We begin with some notions that will be useful in the following. Let \((\Omega,\Sigma, \mu)\) be a non-atomic, complete and σ-finite measure space. We say that functions in a set A contained in the Musielak-Orlicz space \(L^{\Phi}(\Omega)\) have equi-absolutely continuous norms if for any real number \(\varepsilon>0\) there exist a set \(B_{\varepsilon}\in\Sigma\) with \(\mu(B_{\varepsilon})<\infty\) and a real number \(\delta=\delta(\varepsilon)>0\) such that for any function \(x\in A\) we have \(\|x \chi_{\Omega\setminus B_{\varepsilon}}\|_{\Phi}<\varepsilon \) and \(\|x\chi_{B}\|_{\Phi}<\varepsilon\) whenever \(B\in\Sigma\cap B_{\varepsilon}\) and \(\mu(B)< \delta\).
Let \(L^{\Phi}(\Omega)=L^{\Phi}(\Omega, \Sigma, \mu)\) and \(L^{\Psi}(\Omega )=L^{\Psi}(\Omega, \Sigma, \mu)\) be distinct Orlicz spaces. We say that the operator \(T: L^{\Phi}(\Omega) \rightarrow L^{\Psi}(\Omega)\) is equi-absolutely continuous if for any bounded set \(A \subset L^{\Phi}(\Omega)\) all functions of the set \(T (A)\subset L^{\Psi}(\Omega)\) have equi-absolutely continuous norms.
We will make use of the following theorem which gives necessary and sufficient conditions for the relative compactness of a set of functions in a Musielak-Orlicz space.
Theorem 4.1
(Theorem 1.2 in [57])
Let
\((\Omega,\Sigma,\mu)\)
be a non-atomic
σ-finite measure space and let
φ
be a Musielak-Orlicz function. If the functions in a set
\(A\subset L^{\varphi}(\Omega)\)
all have equi-absolutely continuous norms and
A
is relatively compact with respect to local convergence in measure, then
A
is relatively compact in
\(E^{\varphi}(\Omega)\), the subspace of absolutely continuous functions in
\(L^{\varphi}(\Omega)\).
Conversely, if a set
\(A\subset E^{\varphi}(\Omega)\)
is relatively compact, then all the functions in
A
have equi-absolutely continuous norms and
A
is relatively compact with respect to local convergence in measure.
In the proof of the forthcoming theorem we will need the following.
Lemma 4.1
(Lemma 8.3 in [51])
Let the measure
μ
be atomless and let a sequence
\(\{\alpha_{i}\}\)
of positive numbers and a sequence
\(\{a_{i}\}\)
of measurable, finite, non-negative functions in Ω be given, satisfying the inequalities
$$\int_{\Omega}a_{i} (t)\,d\mu\geq2^{i} \alpha_{i} \quad\textit{for } i=1,2,\ldots. $$
Then there exist an increasing sequence
\(\{i_{k}\}\)
of positive integers and a sequence
\(\{A_{k}\}\)
of pairwise disjoint sets from Σ such that
$$\int_{A_{k}} a_{i_{k}} (t)\,d\mu=\alpha_{i_{k}} \quad \textit{for } k=1,2\ldots. $$
The following theorem will be of great importance in proving necessary and sufficient conditions for the compactness of the composition operator \(c_{\tau}: L^{\Phi}(\Omega) \rightarrow L^{\Psi}(\Omega)\).
Theorem 4.2
Let
\((\Omega,\Sigma,\mu)\)
be a finite or infinite but
σ-finite non-atomic and complete measure space and
τ
be such that
\(\mu(\tau ^{-1}(A))<\infty\)
whenever
\(\mu(A)<\infty\)
for any
\(A\in\Sigma\cap \tau(\Omega)\). Then the composition operator
\(c_{\tau}: L^{\Phi}(\Omega )\rightarrow L^{\Psi}(\Omega)\)
is equi-absolutely continuous whenever the following condition is satisfied:
$$ \underset{\sigma>0}{\forall}\ \underset{\substack{A\in \Sigma\cap \tau(\Omega) \\ \mu(A)=0}}{\exists}\ \underset{g_{\sigma}\in L^{1}_{+}(\tau( \Omega))}{\exists}\ \underset{s\in\tau(\Omega)\backslash A}{\forall }\ \underset{u\geq0}{\forall} \Psi(u)h(s)\leq\Phi(\sigma u)+g_{\sigma}(s). $$
(8)
Condition (8) is necessary for the equi-absolute continuity of
\(c_{\tau}\)
if
\(\mu(\Omega)<\infty\).
Proof
Sufficiency. First we prove that for any \(\varepsilon>0\) there exists a set \(D\in\Sigma\) with \(\mu(\Omega\setminus D)<\infty\) such that all the functions in the set \(\{c_{\tau}x: x\in S(L^{\Phi})\}\), where \(S(L^{\Phi})\) is the unit sphere of \(L^{\Phi}\), satisfy the condition \(\|(c_{\tau}x)\chi_{D}\|_{\Psi}<\varepsilon\).
Let \(\sigma>0\) be a number such that \((1+\sigma)\sigma<\varepsilon\) and let \(g_{\sigma}\) be a function from condition (8) corresponding to σ. Since \(g_{\sigma}\in L^{1}_{+}(\tau(\Omega))\), there exists a set \(C\in\Sigma\cap\tau(\Omega)\) such that \(\|g_{\sigma}\chi_{C}\|_{\Psi}<\sigma\) and \(\mu(\tau(\Omega)\setminus C)<\infty\). Defining \(D=\tau^{-1}(C)\), we have for any function \(x\in S(L^{\Phi})\)
$$\begin{aligned} I_{\Psi}\biggl(\frac{c_{\tau}x}{\sigma}\chi_{D} \biggr)&= \int_{\Omega}\Psi\biggl(\frac{c_{\tau}x(t)}{\sigma}\chi_{D} (t) \biggr)\,d\mu(t) \\ &= \int_{D} \Psi\biggl(\frac{x(\tau(t))}{\sigma} \biggr)\,d\mu(t) \\ &= \int_{\tau(D)}\Psi\biggl(\frac{x(s)}{\sigma} \biggr)\,d\mu\circ\tau ^{-1}(s) \\ &= \int_{C}\Psi\biggl(\frac{x(s)}{\sigma} \biggr)h(s)\,d\mu(s) \\ &\leq \int_{C} \Phi\bigl(x(s)\bigr)\,d\mu(s)+ \int_{C} g_{\sigma}(s)\,d\mu(s) \\ &\leq1+\sigma. \end{aligned}$$
(9)
By (9) we get
$$I_{\Psi}\biggl(\frac{(c_{\tau}x)\chi_{D}}{(1+\sigma)\sigma} \biggr)\leq\frac {1}{1+\sigma}I_{\Psi}\biggl(\frac{c_{\tau}x}{\sigma}\chi_{D} \biggr)\leq\frac {1}{1+\sigma}(1+ \sigma)=1 $$
for any \(x\in S(L^{\Phi})\). Consequently, \(\|(c_{\tau}x)\chi_{D}\|_{\Psi}\leq (1+\sigma)\sigma<\varepsilon\) for any \(x\in S(L^{\Phi})\), which finishes the first part of the proof.
Now we show the following implication:
$$\underset{\sigma>0}{\forall}\ \underset{\delta=\delta(\varepsilon)}{\exists}\ \underset{B\subset\tau(\Omega) \setminus D}{\forall}\ \underset{x\in S (L^{\Phi})}{\forall} \mu(B)< \delta\quad\implies\quad \bigl\| (c_{\tau}x)\chi_{B}\bigr\| _{\Psi}< \varepsilon, $$
where D is the set from the first part of this proof.
Take any \(x\in S(L^{\Phi})\). Since \(g_{\sigma}\in L^{1}_{+}(\tau(\Omega))\), it is obvious that there is \(\delta=\delta(\sigma)>0\) such that, if \(C\in \Sigma\cap(\tau(\Omega)\setminus D)\) and \(\mu(C)<\delta\), then \(\int_{C} g_{\sigma}(s)\,d\mu(s)<\sigma\). Let \(B=\tau^{-1}(C)\). Then applying condition (8), we get
$$\begin{aligned} I_{\Psi}\biggl(\frac{c_{\tau}x}{\sigma}\chi_{B} \biggr)&= \int_{\Omega}\Psi\biggl(\frac{c_{\tau}x(t)}{\sigma}\chi_{B} (t) \biggr)\,d\mu(t) \\ &= \int_{B} \Psi\biggl(\frac{x(\tau(t))}{\sigma} \biggr)\,d\mu(t) \\ &= \int_{\tau(B)}\Psi\biggl(\frac{x(s)}{\sigma} \biggr)\,d\mu\circ\tau ^{-1}(s) \\ &= \int_{C}\Psi\biggl(\frac{x(s)}{\sigma} \biggr)h(s)\,d\mu(s) \\ &\leq \int_{C} \Phi\bigl(x(s)\bigr)\,d\mu(s)+ \int_{C} g_{\sigma}(s)\,d\mu(s) \\ &\leq1+\sigma. \end{aligned}$$
(10)
Now, by convexity of the modular \(I_{\Psi}\) and the fact that \(I_{\Psi}\) vanishes at zero we have \(I_{\Psi}(\lambda x)\leq\lambda I_{\Psi}(x)\) for any \(x\in L^{\Psi}(\Omega)\) and \(\lambda\in[0,1]\). From this fact and from (10), we get
$$I_{\Psi}\biggl(\frac{(c_{\tau}x)\chi_{B}}{(1+\sigma)\sigma} \biggr)\leq\frac {1}{1+\sigma}I_{\Psi}\biggl(\frac{c_{\tau}x}{\sigma}\chi_{B} \biggr)\leq\frac {1}{1+\sigma}(1+ \sigma)=1. $$
Hence \(\|(c_{\tau}x)\chi_{B}\|_{\Psi}\leq(1+\sigma)\sigma<\varepsilon\). Since σ depends only on ε, \(\delta=\delta(\sigma )\) depends only on ε, and so the proof of sufficiency is finished.
Necessity. Assume that \(\mu(\Omega)<\infty\) and for any \(\sigma>0\) define the function
$$h_{\sigma}(s)=\sup_{u\geq0}\bigl\{ \Psi(u)h(s)-\Phi(\sigma u)\bigr\} . $$
\(h_{\sigma}(s)\) is a non-negative (since for \(u=0\), we have \(\Psi(0)h(s)-\Phi (\sigma0)=0\)) measurable function.
Suppose condition (8) is not satisfied. Then there is \(\sigma _{0}>0\) such that
$$\int_{\tau(\Omega)} h_{\sigma_{0}}(s)\,d\mu(s)=+\infty. $$
Let \(\{r_{i}\}_{i=0}^{\infty}\) be a sequence of all non-negative rational numbers with \(r_{0} =0\). By the continuity of the Orlicz functions Φ and Ψ, we have
$$h_{\sigma_{0}} (s)=\sup_{i=0,1,2 \dots}\bigl\{ \Psi(r_{i})h(s)- \Phi(\sigma_{0} r_{i})\bigr\} . $$
Let us write
$$h_{\sigma_{0},n} (s)=\max_{0\leq i \leq n}\bigl\{ \Psi(r_{i})h(s)- \Phi(\sigma_{0} r_{i})\bigr\} . $$
It is obvious that \(h_{\sigma_{0},n}\geq0\) and \(h_{\sigma_{0},n}\) are measurable functions such that \(h_{\sigma_{0},n} \nearrow h_{\sigma _{0}}\) as \(n\rightarrow+\infty\)
μ-a.e. in \(\tau(\Omega)\). By Beppo Levi’s theorem, we have
$$\int_{\tau(\Omega)} h_{\sigma_{0},n} (s)\,d\mu(s) \nearrow \int_{\tau(\Omega)} h_{\sigma_{0}} (s)\,d\mu(s). $$
Hence there exists a subsequence \(\{h_{\sigma_{0},n_{k}} \}\subset\{ h_{\sigma_{0},n}\}\) satisfying
$$\int_{\tau(\Omega)} h_{\sigma_{0},n_{k}} (s)\,d\mu(s) \geq2^{k}. $$
Without loss of generality we may assume that \(\int_{\tau(\Omega)} h_{\sigma _{0},n} (s)\,d\mu(s) \geq2^{n}\) for each \(n\in\mathbb{N}\cup\{0\}\). It is clear that for each \(s\in\tau(\Omega)\) and each \(n\in\mathbb{N}\cup\{ 0\}\) there exists \(\widetilde{r}_{n} (s)\in\{r_{0},r_{1},r_{2}, \ldots, r_{n}\}\) such that
$$h_{\sigma_{0},n} (s)=\Psi\bigl(\widetilde{r}_{n} (s)\bigr)h(s)-\Phi \bigl(\sigma_{0} \widetilde{r}_{n} (s)\bigr). $$
Hence
$$\begin{aligned} \int_{\tau(\Omega)}\Psi\bigl(\widetilde{r}_{n} (s)\bigr)h(s)\,d\mu(s)&= \int_{\tau(\Omega)} h_{\sigma _{o},n} (s)\,d\mu(s)+ \int_{\tau(\Omega)}\Phi\bigl(\sigma_{0} \widetilde{r}_{n} (s)\bigr)\,d\mu(s) \geq2^{n}. \end{aligned}$$
Applying Lemma 4.1, we conclude that there is a sequence of sets \(\{\Omega_{k}\} \subset \tau(\Omega)\) with \(\Omega_{i} \cap\Omega_{j} =\emptyset\) for \(i\neq j\) such that
$$\int_{\Omega_{n}} \Psi\bigl(\widetilde{r}_{n} (s)\bigr)h(s)\,d\mu(s)=1. $$
Let \(\bar{r}_{n} (s)=\widetilde{r}_{n} (s) \chi_{\Omega_{n}} (s)\). Since \(\bar {r}_{n}\) are bounded measurable functions and \(h\in L^{1} (\operatorname {supp}\bar {r}_{n})\), we get \(\bar{r}_{n} \in E^{\Psi}_{h} (\Omega)\) with \(\|\bar{r}_{n}\| _{\Psi,h} =1\) and \(\bar{r}_{n} \in L^{\Phi}(\Omega)\) for any \(n\in\mathbb {N}\cup\{0\}\).
Since, by assumption, \(\mu(\Omega)<\infty\), we have \(\sum_{n=0}^{\infty } \mu(\Omega_{n})=\mu(\bigcup_{n=0}^{\infty}\Omega _{n})\leq\mu(\Omega)<\infty\), whence \(\lim_{n\rightarrow\infty} \mu(\Omega_{n})=0\). This means that \(\{\bar{r}_{n}\}_{n=1}^{+\infty}\) does not have equi-absolutely continuous norms in \(E^{\Psi}_{h} (\Omega)\). Yet, using the fact that \(g_{\sigma_{0},n}\geq0 \) (\(n\in\mathbb{N}\cup\{0\}\)), we get the following:
$$\Psi\bigl(\bar{r}_{n} (s)\bigr)h(s)\geq\Phi\bigl(\sigma_{0} \bar{r}_{n} (s)\bigr), \quad n\in\mathbb{N}\cup\{0\}. $$
So \(I_{\Phi}(\sigma_{0} \bar{r}_{n})\leq1\), that is, \(\|\bar{r}_{n}\|_{\Phi}\leq\frac{1}{\sigma_{0}}\), which means that the operator \(c_{\tau}: L^{\Phi}(\Omega)\rightarrow L^{\Psi}(\Omega)\) is not equi-absolutely continuous. Hence we proved that condition (8) is necessary for the equi-absolutely continuity of \(c_{\tau}\). □
From Theorem 4.2, applying Theorem 4.1 and the definition of a compact operator, we directly get the following.
Theorem 4.3
If
\((\Omega,\Sigma, \mu)\)
is a non-atomic complete finite or infinite but
σ-finite measure space and
τ
satisfies the assumption from Theorem
4.2, then the composition operator
\(c_{\tau}\)
from
\(L^{\Phi}(\Omega)\)
into
\(L^{\Psi}(\Omega)\)
is compact whenever the set
\(c_{\tau}(S(L^{\Phi}))\)
is relatively compact with respect to local convergence in measure and condition (8) from Theorem
4.2
is satisfied.
Under the assumption that
\(\mu(\Omega)<\infty\), if the composition operator
\(c_{\tau}\)
from
\(L^{\Phi}(\Omega)\)
into
\(E^{\Psi}(\Omega)\)
is compact then the set
\(c_{\tau}(S(L^{\Phi}))\)
is relatively compact with respect to convergence in measure and condition (8) is satisfied.
In the case when Ω has infinite measure, we were unable to show that (8) is a necessary condition for the equi-absolute continuity of the composition operator \(c_{\tau}\). Instead, we can deduce a slightly different (and weaker) condition, as the following theorem states.
Theorem 4.4
Assume that
\(\mu(\Omega)=\infty\)
and
\(\mu(\Omega\backslash\tau(\Omega))=0\). If the composition operator
\(c_{\tau}: L^{\Phi}(\Omega)\rightarrow L^{\Psi}(\Omega)\)
is equi-absolutely continuous, then the condition
$$ \underset{\lambda>0}{\forall}\ \underset{\substack{A\in \Sigma \\ \mu(A)=0}}{\exists}\ \underset{K_{\lambda}>0}{\exists}\ \underset{g_{\lambda}\in L^{1}_{+}(\Omega)}{\exists}\ \underset{s \in\Omega\backslash A}{\forall}\ \underset{u\geq0}{\forall} \Psi (\lambda u)h(s) \leq K_{\lambda}\Phi(u)+g_{\lambda}(s) $$
(11)
is satisfied.
Proof
We can assume without loss of generality that \(\tau(\Omega)=\Omega\). First, notice that if Φ, Ψ, and h satisfy condition (11) then \(L^{\Phi}(\Omega) \subset E^{\Psi}_{h} (\Omega)\), where
$$E^{\Psi}_{h} (\Omega)=\biggl\{ x\in L^{0} (\Omega): \underset{\lambda>0}{\forall} I_{\Psi,h} (\lambda x)= \int_{\Omega}\Psi\bigl(\lambda x(t)\bigr)h(t)\,d\mu(t)< \infty\biggr\} $$
is the subspace of absolutely continuous elements of \(L^{\Psi}_{h} (\Omega)\). The inclusion results from the assumption that \(c_{\tau}\) is an equi-absolutely continuous operator, i.e., for any bounded set \(A \subset L^{\Phi}(\Omega)\), the functions of the set \(c_{\tau}(A)\subset L^{\Psi}(\Omega)\) all have equi-absolutely continuous norms, and the observation that, for any \(x\in L^{\Phi}(\Omega)\), the singleton set \(\{x\}\) is bounded, and thus x has an equi-absolutely continuous norm, which means that x is an absolutely continuous element of \(L^{\Psi}_{h} (\Omega)\), i.e., \(x\in E^{\Psi}_{h} (\Omega)\).
Further, observe that
$$\begin{aligned}& L^{\Phi}(\Omega)=\bigcup_{\lambda>0} L^{{\Phi_{\lambda}},*} (\Omega),\qquad E^{\Psi}_{h} (\Omega)=\bigcap_{\lambda>0} L^{{\Psi_{\lambda}},*}_{h} (\Omega), \end{aligned}$$
where for any \(\lambda>0\) we define \(\Phi_{\lambda}(u)=\Phi(u)\) and \(L^{\Phi,*} (\Omega)=\{x\in L^{0} (\Omega): I_{\Phi}(x)<\infty\}\) is a Musielak-Orlicz class (\(\Psi_{\lambda}\) and \(L^{\Psi,*}_{h} (\Omega)\) are defined analogously). Hence the inclusion \(L^{\Phi}(\Omega) \subset E^{\Psi}_{h} (\Omega)\) can be expressed as
$$\bigcup_{\lambda>0} L^{{\Phi_{\lambda}},*} (\Omega) \subset \bigcap_{\lambda>0} L^{{\Psi_{\lambda}},*}_{h} ( \Omega). $$
This means that, for any \(\lambda_{1}>0\), the Musielak-Orlicz class \(L^{\Phi_{\lambda_{1}},*} (\Omega)\) is contained in all of \(L^{\Psi _{\lambda},*}_{h} (\Omega) \) (\(\lambda>0\)). In particular, taking \(\lambda _{1} =1\), we see that \(L^{\Phi,*} (\Omega)\) is contained in all the Musielak-Orlicz classes \(L^{\Psi_{\lambda},*}_{h} (\Omega) \) (\(\lambda>0\)). By Theorem 8.4 in [51], this is equivalent to
$$ \underset{\lambda>0}{\forall}\ \underset{\substack{A\in\Sigma \\ \mu(A)=0}}{\exists}\ \underset{K_{\lambda}>0}{\exists}\ \underset{g_{\lambda}\in L^{1}_{+}(\Omega)}{\exists}\ \underset{s \in\Omega\backslash A}{\forall}\ \underset{u\geq0}{\forall} \Psi (\lambda u)h(s) \leq K_{\lambda}\Phi(u)+g_{\lambda}(s), $$
which finishes the proof. □
From the preceding theorem, applying Theorem 4.1, we can deduce the following necessary condition for the compactness of the composition operator \(c_{\tau}\):
Theorem 4.5
Assume that
\(\mu(\Omega)=\infty\)
and
\(\mu(\Omega\backslash\tau(\Omega))=0\). If the composition operator
\(c_{\tau}: L^{\Phi}(\Omega)\rightarrow E^{\Psi}(\Omega)\)
is compact then the following conditions are jointly satisfied:
-
(1)
the set
\(c_{\tau}(S(L^{\Phi}))\)
is relatively compact with respect to local convergence in measure;
-
(2)
the functions Φ and Ψ satisfy condition (11).
Remark 4.1
If there exists \(\varepsilon>0\) such that the set
$$B_{\varepsilon}:= \Bigl\{ t\in\tau(\Omega)\colon\underset{u\geq0}{\forall} \Psi(u)h(t)>\Phi(\varepsilon u) \Bigr\} $$
has positive measure, then no composition operator \(c_{\tau}:L^{\Phi}(\Omega)\rightarrow L^{\Psi}(\Omega)\) over a non-atomic measure space \((\Omega,\Sigma,\mu)\) is compact.
Proof of Remark 4.1
Assume that there exists \(\varepsilon >0\) such that the measure of the set \(B_{\varepsilon }\) is positive. Then in the set \(B_{\varepsilon }\) we can find a sequence of measurable and pairwise disjoint sets \(\{B_{n} \}\) in \(\tau(\Omega)\) having positive and finite measure. Define
$$x_{n}=\frac{\chi_{B_{n}}}{\Vert \chi_{B_{n}}\Vert _{\Phi}}=\Phi^{-1} \biggl(\frac{1}{\mu(B_{n})} \biggr)\chi_{B_{n}}. $$
Obviously, \(I_{\Phi}(x_{n})=1\), so \(\|x_{n}\|_{\Phi}=1\) for all \(n\in\mathbb{N}\). We have
$$\begin{aligned} I_{\Phi}\biggl(\frac{\varepsilon x_{n}}{\Vert x_{n}\circ\tau \Vert _{\Psi}}, \Omega \biggr) =& \int_{\tau(\Omega)}\Phi\biggl(\frac{\varepsilon x_{n}(t)}{\Vert x_{n}\circ\tau \Vert _{\Psi}} \biggr)\,d\mu(t)\\ < & \int_{\tau(\Omega)}\Psi\biggl(\frac{x_{n}(t)}{\Vert x_{n}\circ\tau \Vert _{\Psi}} \biggr)h(t)\,d\mu(t) \\ =& \int_{\Omega}\Psi\biggl(\frac{(x_{n}\circ\tau)(t)}{\Vert x_{n}\circ\tau \Vert _{\Psi}} \biggr)\,d\mu(t) =I_{\Psi}\biggl(\frac{c_{\tau}x_{n}}{\Vert c_{\tau}x_{n} \Vert _{\Psi}},\Omega\biggr)\leq1. \end{aligned}$$
Therefore,
$$\biggl\Vert \frac{\varepsilon x_{n}}{\Vert x_{n}\circ\tau \Vert _{\Psi}}\biggr\Vert _{\Phi}\leq1, $$
i.e., \(\Vert c_{\tau}x_{n}\Vert _{\Psi}\geq \varepsilon \Vert x_{n}\Vert _{\Phi}=\varepsilon \) because \(\|x_{n}\|_{\Phi}=1\). Since the supports of \(x_{n}\) are pairwise disjoint, for all \(n,m\in \mathbb {N}\), \(m\neq n\) we get
$$\Vert c_{\tau}x_{n}-c_{\tau}x_{m} \Vert _{\Psi}=\bigl\Vert c_{\tau}(x_{n}-x_{m}) \bigr\Vert _{\Psi}\geq\max\bigl\{ \Vert c_{\tau}x_{n}\Vert _{\Psi}, \Vert c_{\tau}x_{m} \Vert _{\Psi}\bigr\} \geq \varepsilon . $$
Hence the sequence \(\{c_{\tau}x_{n} \}_{n=1}^{\infty}\) has no Cauchy subsequence, that is, \(c_{\tau}(S (L^{\Phi}(\Omega) ) )\) is not relatively compact. Consequently, no composition operator \(c_{\tau}\) from \(L^{\Phi}(\Omega)\) into \(L^{\Psi}(\Omega)\) over a non-atomic measure space \((\Omega,\Sigma ,\mu)\) is compact. □