}}_n)/c$ in this inequality, we get

$$\begin{aligned} \Phi _2^{-1}\left( \frac{1}{h({\mathfrak {A}}_n)\mu ({\mathfrak {A}}_n)}\right) \le \frac{\Phi _2^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) }{\Phi _2^{-1}\left( \frac{h({\mathfrak {A}}_n)}{c}\right) }. \end{aligned}$$

Combining inequality (4.2) with this, yields

$$\begin{aligned} \Phi _3^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) \Phi _2^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) \le 2M\Phi _2^{-1}\left( \frac{1}{h({\mathfrak {A}}_n)\mu ({\mathfrak {A}}_n)}\right) \le 2M \frac{\Phi _2^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) }{\Phi _2^{-1}\left( \frac{h({\mathfrak {A}}_n)}{c}\right) }. \end{aligned}$$

Therefore

$$\begin{aligned} \Phi _2^{-1}\left( \frac{h({\mathfrak {A}}_n)}{c}\right) \,\Phi _3^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) \le 2M. \end{aligned}$$

Since $\Phi _2^{-1}$ is a concave function and $c\ge 1$, we have

$$\begin{aligned} \frac{1}{c}\Phi _2^{-1}(h({\mathfrak {A}}_n)) \,\Phi _3^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) \le 2M, \end{aligned}$$

hence $\sup \nolimits _{n\in {\mathbb {N}}, \, {\mathfrak {A}}_n\subset T(\Omega )} \Phi _2^{-1}(h({\mathfrak {A}}_n))\,\Phi _3^{-1}(1/\mu ({\mathfrak {A}}_n))\le 2M\,c<\infty$, and this completes the proof. $\square$

Now we present two lemmas on the relationship between multiplication operators and composition operators, which will be needed later on.

Lemma 4.5

Let $T:\Omega \rightarrow \Omega$ be a non-singular measurable transformation, and let $\Phi _1, \Phi _2$ be Young functions vanishing only at zero and taking only finite values, with $\Phi _2\in \nabla '$. For any function $f\in L^{\Phi _1}(\Omega )$ such that $M_{\Phi _2^{-1}\circ h}f\in L^{\Phi _2}(T(\Omega ))$, we have $C_T f \in L^{\Phi _2}(\Omega )$ and the following inequality holds

$$\begin{aligned} \Vert C_T f\Vert _{\Phi _2}\le b\Vert M_{\Phi _2^{-1}\circ h}f\Vert _{\Phi _2}, \end{aligned}$$

where b is some positive constant.

Proof

Let $f\in L^{\Phi _1}(\Omega )$ be such that $M_{\Phi _2^{-1}\circ h}f\in L^{\Phi _2}(T(\Omega ))$. By the definition of the norm $\Vert \cdot \Vert _{\Phi _2}$ and the assumption that $\Phi _2 \in \nabla '$, for $b>0$ such that $\Phi _2(bxy)\ge \Phi _2(x)\Phi _2(y) \quad ( x,y\ge 0)$ we have

$$\begin{aligned} \Vert C_T f\Vert _{\Phi _2}= & {} \inf \left\{ k:\int _\Omega \Phi _2\left( \frac{f(T(t))}{k}\right) d\mu \le 1\right\} \\= & {} \inf \left\{ k:\int _{T(\Omega )} h(t)\Phi _2\left( \frac{f(t)}{k}\right) d\mu \le 1\right\} \\= & {} \inf \left\{ k:\int _{T(\Omega )} \Phi _2(\Phi _2^{-1}(h(t)))\Phi _2\left( \frac{f(t)}{k}\right) d\mu \le 1\right\} \\\le & {} \inf \left\{ k:\int _{T(\Omega )} \Phi _2\left( \frac{b\Phi _2^{-1}(h(t))f(t)}{k}\right) d\mu \le 1\right\} \\= & {} b\inf \left\{ k/b:\int _{T(\Omega )} \Phi _2\left( \frac{\Phi _2^{-1}(h(t))f(t)}{k/b}\right) d\mu \le 1\right\} \\= & {} b\Vert M_{\Phi _2^{-1}\circ h}f\Vert _{\Phi _2}. \end{aligned}$$

$\square$

Corollary 4.6

Under the assumptions of Lemma 4.5, if $M_{\Phi _2^{-1}\circ h}$ is a bounded multiplication operator from $L^{\Phi _1}(T(\Omega ))$ into $L^{\Phi _2}(T(\Omega ))$, then $C_T$ is a bounded composition operator from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$.

Lemma 4.7

Let $T:\Omega \rightarrow \Omega$ be a non-singular measurable transformation and let $\Phi _1, \Phi _2$ be Young functions vanishing only at zero and taking only finite values, with $\Phi _2\in \Delta '$. For any function $f\in L^{\Phi _1}(\Omega )$ such that $C_T f\in L^{\Phi _2}(\Omega )$, we have $M_{\Phi _2^{-1}\circ h}f\in L^{\Phi _2}(T(\Omega ))$ and the following inequality holds

$$\begin{aligned} \Vert M_{\Phi _2^{-1}\circ h}f\Vert _{\Phi _2}\le c\Vert C_T f\Vert _{\Phi _2}, \end{aligned}$$

where c is some positive constant.

Proof

Let $f\in L^{\Phi _1}(\Omega )$ be such that $C_T f\in L^{\Phi _2}(\Omega )$, and let $c\ge 1$ be a constant in the inequality $\Phi _2(xy)\le c\Phi _2(x)\Phi _2(y)$ for all $x,y\ge 0$. Then

$$\begin{aligned} \Vert M_{\Phi _2^{-1}\circ h}f\Vert _{\Phi _2}= & {} \inf \left\{ k:\int _{T(\Omega )} \Phi _2\left( \frac{\Phi _2^{-1}(h(t))f(t)}{k}\right) d\mu \le 1\right\} \\\le & {} \inf \left\{ k:\int _{T(\Omega )} c\,\Phi _2(\Phi _2^{-1}(h(t)))\Phi _2\left( \frac{f(t)}{k}\right) d\mu \le 1\right\} \\= & {} \inf \left\{ k:\int _{T(\Omega )} c\,h(t)\Phi _2\left( \frac{f(t)}{k}\right) d\mu \le 1\right\} \\= & {} \inf \left\{ k:\int _\Omega c\,\Phi _2\left( \frac{f(T(t))}{k}\right) d\mu \le 1\right\} \\\le & {} \inf \left\{ k:\int _\Omega \Phi _2\left( \frac{f(T(t))}{k/c}\right) d\mu \le 1\right\} \\= & {} c\,\inf \left\{ k/c:\int _\Omega \Phi _2\left( \frac{f(T(t))}{k/c}\right) d\mu \le 1\right\} \\= & {} c\,\Vert C_T f\Vert _{\Phi _2}, \end{aligned}$$

as required. $\square$

Corollary 4.8

Under the assumptions of Lemma 4.7, if $C_T$ is a bounded composition operator from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$, then $M_{\Phi _2^{-1}\circ h}$ is a bounded multiplication operator from $L^{\Phi _1}(T(\Omega ))$ into $L^{\Phi _2}(T(\Omega ))$.

Following [38], for any F in the $\sigma$-algebra $\Sigma \cap T(\Omega )$, we define

$$\begin{aligned} Q_T(F):=\inf \{b\ge 0:\mu \circ T^{-1}(E)\le b\,\mu (E),\, E\in \Sigma \cap T(\Omega ), E\subset F\}. \end{aligned}$$

To prove our next results, we will need the following two lemmas.

Lemma 4.9

[38, Lemma 3.6] For any $F\in \Sigma \cap T(\Omega )$, we have $Q_T(F)=\mathrm{ess\,sup}_{t\in F}h(t)$, where $h=d(\mu \circ T^{-1})/d\mu$.

Lemma 4.10

Let T be a measurable transformation of the measure space $(\Omega , \Sigma , \mu )$ into itself and let $\Phi$ be a Young function. Then

$$\begin{aligned} \displaystyle \int _{T(\Omega )} \Phi \circ h\,d\mu =\inf \left\{ \sum ^{\infty }_{j=1}\Phi (Q_T(F_j))\mu (F_j): \{F_j\}\in {\mathcal {P}}_{T(\Omega )}\right\} , \end{aligned}$$

where ${\mathcal {P}}_{T(\Omega )}$ is the family of all measurable partitions of $T(\Omega )$ and the case $\infty =\infty$ is admissible.

Proof

Let $I:=\inf \{\sum \nolimits ^{\infty }_{j=1}\Phi (Q_T(F_j))\mu (F_j):\{F_j\}\in {\mathcal {P}}_{T(\Omega )}\}$. By Lemma 4.9, for the partition $\{F_j\}$ of $T(\Omega )$, we have

$$\begin{aligned} \displaystyle \int _{T(\Omega )} \Phi \circ hd\mu= & {} \sum ^{\infty }_{j=1}\displaystyle \int _{F_j}\Phi (h(t))d\mu \le \sum ^{\infty }_{j=1}\Phi (\mathrm{ess\,sup}_{t\in F_j} h(t))\mu (F_j)\\= & {} \sum ^{\infty }_{j=1}\Phi (Q_T(F_j))\mu (F_j). \end{aligned}$$

Hence

$$\begin{aligned} \int _{T(\Omega )}\Phi \circ h\,d\mu \le I. \end{aligned}$$

To prove the reverse inequality, let $a>1$ be an arbitrary number. For each $m\in {\mathbb {Z}}$ define the set

$$\begin{aligned} G_m:=\{t\in T(\Omega ) : a^{m-1}\le \Phi (h(t))<a^m\}. \end{aligned}$$

Let $\{F_j\}_{j=1}^\infty$ be any rearrangement of the sets $\{G_m\}_{m=-\infty }^\infty$ and the set $\{t\in T(\Omega ): h(t)=0\}$. Clearly, $\{F_j\}_{j=1}^\infty$ is a partition of $T(\Omega )$. By Lemma 4.9, we have

$$\begin{aligned} I\le & {} \sum ^{\infty }_{j=1}\Phi (Q_T(F_j))\mu (F_j)=\sum ^{\infty }_{j=1}\Phi (\mathrm{ess\,sup}_{t\in F_j}h(t))\mu (F_j)\\= & {} \sum ^{\infty }_{m=-\infty }\Phi (\mathrm{ess\,sup}_{t\in G_m}h(t))\mu (G_m)\le \sum ^{\infty }_{m=-\infty }a^m\mu (G_m)\\= & {} a\sum ^{\infty }_{m=-\infty } a^{m-1}\mu (G_m)= a\sum ^{\infty }_{m=-\infty }\displaystyle \int _{G_m}a^{m-1}\,d\mu \\\le & {} a\sum ^{\infty }_{m=-\infty }\displaystyle \int _{G_m}\Phi \circ h\,d\mu =a\sum ^{\infty }_{j=1}\displaystyle \int _{F_j}\Phi \circ h\,d\mu \\= & {} a\int _{T(\Omega )}\Phi \circ h\,d\mu . \end{aligned}$$

Since this holds for any $a>1$, the result follows. $\square$

Proposition 3.2 and Lemma 4.7 allow us to give a necessary condition for the boundedness of the composition operator $C_T$. Statement (ii) of the next theorem is interesting, because under some assumptions on the Young functions $\Phi _1, \Phi _2$ it gives the information that if $C_T$ acts continuously from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$, then the Radon–Nikodym derivative h must be from a concrete Orlicz space $L^\Phi (\Omega )$, where $\Phi$ depends on $\Phi _1$ and $\Phi _2$.

Theorem 4.11

Let T be a non-singular measurable transformation of the measure space $(\Omega , \Sigma , \mu )$ into itself. Then the following assertions hold.

• (i) If $\Phi$ is a Young function, then the Radon–Nikodym derivative $h=d(\mu \circ T^{-1})/d\mu$ belongs to $L^{\Phi }(T(\Omega ))$ if and only if there exists a partition $\{F_j\}_{j=1}^\infty$ of $T(\Omega )$ and $\lambda >0$ such that $\sum ^{\infty }_{j=1}\Phi (\lambda \, Q_T(F_j))\mu (F_j)<\infty$.

• (ii) Let $\Phi _1$ be an N-function, $\Phi _2$ a Young function vanishing only at zero and taking only finite values, such that $\Phi _1\in \Delta '$, $\Phi _2 \in \Delta _2$, and $\Phi _3:=\Psi _2\circ \Psi _1^{-1}$ is a Young function. If T induces a bounded composition operator $C_T:L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$, then the Radon–Nikodym derivative $h=d(\mu \circ T^{-1})/d\mu$ belongs to $L^{\Psi _3\circ \Psi _1\circ \Phi _2^{-1}}(T(\Omega ))$.

Proof

• (i) It is an easy consequence of Lemma 4.10 with the function $\Phi _{\lambda } (x):=\Phi (\lambda \,x)$ in place of the Young function $\Phi$.

• (ii) Since, by assumption, $C_T$ is a bounded composition operator from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$, Lemma 4.7 implies that

  $$\begin{aligned} M_{\Phi _2^{-1}\circ h}:L^{\Phi _1}(T(\Omega ))\rightarrow L^{\Phi _2}(T(\Omega )) \end{aligned}$$

is a bounded multiplication operator. Thus, by Proposition 3.2, for the Young function $\Phi _3=\Psi _2 \circ \Psi _1^{-1}$, we have ${\Phi _2^{-1}\circ h}\in L^{\Psi _3\circ \Psi _1}(T(\Omega ))$, and so

$$\begin{aligned} \int _{T(\Omega )} \Psi _3\circ \Psi _1\circ \lambda \Phi _2^{-1}\circ h\,d\mu <\infty , \end{aligned}$$

for some constant $\lambda >0$. Since $\Psi _3\circ \Psi _1$ is a nondecreasing function on the interval $[0,\infty )$, we may assume that $0<\lambda <1$. It can be easily shown that the $\Delta _2$-condition assumed for the function $\Phi _2$ is equivalent to the condition: for all $\lambda \in (0,1)$ there is a constant $K>0$ such that $\Phi _2^{-1}(Kx) \le \lambda \Phi _2^{-1}(x)$  $(x\ge 0)$. Since $\Psi _3\circ \Psi _1$ is a nondecreasing function on the interval $[0,\infty )$, we then get

$$\begin{aligned} \int _{T(\Omega )} \Psi _3\circ \Psi _1 \circ \Phi _2^{-1}\circ K h\,d\mu <\infty . \end{aligned}$$

Thus $h\in L^{\Psi _3\circ \Psi _1\circ \Phi _2^{-1}}(T(\Omega ))$. $\square$

Finally, we give a sufficient condition for the composition operator $C_T$ to be bounded from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$ when $\Phi _1\circ \Phi ^{-1}_2$ is a Young function.

Theorem 4.12

Let $\Phi _1, \Phi _2$ be Young functions vanishing only at zero and taking only finite values. If $\Phi _3=\Phi _1\circ \Phi ^{-1}_2$ is a Young function and $h\in L^{\Psi _3}(T(\Omega ))$, then T induces a bounded composition operator $C_T:L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$.

Proof

Since

$$\begin{aligned} I_{\Phi _3}(\Phi _2\circ f) =\int _\Omega \Phi _3\circ \Phi _2\circ f\, d\mu= & {} \int _\Omega \Phi _1\circ \Phi _2^{-1}\circ \Phi _2\circ f\, d\mu \\= & {} \int _\Omega \Phi _1\circ f \,d\mu =I_{\Phi _1}(f), \end{aligned}$$

we have $f\in L^{\Phi _1}(\Omega )$ if and only if $\Phi _2\circ f\in L^{\Phi _3}(\Omega )$, and furthermore that $\Vert \Phi _2 \circ f\Vert _{\Phi _3}=\Vert f\Vert _{\Phi _1}$. Hence an application of Hölder’s inequality yields

$$\begin{aligned} I_{\Phi _2}(C_T f)= & {} \int _\Omega \Phi _2\circ f\circ T\,d\mu = \int _{T(\Omega )} h \,\Phi _2 \circ f\,d\mu \\\le & {} 2\Vert h\Vert _{\Psi _3} \Vert \Phi _2 \circ f\Vert _{\Phi _3}= 2\Vert h\Vert _{\Psi _3} \Vert f\Vert _{\Phi _1}. \end{aligned}$$

Thus $\Vert C_T\Vert \le \max (2\Vert h\Vert _{\Psi _3},1)$, and this completes the proof. $\square$

Remark 4.13

We conclude this section with some applications of the theorems proved.

• 1. Taking $\Phi _1(x)=|x|^p/p$ and $\Phi _2(x)=|x|^q/q$, where $1<p<q<\infty$, by Theorems 4.1 and 4.4, we obtain that if $C_{T}$ is a composition operator induced by the non-singular measurable transformation $T:\Omega \rightarrow \Omega$, then the following statements are equivalent:

  • (a) $C_T$ is bounded from $L^p(\Omega )$ into $L^q(\Omega )$.

  • (b) $h(t)=0$ for $\mu$-a.e. $t\in T({\mathfrak {B}})$ and $\sup \nolimits _{n\in {\mathbb {N}}, \, {\mathfrak {A}}_n\subset T(\Omega )}\frac{h({\mathfrak {A}}_n)^p}{\mu ({\mathfrak {A}}_n)^{q-p}}<\infty$.

  • (c) $\mu ({\mathfrak {B}})=0$ and there is a constant $k>0$ such that $\mu \circ T^{-1}({\mathfrak {A}}_n)^p\le k\mu ({\mathfrak {A}}_n)^q$ for all $n\in {\mathbb {N}}$, ${\mathfrak {A}}_n\subset T(\Omega )$.

• 2. Similarly, taking $\Phi _1(x)=|x|^p/p$ and $\Phi _2(x)=|x|^q/q$, where $1<q<p<\infty$, by Theorems 4.11 and 4.12, we obtain that if $C_{T}$ is a composition operator induced by the non-singular measurable transformation $T:\Omega \rightarrow \Omega$, then the following statements are equivalent:

  • (a) $C_T$ is a bounded operator from $L^{p}(\Omega )$ into $L^{q}(\Omega )$.

  • (b) $h\in L^{\frac{r}{q}}(T(\Omega ))$, where $p^{-1}+r^{-1}=q^{-1}$.

  • (c) There exists a partition $\{F_j\}^{\infty }_{j=1}$ of $T(\Omega )$ such that $\sum \nolimits ^{\infty }_{j=1}Q_T(F_j)^{\frac{r}{q}}\mu (F_j)<\infty$.

These characterizations are due to Takagi and Yokouchi [38].

5 Multiplication and composition operators with closed-range and/or finite rank

In this section we are going to investigate closed-range multiplication and composition operators between distinct Orlicz spaces.

We start with a basic observation concerning Young functions.

Lemma 5.1

Let $\Phi _1,\Phi _2$ be Young functions such that $\Phi _1$ is an N-function and $\Phi _3:=\Psi _2\circ \Psi _1^{-1}$ is a Young function. If $\Psi _1\in \Delta '$, then there is a constant $b>0$ such that $b\,\Psi _1(xy)\le \Psi _2(x)+\Psi _3(\Psi _1(y))$ for all $x,y\ge 0$.

Proof

By the definition of the function $\Phi _3$, we have

$$\begin{aligned} \Psi _3(y)=\sup \{x|y|-\Phi _3(x): \ x\ge 0\}=\sup \{x|y|-\Psi _2\circ \Psi _1^{-1}(x): \ x\ge 0\}. \end{aligned}$$

Since $\Phi _1$ is an N-function, $\Psi _1$ is again an N-function, and $a_{\Psi _1}=0$, $b_{\Psi _1}=\infty$, So the usual inverse function $\Psi _1^{-1}$ exists. By substitution, we have

$$\begin{aligned} \Psi _3(\Psi _1(y))=\sup \{x\Psi _1(y)-\Psi _2\circ \Psi _1^{-1}(x): \ x\ge 0\}, \end{aligned}$$

hence

$$\begin{aligned} \Psi _3(\Psi _1(y))=\sup \{\Psi _1(x)\Psi _1(y)-\Psi _2(x): \ x\ge 0\}. \end{aligned}$$

(5.1)

Since $\Psi _1\in \Delta '$, i.e. $\Psi _1(xy)\le c\Psi _1(x)\Psi _1(y)$ for some $c>0$ and all $x,y\ge 0$, we get from (5.1) that

$$\begin{aligned} \Psi _3(\Psi _1(y))\ge \frac{1}{c}\Psi _1(xy)-\Psi _2(x), \end{aligned}$$

for all $x,y\ge 0$, whence the desired inequality with $b=1/c$ follows. $\square$

Now we characterize closed-range multiplication operators $M_u:L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$ under the assumption $\Phi _2(xy)\le \Phi _1(x)+\Phi _3(y)$ for all $x, y\ge 0$.

Theorem 5.2

Let $\Phi _1, \Phi _2, \Phi _3,$ be Young functions vanishing only at zero, taking only finite values, and such that $\Phi _2,\Phi _3\in \Delta _2$ and $\Phi _2(xy)\le \Phi _1(x)+\Phi _3(y)$ for all $x,y\ge 0$. If $u\in L_+^{\Phi _3}(\Omega )$, then $M_u$ is a bounded multiplication operator from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$ and the following assertions are equivalent:

• (a) $u(t)=0$ for $\mu$-a.e. $t\in {\mathfrak {B}}$ and the set $E=\{n\in {\mathbb {N}}: u({\mathfrak {A}}_n)\ne 0\}$ is finite.

• (b) $M_u$ has finite rank.

• (c) $M_u$ has closed-range.

Proof

Let $u\in L^{\Phi _3}(\Omega )$ and $f\in L^{\Phi _1}(\Omega )$, i.e. there are constants $\lambda , \alpha >0$ such that $I_{\Phi _3}(\lambda u)<\infty$ and $I_{\Phi _1}(\alpha f)<\infty$. Therefore, by the inequality $\Phi _2(xy)\le \Phi _1(x)+\Phi _3(y)$   ($x,y\ge 0$), we have

$$\begin{aligned} I_{\Phi _2}(\alpha \lambda uf)\le I_{\Phi _1}(\alpha f)+I_{\Phi _3}(\lambda u)<\infty , \end{aligned}$$

which implies that $u f\in L^{\Phi _2}(\Omega )$.

Let S be the support of u. We may assume that $\mu (S)>0$ since otherwise $M_u$ is a zero operator and there is nothing to prove.

We prove the implication (a) $\Rightarrow$ (b). Assume that (a) holds. Hence there is $r\in {\mathbb {N}}$ such that

$$\begin{aligned} S=\bigcup \limits _{n\in E}{\mathfrak {A}}_n={\mathfrak {A}}_{n_1}\cup \cdots \cup {\mathfrak {A}}_{n_r}. \end{aligned}$$

Since $\Phi _2 \in \Delta _2$, the set $\{\chi _{{\mathfrak {A}}_{n_1}},\ldots ,\chi _{{\mathfrak {A}}_{n_r}}\}$ of characteristic functions generates the subspace

$$\begin{aligned} \{g\in L^{\Phi _2}(\Omega ): g(x)=0 \quad {\text {for}}\quad \mu -{\text {a.e. t}}\;\in \Omega {\setminus } S\}\cong L^{\Phi _2}(S). \end{aligned}$$

The range of $M_u$ is contained in the r-dimensional subspace $L^{\Phi _2}(S)$, hence $M_u$ has finite rank.

(b) $\Rightarrow$ (c):

. If the range of the operator $M_u$ in $L^{\Phi _2}(\Omega )$ is finite-dimensional, then it is also closed, as any finite-dimensional subspace of a Banach space is a closed subspace of this space.

(c) $\Rightarrow$ (a):

. Let $M_u$ have closed-range and assume that $\mu \{t\in {\mathfrak {B}}: u(t)\ne 0\}>0$. Then there is $\delta >0$ such that the set $G=\{t\in {\mathfrak {B}}: u(t)\ge \delta \}$ has positive measure. It is easy to see that the restriction $u_{|_G}$ induces a bounded multiplication operator $M_{u_{|_{G}}}$ from $L^{\Phi _1}(G)$ into $L^{\Phi _2}(G)$, and that if $M_u$ has closed-range, then $M_{u_{|_{G}}}$ has closed-range as well.

We will show that $M_{u_{|_{G}}}(L^{\Phi _1}(G))=L^{\Phi _2}(G)$. Let A be any measurable subset of G with $\mu (A)<\infty$, and define the function $f_A:=\frac{1}{u_{|_{G}}}\chi _{A}$. We get

$$\begin{aligned} I_{\Phi _1}(f_A)=\int _G \Phi _1\circ f_A\,d\mu =\int _A \Phi _1\circ \frac{1}{u_{|_{G}}}\,d\mu \le \Phi _1(1/\delta )\mu (A)<\infty , \end{aligned}$$

and so $f_A\in L^{\Phi _1}(G)$. Moreover, $M_{u_{|_{G}}} f_A =\chi _{A}$, which implies that the linear space $M_{u_{|_{G}}}(L^{\Phi _1}(G))$ contains the set ${\mathcal {F}}$ of all linear combinations of characteristic functions of measurable subsets of G with positive and finite measure.

Now ${\mathcal {F}}$ is a dense subset of $L^{\Phi _2}(G)$ ([26, Lemma 4.1]) and, by assumption, $M_{u_{|_{G}}}(L^{\Phi _1}(G))$ is a closed subspace of $L^{\Phi _2}(G)$, therefore

$$\begin{aligned} L^{\Phi _2}(G)=\overline{{\mathcal {F}}}\subset M_{u_{|_{G}}}(L^{\Phi _1}(G))\subset L^{\Phi _2}(G), \end{aligned}$$

and so $M_{u_{|_{G}}}(L^{\Phi _1}(G))= L^{\Phi _2}(G),$ as claimed.

Consequently, we can define the inverse multiplication operator

$$\begin{aligned} M_{\frac{1}{u_{|G}}}:L^{\Phi _2}(G)\rightarrow L^{\Phi _1}(G),\quad M_{\frac{1}{u_{|G}}} f:=\frac{1}{u_{|_{G}}}f. \end{aligned}$$

The operator $M_{\frac{1}{u_{|G}}}$ is bounded and, since $\Phi _2(xy)\le \Phi _1(x)+\Phi _3(y)$   ($x,y\ge 0$), we can apply Theorem 3.4 to conclude that $\frac{1}{u(t)}=0$ for $\mu$-a.e. $t\in G$, which is absurd. This contradiction shows that $u(t)=0$ for $\mu$-a.e. $t\in {\mathfrak {B}}$.

Next we show that the set $E=\{n\in {\mathbb {N}}: u({\mathfrak {A}}_n)\ne 0\}$ is finite if $M_u$ has closed range. If $E=\emptyset$, we have nothing to prove. So let us assume that $E\ne \emptyset$. Define $S=\bigcup \nolimits _{n\in E}{\mathfrak {A}}_n$.

Analogously as above, we can show that $M_{u_{|_{S}}}(L^{\Phi _1}(S))=L^{\Phi _2}(S)$. Indeed, let A be a subset of S with $\mu (A)<\infty$. Define the function $f_A:=\frac{1}{u_{|_{S}}}\chi _{A}$. The set A having finite measure, we get

$$\begin{aligned} I_{\Phi _1}(f_A)=\int _A \Phi _1\circ \frac{1}{u_{|_{S}}}\,d\mu =\sum _{\mathfrak {A_n}\subset A}\Phi _1(1/u(\mathfrak {A_n}))\mu (\mathfrak {A_n})<\infty . \end{aligned}$$

Hence $f_A\in L^{\Phi _1}(S)$. Since $M_{u_{|_{S}}} f_A =\chi _{A}$, we conclude that $M_{u_{|_{S}}}(L^{\Phi _1}(S))$ contains the set $\ell _f^0$ of all linear combinations of characteristic functions of subsets of S with positive and finite measure.

Now $\ell _f^0$ is a dense subset of $L^{\Phi _2}(S)$, and $M_{u_{|_{S}}}(L^{\Phi _1}(S))$ is a closed subspace of $L^{\Phi _2}(S)$, which implies that

$$\begin{aligned} L^{\Phi _2}(S)=\overline{\ell _f^0}\subset M_{u_{|_{S}}}(L^{\Phi _1}(S))\subset L^{\Phi _2}(S), \end{aligned}$$

and so $M_{u_{|_{S}}}(L^{\Phi _1}(S))= L^{\Phi _2}(S).$

We can thus define a bounded multiplication operator $M_{\frac{1}{u_{|_S}}}$ from $L^{\Phi _2}(S)$ into $L^{\Phi _1}(S)$. Applying Theorem 3.4 to the operator $M_{\frac{1}{u_{|_S}}}$, we obtain

$$\begin{aligned} \sup \limits _{n\in E} \frac{1}{u({\mathfrak {A}}_n)}\,\Phi _3^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) < \infty . \end{aligned}$$

Let $C=\sup \nolimits _{n\in E} \frac{1}{u({\mathfrak {A}}_n)}\,\Phi _3^{-1}(\frac{1}{\mu ({\mathfrak {A}}_n)})>0$. Since $E\ne \varnothing$ and $1\le \Phi _3(Cu({\mathfrak {A}}_n))\,\mu ({\mathfrak {A}}_n)$ for all $n\in E$, we have

$$\begin{aligned} \sum _{n\in E}1 \le \sum _{n\in E}\Phi _3 (Cu({\mathfrak {A}}_n))\,\mu ({\mathfrak {A}}_n)&=\sum _{n\in E}\int _{{\mathfrak {A}}_n}\Phi _3\circ Cu\,d\mu \le \int _{\Omega }\Phi _3\circ Cu\,d\mu <\infty , \end{aligned}$$

where the final inequality follows from the assumption that $\Phi _3\in \Delta _2$. Thus E must be finite. $\square$

In the next theorem we characterize closed-range multiplication operators $M_u:L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$ under the condition that $\Phi _1(xy)\le \Phi _2(x)+\Phi _3(y)$ for all $x, y\ge 0$.

Theorem 5.3

Let $\Phi _1, \Phi _2, \Phi _3$ be Young functions vanishing only at zero, taking only finite values, and such that $\Phi _2,\Phi _3\in \Delta _2$ and $\Phi _1(xy)\le \Phi _2(x)+\Phi _3(y)$ for all $x,y\ge 0$. If $M_u$ is a bounded multiplication operator from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$ and $\frac{1}{u}\in L_+^{\Phi _3}(\Omega )$, then the following statements are equivalent:

• (a) The set $E=\{n\in {\mathbb {N}}: u({\mathfrak {A}}_n)\ne 0\}$ is finite.

• (b) $M_u$ has finite rank.

• (c) $M_u$ has closed-range.

Proof

By Theorem 3.4, we have that $u(t)=0$ for $\mu$-a.e. $t\in {\mathfrak {B}}$. The proofs of implications (a) $\Rightarrow$ (b) and (b) $\Rightarrow$ (c) are as in the proof of Theorem 5.2.

We prove the implication (c) $\Rightarrow$ (a). Using the same notation as in the proof of Theorem 5.2, let $S=\bigcup \nolimits _{n\in E}{\mathfrak {A}}_n$ and $E\ne \varnothing$. Since $M_u:L^{\Phi _1}(S)\rightarrow L^{\Phi _2}(S)$ is bounded, by Theorem 3.4, we have

$$\begin{aligned} \sup \limits _{n\in E} u({\mathfrak {A}}_n)\,\Phi _3^{-1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) < \infty \end{aligned}$$

Let $C=\sup \nolimits _{n\in E} u({\mathfrak {A}}_n)\,\Phi _3^{-1}(\frac{1}{\mu ({\mathfrak {A}}_n)})>0$. Since $E\ne \varnothing$, $1\le \Phi _3\left( \frac{C}{u({\mathfrak {A}}_n)}\right) \,\mu ({\mathfrak {A}}_n)$ for all $n\in E$, and $\Phi _3 \in \Delta _2$, we may write

$$\begin{aligned} \sum _{n\in E}1\le \sum _{n\in E}\Phi _3\left( \frac{C}{u({\mathfrak {A}}_n)}\right) \,\mu ({\mathfrak {A}}_n)&=\sum _{n\in E}\int _{{\mathfrak {A}}_n}\Phi _3\circ \frac{C}{u}\,d\mu \le \int _{\Omega } \Phi _3\circ \frac{C}{u}\,d\mu <\infty . \end{aligned}$$

Thus E is finite. $\square$

Now we characterize closed-range composition operators $C_T:L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$ under the assumption that $\Phi _2(xy)\le \Phi _1(x)+\Phi _3(y)$ for all $x, y\ge 0$.

We will need an elementary lemma with an easy proof which we omit.

Lemma 5.4

Let $\Phi _1, \Phi _2$ be Young functions vanishing only at zero and taking only finite values, and let T be a non-singular measurable transformation of $\Omega$ such that $C_T$ is a bounded composition operator from $L^{\Phi _1}(\Omega )$ into $L^{\Phi _2}(\Omega )$. If T is surjective, then $C_T$ is injective.

Theorem 5.5

Let $\Phi _1, \Phi _2, \Phi _3$ be Young functions vanishing only at zero, taking only finite values, and such that $\Phi _2\in \nabla '\cap \Delta _2$, $\Phi _3\in \Delta _2$, and $\Phi _2(xy)\le \Phi _1(x)+\Phi _3(y)$ for all $x,y\ge 0$. Let T be a surjective non-singular measurable transformation of $\Omega$ such that $\Phi _2^{-1}\circ h \in L^{\Phi _3}(T(\Omega ))$. If $C_T: L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$ is a bounded composition operator, then the following assertions are equivalent:

• (a) $C_T$ has closed-range.

• (b) $h(t)=0$ for $\mu$-a.e. $t\in T({\mathfrak {B}})$ and the set $\{n\in {\mathbb {N}}: {\mathfrak {A}}_n\subset T(\Omega ),\,h({\mathfrak {A}}_n)\ne 0\}$ is finite.

• (c) $\mu ({\mathfrak {B}})=0$ and the set $\{n\in {\mathbb {N}}: {\mathfrak {A}}_n\subset T(\Omega ),\,\mu \circ T^{-1}({\mathfrak {A}}_n)\ne 0\}$ is finite.

• (d) $C_T$ has finite rank.

Proof

The implications (b) $\Rightarrow$ (c) and (d) $\Rightarrow$ (a) are obvious.

We prove the implication (a) $\Rightarrow$ (b). Assume that $C_T$ has closed-range. Since T is surjective, by Lemma 5.4, $C_T$ is injective. It is a well-known fact that an injective operator has closed-range if and only if it is bounded away from zero (see [4], III.12.5), hence $C_T$ is bounded away from zero. This, combined with Lemma 4.5, yields that the multiplication operator $M_{\Phi _2^{-1}\circ h}$ is bounded away from zero and therefore (being also injective) has closed-range. Now an application of Theorem 5.2 shows that $\Phi _2^{-1}\circ h(t)=0$ for $\mu$-a.e. $t\in T({\mathfrak {B}})$ and the set $\{n\in {\mathbb {N}}: {\mathfrak {A}}_n\subset T(\Omega ),\,\Phi _2^{-1}\circ h({\mathfrak {A}}_n)\ne 0\}$ is finite. We conclude that $h(t)=0$ for $\mu$-a.e. $t\in T({\mathfrak {B})}$ and the set $\{n\in {\mathbb {N}}: {\mathfrak {A}}_n\subset T(\Omega ),\,h({\mathfrak {A}}_n)\ne 0\}$ is finite.

Finally, we prove implication (c) $\Rightarrow$ (d). Assume (c) holds. It is easy to see that the range of the operator $C_T$ is then contained in a subspace generated by the functions $\{\chi _{T^{-1}({\mathfrak {A}}_n)}: {\mathfrak {A}}_n\subset T(\Omega ),\,\mu \circ T^{-1}({\mathfrak {A}}_n)\ne 0\}$. Since the set $\{n\in {\mathbb {N}}: {\mathfrak {A}}_n\subset T(\Omega ),\,\mu \circ T^{-1}({\mathfrak {A}}_n)\ne 0\}$ is finite, the subspace $C_T(L^{\Phi _1}(\Omega ))$ is finite-dimensional and so $C_T$ has finite rank. $\square$

As an easy consequence of this theorem we state

Corollary 5.6

If $\Omega$ is non-atomic, then, under the assumptions of Theorem 5.5, there are no non-zero closed-range composition operators from $L^{\Phi _{1}}(\Omega )$ into $L^{\Phi _{2}}(\Omega )$.

In the next theorem we characterize closed-range composition operators $C_T:L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$ in the case when $\Phi _1(xy)\le \Phi _2(x)+\Phi _3(y)$ for all $x,y\ge 0$ (note that compared to Theorem 5.5 the roles of the functions $\Phi _1$ and $\Phi _2$ are reversed).

Theorem 5.7

Let $\Phi _1, \Phi _2, \Phi _3$ be Young functions vanishing only at zero, taking only finite values, and such that $\Phi _2\in \nabla '\cap \Delta _2$, $\Phi _3\in \Delta _2$, and $\Phi _1(xy)\le \Phi _2(x)+\Phi _3(y)$ for all $x,y\ge 0$. Let T be a surjective non-singular measurable transformation of $\Omega$ such that $\frac{1}{\Phi _2^{-1}\circ h} \in L^{\Phi _3}(\Omega )$. If $C_T: L^{\Phi _1}(\Omega )\rightarrow L^{\Phi _2}(\Omega )$ is a bounded composition operator, then the following assertions are equivalent:

• (a) $C_T$ has closed-range.

• (b) The set $\{n\in {\mathbb {N}}: {\mathfrak {A}}_n\subset T(\Omega ),\,h({\mathfrak {A}}_n)\ne 0\}$ is finite.

• (c) The set $\{n\in {\mathbb {N}}: {\mathfrak {A}}_n\subset T(\Omega ),\,\mu \circ T^{-1}({\mathfrak {A}}_n)\ne 0\}$ is finite.

• (d) $C_T$ has finite rank.

Proof

Using Lemma 4.5 and Theorem 5.3 and a method similar as in the proof of Theorem 5.5 proves the theorem. $\square$

Remark 5.8

As an applications of our results we derive characterizations of bounded and closed-range multiplication and composition operators in the special case of $L^p$-spaces. These results recover the main theorems in [38].

• 1. If for $1<p<q<\infty$ the multiplication operator $M_{u}$ is bounded from $L^{p}(\Omega )$ into $L^{q}(\Omega )$, then the following assertions are equivalent:

  • (a) $M_u$ has closed-range.

  • (b) $M_u$ has finite rank.

  • (c) The set $\{n\in {\mathbb {N}}: u({\mathfrak {A}}_n)\ne 0\}$ is finite.

• 2. If for $1<q<p<\infty$ the multiplication operator $M_{u}$ is bounded from $L^{p}(\Omega )$ into $L^{q}(\Omega )$, then the following assertions are equivalent:

  • (a) $M_u$ has closed-range.

  • (b) $M_u$ has finite rank.

  • (c) $u(t)=0$ for $\mu$-almost all $t\in {\mathfrak {B}}$, and the set $\{n\in {\mathbb {N}}: u({\mathfrak {A}}_n)\ne 0\}$ is finite.

• 3. If for $1<p<q<\infty$ the composition operator $C_{T}$ is bounded from $L^{p}(\Omega )$ into $L^{q}(\Omega )$, then the following assertions are equivalent:

  • (a) $C_T$ has closed-range.

  • (b) $C_T$ has finite rank.

  • (c) The set $\{n\in {\mathbb {N}}: h({\mathfrak {A}}_n)\ne 0\}$ is finite.

  • (d) The set $\{n\in {\mathbb {N}}: \mu \circ T^{-1}({\mathfrak {A}}_n)\ne 0\}$ is finite.

• 4. If for $1<q<p<\infty$ the composition operator $C_{T}$ is bounded from $L^{p}(\Omega )$ into $L^{q}(\Omega )$, then the following assertions are equivalent:

  • (a) $C_T$ has closed-range.

  • (b) $C_T$ has finite rank.

  • (c) $h(t)=0$ for $\mu$-almost all $t\in {\mathfrak {B}}$, and the set $\{n\in {\mathbb {N}}: h({\mathfrak {A}}_n)\ne 0\}$ is finite.

  • (d) $\mu \circ T^{-1}({\mathfrak {B}})=0$ and the set $\{n\in {\mathbb {N}}: \mu \circ T^{-1}({\mathfrak {A}}_n)\ne 0\}$ is finite.

6 Simultaneously necessary and sufficient conditions for continuity of the multiplication and composition operators beween distinct Orlicz spaces

Up to this point no conditions that would be simultaneously necessary and sufficient for the continuity of the operators $M_u$ and $C_T$ have been given; the provided necessary conditions and sufficient conditions though have the merit of being quite easy to verify for a pair or a triple of Young functions involved. In order to obtain simultaneously necessary and sufficient conditions for continuity of the operators $M_u$ and $C_T$, it is convenient to take advantage of the theory of embeddings between Musielak–Orlicz spaces. In the case of non-atomic measure spaces such conditions were presented in [3]; now we will present them for purely atomic and mixed measure spaces. However, there is a price: the simultaneously necessary and sufficient conditions for the continuity of the operators $M_u$ and $C_T$ are more complicated and not as easy to verify as those presented in the preceding part of this paper.

A function $\Phi$ acting from $\Omega \times {\mathbb {R}}$ into $[0,\infty )$ such that $\Phi (t,.)$ is an Orlicz function (i.e. a finite-valued continuous Young function) for $\mu$-a.e. $t\in \Omega$ and $\Phi (., x)$ is a $\Sigma$-measurable function for every $x\in {\mathbb {R}}$, is called a generalized Orlicz function or a Musielak–Orlicz function. The Musielak–Orlicz space $L^\Phi = L^\Phi (\Omega ,\Sigma ,\mu )$ is the space of all (equivalence classes of) $\Sigma$-measurable functions $f:\Omega \rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} I_\Phi (k f):=\int _\Omega \Phi (t,k f(t))d\mu <\infty \end{aligned}$$

for some $k>0$ depending on f. The Musielak–Orlicz space furnished with the Luxemburg norm

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

is a Banach space (cf. [26, 28]). It is obvious that Orlicz functions are Musielak–Orlicz functions and, consequently, Orlicz spaces are Musielak–Orlicz spaces. Moreover, an Orlicz weighted space $L^\Phi _w(\Omega )$ is a Musielak–Orlicz space $L^{\Phi _w}(\Omega )$ with $\Phi _w(t,x)=\Phi (x)w(t)$ for any $t\in \Omega$ and any $x\in {\mathbb {R}}$.

When the measure space $(\Omega ,\Sigma ,\mu )$ is purely atomic with infinite number of atoms, we may restrict ourselves to counting measure. Since $(\Omega ,\Sigma ,\mu )$ is $\sigma$-finite, it has countably many atoms, and so, without loss of generality, it may be viewed as having the form $({\mathbb {N}}, 2^{\mathbb {N}}, \mu _c)$, where $\mu _c$ is the counting measure on $2^{{\mathbb {N}}}$. Indeed, considering atoms as singletons, we write $\mu (\{n\})=a_n$ for any $n\in {\mathbb {N}}$, where $\{a_n\}_{n\in {\mathbb {N}}}$ is a sequence of positive real numbers. If $\ell ^0$ denotes the space of all real sequences, then, given the Musielak–Orlicz function $\Phi =(\Phi _n)_{n=1}^\infty$, i.e. a sequence of Young functions $\Phi _n$, we can define on $\ell ^0$ the absolutely convex modular

$$\begin{aligned} I_\Phi (f)=\sum _{n=1}^\infty \Phi _n(f_n)a_n\quad (f=(f_n)\in \ell ^0). \end{aligned}$$

Defining

$$\begin{aligned} B_\Phi =\{f\in \ell ^0 : I_{\Phi }(f)\le 1 \} \end{aligned}$$

and

$$\begin{aligned} \ell ^\Phi ({\mathbb {N}})=\{f\in \ell ^0 : I_{\Phi }(k f)< \infty \text { for some } k>0 \}, \end{aligned}$$

we easily see that $B_{\Phi }$ is an absolutely convex set in $\ell ^\Phi ({\mathbb {N}})$ and that all elements of $\ell ^\Phi ({\mathbb {N}})$ are absorbed by $B_\Phi$. Consequently, the Minkowski functional generated by $B_\Phi$

$$\begin{aligned} m_\Phi (f):= \inf \{\lambda >0 : f/\lambda \in B_\Phi \} \end{aligned}$$

is a seminorm on $\ell ^\Phi ({\mathbb {N}})$. If we assume that for all $n\in {\mathbb {N}}$ the functions $\Phi _n$ are not identically equal to zero, then it is easy to prove that $m_\Phi (f)=0$ if and only if $f=0$. Of course,

$$\begin{aligned} m_\Phi (f)=\inf \{\lambda >0 : I_\Phi (f/\lambda )\le 1 \}. \end{aligned}$$

This is the Luxemburg norm $\Vert f\Vert _\Phi =m_\Phi (f)$.

Let us define a new Musielak–Orlicz function $\Psi =(\Psi _n)_{n=1}^{\infty }$ with

$$\begin{aligned} \Psi _n(x)=\Phi _n(x)\,\mu (\{n\})=\Phi _n(x)\,a_n, \end{aligned}$$

where $\{n\}$ is identified with the nth atom of the measure $\mu$. Then the Musielak–Orlicz space generated by the original Musielak–Orlicz function $\Phi$ and the original purely atomic measure with $\mu (\{n\})=a_n$ is isometric to the Musielak–Orlicz space generated by the Musielak–Orlicz function $\Psi$ defined above and corresponding to counting measure with $\mu _c(\{n\})=1$ for all $n\in {\mathbb {N}}$, because of the equality

$$\begin{aligned} I_\Psi (f,\mu _c)=\sum _{n=1}^\infty \Psi _n(f_n)=\sum _{n=1}^\infty \Phi _n(f_n)\,a_n=I_\Phi (f,\mu ). \end{aligned}$$

For these reasons Musielak–Orlicz sequence spaces, without loss of generality, may be considered (and usually are) over the counting measure space $({\mathbb {N}}, 2^{\mathbb {N}}, \mu _c)$.

Assume that $({\mathbb {N}}, 2^{{\mathbb {N}}}, \mu )$ is a purely atomic measure space with $\mu (\{n\})>0$ for all $n\in {\mathbb {N}}$. If $T:{\mathbb {N}}\rightarrow {\mathbb {N}}$ is a non-singular measurable transformation and $u:{\mathbb {N}}\rightarrow [0,\infty )$ a measurable function, then we can define a composition operator $C_T$ by

$$\begin{aligned} C_Tf(n):=f(T(n))\quad (f\in \ell ^0, n\in {\mathbb {N}}), \end{aligned}$$

and a multiplication operator $M_u$ by

$$\begin{aligned} M_u f(n):= u(n)f(n)\quad (f\in \ell ^0, n\in {\mathbb {N}}). \end{aligned}$$

Given Orlicz functions $\Phi$ and $\Psi$, we can look for conditions on the triples $(\Phi , \Psi , T)$ and $(\Phi , \Psi , u)$, so that $C_T$ or $M_u$ act continuously from $\ell ^\Phi ({\mathbb {N}})$ into $\ell ^\Psi ({\mathbb {N}})$. Note that for every function f on ${\mathbb {N}}$ we have

$$\begin{aligned} I_\Psi (C_T f,\mu )= & {} \sum _{n=1}^\infty \Psi (f(T(n)))\mu (\{n\})=\sum _{m\in T({\mathbb {N}})} \sum _{s\in T^{-1}(m)}\Psi (f(m))\mu (\{s\})\nonumber \\= & {} \sum _{m\in T({\mathbb {N}})} \Psi (f(m))\sum _{s\in T^{-1}(m)}\mu (\{s\})=\sum _{m\in T({\mathbb {N}})}\Psi (f(m))\mu \circ T^{-1}(\{m\})\nonumber \\= & {} I_\Psi (f_{|_{T({\mathbb {N}})}},\mu \circ T^{-1})=I_\Psi ^{\omega }(f_{|_{T({\mathbb {N}})}},\mu ), \end{aligned}$$

(6.1)

where $\omega (n)=\frac{\mu \circ T^{-1}(\{n\})}{\mu (\{n\})}$. This means, of course, that $C_T f\in \ell ^\Psi ({\mathbb {N}})$ if and only if $f_{|_{T({\mathbb {N}})} }\in \ell ^\Psi _\omega (T({\mathbb {N}})).$

Similarly, for every function f on ${\mathbb {N}}$ we have

$$\begin{aligned} I_\Psi (M_u f,\mu )=\sum _{n=1}^\infty \Psi (u(n)f(n))\mu (\{n\})=I_\Psi ^u(f,\mu ), \end{aligned}$$

(6.2)

which implies that $M_u f \in \ell ^\Psi ({\mathbb {N}})$ if and only if $f\in \ell ^{\Psi _u}({\mathbb {N}})$, where $\Psi _u(n,x):= \Psi (u(n)\, x)$.

To work with purely atomic measures, we need to use the discrete version of Ishii’s theorem due to Shragin [37] (see also [15, 17]).

Theorem 6.1

(Shragin 1976) Let $({\mathbb {N}}, 2^{{\mathbb {N}}}, \mu _c)$ be the counting measure space. Let $\Phi =(\Phi _n)_{n=1}^\infty$ and $\Psi =(\Psi _n)_{n=1}^\infty$ be Musielak–Orlicz functions, and let $\ell ^\Phi ({\mathbb {N}})$ and $\ell ^\Psi ({\mathbb {N}})$ be the Musielak–Orlicz spaces corresponding to $\Phi$, $\Psi$, and the counting measure $\mu _c$. Then $\ell ^\Phi ({\mathbb {N}})\subset \ell ^\Psi ({\mathbb {N}})$ if and only if there exist $K,\delta >0$ and a sequence of non-negative numbers $(c_n)_{n=1}^\infty \in \ell ^1({\mathbb {N}})$ such that

$$\begin{aligned} \Psi _n(Kx)\le \Phi _n(x)+ c_n \end{aligned}$$

(6.3)

for all $x\in {\mathbb {R}}$ and $n\in {\mathbb {N}}$ satisfying $\Phi _n(x)\le \delta$.

We use Shragin’s result to prove the necessity part of the following theorem.

Theorem 6.2

Let $({\mathbb {N}}, 2^{{\mathbb {N}}}, \mu )$ be a purely atomic measure space, and let $\Phi$ and $\Psi$ be Young functions vanishing only at zero and taking only finite values. The composition operator $C_T$ acts continuously from $\ell ^{\Phi } ({\mathbb {N}})$ into $\ell ^{\Psi } ({\mathbb {N}})$ if and only if there exist $K,\delta >0$ and a sequence of non-negative numbers $(c_n)_{n=1}^\infty \in \ell ^1(T({\mathbb {N}}))$ such that

$$\begin{aligned} \Psi (Kx)\,\mu \circ T^{-1}(\{n\})\le \Phi (x)\mu (\{n\})+ c_n \end{aligned}$$

(6.4)

for all $x\in {\mathbb {R}}$ and $n\in T({\mathbb {N}})$ satisfying $\Phi (x)\mu (\{n\})\le \delta$.

Remark 6.3

In the preceding two theorems $\delta$ can be taken in the interval (0, 1).

Proof (of Theorem 6.2)

Assume that there are $K>0$, $1>\delta >0$, and a sequence of non-negative numbers $(c_n)_{n=1}^\infty \in \ell ^1(T({\mathbb {N}}))$ such that condition (6.4) is satisfied whenever $\Phi (x)\mu (\{n\})\le \delta$. Let f be a function in the unit sphere of $\ell ^{\Phi } ({\mathbb {N}})$. Then $I_\Phi (f)\le 1$ and further $I_\Phi (\delta f)\le \delta \,I_\Phi (f)\le \delta$, hence $\Phi (\delta f(n))\mu (\{n\})\le \delta$ for all $n\in {\mathbb {N}}$. Now setting

$$\begin{aligned} b:=K\,\delta \quad and \quad c:=\sum _{n\in T({\mathbb {N}})}c_n, \end{aligned}$$

by (6.4) we obtain

$$\begin{aligned} I_\Psi (b\,C_T f)= & {} \sum _{n=1}^\infty \Psi (b\,f(T(n)))\mu (\{n\})=\sum _{n\in T({\mathbb {N}})}\Psi (b\,f(n))\mu \circ T^{-1}(\{n\})\\\le & {} \sum _{n\in T({\mathbb {N}})}\Phi (\delta \,f(n))\mu (\{n\})+\sum _{n\in T({\mathbb {N}})}c_n \le \delta \sum _{n\in T({\mathbb {N}})}\Phi (f(n))\mu (\{n\})+c\\\le & {} \delta +c \end{aligned}$$

Denoting $d=\max (\delta +c, 1)$, we get from this that

$$\begin{aligned} I_\Psi \left( \frac{b}{d}\,C_T f\right) \le \frac{1}{d}I_\Psi (b\,C_T f)\le 1. \end{aligned}$$

Hence, for all functions f in the unit sphere of $\ell ^{\Phi } ({\mathbb {N}})$ we have

$$\begin{aligned} \left\| \frac{b}{d}\,C_T f\right\| _{\ell ^{\Psi } ({\mathbb {N}})}\le 1, \end{aligned}$$

or

$$\begin{aligned} \Vert C_T f\Vert _{\ell ^{\Psi } ({\mathbb {N}})}\le \frac{d}{b}, \end{aligned}$$

i.e. the composition operator $C_T$ acts continuously from $\ell ^\Phi ({\mathbb {N}})$ into $\ell ^\Psi ({\mathbb {N}})$.

Now we shall prove that condition (6.4) follows from the continuity of the composition operator $C_T:\ell ^\Phi ({\mathbb {N}})\rightarrow \ell ^\Psi ({\mathbb {N}})$. To this end, first note that, by Shragin’s result, condition (6.4) is equivalent to the inclusion $\ell ^\Phi (T({\mathbb {N}}))\subset \ell _\omega ^\Psi (T({\mathbb {N}}))$. Indeed, it is sufficient to plug $\Phi _n (x)=\Phi (x)\mu (\{n\})$ and $\Psi _n (x)=\Psi (x)\mu \circ T^{-1}(\{n\})$ in the inequality (6.3). Hence, we only have to show that the inclusion $\ell ^\Phi (T({\mathbb {N}}))\subset \ell _\omega ^\Psi (T({\mathbb {N}}))$ follows from the continuity of $C_T$.

Let $f\in \ell ^\Phi (T({\mathbb {N}}))$. Then ${\tilde{f}}\in \ell ^\Phi ({\mathbb {N}})$, where ${\tilde{f}}$ is the extension of f to ${\mathbb {N}}$ with zero values outside of $T({\mathbb {N}})$. Since $C_T$ is continuous, we have $C_T{\tilde{f}}\in \ell ^\Psi ({\mathbb {N}})$, but, as we noticed above, this means that $f\in \ell ^\Psi _\omega (T({\mathbb {N}}))$. $\square$

Remark 6.4

We could have proved the sufficiency part of Theorem 6.2 via Shragin’s theorem and the fact that ($f\in \ell ^\Phi ({\mathbb {N}}) \implies C_T f\in \ell ^\Psi ({\mathbb {N}})$) implies the continuity of the operator $C_T$ since any non-negative linear operator ($C_T$ is such an operator) between Banach lattices is continuous. Let $\ell ^\Phi (T({\mathbb {N}}))\subset \ell _\omega ^\Psi (T({\mathbb {N}}))$ [which, as we have pointed out, is equivalent to condition (6.4)] and let $f\in \ell ^\Phi ({\mathbb {N}})$. Then $f_{|_{T({\mathbb {N}})}}\in \ell ^\Phi (T({\mathbb {N}}))$ and so $f_{|_{T({\mathbb {N}})}}\in \ell _\omega ^\Psi (T({\mathbb {N}}))$, which means that indeed $C_T f\in \ell ^\Psi ({\mathbb {N}})$. This method of proof, however, does not provide any estimate of the norm of the operator $C_T$.

A similar theorem can be stated for the multiplication operator.

Theorem 6.5

Let $({\mathbb {N}}, \Sigma , \mu )$ be a purely atomic measure space, and let $\Phi$ and $\Psi$ be Young functions vanishing only at zero and taking only finite values. The multiplication operator $M_u$ acts continuously from $\ell ^{\Phi } ({\mathbb {N}})$ into $\ell ^{\Psi } ({\mathbb {N}})$ if and only if there exist $K,\delta >0$ and a sequence of non-negative numbers $(c_n)_{n=1}^\infty \in \ell ^1({\mathbb {N}})$ such that

$$\begin{aligned} \Psi (Ku(n)x)\le \Phi (x)+ c_n \end{aligned}$$

(6.5)

for all $x\in {\mathbb {R}}$ and $n\in {\mathbb {N}}$ satisfying $\Phi (x)\mu (\{n\})\le \delta$ .

Proof

The proof proceeds along the same lines as the proof of Theorem 6.2, and we will omit it. $\square$

Remark 6.6

As in the proof of Theorem 6.2, the sufficiency part can also be proved via Shragin’s theorem with the substitutions $\Phi _n(x)=\Phi (x)$ and $\Psi _n (x)= \Psi (u(n) x)$ in condition 6.3, which then becomes equivalent to the inclusion $\ell ^\Phi ({\mathbb {N}})\subset \ell ^{\Psi _u}({\mathbb {N}})$.

The same techniques can be applied in all theorems in which $h_{\mid _T({{\mathfrak {B}}})}=0$ $\mu$-a.e., where $h=d(\mu \circ T^{-1})/d\mu$, and/or $u_{\mid _{{\mathfrak {B}}}}=0$ $\mu$-a.e.

In the case of a general measure space $\Omega$, the non-atomic part of $\Omega$ and the purely atomic part of $\Omega$ can be treated separately, and this allows us to state the following.

Theorem 6.7

Let $(\Omega ,\Sigma ,\mu )$ be a measure space, and let $\Phi$ and $\Psi$ be Young functions vanishing only at zero and assuming only finite values.

• (a) Assume that a non-singular measurable transformation T maps ${\mathfrak {B}}$ into itself and $\Omega {\setminus }{\mathfrak {B}}$ into itself. The composition operator $C_T$ acts continuously from $L^{\Phi } (\Omega )$ into $L^{\Psi } (\Omega )$ if and only if the following conditions are jointly satisfied:

  • (i) there exist a constant $K>0$, a set $A\in \Sigma \cap T({\mathfrak {B}})$ with $\mu (A)=0$, and a function $g\in L^1_+(T({\mathfrak {B}}))$, such that

    $$\begin{aligned} \Psi (Kx)h(t)\le \Phi (x)+g(t) \end{aligned}$$

    for all $x\ge 0$ and all $t \in T({\mathfrak {B}}){\setminus } A$;

  • (ii) condition (6.4) (we identify ${\mathbb {N}}$ with $\Omega {\setminus }{\mathfrak {B}}).$

• (b) The multiplication operator $M_u$ acts continuously from $L^{\Phi } (\Omega )$ into $L^{\Psi } (\Omega )$ if and only if the following conditions are jointly satisfied:

  • (i) there exist a constant $K>0$, a set $A\in \Sigma \cap {\mathfrak {B}}$ with $\mu (A)=0$, and a function $g\in L^1_+({\mathfrak {B}})$, such that

    $$\begin{aligned} \Psi (u(t)x)\le \Phi (x)+g(t) \end{aligned}$$

    for all $x\ge 0$ and all $t \in {\mathfrak {B}}{\setminus } A$;

  • (ii) condition (6.5) (again, we identify ${\mathbb {N}}$ with $\Omega {\setminus }{\mathfrak {B}}).$

Proof

We give the proof for both operators concurrently. We can represent $L^{\Phi }(\Omega )$ and $L^{\Psi }(\Omega )$ as the direct sums of the spaces over the non-atomic part of $\Omega$ and the purely atomic part of $\Omega$, i.e. $L^{\Phi }(\Omega )=L^{\Phi }({\mathfrak {B}})\oplus L^{\Phi }(\Omega {\setminus } {\mathfrak {B}})$ and $L^{\Psi }(\Omega )=L^{\Psi }({\mathfrak {B}})\oplus L^{\Psi }(\Omega {\setminus } {\mathfrak {B}})$. Therefore, every function $f\in L^{\Phi }(\Omega )$ can be uniquely written as $f=f\chi _{{\mathfrak {B}}}+ f\chi _{\Omega {\setminus } {\mathfrak {B}}}$, and we have

$$\begin{aligned} C_T f (t)&= C_T (f\chi _{{\mathfrak {B}}}+ f\chi _{\Omega {\setminus } {\mathfrak {B}}})(t)\\&= (f\chi _{{\mathfrak {B}}}+ f\chi _{\Omega {\setminus } {\mathfrak {B}}})(T(t))\\&= f(T(t))\chi _ {T^{-1}{({\mathfrak {B}})}}(t)+ f(T(t))\chi _{T^{-1}{(\Omega {\setminus } {\mathfrak {B}})}}(t)\\&= f(T(t))\chi _{{\mathfrak {B}}}{(t)}+f(\chi (t))\chi _{\Omega {\setminus }\mathfrak {B}}{(t)} \end{aligned}$$

and

$$\begin{aligned} M_u f (t)&= M_u (f\chi _{{\mathfrak {B}}}+ f\chi _{\Omega {\setminus } {\mathfrak {B}}})(t)=u( f\chi _{{\mathfrak {B}}}+ f\chi _{\Omega {\setminus } {\mathfrak {B}}})(t)\\&=u(t)f(t)\chi _{{\mathfrak {B}}}(t)+ u(t)f(t)\chi _{\Omega {\setminus } {\mathfrak {B}}}(t). \end{aligned}$$

Hence, if we define the operators $C_{T_{|_{\mathfrak {B}}}}$, $M_{u_{|_{\mathfrak {B}}}}$ on $L^{\Phi }({\mathfrak {B}})$ and the operators $C_{T_{|_{\Omega {\setminus } {\mathfrak {B}}}}}$, $M_{u_{|_{\Omega {\setminus } {\mathfrak {B}}}}}$ on $L^{\Phi }(\Omega {\setminus } {\mathfrak {B}})$ by the obvious formulas, we are justified to write

$$\begin{aligned} C_T=C_{T_{|_{\mathfrak {B}}}}\oplus C_{T_{|_{\Omega {\setminus } {\mathfrak {B}}}}} \text { and } M_u=M_{u_{|_{\mathfrak {B}}}}\oplus M_{u_{|_{\Omega {\setminus } {\mathfrak {B}}}}}. \end{aligned}$$

Of course, we have the following inequalities for the norms

$$\begin{aligned} \max \{\Vert C_{T_{|_{\mathfrak {B}}}} \Vert ,\Vert C_{T_{|_{\Omega {\setminus } {\mathfrak {B}}}}}\Vert \}&\le \Vert C_T\Vert \le \Vert C_{T_{|_{\mathfrak {B}}}} \Vert +\Vert C_{T_{|_{\Omega {\setminus } {\mathfrak {B}}}}}\Vert \\ \max \{\Vert M_{u_{|_{\mathfrak {B}}}}\Vert , \Vert M_{u_{|_{\Omega {\setminus } {\mathfrak {B}}}}}\Vert \}&\le \Vert M_u\Vert \le \Vert M_{u_{|_{\mathfrak {B}}}}\Vert + \Vert M_{u_{|_{\Omega {\setminus } {\mathfrak {B}}}}}\Vert . \end{aligned}$$

Hence it follows that the operator $C_T$ acts continuously from $L^{\Phi }(\Omega )$ into $L^{\Psi }(\Omega )$ if and only if the operator $C_{T_{|_{{\mathfrak {B}}}}}$ acts continuously from $L^{\Phi }({\mathfrak {B}})$ into $L^{\Psi }({\mathfrak {B}})$ and the operator $C_{T_{|_{\Omega {\setminus } {\mathfrak {B}}}}}$ acts continuously from $L^{\Phi }(\Omega {\setminus }{\mathfrak {B}})$ into $L^{\Psi }(\Omega {\setminus }{\mathfrak {B}})$. The same applies to the multiplication operator: $M_u$ acts continuously from $L^{\Phi }(\Omega )$ into $L^{\Psi }(\Omega )$ if and only if $M_{u_{|_{{\mathfrak {B}}}}}$ acts continuously from $L^{\Phi }({\mathfrak {B}})$ into $L^{\Psi }({\mathfrak {B}})$ and $M_{u_{|_{\Omega {\setminus } {\mathfrak {B}}}}}$ acts continuously from $L^{\Phi }(\Omega {\setminus }{\mathfrak {B}})$ into $L^{\Psi }(\Omega {\setminus }{\mathfrak {B}})$. Therefore, the condition for boundedness of the operator $C_T$ is exactly the conjunction of the conditions for the boundedness of the operators $C_{T_{|_{{\mathfrak {B}}}}}$ and $C_{T_{|_{\Omega {\setminus } {\mathfrak {B}}}}}$. Analogously, the condition for boundedness of the operator $M_u$ is the conjunction of conditions for the boundedness of the operators $M_{u_{|{_{\mathfrak {B}}}}}$ and $M_{u_{|_{\Omega {\setminus } {\mathfrak {B}}}}}$. Recall that in [3] it was shown that (a)(i) is a sufficient condition for the continuity of $C_T$ and that (b)(i) is a necessary and sufficient condition for the continuity of $M_u$ in the case of a non-atomic measure space. By analogous arguments to the ones we have used in the proof of the necessity part of Theorem 6.2 (applying Ishii’s theorem rather than Shragin’s) it can be shown that (a)(i) is also a necessary condition for the continuity of $C_T$ in the case of a non-atomic measure space, which generalizes [3], where (a)(i) was shown to be a necessary condition for the continuity of $C_T$ in the non-atomic case under the additional assumption that the mapping T is surjective up to sets of measure zero. Moreover, Theorems 6.2 and 6.5 provide parallel sufficient and necessary conditions (a)(ii) and (b)(ii) in the case of a purely atomic measure space. This completes the proof of Theorem 6.7 $\square$