Ly Van An
Faculty of Mathematics Teacher Education, Tay Ninh University, Tay Ninh, Vietnam.
DOI: 10.4236/oalib.1109821   PDF    HTML   XML   161 Downloads   1,130 Views   Citations

Abstract

In this paper, we study to solve the quadratic type λ-functional equation with 3k variables. First, we investigated in non-Archimedean Banach spaces with a fixed point method, next, we investigated in non-Archimedean Banach spaces with a direct method and finally we do research in non-Archimedean random spaces. I will show that the solutions of the quadratic type λ-functional equation are quadratic type mappings. These are the main results of this paper.

Keywords

Quadratic λ-Functional Equation, Non-Archimedean Normed Space, Non-Archimedean Banach Space, Fixed Point Method, Direct Method, Hyers-Ulam Stability, Random Normed Spaces, Non-Archimedean Random Normed Space

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1. Introduction

Let $X$ and $Y$ be a 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 norm on both $X$ and $Y$ . In this paper, I study and expand the $\lambda$ -function equation from non-Archimedean normed space to non-Archimedean random normed space.

In fact, when $X$ is non-Archimedean normed space and $Y$ is non-Archimedean Banach spaces.

Or $X$ is a vector over field $\mathbb{K}$ and $\left(Y\mathrm{,}\Gamma \mathrm{,}T\right)$ be a non-Archimedean random Banach space over field $\mathbb{K}$ . We solve and prove the Hyers-Ulam-Rassisa type stability of forllowing quadratic λ-functional equation.

$\begin{array}{l}2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j+y_j\right)\\ =f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}x+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\end{array}$ (1)

where: Let $\left|2k\right|\ne 1$ , λ is a fixed non-Archimedean number with ${\lambda }^{-2m}\ne 4k-1$ and $k\mathrm{,}m$ is a positive integer. The notions of non-Archimedean normed space and non-Archimedean Banach spaces and non-Archimedean random Banach space over field $\mathbb{K}$ will remind in the next section. The study the stability of generalized stability of the quadratic type λ-functional equation with variables in non-Archimedean Banach space and non-Archimedean Random normed space originated from a question of S.M. Ulam [1] , concerning the stability of group homomorphisms. Let $\left(G\mathrm{,}\ast \right)$ be a group and let $\left(G^{\prime }\mathrm{,}\circ \mathrm{,}d\right)$ be a metric group with metric $d\left(\cdot \mathrm{,}\cdot \right)$ . Geven $\epsilon >0$ , does there exist a $\delta >0$ such that if $f\mathrm{<mo>\; :\; </mo>}G\to G^{\prime }$ satisfies

$d\left(f\left(x\ast y\right)\mathrm{,}f\left(x\right)\circ f\left(y\right)\right)<\delta \mathrm{,}\forall x\in G$

then there is a homomorphism $h\mathrm{<mo>\; :\; </mo>}G\to G^{\prime }$ with

$d\left(f\left(x\right)\mathrm{,}h\left(x\right)\right)<\epsilon \mathrm{,}\forall x\in G$

The Hyers [2] gave firts 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 difference. Gajda following the same approach as in Rassias gave an affirmative solution to this question for $p>1$ . It was shown by Gajda [5] , as well as by Rassias and Semr [6] that one cannot prove a Rassias, type theorem when $p=1$ . The counterexamples of Gajda, as well as of Rassias and Semr have stimulated several matematicians to invent new definition of approximately additive or approximately linear mappings, was obtained by Găvruta [7] .

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 functonal equation

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

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

The stability the quadratic functional equation was proved by Skof [8] for mappings $f\mathrm{<mo>\; :\; </mo>}{E}_1\to E_2$ , where $E_1$ is a normed space and $E_2$ is a Banach space.

Recently the author studied the Hyers-Ulam stability for the following α-functional equation.

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

in Non-Archimedean Banach spaces and non-Archimedean Random normed space.

In this paper, we solve and proved the Hyers-Ulam stability for λ-functional Equation (1.1), i.e. the λ-functional equation with 3k-variables. Under suitable assumptions on spaces $X$ and $Y$ , we will prove that the mappings satisfying the λ-functional Equation (1.1). Thus, the results in this paper are generalization of those in [9] for λ-functional equation with 3k-variables.

In this paper, based on the work of world mathematicians [1] - [33] , I introduce a new generalized quadratic function equation with 3k-variables to improve the classical form, which is a new breakthrough for the development of this field functional equation.

The paper is organized as followns: In section preliminarier we remind some basic notations in [10] [11] [12] [13] [14] such as non-Archimedean field, Non-Archimedean normed space and non-Archimedean Banach space, Random normed spaces, Non-Archimedean random normed space.

Section 3: Establishing the solution for (1.1) by the fixed point method in Non-Archimedean Banach space.

+ Condition for existence of solutions for Equation (1.1)

+ Constructing a solution for (1.1).

Section 4: Establishing the solution for (1.1) by the direct method in Non-Archimedean Banach space

Section 5: Construct a solution for (1.1) on non-Archimedean Random normed space.

2. Preliminaries

2.1. Non-Archimedean Normed and Banach Spaces

A valuation is a function $\left|\cdot \right|$ from a field $\mathbb{K}$ into $\left[\mathrm{0,}\infty \right)$ such that 0 is the unique element having the 0 valuation,

$\left|r\cdot s\right|=\left|r\right|\cdot \left|s\right|\mathrm{,}\forall r\mathrm{,}s\in \mathbb{K}$

and the triangle inequality holds, i.e.;

$\left|r+s\right|\le \left|r\right|+\left|s\right|\mathrm{,}\forall r\mathrm{,}s\in \mathbb{K}$

A field $\mathbb{K}$ is called a valued filed if $\mathbb{K}$ carries a valuation. The usual absolute values of $ℝ$ and $ℂ$ are examples of valuation. Let us consider a vavluation which satisfies a stronger condition than the triangle inaquality. If the tri triangle inequality is replaced by

$\left|r+s\right|\le \max \left\{\left|r\right|,\left|s\right|\right\},\forall r,s\in \mathbb{K}$

then the function $\left|\cdot \right|$ is called a norm-Archimedean valuational, and filed. Clearly $\left|1\right|=\left|-1\right|=1$ and $\left|n\right|\le \mathrm{1,}\forall n\in N$ . A trivial example of a non-Archimedean valuation is the function $\left|\cdot \right|$ talking everything except for 0 into 1 and $\left|0\right|=0$ this paper, we assume that the base field is a non-Archimedean filed, hence call it simply a filed. Let be a vecter space over a filed $\mathbb{K}$ with a non-Archimedean $\left|\cdot \right|$ . A function $\Vert \cdot \Vert \mathrm{<mo>\; :\; </mo>}X\to \left[\mathrm{0,}\infty \right)$ is said a non-Archimedean norm if it satisfies the follwing conditions:

1) $\Vert x\Vert =0$ if and only if $x=0$ ;

2) $\Vert rx\Vert =\left|r\right|\Vert x\Vert \left(r\in \mathbb{K},x\in X\right)$ ;

3) the strong triangle inequlity

$\Vert x+y\Vert \le \max \left\{\Vert x\Vert ,\Vert y\Vert \right\},x,y\in X$

hold. Then $\left(X,\Vert \cdot \Vert \right)$ is called a norm-Archimedean norm space.

1) Let $\left\{x_n\right\}$ be a sequence in a non-Archimedean normed space X. Then sequence $\left\{x_n\right\}$ is called cauchy if for a given $\epsilon >0$ there a positive integer N such that

$\Vert x_n-x\Vert \le \epsilon$

for all $n,m\ge N$

2) Let $\left\{x_n\right\}$ be a sequence in a norm-Archimedean normed space X. Then sequence $\left\{x_n\right\}$ is called cauchy if for a given $\epsilon >0$ there a positive integer N such that

$\Vert x_n-x\Vert \le \epsilon$

for all $n,m\ge N$ . The we call $x\in X$ a limit of sequence $x_n$ and denote ${\lim }_{n\to \infty }{x}_n=x$ .

3) If every sequence Cauchy in X converger, then the norm-Archimedean normed space X is called a norm-Archimedean Bnanch space.

2.2. Random Normed Spaces

A random normed space is triple $\left(X\mathrm{,}\Gamma \mathrm{,}T\right)$ , where $X$ is a vector space, T is a is a continuous t-norm, and $\Gamma$ is a mapping from $X$ into $D^{+}$ such that, the following conditions hold:

1) (RN₁) ${\Gamma }_x\left(t\right)$ for all $t>0$ if and only if $x=0$ ;

2) (RN₂) ${\Gamma }_{\alpha x}\left(t\right)={\Gamma }_x\left(\frac{t}{\left|\alpha \right|}\right)$ for all $x\in X$ , $\alpha \ne 0$ ;

3) (RN₃) ${\Gamma }_{x+y}\left(t+s\right)\ge T\left({\Gamma }_x\left(t\right)\mathrm{,}{\Gamma }_y\left(s\right)\right)$ for all $x\mathrm{,}y\in X$ , $t\mathrm{,}s\ge 0$ ;

Note: If $\left(X\mathrm{,}\Gamma \mathrm{,}T\right)$ is a random normed space an $\left\{x_n\right\}$ is a sequence such that $x_n\to x$ then ${\lim }_{n\to \infty }{\Gamma }_{x_n}\left(t\right)={\Gamma }_x\left(t\right)$ almost everywhere.

2.3. Non-Archimedean Random Normed Space

A non-Archimedean random normed space is triple $\left(X\mathrm{,}\Gamma \mathrm{,}T\right)$ , where $X$ is a linear space over a non-Archimedean filed $\mathbb{K}$ , T is a is a continuous t-norm, and $\Gamma$ is a mapping from $X$ into $D^{+}$ such that, the following conditions hold:

1) (NA-RN₁) ${\Gamma }_x\left(t\right)={\epsilon }_0\left(t\right)$ for all $t>0$ if and only if $x=0$ ;

2) (NA-RN₂) ${\Gamma }_{\alpha x}\left(t\right)={\Gamma }_x\left(\frac{t}{\left|\alpha \right|}\right)$ for all $x\in X$ , $t>0$ , $\alpha \ne 0$ ;

3) (NA-RN₃) ${\Gamma }_{x+y}\left(\max \left\{t,s\right\}\right)\ge T\left({\Gamma }_x\left(t\right),{\Gamma }_y\left(s\right)\right)$ for all $x\mathrm{,}y\in X$ , $t\mathrm{,}s\ge 0$ ;

It is easy to see that if (NA-RN₃) hold then so is (RN₃) ${\Gamma }_{x+y}\left(\max \left\{t,s\right\}\right)\ge T\left({\Gamma }_x\left(t\right),{\Gamma }_y(\; s\; )\right)$

Let $\left(X\mathrm{,}\Gamma \mathrm{,}T\right)$ is a non-Archimedean random normed space. Suppose $\left\{x_n\right\}$ is a sequence in $X$ . Then $\left\{x_n\right\}$ is said to be convergent if there exists $x\in X$ such that

$\underset{n\to \infty }{lim}{\Gamma }_{x_n-x}\left(t\right)=1$

for all $t>0$ . In that case, x is called the limit of sequence $\left\{x_n\right\}$

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

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

2.4. Solutions of the Equation

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.

The functional equation

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

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

The functional equation

$2f\left(\frac{x+y}{2}\right)+2f\left(\frac{x-y}{2}\right)=f\left(x\right)+f(\; y\; )$

is called a Jensen type the quadratic functional equation

3. Establishing the of (1.1) in Non-Archimedean Banach Space

3.1. Condition for Existence of Solutions for Equation (1.1)

Note that for Quadratic λ-functional equation, $\mathbb{X}$ and $\mathbb{Y}$ is be vector space.

Lemma 2. Suppose $\mathbb{X}$ and $\mathbb{Y}$ be vector space. If mapping $f\mathrm{<mo>\; :\; </mo>}\mathbb{X}\to \mathbb{Y}$ satisfying

$\begin{array}{l}2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j+y_j\right)\\ =f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\end{array}$ (2)

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

Proof. Assume that $f\mathrm{<mo>\; :\; </mo>}\mathbb{X}\to \mathbb{Y}$ satisfies (2)

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 (2), we get

$\left(4k-1\right)f\left(0\right)={\lambda }^{-2m}f\left(0\right)$ (3)

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

$f\left(x\right)={\lambda }^{-2m}f\left({\lambda }^{m}x\right)$ (4)

and so $f\left({\lambda }^{m}x\right)={\lambda }^{2m}f\left(x\right)$ for all $x\in \mathbb{X}$ . Thus from (2)

$\begin{array}{l}2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j+y_j\right)\\ =f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\\ =f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\end{array}$ (5)

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

Next now 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 $\left(x,\mathrm{0,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,0}\right)$ in (2), we have

$f\left(2x\right)=2^{2}f\left(x\right)$ (6)

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

Next we replace x by 2x, we get

$f\left(2^{2}x\right)=2^{4}f\left(x\right)$ (7)

for all $x\in \mathbb{X}$ . for all $x\in \mathbb{X}$ , So from (6) and (7) we have the general case for every m being a positive integer, we have

$f\left(2^{m}x\right)=2^{2m}f\left(x\right)$ (8)

for all $x\in \mathbb{X}$ , So we get the desired result.

Notice now 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 $\left(x\mathrm{,}\cdots \mathrm{,0,0,}\cdots \mathrm{,0,}y\mathrm{,}\cdots \mathrm{,0}\right)$ in (5) we have

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

So, the function f is quadratic. □

3.2. Constructing a Solution for (1.1)

Now, we first study the solutions of (1.1). Note that for Quadratic λ-functional equation, $\mathbb{X}$ is a non-Archimedean normed space and $\mathbb{Y}$ is a non-Archimedean Banach spacebe then use fixed point method, we prove the Hyers-Ulam stability of the Quadratic λ-functional equation in Non-Archimedean Banach space. Under this setting, we can show that the mapping satisfying (1.1) is quadratic. These results are give in the following.

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

$\begin{array}{l}\phi \left(\frac{x_1}{2k}\mathrm{,}\frac{x_2}{2k}\mathrm{,}\cdots \mathrm{,}\frac{x_k}{2k}\mathrm{,}\frac{y_1}{2k}\mathrm{,}\frac{y_2}{2k}\mathrm{,}\cdots \mathrm{,}\frac{y_k}{2k}\mathrm{,}\frac{z_1}{2k}\mathrm{,}\frac{z_2}{2k}\mathrm{,}\cdots \mathrm{,}\frac{z_k}{2k}\right)\\ \le \frac{L}{\left|4k\right|}\phi \left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}{z}_2\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (9)

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

$\begin{array}{l}\Vert 2{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(z_j\right)+2{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(x_j+y_j\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert \\ \le \phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (10)

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

$\Vert f\left(x\right)-H\left(x\right)\Vert \le \frac{L}{\left|4k\right|\left(1-L\right)}\phi \left(x\mathrm{,}\cdots ,x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)$ (11)

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

Proof. 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 $\left(x\mathrm{,}\cdots ,x\mathrm{,0,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,}x\right)$ in (10), we get

$\Vert f\left(2kx\right)-4kf\left(x\right)\Vert \le \phi \left(x\mathrm{,}\cdots ,x\mathrm{,0,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,}x\right)$ (12)

for all $x\in \mathbb{X}$ for all $j=1\to k$ .

Now we consider the set

$\mathbb{M}:=\left\{h:\mathbb{X}\to \mathbb{Y},h\left(0\right)=0\right\}$

and introduce the generalized metric on S as follows:

$d\left(g,h\right):=\inf \left\{\beta \in ℝ:\Vert g\left(x\right)-h\left(x\right)\Vert \le \beta \phi \left(x,\cdots ,x,0,\cdots ,0,x,\cdots ,x\right),\forall x\in \mathbb{X}\right\},$

where, as usual, $\inf \varphi =+\infty$ . That has been proven by mathematicians $\left(\mathbb{M}\mathrm{,}d\right)$ is complete see [14]

Now we cosider the linear mapping $T\mathrm{<mo>\; :\; </mo>}\mathbb{M}\to \mathbb{M}$ such that

$Tg\left(x\right):=4kg\left(\frac{x}{2k}\right)$

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

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

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

Hence

$\begin{array}{c}\Vert Tg\left(x\right)-Th\left(x\right)\Vert =\Vert 4kg\left(\frac{x}{2k}\right)-4khf\left(\frac{x}{2k}\right)\Vert \\ \le \left|4k\right|\epsilon \phi \left(\frac{x}{2k}\mathrm{,}\frac{x}{2k}\mathrm{,}\cdots \mathrm{,}\frac{x}{2k}\mathrm{,0,0,}\cdots \mathrm{,0,}\frac{x}{2k}\mathrm{,}\frac{x}{2k}\mathrm{,}\cdots \mathrm{,}\frac{x}{2k}\right)\\ \le \left|4k\right|\epsilon \frac{L}{\left|4k\right|}\phi \left(x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,0,0,}\cdots \mathrm{,0,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)\\ \le L\epsilon \phi \left(x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,0,0,}\cdots \mathrm{,0,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)\end{array}$

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

$d\left(Tg\mathrm{,}Th\right)\le Ld\left(g\mathrm{,}h\right)$

for all $g\mathrm{,}h\in \mathbb{M}$ . It folows from (12) that

$\begin{array}{c}\Vert f\left(x\right)-4kf\left(\frac{x}{2k}\right)\Vert \le \phi \left(\frac{x}{2k}\mathrm{,}\frac{x}{2k}\mathrm{,}\cdots \mathrm{,}\frac{x}{2k}\mathrm{,0,0,}\cdots \mathrm{,0,}\frac{x}{2k}\mathrm{,}\frac{x}{2k}\mathrm{,}\cdots \mathrm{,}\frac{x}{2k}\right)\\ \le \frac{L}{\left|4k\right|}\phi \left(x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,0,0,}\cdots \mathrm{,0,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)\end{array}$

for all $x\in \mathbb{X}$ . So $d\left(f\mathrm{,}Tf\right)\le \frac{L}{\left|4k\right|}$ for all $x\in \mathbb{X}$ By Theorem 1.2, there exists a mapping $H\mathrm{<mo>\; :\; </mo>}\mathbb{X}\to \mathbb{Y}$ satisfying the fllowing:

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

$H\left(x\right)=4kH\left(\frac{x}{2k}\right)$ (13)

for all $x\in \mathbb{X}$ . The mapping H is a unique fixed point T in the set

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

This implies that H is a unique mapping satisfying (13) such that there exists a $\beta \in \left(\mathrm{0,}\infty \right)$ satisfying

$\Vert f\left(x\right)-H\left(x\right)\Vert \le \beta \phi \left(x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,0,0,}\cdots \mathrm{,0,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)$

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

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

$\underset{l\to \infty }{lim}{\left(4k\right)}^{n}f\left(\frac{x}{{\left(2k\right)}^n}\right)=H(\; x\; )$

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

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

$\Vert f\left(x\right)-H\left(x\right)\Vert \le \frac{L}{\left|4k\right|\left(1-L\right)}\phi \left(x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,0,0,}\cdots \mathrm{,0,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)$

for all $x\in \mathbb{X}$ . It follows (9) and (10) that

$\begin{array}{l}\Vert 2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}H\left(x_j+y_j\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}H\left(z_j\right)-H\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{\lambda }^{-2m}H\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert \\ =\underset{n\to \infty }{\lim }{\left|4k\right|}^n\Vert 2{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(\frac{x_j+y_j}{{\left(2k\right)}^n}\right)+2{\displaystyle \underset{j\mathrm{<mo>\; =\; </mo>1}}{\overset{k}{\sum }}}f\left(\frac{z_j}{{\left(2k\right)}^n}\right)\\ -f\left(\frac{{\displaystyle {\sum }_{j=1}^k{x}_j}+{\displaystyle {\sum }_{j=1}^k{y}_j}+{\displaystyle {\sum }_{j=1}^k{z}_j}}{{\left(2k\right)}^n}\right)\\ -{\lambda }^{-2m}f\left({\lambda }^m\left(\frac{{\displaystyle {\sum }_{j=1}^k{x}_j}+{\displaystyle {\sum }_{j=1}^k{y}_j}+{\displaystyle {\sum }_{j=1}^k{z}_j}}{{\left(2k\right)}^n}\right)\right)\Vert \end{array}$

$\begin{array}{l}\le \underset{n\to \infty }{\lim }{\left|4k\right|}^n\phi (\frac{x_1}{{\left(2k\right)}^n},\frac{x_2}{{\left(2k\right)}^n},\cdots ,\frac{x_k}{{\left(2k\right)}^n},\frac{y_1}{{\left(2k\right)}^n},\frac{y_2}{{\left(2k\right)}^n},\cdots ,\frac{y_k}{{\left(2k\right)}^n},\\ \frac{z_1}{{\left(2k\right)}^n},\frac{z_2}{{\left(2k\right)}^n},\cdots ,\frac{z_k}{{\left(2k\right)}^n})\\ =0\end{array}$

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

$\begin{array}{l}2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}H\left(x_j+y_j\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}H\left(z_j\right)-H\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{\lambda }^{-2m}H\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)=0\end{array}$

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

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

$\begin{array}{l}\phi \left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}{z}_2\mathrm{,}\cdots \mathrm{,}{z}_k\right)\\ \le \left|4k\right|K\phi \left(\frac{x_1}{2k}\mathrm{,}\frac{x_2}{2k}\mathrm{,}\cdots \mathrm{,}\frac{x_k}{2k}\mathrm{,}\frac{y_1}{2k}\mathrm{,}\frac{y_2}{2k}\mathrm{,}\cdots \mathrm{,}\frac{y_k}{2k}\mathrm{,}\frac{z_1}{2k}\mathrm{,}\frac{z_2}{2k}\mathrm{,}\cdots \mathrm{,}\frac{z_k}{2k}\right)\end{array}$ (14)

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

$\begin{array}{l}\Vert 2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j+y_j\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert \\ \le \phi \left(x_1,\cdots ,x_k,y_1,\cdots ,y_k,z_1,\cdots ,z_k\right)\end{array}$ (15)

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

$\Vert f\left(x\right)-H\left(x\right)\Vert \le \frac{L}{\left|4k\right|\left(1-L\right)}\phi \left(x\mathrm{,}\cdots ,x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,}x\mathrm{,}\cdots \mathrm{,}x\right)$ (16)

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

The rest of the proof is similar to the proof of theorem 3.2 with note that mapping $T\mathrm{<mo>\; :\; </mo>}\mathbb{M}\to \mathbb{M}$ , $Tg\left(x\right)\mathrm{<mo>\; :\; </mo>}=\frac{1}{4k}g\left(2kx\right)$ .

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

$\begin{array}{l}\Vert 2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j+y_j\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert \\ \le \theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (17)

for all $x\in X$ . Then there exists a unique quadratic type mapping $H\mathrm{<mo>\; :\; </mo>}\mathbb{X}\to \mathbb{Y}$ such that

$\Vert f\left(x\right)-H\left(x\right)\Vert \le \frac{2\theta }{{\left|2k\right|}^r-\left|4k\right|}{\Vert x\Vert }^r$ (18)

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

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

$\begin{array}{l}\Vert 2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j+y_j\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert \\ \le \theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (19)

for all $x\in \mathbb{X}$ . Then there exists a unique quadratic type mapping $H\mathrm{<mo>\; :\; </mo>}\mathbb{X}\to \mathbb{Y}$ such that

$\Vert f\left(x\right)-H\left(x\right)\Vert \le \frac{2\theta }{\left|4k\right|-{\left|2k\right|}^r}{\Vert x\Vert }^r$ (20)

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

4. Establishing a Solution to the Quadratic λ-Functional Equation Using the Direct Methoduse in Non-Archimedean Banach Space

Next, we are going to study the solutions of (1.1) for Quadratic λ-functional equation use direct method, we prove the Hyers-Ulam stability of the Quadratic λ-functional equation, the $\mathbb{X}$ is a Non-Archimedean normed space and $\mathbb{Y}$ is a Non-Archimedean Banach space, and the field $\mathbb{K}$ satisfy $\left|2k\right|\ne \mathrm{1,}{\lambda }^{-2m}\ne 4k-1$ . Under this setting, we can show that the mapping satisfying (1.1) is quadratic. These results are give in the following

Theorem 5. Let $\phi \mathrm{<mo>\; :\; </mo>}{\mathbb{X}}^{3k}\to \left[\mathrm{0,}\infty \right)$ be a function and let $f\mathrm{<mo>\; :\; </mo>}\mathbb{X}\to \mathbb{Y}$ be a mapping satisfying $f\left(0\right)=0$ and

$\begin{array}{l}\underset{j\to \infty }{\lim }{\left|4k\right|}^j\phi (\frac{x_1}{{\left(2k\right)}^j}\mathrm{,}\frac{x_2}{{\left(2k\right)}^j}\mathrm{,}\cdots \mathrm{,}\frac{x_k}{{\left(2k\right)}^j}\mathrm{,}\frac{y_1}{{\left(2k\right)}^j}\mathrm{,}\frac{y_2}{{\left(2k\right)}^j}\mathrm{,}\cdots \mathrm{,}\\ \frac{y_k}{{\left(2k\right)}^j}\mathrm{,}\frac{z_1}{{\left(2k\right)}^j}\mathrm{,}\frac{z_2}{{\left(2k\right)}^j}\mathrm{,}\cdots \mathrm{,}\frac{z_k}{{\left(2k\right)}^j})=0\end{array}$ (21)

$\begin{array}{l}\Vert 2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j+y_j\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert \\ \le \phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (22)

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

$\begin{array}{l}\Vert f\left(x\right)-H\left(x\right)\Vert \\ \le {\sup }_{j\in \mathbb{N}}\left\{{\left|4k\right|}^{j-1}\phi \left(\frac{x}{{\left(2k\right)}^j},\cdots ,\frac{x}{{\left(2k\right)}^j},\frac{x}{{\left(2k\right)}^j},\cdots ,\frac{x}{{\left(2k\right)}^j},\frac{x}{{\left(2k\right)}^j},\cdots ,\frac{x}{{\left(2k\right)}^j}\right)\right\}\end{array}$ (23)

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

Proof. 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 $\left(x\mathrm{,}\cdots ,x\mathrm{,0,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,}x\right)$ in (22), we have

$\Vert f\left(2kx\right)-4kf\left(x\right)\Vert \le \phi \left(x\mathrm{,}\cdots ,x\mathrm{,0,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,}x\right)$ (24)

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

$\Vert f\left(x\right)-4kf\left(\frac{x}{2k}\right)\Vert \le \phi \left(\frac{x}{2k}\mathrm{,}\cdots ,\frac{x}{2k}\mathrm{,0,}\cdots \mathrm{,0,}\frac{x}{2k}\mathrm{,}\cdots \mathrm{,}\frac{x}{2k}\right)$ (25)

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

Hence

$\begin{array}{l}\Vert {\left(4k\right)}^{l}f\left(\frac{x}{{\left(2k\right)}^l}\right)-{\left(4k\right)}^{m}f\left(\frac{x}{{\left(2k\right)}^m}\right)\Vert \\ \le \max \{\Vert {\left(4k\right)}^{l}f\left(\frac{x}{{\left(2k\right)}^l}\right)-{\left(4k\right)}^{l+1}f\left(\frac{x}{{\left(2k\right)}^{l+1}}\right)\Vert ,\cdots ,\\ \Vert {\left(4k\right)}^{m-1}f\left(\frac{x}{{\left(2k\right)}^{m-1}}\right)-{\left(4k\right)}^{m}f\left(\frac{x}{{\left(2k\right)}^m}\right)\Vert \}\end{array}$

$\begin{array}{l}\le \max \{{\left|4k\right|}^l\Vert f\left(\frac{x}{{\left(2k\right)}^l}\right)-4kf\left(\frac{x}{{\left(2k\right)}^{l+1}}\right)\Vert ,\cdots ,\\ {\left|4k\right|}^{m-1}\Vert f\left(\frac{x}{{\left(2k\right)}^{m-1}}\right)-4kf\left(\frac{x}{{\left(2k\right)}^m}\right)\Vert \}\\ \le {\sup }_{j\in \left\{l,l+1,\cdot \cdot \cdot \right\}}\{{\left|4k\right|}^j\phi (\frac{x_1}{{\left(2k\right)}^{j+1}},\cdots ,\frac{x_k}{{\left(2k\right)}^{j+1}},\frac{y_1}{{\left(2k\right)}^{j+1}},\cdots ,\\ \frac{y_k}{{\left(2k\right)}^{j+1}},\frac{z_1}{{\left(2k\right)}^{j+1}},\cdots ,\frac{z_k}{{\left(2k\right)}^{j+1}})\}\end{array}$ (26)

for all nonnegative integers m and l with $m>l$ and all $x\in X$ . It follows (26)

that the sequence $\left\{{\left(4k\right)}^{n}f\left(\frac{x}{{\left(2k\right)}^n}\right)\right\}$ is a Cauchy sequence for all $x\in X$ . Since $Y$ is complete, the sequence $\left\{{\left(4k\right)}^{n}f\left(\frac{x}{{\left(2k\right)}^n}\right)\right\}$ converger so one can define the mapping $H\mathrm{<mo>\; :\; </mo>}X\to Y$ by

$H\left(x\right)\mathrm{<mo>\; :\; </mo>}=\underset{n\to \infty }{\lim }{\left(4k\right)}^{n}f\left(\frac{x}{{\left(2k\right)}^n}\right)$

for all $x\in X$ . Moreover, letting $l=0$ and passing the limit $m\to \infty$ in (26), we get (23). It follows from (21) and (22) that

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

$\begin{array}{l}\le \underset{j\to \infty }{\lim }{\left|4k\right|}^n\phi \left(\frac{x}{{\left(2k\right)}^j}\mathrm{,}\cdots \mathrm{,}\frac{x}{{\left(2k\right)}^j}\mathrm{,}\frac{x}{{\left(2k\right)}^j}\mathrm{,}\cdots \mathrm{,}\frac{x}{{\left(2k\right)}^j}\mathrm{,}\frac{x}{{\left(2k\right)}^j}\mathrm{,}\cdots \mathrm{,}\frac{x}{{\left(2k\right)}^j}\right)\\ =0\end{array}$ (27)

for all $x\in X$ .

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

$-{\lambda }^{-2m}H\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)=0$

for all $x\in X$ . By Lemma 3.1, the mapping $H\mathrm{<mo>\; :\; </mo>}X\to Y$ is quadratic. Now, let $T\mathrm{<mo>\; :\; </mo>}X\to Y$ be another quadratic mapping satisfying (23). Then we have

$\begin{array}{c}\Vert H\left(x\right)-T\left(x\right)\Vert =\Vert {\left(4k\right)}^{q}H\left(\frac{x}{{\left(2k\right)}^q}\right)-{\left(4k\right)}^{q}T\left(\frac{x}{{\left(2k\right)}^q}\right)\Vert \\ \le \max \{\Vert {\left(4k\right)}^{q}H\left(\frac{x}{{\left(2k\right)}^q}\right)-{\left(4k\right)}^{q}f\left(\frac{x}{{\left(2k\right)}^q}\right)\Vert ,\\ \Vert {\left(4k\right)}^{q}T\left(\frac{x}{{\left(2k\right)}^q}\right)-{\left(4k\right)}^{q}f\left(\frac{x}{{\left(2k\right)}^q}\right)\Vert \}\\ \le {\sup }_{j\in \mathbb{N}}\{{\left|4k\right|}^{q+j-1}\phi (\frac{x_1}{{\left(2k\right)}^{j+1}},\cdots ,\frac{x_k}{{\left(2k\right)}^{j+1}},\frac{y_1}{2^{j+1}},\cdots ,\\ \frac{y_k}{{\left(2k\right)}^{j+1}},\frac{z_1}{{\left(2k\right)}^{j+1}},\cdots ,\frac{z_k}{2^{j+1}})\}\end{array}$

which tends to zero as $q\to \infty$ for all $x\in X$ . So we can conclude that

$H\left(x\right)=T\left(x\right)$ for all $x\in X$ . This proves the uniqueness of H. Thus the mapping $H\mathrm{<mo>\; :\; </mo>}X\to Y$ is a unique quadratic mapping satisfying (23) □

Theorem 6. Let $\phi \mathrm{<mo>\; :\; </mo>}{X}^{3k}\to \left[\mathrm{0,}\infty \right)$ be a function and let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfying $f\left(0\right)=0$ and

$\begin{array}{l}\underset{j\to \infty }{\lim }\{\frac{1}{{\left|4k\right|}^j}\phi ({\left(2k\right)}^{j-1}{x}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{j-1}{x}_k\mathrm{,}{\left(2k\right)}^{j-1}{y}_1\mathrm{,}\cdots \mathrm{,}\\ \underset{\stackrel{}{}}{\overset{}{}}{\left(2k\right)}^{j-1}{y}_k\mathrm{,}{\left(2k\right)}^{j-1}{z}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{j-1}{z}_k)\}=0\end{array}$ (28)

$\begin{array}{l}\Vert 2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j+y_j\right)-f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ -{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert \\ \le \phi \left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}{z}_2\mathrm{,}\cdots \mathrm{,}{z}_k\right)\end{array}$ (29)

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

$\begin{array}{c}\Vert f\left(x\right)-H\left(x\right)\Vert \le {\sup }_{j\in \mathbb{N}}\{\frac{1}{{\left|4k\right|}^{j-1}}\phi ({\left(2k\right)}^{j-1}x,\cdots ,{\left(2k\right)}^{j-1}x,{\left(2k\right)}^{j-1}x,\cdots ,\\ \underset{}{\overset{}{\stackrel{}{}}}{\left(2k\right)}^{j-1}x,{\left(2k\right)}^{j-1}x,\cdots ,{\left(2k\right)}^{j-1}x)\}\end{array}$ (30)

for all $x\in X$ .

The rest of the proof is similar to the proof of theorem 4.1.

Corollary 3. Let $r<2$ and $\theta$ be nonegative real numbers and let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfying $f\left(0\right)=0$ and

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

for all $x\in X$ . Then there exists a unique quadratic mapping $H\mathrm{<mo>\; :\; </mo>}X\to Y$ such that

$\Vert f\left(x\right)-H\left(x\right)\Vert \le \frac{2k\theta }{{\left|2k\right|}^r}{\Vert x\Vert }^r$

for all $x\in X$ .

Corollary 4. Let $r>2$ , and $\theta$ be nonegative real numbers and let $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be a mapping satisfying $f\left(0\right)=0$ and

$\begin{array}{l}\Vert 2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j+y_j\right)=f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\\ +{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\Vert \\ \le \theta \left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert x_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert y_j\Vert }^r+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{\Vert z_j\Vert }^r\right)\end{array}$ (32)

for all $x\in X$ . Then there exists a unique quadratic mapping $H\mathrm{<mo>\; :\; </mo>}X\to Y$ such that

$\Vert f\left(x\right)-H\left(x\right)\Vert \le \frac{2k\theta }{\left|4k\right|}{\Vert x\Vert }^r$

for all $x\in X$ .

5. Construct a Solution for (1.1) on Non-Archimedean Random Normed Space

In this section, $K$ be a non-Archimedean field, $X$ is a vector space over $K$ and let $\left(X\mathrm{,}\Gamma \mathrm{,}T\right)$ be a non-Archimedean random Banach space over $K$

We investigate the stability of the quadratic functional equation

$\begin{array}{l}2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(z_j\right)+2{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left(x_j+y_j\right)\\ =f\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)+{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\end{array}$ (33)

where $f\mathrm{<mo>\; :\; </mo>}X\to Y$ and $f\left(0\right)=0$ .

Next, we define a random approximately quadrtic function. Let $\phi \mathrm{<mo>\; :\; </mo>}{X}^{3k+1}\to \left[\mathrm{0,}\infty \right)$ be a distribution function such that $\phi {\left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}{z}_2\mathrm{,}\cdots \mathrm{,}z\right)}_k$ is symmetric, nondecreasing and

$\phi \left(x\mathrm{,}\cdots \mathrm{,}x\mathrm{,0,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,}\frac{t}{\left|\lambda \right|}\right)\le \phi \left(\lambda x\mathrm{,}\cdots \mathrm{,}\lambda x\mathrm{,0,}\cdots \mathrm{,0,}\lambda x\mathrm{,}\cdots \mathrm{,}\lambda x\mathrm{,}t\right)$ (34)

For $x\in X$ , $\lambda \ne 0$ .

Next, we define:

A mapping $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is said to be φ-approximately quadratic mapping if

$\begin{array}{l}{\Gamma }_{f\left({\displaystyle {\sum }_{j=1}^k{x}_j}+{\displaystyle {\sum }_{j=1}^k{y}_j}+{\displaystyle {\sum }_{j=1}^k{z}_j}\right)+{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle {\sum }_{j=1}^k{x}_j}+{\displaystyle {\sum }_{j=1}^k{y}_j}-{\displaystyle {\sum }_{j=1}^k{z}_j}\right)\right)-2{\displaystyle {\sum }_{j=1}^{k}f\left(z_j\right)}-2{\displaystyle {\sum }_{j=1}^{k}f\left(x_j+y_j\right)}}\\ \le \phi \left(x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}{y}_2\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}{z}_2\mathrm{,}\cdots \mathrm{,}{z}_k\mathrm{,}t\right)\end{array}$ (35)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ , for all $j=1\to k$ , $t>0$ .

* Note: We assume that $2k\ne 0$ in $\mathbb{K}$

Theorem 7 For $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be a φ-approximately quadratic mapping if there exist an $\beta \in ℝ\left(\beta >0\right)$ and an integer h, $h\ge 2$ with $\beta >\left|{\left(2k\right)}^h\right|$ and $\left|2k\right|\ne 0$ such that

$\begin{array}{l}\phi \left({\left(2k\right)}^{-h}{x}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{x}_k\mathrm{,}{\left(2k\right)}^{-h}{y}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{y}_k\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{z}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{z}_k\mathrm{,}t\right)\\ \ge \phi \left({\left(2k\right)}^{-h}{x}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{x}_k\mathrm{,}{\left(2k\right)}^{-h}{y}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{y}_k\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{z}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{z}_k\mathrm{,}\beta t\right)\end{array}$ (36)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for all $j=1\to k$ , $t>0$ and

$\underset{n\to \infty }{\lim }{T}_{j=n}^{\infty }M\left(x,\frac{{\beta }^{j}t}{\left|{\left(2k\right)}^{hj}\right|}\right)=1$ (37)

for all $x\in X$ and $t>0$ .

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

${\Gamma }_{f\left(x\right)-Q\left(x\right)}\left(t\right)\ge T_{i=1}^{\infty }M\left(x,\frac{{\beta }^{i+1}t}{\left|{\left(2k\right)}^{hi}\right|}\right)=1$ (38)

In there

$\begin{array}{l}M\left(x,t\right)=Q(\phi \left(x,\cdots ,x,0,\cdots ,0,x,\cdots ,x,t\right),\phi \left(2kx,\cdots ,2kx,0,\cdots ,0,2kx,\cdots ,2kx,t\right),\\ \cdots ,\phi \left({\left(2k\right)}^{h-1}x,\cdots ,{\left(2k\right)}^{h-1}x,0,\cdots ,0,{\left(2k\right)}^{h-1}x,\cdots ,{\left(2k\right)}^{h-1}x,t\right)\end{array}$ (39)

for all $x\in X$ and $\forall t>0$ .

Proof. First, we show by induction on j that for each $x\in X$ , $t>0$ and $j\ge 1$ ,

$\begin{array}{l}{\Gamma }_{f\left({\left(2k\right)}^{j}x\right)-{\left(4k\right)}^{j}f\left(x\right)}\left(t\right)\ge M_j\left(x\mathrm{,}t\right)\\ \mathrm{<mo>\; :\; </mo>}=T(\phi \left(x\mathrm{,}\cdots \mathrm{,}x\mathrm{,0,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,}t\right)\mathrm{,}\phi \left(2kx\mathrm{,}\cdots \mathrm{,2}kx\mathrm{,0,}\cdots \mathrm{,0,2}kx\mathrm{,}\cdots \mathrm{,2}kx\mathrm{,}t\right)\mathrm{,}\cdots \mathrm{,}\\ \phi \left({\left(2k\right)}^{h-1}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{h-1}x\mathrm{,0,}\cdots \mathrm{,0,}{\left(2k\right)}^{h-1}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{h-1}x\mathrm{,}t\right)\end{array}$ (40)

we replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\mathrm{,}t\right)$ by $\left(x\mathrm{,}\cdots \mathrm{,}x\mathrm{,0,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,}t\right)$ in (35), we obtain

${\Gamma }_{f\left(\left(2k\right)x\right)-\left(4k\right)f\left(x\right)}\left(t\right)\ge \phi \left(x\mathrm{,}\cdots \mathrm{,}x\mathrm{,0,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,}t\right)$ (41)

$x\in X$ , $t>0$ . This proves (40) for $j=1$ . We now assume that (40) holds for some $j\ge 1$ Next we replacing $\left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\mathrm{,}t\right)$ by $\left({\left(2k\right)}^{j}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{j}x\mathrm{,0,}\cdots \mathrm{,0,}{\left(2k\right)}^{j}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{j}x\mathrm{,}t\right)$ in (35) we have

${\Gamma }_{f\left({\left(2k\right)}^{j+1}x\right)-\left(4k\right)f\left({\left(2k\right)}^{j}x\right)}\left(t\right)\ge \phi \left({\left(2k\right)}^{j}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{j}x\mathrm{,0,}\cdots \mathrm{,0,}{\left(2k\right)}^{j}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{j}x\mathrm{,}t\right)$ (42)

Since $\left|4k\right|\le 1$

$\begin{array}{l}{\Gamma }_{f\left({\left(2k\right)}^{j+1}x\right)-{\left(4k\right)}^{j+1}f\left(x\right)}\left(t\right)\\ \ge T\left({\Gamma }_{f\left({\left(2k\right)}^{j+1}x\right)-\left(4k\right)f\left({\left(2k\right)}^{j}x\right)}\left(t\right),{\Gamma }_{\left(4k\right)f\left({\left(2k\right)}^{j}x\right)-{\left(4k\right)}^{j+1}f\left(x\right)}\left(t\right)\right)\\ =T\left({\Gamma }_{f\left({\left(2k\right)}^{j+1}x\right)-\left(4k\right)f\left({\left(2k\right)}^{j}x\right)}\left(t\right),{\Gamma }_{f\left({\left(2k\right)}^{j}x\right)-{\left(4k\right)}^{j}f\left(x\right)}\left(\frac{t}{\left|4k\right|}\right)\right)\\ =T\left({\Gamma }_{f\left({\left(2k\right)}^{j+1}x\right)-\left(4k\right)f\left({\left(2k\right)}^{j}x\right)}\left(t\right),{\Gamma }_{f\left({\left(2k\right)}^{j}x\right)-{\left(4k\right)}^{j}f\left(x\right)}\left(t\right)\right)\\ =T\left(\phi \left({\left(2k\right)}^{j}x,\cdots ,{\left(2k\right)}^{j}x,0,\cdots ,0,{\left(2k\right)}^{j}x,\cdots ,{\left(2k\right)}^{j}x,t\right),M_j\left(x,t\right)\right)\\ =M_{j+1}\left(x,t\right)\end{array}$ (43)

for all $x\in X$ . So in (40) holds for all $j\ge 1$ .

Other way

${\Gamma }_{f\left({\left(2k\right)}^{h}x\right)-{\left(4k\right)}^{h}f\left(x\right)}\left(t\right)\ge M\left(x\mathrm{,}t\right)\mathrm{,}\forall x\in X\mathrm{,}t>0.$ (44)

Next we replacing x by ${\left(2k\right)}^{-\left(hn+h\right)}x$ in (44) and using inequality (36), we have

$\begin{array}{l}{\Gamma }_{f\left(\frac{x}{{\left(2k\right)}^{hn}}\right)-{\left(4k\right)}^{h}f\left(\frac{x}{{\left(2k\right)}^{hn+h}}\right)}\left(t\right)\\ \ge M\left(\frac{x}{{\left(2k\right)}^{hn+h}}\mathrm{,}t\right)\ge M\left(x\mathrm{,}{\beta }^{n+1}t\right)\mathrm{,}\forall x\in X\mathrm{,}t>\mathrm{0,}n\in \mathbb{N}\mathrm{.}\end{array}$ (45)

Then

${\Gamma }_{{\left(4k\right)}^{nh}f\left(\frac{x}{{\left(2k\right)}^{hn}}\right)-{\left(4k\right)}^{h+1}f\left(\frac{x}{{\left(2k\right)}^{hn+h}}\right)}\left(t\right)\ge M\left(x\mathrm{,}\frac{{\beta }^{n+1}}{{\left|4k\right|}^{hn}}t\right),\forall x\in X\mathrm{,}t>\mathrm{0,}n\in \mathbb{N}\mathrm{.}$ (46)

Hence,

$\begin{array}{l}{\Gamma }_{{\left(4k\right)}^{hn}f\left(\frac{x}{{\left(2k\right)}^{hn}}\right)-{\left(4k\right)}^{h\left(n+p\right)}f\left(\frac{x}{{\left(2k\right)}^{h\left(n+p\right)}}\right)}\left(t\right)\\ \ge T_{j=n}^{n+p}\left({\Gamma }_{{\left(4k\right)}^{hj}f\left(\frac{x}{{\left(2k\right)}^{hj}}\right)-{\left(4k\right)}^{h\left(n+j\right)}f\left(\frac{x}{{\left(2k\right)}^{h\left(n+j\right)}}\right)}\left(t\right)\right)\\ \ge T_{j=n}^{n+p}M\left(x\mathrm{,}\frac{{\beta }^{j+1}}{{\left|4k\right|}^{hj}}t\right)\\ \ge T_{j=n}^{n+p}M\left(x\mathrm{,}\frac{{\beta }^{j+1}}{{\left|4k\right|}^j}t\right)\mathrm{,}\forall x\in X\mathrm{,}t>\mathrm{0,}n\in \mathbb{N}\mathrm{.}\end{array}$ (47)

Since

$\underset{n\to \infty }{\lim }{T}_{j=n}^{n+p}M\left(x\mathrm{,}\frac{{\beta }^{j+1}}{{\left|4k\right|}^{hj}}t\right)=\mathrm{1,}\forall x\in X\mathrm{,}t>\mathrm{0,}n\in \mathbb{N}\mathrm{,}$

$\left\{{\left(4k\right)}^{hn}f\left(\frac{x}{{\left(2k\right)}^{hn}}\right)\right\}$ is a Cauchy sequence in the non-Archimedean random Banach space $\left(Y\mathrm{,}\Gamma \mathrm{,}T\right)$ . Hence, we can define a mapping $Q\mathrm{<mo>\; :\; </mo>}X\to Y$ such that

$\underset{n\to \infty }{\lim }{\Gamma }_{{\left(4k\right)}^{hn}f\left(\frac{x}{{\left(2k\right)}^{hn}}\right)-Q\left(x\right)}\left(t\right)=\mathrm{1,}\forall x\in X\mathrm{,}t>\mathrm{0,}$ (48)

Next for each $n\ge 1$ , $\forall x\in X$ and $t>0$ .

$\begin{array}{c}{\Gamma }_{f\left(x\right)-{\left(4k\right)}^{hn}f\left(\frac{x}{{\left(2k\right)}^{hn}}\right)}\left(t\right)={\Gamma }_{{\displaystyle {\sum }_{i=0}^{n-1}{\left(4k\right)}^{hi}f\left(\frac{x}{{\left(2k\right)}^{hi}}\right)-{\left(4k\right)}^{h\left(i+1\right)}f\left(\frac{x}{{\left(2k\right)}^{h\left(i+1\right)}}\right)}}\left(t\right)\\ \ge T_{i=0}^{n+p}\left({\Gamma }_{{\displaystyle {\sum }_{i=0}^{n-1}{\left(4k\right)}^{hi}f\left(\frac{x}{{\left(2k\right)}^{hi}}\right)-{\left(4k\right)}^{h\left(i+1\right)}f\left(\frac{x}{{\left(2k\right)}^{h\left(i+1\right)}}\right)}}\left(t\right)\right)\\ \ge T_{i=0}^{n-1}M\left(x,\frac{{\beta }^{i+1}t}{{\left|4k\right|}^{hi}}\right)\end{array}$ (49)

Therefore,

$\begin{array}{c}{\Gamma }_{f\left(x\right)-Q\left(x\right)}\left(t\right)\ge T\left({\Gamma }_{f\left(x\right)-{\left(4k\right)}^{hn}f\left(\frac{x}{{\left(2k\right)}^{hn}}\right)}\left(t\right),{\Gamma }_{{\left(4k\right)}^{hn}f\left(\frac{x}{{\left(2k\right)}^{hn}}\right)-Q\left(x\right)}\left(t\right)\right)\\ \ge T\left(T_{i=0}^{n-1}M\left(x,\frac{{\beta }^{i+1}t}{{\left|4k\right|}^{hi}}\right),{\Gamma }_{{\left(4k\right)}^{hn}f\left(\frac{x}{{\left(2k\right)}^{hn}}\right)-Q\left(x\right)}\left(t\right)\right)\end{array}$ (50)

By letting $n\to \infty$ , we obtain

${\Gamma }_{f\left(x\right)-Q\left(x\right)}\left(t\right)\ge T_{i=0}^{n-1}M\left(x,\frac{{\beta }^{i+1}t}{{\left|4k\right|}^{hi}}\right)$ (51)

As T is continuous, from a well-known result in probabilistic metric space see [12] .

Now we put

$\begin{array}{c}\Delta x=2{\left(2k\right)}^{hn}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^{-hn}{z}_j\right)+2{\left(2k\right)}^{hn}{\displaystyle \underset{j=1}{\overset{k}{\sum }}}f\left({\left(2k\right)}^{-hn}\left(x_j+y_j\right)\right)\\ -{\left(2k\right)}^{hn}f\left({\left(2k\right)}^{hn}\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right))\\ +{\lambda }^{-2m}{\left(2k\right)}^{hn}f\left({\left(2k\right)}^{-hn}{\lambda }^m\left({\displaystyle \underset{j=1}{\overset{k}{\sum }}}{x}_j+{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{y}_j-{\displaystyle \underset{j=1}{\overset{k}{\sum }}}{z}_j\right)\right)\end{array}$ (52)

it follows that

$\underset{n\to \infty }{\lim }{\Gamma }_{\Delta x}={\Gamma }_{f\left({\displaystyle {\sum }_{j=1}^k{x}_j}+{\displaystyle {\sum }_{j=1}^k{y}_j}+{\displaystyle {\sum }_{j=1}^k{z}_j}\right)+{\lambda }^{-2m}f\left({\lambda }^m\left({\displaystyle {\sum }_{j=1}^k{x}_j}+{\displaystyle {\sum }_{j=1}^k{y}_j}-{\displaystyle {\sum }_{j=1}^k{z}_j}\right)\right)-2{\displaystyle {\sum }_{j=1}^{k}f\left(z_j\right)}-2{\displaystyle {\sum }_{j=1}^{k}f\left(x_j+y_j\right)}}\left(t\right)$ (53)

for almost all $t>0$ , □

On the other hand, replacing $x_j\mathrm{,}{y}_j$ by ${\left(2k\right)}^{hn}{x}_j\mathrm{,}{\left(2k\right)}^{hn}{y}_j$ , respectively, in (35) and suing (NA-RN2) and (36), we have

$\begin{array}{c}{\Gamma }_{\Delta x}\ge \phi \left({\left(2k\right)}^{-hn}{x}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-hn}{x}_k\mathrm{,0,}\cdots \mathrm{,0,}{\left(2k\right)}^{-hn}{z}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-hn}{z}_k\mathrm{,}\frac{t}{{\left|2k\right|}^{hn}}\right)\\ \ge \phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,0,}\cdots \mathrm{,0,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\mathrm{,}\frac{{\beta }^{n}t}{{\left|2k\right|}^{hn}}\right)\end{array}$ (54)

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

$\underset{n\to \infty }{\lim }\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,0,}\cdots \mathrm{,0,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\mathrm{,}\frac{{\beta }^{n}t}{{\left|2k\right|}^{hn}}\right)=\mathrm{1,}$

We infer that Q is a quadratic function.

Finally we have to prove that Q is a unique quadratic mapping.

Let $Q^{\prime }\mathrm{<mo>\; :\; </mo>}X\to Y$ is another quadratic mapping such that

${\Gamma }_{Q^{\prime }\left(x\right)-f\left(x\right)}\left(t\right)\ge M\left(x\mathrm{,}t\right)$ (55)

for all $x\in X$ and $t>0$ , then for each $n\in \mathbb{N}\mathrm{,}x\in X\mathrm{,}t>0$

${\Gamma }_{Q\left(x\right)-Q^{\prime }\left(x\right)}\left(t\right)\ge T\left({\Gamma }_{Q\left(x\right)-{\left(4k\right)}^{hn}f\left(\frac{x}{{\left(2k\right)}^{hn}}\right)}\left(t\right)\mathrm{,}{\Gamma }_{{\left(4k\right)}^{hn}f\left(\frac{x}{{\left(2k\right)}^{hn}}\right)-Q^{\prime }\left(x\right)}\left(t\right)\mathrm{,}t\right)\mathrm{.}$ (56)

Form (48), we infer that $Q^{\prime }=Q$ .

From the theorem 5.1 we get the following corollary:

Corollary 5. For $f\mathrm{<mo>\; :\; </mo>}X\to Y$ be a φ-approximately quadratic mapping if there exist an $\beta \in ℝ\left(\beta >0\right)$ and an integer $h\mathrm{,}h\ge 2$ with $\beta >\left|{\left(2k\right)}^h\right|$ and $\left|2k\right|\ne 0$ such that

$\begin{array}{l}\phi \left({\left(2k\right)}^{-h}{x}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{x}_k\mathrm{,}{\left(2k\right)}^{-h}{y}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{y}_k\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{z}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{z}_k\mathrm{,}t\right)\\ \ge \phi \left({\left(2k\right)}^{-h}{x}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{x}_k\mathrm{,}{\left(2k\right)}^{-h}{y}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{y}_k\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{z}_1\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{-h}{z}_k\mathrm{,}\beta t\right)\end{array}$ (57)

for all $x_j\mathrm{,}{y}_j\mathrm{,}{z}_j\in X$ for all $j=1\to k$ , $t>0$ , then there exists a unique quadratic type mapping $Q\mathrm{<mo>\; :\; </mo>}X\to Y$ such that

${\Gamma }_{f\left(x\right)-Q\left(x\right)}\left(t\right)\ge T_{i=1}^{\infty }M\left(x,\frac{{\beta }^{i+1}t}{\left|{\left(2k\right)}^{hi}\right|}\right)$ (58)

for all $x\in X$ and $\forall t>0$ . In there

$\begin{array}{l}M\left(x\mathrm{,}t\right)=Q(\phi \left(x\mathrm{,}\cdots \mathrm{,}x\mathrm{,0,}\cdots \mathrm{,0,}x\mathrm{,}\cdots \mathrm{,}x\mathrm{,}t\right)\mathrm{,}\phi \left(2kx\mathrm{,}\cdots \mathrm{,2}kx\mathrm{,0,}\cdots \mathrm{,0,2}kx\mathrm{,}\cdots \mathrm{,2}kx\mathrm{,}t\right)\mathrm{,}\\ \cdots \mathrm{,}\phi \left({\left(2k\right)}^{h-1}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{h-1}x\mathrm{,0,}\cdots \mathrm{,0,}{\left(2k\right)}^{h-1}x\mathrm{,}\cdots \mathrm{,}{\left(2k\right)}^{h-1}x\mathrm{,}t\right)\end{array}$ (59)

for all $x\in X$ and $\forall t>0$ .

Application Example: For $\left(X\mathrm{,}\Gamma \mathrm{,}{T}_M\right)$ non-Archimedean random normed space in which

${\Gamma }_x\left(t\right)=\frac{t}{t+\Vert t\Vert }\mathrm{,}\forall x\in X\mathrm{,}t>0$

and assuming that $\left(Y\mathrm{,}\Gamma \mathrm{,}{T}_M\right)$ complete non-Archimedean random normed space.

Now we define

$\phi \left(x_1\mathrm{,}\cdots \mathrm{,}{x}_k\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_k\mathrm{,}{z}_1\mathrm{,}\cdots \mathrm{,}{z}_k\mathrm{,}t\right)=\frac{t}{1+t}\mathrm{.}$

It is easy to see that for $\beta =1$ then (36) holds, sence

$M\left(x,t\right)=\frac{t}{1+t},$

We have

$\begin{array}{c}\underset{n\to \infty }{\lim }{T}_{j=n}^{\infty }M\left(x,\frac{{\beta }^j}{{\left|4k\right|}^{hj}}t\right)=\underset{n\to \infty }{\lim }\left(\underset{m\to \infty }{\lim }{T}_{j=n}^{m}M\left(x,\frac{t}{{\left|4k\right|}^{hj}}t\right)\right)\\ =\underset{n\to \infty }{\lim }\cdot \underset{m\to \infty }{\lim }\left(\frac{t}{t+{\left|{\left(4k\right)}^h\right|}^n}\right)=1\end{array}$ ,

$\forall x\in X$ , $t>0$ .

6. Conclusion

In this paper, I have built the condition for existence of a solution for a functional equation of general form and then I have used two fixed point methods and a direct method to show their solutions on non-Archimedean space and finally establish their solution on the non-Archimedean Random normed space.