Outstanding Development of the Quadratic <i>&Phi;</i>(<i>&mu;</i>₁,<i>&mu;</i>₂)-Functional Inequatities with 2k-Variables in Fuzzy Banach Space

In this paper, I work on expanding the Quadratic Φ( μ ₁, μ ₂)-function inequalities by relying on the general quadratic functional equation with 2k-variables on the fuzzy Banach space. That’s the main result of this.

• Generalized Quadratic Type <i>&Phi;</i>(<i>&mu;</i>₁
• <i>&mu;</i>₂)-Functional Inequality
• Generalized Quadratic Type Functional Equations
• Fuzzy Banach Space
• Fuzzy Normed Vector Spaces

1. Introduction

Let X and Y are fuzzy normed spaces on the same field K , and f : X → Y be a mapping. I use the notation N are the norm on X and on Y respectively. In this paper, I study the relationship between Quadratic-type functional equations and Quadratic ϕ ( μ 1 , μ 2 ) -function inequalities when ( X , N ) is a fuzzy normed space and ( Y , N ) is a fuzzy Banach space.

In fact, when X is a fuzzy normed space and Y is a fuzzy Banach space we solve and prove the Hyers-Ulam stability of the following relationship between quadratic ϕ ( μ 1 , μ 2 ) -function inequalities and quadratic-type functional equations:

N ( 2 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + 2 k f ( ∑ i = 1 k x i − ∑ i = 1 k y i 2 k ) − ∑ i = 1 k     f ( x i ) − ∑ i = 1 k     f ( y i ) , t ) ≤ min ( N ( μ 1 ( f ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ i = 1 k     f ( y i ) ) , t ) ,   N ( μ 2 ( 4 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ i = 1 k     f ( y i ) ) , t ) ) (1)

based on following Generalized Quadratic functional equations with 2k-variable

f ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) = 2 ∑ i = 1 k     f ( x i ) + 2 ∑ k = 1 k     f ( y i )

A 0 = { h : ℝ → ℝ : g ( μ 1 , μ 2 ) = 1 μ 1 + 1 μ 2 < 1 , μ 1 , μ 2 ∈ ℝ } .

Note that: With k is a positive integer and h ∈ A 0 .

The study of the functional equation stability originated from a question of S.M. Ulam [1] , concerning the stability of group homomorphisms. Let ( G , ∗ ) be a group and let ( G ′ , ∘ , d ) be a metric group with metric d ( ⋅ , ⋅ ) . Geven ε > 0 , does there exist a δ > 0 such that if f : G → G ′ satisfy the condition d ( f ( x ∗ y ) , f ( x ) ∘ f ( y ) ) < δ , for all x , y ∈ G then there is a homomorphism h : G → G ′ with d ( f ( x ) , h ( x ) ) < ε , for all x ∈ G , if the answer, is affirmative, we would say that equation of homomophism h ( x ∗ y ) = h ( y ) ∘ h ( y ) is stable. The concept of stability for a functional equation arises when we replace a functional equation with an inequality which acts as a perturbation of the equation. Thus the stability question of functional equations is that how the solutions of the inequality differ from those of the given function equation.

Hyers [2] gave a first affirmative partial answer to the question of Ulam for Banach spaces. Hyers’ Theorem was generalized by Aoki [3] for additive mappings and by Th.M. Rassias [4] for linear mappings by considering an unbounded Cauchy difference. A generalization of the Th.M. Rassias theorem was obtained by Găvrut [5] by replacing the unbounded Cauchy difference with a general control function in the spirit of Th.M. Rassias’ approach. The stability problems of several functional equations have been extensive.

Through the process of studying the works of mathematicians see ( [6] [7] [8] [9] [10] [11] ) in 2020, I set up a general quadratic equation with 2k-variables on the space Non-Archimedean Banach.

f ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ k = 1 k     f ( y i ) (2)

Next in 2020, I build quadratic inequalities on the application of groups and rings,

‖ f ( ∑ j = 1 n     x j + 1 n ∑ j = 1 n     x n + j ) + f ( ∑ j = 1 n     x j − 1 n ∑ j = 1 n     x n + j ) − 2 ∑ j = 1 n     f ( x j ) − 2 ∑ j = 1 n     f ( x n + j n ) ‖ Y ≤ ε , (3)

for all ε ≥ 0 and

‖ f ( ∏ j = 1 n     x j + 1 n ∏ j = 1 n     x n + j ) + f ( ∏ j = 1 n     x j − 1 n ∏ j = 1 n     x n + j ) − 2 ∏ j = 1 n     f ( x j ) − 2 ∏ j = 1 n     f ( x n + j n ) ‖ Y ≤ δ , (4)

for all δ ≥ 0 .

Next in 2021, Ly Van An construct the quadratic inequality functional inequalities in non-Archimedean Banach spaces and Banach spaces,

‖ F ( 1 k ∑ j = 1 k     x k + j + ∑ j = 1 k     x j ) + F ( 1 k ∑ j = 1 k     x k + j − ∑ j = 1 k     x j ) − 2 ∑ j = 1 k     F ( x k + j k ) − 2 ∑ j = 1 k     F ( x j ) ‖ X 2 ≤ ‖ F ( 1 k 2 ∑ j = 1 k     x k + j + 1 k ∑ j = 1 k     x j ) + F ( 1 k 2 ∑ j = 1 k     x k + j − 1 k ∑ j = 1 k     x j ) − 2 k ∑ j = 1 k     F ( x k + j k ) − 2 k ∑ j = 1 k     F ( x j ) ‖ X 2 (5)

and

‖ F ( 1 k 2 ∑ j = 1 k     x k + j + 1 k ∑ j = 1 k     x j ) + F ( 1 k 2 ∑ j = 1 k     x k + j − 1 k ∑ j = 1 k     x j ) − 2 k ∑ j = 1 k     F ( x k + j k ) − 2 k ∑ j = 1 k     F ( x j ) ‖ X 2 ≤ ‖ F ( 1 k ∑ j = 1 k     x k + j + ∑ j = 1 k     x j ) + F ( 1 k ∑ j = 1 k     x k + j − ∑ j = 1 k     x j ) − 2 ∑ j = 1 k     F ( x k + 1 k ) − 2 ∑ j = 1 k     F ( x j ) ‖ X 2 , (6)

Continuing into 2021, Ly Van An construct the quadratic inequality on γ-homogeneous complex Banach space,

‖ f ( ∑ j = 1 k x k + j k + ∑ j = 1 k     x j ) + f ( ∑ j = 1 k x k + j k − ∑ j = 1 k     x j ) − 2 ∑ j = 1 k     f ( x k + j k ) − 2 ∑ j = 1 k     f ( x j ) ‖ Y ≤ ‖ β ( k f ( ∑ j = 1 k x k + j k 2 + 1 k ∑ j = 1 k     x j ) + k f ( ∑ j = 1 k x k + j k 2 − 1 k ∑ j = 1 k     x j ) − 2 ∑ j = 1 k     f ( x k + j k ) − 2 ∑ j = 1 k     f ( x j ) ) ‖ Y (7)

and

‖ k f ( ∑ j = 1 k x k + j k 2 + 1 k ∑ j = 1 k     x j ) + k f ( ∑ j = 1 k x k + j k 2 − 1 k ∑ j = 1 k     x j ) − 2 ∑ j = 1 k     f ( x k + j k ) − 2 ∑ j = 1 k     f ( x j ) ‖ Y ≤ ‖ β ( f ( ∑ j = 1 k x k + j k + ∑ j = 1 k     x j ) + f ( ∑ j = 1 k x k + j k − ∑ j = 1 k     x j ) − 2 ∑ j = 1 k     f ( x k + j k ) − 2 ∑ j = 1 k     f ( x j ) ) ‖ Y (8)

Next in 2023, Ly Van An generalized stability of functional inequalities with 3k-variables associated for Jordan-von Neumann-type additive functional equation,

‖ ∑ j = 1 k     f ( x j ) + ∑ j = 1 k     f ( y j ) + ∑ j = 1 k     f ( z j ) ‖ Y ≤ ‖ 2 k f ( ∑ j = 1 k x j + ∑ j = 1 k y j + ∑ j = 1 k z j 2 k ) ‖ Y , (9)

and

‖ ∑ j = 1 k     f ( x j ) + ∑ j = 1 k     f ( y j ) + ∑ j = 1 k     f ( z j ) ‖ Y ≤ ‖ f ( ∑ j = 1 k     x j + ∑ j = 1 k     y j + ∑ j = 1 k     z j ) ‖ Y , (10)

final

‖ ∑ j = 1 k     f ( x j ) + ∑ j = 1 k     f ( y j ) + 2 k ∑ j = 1 k     f ( z j ) ‖ Y ≤ ‖ 2 k f ( ∑ j = 1 k x j + ∑ j = 1 k y j 2 k + ∑ j = 1 k     z j ) ‖ Y . (11)

Continuing into 2023, Ly Van An construct the broadly derivation on fuzzy Banach algebra involving functional equations and general Cauchy-Jensen functional inequalities,

‖ ∑ j = 1 k     f ( x j ) + ∑ j = 1 k     f ( y j ) + f ( 2 k ∑ j = 1 k     z j ) ‖ ≤ ‖ 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) ‖ (12)

The paper is organized as followings:

In section preliminary, we remind some basic notations in [12] - [18] such as Fuzzy normed spaces, Extended metric space theorem and solutions of the Jensen function equation.

Section 3: Setting up quadratic ϕ ( μ 1 , μ 2 ) -function inequalities (1) based on quadratic Equation (2).

3.1: Condition for existence of solution of (1).

3.2: Establishing a solution for the quadratic h ( μ 1 , μ 2 ) -function inequality (1). So that we solve and proved the Hyers-Ulam type stability for functional Equation (1) i.e. the functional equations with 2k-variables. Under suitable assumptions on spaces X and Y , we will prove that the mappings satisfying the functional Equations (1).

Thus, the results in this paper are generalization of those in [19] - [65] .

2. Preliminaries

2.1. Fuzzy Normed Spaces

Let X be a real vector space. Afunction N : X × R → [ 0,1 ] is called a fuzzy norm on X if for all x , y ∈ X and all s , t ∈ ℝ ,

1) (N1) N ( x , t ) = 0 for t ≤ 0 ;

2) (N2) x = 0 if and only if N ( x , t ) = 1 for all t > 0 ;

3) (N3) N ( c x , t ) = N ( x , t | c | ) if c ≠ 0 ;

4) (N4) N ( x + y , s + t ) ≥ min { N ( x , s ) , N ( y , t ) } ;

5) (N5) N ( x , ⋅ ) is a non-decreasing function of ℝ and lim t → ∞ N ( x , t ) = 1 ;

6) (N6) for x ≠ 0 , N ( x , ⋅ ) is continuous on ℝ .

The pair ( X , N ) is called a fuzzy normed vector space:

1) Let ( X , N ) be a fuzzy normed vector space. A sequence { x n } in X is said to be convergent or converge if there exists an x ∈ X such that lim n → ∞ N ( x n − x , t ) = 1 for all t > 0 . In this case, x is called the limit of the sequence { x n } and we denote it by N − lim n → ∞ x n = x .

2) Let ( X , N ) be a fuzzy normed vector space. A sequence { x n } in X is called Cauchy if for each ε > 0 and each t > 0 there exists an n 0 ∈ N such that for all n = n 0 and all p > 0 , we have N ( x n + p − x n , t ) > 1 − ε .

It is well-known that every convergent sequence in a fuzzy normedvector space is Cauchy. If each Cauchy sequence is convergent, then the fuzzy norm is said to be complete and the fuzzy normed vector space is called a fuzzy Banach space. We say that a mapping f : X → Y between fuzzy normed vector spaces X and Y is continuous at a point x 0 ∈ X if for each sequence { x n } converging to x 0 in X, then the sequence { f ( x n ) } converges to f ( x 0 ) . If f : X → Y is continuous at each x ∈ X , then f : X → Y is said to be continuous on X.

Let X be an algebra and ( X , N ) a fuzzy normed space.

1) The fuzzy normed space ( X , N ) is called a fuzzy normed algebra if N ( x y , s t ) ≥ N ( x , s ) ⋅ N ( y , t ) , for all x , y ∈ X and all positive real numbers s and t.

2) A complete fuzzy normed algebra is called a fuzzy Banach algebra.

Let ( X , N X ) and ( Y , N ) be fuzzy normed algebras. Then a multiplicative ℝ -linear mapping H : ( X , N X ) → ( Y , N ) is called a fuzzy algebra homomorphism. Example:

Let ( X , ‖   ⋅   ‖ ) be a normed algebra. Let N ( x , t ) = { t t + ‖ x ‖ t > 0 0 t ≤ 0     x ∈ X . Then N ( x , t ) is a fuzzy norm on X and ( X , N ( x , t ) ) is a fuzzy normed algebra. Let ( X , N X ) and ( Y , N ) be fuzzy normed algebras. Then a multiplicative ℝ -linear mapping H : ( X , N X ) → ( Y , N ) is called a fuzzy algebra homomorphism.

2.2. Extended Metric Space Theorem

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 nonnegative 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 .

2.3. Solutions of the Equation

The functional equation f ( x + y ) + f ( x − y ) = 2 f ( x ) + 2 f ( y ) is called the Qquadratic equation. In particular, every solution of the quadratic equation is said to be a quadratic mapping.

2.4. Solutions of the Inequalities

The solution of the quadratic function inequalities is called the quadratic mapping.

3. Setting up Quadratic ( μ 1 , μ 2 ) -Function Inequalities (1) Based on Quadratic Equation (2)

3.1. Condition for Existence of Solution of (1)

In this section, assume that X and Y be a fuzzy normed vector spaces Under this setting, we can show that the mappings satisfying (1) is quadratic and h ∈ A .

Lemma 2. Suppose that ( Y , N ) be a fuzzy normed vector space and let f : X → Y be a mapping and it satisfies the functional inequality

N ( 2 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + 2 k f ( ∑ i = 1 k x i − ∑ i = 1 k y i 2 k ) − ∑ i = 1 k     f ( x i ) − ∑ i = 1 k     f ( y i ) , t ) ≤ min ( N ( μ 1 ( f ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ i = 1 k     f ( y i ) ) , t ) ,   N ( μ 2 ( 4 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ i = 1 k     f ( y i ) ) , t ) ) (13)

For all x i , y i ∈ X , i = 1 → k and all t > 0 then f is quadratic.

Proof. I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) by ( 0, ⋯ ,0,0, ⋯ ,0 ) in (13), we have

N ( − 3 k μ 1 f ( 0 ) , t ) ≥ N ( 0, t ) = 1 (14)

Thus f ( 0 ) = 0 .

Next I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) by ( x , ⋯ , x , x , ⋯ , x ) in (13), we have

1 ≤ N ( μ 1 ( f ( 2 k x ) − 4 k f ( x ) , t ) ) (15)

So

f ( 2 k x ) = 4 k f ( x ) (16)

For all x ∈ X .

Now I consider G : X → Y . That

G ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) = 2 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + 2 k f ( ∑ i = 1 k x i − ∑ i = 1 k y i 2 k ) − ∑ i = 1 k     f ( x i ) − ∑ i = 1 k     f ( y i ) . (17)

It follows from (13) and (14)

N ( 1 2 k G ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) , t ) ≤ min ( N ( μ 1 G ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) , t ) , N ( μ 2 G ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) , t ) ) . (18)

Next I put v = 2 t (18) I have

N ( G ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) , v ) ≤ min ( N ( μ 1 G ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) , v 2 ) , N ( μ 2 G ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) , v 2 ) ) = min ( N ( 1 2 k G ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) , v 4 k | μ 1 | ) , N ( 1 2 k G ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) , v 4 k | μ 2 | ) ) ≤ N ( G ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) , 1 4 k h ( μ 1 , μ 2 ) v ) ≤ N ( G ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) , h ( μ 1 , μ 2 ) v ) (19)

for all v > 0 . By (N₅) and (N₆) I have

G ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) = f ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) = 2 ∑ i = 1 k     f ( x i ) + 2 ∑ k = 1 k     f ( y i ) (20)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k ∈ X , since h ( μ 1 , μ 2 ) ∈ A 0 .

Hence f is quadratic mapping as we expected. □

3.2. Establishing a Solution for the Quadratic h ( μ 1 , μ 2 ) -Function Inequality (1)

In this section, assume that ( X , N ) is a fuzzy normed space and ( Y , N ) is a fuzzy Banach space. Under this setting, we can show that the mappings satisfying (1) is quadratic and h ∈ A 0 .

Theorem 3. Let ψ : X 2 k → [ 0, ∞ ) be a function such that there exists an L < 1 2 k ,

ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) ≤ 4 k L ψ ( x 1 2 k , ⋯ , x 1 2 k , y 1 2 k , ⋯ , y k 2 k ) (21)

for all x j , y j ∈ X for j = 1 → k .

Let f : X → Y be a mapping satisfying

min ( N ( 2 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + 2 k f ( ∑ i = 1 k x i − ∑ i = 1 k y i 2 k ) − ∑ i = 1 k     f ( x i ) − ∑ i = 1 k     f ( y i ) , t ) , t t + ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) ) ≤ min ( N ( μ 1 ( f ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ i = 1 k     f ( y i ) ) , t ) ,   N ( μ 2 ( 4 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ i = 1 k     f ( y i ) ) , t ) ) (22)

for all x j , y j ∈ X for j = 1 → k , for all t > 0 . Then

A ( x ) = N − lim n → ∞ 1 ( 4 k ) n f ( ( 2 k ) n x ) (23)

exists each x ∈ X and defines a quadratic mapping A : X → Y such that

N ( f ( x ) − A ( x ) , t ) ≥ 4 k | μ 1 | ( 1 − L ) t 4 k | μ 1 | ( 1 − L ) t + ψ ( x , ⋯ , x , x , ⋯ , x ) (24)

for all x ∈ X and t > 0 .

Proof. I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) by ( 0, ⋯ ,0,0, ⋯ ,0 ) in (22), I have

N ( − 3 k μ 1 f ( 0 ) , t ) ≥ t t + φ ( 0, ⋯ ,0,0, ⋯ ,0 ) = 1 (25)

Thus f ( 0 ) = 0 .

Next I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) by ( x , ⋯ , x , x , ⋯ , x ) in (22), we get

t t + φ ( x , ⋯ , x , x , ⋯ , x ) ≤ N ( μ 1 ( f ( 2 k x ) − 4 k f ( x ) ) , t ) ≤ N ( f ( x ) − 1 4 k f ( 2 k x ) , t 4 k | μ 1 | ) (26)

for all x ∈ X . Now we consider the set M : = { h : X → Y } , and introduce the generalized metric on M as follows:

d ( g , h ) : = inf { β ∈ ℝ + : N ( g ( x ) − h ( x ) , β t )                                   ≥ t t + φ ( x , ⋯ , x , x , ⋯ , x ) , ∀ x ∈ X , ∀ t > 0 } , (27)

where, as usual, inf ϕ = + ∞ . That has been proven by mathematicians ( M , d ) is complete (see [47] ).

Now we cosider the linear mapping T : M → M such that T g ( x ) : = 1 4 k g ( 2 k x ) , for all x ∈ X . Let g , h ∈ M be given such that d ( g , h ) = ε then N ( g ( x ) − h ( x ) , ε t ) ≥ t t + φ ( x , ⋯ , x , x , ⋯ , x ) , ∀ x ∈ X , ∀ t > 0. Hence

N ( g ( x ) − h ( x ) , ε L t ) = N ( 1 4 k g ( 2 k x ) − 1 4 k h ( 2 k x ) , L ε t ) = N ( g ( 2 k x ) − h ( 2 k x ) ,4 L ε t ) ≥ 4 L t 4 L t + φ ( 2 k x , ⋯ ,2 k x ,2 k x , ⋯ ,2 k x ) ≥ 4 L t 4 L t + 4 L φ ( x , ⋯ , x , x , ⋯ , x ) = t t + φ ( x , ⋯ , x , x , ⋯ , x ) , ∀ x ∈ X , ∀ t > 0. (28)

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 (38) that

t t + φ ( x , ⋯ , x , x , ⋯ , x ) ≤ N ( f ( x ) − 1 4 k f ( 2 k x ) , t 4 k | μ 1 | ) (29)

for all x ∈ X . So d ( f , T f ) ≤ 1 4 k | μ 1 | . By Theorem 1, there exists a mapping A : X → Y satisfying the following:

1) A is a fixed point of T, i.e.,

A ( 2 k x ) = 4 k A ( x ) (30)

for all x ∈ X . The mapping A is a unique fixed point T in the set ℚ = { g ∈ M : d ( f , g ) < ∞ } . This implies that A is a unique mapping satisfying (38) such that there exists a β ∈ ( 0, ∞ ) satisfying

N ( f ( x ) − A ( x ) , β t ) ≥ t t + φ ( x , ⋯ , x , x , ⋯ , x ) , ∀ x ∈ X .

2) d ( T l f , H ) → 0 as l → ∞ . This implies equality

N − lim l → ∞ 1 ( 4 k ) l f ( ( 2 k ) l x ) = A ( x ) , for all x ∈ X .

3) d ( f , A ) ≤ 1 1 − L d ( f , T f ) , which implies the inequality.

4) d ( f , A ) ≤ 1 | 4 k | ( 1 − L ) .

This implies that the inequality (24) holds.

By (22)

min ( N ( 1 ( 4 k ) n ( 2 k f ( ( 2 k ) n − 1 ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) ) + 2 k f ( ( 2 k ) n − 1 ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) ) − ∑ i = 1 k     f ( ( 2 k ) n x i ) − ∑ i = 1 k     f ( ( 2 k ) n y i ) , t ( 4 k ) n ) , t t + ψ ( ( 2 k ) n x 1 , ⋯ , ( 2 k ) n x k , ( 2 k ) n y 1 , ⋯ , ( 2 k ) n y k ) )

≤ min ( N ( μ 1 ( 4 k ) n ( f ( ( 2 k ) n ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) ) + f ( ( 2 k ) n ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) )     − 2 ∑ i = 1 k     f ( ( 2 k ) n x i ) − 2 ∑ i = 1 k     f ( ( 2 k ) n y i ) ) , t ( 4 k ) n ) ,

    N ( μ 2 ( 4 k ) n ( 4 k f ( ( 2 k ) n − 1 ( ∑ i = 1 k     x i + ∑ i = 1 k     y i 2 k ) ) + f ( ( 2 k ) n ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) )     − 2 ∑ i = 1 k     f ( ( 2 k ) n x i ) − 2 ∑ i = 1 k     f ( ( 2 k ) n y i ) ) , t ( 4 k ) n ) ) (31)

for all x j , y j ∈ X for j = 1 → k , for all t > 0 and for all n ∈ ℕ . So

min ( N ( 1 ( 4 k ) n ( 2 k f ( ( 2 k ) n − 1 ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) ) + 2 k f ( ( 2 k ) n − 1 ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) ) − ∑ i = 1 k     f ( ( 2 k ) n x i ) − ∑ i = 1 k     f ( ( 2 k ) n y i ) , t ) , ( 4 k ) k t ( 4 k ) k t + ( 4 k ) k ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) )

≤ min ( N ( μ 1 ( 4 k ) n ( f ( ( 2 k ) n ( ∑ i = 1 k x i + ∑ i = 1 k y i ) ) + f ( ( 2 k ) n ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) )     − 2 ∑ i = 1 k     f ( ( 2 k ) n x i ) − 2 ∑ i = 1 k     f ( ( 2 k ) n y i ) ) , t ) ,     N ( μ 2 ( 4 k ) n ( 4 k f ( ( 2 k ) n − 1 ( ∑ i = 1 k     x i + ∑ i = 1 k     y i 2 k ) ) + f ( ( 2 k ) n ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) )     − 2 ∑ i = 1 k     f ( ( 2 k ) n x i ) − 2 ∑ i = 1 k     f ( ( 2 k ) n y i ) ) , t ) ) (32)

for all x j , y j ∈ X for j = 1 → k , for all t > 0 and for all n ∈ ℕ . So since lim n → ∞ ( 4 k ) n t ( 4 k ) n t + ( 4 k ) n L n ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) = 1 , for all x j , y j , z j ∈ X for all j → k , ∀ t > 0 , q ∈ ℝ . So

N ( 2 k A ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + 2 k A ( ∑ i = 1 k x i − ∑ i = 1 k y i 2 k ) − ∑ i = 1 k     A ( x i ) − ∑ i = 1 k     A ( y i ) , t ) ≤ min ( N ( γ 1 ( A ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) + A ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     A ( x i ) − 2 ∑ i = 1 k     A ( y i ) ) , t ) ,   N ( γ 2 ( 4 k A ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + A ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     A ( x i ) − 2 ∑ i = 1 k     A ( y i ) ) , t ) ) (33)

So the mapping A : X → X is a Quadratic mapping, as I desired. □

Theorem 4. Let ψ : X 2 k → [ 0, ∞ ) be a function such that there exists an L < 1 2 k ,

ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) ≤ 1 4 k L ψ ( 2 k x 1 , ⋯ ,2 k x 1 ,2 k y 1 , ⋯ ,2 k y k ) (34)

for all x j , y j ∈ X for j = 1 → k .

Let f : X → Y be a mapping satisfying

min ( N ( 2 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + 2 k f ( ∑ i = 1 k x i − ∑ i = 1 k y i 2 k ) − ∑ i = 1 k     f ( x i ) − ∑ i = 1 k     f ( y i ) , t ) , t t + ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) ) ≤ min ( N ( μ 1 ( f ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ i = 1 k     f ( y i ) ) , t ) ,   N ( μ 2 ( 4 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ i = 1 k     f ( y i ) ) , t ) ) (35)

for all x j , y j ∈ X for j = 1 → k , for all t > 0 . Then

A ( x ) = N − lim n → ∞ ( 4 k ) n f ( x ( 2 k ) n ) (36)

exists each x ∈ X and defines a quadratic mapping A : X → Y such that

N ( f ( x ) − A ( x ) , t ) ≥ 4 k | μ 1 | ( 1 − L ) t 4 k | μ 1 | ( 1 − L ) + L ψ ( x , ⋯ , x , x , ⋯ , x ) (37)

for all x ∈ X and t > 0 .

Proof. Suppose that ( M , d ) be the generalized metric space defined in the proof of theorem 3.

From (35) I have

t t + φ ( x , ⋯ , x , x , ⋯ , x ) ≤ N ( f ( x ) − 4 k f ( x ( 2 k ) n ) , L t 4 k | μ 1 | ) (38)

for all x ∈ X , and for all t > 0.

Now we cosider the linear mapping T : M → M such that T g ( x ) : = 4 k g ( x 2 k ) , for all x ∈ X . So d ( f , T f ) ≤ L 4 k | μ 1 | . Thus d ( f , A ) ≤ L 4 k | μ 1 | ( 1 − L ) . which implies that the inequality (37) Satisfied. The rest of the proof is similar to the proof of Theorem 3. □

From the above theorems we have the following corollary:

Corollary 1. Suppose θ ≥ 0 and let p be a real number with 0 < p < 2 . Let X be a normed vector space with norm ‖   ⋅   ‖ Let f : X → Y be a mapping satisfying

min ( N ( 2 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + 2 k f ( ∑ i = 1 k x i − ∑ i = 1 k y i 2 k )

− ∑ i = 1 k     f ( x i ) − ∑ i = 1 k     f ( y i ) , t ) , t t + θ ( ∑ i = 1 k ‖ x i ‖ p + ∑ i = 1 k ‖ y i ‖ p ) ) ≤ min ( N ( μ 1 ( f ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ i = 1 k     f ( y i ) ) , t ) ,   N ( μ 2 ( 4 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ i = 1 k     f ( y i ) ) , t ) ) (39)

for all x j , y j ∈ X for j = 1 → k , for all t > 0 . Then

A ( x ) = N − lim n → ∞ 1 ( 4 k ) n f ( ( 2 k ) n x ) (40)

exists each x ∈ X and defines a quadratic mapping A : X → Y such that

N ( f ( x ) − A ( x ) , t ) ≥ | μ 1 | ( 4 k − ( 2 k ) p ) t 4 k | μ 1 | ( 4 k − ( 2 k ) p ) t + θ ∑ i = 1 k ‖ 2 k x i ‖ p (41)

for all x ∈ X and t > 0 .

Corollary 2. Suppose θ ≥ 0 and let p be a real number with p > 2 .Let X be a normed vector space with norm ‖   ⋅   ‖ Let f : X → Y be a mapping satisfying

min ( N ( 2 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + 2 k f ( ∑ i = 1 k x i − ∑ i = 1 k y i 2 k ) − ∑ i = 1 k     f ( x i ) − ∑ i = 1 k     f ( y i ) , t ) , t t + θ ( ∑ i = 1 k ‖ x i ‖ p + ∑ i = 1 k ‖ y i ‖ p ) ) ≤ min ( N ( μ 1 ( f ( ∑ i = 1 k     x i + ∑ i = 1 k     y i ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ i = 1 k     f ( y i ) ) , t ) ,   N ( μ 2 ( 4 k f ( ∑ i = 1 k x i + ∑ i = 1 k y i 2 k ) + f ( ∑ i = 1 k     x i − ∑ i = 1 k     y i ) − 2 ∑ i = 1 k     f ( x i ) − 2 ∑ i = 1 k     f ( y i ) ) , t ) ) (42)

for all x j , y j ∈ X for j = 1 → k , for all t > 0 . Then

A ( x ) = N − lim n → ∞ ( 4 k ) n f ( 1 ( 2 k ) n x ) (43)

exists each x ∈ X and defines a quadratic mapping A : X → Y such that

N ( f ( x ) − A ( x ) , t ) ≥ 4 k | μ 1 | ( 4 k − ( 2 k ) p ) t 4 k | μ 1 | ( 4 k − ( 2 k ) p ) t + θ ∑ i = 1 k ‖ 4 k x i ‖ p (44)

for all x ∈ X and t > 0 .

4. Conclusion

In this paper, I construct the ϕ ( μ 1 , μ 2 ) -function inequality on fuzzy space, which is a great idea for the field of functional equations. Then I show how to find their solutions in spaces constructed by Mathematicians.

Conflicts of Interest

The author declares no conflicts of interest.

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