Orlicz-Lorentz Spaces and their Composition Operators

In a self-contained presentation, we discuss the Orlicz-Lorentz space. Also the boundedness of composition operators on Orlicz-Lorentz spaces are characterized in this paper. 2010 Mathematics Subject Clasi ﬁ cation : Primary 47B33, 47B38, secondary 46E30.


Introduction
Let f a complex-valued measurable function defined on a σ-finite measure space (X, A, µ). For λ ≥ 0, define D f (λ) the distribution function of f as D f (λ) = µ ({x ∈ X : |f (x)| > λ}) . (1.1) Observe that D f depends only on the absolute value |f | of the function f and D f may assume the value +∞.
The distribution function D f provides information about the size of f but not about the behavior of f itself near any given point. For instance, a function on R n and each of its translates have the same distribution function. It follows from (1.1) that D f is a decreasing function of λ (not necessarily strictly) and continuous from the right.
Let (X, µ) be a measurable space and f and g be a measurable functions on (X, µ) then D f enjoy the following properties for all λ 1 , λ 2 ≥ 0: This fact demonstrates that f * is the inverse function of the distribution function D f . Let F(X, A) denote the set of all A-measurable functions on X. Let (X, A 0 , µ) and (Y, A 1 , ν) be two measure spaces.
(1.2) So then there exists only one right-continuous decreasing function f * equimeasurable with f . Hence the decreasing rearrangement is unique.
In what follows, we gather some useful properties of the decreasing rearrangement function: c) f and f * are equimeasurables, that is D f (λ) = D f * (λ) for all λ ≥ 0. d) If |f | ≤ lim inf n→∞ |f n | then f * ≤ lim inf n→∞ f * n .
A weight is a nonnegative locally integrable function on R n that takes values in (0, ∞) almost everywhere. Therefore, weights are allowed to be zero or infinite only on a set of Lebesgue measure zero.
Let ϕ : [0, ∞) → [0, ∞) be a convex function such that Such as function is known as a Young function. A Young function is strictly increasing, in fact, let 0 < x < y then 0 < x y < 1 and hence, we might write Since ϕ is convex, we have A Young function is said to satisfy the ∆ 2 -condition if there exists a nonnegative constant x 0 and k such that If x 0 = 0, we say that ϕ satisfy globally the ∆ 2 -condition. The smaller constant k which satisfy (1.3) is denoted by k ∆ . Claim 1.1. If ϕ is a Young function such that satisfy the ∆ 2 -condition, then for each r ≥ 0 there exists a constant k ∆ (r) such that Proof.
[Proof of the claim.] If r > 0, we can choose n ∈ N such that r ≤ 2 n . Then we can applied (1.3) n-times and use the fact that ϕ is increasing to obtain ϕ(rx) ≤ ϕ(2 n x) ≤ k n ϕ(x), and hence we have (1.4). 2 Example 1.2. The function ϕ 1 (x) = x p p with p > 1 is a Young function which satisfy globally the ∆ 2 -condition with k ∆ = 2 p p .
Example 1.3. The function ϕ 2 (t) = t p log(1 + t) with p ≥ 1 and t ≥ 0 is a Young function which satisfy the ∆ 2 -condition, indeed, since Also, ϕ 2 satisfy globally the ∆ 2 -condition. In fact, since for each t ≥ 0 we have (1 + t) 2 ≥ 1 + 2t, then for all t ≥ t 0 , where p is the right derivate of ϕ.

Proof.
Suppose that ϕ satisfy the ∆ 2 -condition, then there exists a constant k > 0 such that for t large enough, since p is increasing, then we have hence, for t large enough, we obtain Since p(s) = ϕ 0 (s), we have log µ ϕ(2t) which implies that ϕ(2t) < 2 λ ϕ(t). 2 The following result show us that the Young functions which satisfy the ∆ 2 -condition have a cross rate less than the function t p for some p > 1.
Theorem 1.5. If ϕ is a Young function which satisfy the ∆ 2 -condition, then there exists constants λ > 1 and C > 0 such that for t large enough.
Proof. By (1.4) we can write And the proof is complete. 2 Example 1.6. The following are Young functions: Related with the Young function ϕ, we define, for t ≥ 0 the complementary function of Young function as next, for t > 0 fixed, we can consider the function It is not hard to check that g achieves its maximum at s = t 1 p−1 which is given by Hence Proposition 1.8. If ϕ is a Young function, then its complementary function ψ is also a Young function.
Proof. It is clear that ψ(0) = 0 if and only if x = 0. Now, we just need to show that ψ is a convex function. To this end, let us choose t 1 , t 2 ∈ [0, +∞) and λ ∈ [0, 1]. Then, by definition of ψ we have On the other hand From the last two inequalities, we have for all s ≥ 0. Which means that λψ(t 1 ) and so ψ is convex. 2 Theorem 1.9 (Young's Inequality). Let ψ be the complementary function of ϕ. Then where t, s ∈ [0, +∞).
and the proof is complete. 2 For more details on Young functions see [13].

Weighted Lorentz-Orlicz Spaces
The aim of this section is to present basic results about Lorentz-Orlicz spaces. We have tried to make the proofs as self-contained and synthetic as possible.
Definition 2.1 (Luxemburg norm). Let ϕ be a Young function. For any measurable function f on X, where it is understood that inf(∅) = +∞.
Remark 2.2. In this article, we will not always require that the Luxemburg norm actually be a norm. k · k ϕ,w is indeed a quasinorm. A quasinorm is a functional that is like a norm except that it does only satisfy the triangle inequality with a constant C ≥ 1, that is, kf Proof. Clearly kf k ϕ,w = 0 if and only if if and only if f = 0 µ − a.e.

2
Identification of almost everywhere equal functions. As with L p spaces, one identifies the function which are µ-almost everywhere equal. This means that one works with the equivalence classes of the equivalence relation defined by the µ-almost everywhere equality. From now on, this will be done without further mention. Consequently, one write: kf kϕ,w´w (t) dt ≤ 1. In particular, kf k ϕ,w ≤ 1 is equivalent to Proof. For all b > kf k ϕ,w , we have Letting b decrease to kf k ϕ,w , one obtains the first result by monotone convergence. The second statement follows from this and lemma 2.8. 2 Proposition 2.5. The gauge k · k ϕ,w is a quasinorm on the vector space of all the measurable functions f such that kf k ϕ,w < ∞.

Proof.
It is already seen that (2.1) holds under identification of a.e. equal functions.
It is clear that for all real λ, kλf k ϕ,w = |λ|kf k ϕ,w . It remains to prove the triangle inequality. Let f and g be two measurable functions such that 0 < kf k ϕ,w + kgk ϕ,w < ∞. Then where the last but one inequality follows from the convexity of ϕ and the fact that w is nonincreasing and the last inequality from lemma 2.4. Therefore kf + gk ϕ,w ≤ 2 (kf k ϕ,w + kgk ϕ,w ) .
As a consequence, the set of all measurable functions f such that kf k ϕ,w < ∞ is a vector space. 2 Definition 2.6. Let ϕ be a Young function. We define the weighted Lorenz-Orlicz spaces It follows from proposition 1.8 that if L ϕ,w is a weighted Lorentz-Orlicz space, then L ψ,w is also a weighted Lorenz-Orlicz space.
2 Lemma 2.8. Let {f n } n∈N be a sequence in L ϕ,w . Then, the following assertions are equivalent:

Proof.
The equivalence (a) ⇔ (b) is a direct consequence of the definition of k·k ϕ,w . Off course (c) ⇒ (b) is obvious. As ϕ is convex and ϕ(0) = 0 for all t ≥ 0 and 0 < ε ≤ 1, we have From which (b) ⇒ (c) follows easily. 2 Theorem 2.9. The space L ϕ,w is a quasi-Banach space.
If n, m ≥ n 0 . By the definition of the Luxemburg quasi-norm we can use k 0 > 0 in such a way that k 0 <ε and Then Since w > 0, we must have lim n,m→∞ D fn−fm (ε) = 0 which means that {f n } n∈N is a Cauchy sequence in measure. Then some subsequence {f n k } k∈N converges almost everywhere to a measurable function f , that is, Let α > 0. By lemma 2.8 there exists a large enough integer n(α) such that

With Fatou's lemma this gives
Moreover, as lim sup m→∞ This proves that L ϕ,w is complete. Proof.
Next, choose n ∈ N such that n ≥ 1 ε , then since s n ≤ f , then s * n (t) ≤ f * (t), for each t > 0. Since n > 1 ε , we have

Composition Operator
Let (X, A, µ) be a σ-finite complete measure space and let T : X → X be a measurable transformation, that is, = 0 for all A ∈ A with µ(A) = 0, then T is said to be nonsingular. This condition means that the measure µ • T −1 , defined by for A ∈ A is absolutely continuous with respect to µ (it is usually denoted µ • T −1 ¿ µ). Then the Radon-Nikodym theorem ensure the existence of a non-negative locally integrable function f T on X such that Any measurable nonsingular transformation T induces a linear operator (composition operator) C T from F (X, A, µ) into itself defined by where F (X, A, µ) denotes the linear space of all equivalence classes of Ameasurable functions on X, where we identify any two functions that are equal µ-almost everywhere on X.
Here the nonsingularty of T guarantees that the operator C T is well defined as a mapping of equivalence classes of functions into itself since f = g µ-a.e. implies C T (f ) = C T (g) µ-a.e.
It is not hard to see that T is B-measurable, also, observe that T is not nonsigular, indeed hence m(T −1 ({1})) = 1 2 but m({1}) = 0. Now, let us consider f = χ [0,1) and g = χ [0,1] note f = g µ-a.e., but Then C T (f ) 6 = C T (g), which means that C T is not well defined. In other words, the nonsingularity of T is a necessary condition in order to T induces a composition operator on F (X, A, µ).
Composition operators are relatively simple operators with a wide range of applications in areas such a partial differential equations, group representation theory, ergodic theory or dynamical systems, etc. For details on composition operator see [7,10,11,12,14,15] and the references given therein.
In what follows, we will consider the transformation C T from L ϕ,w into the space of all complex-valued measurable functions on X as Next, a necessary and sufficient condition for the boundedness of composition mapping C T is given.
If (X, A, µ) is a σ-finite measure space and T : X → X is a non-singular measurable transformation and w is a weight function, define a measure ν on the σ-algebra A as Next, for A ∈ A, kχ A k ϕ,w = inf n k > 0 : Therefore kχ A k ϕ,w = 1 Theorem 3.2. Let T : X → X be a non-singular measurable transformation. Then C T induced by T is bounded on L ϕ,w if and only if there exists M ≥ 1 such that Proof.
Let C T be a bounded transformation on L ϕ,w , then we can find M ≥ 1 such that Hence Then that is kC T f k ϕ,w ≤ M kf k ϕ,w .
On the one hand, let us prove (3.2). Indeed, let and thus kC T f k ϕ,w ≤ N kf k ϕ,w , ∀ f ∈ L ϕ,w hence kC T f k ϕ,w kf k ϕ,w ≤ N, for all 0 6 = f ∈ L ϕ,w . Therefore That is kC T (f )k ϕ,w kf k ϕ,w , then kC T (f )k ϕ,w kf k ϕ,w ≤ M ∀ 0 6 = f ∈ L ϕ,w .