Lower Local Uniform Monotonicity in F-Normed Musielak–Orlicz Spaces

: Lower strict monotonicity points and lower local uniform monotonicity points are considered in the case of Musielak–Orlicz function spaces L Φ endowed with the Mazur–Orlicz F-norm. The findings outlined in this study extend the scope of geometric characteristics observed in F-normed Orlicz spaces, as well as monotonicity properties within specific F-normed lattices. They are suitable for the Orlicz spaces of ordered continuous elements, specifically in relation to the Mazur–Orlicz F-norm. In addition, in this paper presents results that can be used to derive certain monotonicity properties in F-normed Musielak–Orlicz spaces.


Introduction and Preliminaries
It is worth noting that quasi-Banach spaces have been extensively studied over the last century (see [1][2][3][4][5][6]).As we know, in the realm of quasi-Banach spaces, the geometry is heavily influenced by the significant role played by monotonicity properties.Therefore, it is essential to characterize different points of monotonicity in classical quasi-Banach spaces.
This study aims to examine the basic properties in Musielak-Orlicz function spaces that equipped with the Mazur-Orlicz F-norm.Due to the parameterization of generating functions in Musielak-Orlicz function spaces, proving monotonicity in this space is much more complicated than in Orlicz function spaces.We provide several methods for determining lower monotonicity.Some proof methods or ideas mentioned in the paper, such as [3,4,[7][8][9], have reference value.
In this document, we define the set N to represent all natural numbers, and the set R to represent all real numbers.Additionally, we denote R + := [0, ∞).
Definition 1 (see [3]).In a real vector space X, an F-norm is a function ∥•∥ :X → R + that fulfills the following requirements.
(i) The F-norm of x is equal to zero if and only if x equals zero; (ii) For all x ∈ X, the F-norm of x is equal to the F-norm of −x; (iii) For any y, x ∈ X, the F-norm of their sum, ∥x + y∥ F , is always less than or equal to the sum of their individual F-norms, ∥x∥ F + ∥y∥ F ; (iv) For all x ∈ X, λ ∈ R, and λ m limit λ, ∥λ m x m − λx∥ F tends to zero as ∥x m − x∥ F approaches zero, where (x m ) ∞ m=1 is a sequence belongs to X, and (λ m ) ∞ m=1 is a sequence belongs to R.
If a space X = (X, ∥•∥ F ) with the F-norm is topologically complete, we can refer to it as an F-space.A lattice Z = (Z, ≤, ∥ • ∥ F ) is referred to as an F-lattice where the complete and "≤" represents the partial order relation.
We also suppose that (T, Σ, m) is a space that possesses and non-atomic measure, and finite and complete characteristics.L 0 = L 0 (T, Σ, m) is a space that possesses the set of measurable and real-valued functions.Similarly, L 1 = L 1 (T, Σ, m) is a space that possesses Σ−integrable and real-valued functions.
Definition 2 (see [3]).If an F-space(X, ∥•∥ F ) has a linear subspace L 0 , which satisfies the following requirements, the F-space is referred to as a K öthe space endowed with an F-norm.
(i) If y ∈ X, x ∈ L 0 , and |y| ≥ |x|, then x ∈ X and ∥y∥ F ≥ ∥x∥ F ; (ii) There is a positive strictly x ∈ X.
It is important to mention that, in the case where m is non-atomic, X is an F-normed K öthe function space.
The set is defined for a function x(t) that can be measured.
Definition 3 (see [3]).If x belongs to the F-normed K öthe space, for any y ∈ X satisfying the inequality x ̸ = y, and x ≥ y ≥ 0, then the inequality ∥x∥ F > ∥y∥ F holds (equivalently, if y ̸ = 0 and x ≥ y ≥ 0, then ∥x∥ F > ∥x − y∥ F ).We consider x as a lower strict monotonicity point (abbreviated as LSM point).If every point in X has this characteristic, the spaces X is said to be lower strictly monotone.
Definition 4 (see [3]).If x belongs to the F-normed Köthe space, for any sequence, {x m } ∞ m=1 belongs to X, and if the inequality x ≥ x m ≥ 0 holds for all natural numbers m, and lim m→∞ ∥x m ∥ F = ∥x∥ F , then ∥ − x∥ F = 0 holds.In this case, we consider x as a lower local uniform monotonicity point (abbreviated as LLUM point).If every point in X has this characteristic, we can classify X as having lower local uniform monotone.Definition 5. Φ : T × [0, +∞) → [0, +∞] is a function that satisfies the following conditions, which are referred to as a monotone Musielak-Orlicz function.
In addition to that, we also define As we know, b Φ (.) and a Φ (.), as mentioned above, are Σ−measurable functions.The methods used to prove this statement are similar to [7] or [5].Definition 6 (see [5]).If there exists a set T 1 ⊂ Σ with measure m(T 1 ) = 0, a positive constant K and a non-negative function h(t) in the Lebesgue space L 1 (T, Σ, m) for which the inequality Φ(t, 2u) ≤ KΦ(t, u) + h(t) holds for all t ∈ T\T 1 , then we say that the monotone Musielak-Orlicz function Φ is said to satisfy the ∆ 2 −condition (Φ ∈ ∆ 2 for short).
Otherwise, there is a non-empty set T 1 ∈ Σ with a positive measure, and the function b Φ (t) less than positive infinity for t ∈ T 1 .We thus see that < +∞ and t ∈ T 1 , a contradiction.The mapping I Φ : L 0 → [0, ∞] is a modular in L 0 , which can be computed by the integral expression The Musielak-Orlicz space L Φ , its subspace E Φ , and the Mazur-Orlicz F-norm are defined with the above module.
For any x ∈ L Φ , the Mazur-Orlicz F-norm is defined as follows (see [10,11]): Lemma 1 (see [7], Theorem 5.5).If Φ does not satisfy △ 2 , then the set D Φ = {t ∈ T : b Φ (t) < ∞} is non-empty, and this holds true for any sequence of natural numbers in Σ, there exist mutually disjoint sets {F m } ∞ m=1 and measurable functions {x m (t)} ∞ m=1 such that, for natural number m, 0 ≤ x m (t) < ∞ on the set F m , and Lemma 2. For a non-zero element x ∈ L Φ and a monotone Musielak-Orlicz function Φ, all the statements mentioned below hold true.
(ii) Whenever there exists some λ > 1 such that Proof.The evidence follows a similar pattern as the evidences presented in [3,4]; for convenience, we only prove the statement (i).
The necessity is obvious.Now, we will show the sufficiency.Let f (λ) = I Φ ( x λ ), using the definition of the F-norm and for a non-zero x ∈ L Φ , in the interval (0, ∞), I Φ ( x λ ) is non-increasing, we can establish the following inequality for any positive real number ε We can find a sequence (ε m ) ∞ m=1 that satisfies ε m = 1 m , and for any natural number m.According to Beppo Levi's theorem, the inequality Lemma 3.For any positive value of λ, It is clear, so we omit the proof in here.

Conclusions in Musielak-Orlicz Space
Theorem 1.A non-zero element x ∈ L Φ is an LSM point if and only if it satisfies the following conditions.
We will divided the proof in following into two cases.Case 1: There exists a positive constant A for which Take disjoint sets T 1 , T 2 such that supp x = T 1 ∪ T 2 , where both T 1 and T 2 have positive measures.Then, it holds true that, for any λ in the interval (0, 1), we have Suppose that there is a sequence λ m ∈ (0, 1) with λ m → 1 for which and put y(t) = xχ T 1 (t).Next, we have ∥x∥ F ≥ ∥y∥ F .Thanks to the equality I Φ ( y Case 2: For any positive constant A, m({t ∈ supp(x) : Take a positive constant c > 0 that satisfies this condition where Using the equality I Φ ( x λ∥x∥ F ) = ∞, for any λ in the interval (0, 1), we can obtain λ∥x∥ F ≤ ∥y∥ F .As we let λ → 1, so ∥x∥ F ≤ ∥y∥ F holds.Obviously, ∥x∥ F ≥ ∥y∥ F .Therefore, we have ∥x∥ F = ∥y∥ F .This contradicts that x is an LSM point.
Let us demonstrate the validity of condition (ii).Suppose that Denote by ∥x∥ F > 0} and put y(t) = xχ T\T 0 (t).Then, ∥x∥ F ≥ ∥y∥ F and According to Lemma 2 (iii), we can conclude that the F-norm of x is equal to the F-norm of y.This contradicts that x is an LSM point.Now, we will provide evidence to validate condition (iii).Case 1: There is a subset A ⊂ supp x, and it has positive measure such that x(t) For any λ in the interval (0, 1), we have )m(dt) = +∞, which implies λ∥x∥ F ≤ ∥y∥ F .Because λ can take any value of (0, 1), we have that ∥x∥ F ≤ ∥y∥ F holds.Obviously, ∥x∥ F ≥ ∥y∥ F .Hence, ∥x∥ F = ∥y∥ F .This contradicts that x is an LSM point.
Case 2: For any subset A ⊂ supp x, we have b • • • and y(t) = ∑ ∞ m=1 xχ e m (t), without sacrificing the generalizability, it is reasonable to assume that m(e m ) > 0. We obtain that )m(dt).
For any k ∈ N, we take m ∈ N with k < m.Further, we obtain 1 )m(dt) = +∞.
We can yield that (1 Let k → ∞, we observe that the inequality ∥x∥ F ≤ ∥y∥ F is satisfied.Hence, the F-norm of x is equal to the F-norm of y, a contradiction.
We aim to demonstrate the indispensability of condition (iv).Assuming that there exists m({t ∈ supp x : we will establish the existence of a, b ∈ R + ,where b < a, such that where Since positive rational numbers are countable sets, we denote them as {r 1 , r 2 , • • • } and put Hence, By m(A) > 0 and m(A) ≤ ∑ ∞ m,n=1 m(A m,n ∩ A), there exist r m 0 , r n 0 such that m(A m 0 ,n 0 ∩ A) > 0. Let us set a = r m 0 , b = r n 0 , with the assumption that a > b. Then, Put y(t) = xχ T\T b,a (t) + b∥x∥ F χ T b,a (t).
We have Thus, the F-norm of x is equal to the F-norm of y, a contradiction.Sufficiency: Assume x(t) ≥ y(t) ≥ 0 and ∃ e ⊂ T with positive measure, where for all t ∈ e, the inequality y(t) < x(t) holds.We need to prove ∥x∥ F > ∥y∥ F .Assuming that it is false.Under condition (i), there is value λ > 1 such that According to condition (ii) stated in Lemma 2, it is evident that the equation Then,

Corollary 1.
x ∈ E Φ is an LSM point only when these conditions are satisfied.
Denoted by Thus, that is, the F-norm of x equals 1.For any λ within the range of (0, 1), there exists n ∈ N, n ≥ 2, such that, for all m ≥ n, the inequality 1 λ > 1 + 1 m holds.Then, Hence, we have λ ≤ ∥y∥ F .Due to the arbitrariness of λ, it can be inferred that 1 ≤ ∥y∥ F , the F-norms of both x and y are equal to 1, which does not qualify as an LSM point.Therefore, it can be concluded that L Φ does not exhibit strict monotonicity.(iii) If there is a non-empty subset T 0 ⊂ T with a positive measure, such that Φ(t, u) does not strictly monotonically increase, then there exists a subset T 1 ⊂ T, where m(T) > m(T 1 ) > 0 and b > a > 0, such that Φ(t, u) remains constant for all (t, u) To ensure generality, for all t ∈ T, it can be assumed that lim u→∞ Φ(t, u) = +∞ holds.Take a positive number M satisfying For t ∈ T, let δ(t) = inf{u ≥ 0 : M ≤ Φ(t, u)}.On the basis of the limit lim u→∞ Φ(t, u) tends to infinity, we see that δ(t) is well-defined and δ(t) is a measurable function.
Using the condition lim .
So, we have It is clear that Φ(t, m 0 )χ T 3 is a measurable function with finite values almost everywhere.Hence, a subset T 4 is present within T 3 such that and Φ(t, m 0 )χ T 4 is an integrable function.So, we posses Since (T, Σ, m) is a measure space without atoms, there exists a subset T 5 is present within This is denoted by x(t) = bχ T 1 + m 0 χ T 5 .Then, I Φ (x) equals 1, indicating that ∥x∥ F is equal to 1. Based on the given condition we are aware that x does not qualify as a point of strict monotonicity.Hence, the strict monotonicity of L Φ is not established.Sufficiency: For any ) < ∞ for any λ > 0 and b Φ (t) is equal to positive infinity for nearly all values of t ∈ T.Then, x is an LSM point, further L Φ has strict monotonicity.
With the utilization of the evidence provided in Corollary 2, it becomes feasible to derive the subsequent outcomes effortlessly.Corollary 3. E Φ has LSM property when all of the following criteria are satisfied.
Theorem 2. x ∈ L Φ \{0} is an LLUM point when all of the following criteria are met.
(i) For a.e.t ∈ supp x, x(t) Proof.Necessity: We only need to prove condition (iii).Suppose x does not belong to the set where T m = {t ∈ T : m ≥ x(t)}, x m (t) = xχ T m (t).
In fact, according to θ(x), we can achieve this for any positive value of ε. So, hold true.By the condition m(T\T m ) → 0, we have lim m→∞ T\T m Φ(t, x(t) θ(x)+ε )dm(t) = 0. Thus, there exist a natural number m 0 such that Similarly, by the condition I Φ ( x−x m θ(x)−ε ) = +∞, we can deduce that ∥x − x m ∥ F ≥ θ(x) − ε for any natural number m.Therefore, for all positive ε, there exists a m 0 ∈ N, and we have . Then, we can easily obtain that if x / ∈ E Φ , then θ(x) > 0. Next, we will prove that lim m→∞ ∥x m ∥ F = ∥x∥ F .
By the conditions 0 ≤ x m ≤ x and x being a point with strict monotonicity, there exists As a matter of fact, Φ(t, for each t ∈ T. Since, for a.e.t ∈ T, x(t) ∥x∥ F ∈ S − Φ (t) holds, there exists e 0 ⊂ supp x with m(e 0 ) = 0, such that x(t) ∥x∥ F ∈ S − Φ (t) for any t ∈ supp x\e 0 .Hence, we have if u < x(t 0 ) ∥x∥ F , then for any t 0 ∈ supp x\e 0 .
Next, we will prove that lim for any t 0 ∈ supp x\e 0 .If not, according to the Density's theorem, let us assume that There exists an interval [b, a], which is strictly monotonicity interval of Φ(t, u) such that The following two cases are being considered in the next proof.Case 1.There exists We will next prove that there is a positive value d, such that whenever u ∈ (b, a], u − v ≥ c 2 .Otherwise, there are sequences {u m } ∞ m=1 and {v m } ∞ m=1 , where u m ∈ (b, a] and v m ∈ R, with u m − v m ≥ c 2 for which We can make the assumption that u m → u 0 and v m → v 0 , as {u m } ∞ m=1 is a bounded sequence and using the continuity of Φ(t 0 , u), we obtain the following equality: Φ(t 0 , v 0 ) = Φ(t 0 , u 0 ).
Thanks to u 0 ∈ (b, a] and u 0 > v 0 , we have the following inequality: Φ(t 0 , v 0 ) < Φ(t 0 , u 0 ).This is a contradiction.Ultimately, by utilizing the double extraction subsequence theorem, we can prove this.
Levi's theorem, we have the following inequality:lim m→∞ I Φ ( x m λ∥x∥ F ) = I Φ ( x λ∥x∥ F ) ≥ λ∥x∥ F .Hence, lim m→∞ ∥x m ∥ F ≥ λ∥x∥ F .Let λ → 1,we have that the equality lim m→∞ ∥x m ∥ F = ∥x∥ F holds.However, lim m→∞ ∥x − x m ∥ F = θ(x) > 0, a contradiction.Sufficiency: For any x ∈ Ł Φ and a sequence {x m } contained in Ł Φ with 0 ≤ x m ≤ x, if lim m→∞ ∥x m ∥ F = ∥x∥ F .We want to demonstrate that lim m→∞ ∥x − x m ∥ F = 0.In virtue of the property of Φ ∈ ∆ 2 , the following equality lim m→∞ I Φ ( x ∥x m ∥ F ) = ∥x∥ F holds.Hence, lim m→∞ T 0 ) ∥x m k ∥ F − x m k (t 0 ) ∥x m k ∥ F = c,we are able to determine a positive integer k 2 such that lim k→∞ I Φ (λ(x − x m k )) = 0 for any positive value of λ.According to Lemma 3, we can conclude that lim k→∞ ∥x − x m k ∥ F = 0.
By considering x as a member of the set E Φ , it follows that for any positive λ, the value of I Φ (λx) is finite.Therefore, condition (i) stated in Theorem 1 remains valid.Based on the definition of b Φ (t), it can be observed that for a.e.t ∈ suppx, b Φ (t) equals positive infinity.Therefore, condition (iii) stated in Theorem 1 is satisfied.
measurable functions, and mutually disjoint sets {F m } in ∑, such that ≥ k 2 .In virtue of the inequality (2), there is a positive value d that satisfies the inequalityΦ(t 0 , x m k (t 0 ) ∥x m k ∥ F ) + d ≤ Φ(t 0 , x(t 0 ) ∥x m k ∥ F ) holds.The above inequality contradicts with Equality (1).So, in this case, we havelim k→∞ x m k (t 0 ) ∥x m k ∥ F = x(t 0 ) ∥x∥ F .Case 2. There isk 3 > 0, such that x(t 0 ) ∥x m k ∥ F > a whenever k 3 ≤ k. (t 0 ) ∥x m k ∥ F = c > 0, there exists k 4 > 0 and b 1 ∈ (b, a] such that ( x m k (t 0 ) ∥x m k ∥ F , x(t 0 ) ∥x m k ∥ F ) ⊇ (b 1 , a]whenever k ≥ k 4 .Using the proof as in Case 1, there is a positive number d 1 such that F = ∥x∥ F , it can be concluded that as k → ∞, x m k (t) → x(t), for a.e.t ∈ T. Therefore, for every positive value of λ and almost every t ∈ T, lim −x m k (t))) equals to 0. Hence,Φ(t, λ(x(t) − x m k (t))) ≤ Φ(t, λx(t)) ∈ L 1for all t ∈ T. The convergence theorem of Lebesgue dominated implies that k→∞ ∥x m k ∥ k→∞ Φ(t, λ(x(t)