Basic properties of multiplication and composition operators between distinct Orlicz spaces

First, we present some simple (and easily verifiable) necessary conditions and sufficient conditions for boundedness of the multiplication operator Mu and composition operator CT acting from Orlicz space L 1( ) into Orlicz space L 2( ) over arbitrary complete, σ -finite measure space ( , ,μ). Next, we investigate the problem of conditions on the generating Young functions, the function u, and/or the function h = d(μ ◦ T−1)/dμ, under which the operators Mu and CT are of closed range or finite rank. Finally, we give necessary and sufficient conditions for boundedness of the operators Mu and CT in terms of techniques developed within the theory of Musielak–Orlicz spaces. B H. Hudzik hudzik@amu.edu.pl T. Chawziuk tchawz@gmail.com Y. Estaremi yestaremi@pnu.ac.ir S. Maghsoudi s_maghsodi@znu.ac.ir I. Rahmani i_rahmani@znu.ac.ir 1 Faculty of Mathematics and Computer Science, Adam Mickiewicz University in Poznań, Umultowska 87, 61-614 Poznan, Poland 2 Department of Mathematics, Payame Noor University (PNU), P. O. Box: 19395-3697, Tehran, Iran 3 Faculty of Eonomics and Information Technology, The State University of Applied Sciences in Płock, Nowe Trzepowo 55, 09-402 Płock, Poland 4 Department of Mathematics, University of Zanjan, P.O. Box: 45195-313, Zanjan, Iran


Introduction
Given the functions f and T ( f : → R, T : → ), 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 • T , and the operator f → f • 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 into 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 • T ). Composition and multiplication operators received much attention over the past several decades, especially in spaces of measurable functions such as L p -spaces, Bergman spaces, and, to a lesser degree, Orlicz spaces. They also play an important role in the study of operators on Hilbert spaces. Consequently, by now there exists a vast literature on the properties of these transformations in various function spaces; we refer the interested reader to the beautiful books [5,35,36], and, for more recent studies, to [9][10][11]14,23,25], for instance.
As said, composition and multiplication operators between classical L p -spaces have been the subject-matter of intensive and extensive study and they feature prominently in operator theory, operator algebras, dynamical systems, and endomorphisms of Banach algebras. Composition operators were used by Banach [2] in his study of isometries between function spaces. In this paper, we aim at a generalization to Orlicz spaces of some results of the paper [38] concerning the classical Lebesgue spaces. Apparently, the paper [24] by Kumar in 1997 was the first attempt at the study of composition operators between Orlicz spaces. In 2004, Cui, Hudzik, Kumar, and Maligranda in [6] considered the composition operator between Orlicz spaces induced by a non-singular measurable transformation and studied its boundedness and compactness. Some other results on the boundedness and compactness of the composition operator between Orlicz spaces were published in [6][7][8]12,13,16,21,25,33].
Our concern in this paper is to state and prove some necessary conditions, sufficient conditions, and some simultaneously necessary and sufficient conditions for the composition and multiplication operator between distinct Orlicz spaces to be bounded or to have closed-range or finite rank. Our results generalize and improve on some recent results to be found in the literature.

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].
To each Young function is associated another convex function : R → [0, ∞) with similar properties, defined by for all x ≥ 0 (Proposition 2.1.1(ii) in [34]). By an N -function we mean a Young function vanishing only at zero, taking only finite values, and such that lim x→∞ (x)/x = ∞ and lim x→0 + (x)/x = 0. Note that then a = 0, b = ∞, and, as we said above, is continuous and strictly increasing on [0, ∞). Moreover, a function complementary to an N -function is again an N -function.
A Young function is said to satisfy the 2 A Young function is said to satisfy the -condition (respectively, the ∇condition) at ∞, if there exist c > 0 (respectively, b > 0) and If x 0 = 0, these conditions are said to hold globally. Notice that if ∈ , then ∈ 2 (both at ∞ and globally). Let 1 , 2 be Young functions. Then 1 is called stronger than 2 at ∞, which is denoted by 1 2 for some a ≥ 0 and Proof The inequality in the assumptions of the lemma implies that 1 for all x ≥ 0, and this, together with subadditivity of −1 1 , leads to Now suppose, towards a contradiction, that there are a > 0 and x 0 > 0 such that Then, since −1 1 is increasing on [0, ∞), we have −1 1 ( 2 (x)) < −1 1 ( 1 (ax)) = ax, and so, replacing x with 2 (x) in (2.2), we get

This means, since 3 is increasing on
Let ( , , μ) be a complete σ -finite measure space and let L 0 ( ) be the linear space of equivalence classes of -measurable real-valued functions on , that is, we identify functions equal μ-almost everywhere on . The support S of a measurable function f is defined by S( f ) := {t ∈ : f (t) = 0}. For a Young function , the space where The couple (L ( ), · ) is called the Orlicz space generated by a Young function . For any f ∈ L 0 ( ), we have Let (x) = |x| p / p with 1 < p < ∞; is then a Young function and (x) = |x| p / p , with 1/ p + 1/ p = 1, is the Young function complementary to . Thus, with this function we retrieve the classical Lebesgue space L p ( ), i.e. L ( ) = L p ( ).
Recall that an atom of the measure space ( , , μ) is a set A ∈ with μ(A) > 0 such that if F ∈ and F ⊂ A, then either μ(F) = 0 or μ(F) = μ(A). A measure space ( , , μ) with no atoms is called a non-atomic measure space [29]. Remark 2.2 By saying that ∈ 2 we will always mean that the Young function satisfies the 2 -condition globally, since we place no restrictions on the underlying measure space ( , , μ), which can be finite or infinite, non-atomic, purely atomic, or mixed.
The following lemma is a key tool in some of our investigations.

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 ∈ L 0 + ( ) the mapping f → u f is a linear transformation on L 0 ( ) called the multiplication operator induced by u and denoted M u .
Let T : → be a measurable transformation. The transformation T is said to be non-singular if μ(T −1 (A)) = 0 for all A ∈ ∩ T ( ) with μ(A) = 0. The condition ensures that the measure μ • T −1 defined by μ • T −1 (A) = μ(T −1 (A)) for all A ∈ ∩ T ( ) is absolutely continuous with respect to the measure μ (this fact is usually denoted by μ • T −1 μ). Then the Radon-Nikodym theorem guarantees the existence of a non-negative locally integrable function h : T ( ) → R + , called the Radon-Nikodym derivative of μ • T −1 with respect to μ, such that We do not necessarily assume that the transformation T is surjective or injective.
The non-singular measurable transformation T induces a linear operator C T on L 0 ( ), called the composition operator, defined by Finally, we recall two well-known facts from measure theory; for proofs see, for example, Section 11.9 in [29]: (i) If ( , , μ) be a σ -finite measure space, then for every measurable real-valued function f on and every atom A, there is a real number, denoted by f (A), such that f = f (A) μ-a.e. on A. (ii) If ( , , μ) is a σ -finite measure space that fails to be non-atomic, there is a non-empty countable set of pairwise disjoint atoms {A n } n∈N with the property that B := \ n∈N 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 ( ). We keep the above notation throughout the paper.

Boundedness of the multiplication operator
In this section we state various necessary conditions and sufficient conditions under which the multiplication operator M u between distinct Orlicz spaces is bounded.
Recall the well-known theorem due to O'Neil [32] on the multiplication operator between Orlicz spaces (see also [1] and Theorem 5.4.1 in [33]): suppose that 1 , 2 , 3 are Young functions such that either (a) there exist c > 0 and x 0 > 0 with 2 for all x, y > 0 in the case when μ( ) = ∞. Let contain a non-atomic set of positive measure. Then, whenever u ∈ L 3 ( ), the multiplication operator M u is bounded from L 1 ( ) into L 2 ( ).
Before going any further, let us consider some related examples.
Maligranda and Persson in [27] proved a theorem which can be stated in the following way: assume that 1 , 2 , 3 are Young functions with values in [0,∞) such for all x ≥ 0, and assume further that either (A) is a non-atomic measure space or (B) 2 (x) = x r and x −r 1 (x) is nondecreasing. Then M u is a bounded multiplication operator from L 1 ( ) into L 2 ( ) if and only if u ∈ L 3 ( ).
These results have been generalized and extended to the more general setting of Calderón-Lozanovskiȋ 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 1 and 2 be Young functions vanishing only at zero, taking only finite values, and such that 1 is an N-function, 1 ∈ , 2 ∈ 2 , and 3 : Proof Let u ∈ L 0 + ( ) and let M u be a bounded multiplication operator from L 1 ( ) into L 2 ( ) induced by the function u. Since 1 ∈ 2 (as a consequence of the fact that 1 ∈ ) and 2 ∈ 2 , the dual spaces of L 1 ( ) and L 2 ( ) are equal to L 1 ( ) and L 2 ( ), respectively. Hence the adjoint operator M * u to the bounded operator M u is bounded and acts from L 2 ( ) into L 1 ( ).
It is easy to check that 1 ∈ implies that 1 ∈ ∇ , i.e. there is b > 0 such that Let f be any function from L 3 ( ). By the definition of 3 , we have that the This means that 1 • λ b u belongs to the Köthe dual of L 3 ( ), which is equal to In the case of a non-atomic complete and σ -finite measure space, some necessary and sufficient conditions for the boundedness of M u and C T from L 1 ( ) into L 2 ( ) were established in [3]. It turns out that these are equivalent to the conditions for the inclusion of the Orlicz space L 1 ( ) into the Musielak-Orlicz space L 2 ,u ( ) and of the Orlicz space h (T ( )) are generated over the measure spaces ( , , μ) and (T ( ), ∩T ( ), μ | ∩T ( ) ) by the Musielak-Orlicz functions 2 (xu(t)) (t ∈ and x ∈ R) and 2 (x)h(t) (t ∈ T ( ) and x ∈ R), respectively, t ∈ and x ∈ R.
We shall now prove a theorem from which it follows that, if the measure space ( , , μ) is non-atomic, for the above inclusions to hold it is necessary that 2 ≺ 1 .
This might come as a surprise since it means that the simple condition 2 ≺ 1 , which does not involve the Radon-Nikodym derivative h, is necessary for the inclusion L 1 (T ( )) ⊂ L 2 h (T ( )), which does. Proof Suppose, to the contrary, that C T M u is a non-zero bounded linear operator from L 1 ( ) into L 2 ( ) and let We assume without loss of generality that μ(E n ) > 0 for all n ∈ N. Let The assumption that 2 ⊀ 1 implies that an increasing sequence of positive numbers {y n } can be found such that 2 (y n ) > 1 (2 n n 3 y n ). Since the measure space ( , , μ) is non-atomic, we can choose a sequence {F n } of pairwise disjoint measurable subsets of F such that F n ⊂ E n and μ(F n ) = 1 (y 1 )μ(F) 2 n 1 (n 3 y n ) For any n ∈ N. This is possible because ν is non-atomic and Define the function f := ∞ n=1 b n χ F n , where b n := n 2 y n , and take arbitrary α > 0. Then for a natural number n 0 > α we have This implies that f ∈ L 1 ( ). But for m 0 > 0 such that 1 m 0 < α, we obtain which, by the arbitratiness of α > 0, contradicts the boundedness of C T M u . (In fact, we even proved that C T M u does not act from L 1 ( ) into L 2 ( )).
We provide some necessary conditions for the boundedness of the multiplication operator M u from L 1 ( ) into L 2 ( ) under the assumption that 1 (x y) ≤ 2 (x)+ 3 (y) for some Young function 3 and for all x, y ≥ 0.
Proof Let the assumptions about i (i=1,2,3) be satisfied and M u be bounded. First which is a contradiction. Thus (i) holds. Now we prove (ii). We may assume that the function u is not identically zero. For each n ∈ N, put f n = −1 Therefore This completes the proof.
We have a straightforward consequence.

Corollary 3.5 Under the assumptions of Theorem 3.4, if ( , , μ) is a non-atomic measure space, then the multiplication operator M u is bounded from L
Now we present some sufficient conditions for the continuity of the operator M u from one Orlicz space into another. Theorem 3.6 Let 1 , 2 be Young functions vanishing only at zero, taking only finite values, and such that 1 , 2 ∈ and 2 • −1 1 is a Young function. Then u ∈ L 0 + ( ) induces a bounded multiplication operator M u : Proof Suppose that (i) and (ii) hold and set M : Then where c 1 , c 2 are positive constants related to conditions 1 ∈ and 2 ∈ , respectively, and the last inequality follows from the superadditivity of the convex function 2 • −1 1 on the interval [0, ∞). Now let f 1 ≤ 1. We get This implies that M u f 2 ≤ max(c 2 M 2 • −1 1 (c 1 ), 1), and so M u is bounded.

Remark 3.7
The following results on the classical Lebesgue spaces, presented in [38] (see also [18]), are an immediate consequence of Theorem 3.6.
1. Taking 1 (x) = |x| p / p and 2 (x) = |x| q /q, where 1 < p < q < ∞, by Theorems 3.4 and 3.6, we obtain that the multiplication operator M u induced by a function u ∈ L 0 + ( ) is bounded from L p ( ) into L q ( ) if and only if the following conditions hold: 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 ∈ L 0

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.
, we obtain that C T f = 0 (otherwise we would have f | E ∈ L 2 (E)). Together with the fact that C T is bounded, the above gives which is a contradiction finishing the proof.
Proof Assume that (i) and (ii) hold and put Then for each f ∈ L 1 ( ) we have where c 1 , c 2 are positive constants related to conditions 1 ∈ and 2 ∈ , respectively, and the last inequality follows from the superadditivity of the convex function 2 • −1 1 on the interval [0, ∞). Now let f 1 ≤ 1. We obtain Hence 1), and so C T is bounded.
Proof To prove that (i) ⇒ (ii), assume that C T is a bounded operator from L 1 ( ) into L 2 ( ). It implies that there is a constant M > 0 such that for every n ∈ N, where χ A n is the characteristic function of the atom A n ⊂ T ( ). Calculating the norm of the characteristic function χ A n , we obtain , which yields the desired inequality. By . (4.1) in this inequality and using (4.1), we have

Lemma 4.5 Let T : → be a non-singular measurable transformation, and let
1 , 2 be Young functions vanishing only at zero and taking only finite values, with 2 ∈ ∇ . For any function f ∈ L 1 ( ) such that M −1 2 •h f ∈ L 2 (T ( )), we have C T f ∈ L 2 ( ) and the following inequality holds . By the definition of the norm · 2 and the assumption that 2 ∈ ∇ , for b > 0 such that 2 (bx y) ≥ 2 (x) 2 (y) (x, y ≥ 0) we have

T ( )) and the following inequality holds
where c is some positive constant.
Proof Let f ∈ L 1 ( ) be such that C T f ∈ L 2 ( ), and let c ≥ 1 be a constant in the inequality 2 (x y) ≤ c 2 (x) 2 (y) for all x, y ≥ 0. Then as required.

Corollary 4.8 Under the assumptions of Lemma 4.7, if C T is a bounded composition operator from L 1 ( ) into L 2 ( ), then M −1 2 •h is a bounded multiplication operator from L 1 (T ( )) into L 2 (T ( )).
Following [38], for any F in the σ -algebra ∩ T ( ), we define To prove our next results, we will need the following two lemmas.

Lemma 4.10 Let T be a measurable transformation of the measure space ( , , μ) into itself and let be a Young function. Then
where P T ( ) is the family of all measurable partitions of T ( ) and the case ∞ = ∞ is admissible.
To prove the reverse inequality, let a > 1 be an arbitrary number. For each m ∈ Z define the set Since this holds for any a > 1, the result follows.
is a bounded multiplication operator. Thus, by Proposition 3.2, for the Young function for some constant λ > 0. Since 3 • 1 is a nondecreasing function on the interval [0, ∞), we may assume that 0 < λ < 1. It can be easily shown that the 2 -condition assumed for the function 2 is equivalent to the condition: for all λ ∈ (0, 1) there is a constant K > 0 such that −1 2 (K x) ≤ λ −1 2 (x) (x ≥ 0). Since 3 • 1 is a nondecreasing function on the interval [0, ∞), we then get Finally, we give a sufficient condition for the composition operator C T to be bounded from L 1 ( ) into L 2 ( ) when 1 • −1 2 is a Young function.

Proof
Since we have f ∈ L 1 ( ) if and only if 2 • f ∈ L 3 ( ), and furthermore that 2 • f 3 = f 1 . Hence an application of Hölder's inequality yields Thus C T ≤ max(2 h 3 , 1), and this completes the proof.

Remark 4.13
We conclude this section with some applications of the theorems proved.
1. Taking 1 (x) = |x| p / p and 2 (x) = |x| q /q, where 1 < p < q < ∞, by Theorems 4.1 and 4.4, we obtain that if C T is a composition operator induced by the non-singular measurable transformation T : → , then the following statements are equivalent: for all n ∈ N, A n ⊂ T ( ). 2. Similarly, taking 1 (x) = |x| p / p and 2 (x) = |x| q /q, where 1 < q < p < ∞, by Theorems 4.11 and 4.12, we obtain that if C T is a composition operator induced by the non-singular measurable transformation T : → , then the following statements are equivalent: (a) C T is a bounded operator from L p ( ) into L q ( ).
These characterizations are due to Takagi and Yokouchi [38].

Multiplication and composition operators with closed-range and/or finite rank
In this section we are going to investigate closed-range multiplication and composition operators between distinct Orlicz spaces. We start with a basic observation concerning Young functions.

Lemma 5.1 Let 1 , 2 be Young functions such that 1 is an N -function and
for all x, y ≥ 0, whence the desired inequality with b = 1/c follows. Now we characterize closed-range multiplication operators M u : L 1 ( ) → L 2 ( ) under the assumption 2 (x y) ≤ 1 (x) + 3 (y) for all x, y ≥ 0.  Proof Let u ∈ L 3 ( ) and f ∈ L 1 ( ), i.e. there are constants λ, α > 0 such that I 3 (λu) < ∞ and I 1 (α f ) < ∞. Therefore, by the inequality 2 (x y) ≤ 1 (x) + 3 (y) (x, y ≥ 0), we have which implies that u f ∈ L 2 ( ). Let S be the support of u. We may assume that μ(S) > 0 since otherwise M u is a zero operator and there is nothing to prove.
We prove the implication (a) ⇒ (b). Assume that (a) holds. Hence there is r ∈ N such that S = n∈E A n = A n 1 ∪ · · · ∪ A n r .
Since 2 ∈ 2 , the set {χ A n 1 , . . . , χ A nr } of characteristic functions generates the subspace The range of M u is contained in the r -dimensional subspace L 2 (S), hence M u has finite rank.
(b) ⇒ (c) . If the range of the operator M u in L 2 ( ) is finite-dimensional, then it is also closed, as any finite-dimensional subspace of a Banach space is a closed subspace of this space. (c) ⇒ (a) . Let M u have closed-range and assume that μ{t ∈ B : u(t) = 0} > 0.
Then there is δ > 0 such that the set G = {t ∈ B : u(t) ≥ δ} has positive measure. It is easy to see that the restriction u | G induces a bounded multiplication operator M u | G from L 1 (G) into L 2 (G), and that if M u has closed-range, then M u | G has closed-range as well.
We will show that M u | G (L 1 (G)) = L 2 (G). Let A be any measurable subset of G with μ(A) < ∞, and define the function f A := 1 u | G χ A . We get and so f A ∈ L 1 (G). Moreover, M u | G f A = χ A , which implies that the linear space M u | G (L 1 (G)) contains the set F of all linear combinations of characteristic functions of measurable subsets of G with positive and finite measure. Now F is a dense subset of L 2 (G) ([26, Lemma 4.1]) and, by assumption, M u | G (L 1 (G)) is a closed subspace of L 2 (G), therefore and so M u | G (L 1 (G)) = L 2 (G), as claimed.
Consequently, we can define the inverse multiplication operator The operator M 1 u |G is bounded and, since 2 (x y) ≤ 1 (x) + 3 (y) (x, y ≥ 0), we can apply Theorem 3.4 to conclude that 1 u(t) = 0 for μ-a.e. t ∈ G, which is absurd. This contradiction shows that u(t) = 0 for μ-a.e. t ∈ B.
Next we show that the set E = {n ∈ N : u(A n ) = 0} is finite if M u has closed range. If E = ∅, we have nothing to prove. So let us assume that E = ∅. Define S = n∈E A n .
Analogously as above, we can show that M u | S (L 1 (S)) = L 2 (S). Indeed, let A be a subset of S with μ(A) < ∞. Define the function f A := 1 u | S χ A . The set A having finite measure, we get Hence f A ∈ L 1 (S). Since M u | S f A = χ A , we conclude that M u | S (L 1 (S)) contains the set 0 f of all linear combinations of characteristic functions of subsets of S with positive and finite measure. Now 0 f is a dense subset of L 2 (S), and M u | S (L 1 (S)) is a closed subspace of L 2 (S), which implies that and so M u | S (L 1 (S)) = L 2 (S). We can thus define a bounded multiplication operator M 1 Let C = sup n∈E where the final inequality follows from the assumption that 3 ∈ 2 . Thus E must be finite.
In the next theorem we characterize closed-range multiplication operators M u : We prove the implication (c) ⇒ (a). Using the same notation as in the proof of Theorem 5.2, let S = n∈E A n and E = ∅. Since M u : for all n ∈ E, and 3 ∈ 2 , we may write Thus E is finite.
We will need an elementary lemma with an easy proof which we omit. • h ∈ L 3 (T ( )). If C T : L 1 ( ) → L 2 ( ) is a bounded composition operator, then the following assertions are equivalent: (a) C T has closed-range. We prove the implication (a) ⇒ (b). Assume that C T has closed-range. Since T is surjective, by Lemma 5.4, C T is injective. It is a well-known fact that an injective operator has closed-range if and only if it is bounded away from zero (see [4], III.12.5), hence C T is bounded away from zero. This, combined with Lemma 4.5, yields that the multiplication operator M −1 2 •h is bounded away from zero and therefore (being also injective) has closed-range. Now an application of Theorem 5.2 shows that −1 2 • h(t) = 0 for μ-a.e. t ∈ T (B) and the set {n ∈ N : Finally, we prove implication (c) ⇒ (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 is finite-dimensional and so C T has finite rank.
As an easy consequence of this theorem we state In the next theorem we characterize closed-range composition operators C T : L 1 ( ) → L 2 ( ) in the case when 1 (x y) ≤ 2 (x) + 3 (y) for all x, y ≥ 0 (note that compared to Theorem 5.5 the roles of the functions 1 and 2 are reversed).
Theorem 5.7 Let 1 , 2 , 3 be Young functions vanishing only at zero, taking only finite values, and such that 2 ∈ ∇ ∩ 2 , 3 ∈ 2 , and 1 (x y) ≤ 2 (x) + 3 (y) for all x, y ≥ 0. Let T be a surjective non-singular measurable transformation of such that 1  3. If for 1 < p < q < ∞ the composition operator C T is bounded from L p ( ) into L q ( ), then the following assertions are equivalent: (a) C T has closed-range.

Simultaneously necessary and sufficient conditions for continuity of the multiplication and composition operators beween distinct Orlicz spaces
Up to this point no conditions that would be simultaneously necessary and sufficient for the continuity of the operators M u and C T have been given; the provided necessary conditions and sufficient conditions though have the merit of being quite easy to verify for a pair or a triple of Young functions involved. In order to obtain simultaneously necessary and sufficient conditions for continuity of the operators M u and C T , it is convenient to take advantage of the theory of embeddings between Musielak-Orlicz spaces. In the case of non-atomic measure spaces such conditions were presented in [3]; now we will present them for purely atomic and mixed measure spaces. However, there is a price: the simultaneously necessary and sufficient conditions for the continuity of the operators M u and C T are more complicated and not as easy to verify as those presented in the preceding part of this paper. A function acting from × R into [0, ∞) such that (t, .) is an Orlicz function (i.e. a finite-valued continuous Young function) for μ-a.e. t ∈ and (., x) is a -measurable function for every x ∈ R, is called a generalized Orlicz function or a Musielak-Orlicz function. The Musielak-Orlicz space L = L ( , , μ) is the space of all (equivalence classes of) -measurable functions f : → R such that is a Banach space (cf. [26,28] for any t ∈ and any x ∈ R. When the measure space ( , , μ) is purely atomic with infinite number of atoms, we may restrict ourselves to counting measure. Since ( , , μ) is σ -finite, it has countably many atoms, and so, without loss of generality, it may be viewed as having the form (N, 2 N , μ c ), where μ c is the counting measure on 2 N . Indeed, considering atoms as singletons, we write μ({n}) = a n for any n ∈ N, where {a n } n∈N is a sequence of positive real numbers. If 0 denotes the space of all real sequences, then, given the Musielak-Orlicz function = ( n ) ∞ n=1 , i.e. a sequence of Young functions n , we can define on 0 the absolutely convex modular we easily see that B is an absolutely convex set in (N) and that all elements of (N) are absorbed by B . Consequently, the Minkowski functional generated by B is a seminorm on (N). If we assume that for all n ∈ N the functions n are not identically equal to zero, then it is easy to prove that m ( f ) = 0 if and only if f = 0. Of course, This is the Luxemburg norm f = m ( f ).

Let us define a new Musielak-Orlicz function
where {n} is identified with the nth atom of the measure μ. Then the Musielak-Orlicz space generated by the original Musielak-Orlicz function and the original purely atomic measure with μ({n}) = a n is isometric to the Musielak-Orlicz space generated by the Musielak-Orlicz function defined above and corresponding to counting measure with μ c ({n}) = 1 for all n ∈ N, because of the equality For these reasons Musielak-Orlicz sequence spaces, without loss of generality, may be considered (and usually are) over the counting measure space (N, 2 N , μ c ).
In the case of a general measure space , the non-atomic part of and the purely atomic part of can be treated separately, and this allows us to state the following.  (T (B)), such that for all x ≥ 0 and all t ∈ T (B)\A; (ii) condition (6.4) (we identify N with \B). for all x ≥ 0 and all t ∈ B\A; (ii) condition (6.5) (again, we identify N with \B).
Proof We give the proof for both operators concurrently. We can represent L ( ) and L ( ) as the direct sums of the spaces over the non-atomic part of and the purely atomic part of , i.e. L ( ) = L (B) ⊕ L ( \B) and L ( ) = L (B) ⊕ L ( \B). Therefore, every function f ∈ L ( ) can be uniquely written as f = f χ B + f χ \B , and we have and Hence, if we define the operators C T | B , M u | B on L (B) and the operators C T | \B , M u | \B on L ( \B) by the obvious formulas, we are justified to write Of course, we have the following inequalities for the norms Hence it follows that the operator C T acts continuously from L ( ) into L ( ) if and only if the operator C T | B acts continuously from L (B) into L (B) and the operator C T | \B acts continuously from L ( \B) into L ( \B). The same applies to the multiplication operator: M u acts continuously from L ( ) into L ( ) if and only if M u | B acts continuously from L (B) into L (B) and M u | \B acts continuously from L ( \B) into L ( \B). Therefore, the condition for boundedness of the operator C T is exactly the conjunction of the conditions for the boundedness of the operators C T | B and C T | \B . Analogously, the condition for boundedness of the operator M u is the conjunction of conditions for the boundedness of the operators M u | B and M u | \B . Recall that in [3] it was shown that (a)(i) is a sufficient condition for the continuity of C T and that (b)(i) is a necessary and sufficient condition for the continuity of M u in the case of a non-atomic measure space. By analogous arguments to the ones we have used in the proof of the necessity part of Theorem 6.2 (applying Ishii's theorem rather than Shragin's) it can be shown that (a)(i) is also a necessary condition for the continuity of C T in the case of a non-atomic measure space, which generalizes [3], where (a)(i) was shown to be a necessary condition for the continuity of C T in the non-atomic case under the additional assumption that the mapping T is surjective up to sets of measure zero. Moreover, Theorems 6.2 and 6.5 provide parallel sufficient and necessary conditions (a)(ii) and (b)(ii) in the case of a purely atomic measure space. This completes the proof of Theorem 6.7