Development of the Additive-Quadratic η-Function Inequality with 3k-Variables Based on a General Quadratic Function Variables on a Complex Banach Spaces

Ly Van An

Faculty of Mathematics Teacher Education, Tay Ninh University, Tay Ninh, Vietnam.

DOI: 10.4236/oalib.1110777                

Abstract

In this article, I study the establishment of the quadratic-additive η-function inequality with 3k-variables on the homogeneous complex Banach space and prove the quadratic-additive η-function equation related to the additive and quadratic η-functional inequalities in (α₁,α₂)-homogeneous Banach complex space.

Keywords

Additive-Quadratic η-Functional Inequalities, (α₁, α₂)-Homogeneous Complex Banach Spaces, Hyers-Ulam-Rassias Stability

1. Introduction

Let $X$ and $Y$ be normed spaces on the same field $\mathbb{K}$ , and $f\mathrm{<mo>\; :\; </mo>}X\to Y$ . I use the notations ${\Vert \cdot \Vert }_X$ , ${\Vert \cdot \Vert }_Y$ as the normals on $X$ and the normals on $Y$ , respectively. In this paper, I investigate some additive-quadratic η-functional inequalities in $\left({\alpha }_1\mathrm{,}{\alpha }_2\right)$ -homogeneous complex Banach spaces.

In fact, when $X$ is a ${\alpha }_1$ -homogeneous real or complex normed spaces ${\Vert \cdot \Vert }_X$ and that $Y$ is a ${\alpha }_2$ -homogeneous real or complex Banach spaces ${\Vert \cdot \Vert }_Y$

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

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert h\left(\eta \right)(2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\end{array}$ (1)

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

$\begin{array}{l}\Vert 2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert h\left(\eta \right)(f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\end{array}$ (2)

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

$f\left({\displaystyle \underset{i=1}{\overset{k}{\sum }}}{x}_i+{\displaystyle \underset{i=1}{\overset{k}{\sum }}}{y}_i\right)+f\left({\displaystyle \underset{i=1}{\overset{k}{\sum }}}{x}_i-{\displaystyle \underset{i=1}{\overset{k}{\sum }}}{y}_i\right)=2{\displaystyle \underset{i=1}{\overset{k}{\sum }}}f\left(x_i\right)+2{\displaystyle \underset{k=1}{\overset{k}{\sum }}}f\left(y_i\right)$ (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.

$\Vert 2f\left(x\right)+2f\left(y\right)-f\left(x-y\right)\Vert \le \Vert f\left(x+y\right)\Vert$ (4)

Then f satisfies the Jordan-von Newman functional equation.

$2f\left(x\right)+2f\left(y\right)=f\left(x+y\right)+f\left(x-y\right)$ (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 $F^{\mathrm{<mo>\; *\; </mo>}}$ -space.

Section 3: Constructing solution to the quadratic η-functional inequalities (1) in $\left({\alpha }_1\mathrm{,}{\alpha }_2\right)$ -homogeneous complex Banach spaces.

Section 4: Constructing solution to the quadratic η-functional inequalities (2) in $\left({\alpha }_1\mathrm{,}{\alpha }_2\right)$ -homogeneous complex Banach spaces.

Section 5: Constructing solution to the additive η-functional inequalities (1) in $\left({\alpha }_1\mathrm{,}{\alpha }_2\right)$ -homogeneous complex Banach spaces.

Section 6: Constructing solution to the additive η-functional inequalities (2) in $\left({\alpha }_1\mathrm{,}{\alpha }_2\right)$ -homogeneous complex Banach spaces.

2. Preliminaries

2.1. $F^{\ast }$ -Spaces

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

1. $\Vert x\Vert =0$ if and only if $x=0$ ;

2. $\Vert \lambda x\Vert =\Vert x\Vert$ for all $x\in X$ and all $\lambda$ with $\left|\lambda \right|=1$ ;

3. $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$ for all $x\mathrm{,}y\in X$ ;

4. $\Vert {\lambda }_{n}x\Vert \to 0$ , ${\lambda }_n\to 0$ ;

5. $\Vert {\lambda }_{n}x\Vert \to 0$ , $x_n\to 0$ .

Then $\left(X,\Vert \cdot \Vert \right)$ is called an $F^{\mathrm{<mo>\; *\; </mo>}}$ -space. An F-space is a complete $F^{\mathrm{<mo>\; *\; </mo>}}$ -space. An F-norm is called β-homgeneous ( $\beta >0$ ) if $\Vert tx\Vert ={\left|t\right|}^{\beta }\Vert x\Vert$ for all $x\in X$ and for all $t\in ℂ$ and $\left(X,\Vert \cdot \Vert \right)$ is called α-homogeneous F-space.

2.2. Solutions of the Inequalities

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

The functional equation: $f\left(x+y\right)=f\left(x\right)+f\left(y\right)$ is called the Cauchy equation. In particular, every solution of the Cauchy equation is said to be an additive mapping.

The functional equation: $f\left(\frac{x+y}{2}\right)=\frac{1}{2}f\left(x\right)+\frac{1}{2}f\left(y\right)$ is called the Jensen equation. In particular, every solution of the Jensen equation is said to be a Jensen mapping.

The functional equation: $f\left(\frac{x+y}{2}\right)+f\left(\frac{x-y}{2}\right)=\frac{1}{2}f\left(x\right)+\frac{1}{2}f\left(y\right)$ is called the Jensen type qudratic equation. In particular, every solution of the quadratic equation is said to be a Jensen type quadratic mapping.

$D=\left\{\phi :ℂ\to ℂ:g\left(\eta \right)=\eta ,\left|g\left(\eta \right)\right|=\left|\eta \right|\le \frac{1}{2}\right\}$ (6)

Note: With k is a positive integer and $h\in A$ ${\alpha }_1\mathrm{,}{\alpha }_2\in {ℝ}^{+}$ , ${\alpha }_1\mathrm{,}{\alpha }_2\le 1$ .

3. Constructing Solution to the η-Functional Inequalities (2) in $\left({\alpha }_1,{\alpha }_2\right)$ -Homogeneous Complex Banach Spaces

Now, I first study the solutions of (1). Note that for these inequalities, when $X$ is a ${\alpha }_1$ -homogeneous real or complex normed spaces ${\Vert \cdot \Vert }_X$ and that $Y$ is a ${\alpha }_2$ -homogeneous real or complex Banach spaces ${\Vert \cdot \Vert }_Y$ . Under this setting, I can show that the mapping satisfying (1) is quadratic. These results are given in the following.

Lemma 1 Let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an even mapping satilies:

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\end{array}$ (7)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ if and only if $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is quadratic.

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (7).

I replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (7), I have: ${\Vert \left(4k-2\right)f\left(0\right)\Vert }_Y\le {\Vert \eta f\left(0\right)\Vert }_Y\le 0$ therefore, $\left({\left|4k-2\right|}^{{\alpha }_2}-{\left|\eta \right|}^{{\alpha }_2}\right){\Vert f\left(0\right)\Vert }_Y\le 0$

So $f\left(0\right)=0$ .

Next replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(kx\mathrm{,}\cdots \mathrm{,}kx\mathrm{,}kx\mathrm{,}\cdots \mathrm{,}kx\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)$ in (7), I have.

Thus ${\Vert f\left(2kx\right)-4kf\left(x\right)\Vert }_Y\le 0$

$f\left(\frac{x}{2k}\right)=\frac{1}{4k}f\left(x\right)$ (8)

for all $x\in X$ .

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

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\\ =\frac{{\left|\eta \right|}^{{\alpha }_2}}{{\left|2k\right|}^{{\alpha }_2}}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\end{array}$ (9)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ and so, $\begin{array}{l}f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ =2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\end{array}$ for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ , as I expected. The couverse is obviously true. □

Corollary 1 Let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an even mapping satilies:

$\begin{array}{l}f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\\ =\eta (2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\end{array}$ (10)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ if and only if $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is quadratic.

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

Theorem 2 Assume for $r>\frac{2{\alpha }_2}{{\alpha }_1}$ , $\theta$ be nonngative real number, and suppose $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an even mapping such that:

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\\ -{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y+\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }_X^r\right)\end{array}$ (11)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ . Then there exists a unique quadratic mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ such that:

${\Vert f\left(x\right)-\varphi \left(x\right)\Vert }_Y\le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_{1}r}-{\left(4k\right)}^{{\alpha }_2}}\theta {\Vert x\Vert }_X^r$ (12)

for all $x\in X$ .

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (11).

I replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (11), I have: ${\Vert \left(4k-2\right)f\left(0\right)\Vert }_Y\le {\Vert 2\eta f\left(0\right)\Vert }_Y\le 0$ therefore, $\left({\left|4k-2\right|}^{{\alpha }_2}-{\left|2\eta \right|}^{{\alpha }_2}\right){\Vert f\left(0\right)\Vert }_Y\le 0$

So $f\left(0\right)=0$ .

Next replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(kx\mathrm{,}\cdots \mathrm{,}kx\mathrm{,}kx\mathrm{,}\cdots \mathrm{,}kx\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)$ in (11) I have:

${\Vert f\left(2x\right)-4kf\left(x\right)\Vert }_Y\le \left(2k^{{\alpha }_{1}r+1}+1\right)\theta {\Vert x\Vert }_X^r$ (13)

for all $x\in X$ . Thus,

${\Vert f\left(x\right)-4kf\left(\frac{x}{2k}\right)\Vert }_Y\le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\Vert x\Vert }_X^r$ (14)

for all $x\in X$ .

$\begin{array}{l}{\Vert {\left(4k\right)}^{l}f\left(\frac{x}{{\left(2k\right)}^l}\right)-{\left(4k\right)}^{m}f\left(\frac{x}{{\left(2k\right)}^m}\right)\Vert }_Y\\ \le {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}{\Vert {\left(4k\right)}^{j}f\left(\frac{x}{{\left(2k\right)}^j}\right)-{\left(4k\right)}^{j+1}f\left(\frac{x}{{\left(2k\right)}^{j+1}}\right)\Vert }_Y\\ \le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\frac{{\left(4k\right)}^{{\alpha }_{2}j}}{{\left(2k\right)}^{{\alpha }_{1}rj}}{\Vert x\Vert }_X^r\end{array}$ (15)

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

So one can define the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ by, $\varphi \left(x\right)\mathrm{<mo>\; :\; </mo>}=\underset{n\to \infty }{\lim }{\left(4k\right)}^{n}f\left(\frac{x}{{\left(2k\right)}^n}\right)$ for all $x\in X$ . Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (15), I get (12).

Form $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is even, the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is even.

It follows from (11) that:

$\begin{array}{l}\Vert \varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-z_j\right)\Vert }_Y\\ =\underset{n\to \infty }{\lim }{\left(4k\right)}^{{\alpha }_{2}n}\Vert f\left(\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ +f(\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{1}{{\left(2k\right)}^n}\frac{x_j+y_j}{2k}\right)\\ -{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{1}{{\left(2k\right)}^n}{z}_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{1}{{\left(2k\right)}^n}{z}_j\right)\Vert }_Y\end{array}$

$\begin{array}{l}\le \underset{n\to \infty }{\lim }{\left(4k\right)}^{{\alpha }_{2}n}{\left|\eta \right|}^{{\alpha }_2}\Vert 2f\left(\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ +2f\left(\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{1}{{\left(2k\right)}^n}\frac{x_j+y_j}{2k}\right)\\ +\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{1}{{\left(2k\right)}^n}\frac{x_j+y_j}{2k}\right)-\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{1}{{\left(2k\right)}^n}{z}_j\right)-{\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{1}{{\left(2k\right)}^n}{z}_j\right)\Vert }_Y\\ +\underset{n\to \infty }{\lim }\frac{{\left(4k\right)}^{{\alpha }_{2}n}}{{\left(2k\right)}^{{\alpha }_{1}nr}}\theta \left({\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{\Vert x_j\Vert }_X^r+{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{\Vert y_j\Vert }_X^r+{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{\Vert z_j\Vert }_X^r\right)\end{array}$

$\begin{array}{l}={\left|\eta \right|}^{{\alpha }_2}\Vert 2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\end{array}$ (16)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for all $j=1\to n$ .

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\end{array}$ (17)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for $j=1\to n$ , So by lemma 3.1, it follows that the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is additive. Now I need to prove uniqueness, Suppose ${\varphi }^{\prime }\mathrm{<mo>\; :\; </mo>}X\to Y$ is also a quadratic mapping that satisfies (12). Then I have:

$\begin{array}{l}{\Vert \varphi \left(x\right)-{\varphi }^{\prime }\left(x\right)\Vert }_Y={\left(4k\right)}^{{\alpha }_{2}n}{\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-{\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert }_Y\\ \le {\left(4k\right)}^{{\alpha }_{2}n}\left({\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert }_Y+{\Vert {\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert }_Y\right)\\ \le \frac{2\cdot {\left(4k\right)}^{{\alpha }_{2}n}\cdot \left(2k^{{\alpha }_{1}r+1}+1\right)}{{\left(2k\right)}^{{\alpha }_{1}nr}\left({\left(2k\right)}^{{\alpha }_{1}r}-{\left(4k\right)}^{{\alpha }_2}\right)}\theta {\Vert x\Vert }_X^r\end{array}$ (18)

which tends to zero as $n\to \infty$ for all $x\in X$ . So I can conclude that $\varphi \left(x\right)={\varphi }^{\prime }\left(x\right)$ for all $x\in X$ . This proves thus the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is a unique mapping satisfying(12) as I expected.

Theorem 3 Assume for $r<\frac{2{\alpha }_2}{{\alpha }_1}$ , $\theta$ be nonngative real number, and Suppose $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an even mapping satisfiying (11). Then there exists a unique quadratic mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ such that:

$\Vert f\left(x\right)-\varphi \left(x\right)\Vert \le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(4k\right)}^{{\alpha }_2}-{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\Vert x\Vert }^r$ (19)

for all $x\in X$ .

The proof is similar to the proof of theorem 3.3.

4. Constructing Solution to the η-Functional Inequalities (2) in $\left({\alpha }_1,{\alpha }_2\right)$ -Homogeneous Complex Banach Spaces

Now, I study the solutions of (2). Note that for these inequalities, when $X$ is a ${\alpha }_1$ -homogeneous complex Banach spaces and that $Y$ is a ${\alpha }_2$ -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 $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an even mapping satilies $f\left(0\right)=0$ and:

$\begin{array}{l}\Vert 2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\end{array}$ (20)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ if and only if $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is quadratic.

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (20).

Replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(2kx\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (20), I have.

Thus ${\Vert 4f\left(\frac{x}{2k}\right)-\frac{1}{k}f\left(x\right)\Vert }_Y\le 0$

$f\left(\frac{x}{2k}\right)=\frac{1}{4k}f\left(x\right)$ (21)

for all $x\in X$ .

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

$\begin{array}{l}\frac{1}{{\left(2k\right)}^{{\alpha }_2}}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ =\Vert 2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le {\left|\eta \right|}^{{\alpha }_2}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\end{array}$ (22)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ and so: $f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)=2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)$ for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ , as I expected. The couverse is obviously true.

Let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an even mapping satilies,

$\begin{array}{l}2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\\ =\eta (f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\end{array}$ (23)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ if and only if $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is quadratic Note! The functional Equation (23) is called an quadratic λ-functional equation.

Theorem 5 Assume for $r>\frac{2{\alpha }_2}{{\alpha }_1}$ , $\theta$ be nonngative real number, and suppose $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping such that $f\left(0\right)=0$ and

$\begin{array}{l}\Vert 2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)\\ -{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right){-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y+\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }_X^r\right)\end{array}$ (24)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ . Then there exists a unique quadratic mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ such that:

${\Vert f\left(x\right)-\varphi \left(x\right)\Vert }_Y\le \frac{{\left(2k\right)}^{{\alpha }_{1}r}}{{\left(2k\right)}^{{\alpha }_{1}r}-{\left(4k\right)}^{{\alpha }_2}}\theta {\Vert x\Vert }_X^r$ (25)

for all $x\in X$ .

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (24).

Replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(2kx\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (24) I have:

${\Vert 4f\left(\frac{x}{2k}\right)-\frac{1}{k}f\left(x\right)\Vert }_Y\le {\left(2k\right)}^{{\alpha }_{1}r}\theta {\Vert x\Vert }_X^r$ (26)

for all $x\in X$ . Thus

$\Vert 4kf\left(\frac{x}{2k}\right)-f\left(x\right)\Vert \le {\left(2k\right)}^{{\alpha }_{1}r}{k}^{{\alpha }_2}\theta {\Vert x\Vert }_X^r$ (27)

for all $x\in X$ .

$\begin{array}{l}\Vert {\left(4k\right)}^{l}f\left(\frac{x}{{\left(2k\right)}^l}\right)-{\left(4k\right)}^{m}f\left(\frac{x}{{\left(2k\right)}^m}\right)\Vert \\ \le {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}{\Vert {\left(4k\right)}^{j}f\left(\frac{x}{{\left(2k\right)}^j}\right)-{\left(4k\right)}^{j+1}f\left(\frac{x}{{\left(2k\right)}^{j+1}}\right)\Vert }_Y\\ \le {\left(2k\right)}^{{\alpha }_{1}r}{k}^{{\alpha }_2}\theta {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\frac{{\left(4k\right)}^{{\alpha }_{2}j}}{{\left(2k\right)}^{{\alpha }_{1}rj}}{\Vert x\Vert }_X^r\end{array}$ (28)

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

So one can define the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ by $\varphi \left(x\right)\mathrm{<mo>\; :\; </mo>}=\underset{n\to \infty }{\lim }{\left(4k\right)}^{n}f\left(\frac{x}{{\left(2k\right)}^n}\right)$ for all $x\in X$ . Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (28), I get (25). Form $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is even, the mapping: $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is even. It follows from (24) I have:

$\begin{array}{l}\Vert 2\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-z_j\right)\Vert }_Y\\ =\underset{n\to \infty }{\lim }{\left(4k\right)}^{{\alpha }_{2}n}\Vert 2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^{n+2}}+\frac{1}{{\left(2k\right)}^{n+1}}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ +2f({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^{n+2}}-\frac{1}{{\left(2k\right)}^{n+1}}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j)-\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{{\left(2k\right)}^{n+1}}\right)\\ +\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{{\left(2k\right)}^{n+1}}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{z_j}{{\left(2k\right)}^n}\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{-z_j}{{\left(2k\right)}^n}\right)\Vert }_Y\end{array}$

$\begin{array}{l}\le \underset{n\to \infty }{\lim }{\left(4k\right)}^{{\alpha }_{2}n}{\left|\eta \right|}^{{\alpha }_2}\Vert 2f\left(\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ +2f\left(\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{1}{{\left(2k\right)}^n}\frac{x_j+y_j}{2k}\right)\\ -\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{1}{{\left(2k\right)}^n}{z}_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{1}{{\left(2k\right)}^n}{z}_j\right)\Vert }_Y\\ +\underset{n\to \infty }{\lim }\frac{{\left(4k\right)}^{{\alpha }_{2}n}}{{\left(2k\right)}^{{\alpha }_{1}nr}}\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }_X^r\right)\end{array}$

$\begin{array}{l}=\Vert \varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-z_j\right)\Vert }_Y\end{array}$ (29)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ .

$\begin{array}{l}\Vert 2\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-z_j\right))\Vert }_Y\end{array}$

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ , So by lemma 4.1 it follows that the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is quadratic. Now I need to prove uniqueness, Suppose ${\varphi }^{\prime }\mathrm{<mo>\; :\; </mo>}X\to Y$ is also a quadratic mapping that satisfies (25). Then I have:

$\begin{array}{l}{\Vert \varphi \left(x\right)-{\varphi }^{\prime }\left(x\right)\Vert }_Y={\left(4k\right)}^{{\alpha }_{2}n}{\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-{\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert }_Y\\ \le {\left(4k\right)}^{{\alpha }_{2}n}\left({\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert }_Y+{\Vert {\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert }_Y\right)\\ \le \frac{2\cdot {\left(4k\right)}^{{\alpha }_{2}n}\cdot {\left(2k\right)}^{{\alpha }_{1}r}}{{\left(2k\right)}^{{\alpha }_{1}nr}\left({\left(2k\right)}^{{\alpha }_{1}r}-{\left(4k\right)}^{{\alpha }_2}\right)}\theta {\Vert x\Vert }_X^r\end{array}$ (30)

which tends to zero as $n\to \infty$ for all $x\in X$ . So I can conclude that $\varphi \left(x\right)={\varphi }^{\prime }\left(x\right)$ for all $x\in X$ .This proves thus the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is a unique mapping satisfying(25) as I expected. □

Theorem 6 Assume for $r<\frac{2{\alpha }_2}{{\alpha }_1}$ , $\theta$ be nonngative real number, $f\left(0\right)=0$ and suppose $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an odd mapping (24). Then there exists a unique quadratic mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ such that:

${\Vert f\left(x\right)-\varphi \left(x\right)\Vert }_Y\le \frac{{\left(2k\right)}^{{\alpha }_{1}r}}{{\left(4k\right)}^{{\alpha }_2}-{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\Vert x\Vert }_X^r$ (31)

for all $x\in X$ .

The proof is similar to the proof of theorem 4.3.

5. Constructing Solution to the Additive η-Functional Inequalities (1) in $\left({\alpha }_1,{\alpha }_2\right)$ -Homogeneous Complex Banach Spaces

Now, I first study the solutions of (1). Note that for these inequalities, when $X$ is a ${\alpha }_1$ -homogeneous complex Banach spaces and that $Y$ is a ${\alpha }_2$ -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 $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an odd mapping satilies:

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\end{array}$ (32)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ if and only if $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is additive.

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (32).

I replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (32), I have: ${\Vert \left(4k-2\right)f\left(0\right)\Vert }_Y\le {\left|\eta \right|}^{{\alpha }_2}{\Vert 5f\left(0\right)\Vert }_Y\le 0$ therefore $f\left(0\right)=0$ .

Next replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(kx\mathrm{,}\cdots \mathrm{,}kx\mathrm{,}kx\mathrm{,}\cdots \mathrm{,}kx\mathrm{,}x\mathrm{,}\cdots ,x\right)$ in (32), I have.

Thus ${\Vert f\left(2kx\right)-2kf\left(x\right)\Vert }_Y\le 0$

$f\left(\frac{x}{2k}\right)=\frac{1}{2k}f\left(x\right)$ (33)

for all $x\in X$ From (32) and (33) I infer that:

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\\ =\frac{{\left|\eta \right|}^{{\alpha }_2}}{{\left|k\right|}^{{\alpha }_2}}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\end{array}$ (34)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ and so.

$f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)=2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)$ (35)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ .

Next I replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(kx\mathrm{,}\cdots \mathrm{,}kx\mathrm{,}kx\mathrm{,}\cdots \mathrm{,}kx\mathrm{,}z\mathrm{,}\cdots ,z\right)$ in (35), I have

$f\left(kx+kz\right)+f\left(kx-kz\right)=2kf\left(x\right)$ (36)

for all $x\mathrm{,}z\in X$ Now letting $p=kx+kz,q=kx-kz$ when that in (36), I get

$f\left(p\right)+f\left(q\right)=2kf\left(\frac{p+q}{2k}\right)=2k\cdot \frac{1}{2k}f\left(p+q\right)=f\left(p+q\right)$ (37)

for all $p\mathrm{,}q\in X$ . So f is an additive mapping. as I expected. The couverse is obviously true. □

Corollary 2 Let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an even mapping satilies:

$\begin{array}{l}f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\\ =\eta (2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\end{array}$ (38)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ if and only if $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is additive.

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

Theorem 8 Assume for $r>\frac{{\alpha }_2}{{\alpha }_1}$ , $\theta$ be nonngative real number, and suppose $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an odd mapping such that:

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\\ -{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y+\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }_X^r\right)\end{array}$ (39)

for all $x_1,\cdots ,x_k,y_1,\cdots ,y_k,z_1,\cdots ,z_k\in X$ . Then there exists a unique additive mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ such that:

${\Vert f\left(x\right)-\varphi \left(x\right)\Vert }_Y\le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_{1}r}-{\left(2k\right)}^{{\alpha }_2}}\theta {\Vert x\Vert }_X^r\mathrm{.}$ (40)

for all $x\in X$ .

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (39). I replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (39), I have: $O{\Vert \left(4k-2\right)f\left(0\right)\Vert }_Y\le {\Vert 5\lambda f\left(0\right)\Vert }_Y$ therefore, $\left({\left|4k-2\right|}^{{\alpha }_2}-{\left|5\lambda \right|}^{{\alpha }_2}\right){\Vert f\left(0\right)\Vert }_Y\le 0$

So $f\left(0\right)=0$ . Next replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(kx\mathrm{,}\cdots \mathrm{,}kx\mathrm{,}kx\mathrm{,}\cdots \mathrm{,}kx\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)$ in (39) I have:

${\Vert f\left(2kx\right)-2kf\left(x\right)\Vert }_Y\le \left(2k^{{\alpha }_{1}r+1}+1\right)\theta {\Vert x\Vert }_X^r$ (41)

for all $x\in X$ . Thus

${\Vert f\left(x\right)-2kf\left(\frac{x}{2k}\right)\Vert }_Y\le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\Vert x\Vert }_X^r$ (42)

for all $x\in X$ .

$\begin{array}{l}{\Vert {\left(2k\right)}^{l}f\left(\frac{x}{{\left(2k\right)}^l}\right)-{\left(2k\right)}^{m}f\left(\frac{x}{{\left(2k\right)}^m}\right)\Vert }_Y\\ \le {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}{\Vert {\left(2k\right)}^{j}f\left(\frac{x}{{\left(2k\right)}^j}\right)-{\left(2k\right)}^{j+1}f\left(\frac{x}{{\left(2k\right)}^{j+1}}\right)\Vert }_Y\\ \le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{m-1}{\sum }}}\frac{{\left(2k\right)}^{{\alpha }_{2}j}}{{\left(2k\right)}^{{\alpha }_{1}rj}}{\Vert x\Vert }_X^r\end{array}$ (43)

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

So one can define the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ by $\varphi \left(x\right)\mathrm{<mo>\; :\; </mo>}=\underset{n\to \infty }{\lim }{\left(2k\right)}^{n}f\left(\frac{x}{{\left(2k\right)}^n}\right)$ for all $x\in X$ . Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (15), I get (40).

Form $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is even, the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is even.

It follows from (39) I have:

$\begin{array}{l}\Vert \varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-z_j\right)\Vert }_Y\\ =\underset{n\to \infty }{\lim }{\left(2k\right)}^{{\alpha }_{2}n}\Vert f\left(\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ +f\left(\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{1}{{\left(2k\right)}^n}\frac{x_j+y_j}{2k}\right)\\ -{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{1}{{\left(2k\right)}^n}{z}_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{1}{{\left(2k\right)}^n}{z}_j\right)\Vert }_Y\end{array}$

$\begin{array}{l}\le \underset{n\to \infty }{\lim }{\left(2k\right)}^{{\alpha }_{2}n}{\left|\lambda \right|}^{{\alpha }_2}\Vert 2f\left(\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{{\left(2k\right)}^{n+1}}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ +2f\left(\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{{\left(2k\right)}^{n+1}}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{1}{{\left(2k\right)}^n}\frac{x_j+y_j}{2k}\right)\\ -\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{1}{{\left(2k\right)}^n}\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{1}{{\left(2k\right)}^n}{z}_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{1}{{\left(2k\right)}^n}{z}_j\right)\Vert }_Y\\ +\underset{n\to \infty }{\lim }\frac{{\left(2k\right)}^{{\alpha }_{2}n}}{{\left(2k\right)}^{{\alpha }_{1}nr}}\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }_X^r\right)\end{array}$

$\begin{array}{l}={\left|\lambda \right|}^{{\alpha }_2}\Vert 2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\end{array}$ (44)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for all $j=1\to n$ .

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\end{array}$

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for $j=1\to n$ , So by lemma 5.1, it follows that the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is additive. Now I need to prove uniqueness, suppose ${\varphi }^{\prime }\mathrm{<mo>\; :\; </mo>}X\to Y$ is also an additive mapping that satisfies (40). Then I have:

$\begin{array}{l}{\Vert \varphi \left(x\right)-{\varphi }^{\prime }\left(x\right)\Vert }_Y={\left(2k\right)}^{{\alpha }_{2}n}{\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-{\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert }_Y\\ \le {\left(2k\right)}^{{\alpha }_{2}n}\left({\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert }_Y+{\Vert {\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert }_Y\right)\\ \le \frac{2\cdot {\left(2k\right)}^{{\alpha }_{2}n}\cdot \left(2k^{{\alpha }_{1}r+1}+1\right)}{{\left(2k\right)}^{{\alpha }_{1}nr}\left({\left(2k\right)}^{{\alpha }_{1}r}-{\left(2k\right)}^{{\alpha }_2}\right)}\theta {\Vert x\Vert }_X^r\end{array}$ (45)

which tends to zero as $n\to \infty$ for all $x\in X$ . So I can conclude that $\varphi \left(x\right)={\varphi }^{\prime }\left(x\right)$ for all $x\in X$ . This proves thus the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is a unique mapping satisfying(40) as I expected. □

Theorem 9 Assume for $r<\frac{{\alpha }_2}{{\alpha }_1}$ , $\theta$ be nonngative real number, and suppose $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an odd mapping satisfying (1). Then there exists a unique additive mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ such that:

${\Vert f\left(x\right)-\varphi \left(x\right)\Vert }_Y\le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_2}-{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\Vert x\Vert }_X^r$ (46)

for all $x\in X$ .

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 $\left({\alpha }_1,{\alpha }_2\right)$ -Homogeneous Complex Banach Spaces

Now, I first study the solutions of (2). Note that for these inequalities, when $X$ is a ${\alpha }_1$ -homogeneous complex Banach spaces and that $Y$ is a ${\alpha }_2$ -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 $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an odd mapping satilies:

$\begin{array}{l}\Vert 2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right){-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\end{array}$ (47)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for $j=1\to n$ , if and only if $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is additive.

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (47).

I replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (20), I have: ${\Vert 2kf\left(0\right)\Vert }_Y\le {\left|\eta \right|}^{{\alpha }_2}{\Vert \left(4k-2\right)f\left(0\right)\Vert }_Y$

So $f\left(0\right)=0$ .

Replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(2kx\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (47), I have.

Thus ${\Vert 4kf\left(\frac{x}{2k}\right)-2f\left(x\right)\Vert }_Y\le 0$

$f\left(\frac{x}{2k}\right)=\frac{1}{2k}f\left(x\right)$ (48)

for all $x\in X$ . From (47) and (48) I infer that:

$\begin{array}{l}\frac{1}{{\left|k\right|}^{{\alpha }_2}}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ =\Vert 2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le {\left|\eta \right|}^{{\alpha }_2}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\end{array}$ (49)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for $j=1\to n$ , and so: $f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)=2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)$ for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for $j=1\to n$ , as I expected. The couverse is obviously true. □

Let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an even mapping satilies.

Theorem 11 Assume for $r>\frac{{\alpha }_2}{{\alpha }_1}$ , $\theta$ be nonngative real number, and suppose $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping such that $f\left(0\right)=0$ and:

$\begin{array}{l}\Vert 2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \eta (f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)\\ -{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right){-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y+\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }_X^r\right)\end{array}$ (50)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for all $j=1\to n$ . Then there exists a unique additive mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ such that:

${\Vert f\left(x\right)-\varphi \left(x\right)\Vert }_Y\le \frac{{\left(2k\right)}^{{\alpha }_{1}r}}{{\left(2k\right)}^{{\alpha }_{1}r}-{\left(4k\right)}^{{\alpha }_2}}\theta {\Vert x\Vert }_X^r$ (51)

for all $x\in X$ .

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (50).

I replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (50), I have: ${\Vert 2f\left(0\right)\Vert }_Y\le {\left|\eta \right|}^{{\alpha }_2}{\Vert \left(4k-2\right)f\left(0\right)\Vert }_Y$ therefore, $\left({\left|4k-2\right|}^{{\alpha }_2}-{\left|2\eta \right|}^{{\alpha }_2}\right){\Vert f\left(0\right)\Vert }_Y\le 0$

So $f\left(0\right)=0$ .

Replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(2kx\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (50) I have:

${\Vert 4f\left(\frac{x}{2k}\right)-\frac{1}{k}f\left(x\right)\Vert }_Y\le {\left(2k\right)}^{{\alpha }_{1}r}\theta {\Vert x\Vert }_X^r$ (52)

for all $x\in X$ . Thus

${\Vert 4kf\left(\frac{x}{2k}\right)-f\left(x\right)\Vert }_Y\le {\left(2k\right)}^{{\alpha }_{1}r}{k}^{{\alpha }_2}\theta {\Vert x\Vert }_X^r$ (53)

for all $x\in X$ .

$\begin{array}{l}{\Vert {\left(4k\right)}^{l}f\left(\frac{x}{{\left(2k\right)}^l}\right)-{\left(4k\right)}^{m}f\left(\frac{x}{{\left(2k\right)}^m}\right)\Vert }_Y\\ \le {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}{\Vert {\left(4k\right)}^{j}f\left(\frac{x}{{\left(2k\right)}^j}\right)-{\left(4k\right)}^{j+1}f\left(\frac{x}{{\left(2k\right)}^{j+1}}\right)\Vert }_Y\\ \le {\left(2k\right)}^{{\alpha }_{1}r}{k}^{{\alpha }_2}\theta {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\frac{{\left(4k\right)}^{{\alpha }_{2}j}}{{\left(2k\right)}^{{\alpha }_{1}rj}}{\Vert x\Vert }_X^r\end{array}$ (54)

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

So one can define the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ by $\varphi \left(x\right)\mathrm{<mo>\; :\; </mo>}=\underset{n\to \infty }{\lim }{\left(4k\right)}^{n}f\left(\frac{x}{{\left(2k\right)}^n}\right)$ for all $x\in X$ . Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (28), I get (51). Form $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is even, the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is even. It follows from (50) I have:

$\begin{array}{l}\Vert 2\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-z_j\right))\Vert }_Y\\ =\underset{n\to \infty }{\lim }{\left(4k\right)}^{{\alpha }_{2}n}\Vert 2f\left({\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^{n+2}}+\frac{1}{{\left(2k\right)}^{n+1}}{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{z}_j\right)\\ +2f({\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^{n+2}}-\frac{1}{{\left(2k\right)}^{n+1}}{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{z}_j)-\frac{3}{2k}{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{{\left(2k\right)}^{n+1}}\right)\\ +\frac{1}{2k}{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{{\left(2k\right)}^{n+1}}\right)-\frac{1}{2k}{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(\frac{z_j}{{\left(2k\right)}^n}\right)-{\frac{1}{2k}{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(\frac{-z_j}{{\left(2k\right)}^n}\right)\Vert }_Y\end{array}$

$\begin{array}{l}\le \underset{n\to \infty }{\lim }{\left(4k\right)}^{{\alpha }_{2}n}{\left|\eta \right|}^{{\alpha }_2}\Vert 2f\left(\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ +2f\left(\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{1}{{\left(2k\right)}^n}\frac{x_j+y_j}{2k}\right)\\ -\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{1}{{\left(2k\right)}^n}{z}_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{1}{{\left(2k\right)}^n}{z}_j\right)\Vert }_Y\\ +\underset{n\to \infty }{\lim }\frac{{\left(4k\right)}^{{\alpha }_{2}n}}{{\left(2k\right)}^{{\alpha }_{1}nr}}\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }_X^r\right)\end{array}$

$\begin{array}{l}=\Vert \varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-z_j\right))\Vert }_Y\end{array}$ (55)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for all $j=1\to n$ .

$\begin{array}{l}\Vert 2\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-z_j\right))\Vert }_Y\\ \le \Vert \lambda (\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)-{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-z_j\right))\Vert }_Y\end{array}$

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for $j=1\to n$ , So by lemma 6.1, it follows that the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is quadratic. Now I need to prove uniqueness, suppose ${\varphi }^{\prime }\mathrm{<mo>\; :\; </mo>}X\to Y$ is also a quadratic mapping that satisfies (50). Then I have:

$\begin{array}{l}{\Vert \varphi \left(x\right)-{\varphi }^{\prime }\left(x\right)\Vert }_Y={\left(4k\right)}^{{\alpha }_{2}n}{\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-{\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert }_Y\\ \le {\left(4k\right)}^{{\alpha }_{2}n}\left({\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert }_Y+{\Vert {\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert }_Y\right)\\ \le \frac{2\cdot {\left(4k\right)}^{{\alpha }_{2}n}\cdot {\left(2k\right)}^{{\alpha }_{1}r}}{{\left(2k\right)}^{{\alpha }_{1}nr}\left({\left(2k\right)}^{{\alpha }_{1}r}-{\left(4k\right)}^{{\alpha }_2}\right)}\theta {\Vert x\Vert }_X^r\end{array}$ (56)

which tends to zero as $n\to \infty$ for all $x\in X$ . So I can conclude that $\varphi \left(x\right)={\varphi }^{\prime }\left(x\right)$ for all $x\in X$ . This proves thus the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is a unique mapping satisfying(51) as I expected.

Theorem 12 Assume for $r<\frac{{\alpha }_2}{{\alpha }_1}$ , $\theta$ be nonngative real number, $f\left(0\right)=0$ and suppose $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an odd mapping satisfying (50). Then there exists a unique quadratic mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ such that:

${\Vert f\left(x\right)-\varphi \left(x\right)\Vert }_Y\le \frac{{\left(2k\right)}^{{\alpha }_{1}r}}{{\left(4k\right)}^{{\alpha }_2}-{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\Vert x\Vert }_X^r\mathrm{.}$ (57)

for all $x\in X$ .

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 $\left({\alpha }_1\mathrm{,}{\alpha }_2\right)$ -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.

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