Surjectivity, Closed Range, and Fredholmness of the Composition and Multiplication Operators Between Possibly Distinct Orlicz Spaces

Abstract

We give criteria for the closely related concepts of surjectivity, closed range, and Fredholmness of the composition and multiplication operators between possibly different Orlicz spaces over non-atomic measure spaces.

1 Introduction

We continue our investigation of the composition and multiplication operators between possibly different Orlicz spaces over non-atomic measure spaces. It was initiated in the papers [1, 2] (see also [16]). For results on the composition and multiplication operators between the same Orlicz spaces, see, e.g. [3, 4, 6, 8].

In this paper we consider the closely related concepts of surjectivity, closed range, and Fredholmness of these operators. Here the criterion for the continuity of these operators, based on Ishii’s inclusion theorem (see [5]), again proves its usefulness. Some results on the Fredholm property of the multiplication operator between Lebesgue spaces were published in [15] and between Orlicz spaces in [6].

The criteria for surjectivity of the composition and multiplication operators given below extend and/or partly correct the criteria found in the papers [1, 4].

2 Preliminaries

We explain some of the terms used in this paper.

We will call an Orlicz function a function $\Phi :\mathbb {R}\rightarrow [0,\infty )$ that is convex, even, vanishes at 0 and only at 0.

The function $\Phi ^*:\mathbb {R}\rightarrow [0,\infty ]$ defined by

$$\begin{aligned} \Phi ^*(x):=\sup \{y|x|-\Phi (y):y\ge 0\} \qquad (x\in \mathbb {R}) \end{aligned}$$

is called the function complementary to $\Phi$in the sense of Young.

An Orlicz function $\Phi$ is said to satisfy the $\Delta _{2}$-condition at $\infty$ if for every $a>1$ there are constants $K>0$ and $x_0>0$ such that

$$\begin{aligned} \Phi (ax)\le K\Phi (x) \qquad ( x\ge x_{0}). \end{aligned}$$

If $x_0=0$, then the Orlicz function $\Phi$ is said to satisfy the $\Delta _{2}$-condition globally.

Let $\Phi , \Psi$ be Orlicz functions. Then $\Phi$ is called stronger than $\Psi$ at $\infty$ if

$$\begin{aligned} \Psi (x)\le \Phi (ax)\qquad (x\ge x_0) \end{aligned}$$

for some $a\ge 0$ and some $x_0>0$.

Let $(\Omega , \mathcal M, \mu )$ be a $\sigma$-finite non-atomic complete measure space and let $L^0(\Omega , \mathcal M, \mu )$ be the linear space of equivalence classes of $\mathcal M$-measurable real-valued functions on $\Omega$. We often write $\Omega$ instead of $(\Omega ,\mathcal M, \mu )$, for brevity. Throughout the paper all measures are assumed to be positive.

If $\Phi$ is an Orlicz function, then the space

$$\begin{aligned} L^{\Phi }(\Omega ):=\left\{ f\in L^0(\Omega , \mathcal M, \mu ):\exists \,\lambda >0\;\; I_{\Phi }(\lambda \,f)<\infty \right\} \end{aligned}$$

equipped with the norm

$$\begin{aligned} \Vert f\Vert _{L^{\Phi }(\Omega )}:=\inf \Big \{\lambda >0:I_{\Phi }\Big (\frac{f}{\lambda }\Big )\le 1\Big \}, \end{aligned}$$

where $I_\Phi (f):=\int _{\Omega }\Phi \big (f(\omega )\big )d\mu (\omega )$, is a Banach space called Orlicz space. The functional $I_\Phi$ is called a modular (see [7, 9]).

If $A\in \mathcal M$, by $I_\Phi (f,A)$ we mean the value of the modular $I_\Phi$ at f in the space $L^\Phi (A,\mathcal M\cap A,\mu _{|A})$.

For $f_n,f\in L^{\Phi }(\Omega )\;(n\in \mathbb {N})$, we have

$$\begin{aligned} \Vert f_n-f\Vert _{L^{\Phi }(\Omega )}\rightarrow 0\quad {\text {iff}}\quad I_\Phi \big (\lambda (f_n-f)\big )\rightarrow 0 \quad \text {for every}\quad \lambda >0 \end{aligned}$$

(cf. [10, Theorem 1.3]).

If $\Phi$ satisfies the $\Delta _2$-condition globally, then F is a bounded linear functional on $L^\Phi (\Omega )$ if and only if it has the form $F(g)=\int _\Omega g(\omega )\tilde{g}(\omega )d\mu (\omega )$, for some $\tilde{g}\in L^{\Phi ^*}(\Omega )$ and every $g\in L^\Phi (\Omega )$ (cf. [10, Theorem 9.3]).

Let $(\Omega ,\mathcal M,\mu )$ and $(\Xi ,\mathcal N,\nu )$ be general measure spaces and let $T:\Xi \rightarrow \Omega$ be a measurable transformation in the sense that the set $T^{-1}(A)$ is $\mathcal N$-measurable for any $\mathcal M$-measurable set A. The transformation T is said to be nonsingular if $\nu (T^{-1}(F))=0$ for any $F\in \mathcal M$ with $\mu (F)=0$. This property ensures that the measure $\nu \circ T^{-1}$ defined by $\nu \circ T^{-1}(A):=\nu (T^{-1}(A))$ for any $A\in \mathcal M$ is absolutely continuous with respect to the measure $\mu$. Under this condition, the Radon–Nikodym theorem guarantees the existence of a non-negative locally integrable function $h:\Omega \rightarrow \mathbb {R}_+$, called the Radon–Nikodym derivative of $\nu \circ T^{-1}$ with respect to $\mu$, such that

$$\begin{aligned} \nu \circ T^{-1}(A)=\int _{A}h(\omega )\, d\mu (\omega ) \quad (A\in \mathcal M). \end{aligned}$$

Upon extending it, we may and shall assume that the Radon–Nikodym derivative h is 0 on $\Omega {\setminus } T(\Xi )$.

In case $T:\Xi \rightarrow \Omega$ is measurable, nonsingular, and injective, we can define the function $T^{-1}:T(\xi )\mapsto \xi$. If, additionally, we assume that the function $T^{-1}$ measurable and nonsingular, then $h(\omega )>0$ for a.e. $\omega \in T(\Xi )$. Indeed, suppose that $h=0$ on a $\mathcal {M}$-measurable set $A\subset T(\Xi )$ with $\mu (A)>0$. Then $\nu \circ T^{-1}(A)=\int _{A}h(\omega )\, d\mu (\omega )=0$. Hence, for the $\mathcal {N}$-measurable set $B:=T^{-1}(A)$, we have $\nu (B)=0$ and also $\mu \big ((T^{-1})^{-1}(B)\big )=\mu (T(B))=\mu (A)>0$, which means that $T^{-1}$ would not be nonsingular.

Instead of $T:\Xi \rightarrow \Omega$ being injective (as we will be doing all along), we might assume that $T:\Xi \rightarrow \Omega$ is essentially injective in the sense that $\nu \big (\{\xi \in \Xi : \text {there is } \tilde{\xi }\in \Xi \text { such that } \xi \ne \tilde{\xi } \text { and } T(\xi )=T(\tilde{\xi })\}\big )=0$. Then a function $T^{-1}:T(\Xi )\rightarrow \Xi$ may be defined, for any $\xi \in \Xi$, by $T(\xi )\mapsto \tilde{\xi }$ with $\tilde{\xi }\in \Xi$ such that $T(\xi )=T(\tilde{\xi })$ chosen arbitrarily. Moreover, if any such function $T^{-1}$ is measurable and nonsingular, then all of them are.

Let $(\Omega ,\mathcal M,\mu )$ and $(\Xi ,\mathcal N,\nu )$ be measure spaces and let $T:\Xi \rightarrow \Omega$ be a nonsingular measurable transformation. The linear transformation $C_T$ defined by

$$\begin{aligned} (C_{T}f)(\xi ):=f\big (T(\xi )\big )\quad \big (\xi \in \Xi ,\;\,f\in L^0(\Omega ,\mathcal M,\mu )\big ) \end{aligned}$$

is called a composition operator induced by the transformation T.

The nonsingularity of the transformation $T:\Xi \rightarrow \Omega$ ensures that the composition operator $C_T$ is well defined, i.e. if $f=g$$\mu$-a.e. on $\Omega$ then $C_T\,f=C_T\,g$$\nu$-a.e. on $\Xi$ (see [14, p. 18]).

Let $(\Omega ,\mathcal M,\mu )$ be a measure space. Given a measurable function $u\in L_+^0 (\Omega ,\mathcal M,\mu )$ (with non-negative values), the linear transformation $M_u$ defined by

$$\begin{aligned} (M_uf)(\omega ):=u(\omega )\,f(\omega )\quad \big (f\in L^0 (\Omega ,\mathcal M,\mu )\big ), \end{aligned}$$

is called a multiplication operator induced by the function u.

A function $\varphi :\Omega \times \mathbb {R}\rightarrow [0,\infty ]$ such that $\varphi (\omega ,\cdot )$ is an Orlicz function for $\mu$-a.e. $\omega \in \Omega$ and $\varphi (\cdot , x)$ is a $\mathcal M$-measurable function for every $x\in \mathbb {R}$ is called a Musielak–Orlicz function.

The space $L^\varphi (\Omega ,\mathcal M,\mu )$ of all equivalence classes of $\mathcal M$-measurable functions $f:\Omega \rightarrow \mathbb {R}$ such that

$$\begin{aligned} I_\varphi (\lambda f):=\int _\Omega \varphi \big (\omega ,\lambda f(\omega )\big )d\mu (\omega )<\infty \end{aligned}$$

for some $\lambda >0$, equipped with the Luxemburg norm $\Vert f\Vert _\varphi$ defined analogously as for the Orlicz space, is called a Musielak–Orlicz space generated by the Musielak–Orlicz function $\varphi$. The Musielak–Orlicz space is a Banach space (cf. [11, 12]).

We are only interested in a special class of Musielak–Orlicz spaces. Firstly, the Orlicz weighted space $L^\Psi _h(\Omega )$ is the Musielak–Orlicz space $L^\psi (\Omega )$ generated by the Musielak–Orlicz function $\psi (\omega ,x):=\Psi (x)h(\omega )$ for $\omega \in \Omega$ and $x\in \mathbb {R}$, where $h\in L^0_+(\Omega )$ is the Radon–Nikodym derivative related to the transformation T inducing the composition operator $C_T$.

Secondly, the Orlicz weighted space $L^{\Psi _u}(\Omega )$ is the Musielak–Orlicz space $L^\psi (\Omega )$ generated by the Musielak–Orlicz function $\psi (\omega ,x):=\Psi _u (\omega ,x):=\Psi \left( u(\omega )x\right)$ for $\omega \in \Omega$ and $x\in \mathbb {R}$, where $u\in L^0_+(\Omega )$ is the function inducing the multiplication operator $M_u$.

The consideration of the composition and multiplication operators between Orlicz spaces rather naturally leads to Orlicz weighted spaces.

Recall that a continuous linear operator between Banach spaces is said to be Fredholm if the codimension of its range and the dimension of its kernel are finite and the range is closed. It is a consequence of the Open Mapping Theorem that, actually, finite codimension of the range of a continuous linear operator between Banach spaces already implies that its range is closed (cf. [13, p. 267]).

We will take advantage of the following two theorems proved elsewhere. For the following theorem on the composition operator, see Theorem 2.1 and part of the proof of Theorem 2.2 in [1] (when the Radom–Nikodym derivative h is assumed to vanish outside the image of T, as we do here, the sufficient conditions in [1, Theorem 2.1] are also necessary).

Theorem 1

Let $T:\Xi \rightarrow \Omega$ be a nonsingular measurable transformation. The composition operator $C_T$ acts continuously from $L^\Phi (\Omega )$ to $L^\Psi (\Xi )$ if and only if one of the following equivalent conditions is satisfied.

• (1) $L^\Phi (\Omega )\subset L^\Psi _h(\Omega )$.

• (2) There are $K>0$ and $g\in L^1_+(\Omega )$ such that $\Psi (x)h(\omega )\le \Phi (Kx)+g(\omega )$ for every $x\ge 0$ and a.e. $\omega \in \Omega$.

• (3) The function $\sup _{x\ge 0} \{\Psi (x)h(\omega )-\Phi (Kx)\}$ is integrable over $\Omega$ for some $K>0$.

For the following corresponding theorem on the multiplication operator, see the proof of Theorem 3.1 in [1].

Theorem 2

Let $u:\Omega \rightarrow \mathbb R_+$ be a measurable function. The multiplication operator $M_u$ acts continuously from $L^\Phi (\Omega )$ to $L^\Psi (\Omega )$ if and only if one of the following equivalent conditions is satisfied.

• (1) $L^\Phi (\Omega )\subset L^{\Psi _u}(\Omega )$.

• (2) There are $K>0$ and $g\in L^1_+(\Omega )$ such that $\Psi \big (u(\omega )x\big )\le \Phi (Kx)+g(\omega )$ for every $x\ge 0$ and a.e. $\omega \in \Omega$.

• (3) The function $\sup _{x\ge 0} \{\Psi \big (u(\omega )x\big )-\Phi (Kx)\}$ is integrable over $\Omega$ for some $K>0$.

3 Surjectivity

3.1 Surjectivity of $C_T$

We would like to give necessary and sufficient conditions for a composition operator to be surjective. The following theorem extends Theorem 4.1. from [4], and Theorems 2.4. and 2.5. from [1], as well as corrects their points (2), (ii), and (2), respectively. More precisely, the surjectivity of the operator $C_T$ is in fact not equivalent to $L^\Phi (\Omega ) = L^\Psi _h(\Omega )$, as it was claimed in [1, Fact 2.1] and used in the proof of Theorems 2.4 (ii) and 2.5 (2) in [1].

Theorem 3

Let $T:\Xi \rightarrow \Omega$ be a nonsingular measurable transformation which is injective and such that the mapping $T^{-1}:T(\xi )\mapsto \xi$ is measurable and nonsingular. Then a continuous composition operator from $C_T:L^\Phi (\Omega )$ to $L^\Psi (\Xi )$ is surjective if and only if one of the following equivalent conditions is satisfied.

• (1) $L^\Psi (\Xi )\subset L^\Phi _{\frac{1}{h\circ T}}(\Xi )$.

• (2) There is $K>0$ and $g\in L^1_+(\Xi )$ such that

  $$\begin{aligned} \Phi (x)\frac{1}{h\circ T(\xi )}\le \Psi (Kx)+g(\xi ) \end{aligned}$$

  for all $x\ge 0$ and a.e. $\xi \in \Xi$.

• (3) There is $K>0$ for which the function

  $$\begin{aligned} \sup _{x\ge 0}\Big \{ \Phi (x)\frac{1}{h\circ T(\xi )}-\Psi (Kx)\Big \} \end{aligned}$$

  is integrable over $\Xi$.

Proof

A continuous composition operator $C_T$ from $L^\Phi (\Omega )$ to $L^\Psi (\Xi )$ is surjective if and only if for every $g\in L^\Psi (\Xi )$ there is $f\in L^\Phi (\Omega )$ such that $C_Tf=g$, i.e $f(T(\xi ))=g(\xi )$ for a.e. $\xi \in \Xi$. Since T is injective and $T^{-1}$ is nonsingular, we have $f(\omega )=g(T^{-1}(\omega ))$ for a.e. $\omega \in T(\Xi )$, i.e. $f=C_{T^{-1}}g$. Therefore $C_T$ is surjective if and only if $C_{T^{-1}}$ acts continuously from $L^\Psi (\Xi )$ to $L^\Phi (\Omega )$, supposing $C_{T^{-1}}$ is well defined. But the mapping $T^{-1}:T(\Xi )\rightarrow \Xi$ is measurable and nonsingular, hence $C_{T^{-1}}$ is indeed well defined. By the Radon–Nikodym theorem, there is a locally integrable function $\tilde{h}:\Xi \rightarrow R_+$ such that for any measurable set $A\subset \Xi$,

$$\begin{aligned} \mu \circ T(A)=\int _A \tilde{h}(\xi )d\nu (\xi ). \end{aligned}$$

In fact, $\tilde{h}=\frac{1}{h\circ T}$ a.e. on $\Xi$ because

$$\begin{aligned} \int _A \frac{1}{h\big (T(\xi )\big )}d\nu (\xi )&=\int _{T(A)} \frac{1}{h(\omega )}d\nu \circ T^{-1}(\omega ) =\int _{T(A)} \frac{1}{h(\omega )}h(\omega )d\mu (\omega )\\&=\mu (T(A))=\int _A \tilde{h}(\xi )d\nu (\xi ), \end{aligned}$$

for any measurable $A\subset \Xi$. Now by Theorem 1 (substituting $T^{-1}:T(\Xi )\rightarrow \Xi$ for $T:\Xi \rightarrow \Omega$ and $\frac{1}{h\circ T}$ for h) we infer that $C_T$ is surjective if and only if $L^\Psi (\Xi )\subset L^\Phi _{\frac{1}{h\circ T}}(\Xi )$. The same theorem yields the other two equivalent conditions for the surjectivity of $C_T$.

Remark 1

It is possible to prove that if a continuous nonzero composition operator $C_T:L^\Phi (\Omega )\rightarrow L^\Psi (\Xi )$ is surjective, then the functions $\Phi$ and $\Psi$ are equivalent at $\infty$; more precisely, there are constants $K,L,a, x_0>0$ such that $\Psi (Kx)\le a\Phi (x)\le \Psi (Lx)$ for all $x\ge x_0$.

In the special case when $\Psi =\Phi$, a much simpler condition for the surjectivity of a continuous composition operator can be derived.

Corollary 1

Let $T:\Xi \rightarrow \Omega$ be a nonsingular measurable transformation which is injective and such that the mapping $T^{-1}:T(\xi )\mapsto \xi$ is measurable and nonsingular. Then a continuous composition operator $C_T:L^\Phi (\Omega )\rightarrow L^\Phi (\Xi )$ is surjective if there is $\varepsilon >0$ such that $h(\omega )\ge \varepsilon$ for a.e. $\omega \in T(\Xi )$. Moreover, assuming that $\Phi$ satisfies the $\Delta _2$-condition at $\infty$, if a continuous composition operator $C_T:L^\Phi (\Omega )\rightarrow L^\Phi (\Xi )$ is surjective, then this condition is also necessary.

Proof

Sufficiency. Let $h(\omega )\ge \varepsilon$ for a.e. $\omega \in T(\Xi )$ and some $\varepsilon >0$. We may assume that $\varepsilon \le 1$. By the convexity of $\Phi$ and the fact that $\Phi (0)=0$, we have

$$\begin{aligned} \Phi (x)\frac{1}{h\circ T(\xi )}\le \Phi (x)\frac{1}{\varepsilon }\le \Phi \Big (\frac{1}{\varepsilon }x\Big ) \end{aligned}$$

for all $x\ge 0$ and a.e. $\xi \in \Xi$. Hence, by the preceding theorem (with $K=\frac{1}{\varepsilon }$ and $g=0$), the operator $C_T$ is surjective.

Necessity. Let $C_T:L^\Phi (\Omega )\rightarrow L^\Phi (\Xi )$ be surjective and let $K>0$ be arbitrary. Since $\Phi$ satisfies the $\Delta _2$-condition at $\infty$, there is $a>1$ and $x_0\ge 0$ such that $\Phi (Kx)\le a\Phi (x)$ for $x\ge x_0$. Suppose that the set $\{\xi \in \Xi : h\circ T(\xi )<\varepsilon \}$ has positive measure, for every $\varepsilon >0$. Let $\varepsilon _0:=\frac{1}{2a}$ and set $A:=\{\xi \in \Xi : h\circ T(\xi )<\varepsilon _0\}$. Notice that

$$\begin{aligned} \Phi (x)\frac{1}{h\circ T(\xi )}-\Phi (Kx)> \Phi (x)\frac{1}{\varepsilon _0}-\Phi (Kx)\ge 2a\Phi (x)-a\Phi (x)=a\Phi (x) \end{aligned}$$

for $x\ge x_0$ and $\xi \in A$. Therefore, for every $K>0$,

$$\begin{aligned} \int _\Xi \;\sup _{x\ge 0}\Big \{ \Phi (x)\frac{1}{h\circ T(\xi )}-\Phi (Kx)\Big \}\,d\nu (\xi )\ge \int _{A} \sup _{x\ge x_0}a \Phi (x)\,d\nu (\xi )\;=\;\infty , \end{aligned}$$

which, by Theorem 3, implies that $C_T$ is not surjective. This contradiction shows that there is $\varepsilon >0$ such that $h(\omega )\ge \varepsilon$ for a.e. $\omega \in T(\Xi )$. $\square$

Proceeding in exactly the same manner one can prove that if there is $M>0$ such that $h(\omega )\le M$ for a.e. $\omega \in T(\Xi )$ then $C_T$ is continuous from $L^\Phi (\Omega )$ to $L^\Phi (\Xi )$, and that if $\Phi$ satisfies the $\Delta _2$-condition at $\infty$, then the same condition is also necessary for the continuity of $C_T:L^\Phi (\Omega )\rightarrow L^\Phi (\Xi )$. Hence the last corollary may be written in a more specific way.

Corollary 2

Let $T:\Xi \rightarrow \Omega$ be a nonsingular measurable transformation which is injective and such that the mapping $T^{-1}:T(\xi )\mapsto \xi$ is measurable and nonsingular. Then the composition operator $C_T:L^\Phi (\Omega )\rightarrow L^\Phi (\Xi )$ is continuous and surjective if there are $\varepsilon ,M>0$ such that $\varepsilon \le h(\omega )\le M$ for a.e. $\omega \in T(\Xi )$. Moreover, assuming that $\Phi$ satisfies the $\Delta _2$-condition at $\infty$, if the composition operator $C_T$ is continuous from $L^\Phi (\Omega )$ to $L^\Phi (\Xi )$ and surjective, then this condition is also necessary.

3.2 Surjectivity of $M_u$

We proceed to the surjectivity of the multiplication operator. The following theorem corrects point (ii) of Theorem 3.2. from [1]. The surjectivity of $M_u$ is a different property from the equality $L^\Phi (\Omega )= L^{\Psi _u}(\Omega )$, as was claimed there.

Theorem 4

Let $u:\Omega \rightarrow \mathbb R_+$ be a measurable function such that $u(\omega )>0$ for a.e. $\omega \in \Omega$. Then a continuous multiplication operator from $M_u:L^\Phi (\Omega )$ to $L^\Psi (\Omega )$ is surjective if and only if one of the following equivalent conditions is satisfied.

• (1) $L^\Psi (\Omega )\subset L^{\Phi _{\frac{1}{u}}}(\Omega )$.

• (2) There is $K>0$ and $g\in L^1_+(\Omega )$ such that

  $$\begin{aligned} \Phi \Big (\frac{x}{u(\omega )}\Big )\le \Psi (Kx)+g(\omega ) \end{aligned}$$

  for all $x\ge 0$ and a.e. $\omega \in \Omega$.

• (3) There is $K>0$ for which the function

  $$\begin{aligned} \sup _{x\ge 0}\Big \{ \Phi \Big (\frac{x}{u(\omega )}\Big )-\Psi (Kx)\Big \} \end{aligned}$$

  is integrable over $\Omega$.

Proof

A continuous multiplication operator $M_u$ from $L^\Phi (\Omega )$ to $L^\Psi (\Omega )$ is surjective if and only if for every $g\in L^\Psi (\Omega )$ there is $f\in L^\Phi (\Omega )$ such that $M_u f=g$. But, since $u(\omega )>0$ for a.e. $\omega \in \Omega$, we may write, equivalently, $f=M_{\frac{1}{u}}g$, which means that a continuous multiplication operator $M_u$ from $L^\Phi (\Omega )$ to $L^\Psi (\Omega )$ is surjective if and only if $M_{\frac{1}{u}}$ acts continuously from $L^\Psi (\Omega )$ to $L^\Phi (\Omega )$.

Now by Theorem 2 (substituting $\frac{1}{u}$ for u) we infer that $M_u$ is surjective if and only if $L^\Psi (\Omega )\subset L^{\Phi _{\frac{1}{u}}}(\Omega )$, and the same theorem also yields the other two equivalent conditions for the surjectivity of $M_u$, which finishes the proof.

Remark 2

Analogously to Remark 1, the surjective continuous multiplication operator $M_u:L^\Phi (\Omega )\rightarrow L^\Psi (\Omega )$ implies the equivalence of the functions $\Phi$ and $\Psi$ at $\infty$.

A simpler condition for the surjectivity of a continuous multiplication operator can be obtained in the case when $\Psi =\Phi$.

Corollary 3

Let $u:\Omega \rightarrow \mathbb R_+$ be a measurable function such that $u(\omega )>0$ for a.e. $\omega \in \Omega$. Then a continuous multiplication operator $M_u:L^\Phi (\Omega )\rightarrow L^\Phi (\Omega )$ is surjective if there is $\varepsilon >0$ such that $u(\omega )\ge \varepsilon$ for a.e. $\omega \in \Omega$. Moreover, assuming that $\Phi$ satisfies the $\Delta _2$-condition at $\infty$, if a continuous multiplication operator $M_u:L^\Phi (\Omega )\rightarrow L^\Phi (\Omega )$ is surjective, then there is $\varepsilon >0$ such that $u(\omega )\ge \varepsilon$ for a.e. $\omega \in \Omega$.

We omit the proof of this corollary as it is analogous to the proof of Corollary 1.

4 Closed Range

We have just found necessary and sufficient conditions for the surjectivity of a continuous composition operator from $L^\Phi (\Omega )$ to $L^\Phi (\Xi )$ and a continuous multiplication operator from $L^\Phi (\Omega )$ to $L^\Phi (\Omega )$. It turns out that basically the same conditions are necessary and sufficient for the range of these operators to be closed when the generating functions are possibly different. As is to be expected, they look somewhat stronger than the conditions we obtained for the surjectivity of $C_T:L^\Phi (\Omega )\rightarrow L^\Psi (\Xi )$ and $M_u:L^\Phi (\Omega )\rightarrow L^\Psi (\Omega )$.

4.1 $C_T$ with Closed Range

The sufficient condition for closedness of the range of a continuous composition operator $C_T$ we give below requires that $C_T$ acts between Orlicz spaces over the same measure space.

Theorem 5

Suppose that $T:\Omega \rightarrow \Omega$ is an injective, nonsingular measurable transformation, the composition operator $C_T$ is continuous from $L^\Phi (\Omega )$ to $L^\Psi (\Omega )$, and $L^\Phi (T(\Omega ))\cap L^\Psi (T(\Omega ))$ is a closed subset of $L^\Psi (T(\Omega ))$. If there is $\varepsilon >0$ such that $h(\omega )\ge \varepsilon$ for a.e. $\omega \in T(\Omega )$, then the range of $C_T$ is closed.

Proof

Suppose that $C_T$ is continuous from $L^\Phi (\Omega )$ to $L^\Psi (\Omega )$ and there is $\varepsilon >0$ such that $h(\omega )\ge \varepsilon$ for a.e. $\omega \in T(\Omega )$ but the range of $C_T$ is not closed. Then there is a sequence $(f_n)_{n=1}^\infty \subset L^\Phi (\Omega )$ and a function $g\in L^\Psi (\Omega )$ such that

$$\begin{aligned} (C_T f_n)_{n=1}^\infty \subset \mathcal R(C_T),\quad C_T f_n\rightarrow g~\text {in}~L^\Psi (\Omega ) ~\text {as}~n\rightarrow \infty \quad \text {but}\quad g\notin \mathcal R(C_T). \end{aligned}$$

The fact that $g\notin \mathcal R(C_T)$ implies that there is no $f\in L^\Phi (\Omega )$ for which $C_T f= g$, or, equivalently, $f(T(\omega ))=g(\omega ) \;\big (\omega \in \Omega \big )$. Hence there is no $f\in L^\Phi (\Omega )$ such that $f(\omega )=g\big (T^{-1}(\omega )\big )\; \big (\omega \in T(\Omega )\big )$ since, by assumption, T is injective. In other words, $g\notin \mathcal R(C_T)$ is equivalent to $g\circ T^{-1}\notin L^\Phi (T(\Omega ))$.

Since $C_T$ is continuous from $L^\Phi (\Omega )$ to $L^\Psi (\Omega )$, Theorem 1 implies that there are $K>0$ and $\tilde{g}\in L_+^1(T(\Omega ))$ such that

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

for all $x\ge 0$ and a.e. $\omega \in T(\Omega )$. Now the condition $h(\omega )\ge \varepsilon >0$ for a.e. $\omega \in T(\Omega )$ yields that

$$\begin{aligned} \varepsilon \Psi (x)\le \Phi (Kx)+\tilde{g}(\omega ) \end{aligned}$$

for all $x\ge 0$ and a.e. $\omega \in T(\Omega )$. Thus, if $f\in L^\Phi (T(\Omega ))$, then for $\lambda >0$ such that $I_\Phi \big (\lambda f,T(\Omega )\big )<\infty$ we have

$$\begin{aligned} \varepsilon I_\Psi \big (\frac{\lambda }{K}f, T(\Omega )\big )\le I_\Phi \big (\lambda f, T(\Omega )\big )+\int _{T(\Omega )} \tilde{g}(\omega )d\mu (\omega )\;<\;\infty , \end{aligned}$$

i.e. $f\in L^\Psi (T(\Omega ))$. Hence $L^\Phi (T(\Omega ))\subset L^\Psi (T(\Omega ))$. In particular, $(f_n)_{n=1}^\infty \subset L^\Phi (\Omega )\subset L^\Phi (T(\Omega ))\subset L^\Psi (T(\Omega ))$. Further, since we assume that $L^\Phi (T(\Omega ))\cap L^\Psi (T(\Omega ))$ is a closed subset of $L^\Psi (T(\Omega ))$, we conclude that $L^\Phi (T(\Omega ))$ is a closed subspace of $L^\Psi (T(\Omega ))$. Consequently, there is no sequence in $L^\Phi (T(\Omega ))$ that converges to $g\circ T^{-1}\notin L^\Phi (T(\Omega ))$ in $L^\Psi (T(\Omega ))$. However, applying the fact that $C_T f_n\rightarrow g \text { in }L^\Psi (\Omega )$, which is equivalent to $I_\Psi \big (\lambda (f_n\circ T-g), \Omega \big )\rightarrow 0$ for every $\lambda >0$, and the fact that $h(\omega )\ge \varepsilon >0$ for a.e. $\omega \in T(\Omega )$, we obtain

$$\begin{aligned} \varepsilon I_\Psi \Big (\lambda (f_n-g\circ T^{-1}), T(\Omega )\Big )&= \varepsilon \int _{T(\Omega )}\Psi \Big (\lambda \big (f_n(\omega )-g\circ T^{-1}(\omega )\big )\Big ) d\mu (\omega )\\&\le \int _{T(\Omega )}\Psi \Big (\lambda \big (f_n(\omega )-g\circ T^{-1}(\omega )\big )\Big )h(\omega ) d\mu (\omega )\\&= \int _{T(\Omega )}\Psi \Big (\lambda \big (f_n(\omega )-g\circ T^{-1}(\omega )\big )\Big ) d\mu \circ T^{-1}(\omega )\\&= \int _{\Omega }\Psi \Big (\lambda \big (f_n\circ T(\omega )-g(\omega )\big )\Big ) d\mu (\omega )\\&=I_\Psi \Big (\lambda (f_n\circ T-g), \Omega \Big )\;\rightarrow \;0, \end{aligned}$$

for every $\lambda >0$. Therefore $f_n\rightarrow g\circ T^{-1} \text { in }L^\Psi (T(\Omega ))$. This contradiction finishes the proof. $\square$

The following theorem gives a necessary condition for the closedness of the range of a continuous composition operator $C_T$ from $L^\Phi (\Omega )$ to $L^\Psi (\Xi )$. (Note that here measure spaces may be different.)

Theorem 6

Suppose that $T:\Xi \rightarrow \Omega$ is a nonsingular measurable transformation such that $h(\omega )>0$ for a.e. $\omega \in T(\Xi )$, and the composition operator $C_T$ is continuous from $L^\Phi (\Omega )$ to $L^\Psi (\Xi )$. If the range of $C_T$ is closed and $\Psi$ satisfies the $\Delta _2$-condition at $\infty$, then there is $\varepsilon >0$ such that $h(\omega )\ge \varepsilon$ for a.e. $\omega \in T(\Xi )$.

Proof

Let $C_T$ be continuous from $L^\Phi (\Omega )$ to $L^\Psi (\Xi )$ and suppose that for every $\varepsilon >0$ we have $\mu \big (\{\omega \in T(\Xi ): h(\omega )<\varepsilon \}\big )>0$.

We will show that there is a sequence $(g_n)_{n=1}^\infty \subset \mathcal R(C_T)\subset L^\Psi (\Xi )$ and a function $g\in L^\Psi (\Xi )$ such that $g_n\rightarrow g$ in $L^\Psi (\Xi )$ and $g\notin \mathcal R(C_T)$. This will imply that the range of $C_T$ is not closed, proving the theorem.

For each $k\in \mathbb {N}$ define the set

$$\begin{aligned} A_k:=\Big \{\omega \in T(\Xi ):\frac{1}{(k+1)^2}\le h(\omega )< \frac{1}{k^2}\Big \}. \end{aligned}$$

By our assumptions, infinitely many sets $A_k$ have nonzero measure and, upon taking a subsequence, one may and shall assume that the sequence $(\mu (A_{k})_k$ is decreasing to 0.

For each $n\in \mathbb {N}$, define the function

$$\begin{aligned} f_n:=\sum _{k=1}^n\Psi ^{-1}\Big (\frac{\chi _{A_k}}{\mu (A_k)}\Big )\in L^\Phi (\Omega ). \end{aligned}$$

Set $g_n:=C_T f_n\in L^\Psi (\Xi )$, for each $n\in \mathbb {N}$, and

$$\begin{aligned} g:=\sum _{k=1}^\infty \Psi ^{-1}\Big (\frac{\chi _{A_k}\circ T}{\mu (A_k)}\Big ). \end{aligned}$$

Note that, since T is injective, the function

$$\begin{aligned} f=\sum _{k=1}^\infty \Psi ^{-1}\Big (\frac{\chi _{A_k}}{\mu (A_k)}\Big ) \end{aligned}$$

is the only function measurable on $\Omega$ such that $g=C_T f$. Thus the theorem will be proved once we have shown that $g\in L^\Psi (\Xi )$ but $f\notin L^\Phi (\Omega )$.

Let $\lambda \ge 1$. Since $\Psi$ satisfies the $\Delta _2$-condition at $\infty$, there are $b,x_0>0$ such that $\Psi (\lambda x)\le b\Psi (x)$ for all $x\ge x_0$. Then for every $n\ge n_0$, where $n_0$ is the smallest positive integer such that $\Psi ^{-1}\Big (\frac{1}{\mu (A_{n_0})}\Big )\ge x_0$, we have

$$\begin{aligned} I_\Psi \big (\lambda (g-g_n)\big )&=\sum _{k=n+1}^\infty \int _\Xi \Psi \Bigg (\lambda \Psi ^{-1}\Big (\frac{\chi _{A_k}\circ T(\xi )}{\mu (A_k)}\Big )\Bigg )d\nu (\xi )\\&= \sum _{k=n+1}^\infty \int _{T(\Xi )}\Psi \Bigg (\lambda \Psi ^{-1}\Big (\frac{\chi _{A_k}(\omega )}{\mu (A_k)}\Big )\Bigg ) h(\omega )d\mu (\omega )\\&\le \sum _{k=n+1}^\infty \int _{T(\Xi )}b\frac{\chi _{A_k}(\omega )}{\mu (A_k)} h(\omega )d\mu (\omega )\\&\le b\sum _{k=n+1}^\infty \frac{1}{k^2}. \end{aligned}$$

Hence $I_\Psi \big (\lambda (g-g_n)\big )\rightarrow 0$. Since the function $x\mapsto \Psi (x)$ is nondecreasing for $x> 0$, the above convergence holds for all $\lambda >0$. By [10, Theorem 1.3] referred to in the Preliminaries, this implies that $g\in L^\Psi (\Xi )$ and $g_n\rightarrow g$ in $L^\Psi (\Xi )$. But the continuity of $C_T$ from $L^\Phi (\Omega )$ to $L^\Psi (\Xi )$ easily implies that there are $a,x_1\ge 0$ such that $\Psi (x)\le \Phi (ax)$ for $x\ge x_1$ (i.e. $\Psi$ is stronger than $\Phi$ at $\infty$). Let $\lambda \in (0,1)$. Since $\Psi$ satisfies the $\Delta _2$-condition at $\infty$, there are $b,x_2>0$ such that $\Psi (x)\le \frac{1}{b}\Psi (\lambda x)$ for all $x\ge x_2$. Therefore, for $N\in \mathbb {N}$ so large that $\lambda \Psi ^{-1}\Big (\frac{1}{\mu (A_N)}\Big )\ge x_1$ and $\Psi ^{-1}\Big (\frac{1}{\mu (A_N)}\Big )\ge x_2$, we have

$$\begin{aligned} I_\Phi ( \lambda a f)&=\sum _{k=1}^\infty \int _\Omega \Phi \Bigg (a \lambda \Psi ^{-1}\Big (\frac{\chi _{A_k}(\omega )}{\mu (A_k)}\Big )\Bigg )d\mu (\omega )\\&\ge \sum _{k=N}^\infty \int _\Omega \Psi \Bigg (\lambda \Psi ^{-1}\Big (\frac{\chi _{A_k}(\omega )}{\mu (A_k)}\Big )\Bigg )d\mu (\omega )\\&\ge \sum _{k=N}^\infty \int _\Omega b \frac{\chi _{A_k}(\omega )}{\mu (A_k)}d\mu (\omega )=b\sum _{k=N}^\infty 1\;=\;\infty , \end{aligned}$$

and the proof is complete. $\square$

Combining Theorem 6 and Corollary 1, we obtain the following corollary.

Corollary 4

Suppose that $T:\Xi \rightarrow \Omega$ is an injective, nonsingular measurable transformation, $T^{-1}$ is measurable and nonsingular, $\Phi$ satisfies the $\Delta _2$-condition at $\infty$, and the composition operator $C_T$ is continuous from $L^\Phi (\Omega )$ to $L^\Phi (\Xi )$. Then $C_T$ is surjective if and only if $C_T$ has closed range.

4.2 $M_u$ with Closed Range

Now we provide a necessary and sufficient condition for a continuous multiplication operator $M_u$ from $L^\Phi (\Omega )$ to $L^\Psi (\Omega )$ to have closed range.

Theorem 7

Suppose that $u:\Omega \rightarrow \mathbb {R}$ is a measurable function, the multiplication operator $M_u$ is continuous from $L^\Phi (\Omega )$ to $L^\Psi (\Omega )$, and $L^\Phi (\Omega )\cap L^\Psi (\Omega )$ is a closed subset of $L^\Psi (\Omega )$. If there is $\varepsilon >0$ such that $u(\omega )\ge \varepsilon$ for a.e. $\omega \in \Omega$, then the range of $M_u$ is closed.

Proof

Assume that $M_u$ is continuous from $L^\Phi (\Omega )$ to $L^\Psi (\Omega )$, and let $\varepsilon >0$ be such that $u(\omega )\ge \varepsilon$ for a.e. $\omega \in \Omega$ but suppose that the range of $M_u$ is not closed. Then there is a sequence $(f_n)_{n=1}^\infty \subset L^\Phi (\Omega )$ and a function $g\in L^\Psi (\Omega )$ such that

$$\begin{aligned} (M_u f_n)_{n=1}^\infty \subset \mathcal R(M_u),\quad M_u f_n\rightarrow g ~\text {in}~L^\Psi (\Omega )~\text {as}~n\rightarrow \infty ,\quad \text {but}\quad g\notin \mathcal R(M_u). \end{aligned}$$

The fact that $g\notin \mathcal R(M_u)$ means that there is no $f\in L^\Phi (\Omega )$ for which $M_u f= g$, i.e. $f=\frac{g}{u}$. Hence $g\notin \mathcal R(M_u)$ is equivalent to $\frac{g}{u}\notin L^\Phi (\Omega )$.

Since $M_u$ is continuous from $L^\Phi (\Omega )$ to $L^\Psi (\Omega )$, Theorem 2 implies that there are $K>0$ and $\tilde{g}\in L_+^1(\Omega )$ such that

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

for all $x\ge 0$ and a.e. $\omega \in \Omega$. The assumption that $u(\omega )\ge \varepsilon >0$ for a.e. $\omega \in \Omega$ yields

$$\begin{aligned} \Psi (\varepsilon x)\le \Phi (Kx)+\tilde{g}(\omega ) \end{aligned}$$

for all $x\ge 0$ and a.e. $\omega \in \Omega$. Thus, if $f\in L^\Phi (\Omega )$, then for $\lambda >0$ such that $I_\Phi (\lambda f)<\infty$ we have

$$\begin{aligned} I_\Psi \Big (\frac{\lambda \varepsilon }{K}f\Big )\le I_\Phi (\lambda f)+\int _{\Omega } \tilde{g}(\omega )d\mu (\omega )\;<\;\infty , \end{aligned}$$

hence $f\in L^\Psi (\Omega )$. Therefore $L^\Phi (\Omega )\subset L^\Psi (\Omega )$. In particular, $(f_n)_{n=0}^{\infty }\subset L^\Psi (\Omega )$. Since $L^\Phi (\Omega )\cap L^\Psi (\Omega )$ is a closed subset of $L^\Psi (\Omega )$, we conclude that $L^\Phi (\Omega )$ is a closed subspace of $L^\Psi (\Omega )$. Consequently, no sequence in $L^\Phi (\Omega )$ converges to $\frac{g}{u}\notin L^\Phi (\Omega )$. However, applying the fact that $M_u f_n\rightarrow g \text { in }L^\Psi (\Omega )$, which is equivalent to $I_\Psi \big (\lambda (u f_n-g)\big )\rightarrow 0$ for every $\lambda >0$, and the fact that $u(\omega )\ge \varepsilon >0$ for a.e. $\omega \in \Omega$, we obtain

$$\begin{aligned} I_\Psi \Big (\lambda \varepsilon \big (f_n-\frac{g}{u}\big )\Big )&\le I_\Psi \Big (\lambda u \big (f_n-\frac{g}{u}\big )\Big ) \le I_\Psi \big (\lambda (u f_n-g)\big )\;\rightarrow \;0 \end{aligned}$$

for every $\lambda >0$. Therefore $f_n\rightarrow \frac{g}{u} \text { in }L^\Psi (\Omega )$, and this contradiction finishes the proof. $\square$

Theorem 8

Suppose that $u(\omega )>0$ for a.e. $\omega \in \Omega$ and the multiplication operator $M_u$ is continuous from $L^\Phi (\Omega )$ to $L^\Psi (\Omega )$. Assume, additionally, $\Psi$ satisfies the $\Delta _2$-condition at $\infty$. If the range of $M_u$ is closed, then there is $\varepsilon >0$ such that $u(\omega )\ge \varepsilon$ for a.e. $\omega \in \Omega$.

Proof

Suppose that $\mu \big (\{\omega \in \Omega : u(\omega )<\varepsilon \}\big )>0$ for every $\varepsilon >0$. We claim that there is a sequence $(g_n)_{n=1}^\infty \subset \mathcal R(M_u)\subset L^\Psi (\Omega )$ and a function $g\in L^\Psi (\Omega )$ such that $g_n\rightarrow g$ in $L^\Psi (\Omega )$ and $g\notin \mathcal R(M_u)$. This will imply that the range of $M_u$ is not closed, proving the necessity of the condition $u(\omega )\ge \varepsilon >0$ for a.e. $\omega \in \Omega$

For each $k\in \mathbb {N}$ define the set

$$\begin{aligned} A_k:=\Big \{\omega \in \Omega :\frac{1}{(k+1)^2}\le u(\omega )< \frac{1}{k^2}\Big \}. \end{aligned}$$

As in the proof of Theorem 6, upon taking a subsequence, we may and shall assume that the sequence $(\mu (A_{k})_k$ is decreasing to 0.

For each $n\in \mathbb {N}$, define the function

$$\begin{aligned} f_n:=\sum _{k=1}^n\Psi ^{-1}\Big (\frac{1}{\mu (A_k)}\Big )\chi _{A_k}\in L^\Phi (\Omega ). \end{aligned}$$

Let $g_n:=M_u f_n\in L^\Psi (\Omega )$ for each $n\in \mathbb {N}$ and

$$\begin{aligned} g:=\sum _{k=1}^\infty u \Psi ^{-1}\Big (\frac{1}{\mu (A_k)}\Big )\chi _{A_k}. \end{aligned}$$

Since $u>0$, the function

$$\begin{aligned} f=\sum _{k=1}^\infty \Psi ^{-1}\Big (\frac{1}{\mu (A_k)}\Big )\chi _{A_k} \end{aligned}$$

is the only function measurable on $\Omega$ such that $g=M_u f$. We will show that $g\in L^\Psi (\Omega )$ but $f\notin L^\Phi (\Omega )$.

For every $\lambda > 0$ and all $n\ge n_0$, where $n_0$ is the smallest positive integer satisfying $\frac{\lambda }{n_0^2}\le 1$, we have

$$\begin{aligned} I_\Psi \big (\lambda (g-g_n)\big )&=\sum _{k=n+1}^\infty \int _\Omega \Psi \Bigg (\lambda u(\omega ) \Psi ^{-1}\Big (\frac{1}{\mu (A_k)}\Big )\Bigg )\chi _{A_k}(\omega )d\mu (\omega )\\&\le \sum _{k=n+1}^\infty \Psi \Bigg (\frac{\lambda }{k^2} \Psi ^{-1}\Big (\frac{1}{\mu (A_k)}\Big )\Bigg )\mu (A_k)\\&\le \lambda \sum _{k=n+1}^\infty \frac{1}{k^2}. \end{aligned}$$

Hence $I_\Psi \big (\lambda (g-g_n)\big )\rightarrow 0$ for every $\lambda > 0$. By [10, Theorem 1.3] cited in the Preliminaries, this implies that $g\in L^\Psi (\Omega )$ and $g_n\rightarrow g$ in $L^\Psi (\Omega )$. But the continuity of $M_u$ from $L^\Phi (\Omega )$ to $L^\Psi (\Omega )$ implies there is $a>0$ and $x_0\ge 0$ such that $\Psi (x)\le \Phi (ax)$ for $x\ge x_0$. Let $\lambda \in (0,1)$. Since $\Psi$ satisfies the $\Delta _2$-condition at $\infty$, there are $b,x_1>0$ such that $\Psi (x)\le \frac{1}{b}\Psi (\lambda x)$ for all $x\ge x_1$. Therefore, for $N\in \mathbb {N}$ so large that $\lambda \Psi ^{-1}\Big (\frac{1}{\mu (A_N)}\Big )\ge x_0$ and $\Psi ^{-1}\Big (\frac{1}{\mu (A_N)}\Big )\ge x_1$, we have

$$\begin{aligned} I_\Phi ( \lambda a f)&=\sum _{k=1}^\infty \int _\Omega \Phi \Bigg (a \lambda \Psi ^{-1}\Big (\frac{1}{\mu (A_k)}\Big )\Bigg )\chi _{A_k}(\omega )d\mu (\omega )\\&\ge \sum _{k=N}^\infty \Psi \Bigg (\lambda \Psi ^{-1}\Big (\frac{1}{\mu ({A_k})}\Big )\Bigg )\mu (A_k)\\&\ge b \sum _{k=N}^\infty 1\;=\;\infty , \end{aligned}$$

which completes the proof of the theorem. $\square$

From Theorem 8 and Corollary 3 we derive the following corollary.

Corollary 5

Suppose that $u(\omega )>0$ for a.e. $\omega \in \Omega$, $\Phi$ satisfies the $\Delta _2$-condition at $\infty$, and the multiplication operator $M_u$ is continuous from $L^\Phi (\Omega )$ to $L^\Phi (\Omega )$. Then $M_u$ is surjective if and only if $M_u$ has closed range.

5 Fredholm Property

Finally, we present, partially based on the preceding results, necessary and sufficient conditions for a continuous composition and a multiplication operator to be Fredholm. The conditions will be reasonably simple if in the respective theorems we limit ourselves to the case $\Psi =\Phi$.

5.1 Fredholmness of $C_T$

In the following lemma we show that a finite codimension of the range of $C_T$ implies the surjectivity of this operator.

Lemma 1

Let $\Psi$ satisfy the $\Delta _2$-condition globally, and let the composition operator $C_T$ act continuously from $L^\Phi (\Omega )$ to $L^\Psi (\Xi )$. Suppose, further, that for any measurable set $B\subset \Xi$ of positive measure, there is a measurable set $A\subset T(\Xi )$ such that $T^{-1}(A)\subset B$ and $\nu (T^{-1}(A))>0$. If the codimension of the range of $C_T$ is finite, then $C_T$ is surjective.

Proof

From $codim\, \mathcal {R}(C_T)<\infty$ it follows that $\mathcal {R}(C_T)$ is closed. Suppose that $C_T$ is not surjective. Then there is a function $g\in L^{\Psi }(\Xi )$ which does not belong to $\mathcal {R}(C_T)$. Since $\Psi$ satisfies the $\Delta _2$-condition globally and $\mathcal {R}(C_T)$ is a closed subspace of $L^\Psi (\Xi )$, there is $\tilde{g}\in L^{\Psi ^*}(\Xi )$ such that

$$\begin{aligned} \big \langle g, \tilde{g}\big \rangle =\int _\Xi g(\xi )\tilde{g}(\xi )\, d\nu (\xi )\,=\,1 \end{aligned}$$

(1)

and

$$\begin{aligned} \big \langle C_T f,\tilde{g} \big \rangle =\int _\Xi f\big (T(\xi )\big ) \tilde{g}(\xi )\, d\nu (\xi )\,=\,0\;\; \text { for every }\; f\in L^{\Phi }(\Omega ). \end{aligned}$$

(2)

In virtue of (1), there is $\varepsilon >0$ such that the set $B:=\{\xi \in \Xi : g(\xi ) \tilde{g}(\xi )>\varepsilon \}$ has positive measure. Note that $\tilde{g}$ is nonzero on B. Since $\Xi$ is non-atomic, a sequence $(B_n)_{n=1}^\infty$ of pairwise disjoint subsets of B can be found such that $0<\nu (B_n)<\infty$ for $n\in \mathbb {N}$. By assumption, for each $n\in \mathbb {N}$, there is a measurable subset $A_n\subset T(\Xi )$ such that $T^{-1}({A_n})\subset B_n$ and $T^{-1}({A_n})$ has positive measure. Define the functions $\tilde{g}_n:=\tilde{g} \chi _{T^{-1}({A_n})}$, for $n\in \mathbb {N}$; they are all nonzero, linearly independent, and belong to $L^{\Psi ^*}(\Xi )$. Now, by (2), for every $f\in L^\Phi (\Omega )$ and each $n\in \mathbb {N}$, we have

$$\begin{aligned} \big \langle f,C_T^* \tilde{g}_n\big \rangle&=\big \langle C_T f,\tilde{g}_n\big \rangle =\big \langle C_T f,\tilde{g}\chi _{T^{-1}(A_n)}\big \rangle =\big \langle \chi _{T^{-1}(A_n)} C_T f,\tilde{g}\big \rangle \\&=\big \langle (\chi _{A_n}\circ T)(f\circ T) ,\tilde{g}\big \rangle =\big \langle C_T(f\chi _{A_n}),\tilde{g}\big \rangle \,=\,0. \end{aligned}$$

Therefore, each $\tilde{g}_n$ belongs to $\mathcal {N}(C^*_T)$, the null space of the adjoint operator of $C_T$. But

$$\begin{aligned} \mathcal {N}(C^*_T)=\mathcal {R}(C_T)^\perp =\big \{\tilde{g}\in L^{\Psi ^*}(\Xi ): \langle g,\tilde{g}\rangle =0\; \text { for all } g\in \mathcal {R}(C_T)\big \}. \end{aligned}$$

Hence $\mathcal {R}(C_T)^{\perp }$ contains an infinite number of linearly independent functions. Consequently, the quotient space $L^\Psi (\Xi )/\mathcal {R}(C_T)$ also contains an infinite number of linearly independent functions. Since $codim\,\mathcal {R}(C_T)=dim\, L^\Psi (\Xi )/\mathcal {R}(C_T)$, this contradicts the assumption that ${codim}\,\mathcal {R}(C_T)<\infty$. Therefore, $C_T$ is surjective. $\square$

We will use the following lemma in the proof of the next theorem. The notation ’$T(\Xi )= \Omega$ essentially’ means that $\mu (\Omega {\setminus } T(\Xi ))=0$.

Lemma 2

Let $T:\Xi \rightarrow \Omega$ be a nonsingular measurable transformation.

• (1) If there is $\varepsilon >0$ such that $h(\omega )\ge \varepsilon$ for a.e. $\omega \in \Omega$, then $Ker\,( C_T)=\{0\}$.

• (2) If $dim\, Ker\, (C_T)<\infty$, then $T(\Xi )=\Omega$ essentially.

Proof

• (1) First note that if $h\ge \varepsilon >0$ on $\Omega$, then $T(\Xi )=\Omega$ essentially. Indeed, if not, we can take a measurable $B\subset \Omega {\setminus } T(\Xi )$ such that $\mu (B)>0$. Then

  $$\begin{aligned} 0=\nu (\emptyset )=\nu \circ T^{-1}(B)=\int _B h(\omega )d\mu (\omega )\ge \int _B \varepsilon \, d\mu (\omega )=\varepsilon \, \mu (B)>0. \end{aligned}$$

  Now if $C_T f=0$, then

  $$\begin{aligned} 0=I_\Psi (C_T f)=\int _\Xi \Psi \big (f\circ T(\xi )\big )d\nu (\xi )&=\int _{T(\Xi )}\Psi \big (f(\omega )\big )h(\omega )d\mu (\omega )\\&\ge \varepsilon \int _{\Omega }\Psi \big (f(\omega )\big )d\mu (\omega ). \end{aligned}$$

  Hence $\mu (\text {supp} f)=0$, which implies that $f=0$ (as usual, up to sets of measure zero). Thus $Ker\,( C_T)=\{0\}$.

• (2) If it is not true that $T(\Xi )= \Omega$ essentially, then there is an infinite sequence $(A_n)_{n=1}^{\infty }$ of pairwise disjoint subsets of $\Omega {\setminus } T(\Xi )$ with nonzero measure. Since each function $f_n:=\chi _{A_n}$ is nonzero and $C_T f_n=\chi _{T^{-1}(A_n)}=0$, we conclude that $dim\, Ker\, (C_T)=\infty$. $\square$

Theorem 9

Suppose that $T:\Xi \rightarrow \Omega$ is a nonsingular measurable transformation which is injective and such that the mapping $T^{-1}:T(\xi )\mapsto \xi$ is nonsingular. Suppose, further, that for any measurable set $B\subset \Xi$ of positive measure, there is a measurable set $A\subset T(\Xi )$ such that $T^{-1}(A)\subset B$ and $\nu (T^{-1}(A))>0$. If $\Phi$ satisfies the $\Delta _2$-condition globally and $C_T$ is a continuous operator from $L^\Phi (\Omega )$ to $L^\Phi (\Xi )$, then the following conditions are equivalent.

• (1) $C_T$ is a bijection.

• (2) $C_T$ is Fredholm.

• (3) There is $\varepsilon >0$ such that $h(\omega )\ge \varepsilon$ for a.e. $\omega \in \Omega$.

Proof

The implication $(a)\Rightarrow (b)$ is obvious.

Let $C_T$ be Fredholm. Since $dim\, Ker\, (C_T)<\infty$, by Lemma 2, we get that $T(\Xi )=\Omega$ essentially. Since $codim\, \mathcal {R}(C_T)<\infty$, Lemma 1 implies that $C_T$ is surjective. Applying Corollary 1 or Theorem 6 (with $\Psi =\Phi$), we infer that there is $\varepsilon >0$ such that $h(\omega )\ge \varepsilon$ for a.e. $\omega \in T(\Xi )=\Omega$. Hence $(b)\Rightarrow (c)$ holds.

Finally we prove $(c)\Rightarrow (a)$. If there is $\varepsilon >0$ such that $h(\omega )\ge \varepsilon$ for a.e. $\omega \in \Omega$, then, by Corollary 1, the continuous composition operator $C_T$ is surjective. This immediately implies that $\mathcal {R}(C_T)$ is closed and $codim\, \mathcal {R}(C_T)=0$. Moreover, Lemma 2 yields that $dim\, Ker\,(C_T)=0$. Therefore $C_T$ is a bijection.

$\square$

5.2 Fredholmness of $M_u$

Lemma 3

Suppose that $\Phi$ satisfies the $\Delta _2$-condition globally and the multiplication operator $M_u$ acts continuously from $L^\Phi (\Omega )$ to $L^\Phi (\Omega )$. If the codimension of $M_u$ is finite, then the operator $M_u$ is surjective.

The proof of this lemma is so similar to the proof of Lemma 1 that we omit it.

To complete the picture we prove necessary and sufficient conditions for Fredholmness of the multiplication operator. This theorem was proved in [6] by essentially different methods.

Theorem 10

Suppose that $u:\Omega \rightarrow \mathbb {R}_+$ is a measurable function such that $u(\omega )>0$ for a.e. $\omega \in \Omega$. If $\Phi$ satisfies the $\Delta _2$-condition globally and $M_u$ is a continuous operator from $L^\Phi (\Omega )$ to $L^\Phi (\Omega )$, then the following conditions are equivalent.

• (1) $M_u$ is a bijection.

• (2) $M_u$ is Fredholm.

• (3) The codimension of $M_u$ is finite.

• (4) There is $\varepsilon >0$ such that $u(\omega )\ge \varepsilon$ for a.e. $\omega \in \Omega$.

Proof

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

If $codim\, \mathcal {R}(M_u)<\infty$, then $\mathcal {R}(M_u)$ is closed. Lemma 3 implies that $M_u$ is surjective. Therefore, by Corollary 3 or Theorem 8 (with $\Psi =\Phi$), we get that there is $\varepsilon >0$ such that $u(\omega )\ge \varepsilon$ for a.e. $\omega \in \Omega$. Hence $(c)\Rightarrow (d)$ holds.

We only need to show $(d)\Rightarrow (a)$. If there is $\varepsilon >0$ such that $u(\omega )\ge \varepsilon$ for a.e. $\omega \in \Omega$, by Corollary 3, the continuous multiplication operator $M_u$ is surjective. This immediately implies that $\mathcal {R}(M_u)$ is closed and $codim\, \mathcal {R}(M_u)<\infty$. It is obvious that $dim\, Ker\, (M_u)=0$. Therefore $M_u$ is a bijection. $\square$