Basic properties of multiplication and composition operators between distinct Orlicz spaces
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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.
Keywords
Multiplication operator Composition operator Continuous operators Closedrange operators Finiterank operators Orlicz space Musielak–Orlicz space Lebesgue space Measurable transformation Radon–Nikodym derivativeMathematics Subject Classification
47B38 47B33 46E301 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 subjectmatter 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 nonsingular 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 closedrange 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 leftcontinuity of the function \(\Phi \) at \(b_\Phi \), i.e. \(\lim _{x\rightarrow b_\Phi ^} \Phi (x)=\Phi (b_\Phi )\in [0,\infty ]\).
By an Nfunction 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 Nfunction is again an Nfunction.
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 \(K>0\).
Lemma 2.1
Proof
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 nonatomic 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, nonatomic, 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 nonatomic \(\Sigma \)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 wellknown that if \(L^{\Phi _{1}}(\Omega , \Sigma ,\mu )\subset L^{\Phi _{2}}(\Omega , \Sigma ,\mu )\) for a nonatomic 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 \(M_u\).
 (i)
If \((\Omega , \Sigma , \mu )\) be a \(\sigma \)finite measure space, then for every measurable realvalued 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 nonatomic, there is a nonempty 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 \(T(\Omega )\).
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 wellknown 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 nonatomic 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 \(\frac{x^p}{p}\le \frac{1}{p}(\Phi (x)+\Psi (x^{p1}))\) and \(\frac{x^p}{p}\le \frac{1}{p}(\Phi (x^{p1})+\Psi (x))\), for \(x\ge 0\) and \(p>2\), by O’Neil’s theorem we get that if \(u^{\frac{1}{p1}}\in L^{\Psi }(\Omega )\) then the operator \(M_u\) is bounded from \(L^{\Phi }(\Omega )\) into \(L^p(\Omega )\), and if \(v^{\frac{1}{p1}}\in L^{\Phi }(\Omega )\) 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 \(\Phi _1(x)=e^{x^p}x^p1\), \(\Phi _2(x)=\frac{x^p}{p}\), 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 \(u(t)=\root p \of {t^p1}\), 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 \(\Phi _1(x)=e^{x}x1\), \(\Phi _2(x)=(1+x)\ln (1+x)x\), 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 havewhere we used the following calculation$$\begin{aligned} I_{\Phi _1}(f)= & {} \int _\Omega (\Phi _1\circ f)\,d\mu =\int _\Omega \left( e^tt1\right) \,d\mu \\= & {} \int _A \left( e^tt1\right) \,d\mu +\int _B\left( e^tt1\right) \,d\mu <\infty , \end{aligned}$$Hence \(f\in L^{\Phi _1}(\Omega )\). But for any \(\lambda >0\) we have$$\begin{aligned} \int _B\left( e^tt1\right) \,d\mu&=\sum \limits _{n>a}(e^{\ln n}\ln n1)\frac{1}{n^3}<\sum \limits _{n>a}\frac{1}{n^2}<\infty . \end{aligned}$$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 \(L^{\Phi _3}(\Omega )\).$$\begin{aligned} I_{\Phi _2}(\lambda M_u f)= & {} \int _\Omega (\Phi _2\circ \lambda M_u f)d\mu =\int _\Omega \Phi _2\left( \frac{\lambda }{t}\right) d\mu \\= & {} \int _\Omega \left( \left( 1+\frac{\lambda }{t}\right) \ln \left( 1+\frac{\lambda }{t}\right) \frac{\lambda }{t}\right) d\mu \\= & {} \int _A\left( \left( 1+\frac{\lambda }{t}\right) \ln \left( 1+\frac{\lambda }{t}\right) \frac{\lambda }{t}\right) d\mu \\&+\int _B\left( \left( 1+\frac{\lambda }{t}\right) \ln \left( 1+\frac{\lambda }{t}\right) \frac{\lambda }{t}\right) d\mu \\> & {} \int _0^a\left( \left( 1+\frac{\lambda }{t}\right) \ln \left( 1+\frac{\lambda }{t}\right) \frac{\lambda }{t}\right) dt\\ {}= & {} \lambda \int _{\lambda /a}^\infty \left( (1+s)\ln (1+s)s\right) \frac{1}{s^2}ds\\= & {} \lambda \int _{\lambda /a}^\infty \left( \frac{\ln (1+s)}{s^2}+\frac{\ln (1+s)1}{s}\right) ds\\ {}> & {} \lambda \int _{\lambda /a}^\infty \left( \frac{\ln (1+s)1}{s}\right) ds =\infty , \end{aligned}$$
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 nonatomic 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 Nfunction, \(\Phi _1\in \Delta '\), \(\Phi _2\in \Delta _2\), 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 \( x,y\ge 0\).
In the case of a nonatomic 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 nonatomic, 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 nonatomic measure space. Then there are no nonzero bounded operators \(C_TM_u\) from \(L^{\Phi _1}(\Omega )\) into \(L^{\Phi _2}(\Omega )\).
Proof
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
 (i)
\(u(t)=0\) for \(\mu \)a.e. \(t\in {\mathfrak {B}}\), the nonatomic 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
We have a straightforward consequence.
Corollary 3.5
Under the assumptions of Theorem 3.4, if \((\Omega , \Sigma , \mu )\) is a nonatomic 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
 (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
Remark 3.7
 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 \(q^{1}+r^{1}=p^{1}\).
 (i)
 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 nonsingular 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
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
 (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
Theorem 4.4
 (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. \(t\in T({\mathfrak {B}})\) andThen (i) \(\Rightarrow \) (ii) \(\Rightarrow \) (iii). Moreover, if \(\Phi _2\in \Delta '\), then (ii) \(\Rightarrow \) (iv).$$\begin{aligned} \sup \limits _{n\in {\mathbb {N}},\,{\mathfrak {A}}_n\subset T(\Omega )} \Phi _2^{1}(h({\mathfrak {A}}_n))\,\Phi _3^{1}\left( \frac{1}{\mu ({\mathfrak {A}}_n)}\right) <\infty . \end{aligned}$$
Proof
By Theorem 4.1, we have \(h(t)=0\) for \(\mu \)a.e. \(t\in T({\mathfrak {B}})\), hence \(\mu (\mathfrak {B})=\mu \circ T^{1}({T(\mathfrak {B}}))=\int _{T(\mathfrak {B})}h(t)d\mu (t)=0\).
Now we present two lemmas on the relationship between multiplication operators and composition operators, which will be needed later on.
Lemma 4.5
Proof
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
Proof
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 \(L^{\Phi _2}(T(\Omega ))\).
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
Proof
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
 (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 Nfunction, \(\Phi _2\) 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$$\begin{aligned} M_{\Phi _2^{1}\circ h}:L^{\Phi _1}(T(\Omega ))\rightarrow L^{\Phi _2}(T(\Omega )) \end{aligned}$$
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
Remark 4.13
 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 nonsingular 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 \(\sup \nolimits _{n\in {\mathbb {N}}, \, {\mathfrak {A}}_n\subset T(\Omega )}\frac{h({\mathfrak {A}}_n)^p}{\mu ({\mathfrak {A}}_n)^{qp}}<\infty \).
 (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 \(n\in {\mathbb {N}}\), \({\mathfrak {A}}_n\subset T(\Omega )\).
 (a)
 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 nonsingular 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 \(\sum \nolimits ^{\infty }_{j=1}Q_T(F_j)^{\frac{r}{q}}\mu (F_j)<\infty \).
 (a)
5 Multiplication and composition operators with closedrange and/or finite rank
In this section we are going to investigate closedrange 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 Nfunction 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
Now we characterize closedrange 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
 (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 closedrange.
Proof
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.
 (b) \(\Rightarrow \) (c)

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

. Let \(M_u\) have closedrange 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 closedrange, then \(M_{u_{_{G}}}\) has closedrange as well.
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 \(S=\bigcup \nolimits _{n\in E}{\mathfrak {A}}_n\).
In the next theorem we characterize closedrange 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
 (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 closedrange.
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.
Now we characterize closedrange 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 nonsingular 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
 (a)
\(C_T\) has closedrange.
 (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 closedrange. Since T is surjective, by Lemma 5.4, \(C_T\) is injective. It is a wellknown fact that an injective operator has closedrange 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 closedrange. 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 finitedimensional and so \(C_T\) has finite rank. \(\square \)
As an easy consequence of this theorem we state
Corollary 5.6
If \(\Omega \) is nonatomic, then, under the assumptions of Theorem 5.5, there are no nonzero closedrange composition operators from \(L^{\Phi _{1}}(\Omega )\) into \(L^{\Phi _{2}}(\Omega )\).
In the next theorem we characterize closedrange 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
 (a)
\(C_T\) has closedrange.
 (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
 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 closedrange.
 (b)
\(M_u\) has finite rank.
 (c)
The set \(\{n\in {\mathbb {N}}: u({\mathfrak {A}}_n)\ne 0\}\) is finite.
 (a)
 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 closedrange.
 (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.
 (a)
 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 closedrange.
 (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.
 (a)
 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 closedrange.
 (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.
 (a)
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 nonatomic 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.
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
We use Shragin’s result to prove the necessity part of the following theorem.
Theorem 6.2
Remark 6.3
In the preceding two theorems \(\delta \) can be taken in the interval (0, 1).
Proof (of Theorem 6.2)
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 nonnegative 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
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 nonatomic 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
 (a)Assume that a nonsingular 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 thatfor all \(x\ge 0\) and all \(t \in T({\mathfrak {B}}){\setminus } A\);$$\begin{aligned} \Psi (Kx)h(t)\le \Phi (x)+g(t) \end{aligned}$$
 (ii)
condition (6.4) (we identify \({\mathbb {N}}\) with \(\Omega {\setminus }{\mathfrak {B}}).\)
 (i)
 (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 thatfor all \(x\ge 0\) and all \(t \in {\mathfrak {B}}{\setminus } A\);$$\begin{aligned} \Psi (u(t)x)\le \Phi (x)+g(t) \end{aligned}$$
 (ii)
condition (6.5) (again, we identify \({\mathbb {N}}\) with \(\Omega {\setminus }{\mathfrak {B}}).\)
 (i)
Proof
Notes
Acknowledgements
The authors wish to thank the anonymous referee for providing important corrections and insightful suggestions that helped to improve the text significantly.
References
 1.Ando, T.: On products of Orlicz spaces. Math. Ann. 140, 174–186 (1960)MathSciNetCrossRefzbMATHGoogle Scholar
 2.Banach, S.: Théorie des opérations linéaires. Monografie Matematyczne, Warsaw (1932)zbMATHGoogle Scholar
 3.Chawziuk, T., Cui, Y., Estaremi, Y., Hudzik, H., Kaczmarek, R.: Composition and multiplication operators between Orlicz function spaces. J. Inequal. Appl. 2016, 52 (2016). doi: 10.1186/s1366001609729
 4.Conway, J.B.: A course in functional analysis. Springer, New York (1990)zbMATHGoogle Scholar
 5.Cowen, C.C., MacCluer, B.D.: Composition Operators on Spaces of Analytic Functions. Studies in Advanced Mathematics. CRC Press, Boca Raton (1995)zbMATHGoogle Scholar
 6.Cui, Y., Hudzik, H., Kumar, R., Maligranda, L.: Composition operators in Orlicz spaces. J. Aust. Math. Soc. 76, 189–206 (2004)MathSciNetCrossRefzbMATHGoogle Scholar
 7.Cui, Y., Hudzik, H., Kumar, R., Kumar, R.: Compactness and essential norms of composition operators on OrliczLorentz spaces. Appl. Math. Lett. 25(11), 1778–1783 (2012)MathSciNetCrossRefzbMATHGoogle Scholar
 8.Cui, Y., Hudzik, H., Kumar, R., Kumar, R.: Erratum and addendum to the paper Compactness and essential norms of composition operators on OrliczLorentz spaces [Applied Mathematics Letters 25 (2012) 1778–1783]. Comment. Math. 54(2), 259–260 (2014)MathSciNetzbMATHGoogle Scholar
 9.del Campo, R., Fernandez, A., Ferrando, I., Mayoral, F., Naranjo, F.: Multiplication operators on spaces on integrable functions with respect to a vector measure. J. Math. Anal. Appl. 343, 514–524 (2008)MathSciNetCrossRefzbMATHGoogle Scholar
 10.Domański, P., Langenbruch, M.: Composition operators on spaces of real analytic functions. Math. Nachr. 254(255), 68–86 (2003)MathSciNetCrossRefzbMATHGoogle Scholar
 11.Eryilmaz, I.: Weighted composition operators on weighted Lorentz spaces. Colloq. Math. 128, 143–151 (2012)MathSciNetCrossRefzbMATHGoogle Scholar
 12.Estaremi, Y.: Multiplication and composition operators between two diferent Orlicz spaces. arXiv:1301.4830v2. 20 Oct 2013
 13.Gupta, S., Komal, B.S., Suri, N.: Weighted composition operators on Orlicz spaces. Int. J. Contemp. Math. Sci. 1, 11–20 (2010)MathSciNetzbMATHGoogle Scholar
 14.Hencl, S., Kleprlík, L., Malý, J.: Composition operator and SobolevLorentz spaces \(WL^{n, q}\). Studia Math. 221, 197–208 (2014)MathSciNetCrossRefzbMATHGoogle Scholar
 15.Hudzik, H., Kamińska, A.: On uniformly convexifiable and Bconvex MusielakOrlicz spaces. Comment. Math. Prace Mat. 25(1), 59–75 (1985)MathSciNetzbMATHGoogle Scholar
 16.Hudzik, H., Krbec, M.: On noneffective weights in Orlicz spaces. Indag. Mathem. (N.S.) 18(2), 215–231 (2007)Google Scholar
 17.Ishii, J.: On equivalence of modular function spaces. Proc. Jpn. Acad. 35, 551–556 (1959)MathSciNetCrossRefzbMATHGoogle Scholar
 18.Jabbarzadeh, M.R.: Weighted composition operators between \(L^p\)spaces. Bull. Korean Math. Soc. 42, 369–378 (2005)MathSciNetCrossRefzbMATHGoogle Scholar
 19.Kolwicz, P., Leśnik, K., Maligranda, L.: Pointwise multipliers of CalderónLozanovskiĭ spaces. Math. Nachr. 286, 876–907 (2013)MathSciNetCrossRefzbMATHGoogle Scholar
 20.Kolwicz, P., Leśnik, K., Maligranda, L.: Pointwise product of some Banach function spaces and factorization. J. Funct. Anal. 266, 616–659 (2014)MathSciNetCrossRefzbMATHGoogle Scholar
 21.Komal, B.S., Gupta, S.: Multiplication operators between Orlicz spaces. Integr. Equ. Oper. Theory 41, 324–330 (2001)MathSciNetCrossRefzbMATHGoogle Scholar
 22.Krasnoselskii, M.A., Rutickii, B.: Convex functions and Orlicz spaces. Noordhoff, Netherlands (1961) [Translated from Russian]Google Scholar
 23.Kumar, R.: Weighted composition operators between two \(L^p\)spaces. Mat. Vesnik 61, 111–118 (2009)MathSciNetzbMATHGoogle Scholar
 24.Kumar, R.: Composition operators on Orlicz spaces. Integr. Equ. Oper. Theory 29, 17–22 (1997)MathSciNetCrossRefzbMATHGoogle Scholar
 25.Kumar, R., Kumar, R.: Composition operators on OrliczLorentz spaces. Integr. Equ. Oper. Theory 60, 79–88 (2008)MathSciNetCrossRefzbMATHGoogle Scholar
 26.Maligranda, L.: Orlicz spaces and interpolation, Seminars in Math. 5, Univ. Estadual de Campinas, Campinas SP, Brazil (1989)Google Scholar
 27.Maligranda, L., Persson, L.E.: Generalized duality of some Banach function spaces. Indag. Math. 92(3), 323–338 (1989)MathSciNetCrossRefzbMATHGoogle Scholar
 28.Musielak, J.: Orlicz spaces and modular spaces. Lecture Notes in Math, vol. 1034. Springer, Berlin (1983)Google Scholar
 29.Nielsen, O.A.: An introduction to integration and measure theory. WileyInterscience, New York (1997)zbMATHGoogle Scholar
 30.Nowak, M.: On inclusions between Orlicz spaces. Comment. Math. Prace Mat. 26, 265–277 (1986)MathSciNetzbMATHGoogle Scholar
 31.Nowak, M.: Some equalities among Orlicz spaces. I. Comment. Math. Prace Mat. 29, 255–275 (1990)MathSciNetzbMATHGoogle Scholar
 32.O’Neil, R.: Fractional integration in Orlicz spaces. I. Trans. Am. Math. Soc. 115, 300–328 (1965)MathSciNetCrossRefzbMATHGoogle Scholar
 33.Rao, M.M.: Convolutions of vector fields II: random walk models. Nonlinear Anal. 47, 3599–3615 (2001)MathSciNetCrossRefzbMATHGoogle Scholar
 34.Rao, M.M., Ren, Z.D.: Theory of Orlicz spaces. Marcel Dekker, New York (1991)zbMATHGoogle Scholar
 35.Shapiro, J.H.: Composition Operators and Classical Function Theory. Universitext Tracts in Mathematics. Springer, New York (1993)CrossRefzbMATHGoogle Scholar
 36.Singh, R.K., Manhas, J.S.: Composition operators on function spaces, NorthHolland Mathematics Studies, 179. NorthHolland Publishing Co., Amsterdam (1993)Google Scholar
 37.Shragin, I.V.: The conditions on embeddings of some classes and conclusions from them. Mat. Zam. 20, 681–692 (1976). (in Russian)Google Scholar
 38.Takagi, H., Yokouchi, K.: Multiplication and composition operators between two \(L^p\)spaces. Contemp. Math. 232, 321–338 (1999)MathSciNetCrossRefzbMATHGoogle Scholar
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