Generalized Hyers-Ulam-Rassisa Type Stability of a Cauchy Additive (ξ12)-Functional Inequalities with 3k-Variables in Complex Banach Space

Abstract

In this paper, we study to solve two additive (ξ12)-functional inequalities with 3k-variables and their Hyers-Ulam stability: First are investigated in complex Banach spaces with a fixed point method and last are investigated in complex Banach spaces with a direct method: These are the main results of this paper.

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An, L.V. (2022) Generalized Hyers-Ulam-Rassisa Type Stability of a Cauchy Additive (ξ1,ξ2)-Functional Inequalities with 3k-Variables in Complex Banach Space. Open Access Library Journal, 9, 1-24. doi: 10.4236/oalib.1109480.

Mathematics Subject Classification

46S10, 39B62, 39B52, 47H10

1. Introduction

Let X and Y be normed spaces on the same field K , and f : X Y . We use the notation for all the norms on both X and Y . In this paper, we investisgate additive (ξ12)-functional inequalitíes when X be a real or complex normed space and Y a complex Banach spaces. We solve and prove the Hyers-Ulam-Rassisa type stability of following Cauchy additive (ξ12)-functional inequalities.

2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y (1)

and when we change the role of the function inequality (1), we continue to prove the following function inequality

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( 2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y (2)

So (1) and (2) are equivalent propositions, which ξ 1 , ξ 2 are fixed nonzero complex numbers with G ( ξ 1 , ξ 2 ) -functional inequality. Note that in the preliminaries, we just recap some of the essential properties for the above problem and for the specific problem, please see the document. The Hyers-Ulam stability was first investigated for the functional equation of Ulam in [1] concerning the stability of group homomorphisms.

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 Hyers [2] gave the first affirmative partial answer to the equation of Ulam in Banach spaces. After that, Hyers’ Theorem was generalized by Aoki [3] additive mappings and by Rassias [4] for linear mappings considering an unbounded Cauchy difference. A generalization of the Rassias theorem was obtained by Găvruta [5] by replacing the unbounded Cauchy difference with a general control function in the spirit of Rassias’ approach.

The stability of the quadratic functional equation was proved by Skof [6] for mappings f : X Y , where X is a normed space and Y is a Banach space. Park [7] [8] defined additive γ -functional inequalities and proved the Hyers-Ulam stability of the additive γ -functional inequalities in Banach spaces and non-archimedean Banach spaces. The stability problems of various functional equations have been extensively investigated by a number of authors on the world even term [4] - [29]. We recall a fundamental result in fixed point theory. The authors studied the Hyers-Ulam stability for the following functional inequalities

f ( x + y 2 + z ) f ( x + y 2 ) f ( z ) f ( x + y 2 2 + z 2 ) 1 2 f ( x + y 2 ) 1 2 f ( z ) (3)

f ( x + y 2 2 + z 2 ) 1 2 f ( x + y 2 ) 1 2 f ( z ) f ( x + y 2 + z ) f ( x + y 2 ) f ( z ) (4)

f ( x + y ) f ( x ) f ( y ) ρ ( 2 f ( x + y 2 ) f ( x ) f ( y ) ) (5)

2 f ( x + y 2 ) f ( x ) f ( y ) ρ ( f ( x + y ) f ( x ) f ( y ) ) (6)

and

f ( x + y 2 + z ) + f ( x + y 2 z ) 2 f ( x + y 2 ) 2 f ( z ) β ( 2 f ( x + y 2 2 + z 2 ) + 2 f ( x + y 2 2 z 2 ) f ( x + y 2 ) f ( z ) ) (7)

2 f ( x + y 2 2 + z 2 ) + 2 f ( x + y 2 2 z 2 ) f ( x + y 2 ) f ( z ) β ( f ( x + y 2 + z ) + f ( x + y 2 z ) 2 f ( x + y 2 ) 2 f ( z ) ) (8)

finaly

f ( x + y ) f ( x ) f ( y ) β 1 ( f ( x + y ) + f ( x y ) 2 f ( x ) ) + β 2 ( 2 f ( x + y 2 ) f ( x ) f ( y ) ) (9)

next

f ( x 1 + x 2 + + x n ) f ( x 1 ) f ( x 2 + + x n ) Y β 1 ( f ( x 1 + x 2 + + x n ) f ( x 1 x 2 x n ) 2 f ( x 1 ) ) Y + β 2 ( 2 f ( x 1 + x 2 + + x n 2 ) f ( x 1 ) f ( x 2 + + x n ) ) Y (10)

final

2 f ( x 1 + x 2 2 + x 3 + x 4 + + x k 4 ) f ( x 1 ) f ( x 2 + x 3 + x 4 + + x k 2 ) Y β 1 ( f ( x 1 + x 2 + x 3 + x 4 + + x k 2 ) + f ( x 1 x 2 x 3 + x 4 + + x k 2 ) 2 f ( x 1 ) ) Y + β 2 ( f ( x 1 + x 2 + x 3 + x 4 + + x k 2 ) f ( x 1 ) f ( x 2 + x 3 + x 4 + + x k 2 ) ) Y (11)

in complex Banach spaces.

In this paper, we solve and prove the Hyers-Ulam stability for (ξ12)-functional inequalities (1) and (2), i.e., the (ξ12)-functional inequalities with n-variables. Under suitable assumptions on spaces X and Y , we will prove that the mappings satisfy the (ξ12)-functional inequalities (1) and (2). Thus, the results in this paper are the generalization of those in [13] [14] [15] [16] [26] [27] [28] [29] for (ξ12)-functional inequalities with n-variables.

The goal of the paper is to develop functional inequalities with a higher number of variables to solve problems of general nonlinear functional equations in order to develop the field of nonlinear analysis.

The paper is organized as follows: In the section preliminaries, we remind some basic notations in [13] [14] [17], such as complete generalized metric space and Solutions of the inequalities.

Section 3: In this section, I use the method of the fixed to prove the Hyers-Ulam stability of the additive (ξ12)-functional inequalities (1) when X be a real or complete normed space and Y complex Banach space.

Section 4: In this section, I use the method of directly determining the solution for (1) when X be a real or complete normed space and Y complex Banach space.

Section 5: In this section, I use the method of the fixed to prove the Hyers-Ulam stability of the additive (ξ12)-functional inequalities (2) when X be a real or complete normed space and Y complex Banach space.

Section 6: In this section, I use the method of directly determining the solution for (2) when X be a real or complete normed space and Y complex Banach space.

2. Preliminaries

2.1. Complete Generalized Metric Space and Solutions of the Inequalities

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 .

2.2. Solutions of the Inequalities

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.

3. Establish the Solution of the Additive (ξ12)-Function Inequalities Using a Fixed Point Method

Now, we first study the solutions of (1). Note that for these inequalities, when X be a real or complete normed space and Y complex Banach space.

Lemma 2. Suppose mapping Γ : X Y satisfies Γ ( 0 ) = 0 and

2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y (12)

for all x j , y j , z j X , j = 1 k , then Γ : X Y is Cauchy additive

Proof. Assume that Γ : X Y satisfies (12)

We replacing ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) by ( x , ,0, x , ,0,0, ,0 ) in (12) we have

2 Γ ( x 2 ) Γ ( x ) Y 0

Thus

Γ ( x 2 ) = 1 2 f ( x ) (13)

for all x X .

It follows from (12) and (13) that

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y = 2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 k j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y (14)

and so

( 1 | ξ 2 | ) Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y | ξ 1 | Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) Y (15)

we let u = j = 1 k x j + y j 2 + j = 1 k z j , v = j = 1 k x j + y j 2 j = 1 k z j , for all j = 1 k ,

we get

( 1 | ξ 2 | ) Γ ( u ) f ( u + v 2 ) Γ ( u v 2 ) Y | ξ 1 | Γ ( u ) + Γ ( v ) 2 Γ ( u + v 2 ) Y (16)

for all u , v X

and so

1 2 ( 1 | ξ 2 | ) Γ ( u + v ) + Γ ( u v ) 2 Γ ( u ) Y | ξ 1 | Γ ( u + v ) f ( u ) Γ ( v ) Y (17)

for all u , v X . It follows from (15) and (17) that

1 2 ( 1 | ξ 2 | ) 2 Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) f ( j = 1 k z j ) Y | ξ 1 | 2 Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y (18)

Since 2 | ξ 1 | + | ξ 2 | < 1

and so

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) = Γ ( j = 1 k x j + y j 2 ) + Γ ( j = 1 k z j )

for all x j , y j , z j X , j = 1 k . Thus Γ is Cauchy additive. □

Theorem 3. Suppose φ : X 3 k [ 0, ) be a function such that there exists an L < 1 with

φ ( x 1 2 , , x k 2 , y 1 2 , , y k 2 , z 1 2 , , z k 2 ) L 2 φ ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) (19)

for all x j , y j , z j X , j = 1 k . If Γ : X Y be a mapping satisfy Γ ( 0 ) = 0 and

2 Γ ( j = 1 n x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 f ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + φ ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) (20)

for all x j , y j , z j X , j = 1 k .

Then there exists a unique mapping ψ : X Y such that

Γ ( x ) ψ ( x ) Y 1 1 L φ ( x , ,0, x , ,0,0, ,0 ) (21)

for all x X .

Proof. Replacing ( x 1 , x 2 , , x k , y 1 , y 2 , , y k , z 1 , z 2 , , z k ) by ( x , ,0, x , ,0,0, ,0 ) in (20), we get

2 Γ ( x 2 ) Γ ( x ) Y φ ( x ,0, ,0, x , ,0,0, ,0 ) (22)

for all x X .

Consider the set

S : = { h : X Y , h ( 0 ) = 0 }

and introduce the generalized metric on S :

d ( g , h ) : = i n f { λ : g ( x ) h ( x ) λ φ ( x ,0, ,0, x , ,0,0, ,0 ) , x X } ,

where, as usual, i n f ϕ = + . It easy to show that ( S , d ) is complete [17] Now we cosider the linear mapping J : S S such that

J g ( x ) : = 2 g ( x 2 )

for all x X . Let g , h S be given such that d ( g , h ) = ε . Then

g ( x ) h ( x ) ε φ ( x ,0 , ,0, x , ,0,0, ,0 )

for all x X .

Hence

J g ( x ) J h ( x ) = 2 g ( x 2 ) 2 h Γ ( x 2 ) 2 ε φ ( x 2 ,0, ,0, x 2 ,0, ,0,0 , ,0 ) 2 ε L 2 φ ( x ,0 , ,0, x , ,0,0, ,0 ) L ε φ ( x ,0 , ,0, x , ,0,0, ,0 )

for all x X . So d ( g , h ) = ε implies that d ( J g , J h ) L ε . This means that

d ( J g , J h ) L d ( g , h ) .

for all g , h S It folows from (22) that

d ( Γ , J Γ ) 1.

By Theorem 2.1, there exists a mapping ψ : X Y satisfying the fllowing:

1) ψ is a fixed point of J, i.e.,

ψ ( x ) = 2 ψ ( x 2 ) (23)

for all x X . The mapping ψ is a unique fixed point J in the set

M = { g S : d ( Γ , g ) < }

This implies that ψ is a unique mapping satisfying (23) such that there exists a λ ( 0, ) satisfying

Γ ( x ) ψ ( x ) λ φ ( x ,0 , ,0, x , ,0,0, ,0 )

for all x X

2) d ( J l f , ψ ) 0 as l . This implies equality

l i m l 2 n f ( x 2 n ) = ψ ( x )

for all x X

3) d ( Γ , ψ ) 1 1 L d ( Γ , J Γ ) . which implies

Γ ( x ) ψ ( x ) 1 1 L φ ( x ,0 , ,0, x , ,0,0, ,0 )

for all x X . It follows (19) and (20) that

2 ψ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) Y = l i m n 2 n 2 Γ ( j = 1 k x j + y j 2 n + 2 + 1 2 n + 1 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 n + 1 ) Γ ( 1 2 n j = 1 k z j ) Y l i m n 2 n | ξ 1 | Γ ( j = 1 k x j + y j 2 n + 1 + 1 2 n j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 n + 1 1 2 n j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 n + 1 ) Y + l i m n 2 n | ξ 2 | Γ ( j = 1 k x j + y j 2 n + 1 + 1 2 n j = 1 k z j ) Γ ( j = 1 k x j + y j 2 n + 1 ) Γ ( 1 2 n j = 1 k z j ) Y + l i m n 2 n φ ( x 1 2 n , , x n 2 n , y 1 2 n , , y n 2 n , z 1 2 n , , z n 2 n )

= ξ 1 ( ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) + ψ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 ψ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) ) Y (24)

for all x j , y j , z j X , j = 1 k . So

2 ψ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) Y β 1 ( ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) + ψ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 ψ ( j = 1 k x j + y j 2 ) ) Y + β 2 ( ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) ) Y + φ ( x 1 , , x k , y 1 , , y k , z 1 , , z k )

for all x j , y j , z j X , j = 1 k . By Lemma 3.1, the mapping ψ : X Y is additive. Ei

ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) = 0

Theorem 4. Let φ : X 3 k [ 0, ) be a function such that there exists an L < 1 with

φ ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) 2 L φ ( x 1 2 , , x k 2 , y 1 2 , , y k 2 , z 1 2 , , z k 2 ) (25)

for all x , y , z X . Let Γ : X Y be a mapping satisfy Γ ( 0 ) = 0 and

2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + φ ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) (26)

for all x j , y j , z j X , j = 1 k .

Then there exists a unique mapping ψ : X Y such that

Γ ( x ) ψ ( x ) Y L 1 L φ ( x , ,0, x , ,0,0, ,0 ) (27)

for all x X .

The rest of the proof is similar to the proof of Theorem 3.

From proving the theorems we have consequences:

Corollary 1. Let r > 1 and θ be nonnegative real numbers and let Γ : X Y be a mapping satisfy Γ ( 0 ) = 0 and

2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 f ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + θ ( j = 1 k x j r + j = 1 k y j r + j = 1 k z j r ) (28)

for all x j , y j , z j X , j = 1 k .

Then there exists a unique mapping ψ : X Y such that

f ( x ) ψ ( x ) Y 2.2 r θ 2 r 2 x X r (29)

for all x X .

Corollary 2. Let r < 1 and θ be nonnegative real numbers and let Γ : X Y be a mapping satisfy Γ ( 0 ) = 0 and

2 f ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + θ ( j = 1 k x j r + j = 1 k y j r + j = 1 k z j r ) (30)

for all x j , y j , z j X , j k .

Then there exists a unique mapping ψ : X Y such that

Γ ( x ) ψ ( x ) Y 2.2 r θ 2 2 r x X r (31)

for all x X .

4. Establish the Solution of the Additive (ξ12)-Function Inequalities Using a Direect Method

Next, we study the solutions of (1). Note that for these inequalities, when X be a real or complete normed space and Y complex Banach space.

Theorem 5. Suppose φ : X 3 k [ 0, ) be a function such that

ϕ ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) : = j = 1 2 j φ ( x 1 2 j , , x k 2 j , y 1 2 j , , y k 2 j , z 1 2 j , , z k 2 j ) < (32)

for all x j , y j , z j X , j = 1 k and let Γ : X Y be a mapping satisfies Γ ( 0 ) = 0 and

2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + φ ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) (33)

for all x j , y j , z j X , j = 1 k .

Then there exists a unique mapping ψ : X Y such that

Γ ( x ) ψ ( x ) Y ϕ ( x , ,0, x , ,0,0, ,0 ) (34)

for all x X

Proof. Replacing ( x 1 , x 2 , , x k , y 1 , y 2 , , y k , z 1 , z 2 , , z k ) by ( x , ,0, x , ,0,0, ,0 ) in (33), we get

2 Γ ( x 2 ) Γ ( x ) Y φ ( x , 0 , ,0, x , ,0,0, ,0 ) (35)

for all x X .

Hence

2 l Γ ( x 2 l ) 2 m Γ ( x 2 m ) Y j = l m 1 2 j Γ ( x 2 j ) 2 j + 1 Γ ( x 2 j + 1 ) Y j = l m 1 2 j φ ( x 2 j + 1 ,0, , 0, x 2 j + 1 ,0, ,0,0, ,0 ) (36)

for all nonnegative integers m and l with m > l and all x X . It follows from (36) that the sequence { 2 n Γ ( x 2 n ) } is a Cauchy sequence for all x X . Since Y is complete, the sequence { 2 n Γ ( x 2 n ) } coverges. So one can define the mapping ψ : X Y by

ψ ( x ) : = lim n 2 n Γ ( x 2 n x ) (37)

for all x X . Moreover, letting l = 0 and passing the limit m in (37), we get (34) It follows from (32) and (33) that

2 ψ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) Y = l i m n 2 n 2 Γ ( j = 1 k x j + y j 2 n + 2 + 1 2 n + 1 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 n + 1 ) f ( 1 2 n j = 1 k z j ) Y l i m n 2 n | ξ 1 | Γ ( j = 1 k x j + y j 2 n + 1 + 1 2 n j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 n + 1 1 2 n j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 n + 1 ) Y + l i m n 2 n | ξ 2 | Γ ( j = 1 k x j + y j 2 n + 1 + 1 2 n j = 1 k z j ) Γ ( j = 1 k x j + y j 2 n + 1 ) Γ ( 1 2 n j = 1 k z j ) Y + l i m n 2 n φ ( x 1 2 n , , x k 2 n , y 1 2 n , , y k 2 n , z 1 2 n , , z k 2 n )

= β 1 ( ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) + ψ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 ψ ( j = 1 k x j + y j 2 ) ) Y + β 2 ( ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) ) Y (38)

for all x j , y j , z j X , j = 1 k . So

2 ψ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) Y ξ 1 ( ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) + ψ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 ψ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) ) Y + φ ( x 1 , , x k , y 1 , , y k , z 1 , , z k )

for all x j , y j , z j X , j = 1 n . By Lemma 3.1, the mapping ψ : X Y is additive. Ei

ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) = 0

Now, let ψ : X Y be another additive mapping satisfying (34). Then we have

ψ ( x ) ψ ( x ) = 2 q ψ ( x 2 q ) 2 q ψ ( x 2 q ) Y 2 q ψ ( x 2 q ) 2 q f ( x 2 q ) + 2 q ψ ( x 2 q ) 2 q f ( x 2 q ) Y 2 q ϕ ( x 2 q ,0, ,0, x 2 q ,0, ,0,0, ,0 )

which tends to zero as q for all x X . So we can conclude that ψ ( x ) = ψ ( x ) for all x X . This proves the uniqueness of ψ .

Theorem 6. Suppose φ : X 3 k [ 0, ) be a function such that

ψ ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) : = j = 0 1 2 j φ ( 2 j x 1 , , 2 j x k , 2 j y 1 , , 2 j y k , 2 j z 1 , , 2 j z k ) < (39)

for all x j , y j , z j X , j = 1 k and let Γ : X Y be a mapping satisfies Γ ( 0 ) = 0 and

2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + φ ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) (40)

for all x j , y j , z j X , j = 1 k .

Then there exists a unique mapping ψ : X Y such that

Γ ( x ) ψ ( x ) Y ϕ ( x , 0 , ,0, x , ,0,0, ,0 ) (41)

for all x X

The rest of the proof is similar to the proof of theorem 5.

From proving the theorems we have consequences:

Corollary 3. Let r > 1 and θ be nonnegative real numbers and let Γ : X Y be a mapping satisfy Γ ( 0 ) = 0 and

2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) f ( j = 1 k x j + y j 2 ) f ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + θ ( j = 1 k x j r + j = 1 k y j r + j = 1 k z j r ) (42)

for all x j X .

Then there exists a unique mapping ψ : X Y such that

f ( x ) ψ ( x ) Y 2.2 r θ 2 r 2 x X r (43)

for all x X

Corollary 4. Let r < 1 and θ be non-negative real numbers and let Γ : X Y be a mapping satisfy f ( 0 ) = 0 and

2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) f ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + θ ( j = 1 k x j r + j = 1 k y j r + j = 1 k z j r ) (44)

for all x j , y j , z j X .

Then there exists a unique mapping ψ : X Y such that

Γ ( x ) ψ ( x ) Y 2.2 r θ 2 2 r x X r (45)

for all x X

5. Establish the Solution of the Cauchy Additive (ξ12)-Function Inequalities Using a Fixed Point Method

Now, we first study the solutions of (2). Note that for these inequalities, when X be a real or complete normed space and Y complex Banach space.

Lemma 7. Suppose mapping Γ : X Y satisfies Γ ( 0 ) = 0 and

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) f ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( 2 f ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y (46)

for all x j , y j , z j X , j = 1 k if and only if Γ : X Y is Cauchy additive

Proof. Assume that Γ : X Y satisfies (46)

( x 1 , , x k , y 1 , , y k , z 1 , , z k ) by ( x , ,0, x , ,0, x , ,0 ) in (46) we have

Γ ( 2 x ) 2 Γ ( x ) Y | β 1 | Γ ( 2 x ) 2 Γ ( x ) Y

and so Γ ( 2 x ) = 2 Γ ( x ) for all x X .

Thus

Γ ( x 2 ) = 1 2 Γ ( x ) (47)

for all x X

It follows from (46) and (47) that

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 k j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( 2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) f ( j = 1 k z j ) ) Y = ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 k j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y (48)

and so

( 1 | ξ 2 | ) Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y | ξ 1 | Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) Y (49)

we let u = j = 1 k x j + y j 2 + j = 1 k z j , v = j = 1 k x j + y j 2 j = 1 k z j , for all j = 1 k ,

we get

( 1 | ξ 2 | ) Γ ( u ) Γ ( u + v 2 ) Γ ( u v 2 ) | ξ 1 | Γ ( u ) + Γ ( v ) 2 Γ ( u + v 2 ) Y (50)

for all u , v X

and so

1 2 ( 1 | ξ 2 | ) Γ ( u + v ) + Γ ( u v ) 2 Γ ( u ) Y | ξ 1 | Γ ( u + v ) Γ ( u ) f ( v ) Y (51)

for all u , v X It follows from (49) and (51) that

1 2 ( 1 | ξ 2 | ) 2 Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) f ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y | ξ 1 | 2 Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y (52)

Since 2 | ξ 1 | + | ξ 2 | < 1

and so

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) = Γ ( j = 1 k x j + y j 2 ) + Γ ( j = 1 k z j )

for all x j , y j , z j X , j = 1 k . Thus f is Cauchy additive. □

The rest of the proof is similar to the proof of Lemma 2.

Theorem 8. Suppose φ : X 3 n [ 0, ) be a function such that there exists an L < 1 with

φ ( x 1 2 , , x n 2 , y 1 2 , , y n 2 , z 1 2 , , z n 2 ) L 2 φ ( x 1 , , x n , y 1 , , y n , z 1 , , z n ) (53)

for all x j , y j , z j X , j = 1 k . If f : X Y be a mapping satisfy Γ ( 0 ) = 0 and

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + f ( j = 1 k x j + y j 2 j = 1 k z j ) 2 f ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( 2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + φ ( x 1 , , x n , y 1 , , y n , z 1 , , z n ) (54)

for all x j , y j , z j X , j = 1 k .

Then there exists a unique mapping ψ : X Y such that

f ( x ) ψ ( x ) Y L 2 ( 1 L ) ( 1 | ξ 1 | ) φ ( x , 0 , ,0, x , ,0, x , ,0 ) (55)

for all x X

Proof. Replacing ( x 1 , x 2 , , x n , y 1 , y 2 , , y n , z 1 , z 2 , , z n ) by ( x , 0 , ,0, x , ,0,0, ,0 ) in (54), we get

( 1 | ξ 1 | ) Γ ( 2 x ) 2 Γ ( x ) Y φ ( x , 0 , ,0, x , ,0, x , ,0 ) (56)

for all x X .

Consider the set

S : = { h : X Y , h ( 0 ) = 0 }

and introduce the generalized metric on S :

d ( g , h ) : = i n f { λ : g ( x ) h ( x ) λ φ ( x , 0 , ,0, x , ,0, x , ,0 ) , x X } ,

where, as usual, i n f ϕ = + . It easy to show that ( S , d ) is complete [17] Now we cosider the linear mapping J : S S such that

J g ( x ) : = 2 g ( x 2 )

for all x X . Let g , h S be given such that d ( g , h ) = ε then

g ( x ) h ( x ) ε φ ( x , 0 , ,0, x , ,0, x , ,0 )

for all x X .

Hence

J g ( x ) J h ( x ) = 2 g ( x 2 ) 2 h Γ ( x 2 ) 2 ε φ ( x 2 ,0, ,0, x 2 , ,0, x 2 , ,0 ) 2 ε L 2 φ ( x , 0 , ,0, x , ,0, x , ,0 ) L ε φ ( x , 0 , ,0, x , ,0, x , ,0 )

for all x X . So d ( g , h ) = ε implies that d ( J g , J h ) L ε . This means that

d ( J g , J h ) L d ( g , h )

for all g , h X It follows from (56) that

Γ ( x ) 2 Γ ( x 2 ) 1 1 | ξ 1 | φ ( x 2 ,0, ,0, x 2 , ,0, x 2 , ,0 ) L 2 ( 1 | ξ 1 | ) φ ( x , 0 , ,0, x , ,0, x , ,0 )

for all x X . So d ( Γ , J Γ ) L 2 ( 1 | ξ 1 | ) for all x X By Theorem 2.3, there exists a mapping ψ : X Y satisfying the following:

1) ψ is a fixed point of J, i.e.,

ψ ( x ) = 2 ψ ( x 2 ) (57)

for all x X . The mapping ψ is a unique fixed point J in the set

M = { g S : d ( f , g ) < }

This implies that ψ is a unique mapping satisfying (57) such that there exists a λ ( 0, ) satisfying

Γ ( x ) ψ ( x ) λ φ ( x , 0 , ,0, x , ,0, x , ,0 )

for all x X

2) d ( J l Γ , ψ ) 0 as l . This implies equality

l i m l 2 n Γ ( x 2 n ) = ψ ( x )

for all x X

3) d ( Γ , ψ ) 1 1 L d ( Γ , J Γ ) . which implies

Γ ( x ) ψ ( x ) L 2 ( 1 L ) ( 1 | ξ 1 | ) φ ( x , 0 , ,0, x , ,0, x , ,0 )

for all x X . □

The rest of the proof is similar to the proof of Theorem 3.

Theorem 9. Let φ : X 3 n [ 0, ) be a function such that there exists an L < 1 with

φ ( x 1 , , x n , y 1 , , y n , z 1 , , z n ) 2 L φ ( x 1 2 , , x n 2 , y 1 2 , , y n 2 , z 1 2 , , z n 2 ) (58)

for all x , y , z X . Let Γ : X Y be a mapping satisfy Γ ( 0 ) = 0 and

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( 2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + φ ( x 1 , , x n , y 1 , , y n , z 1 , , z n ) (59)

for all x j , y j , z j X , j = 1 k .

Then there exists a unique mapping ψ : X Y such that

Γ ( x ) ψ ( x ) Y L 2 ( 1 L ) ( 1 | ξ 1 | ) φ ( x , 0 , ,0, x , ,0, x , ,0 ) (60)

for all x X

The rest of the proof is similar to the proof of Theorem 8.

From proving the theorems we have consequences:

Corollary 5. Let r > 1 and θ be nonnegative real numbers and let Γ : X Y be a mapping satisfy Γ ( 0 ) = 0 and

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( 2 f ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y x i + θ ( j = 1 k x j X r + j = 1 k y j X r + j = 1 k z j X r ) (61)

for all x j , y j , z j X , j = 1 k .

Then there exists a unique mapping ψ : X Y such that

Γ ( x ) ψ ( x ) Y 3 θ ( 2 r 2 ) ( 1 | ξ 1 | ) x X r (62)

for all x X

Corollary 6. Let r < 1 and θ be nonnegative real numbers and let Γ : X Y be a mapping satisfy Γ ( 0 ) = 0 and

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( 2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + θ ( j = 1 k x j X r + j = 1 k y j X r + j = 1 k z j X r ) (63)

for all x j , y j , z j X , j = 1 k .

Then there exists a unique mapping ψ : X Y such that

f ( x ) ψ ( x ) Y 3 θ ( 2 2 r ) ( 1 | ξ 1 | ) x X r (64)

for all x X .

6. Establish the Solution of the Additive (ξ12)-Function Inequalities Using a Direect Method

Next, we study the solutions of (2). Note that for these inequalities, when X be a real or complete normed space and Y complex Banach space.

Theorem 10. Suppose φ : X 3 k [ 0, ) be a function such that

ϕ ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) : = j = 1 2 j φ ( x 1 2 j , , x k 2 j , y 1 2 j , , y k 2 j , z 1 2 j , , z k 2 j ) < (65)

for all x j , y j , z j X , j = 1 k and let Γ : X Y be a mapping satisfies Γ ( 0 ) = 0 and

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( 2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + φ ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) (66)

for all x j , y j , z j X , j = 1 k .

Then there exists a unique mapping ψ : X Y such that

Γ ( x ) ψ ( x ) Y ϕ ( x , ,0, x , ,0, x , ,0 ) (67)

for all x X

Proof. Replacing ( x 1 , x 2 , , x k , y 1 , y 2 , , y k , z 1 , z 2 , , z k ) by ( x , ,0, x , ,0, x , 0, ,0 ) in (66), we get

( 1 | ξ 1 | ) Γ ( 2 x ) 2 Γ ( x ) Y φ ( x , 0 , ,0, x , ,0, x ,0 , ,0 ) (68)

for all x X .

So

Γ ( x ) 1 2 Γ ( 2 x ) Y 1 2 ( 1 | ξ 1 | ) φ ( x , 0 , ,0, x , ,0, x , ,0 ) (69)

for all x X .

Hence

1 2 l Γ ( 2 l x ) 1 2 m Γ ( 2 m x ) Y j = l m 1 2 j f ( x 2 j ) 2 j + 1 Γ ( x 2 j + 1 ) Y j = l m 1 2 j + 1 2 ( 1 | ξ 1 | ) φ ( x 2 j + 1 ,0, , 0, x 2 j + 1 ,0, ,0, x 2 j + 1 , ,0 ) (70)

for all nonnegative integers m and l with m > l and all x X . It follows from (70) that the sequence { 2 n Γ ( x 2 n ) } is a Cauchy sequence for all x X . Since Y is complete, the sequence { 2 n Γ ( x 2 n ) } converges. So one can define the mapping ψ : X Y by

ψ ( x ) : = l i m n 2 n Γ ( x 2 n x ) (71)

for all x X . Moreover, letting l = 0 and passing the limit m in (37), we get (67) It follows from (65) and (66) that

ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) Y = l i m n 2 n Γ ( j = 1 k x j + y j 2 n + 1 + 1 2 n j = 1 k z j ) Γ ( j = 1 k x j + y j 2 n + 1 ) Γ ( 1 2 n j = 1 k z j ) Y l i m n 2 n | ξ 1 | Γ ( j = 1 k x j + y j 2 n + 1 + 1 2 n j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 n + 1 1 2 n j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 n + 1 ) Y + l i m n 2 n | ξ 2 | 2 Γ ( j = 1 k x j + y j 2 n + 2 + 1 2 n + 1 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 n + 1 ) Γ ( 1 2 n j = 1 k z j ) Y + l i m n 2 n φ ( x 1 2 n , , x k 2 n , y 1 2 n , , y k 2 n , z 1 2 n , , z k 2 n )

= ξ 1 ( ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) + ψ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 ψ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( 2 ψ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) ) Y (72)

for all x j , y j , z j X , j = 1 k . So

ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) Y ξ 1 ( ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) + ψ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 ψ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( 2 ψ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) ψ ( j = 1 k x j + y j 2 ) ψ ( j = 1 k z j ) ) Y

for all x j , y j , z j X , j = 1 n . By Lemma 5.1, the mapping ψ : X Y is additive. Ei

ψ ( j = 1 k x j + y j 2 + j = 1 k z j ) = ψ ( j = 1 k x j + y j 2 ) + ψ ( j = 1 k z j )

Now, let ψ : X Y be another additive mapping satisfying (67). Then we have

ψ ( x ) ψ ( x ) = 2 q ψ ( x 2 q ) 2 q ψ ( x 2 q ) Y 2 q ψ ( x 2 q ) 2 q Γ ( x 2 q ) + 2 q ψ ( x 2 q ) 2 q Γ ( x 2 q ) Y 2 q ϕ ( x 2 q ,0, ,0, x 2 q ,0, ,0, x 2 q 0, ,0 )

which tends to zero as q for all x X . So we can conclude that ψ ( x ) = ψ ( x ) for all x X . This proves the uniqueness of ψ . □

Theorem 11. Suppose φ : X 3 k [ 0, ) be a function such that

ψ ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) : = j = 0 1 2 j φ ( 2 j x 1 , , 2 j x k , 2 j y 1 , , 2 j y k , 2 j z 1 , , 2 j z k ) < (73)

for all x j , y j , z j X , j = 1 k and let Γ : X Y be a mapping satisfies Γ ( 0 ) = 0 and

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y β 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + β 2 ( 2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + φ ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) (74)

for all x j , y j , z j X , j = 1 k .

Then there exists a unique mapping ψ : X Y such that

Γ ( x ) ψ ( x ) Y ϕ ( x , 0 , ,0, x , ,0, x ,0 , ,0 ) (75)

for all x X

Proof. Replacing ( x 1 , x 2 , , x k , y 1 , y 2 , , y k , z 1 , z 2 , , z k ) by ( x , ,0, x , ,0, x , ,0 ) in (74), we get

( 1 | ξ 1 | ) Γ ( 2 x ) 2 Γ ( x ) Y φ ( x , 0 , ,0, x , ,0, x , ,0 ) (76)

for all x X .

So Replacing ( x 1 , , x k , y 1 , , y k , z 1 , , z k ) by ( x , 0 , ,0, x ,0 , ,0, x ,0 , ,0 ) in (74), we get

2 Γ ( x 2 ) Γ ( x ) Y φ ( x , 0 , ,0, x ,0 , ,0, x ,0 , ,0 ) (77)

for all x X . So

Γ ( x ) 1 2 Γ ( 2 x ) Y 1 2 φ ( 2 x ,0, ,0,2 x ,0, ,0,2 x ,0, ,0 ) (78)

for all x X . Hence

1 2 l Γ ( 2 l x ) 1 2 m Γ ( 2 m x ) Y j = l m 1 1 2 j Γ ( 2 j x ) 1 2 j + 1 Γ ( 2 j + 1 x ) Y j = l m 1 1 2 j + 1 φ ( 2 j + 1 x ,0, ,0,2 j + 1 x ,0, ,0,2 j + 1 x ,0, ,0 ) (79)

for all nonnegative integers m and l with m > l and all x X . It follows from (79) that the sequence { 1 2 n Γ ( 2 n x ) } is a Cauchy sequence for all x X . Since Y is complete, the sequence { 1 2 n Γ ( 2 n x ) } converges. So one can define the mapping ψ : X Y by

ψ ( x ) : = lim n 1 2 n Γ ( 2 n x ) (80)

for all x X . Moreover, letting l = 0 and passing the limit m in (79), we get (75).

The rest of the proof is similar to the proof of theorem 10. □

From proving the theorems we have consequences:

Corollary 7. Let r > 1 and θ be nonnegative real numbers and let Γ : X Y be a mapping satisfy Γ ( 0 ) = 0 and

Γ ( j = 1 k x j + y j 2 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( 2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + θ ( j = 1 k x j r + j = 1 k y j r + j = 1 k z j r ) (81)

for all x j X .

Then there exists a unique mapping ψ : X Y such that

Γ ( x ) ψ ( x ) Y 3 θ ( 2 r 2 ) ( 1 | ξ 1 | ) x X r (82)

for all x X

Corollary 8. Let r < 1 and θ be nonnegative real numbers and let Γ : X Y be a mapping satisfy Γ ( 0 ) = 0 and

Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) Y ξ 1 ( Γ ( j = 1 k x j + y j 2 + j = 1 k z j ) + Γ ( j = 1 k x j + y j 2 j = 1 k z j ) 2 Γ ( j = 1 k x j + y j 2 ) ) Y + ξ 2 ( 2 Γ ( j = 1 k x j + y j 4 + 1 2 j = 1 k z j ) Γ ( j = 1 k x j + y j 2 ) Γ ( j = 1 k z j ) ) Y + θ ( j = 1 k x j r + j = 1 k y j r + j = 1 k z j r ) (83)

for all x j , y j , z j X .

Then there exists a unique mapping ψ : X Y such that

Γ ( x ) ψ ( x ) Y 3 θ ( 2 2 r ) ( 1 | ξ 1 | ) x X r (84)

for all x X

7. Conclusion

In this paper, I have given two functional inequalities with 3k variables and fully solved complex Banach space by fixed point methods and direct methods. This result is based on special results such as [25] [26].

Conflicts of Interest

The author declares no conflicts of interest.

References

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