Ly Van An
Faculty of Mathematics Teacher Education, Tay Ninh University, Tay Ninh, Vietnam.
DOI: 10.4236/oalib.1109480   PDF    HTML   XML   121 Downloads   1,037 Views   Citations

Abstract

In this paper, we study to solve two additive (ξ₁,ξ₂)-functional inequalities with 3k-variables and their Hyers-Ulam stability: First are investigated in complex Banach spaces with a fixed point method and last are investigated in complex Banach spaces with a direct method: These are the main results of this paper.

Keywords

Additive (ξ1, ξ2)-Functional Inequality, Fixed Point Method, Direct Method, Banach Space, Hyers-Ulam Stability

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Mathematics Subject Classification

46S10, 39B62, 39B52, 47H10

1. Introduction

Let $\mathbb{X}$ and $Y$ be normed spaces on the same field $\mathbb{K}$, and $f\mathrm{<mo>\; :\; </mo>}\mathbb{X}\to Y$. We use the notation $\Vert \cdot \Vert$ for all the norms on both $\mathbb{X}$ and $Y$. In this paper, we investisgate additive (ξ₁,ξ₂)-functional inequalitíes when $X$ be a real or complex normed space and $Y$ a complex Banach spaces. We solve and prove the Hyers-Ulam-Rassisa type stability of following Cauchy additive (ξ₁,ξ₂)-functional inequalities.

$\begin{array}{l}{\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\end{array}$ (1)

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

$\begin{array}{l}{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\end{array}$ (2)

So (1) and (2) are equivalent propositions, which ${\xi }_1,{\xi }_2$ are fixed nonzero complex numbers with $G\left({\xi }_1\mathrm{,}{\xi }_2\right)$ -functional inequality. Note that in the preliminaries, we just recap some of the essential properties for the above problem and for the specific problem, please see the document. The Hyers-Ulam stability was first investigated for the functional equation of Ulam in [1] concerning the stability of group homomorphisms.

The functional equation

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

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

The Hyers [2] gave the first affirmative partial answer to the equation of Ulam in Banach spaces. After that, Hyers’ Theorem was generalized by Aoki [3] additive mappings and by Rassias [4] for linear mappings considering an unbounded Cauchy difference. A generalization of the Rassias theorem was obtained by Găvruta [5] by replacing the unbounded Cauchy difference with a general control function in the spirit of Rassias’ approach.

The stability of the quadratic functional equation was proved by Skof [6] for mappings $f\mathrm{<mo>\; :\; </mo>}X\to \mathbb{Y}$, where $X$ is a normed space and $Y$ is a Banach space. Park [7] [8] defined additive $\gamma$ -functional inequalities and proved the Hyers-Ulam stability of the additive $\gamma$ -functional inequalities in Banach spaces and non-archimedean Banach spaces. The stability problems of various functional equations have been extensively investigated by a number of authors on the world even term [4] - [29]. We recall a fundamental result in fixed point theory. The authors studied the Hyers-Ulam stability for the following functional inequalities

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

$\Vert f\left(\frac{x+y}{2^2}+\frac{z}{2}\right)-\frac{1}{2}f\left(\frac{x+y}{2}\right)-\frac{1}{2}f\left(z\right)\Vert \le \Vert f\left(\frac{x+y}{2}+z\right)-f\left(\frac{x+y}{2}\right)-f\left(z\right)\Vert$ (4)

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

$\Vert 2f\left(\frac{x+y}{2}\right)-f\left(x\right)-f\left(y\right)\Vert \le \Vert \rho \left(f\left(x+y\right)-f\left(x\right)-f\left(y\right)\right)\Vert$ (6)

and

$\begin{array}{l}\Vert f\left(\frac{x+y}{2}+z\right)+f\left(\frac{x+y}{2}-z\right)-2f\left(\frac{x+y}{2}\right)-2f\left(z\right)\Vert \\ \le \Vert \beta \left(2f\left(\frac{x+y}{2^2}+\frac{z}{2}\right)+2f\left(\frac{x+y}{2^2}-\frac{z}{2}\right)-f\left(\frac{x+y}{2}\right)-f\left(z\right)\right)\Vert \end{array}$ (7)

$\begin{array}{l}\Vert 2f\left(\frac{x+y}{2^2}+\frac{z}{2}\right)+2f\left(\frac{x+y}{2^2}-\frac{z}{2}\right)-f\left(\frac{x+y}{2}\right)-f\left(z\right)\Vert \\ \le \Vert \beta \left(f\left(\frac{x+y}{2}+z\right)+f\left(\frac{x+y}{2}-z\right)-2f\left(\frac{x+y}{2}\right)-2f\left(z\right)\right)\Vert \end{array}$ (8)

finaly

$\begin{array}{l}\Vert f\left(x+y\right)-f\left(x\right)-f\left(y\right)\Vert \\ \le \Vert {\beta }_1\left(f\left(x+y\right)+f\left(x-y\right)-2f\left(x\right)\right)\Vert +\Vert {\beta }_2\left(2f\left(\frac{x+y}{2}\right)-f\left(x\right)-f\left(y\right)\right)\Vert \end{array}$ (9)

next

$\begin{array}{l}{\Vert f\left(x_1+x_2+\cdots +x_n\right)-f\left(x_1\right)-f\left(x_2+\cdots +x_n\right)\Vert }_{\mathbb{Y}}\\ \le {\Vert {\beta }_1\left(f\left(x_1+x_2+\cdots +x_n\right)-f\left(x_1-x_2-\cdots -x_n\right)-2f\left(x_1\right)\right)\Vert }_{\mathbb{Y}}\\ +{\Vert {\beta }_2\left(2f\left(\frac{x_1+x_2+\cdots +x_n}{2}\right)-f\left(x_1\right)-f\left(x_2+\cdots +x_n\right)\right)\Vert }_{\mathbb{Y}}\end{array}$ (10)

final

$\begin{array}{l}{\Vert 2f\left(\frac{x_1+x_2}{2}+\frac{x_3+x_4+\cdots +x_k}{4}\right)-f\left(x_1\right)-f\left(x_2+\frac{x_3+x_4+\cdots +x_k}{2}\right)\Vert }_Y\\ \le {\Vert {\beta }_1\left(f\left(x_1+x_2+\frac{x_3+x_4+\cdots +x_k}{2}\right)+f\left(x_1-x_2-\frac{x_3+x_4+\cdots +x_k}{2}\right)-2f\left(x_1\right)\right)\Vert }_Y\\ +{\Vert {\beta }_2\left(f\left(x_1+x_2+\frac{x_3+x_4+\cdots +x_k}{2}\right)-f\left(x_1\right)-f\left(x_2+\frac{x_3+x_4+\cdots +x_k}{2}\right)\right)\Vert }_Y\end{array}$ (11)

in complex Banach spaces.

In this paper, we solve and prove the Hyers-Ulam stability for (ξ₁,ξ₂)-functional inequalities (1) and (2), i.e., the (ξ₁,ξ₂)-functional inequalities with n-variables. Under suitable assumptions on spaces $X$ and $Y$, we will prove that the mappings satisfy the (ξ₁,ξ₂)-functional inequalities (1) and (2). Thus, the results in this paper are the generalization of those in [13] [14] [15] [16] [26] [27] [28] [29] for (ξ₁,ξ₂)-functional inequalities with n-variables.

The goal of the paper is to develop functional inequalities with a higher number of variables to solve problems of general nonlinear functional equations in order to develop the field of nonlinear analysis.

The paper is organized as follows: In the section preliminaries, we remind some basic notations in [13] [14] [17], such as complete generalized metric space and Solutions of the inequalities.

Section 3: In this section, I use the method of the fixed to prove the Hyers-Ulam stability of the additive (ξ₁,ξ₂)-functional inequalities (1) when $\mathbb{X}$ be a real or complete normed space and $\mathbb{Y}$ complex Banach space.

Section 4: In this section, I use the method of directly determining the solution for (1) when $X$ be a real or complete normed space and $Y$ complex Banach space.

Section 5: In this section, I use the method of the fixed to prove the Hyers-Ulam stability of the additive (ξ₁,ξ₂)-functional inequalities (2) when $X$ be a real or complete normed space and $Y$ complex Banach space.

Section 6: In this section, I use the method of directly determining the solution for (2) when $X$ be a real or complete normed space and $Y$ complex Banach space.

2. Preliminaries

2.1. Complete Generalized Metric Space and Solutions of the Inequalities

Theorem 1. Let $\left(\mathbb{X}\mathrm{,}d\right)$ be a complete generalized metric space and let $J\mathrm{<mo>\; :\; </mo>}\mathbb{X}\to \mathbb{X}$ be a strictly contractive mapping with Lipschitz constant $L<1$. Then for each given element $x\in \mathbb{X}$, either

$d\left(J^n\mathrm{,}{J}^{n+1}\right)=\infty$

for all nonegative integers n or there exists a positive integer $n_0$ such that

1). $d\left(J^n,J^{n+1}\right)<\infty$, $\forall n\ge n_0$ ;

2). The sequence $\left\{J^{n}x\right\}$ converges to a fixed point $y^{\mathrm{<mo>\; *\; </mo>}}$ of J;

3). $y^{\mathrm{<mo>\; *\; </mo>}}$ is the unique fixed point of J in the set $Y=\left\{y\in \mathbb{X}|d\left(J^n,J^{n+1}\right)<\infty \right\}$ ;

4). $d\left(y\mathrm{,}{y}^{\mathrm{<mo>\; *\; </mo>}}\right)\le \frac{1}{1-l}d\left(y\mathrm{,}Jy\right)$ $\forall y\in \mathbb{Y}$.

2.2. Solutions of the Inequalities

The functional equation

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

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

3. Establish the Solution of the Additive (ξ₁,ξ₂)-Function Inequalities Using a Fixed Point Method

Now, we first study the solutions of (1). Note that for these inequalities, when $X$ be a real or complete normed space and $Y$ complex Banach space.

Lemma 2. Suppose mapping $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\end{array}$ (12)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}\forall j=1\to k$, then $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ is Cauchy additive

Proof. Assume that $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (12)

We replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (12) we have

${\Vert 2\Gamma \left(\frac{x}{2}\right)-\Gamma \left(x\right)\Vert }_Y\le 0$

Thus

$\Gamma \left(\frac{x}{2}\right)=\frac{1}{2}f\left(x\right)$ (13)

for all $x\in X$.

It follows from (12) and (13) that

$\begin{array}{l}{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ ={\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\end{array}$ (14)

and so

$\begin{array}{l}\left(1-\left|{\xi }_2\right|\right){\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le \left|{\xi }_1\right|{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\Vert }_Y\end{array}$ (15)

we let $u={\displaystyle {\sum }_{j=1}^k}\frac{x_j+y_j}{2}+{\displaystyle {\sum }_{j=1}^k}{z}_j,v={\displaystyle {\sum }_{j=1}^k}\frac{x_j+y_j}{2}-{\displaystyle {\sum }_{j=1}^k}{z}_j$, for all $j=1\to k$,

we get

$\begin{array}{l}\left(1-\left|{\xi }_2\right|\right){\Vert \Gamma \left(u\right)-f\left(\frac{u+v}{2}\right)-\Gamma \left(\frac{u-v}{2}\right)\Vert }_Y\\ \le \left|{\xi }_1\right|{\Vert \Gamma \left(u\right)+\Gamma \left(v\right)-2\Gamma \left(\frac{u+v}{2}\right)\Vert }_Y\end{array}$ (16)

for all $u\mathrm{,}v\in X$

and so

$\begin{array}{l}\frac{1}{2}\left(1-\left|{\xi }_2\right|\right){\Vert \Gamma \left(u+v\right)+\Gamma \left(u-v\right)-2\Gamma \left(u\right)\Vert }_Y\\ \le \left|{\xi }_1\right|{\Vert \Gamma \left(u+v\right)-f\left(u\right)-\Gamma \left(v\right)\Vert }_Y\end{array}$ (17)

for all $u\mathrm{,}v\in X$. It follows from (15) and (17) that

$\begin{array}{l}\frac{1}{2}{\left(1-\left|{\xi }_2\right|\right)}^2{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\left|{\xi }_1\right|}^2{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\end{array}$ (18)

Since $\sqrt{2}\left|{\xi }_1\right|+\left|{\xi }_2\right|<1$

and so

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

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}\forall j=1\to k$. Thus $\Gamma$ is Cauchy additive. □

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

$\phi \left(\frac{x_1}{2}\mathrm{,}\cdots \mathrm{,}\frac{x_k}{2}\mathrm{,}\frac{y_1}{2}\mathrm{,}\cdots \mathrm{,}\frac{y_k}{2}\mathrm{,}\frac{z_1}{2}\mathrm{,}\cdots \mathrm{,}\frac{z_k}{2}\right)\le \frac{L}{2}\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ (19)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}\forall j=1\to k$. If $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfy $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{n}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (20)

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

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

${\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert }_Y\le \frac{1}{1-L}\phi \left(x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ (21)

for all $x\in X$.

Proof. Replacing $\left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}{z}_2\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (20), we get

${\Vert 2\Gamma \left(\frac{x}{2}\right)-\Gamma \left(x\right)\Vert }_Y\le \phi \left(x\mathrm{,0,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ (22)

for all $x\in \mathbb{X}$.

Consider the set

$\mathbb{S}\mathrm{<mo>\; :\; </mo>}=\left\{h\mathrm{<mo>\; :\; </mo>}X\to Y\mathrm{,}h\left(0\right)=0\right\}$

and introduce the generalized metric on $\mathbb{S}$ :

$d\left(g\mathrm{,}h\right)\mathrm{<mo>\; :\; </mo>}=inf\left\{\lambda \in ℝ\mathrm{<mo>\; :\; </mo>}\Vert g\left(x\right)-h\left(x\right)\Vert \le \lambda \phi \left(x\mathrm{,0,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)\mathrm{,}\forall x\in X\right\}\mathrm{,}$

where, as usual, $inf\varphi =+\infty$. It easy to show that $\left(\mathbb{S}\mathrm{,}d\right)$ is complete [17] Now we cosider the linear mapping $J\mathrm{<mo>\; :\; </mo>}\mathbb{S}\to \mathbb{S}$ such that

$Jg\left(x\right)\mathrm{<mo>\; :\; </mo>}=2g(\; x\; 2\; )$

for all $x\in X$. Let $g\mathrm{,}h\in \mathbb{S}$ be given such that $d\left(g\mathrm{,}h\right)=\epsilon$. Then

$\Vert g\left(x\right)-h\left(x\right)\Vert \le \epsilon \phi \left(x\mathrm{,0},\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$

for all $x\in X$.

Hence

$\begin{array}{c}\Vert Jg\left(x\right)-Jh\left(x\right)\Vert =\Vert 2g\left(\frac{x}{2}\right)-2h\Gamma \left(\frac{x}{2}\right)\Vert \\ \le 2\epsilon \phi \left(\frac{x}{2}\mathrm{,0,}\cdots \mathrm{,0,}\frac{x}{2}\mathrm{,0,}\cdots \mathrm{,0,0},\cdots \mathrm{,0}\right)\\ \le 2\epsilon \frac{L}{2}\phi \left(x\mathrm{,0},\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)\\ \le L\epsilon \phi \left(x\mathrm{,0},\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)\end{array}$

for all $x\in X$. So $d\left(g\mathrm{,}h\right)=\epsilon$ implies that $d\left(Jg\mathrm{,}Jh\right)\le L\cdot \epsilon$. This means that

$d\left(Jg\mathrm{,}Jh\right)\le Ld\left(g\mathrm{,}h\right)$.

for all $g\mathrm{,}h\in \mathbb{S}$ It folows from (22) that

$d\left(\Gamma \mathrm{,}J\Gamma \right)\le 1.$

By Theorem 2.1, there exists a mapping $\psi \mathrm{<mo>\; :\; </mo>}X\to Y$ satisfying the fllowing:

1) $\psi$ is a fixed point of J, i.e.,

$\psi \left(x\right)=2\psi \left(\frac{x}{2}\right)$ (23)

for all $x\in X$. The mapping $\psi$ is a unique fixed point J in the set

$\mathbb{M}=\left\{g\in \mathbb{S}\mathrm{<mo>\; :\; </mo>}d\left(\Gamma \mathrm{,}g\right)<\infty \right\}$

This implies that $\psi$ is a unique mapping satisfying (23) such that there exists a $\lambda \in \left(\mathrm{0,}\infty \right)$ satisfying

$\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert \le \lambda \phi \left(x\mathrm{,0},\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$

for all $x\in X$

2) $d\left(J^{l}f\mathrm{,}\psi \right)\to 0$ as $l\to \infty$. This implies equality

$\underset{l\to \infty }{lim}{2}^{n}f\left(\frac{x}{2^n}\right)=\psi (\; x\; )$

for all $x\in \mathbb{X}$

3) $d\left(\Gamma \mathrm{,}\psi \right)\le \frac{1}{1-L}d\left(\Gamma \mathrm{,}J\Gamma \right)$. which implies

$\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert \le \frac{1}{1-L}\phi \left(x\mathrm{,0},\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$

for all $x\in X$. It follows (19) and (20) that

$\begin{array}{l}{\Vert 2\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ =\underset{n\to \infty }{lim}{2}^n{\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+2}}+\frac{1}{2^{n+1}}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}\right)-\Gamma \left(\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le \underset{n\to \infty }{lim}{2}^n\left|{\xi }_1\right|{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}+\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}-\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}\right)\Vert }_Y\\ +\underset{n\to \infty }{lim}{2}^n\left|{\xi }_2\right|{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}+\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}\right)-\Gamma \left(\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ +\underset{n\to \infty }{lim}{2}^n\phi \left(\frac{x_1}{2^n}\mathrm{,}\cdots \mathrm{,}\frac{x_n}{2^n}\mathrm{,}\frac{y_1}{2^n}\mathrm{,}\cdots \mathrm{,}\frac{y_n}{2^n}\mathrm{,}\frac{z_1}{2^n}\mathrm{,}\cdots \mathrm{,}\frac{z_n}{2^n}\right)\end{array}$

$\begin{array}{l}={\Vert {\xi }_1\left(\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\end{array}$ (24)

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

$\begin{array}{l}{\Vert 2\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\beta }_1\left(\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\beta }_2\left(\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}j=1\to k$. By Lemma 3.1, the mapping $\psi \mathrm{<mo>\; :\; </mo>}X\to Y$ is additive. Ei

$\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)=0$

□

Theorem 4. Let $\phi \mathrm{<mo>\; :\; </mo>}{X}^{3k}\to \left[\mathrm{0,}\infty \right)$ be a function such that there exists an $L<1$ with

$\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\le 2L\phi \left(\frac{x_1}{2}\mathrm{,}\cdots \mathrm{,}\frac{x_k}{2}\mathrm{,}\frac{y_1}{2}\mathrm{,}\cdots \mathrm{,}\frac{y_k}{2}\mathrm{,}\frac{z_1}{2}\mathrm{,}\cdots \mathrm{,}\frac{z_k}{2}\right)$ (25)

for all $x\mathrm{,}y\mathrm{,}z\in \mathbb{X}$. Let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfy $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (26)

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

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

${\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert }_Y\le \frac{L}{1-L}\phi \left(x,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ (27)

for all $x\in X$.

The rest of the proof is similar to the proof of Theorem 3.

From proving the theorems we have consequences:

Corollary 1. Let $r>1$ and $\theta$ be nonnegative real numbers and let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfy $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (28)

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

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

${\Vert f\left(x\right)-\psi \left(x\right)\Vert }_{\mathbb{Y}}\le \frac{{2.2}^r\theta }{2^r-2}{\Vert x\Vert }_X^r$ (29)

for all $x\in X$.

Corollary 2. Let $r<1$ and $\theta$ be nonnegative real numbers and let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfy $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert 2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (30)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}\forall j\to k$.

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

${\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert }_Y\le \frac{{2.2}^r\theta }{2-2^r}{\Vert x\Vert }_X^r$ (31)

for all $x\in X$.

4. Establish the Solution of the Additive (ξ₁,ξ₂)-Function Inequalities Using a Direect Method

Next, we study the solutions of (1). Note that for these inequalities, when $X$ be a real or complete normed space and $Y$ complex Banach space.

Theorem 5. Suppose $\phi \mathrm{<mo>\; :\; </mo>}{X}^{3k}\to \left[\mathrm{0,}\infty \right)$ be a function such that

$\begin{array}{l}\varphi \left(x_1,\cdots ,x_k,y_1,\cdots ,y_k,z_1,\cdots ,z_k\right)\\ :={\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}{2}^j\phi \left(\frac{x_1}{2^j},\cdots ,\frac{x_k}{2^j},\frac{y_1}{2^j},\cdots ,\frac{y_k}{2^j},\frac{z_1}{2^j},\cdots ,\frac{z_k}{2^j}\right)<\infty \end{array}$ (32)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}j=1\to k$ and let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfies $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (33)

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

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

${\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert }_{\mathbb{Y}}\le \varphi \left(x,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ (34)

for all $x\in X$

Proof. Replacing $\mathrm{<mo\; stretchy="false">\; (\; </mo>}{x}_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}{z}_2\mathrm{,}\cdots \mathrm{,}{z}_k\mathrm{<mo\; stretchy="false">\; )\; </mo>}$ by $\left(x,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (33), we get

${\Vert 2\Gamma \left(\frac{x}{2}\right)-\Gamma \left(x\right)\Vert }_Y\le \phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ (35)

for all $x\in X$.

Hence

$\begin{array}{l}{\Vert 2^l\Gamma \left(\frac{x}{2^l}\right)-2^m\Gamma \left(\frac{x}{2^m}\right)\Vert }_Y\\ \le {\displaystyle \underset{j=l}{\overset{m-1}{\sum }}}{\Vert 2^j\Gamma \left(\frac{x}{2^j}\right)-2^{j+1}\Gamma \left(\frac{x}{2^{j+1}}\right)\Vert }_Y\\ \le {\displaystyle \underset{j=l}{\overset{m-1}{\sum }}}{2}^j\phi \left(\frac{x}{2^{j+1}}\mathrm{,0,}\cdots ,\mathrm{0,}\frac{x}{2^{j+1}}\mathrm{,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)\end{array}$ (36)

for all nonnegative integers m and l with $m>l$ and all $x\in \mathbb{X}$. It follows from (36) that the sequence $\left\{2^n\Gamma \left(\frac{x}{2^n}\right)\right\}$ is a Cauchy sequence for all $x\in \mathbb{X}$. Since $\mathbb{Y}$ is complete, the sequence $\left\{2^n\Gamma \left(\frac{x}{2^n}\right)\right\}$ coverges. So one can define the mapping $\psi \mathrm{<mo>\; :\; </mo>}X\to Y$ by

$\psi \left(x\right):=\underset{n\to \infty }{\lim }{2}^n\Gamma \left(\frac{x}{2^n}x\right)$ (37)

for all $x\in \mathbb{X}$. Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (37), we get (34) It follows from (32) and (33) that

$\begin{array}{l}{\Vert 2\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ =\underset{n\to \infty }{lim}{2}^n{\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+2}}+\frac{1}{2^{n+1}}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}\right)-f\left(\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le \underset{n\to \infty }{lim}{2}^n\left|{\xi }_1\right|{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}+\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}-\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}\right)\Vert }_Y\\ +\underset{n\to \infty }{lim}{2}^n\left|{\xi }_2\right|{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}+\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}\right)-\Gamma \left(\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ +\underset{n\to \infty }{lim}{2}^n\phi \left(\frac{x_1}{2^n}\mathrm{,}\cdots \mathrm{,}\frac{x_k}{2^n}\mathrm{,}\frac{y_1}{2^n}\mathrm{,}\cdots \mathrm{,}\frac{y_k}{2^n}\mathrm{,}\frac{z_1}{2^n}\mathrm{,}\cdots \mathrm{,}\frac{z_k}{2^n}\right)\end{array}$

$\begin{array}{l}={\Vert {\beta }_1\left(\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\beta }_2\left(\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\end{array}$ (38)

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

$\begin{array}{l}{\Vert 2\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}j=1\to n$. By Lemma 3.1, the mapping $\psi \mathrm{<mo>\; :\; </mo>}X\to Y$ is additive. Ei

$\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)=0$

Now, let ${\psi }^{\prime }\mathrm{<mo>\; :\; </mo>}\mathbb{X}\to \mathbb{Y}$ be another additive mapping satisfying (34). Then we have

$\begin{array}{c}\Vert \psi \left(x\right)-{\psi }^{\prime }\left(x\right)\Vert ={\Vert 2^q\psi \left(\frac{x}{2^q}\right)-2^q{\psi }^{\prime }\left(\frac{x}{2^q}\right)\Vert }_Y\\ \le \Vert 2^q\psi \left(\frac{x}{2^q}\right)-2^{q}f\left(\frac{x}{2^q}\right)\Vert +{\Vert 2^q{\psi }^{\prime }\left(\frac{x}{2^q}\right)-2^{q}f\left(\frac{x}{2^q}\right)\Vert }_{\mathbb{Y}}\\ \le 2^q\varphi \left(\frac{x}{2^q}\mathrm{,0,}\cdots \mathrm{,0,}\frac{x}{2^q}\mathrm{,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)\end{array}$

which tends to zero as $q\to \infty$ 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 uniqueness of $\psi$.

Theorem 6. Suppose $\phi \mathrm{<mo>\; :\; </mo>}{X}^{3k}\to \left[\mathrm{0,}\infty \right)$ be a function such that

$\begin{array}{l}\psi \left(x_1,\cdots ,x_k,y_1,\cdots ,y_k,z_1,\cdots ,z_k\right)\\ :={\displaystyle \underset{j=0}{\overset{\infty }{\sum }}}\frac{1}{2^j}\phi \left(2^j{x}_1,\cdots ,2^j{x}_k,2^j{y}_1,\cdots ,2^j{y}_k,2^j{z}_1,\cdots ,2^j{z}_k\right)<\infty \end{array}$ (39)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}j=1\to k$ and let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfies $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (40)

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

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

$\begin{array}{l}{\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert }_Y\\ \le \varphi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)\end{array}$ (41)

for all $x\in \mathbb{X}$

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

From proving the theorems we have consequences:

Corollary 3. Let $r>1$ and $\theta$ be nonnegative real numbers and let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfy $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (42)

for all $x_j\in X$.

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

${\Vert f\left(x\right)-\psi \left(x\right)\Vert }_{\mathbb{Y}}\le \frac{{2.2}^r\theta }{2^r-2}{\Vert x\Vert }_X^r$ (43)

for all $x\in X$

Corollary 4. Let $r<1$ and $\theta$ be non-negative real numbers and let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfy $f\left(0\right)=0$ and

$\begin{array}{l}{\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (44)

for all $x_j\mathrm{,}{y}_j\mathrm{,}z-j\in X$.

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

${\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert }_Y\le \frac{{2.2}^r\theta }{2-2^r}{\Vert x\Vert }_X^r$ (45)

for all $x\in X$

5. Establish the Solution of the Cauchy Additive (ξ₁,ξ₂)-Function Inequalities Using a Fixed Point Method

Now, we first study the solutions of (2). Note that for these inequalities, when $X$ be a real or complete normed space and $Y$ complex Banach space.

Lemma 7. Suppose mapping $\Gamma \mathrm{<mo>\; :\; </mo>}\mathbb{X}\to Y$ satisfies $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\end{array}$ (46)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in \mathbb{X}\mathrm{,}\forall j=1\to k$ if and only if $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ is Cauchy additive

Proof. Assume that $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (46)

$\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(x,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)$ in (46) we have

${\Vert \Gamma \left(2x\right)-2\Gamma \left(x\right)\Vert }_Y\le \left|{\beta }_1\right|{\Vert \Gamma \left(2x\right)-2\Gamma \left(x\right)\Vert }_Y$

and so $\Gamma \left(2x\right)=2\Gamma \left(x\right)$ for all $x\in X$.

Thus

$\Gamma \left(\frac{x}{2}\right)=\frac{1}{2}\Gamma \left(x\right)$ (47)

for all $x\in X$

It follows from (46) and (47) that

$\begin{array}{l}{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_{\mathbb{Y}}\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_{\mathbb{Y}}\\ +{\Vert {\xi }_2\left(2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_{\mathbb{Y}}\\ ={\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_{\mathbb{Y}}\\ +{\Vert {\xi }_2\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_{\mathbb{Y}}\end{array}$ (48)

and so

$\begin{array}{l}\left(1-\left|{\xi }_2\right|\right){\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le \left|{\xi }_1\right|{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\Vert }_Y\end{array}$ (49)

we let $u={\displaystyle {\sum }_{j=1}^k}\frac{x_j+y_j}{2}+{\displaystyle {\sum }_{j=1}^k}{z}_j,v={\displaystyle {\sum }_{j=1}^k}\frac{x_j+y_j}{2}-{\displaystyle {\sum }_{j=1}^k}{z}_j$, for all $j\mathrm{<mo>\; =\; </mo>1}\to k$,

we get

$\begin{array}{l}\left(1-\left|{\xi }_2\right|\right)\Vert \Gamma \left(u\right)-\Gamma \left(\frac{u+v}{2}\right)-\Gamma \left(\frac{u-v}{2}\right)\Vert \\ \le \left|{\xi }_1\right|{\Vert \Gamma \left(u\right)+\Gamma \left(v\right)-2\Gamma \left(\frac{u+v}{2}\right)\Vert }_Y\end{array}$ (50)

for all $u\mathrm{,}v\in X$

and so

$\begin{array}{l}\frac{1}{2}\left(1-\left|{\xi }_2\right|\right){\Vert \Gamma \left(u+v\right)+\Gamma \left(u-v\right)-2\Gamma \left(u\right)\Vert }_{\mathbb{Y}}\\ \le \left|{\xi }_1\right|{\Vert \Gamma \left(u+v\right)-\Gamma \left(u\right)-f\left(v\right)\Vert }_Y\end{array}$ (51)

for all $u\mathrm{,}v\in X$ It follows from (49) and (51) that

$\begin{array}{l}\frac{1}{2}{\left(1-\left|{\xi }_2\right|\right)}^2{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\left|{\xi }_1\right|}^2{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\end{array}$ (52)

Since $\sqrt{2}\left|{\xi }_1\right|+\left|{\xi }_2\right|<1$

and so

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

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}\forall j=1\to k$. Thus f is Cauchy additive. □

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

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

$\phi \left(\frac{x_1}{2}\mathrm{,}\cdots \mathrm{,}\frac{x_n}{2}\mathrm{,}\frac{y_1}{2}\mathrm{,}\cdots \mathrm{,}\frac{y_n}{2}\mathrm{,}\frac{z_1}{2}\mathrm{,}\cdots \mathrm{,}\frac{z_n}{2}\right)\le \frac{L}{2}\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_n\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_n\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_n\right)$ (53)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}j=1\to k$. If $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfy $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_n\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_n\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_n\right)\end{array}$ (54)

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

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

${\Vert f\left(x\right)-\psi \left(x\right)\Vert }_Y\le \frac{L}{2\left(1-L\right)\left(1-\left|{\xi }_1\right|\right)}\phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)$ (55)

for all $x\in X$

Proof. Replacing $\left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_n\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}\cdots \mathrm{,}{y}_n\mathrm{,}{z}_1\mathrm{,}{z}_2\mathrm{,}\cdots \mathrm{,}{z}_n\right)$ by $\left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (54), we get

$\left(1-\left|{\xi }_1\right|\right){\Vert \Gamma \left(2x\right)-2\Gamma \left(x\right)\Vert }_Y\le \phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)$ (56)

for all $x\in X$.

Consider the set

$\mathbb{S}\mathrm{<mo>\; :\; </mo>}=\left\{h\mathrm{<mo>\; :\; </mo>}X\to Y\mathrm{,}h\left(0\right)=0\right\}$

and introduce the generalized metric on $\mathbb{S}$ :

$d\left(g\mathrm{,}h\right)\mathrm{<mo>\; :\; </mo>}=inf\left\{\lambda \in ℝ\mathrm{<mo>\; :\; </mo>}\Vert g\left(x\right)-h\left(x\right)\Vert \le \lambda \phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)\mathrm{,}\forall x\in X\right\}\mathrm{,}$

where, as usual, $inf\varphi =+\infty$. It easy to show that $\left(\mathbb{S}\mathrm{,}d\right)$ is complete [17] Now we cosider the linear mapping $J\mathrm{<mo>\; :\; </mo>}\mathbb{S}\to \mathbb{S}$ such that

$Jg\left(x\right)\mathrm{<mo>\; :\; </mo>}=2g(\; x\; 2\; )$

for all $x\in X$. Let $g\mathrm{,}h\in \mathbb{S}$ be given such that $d\left(g\mathrm{,}h\right)=\epsilon$ then

$\Vert g\left(x\right)-h\left(x\right)\Vert \le \epsilon \phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)$

for all $x\in X$.

Hence

$\begin{array}{l}\Vert Jg\left(x\right)-Jh\left(x\right)\Vert =\Vert 2g\left(\frac{x}{2}\right)-2h\Gamma \left(\frac{x}{2}\right)\Vert \le 2\epsilon \phi \left(\frac{x}{2}\mathrm{,0,}\cdots \mathrm{,0,}\frac{x}{2}\mathrm{,}\cdots \mathrm{,0,}\frac{x}{2}\mathrm{,}\cdots \mathrm{,0}\right)\\ \le 2\epsilon \frac{L}{2}\phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)\le L\epsilon \phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)\end{array}$

for all $x\in X$. So $d\left(g\mathrm{,}h\right)=\epsilon$ implies that $d\left(Jg\mathrm{,}Jh\right)\le L\cdot \epsilon$. This means that

$d\left(Jg\mathrm{,}Jh\right)\le Ld\left(g\mathrm{,}h\right)$

for all $g\mathrm{,}h\in X$ It follows from (56) that

$\begin{array}{c}\Vert \Gamma \left(x\right)-2\Gamma \left(\frac{x}{2}\right)\Vert \le \frac{1}{1-\left|{\xi }_1\right|}\phi \left(\frac{x}{2}\mathrm{,0,}\cdots \mathrm{,0,}\frac{x}{2}\mathrm{,}\cdots \mathrm{,0,}\frac{x}{2}\mathrm{,}\cdots \mathrm{,0}\right)\\ \le \frac{L}{2\left(1-\left|{\xi }_1\right|\right)}\phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)\end{array}$

for all $x\in X$. So $d\left(\Gamma \mathrm{,}J\Gamma \right)\le \frac{L}{2\left(1-\left|{\xi }_1\right|\right)}$ for all $x\in X$ By Theorem 2.3, there exists a mapping $\psi \mathrm{<mo>\; :\; </mo>}X\to \mathbb{Y}$ satisfying the following:

1) $\psi$ is a fixed point of J, i.e.,

$\psi \left(x\right)=2\psi \left(\frac{x}{2}\right)$ (57)

for all $x\in X$. The mapping $\psi$ is a unique fixed point J in the set

$\mathbb{M}=\left\{g\in \mathbb{S}:d\left(f,g\right)<\infty \right\}$

This implies that $\psi$ is a unique mapping satisfying (57) such that there exists a $\lambda \in \left(\mathrm{0,}\infty \right)$ satisfying

$\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert \le \lambda \phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)$

for all $x\in X$

2) $d\left(J^l\Gamma \mathrm{,}\psi \right)\to 0$ as $l\to \infty$. This implies equality

$\underset{l\to \infty }{lim}{2}^n\Gamma \left(\frac{x}{2^n}\right)=\psi (\; x\; )$

for all $x\in X$

3) $d\left(\Gamma \mathrm{,}\psi \right)\le \frac{1}{1-L}d\left(\Gamma \mathrm{,}J\Gamma \right)$. which implies

$\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert \le \frac{L}{2\left(1-L\right)\left(1-\left|{\xi }_1\right|\right)}\phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)$

for all $x\in X$. □

The rest of the proof is similar to the proof of Theorem 3.

Theorem 9. Let $\phi \mathrm{<mo>\; :\; </mo>}{X}^{3n}\to \left[\mathrm{0,}\infty \right)$ be a function such that there exists an $L<1$ with

$\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_n\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_n\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_n\right)\le 2L\phi \left(\frac{x_1}{2}\mathrm{,}\cdots \mathrm{,}\frac{x_n}{2}\mathrm{,}\frac{y_1}{2}\mathrm{,}\cdots \mathrm{,}\frac{y_n}{2}\mathrm{,}\frac{z_1}{2}\mathrm{,}\cdots \mathrm{,}\frac{z_n}{2}\right)$ (58)

for all $x\mathrm{,}y\mathrm{,}z\in X$. Let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfy $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_n\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_n\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_n\right)\end{array}$ (59)

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

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

${\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert }_Y\le \frac{L}{2\left(1-L\right)\left(1-\left|{\xi }_1\right|\right)}\phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)$ (60)

for all $x\in X$

The rest of the proof is similar to the proof of Theorem 8.

From proving the theorems we have consequences:

Corollary 5. Let $r>1$ and $\theta$ be nonnegative real numbers and let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfy $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(2f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ xi+\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }_X^r\right)\end{array}$ (61)

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

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

${\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert }_Y\le \frac{3\theta }{\left(2^r-2\right)\left(1-\left|{\xi }_1\right|\right)}{\Vert x\Vert }_X^r$ (62)

for all $x\in X$

Corollary 6. Let $r<1$ and $\theta$ be nonnegative real numbers and let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfy $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }_{\mathbb{X}}^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }_{\mathbb{X}}^r\right)\end{array}$ (63)

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

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

${\Vert f\left(x\right)-\psi \left(x\right)\Vert }_Y\le \frac{3\theta }{\left(2-2^r\right)\left(1-\left|{\xi }_1\right|\right)}{\Vert x\Vert }_X^r$ (64)

for all $x\in X$.

6. Establish the Solution of the Additive (ξ₁,ξ₂)-Function Inequalities Using a Direect Method

Next, we study the solutions of (2). Note that for these inequalities, when $X$ be a real or complete normed space and $Y$ complex Banach space.

Theorem 10. Suppose $\phi \mathrm{<mo>\; :\; </mo>}{X}^{3k}\to \left[\mathrm{0,}\infty \right)$ be a function such that

$\begin{array}{l}\varphi \left(x_1,\cdots ,x_k,y_1,\cdots ,y_k,z_1,\cdots ,z_k\right)\\ :={\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}{2}^j\phi \left(\frac{x_1}{2^j},\cdots ,\frac{x_k}{2^j},\frac{y_1}{2^j},\cdots ,\frac{y_k}{2^j},\frac{z_1}{2^j},\cdots ,\frac{z_k}{2^j}\right)<\infty \end{array}$ (65)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}j=1\to k$ and let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfies $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\phi \left(x_1,\cdots ,x_k,y_1,\cdots ,y_k,z_1,\cdots ,z_k\right)\end{array}$ (66)

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

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

${\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert }_{\mathbb{Y}}\le \varphi \left(x,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)$ (67)

for all $x\in X$

Proof. Replacing $\left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}{z}_2\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(x,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x,\mathrm{0,}\cdots \mathrm{,0}\right)$ in (66), we get

$\left(1-\left|{\xi }_1\right|\right){\Vert \Gamma \left(2x\right)-2\Gamma \left(x\right)\Vert }_Y\le \phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,0},\cdots \mathrm{,0}\right)$ (68)

for all $x\in X$.

So

${\Vert \Gamma \left(x\right)-\frac{1}{2}\Gamma \left(2x\right)\Vert }_Y\le \frac{1}{2\left(1-\left|{\xi }_1\right|\right)}\phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)$ (69)

for all $x\in X$.

Hence

$\begin{array}{l}{\Vert \frac{1}{2^l}\Gamma \left(2^{l}x\right)-\frac{1}{2^m}\Gamma \left(2^{m}x\right)\Vert }_Y\\ \le {\displaystyle \underset{j=l}{\overset{m-1}{\sum }}}{\Vert 2^{j}f\left(\frac{x}{2^j}\right)-2^{j+1}\Gamma \left(\frac{x}{2^{j+1}}\right)\Vert }_Y\\ \le {\displaystyle \underset{j=l}{\overset{m-1}{\sum }}}\frac{2^{j+1}}{2\left(1-\left|{\xi }_1\right|\right)}\phi \left(\frac{x}{2^{j+1}}\mathrm{,0,}\cdots ,\mathrm{0,}\frac{x}{2^{j+1}}\mathrm{,0,}\cdots \mathrm{,0,}\frac{x}{2^{j+1}}\mathrm{,}\cdots \mathrm{,0}\right)\end{array}$ (70)

for all nonnegative integers m and l with $m>l$ and all $x\in X$. It follows from (70) that the sequence $\left\{2^n\Gamma \left(\frac{x}{2^n}\right)\right\}$ is a Cauchy sequence for all $x\in \mathbb{X}$. Since $Y$ is complete, the sequence $\left\{2^n\Gamma \left(\frac{x}{2^n}\right)\right\}$ converges. So one can define the mapping $\psi \mathrm{<mo>\; :\; </mo>}X\to Y$ by

$\psi \left(x\right)\mathrm{<mo>\; :\; </mo>}=\underset{n\to \infty }{lim}{2}^n\Gamma \left(\frac{x}{2^n}x\right)$ (71)

for all $x\in \mathbb{X}$. Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (37), we get (67) It follows from (65) and (66) that

$\begin{array}{l}{\Vert \psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ =\underset{n\to \infty }{lim}{2}^n{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}+\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}\right)-\Gamma \left(\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le \underset{n\to \infty }{lim}{2}^n\left|{\xi }_1\right|{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}+\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}-\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}\right)\Vert }_Y\\ +\underset{n\to \infty }{lim}{2}^n\left|{\xi }_2\right|{\Vert 2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+2}}+\frac{1}{2^{n+1}}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2^{n+1}}\right)-\Gamma \left(\frac{1}{2^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ +\underset{n\to \infty }{lim}{2}^n\phi \left(\frac{x_1}{2^n}\mathrm{,}\cdots \mathrm{,}\frac{x_k}{2^n}\mathrm{,}\frac{y_1}{2^n}\mathrm{,}\cdots \mathrm{,}\frac{y_k}{2^n}\mathrm{,}\frac{z_1}{2^n}\mathrm{,}\cdots \mathrm{,}\frac{z_k}{2^n}\right)\end{array}$

$\begin{array}{l}={\Vert {\xi }_1\left(\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(2\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\end{array}$ (72)

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

$\begin{array}{l}{\Vert \psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(2\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\psi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\end{array}$

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}j=1\to n$. By Lemma 5.1, the mapping $\psi \mathrm{<mo>\; :\; </mo>}X\to Y$ is additive. Ei

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

Now, let ${\psi }^{\prime }\mathrm{<mo>\; :\; </mo>}X\to Y$ be another additive mapping satisfying (67). Then we have

$\begin{array}{c}\Vert \psi \left(x\right)-{\psi }^{\prime }\left(x\right)\Vert ={\Vert 2^q\psi \left(\frac{x}{2^q}\right)-2^q{\psi }^{\prime }\left(\frac{x}{2^q}\right)\Vert }_Y\\ \le \Vert 2^q\psi \left(\frac{x}{2^q}\right)-2^q\Gamma \left(\frac{x}{2^q}\right)\Vert +{\Vert 2^q{\psi }^{\prime }\left(\frac{x}{2^q}\right)-2^q\Gamma \left(\frac{x}{2^q}\right)\Vert }_Y\\ \le 2^q\varphi \left(\frac{x}{2^q}\mathrm{,0,}\cdots \mathrm{,0,}\frac{x}{2^q}\mathrm{,0,}\cdots \mathrm{,0,}\frac{x}{2^q}\mathrm{0,}\cdots \mathrm{,0}\right)\end{array}$

which tends to zero as $q\to \infty$ 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 uniqueness of $\psi$. □

Theorem 11. Suppose $\phi \mathrm{<mo>\; :\; </mo>}{X}^{3k}\to \left[\mathrm{0,}\infty \right)$ be a function such that

$\begin{array}{l}\psi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\\ :={\displaystyle \underset{j=0}{\overset{\infty }{\sum }}}\frac{1}{2^j}\phi \left(2^j{x}_1,\cdots ,2^j{x}_k,2^j{y}_1,\cdots ,2^j{y}_k,2^j{z}_1,\cdots ,2^j{z}_k\right)<\infty \end{array}$ (73)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X\mathrm{,}j=1\to k$ and let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfies $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\beta }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\beta }_2\left(2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (74)

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

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

${\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert }_Y\le \varphi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,0},\cdots \mathrm{,0}\right)$ (75)

for all $x\in \mathbb{X}$

Proof. Replacing $\left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}{z}_2\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(x,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)$ in (74), we get

$\left(1-\left|{\xi }_1\right|\right){\Vert \Gamma \left(2x\right)-2\Gamma \left(x\right)\Vert }_{\mathbb{Y}}\le \phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)$ (76)

for all $x\in X$.

So 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 $\left(x,0,\cdots \mathrm{,0,}x\mathrm{,0},\cdots \mathrm{,0,}x\mathrm{,0},\cdots \mathrm{,0}\right)$ in (74), we get

${\Vert 2\Gamma \left(\frac{x}{2}\right)-\Gamma \left(x\right)\Vert }_Y\le \phi \left(x,0,\cdots \mathrm{,0,}x\mathrm{,0},\cdots \mathrm{,0,}x\mathrm{,0},\cdots \mathrm{,0}\right)$ (77)

for all $x\in X$. So

${\Vert \Gamma \left(x\right)-\frac{1}{2}\Gamma \left(2x\right)\Vert }_Y\le \frac{1}{2}\phi \left(2x\mathrm{,0,}\cdots \mathrm{,0,2}x\mathrm{,0,}\cdots \mathrm{,0,2}x\mathrm{,0,}\cdots \mathrm{,0}\right)$ (78)

for all $x\in X$. Hence

$\begin{array}{l}{\Vert \frac{1}{2^l}\Gamma \left(2^{l}x\right)-\frac{1}{2^m}\Gamma \left(2^{m}x\right)\Vert }_Y\\ \le {\displaystyle \underset{j=l}{\overset{m-1}{\sum }}}{\Vert \frac{1}{2^j}\Gamma \left(2^{j}x\right)-\frac{1}{2^{j+1}}\Gamma \left(2^{j+1}x\right)\Vert }_Y\\ \le {\displaystyle \underset{j=l}{\overset{m-1}{\sum }}}\frac{1}{2^{j+1}}\phi \left(2^{j+1}x\mathrm{,0,}\cdots {\mathrm{,0,2}}^{j+1}x\mathrm{,0,}\cdots {\mathrm{,0,2}}^{j+1}x\mathrm{,0,}\cdots \mathrm{,0}\right)\end{array}$ (79)

for all nonnegative integers m and l with $m>l$ and all $x\in X$. It follows from (79) that the sequence $\left\{\frac{1}{2^n}\Gamma \left(2^{n}x\right)\right\}$ is a Cauchy sequence for all $x\in X$. Since $Y$ is complete, the sequence $\left\{\frac{1}{2^n}\Gamma \left(2^{n}x\right)\right\}$ converges. So one can define the mapping $\psi \mathrm{<mo>\; :\; </mo>}X\to Y$ by

$\psi \left(x\right):=\underset{n\to \infty }{\lim }\frac{1}{2^n}\Gamma \left(2^{n}x\right)$ (80)

for all $x\in X$. Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (79), we get (75).

The rest of the proof is similar to the proof of theorem 10. □

From proving the theorems we have consequences:

Corollary 7. Let $r>1$ and $\theta$ be nonnegative real numbers and let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfy $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (81)

for all $x_j\in X$.

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

${\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert }_Y\le \frac{3\theta }{\left(2^r-2\right)\left(1-\left|{\xi }_1\right|\right)}{\Vert x\Vert }_X^r$ (82)

for all $x\in X$

Corollary 8. Let $r<1$ and $\theta$ be nonnegative real numbers and let $\Gamma \mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfy $\Gamma \left(0\right)=0$ and

$\begin{array}{l}{\Vert \Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\Vert }_Y\\ \le {\Vert {\xi }_1\left(\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)\right)\Vert }_Y\\ +{\Vert {\xi }_2\left(2\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{4}+\frac{1}{2}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2}\right)-\Gamma \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (83)

for all $x_j\mathrm{,}{y}_j\mathrm{,}z-j\in X$.

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

${\Vert \Gamma \left(x\right)-\psi \left(x\right)\Vert }_Y\le \frac{3\theta }{\left(2-2^r\right)\left(1-\left|{\xi }_1\right|\right)}{\Vert x\Vert }_X^r$ (84)

for all $x\in X$

7. Conclusion

In this paper, I have given two functional inequalities with 3k variables and fully solved complex Banach space by fixed point methods and direct methods. This result is based on special results such as [25] [26].