Tremendous Development of Functional Inequalities and Cauchy-Jensen Functional Equations with 3k-Variables on Banach Space and Stability Derivation on Fuzzy-Algebras

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.

Share and Cite:

An, L. (2024) Tremendous Development of Functional Inequalities and Cauchy-Jensen Functional Equations with 3k-Variables on Banach Space and Stability Derivation on Fuzzy-Algebras. Open Access Library Journal, 11, 1-23. doi: 10.4236/oalib.1111241.

1. Introduction

Let $G$ be an m-divisible group where $m\in ℕ\\left\{0\right\}$ and $X$ , $Y$ be a normed space on the same field $\mathbb{K}$ , and $f:G\to X$ ( $f:G\to Y$ ) be a mapping. I use the notation ${‖\text{ }\cdot \text{ }‖}_{X}$ ( ${‖\text{ }\cdot \text{ }‖}_{Y}$ ) 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 $m\in ℕ$ and $X$ is a normed space with norm ${‖\text{ }\cdot \text{ }‖}_{X}$ and that $Y$ is a Banach space with norm ${‖\text{ }\cdot \text{ }‖}_{Y}$ .

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

${‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)‖}_{Y}\le {‖2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}$ (1)

and

$\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)=2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)$ (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 $\left(G,\ast \right)$ be a group and let $\left({G}^{\prime },\circ ,d\right)$ be a metric group with metric $d\left(\cdot ,\cdot \right)$ . Geven $\epsilon >0$ , does there exist a $\delta >0$ such that if $f:G\to {G}^{\prime }$ satisfies: $d\left(f\left(x\ast y\right),f\left(x\right)\circ f\left(y\right)\right)<\delta$ for all $x,y\in G$ then there is a homomorphism $h:G\to {G}^{\prime }$ with $d\left(f\left(x\right),h\left(x\right)\right)<\epsilon$ , 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 $f:{E}_{1}\to {E}_{2}$ be a mapping between Banach space such that $‖f\left(x+y\right)-f\left(x\right)-f\left(y\right)‖\le \epsilon ,$ for all $x,y\in {E}_{1}$ and some $\epsilon \ge 0$ . It was shown that the limit $T\left(x\right)=\underset{n\to \infty }{\mathrm{lim}}\frac{f{\left(2}^{n}x\right)}{{2}^{n}}$ exists for all $x\in {E}_{1}$ and that $T:{E}_{1}\to {E}_{2}$ is that unique additive mapping satisfying $‖f\left(x\right)-T\left(x\right)‖\le \epsilon ,\forall x\in {E}_{1}.$

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

Consider $E,{E}^{\prime }$ to be two Banach spaces, and let $f:E\to {E}^{\prime }$ 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 $p\in \left[0,1\right)$ such that $‖f\left(x+y\right)-f\left(x\right)-f\left(y\right)‖\le \epsilon \left({‖x‖}^{p}+{‖y‖}^{p}\right),\forall x,y\in \mathbb{E}.$ then there exists a unique linear $L:E\to {E}^{\prime }$ satifies $‖f\left(x\right)-L\left(x\right)‖\le \frac{2\theta }{2-{2}^{p}}‖x‖,x\in E.$

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 ${‖x‖}^{p}+{‖y‖}^{p}$ by ${‖x‖}^{p}{‖y‖}^{p}$ for $p,q\in ℝ$ with $p+q\ne 1$ .

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

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

Denote by $\varphi :G×G\to \left[0,\infty \right)$ a function such that $\stackrel{˜}{\varphi }\left(x,y\right)=\underset{n=0}{\overset{\infty }{\sum }}\text{ }\text{ }{2}^{-n}\varphi \left({2}^{n}x,{2}^{n}y\right)<\infty$ for all $x,y\in G$ . Suppose that $f:G\to E$ is a mapping satisfying $‖f\left(x+y\right)-f\left(x\right)-f\left(y\right)‖\le \epsilon$ , $\forall x,y\in G$ . There exists a unique additive mapping $T:G\to E$ such that $‖f\left(x\right)-T\left(x\right)‖\le \stackrel{˜}{\varphi }\left(x,x\right)$ , $\forall x,y\in G$ .

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 $\left({E}_{1},{E}_{2}\right)$ with ${E}_{1}$ and ${E}_{2}$ are Banach spaces if for each $f:{E}_{1}\to {E}_{2}$ satisfying $‖f\left(x+y\right)-f\left(x\right)-f\left(y\right)‖\le \epsilon$ for all $x,y\in {E}_{1}$ for some $\epsilon >0$ , there exists an additive $h:{E}_{1}\to {E}_{2}$ 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

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

$‖f\left(x\right)+f\left(y\right)+2f\left(y\right)‖\le ‖2f\left(\frac{x+y}{2}+z\right)‖$ (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 $n\in ℕ$ there is some $y\in G$ so that $x=ny$ . divisible by every positive integer. Let $G$ be an n-divisible abelian group where $n\in ℕ$ (i.e., $a\to na:G\to G$ is a surjection).

Denote by

$M\left(G,X\right)=\left\{f|f:G\to X\right\}$

${L}^{\infty }\left(G,X\right)=\left\{f:G\to X|{‖f‖}_{\infty }:={\mathrm{sup}}_{x\in G}{‖f‖}_{X}<\infty \right\}$

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

${G}^{r}=\left\{\left({x}_{1},{x}_{2},\cdots ,{x}_{r}\right):{x}_{j}\in G,j=1,\cdots ,k\right\}$

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

Given mappings $E:M\left(G,X\right)\to M\left({G}^{r},{ℝ}^{+}\right)$ , $\phi :{G}^{r}\to ℝ$ and $\psi :G\to {ℝ}^{+}$ . If $E\left(f\right)\left({x}_{1},{x}_{2},\cdots ,{x}_{r}\right)\le \phi \left({x}_{1},{x}_{2},\cdots ,{x}_{r}\right)$ for all ${x}_{1},{x}_{2},\cdots ,{x}_{r}\in G$ implies that there exists $g\in M\left(G,X\right)$ such that $E\left(g\right)\le 0$ and ${‖f\left(x\right)-g\left(x\right)‖}_{\infty }\le \psi \left(x\right)$ , for all $x\in G$ , then we say that the inequality $E\left(f\right)\le 0$ is $\left(\phi ,\psi \right)$ -stable in $M\left(G,X\right)$ . In this case, we also say that the solutions of the inequality $E\left(f\right)\le 0$ is $\left(\phi ,\psi \right)$ -stable in $M\left(G,X\right)$ . Given mappings $E:M\left(G,X\right)\to M\left({G}^{r},{ℝ}^{+}\right)$ , $\phi :{G}^{r}\to ℝ$ and $\psi :G\to {ℝ}^{+}$ if ${‖E\left(f\right)\left({x}_{1},{x}_{2},\cdots ,{x}_{r}\right)‖}_{\infty }\le \phi \left({x}_{1},{x}_{2},\cdots ,{x}_{r}\right)$ for all ${x}_{1},{x}_{2},\cdots ,{x}_{r}\in G$ , implies that there exists $g\in M\left(G,X\right)$ such that $E\left(g\right)=0$ and ${‖f\left(x\right)-g\left(x\right)‖}_{\infty }\le \psi \left(x\right)$ , for all $x\in G$ , then we say that the inequality $E\left(f\right)\le 0$ is $\left(\phi ,\psi \right)$ -stable in $M\left(G,X\right)$ . In this case, we also say that the solutions of the inequality $E\left(f\right)=0$ is $\left(\phi ,\psi \right)$ -stable in $M\left(G,X\right)$ .

It is well known that if an additive function $f:ℝ\to ℝ$ 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 $‖f\left(x\right)+f\left(y\right)+2f\left(z\right)‖\le ‖2f\left(\frac{x+y}{2}+z\right)‖$ 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 $f:G\to X$ , the equation

$\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+m\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)=mf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{m}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)$ (7)

is said to a generalized Cauchy-Jensen equation.

And function inequalities

$\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+m\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)\le mf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{m}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)$ (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 $m\in ℕ\\left\{0\right\}$ 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 $f:G\to X$ be a mapping such that satisfies

${‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+m\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)‖}_{X}\le {‖mf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{m}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{X}$ (9)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ if and only if $f:G\to X$ is additive.

Proof. Prerequisites

Assume that $f:G\to Y$ satisfies (9) Replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left(0,\cdots ,0,0,\cdots ,0,0,\cdots ,0\right)$ in (9), I get $|2k+m|{‖f\left(0\right)‖}_{X}\le |m|{‖f\left(0\right)‖}_{X}$ $\left(|2k+m|-|m|\right){‖f\left(0\right)‖}_{X}\le 0$

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

Next I replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left(-mz,\cdots ,-mz,0,\cdots ,0,z,\cdots ,z\right)$ in (9), I get $‖kf\left(-mz\right)+kmf\left(z\right)‖\le 0$ and so

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

for all $z\in G$ .

Next I replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},-\frac{{x}_{j}+{y}_{j}}{m},\cdots ,-\frac{{x}_{j}+{y}_{j}}{m}\right)$ in (9) and (10) I have

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}f\left({y}_{j}\right)+m\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)‖}_{X}\\ ={‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)-\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}+{y}_{j}\right)‖}_{X}\\ \le {‖mf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{m}-\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{m}\right)‖}_{X}={‖f\left(0\right)‖}_{X}=0\end{array}$ (11)

Therefore

$\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)=\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}+{y}_{j}\right)$ (12)

Finally we replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k}\right)$ by $\left(u,\cdots ,u,v,\cdots ,v\right)$ in (12) so $f\left(u\right)+f\left(v\right)=f\left(u+v\right)$ .

Sufficient conditions:

Suppose $f:G\to Y$ is additive. Then

$f\left(\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{x}_{j}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{y}_{j}\right)=\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)$ (13)

and so $f\left(p\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{x}_{j}\right)=p\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)$ for all $p\in ℚ$ and ${x}_{1},{x}_{2},\cdots ,{x}_{r}\in G$ .

Therefore

$\begin{array}{l}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+m\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)\\ =mf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{m}\right)+m\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)=mf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{m}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)\end{array}$ (14)

So I have something to prove

${‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+m\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)‖}_{Y}\le {‖mf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{m}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}$ (15)

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

Corollary 1. Suppose a mapping $f:G\to X$ , The following clauses are equivalent

2) ${\sum }_{j=1}^{k}\text{ }\text{ }f\left({x}_{j}\right)+{\sum }_{j=1}^{k}\text{ }\text{ }f\left({y}_{j}\right)+m{\sum }_{j=1}^{k}\text{ }\text{ }f\left({z}_{j}\right)=mf\left({\sum }_{j=1}^{k}\frac{{x}_{j}+{y}_{j}}{m}+{\sum }_{j=1}^{k}\text{ }\text{ }{z}_{j}\right)$ ,

$\forall {x}_{j},{y}_{j},{z}_{j}\in G$ , $j=1,\cdots ,k$ .

3) $‖{\sum }_{j=1}^{k}\text{ }\text{ }f\left({x}_{j}\right)+{\sum }_{j=1}^{k}\text{ }\text{ }f\left({y}_{j}\right)+m{\sum }_{j=1}^{k}\text{ }\text{ }f\left({z}_{j}\right)‖\le ‖mf\left({\sum }_{j=1}^{k}\frac{{x}_{j}+{y}_{j}}{m}+{\sum }_{j=1}^{k}\text{ }\text{ }{z}_{j}\right)‖$

$\forall {x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

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 $m\in ℕ\\left\{0\right\}$ 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 $\varphi :{G}^{3k}\to {ℝ}^{+}$ be a function such that

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

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

And

$\begin{array}{l}\stackrel{˜}{\varphi }\left({x}_{1},\cdots ,{x}_{k},{z}_{1},\cdots ,{z}_{k}\right)\\ =\underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(2k\right)}^{n+1}}\varphi \left({\left(2k\right)}^{n+1}{x}_{1},\cdots ,{\left(2k\right)}^{n+1}{x}_{k},0,\cdots ,0,{\left(2k\right)}^{n}{z}_{1},\cdots ,{\left(2k\right)}^{n}{z}_{k}\right)<\infty \end{array}$ (17)

for all ${x}_{1},\cdots ,{x}_{k},{z}_{1},\cdots ,{z}_{k},{z}_{j}\in G$ . Suppose that an odd mapping $f:G\to Y$ satisfies

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)‖}_{Y}\\ \le {‖2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}+\varphi \left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)\end{array}$ (18)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

Then there exists a unique additive mapping $\psi :G\to Y$ such that

${‖f\left(x\right)-\psi \left(x\right)‖}_{Y}\le \stackrel{˜}{\varphi }\left(x,\cdots ,x,x,\cdots ,x\right)$ (19)

for all $x\in G$ .

Proof. Replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left(0,\cdots ,0,0,\cdots ,0,0,\cdots ,0\right)$ in (18), we get

$\left(|2{k}^{2}+k|-|2k|\right){‖f\left(0\right)‖}_{Y}\le 0.$ (20)

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

Next I replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left(2kx,\cdots ,2kx,0,\cdots ,0,-x,\cdots ,-x\right)$ in (18), I get

${‖kf\left(2kx\right)-2{k}^{2}f\left(x\right)‖}_{Y}\le \varphi \left(2kx,2kx,\cdots ,2kx,0,0,\cdots ,0,-x,-x,\cdots ,-x\right)$ (21)

${‖f\left(x\right)-\frac{1}{2k}f\left(2kx\right)‖}_{Y}\le \frac{1}{2{k}^{2}}\varphi \left(2kx,2kx,\cdots ,2kx,0,0,\cdots ,0,-x,-x,\cdots ,-x\right)$

Hence

$\begin{array}{l}{‖\frac{1}{{\left(2k\right)}^{l}}f\left({\left(2k\right)}^{l}x\right)-\frac{1}{{\left(2k\right)}^{m}}f\left({\left(2k\right)}^{m}x\right)‖}_{Y}\\ \le \underset{j=l}{\overset{m-1}{\sum }}{‖\frac{1}{{\left(2k\right)}^{j}}f\left({\left(2k\right)}^{j}x\right)-\frac{1}{{\left(2k\right)}^{j+1}}f\left({\left(2k\right)}^{j+1}x\right)‖}_{Y}\\ \le \frac{1}{2{k}^{2}}\underset{j=l+1}{\overset{m}{\sum }}\frac{1}{{\left(2k\right)}^{j}}\varphi \left({\left(2k\right)}^{j+1}x,\cdots ,{\left(2k\right)}^{j+1}x,0,0,\cdots ,0,-{\left(2k\right)}^{j}x,\cdots ,-{\left(2k\right)}^{j}x\right)\\ =0\end{array}$ (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 $\psi :G\to Y$ by $\psi \left(x\right):=\underset{n\to \infty }{\mathrm{lim}}\frac{1}{{\left(2k\right)}^{n}}f\left({\left(2k\right)}^{n}x\right)$ 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

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }\psi \left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }\psi \left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }\psi \left({z}_{j}\right)‖}_{Y}\\ =\underset{n\to \infty }{\mathrm{lim}}{‖\frac{1}{{\left(2k\right)}^{n}}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({\left(2k\right)}^{n}{x}_{j}\right)+\frac{1}{{\left(2k\right)}^{n}}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({\left(2k\right)}^{n}{y}_{j}\right)+2k\frac{1}{{\left(2k\right)}^{n}}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({\left(2k\right)}^{n}{z}_{j}\right)‖}_{Y}\\ =\underset{n\to \infty }{\mathrm{lim}}\frac{1}{{\left(2k\right)}^{n}}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({\left(2k\right)}^{n}{x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({\left(2k\right)}^{n}{y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({\left(2k\right)}^{n}{z}_{j}\right)‖}_{Y}\\ \le \underset{n\to \infty }{\mathrm{lim}}\frac{1}{{\left(2k\right)}^{n}}\left({‖2kf\left({\left(2k\right)}^{n}\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+{\left(2k\right)}^{n}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}\\ \text{\hspace{0.17em}}\stackrel{\stackrel{}{\text{ }}}{}+\varphi \left({\left(2k\right)}^{n}{x}_{1},\cdots ,{\left(2k\right)}^{n}{x}_{k},{\left(2k\right)}^{n}{y}_{1},\cdots ,{\left(2k\right)}^{n}{y}_{k},{\left(2k\right)}^{n}{z}_{1},\cdots ,{\left(2k\right)}^{n}{z}_{k}\right)\right)\\ ={‖2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}\end{array}$ (23)

So I have

${‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }\psi \left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }\psi \left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }\psi \left({z}_{j}\right)‖}_{Y}\le {‖2k\psi \left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}$ (24)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

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 ${\psi }^{\prime }:G\to Y$ be another generalized Cauchy-Jensen additive mapping satisfying (19). Then I have

$\begin{array}{l}{‖\psi \left(x\right)-{\psi }^{\prime }\left(x\right)‖}_{Y}=\frac{1}{{\left(2k\right)}^{n}}{‖\psi \left({\left(2k\right)}^{n}x\right)-{\psi }^{\prime }\left({\left(2k\right)}^{n}x\right)‖}_{Y}\\ \le \frac{1}{{\left(2k\right)}^{n}}\left({‖f\left({\left(2k\right)}^{n}x\right)-\psi \left({\left(2k\right)}^{n}x\right)‖}_{Y}+{‖f\left({\left(2k\right)}^{n}x\right)-{\psi }^{\prime }\left(\frac{x}{{2}^{n}}\right)‖}_{Y}\right)\\ \le 2\frac{1}{{\left(2k\right)}^{n}}\stackrel{˜}{\varphi }\left({\left(2k\right)}^{n}x,\cdots ,{\left(2k\right)}^{n}x,0,\cdots ,0,{\left(2k\right)}^{n}x,\cdots ,{\left(2k\right)}^{n}x\right)\end{array}$ (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 $p,r\ne 0,\text{\hspace{0.17em}}q>0,\text{\hspace{0.17em}}\theta >0$ . Suppose $f:G\to Y$ be a function such that

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)‖}_{Y}\\ \le {‖2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}+\theta \cdot \underset{j=1}{\overset{k}{\prod }}{‖{x}_{j}‖}^{p}\cdot \underset{j=1}{\overset{k}{\prod }}{‖{y}_{j}‖}^{q}\cdot \underset{j=1}{\overset{k}{\prod }}{‖{z}_{j}‖}^{r}\end{array}$ (26)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ then f í a additive mapping.

Corollary 3. For $G$ is a normed space and $00$ . Suppose $f:G\to Y$ be a function such that

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)‖}_{Y}\\ \le {‖2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}+\theta \left(\underset{j=1}{\overset{k}{\sum }}{‖{x}_{j}‖}^{p}+\underset{j=1}{\overset{k}{\sum }}{‖{y}_{j}‖}^{q}+\underset{j=1}{\overset{k}{\sum }}{‖{z}_{j}‖}^{r}\right)\end{array}$ (27)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ . Then there exists a unique additive mapping $\psi :G\to Y$ such that

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

for all $x\in G$ .

Theorem 3. For $\varphi :{G}^{3k}\to {ℝ}^{+}$ be a function such that

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

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ , and

$\begin{array}{l}\stackrel{˜}{\varphi }\left({x}_{1},\cdots ,{x}_{k},{z}_{1},\cdots ,{z}_{k}\right)\\ =\underset{n=0}{\overset{\infty }{\sum }}\text{ }\text{ }\varphi {\left(2k\right)}^{n}\varphi \left(\frac{1}{{\left(2k\right)}^{n}}{x}_{1},\cdots ,\frac{1}{{\left(2k\right)}^{n}}{x}_{k},0,0,\cdots ,0,\cdots ,\frac{1}{{\left(2k\right)}^{n+1}}{z}_{1},\cdots ,\frac{1}{{\left(2k\right)}^{n+1}}{z}_{k}\right)<\infty \end{array}$ (30)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

Suppose that be an odd mapping $f:G\to Y$ satisfies

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)‖}_{Y}\\ \le {‖2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}+\varphi \left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)\end{array}$ (31)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

Then there exists a unique additive mapping $\psi :G\to Y$ such that

${‖f\left(x\right)-\psi \left(x\right)‖}_{Y}\le \stackrel{˜}{\varphi }\left(x,\cdots ,x,x,\cdots ,x\right)$ (32)

for all $x\in G$ .

Proof. Replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left(0,\cdots ,0,0,\cdots ,0,0,\cdots ,0\right)$ in (31), I get

$\left(|2{k}^{2}+k|-|2k|\right){‖f\left(0\right)‖}_{Y}\le 0.$ (33)

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

Replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left(2kx,\cdots ,2kx,0,\cdots ,0,-x,\cdots ,-x\right)$ in (31), I get

${‖kf\left(2kx\right)-2{k}^{2}f\left(x\right)‖}_{Y}\le \varphi \left(2kx,2kx,\cdots ,2kx,0,0,\cdots ,0,-x,-x,\cdots ,-x\right)$ (34)

${‖f\left(x\right)-2kf\left(\frac{x}{2k}\right)‖}_{Y}\le \frac{1}{k}\varphi \left(x,x,\cdots ,x,0,0,\cdots ,0,-\frac{x}{2k},-\frac{x}{2k},\cdots ,-\frac{x}{2k}\right)$

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 $p,r\ne 0,\text{\hspace{0.17em}}q>0,\text{\hspace{0.17em}}\theta >0$ . Suppose $f:G\to Y$ be a function such that

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)‖}_{Y}\\ \le {‖2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}+\theta \cdot \underset{j=1}{\overset{k}{\prod }}{‖{x}_{j}‖}^{p}\cdot \underset{j=1}{\overset{k}{\prod }}{‖{y}_{j}‖}^{q}\cdot \underset{j=1}{\overset{k}{\prod }}{‖{z}_{j}‖}^{r}\end{array}$ (35)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ , then f is a additive mapping.

Corollary 5. For $G$ is a normed space and $00$ . Suppose $f:G\to Y$ be a function such that

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)‖}_{Y}\\ \le {‖2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}+\theta \left(\underset{j=1}{\overset{k}{\sum }}{‖{x}_{j}‖}^{p}+\underset{j=1}{\overset{k}{\sum }}{‖{y}_{j}‖}^{q}+\underset{j=1}{\overset{k}{\sum }}{‖{z}_{j}‖}^{r}\right)\end{array}$ (36)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ . Then there exists a unique additive mapping $\psi :G\to Y$ such that

${‖f\left(x\right)-\psi \left(x\right)‖}_{Y}\le \theta k\left(\frac{{\left(2k\right)}^{p}}{{\left(2k\right)}^{p}-2k}{‖x‖}^{p}+\frac{1}{{\left(2k\right)}^{k}-2k}{‖x‖}^{r}\right)$ (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 $E:M\left(G,Y\right)\to M\left({G}^{r},{ℝ}^{*}\right)$ as $\begin{array}{l}E\left(f\right)\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)\\ =‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)‖-‖2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖\end{array}$ then the inequalities $Ef\le 0$ is $\left(\varphi ,\stackrel{˜}{\varphi }\right)$ -stable in $M\left(G,Y\right)$ where $\left(\varphi ,\stackrel{˜}{\varphi }\right)$ is as Theorem 2 and Theorem 3.

Note that for inequalities, $G$ be a m-divisible group where $m\in ℕ\\left\{0\right\}$ 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 $\varphi :{G}^{3k}\to {ℝ}^{+}$ be a function such that

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

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ , and

$\begin{array}{l}\stackrel{˜}{\varphi }\left({x}_{1},\cdots ,{x}_{k},{z}_{1},\cdots ,{z}_{k}\right)\\ =\underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(2k\right)}^{n+1}}\left(\varphi \left({\left(2k\right)}^{n+1}{x}_{1},\cdots ,{\left(2k\right)}^{n+1}{x}_{k},0,\cdots ,0,-{\left(2k\right)}^{n}{z}_{1},\cdots ,-{\left(2k\right)}^{n}{z}_{k}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+\varphi \left(-{\left(2k\right)}^{n+1}{x}_{1},\cdots ,-{\left(2k\right)}^{n+1}{x}_{k},0,\cdots ,0,{\left(2k\right)}^{n}{z}_{1},\cdots ,{\left(2k\right)}^{n}{z}_{k}\right)\right)<\infty \end{array}$ (39)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

Suppose that a mapping $f:G\to Y$ satisfies $f\left(0\right)=0$ for all $x\in G$ , and

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)-2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}\\ \le \varphi \left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)\end{array}$ (40)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

Then there exists a unique additive mapping $\psi :G\to Y$ such that

${‖f\left(x\right)-\psi \left(x\right)‖}_{Y}\le \stackrel{˜}{\varphi }\left(x,\cdots ,x,x,\cdots ,x\right)$ (41)

for all $x\in G$ .

Proof. I replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left(2kx,\cdots ,2kx,0,\cdots ,0,-x,\cdots ,-x\right)$ in (40), I get

${‖kf\left(2kx\right)+2{k}^{2}f\left(-x\right)‖}_{Y}\le \varphi \left(2kx,2kx,\cdots ,2kx,0,0,\cdots ,0,-x,-x,\cdots ,-x\right)$ (42)

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

${‖kf\left(-2kx\right)+2{k}^{2}f\left(x\right)‖}_{Y}\le \varphi \left(-2kx,-2kx,\cdots ,-2kx,0,0,\cdots ,0,x,x,\cdots ,x\right)$ (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

$\begin{array}{c}{‖f\left(x\right)-\frac{1}{2k}f\left(2kx\right)‖}_{Y}\le \frac{1}{2{k}^{2}}\left(\varphi \left(2kx,2kx,\cdots ,2kx,0,0,\cdots ,0,-x,-x,\cdots ,-x\right)\\ \text{\hspace{0.17em}}+\varphi \left(-2kx,-2kx,\cdots ,-2kx,0,0,\cdots ,0,x,x,\cdots ,x\right)\right)\end{array}$ (45)

Hence

$\begin{array}{l}{‖\frac{1}{{\left(2k\right)}^{l}}f\left({\left(2k\right)}^{l}x\right)-\frac{1}{{\left(2k\right)}^{m}}f\left({\left(2k\right)}^{m}x\right)‖}_{Y}\\ \le \underset{j=l}{\overset{m-1}{\sum }}{‖\frac{1}{{\left(2k\right)}^{j}}f\left({\left(2k\right)}^{j}x\right)-\frac{1}{{\left(2k\right)}^{j+1}}f\left({\left(2k\right)}^{j+1}x\right)‖}_{Y}\\ \le \frac{1}{2{k}^{2}}\underset{j=l+1}{\overset{m}{\sum }}\frac{1}{{\left(2k\right)}^{j}}\left(\varphi \left({\left(2k\right)}^{j+1}x,\cdots ,{\left(2k\right)}^{j+1}x,0,0,\cdots ,0,-{\left(2k\right)}^{j}x,\cdots ,-{\left(2k\right)}^{j}x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+\varphi \left(-2kx,-2kx,\cdots ,-2kx,0,0,\cdots ,0,x,x,\cdots ,x\right)\right)\\ =0\end{array}$ (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 $\psi :G\to Y$ by $\psi \left(x\right):=\underset{n\to \infty }{\mathrm{lim}}\frac{1}{{\left(2k\right)}^{n}}f\left({\left(2k\right)}^{n}x\right)$ 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

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }\psi \left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }\psi \left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }\psi \left({z}_{j}\right)-2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}\\ =\underset{n\to \infty }{\mathrm{lim}}‖\frac{1}{{\left(2k\right)}^{n}}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({\left(2k\right)}^{n}{x}_{j}\right)+\frac{1}{{\left(2k\right)}^{n}}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({\left(2k\right)}^{n}{y}_{j}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+2k\frac{1}{{\left(2k\right)}^{n}}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({\left(2k\right)}^{n}{z}_{j}\right)-{2kf\left({\left(2k\right)}^{n}\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+{\left(2k\right)}^{n}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}\\ =\underset{n\to \infty }{\mathrm{lim}}\frac{1}{{\left(2k\right)}^{n}}‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({\left(2k\right)}^{n}{x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({\left(2k\right)}^{n}{y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({\left(2k\right)}^{n}{z}_{j}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }-{2kf\left({\left(2k\right)}^{n}\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+{\left(2k\right)}^{n}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}\\ \le \varphi \left({\left(2k\right)}^{n}{x}_{1},\cdots ,{\left(2k\right)}^{n}{x}_{k},{\left(2k\right)}^{n}{y}_{1},\cdots ,{\left(2k\right)}^{n}{y}_{k},{\left(2k\right)}^{n}{z}_{1},\cdots ,{\left(2k\right)}^{n}{z}_{k}\right)=0\end{array}$ (47)

So I have

$\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }\psi \left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }\psi \left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }\psi \left({z}_{j}\right)=2k\psi \left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)$ (48)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

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 ${\psi }^{\prime }:G\to Y$ be another generalized Cauchy-Jensen additive mapping satisfying (41). Then we have

$\begin{array}{l}{‖\psi \left(x\right)-{\psi }^{\prime }\left(x\right)‖}_{Y}=\frac{1}{{\left(2k\right)}^{n}}{‖\psi \left({\left(2k\right)}^{n}x\right)-{\psi }^{\prime }\left({\left(2k\right)}^{n}x\right)‖}_{Y}\\ \le \frac{1}{{\left(2k\right)}^{n}}\left({‖f\left({\left(2k\right)}^{n}x\right)-\psi \left({\left(2k\right)}^{n}x\right)‖}_{Y}+{‖f\left({\left(2k\right)}^{n}x\right)-{\psi }^{\prime }\left(\frac{x}{{2}^{n}}\right)‖}_{Y}\right)\\ \le 2\frac{1}{{\left(2k\right)}^{n}}\stackrel{˜}{\varphi }\left({\left(2k\right)}^{n}x,\cdots ,{\left(2k\right)}^{n}x,0,\cdots ,0,{\left(2k\right)}^{n}x,\cdots ,{\left(2k\right)}^{n}x\right)\\ =\underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(2k\right)}^{n+1}}\left(\varphi \left({\left(2k\right)}^{n+1}{x}_{1},\cdots ,{\left(2k\right)}^{n+1}{x}_{k},0,\cdots ,0,-{\left(2k\right)}^{n}{z}_{1},\cdots ,-{\left(2k\right)}^{n}{z}_{k}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+\varphi \left(-{\left(2k\right)}^{n+1}{x}_{1},\cdots ,-{\left(2k\right)}^{n+1}{x}_{k},0,\cdots ,0,{\left(2k\right)}^{n}{z}_{1},\cdots ,{\left(2k\right)}^{n}{z}_{k}\right)\right)<\infty \end{array}$ (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 $p,r\ne 0,\text{\hspace{0.17em}}q>0,\text{\hspace{0.17em}}\theta >0$ . Suppose $f:G\to Y$ be a function such that $f\left(0\right)=0$ and

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)-2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}\\ \le \theta \cdot \underset{j=1}{\overset{k}{\prod }}{‖{x}_{j}‖}^{p}\cdot \underset{j=1}{\overset{k}{\prod }}{‖{y}_{j}‖}^{q}\cdot \underset{j=1}{\overset{k}{\prod }}{‖{z}_{j}‖}^{r}\end{array}$ (50)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ then f is an additive mapping.

Corollary 7. For $G$ is a normed space and $00$ . Suppose $f:G\to Y$ be a function such that $f\left(0\right)=0$ and

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)-2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}\\ \le \theta \left(\underset{j=1}{\overset{k}{\sum }}{‖{x}_{j}‖}^{p}+\underset{j=1}{\overset{k}{\sum }}{‖{y}_{j}‖}^{q}+\underset{j=1}{\overset{k}{\sum }}{‖{z}_{j}‖}^{r}\right)\end{array}$ (51)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ . Then there exists a unique additive mapping $\psi :G\to Y$ such that

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

for all $x\in G$ .

Theorem 5. For $\varphi :{G}^{3k}\to {ℝ}^{+}$ be a function such that

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

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

And

$\begin{array}{l}\stackrel{˜}{\varphi }\left({x}_{1},\cdots ,{x}_{k},{z}_{1},\cdots ,{z}_{k}\right)\\ =\underset{n=0}{\overset{\infty }{\sum }}{\left(2k\right)}^{n-1}\left(\varphi \left({\left(2k\right)}^{-n}{x}_{1},\cdots ,{\left(2k\right)}^{-\left(n+1\right)}{x}_{k},0,\cdots ,0,-{\left(2k\right)}^{n+1}{z}_{1},\cdots ,-{\left(2k\right)}^{n+1}{z}_{k}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+\varphi \left(-{\left(2k\right)}^{-n}{x}_{1},\cdots ,-{\left(2k\right)}^{-n}{x}_{k},0,\cdots ,0,{\left(2k\right)}^{n+1}{z}_{1},\cdots ,{\left(2k\right)}^{n+1}{z}_{k}\right)\right)<\infty \end{array}$ (54)

for all ${x}_{1},\cdots ,{x}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

Suppose that a mapping $f:G\to Y$ satisfies $f\left(0\right)=0$ for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

And

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)-2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}\\ \le \varphi \left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)\end{array}$ (55)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ .

Then there exists a unique additive mapping $\psi :G\to Y$ such that

${‖f\left(x\right)-\psi \left(x\right)‖}_{Y}\le \stackrel{˜}{\varphi }\left(x,\cdots ,x,x,\cdots ,x\right)$ (56)

for all $x\in G$ .

The proof is similar to theorem 4.

Corollary 8. For $G$ is a normed space and $p,r\ne 0,\text{\hspace{0.17em}}q>0,\text{\hspace{0.17em}}\theta >0$ . Suppose $f:G\to Y$ be a function such that $f\left(0\right)=0$ and

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)-2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}\\ \le \theta \cdot \underset{j=1}{\overset{k}{\prod }}{‖{x}_{j}‖}^{p}\cdot \underset{j=1}{\overset{k}{\prod }}{‖{y}_{j}‖}^{q}\cdot \underset{j=1}{\overset{k}{\prod }}{‖{z}_{j}‖}^{r}\end{array}$ (57)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ then f í a additive mapping.

Corollary 9. For $G$ is a normed space and $00$ . Suppose $f:G\to Y$ be a function such that $f\left(0\right)=0$ and

$\begin{array}{l}{‖\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right)-2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right)‖}_{Y}\\ \le \theta \left(\underset{j=1}{\overset{k}{\sum }}{‖{x}_{j}‖}^{p}+\underset{j=1}{\overset{k}{\sum }}{‖{y}_{j}‖}^{q}+\underset{j=1}{\overset{k}{\sum }}{‖{z}_{j}‖}^{r}\right)\end{array}$ (58)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in G$ . Then there exists a unique additive mapping $\psi :G\to Y$ such that

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

for all $x\in G$ .

6. The Stability of Derivation on Fuzzy-Algebras

Lemma 6. Let $\left(Y,ℕ\right)$ be a fuzzy normed vector space and $f:X\to Y$ be a mapping such that

$N\left(\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({x}_{j}\right)+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({y}_{j}\right)+2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }f\left({z}_{j}\right),t\right)\ge N\left(2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{{x}_{j}+{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }{z}_{j}\right),\frac{t}{2k}\right)$ (60)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in Y$ and all $t>0$ . Then f is Cauchy additive.

Proof. I replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left(0,\cdots ,0,0,\cdots ,0,0,\cdots ,0\right)$ in (60), I have

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

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

Next I replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left(-y,\cdots ,-y,y,\cdots ,y,0,\cdots ,0\right)$ in (60), I have

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

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

So

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

Next I replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left(-2z,\cdots ,-2z,0,\cdots ,0,z,0,\cdots ,0\right)$ in (60), we have

$N\left(-kf\left(2z\right)+2kf\left(z\right),t\right)=N\left(f\left(-2z\right)+2f\left(z\right),\frac{t}{k}\right)\ge N\left(2kf\left(0\right),\frac{t}{2{k}^{2}+2k}\right)$ (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 $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left(x,\cdots ,x,y,\cdots ,y,{z}_{1}=-\frac{x+y}{2},{z}_{2}=0,\cdots ,0\right)$ in (60), we have

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

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

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

$\begin{array}{l}\psi \left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)\\ \le \frac{L}{2k}\psi \left(2k{x}_{1},\cdots ,2k{x}_{k},2k{y}_{1},\cdots ,2k{y}_{k},2k{z}_{1},\cdots ,2k{z}_{k}\right)\end{array}$ (65)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in X$ and $f\left(0\right)=0$ .

Let $f:X\to X$ be a mapping sattisfying

$\begin{array}{l}N\left(2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{q{x}_{j}+q{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }q{z}_{j}\right)-\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qf\left({x}_{j}\right)-\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qf\left({y}_{j}\right)-2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qf\left({z}_{j}\right),t\right)\\ \ge \frac{t}{t+\psi \left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)}\end{array}$ (66)

$\begin{array}{l}N\left(f\left(\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{x}_{j}\cdot {y}_{j}\right)-\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }f\left({x}_{j}\right)\cdot \underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{y}_{j}-\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{x}_{j}\cdot \underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }f\left({y}_{j}\right),t\right)\\ \ge \frac{t}{t+\psi \left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},0,\cdots ,0\right)}\end{array}$ (67)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k}\in X$ , for all $t>0$ and for all $q>0$ . Then

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

exists each $x\in X$ and defines a fuzzy derivation $H:X\to X$ , such that

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

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

Proof. Letting $q=1$ and I replacing $\left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)$ by $\left(x,0,\cdots ,0,0,\cdots ,0,0,\cdots ,0\right)$ in (84), I get

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

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

$\begin{array}{l}d\left(g,h\right):=\mathrm{inf}\left\{\beta \in {ℝ}_{+}:N\left(g\left(x\right)-h\left(x\right),\beta t\right)\begin{array}{c}\text{ }\\ \text{ }\end{array}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \frac{t}{t+\phi \left(x,0,\cdots ,0,0,\cdots ,0,0,\cdots ,0\right)},\forall x\in X,\forall t>0\right\},\end{array}$ (71)

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

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

Hence

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

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

For all $x\in \mathbb{X}$ . So $d\left(f,Tf\right)\le 1$ . By Theorem 1.2, there exists a mapping $H:X\to Y$ 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 $ℚ=\left\{g\in \mathbb{M}:d\left(f,g\right)<\infty \right\}$ .

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

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

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

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

This follows that the inequality (70) is satisfied.

By (85)

$\begin{array}{l}N\left({\left(2k\right)}^{p+1}f\left(\underset{j=1}{\overset{k}{\sum }}\frac{q{x}_{j}+q{y}_{j}}{{\left(2k\right)}^{p+1}}+\underset{j=1}{\overset{k}{\sum }}\frac{q{z}_{j}}{{\left(2k\right)}^{p}}\right)-{\left(2k\right)}^{p}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qf\left(\frac{{x}_{j}}{{\left(2k\right)}^{p}}\right)\\ -{\left(2k\right)}^{p}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qf\left(\frac{{y}_{j}}{{\left(2k\right)}^{p}}\right)-{\left(2k\right)}^{p}2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qf\left(\frac{{z}_{j}}{{\left(2k\right)}^{p}}\right),t\right)\\ \ge \frac{t}{t+\psi \left(\frac{{x}_{1}}{{\left(2k\right)}^{p}},\cdots ,\frac{{x}_{k}}{{\left(2k\right)}^{p}},\frac{{y}_{1}}{{\left(2k\right)}^{p}},\cdots ,\frac{{y}_{k}}{{\left(2k\right)}^{p}},\frac{{z}_{1}}{{\left(2k\right)}^{p}},\cdots ,\frac{{z}_{k}}{{\left(2k\right)}^{p}}\right)}\end{array}$ (74)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in X$ , for all $t>0$ and for all $q\in ℝ$ . So

$\begin{array}{l}N\left({\left(2k\right)}^{p+1}f\left(\underset{j=1}{\overset{k}{\sum }}\frac{q{x}_{j}+q{y}_{j}}{{\left(2k\right)}^{p+1}}+\underset{j=1}{\overset{k}{\sum }}\frac{q{z}_{j}}{{\left(2k\right)}^{p}}\right)-{\left(2k\right)}^{p}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qf\left(\frac{{x}_{j}}{{\left(2k\right)}^{p}}\right)\\ -{\left(2k\right)}^{p}\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qf\left(\frac{{y}_{j}}{{\left(2k\right)}^{p}}\right)-{\left(2k\right)}^{p}2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qf\left(\frac{{z}_{j}}{{\left(2k\right)}^{p}}\right),t\right)\\ \ge \frac{\frac{t}{{\left(2k\right)}^{p}}}{\frac{t}{{\left(2k\right)}^{p}}+\frac{{L}^{p}}{{\left(2k\right)}^{p}}\psi \left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)}\end{array}$ (75)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in X$ , for all $t>0$ and for all $q\in ℝ$ .

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

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

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in X$ , $\forall t>0$ , $q\in ℝ$ . So

$2kH\left(\underset{j=1}{\overset{k}{\sum }}\frac{q{x}_{j}+q{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }q{z}_{j}\right)-\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qH\left({x}_{j}\right)-\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qH\left({y}_{j}\right)-2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qH\left({z}_{j}\right)=0$ (77)

Thus the mapping $H:X\to X$ is additive and $R$ -linear by (85) I have

$\begin{array}{l}N\left({\left(2k\right)}^{2p}f\left(\underset{j=1}{\overset{k}{\prod }}\frac{{x}_{j}\cdot {y}_{j}}{{\left(2k\right)}^{2p}}\right)-{\left(2k\right)}^{p}\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }f\left(\frac{{x}_{j}}{{\left(2k\right)}^{p}}\right)\cdot \underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{y}_{j}\\ -\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{x}_{j}\cdot {\left(2k\right)}^{p}\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }f\left(\frac{{y}_{j}}{{\left(2k\right)}^{p}}\right),t\right)\\ \ge \frac{t}{t+\psi \left(\frac{{x}_{1}}{{\left(2k\right)}^{p}},\cdots ,\frac{{x}_{k}}{{\left(2k\right)}^{p}},\frac{{y}_{1}}{{\left(2k\right)}^{p}},\cdots ,\frac{{y}_{k}}{{\left(2k\right)}^{p}},0,\cdots ,0\right)}\end{array}$ (78)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k}\in X$ , for all $t>0$ .

$\begin{array}{l}N\left({\left(2k\right)}^{2p}f\left(\underset{j=1}{\overset{k}{\prod }}\frac{{x}_{j}\cdot {y}_{j}}{{\left(2k\right)}^{2p}}\right)-{\left(2k\right)}^{p}\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }f\left(\frac{{x}_{j}}{{\left(2k\right)}^{p}}\right)\cdot \underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{y}_{j}\\ -\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{x}_{j}\cdot {\left(2k\right)}^{p}\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }f\left(\frac{{y}_{j}}{{\left(2k\right)}^{p}}\right),t\right)\\ \ge \frac{\frac{t}{{\left(2k\right)}^{2p}}}{\frac{t}{{\left(2k\right)}^{2p}}+\frac{{L}^{p}}{{\left(2k\right)}^{p}}\psi \left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},0,\cdots ,0\right)}\end{array}$ (79)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k}\in X$ , for all $t>0$ Since

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

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k}\in X$ , for all $t>0$ Thus

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

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k}\in X$ , for all $t>0$ Thus

$f\left(\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{x}_{j}\cdot {y}_{j}\right)-\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }f\left({x}_{j}\right)\cdot \underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{y}_{j}-\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{x}_{j}\cdot \underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }f\left({y}_{j}\right)=0$ (82)

So the mapping $H:X\to X$ is a fuzzy derivation, as desired.

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

$\psi \left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)\le 2k\psi \left(\frac{{x}_{1}}{2k},\cdots ,\frac{{x}_{k}}{2k},\frac{{y}_{1}}{2k},\cdots ,\frac{{y}_{k}}{2k},\frac{{z}_{1}}{2k},\cdots ,\frac{{z}_{k}}{2k}\right)$ (83)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\in X$ and $f\left(0\right)=0$ .

Let $f:X\to X$ be a mapping sattisfying

$\begin{array}{l}N\left(2kf\left(\underset{j=1}{\overset{k}{\sum }}\frac{q{x}_{j}+q{y}_{j}}{2k}+\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }q{z}_{j}\right)-\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qf\left({x}_{j}\right)-\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qf\left({y}_{j}\right)-2k\underset{j=1}{\overset{k}{\sum }}\text{ }\text{ }qf\left({z}_{j}\right),t\right)\\ \ge \frac{t}{t+\psi \left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},{z}_{1},\cdots ,{z}_{k}\right)}\end{array}$ (84)

$\begin{array}{l}N\left(f\left(\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{x}_{j}\cdot {y}_{j}\right)-\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }f\left({x}_{j}\right)\cdot \underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{y}_{j}-\underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }{x}_{j}\cdot \underset{j=1}{\overset{k}{\prod }}\text{ }\text{ }f\left({y}_{j}\right),t\right)\\ \ge \frac{t}{t+\psi \left({x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k},0,\cdots ,0\right)}\end{array}$ (85)

for all ${x}_{1},\cdots ,{x}_{k},{y}_{1},\cdots ,{y}_{k}\in X$ , for all $t>0$ and for all $q>0$ . Then

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

exists each $x\in X$ and defines a fuzzy derivation $H:X\to X$ .

Such that

$N\left(f\left(x\right)-H\left(x\right),t\right)\ge \frac{\left(1-L\right)t}{\left(1-L\right)+L\psi \left({x}_{1},\cdots ,{x}_{k},0,\cdots ,0\right)}$ (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.

Conflicts of Interest

The authors declare no conflicts of interest.