1. Introduction

Let X and Y be normed spaces on the same field K , and f : X → Y . I use the notations ‖ ⋅ ‖ X , ‖ ⋅ ‖ Y as the normals on X and the normals on Y , respectively. In this paper, I investigate some additive-quadratic η-functional inequalities in ( α 1 , α 2 ) -homogeneous complex Banach spaces.

In fact, when X is a α 1 -homogeneous real or complex normed spaces ‖ ⋅ ‖ X and that Y is a α 2 -homogeneous real or complex Banach spaces ‖ ⋅ ‖ Y

I solve and prove the Hyers-Ulam-Rassias type stability of two following additive-quadratic η-functional inequalities.

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ‖ Y ≤ ‖ h ( η ) ( 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) ) ‖ Y (1)

and when I change the role of the function inequality (1.1), I continue to prove the following function inequality.

‖ 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) ‖ Y ≤ ‖ h ( η ) ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ) ‖ Y (2)

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 ) (3)

The Hyers-Ulam stability was the first investigated for the functional equation of Ulam in [1] concerning the stability of group homomorphisms.

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 ageneralization of the Rassias theorem was obtained by Găvruta [5] with replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias’ approach.

The Hyers-Ulam stability for functional inequalities has been investigated such as Gilányi [6] showed that is if satisfies the functional inequality.

‖ 2 f ( x ) + 2 f ( y ) − f ( x − y ) ‖ ≤ ‖ f ( x + y ) ‖ (4)

Then f satisfies the Jordan-von Newman functional equation.

2 f ( x ) + 2 f ( y ) = f ( x + y ) + f ( x − y ) (5)

Gilányi [7] and Fechner [8] proved the Hyers-Ulam stability of the functional inequality (4).

Next Chookil [9] and [10] proved the of additive β-functional inequalities in non-Archimedean Banach spaces and in complex Banach spaces, and Harin Lee^a [11] [12] [13] proved the Hyers-Ulam stability of additive β-functional inequalities in ρ-homogeneous F space.

Recently, the author has studied the additive-quadratic functional inequalities of mathematicians around the world, on spaces complex Banach spaces, non-Archimedan Banach spaces or additive β-functional inequalities in p-homogeneous F-space.... See [14] - [19] .

So in this paper, I solve and prove the Hyers-Ulam stability for two additive-quadratic η-functional inequalities (1)-(2), i.e. the additive-quadratic η-functional inequalities with 3k-variables. Under suitable assumptions on spaces X and Y, I will prove that the mappings satisfy the additive-quadratic η-functional inequalities (1) or (2). Thus, the results in this paper are a generalization of those in [14] - [20] for additive-quadratic η-functional inequalities with 3k-variables.

In this paper, I have constructed a general quadratic linear functional inequality to improve the classical linear linear inequality. This problem I think is one outstanding development for the mathematics industry modern studies in the field of functional equations in particular and mathematics in general. I would like to express my gratitude to the senior mathematicians [1] - [24] who have inspired today’s mathematics researchers.

The paper is organized as follows: In section preliminariers, I remind a basic property such as I only redefine the solution definition of the equations of the additive function, the equations of the quadratic function and F * -space.

Section 3: Constructing solution to the quadratic η-functional inequalities (1) in ( α 1 , α 2 ) -homogeneous complex Banach spaces.

Section 4: Constructing solution to the quadratic η-functional inequalities (2) in ( α 1 , α 2 ) -homogeneous complex Banach spaces.

Section 5: Constructing solution to the additive η-functional inequalities (1) in ( α 1 , α 2 ) -homogeneous complex Banach spaces.

Section 6: Constructing solution to the additive η-functional inequalities (2) in ( α 1 , α 2 ) -homogeneous complex Banach spaces.

2. Preliminaries2.1. F ∗ -Spaces

Let X be a (complex) linear space. A nonnegative valued function ‖ ⋅ ‖ is an F-norm if it satisfies the following conditions:

1. ‖ x ‖ = 0 if and only if x = 0 ;

2. ‖ λ x ‖ = ‖ x ‖ for all x ∈ X and all λ with | λ | = 1 ;

3. ‖ x + y ‖ ≤ ‖ x ‖ + ‖ y ‖ for all x , y ∈ X ;

4. ‖ λ n x ‖ → 0 , λ n → 0 ;

5. ‖ λ n x ‖ → 0 , x n → 0 .

Then ( X , ‖ ⋅ ‖ ) is called an F * -space. An F-space is a complete F * -space. An F-norm is called β-homgeneous ( β > 0 ) if ‖ t x ‖ = | t | β ‖ x ‖ for all x ∈ X and for all t ∈ ℂ and ( X , ‖ ⋅ ‖ ) is called α-homogeneous F-space.

2.2. Solutions of the Inequalities

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

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 functional equation: f ( x + y 2 ) = 1 2 f ( x ) + 1 2 f ( y ) is called the Jensen equation. In particular, every solution of the Jensen equation is said to be a Jensen mapping.

The functional equation: f ( x + y 2 ) + f ( x − y 2 ) = 1 2 f ( x ) + 1 2 f ( y ) is called the Jensen type qudratic equation. In particular, every solution of the quadratic equation is said to be a Jensen type quadratic mapping.

D = { φ : ℂ → ℂ : g ( η ) = η , | g ( η ) | = | η | ≤ 1 2 } (6)

Note: With k is a positive integer and h ∈ A α 1 , α 2 ∈ ℝ + , α 1 , α 2 ≤ 1 .

3. Constructing Solution to the η-Functional Inequalities (2) in ( α 1 , α 2 ) -Homogeneous Complex Banach Spaces

Now, I first study the solutions of (1). Note that for these inequalities, when X is a α 1 -homogeneous real or complex normed spaces ‖ ⋅ ‖ X and that Y is a α 2 -homogeneous real or complex Banach spaces ‖ ⋅ ‖ Y . Under this setting, I can show that the mapping satisfying (1) is quadratic. These results are given in the following.

Lemma 1 Let f : X → Y be an even mapping satilies:

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X if and only if f : X → Y is quadratic.

Proof. Assume that f : X → Y satisfies (7).

I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (7), I have: ‖ ( 4 k − 2 ) f ( 0 ) ‖ Y ≤ ‖ η f ( 0 ) ‖ Y ≤ 0 therefore, ( | 4 k − 2 | α 2 − | η | α 2 ) ‖ f ( 0 ) ‖ Y ≤ 0

So f ( 0 ) = 0 .

Next replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , ⋯ , k x , k x , ⋯ , k x , x , ⋯ , x ) in (7), I have.

Thus ‖ f ( 2 k x ) − 4 k f ( x ) ‖ Y ≤ 0

f ( x 2 k ) = 1 4 k f ( x ) (8)

for all x ∈ X .

From (7) and (8) I infer that:

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ‖ Y ≤ ‖ η ( 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) ) ‖ Y = | η | α 2 | 2 k | α 2 ‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ‖ Y (9)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X and so, f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) = 2 ∑ j = 1 k f ( x j + y j 2 k ) + 2 ∑ j = 1 k f ( z j ) for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X , as I expected. The couverse is obviously true. □

Corollary 1 Let f : X → Y be an even mapping satilies:

f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) = η ( 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) ) (10)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X if and only if f : X → Y is quadratic.

Note! The functional Equation (10) is called an quadratic η-functional equation.

Theorem 2 Assume for r > 2 α 2 α 1 , θ be nonngative real number, and suppose f : X → Y be an even mapping such that:

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ‖ Y ≤ ‖ η ( 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − 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 ) (11)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X . Then there exists a unique quadratic mapping ϕ : X → Y such that:

‖ f ( x ) − ϕ ( x ) ‖ Y ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 1 r − ( 4 k ) α 2 θ ‖ x ‖ X r (12)

for all x ∈ X .

Proof. Assume that f : X → Y satisfies (11).

I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (11), I have: ‖ ( 4 k − 2 ) f ( 0 ) ‖ Y ≤ ‖ 2 η f ( 0 ) ‖ Y ≤ 0 therefore, ( | 4 k − 2 | α 2 − | 2 η | α 2 ) ‖ f ( 0 ) ‖ Y ≤ 0

So f ( 0 ) = 0 .

Next replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , ⋯ , k x , k x , ⋯ , k x , x , ⋯ , x ) in (11) I have:

‖ f ( 2 x ) − 4 k f ( x ) ‖ Y ≤ ( 2 k α 1 r + 1 + 1 ) θ ‖ x ‖ X r (13)

for all x ∈ X . Thus,

‖ f ( x ) − 4 k f ( x 2 k ) ‖ Y ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 1 r θ ‖ x ‖ X r (14)

for all x ∈ X .

‖ ( 4 k ) l f ( x ( 2 k ) l ) − ( 4 k ) m f ( x ( 2 k ) m ) ‖ Y ≤ ∑ j = 1 m − 1 ‖ ( 4 k ) j f ( x ( 2 k ) j ) − ( 4 k ) j + 1 f ( x ( 2 k ) j + 1 ) ‖ Y ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 1 r θ ∑ j = 1 m − 1 ( 4 k ) α 2 j ( 2 k ) α 1 r j ‖ x ‖ X r (15)

for all nonnegative integers p, l with p > l and all x ∈ X . It follows from (15) that the sequence { ( 4 k ) n f ( x ( 2 k ) n ) } is a cauchy sequence for all x ∈ X . Since Y is complete, the sequence { ( 4 k ) n f ( x ( 2 k ) n ) } coverges.

So one can define the mapping ϕ : X → Y by, ϕ ( x ) : = lim n → ∞ ( 4 k ) n f ( x ( 2 k ) n ) for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (15), I get (12).

Form f : X → Y is even, the mapping ϕ : X → Y is even.

It follows from (11) that:

‖ ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + ϕ ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k ϕ ( x j + y j 2 k ) − ∑ j = 1 k ϕ ( z j ) − ∑ j = 1 k ϕ ( − z j ) ‖ Y = lim n → ∞ ( 4 k ) α 2 n ‖ f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k + 1 ( 2 k ) n ∑ j = 1 k z j ) + f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k − 1 ( 2 k ) n ∑ j = 1 k z j ) − ∑ j = 1 k f ( 1 ( 2 k ) n x j + y j 2 k ) − ∑ j = 1 k f ( 1 ( 2 k ) n z j ) − ∑ j = 1 k f ( − 1 ( 2 k ) n z j ) ‖ Y

≤ lim n → ∞ ( 4 k ) α 2 n | η | α 2 ‖ 2 f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k + 1 ( 2 k ) n ∑ j = 1 k z j ) + 2 f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k − 1 ( 2 k ) n ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( 1 ( 2 k ) n x j + y j 2 k ) + 1 2 k ∑ j = 1 k f ( − 1 ( 2 k ) n x j + y j 2 k ) − 1 2 ∑ j = 1 k f ( 1 ( 2 k ) n z j ) − 1 2 ∑ j = 1 k f ( − 1 ( 2 k ) n z j ) ‖ Y + lim n → ∞ ( 4 k ) α 2 n ( 2 k ) α 1 n r θ ( ∑ j = 1 k ‖ x j ‖ X r + ∑ j = 1 k ‖ y j ‖ X r + ∑ j = 1 k ‖ z j ‖ X r )

= | η | α 2 ‖ 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) ‖ Y (16)

for all x j , y j , z j ∈ X for all j = 1 → n .

for all x j , y j , z j ∈ X for j = 1 → n , So by lemma 3.1, it follows that the mapping ϕ : X → Y is additive. Now I need to prove uniqueness, Suppose ϕ ′ : X → Y is also a quadratic mapping that satisfies (12). Then I have:

‖ ϕ ( x ) − ϕ ′ ( x ) ‖ Y = ( 4 k ) α 2 n ‖ ϕ ( x ( 2 k ) n ) − ϕ ′ ( x ( 2 k ) n ) ‖ Y ≤ ( 4 k ) α 2 n ( ‖ ϕ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ Y + ‖ ϕ ′ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ Y ) ≤ 2 ⋅ ( 4 k ) α 2 n ⋅ ( 2 k α 1 r + 1 + 1 ) ( 2 k ) α 1 n r ( ( 2 k ) α 1 r − ( 4 k ) α 2 ) θ ‖ x ‖ X r (18)

which tends to zero as n → ∞ for all x ∈ X . So I can conclude that ϕ ( x ) = ϕ ′ ( x ) for all x ∈ X . This proves thus the mapping ϕ : X → Y is a unique mapping satisfying(12) as I expected.

Theorem 3 Assume for r < 2 α 2 α 1 , θ be nonngative real number, and Suppose f : X → Y be an even mapping satisfiying (11). Then there exists a unique quadratic mapping ϕ : X → Y such that:

‖ f ( x ) − ϕ ( x ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 4 k ) α 2 − ( 2 k ) α 1 r θ ‖ x ‖ r (19)

for all x ∈ X .

The proof is similar to the proof of theorem 3.3.

4. Constructing Solution to the η-Functional Inequalities (2) in ( α 1 , α 2 ) -Homogeneous Complex Banach Spaces

Now, I study the solutions of (2). Note that for these inequalities, when X is a α 1 -homogeneous complex Banach spaces and that Y is a α 2 -homogeneous complex Banach spaces.

Under this setting, I can show that the mapping satisfying (2) is quadratic. These results are given in the following.

Lemma 4 Let f : X → Y be an even mapping satilies f ( 0 ) = 0 and:

‖ 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) ‖ Y ≤ ‖ η ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ) ‖ Y (20)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X if and only if f : X → Y is quadratic.

Proof. Assume that f : X → Y satisfies (20).

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 2 k x , ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (20), I have.

Thus ‖ 4 f ( x 2 k ) − 1 k f ( x ) ‖ Y ≤ 0

f ( x 2 k ) = 1 4 k f ( x ) (21)

for all x ∈ X .

From (20) and (21) I infer that:

1 ( 2 k ) α 2 ‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ‖ Y = ‖ 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) ‖ Y ≤ | η | α 2 ‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ‖ Y (22)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X and so: f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) = 2 ∑ j = 1 k f ( x j + y j 2 k ) + 2 ∑ j = 1 k f ( z j ) for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X , as I expected. The couverse is obviously true.

Let f : X → Y be an even mapping satilies,

2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) = η ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ) (23)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X if and only if f : X → Y is quadratic Note! The functional Equation (23) is called an quadratic λ-functional equation.

Theorem 5 Assume for r > 2 α 2 α 1 , θ be nonngative real number, and suppose f : X → Y be a mapping such that f ( 0 ) = 0 and

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X . Then there exists a unique quadratic mapping ϕ : X → Y such that:

‖ f ( x ) − ϕ ( x ) ‖ Y ≤ ( 2 k ) α 1 r ( 2 k ) α 1 r − ( 4 k ) α 2 θ ‖ x ‖ X r (25)

for all x ∈ X .

Proof. Assume that f : X → Y satisfies (24).

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 2 k x , ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (24) I have:

‖ 4 f ( x 2 k ) − 1 k f ( x ) ‖ Y ≤ ( 2 k ) α 1 r θ ‖ x ‖ X r (26)

for all x ∈ X . Thus

‖ 4 k f ( x 2 k ) − f ( x ) ‖ ≤ ( 2 k ) α 1 r k α 2 θ ‖ x ‖ X r (27)

for all x ∈ X .

‖ ( 4 k ) l f ( x ( 2 k ) l ) − ( 4 k ) m f ( x ( 2 k ) m ) ‖ ≤ ∑ j = 1 m − 1 ‖ ( 4 k ) j f ( x ( 2 k ) j ) − ( 4 k ) j + 1 f ( x ( 2 k ) j + 1 ) ‖ Y ≤ ( 2 k ) α 1 r k α 2 θ ∑ j = 1 m − 1 ( 4 k ) α 2 j ( 2 k ) α 1 r j ‖ x ‖ X r (28)

for all nonnegative integers p, l with p > l and all x ∈ X . It follows from (28) that the sequence { ( 4 k ) n f ( x ( 2 k ) n ) } is a cauchy sequence for all x ∈ X . Since Y is complete, the sequence { ( 4 k ) n f ( x ( 2 k ) n ) } coverges.

So one can define the mapping ϕ : X → Y by ϕ ( x ) : = lim n → ∞ ( 4 k ) n f ( x ( 2 k ) n ) for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (28), I get (25). Form f : X → Y is even, the mapping: ϕ : X → Y is even. It follows from (24) I have:

‖ 2 ϕ ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 ϕ ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k ϕ ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k ϕ ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k ϕ ( z j ) − 1 2 k ∑ j = 1 k ϕ ( − z j ) ‖ Y = lim n → ∞ ( 4 k ) α 2 n ‖ 2 f ( ∑ j = 1 k x j + y j ( 2 k ) n + 2 + 1 ( 2 k ) n + 1 ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) n + 2 − 1 ( 2 k ) n + 1 ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j ( 2 k ) n + 1 ) + 1 2 k ∑ j = 1 k f ( − x j + y j ( 2 k ) n + 1 ) − 1 2 k ∑ j = 1 k f ( z j ( 2 k ) n ) − 1 2 k ∑ j = 1 k f ( − z j ( 2 k ) n ) ‖ Y

≤ lim n → ∞ ( 4 k ) α 2 n | η | α 2 ‖ 2 f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k + 1 ( 2 k ) n ∑ j = 1 k z j ) + 2 f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k − 1 ( 2 k ) n ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( 1 ( 2 k ) n x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( 1 ( 2 k ) n z j ) − 1 2 k ∑ j = 1 k f ( − 1 ( 2 k ) n z j ) ‖ Y + lim n → ∞ ( 4 k ) α 2 n ( 2 k ) α 1 n r θ ( ∑ j = 1 k ‖ x j ‖ X r + ∑ j = 1 k ‖ y j ‖ X r + ∑ j = 1 k ‖ z j ‖ X r )

= ‖ ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + ϕ ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k ϕ ( x j + y j 2 k ) − ∑ j = 1 k ϕ ( z j ) − ∑ j = 1 k ϕ ( − z j ) ‖ Y (29)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X .

‖ 2 ϕ ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 ϕ ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k ϕ ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k ϕ ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k ϕ ( z j ) − 1 2 k ∑ j = 1 k ϕ ( − z j ) ‖ Y ≤ ‖ η ( ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + ϕ ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k ϕ ( x j + y j 2 k ) − ∑ j = 1 k ϕ ( z j ) − ∑ j = 1 k ϕ ( − z j ) ) ‖ Y

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X , So by lemma 4.1 it follows that the mapping ϕ : X → Y is quadratic. Now I need to prove uniqueness, Suppose ϕ ′ : X → Y is also a quadratic mapping that satisfies (25). Then I have:

‖ ϕ ( x ) − ϕ ′ ( x ) ‖ Y = ( 4 k ) α 2 n ‖ ϕ ( x ( 2 k ) n ) − ϕ ′ ( x ( 2 k ) n ) ‖ Y ≤ ( 4 k ) α 2 n ( ‖ ϕ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ Y + ‖ ϕ ′ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ Y ) ≤ 2 ⋅ ( 4 k ) α 2 n ⋅ ( 2 k ) α 1 r ( 2 k ) α 1 n r ( ( 2 k ) α 1 r − ( 4 k ) α 2 ) θ ‖ x ‖ X r (30)

which tends to zero as n → ∞ for all x ∈ X . So I can conclude that ϕ ( x ) = ϕ ′ ( x ) for all x ∈ X .This proves thus the mapping ϕ : X → Y is a unique mapping satisfying(25) as I expected. □

Theorem 6 Assume for r < 2 α 2 α 1 , θ be nonngative real number, f ( 0 ) = 0 and suppose f : X → Y be an odd mapping (24). Then there exists a unique quadratic mapping ϕ : X → Y such that:

‖ f ( x ) − ϕ ( x ) ‖ Y ≤ ( 2 k ) α 1 r ( 4 k ) α 2 − ( 2 k ) α 1 r θ ‖ x ‖ X r (31)

for all x ∈ X .

The proof is similar to the proof of theorem 4.3.

5. Constructing Solution to the Additive η-Functional Inequalities (1) in ( α 1 , α 2 ) -Homogeneous Complex Banach Spaces

Now, I first study the solutions of (1). Note that for these inequalities, when X is a α 1 -homogeneous complex Banach spaces and that Y is a α 2 -homogeneous complex Banach spaces. Under this setting, I can show that the mapping satisfying (1) is additive. These results are given in the following.

Lemma 7 Let f : X → Y be an odd mapping satilies:

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X if and only if f : X → Y is additive.

Proof. Assume that f : X → Y satisfies (32).

I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (32), I have: ‖ ( 4 k − 2 ) f ( 0 ) ‖ Y ≤ | η | α 2 ‖ 5 f ( 0 ) ‖ Y ≤ 0 therefore f ( 0 ) = 0 .

Next replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , ⋯ , k x , k x , ⋯ , k x , x , ⋯ , x ) in (32), I have.

Thus ‖ f ( 2 k x ) − 2 k f ( x ) ‖ Y ≤ 0

f ( x 2 k ) = 1 2 k f ( x ) (33)

for all x ∈ X From (32) and (33) I infer that:

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ‖ Y ≤ ‖ η ( 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) ) ‖ Y = | η | α 2 | k | α 2 ‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ‖ Y (34)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X and so.

f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) = 2 ∑ j = 1 k f ( x j + y j 2 k ) (35)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X .

Next I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , ⋯ , k x , k x , ⋯ , k x , z , ⋯ , z ) in (35), I have

f ( k x + k z ) + f ( k x − k z ) = 2 k f ( x ) (36)

for all x , z ∈ X Now letting p = k x + k z , q = k x − k z when that in (36), I get

f ( p ) + f ( q ) = 2 k f ( p + q 2 k ) = 2 k ⋅ 1 2 k f ( p + q ) = f ( p + q ) (37)

for all p , q ∈ X . So f is an additive mapping. as I expected. The couverse is obviously true. □

Corollary 2 Let f : X → Y be an even mapping satilies:

f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) = η ( 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) ) (38)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X if and only if f : X → Y is additive.

Note! The functional Equation (38) is called an additive η-functional equation.

Theorem 8 Assume for r > α 2 α 1 , θ be nonngative real number, and suppose f : X → Y be an odd mapping such that:

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ‖ Y ≤ ‖ η ( 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − 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 ) (39)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X . Then there exists a unique additive mapping ϕ : X → Y such that:

‖ f ( x ) − ϕ ( x ) ‖ Y ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 1 r − ( 2 k ) α 2 θ ‖ x ‖ X r . (40)

for all x ∈ X .

Proof. Assume that f : X → Y satisfies (39). I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (39), I have: O ‖ ( 4 k − 2 ) f ( 0 ) ‖ Y ≤ ‖ 5 λ f ( 0 ) ‖ Y therefore, ( | 4 k − 2 | α 2 − | 5 λ | α 2 ) ‖ f ( 0 ) ‖ Y ≤ 0

So f ( 0 ) = 0 . Next replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , ⋯ , k x , k x , ⋯ , k x , x , ⋯ , x ) in (39) I have:

‖ f ( 2 k x ) − 2 k f ( x ) ‖ Y ≤ ( 2 k α 1 r + 1 + 1 ) θ ‖ x ‖ X r (41)

for all x ∈ X . Thus

‖ f ( x ) − 2 k f ( x 2 k ) ‖ Y ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 1 r θ ‖ x ‖ X r (42)

for all x ∈ X .

‖ ( 2 k ) l f ( x ( 2 k ) l ) − ( 2 k ) m f ( x ( 2 k ) m ) ‖ Y ≤ ∑ j = 1 m − 1 ‖ ( 2 k ) j f ( x ( 2 k ) j ) − ( 2 k ) j + 1 f ( x ( 2 k ) j + 1 ) ‖ Y ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 1 r θ ∑ j = 1 m − 1 ( 2 k ) α 2 j ( 2 k ) α 1 r j ‖ x ‖ X r (43)

for all nonnegative integers p, l with p > l and all x ∈ X . It follows from (15) that the sequence { ( 2 k ) n f ( x ( 2 k ) n ) } is a cauchy sequence for all x ∈ X . Since Y is complete, the sequence { ( 2 k ) n f ( x ( 2 k ) n ) } coverges.

So one can define the mapping ϕ : X → Y by ϕ ( x ) : = lim n → ∞ ( 2 k ) n f ( x ( 2 k ) n ) for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (15), I get (40).

Form f : X → Y is even, the mapping ϕ : X → Y is even.

It follows from (39) I have:

‖ ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + ϕ ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k ϕ ( x j + y j 2 k ) − ∑ j = 1 k ϕ ( z j ) − ∑ j = 1 k ϕ ( − z j ) ‖ Y = lim n → ∞ ( 2 k ) α 2 n ‖ f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k + 1 ( 2 k ) n ∑ j = 1 k z j ) + f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k − 1 ( 2 k ) n ∑ j = 1 k z j ) − ∑ j = 1 k f ( 1 ( 2 k ) n x j + y j 2 k ) − ∑ j = 1 k f ( 1 ( 2 k ) n z j ) − ∑ j = 1 k f ( − 1 ( 2 k ) n z j ) ‖ Y

≤ lim n → ∞ ( 2 k ) α 2 n | λ | α 2 ‖ 2 f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 ( 2 k ) n + 1 ∑ j = 1 k z j ) + 2 f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 ( 2 k ) n + 1 ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( 1 ( 2 k ) n x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( − 1 ( 2 k ) n x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( 1 ( 2 k ) n z j ) − 1 2 k ∑ j = 1 k f ( − 1 ( 2 k ) n z j ) ‖ Y + lim n → ∞ ( 2 k ) α 2 n ( 2 k ) α 1 n r θ ( ∑ j = 1 k ‖ x j ‖ X r + ∑ j = 1 k ‖ y j ‖ X r + ∑ j = 1 k ‖ z j ‖ X r )

= | λ | α 2 ‖ 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) ‖ Y (44)

for all x j , y j , z j ∈ X for all j = 1 → n .

for all x j , y j , z j ∈ X for j = 1 → n , So by lemma 5.1, it follows that the mapping ϕ : X → Y is additive. Now I need to prove uniqueness, suppose ϕ ′ : X → Y is also an additive mapping that satisfies (40). Then I have:

‖ ϕ ( x ) − ϕ ′ ( x ) ‖ Y = ( 2 k ) α 2 n ‖ ϕ ( x ( 2 k ) n ) − ϕ ′ ( x ( 2 k ) n ) ‖ Y ≤ ( 2 k ) α 2 n ( ‖ ϕ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ Y + ‖ ϕ ′ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ Y ) ≤ 2 ⋅ ( 2 k ) α 2 n ⋅ ( 2 k α 1 r + 1 + 1 ) ( 2 k ) α 1 n r ( ( 2 k ) α 1 r − ( 2 k ) α 2 ) θ ‖ x ‖ X r (45)

which tends to zero as n → ∞ for all x ∈ X . So I can conclude that ϕ ( x ) = ϕ ′ ( x ) for all x ∈ X . This proves thus the mapping ϕ : X → Y is a unique mapping satisfying(40) as I expected. □

Theorem 9 Assume for r < α 2 α 1 , θ be nonngative real number, and suppose f : X → Y be an odd mapping satisfying (1). Then there exists a unique additive mapping ϕ : X → Y such that:

‖ f ( x ) − ϕ ( x ) ‖ Y ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 2 − ( 2 k ) α 1 r θ ‖ x ‖ X r (46)

for all x ∈ X .

The rest of the proof is similar to the proof of Theorem 5.3.

6. Constructing Solution to the Additive η-Functional Inequalities (2) in ( α 1 , α 2 ) -Homogeneous Complex Banach Spaces

Now, I first study the solutions of (2). Note that for these inequalities, when X is a α 1 -homogeneous complex Banach spaces and that Y is a α 2 -homogeneous complex Banach spaces. Under this setting, I can show that the mapping satisfying (2) is additive. These results are given in the following.

Lemma 10 Let f : X → Y be an odd mapping satilies:

‖ 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) ‖ Y ≤ ‖ η ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ) ‖ Y (47)

for all x j , y j , z j ∈ X for j = 1 → n , if and only if f : X → Y is additive.

Proof. Assume that f : X → Y satisfies (47).

I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (20), I have: ‖ 2 k f ( 0 ) ‖ Y ≤ | η | α 2 ‖ ( 4 k − 2 ) f ( 0 ) ‖ Y

So f ( 0 ) = 0 .

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 2 k x , ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (47), I have.

Thus ‖ 4 k f ( x 2 k ) − 2 f ( x ) ‖ Y ≤ 0

f ( x 2 k ) = 1 2 k f ( x ) (48)

for all x ∈ X . From (47) and (48) I infer that:

1 | k | α 2 ‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ‖ Y = ‖ 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k f ( z j ) − 1 2 k ∑ j = 1 k f ( − z j ) ‖ Y ≤ | η | α 2 ‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k f ( z j ) − ∑ j = 1 k f ( − z j ) ‖ Y (49)

for all x j , y j , z j ∈ X for j = 1 → n , and so: f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) = 2 ∑ j = 1 k f ( x j + y j 2 k ) for all x j , y j , z j ∈ X for j = 1 → n , as I expected. The couverse is obviously true. □

Let f : X → Y be an even mapping satilies.

Theorem 11 Assume for r > α 2 α 1 , θ be nonngative real number, and suppose f : X → Y be a mapping such that f ( 0 ) = 0 and:

for all x j , y j , z j ∈ X for all j = 1 → n . Then there exists a unique additive mapping ϕ : X → Y such that:

‖ f ( x ) − ϕ ( x ) ‖ Y ≤ ( 2 k ) α 1 r ( 2 k ) α 1 r − ( 4 k ) α 2 θ ‖ x ‖ X r (51)

for all x ∈ X .

Proof. Assume that f : X → Y satisfies (50).

I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (50), I have: ‖ 2 f ( 0 ) ‖ Y ≤ | η | α 2 ‖ ( 4 k − 2 ) f ( 0 ) ‖ Y therefore, ( | 4 k − 2 | α 2 − | 2 η | α 2 ) ‖ f ( 0 ) ‖ Y ≤ 0

So f ( 0 ) = 0 .

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 2 k x , ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (50) I have:

‖ 4 f ( x 2 k ) − 1 k f ( x ) ‖ Y ≤ ( 2 k ) α 1 r θ ‖ x ‖ X r (52)

for all x ∈ X . Thus

‖ 4 k f ( x 2 k ) − f ( x ) ‖ Y ≤ ( 2 k ) α 1 r k α 2 θ ‖ x ‖ X r (53)

for all x ∈ X .

‖ ( 4 k ) l f ( x ( 2 k ) l ) − ( 4 k ) m f ( x ( 2 k ) m ) ‖ Y ≤ ∑ j = 1 m − 1 ‖ ( 4 k ) j f ( x ( 2 k ) j ) − ( 4 k ) j + 1 f ( x ( 2 k ) j + 1 ) ‖ Y ≤ ( 2 k ) α 1 r k α 2 θ ∑ j = 1 m − 1 ( 4 k ) α 2 j ( 2 k ) α 1 r j ‖ x ‖ X r (54)

for all nonnegative integers p, l with p > l and all x ∈ X . It follows from (54) that the sequence { ( 4 k ) n f ( x ( 2 k ) n ) } is a cauchy sequence for all x ∈ X . Since Y is complete, the sequence { ( 4 k ) n f ( x ( 2 k ) n ) } coverges.

So one can define the mapping ϕ : X → Y by ϕ ( x ) : = lim n → ∞ ( 4 k ) n f ( x ( 2 k ) n ) for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (28), I get (51). Form f : X → Y is even, the mapping ϕ : X → Y is even. It follows from (50) I have:

‖ 2 ϕ ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 ϕ ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k ϕ ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k ϕ ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k ϕ ( z j ) − 1 2 k ∑ j = 1 k ϕ ( − z j ) ) ‖ Y = lim n → ∞ ( 4 k ) α 2 n ‖ 2 f ( ∑ j = 1 k x j + y j ( 2 k ) n + 2 + 1 ( 2 k ) n + 1 ∑ j = 1 k z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) n + 2 − 1 ( 2 k ) n + 1 ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k f ( x j + y j ( 2 k ) n + 1 ) + 1 2 k ∑ j = 1 k f ( − x j + y j ( 2 k ) n + 1 ) − 1 2 k ∑ j = 1 k f ( z j ( 2 k ) n ) − 1 2 k ∑ j = 1 k f ( − z j ( 2 k ) n ) ‖ Y

= ‖ ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + ϕ ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k ϕ ( x j + y j 2 k ) − ∑ j = 1 k ϕ ( z j ) − ∑ j = 1 k ϕ ( − z j ) ) ‖ Y (55)

for all x j , y j , z j ∈ X for all j = 1 → n .

‖ 2 ϕ ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k z j ) + 2 ϕ ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k z j ) − 3 2 k ∑ j = 1 k ϕ ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k ϕ ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k ϕ ( z j ) − 1 2 k ∑ j = 1 k ϕ ( − z j ) ) ‖ Y ≤ ‖ λ ( ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) + ϕ ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k z j ) − 2 ∑ j = 1 k ϕ ( x j + y j 2 k ) − ∑ j = 1 k ϕ ( z j ) − ∑ j = 1 k ϕ ( − z j ) ) ‖ Y

for all x j , y j , z j ∈ X for j = 1 → n , So by lemma 6.1, it follows that the mapping ϕ : X → Y is quadratic. Now I need to prove uniqueness, suppose ϕ ′ : X → Y is also a quadratic mapping that satisfies (50). Then I have:

‖ ϕ ( x ) − ϕ ′ ( x ) ‖ Y = ( 4 k ) α 2 n ‖ ϕ ( x ( 2 k ) n ) − ϕ ′ ( x ( 2 k ) n ) ‖ Y ≤ ( 4 k ) α 2 n ( ‖ ϕ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ Y + ‖ ϕ ′ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ Y ) ≤ 2 ⋅ ( 4 k ) α 2 n ⋅ ( 2 k ) α 1 r ( 2 k ) α 1 n r ( ( 2 k ) α 1 r − ( 4 k ) α 2 ) θ ‖ x ‖ X r (56)

which tends to zero as n → ∞ for all x ∈ X . So I can conclude that ϕ ( x ) = ϕ ′ ( x ) for all x ∈ X . This proves thus the mapping ϕ : X → Y is a unique mapping satisfying(51) as I expected.

Theorem 12 Assume for r < α 2 α 1 , θ be nonngative real number, f ( 0 ) = 0 and suppose f : X → Y be an odd mapping satisfying (50). Then there exists a unique quadratic mapping ϕ : X → Y such that:

‖ f ( x ) − ϕ ( x ) ‖ Y ≤ ( 2 k ) α 1 r ( 4 k ) α 2 − ( 2 k ) α 1 r θ ‖ x ‖ X r . (57)

for all x ∈ X .

The proof is similar to theorem 6.2.

7. Conclusion

In the article, I developed the quadratic additivity η-function inequality with many variables on the complex ( α 1 , α 2 ) -homogeneous Banach space and showed that their solution is a quadratic additivity map. This is a remarkable idea for modern mathematics.

Conflicts of Interest

The author declares no conflicts of interest.

Cite this paper

An, L.V. (2023) Development the Additive-Quadratic η-Function Inequality with 3k-Variables Based on a General Quadratic Function Variables on a Complex Banach Spaces. Open Access Library Journal, 10: e10777. https://doi.org/10.4236/oalib.1110777

References1

Ulam, S.M. (1960) A Collection of the Mathematical Problems. Interscience Publ., New York.

Hyers, D.H. (1941) On the Stability of the Linear Functional Equation. Proceedings of the National Academy of Sciences of the United States of America, 27, 222-224. https://doi.org/10.1073/pnas.27.4.222

Aoki, T. (1950) On the Stability of the Linear Transformation in Banach Spaces. Journal of the Mathematical Society of Japan, 2, 64-66. https://doi.org/10.2969/jmsj/00210064

Rassias, Th.M. (1978) On the Stability of the Linear Mapping in Banach Spaces. Proceedings of the AMS, 72, 297-300. https://doi.org/10.1090/S0002-9939-1978-0507327-1

G&abreve;vruta, P. (1994) A Generalization of the Hyers-Ulam-Rassias Stability of Approximately Additive Mappings. Journal of Mathematical Analysis and Applications, 184, 431-436. https://doi.org/10.1006/jmaa.1994.1211

Gilányi, A. (2002) On a Problem by K. Nikodem. Mathematical Inequalities & Applications, 5, 707-710. https://doi.org/10.7153/mia-05-71

Gilányi, A. (2002) Eine zur parallelogrammleichung äquivalente ungleichung. Aequations, 5, 707-710. https://doi.org/10.7153/mia-05-71

Fechner, W. (2006) Stability of a Functional Inequalities Associated with the Jordan-von Neumann Functional Equation. Aequationes Mathematicae, 71, 149-161. https://doi.org/10.1007/s00010-005-2775-9

Lee, J.R., Park, C. and Shin, D.Y. (2014) Additive and Quadratic Functional in Equalities in Non-Archimedean Normed Spaces. International Journal of Mathematical Analysis, 8, 1233-1247. https://doi.org/10.12988/ijma.2014.44113

Lee, H., Cha, J.Y., Cho, M.W. and Kwon, M. (2016) Additive ρ-Functional Inequalities in β-Homogeneous F-Spaces. Journal of the Korean Society of Mathematical Education Series B-Pure and Applied Mathematics, 23, 319-328. https://doi.org/10.7468/jksmeb.2016.23.3.319

Park, C., Cho, Y. and Han, M. (2007) Functional Inequalities Associated with Jordan-von Newman-Type Additive Functional Equations. Journal of Inequalities and Applications, 2007, Article ID: 041820. https://doi.org/10.1155/2007/41820

Prager, W. and Schwaiger, J. (2013) A System of Two in Homogeneous Linear Functional Equations. Acta Mathematica Hungarica, 140, 377-406. https://doi.org/10.1007/s10474-013-0315-y

Park, C. (2014) Additive β-Functional Inequalities. Journal of Nonlinear Sciences and Applications, 7, 296-310. https://doi.org/10.22436/jnsa.007.05.02

Van An

,et al. (2020)Generalized Hyers-Ulam Type Stability of the Type Functional Equation with 2k-Variable in Non-Archimedean Space Asia Mathematika 5, 69- 83.

Van An

,et al. (2020)Generalized Hyers-Ulam Type Stability of the Additive Functional Equation Inequalities with 2n-Variables on an Approximate Group and Ring Homomorphism Asia Mathematika 4, 161-175.

Van An, L. (2021) Generalized Hyers-Ulam-Rassias Type Stability of the with 2k-Variable Quadratic Functional Inequalities in Non-Archimedean Banach Spaces and Banach Spaces. Asia Mathematika, 5, 69-83. https://doi.org/10.14445/22315373/IJMTT-V67I9P521

Van An

,et al. (2021)Generalized Hyers-Ulam Type Stability of the 2k-Variables Quadratic β-Functional Inequalities and Function in γ-Homogeneous Normed Space International Journal of Mathematics and Its Applications 9, 81-93.

Van An, L. (2023) Generalized Stability of Functional Inequalities with 3k-Variables Associated for Jordan-von Neumann-Type Additive Functional Equation. Open Access Library Journal, 10, e9681.

Van An, L. (2020) Generalized Hyers-Ulam Type Stability of the 2k-Variable Additive β-Functional Inequalities and Equations in Complex Banach Space. International Journal of Mathematics Trends and Technology, 66, 134-147. https://doi.org/10.14445/22315373/IJMTT-V66I7P518

Van An, L. (2022) Generalized Stability Additive λ-Functional Inequalities with 3k-Variable in α-Homogeneous F-Spaces. International Journal of Analysis and Applications, 20, 43. https://doi.org/10.28924/2291-8639-20-2022-43

Skof, F. (1983) Proprieta’ locali e approssimazione di operatori. Rendiconti del Seminario Matematico e Fisico di Milano, 53, 113-129. https://doi.org/10.1007/BF02924890

Cholewa, P.W. (1984) Remarks on the Stability of Functional Equations. Aequationes Mathematicae, 27, 76-86. https://doi.org/10.1007/BF02192660

Bahyrycz, A. and Piszczek, M. (2014) Hyers Stability of the Jensen Function Equation. Acta Mathematica Hungarica, 142, 353-365. https://doi.org/10.1007/s10474-013-0347-3

Balcerowski, M. (2013) On the Functional Equations Related to a Problem of Z Boros and Z. Dróczy. Acta Mathematica Hungarica, 138, 329-340. https://doi.org/10.1007/s10474-012-0278-4