Considerable Development of the Type Additive-Quadratic g(λ)-Functional Inequalities with 3k-Variable in (α₁,α₂)-Homogeneous F-Spaces

Ly Van An

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

DOI: 10.4236/oalib.1110970                

Abstract

In this article, I use the direct method to study two general functional inequalities with multivariables. First, I prove that the g(λ)-function inequalities (1) and (2) are additive in (α₁;α₂)-homogeneous F-spaces. After that, I continue to prove that the g(λ)-function inequality (1) and (2) are quadratic in the (α₁;α₂)-homogeneous F-space. That is the main result in this paper.

Keywords

Additive g(λ)-Functional Inequality, (α₁;α₂)-Homogeneous F-Space, Additive-Quadratic g(λ)-Functional Inequality, (α₁;α₂)-Homogeneous F-Space

1. Introduction

Let $X$ and $Y$ be a normed spaces on the same field $\mathbb{K}$ , and $f\mathrm{<mo>\; :\; </mo>}X\to Y$ . I use the notation $\Vert \cdot \Vert$ for all the norm on both $X$ and $Y$ . In this paper, I investisgate some additive-quadraic $\lambda$ -functional inequality in $\left({\alpha }_1\mathrm{<mo>\; ;\; </mo>}{\alpha }_2\right)$ -homogeneous F-spaces.

In fact, when $X$ is a ${\alpha }_1$ -homogeneous F-spaces and that $Y$ is a ${\alpha }_2$ -homogeneous F-spaces, I solve and prove the Hyers-Ulam-Rassias type stability of two forllowing additive-quadratic $g\left(\lambda \right)$ -functional inequality.

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

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

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

$H=\left\{h\mathrm{<mo>\; :\; </mo>}ℂ\backslash \left\{0\right\}\to ℂ\mathrm{,}h\left(\lambda \right)=\lambda \right\}$ (3)

where $g\in H$ .

The stability problem of functional equations originated from a question of Ulam [1] concerning the stability of group homomorphisms.

The functional equation

$f\left(x+y\right)=f\left(x\right)+f\left(y\right)$ (4)

is called the Cauchy equation.

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

The functional equation

$f\left(x+y\right)+f\left(x-y\right)=2f\left(x\right)+2f\left(y\right)$ (5)

is called the quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic mapping. The stability of quadratic functional equation was proved by Skof [6] for mappings $f\mathrm{<mo>\; :\; </mo>}{E}_1\to E_2$ , where $E_1$ is a normed space and $E_2$ is a Banach space. Cholewa [7] noticed that the theorem of Skof is still true if the relevant domain E₁ is replaced by an Abelian group.

Recently, the I has studied the additive function inequalities or quadratic function inequalities of mathematicians around the world see [1] - [24] , on spaces as complex Banach spaces, non-Archimedan Banach spaces or homogeneous F-space let me give two general additive-quadratic functional inequalities and show their solutions exist on $\left({\alpha }_1\mathrm{,}{\alpha }_2\right)$ -homogeneous F-space.

In this article, I successfully built quadratic functional inequalities with the number of variables more than 3 on F-homogeneous space and I showed their solutions. This is a great step forward in the field of functional equations. Application to solve problems in many spaces with no limit on the number of variables.

The paper is organized as followns: In section preliminarier I remind a basic property such as I only redefine the solution definition of the equation of the additive function and F*-space.

Section 3: is devoted to prove the Hyers-Ulam stability of the addive $g\left(\lambda \right)$ -functional inequalities (1) when when $X$ is a ${\alpha }_1$ -homogeneous F-spaces and that $Y$ is a ${\alpha }_2$ -homogeneous F-spaces.

Section 4: is devoted to prove the Hyers-Ulam stability of the addive $g\left(\lambda \right)$ -functional inequalities (2) when when $X$ is a ${\alpha }_1$ -homogeneous F-spaces and that $Y$ is a ${\alpha }_2$ -homogeneous F-spaces.

Section 5: is devoted to prove the Hyers-Ulam stability of the quadratic $g\left(\lambda \right)$ -functional inequalities (1) when when $X$ is a ${\alpha }_1$ -homogeneous F-spaces and that $Y$ is a ${\alpha }_2$ -homogeneous F-spaces.

Section 6: is devoted to prove the Hyers-Ulam stability of the quadratic $g\left(\lambda \right)$ -functional inequalities (2) when when $X$ is a ${\alpha }_1$ -homogeneous F-spaces and that $Y$ is a ${\alpha }_2$ -homogeneous F-spaces.

2. Preliminaries

2.1. F*-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 $\left(X\mathrm{,}\Vert \cdot \Vert \right)$ is called an F*-space. An F-space is a complete F*-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 $\left(X\mathrm{,}\Vert \cdot \Vert \right)$ is called α-homogeneous F-space.

2.2. Solutions of the Inequalities

The functional equation The functional equation

$f\left(x+y\right)=f\left(x\right)+f\left(y\right)$ (6)

is called the Cauchy equation. In particular, every solution of the Cauchy equation is said to be an additive mapping.

The functional equation

$f\left(x+y\right)+f\left(x-y\right)=2f\left(x\right)+2f\left(y\right)$ (7)

is called the quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic mapping.

3. Hyers-Ulam-Rassias Stability Additive $g\left(\lambda \right)$ -Functional Inequalities (1) in α-Homogeneous F-Spaces

Now, I first study the solutions of (1). Note that for these inequalities, when $X$ is a ${\alpha }_1$ -homogeneous F-spaces and that $Y$ is a ${\alpha }_2$ -homogeneous F-spaces. Under this setting, I can show that the mapping satisfying (1) is additive. These results are give in the following.

Where: ${\alpha }_1\mathrm{,}{\alpha }_1\in {ℝ}^{+}$ and ${\alpha }_1\mathrm{,}{\alpha }_1\le 1$ .

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

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \lambda (2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\end{array}$ (8)

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 (8).

We 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 (8), we have

$\Vert \left(4k-2\right)f\left(0\right)\Vert \le {\left|g\left(\lambda \right)\right|}^{{\alpha }_2}\Vert 2kf\left(0\right)\Vert \le 0$

therefore

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 (8), we have

Thus

$\Vert f\left(2kx\right)-2kf\left(x\right)\Vert \le 0$

$f\left(\frac{x}{2k}\right)=\frac{1}{2k}f\left(x\right)$ (9)

for all $x\in X$ .

From (8) and (9) I infer that

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert g\left(\lambda \right)(2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\\ ={\left|g\left(\lambda \right)\right|}^{{\alpha }_2}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\end{array}$ (10)

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

$f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)=2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)$ (11)

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

Next we 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 $\left(kx\mathrm{,}\cdots \mathrm{,}kx\mathrm{,}kx\mathrm{,}\cdots \mathrm{,}kx\mathrm{,}z\mathrm{,}\cdots \mathrm{,}z\right)$ in (11), we have

$f\left(kx+kz\right)+f\left(kx-kz\right)=2kf\left(x\right)$ (12)

for all $x\mathrm{,}z\in X$ .

Now letting $p=kx+kz,q=kx-kz$ when that in (12), we get

$f\left(p\right)+f\left(q\right)=2kf\left(\frac{p+q}{2k}\right)=2k\cdot \frac{1}{2k}f\left(p+q\right)=f\left(p+q\right)$ (13)

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

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

$\begin{array}{l}f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\\ =g\left(\lambda \right)(2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\end{array}$ (14)

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.

Note! The functional equation (14) is called an additive λ-functional equation.

Theorem 2. 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

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert g\left(\lambda \right)(2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (15)

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

$\Vert f\left(x\right)-\varphi \left(x\right)\Vert \le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_{1}r}-{\left(2k\right)}^{{\alpha }_2}}\theta {\Vert x\Vert }^r\mathrm{.}$ (16)

for all $x\in X$ .

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (15).

We 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 (15), we have

$\Vert \left(4k-2\right)f\left(0\right)\Vert \le \Vert 2kg\left(\lambda \right)f\left(0\right)\Vert$

therefore

$\left({\left|4k-2\right|}^{{\alpha }_2}-{\left|2kg\left(\lambda \right)\right|}^{{\alpha }_2}\right)\Vert f\left(0\right)\Vert$

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 (15) we have

$\Vert f\left(2kx\right)-2kf\left(x\right)\Vert \le \left(2k^{{\alpha }_{1}r+1}+1\right)\theta {\Vert x\Vert }^r$ (17)

for all $x\in X$ . Thus

$\Vert f\left(x\right)-2kf\left(\frac{x}{2k}\right)\Vert \le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\Vert x\Vert }^r$ (18)

for all $x\in X$ .

$\begin{array}{l}\Vert {\left(2k\right)}^{l}f\left(\frac{x}{{\left(2k\right)}^l}\right)-{\left(2k\right)}^{m}f\left(\frac{x}{{\left(2k\right)}^m}\right)\Vert \\ \le {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\Vert {\left(2k\right)}^{j}f\left(\frac{x}{{\left(2k\right)}^j}\right)-{\left(2k\right)}^{j+1}f\left(\frac{x}{{\left(2k\right)}^{j+1}}\right)\Vert \\ \le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\frac{{\left(2k\right)}^{{\alpha }_{2}j}}{{\left(2k\right)}^{{\alpha }_{1}rj}}{\Vert x\Vert }^r\end{array}$ (19)

for all nonnegative integers $p\mathrm{,}l$ with $p>l$ and all $x\in X$ . It follows from (19) 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

$\varphi \left(x\right):=\underset{n\to \infty }{\lim }{\left(2k\right)}^{n}f\left(\frac{x}{{\left(2k\right)}^n}\right)$

for all $x\in X$ . Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (19), we get (16).

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

It follows from (15) that

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

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

$\begin{array}{l}={\left|g\left(\lambda \right)\right|}^{{\alpha }_2}\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\end{array}$ (20)

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

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

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for $j=1\to n$ , So by Lemma 1 it follows that the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is additive. Now we need to prove uniqueness, Suppose ${\varphi }^{\prime }\mathrm{<mo>\; :\; </mo>}X\to Y$ is also an additive mapping that satisfies (16). Then we have

$\begin{array}{c}\Vert \varphi \left(x\right)-{\varphi }^{\prime }\left(x\right)\Vert ={\left(2k\right)}^{{\alpha }_{2}n}\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-{\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert \\ \le {\left(2k\right)}^{{\alpha }_{2}n}\left(\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert +\Vert {\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert \right)\\ \le \frac{2\cdot {\left(2k\right)}^{{\alpha }_{2}n}\cdot \left(2k^{{\alpha }_{1}r+1}+1\right)}{{\left(2k\right)}^{{\alpha }_{1}nr}\left({\left(2k\right)}^{{\alpha }_{1}r}-{\left(2k\right)}^{{\alpha }_2}\right)}\theta {\Vert x\Vert }^r\end{array}$ (21)

which tends to zero as $n\to \infty$ for all $x\in X$ . So we 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 (16) as we expected.

Theorem 3. 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

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

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

$\Vert f\left(x\right)-\varphi \left(x\right)\Vert \le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_2}-{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\Vert x\Vert }^r\mathrm{.}$ (23)

for all $x\in X$ .

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (22).

We 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 (22), we have

$\Vert \left(4k-2\right)f\left(0\right)\Vert \le \Vert 2kg\left(\lambda \right)f\left(0\right)\Vert$

therefore

$\left({\left|4k-2\right|}^{{\alpha }_2}-{\left|2kg\left(\lambda \right)\right|}^{{\alpha }_2}\right)\Vert f\left(0\right)\Vert$

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 (22) we have

$\Vert f\left(2kx\right)-2kf\left(x\right)\Vert \le \left(2k^{{\alpha }_{1}r+1}+1\right)\theta {\Vert x\Vert }^r$ (24)

for all $x\in X$ . Thus

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

for all $x\in X$ .

$\begin{array}{l}\Vert \frac{1}{{\left(2k\right)}^l}f\left({\left(2k\right)}^{l}x\right)-\frac{1}{{\left(2k\right)}^m}f\left({\left(2k\right)}^{m}x\right)\Vert \\ \le {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\Vert \frac{1}{{\left(2k\right)}^j}f\left({\left(2k\right)}^{j}x\right)-\frac{1}{{\left(2k\right)}^{j+1}}f\left({\left(2k\right)}^{j+1}x\right)\Vert \\ \le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_2}}\theta {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\frac{{\left(2k\right)}^{{\alpha }_{1}rj}}{{\left(2k\right)}^{{\alpha }_{2}j}}{\Vert x\Vert }^r\end{array}$ (26)

for all nonnegative integers $p\mathrm{,}l$ with $p>l$ and all $x\in X$ . It follows from (26) that the sequence $\left\{\frac{1}{{\left(2k\right)}^n}f\left({\left(2k\right)}^{n}x\right)\right\}$ is a cauchy sequence for all $x\in X$ . Since $Y$ is complete, the sequence $\left\{\frac{1}{{\left(2k\right)}^n}f\left({\left(2k\right)}^{n}x\right)\right\}$ coverges.

So one can define the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ by

$\varphi \left(x\right):=\underset{n\to \infty }{\lim }\frac{1}{{\left(2k\right)}^n}f\left({\left(2k\right)}^{n}x\right)$

for all $x\in X$ . Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (26), we get (23).

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

4. Stability Additive $g\left(\lambda \right)$ -Functional Inequalities (2) in $\left({\alpha }_1,{\alpha }_2\right)$ -Homogeneous F-Spaces

Now, we study the solutions of (2). Note that for these inequalities, when $X$ is a ${\alpha }_1$ -homogeneous F-spaces and that $Y$ is a ${\alpha }_2$ -homogeneous F-spaces. Under this setting, I can show that the mapping satisfying (2) is additive. These results are give in the following.

Lemma 4. Let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an odd mapping satilies

$\begin{array}{l}\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert g\left(\lambda \right)(f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\end{array}$ (27)

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 (27).

We 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 (27), we have

$\Vert 2kf\left(0\right)\Vert \le {\left|g\left(\lambda \right)\right|}^{{\alpha }_2}\Vert \left(4k-2\right)f\left(0\right)\Vert$

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 (27), we have

Thus

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

$f\left(\frac{x}{2k}\right)=\frac{1}{2k}f\left(x\right)$ (28)

for all $x\in X$ .

From (27) and (28) we infer that

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

$\begin{array}{l}-{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le {\left|g\left(\lambda \right)\right|}^{{\alpha }_2}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\end{array}$ (29)

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 we expected. The couverse is obviously true.

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

$\begin{array}{l}2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\\ =g\left(\lambda \right)(f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\end{array}$ (30)

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.

Note! The functional equation (30) is called an additive λ-functional equation.

Theorem 5. 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

$\begin{array}{l}\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \lambda (f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (31)

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

$\Vert f\left(x\right)-\varphi \left(x\right)\Vert \le \frac{{\left(2k\right)}^{{\alpha }_{1}r}}{{\left(2k\right)}^{{\alpha }_{1}r}-{\left(4k\right)}^{{\alpha }_2}}\theta {\Vert x\Vert }^r\mathrm{.}$ (32)

for all $x\in X$ .

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (38).

We 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 (38), we have

$\Vert 2f\left(0\right)\Vert \le {\left|\lambda \right|}^{{\alpha }_2}\Vert \left(4k-2\right)f\left(0\right)\Vert$

therefore

$\left({\left|4k-2\right|}^{{\alpha }_2}-{\left|2\lambda \right|}^{{\alpha }_2}\right)\Vert f\left(0\right)\Vert \le 0$

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 (38) we have

${\Vert 4f\left(\frac{x}{2k}\right)-\frac{1}{k}f\left(x\right)\Vert }_Y\le {\left(2k\right)}^{{\alpha }_{1}r}\theta {\Vert x\Vert }^r$ (33)

for all $x\in X$ . Thus

$\Vert 4kf\left(\frac{x}{2k}\right)-f\left(x\right)\Vert \le {\left(2k\right)}^{{\alpha }_{1}r}{k}^{{\alpha }_2}\theta {\Vert x\Vert }^r$ (34)

for all $x\in X$ .

$\begin{array}{l}\Vert {\left(4k\right)}^{l}f\left(\frac{x}{{\left(2k\right)}^l}\right)-{\left(4k\right)}^{m}f\left(\frac{x}{{\left(2k\right)}^m}\right)\Vert \\ \le {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\Vert {\left(4k\right)}^{j}f\left(\frac{x}{{\left(2k\right)}^j}\right)-{\left(4k\right)}^{j+1}f\left(\frac{x}{{\left(2k\right)}^{j+1}}\right)\Vert \\ \le {\left(2k\right)}^{{\alpha }_{1}r}{k}^{{\alpha }_2}\theta {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\frac{{\left(4k\right)}^{{\alpha }_{2}j}}{{\left(2k\right)}^{{\alpha }_{1}rj}}{\Vert x\Vert }^r\end{array}$ (35)

for all nonnegative integers $p\mathrm{,}l$ with $p>l$ and all $x\in X$ . It follows from (35) 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 (35), we get (39). Form $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is even, the mapping

$\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$

is even. It follows from (38) that

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

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

$\begin{array}{l}=\Vert \varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)\\ -{{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(-z_j\right)\Vert }_Y\end{array}$ (36)

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

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

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for $j=1\to n$ , So by Lemma 4.1 it follows that the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is additive. Now we need to prove uniqueness, Suppose ${\varphi }^{\prime }\mathrm{<mo>\; :\; </mo>}X\to Y$ is also a quadratic mapping that satisfies (39). Then we have

$\begin{array}{l}\Vert \varphi \left(x\right)-{\varphi }^{\prime }\left(x\right)\Vert ={\left(4k\right)}^{{\alpha }_{2}n}\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-{\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert \\ \le {\left(4k\right)}^{{\alpha }_{2}n}\left(\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert +\Vert {\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert \right)\\ \le \frac{2\cdot {\left(4k\right)}^{{\alpha }_{2}n}\cdot {\left(2k\right)}^{{\alpha }_{1}r}}{{\left(2k\right)}^{{\alpha }_{1}nr}\left({\left(2k\right)}^{{\alpha }_{1}r}-{\left(4k\right)}^{{\alpha }_2}\right)}\theta {\Vert x\Vert }^r\end{array}$ (37)

which tends to zero as $n\to \infty$ for all $x\in X$ . So we 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 (39) as we 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 a mapping such that

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

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

$\Vert f\left(x\right)-\varphi \left(x\right)\Vert \le \frac{{\left(2k\right)}^{{\alpha }_{1}r}}{{\left(4k\right)}^{{\alpha }_2}-{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\Vert x\Vert }^r.$ (39)

for all $x\in X$ .

The proof is similar to theorem 5.

5. Hyers-Ulam-Rassias Stability Quadratic $g\left(\lambda \right)$ -Functional Inequalities (1) in $\left({\alpha }_1,{\alpha }_2\right)$ -Homogeneous F-Spaces

Now, we first study the solutions of (1). Note that for these inequalities, when $X$ is a ${\alpha }_1$ -homogeneous F-spaces and that $Y$ is a ${\alpha }_2$ -homogeneous F-spaces. Under this setting, we can show that the mapping satisfying (1) is quadratic. These results are give in the following.

Lemma 7. Let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an even mapping satilies

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert g\left(\lambda \right)(2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\end{array}$ (40)

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 quadratic.

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (40).

We 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 (40), we have

$\Vert \left(4k-2\right)f\left(0\right)\Vert \le {\left|\lambda \right|}^{{\alpha }_2}\Vert 2kf\left(0\right)\Vert \le 0$

therefore

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 (40), we have

Thus

$\Vert f\left(2kx\right)-2kf\left(x\right)\Vert \le 0$

$f\left(\frac{x}{2k}\right)=\frac{1}{2k}f\left(x\right)$ (41)

for all $x\in X$ .

From (40) and (41) we infer that

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert \lambda (2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\\ ={\left|\lambda \right|}^{{\alpha }_2}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f(\frac{x_j+y_j}{2k})-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\end{array}$ (42)

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

$f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)=2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)$ (43)

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

As we expected. The couverse is obviously true.

Corollary 3. Let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an even mapping satilies

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

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 quadratic.

Note! The functional equation (44) is called an quadratic $g\left(\lambda \right)$ -functional equation.

Theorem 8. 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

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert g\left(\lambda \right)(2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (45)

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

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

for all $x\in X$ .

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (45).

We 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 (45), we have

$\Vert \left(4k-2\right)f\left(0\right)\Vert \le \Vert 2kg\left(\lambda \right)f\left(0\right)\Vert$

therefore

$\left({\left|4k-2\right|}^{{\alpha }_2}-{\left|2kg\left(\lambda \right)\right|}^{{\alpha }_2}\right)\Vert f\left(0\right)\Vert$

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 (45) we have

$\Vert f\left(2kx\right)-2kf\left(x\right)\Vert \le \left(2k^{{\alpha }_{1}r+1}+1\right)\theta {\Vert x\Vert }^r$ (47)

for all $x\in X$ . Thus

$\Vert f\left(x\right)-2kf\left(\frac{x}{2k}\right)\Vert \le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\Vert x\Vert }^r$ (48)

for all $x\in X$ .

$\begin{array}{l}\Vert {\left(2k\right)}^{l}f\left(\frac{x}{{\left(2k\right)}^l}\right)-{\left(2k\right)}^{m}f\left(\frac{x}{{\left(2k\right)}^m}\right)\Vert \\ \le {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\Vert {\left(2k\right)}^{j}f\left(\frac{x}{{\left(2k\right)}^j}\right)-{\left(2k\right)}^{j+1}f\left(\frac{x}{{\left(2k\right)}^{j+1}}\right)\Vert \\ \le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\frac{{\left(2k\right)}^{{\alpha }_{2}j}}{{\left(2k\right)}^{{\alpha }_{1}rj}}{\Vert x\Vert }^r\end{array}$ (49)

for all nonnegative integers $p\mathrm{,}l$ with $p>l$ and all $x\in X$ . It follows from (49) 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 (49), we get (46).

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

It follows from (45) that

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

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

$\begin{array}{l}+\underset{n\to \infty }{\lim }\frac{{\left(2k\right)}^{{\alpha }_{2}n}}{{\left(2k\right)}^{{\alpha }_{1}nr}}\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\\ ={\left|g\left(\lambda \right)\right|}^{{\alpha }_2}\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\end{array}$ (50)

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

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

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for $j=1\to n$ , So by Lemma 7 it follows that the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is quadratc. Now we need to prove uniqueness, Suppose ${\varphi }^{\prime }\mathrm{<mo>\; :\; </mo>}X\to Y$ is also an additive mapping that satisfies (46). Then we have

$\begin{array}{l}\Vert \varphi \left(x\right)-{\varphi }^{\prime }\left(x\right)\Vert ={\left(2k\right)}^{{\alpha }_{2}n}\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-{\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert \\ \le {\left(2k\right)}^{{\alpha }_{2}n}\left(\Vert \varphi \left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert +\Vert {\varphi }^{\prime }\left(\frac{x}{{\left(2k\right)}^n}\right)-f\left(\frac{x}{{\left(2k\right)}^n}\right)\Vert \right)\\ \le \frac{2\cdot {\left(2k\right)}^{{\alpha }_{2}n}\cdot \left(2k^{{\alpha }_{1}r+1}+1\right)}{{\left(2k\right)}^{{\alpha }_{1}nr}\left({\left(2k\right)}^{{\alpha }_{1}r}-{\left(2k\right)}^{{\alpha }_2}\right)}\theta {\Vert x\Vert }^r\end{array}$ (51)

which tends to zero as $n\to \infty$ for all $x\in X$ . So we 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 (46) as we 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 a mapping such that

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert g\left(\lambda \right)(2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (52)

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

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

for all $x\in X$ .

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (52).

We 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 (52), we have

$\Vert \left(4k-2\right)f\left(0\right)\Vert \le \Vert 2kg\left(\lambda \right)f\left(0\right)\Vert$

therefore

$\left({\left|4k-2\right|}^{{\alpha }_2}-{\left|2kg\left(\lambda \right)\right|}^{{\alpha }_2}\right)\Vert f\left(0\right)\Vert$

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 (52) we have

$\Vert f\left(2kx\right)-2kf\left(x\right)\Vert \le \left(2k^{{\alpha }_{1}r+1}+1\right)\theta {\Vert x\Vert }^r$ (54)

for all $x\in X$ . Thus

$\Vert f\left(x\right)-\frac{1}{2k}f\left(2kx\right)\Vert \le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_2}}\theta {\Vert x\Vert }^r$ (55)

for all $x\in X$ .

$\begin{array}{l}\Vert \frac{1}{{\left(2k\right)}^l}f\left({\left(2k\right)}^{l}x\right)-\frac{1}{{\left(2k\right)}^m}f\left({\left(2k\right)}^{m}x\right)\Vert \\ \le {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\Vert \frac{1}{{\left(2k\right)}^j}f\left({\left(2k\right)}^{j}x\right)-\frac{1}{{\left(2k\right)}^{j+1}}f\left({\left(2k\right)}^{j+1}x\right)\Vert \\ \le \frac{2k^{{\alpha }_{1}r+1}+1}{{\left(2k\right)}^{{\alpha }_2}}\theta {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\frac{{\left(2k\right)}^{{\alpha }_{1}rj}}{{\left(2k\right)}^{{\alpha }_{2}j}}{\Vert x\Vert }^r\end{array}$ (56)

for all nonnegative integers $p\mathrm{,}l$ with $p>l$ and all $x\in X$ . It follows from (56) that the sequence $\left\{\frac{1}{{\left(2k\right)}^n}f\left({\left(2k\right)}^{n}x\right)\right\}$ is a cauchy sequence for all $x\in X$ . Since $Y$ is complete, the sequence $\left\{\frac{1}{{\left(2k\right)}^n}f\left({\left(2k\right)}^{n}x\right)\right\}$ coverges.

So one can define the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ by

$\varphi \left(x\right):=\underset{n\to \infty }{\lim }\frac{1}{{\left(2k\right)}^n}f\left({\left(2k\right)}^{n}x\right)$

for all $x\in X$ . Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (56), we get (53).

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

6. Stability Quadratic λ-Functional Inequalities (2) in $\left({\alpha }_1,{\alpha }_2\right)$ -Homogeneous F-Spaces

Now, we study the solutions of (2). Note that for these inequalities, when $X$ is a ${\alpha }_1$ -homogeneous F-spaces and that $Y$ is a ${\alpha }_2$ -homogeneous F-spaces. Under this setting, we can show that the mapping satisfying (2) is quadratic. These results are give in the following.

Lemma 10. Let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an even mapping satilies

$\begin{array}{l}\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert g\left(\lambda \right)(f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\end{array}$ (57)

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 quadratic.

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (57).

We 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 (57), we have

$\Vert 2kf\left(0\right)\Vert \le {\left|g\left(\lambda \right)\right|}^{{\alpha }_2}\Vert \left(4k-2\right)f\left(0\right)\Vert$

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 (57), we have

Thus

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

$f\left(\frac{x}{2k}\right)=\frac{1}{2k}f\left(x\right)$ (58)

for all $x\in X$ .

From (57) and (58) we infer that

$\begin{array}{l}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ =\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le {\left|g\left(\lambda \right)\right|}^{{\alpha }_2}\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\end{array}$ (59)

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

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

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for $j=1\to n$ , as we expected. The couverse is obviously true.

Corollary 4. Let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be an even mapping satilies

$\begin{array}{l}2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\\ =g\left(\lambda \right)(f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\end{array}$ (60)

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 quadratic.

Note! The functional equation (60) is called a quadratic $g\left(\lambda \right)$ -functional equation.

Theorem 11. 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 even mapping such that $f\left(0\right)=0$ and

$\begin{array}{l}\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{3{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert g\left(\lambda \right)(f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (61)

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

$\Vert f\left(x\right)-\varphi \left(x\right)\Vert \le \frac{{\left(2k\right)}^{{\alpha }_{1}r}}{{\left(2k\right)}^{{\alpha }_{1}r}-{\left(4k\right)}^{{\alpha }_2}}\theta {\Vert x\Vert }^r$ (62)

for all $x\in X$ .

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (61).

We 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 (61), we have

$\Vert 2f\left(0\right)\Vert \le {\left|g\left(\lambda \right)\right|}^{{\alpha }_2}\Vert \left(4k-2\right)f\left(0\right)\Vert$

therefore

$\left({\left|4k-2\right|}^{{\alpha }_2}-{\left|2g\left(\lambda \right)\right|}^{{\alpha }_2}\right)\Vert f\left(0\right)\Vert \le 0$

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 (61) we have

${\Vert 4f\left(\frac{x}{2k}\right)-\frac{1}{k}f\left(x\right)\Vert }_Y\le {\left(2k\right)}^{{\alpha }_{1}r}\theta {\Vert x\Vert }^r$ (63)

for all $x\in X$ . Thus

$\Vert 4kf\left(\frac{x}{2k}\right)-f\left(x\right)\Vert \le {\left(2k\right)}^{{\alpha }_{1}r}{k}^{{\alpha }_2}\theta {\Vert x\Vert }^r$ (64)

for all $x\in X$ .

$\begin{array}{l}\Vert {\left(4k\right)}^{l}f\left(\frac{x}{{\left(2k\right)}^l}\right)-{\left(4k\right)}^{m}f\left(\frac{x}{{\left(2k\right)}^m}\right)\Vert \\ \le {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\Vert {\left(4k\right)}^{j}f\left(\frac{x}{{\left(2k\right)}^j}\right)-{\left(4k\right)}^{j+1}f\left(\frac{x}{{\left(2k\right)}^{j+1}}\right)\Vert \\ \le {\left(2k\right)}^{{\alpha }_{1}r}{k}^{{\alpha }_2}\theta {\displaystyle \underset{j=1}{\overset{m-1}{\sum }}}\frac{{\left(4k\right)}^{{\alpha }_{2}j}}{{\left(2k\right)}^{{\alpha }_{1}rj}}{\Vert x\Vert }^r\end{array}$ (65)

for all nonnegative integers $p\mathrm{,}l$ with $p>l$ and all $x\in X$ . It follows from (65) 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 (65), we get (62). The rest of the proof is similar to the proof of Theorem 8.

Theorem 12. 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 a mapping such that

$\begin{array}{l}\Vert 2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+2f{\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{{\left(2k\right)}^2}-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}z\right)}_j\\ -\frac{3}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)+\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-\frac{x_j+y_j}{2k}\right)-\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\frac{1}{2k}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right)\Vert }_Y\\ \le \Vert g\left(\lambda \right)(f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(-z_j\right))\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (66)

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

$\Vert f\left(x\right)-\varphi \left(x\right)\Vert \le \frac{{\left(2k\right)}^{{\alpha }_{1}r}}{{\left(4k\right)}^{{\alpha }_2}-{\left(2k\right)}^{{\alpha }_{1}r}}\theta {\Vert x\Vert }^r\mathrm{.}$ (67)

for all $x\in X$ .

The proof is similar to theorem 8 and 9.

7. Conclusion

In this article, I construct two general functional inequalities with multivariables on homogeneous space and show that their solutions are additive-quadratic maps.

Conflicts of Interest

The author declares no conflicts of interest.

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