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 (ξ₁,ξ₂)-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 (ξ₁,ξ₂)-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 (ξ₁,ξ₂)-functional inequalities (1) and (2), i.e., the (ξ₁,ξ₂)-functional inequalities with n-variables. Under suitable assumptions on spaces X and Y , we will prove that the mappings satisfy the (ξ₁,ξ₂)-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 (ξ₁,ξ₂)-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 (ξ₁,ξ₂)-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 (ξ₁,ξ₂)-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.

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 ∈ ℝ : ‖ 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 .

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

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 ∈ ℝ : ‖ 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 .

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

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].

The author declares no conflicts of interest.

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: e9480. https://doi.org/10.4236/oalib.1109480