In this paper, we study to solve the quadratic type λ-functional equation with 3 k variables. First, we investigated in non-Archimedean Banach spaces with a fixed point method, next, we investigated in non-Archimedean Banach spaces with a direct method and finally we do research in non-Archimedean random spaces. I will show that the solutions of the quadratic type λ-functional equation are quadratic type mappings. These are the main results of this paper.
Let X and Y be a normed spaces on the same field K , and f : X → Y . We use the notation ‖ ⋅ ‖ for all the norm on both X and Y . In this paper, I study and expand the λ -function equation from non-Archimedean normed space to non-Archimedean random normed space.
In fact, when X is non-Archimedean normed space and Y is non-Archimedean Banach spaces.
Or X is a vector over field K and ( Y , Γ , T ) be a non-Archimedean random Banach space over field K . We solve and prove the Hyers-Ulam-Rassisa type stability of forllowing quadratic λ-functional equation.
2 ∑ j = 1 k f ( z j ) + 2 ∑ j = 1 k f ( x j + y j ) = f ( ∑ j = 1 k x + ∑ j = 1 k y j + ∑ j = 1 k z j ) + λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) (1)
where: Let | 2 k | ≠ 1 , λ is a fixed non-Archimedean number with λ − 2 m ≠ 4 k − 1 and k , m is a positive integer. The notions of non-Archimedean normed space and non-Archimedean Banach spaces and non-Archimedean random Banach space over field K will remind in the next section. The study the stability of generalized stability of the quadratic type λ-functional equation with variables in non-Archimedean Banach space and non-Archimedean Random normed space originated from a question of S.M. Ulam [
d ( f ( x ∗ y ) , f ( x ) ∘ f ( y ) ) < δ , ∀ x ∈ G
then there is a homomorphism h : G → G ′ with
d ( f ( x ) , h ( x ) ) < ε , ∀ x ∈ G
The Hyers [
The functional equation
f ( x + y ) = f ( x ) + f ( y )
is called the Cauchy equation. In particular, every solution of the Cauchy equation is said to be an additive mapping.
The functonal equation
f ( x + y ) + f ( x − y ) = 2 f ( x ) + 2 f ( y )
is called the quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic functional mapping.
The stability the quadratic functional equation was proved by Skof [
Recently the author studied the Hyers-Ulam stability for the following α-functional equation.
2 f ( x ) + 2 f ( y ) = f ( x + y ) + α − 2 f ( α ( x − y ) )
in Non-Archimedean Banach spaces and non-Archimedean Random normed space.
In this paper, we solve and proved the Hyers-Ulam stability for λ-functional Equation (1.1), i.e. the λ-functional equation with 3k-variables. Under suitable assumptions on spaces X and Y , we will prove that the mappings satisfying the λ-functional Equation (1.1). Thus, the results in this paper are generalization of those in [
In this paper, based on the work of world mathematicians [
The paper is organized as followns: In section preliminarier we remind some basic notations in [
Section 3: Establishing the solution for (1.1) by the fixed point method in Non-Archimedean Banach space.
+ Condition for existence of solutions for Equation (1.1)
+ Constructing a solution for (1.1).
Section 4: Establishing the solution for (1.1) by the direct method in Non-Archimedean Banach space
Section 5: Construct a solution for (1.1) on non-Archimedean Random normed space.
A valuation is a function | ⋅ | from a field K into [ 0, ∞ ) such that 0 is the unique element having the 0 valuation,
| r ⋅ s | = | r | ⋅ | s | , ∀ r , s ∈ K
and the triangle inequality holds, i.e.;
| r + s | ≤ | r | + | s | , ∀ r , s ∈ K
A field K is called a valued filed if K carries a valuation. The usual absolute values of ℝ and ℂ are examples of valuation. Let us consider a vavluation which satisfies a stronger condition than the triangle inaquality. If the tri triangle inequality is replaced by
| r + s | ≤ max { | r | , | s | } , ∀ r , s ∈ K
then the function | ⋅ | is called a norm-Archimedean valuational, and filed. Clearly | 1 | = | − 1 | = 1 and | n | ≤ 1, ∀ n ∈ N . A trivial example of a non-Archimedean valuation is the function | ⋅ | talking everything except for 0 into 1 and | 0 | = 0 this paper, we assume that the base field is a non-Archimedean filed, hence call it simply a filed. Let be a vecter space over a filed K with a non-Archimedean | ⋅ | . A function ‖ ⋅ ‖ : X → [ 0, ∞ ) is said a non-Archimedean norm if it satisfies the follwing conditions:
1) ‖ x ‖ = 0 if and only if x = 0 ;
2) ‖ r x ‖ = | r | ‖ x ‖ ( r ∈ K , x ∈ X ) ;
3) the strong triangle inequlity
‖ x + y ‖ ≤ max { ‖ x ‖ , ‖ y ‖ } , x , y ∈ X
hold. Then ( X , ‖ ⋅ ‖ ) is called a norm-Archimedean norm space.
1) Let { x n } be a sequence in a non-Archimedean normed space X. Then sequence { x n } is called cauchy if for a given ε > 0 there a positive integer N such that
‖ x n − x ‖ ≤ ε
for all n , m ≥ N
2) Let { x n } be a sequence in a norm-Archimedean normed space X. Then sequence { x n } is called cauchy if for a given ε > 0 there a positive integer N such that
‖ x n − x ‖ ≤ ε
for all n , m ≥ N . The we call x ∈ X a limit of sequence x n and denote lim n → ∞ x n = x .
3) If every sequence Cauchy in X converger, then the norm-Archimedean normed space X is called a norm-Archimedean Bnanch space.
A random normed space is triple ( X , Γ , T ) , where X is a vector space, T is a is a continuous t-norm, and Γ is a mapping from X into D + such that, the following conditions hold:
1) (RN1) Γ x ( t ) for all t > 0 if and only if x = 0 ;
2) (RN2) Γ α x ( t ) = Γ x ( t | α | ) for all x ∈ X , α ≠ 0 ;
3) (RN3) Γ x + y ( t + s ) ≥ T ( Γ x ( t ) , Γ y ( s ) ) for all x , y ∈ X , t , s ≥ 0 ;
Note: If ( X , Γ , T ) is a random normed space an { x n } is a sequence such that x n → x then lim n → ∞ Γ x n ( t ) = Γ x ( t ) almost everywhere.
A non-Archimedean random normed space is triple ( X , Γ , T ) , where X is a linear space over a non-Archimedean filed K , T is a is a continuous t-norm, and Γ is a mapping from X into D + such that, the following conditions hold:
1) (NA-RN1) Γ x ( t ) = ε 0 ( t ) for all t > 0 if and only if x = 0 ;
2) (NA-RN2) Γ α x ( t ) = Γ x ( t | α | ) for all x ∈ X , t > 0 , α ≠ 0 ;
3) (NA-RN3) Γ x + y ( max { t , s } ) ≥ T ( Γ x ( t ) , Γ y ( s ) ) for all x , y ∈ X , t , s ≥ 0 ;
It is easy to see that if (NA-RN3) hold then so is (RN3) Γ x + y ( max { t , s } ) ≥ T ( Γ x ( t ) , Γ y ( s ) )
Let ( X , Γ , T ) is a non-Archimedean random normed space. Suppose { x n } is a sequence in X . Then { x n } is said to be convergent if there exists x ∈ X such that
l i m n → ∞ Γ x n − x ( t ) = 1
for all t > 0 . In that case, x is called the limit of sequence { x n }
Theorem 1. Let ( X , d ) be a complete generalized metric space and let J : X → X be a strictly contractive mapping with Lipschitz constant L < 1 . Then for each given element x ∈ X , either
d ( J n , J n + 1 ) = ∞
for all nonegative integers n or there exists a positive integer n 0 such that
1) d ( J n , J n + 1 ) < ∞ , ∀ n ≥ n 0 ;
2) The sequence { J n x } converges to a fixed point y * of J;
3) y * is the unique fixed point of J in the set Y = { y ∈ X | d ( J n , J n + 1 ) < ∞ } ;
4) d ( y , y * ) ≤ 1 1 − l d ( y , J y ) ∀ y ∈ Y
The functional equation
f ( x + y ) = f ( x ) + f ( y )
is called the cauchuy equation. In particular, every solution of the cauchuy equation is said to be an additive mapping.
The functional equation
f ( x + y ) + f ( x − y ) = 2 f ( x ) + 2 f ( y )
is called the quadratic functional equation In particular, every solution of the quadratic functional equation is said to be an quadratic mapping.
The functional equation
2 f ( x + y 2 ) + 2 f ( x − y 2 ) = f ( x ) + f ( y )
is called a Jensen type the quadratic functional equation
Note that for Quadratic λ-functional equation, X and Y is be vector space.
Lemma 2. Suppose X and Y be vector space. If mapping f : X → Y satisfying
2 ∑ j = 1 k f ( z j ) + 2 ∑ j = 1 k f ( x j + y j ) = f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) + λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) (2)
for all x j , y j , z j ∈ X for all j = 1 → k then f : X → Y is quadratic type
Proof. Assume that f : X → Y satisfies (2)
We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (2), we get
( 4 k − 1 ) f ( 0 ) = λ − 2 m f ( 0 ) (3)
So f ( 0 ) = 0 .
Next we replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (2), we have
f ( x ) = λ − 2 m f ( λ m x ) (4)
and so f ( λ m x ) = λ 2 m f ( x ) for all x ∈ X . Thus from (2)
2 ∑ j = 1 k f ( z j ) + 2 ∑ j = 1 k f ( x j + y j ) = f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) + λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) = f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) (5)
for all x j , y j , z j ∈ X for all j = 1 → k
Next now we replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , 0, ⋯ ,0,0, ⋯ ,0, x , ⋯ ,0 ) in (2), we have
f ( 2 x ) = 2 2 f ( x ) (6)
for all v ∈ X .
Next we replace x by 2x, we get
f ( 2 2 x ) = 2 4 f ( x ) (7)
for all x ∈ X . for all x ∈ X , So from (6) and (7) we have the general case for every m being a positive integer, we have
f ( 2 m x ) = 2 2 m f ( x ) (8)
for all x ∈ X , So we get the desired result.
Notice now we replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , ⋯ ,0,0, ⋯ ,0, y , ⋯ ,0 ) in (5) we have
f ( x + y ) + f ( x − y ) = 2 f ( x ) + 2 f ( y )
So, the function f is quadratic. □
Now, we first study the solutions of (1.1). Note that for Quadratic λ-functional equation, X is a non-Archimedean normed space and Y is a non-Archimedean Banach spacebe then use fixed point method, we prove the Hyers-Ulam stability of the Quadratic λ-functional equation in Non-Archimedean Banach space. Under this setting, we can show that the mapping satisfying (1.1) is quadratic. These results are give in the following.
Theorem 3. Suppose φ : X 3 k → [ 0, ∞ ) be a function such that there exists an 0 < L < 1 with
φ ( x 1 2 k , x 2 2 k , ⋯ , x k 2 k , y 1 2 k , y 2 2 k , ⋯ , y k 2 k , z 1 2 k , z 2 2 k , ⋯ , z k 2 k ) ≤ L | 4 k | φ ( x 1 , x 2 , ⋯ , x k , y 1 , y 2 , ⋯ , y k , z 1 , z 2 , ⋯ , z k ) (9)
for all x j , y j , z j ∈ X , for all j = 1 → k . Let f : X → Y be a mapping satisfying f ( 0 ) = 0 and
‖ 2 ∑ j = 1 k f ( z j ) + 2 ∑ j = 1 k f ( x j + y j ) − f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) − λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) ‖ ≤ φ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) (10)
for all x j , y j , z j ∈ X , for all j = 1 → k . Then there exists a unique quadratic type mapping H : X → Y such that
‖ f ( x ) − H ( x ) ‖ ≤ L | 4 k | ( 1 − L ) φ ( x , ⋯ , x , x , ⋯ , x , x , ⋯ , x ) (11)
for all x ∈ X .
Proof. We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , ⋯ , x ,0, ⋯ ,0, x , ⋯ , x ) in (10), we get
‖ f ( 2 k x ) − 4 k f ( x ) ‖ ≤ φ ( x , ⋯ , x ,0, ⋯ ,0, x , ⋯ , x ) (12)
for all x ∈ X for all j = 1 → k .
Now we consider the set
M : = { h : X → Y , h ( 0 ) = 0 }
and introduce the generalized metric on S as follows:
d ( g , h ) : = inf { β ∈ ℝ : ‖ g ( x ) − h ( x ) ‖ ≤ β φ ( x , ⋯ , x , 0 , ⋯ , 0 , x , ⋯ , x ) , ∀ x ∈ X } ,
where, as usual, inf ϕ = + ∞ . That has been proven by mathematicians ( M , d ) is complete see [
Now we cosider the linear mapping T : M → M such that
T g ( x ) : = 4 k g ( x 2 k )
for all x ∈ X . Let g , h ∈ M be given such that d ( g , h ) = ε then
‖ g ( x ) − h ( x ) ‖ ≤ ε φ ( x , ⋯ , x ,0, ⋯ ,0, x , ⋯ , x )
for all x ∈ X .
Hence
‖ T g ( x ) − T h ( x ) ‖ = ‖ 4 k g ( x 2 k ) − 4 k h f ( x 2 k ) ‖ ≤ | 4 k | ε φ ( x 2 k , x 2 k , ⋯ , x 2 k ,0,0, ⋯ ,0, x 2 k , x 2 k , ⋯ , x 2 k ) ≤ | 4 k | ε L | 4 k | φ ( x , x , ⋯ , x ,0,0, ⋯ ,0, x , x , ⋯ , x ) ≤ L ε φ ( x , x , ⋯ , x ,0,0, ⋯ ,0, x , x , ⋯ , x )
for all x ∈ X . So d ( g , h ) = ε implies that d ( T g , T h ) ≤ L ⋅ ε . This means that
d ( T g , T h ) ≤ L d ( g , h )
for all g , h ∈ M . It folows from (12) that
‖ f ( x ) − 4 k f ( x 2 k ) ‖ ≤ φ ( x 2 k , x 2 k , ⋯ , x 2 k ,0,0, ⋯ ,0, x 2 k , x 2 k , ⋯ , x 2 k ) ≤ L | 4 k | φ ( x , x , ⋯ , x ,0,0, ⋯ ,0, x , x , ⋯ , x )
for all x ∈ X . So d ( f , T f ) ≤ L | 4 k | for all x ∈ X By Theorem 1.2, there exists a mapping H : X → Y satisfying the fllowing:
1) H is a fixed point of T, i.e.,
H ( x ) = 4 k H ( x 2 k ) (13)
for all x ∈ X . The mapping H is a unique fixed point T in the set
ℚ = { g ∈ M : d ( f , g ) < ∞ }
This implies that H is a unique mapping satisfying (13) such that there exists a β ∈ ( 0, ∞ ) satisfying
‖ f ( x ) − H ( x ) ‖ ≤ β φ ( x , x , ⋯ , x ,0,0, ⋯ ,0, x , x , ⋯ , x )
for all x ∈ X
2) d ( T l f , H ) → 0 as l → ∞ . This implies equality
l i m l → ∞ ( 4 k ) n f ( x ( 2 k ) n ) = H ( x )
for all x ∈ X
3) d ( f , H ) ≤ 1 1 − L d ( f , T f ) . Which implies
‖ f ( x ) − H ( x ) ‖ ≤ L | 4 k | ( 1 − L ) φ ( x , x , ⋯ , x ,0,0, ⋯ ,0, x , x , ⋯ , x )
for all x ∈ X . It follows (9) and (10) that
‖ 2 ∑ j = 1 k H ( x j + y j ) + 2 ∑ j = 1 k H ( z j ) − H ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) − λ − 2 m H ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) ‖ = lim n → ∞ | 4 k | n ‖ 2 ∑ j = 1 k f ( x j + y j ( 2 k ) n ) + 2 ∑ j = 1 k f ( z j ( 2 k ) n ) − f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ( 2 k ) n ) − λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ( 2 k ) n ) ) ‖
≤ lim n → ∞ | 4 k | n φ ( x 1 ( 2 k ) n , x 2 ( 2 k ) n , ⋯ , x k ( 2 k ) n , y 1 ( 2 k ) n , y 2 ( 2 k ) n , ⋯ , y k ( 2 k ) n , z 1 ( 2 k ) n , z 2 ( 2 k ) n , ⋯ , z k ( 2 k ) n ) = 0
for all x j , y j , z j ∈ X for all j → k . So
2 ∑ j = 1 k H ( x j + y j ) + 2 ∑ j = 1 k H ( z j ) − H ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) − λ − 2 m H ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) = 0
for all x j , y j , z j ∈ X for all j = 1 → k . By Lemma 3.1, the mapping H : X → Y is quadratic type. □
Theorem 4. Suppose φ : X 3 k → [ 0, ∞ ) be a function such that there exists an 0 < L < 1 with
φ ( x 1 , x 2 , ⋯ , x k , y 1 , y 2 , ⋯ , y k , z 1 , z 2 , ⋯ , z k ) ≤ | 4 k | K φ ( x 1 2 k , x 2 2 k , ⋯ , x k 2 k , y 1 2 k , y 2 2 k , ⋯ , y k 2 k , z 1 2 k , z 2 2 k , ⋯ , z k 2 k ) (14)
for all x j , y j , z j ∈ X , for all j = 1 → k . Let f : X → Y be a mapping satisfying f ( 0 ) = 0 and
‖ 2 ∑ j = 1 k f ( z j ) + 2 ∑ j = 1 k f ( x j + y j ) − f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) − λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) ‖ ≤ φ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) (15)
for all x j , y j , z j ∈ X , for all j = 1 → k . Then there exists a unique quadratic type mapping H : X → Y such that
‖ f ( x ) − H ( x ) ‖ ≤ L | 4 k | ( 1 − L ) φ ( x , ⋯ , x , x , ⋯ , x , x , ⋯ , x ) (16)
for all x ∈ X .
The rest of the proof is similar to the proof of theorem 3.2 with note that mapping T : M → M , T g ( x ) : = 1 4 k g ( 2 k x ) .
Corollary 1. Let r < 2 and θ be nonegative real numbers and let f : X → Y be a mapping satisfying f ( 0 ) = 0 and
‖ 2 ∑ j = 1 k f ( z j ) + 2 ∑ j = 1 k f ( x j + y j ) − f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) − λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) ‖ ≤ θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (17)
for all x ∈ X . Then there exists a unique quadratic type mapping H : X → Y such that
‖ f ( x ) − H ( x ) ‖ ≤ 2 θ | 2 k | r − | 4 k | ‖ x ‖ r (18)
for all x ∈ X .
Corollary 2. Let r > 2 and θ be nonegative real numbers and let f : X → Y be a mapping satisfying f ( 0 ) = 0 and
‖ 2 ∑ j = 1 k f ( z j ) + 2 ∑ j = 1 k f ( x j + y j ) − f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) − λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) ‖ ≤ θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (19)
for all x ∈ X . Then there exists a unique quadratic type mapping H : X → Y such that
‖ f ( x ) − H ( x ) ‖ ≤ 2 θ | 4 k | − | 2 k | r ‖ x ‖ r (20)
for all x ∈ X .
Next, we are going to study the solutions of (1.1) for Quadratic λ-functional equation use direct method, we prove the Hyers-Ulam stability of the Quadratic λ-functional equation, the X is a Non-Archimedean normed space and Y is a Non-Archimedean Banach space, and the field K satisfy | 2 k | ≠ 1, λ − 2 m ≠ 4 k − 1 . Under this setting, we can show that the mapping satisfying (1.1) is quadratic. These results are give in the following
Theorem 5. Let φ : X 3 k → [ 0, ∞ ) be a function and let f : X → Y be a mapping satisfying f ( 0 ) = 0 and
lim j → ∞ | 4 k | j φ ( x 1 ( 2 k ) j , x 2 ( 2 k ) j , ⋯ , x k ( 2 k ) j , y 1 ( 2 k ) j , y 2 ( 2 k ) j , ⋯ , y k ( 2 k ) j , z 1 ( 2 k ) j , z 2 ( 2 k ) j , ⋯ , z k ( 2 k ) j ) = 0 (21)
‖ 2 ∑ j = 1 k f ( z j ) + 2 ∑ j = 1 k f ( x j + y j ) − f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) − λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) ‖ ≤ φ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) (22)
for all x j , y j , z j ∈ X for all j = 1 → k . Then there exists a unique quadratic type mapping H : X → Y such that
‖ f ( x ) − H ( x ) ‖ ≤ sup j ∈ ℕ { | 4 k | j − 1 φ ( x ( 2 k ) j , ⋯ , x ( 2 k ) j , x ( 2 k ) j , ⋯ , x ( 2 k ) j , x ( 2 k ) j , ⋯ , x ( 2 k ) j ) } (23)
for all x ∈ X .
Proof. We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , ⋯ , x ,0, ⋯ ,0, x , ⋯ , x ) in (22), we have
‖ f ( 2 k x ) − 4 k f ( x ) ‖ ≤ φ ( x , ⋯ , x ,0, ⋯ ,0, x , ⋯ , x ) (24)
for all x ∈ X . Therefore
‖ f ( x ) − 4 k f ( x 2 k ) ‖ ≤ φ ( x 2 k , ⋯ , x 2 k ,0, ⋯ ,0, x 2 k , ⋯ , x 2 k ) (25)
for all x ∈ X .
Hence
‖ ( 4 k ) l f ( x ( 2 k ) l ) − ( 4 k ) m f ( x ( 2 k ) m ) ‖ ≤ max { ‖ ( 4 k ) l f ( x ( 2 k ) l ) − ( 4 k ) l + 1 f ( x ( 2 k ) l + 1 ) ‖ , ⋯ , ‖ ( 4 k ) m − 1 f ( x ( 2 k ) m − 1 ) − ( 4 k ) m f ( x ( 2 k ) m ) ‖ }
≤ max { | 4 k | l ‖ f ( x ( 2 k ) l ) − 4 k f ( x ( 2 k ) l + 1 ) ‖ , ⋯ , | 4 k | m − 1 ‖ f ( x ( 2 k ) m − 1 ) − 4 k f ( x ( 2 k ) m ) ‖ } ≤ sup j ∈ { l , l + 1 , ⋅ ⋅ ⋅ } { | 4 k | j φ ( x 1 ( 2 k ) j + 1 , ⋯ , x k ( 2 k ) j + 1 , y 1 ( 2 k ) j + 1 , ⋯ , y k ( 2 k ) j + 1 , z 1 ( 2 k ) j + 1 , ⋯ , z k ( 2 k ) j + 1 ) } (26)
for all nonnegative integers m and l with m > l and all x ∈ X . It follows (26)
that the sequence { ( 4 k ) n f ( x ( 2 k ) n ) } is a Cauchy sequence for all x ∈ X . Since Y is complete, the sequence { ( 4 k ) n f ( x ( 2 k ) n ) } converger so one can define the mapping H : X → Y by
H ( x ) : = lim n → ∞ ( 4 k ) n f ( x ( 2 k ) n )
for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (26), we get (23). It follows from (21) and (22) that
‖ 2 ∑ j = 1 k H ( x j + y j ) + 2 ∑ j = 1 k H ( z j ) − H ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) − λ − 2 m H ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) ‖ = lim n → ∞ | 4 k | n ‖ 2 ∑ j = 1 k f ( x j + y j ( 2 k ) n ) + 2 ∑ j = 1 k f ( z j ( 2 k ) n ) − f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ( 2 k ) n ) − λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ( 2 k ) n ) ) ‖
≤ lim j → ∞ | 4 k | n φ ( x ( 2 k ) j , ⋯ , x ( 2 k ) j , x ( 2 k ) j , ⋯ , x ( 2 k ) j , x ( 2 k ) j , ⋯ , x ( 2 k ) j ) = 0 (27)
for all x ∈ X .
2 ∑ j = 1 k H ( x j + y j ) + 2 ∑ j = 1 k H ( z j ) − H ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j )
− λ − 2 m H ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) = 0
for all x ∈ X . By Lemma 3.1, the mapping H : X → Y is quadratic. Now, let T : X → Y be another quadratic mapping satisfying (23). Then we have
‖ H ( x ) − T ( x ) ‖ = ‖ ( 4 k ) q H ( x ( 2 k ) q ) − ( 4 k ) q T ( x ( 2 k ) q ) ‖ ≤ max { ‖ ( 4 k ) q H ( x ( 2 k ) q ) − ( 4 k ) q f ( x ( 2 k ) q ) ‖ , ‖ ( 4 k ) q T ( x ( 2 k ) q ) − ( 4 k ) q f ( x ( 2 k ) q ) ‖ } ≤ sup j ∈ ℕ { | 4 k | q + j − 1 φ ( x 1 ( 2 k ) j + 1 , ⋯ , x k ( 2 k ) j + 1 , y 1 2 j + 1 , ⋯ , y k ( 2 k ) j + 1 , z 1 ( 2 k ) j + 1 , ⋯ , z k 2 j + 1 ) }
which tends to zero as q → ∞ for all x ∈ X . So we can conclude that
H ( x ) = T ( x ) for all x ∈ X . This proves the uniqueness of H. Thus the mapping H : X → Y is a unique quadratic mapping satisfying (23) □
Theorem 6. Let φ : X 3 k → [ 0, ∞ ) be a function and let f : X → Y be a mapping satisfying f ( 0 ) = 0 and
lim j → ∞ { 1 | 4 k | j φ ( ( 2 k ) j − 1 x 1 , ⋯ , ( 2 k ) j − 1 x k , ( 2 k ) j − 1 y 1 , ⋯ , ( 2 k ) j − 1 y k , ( 2 k ) j − 1 z 1 , ⋯ , ( 2 k ) j − 1 z k ) } = 0 (28)
‖ 2 ∑ j = 1 k f ( z j ) + 2 ∑ j = 1 k f ( x j + y j ) − f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) − λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) ‖ ≤ φ ( x 1 , x 2 , ⋯ , x k , y 1 , y 2 , ⋯ , y k , z 1 , z 2 , ⋯ , z k ) (29)
for all x j , y j , z j ∈ X for all j = 1 → n . Then there exists a unique quadratic type mapping H : X → Y such that
‖ f ( x ) − H ( x ) ‖ ≤ sup j ∈ ℕ { 1 | 4 k | j − 1 φ ( ( 2 k ) j − 1 x , ⋯ , ( 2 k ) j − 1 x , ( 2 k ) j − 1 x , ⋯ , ( 2 k ) j − 1 x , ( 2 k ) j − 1 x , ⋯ , ( 2 k ) j − 1 x ) } (30)
for all x ∈ X .
The rest of the proof is similar to the proof of theorem 4.1.
Corollary 3. Let r < 2 and θ be nonegative real numbers and let f : X → Y be a mapping satisfying f ( 0 ) = 0 and
‖ 2 ∑ j = 1 k f ( z j ) + 2 ∑ j = 1 k f ( x j + y j ) = f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) + λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) ‖ ≤ θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (31)
for all x ∈ X . Then there exists a unique quadratic mapping H : X → Y such that
‖ f ( x ) − H ( x ) ‖ ≤ 2 k θ | 2 k | r ‖ x ‖ r
for all x ∈ X .
Corollary 4. Let r > 2 , and θ be nonegative real numbers and let f : X → Y be a mapping satisfying f ( 0 ) = 0 and
‖ 2 ∑ j = 1 k f ( z j ) + 2 ∑ j = 1 k f ( x j + y j ) = f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) + λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) ‖ ≤ θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (32)
for all x ∈ X . Then there exists a unique quadratic mapping H : X → Y such that
‖ f ( x ) − H ( x ) ‖ ≤ 2 k θ | 4 k | ‖ x ‖ r
for all x ∈ X .
In this section, K be a non-Archimedean field, X is a vector space over K and let ( X , Γ , T ) be a non-Archimedean random Banach space over K
We investigate the stability of the quadratic functional equation
2 ∑ j = 1 k f ( z j ) + 2 ∑ j = 1 k f ( x j + y j ) = f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) + λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) (33)
where f : X → Y and f ( 0 ) = 0 .
Next, we define a random approximately quadrtic function. Let φ : X 3 k + 1 → [ 0, ∞ ) be a distribution function such that φ ( x 1 , x 2 , ⋯ , x k , y 1 , y 2 , ⋯ , y k , z 1 , z 2 , ⋯ , z ) k is symmetric, nondecreasing and
φ ( x , ⋯ , x ,0, ⋯ ,0, x , ⋯ , x , t | λ | ) ≤ φ ( λ x , ⋯ , λ x ,0, ⋯ ,0, λ x , ⋯ , λ x , t ) (34)
For x ∈ X , λ ≠ 0 .
Next, we define:
A mapping f : X → Y is said to be φ-approximately quadratic mapping if
Γ f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) + λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) − 2 ∑ j = 1 k f ( z j ) − 2 ∑ j = 1 k f ( x j + y j ) ≤ φ ( x 1 , x 2 , ⋯ , x k , y 1 , y 2 , ⋯ , y k , z 1 , z 2 , ⋯ , z k , t ) (35)
for all x j , y j , z j ∈ X , for all j = 1 → k , t > 0 .
* Note: We assume that 2 k ≠ 0 in K
Theorem 7 For f : X → Y be a φ-approximately quadratic mapping if there exist an β ∈ ℝ ( β > 0 ) and an integer h, h ≥ 2 with β > | ( 2 k ) h | and | 2 k | ≠ 0 such that
φ ( ( 2 k ) − h x 1 , ⋯ , ( 2 k ) − h x k , ( 2 k ) − h y 1 , ⋯ , ( 2 k ) − h y k , ⋯ , ( 2 k ) − h z 1 , ⋯ , ( 2 k ) − h z k , t ) ≥ φ ( ( 2 k ) − h x 1 , ⋯ , ( 2 k ) − h x k , ( 2 k ) − h y 1 , ⋯ , ( 2 k ) − h y k , ⋯ , ( 2 k ) − h z 1 , ⋯ , ( 2 k ) − h z k , β t ) (36)
for all x j , y j , z j ∈ X for all j = 1 → k , t > 0 and
lim n → ∞ T j = n ∞ M ( x , β j t | ( 2 k ) h j | ) = 1 (37)
for all x ∈ X and t > 0 .
Then there exists a unique quadratic type mapping Q : X → Y such that
Γ f ( x ) − Q ( x ) ( t ) ≥ T i = 1 ∞ M ( x , β i + 1 t | ( 2 k ) h i | ) = 1 (38)
In there
M ( x , t ) = Q ( φ ( x , ⋯ , x , 0 , ⋯ , 0 , x , ⋯ , x , t ) , φ ( 2 k x , ⋯ , 2 k x , 0 , ⋯ , 0 , 2 k x , ⋯ , 2 k x , t ) , ⋯ , φ ( ( 2 k ) h − 1 x , ⋯ , ( 2 k ) h − 1 x , 0 , ⋯ , 0 , ( 2 k ) h − 1 x , ⋯ , ( 2 k ) h − 1 x , t ) (39)
for all x ∈ X and ∀ t > 0 .
Proof. First, we show by induction on j that for each x ∈ X , t > 0 and j ≥ 1 ,
Γ f ( ( 2 k ) j x ) − ( 4 k ) j f ( x ) ( t ) ≥ M j ( x , t ) : = T ( φ ( x , ⋯ , x ,0, ⋯ ,0, x , ⋯ , x , t ) , φ ( 2 k x , ⋯ ,2 k x ,0, ⋯ ,0,2 k x , ⋯ ,2 k x , t ) , ⋯ , φ ( ( 2 k ) h − 1 x , ⋯ , ( 2 k ) h − 1 x ,0, ⋯ ,0, ( 2 k ) h − 1 x , ⋯ , ( 2 k ) h − 1 x , t ) (40)
we replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k , t ) by ( x , ⋯ , x ,0, ⋯ ,0, x , ⋯ , x , t ) in (35), we obtain
Γ f ( ( 2 k ) x ) − ( 4 k ) f ( x ) ( t ) ≥ φ ( x , ⋯ , x ,0, ⋯ ,0, x , ⋯ , x , t ) (41)
x ∈ X , t > 0 . This proves (40) for j = 1 . We now assume that (40) holds for some j ≥ 1 Next we replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k , t ) by ( ( 2 k ) j x , ⋯ , ( 2 k ) j x ,0, ⋯ ,0, ( 2 k ) j x , ⋯ , ( 2 k ) j x , t ) in (35) we have
Γ f ( ( 2 k ) j + 1 x ) − ( 4 k ) f ( ( 2 k ) j x ) ( t ) ≥ φ ( ( 2 k ) j x , ⋯ , ( 2 k ) j x ,0, ⋯ ,0, ( 2 k ) j x , ⋯ , ( 2 k ) j x , t ) (42)
Since | 4 k | ≤ 1
Γ f ( ( 2 k ) j + 1 x ) − ( 4 k ) j + 1 f ( x ) ( t ) ≥ T ( Γ f ( ( 2 k ) j + 1 x ) − ( 4 k ) f ( ( 2 k ) j x ) ( t ) , Γ ( 4 k ) f ( ( 2 k ) j x ) − ( 4 k ) j + 1 f ( x ) ( t ) ) = T ( Γ f ( ( 2 k ) j + 1 x ) − ( 4 k ) f ( ( 2 k ) j x ) ( t ) , Γ f ( ( 2 k ) j x ) − ( 4 k ) j f ( x ) ( t | 4 k | ) ) = T ( Γ f ( ( 2 k ) j + 1 x ) − ( 4 k ) f ( ( 2 k ) j x ) ( t ) , Γ f ( ( 2 k ) j x ) − ( 4 k ) j f ( x ) ( t ) ) = T ( φ ( ( 2 k ) j x , ⋯ , ( 2 k ) j x , 0 , ⋯ , 0 , ( 2 k ) j x , ⋯ , ( 2 k ) j x , t ) , M j ( x , t ) ) = M j + 1 ( x , t ) (43)
for all x ∈ X . So in (40) holds for all j ≥ 1 .
Other way
Γ f ( ( 2 k ) h x ) − ( 4 k ) h f ( x ) ( t ) ≥ M ( x , t ) , ∀ x ∈ X , t > 0. (44)
Next we replacing x by ( 2 k ) − ( h n + h ) x in (44) and using inequality (36), we have
Γ f ( x ( 2 k ) h n ) − ( 4 k ) h f ( x ( 2 k ) h n + h ) ( t ) ≥ M ( x ( 2 k ) h n + h , t ) ≥ M ( x , β n + 1 t ) , ∀ x ∈ X , t > 0, n ∈ ℕ . (45)
Then
Γ ( 4 k ) n h f ( x ( 2 k ) h n ) − ( 4 k ) h + 1 f ( x ( 2 k ) h n + h ) ( t ) ≥ M ( x , β n + 1 | 4 k | h n t ) , ∀ x ∈ X , t > 0, n ∈ ℕ . (46)
Hence,
Γ ( 4 k ) h n f ( x ( 2 k ) h n ) − ( 4 k ) h ( n + p ) f ( x ( 2 k ) h ( n + p ) ) ( t ) ≥ T j = n n + p ( Γ ( 4 k ) h j f ( x ( 2 k ) h j ) − ( 4 k ) h ( n + j ) f ( x ( 2 k ) h ( n + j ) ) ( t ) ) ≥ T j = n n + p M ( x , β j + 1 | 4 k | h j t ) ≥ T j = n n + p M ( x , β j + 1 | 4 k | j t ) , ∀ x ∈ X , t > 0, n ∈ ℕ . (47)
Since
lim n → ∞ T j = n n + p M ( x , β j + 1 | 4 k | h j t ) = 1, ∀ x ∈ X , t > 0, n ∈ ℕ ,
{ ( 4 k ) h n f ( x ( 2 k ) h n ) } is a Cauchy sequence in the non-Archimedean random Banach space ( Y , Γ , T ) . Hence, we can define a mapping Q : X → Y such that
lim n → ∞ Γ ( 4 k ) h n f ( x ( 2 k ) h n ) − Q ( x ) ( t ) = 1, ∀ x ∈ X , t > 0, (48)
Next for each n ≥ 1 , ∀ x ∈ X and t > 0 .
Γ f ( x ) − ( 4 k ) h n f ( x ( 2 k ) h n ) ( t ) = Γ ∑ i = 0 n − 1 ( 4 k ) h i f ( x ( 2 k ) h i ) − ( 4 k ) h ( i + 1 ) f ( x ( 2 k ) h ( i + 1 ) ) ( t ) ≥ T i = 0 n + p ( Γ ∑ i = 0 n − 1 ( 4 k ) h i f ( x ( 2 k ) h i ) − ( 4 k ) h ( i + 1 ) f ( x ( 2 k ) h ( i + 1 ) ) ( t ) ) ≥ T i = 0 n − 1 M ( x , β i + 1 t | 4 k | h i ) (49)
Therefore,
Γ f ( x ) − Q ( x ) ( t ) ≥ T ( Γ f ( x ) − ( 4 k ) h n f ( x ( 2 k ) h n ) ( t ) , Γ ( 4 k ) h n f ( x ( 2 k ) h n ) − Q ( x ) ( t ) ) ≥ T ( T i = 0 n − 1 M ( x , β i + 1 t | 4 k | h i ) , Γ ( 4 k ) h n f ( x ( 2 k ) h n ) − Q ( x ) ( t ) ) (50)
By letting n → ∞ , we obtain
Γ f ( x ) − Q ( x ) ( t ) ≥ T i = 0 n − 1 M ( x , β i + 1 t | 4 k | h i ) (51)
As T is continuous, from a well-known result in probabilistic metric space see [
Now we put
Δ x = 2 ( 2 k ) h n ∑ j = 1 k f ( ( 2 k ) − h n z j ) + 2 ( 2 k ) h n ∑ j = 1 k f ( ( 2 k ) − h n ( x j + y j ) ) − ( 2 k ) h n f ( ( 2 k ) h n ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) ) ) + λ − 2 m ( 2 k ) h n f ( ( 2 k ) − h n λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) (52)
it follows that
lim n → ∞ Γ Δ x = Γ f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j ) + λ − 2 m f ( λ m ( ∑ j = 1 k x j + ∑ j = 1 k y j − ∑ j = 1 k z j ) ) − 2 ∑ j = 1 k f ( z j ) − 2 ∑ j = 1 k f ( x j + y j ) ( t ) (53)
for almost all t > 0 , □
On the other hand, replacing x j , y j by ( 2 k ) h n x j , ( 2 k ) h n y j , respectively, in (35) and suing (NA-RN2) and (36), we have
Γ Δ x ≥ φ ( ( 2 k ) − h n x 1 , ⋯ , ( 2 k ) − h n x k ,0, ⋯ ,0, ( 2 k ) − h n z 1 , ⋯ , ( 2 k ) − h n z k , t | 2 k | h n ) ≥ φ ( x 1 , ⋯ , x k ,0, ⋯ ,0, z 1 , ⋯ , z k , β n t | 2 k | h n ) (54)
for all x j , y j , z j ∈ X , j = 1 → k . Sence
lim n → ∞ φ ( x 1 , ⋯ , x k ,0, ⋯ ,0, z 1 , ⋯ , z k , β n t | 2 k | h n ) = 1,
We infer that Q is a quadratic function.
Finally we have to prove that Q is a unique quadratic mapping.
Let Q ′ : X → Y is another quadratic mapping such that
Γ Q ′ ( x ) − f ( x ) ( t ) ≥ M ( x , t ) (55)
for all x ∈ X and t > 0 , then for each n ∈ ℕ , x ∈ X , t > 0
Γ Q ( x ) − Q ′ ( x ) ( t ) ≥ T ( Γ Q ( x ) − ( 4 k ) h n f ( x ( 2 k ) h n ) ( t ) , Γ ( 4 k ) h n f ( x ( 2 k ) h n ) − Q ′ ( x ) ( t ) , t ) . (56)
Form (48), we infer that Q ′ = Q .
From the theorem 5.1 we get the following corollary:
Corollary 5. For f : X → Y be a φ-approximately quadratic mapping if there exist an β ∈ ℝ ( β > 0 ) and an integer h , h ≥ 2 with β > | ( 2 k ) h | and | 2 k | ≠ 0 such that
φ ( ( 2 k ) − h x 1 , ⋯ , ( 2 k ) − h x k , ( 2 k ) − h y 1 , ⋯ , ( 2 k ) − h y k , ⋯ , ( 2 k ) − h z 1 , ⋯ , ( 2 k ) − h z k , t ) ≥ φ ( ( 2 k ) − h x 1 , ⋯ , ( 2 k ) − h x k , ( 2 k ) − h y 1 , ⋯ , ( 2 k ) − h y k , ⋯ , ( 2 k ) − h z 1 , ⋯ , ( 2 k ) − h z k , β t ) (57)
for all x j , y j , z j ∈ X for all j = 1 → k , t > 0 , then there exists a unique quadratic type mapping Q : X → Y such that
Γ f ( x ) − Q ( x ) ( t ) ≥ T i = 1 ∞ M ( x , β i + 1 t | ( 2 k ) h i | ) (58)
for all x ∈ X and ∀ t > 0 . In there
M ( x , t ) = Q ( φ ( x , ⋯ , x ,0, ⋯ ,0, x , ⋯ , x , t ) , φ ( 2 k x , ⋯ ,2 k x ,0, ⋯ ,0,2 k x , ⋯ ,2 k x , t ) , ⋯ , φ ( ( 2 k ) h − 1 x , ⋯ , ( 2 k ) h − 1 x ,0, ⋯ ,0, ( 2 k ) h − 1 x , ⋯ , ( 2 k ) h − 1 x , t ) (59)
for all x ∈ X and ∀ t > 0 .
Application Example: For ( X , Γ , T M ) non-Archimedean random normed space in which
Γ x ( t ) = t t + ‖ t ‖ , ∀ x ∈ X , t > 0
and assuming that ( Y , Γ , T M ) complete non-Archimedean random normed space.
Now we define
φ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k , t ) = t 1 + t .
It is easy to see that for β = 1 then (36) holds, sence
M ( x , t ) = t 1 + t ,
We have
lim n → ∞ T j = n ∞ M ( x , β j | 4 k | h j t ) = lim n → ∞ ( lim m → ∞ T j = n m M ( x , t | 4 k | h j t ) ) = lim n → ∞ ⋅ lim m → ∞ ( t t + | ( 4 k ) h | n ) = 1 ,
∀ x ∈ X , t > 0 .
In this paper, I have built the condition for existence of a solution for a functional equation of general form and then I have used two fixed point methods and a direct method to show their solutions on non-Archimedean space and finally establish their solution on the non-Archimedean Random normed space.
The author declares no conflicts of interest.
An, L.V. (2023) Generalized Stability of the Quadratic Type λ-Functional Equation with 3k-Variables in Non-Archimedean Banach Space and Non-Archimedean Random Normed Space. Open Access Library Journal, 10: e9821. https://doi.org/10.4236/oalib.1109821