HUR stability of a generalized Apollonius type quadratic functional equation in non-Archimedean Banach spaces

Abstract

Using the fixed point and direct methods, we prove the generalized Hyers-Ulam stability of the following generalized Apollonius type quadratic functional equation

$$f\left(\sum _{i=1}^m{z}_i-\sum _{i=1}^m{x}_i\right)+f\left(\sum _{i=1}^m{z}_i-\sum _{i=1}^m{y}_i\right)=\frac{1}{2}f\left(\sum _{i=1}^m{x}_i-\sum _{i=1}^m{y}_i\right)+2f\left(\sum _{i=1}^m{z}_i-\frac{\sum _{i=1}^m{x}_i+\sum _{i=1}^m{y}_i}{2}\right)$$

in non-Archimedean Banach spaces.

Introduction

The stability problem of functional equations originates from a question of Ulam [1] concerning the stability of group homomorphisms. Hyers [2] gave the first affirmative partial answer to the question of Ulam for Banach spaces. Hyers’ theorem was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference.

Theorem 1

Let f be an approximately additive mapping from a normed vector space E into a Banach space E’, i.e., f satisfies the inequality ∥f(x+y)−f(x)−f(y)∥≤ε(∥x∥^r+∥y∥^r) for all x,y∈E, where ε and r are constants with ε>0 and 0≤r<1. Then, the mapping L:E→E^′ defined by L(x)= limn→∞ 2⁻ⁿf(2ⁿx) is the unique additive mapping which satisfies

$$||f(x+y)-L(x)||\le \frac{2\epsilon }{2-2^r}||x{||}^r$$

for all x∈E.

However, the following example shows that the same result of Theorem 1 is not true in non-Archimedean normed spaces.

Example 1

Let p>2 and let $f:{\mathbb{Q}}_p\to {\mathbb{Q}}_p$ be defined by f(x)=2. Then for ε=1,

$$|f(x+y)-f(x)-f(y)|=1\le \epsilon$$

for all $x,y\in {\mathbb{Q}}_p$. However, the sequences ${\left\{\frac{f(2^{n}x)}{2^n}\right\}}_{n=1}^{\infty }$ and ${\left\{2^{n}f\left(\frac{x}{2^n}\right)\right\}}_{n=1}^{\infty }$ are not Cauchy. In fact, by using the fact that |2|=1, we have

$$\left|\frac{f(2^{n}x)}{2^n}-\frac{f(2^{n+1}x)}{2^{n+1}}\right|=|2^{-n}·2-2^{-(n+1)}·2|=\left|2^{-n}\right|=1$$

and

$$\left|2^{n}f\left(\frac{x}{2^n}\right)-2^{n+1}f\left(\frac{x}{2^{n+1}}\right)\right|=|2^n.2-2^{(n+1)}.2|=\left|2^{n+1}\right|=1$$

for all $x,y\in {\mathbb{Q}}_p$ and $n\in \mathbb{N}$. Hence, these sequences are not convergent in ${\mathbb{Q}}_p$.

The paper of Rassias [4] has provided a lot of influence on the development of what we call the ‘Hyers-Ulam stability’ or ‘Hyers-Ulam-Rassias stability’ of functional equations. A generalization of the Th.M. Rassias theorem was obtained by Gǎvruta [5] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias’ approach.

The functional equation f(x+y)+f(x−y)=2f(x)+2f(y) is called a ‘quadratic functional equation’. In particular, every solution of the quadratic functional equation is said to be a ‘quadratic mapping’. A Hyers-Ulam stability problem for the quadratic functional equation was proven by Skof [6] for mappings f:X→Y, where X is a normed space and Y is a Banach space. Cholewa [7] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group. Czerwik [8] proved the Hyers-Ulam stability of the quadratic functional equation. The stability problems of several functional equations have been extensively investigated by a number of authors, and there are many interesting results concerning this problem (see [3–47]).

In 1897, Hensel [15] introduced a normed space which does not have the Archimedean property. It turned out that non-Archimedean spaces have many nice applications [17, 18, 22, 48].

In this paper, we prove the Hyers-Ulam-Rassias (or generalized Hyers-Ulam) stability of the following generalized Apollonius type quadratic functional equation:

$$\begin{array}{l}f\left(\sum _{i=1}^m{z}_i-\sum _{i=1}^m{x}_i\right)+f\left(\sum _{i=1}^m{z}_i-\sum _{i=1}^m{y}_i\right)\\ =\frac{1}{2}f\left(\sum _{i=1}^m{x}_i-\sum _{i=1}^m{y}_i\right)\\ +2f\left(\sum _{i=1}^m{z}_i-\frac{\sum _{i=1}^m{x}_i-\sum _{i=1}^m{y}_i}{2}\right)\end{array}$$

(1)

in non-Archimedean Banach spaces. It is easy to show that the function f(x)=x² satisfies the functional Equation (1), which is called a quadratic functional equation, and every solution of the quadratic functional equation is said to be a quadratic mapping.

Definition 1

By a non-Archimedean field we mean a field $\mathbb{K}$ equipped with a function (valuation) $|·|:\mathbb{K}\to [0,\infty )$ such that for all $r,s\in \mathbb{K}$, the following conditions hold: (a) |r|=0 if and only if r=0; (b) |r s|=|r||s|; and (c) |r+s|≤m a x{|r|,|s|}.

Remark 1

Clearly, |1|=|−1|=1 and |n|≤1 for all $n\in \mathbb{N}$.

Definition 2

Let X be a vector space over a scalar field $\mathbb{K}$ with a non-Archimedean, non-trivial valuation |·|. A function $\left|\right|·\left|\right|:X\to \mathbb{R}$ is a non-Archimedean norm (valuation) if it satisfies the following conditions: (a) ||x||=0 if and only if x=0; $(b)\left|\right|\mathit{\text{rx}}\left|\right|=\left|r\right|\left|\right|x\left|\right|(r\in \mathbb{K},x\in X$); and (c) the strong triangle inequality (ultrametric), namely

$$\left|\right|x+y\left|\right|\le \mathit{\text{max}}\left\{\right|\left|x\right||,|\left|y\right|\left|\right\},x,y\in \mathrm{X.}$$

Then, (X,||·||) is called a non-Archimedean space.

Definition 3

A sequence {x _n } is Cauchy if and only if {x_n+1−x _n } converges to zero in a non-Archimedean space. By a complete non-Archimedean space we mean one in which every Cauchy sequence is convergent.

The most important examples of non-Archimedean spaces are p-adic numbers. A key property of p-adic numbers is that they do not satisfy the Archimedean axiom: ‘for x,y>0, there exists $n\in \mathbb{N}$ such that x<n y’.

Example 2

Fix a prime number p. For any nonzero rational number x, there exists a unique integer $n_x\in \mathbb{Z}$ such that $x=\frac{a}{b}{p}^{n_x}$, where a and b are integers not divisible by p. Then, $|x{|}_p:=p^{-n_x}$ defines a non-Archimedean norm on $\mathbb{Q}$. The completion of $\mathbb{Q}$ with respect to the metric d(x,y)=|x−y| _p is denoted by ${\mathbb{Q}}_p$, which is called the p-adic number field. In fact, ${\mathbb{Q}}_p$ is the set of all formal series $x=\sum _{k\ge n_x}^{\infty }{a}_k{p}^k$ where |a _k |≤p−1 are integers. The addition and multiplication between any two elements of ${\mathbb{Q}}_p$ are defined naturally. The norm $|\sum _{k\ge n_x}^{\infty }{a}_k{p}^k{|}_p=p^{-n_x}$ is a non-Archimedean norm on ${\mathbb{Q}}_p$, and it makes ${\mathbb{Q}}_p$ a locally compact field.

Definition 4

Let X be a set. A function d:X×X→[0,∞] is called a generalized metric on X if d satisfies the following conditions: (a) d(x,y)=0 if and only if x=y for all x,y∈X; (b) d(x,y)=d(y,x) for all x,y∈X; and (c) d(x,z)≤d(x,y)+d(y,z) for all x,y,z∈X.

Theorem 2

Let (X,d) be a complete generalized metric space and J:X→X be a strictly contractive mapping with Lipschitz constant L<1. Then, for all x∈X, either d(Jⁿx,Jⁿ⁺¹x)=∞ for all nonnegative integers n or there exists a positive integer n₀ such that (a) d(Jⁿx,Jⁿ⁺¹x)<∞ for all n₀≥n₀; (b) the sequence {Jⁿx} converges to a fixed point y^∗ of J; and (c) y^∗ is the unique fixed point of J in the set $Y=\{y\in X:d(J^{n_0}x,y)<\infty \}$; (d) $d(y,y^{\ast })\le \frac{1}{1-L}d(y,\mathit{\text{Jy}})$ for all y∈Y.

Arriola and Beyer [49] investigated the Hyers-Ulam stability of approximate additive functions $f:{\mathbb{Q}}_p\to \mathbb{R}$. They showed that if $f:{\mathbb{Q}}_p\to \mathbb{R}$ is a continuous function for which there exists a fixed ε: |f(x+y)−f(x)−f(y)|≤ε for all $x,y\in {\mathbb{Q}}_p$, then there exists a unique additive function $T:{\mathbb{Q}}_p\to \mathbb{R}$ such that |f(x)−T(x)|≤ε for all $x\in {\mathbb{Q}}_p$. In this paper, using the fixed point and direct method, we prove the generalized Hyers-Ulam stability of the functional equation (1) in non-Archimedean normed spaces.

Methods

Non-archimedean stability of Equation 1: fixed point method

Throughout this section, using the fixed point alternative approach, we prove the generalized Hyers-Ulam stability of functional Equation 1 in non-Archimedean normed spaces. Let X be a non-Archimedean normed space and Y be a non-Archimedean Banach space.

Remark 2

Let $x:=\sum _{i=1}^m{x}_i$,

$$y:=\sum _{i=1}^m{y}_i$$

, $z:=\sum _{i=1}^m{z}_i$ and |4|≠1.

Theorem 3

Let ζ:X²→[0,∞) be a function such that there exists L<1 with

$$\zeta \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right)\le \frac{\mathrm{L\zeta }(x,y,z)}{\left|4\right|}$$

(2)

for all x,y,z∈X. If f:X→Y is a mapping with f(0)=0 and satisfying

$$\begin{array}{l}∥f(z-x)+f(z-y)-\frac{1}{2}f(x-y)-2f\left(z-\frac{x+y}{2}\right)∥\\ \le \zeta (x,y,z)\end{array}$$

(3)

for all x,y,z∈X, then the limit $Q(x)=\underset{n\to \infty }{\text{lim}}{4}^{n}f\left(\frac{x}{2^n}\right)$ exists for all x∈X and defines a unique quadratic mapping Q:X→Y such that

$$||f(x)-Q(x)||\le \frac{\mathrm{L\zeta }(x,-x,x)}{\left|2\right|-\left|2\right|L}.$$

(4)

Proof

Putting z=x and y=−x in Equation 3, we have

$$∥\frac{1}{2}f(2x)-2f(x)∥\le \zeta (x,-x,x).$$

(5)

Replacing x by $\frac{x}{2}$ in the above inequality, we obtain

$$∥4f\left(\frac{x}{2}\right)-f(x)∥\le \left|2\right|\zeta \left(\frac{x}{2},\frac{-x}{2},\frac{x}{2}\right)$$

(6)

for all x∈X. Consider the set S:={g:X→Y; g(0)=0} and the generalized metric d in S defined by

$$\begin{array}{c}d(f,g)=inf\left\{\mu \in {\mathbb{R}}^{+}:||g(x)-h(x)||\right.\\ \left.\le \mathrm{\mu \zeta }(x,-x,x),\forall x\in X\right\},\end{array}$$

(7)

where inf ∅=+∞. It is easy to show that (S,d) is complete (see Lemma 2.1 in [20]). Now, we consider a linear mapping J:S→S such that $\mathit{\text{Jh}}(x):=4h\left(\frac{x}{2}\right)$ for all x∈X. Let g,h∈S be such that d(g,h)=ε. Then, we have ∥g(x)−h(x)∥≤ε ζ(x,−x,x) for all x∈X, and so,

$$\begin{array}{cc}||\mathit{\text{Jg}}(x)-\mathit{\text{Jh}}(x)||& =∥4g\left(\frac{x}{2}\right)-4h\left(\frac{x}{2}\right)∥\le \left|4\right|\mathrm{\epsilon \zeta }\left(\frac{x}{2},\frac{-x}{2},\frac{x}{2}\right)\\ \le \frac{\left|4\right|\mathrm{L\epsilon \zeta }(x,-x,x)}{\left|4\right|}\end{array}$$

for all x∈X. Thus, d(g,h)=ε implies that d(J g,J h)≤L ε. This means that d(J g,J h)≤L d(g,h) for all g,h∈S. It follows from Equation 6 that $d(f,\mathit{\text{Jf}})\le \frac{L}{\left|2\right|}.$ By Theorem 2, there exists a mapping Q:X→Y satisfying the following: (1) Q is a fixed point of J, that is,

$$Q\left(\frac{x}{2}\right)=\frac{1}{4}Q(x)$$

(8)

for all x∈X. The mapping Q is a unique fixed point of J in the set Ω={h∈S:d(g,h)<∞}. This implies that Q is a unique mapping satisfying Equation 8 such that there exists μ∈(0,∞) satisfying ∥f(x)−Q(x)∥≤μ ζ(x,−x,x) for all x∈X. (2) d(Jⁿf,Q)→0 as n→∞. This implies the equality $\underset{n\to \infty }{\text{lim}}{4}^{n}f\left(\frac{x}{2^n}\right)=Q(x)$ for all x∈X. (3) $d(f,Q)\le \frac{d(f,\mathit{\text{Jf}})}{1-L}$ with f∈Ω, which implies the inequality $d(f,Q)\le \frac{L}{\left|2\right|-\left|2\right|L}.$ This implies that the inequality (Equation 4) holds. By Equation 3, we have

$$\begin{array}{c}∥4^{n}f\left(\frac{z-x}{2^n}\right)+4^{n}f\left(\frac{z-y}{2^n}\right)-\frac{4^n}{2}f\left(\frac{x-y}{2^n}\right)\\ -2.4^{n}f\left(\frac{z}{2^n}-\frac{x+y}{2^{n+1}}\right)∥\le |4{|}^n\zeta \left(\frac{x}{2^n},\frac{y}{2^n},\frac{z}{2^n}\right)\\ \le \frac{|4{|}^n{L}^n\zeta (x,y,z)}{|4{|}^n}\end{array}$$

for all x,y∈X and n≥1, and so, $∥Q(z-x)+Q(z-y)-\frac{1}{2}Q(x-y)-2Q\left(z-\frac{x+y}{2}\right)∥=0$ for all x,y∈X. Therefore, the mapping Q:X→Y satisfies Equation 1. On the other hand,

$$\begin{array}{l}Q(2x)-4Q(x)=\underset{n\to \infty }{\text{lim}}{4}^{n}f\left(\frac{x}{2^{n-1}}\right)-4\underset{n\to \infty }{\text{lim}}{4}^{n}f\left(\frac{x}{2^n}\right)\\ =4\left[\underset{n\to \infty }{\text{lim}}{4}^{n-1}f\left(\frac{x}{2^{n-1}}\right)-\underset{n\to \infty }{\text{lim}}{4}^{n}f\left(\frac{x}{2^n}\right)\right]\\ =0\end{array}$$

So, Q:X→Y is quadratic. This completes the proof. □

Corollary 1

Let θ₁,θ₂≥0 and r be a real number with r∈(1,+∞). Let f:X→Y be a mapping with f(0)=0 and satisfying

$$\begin{array}{l}∥f(z-x)+f(z-y)-\frac{1}{2}f(x-y)-2f\left(z-\frac{x+y}{2}\right)∥\\ \le {\theta }_1(||x{||}^r+||y{||}^r+||z{||}^r)+{\theta }_2||x{||}^{\frac{r}{3}}.||y{||}^{\frac{r}{3}}||z{||}^{\frac{r}{3}}\end{array}$$

(9)

for all x,y,z∈X. Then, the limit $Q(x)=\underset{n\to \infty }{\text{lim}}{4}^{n}f\left(\frac{x}{2^n}\right)$ exists for all x∈X, and Q:X→Y is a unique quadratic mapping such that

$$||f(x)-Q(x)||\le \frac{|4{|}^r(3{\theta }_1+{\theta }_2)||x{||}^r}{|2|(1-|4{|}^r)}$$

for all x∈X.

Proof

The proof follows from Theorem 3 if we take

$$\zeta (x,y,z)={\theta }_1(||x{||}^r+||y{||}^r+||z{||}^r)+{\theta }_2||x{||}^{\frac{r}{3}}.||y{||}^{\frac{r}{3}}||z{||}^{\frac{r}{3}}$$

for all x,y,z∈X. In fact, if we choose L=|4|^r, we then get the desired result. □

Theorem 4

Let ζ:X²→[0,∞) be a function such that there exists an L<1 with ζ(2x,2y,2z)≤|4|L ζ(x,y,z) for all x,y,z∈X. Let f:X→Y be mapping with f(0)=0 and satisfying Equation 3. Then, the limit $Q(x)\underset{n\to \infty }{\text{lim}}\frac{f(2^{n}x)}{4^n}$ exists for all x∈X and defines a unique quadratic mapping Q:X→Y such that

$$||f(x)-Q(x)||\le \frac{\zeta (x,-x,x)}{|2|-|2|L}.$$

Proof

It follows from Equation 5 that $∥f(x)-\frac{1}{4}f(2x)∥\le \frac{\zeta (x,-x,x)}{\left|2\right|}$ for all x∈X. The rest of the proof is similar to the proof of Theorem 3. □

Corollary 2

Let θ₁,θ₂≥0 and r be a real number with r∈(0,1). Let f:X→Y be a mapping with f(0)=0 and satisfying Equation 9. Then, the limit $Q(x)=\underset{n\to \infty }{\text{lim}}\frac{f(2^{n}x)}{4^n}$ exists for all x∈X, and Q:X→Y is a unique quadratic mapping such that

$$||f(x)-Q(x)||\le \frac{(3{\theta }_1+{\theta }_2)||x{||}^r}{\left|2\right|(1-|4{|}^{1-r})}$$

for all x∈X.

Proof

The proof follows from Theorem 4 if we take

$$\zeta (x,y,z)={\theta }_1(||x{||}^r+||y{||}^r+||z{||}^r)+{\theta }_2||x{||}^{\frac{r}{3}}.||y{||}^{\frac{r}{3}}||z{||}^{\frac{r}{3}}$$

for all x,y,z∈X. In fact, if we choose L=|4|^1−r, we then get the desired result. □

Non-archimedean stability of Equation 1: direct method

In this section, using the direct method, we prove the generalized Hyers-Ulam stability of functional Equation 1 in non-Archimedean normed spaces. Throughout this section, let G be 2-divisible.

Theorem 5

Let G be an additive semigroup and X be a complete non-Archimedean space. Assume that ζ:G³→[0,+∞) is a function such that

$$\underset{n\to \infty }{\text{lim}}\frac{\zeta (2^{n}x,2^{n}y,2^{n}z)}{|4{|}^n}=0$$

(10)

for all x,y,z∈G. Let, for each x∈G, the limit

$$\$(x)=\underset{n\to \infty }{\text{lim}}\mathit{\text{max}}\left\{\frac{\zeta (2^{k}x,-2^{k}x,2^{k}x)}{|4{|}^k}:0\le k<n\right\}$$

(11)

exists for all x∈G. Suppose that f:G→X is a mapping with f(0)=0 and satisfying the inequality

$$\begin{array}{l}∥f(z-x)+f(z-y)-\frac{1}{2}f(x-y)-2f\left(z-\frac{x+y}{2}\right)∥\\ \le \zeta (x,y,z)\end{array}$$

(12)

for all x,y,z∈G. Then, the limit $\alpha (x):=\underset{n\to \infty }{\text{lim}}\frac{f(2^{n}x)}{4^n}$ exists for all x∈G, and α(x):G→X is a quadratic mapping satisfying

$$\left|\right|f(x)-\alpha (x)\left|\right|\le |2{|}^{-1}\$(x)$$

(13)

for all x∈G. Moreover, if

$$\underset{j\to \infty }{\text{lim}}\underset{n\to \infty }{\text{lim}}\mathit{\text{max}}\left\{\frac{\zeta (2^{k}x,-2^{k}x,2^{k}x)}{|4{|}^k}:j\le k<j+n\right\}=0$$

(14)

then, α(x) is the unique mapping satisfying Equation 13.

Proof

Putting z = x and y =-x in Equation 12, we have

$$∥f(x)-\frac{f(2x)}{4}∥\le \frac{\zeta (x,-x,x)}{\left|2\right|}.$$

(15)

for all x ε G. Replacing x by 2ⁿx in Equation 15, we get

$$∥\frac{f(2^{n+1}x)}{4^{n+1}}-\frac{f(2^{n}x)}{4^n}∥\le \frac{\zeta (2^{n}x,-2^{n}x,2^{n}x)}{\left|2\right|.|4{|}^n}.$$

(16)

It follows from Equations 10 and 16 that the sequence ${\left\{\frac{f(2^{n}x)}{4^n}\right\}}_{n=1}^{\infty }$ is a Cauchy sequence. Since X is complete, ${\left\{\frac{f(2^{n}x)}{4^n}\right\}}_{n=1}^{\infty }$ is convergent. Set $\alpha (x):=\underset{n\to \infty }{\text{lim}}\frac{f(2^{n}x)}{4^n}.$ Using induction, we see that

$$∥\frac{f(2^{n}x)}{4^n}-f(x)∥\le \frac{\mathit{\text{max}}\left\{\frac{\zeta (2^{k}x,-2^{k}x,2^{k}x)}{|4{|}^k}:0\le k<n\right\}}{\left|2\right|}.$$

(17)

Indeed, Equation 17 holds for n=1 by Equation 15. Now, if Equation 17 holds for n, then by Equation 16, we obtain

$$\begin{array}{l}∥\frac{f(2^{n+1}x)}{4^{n+1}}-f(x)∥=∥\frac{f(2^{n+1}x)}{4^{n+1}}\pm \frac{f(2^{n}x)}{4^n}-f(x)∥\\ \le \mathit{\text{max}}\left\{∥\frac{f(2^{n+1}x)}{4^{n+1}}-\frac{f(2^{n}x)}{4^n}∥,\right.\\ \left.∥\frac{f(2^{n}x)}{4^n}-f(x)∥\right\}\le \frac{1}{\left|2\right|}\mathit{\text{max}}\left\{\frac{\zeta (2^{n}x,-2^{n}x,2^{n}x)}{|4{|}^n},\right.\\ \mathit{\text{max}}\left\{\frac{\zeta (2^{k}x,-2^{k}x,2^{k}x)}{|4{|}^k}:0\right.\le \left.\left.k<n\right\}\right\}\\ =\frac{1}{\left|2\right|}\mathit{\text{max}}\left\{\frac{\zeta (2^{k}x,-2^{k}x,2^{k}x)}{|4{|}^k}:0\le k<n+1\right\}.\end{array}$$

(18)

So for all $n\in \mathbb{N}$ and all x∈G, Equation 17 holds. By taking n to approach infinity in Equation 17, one obtains Equation 13. If β(x) is another mapping that satisfies Equation 13, then for all x∈G, we get

$$\begin{array}{l}\left|\right|\alpha (x)-\beta (x)\left|\right|=\underset{k\to \infty }{\text{lim}}∥\frac{\alpha (2^{k}x)}{4^k}-\frac{\beta (2^{k}x)}{4^k}∥\\ \le \underset{k\to \infty }{\text{lim}}\mathit{\text{max}}\left\{∥\frac{\alpha (2^{k}x)}{4^k}-\frac{f(2^{k}x)}{4^k}∥,\right.\\ \left.∥\frac{f(2^{k}x)}{4^k}-\frac{\beta (2^{k}x)}{4^k}∥\right\}\\ \le \underset{j\to \infty }{\text{lim}}\underset{n\to \infty }{\text{lim}}\mathit{\text{max}}\left\{\frac{\zeta (2^{k}x,-2^{k}x,2^{k}x)}{|4{|}^k}:\right.\\ \left.j\le k<j+n\right\}=0.\end{array}$$

Therefore, for all x∈G, we obtain α(x)=β(x). □

Corollary 3

Let ξ:[0,∞)→[0,∞) be a function satisfying

$$\xi (|2|t)\le \xi (|2|)\xi (t)(t\ge 0),\xi (|2|)<\left|4\right|.$$

Let κ>0 and f:G→X be a mapping with f(0)=0 and satisfying the inequality

$$\begin{array}{c}∥f(z-x)+f(z-y)-\frac{1}{2}f(x-y)-2f\left(z-\frac{x+y}{2}\right)∥\\ \le \kappa \left(\xi (|x|)+\xi (|y|)+\xi (|z|)\right)\end{array}$$

for all x,y,z∈G. Then the limit $\alpha (x):=\underset{n\to \infty }{\text{lim}}\frac{f(2^{n}x)}{4^n}$ exists for all x∈G, and α(x):G→X is a unique quadratic mapping satisfying

$$||f(x)-\alpha (x)||\le \frac{3\mathrm{\kappa \xi }(|x|)}{|2|}$$

for all x∈G.

Proof

Define ζ:G³→[0,∞) by ζ(x,y,z):=κ(ξ(|x|)+ξ(|y|)+ξ(|z|)). Since $\frac{\xi (|2|)}{\left|4\right|}<1$, we have $\underset{n\to \infty }{\text{lim}}\frac{\zeta (2^{n}x,2^{n}y,2^{n}z)}{|4{|}^n}\le \underset{n\to \infty }{\text{lim}}{\left(\frac{\xi (|2|)}{\left|4\right|}\right)}^n\zeta (x,y,z)=0$

for all x,y,z∈G. Also, for all x∈G

$$\begin{array}{c}\$(x)=\underset{n\to \infty }{\text{lim}}\mathit{\text{max}}\left\{\frac{\zeta (2^{k}x,-2^{k}x,2^{k}x)}{|4{|}^k}:0\le k<n\right\}\\ =3\mathrm{\kappa \xi }(|x|)\end{array}$$

exists for all x∈G. Moreover, $\underset{j\to \infty }{\text{lim}}\underset{n\to \infty }{\text{lim}}\mathit{\text{max}}\left\{\frac{\zeta (2^{k}x,-2^{k}x,2^{k}x)}{|4{|}^k}:j\le k<j+n\right\}=\underset{j\to \infty }{\text{lim}}\frac{\zeta (2^{j}x,-2^{j}x,2^{j}x)}{|4{|}^j}=0$ for all x∈G. Applying Theorem 5, we get the desired results. □

Theorem 6

Let ζ:G³→[0,+∞) be a function such that

$$\underset{n\to \infty }{\text{lim}}|4{|}^n\zeta \left(\frac{x}{2^n},\frac{y}{2^n},\frac{z}{2^n}\right)=0$$

(19)

for all x,y,z∈G. Let the limit

$$\$(x)=\underset{n\to \infty }{\text{lim}}\mathit{\text{max}}\left\{|4{|}^k\zeta \left(\frac{x}{2^{k+1}},\frac{-x}{2^{k+1}},\frac{x}{2^{k+1}}\right):0\le k<n\right\}$$

(20)

exist for each x∈G. Suppose that f:G→X is a mapping with f(0)=0 and satisfying the inequality

$$\begin{array}{cc}∥f(z-x)& +f(z-y)-\frac{1}{2}f(x-y)-2f\left(z-\frac{x+y}{2}\right)∥\\ \le \zeta (x,y,z)\end{array}$$

(21)

for all x,y,z∈G. Then the limit $\alpha (x):=\underset{n\to \infty }{\text{lim}}{4}^{n}f\left(\frac{x}{2^n}\right)$ exists for all x∈G, and α:G→X is a quadratic mapping satisfying

$$\left|\right|f(x)-\alpha (x)\left|\right|\le \left|2\right|\$(x)$$

(22)

for all x∈G. Moreover, if

$$\begin{array}{cc}\underset{k\to \infty }{\text{lim}}& \underset{n\to \infty }{\text{lim}}\mathit{\text{max}}\left\{|4{|}^k\zeta \left(\frac{x}{2^{k+1}},\frac{-x}{2^{k+1}},\frac{x}{2^{k+1}}\right):j\le k<n+j\right\}\\ =0\end{array}$$

then α(x) is the unique mapping satisfying Equation 22.

Proof

Proof. By Equation 6, we know that

$$∥4f\left(\frac{x}{2}\right)-f(x)∥\le \left|2\right|\zeta \left(\frac{x}{2},\frac{-x}{2},\frac{x}{2}\right)$$

(23)

for all x ∈G. Replacing x by $\frac{x}{2^n}$ in Equation 23, we get

$$∥4^{n+1}f\left(\frac{x}{2^{n+1}}\right)-4^{n}f\left(\frac{x}{2^n}\right)∥\le \left|2\right|.|4{|}^n\zeta \left(\frac{x}{2^{n+1}},\frac{-x}{2^{n+1}},\frac{x}{2^{n+1}}\right).$$

(24)

for all x ∈G. It follows from Equations 19 and 24 that the sequence ${\left\{4^{n}f\left(\frac{x}{2^n}\right)\right\}}_{n=1}^{\infty }$ is a Cauchy sequence. Since X is complete, ${\left\{4^{n}f\left(\frac{x}{2^n}\right)\right\}}_{n=1}^{\infty }$ is convergent. It follows from Equation 24 that

$$\begin{array}{l}∥4^{n}f\left(\frac{x}{2^n}\right)-4^{p}f\left(\frac{x}{2^p}\right)∥=∥\sum _{k=p}^n{4}^{k+1}f\left(\frac{x}{2^{k+1}}\right)-4^{k}f\left(\frac{x}{2^k}\right)∥\\ \le \mathit{\text{max}}\left\{∥4^{k+1}f\left(\frac{x}{2^{k+1}}\right)-4^{k}f\left(\frac{x}{2^k}\right)∥:\right.\\ \left.p\le k<n\right\}\\ \le \left|2\right|\mathit{\text{max}}\left\{4{|}^k\zeta \left(\frac{x}{2^{k+1}},\frac{-x}{2^{k+1}},\frac{x}{2^{k+1}}\right):\right.\\ \left.p\le k<n\right\}\end{array}$$

for all x∈G and all nonnegative integers n,p with n>p≥0. Letting p=0 and passing the limit n→∞ in the last inequality, we obtain Equation 22. The rest of the proof is similar to the proof of Theorem 5. □

Corollary 4

Let ξ:[0,∞)→[0,∞) be a function satisfying

$$\xi (|2{|}^{-1}t)\le \xi (|2{|}^{-1})\xi (t)(t\ge 0),\xi (|2{|}^{-1})<|4{|}^{-1}.$$

Let κ>0 and f:G→X be a mapping with f(0)=0 and satisfying the inequality

$$\begin{array}{l}∥f(z-x)+f(z-y)-\frac{1}{2}f(x-y)-2f\left(z-\frac{x+y}{2}\right)∥\\ \le \kappa \left(\xi (|x|)\mathrm{.\xi }(|y|)\mathrm{.\xi }(|z|)\right)\end{array}$$

for all x,y,z∈G. Then the limit $\alpha (x):=\underset{n\to \infty }{\text{lim}}{4}^{n}f\left(\frac{x}{2^n}\right)$ exists for all x∈G, and α:G→X is a unique quadratic mapping satisfying

$$\left|\right|f(x)-\alpha (x)\left|\right|\le \frac{\left|2\right|\kappa {\xi }^3(|x|)}{|4{|}^3}$$

for all x∈G.

Proof

Define ζ:G³→[0,∞) by ζ(x,y,z):=κ(ξ(|x|).ξ(|y|).ξ(|z|)). The rest of the proof is similar to the proof of Corollary 3. □

Results and discussion

We linked here four different disciplines, namely, non- Archimedean Banach spaces, functional equations, direct method and fixed point theory. We established the Hyers-Ulam-Rassias stability of the functional Equation 1 in Archimedean Banach spaces by using direct and fixed point methods.

Conclusions

Throughout this paper, using the fixed point and direct method we proved the Hyers-Ulam-Rassias stability of a generalized Apollonius type quadratic functional equation in non-Archimedean Banach spaces.