Ly Van An
Faculty of Mathematics Teacher Education, Tay Ninh University, Tay Ninh, Vietnam.
DOI: 10.4236/oalib.1111241   PDF    HTML   XML   85 Downloads   669 Views   Citations

Abstract

In this paper, I study to solve functional inequalities and equations of type Cauchy-Jensen with 3k-variables in a general form. I first introduce the con-cept of the general Cauchy-Jensen equation and next, I use the direct method of proving the solutions of the Jensen-Cauchy functional inequalities relative to the general Cauchy-Jensen equations and then I show that their solutions are mappings that are additive mappings calculated and finally apply the de-rivative setup on fuzzy algebra also the results of the paper.

Keywords

Functional Equation, Functional Inequality Additivity, Banach Space, Derivation on Fuzzy-Algebras

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1. Introduction

Let $G$ be an m-divisible group where $m\in \mathbb{N}\backslash \left\{0\right\}$ and $X$ , $Y$ be a normed space on the same field $\mathbb{K}$ , and $f\mathrm{<mo>\; :\; </mo>}G\to X$ ( $f\mathrm{<mo>\; :\; </mo>}G\to Y$ ) be a mapping. I use the notation ${\Vert \cdot \Vert }_X$ ( ${\Vert \cdot \Vert }_Y$ ) for corresponding the norms on $X$ and $Y$ . In this paper, I investigate functional inequalities and equations when when $G$ be an m-divisible group where $m\in \mathbb{N}$ and $X$ is a normed space with norm ${\Vert \cdot \Vert }_X$ and that $Y$ is a Banach space with norm ${\Vert \cdot \Vert }_Y$ .

In fact, when $G$ be an m-divisible group where $m\in \mathbb{N}$ and $X$ is a normed space with norm ${\Vert \cdot \Vert }_X$ and that $Y$ is a Banach space with norm ${\Vert \cdot \Vert }_Y$ I solve and prove the Hyers-Ulam-Rassias type stability of following functional inequalities and equations.

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

and

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

Where k is a positive integer.

The study of the functional equation stability originated from a question of S. M. Ulam [1] , concerning the stability of group homomorphisms. Let $\left(G\mathrm{,}\ast \right)$ be a group and let $\left(G^{\prime }\mathrm{,}\circ \mathrm{,}d\right)$ be a metric group with metric $d\left(\cdot \mathrm{,}\cdot \right)$ . Geven $\epsilon >0$ , does there exist a $\delta >0$ such that if $f\mathrm{<mo>\; :\; </mo>}G\to G^{\prime }$ satisfies: $d\left(f\left(x\ast y\right),f\left(x\right)\circ f\left(y\right)\right)<\delta$ for all $x\mathrm{,}y\in G$ then there is a homomorphism $h\mathrm{<mo>\; :\; </mo>}G\to G^{\prime }$ with $d\left(f\left(x\right)\mathrm{,}h\left(x\right)\right)<\epsilon$ , for all $x\in G$ , if the answer, is affirmative, I would say that equation of homomophism $h\left(x\ast y\right)=h\left(y\right)\circ h\left(y\right)$ is stable. The concept of stability for a functional equation arises when we replace a functional equation with an inequality which acts as a perturbation of the equation. Thus the stability question of functional equations is how the solutions of the inequality differ from those of the given function equation. Hyers gave a first affirmative answer the question Ulam as follows:

In 1941 D. H. Hyers [2] Let $\epsilon \ge 0$ and let $f\mathrm{<mo>\; :\; </mo>}{E}_1\to E_2$ be a mapping between Banach space such that $\Vert f\left(x+y\right)-f\left(x\right)-f\left(y\right)\Vert \le \epsilon \mathrm{,}$ for all $x\mathrm{,}y\in E_1$ and some $\epsilon \ge 0$ . It was shown that the limit $T\mathrm{<mo\; stretchy="false">\; (\; </mo>}x\mathrm{<mo\; stretchy="false">\; )\; </mo>\; <mo>\; =\; </mo>}\underset{n\to \infty }{{\displaystyle \lim }}\frac{f{\mathrm{<mo\; stretchy="false">\; (\; </mo>2}}^{n}x\mathrm{<mo\; stretchy="false">\; )\; </mo>}}{2^n}$ exists for all $x\in E_1$ and that $T\mathrm{<mo>\; :\; </mo>}{E}_1\to E_2$ is that unique additive mapping satisfying $\Vert f\left(x\right)-T\left(x\right)\Vert \le \epsilon \mathrm{,}\forall x\in E_1\mathrm{.}$

Next in 1978 Th. M. Rassias [3] provided a generalization of Hyers’ Theorem which allows the Cauchy difference to be unbounded:

Consider $E\mathrm{,}{E}^{\prime }$ to be two Banach spaces, and let $f\mathrm{<mo>\; :\; </mo>}E\to E^{\prime }$ be a mapping such that $f\left(tx\right)$ is continous in t for each fixed x. Assume that there exist $\theta \ge 0$ and $p\in \left[\mathrm{0,1}\right)$ such that $\Vert f\left(x+y\right)-f\left(x\right)-f\left(y\right)\Vert \le \epsilon \left({\Vert x\Vert }^p+{\Vert y\Vert }^p\right)\mathrm{,}\forall x\mathrm{,}y\in \mathbb{E}\mathrm{.}$ then there exists a unique linear $L\mathrm{<mo>\; :\; </mo>}E\to E^{\prime }$ satifies $\Vert f\left(x\right)-L\left(x\right)\Vert \le \frac{2\theta }{2-2^p}\Vert x\Vert \mathrm{,}x\in E\mathrm{.}$

Next J. M. Rassias [4] following the spirit of the innovative approach of Th. M. Rassias for the unbounded Cauchy difference proved a similar stability theorem in which he replaced the factor ${\Vert x\Vert }^p+{\Vert y\Vert }^p$ by ${\Vert x\Vert }^p{\Vert y\Vert }^p$ for $p\mathrm{,}q\in ℝ$ with $p+q\ne 1$ .

Next in 1992, a generalized of Rassias’ Theorem was obtained by Găvruta [5] .

Let $\left(G\mathrm{,}+\right)$ be a group Abelian and $E$ a Banach space.

Denote by $\varphi \mathrm{<mo>\; :\; </mo>}G\times G\to \left[\mathrm{0,}\infty \right)$ a function such that $\tilde{\varphi }\left(x,y\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}{2}^{-n}\varphi \left(2^{n}x,2^{n}y\right)<\infty$ for all $x\mathrm{,}y\in G$ . Suppose that $f\mathrm{<mo>\; :\; </mo>}G\to E$ is a mapping satisfying $\Vert f\left(x+y\right)-f\left(x\right)-f\left(y\right)\Vert \le \epsilon$ , $\forall x,y\in G$ . There exists a unique additive mapping $T\mathrm{<mo>\; :\; </mo>}G\to E$ such that $\Vert f\left(x\right)-T\left(x\right)\Vert \le \tilde{\varphi }\left(x\mathrm{,}x\right)$ , $\forall x,y\in G$ .

Generally speaking for a more specific problem, when considering this famous result, the additive Cauchy equation $f\left(x+y\right)=f\left(x\right)+f\left(y\right)$ is said to have the Hyers-Ulam stability on $\left(E_1\mathrm{,}{E}_2\right)$ with $E_1$ and $E_2$ are Banach spaces if for each $f\mathrm{<mo>\; :\; </mo>}{E}_1\to E_2$ satisfying $\Vert f\left(x+y\right)-f\left(x\right)-f\left(y\right)\Vert \le \epsilon$ for all $x\mathrm{,}y\in E_1$ for some $\epsilon >0$ , there exists an additive $h\mathrm{<mo>\; :\; </mo>}{E}_1\to E_2$ such that $f-h$ is bounded on $E_1$ . The method which was provided by Hyers, and which produces the additive h, was called a direct method.

Afterward, Gilány showed that if satisfies the functional inequality

$\Vert 2f\left(x\right)+2f\left(y\right)-f\left(x{y}^{-1}\right)\Vert \le \Vert f\left(xy\right)\Vert$ (3)

Then f satisfies the Jordan-von Newman functional equation

$2f\left(x\right)+2f\left(y\right)=f\left(xy\right)+f\left(x{y}^{-1}\right)$ (4)

Gilányi [6] and Fechner [7] proved the Hyers-Ulam stability of the functional inequality.

Recently, the authors studied the Hyers-Ulam stability for the following functional inequalities and equation

$\Vert f\left(x\right)+f\left(y\right)+2f\left(y\right)\Vert \le \Vert 2f\left(\frac{x+y}{2}+z\right)\Vert$ (5)

$f\left(x\right)+f\left(y\right)+2f\left(y\right)=2f\left(\frac{x+y}{2}+z\right)$ (6)

in Banach spaces.

In this paper, I solve and prove the Hyers-Ulam stability for inequality (1.1) is related to Equation (1.2), ie the functional inequalities and equation with 3k variables. Under suitable assumptions on spaces $G$ and $X$ or $G$ and $Y$ , I will prove that the mappings satisfy the (1.1) - (1.2). Thus, the results in this paper are generalization of those in [1] - [33] for inequality (1.1) is related to Equation (1.2) with 3k variables.

The paper is organized as follows:

In the section preliminary, I remind some basic notations such as:

Concept of the divisible group, definition of the stability of Cauchy-Jenen functional inequalities and functional equation, Solutions of the equation, functional inequalities and functional equation, the crucial problem when constructing solutions for Cauchy-Jensen inequalities.

Section 3: Establish a solution to the generalized Cauchy-Jensen functional inequalities (2.2) when I assume that G be a m-divisible abelian group and X is a normed space.

Section 4: Stability of functional inequalities (1.1) related to the Cauchy-Jensen equation when I assume that G be a m-divisible abelian group and Y is a Banach space.

Section 5: Establish solutions to functional inequalities (1.1) based on the definition when I assume that G be a m-divisible abelian group and Y is a Banach space.

Section 6: The stability of derivation on fuzzy-algebras.

2. Preliminaries

2.1. Concept of Divisible Group

A group $G$ is called divisible if for every $x\in G$ and every positive integer n there is a $y\in G$ so that $ny=x$ , i.e., every element of $G$ is divisible by every positive integer. A abelian group $G$ is called divisible if for every $x\in G$ and every $n\in \mathbb{N}$ there is some $y\in G$ so that $x=ny$ . divisible by every positive integer. Let $G$ be an n-divisible abelian group where $n\in \mathbb{N}$ (i.e., $a\to na\mathrm{<mo>\; :\; </mo>}G\to G$ is a surjection).

Denote by

$M\left(G\mathrm{,}X\right)=\left\{f\mathrm{<mo>\; |\; </mo>}f\mathrm{<mo>\; :\; </mo>}G\to X\right\}$

$L^{\infty }\left(G\mathrm{,}X\right)=\left\{f\mathrm{<mo>\; :\; </mo>}G\to X\mathrm{<mo>\; |\; </mo>}{\Vert f\Vert }_{\infty }\mathrm{<mo>\; :\; </mo>}={\sup }_{x\in G}{\Vert f\Vert }_X<\infty \right\}$

The sets $M\left(G\mathrm{,}Y\right)\mathrm{,}M\left(G^r\mathrm{,}X\right)$ and $M\left(G^r\mathrm{,}{ℝ}^{+}\right)$ can be defined similarly where

$G^r=\left\{\left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_r\right)\mathrm{<mo>\; :\; </mo>}{x}_j\in G\mathrm{,}j=\mathrm{1,}\cdots \mathrm{,}k\right\}$

2.2. Definition of the Stability of Functional Inequalities and Functional Equation

Given mappings $E\mathrm{<mo>\; :\; </mo>}M\left(G\mathrm{,}X\right)\to M\left(G^r\mathrm{,}{ℝ}^{+}\right)$ , $\phi \mathrm{<mo>\; :\; </mo>}{G}^r\to ℝ$ and $\psi \mathrm{<mo>\; :\; </mo>}G\to {ℝ}^{+}$ . If $E\left(f\right)\left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_r\right)\le \phi \left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_r\right)$ for all $x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_r\in G$ implies that there exists $g\in M\left(G\mathrm{,}X\right)$ such that $E\left(g\right)\le 0$ and ${\Vert f\left(x\right)-g\left(x\right)\Vert }_{\infty }\le \psi \left(x\right)$ , for all $x\in G$ , then we say that the inequality $E\left(f\right)\le 0$ is $\left(\phi \mathrm{,}\psi \right)$ -stable in $M\left(G\mathrm{,}X\right)$ . In this case, we also say that the solutions of the inequality $E\left(f\right)\le 0$ is $\left(\phi \mathrm{,}\psi \right)$ -stable in $M\left(G\mathrm{,}X\right)$ . Given mappings $E\mathrm{<mo>\; :\; </mo>}M\left(G\mathrm{,}X\right)\to M\left(G^r\mathrm{,}{ℝ}^{+}\right)$ , $\phi \mathrm{<mo>\; :\; </mo>}{G}^r\to ℝ$ and $\psi \mathrm{<mo>\; :\; </mo>}G\to {ℝ}^{+}$ if ${\Vert E\left(f\right)\left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_r\right)\Vert }_{\infty }\le \phi \left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_r\right)$ for all $x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_r\in G$ , implies that there exists $g\in M\left(G\mathrm{,}X\right)$ such that $E\left(g\right)=0$ and ${\Vert f\left(x\right)-g\left(x\right)\Vert }_{\infty }\le \psi \left(x\right)$ , for all $x\in G$ , then we say that the inequality $E\left(f\right)\le 0$ is $\left(\phi \mathrm{,}\psi \right)$ -stable in $M\left(G\mathrm{,}X\right)$ . In this case, we also say that the solutions of the inequality $E\left(f\right)=0$ is $\left(\phi \mathrm{,}\psi \right)$ -stable in $M\left(G\mathrm{,}X\right)$ .

It is well known that if an additive function $f\mathrm{<mo>\; :\; </mo>}ℝ\to ℝ$ satisfies one of the following conditions:

1) f is continuous at a point;

2) f is monotonic on an interval of positive length;

3) f is bounded on an interval of positive length;

4) f is integrable;

5) f is measurable;

then f is of the form $f\left(x\right)=cx$ with a real constant c.

2.3. Solutions of the Equation

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 an Jensen additive mapping.

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

2.4. Solutions of the Functional Inequalities

The functional inequalities $\Vert f\left(x\right)+f\left(y\right)+2f\left(z\right)\Vert \le \Vert 2f\left(\frac{x+y}{2}+z\right)\Vert$ is called the Cauchy-Jensen inequalities. In particular, every solution of the inequalities is said to be an additive mapping

2.5. The Crucial Problem When Constructing Solutions for Cauchy-Jensen Inequalities

Suppose a mapping $f\mathrm{<mo>\; :\; </mo>}G\to X$ , the equation

${\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+m{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)=mf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{m}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)$ (7)

is said to a generalized Cauchy-Jensen equation.

And function inequalities

${\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+m{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\le mf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{m}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)$ (8)

is said to a generalized Cauchy-Jensen function inequalitiess Note: case $m=2$ and $k=1$ so (7) it is called a classical Cauchy-Jensen equation, (8) it is called a Cauchy-Jensen function inequalities.

3. Establish a Solution to the Generalized Cauchy-Jensen Functional Inequality

Now, I first study the solutions of (8). Note that for inequalities, $G$ be a m-divisible group where $m\in \mathbb{N}\backslash \left\{0\right\}$ and $X$ be a normed spaces. Under this setting, I can show that the mapping satisfying (8) is additive. These results are give in the following.

Lemma 1. Let $f\mathrm{<mo>\; :\; </mo>}G\to X$ be a mapping such that satisfies

${\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+m{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\Vert }_X\le {\Vert mf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{m}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_X$ (9)

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

Proof. Prerequisites

Assume that $f\mathrm{<mo>\; :\; </mo>}G\to Y$ satisfies (9) 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 (9), I get $\left|2k+m\right|{\Vert f\left(0\right)\Vert }_X\le \left|m\right|{\Vert f\left(0\right)\Vert }_X$ $\left(\left|2k+m\right|-\left|m\right|\right){\Vert f\left(0\right)\Vert }_X\le 0$

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

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 $\left(-mz\mathrm{,}\cdots \mathrm{,}-mz\mathrm{,0,}\cdots \mathrm{,0,}z,\cdots \mathrm{,}z\right)$ in (9), I get $\Vert kf\left(-mz\right)+kmf\left(z\right)\Vert \le 0$ and so

$f\left(-mz\right)=-mf\left(z\right)$ (10)

for all $z\in G$ .

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 $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}-\frac{x_j+y_j}{m}\mathrm{,}\cdots \mathrm{,}-\frac{x_j+y_j}{m}\right)$ in (9) and (10) I have

$\begin{array}{l}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+m{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\Vert }_X\\ ={\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j+y_j\right)\Vert }_X\\ \le {\Vert mf\left({\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}\frac{x_j+y_j}{m}-{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}\frac{x_j+y_j}{m}\right)\Vert }_X={\Vert f\left(0\right)\Vert }_X=0\end{array}$ (11)

Therefore

${\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)={\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j+y_j\right)$ (12)

Finally we replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\right)$ by $\left(u\mathrm{,}\cdots \mathrm{,}u\mathrm{,}v\mathrm{,}\cdots \mathrm{,}v\right)$ in (12) so $f\left(u\right)+f\left(v\right)=f\left(u+v\right)$ .

Sufficient conditions:

Suppose $f\mathrm{<mo>\; :\; </mo>}G\to Y$ is additive. Then

$f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j\right)={\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)$ (13)

and so $f\left(p{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j\right)=p{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)$ for all $p\in \mathbb{Q}$ and $x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_r\in G$ .

Therefore

$\begin{array}{l}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+m{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\\ =mf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{m}\right)+m{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)=mf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{m}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\end{array}$ (14)

So I have something to prove

${\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+m{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\Vert }_Y\le {\Vert mf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{m}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y$ (15)

From the proof of the lemma 2, I get the following corollary:

Corollary 1. Suppose a mapping $f\mathrm{<mo>\; :\; </mo>}G\to X$ , The following clauses are equivalent

1) f is additive.

2) ${\displaystyle {\sum }_{j=1}^k}f\left(x_j\right)+{\displaystyle {\sum }_{j=1}^k}f\left(y_j\right)+m{\displaystyle {\sum }_{j=1}^k}f\left(z_j\right)=mf\left({\displaystyle {\sum }_{j=1}^k}\frac{x_j+y_j}{m}+{\displaystyle {\sum }_{j=1}^k}{z}_j\right)$ ,

$\forall x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in G$ , $j=\mathrm{1,}\cdots \mathrm{,}k$ .

3) $\Vert {\displaystyle {\sum }_{j=1}^k}f\left(x_j\right)+{\displaystyle {\sum }_{j=1}^k}f\left(y_j\right)+m{\displaystyle {\sum }_{j=1}^k}f\left(z_j\right)\Vert \le \Vert mf\left({\displaystyle {\sum }_{j=1}^k}\frac{x_j+y_j}{m}+{\displaystyle {\sum }_{j=1}^k}{z}_j\right)\Vert$

$\forall x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in G$ .

Note: Clearly, a vector space is a m-divisible abelian group, so Corollary 3.2 is right when $G$ is a vector space.

Through the Lemma 2 proof, I have the remark:

Remark: When the letting m = 2k (means that m is always even) and $G$ is an m-divisible abelian gourp then $G$ must be a 2-divisible abelian gourp.

4. Stability of Functional Inequalities Related to the Cauchy-Jensen Equation

Now, I first study the solutions of (1.1). Note that for inequalities, $G$ be a m-divisible group where $m\in \mathbb{N}\backslash \left\{0\right\}$ and $Y$ be a Banach spaces. Under this setting, I can show that the mapping satisfying (1.1) is additive. These results are give in the following.

Theorem 2. For $\varphi \mathrm{<mo>\; :\; </mo>}{G}^{3k}\to {ℝ}^{+}$ be a function such that

$\underset{n\to \infty }{\lim }\frac{1}{{\left(2k\right)}^n}\varphi \left({\left(2k\right)}^n{x}_1,\cdots ,{\left(2k\right)}^n{x}_k,{\left(2k\right)}^n{y}_1,\cdots ,{\left(2k\right)}^n{y}_k,\cdots ,{\left(2k\right)}^n{z}_1,\cdots ,{\left(2k\right)}^n{z}_k\right)=0$ (16)

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

And

$\begin{array}{l}\tilde{\varphi }\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\\ ={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{1}{{\left(2k\right)}^{n+1}}\varphi \left({\left(2k\right)}^{n+1}{x}_1,\cdots ,{\left(2k\right)}^{n+1}{x}_k,0,\cdots ,0,{\left(2k\right)}^n{z}_1,\cdots ,{\left(2k\right)}^n{z}_k\right)<\infty \end{array}$ (17)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\mathrm{,}{z}_j\in G$ . Suppose that an odd mapping $f\mathrm{<mo>\; :\; </mo>}G\to Y$ satisfies

$\begin{array}{l}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\Vert }_Y\\ \le {\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y+\varphi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (18)

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

Then there exists a unique additive mapping $\psi \mathrm{<mo>\; :\; </mo>}G\to Y$ such that

${\Vert f\left(x\right)-\psi \left(x\right)\Vert }_Y\le \tilde{\varphi }\left(x\mathrm{,}\cdots \mathrm{,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)$ (19)

for all $x\in G$ .

Proof. 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 (18), we get

$\left(\left|2k^2+k\right|-\left|2k\right|\right){\Vert f\left(0\right)\Vert }_Y\le 0.$ (20)

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

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 $\left(2kx\mathrm{,}\cdots \mathrm{,2}kx\mathrm{,0,}\cdots \mathrm{,0,}-x\mathrm{,}\cdots \mathrm{,}-x\right)$ in (18), I get

${\Vert kf\left(2kx\right)-2k^{2}f\left(x\right)\Vert }_Y\le \varphi \left(2kx\mathrm{,2}kx\mathrm{,}\cdots \mathrm{,2}kx\mathrm{,0,0,}\cdots \mathrm{,0,}-x\mathrm{,}-x\mathrm{,}\cdots \mathrm{,}-x\right)$ (21)

${\Vert f\left(x\right)-\frac{1}{2k}f\left(2kx\right)\Vert }_Y\le \frac{1}{2k^2}\varphi \left(2kx\mathrm{,2}kx\mathrm{,}\cdots \mathrm{,2}kx\mathrm{,0,0,}\cdots \mathrm{,0,}-x\mathrm{,}-x\mathrm{,}\cdots \mathrm{,}-x\right)$

Hence

$\begin{array}{l}{\Vert \frac{1}{{\left(2k\right)}^l}f\left({\left(2k\right)}^{l}x\right)-\frac{1}{{\left(2k\right)}^m}f\left({\left(2k\right)}^{m}x\right)\Vert }_Y\\ \le {\displaystyle \underset{j=l}{\overset{m-1}{\sum }}}{\Vert \frac{1}{{\left(2k\right)}^j}f\left({\left(2k\right)}^{j}x\right)-\frac{1}{{\left(2k\right)}^{j+1}}f\left({\left(2k\right)}^{j+1}x\right)\Vert }_Y\\ \le \frac{1}{2k^2}{\displaystyle \underset{j=l+1}{\overset{m}{\sum }}}\frac{1}{{\left(2k\right)}^j}\varphi \left({\left(2k\right)}^{j+1}x,\cdots ,{\left(2k\right)}^{j+1}x,0,0,\cdots ,0,-{\left(2k\right)}^{j}x,\cdots ,-{\left(2k\right)}^{j}x\right)\\ =0\end{array}$ (22)

for all nonnegative integers m and l with $m>l$ and all $x\in G$ . It follows from (22) 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 $x\in G$ . Since $Y$ is complete space, 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 $\psi \mathrm{<mo>\; :\; </mo>}G\to Y$ by $\psi \left(x\right)\mathrm{<mo>\; :\; </mo>}=\underset{n\to \infty }{\lim }\frac{1}{{\left(2k\right)}^n}f\left({\left(2k\right)}^{n}x\right)$ for all $x\in G$ . Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (22), I get (19).

Now, It follows from (18) I have

$\begin{array}{l}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}\psi \left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\psi \left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\psi \left(z_j\right)\Vert }_Y\\ =\underset{n\to \infty }{\lim }{\Vert \frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^n{x}_j\right)+\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^n{y}_j\right)+2k\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^n{z}_j\right)\Vert }_Y\\ =\underset{n\to \infty }{\lim }\frac{1}{{\left(2k\right)}^n}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^n{x}_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^n{y}_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^n{z}_j\right)\Vert }_Y\\ \le \underset{n\to \infty }{\lim }\frac{1}{{\left(2k\right)}^n}({\Vert 2kf\left({\left(2k\right)}^n{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\left(2k\right)}^n{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \stackrel{\stackrel{}{}}{}+\varphi \left({\left(2k\right)}^n{x}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^n{x}_k\mathrm{,}{\left(2k\right)}^n{y}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^n{y}_k\mathrm{,}{\left(2k\right)}^n{z}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^n{z}_k\right))\\ ={\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\end{array}$ (23)

So I have

${\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}\psi \left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\psi \left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\psi \left(z_j\right)\Vert }_Y\le {\Vert 2k\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y$ (24)

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

Hence from Lemma 1 and corollary 1 it follows that $\psi$ is an additive mapping.

Finally I have to prove that $\psi$ is a unique additive mapping.

Now, let ${\psi }^{\prime }\mathrm{<mo>\; :\; </mo>}G\to Y$ be another generalized Cauchy-Jensen additive mapping satisfying (19). Then I have

$\begin{array}{l}{\Vert \psi \left(x\right)-{\psi }^{\prime }\left(x\right)\Vert }_Y=\frac{1}{{\left(2k\right)}^n}{\Vert \psi \left({\left(2k\right)}^{n}x\right)-{\psi }^{\prime }\left({\left(2k\right)}^{n}x\right)\Vert }_Y\\ \le \frac{1}{{\left(2k\right)}^n}\left({\Vert f\left({\left(2k\right)}^{n}x\right)-\psi \left({\left(2k\right)}^{n}x\right)\Vert }_Y+{\Vert f\left({\left(2k\right)}^{n}x\right)-{\psi }^{\prime }\left(\frac{x}{2^n}\right)\Vert }_Y\right)\\ \le 2\frac{1}{{\left(2k\right)}^n}\tilde{\varphi }\left({\left(2k\right)}^{n}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{n}x\mathrm{,0,}\cdots \mathrm{,0,}{\left(2k\right)}^{n}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{n}x\right)\end{array}$ (25)

which tends to zero as $n\to \infty$ for all $x\in X$ . So we can conclude that $\psi \left(x\right)={\psi }^{\prime }\left(x\right)$ for all $x\in G$ . This proves the uniquence of ${\psi }^{\prime }$ .

From Theorem 2 I have the following corollarys.

Corollary 2. For $G$ is a normed space and $p,r\ne 0,q>0,\theta >0$ . Suppose $f\mathrm{<mo>\; :\; </mo>}G\to Y$ be a function such that

$\begin{array}{l}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\Vert }_Y\\ \le {\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y+\theta \cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}{\Vert x_j\Vert }^p\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}{\Vert y_j\Vert }^q\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}{\Vert z_j\Vert }^r\end{array}$ (26)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in G$ then f í a additive mapping.

Corollary 3. For $G$ is a normed space and $0<p,r<1,q\ne 0,\theta >0$ . Suppose $f\mathrm{<mo>\; :\; </mo>}G\to Y$ be a function such that

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

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

${\Vert f\left(x\right)-\psi \left(x\right)\Vert }_Y\le \theta k\left(\frac{{\left(2k\right)}^p}{2k-{\left(2k\right)}^p}{\Vert x\Vert }^p+\frac{1}{2k-{\left(2k\right)}^k}{\Vert x\Vert }^r\right)$ (28)

for all $x\in G$ .

Theorem 3. For $\varphi \mathrm{<mo>\; :\; </mo>}{G}^{3k}\to {ℝ}^{+}$ be a function such that

$\underset{n\to \infty }{\lim }{\left(2k\right)}^n\varphi \left(\frac{1}{{\left(2k\right)}^n}{x}_1\mathrm{,}\cdots \mathrm{,}\frac{1}{{\left(2k\right)}^n}{x}_k\mathrm{,}\frac{1}{{\left(2k\right)}^n}{y}_1\mathrm{,}\cdots \mathrm{,}\frac{1}{{\left(2k\right)}^n}{y}_k\mathrm{,}-\frac{1}{{\left(2k\right)}^n}{z}_1\mathrm{,}\cdots \mathrm{,}-\frac{1}{{\left(2k\right)}^n}{z}_k\right)=0$ (29)

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

$\begin{array}{l}\tilde{\varphi }\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\\ ={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\varphi {\left(2k\right)}^n\varphi \left(\frac{1}{{\left(2k\right)}^n}{x}_1,\cdots ,\frac{1}{{\left(2k\right)}^n}{x}_k,0,0,\cdots ,0,\cdots ,\frac{1}{{\left(2k\right)}^{n+1}}{z}_1,\cdots ,\frac{1}{{\left(2k\right)}^{n+1}}{z}_k\right)<\infty \end{array}$ (30)

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

Suppose that be an odd mapping $f\mathrm{<mo>\; :\; </mo>}G\to Y$ satisfies

$\begin{array}{l}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\Vert }_Y\\ \le {\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y+\varphi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (31)

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

Then there exists a unique additive mapping $\psi \mathrm{<mo>\; :\; </mo>}G\to Y$ such that

${\Vert f\left(x\right)-\psi \left(x\right)\Vert }_Y\le \tilde{\varphi }\left(x\mathrm{,}\cdots \mathrm{,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)$ (32)

for all $x\in G$ .

Proof. 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 (31), I get

$\left(\left|2k^2+k\right|-\left|2k\right|\right){\Vert f\left(0\right)\Vert }_Y\le 0.$ (33)

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 $\left(2kx\mathrm{,}\cdots \mathrm{,2}kx\mathrm{,0,}\cdots \mathrm{,0,}-x\mathrm{,}\cdots \mathrm{,}-x\right)$ in (31), I get

${\Vert kf\left(2kx\right)-2k^{2}f\left(x\right)\Vert }_Y\le \varphi \left(2kx\mathrm{,2}kx\mathrm{,}\cdots \mathrm{,2}kx\mathrm{,0,0,}\cdots \mathrm{,0,}-x\mathrm{,}-x\mathrm{,}\cdots \mathrm{,}-x\right)$ (34)

${\Vert f\left(x\right)-2kf\left(\frac{x}{2k}\right)\Vert }_Y\le \frac{1}{k}\varphi \left(x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,0,0,}\cdots \mathrm{,0,}-\frac{x}{2k}\mathrm{,}-\frac{x}{2k}\mathrm{,}\cdots \mathrm{,}-\frac{x}{2k}\right)$

The remainder is similar to the proof of Theorem 2. This completes the proof.

From Theorem 2 andTheorem 2. I have the following corollarys.

Corollary 4. For $G$ is a normed space and $p,r\ne 0,q>0,\theta >0$ . Suppose $f\mathrm{<mo>\; :\; </mo>}G\to Y$ be a function such that

$\begin{array}{l}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\Vert }_Y\\ \le {\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y+\theta \cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}{\Vert x_j\Vert }^p\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}{\Vert y_j\Vert }^q\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}{\Vert z_j\Vert }^r\end{array}$ (35)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in G$ , then f is a additive mapping.

Corollary 5. For $G$ is a normed space and $0<p,r<1,q\ne 0,\theta >0$ . Suppose $f\mathrm{<mo>\; :\; </mo>}G\to Y$ be a function such that

$\begin{array}{l}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\Vert }_Y\\ \le {\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y+\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^p+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^q+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (36)

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

${\Vert f\left(x\right)-\psi \left(x\right)\Vert }_Y\le \theta k\left(\frac{{\left(2k\right)}^p}{{\left(2k\right)}^p-2k}{\Vert x\Vert }^p+\frac{1}{{\left(2k\right)}^k-2k}{\Vert x\Vert }^r\right)$ (37)

for all $x\in G$ .

5. Establish Solutions to Functional Inequalities Based on the Definition

Now, I first study the solutions of (1). We first consider the mapping $E\mathrm{<mo>\; :\; </mo>}M\left(G\mathrm{,}Y\right)\to M\left(G^r\mathrm{,}{ℝ}^{\mathrm{<mo>\; *\; </mo>}}\right)$ as $\begin{array}{l}E\left(f\right)\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\\ =\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\Vert -\Vert 2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert \end{array}$ then the inequalities $Ef\le 0$ is $\left(\varphi \mathrm{,}\tilde{\varphi }\right)$ -stable in $M\left(G\mathrm{,}Y\right)$ where $\left(\varphi \mathrm{,}\tilde{\varphi }\right)$ is as Theorem 2 and Theorem 3.

Note that for inequalities, $G$ be a m-divisible group where $m\in \mathbb{N}\backslash \left\{0\right\}$ and $Y$ be a Banach spaces. Under this setting, we can show that the mapping satisfying (1) is additive. These results are give in the following.

Theorem 4. For $\varphi \mathrm{<mo>\; :\; </mo>}{G}^{3k}\to {ℝ}^{+}$ be a function such that

$\underset{n\to \infty }{\lim }\frac{1}{{\left(2k\right)}^n}\varphi \left({\left(2k\right)}^n{x}_1,\cdots ,{\left(2k\right)}^n{x}_k,{\left(2k\right)}^n{y}_1,\cdots ,{\left(2k\right)}^n{y}_k,\cdots ,{\left(2k\right)}^n{z}_1,\cdots ,{\left(2k\right)}^n{z}_k\right)=0$ (38)

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

$\begin{array}{l}\tilde{\varphi }\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\\ ={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{1}{{\left(2k\right)}^{n+1}}(\varphi \left({\left(2k\right)}^{n+1}{x}_1,\cdots ,{\left(2k\right)}^{n+1}{x}_k,0,\cdots ,0,-{\left(2k\right)}^n{z}_1,\cdots ,-{\left(2k\right)}^n{z}_k\right)\\ +\varphi \left(-{\left(2k\right)}^{n+1}{x}_1,\cdots ,-{\left(2k\right)}^{n+1}{x}_k,0,\cdots ,0,{\left(2k\right)}^n{z}_1,\cdots ,{\left(2k\right)}^n{z}_k\right))<\infty \end{array}$ (39)

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

Suppose that a mapping $f\mathrm{<mo>\; :\; </mo>}G\to Y$ satisfies $f\left(0\right)=0$ for all $x\in G$ , and

$\begin{array}{l}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le \varphi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (40)

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

Then there exists a unique additive mapping $\psi \mathrm{<mo>\; :\; </mo>}G\to Y$ such that

${\Vert f\left(x\right)-\psi \left(x\right)\Vert }_Y\le \tilde{\varphi }\left(x\mathrm{,}\cdots \mathrm{,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)$ (41)

for all $x\in G$ .

Proof. 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 $\left(2kx\mathrm{,}\cdots \mathrm{,2}kx\mathrm{,0,}\cdots \mathrm{,0,}-x\mathrm{,}\cdots \mathrm{,}-x\right)$ in (40), I get

${\Vert kf\left(2kx\right)+2k^{2}f\left(-x\right)\Vert }_Y\le \varphi \left(2kx\mathrm{,2}kx\mathrm{,}\cdots \mathrm{,2}kx\mathrm{,0,0,}\cdots \mathrm{,0,}-x\mathrm{,}-x\mathrm{,}\cdots \mathrm{,}-x\right)$ (42)

continue I replace x by −x in (42), I have

${\Vert kf\left(-2kx\right)+2k^{2}f\left(x\right)\Vert }_Y\le \varphi \left(-2kx\mathrm{,}-2kx\mathrm{,}\cdots \mathrm{,}-2kx\mathrm{,0,0,}\cdots \mathrm{,0,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)$ (43)

put

$g\left(x\right)=\frac{f\left(x\right)-f\left(-x\right)}{2}$ (44)

So since (45), (43) and (44), I have

$\begin{array}{c}{\Vert f\left(x\right)-\frac{1}{2k}f\left(2kx\right)\Vert }_Y\le \frac{1}{2k^2}(\varphi \left(2kx\mathrm{,2}kx\mathrm{,}\cdots \mathrm{,2}kx\mathrm{,0,0,}\cdots \mathrm{,0,}-x\mathrm{,}-x\mathrm{,}\cdots \mathrm{,}-x\right)\\ +\varphi \left(-2kx\mathrm{,}-2kx\mathrm{,}\cdots \mathrm{,}-2kx\mathrm{,0,0,}\cdots \mathrm{,0,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right))\end{array}$ (45)

Hence

$\begin{array}{l}{\Vert \frac{1}{{\left(2k\right)}^l}f\left({\left(2k\right)}^{l}x\right)-\frac{1}{{\left(2k\right)}^m}f\left({\left(2k\right)}^{m}x\right)\Vert }_Y\\ \le {\displaystyle \underset{j=l}{\overset{m-1}{\sum }}}{\Vert \frac{1}{{\left(2k\right)}^j}f\left({\left(2k\right)}^{j}x\right)-\frac{1}{{\left(2k\right)}^{j+1}}f\left({\left(2k\right)}^{j+1}x\right)\Vert }_Y\\ \le \frac{1}{2k^2}{\displaystyle \underset{j=l+1}{\overset{m}{\sum }}}\frac{1}{{\left(2k\right)}^j}(\varphi \left({\left(2k\right)}^{j+1}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{j+1}x\mathrm{,0,0,}\cdots \mathrm{,0,}-{\left(2k\right)}^{j}x\mathrm{,}\cdots \mathrm{,}-{\left(2k\right)}^{j}x\right)\\ +\varphi \left(-2kx\mathrm{,}-2kx\mathrm{,}\cdots \mathrm{,}-2kx\mathrm{,0,0,}\cdots \mathrm{,0,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right))\\ =0\end{array}$ (46)

for all nonnegative integers m and l with $m>l$ and all $x\in G$ . It follows from (46) 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 $x\in G$ . Since $Y$ is complete space, 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 $\psi \mathrm{<mo>\; :\; </mo>}G\to Y$ by $\psi \left(x\right)\mathrm{<mo>\; :\; </mo>}=\underset{n\to \infty }{\lim }\frac{1}{{\left(2k\right)}^n}f\left({\left(2k\right)}^{n}x\right)$ for all $x\in G$ . Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (46), I get (41).

Now, It follows from (40)we have

$\begin{array}{l}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}\psi \left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\psi \left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\psi \left(z_j\right)-2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ =\underset{n\to \infty }{\lim }\Vert \frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^n{x}_j\right)+\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^n{y}_j\right)\\ +2k\frac{1}{{\left(2k\right)}^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^n{z}_j\right)-{2kf\left({\left(2k\right)}^n{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\left(2k\right)}^n{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ =\underset{n\to \infty }{\lim }\frac{1}{{\left(2k\right)}^n}\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^n{x}_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^n{y}_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^n{z}_j\right)\\ -{2kf\left({\left(2k\right)}^n{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\left(2k\right)}^n{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le \varphi \left({\left(2k\right)}^n{x}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^n{x}_k\mathrm{,}{\left(2k\right)}^n{y}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^n{y}_k\mathrm{,}{\left(2k\right)}^n{z}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^n{z}_k\right)=0\end{array}$ (47)

So I have

${\displaystyle \underset{j=1}{\overset{k}{\sum }}}\psi \left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\psi \left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\psi \left(z_j\right)=2k\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)$ (48)

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

Hence from Lemma 2 and corollary 1, it follows that $\psi$ is an additive mapping.

Finally I have to prove that $\psi$ is a unique additive mapping.

Now, let ${\psi }^{\prime }\mathrm{<mo>\; :\; </mo>}G\to Y$ be another generalized Cauchy-Jensen additive mapping satisfying (41). Then we have

$\begin{array}{l}{\Vert \psi \left(x\right)-{\psi }^{\prime }\left(x\right)\Vert }_Y=\frac{1}{{\left(2k\right)}^n}{\Vert \psi \left({\left(2k\right)}^{n}x\right)-{\psi }^{\prime }\left({\left(2k\right)}^{n}x\right)\Vert }_Y\\ \le \frac{1}{{\left(2k\right)}^n}\left({\Vert f\left({\left(2k\right)}^{n}x\right)-\psi \left({\left(2k\right)}^{n}x\right)\Vert }_Y+{\Vert f\left({\left(2k\right)}^{n}x\right)-{\psi }^{\prime }\left(\frac{x}{2^n}\right)\Vert }_Y\right)\\ \le 2\frac{1}{{\left(2k\right)}^n}\tilde{\varphi }\left({\left(2k\right)}^{n}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{n}x\mathrm{,0,}\cdots \mathrm{,0,}{\left(2k\right)}^{n}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{n}x\right)\\ ={\displaystyle \underset{n\mathrm{<mo>\; =\; </mo>0}}{\overset{\infty }{\sum }}}\frac{1}{{\left(2k\right)}^{n+1}}(\varphi \left({\left(2k\right)}^{n+1}{x}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{n+1}{x}_k\mathrm{,0,}\cdots \mathrm{,0,}-{\left(2k\right)}^n{z}_1\mathrm{,}\cdots \mathrm{,}-{\left(2k\right)}^n{z}_k\right)\\ +\varphi \left(-{\left(2k\right)}^{n+1}{x}_1\mathrm{,}\cdots \mathrm{,}-{\left(2k\right)}^{n+1}{x}_k\mathrm{,0,}\cdots \mathrm{,0,}{\left(2k\right)}^n{z}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^n{z}_k\right))<\infty \end{array}$ (49)

which tends to zero as $n\to \infty$ for all $x\in X$ . So we can conclude that $\psi \left(x\right)={\psi }^{\prime }\left(x\right)$ for all $x\in X$ . This proves the uniquence of ${\psi }^{\prime }$ .

From Theorem 4 I have the following corollarys.

Corollary 6. For $G$ is a normed space and $p,r\ne 0,q>0,\theta >0$ . Suppose $f\mathrm{<mo>\; :\; </mo>}G\to Y$ be a function such that $f\left(0\right)=0$ and

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

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in G$ then f is an additive mapping.

Corollary 7. For $G$ is a normed space and $0<p,r<1,q\ne 0,\theta >0$ . Suppose $f\mathrm{<mo>\; :\; </mo>}G\to Y$ be a function such that $f\left(0\right)=0$ and

$\begin{array}{l}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le \theta \left({\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{\Vert x_j\Vert }^p+{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{\Vert y_j\Vert }^q+{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (51)

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

${\Vert f\left(x\right)-\psi \left(x\right)\Vert }_Y\le \theta k\left(\frac{{\left(2k\right)}^p}{2k-{\left(2k\right)}^p}{\Vert x\Vert }^p+\frac{1}{2k-{\left(2k\right)}^k}{\Vert x\Vert }^r\right)$ (52)

for all $x\in G$ .

Theorem 5. For $\varphi \mathrm{<mo>\; :\; </mo>}{G}^{3k}\to {ℝ}^{+}$ be a function such that

$\underset{n\to \infty }{\lim }{\left(2k\right)}^n\varphi \left(\frac{1}{{\left(2k\right)}^n}{x}_1,\cdots ,\frac{1}{{\left(2k\right)}^n}{x}_k,\frac{1}{{\left(2k\right)}^n}{y}_1,\cdots ,\frac{1}{{\left(2k\right)}^n}{y}_k,\cdots ,\frac{1}{{\left(2k\right)}^n}{z}_1,\cdots ,\frac{1}{{\left(2k\right)}^n}{z}_k\right)=0$ (53)

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

And

$\begin{array}{l}\tilde{\varphi }\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\\ ={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}{\left(2k\right)}^{n-1}(\varphi \left({\left(2k\right)}^{-n}{x}_1,\cdots ,{\left(2k\right)}^{-\left(n+1\right)}{x}_k,0,\cdots ,0,-{\left(2k\right)}^{n+1}{z}_1,\cdots ,-{\left(2k\right)}^{n+1}{z}_k\right)\\ +\varphi \left(-{\left(2k\right)}^{-n}{x}_1,\cdots ,-{\left(2k\right)}^{-n}{x}_k,0,\cdots ,0,{\left(2k\right)}^{n+1}{z}_1,\cdots ,{\left(2k\right)}^{n+1}{z}_k\right))<\infty \end{array}$ (54)

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

Suppose that a mapping $f\mathrm{<mo>\; :\; </mo>}G\to Y$ satisfies $f\left(0\right)=0$ for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in G$ .

And

$\begin{array}{l}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le \varphi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (55)

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

Then there exists a unique additive mapping $\psi \mathrm{<mo>\; :\; </mo>}G\to Y$ such that

${\Vert f\left(x\right)-\psi \left(x\right)\Vert }_Y\le \tilde{\varphi }\left(x\mathrm{,}\cdots \mathrm{,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)$ (56)

for all $x\in G$ .

The proof is similar to theorem 4.

Corollary 8. For $G$ is a normed space and $p,r\ne 0,q>0,\theta >0$ . Suppose $f\mathrm{<mo>\; :\; </mo>}G\to Y$ be a function such that $f\left(0\right)=0$ and

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

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in G$ then f í a additive mapping.

Corollary 9. For $G$ is a normed space and $0<p,r<1,q\ne 0,\theta >0$ . Suppose $f\mathrm{<mo>\; :\; </mo>}G\to Y$ be a function such that $f\left(0\right)=0$ and

$\begin{array}{l}{\Vert {\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)-2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le \theta \left({\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{\Vert x_j\Vert }^p+{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{\Vert y_j\Vert }^q+{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (58)

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

${\Vert f\left(x\right)-\psi \left(x\right)\Vert }_Y\le \theta k\left(\frac{{\left(2k\right)}^p}{2k-{\left(2k\right)}^p}{\Vert x\Vert }^p+\frac{1}{2k-{\left(2k\right)}^k}{\Vert x\Vert }^r\right)$ (59)

for all $x\in G$ .

6. The Stability of Derivation on Fuzzy-Algebras

Lemma 6. Let $\left(Y\mathrm{,}\mathbb{N}\right)$ be a fuzzy normed vector space and $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping such that

$N\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j\right)+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(y_j\right)+2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\mathrm{,}t\right)\ge N\left(2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right),\frac{t}{2k}\right)$ (60)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in Y$ and all $t>0$ . Then f is Cauchy additive.

Proof. 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 (60), I have

$N\left(\left(2k^2+2k\right)f\left(0\right),t\right)=N\left(f\left(0\right),\frac{t}{2k^2+2k}\right)\ge N\left(2kf\left(0\right),\frac{t}{2k}\right)=1$ (61)

for all $t>0$ . By N₅ and N₆, $N\left(f\left(0\right)\mathrm{,}\frac{t}{2k}\right)=1$ . It follows N₂ that $f\left(0\right)=0$ .

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 $\left(-y\mathrm{,}\cdots \mathrm{,}-y\mathrm{,}y\mathrm{,}\cdots \mathrm{,}y\mathrm{,0,}\cdots \mathrm{,0}\right)$ in (60), I have

$N\left(kf\left(-y\right)+kf\left(y\right)\mathrm{,}t\right)=N\left(f\left(-y\right)+f\left(y\right)\mathrm{,}\frac{t}{k}\right)\ge N\left(2kf\left(0\right)\mathrm{,}\frac{t}{2k^2+2k}\right)$ (62)

It follows N₂ that $f\left(-y\right)+f\left(y\right)=0$ .

So

$f\left(-y\right)=-f(\; y\; )$

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 $\left(-2z\mathrm{,}\cdots \mathrm{,}-2z\mathrm{,0,}\cdots \mathrm{,0,}z\mathrm{,0},\cdots \mathrm{,0}\right)$ in (60), we have

$N\left(-kf\left(2z\right)+2kf\left(z\right),t\right)=N\left(f\left(-2z\right)+2f\left(z\right),\frac{t}{k}\right)\ge N\left(2kf\left(0\right),\frac{t}{2k^2+2k}\right)$ (63)

It follows N₂ that $f\left(-2z\right)+2f\left(z\right)=0$ .

So $f\left(2z\right)=2f\left(z\right)$ for all $t>0$ and for all $z\in X$ .

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 $\left(x,\cdots ,x,y,\cdots ,y,z_1=-\frac{x+y}{2},z_2=0,\cdots ,0\right)$ in (60), we have

$\begin{array}{c}N\left(f\left(x\right)+f\left(y\right)-f\left(x+y\right)\mathrm{,}\frac{t}{k}\right)=N\left(f\left(x\right)+f\left(y\right)+2f\left(-\frac{x+y}{2}\right)\mathrm{,}\frac{t}{k}\right)\\ \ge N\left(2kf\left(0\right)\mathrm{,}\frac{t}{2k^2+2k}\right)\end{array}$ (64)

for all $t>0$ . and for all $x\mathrm{,}y\in X$ Thus $f\left(x\right)+f\left(y\right)=f\left(x+y\right)$ for all $x\mathrm{,}y\in X$ , as desired.

Theorem 7. Let $\psi \mathrm{<mo>\; :\; </mo>}{X}^{3k}\to \left[\mathrm{0,}\infty \right)$ be a function such that there exists an $L<\frac{1}{2k}$

$\begin{array}{l}\psi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\\ \le \frac{L}{2k}\psi \left(2k{x}_1\mathrm{,}\cdots \mathrm{,2}k{x}_k\mathrm{,2}k{y}_1\mathrm{,}\cdots \mathrm{,2}k{y}_k\mathrm{,2}k{z}_1\mathrm{,}\cdots \mathrm{,2}k{z}_k\right)\end{array}$ (65)

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

Let $f\mathrm{<mo>\; :\; </mo>}X\to X$ be a mapping sattisfying

$\begin{array}{l}N\left(2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{q{x}_j+q{y}_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}q{z}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qf\left(x_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qf\left(y_j\right)-2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qf\left(z_j\right),t\right)\\ \ge \frac{t}{t+\psi \left(x_1,\cdots ,x_k,y_1,\cdots ,y_k,z_1,\cdots ,z_k\right)}\end{array}$ (66)

$\begin{array}{l}N\left(f\left({\displaystyle \underset{j=1}{\overset{k}{\prod }}}{x}_j\cdot y_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\prod }}}f\left(x_j\right)\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\prod }}}{x}_j\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}f\left(y_j\right),t\right)\\ \ge \frac{t}{t+\psi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,0,}\cdots \mathrm{,0}\right)}\end{array}$ (67)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\in X$ , for all $t>0$ and for all $q>0$ . Then

$H\left(x\right)=N-\underset{n\to \infty }{\lim }{\left(2k\right)}^{n}f\left(\frac{x}{{\left(2k\right)}^n}\right)$ (68)

exists each $x\in X$ and defines a fuzzy derivation $H\mathrm{<mo>\; :\; </mo>}X\to X$ , such that

$N\left(f\left(x\right)-H\left(x\right)\mathrm{,}t\right)\ge \frac{\left(1-L\right)t}{\left(1-L\right)+L\psi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,0,}\cdots \mathrm{,0}\right)}$ (69)

for all $t>0$ and for all $q>0$ .

Proof. Letting $q=1$ and 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 $\left(x,\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (84), I get

$N\left(2kf\left(\frac{x}{2k}\right)-f\left(x\right)\mathrm{,}t\right)\ge \frac{t}{1+\phi \left(x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)}$ (70)

for all $x\in X$ . Now I consider the set $\mathbb{M}\mathrm{<mo>\; :\; </mo>}=\left\{h\mathrm{<mo>\; :\; </mo>}X\to Y\right\}$ and introduce the generalized metric on S as follows:

$\begin{array}{l}d\left(g\mathrm{,}h\right)\mathrm{<mo>\; :\; </mo>}=\inf \{\beta \in {ℝ}_{+}\mathrm{<mo>\; :\; </mo>}N\left(g\left(x\right)-h\left(x\right)\mathrm{,}\beta t\right)\begin{array}{c}\\ \end{array}\\ \ge \frac{t}{t+\phi \left(x,\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)}\mathrm{,}\forall x\in X\mathrm{,}\forall t>0\}\mathrm{,}\end{array}$ (71)

where, as usual, $\inf \varphi =+\infty$ . That has been proven by mathematicians $\left(\mathbb{M}\mathrm{,}d\right)$ is complete (see [32] ).

Now I cosider the linear mapping $T\mathrm{<mo>\; :\; </mo>}\mathbb{M}\to \mathbb{M}$ such that $Tg\left(x\right)\mathrm{<mo>\; :\; </mo>}=2kg\left(\frac{x}{2k}\right)$ for all $x\in X$ . Let $g\mathrm{,}h\in \mathbb{M}$ be given such that $d\left(g\mathrm{,}h\right)=\epsilon$ then $N\left(g\left(x\right)-h\left(x\right)\mathrm{,}\epsilon t\right)\ge \frac{t}{t+\phi \left(x,\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)}\mathrm{,}\forall x\in X\mathrm{,}\forall t>0.$

Hence

$\begin{array}{c}N\left(g\left(x\right)-h\left(x\right)\mathrm{,}\epsilon t\right)=N\left(2kg\left(\frac{x}{2k}\right)-2kh\left(\frac{x}{2k}\right)\mathrm{,}L\epsilon t\right)\\ =N\left(g\left(\frac{x}{2k}x\right)-h\left(\frac{x}{2k}x\right)\mathrm{,}\frac{L}{2k}\epsilon t\right)\\ \ge \frac{\frac{Lt}{2k}}{\frac{Lt}{2k}+\phi \left(\frac{x}{2k}\mathrm{,}\cdots ,\mathrm{0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)}\\ \ge \frac{\frac{Lt}{2k}}{\frac{Lt}{2k}+\frac{L}{2k}\phi \left(x\mathrm{,0},\cdots \mathrm{,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)}\\ =\frac{t}{t+\phi \left(x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)}\mathrm{,}\forall x\in X\mathrm{,}\forall t>0.\end{array}$ (72)

So $d\left(g\mathrm{,}h\right)=\epsilon$ implies that $d\left(Tg\mathrm{,}Th\right)\le L\cdot \epsilon$ . This means that $d\left(Tg\mathrm{,}Th\right)\le Ld\left(g\mathrm{,}h\right)$ for all $g\mathrm{,}h\in \mathbb{M}$ . It folows from (70) that I have.

For all $x\in \mathbb{X}$ . So $d\left(f\mathrm{,}Tf\right)\le 1$ . By Theorem 1.2, there exists a mapping $H\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfying the fllowing:

1) H is a fixed point of T, i.e.,

$H\left(\frac{x}{2k}\right)=\frac{1}{2k}H\left(x\right)$ (73)

for all $x\in X$ . The mapping H is a unique fixed point T in the set $\mathbb{Q}=\left\{g\in \mathbb{M}:d\left(f,g\right)<\infty \right\}$ .

This implies that H is a unique mapping satisfying (73) such that there exists a $\beta \in \left(\mathrm{0,}\infty \right)$ satisfying $N\left(f\left(x\right)-H\left(x\right)\mathrm{,}\beta t\right)\ge \frac{t}{t+\phi \left(x,\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)}\mathrm{,}\forall x\in X\mathrm{.}$

2) $d\left(T^{l}f\mathrm{,}H\right)\to 0$ as $l\to \infty$ . This implies equality $N-\underset{l\to \infty }{\lim }{\left(2k\right)}^{l}f\left(\frac{x}{{\left(2k\right)}^l}\right)=H\left(x\right)$ for all $x\in \mathbb{X}$ .

3) $d\left(f\mathrm{,}H\right)\le \frac{1}{1-L}d\left(f\mathrm{,}Tf\right)$ . which implies the inequality.

4) $d\left(f\mathrm{,}H\right)\le \frac{1}{1-L}$ .

This follows that the inequality (70) is satisfied.

By (85)

$\begin{array}{l}N({\left(2k\right)}^{p+1}f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{q{x}_j+q{y}_j}{{\left(2k\right)}^{p+1}}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{q{z}_j}{{\left(2k\right)}^p}\right)-{\left(2k\right)}^p{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qf\left(\frac{x_j}{{\left(2k\right)}^p}\right)\\ -{\left(2k\right)}^p{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qf\left(\frac{y_j}{{\left(2k\right)}^p}\right)-{\left(2k\right)}^{p}2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qf\left(\frac{z_j}{{\left(2k\right)}^p}\right),t)\\ \ge \frac{t}{t+\psi \left(\frac{x_1}{{\left(2k\right)}^p},\cdots ,\frac{x_k}{{\left(2k\right)}^p},\frac{y_1}{{\left(2k\right)}^p},\cdots ,\frac{y_k}{{\left(2k\right)}^p},\frac{z_1}{{\left(2k\right)}^p},\cdots ,\frac{z_k}{{\left(2k\right)}^p}\right)}\end{array}$ (74)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ , for all $t>0$ and for all $q\in ℝ$ . So

$\begin{array}{l}N({\left(2k\right)}^{p+1}f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{q{x}_j+q{y}_j}{{\left(2k\right)}^{p+1}}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{q{z}_j}{{\left(2k\right)}^p}\right)-{\left(2k\right)}^p{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qf\left(\frac{x_j}{{\left(2k\right)}^p}\right)\\ -{\left(2k\right)}^p{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qf\left(\frac{y_j}{{\left(2k\right)}^p}\right)-{\left(2k\right)}^{p}2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qf\left(\frac{z_j}{{\left(2k\right)}^p}\right),t)\\ \ge \frac{\frac{t}{{\left(2k\right)}^p}}{\frac{t}{{\left(2k\right)}^p}+\frac{L^p}{{\left(2k\right)}^p}\psi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)}\end{array}$ (75)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ , for all $t>0$ and for all $q\in ℝ$ .

Since $\underset{n\to \infty }{\lim }\frac{\frac{t}{{\left(2k\right)}^p}}{\frac{t}{{\left(2k\right)}^p}+\frac{L^p}{{\left(2k\right)}^p}\psi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)}=1$ for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in \mathbb{X}$ , $\forall t>0$ , $q\in ℝ$ . So

$N\left(2kH\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{q{x}_j+q{y}_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}q{z}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qH\left(x_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qH\left(y_j\right)-2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qH\left(z_j\right),t\right)=1$ (76)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\in X$ , $\forall t>0$ , $q\in ℝ$ . So

$2kH\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{q{x}_j+q{y}_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}q{z}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qH\left(x_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qH\left(y_j\right)-2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qH\left(z_j\right)=0$ (77)

Thus the mapping $H\mathrm{<mo>\; :\; </mo>}X\to X$ is additive and $R$ -linear by (85) I have

$\begin{array}{l}N({\left(2k\right)}^{2p}f\left({\displaystyle \underset{j=1}{\overset{k}{\prod }}}\frac{x_j\cdot y_j}{{\left(2k\right)}^{2p}}\right)-{\left(2k\right)}^p{\displaystyle \underset{j=1}{\overset{k}{\prod }}}f\left(\frac{x_j}{{\left(2k\right)}^p}\right)\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}{y}_j\\ -{\displaystyle \underset{j=1}{\overset{k}{\prod }}}{x}_j\cdot {\left(2k\right)}^p{\displaystyle \underset{j=1}{\overset{k}{\prod }}}f\left(\frac{y_j}{{\left(2k\right)}^p}\right),t)\\ \ge \frac{t}{t+\psi \left(\frac{x_1}{{\left(2k\right)}^p}\mathrm{,}\cdots \mathrm{,}\frac{x_k}{{\left(2k\right)}^p}\mathrm{,}\frac{y_1}{{\left(2k\right)}^p}\mathrm{,}\cdots \mathrm{,}\frac{y_k}{{\left(2k\right)}^p}\mathrm{,0},\cdots \mathrm{,0}\right)}\end{array}$ (78)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\in X$ , for all $t>0$ .

$\begin{array}{l}N({\left(2k\right)}^{2p}f\left({\displaystyle \underset{j=1}{\overset{k}{\prod }}}\frac{x_j\cdot y_j}{{\left(2k\right)}^{2p}}\right)-{\left(2k\right)}^p{\displaystyle \underset{j=1}{\overset{k}{\prod }}}f\left(\frac{x_j}{{\left(2k\right)}^p}\right)\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}{y}_j\\ -{\displaystyle \underset{j=1}{\overset{k}{\prod }}}{x}_j\cdot {\left(2k\right)}^p{\displaystyle \underset{j=1}{\overset{k}{\prod }}}f\left(\frac{y_j}{{\left(2k\right)}^p}\right),t)\\ \ge \frac{\frac{t}{{\left(2k\right)}^{2p}}}{\frac{t}{{\left(2k\right)}^{2p}}+\frac{L^p}{{\left(2k\right)}^p}\psi \left(x_1,\cdots ,x_k,y_1,\cdots ,y_k,0,\cdots ,0\right)}\end{array}$ (79)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\in X$ , for all $t>0$ Since

$\underset{p\to \infty }{\lim }\frac{\frac{t}{{\left(2k\right)}^{2p}}}{\frac{t}{{\left(2k\right)}^{2p}}+\frac{L^p}{{\left(2k\right)}^p}\psi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,0},\cdots \mathrm{,0}\right)}=1$ (80)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\in X$ , for all $t>0$ Thus

$N\left(f\left({\displaystyle \underset{j=1}{\overset{k}{\prod }}}{x}_j\cdot y_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\prod }}}f\left(x_j\right)\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\prod }}}{x}_j\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}f\left(y_j\right),t\right)=1$ (81)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\in X$ , for all $t>0$ Thus

$f\left({\displaystyle \underset{j=1}{\overset{k}{\prod }}}{x}_j\cdot y_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\prod }}}f\left(x_j\right)\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\prod }}}{x}_j\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}f\left(y_j\right)=0$ (82)

So the mapping $H\mathrm{<mo>\; :\; </mo>}X\to X$ is a fuzzy derivation, as desired.

Theorem 8. Let $\psi \mathrm{<mo>\; :\; </mo>}{X}^{3k}\to \left[\mathrm{0,}\infty \right)$ be a function such that there exists an $L<1$

$\psi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\le 2k\psi \left(\frac{x_1}{2k}\mathrm{,}\cdots \mathrm{,}\frac{x_k}{2k}\mathrm{,}\frac{y_1}{2k}\mathrm{,}\cdots \mathrm{,}\frac{y_k}{2k}\mathrm{,}\frac{z_1}{2k}\mathrm{,}\cdots \mathrm{,}\frac{z_k}{2k}\right)$ (83)

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

Let $f\mathrm{<mo>\; :\; </mo>}X\to X$ be a mapping sattisfying

$\begin{array}{l}N\left(2kf\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{q{x}_j+q{y}_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}q{z}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qf\left(x_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qf\left(y_j\right)-2k{\displaystyle \underset{j=1}{\overset{k}{\sum }}}qf\left(z_j\right),t\right)\\ \ge \frac{t}{t+\psi \left(x_1,\cdots ,x_k,y_1,\cdots ,y_k,z_1,\cdots ,z_k\right)}\end{array}$ (84)

$\begin{array}{l}N\left(f\left({\displaystyle \underset{j=1}{\overset{k}{\prod }}}{x}_j\cdot y_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\prod }}}f\left(x_j\right)\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\prod }}}{x}_j\cdot {\displaystyle \underset{j=1}{\overset{k}{\prod }}}f\left(y_j\right),t\right)\\ \ge \frac{t}{t+\psi \left(x_1,\cdots ,x_k,y_1,\cdots ,y_k,0,\cdots ,0\right)}\end{array}$ (85)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\in X$ , for all $t>0$ and for all $q>0$ . Then

$\beta \left(x\right)=N-\underset{n\to \infty }{\lim }\frac{1}{{\left(2k\right)}^n}f\left({\left(2k\right)}^{n}x\right)$ (86)

exists each $x\in X$ and defines a fuzzy derivation $H\mathrm{<mo>\; :\; </mo>}X\to X$ .

Such that

$N\left(f\left(x\right)-H\left(x\right)\mathrm{,}t\right)\ge \frac{\left(1-L\right)t}{\left(1-L\right)+L\psi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,0,}\cdots \mathrm{,0}\right)}$ (87)

for all $t>0$ and for all $q>0$ .

7. Conclusion

In this article, I introduced the concept of the general Jensen Cauchy functional equation, then I used a direct method to show that the solutions of the Jensen-Cauchy functional inequality are additive maps related to the functional equation, Jensen-Cauchy. Then apply the derivative setup on fuzzy algebra.