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Article

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

Department of Mathematics, Harbin University of Science and Technology, Harbin 150080, China
*
Author to whom correspondence should be addressed.
Axioms 2023, 12(6), 539; https://doi.org/10.3390/axioms12060539
Submission received: 24 April 2023 / Revised: 18 May 2023 / Accepted: 18 May 2023 / Published: 30 May 2023
(This article belongs to the Special Issue Modern Functional Analysis and Related Applications)

Abstract

:
In this paper, the necessary and sufficient conditions for the upper strict monotonicity point and the upper local uniform monotonicity point are given in the case of Musielak–Orlicz spaces equipped with the Mazur–Orlicz F-norm. Moreover, strict monotonicity and upper local uniform monotonicity are easily deduced in the case of Musielak–Orlicz spaces endowed with the Mazur–Orlicz F-norm, and the work by Kaczmarek presented in the references is encompassed by the corollaries presented in this paper.

1. Introduction and Preliminaries

It is widely known that monotonicity properties play an important role in geometric properties. For example, in the best approximation problem of a Banach lattice, monotonicity properties play a similar role to that of rotundity properties in the best approximation problem of a Banach space. Therefore, we can know that various monotonicity points play an analogous role to rotundity points (exposed points, strongly exposed points, etc.) in geometric properties. In recent years, monotonicity points have been extensively studied in Musielak–Orlicz, Orlicz–Lorentz, and Calder o ´ n–Lozanovskiĭ spaces (see [1,2,3,4]).
In this paper, we obtain the necessary and sufficient conditions for the upper strict monotonicity point and the upper local uniform monotonicity point in the case of Musielak–Orlicz spaces equipped with the Mazur–Orlicz F-norm under various conditions. Furthermore, the necessary and sufficient conditions for strict monotonicity and upper local uniform monotonicity in the case of Musielak–Orlicz spaces equipped with the Mazur–Orlicz F-norm are also obtained.
Let us denote N and R as the sets of natural and real numbers, respectively, and R + : = 0 , .
Assume that ( T , Σ , m ) is a complete, finite, and non-atomic measure space. Let L 0 = L 0 ( T , Σ , m ) be the space of all real-valued and Σ -measurable functions on T. L 1 = L 1 ( T , Σ , m ) is the space of all real-valued and Σ -integrable functions on T.
Definition 1. 
A function Φ : T × [ 0 , + ) 0 , + is called a monotone Musielak–Orlicz function if the following conditions are satisfied:
(1) 
Φ ( t , 0 ) = 0 ;
(2) 
Φ ( t , u ) is non-decreasing and continuous with respect to u on [ 0 , b Φ ( t ) ) for almost every t T and left continuous at b Φ ( t ) ;
i.e.,
(i) 
lim u b Φ ( t ) Φ ( t , u ) = Φ ( t , b Φ ( t ) ) ( 0 , + ) whenever b Φ ( t ) < + ;
(ii) 
lim u b Φ ( t ) Φ ( t , u ) = + whenever b Φ ( t ) = + ;
(iii) 
For almost every t T , there is a u t > 0 such that Φ ( t , u t ) > 0 and Φ ( t , u ) is Σ measurable with respect to t for each u R + ,
where
b Φ ( t ) : = sup { u 0 : Φ ( t , u ) < + } .
Define
a Φ ( t ) : = sup { u 0 : Φ ( t , u ) = 0 } ,
supp x = { t T : x ( t ) 0 }
and
S Φ + ( t ) = { u : Φ ( t , u ) < Φ ( t , v ) f o r 0 u < v } .
The functions a Φ ( . ) and b Φ ( . ) are Σ measurable and the proofs are similar to the proof of [5]. The function Φ ( t , u ) is continuous on [ 0 , b Φ ( t ) ) in regard to u for almost every t T .
Definition 2 
(see [6]). We say that a monotone Musielak–Orlicz function Φ satisfies the Δ 2 condition (for brevity, we write Φ Δ 2 ) if there exists a set T 1 Σ with m ( T 1 ) = 0 , a constant K > 0 , and a function 0 h L 1 ( T , Σ , m ) such that Φ ( t , 2 u ) K Φ ( t , u ) + h ( t ) for all t T \ T 1 .
If Φ Δ 2 , then b Φ ( t ) = + for almost every t T .
The function I Φ : L 0 [ 0 , ] is defined by the formula
I Φ ( x ) = T Φ ( t , | x ( t ) | ) d t .
The space
L Φ = { x L 0 : I Φ ( λ x ) < f o r s o m e λ > 0 }
is said to be a Musielak–Orlicz space (see [7,8,9]). Define the subspace E Φ of L Φ by the formula:
E Φ = { x L 0 : I Φ ( λ x ) < f o r a n y λ > 0 } .
The Mazur–Orlicz F-norm is defined by the formula (see [7,8,9]):
x F = inf { λ > 0 : I Φ ( x λ ) λ } , x L Φ .
Definition 3 
(see [10]). Given any real vector space X the functional x x F R + , is called an F-norm if the following conditions are satisfied:
(i) 
x F = 0 if and only if x = 0 ;
(ii) 
x F = x F for all x X ;
(iii) 
x + y F x F + y F for all x , y X ;
(iv) 
λ n x n λ x F 0 whenever x n x F 0 and λ n λ for any x X , x n n = 1 in X, λ R and λ n n = 1 in R .
An F-normed space X = X , · F is an F-space under the condition that the F-normed space X is complete with regard to the F-norm topology. If Z is complete, the lattice Z = ( Z , , · F ) equipped with a monotone F-norm · F is an F-lattice, where ≤ denotes the partial order relation.
Definition 4 
(see [10]). An F-space X , · F is called an F-normed K o ¨ t h e space if it is a linear subspace of L 0 satisfying the following conditions:
(i) 
If x L 0 , y X and x y , then x X and x F y F ;
(ii) 
There exists a strictly positive x X (called a weak unit).
A complete F-normed K o ¨ t h e space is an F-lattice. If the measure m is non-atomic, then X is an F-normed K o ¨ t h e function space.
Definition 5 
(see [10]). A point x in an F-normed K o ¨ t h e space X , · F is said to be an upper strict monotonicity point if for any y X such that 0 x y , the condition x F < y F holds whenever x y (or equivalently x + y F > x F whenever 0 y and y 0 ). If every point in X is an upper strict monotonicity point, then we say that X is upper strictly monotone.
Definition 6 
(see [10]). A point x in an F-normed K o ¨ t h e space X , · F is said to be an upper local uniform monotonicity point if for any x n n = 1 in X such that 0 x x n for all n N and x n F x F as n , the condition x n x F 0 as n holds. If every point in X is an upper local uniform monotonicity point, we then say that X is upper locally uniformly monotone.
If a point x is an upper local uniform monotonicity point, it must be an upper strict monotonicity point. However, an upper strict monotonicity point may not be an upper local uniform monotonicity point.
Lemma 1 
(see [5]). If Φ 2 , then D Φ = { t T : b Φ ( t ) < } is a non-null set or for any
b 1 b 2 1 , 1 < p 1 p 2 , q n > 0 ( n N ) ,
there exist measurable functions { x n ( t ) } n = 1 and mutually disjoint { F n ( t ) } n = 1 in Σ such that 0 x n ( t ) < on F n and
F n Φ ( t , x n ( t ) ) d t = q n , Φ ( t , b n x n ( t ) ) p n Φ ( t , x n ( t ) ) ( t F n , n N ) .
Lemma 2. 
For every monotone Musielak–Orlicz function Φ, and x L Φ \ { 0 } we have:
(1) 
I Φ x x F x F I Φ x x F < + ;
(2) 
I Φ x x F = x F whenever I Φ λ x x F < for some λ > 1 ;
(3) 
If I Φ x λ = λ for λ > 0 , then x F = λ .
Proof. 
The proofs of conditions (1) and (2) are similar to the proof of Lemma 6.1 from [11], and the proof of condition (3) is similar to the proof of Lemma 2.16 from [10], so they are omitted. □
Lemma 3. 
lim n x n F = 0 if and only if lim n I Φ ( λ x n ) = 0 for any λ > 0 .
Proof. 
The proof is similar to the proof of [11], so it is omitted. □

2. Results in F-Normed Musielak–Orlicz Spaces

Theorem 1. 
(a) 
If x ( t ) x F = b Φ ( t ) and b Φ ( t ) < + for t supp x , then x L Φ \ { 0 } is upper strict monotonicity point;
(b) 
If m ( { t supp x : x ( t ) x F < b Φ ( t ) < + } ) > 0 , then x L Φ \ { 0 } is upper strict monotonicity point if and only if the following conditions are satisfied:
(1) 
I Φ ( x x F ) = x F ;
(2) 
x ( t ) x F a Φ ( t ) for almost every t supp x ;
(3) 
x ( t ) x F S Φ + ( t ) for almost every t supp x .
(c) 
If m ( { t supp x : x ( t ) x F < b Φ ( t ) < + } ) = 0 , then x L Φ \ { 0 } is an upper strict monotonicity point if and only if the following conditions are satisfied:
(1) 
x ( t ) x F a Φ ( t ) for almost every t supp x ;
(2) 
x ( t ) x F S Φ + ( t ) for almost every t supp x .
Proof. 
Case (a): Assume 0 x ( t ) y 1 ( t ) for t T and there exists e T , m ( e ) > 0 such that x ( t ) < y 1 ( t ) for t e . Denote e n = { t e : ( 1 + 1 n ) x ( t ) y 1 ( t ) } . Then n = 1 e n = e , and there exists n 0 N such that m ( e n 0 ) > 0 . Hence, we have
I Φ y 1 ( 1 + 1 n 0 2 ) x F e n 0 Φ ( t , ( 1 + 1 n 0 ) x ( t ) ( 1 + 1 n 0 2 ) x F ) d t e n 0 Φ ( t , ( 1 + 1 n 0 ) ( 1 + 1 n 0 2 ) b Φ ( t ) ) d t = + ,
which means that y 1 F ( 1 + 1 n 0 2 ) x F > x F ; hence, x is an upper strict monotonicity point.
Case (b): The necessity.
First let us prove the necessity of condition (1). Assume T Φ ( t , x ( t ) x F ) d t < x F . Put e = { t supp x : x ( t ) x F < b Φ ( t ) } . Because the function Φ ( t , x ( t ) + x F b Φ ( t ) 2 x F ) χ e ( t ) is a measurable function and finite for almost every t e , there exists a subset e 0 e with m ( e 0 ) > 0 such that Φ ( t , x ( t ) + x F b Φ ( t ) 2 x F ) χ e 0 ( t ) is a bounded measurable function. Hence,
e 0 Φ ( t , x ( t ) + x F b Φ ( t ) 2 x F ) d t < + .
By absolute continuity of the Lebesgue integral, there is a subset e ˜ e 0 with m ( e ˜ ) > 0 for which
e ˜ Φ ( t , x ( t ) + x F b Φ ( t ) 2 x F ) d t x F T Φ ( t , x ( t ) x F ) d t .
Set
y 2 ( t ) = x ( t ) χ T e ˜ ( t ) + x ( t ) + x F b Φ ( t ) 2 χ e ˜ ( t ) .
Then, 0 x ( t ) y 2 ( t ) for t T and x ( t ) < y 2 ( t ) for t e ˜ , hence x F < y 2 F . However, the inequality
I Φ ( y 2 x F ) = T e ˜ Φ ( t , x ( t ) x F ) d t + e ˜ Φ ( t , x ( t ) + x F b Φ ( t ) 2 x F ) d t T e ˜ Φ ( t , x ( t ) x F ) d t + x F T Φ ( t , x ( t ) x F ) d t x F ,
we have that the inequality y 2 F x F holds; This is a contradiction.
Next, we are going to prove that condition (2) is true. If m ( { t supp x : x ( t ) x F < a Φ ( t ) } ) > 0 . Denote F 0 = { t supp x : x ( t ) x F < a Φ ( t ) } and y 2 ( t ) = x χ T F 0 ( t ) + x + x F a Φ 2 χ F 0 ( t ) . We obtain that 0 x ( t ) y 2 ( t ) for t T and x ( t ) < y 2 ( t ) for any t F 0 . It is known that
x F = T Φ ( t , x ( t ) x F ) d t = T F 0 Φ ( t , x ( t ) x F ) d t = T F 0 Φ ( t , y 2 ( t ) x F ) d t + F 0 Φ ( t , y 2 ( t ) x F ) d t = T Φ ( t , y 2 ( t ) x F ) d t .
We can discern from the above equality that y 2 F x F . Together with the previous conditions, we can obtain that y 2 F = x F , which contradicts the fact that x is an upper strict monotonicity point.
Finally, we will prove that condition (3) holds. If m ( { t T : x ( t ) x F S Φ + ( t ) } ) > 0 . We want to prove that there exists a , b R + , a < b satisfying
Φ ( t , a ) = Φ ( t , b ) , t T a , b ,
where T a , b = { t T : a x ( t ) x F < b } . As the set of positive rational numbers is countable, assume them to be { r 1 , r 2 , } and
A n , m = { t T : Φ ( t , r n ) = Φ ( t , r m ) } .
We obtain that
A = { t T : x ( t ) x F S Φ + ( t ) } = n , m = 1 ( A A n , m )
hence, m ( A ) n , m = 1 m ( A A n , m ) . In virtue of the condition m ( A ) > 0 , there exist r n 0 , r m 0 Q + such that m ( A A n 0 , m 0 ) > 0 . Let a = r n 0 , b = r m 0 . Suppose that a < b . Thus, m ( { t T : a x ( t ) x F < b } ) > 0 . Denote
y 2 ( t ) = x χ T T a , b ( t ) + b x F χ T a , b ( t ) .
We obtain that
I Φ ( y 2 x F ) = T T a , b Φ ( t , x ( t ) x F ) d t + T a , b Φ ( t , b ) d t = T T a , b Φ ( t , x ( t ) x F ) d t + T a , b Φ ( t , x ( t ) x F ) d t = T Φ ( t , x ( t ) x F ) d t = I Φ ( x x F ) = x F ,
which can further yield that y 2 F = x F ; the equality contradicts the fact that x is upper strict monotonicity point.
The sufficiency.
Suppose 0 x ( t ) y 2 ( t ) for t T and there exist e 1 T and m ( e 1 ) > 0 satisfying x ( t ) < y 2 ( t ) for t e 1 . We will to prove that the inequality x F < y 2 F holds. By the condition (1) we can get that
x F = T Φ ( t , x ( t ) x F ) d t = T e 1 Φ ( t , x ( t ) x F ) d t + e 1 Φ ( t , x ( t ) x F ) d t < T e 1 Φ ( t , y 2 ( t ) x F ) d t + e 1 Φ ( t , y 2 ( t ) x F ) d t = I Φ ( y 2 x F ) .
From the above inequality we can get that I Φ ( y 2 x F ) > x F . By the definition of F-norm we have that I Φ ( y 2 y 2 F ) y 2 F . Thus, we obtain x F < y 2 F .
Case (c): The proof of Case (c) is similar to that of Case (b), so we have omitted the proof. □
Corollary 1. 
x E Φ is upper strict monotonicity point if and only if the following conditions are satisfied:
(1) 
m ( { t T : 0 < x ( t ) x F < a Φ ( t ) } ) = 0 ;
(2) 
x ( t ) x F S Φ + ( t ) for almost every t supp x .
Proof. 
The condition x E Φ implies that b Φ ( t ) = , which means that the statement m ( { t supp x : x ( t ) x F < b Φ ( t ) < + } ) = 0 in Theorem 1 (c) is satisfied. □
Corollary 2. 
L Φ is strictly monotone if and only if the following conditions are satisfied:
(1) 
a Φ ( t ) = 0 for almost every t T ;
(2) 
Φ 2 ;
(3) 
Φ ( t , u ) is strictly monotonically increasing with respect to u for almost every t T .
Proof. 
The necessity.
(1) If there is a set e T satisfying m ( e ) > 0 and a Φ ( t ) > 0 for t e , and let x ( t ) = 1 2 a Φ ( t ) ; this yields that x ( t ) is not an upper strict monotonicity point. Further, L Φ is not strictly monotone.
(2) If Φ 2 , in combination with Lemma 1 we can take b n = 1 + 1 n , p n = 2 n , q n = 1 2 n for n N such that
F n Φ ( t , x n ( t ) ) d t = 1 2 n , Φ ( t , b n x n ( t ) ) p n Φ ( t , x n ( t ) ) ( t F n , n N ) ,
where { x n ( t ) } n = 1 are measurable functions and { F n } in ∑ are mutually disjoint sets satisfying 0 x n ( t ) < on the set { F n } .
Define
x ( t ) = n = 1 x n χ F n ( t ) , y ( t ) = n = 2 x n χ F n ( t ) .
Then
I Φ ( x ) = n = 1 F n Φ ( t , x n ( t ) ) d t = n = 1 1 2 n = 1 .
From the condition (3) in Lemma 2, we can obtain that x , y L Φ and y F x F = 1 . For any λ ( 0 , 1 ) , there exists m N , m 2 such that 1 λ > 1 + 1 n for any n m . Then
I Φ ( y λ ) n = m F n Φ ( t , x n ( t ) λ ) d t n = m F n Φ t , ( 1 + 1 n ) x n ( t ) d t n = m 2 n F n Φ ( t , x n ( t ) ) d t = n = m 1 = .
Consequently, we have that the inequality y F λ holds. Because λ is arbitrary, we obtain y F 1 . Further, we conclude that x F = y F = 1 and x ( t ) is not an upper strict monotonicity point, which means that L Φ is not strictly monotone.
(3) If Φ ( t , u ) is constant function for each t T 1 and a u < b , where T 1 T , 0 < m ( T 1 ) < m ( T ) and 0 < a < b . Suppose lim u Φ ( t , u ) = + for t T . Select M > 0 such that
M · 1 3 m ( T T 1 ) > 1 T 1 Φ ( t , b ) d t .
Define δ ( t ) = inf { u 0 : Φ ( t , u ) M } for t T . Because lim u Φ ( t , u ) = + , we yield that δ ( t ) is measurable. The fact lim n m ( { t T T 1 : δ ( t ) > n } ) = 0 implies that there exists n 0 N satisfying
m ( { t T T 1 : δ ( t ) > n 0 } ) < 1 3 m ( T T 1 ) .
Define T 2 = { t T T 1 : δ ( t ) > n 0 } and T 3 = T ( T 1 T 2 ) . Therefore, we obtain m ( T 3 ) 1 3 m ( T T 1 ) and
M · m ( T 3 ) > 1 T 1 Φ ( t , b ) d t .
Because Φ ( t , n 0 ) χ T 3 is an almost everywhere finite measurable function, we know that there exists T 4 T 3 such that
M · m ( T 4 ) > 1 T 1 Φ ( t , b ) d t
and Φ ( t , n 0 ) χ T 4 is an integrable function. We can yield that
T 4 Φ ( t , n 0 ) d t > 1 T 1 Φ ( t , b ) d t .
There must exist T 5 T 4 such that
T 5 Φ ( t , n 0 ) d t = 1 T 1 Φ ( t , b ) d t
under the condition ( T , Σ , m ) is a non-atomic measure space. Define x ( t ) = b χ T 1 ( t ) + n 0 χ T 5 ( t ) . Thus, I Φ ( x ) = 1 , x F = 1 . According to the condition m ( { t T : x ( t ) x F S Φ + ( t ) } ) > 0 , we have that x is not strict monotonicity point; hence, L Φ is not strictly monotone.
The sufficiency.
The condition Φ 2 implies that m ( { t supp x : x ( t ) x F < b Φ ( t ) < + } ) = 0 . Thus, x is a upper strict monotonicity point, and L Φ is strictly monotone. □
Theorem 2. 
x L Φ { 0 } is upper local uniform monotonicity point if and only if the following conditions are satisfied:
(1) 
m ( t supp x : x ( t ) x F < a Φ ( t ) ) = 0 ;
(2) 
x ( t ) x F S Φ + ( t ) for almost every t supp x ;
(3) 
Φ 2 .
Proof. 
The necessity.
Because the upper local uniform monotonicity point is an upper strict monotonicity point, we only need to prove condition (3).
Case 1: Let e = { t T : b Φ ( t ) < + } . Suppose that m ( e ) > 0 . From the definition of the Musielak–Orlicz function, we know that Φ ( t , b Φ ( t ) ) χ e ( t ) is a finite measurable function. Hence, there exists a e 0 e with m ( e 0 ) > 0 such that
e 0 Φ ( t , b Φ ( t ) ) d t < + .
Let e n = { t T : x ( t ) x F < ( 1 1 n ) b Φ ( t ) } . Thus, there is a n 0 N such that m ( e n 0 ) > 0 . Take G n e n 0 such that m ( G n ) > 0 and m ( G n ) 0 . Denote
x n ( t ) = x ( t ) χ T G n ( t ) + x F b Φ χ G n ( t ) , ( n = 1 , 2 , ) .
Then 0 x ( t ) x n ( t ) for t T and
I Φ ( x n x F ) = T G n Φ ( t , x ( t ) x F ) d t + G n Φ ( t , b Φ ( t ) ) d t x F + G n Φ ( t , b Φ ( t ) ) d t .
Because lim n G n Φ ( t , b Φ ( t ) ) d t = 0 , then x n F x F . Moreover,
I Φ x n ( t ) x ( t ) x F 2 n 0 > G n Φ t , x F n 0 b Φ ( t ) x F 2 n 0 d t = G n Φ ( t , 2 b Φ ( t ) ) d t = .
The above inequality shows that x n x F > x F 2 n 0 ; this is a contradiction.
Case 2: Let b Φ ( t ) = + . Take c > 0 satisfying m ( { t supp x : 0 < x ( t ) c } ) > 0 and denote A = { t supp x : 0 < x ( t ) c } . Let r 1 , r 2 , be the set of rational numbers on the interval [ 0 , 1 ] . According to Φ ( t , 1 r i ) is a measurable function on A that is finite for almost every t A . Then, there exists a Q i A , m ( Q i ) m ( A ) 2 i + 1 satisfying Φ ( t , 1 r i ) is an integrable function on the interval A Q i . Denote Q = n = 1 Q i . Then, Φ ( t , 1 r i ) is an integrable function on the interval A Q and
m ( A Q ) m ( A ) i = 1 m ( Q i ) m ( A ) 1 2 m ( A ) = 1 2 m ( A ) .
Hence, we can assume that Φ ( t , 1 r i ) is an integrable function on A for each i N .
Applying Lemma 1 with b n = 1 + 1 n , p n = 2 n , q n = 1 2 n , where n N . We can find a sequence { x n ( t ) } n = 1 of measurable functions and mutually disjoint measurable subsets { F n } in A such that 0 x n ( t ) < ,
F n Φ ( t , x n ( t ) ) d t = 1 2 n , Φ ( t , b n x n ( t ) ) p n Φ ( t , x n ( t ) ) ( t F n , n N ) .
Let us define
y ( t ) = n = 1 x n ( t ) χ F n ( t ) .
Then, I Φ ( y ) = n = 1 F n Φ ( t , x n ( t ) ) d t = n = 1 1 2 n = 1 , y L Φ and y F 1 . Moreover, for any λ ( 0 , 1 ) , there exists an m N , m 2 such that 1 λ > 1 + 1 n for any n m . Then,
I Φ ( y λ ) n = m F n Φ ( t , x n ( t ) λ ) d t n = m F n Φ t , ( 1 + 1 n ) x n ( t ) d t n = m 2 n F n Φ ( t , x n ( t ) ) d t = n = m 1 = .
By the above inequality we conclude that there exists y ( t ) = n = 1 x n ( t ) χ F n ( t ) satisfying the following conditions:
(a)
supp y A ;
(b)
n = m x n χ F n F = 1 as m ;
(c)
I Φ n = m x n χ F n 0 ;
(d)
I Φ n = m x n χ F n λ = + for any λ ( 0 , 1 ) .
Define
W m ( t ) = x ( t ) χ T n = m F n ( t ) + x F n = m x n ( t ) χ F n ( t )
and
Z m ( t ) = x ( t ) + x F n = m x n ( t ) χ F n ( t ) .
We have 0 x ( t ) Z m ( t ) for t T and want to prove that W m F x F . According to inequality
I Φ ( W m x F ) I Φ n = m x n χ F n + I Φ ( x x F ) x F + I Φ n = m x n χ F n ,
we know that lim m ¯ W m F x F . For any λ ( 0 , 1 ) ,
I Φ ( W m λ x F ) I Φ n = m x n χ F n λ = .
Hence, lim inf n = W m F λ x F . Further by the arbitrariness of λ , we obtain lim inf n W m F x F , thus the equality W m F = x F holds.
Next, we are going to prove lim m Z m W m F = 0 . Because
Z m ( t ) W m ( t ) = n = m x ( t ) χ F n ( t )
and
Z m W m F = x χ n = m F n F c χ n = m F n F ,
we only need to prove the condition lim m χ n = m F n F = 0 . For any i N , we have I Φ χ n = m F n r i = n = m F n Φ ( t , 1 r i ) d t . The condition Φ ( t , 1 r i ) is an integrable function and m ( n = m F n ) 0 imply that there exists m 0 N such that I Φ χ n = m F n r i r i as m m 0 . Hence, χ n = m F n F r i . By the arbitrariness of r i , we obtain lim m χ n = m F n F = 0 . The three-angle inequality implies that
lim m Z m F = x F ,
Z m ( t ) x ( t ) = x F n = m x n ( t ) χ F n ( t ) .
From the above equalities we obtain
I Φ Z m x 1 2 x F = I Φ n = m x n χ F n 1 2 = ,
Z m x F 1 2 x F ,
which contradicts the fact that x is an upper local uniform monotonicity point.
The sufficiency.
Suppose the conditions 0 x x n and lim n x n F = x F are satisfied; then, we only need to prove that the equality lim n x n x F = 0 holds. It is known that
lim n I Φ ( x x n F ) = x F .
Further, we can obtain that
lim n T Φ ( t , x n ( t ) x n F ) d t T Φ ( t , x ( t ) x n F ) d t = 0 ,
lim n T Φ ( t , x n ( t ) x n F ) Φ ( t , x ( t ) x n F ) d t = 0 .
Because Φ ( t , x n ( t ) x n F ) Φ ( t , x ( t ) x n F ) 0 , the equality
lim n T Φ ( t , x n ( t ) x n F ) Φ ( t , x ( t ) x n F ) d t = 0
holds. According to finiteness of the measure, there is a subsequence { x n k } k = 1 satisfying Φ ( t , x n k ( t ) x n k F ) Φ ( t , x ( t ) x n k F ) 0 for almost every t T . Without loss of generality, suppose the equality
lim k Φ ( t , x n k ( t ) x n k F ) Φ ( t , x ( t ) x n k F ) = 0
for any t T is true. Furthermore, there is a sequence { η k } k = 1 ( 0 , 1 ) and y L 1 satisfying lim k η k = 0 and
0 Φ ( t , x n k ( t ) x n k F ) Φ ( t , x ( t ) x n k F ) η k y ( k N )
for any t T , where
Φ ( t , x n k ( t ) x n k F ) η k y + Φ ( t , x ( t ) x n k F ) y + Φ ( t , x ( t ) x F )
for any t T .
By the fact that x ( t ) x n F S Φ + ( t ) for almost every t supp x , there exists a set e 0 supp x and m ( e 0 ) = 0 such that x ( t ) x n F S Φ + ( t ) for t supp x e 0 . Let t 0 supp x e 0 ; we can easily obtain that Φ ( t 0 , u ) > Φ ( t 0 , x ( t ) x F ) for t 0 supp x e 0 under the condition u > x ( t ) x F .
The next step we need to prove is lim k x n k ( t 0 ) x n k F = x ( t 0 ) x F for t 0 supp x e 0 . If the equality ( 1 ) imply that the sequence { x n k ( t 0 ) x n k F } is bounded. Denote lim k x n k ( t 0 ) x n k F x ( t 0 ) x F = c . There is a maximum strict monotonicity interval of Φ ( t , u ) [ a , b ) satisfying x ( t 0 ) x F [ a , b ) . From the fact that lim k x ( t 0 ) x n k F = x ( t 0 ) x F < b , we can obtain that there exists an n 0 N such that x ( t 0 ) x n k F < b whenever n k n 0 .
We have to consider two cases.
Case 1: If there is a k 1 N such that a x ( t 0 ) x n k F < b whenever k k 1 . We want to prove that for some d > 0 , the inequality
Φ ( t 0 , u ) + d Φ ( t 0 , v ) ( )
holds whenever u [ a , b ) , v u c 2 . Suppose to the contrary, there are the sequences { u n } n = 1 [ a , b ) and { v n } n = 1 R with u n v n c 2 satisfying
Φ ( t 0 , u n ) + 1 n Φ ( t 0 , v n ) > Φ ( t 0 , u n ) .
The sequence { u n } n = 1 is bounded, which implies that there are u 0 , v 0 R such that u n u 0 , v n v 0 . Because Φ ( t 0 , u ) is continuous, we have that Φ ( t 0 , u 0 ) = Φ ( t 0 , v 0 ) . However, because u 0 [ a , b ) and u 0 < v 0 , it is easy to see that Φ ( t 0 , u 0 ) < Φ ( t 0 , v 0 ) . We obtain a contradiction.
As lim k x n k ( t 0 ) x n k F x ( t 0 ) x n k F = c , we can find a k 2 N such that x n k ( t 0 ) x n k F x ( t 0 ) x n k F c 2 whenever k k 2 . The inequality ( ) can easily yield that there exists d > 0 such that
Φ ( t 0 , x ( t 0 ) x n k F ) + d Φ ( t 0 , x n k ( t 0 ) x n k F ) .
Combined with equality (2) we also obtain a contradiction. Thus, we have
lim k x n k ( t 0 ) x n k F = x ( t 0 ) x F .
Case 2: If there exists k 3 N such that x ( t 0 ) x n k F < a whenever k k 3 .
There must exist k 4 N and b 1 [ a , b ) such that [ a , b 1 ) ( x ( t 0 ) x n k F , x n k ( t 0 ) x n k F ) . Then, the following proof is similar to the Case 1. There is d > 0 satisfying the inequality:
Φ ( t 0 , x ( t 0 ) x n k F ) + d Φ ( t 0 , x n k ( t 0 ) x n k F ) ,
which contradicts equality (2). Then, we can yield that lim k x n k ( t ) x n k F = x ( t ) x F for almost every t supp x . By the fact that lim k x n k F = x F , we obtain that the equality
lim k x n k ( t ) = x ( t )
for almost every t T holds.
Therefore, by inequality (3), we can conclude that
Φ ( t , x n k ( t ) x n k F x ( t ) x n k F ) Φ ( t , x n k ( t ) x n k F ) y + Φ ( t , x ( t ) x F ) L 1 .
Then, for any λ > 0 , the Lebesgue dominated convergence theorem concludes that I Φ ( λ ( x n k x ) ) 0 as k . Further, Lemma 3 yields that x x n k F 0 as k . The double extract subsequence theorem fulfills the proof. □
Corollary 3. 
E Φ is upper locally uniformly monotone if and only if the following conditions are satisfied:
(1) 
a Φ ( t ) = 0 for almost every t T ;
(2) 
Φ ( t , u ) is strictly monotonically increasing with respect to u for almost every t T .
Corollary 4. 
L Φ is upper locally uniformly monotone if and only if the following conditions are satisfied:
(1) 
a Φ ( t ) = 0 for almost every t T ;
(2) 
Φ ( t , u ) is strictly monotonically increasing with respect to u for almost every t T ;
(3) 
Φ 2 .

Author Contributions

All authors contributed equally and significantly in writing this article. All authors have read and agreed to the published version of the manuscript.

Funding

This work was supported by the National Key Laboratory Foundation Project (Project Number is 6142217210206.

Data Availability Statement

Not applicable.

Acknowledgments

We would like to express our thanks to the anonymous referees and the editor for their constructive comments and suggestions, which greatly improved this article.

Conflicts of Interest

The authors declare no conflict of interest.

References

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Liu, Y.; Xue, Y.; Cui, Y. Upper Local Uniform Monotonicity in F-Normed Musielak–Orlicz Spaces. Axioms 2023, 12, 539. https://doi.org/10.3390/axioms12060539

AMA Style

Liu Y, Xue Y, Cui Y. Upper Local Uniform Monotonicity in F-Normed Musielak–Orlicz Spaces. Axioms. 2023; 12(6):539. https://doi.org/10.3390/axioms12060539

Chicago/Turabian Style

Liu, Yanli, Yangyang Xue, and Yunan Cui. 2023. "Upper Local Uniform Monotonicity in F-Normed Musielak–Orlicz Spaces" Axioms 12, no. 6: 539. https://doi.org/10.3390/axioms12060539

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