Generalized Hyers-Ulam-Rassias Type Stability Additive α-Functional Inequalities with 3k-Variable in Complex Banach Spaces

Ly Van An

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

DOI: 10.4236/oalib.1109373                

Abstract

In this paper we study to solve two-additive α-functional inequality with 3k-variables and their Hyers-Ulam-Rassias type stability. It is investigated in complex Banach spaces. These are the main results of this paper.

Keywords

Additive β-Functional Equation, Additive β-Functional Inequality, Complex Banach Space, Hyers-Ulam-Rassisa Stability

Mathematics Subject Classification

Primary 4610, 4710, 39B62, 39B72, 39B52

1. Introduction

Let $X$ and $Y$ be normed spaces on the same field $\mathbb{K}$, and $f\mathrm{<mo>\; :\; </mo>}X\to Y$. We use the notation $\Vert \cdot \Vert$ for all the norms on both $X$ and $Y$. In this paper, we investigate some additive α-functional inequality when $X$ is a real or complex normed space and $Y$ is a complex Banach space.

In fact, when $X$ is a real or complex normed space and $Y$ is a complex Banach space, we solve and prove the Hyers-Ulam stability of following additive α-functional inequality.

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

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

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

So (1) and (2) are equivalent propositions.

Where $\alpha$ is a fixed complex number with $\left|\alpha \right|<1$ and m be a fixed integer with $m>1$.

The Hyers-Ulam stability was first investigated for 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 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 unbouned Cauchy diffrence. Ageneralization of the Rassias theorem was obtained by Găvruta [5] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias’ approach.

The Hyers-Ulam stability for functional inequalities has been investigated such as in [5] [6] [7]. Gilány showed that if it satisfies the functional inequality

$\Vert 2f\left(x\right)+2f\left(y\right)-f\left(x-y\right)\Vert \le \Vert f\left(x+y\right)\Vert$ (3)

Then f satisfies the Jordan-von Newman functional equation

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

Gilányi [5] and Fechner [8] proved the Hyers-Ulam stability of the functional inequality (3).

Next Choonkil Park [9] proved the Hyers-Ulam stability of additive β-functional inequalities. Recently, the author has studied the addition inequalities of mathematicians in the world as [5] [8] [10] - [24] and I have introduced two general additive function inequalities (1) and (2) based on the $\left({\beta }_1\mathrm{,}{\beta }_2\right)$ -function inequality result, see [25]. When inserting the parameter m this is the opening for modern functional equations. That is, it demonstrates the superiority of the field of functional equations and is also a bright horizon for the special development of functional equations. So in this paper, we solve and proved the Hyers-Ulam stability for two α-functional inequalities (1)-(2), i.e. the α-functional inequalities with 3k-variables. Under suitable assumptions on spaces $X$ and $Y$, we will prove that the mappings satisfying the α-functional inequatilies (1) or (2). Thus, the results in this paper are generalization of those in [7] [9] [17] [25] [26] [27] for α-functional inequatilies with 3k-variables. The paper is organized as followns: In section preliminarier we remind a basic property such as We only redefine the solution definition of the equation of the additive function.

Notice here that we make the general assumption that: $G$ be a k-divisible abelian group.

Section 3: is devoted to prove the Hyers-Ulam stability of the addive α-functional inequalities (1) when $X$ is a real or complex normed space and $Y$ complex Banach space.

Section 4: is devoted to prove the Hyers-Ulam stability of the addive α-functional inequalities (2) when $X$ is a real or complex normed space and $Y$ complex Banach space.

2. Preliminaries

Solutions of the Inequalities

The functional equation

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

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

3. Establish the Solution of the Additive α-Function Inequalities

Now, we first study the solutions of (1). Note that for these inequalities, $G$ be a k-divisible abelian group, $X$ is a real or complex normed space and $Y$ is a complex Banach spaces. Under this setting, we can show that the mapping satisfying (1.1) is additive. These results are give in the following.

Lemma 1. Let $m\in \mathbb{N}$ and a mapping $f\mathrm{<mo>\; :\; </mo>}G\to Y$ satilies

$\begin{array}{l}{\Vert f\left(\left(m+1\right){\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(m\frac{x_j+y_j}{2k}-z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)\Vert }_Y\\ \le {\Vert \alpha \left(f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\right)\Vert }_Y\end{array}$ (5)

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

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}G\to Y$ satisfies (5).

We replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (5), we have

${\Vert \left(2k-1\right)f\left(0\right)\Vert }_Y\le {\Vert \alpha \left(2k-1\right)f\left(0\right)\Vert }_Y\le 0$

therefore

$\left(\left|2k-1\right|-\left|\alpha \left(2k-1\right)\right|\right){\Vert f\left(0\right)\Vert }_Y\le 0$

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

Replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}m\cdot \frac{x_1+y_1}{2k}-v_1\mathrm{,}\cdots \mathrm{,}m\cdot \frac{x_k+y_k}{2k}-v_k\right)$ in (5), we have

$\begin{array}{l}{\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{v}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(v_j\right)\Vert }_Y\\ \le {\Vert \alpha \left(f\left(\left(m+1\right){\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{v}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(m\frac{x_j+y_j}{2k}-v_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)\right)\Vert }_Y\end{array}$ (6)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}\frac{x_1+y_1}{2k}-v_1\mathrm{,}\cdots \mathrm{,}\frac{x_k+y_k}{2k}-v_k\in G$. From (5) and (6) we infer that

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

and so

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

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in G$ for $j=1\to n$, as we expected.

Theorem 2. Let $r>1,m\in \mathbb{Z},m>1$, $\theta$ be nonngative real number, and let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping such that

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

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

${\Vert f\left(x\right)-h\left(x\right)\Vert }_Y\le \frac{{\displaystyle {\sum }_{q=1}^{m-1}\left(q^r+2k^r\right)}}{\left(1-\left|\alpha \right|\right)\left(m^r-m\right)}\theta {\Vert x\Vert }_X^r\mathrm{.}$ (9)

for all $x\in X$.

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (8).

Replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (8), we have

${\Vert \left(2k-1\right)f\left(0\right)\Vert }_Y\le {\Vert \alpha \left(2k-1\right)f\left(0\right)\Vert }_Y\le 0$

therefore

$\left(\left|2k-1\right|-\left|\alpha \left(2k-1\right)\right|\right){\Vert f\left(0\right)\Vert }_Y\le 0$

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

Next we:

Replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(kx,\mathrm{0,}\cdots \mathrm{,0,}kx,\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (8), we get

${\Vert f\left(\left(m+1\right)x\right)-f\left(mx\right)-f\left(x\right)\Vert }_Y\le 2k^r\theta {\Vert x\Vert }_X^r$ (10)

for all $x\in X$. Thus for $q\in \mathbb{N}$.

We replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(kx,\mathrm{0,}\cdots \mathrm{,0,}kx,\mathrm{0,}\cdots \mathrm{,0,}qx,\mathrm{0,}\cdots \mathrm{,0}\right)$ in (8), we have

$\begin{array}{l}{\Vert f\left(\left(m-q+1\right)x\right)-f\left(\left(m-q\right)x\right)-f\left(x\right)\Vert }_Y\\ \le {\Vert \alpha \left(f\left(\left(q+1\right)x\right)-f\left(qx\right)-f\left(x\right)\right)\Vert }_Y+\theta \left(2k^r+q^r\right){\Vert x\Vert }_Y^r\end{array}$ (11)

for all $x\in X$.

For (10) and (11)

$\begin{array}{l}{\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}{\Vert f\left(\left(m-q+1\right)x\right)-f\left(\left(m-q\right)x\right)-f\left(x\right)\Vert }_Y\\ \le {\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}{\Vert \alpha \left(f\left(\left(q+1\right)x\right)-f\left(qx\right)-f\left(x\right)\right)\Vert }_Y+\theta \left({\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}\left(2k^r+q^r\right){\Vert x\Vert }_X^r\right)\end{array}$ (12)

for all $x\in X$.

From (11) and (12) and triangle inequality, we have

$\begin{array}{l}\left(1-\left|\alpha \right|\right){\Vert f\left(mx\right)-mf\left(x\right)\Vert }_Y\\ =\left(1-\left|\alpha \right|\right){\displaystyle \underset{q\mathrm{<mo>\; =\; </mo>1}}{\overset{m-1}{\sum }}}{\Vert f\left(\left(q+1\right)x\right)-f\left(qx\right)-f\left(x\right)\Vert }_Y\\ \le {\displaystyle \underset{q\mathrm{<mo>\; =\; </mo>1}}{\overset{m-1}{\sum }}}\left(1-\left|\alpha \right|\right){\Vert f\left(\left(q+1\right)x\right)-f\left(qx\right)-f\left(x\right)\Vert }_Y\\ \le {\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}{\Vert f\left(\left(q+1\right)x\right)-f\left(qx\right)-f\left(x\right)\Vert }_Y-{\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}{\Vert \alpha \left(f\left(\left(q+1\right)x\right)-f\left(qx\right)-f\left(x\right)\right)\Vert }_Y\\ \le \theta \left({\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}\left(2k^r+q^r\right){\Vert x\Vert }_X^r\right)\end{array}$ (13)

for all $x\in X$. from

$\begin{array}{l}{\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}{\Vert f\left(\left(m-q+1\right)x\right)-f\left(\left(m-q\right)x\right)-f\left(x\right)\Vert }_Y\\ ={\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}{\Vert f\left(\left(q+1\right)x\right)-f\left(qx\right)-f\left(x\right)\Vert }_Y\end{array}$

Since $\left|\alpha \right|<1$, the mapping f satisfies the inequalities

${\Vert f\left(mx\right)-mf\left(x\right)\Vert }_Y\le \frac{\theta \left({\displaystyle {\sum }_{q=1}^{m-1}\left(2k^r+q^r\right){\Vert x\Vert }_X^r}\right)}{1-\left|\alpha \right|}$

for all $x\in X$.

Therefore

${\Vert f\left(x\right)-mf\left(\frac{x}{m}\right)\Vert }_Y\le \frac{\theta \left({\displaystyle {\sum }_{q=1}^{m-1}\left(2k^r+q^r\right){\Vert x\Vert }_X^r}\right)}{\left(1-\left|\alpha \right|\right)m^r}$ (14)

for all $x\in X$. So

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

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

So one can define the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ by $\varphi \left(x\right):={\lim }_{n\to \infty }{m}^{n}f\left(\frac{x}{m^n}\right)$ for all $x\in X$. Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (15), we get (9).

It follows from (8) that

$\begin{array}{l}{\Vert \varphi \left(\left(m+1\right){\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{z}_j\right)-{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}\varphi \left(m\frac{x_j+y_j}{2k}-z_j\right)-{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)\Vert }_Y\\ =\underset{n\to \infty }{lim}{m}^n\Vert f\left(\frac{m+1}{m^n}{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-\frac{1}{m^n}{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}{z}_j\right)-{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(\frac{m}{m^n}\frac{x_j+y_j}{2k}-\frac{1}{m^n}{z}_j\right)\\ -{{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(\frac{1}{m^n}\frac{x_j+y_j}{2k}\right)\Vert }_Y\\ \le \underset{n\to \infty }{lim}{m}^n\Vert \alpha (f\left(\frac{1}{m^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+\frac{1}{m^n}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(\frac{1}{m^n}\frac{x_j+y_j}{2k}\right)\\ -{{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(\frac{1}{m^n}{z}_j\right))\Vert }_Y+\underset{n\to \infty }{lim}\frac{m^n}{m^{nr}}\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }_X^r\right)\\ \le \left|\alpha \right|{\Vert \varphi \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)\Vert }_Y\end{array}$ (16)

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

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

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for all $j=1\to n$. So by lemma 21 it follows that the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is additive. Now we need to prove uniqueness, suppose ${\varphi }^{\prime }\mathrm{<mo>\; :\; </mo>}X\to Y$ is also an additive mapping that satisfies (9). Then we have

$\begin{array}{l}{\Vert \varphi \left(x\right)-{\varphi }^{\prime }\left(x\right)\Vert }_Y\\ =m^n{\Vert \varphi \left(\frac{x}{m^n}\right)-{\varphi }^{\prime }\left(\frac{x}{m^n}\right)\Vert }_Y\\ \le m^n\left({\Vert \varphi \left(\frac{x}{m^n}\right)-f\left(\frac{x}{m^n}\right)\Vert }_Y+{\Vert {\varphi }^{\prime }\left(\frac{x}{m^n}\right)-f\left(\frac{x}{m^n}\right)\Vert }_Y\right)\\ \le \frac{2\cdot m^n\cdot {\displaystyle {\sum }_{q=1}^{m-1}\left(q^r+2k^r\right)}}{\left(1-\left|\alpha \right|\right)m^{nr}\left(m^r-m\right)}\theta {\Vert x\Vert }_X^r\end{array}$ (17)

which tends to zero as $n\to \infty$ for all $x\in X$. So we can conclude that $\varphi \left(x\right)={\varphi }^{\prime }\left(x\right)$ for all $x\in X$.This proves thus the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is a unique mapping satisfying (9) as we expected.

Theorem 3. Let $r>1,m\in \mathbb{Z},m>1$, $\theta$ be nonngative real number, and let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping such that

$\begin{array}{l}{\Vert f\left(\left(m+1\right){\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(m\frac{x_j+y_j}{2k}-z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)\Vert }_Y\\ \le {\Vert \alpha \left(f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)\right)\Vert }_Y\\ +\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }_X^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }_X^r\right)\end{array}$ (18)

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

${\Vert f\left(x\right)-\varphi \left(x\right)\Vert }_Y\le \frac{m^n\cdot {\displaystyle {\sum }_{q=1}^{m-1}\left(q^r+2k^r\right)}}{\left(1-\left|\alpha \right|\right)\left(m-m^r\right)}\theta {\Vert x\Vert }_X^r\mathrm{.}$ (19)

for all $x\in X$.

The rest of the proof is similar to the proof of Theorem 2.2.

4. Establish the Solution of the Additive α-Function Inequalities

Next, we study the solutions of (2). Note that for these inequalities, when $\mathbb{X}$ be a real or complete normed space and $\mathbb{Y}$ complex Banach space. Now, we study the solutions of (2). Note that for these inequalities, $G$ be a k-divisible abelian group, $X$ is a real or complex normed space and $Y$ is complex Banach spaces. Under this setting, we can show that the mapping satisfying (2) is additive. These results are give in the following.

Lemma 4. Let $m\in \mathbb{N}$ and a mapping $f\mathrm{<mo>\; :\; </mo>}G\to Y$ satilies

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

$\le {\Vert \alpha \left(f\left(\left(m+1\right){\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(m\frac{x_j+y_j}{2k}-z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)\right)\Vert }_Y$ (20)

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

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

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

${\Vert \left(2k-1\right)f\left(0\right)\Vert }_Y\le {\Vert \left(2k-1\right)\alpha f\left(0\right)\Vert }_Y\le 0$

therefore

$\left(\left|2k-1\right|-\left|\alpha \left(2k-1\right)\right|\right){\Vert f\left(0\right)\Vert }_Y\le 0$

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

Replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}m\cdot \frac{x_1+y_1}{2k}-v_1\mathrm{,}\cdots \mathrm{,}m\cdot \frac{x_k+y_k}{2k}-v_k\right)$ in (20), we have

$\begin{array}{l}{\Vert f\left(\left(m+1\right){\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{v}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(m\frac{x_j+y_j}{2k}-v_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)\Vert }_Y\\ \le {\Vert \alpha \left(f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{v}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(v_j\right)\right)\Vert }_Y\end{array}$ (21)

for all $x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}\frac{x_1+y_1}{2k}-v_1\mathrm{,}\cdots \mathrm{,}\frac{x_k+y_k}{2k}-v_k\in G$. From (20) and (21) we infer that

$\begin{array}{l}{\Vert f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{v}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(v_j\right)\Vert }_Y\\ \le {\Vert \alpha \left(f\left(\left(m+1\right){\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{v}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(m\frac{x_j+y_j}{2k}-v_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)\right)\Vert }_Y\\ \le {\Vert {\alpha }^2\left(f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{v}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{2k}\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(v_j\right)\right)\Vert }_Y\end{array}$ (22)

and so

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

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in G$ for $j=1\to n$, as we expected.

Theorem 5. Let $r>1,m\in \mathbb{Z},m>1$, $\theta$ be nonngative real number, and let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping such that

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

$+\theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)$ (23)

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

${\Vert f\left(x\right)-h\left(x\right)\Vert }_Y\le \frac{{\displaystyle {\sum }_{q=1}^{m-1}\left(q^r+2k^r\right)}}{\left(1-\left|\alpha \right|\right)\left(m-m^r\right)}\theta {\Vert x\Vert }_X^r\mathrm{.}$ (24)

for all $x\in X$.

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfies (23).

Replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0}\right)$ in (23), we have

$\Vert 2kf\left(0\right)\Vert \le {\Vert \alpha \left(2k-1\right)f\left(0\right)\Vert }_Y\le 0$

therefore

$\left(\left|2k-1\right|-\left|\alpha \left(2k-1\right)\right|\right){\Vert f\left(0\right)\Vert }_Y\le 0$

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

Next we:

Replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(kx,\mathrm{0,}\cdots \mathrm{,0,}kx,\mathrm{0,}\cdots \mathrm{,0},\mathrm{0,}\cdots \mathrm{,0}\right)$ in (23), we get

${\Vert f\left(\left(m+1\right)x\right)-f\left(mx\right)-f\left(x\right)\Vert }_Y\le 2k^r\theta {\Vert x\Vert }_X^r$ (25)

for all $x\in X$. Thus for $q\in \mathbb{N}$.

We replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)$ by $\left(kx,\mathrm{0,}\cdots \mathrm{,0,}kx,\mathrm{0,}\cdots \mathrm{,0,}qx,\mathrm{0,}\cdots \mathrm{,0}\right)$ in (23), we have

$\begin{array}{l}{\Vert f\left(\left(m-q+1\right)x\right)-f\left(\left(m-q\right)x\right)-f\left(x\right)\Vert }_Y\\ \le \Vert \alpha \left(f\left(\left(q+1\right)x\right)-f\left(qx\right)-f\left(x\right)\right)\Vert +\theta \left(2k^r+q^r\right){\Vert x\Vert }_Y^r\end{array}$ (26)

for all $x\in X$.

For (25) and (26)

$\begin{array}{l}{\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}{\Vert f\left(\left(m-q+1\right)x\right)-f\left(\left(m-q\right)x\right)-f\left(x\right)\Vert }_Y\\ \le {\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}{\Vert \alpha \left(f\left(\left(q+1\right)x\right)-f\left(qx\right)-f\left(x\right)\right)\Vert }_Y+\theta \left({\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}\left(2k^r+q^r\right){\Vert x\Vert }^r\right)\end{array}$ (27)

for all $x\in X$.

From (26) and (27) and triangle inequality, we have

$\begin{array}{l}\left(1-\left|\alpha \right|\right){\Vert f\left(mx\right)-mf\left(x\right)\Vert }_Y\\ =\left(1-\left|\alpha \right|\right){\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}{\Vert f\left(\left(q+1\right)x\right)-f\left(qx\right)-f\left(x\right)\Vert }_Y\\ \le {\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}\left(1-\left|\alpha \right|\right){\Vert f\left(\left(q+1\right)x\right)-f\left(qx\right)-f\left(x\right)\Vert }_Y\end{array}$

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

for all $x\in X$. from

$\begin{array}{l}{\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}{\Vert f\left(\left(m-q+1\right)x\right)-f\left(\left(m-q\right)x\right)-f\left(x\right)\Vert }_Y\\ ={\displaystyle \underset{q=1}{\overset{m-1}{\sum }}}{\Vert f\left(\left(q+1\right)x\right)-f\left(qx\right)-f\left(x\right)\Vert }_Y\end{array}$

Since $\left|\alpha \right|<1$, the mapping f satisfies the inequalities

${\Vert f\left(mx\right)-mf\left(x\right)\Vert }_Y\le \frac{\theta \left({\displaystyle {\sum }_{q=1}^{m-1}\left(2k^r+q^r\right){\Vert x\Vert }_X^r}\right)}{1-\left|\alpha \right|}$

for all $x\in X$.

Therefore

${\Vert f\left(x\right)-mf\left(\frac{x}{m}\right)\Vert }_Y\le \frac{\theta \left({\displaystyle {\sum }_{q=1}^{m-1}\left(2k^r+q^r\right){\Vert x\Vert }_X^r}\right)}{\left(1-\left|\alpha \right|\right)m^r}$ (29)

for all $x\in X$. So

$\begin{array}{c}{\Vert m^{l}f\left(\frac{x}{m^n}\right)-m^{p}f\left(\frac{x}{m^h}\right)\Vert }_Y\le {\displaystyle \underset{j=l}{\overset{p-1}{\sum }}}{\Vert m^{j}f\left(\frac{x}{m^j}\right)-m^{j+1}f\left(\frac{x}{m^{j+1}}\right)\Vert }_Y\\ \le \frac{\theta \left({\displaystyle {\sum }_{q=1}^{m-1}\left(2k^r+q^r\right)}\right)}{\left(1-\left|\alpha \right|\right)m^r}{\displaystyle \underset{j=l}{\overset{p-1}{\sum }}}\frac{m^j}{m^{rj}}{\Vert x\Vert }_X^r\end{array}$ (30)

for all nonnegative integers $p\mathrm{,}l$ with $p>l$ and all $x\in X$. It follows from (30) that the sequence $\left\{m^{n}f\left(\frac{x}{m^n}\right)\right\}$ is a Cauchy sequence for all $x\in X$. Since $Y$ is complete, the sequence $\left\{m^{n}f\left(\frac{x}{m^n}\right)\right\}$ coverges.

So one can define the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ by $\varphi \left(x\right)\mathrm{<mo>\; :\; </mo>}={\lim }_{n\to \infty }{m}^{n}f\left(\frac{x}{m^n}\right)$ for all $x\in X$. Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (30), we get (24).

It follows from (23) that

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

$\begin{array}{l}-{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(\frac{m}{m^n}\left(\frac{x_j+y_j}{2k}\right)-\frac{1}{m^n}{z}_j\right)-{{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(\frac{1}{m^n}{z}_j\right)\Vert }_Y\\ \le \left|\alpha \right|{\Vert \varphi \left(\left(m+1\right){\displaystyle \underset{j=1}{\overset{k}{\sum }}}\frac{x_j+y_j}{2k}+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(m\frac{x_j+y_j}{2k}-z_j\right)-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}\varphi \left(z_j\right)\Vert }_Y\end{array}$ (31)

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

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

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for all $j=1\to n$. So by lemma 4.1 it follows that the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is additive. Now we need to prove uniqueness, suppose ${\varphi }^{\prime }\mathrm{<mo>\; :\; </mo>}X\to Y$ is also an additive mapping that satisfies (24). Then we have

$\begin{array}{c}\Vert \varphi \left(x\right)-{\varphi }^{\prime }\left(x\right)\Vert =m^n\Vert \varphi \left(\frac{x}{m^n}\right)-{\varphi }^{\prime }\left(\frac{x}{m^n}\right)\Vert \\ \le m^n\left(\Vert \varphi \left(\frac{x}{m^n}\right)-f\left(\frac{x}{m^n}\right)\Vert +\Vert {\varphi }^{\prime }\left(\frac{x}{m^n}\right)-f\left(\frac{x}{m^n}\right)\Vert \right)\\ \le \frac{2\cdot m^n\cdot {\displaystyle {\sum }_{q=1}^{m-1}\left(q^r+2k^r\right)}}{\left(1-\left|\alpha \right|\right)m^{nr}\left(m^r-m\right)}\theta {\Vert x\Vert }^r\end{array}$ (32)

which tends to zero as $n\to \infty$ for all $x\in X$. So we can conclude that $\varphi \left(x\right)={\varphi }^{\prime }\left(x\right)$ for all $x\in X$.This proves thus the mapping $\varphi \mathrm{<mo>\; :\; </mo>}X\to Y$ is a unique mapping satisfying (24) as we expected.

5. Conclusion

In this article, I have solved two problems posed as establishing the solution of the additive α-function inequality (1) and (2) in complex Banach spaces with 3k variable. So when I develop this result, I rely on the inequality $\left({\beta }_1\mathrm{,}{\beta }_2\right)$ -function.

Conflicts of Interest

The author declares no conflicts of interest.

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