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

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.


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].

Preliminaries
We explain some of the terms used in this paper.
We will call an Orlicz function a function Φ : R → [0, ∞) that is convex, even, vanishes at 0 and only at 0.
The function Φ * : R → [0, ∞] defined by is called the function complementary to Φ in the sense of Young. An Orlicz function Φ is said to satisfy the Δ 2 -condition at ∞ if for every a > 1 there are constants K > 0 and x 0 > 0 such that If x 0 = 0, then the Orlicz function Φ is said to satisfy the Δ 2 -condition globally. Let Φ, Ψ be Orlicz functions. Then Φ is called stronger than Ψ at ∞ if for some a ≥ 0 and some x 0 > 0. Let (Ω, M, μ) be a σ-finite non-atomic complete measure space and let L 0 (Ω, M, μ) be the linear space of equivalence classes of M-measurable realvalued functions on Ω. We often write Ω instead of (Ω, M, μ), for brevity. Throughout the paper all measures are assumed to be positive.
If Φ is an Orlicz function, then the space where , is a Banach space called Orlicz space. The functional I Φ is called a modular (see [7,9]). If A ∈ M, by I Φ (f, A) we mean the value of the modular I Φ at f in the space L Φ (A, M ∩ A, μ |A ).
Let (Ω, M, μ) and (Ξ, N , ν) be general measure spaces and let T : Ξ → Ω be a measurable transformation in the sense that the set T −1 (A) is Nmeasurable for any M-measurable set A. The transformation T is said to be nonsingular if ν(T −1 (F )) = 0 for any F ∈ M with μ(F ) = 0. This property ensures that the measure ν • T −1 defined by ν • T −1 (A) := ν(T −1 (A)) for any A ∈ M is absolutely continuous with respect to the measure μ. Under this condition, the Radon-Nikodym theorem guarantees the existence of a nonnegative locally integrable function h : Ω → R + , called the Radon-Nikodym derivative of ν • T −1 with respect to μ, such that Upon extending it, we may and shall assume that the Radon-Nikodym derivative h is 0 on Ω\T (Ξ).
Let (Ω, M, μ) and (Ξ, N , ν) be measure spaces and let T : Ξ → Ω be a nonsingular measurable transformation. The linear transformation C T defined by is called a composition operator induced by the transformation T . The nonsingularity of the transformation T : Ξ → Ω ensures that the composition operator C T is well defined, i.e. if f = g μ-a.e. on Ω then Let (Ω, M, μ) be a measure space. Given a measurable function u ∈ L 0 + (Ω, M, μ) (with non-negative values), the linear transformation M u defined by is called a multiplication operator induced by the function u.
is an Orlicz function for μ-a.e. ω ∈ Ω and ϕ(·, x) is a M-measurable function for every x ∈ R is called a Musielak-Orlicz function.
The space L ϕ (Ω, M, μ) of all equivalence classes of M-measurable functions f : Ω → R such that for some λ > 0, equipped with the Luxemburg norm f ϕ defined analogously as for the Orlicz space, is called a Musielak-Orlicz space generated by the Musielak-Orlicz function ϕ. 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 Ψ h (Ω) is the Musielak-Orlicz space L ψ (Ω) generated by the Musielak-Orlicz function ψ(ω, x) := Ψ(x)h(ω) for ω ∈ Ω and x ∈ R, where h ∈ L 0 + (Ω) is the Radon-Nikodym derivative related to the transformation T inducing the composition operator C T .
Secondly, the Orlicz weighted space L Ψu (Ω) is the Musielak-Orlicz space L ψ (Ω) generated by the Musielak-Orlicz function ψ(ω, 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).
For the following corresponding theorem on the multiplication operator, see the proof of Theorem 3.1 in [1].

Theorem 2.
Let u : Ω → R + be a measurable function. The multiplication operator M u acts continuously from L Φ (Ω) to L Ψ (Ω) if and only if one of the following equivalent conditions is satisfied. ( (2) There are K > 0 and g ∈ L 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 ( (2) There is K > 0 and g ∈ L 1 + (Ξ) such that for all x ≥ 0 and a.e. ξ ∈ Ξ.
and nonsingular, hence C T −1 is indeed well defined. By the Radon-Nikodym theorem, there is a locally integrable functionh : Ξ → R + such that for any measurable set A ⊂ Ξ, 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
In the special case when Ψ = Φ, a much simpler condition for the surjectivity of a continuous composition operator can be derived.

Corollary 1.
Let T : Ξ → Ω be a nonsingular measurable transformation which is injective and such that the mapping T −1 : T (ξ) → ξ is measurable and nonsingular. Then a continuous composition operator C T : is surjective, then this condition is also necessary.
for x ≥ x 0 and ξ ∈ A. Therefore, for every K > 0, which, by Theorem 3, implies that C T is not surjective. This contradiction shows that there is ε > 0 such that h(ω) ≥ ε for a.e. ω ∈ T (Ξ).
Proceeding in exactly the same manner one can prove that if there is , and that if Φ satisfies the Δ 2 -condition at ∞, then the same condition is also necessary for the continuity of C T : L Φ (Ω) → L Φ (Ξ). Hence the last corollary may be written in a more specific way.

Corollary 2. Let T : Ξ → Ω be a nonsingular measurable transformation which is injective and such that the mapping T −1 : T (ξ) → ξ is measurable and nonsingular. Then the composition operator
and surjective, then this condition is also necessary.

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 Φ (Ω) = L Ψu (Ω), as was claimed there.
is integrable over Ω.
We omit the proof of this corollary as it is analogous to the proof of Corollary 1.

Closed Range
We have just found necessary and sufficient conditions for the surjectivity of a continuous composition operator from L Φ (Ω) to L Φ (Ξ) and a continuous multiplication operator from L Φ (Ω) to L Φ (Ω). 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 :

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.
Proof. Let C T be continuous from L Φ (Ω) to L Ψ (Ξ) and suppose that for every We will show that there is a sequence (g n ) ∞ n=1 ⊂ R(C T ) ⊂ L Ψ (Ξ) and a function g ∈ L Ψ (Ξ) such that g n → g in L Ψ (Ξ) and g / ∈ R(C T ). This will imply that the range of C T is not closed, proving the theorem.
For each k ∈ N define the set By our assumptions, infinitely many sets A k have nonzero measure and, upon taking a subsequence, one may and shall assume that the sequence (μ(A k ) k is decreasing to 0.
For each n ∈ N, define the function Set g n := C T f n ∈ L Ψ (Ξ), for each n ∈ N, and Note that, since T is injective, the function is the only function measurable on Ω such that g = C T f . Thus the theorem will be proved once we have shown that g ∈ L Ψ (Ξ) but f / ∈ L Φ (Ω). Let λ ≥ 1. Since Ψ satisfies the Δ 2 -condition at ∞, there are b, x 0 > 0 such that Ψ(λx) ≤ bΨ(x) for all x ≥ x 0 . Then for every n ≥ n 0 , where n 0 is the smallest positive integer such that Ψ −1 1 μ(An 0 ) ≥ x 0 , we have Vol. 75 (2020) Surjectivity, Closed Range, and Fredholmness of Composition Page 11 of 18 103 Hence I Ψ λ(g − g n ) → 0. Since the function x → Ψ(x) is nondecreasing for x > 0, the above convergence holds for all λ > 0. By [10, Theorem 1.3] referred to in the Preliminaries, this implies that g ∈ L Ψ (Ξ) and g n → g in L Ψ (Ξ). But the continuity of C T from L Φ (Ω) to L Ψ (Ξ) easily implies that there are a, and the proof is complete.
Combining Theorem 6 and Corollary 1, we obtain the following corollary.
Corollary 4. Suppose that T : Ξ → Ω is an injective, nonsingular measurable transformation, T −1 is measurable and nonsingular, Φ satisfies the Δ 2condition at ∞, and the composition operator C T is continuous from L Φ (Ω) to L Φ (Ξ). Then C T is surjective if and only if C T has closed range.

M u with Closed Range
Now we provide a necessary and sufficient condition for a continuous multiplication operator M u from L Φ (Ω) to L Ψ (Ω) to have closed range.
Theorem 7. Suppose that u : Ω → R is a measurable function, the multiplication operator M u is continuous from L Φ (Ω) to L Ψ (Ω), and L Φ (Ω) ∩ L Ψ (Ω) is a closed subset of L Ψ (Ω). If there is ε > 0 such that u(ω) ≥ ε for a.e. ω ∈ Ω, then the range of M u is closed.
Proof. Suppose that μ {ω ∈ Ω : u(ω) < ε} > 0 for every ε > 0. We claim that there is a sequence (g n ) ∞ n=1 ⊂ R(M u ) ⊂ L Ψ (Ω) and a function g ∈ L Ψ (Ω) such that g n → g in L Ψ (Ω) and g / ∈ R(M u ). This will imply that the range of M u is not closed, proving the necessity of the condition u(ω) ≥ ε > 0 for a.e. ω ∈ Ω For each k ∈ N define the set Vol. 75 (2020) Surjectivity, Closed Range, and Fredholmness of Composition Page 13 of 18 103 As in the proof of Theorem 6, upon taking a subsequence, we may and shall assume that the sequence (μ(A k ) k is decreasing to 0.
For each n ∈ N, define the function Let g n := M u f n ∈ L Ψ (Ω) for each n ∈ N and Since u > 0, the function is the only function measurable on Ω such that g = M u f . We will show that For every λ > 0 and all n ≥ n 0 , where n 0 is the smallest positive integer satisfying λ n 2 0 ≤ 1, we have Hence I Ψ λ(g − g n ) → 0 for every λ > 0. By [10,Theorem 1.3] cited in the Preliminaries, this implies that g ∈ L Ψ (Ω) and g n → g in L Ψ (Ω). But the continuity of M u from L Φ (Ω) to L Ψ (Ω) implies there is a > 0 and x 0 ≥ 0 such that Ψ(x) ≤ Φ(ax) for x ≥ x 0 . Let λ ∈ (0, 1). Since Ψ satisfies the Δ 2condition at ∞, there are b, x 1 > 0 such that Ψ(x) ≤ 1 b Ψ(λx) for all x ≥ x 1 . Therefore, for N ∈ N so large that λΨ −1 which completes the proof of the theorem.
From Theorem 8 and Corollary 3 we derive the following corollary.

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 Ψ = Φ.

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. Proof. From codim R(C T ) < ∞ it follows that R(C T ) is closed. Suppose that C T is not surjective. Then there is a function g ∈ L Ψ (Ξ) which does not belong to R(C T ). Since Ψ satisfies the Δ 2 -condition globally and R(C T ) is a closed subspace of L Ψ (Ξ), there isg ∈ L Ψ * (Ξ) such that and In virtue of (1), there is ε > 0 such that the set B := {ξ ∈ Ξ : g(ξ)g(ξ) > ε} has positive measure. Note thatg is nonzero on B. Since Ξ is non-atomic, a sequence (B n ) ∞ n=1 of pairwise disjoint subsets of B can be found such that 0 < ν(B n ) < ∞ for n ∈ N. By assumption, for each n ∈ N, there is a measurable subset A n ⊂ T (Ξ) such that T −1 (A n ) ⊂ B n and T −1 (A n ) has positive measure. Define the functionsg n :=gχ T −1 (An) , for n ∈ N; they are all nonzero, linearly independent, and belong to L Ψ * (Ξ). Now, by (2), for every f ∈ L Φ (Ω) and each n ∈ N, we have Therefore, eachg n belongs to N (C * T ), the null space of the adjoint operator of C T . But Hence R(C T ) ⊥ contains an infinite number of linearly independent functions. Consequently, the quotient space L Ψ (Ξ)/R(C T ) also contains an infinite number of linearly independent functions. Since codim R(C T ) = dim L Ψ (Ξ)/R(C T ), this contradicts the assumption that codim R(C T ) < ∞. Therefore, C T is surjective.
We will use the following lemma in the proof of the next theorem. The notation 'T (Ξ) = Ω essentially' means that μ(Ω\T (Ξ)) = 0.  Indeed, if not, we can take a measurable B ⊂ Ω\T (Ξ) such that μ(B) > 0. Then Hence μ(suppf ) = 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 (Ξ) = Ω essentially, then there is an infinite sequence (A n ) ∞ n=1 of pairwise disjoint subsets of Ω\T (Ξ) with nonzero measure. Since each function f n := χ An is nonzero and C T f n = χ T −1 (An) = 0, we conclude that dim Ker (C T ) = ∞. Theorem 9. Suppose that T : Ξ → Ω is a nonsingular measurable transformation which is injective and such that the mapping T −1 : T (ξ) → ξ is nonsingular. Suppose, further, that for any measurable set B ⊂ Ξ of positive measure, there is a measurable set A ⊂ T (Ξ) such that T −1 (A) ⊂ B and ν(T −1 (A)) > 0. If Φ satisfies the Δ 2 -condition globally and C T is a continuous operator from L Φ (Ω) to L Φ (Ξ), then the following conditions are equivalent.
(1) C T is a bijection.
Finally we prove (c) ⇒ (a). If there is ε > 0 such that h(ω) ≥ ε for a.e. ω ∈ Ω, then, by Corollary 1, the continuous composition operator C T is surjective. This immediately implies that R(C T ) is closed and codim R(C T ) = 0. Moreover, Lemma 2 yields that dim Ker (C T ) = 0. Therefore C T is a bijection.

Fredholmness of M u Lemma 3.
Suppose that Φ satisfies the Δ 2 -condition globally and the multiplication operator M u acts continuously from L Φ (Ω) to L Φ (Ω). 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 : Ω → R + is a measurable function such that u(ω) > 0 for a.e. ω ∈ Ω. If Φ satisfies the Δ 2 -condition globally and M u is a continuous operator from L Φ (Ω) to L Φ (Ω), then the following conditions are equivalent.
(1) M u is a bijection.
(3) The codimension of M u is finite.