Nonlinear approximation of an ACQ-functional equation in nan-spaces

Abstract

In this paper, using the fixed point and direct methods, we prove the generalized Hyers-Ulam stability of an additive-cubic-quartic functional equation in NAN-spaces.

Mathematics Subject Classification (2010)

39B52·47H10·26E30·46S10·47S10

1. Introduction and preliminaries

A classical question in the theory of functional equations is the following: "When is it true that a function which approximately satisfies a functional equation must be close to an exact solution of the equation?" If the problem accepts a solution, we say that the equation is stable. The first stability problem concerning group homomorphisms was raised by Ulam [1] in 1940. In the next year, Hyers [2] gave a positive answer to the above question for additive groups under the assumption that the groups are Banach spaces. In 1978, Rassias [3] proved a generalization of the Hyers' theorem for additive mappings. The result of Rassias has provided a lot of influence during the last three decades in the development of a generalization of the Hyers-Ulam stability concept. This new concept is known as generalized Hyers-Ulam stability or Hyers-Ulam-Rassias stability of functional equations (see [4–8]). Furthermore, in 1994, a generalization of the Rassias' theorem was obtained by Găvruta [9] by replacing the bound ε(||x|| ^p + ||y|| ^p ) by a general control function φ(x, y).

The functional equation

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

is called a quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic mapping. In 1983, a generalized Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [10] for mappings f : X → Y, where X is a normed space and Y is a Banach space. In 1984, Cholewa [11] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group and, in 2002, Czerwik [12] proved the generalized 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 [13–32]).

In 1897, Hensel [33] has introduced a normed space that does not have the Archimedean property. It turned out that non-Archimedean spaces have many nice applications (see [34–37]).

Now, we give some definitions and lemmas for the main results in this paper.

A valuation is a function |·| from a field $K$ into [0, ∞) such that, for all $r,s\in K$, the following conditions hold:

• (a) |r| = 0 if and only if r = 0;

• (b) |rs| = |r||s|;

• (c) |r + s| ≤ |r| + |s|.

A field $K$ is called a valued field if $K$ carries a valuation. The usual absolute values of ℝ and $ℂ$ are examples of valuations.

Let us consider a valuation that satisfies a stronger condition than the triangle inequality. If the triangle inequality is replaced by

$$|r+s|\le \mathsf{\text{max}}\left\{\left|r\right|,\left|s\right|\right\}$$

for all $r,s\in K$, then the function |·| is called a non-Archimedean valuation and the field is called a non-Archimedean field. Clearly, |1| = | -1| = 1 and |n| ≤ 1 for all n ∈ ℕ. A trivial example of a non-Archimedean valuation is the function |·| taking everything except for 0 into 1 and |0| = 0.

Definition 1.1. Let X be a vector space over a field $K$ with a non-Archimedean valuation |·|. A function ||·|| : X → [0, ∞) is called a non-Archimedean norm if the following conditions hold:

• (a) ||x|| = 0 if and only if x = 0 for all x ∈ X;

• (b) ||rx|| = |r| ||x|| for all r ∈ K and x ∈ X;

• (c) the strong triangle inequality holds:

  $$\left|\right|x+y\left|\right|\le \mathsf{\text{max}}\left\{\left|\right|x\left|\right|,\left|\right|y\left|\right|\right\}$$

for all x, y ∈ X.

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

Definition 1.2. Let {x _n } be a sequence in a non-Archimedean normed space X.

• (1) The sequence {x _n } is called a Cauchy sequence if, for any ε > 0, there is a positive integer N such that

  $$\left|\right|x_n-x_m\left|\right|\le \epsilon$$

for all n, m ≥ N.

• (2) The sequence {x _n } is said to be convergent if, for any ε > 0, there are a positive integer N and x ∈ X such that

  $$\left|\right|x_n-x\left|\right|\le \epsilon$$

for all n ≥ N. Then, the point x ∈ X is called the limit of the sequence {x _n }, which is denoted by lim_n→∞x _n = x.

• (3) If every Cauchy sequence in X converges, then the non-Archimedean normed space X is called a non-Archimedean Banach space.

Note that ||x _n - x _m || ≤ max{||x_j+1- x _j || : m ≤ j ≤ n - 1} for all m, n ≥ 1 with n > m.

Definition 1.3. 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;

• (c) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X.

Theorem 1.1. [38, 39]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

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

for all nonnegative integers n or there exists a positive integer n ₀ such that

• (a) d(Jⁿx, J ⁿ⁺¹ x) < ∞ for all n ₀ ≥ n ₀ ;

• (b) the sequence {Jⁿx} converges to a fixed point y* of J;

• (c) y* is the unique fixed point of J in the set $Y=\left\{y\in X:d\left(J^{n_0}x,y\right)<\infty \right\}$;

• (d) $d\left(y,y^{*}\right)\le \frac{1}{1-L}d\left(y,Jy\right)$ for all y ∈ Y.

In this paper, using the fixed point and direct methods, we prove the generalized Hyers-Ulam stability of the following functional equation

$$\begin{array}{ll}\hfill 11f\left(x+2y\right)+11f\left(x-2y\right)& =44\left\{f\left(x+y\right)+f\left(x-y\right)\right\}+12f\left(3y\right)\\ -48f\left(2y\right)+60f\left(y\right)-66f\left(x\right)\end{array}$$

(1.1)

in non-Archimedean normed spaces.

2. Non-Archimedean stability of the equation (1.1): a fixed point method-odd case

Using the fixed point alternative approach, we prove the generalized Hyers-Ulam stability of functional Equation (1.1) in non-Archimedean normed spaces for an odd case.

In [40], Lee et al. considered the following quartic functional equation:

$$f\left(2x+y\right)+f\left(2x-y\right)=4\left\{f\left(x+y\right)+f\left(x-y\right)\right\}+24f\left(x\right)-6f\left(y\right)$$

(2.1)

It is easy to show that the function f(x) = x⁴ satisfies the functional Equation (2.1), which is called a quartic functional equation and every solution of the quartic functional equation is said to be a quartic mapping.

One can easily show that an even mapping f : X → Y satisfies (1.1) if and only if the even mapping f : X → Y is a quartic mapping, that is,

$$f\left(2x+y\right)+f\left(2x-y\right)=4\left\{f\left(x+y\right)+f\left(x-y\right)\right\}+24f\left(x\right)-6f\left(y\right)$$

(2.2)

and an odd mapping f : X → Y satisfies (1.1) if and only if the odd mapping f : X → Y is a additive-cubic mapping, that is,

$$f\left(2x+y\right)+f\left(2x-y\right)=4\left\{f\left(x+y\right)+f\left(x-y\right)\right\}-6f\left(x\right)$$

(2.3)

It was shown in [[41], Lemma 2.2] that g(x) = f(2x) - 2f(x) and h(x) = f(2x) - 8f(x) are cubic and additive, respectively, and that $f\left(x\right):=\frac{1}{16}g\left(x\right)-\frac{1}{16}h\left(x\right)$.

For a given mapping f : X → Y, we define

$$\begin{array}{ll}\hfill {\Phi }_f\left(x,y\right)& =11f\left(x+2y\right)+11f\left(x-2y\right)-44\left\{f\left(x+y\right)+f\left(x-y\right)\right\}\\ -12f\left(3y\right)+48f\left(2y\right)-60f\left(y\right)+66f\left(x\right)\end{array}$$

for all x, y ∈ X.

Using the fixed point method, we prove the generalized Hyers-Ulam stability of the functional equation Φ _f (x, y) = 0 in non-Archimedean normed spaces: an odd case.

Throughout this section, let |8| ≠ 1.

Theorem 2.1. Let X be a non-Archimedean normed space and Y a non-Archimedean Banach space. Assume that γ : X² → [0, ∞) is a function such that there exists an L < 1 with

$$\gamma \left(\frac{x}{2},\frac{y}{2}\right)\le \frac{L}{\left|8\right|}\gamma \left(x,y\right)$$

(2.4)

for all x, y ∈ X. If f : X → Y is an odd mapping satisfying

$$\left|\right|{\Phi }_f\left(x,y\right)\left|\right|\le \gamma \left(x,y\right)$$

(2.5)

for all x, y ∈ X, then the limit

$$C\left(x\right):=\underset{n\to \infty }{lim}{8}^n\left(f\left(\frac{x}{2^{n-1}}\right)-2f\left(\frac{x}{2^n}\right)\right)$$

exists for all x ∈ X and defines a unique cubic mapping C : X → Y such that

$$\left|\right|f\left(2x\right)-2f\left(x\right)-C\left(x\right)\left|\right|\le \frac{L}{\left|8\right|-\left|8\right|L}max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}.$$

(2.6)

Proof. Putting x = 0 in (2.5), we have

$$\left|\right|12f\left(3y\right)-48f\left(2y\right)+60f\left(y\right)\left|\right|\le \gamma \left(y,0\right)$$

(2.7)

for all y ∈ X.

Replacing x by 2y in (2.5), we get

$$\left|\right|11f\left(4y\right)-56f\left(3y\right)+114f\left(2y\right)-104f\left(y\right)\left|\right|\le \gamma \left(2y,y\right)$$

(2.8)

for all y ∈ X. By (2.7) and (2.8), we have

$$\begin{array}{ll}\hfill ∥f\left(4y\right)-10f\left(2y\right)+16f\left(y\right)∥& =∥\frac{1}{11}\left[11f\left(4y\right)-56f\left(3y\right)+114f\left(2y\right)-104f\left(y\right)\right]\\ +\frac{14}{33}\left[12f\left(3y\right)-48f\left(2y\right)+60f\left(y\right)\right]∥\\ \le max\left\{\frac{1}{\left|11\right|}\gamma \left(2y,y\right),\left|\frac{14}{33}\right|\gamma \left(y,0\right)\right\}\end{array}$$

(2.9)

for all y ∈ X. Letting $y:=\frac{x}{2}$ and g(x) := f (2x) - 2f(x) for all x ∈ X, we get

$$∥g\left(x\right)-8g\left(\frac{x}{2}\right)∥\le max\left\{\frac{1}{\left|11\right|}\gamma \left(x,\frac{x}{2}\right),\left|\frac{14}{33}\right|\gamma \left(\frac{x}{2},0\right)\right\}.$$

(2.10)

Consider the set

$$S:=\left\{g:X\to Y\right\}$$

and the generalized metric d in S defined by

$$d\left(f,g\right)=\underset{\mu \in \left(0,+\infty \right)}{inf}\left\{\left|\right|g\left(x\right)-h\left(x\right)\left|\right|\le \mu max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\},\forall x\in X\right\},$$

where inf ∅ = +∞. It is easy to show that (S, d) is complete (see [[42], Lemma 2.1]).

Now, we consider a linear mapping J : S → S such that

$$Jg\left(x\right):=8g\left(\frac{x}{2}\right)$$

(2.11)

for all x ∈ X. Let g, h ∈ S be such that d(g, h) = ε. Then we have

$$\left|\right|g\left(x\right)-h\left(x\right)\left|\right|\le \epsilon max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}$$

for all x ∈ X and so

$$\begin{array}{ll}\hfill \left|\right|Jg\left(x\right)-Jh\left(x\right)\left|\right|& =∥8g\left(\frac{x}{2}\right)-8h\left(\frac{x}{2}\right)∥\\ \le \left|8\right|max\left\{\frac{1}{\left|11\right|}\gamma \left(x,\frac{x}{2}\right),\left|\frac{14}{33}\right|\gamma \left(\frac{x}{2},0\right)\right\}\\ \le \left|8\right|\cdot \frac{L}{\left|8\right|}\epsilon max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}\end{array}$$

for all x ∈ X. Thus d(g, h) = ε implies that d(Jg, Jh) ≤ Lε. This means that

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

for all g, h ∈ S. It follows from (2.10) that

$$d\left(g,Jg\right)\le \frac{L}{\left|8\right|}.$$

(2.12)

By Theorem 1.1, there exists a mapping C : X → Y satisfying the following:

• (1) C is a fixed point of J, that is,

  $$\frac{1}{8}C\left(x\right)=C\left(\frac{x}{2}\right)$$

  (2.13)

for all x ∈ X. The mapping C is a unique fixed point of J in the set

$$\Omega =\left\{h\in S:d\left(g,h\right)<\infty \right\}.$$

This implies that C is a unique mapping satisfying (2.13) such that there exists μ ∈ (0, ∞) satisfying

$$\left|\right|g\left(x\right)-C\left(x\right)\left|\right|\le \mu max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}$$

for all x ∈ X.

• (2) d(Jⁿg, C) → 0 as n → ∞. This implies the equality

  $$\underset{n\to \infty }{lim}{8}^{n}g\left(\frac{x}{2^n}\right)=\underset{n\to \infty }{lim}{8}^n\left(f\left(\frac{x}{2^{n-1}}\right)-2f\left(\frac{x}{2^n}\right)\right)=C\left(x\right)$$

for all x ∈ X.

• (3) $d\left(g,C\right)\le \frac{d\left(g,Jg\right)}{1-L}$ with g ∈ Ω, which implies the inequality

  $$d\left(g,C\right)\le \frac{L}{\left|8\right|-\left|8\right|L}.$$

  (2.14)

This implies that the inequality (2.6) holds.

Since Φ _g (x, y) = Φ _f (2x, 2y) - 2Φ _f (x, y), using (2.4) and (2.5), we have

$$\begin{array}{ll}\hfill \left|\right|{\Phi }_C\left(x,y\right)\left|\right|& =\underset{n\to \infty }{lim}|8{|}^n∥{\Phi }_g\left(\frac{x}{2^n},\frac{y}{2^n}\right)∥\\ =\underset{n\to \infty }{lim}|8{|}^n∥{\Phi }_f\left(\frac{x}{2^{n-1}},\frac{y}{2^{n-1}}\right)-2{\Phi }_f\left(\frac{x}{2^n},\frac{y}{2^n}\right)∥\\ \le \underset{n\to \infty }{lim}|8{|}^{n}max\left\{∥{\Phi }_f\left(\frac{x}{2^{n-1}},\frac{y}{2^{n-1}}\right)∥,\left|2\right|∥{\Phi }_f\left(\frac{x}{2^n},\frac{y}{2^n}\right)∥\right\}\\ \le \underset{n\to \infty }{lim}|8{|}^{n}max\left\{\gamma \left(\frac{x}{2^{n-1}},\frac{y}{2^{n-1}}\right),\left|2\right|\gamma \left(\frac{x}{2^n},\frac{y}{2^n}\right)\right\}\\ \le \underset{n\to \infty }{lim}|8{|}^{n}max\left\{\frac{L^{n-1}}{|8{|}^{n-1}}\gamma \left(x,y\right),\frac{\left|2\right|L^n}{|8{|}^n}\gamma \left(x,y\right)\right\}\\ =0\end{array}$$

for all x, y ∈ X and n ≥ 1 and so ||Φ _C (x, y)|| = 0 for all x, y ∈ X. Therefore, the mapping C : X → Y is cubic. This completes the proof.   □

Corollary 2.1. Let θ ≥ 0 and r be a real number with r > 1. Let f : X → Y be an odd mapping satisfying

$$\left|\right|{\Phi }_f\left(x,y\right)\left|\right|\le \theta \left(\left|\right|x|{|}^r+|\left|y\right|{|}^r\right)$$

(2.15)

for all x, y ∈ X. Then the limit$C\left(x\right)=\underset{n\to \infty }{lim}{8}^n\left(f\left(\frac{x}{2^{n-1}}\right)-2f\left(\frac{x}{2^n}\right)\right)$exists for all x ∈ X and C : X → Y is a unique cubic mapping such that

$$\left|\right|f\left(2x\right)-2f\left(x\right)-C\left(x\right)\left|\right|\le \frac{|8{|}^r}{\left|8\right|-|8{|}^{r+1}}max\left\{\frac{\left(|2{|}^r+1\right)\theta \left|\right|x|{|}^r}{\left|11\right|},\left|\frac{14}{33}\right|\theta \left|\right|x|{|}^r\right\}$$

for all x ∈ X.

Proof. The proof follows from Theorem 2.1 if we take

$$\gamma \left(x,y\right)=\theta \left(\left|\right|x|{|}^r+|\left|y\right|{|}^r\right)$$

for all x, y ∈ X. In fact, if we choose L = |8| ^r , then we get the desired result.   □

Theorem 2.2. Let X be a non-Archimedean normed space and Y a non-Archimedean Banach space. Assume that γ : X² → [0, ∞) is a function such that there exists an L < 1 with

$$\gamma \left(2x,2y\right)\le \left|8\right|L\gamma \left(x,y\right)$$

(2.16)

for all x, y ∈ X. If f : X → Y is an odd mapping satisfying (2.5), then the limit

$$C\left(x\right)=\underset{n\to \infty }{lim}\frac{f\left(2^{n+1}x\right)-2f\left(2^{n}x\right)}{8^n}$$

exists for all x ∈ X and defines a unique cubic mapping C : X → Y such that

$$\left|\right|f\left(2x\right)-2f\left(x\right)-C\left(x\right)\left|\right|\le \frac{1}{\left|8\right|-\left|8\right|L}max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}.$$

(2.17)

Proof. Let (S, d) be the generalized metric space defined in the proof of Theorem 2.1. Consider the mapping J : (S, d) → (S, d) such that

$$Jg\left(x\right):=\frac{1}{8}g\left(2x\right)$$

(2.18)

for all x ∈ X.

Proceeding as in the proof of Theorem 2.1, we find that d(g, h) = ε implies that d(Jg, Jh) ≤ Lε. This means that d(Jg, Jh) ≤ Ld(g, h) for all g, h ∈ S.

It follows from (2.10) that

$$∥\frac{g\left(2x\right)}{8}-g\left(x\right)∥\le \frac{1}{\left|8\right|}max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}$$

for all x ∈ X. So

$$d\left(g,Jg\right)\le \frac{1}{\left|8\right|}.$$

(2.19)

By Theorem 1.1, there exists a mapping C : X → Y satisfying the following:

• (1) C is a fixed point of J, that is,

  $$\mathsf{\text{8}}C\mathsf{\text{(}}x\mathsf{\text{) = }}C\mathsf{\text{(2}}x\mathsf{\text{)}}$$

  (2.20)

for all x ∈ X. The mapping C is a unique fixed point of J in the set

$$\Omega =\left\{h\in S:d\left(g,h\right)<\infty \right\}.$$

This implies that C is a unique mapping satisfying (2.20) such that there exists μ ∈ (0, ∞) satisfying

$$\left|\right|g\left(x\right)-C\left(x\right)\left|\right|\le \mu max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}$$

for all x ∈ X.

• (2) d(Jⁿg, C) → 0 as n → ∞. This implies the equality

  $$\underset{n\to \infty }{lim}\frac{g\left(2^{n}x\right)}{8^n}=\underset{n\to \infty }{lim}\frac{f\left(2^{n+1}x\right)-2f\left(2^{n}x\right)}{8^n}=C\left(x\right)$$

for all x ∈ X.

• (3) $d\left(g,C\right)\le \frac{d\left(g,Jg\right)}{1-L}$ with g ∈ Ω, which implies the inequality

  $$d\left(g,C\right)\le \frac{1}{\left|8\right|-\left|8\right|L}.$$

  (2.21)

This implies that the inequality (2.17) holds. The rest of the proof is similar to the proof of Theorem 2.1.   □

Corollary 2.2. Let θ ≥ 0 and r be a real number with 0 < r < 1. Let f : X → Y be an odd mapping satisfying (2.15). Then the limit$C\left(x\right)=\underset{n\to \infty }{lim}\frac{f\left(2^{n+1}x\right)-2f\left(2^{n}x\right)}{8^n}$exists for all x ∈ X and C : X → Y is a unique cubic mapping such that

$$\left|\right|f\left(2x\right)-2f\left(x\right)-C\left(x\right)\left|\right|\le \frac{|8{|}^r}{|8{|}^{r+1}-|8{|}^2}max\left\{\frac{\left(|2{|}^r+1\right)\theta \left|\right|x|{|}^r}{\left|11\right|},\left|\frac{14}{33}\right|\theta \left|\right|x|{|}^r\right\}$$

for all x ∈ X.

Proof. The proof follows from Theorem 2.2 if we take

$$\gamma \left(x,y\right)=\theta \left(\left|\right|x|{|}^r+|\left|y\right|{|}^r\right)$$

for all x, y ∈ X. In fact, if we choose L = |8|^1-r, then we get the desired result.   □

Theorem 2.3. Let X be a non-Archimedean normed space and Y a non-Archimedean Banach space. Assume that γ : X² → [0, ∞) is a function such that there exists an L < 1 with

$$\gamma \left(\frac{x}{2},\frac{y}{2}\right)\le \frac{L}{\left|2\right|}\gamma \left(x,y\right)$$

(2.22)

for all x, y ∈ X. If f : X → Y is an odd mapping satisfying (2.5), then the limit

$$A\left(x\right):=\underset{n\to \infty }{lim}{2}^n\left(f\left(\frac{x}{2^{n-1}}\right)-8f\left(\frac{x}{2^n}\right)\right)$$

exists for all x ∈ X and defines a unique additive mapping A : X → Y such that

$$\left|\right|f\left(2x\right)-8f\left(x\right)-A\left(x\right)\left|\right|\le \frac{L}{\left|2\right|-\left|2\right|L}max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}.$$

(2.23)

Proof. Let (S, d) be the generalized metric space defined in the proof of Theorem 2.1.

Letting $y:=\frac{x}{2}$ and h(x) := f (2x) - 8f (x) for all x ∈ X in (2.9), we get

$$∥h\left(x\right)-2h\left(\frac{x}{2}\right)∥\le max\left\{\frac{1}{\left|11\right|}\gamma \left(x,\frac{x}{2}\right),\left|\frac{14}{33}\right|\gamma \left(\frac{x}{2},0\right)\right\}.$$

(2.24)

Now, we consider a linear mapping J : S → S such that

$$Jh\left(x\right):=2h\left(\frac{x}{2}\right)$$

(2.25)

for all x ∈ X. Let g, h ∈ S be such that d(g, h) = ε. Then we have

$$∥g\left(x\right)-h\left(x\right)∥\le \epsilon max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}$$

for all x ∈ X and so

$$\begin{array}{ll}\hfill \parallel Jg\left(x\right)-Jh\left(x\right)\parallel =∥2g\left(\frac{x}{2}\right)-2h\left(\frac{x}{2}\right)∥& \le \left|2\right|max\left\{\frac{1}{\left|11\right|}\gamma \left(x,\frac{x}{2}\right),\left|\frac{14}{33}\right|\gamma \left(\frac{x}{2},0\right)\right\}\\ \le \left|2\right|\cdot \frac{L}{\left|2\right|}\epsilon max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}\end{array}$$

for all x ∈ X. Thus, d(g, h) = ε implies that d(Jg, Jh) ≤ Lε. This means that

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

for all g, h ∈ S. It follows from (2.24) that

$$d\left(g,Jg\right)\le \frac{L}{\left|2\right|}.$$

(2.26)

By Theorem 1.1, there exists a mapping A : X → Y satisfying the following:

• (1) A is a fixed point of J, that is,

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

  (2.27)

for all x ∈ X. The mapping A is a unique fixed point of J in the set

$$\Omega =\left\{h\in S:d\left(g,h\right)<\infty \right\}.$$

This implies that A is a unique mapping satisfying (2.27) such that there exists μ ∈ (0, ∞) satisfying

$$\parallel h\left(x\right)-A\left(x\right)\parallel \le \mu max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}$$

for all x ∈ X.

• (2) d(Jⁿh, A) → 0 as n → ∞. This implies the equality

  $$\underset{n\to \infty }{lim}{2}^{n}h\left(\frac{x}{2^n}\right)=\underset{n\to \infty }{lim}{2}^n\left(f\left(\frac{x}{2^{n-1}}\right)-8f\left(\frac{x}{2^n}\right)\right)=A\left(x\right)$$

for all x ∈ X.

• (3) $d\left(h,A\right)\le \frac{d\left(h,Jh\right)}{1-L}$ with h ∈ Ω, which implies the inequality

  $$d\left(h,A\right)\le \frac{L}{\left|2\right|-\left|2\right|L}.$$

  (2.28)

This implies that the inequality (2.23) holds. The rest of the proof is similar to the proof of Theorem 2.1.   □

Corollary 2.3. Let θ ≥ 0 and r be a real number with r > 1. Let f : X → Y be an odd mapping satisfying (2.15). Then, the limit$A\left(x\right)=\underset{n\to \infty }{lim}{2}^n\left(f\left(\frac{x}{2^{n-1}}\right)-8f\left(\frac{x}{2^n}\right)\right)$exists for all x ∈ X and A : X → Y is a unique additive mapping such that

$$\parallel f(2x)-8f(x)-A(x)\parallel \le \frac{{\left|2\right|}^r}{\left|2\right|-{\left|2\right|}^{r+1}}\max \left\{\frac{{(|2|}^r+1)\theta \parallel x{\parallel }^r}{\left|11\right|}\mathrm{,}\left|\frac{14}{33}\right|\theta \parallel x{\parallel }^r\right\}$$

for all x ∈ X.

Proof. The proof follows from Theorem 2.3 if we take

$$\gamma \left(x,y\right)=\theta \left(\parallel x{\parallel }^r+\parallel y{\parallel }^r\right)$$

for all x, y ∈ X. In fact, if we choose L = |2| ^r , then we get the desired result.   □

Theorem 2.4. Let X be a non-Archimedean normed space and Y a non-Archimedean Banach space. Assume that γ : X² → [0, ∞) is a function such that there exists an L < 1 with

$$\gamma \left(2x,2y\right)\le \left|2\right|L\gamma \left(x,y\right)$$

(2.29)

for all x, y ∈ X. If f : X → Y is an odd mapping satisfying (2.5), then the limit

$$A(x)=\underset{n\to \infty }{\lim }\frac{f{(2}^{n+1}x)-8f{(2}^{n}x)}{2^n}$$

exists for all x ∈ X and defines a unique additive mapping A : X → Y such that

$$\parallel f\left(2x\right)-8f\left(x\right)-A\left(x\right)\parallel \le \frac{1}{\left|2\right|-\left|2\right|L}max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}.$$

(2.30)

Proof. Let (S, d) be the generalized metric space defined in the proof of Theorem 2.1. Consider the mapping J : (S, d) → (S, d) such that

$$Jg\left(x\right):=\frac{1}{2}g\left(2x\right)$$

(2.31)

for all x ∈ X. By (2.24), we obtain

$$∥\frac{h\left(2x\right)}{2}-g\left(x\right)∥\le \frac{1}{\left|2\right|}max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}$$

for all x ∈ X. So

$$d\left(g,Jg\right)\le \frac{1}{\left|2\right|}.$$

(2.32)

By Theorem 1.1, there exists a mapping A : X → Y satisfying the following:

• (1) A is a fixed point of J, that is,

  $$2A\left(x\right)=A\left(2x\right)$$

  (2.33)

for all x ∈ X. The mapping A is a unique fixed point of J in the set

$$\Omega =\left\{h\in S:d\left(g,h\right)<\infty \right\}.$$

This implies that A is a unique mapping satisfying (2.33) such that there exists μ ∈ (0, ∞) satisfying

$$\parallel h\left(x\right)-A\left(x\right)\parallel \le \mu max\left\{\frac{1}{\left|11\right|}\gamma \left(2x,x\right),\left|\frac{14}{33}\right|\gamma \left(x,0\right)\right\}$$

for all x ∈ X.

• (2) d(Jⁿh, A) → 0 as n → ∞. This implies the equality

  $$\underset{n\to \infty }{\lim }{2}^{n}h\left(\frac{x}{2^n}\right)=\underset{n\to \infty }{\lim }{2}^n\left(f\left(\frac{x}{2^{n-1}}\right)-8f\left(\frac{x}{2^n}\right)\right)=A(x)$$

for all x ∈ X.

• (3) $d\left(h,A\right)\le \frac{d\left(h,Jh\right)}{1-L}$ with h ∈ Ω, which implies the inequality

  $$d\left(h,A\right)\le \frac{1}{\left|2\right|-\left|2\right|L}.$$

  (2.34)

This implies that the inequality (2.30) holds. The rest of the proof is similar to the proof of Theorem 2.1.   □

Corollary 2.4. Let θ ≥ 0 and r be a real number with 0 < r < 1. Let f : X → Y be an odd mapping satisfying (2.15). Then, the limit$A(x)={\lim }_{n\to \infty }\frac{f{(2}^{n+1}x)-8f{(2}^{n}x)}{2^n}$exists for all x ∈ X and A : X → Y is a unique additive mapping such that

$$\parallel f(2x)-8f(x)-A(x)\parallel \le \frac{1}{\left|2\right|-{\left|2\right|}^{r+2}}\max \left\{\frac{{(|2|}^r+1)\theta \parallel x{\parallel }^r}{\left|11\right|}\mathrm{,}\left|\frac{14}{33}\right|\theta \parallel x{\parallel }^r\right\}$$

for all x ∈ X.

Proof. The proof follows from Theorem 2.4 if we take

$$\gamma \left(x,y\right)=\theta \left(\parallel x{\parallel }^r+\parallel y{\parallel }^r\right)$$

for all x, y ∈ X. In fact, if we choose L = |2|^{r + 1}, then we get the desired result.   □

3. Non-Archimedean stability of the equation (1.1): a fixed point method-even case

Using the fixed point method, we prove the generalized Hyers-Ulam stability of the functional Equation (1.1) in non-Archimedean normed spaces for an even case. Throughout this section, let |16| ≠ 1.

Theorem 3.1. Let X be a non-Archimedean normed space and Y a non-Archimedean Banach space. Assume that γ : X² → [0, ∞) is a function such that there exists an L < 1 with

$$\gamma \left(2x,2y\right)\le \left|16\right|L\gamma \left(x,y\right)$$

(3.1)

for all x, y ∈ X. If f : X → Y is an even mapping with f(0) = 0 satisfying (2.5), then the limit

$$Q(x):=\underset{n\to \infty }{\lim }\frac{f{(2}^{n}x)}{{16}^n}$$

exists for all x ∈ X and defines a unique quartic mapping Q : X → Y such that

$$\parallel f\left(x\right)-Q\left(x\right)\parallel \le \frac{1}{\left|16\right|-\left|16\right|L}max\left\{\frac{1}{\left|22\right|}\gamma \left(0,x\right),\left|\frac{6}{11}\right|\gamma \left(x,x\right)\right\}.$$

(3.2)

Proof. Putting x = 0 in (2.5), we have

$$∥12f\left(3y\right)-70f\left(2y\right)+148f\left(y\right)∥\le \gamma \left(0,y\right)$$

(3.3)

for all y ∈ X.

Substituting x = y in (2.5), we get

$$∥f\left(3y\right)-4f\left(2y\right)-17f\left(y\right)∥\le \gamma \left(y,y\right)$$

(3.4)

for all y ∈ X. By (3.3) and (3.4), we have

$$\begin{array}{ll}\hfill ∥f\left(2y\right)-16f\left(y\right)∥& =∥\frac{-1}{22}\left[12f\left(3y\right)-70f\left(2y\right)+148f\left(y\right)\right]+\frac{6}{11}\left[f\left(3y\right)-4f\left(2y\right)-17f\left(y\right)\right]∥\\ \le max\left\{\frac{1}{\left|22\right|}\gamma \left(0,y\right),\left|\frac{6}{11}\right|\gamma \left(y,y\right)\right\}\end{array}$$

(3.5)

for all y ∈ X. Consider the set

$$S:=\left\{g:X\to Y,g\left(0\right)=0\right\}$$

and the generalized metric d in S defined by

$$d\left(f,g\right)=\underset{\mu \in \left(0,+\infty \right)}{inf}\left\{\parallel g\left(x\right)-h\left(x\right)\parallel \le \mu max\left\{\frac{1}{\left|22\right|}\gamma \left(0,x\right),\left|\frac{6}{11}\right|\gamma \left(x,x\right)\right\},\forall x\in X\right\}$$

where inf ∅ = +∞. It is easy to show that (S, d) is complete (see [[42], Lemma 2.1]).

Now, we consider a linear mapping J : S → S such that

$$Jg\left(x\right):=\frac{1}{16}g\left(2x\right)$$

(3.6)

for all x ∈ X. It follows from (3.5) that

$$d\left(f,Jf\right)\le \frac{1}{\left|16\right|}.$$

(3.7)

By Theorem 1.1, there exists a mapping Q : X → Y satisfying the following:

• (1) Q is a fixed point of J, that is,

  $$16Q\left(x\right)=Q\left(2x\right)$$

  (3.8)

for all x ∈ X. The mapping Q is a unique fixed point of J in the set

$$\Omega =\left\{h\in S:d\left(g,h\right)<\infty \right\}.$$

This implies that Q is a unique mapping satisfying (3.8) such that there exists μ ∈ (0, ∞) satisfying

$$\parallel f\left(x\right)-Q\left(x\right)\parallel \le \mu max\left\{\frac{1}{\left|22\right|}\gamma \left(0,x\right),\left|\frac{6}{11}\right|\gamma \left(x,x\right)\right\}$$

for all x ∈ X.

• (2) d(Jⁿf, Q) → 0 as n → ∞. This implies the equality

  $$\underset{n\to \infty }{\lim }\frac{f{(2}^{n}x)}{{16}^n}=Q(x)$$

for all x ∈ X.

• (3) $d\left(f,Q\right)\le \frac{d\left(f,Jf\right)}{1-L}$ with f ∈ Ω, which implies the inequality

  $$d\left(f,C\right)\le \frac{1}{\left|16\right|-\left|16\right|L}.$$

  (3.9)

This implies that the inequality (3.2) holds. The rest of the proof is similar to the proof of Theorem 2.1.   □

Corollary 3.1. Let θ ≥ 0 and r be a real number with r > 1. Let f : X → Y be an even mapping with f(0) = 0 satisfying (2.15). Then, the limit$Q(x)={\lim }_{n\to \infty }\frac{f{(2}^{n}x)}{{16}^n}$exists for all x ∈ X and Q : X → Y is a unique quartic mapping such that

$$\parallel f\left(x\right)-Q\left(x\right)\parallel \le \frac{1}{\left|16\right|-{\left|16\right|}^{r+1}}max\left\{\frac{\theta \parallel x{\parallel }^r}{\left|22\right|},2\left|\frac{6}{11}\right|\theta \parallel x{\parallel }^r\right\}$$

for all x ∈ X.

Proof. The proof follows from Theorem 3.1 if we take

$$\gamma \left(x,y\right)=\theta \left(\parallel x{\parallel }^r+\parallel y{\parallel }^r\right)$$

for all x, y ∈ X. In fact, if we choose L = |16| ^r , then we get the desired result.   □

Similarly, we can obtain the following. We will omit the proof.

Theorem 3.2. Let X be a non-Archimedean normed space and Y a non-Archimedean Banach space. Assume that γ : X² → [0, ∞) is a function such that there exists an L < 1 with

$$\gamma \left(\frac{x}{2},\frac{y}{2}\right)\le \frac{L}{\left|16\right|}\gamma \left(x,y\right)$$

(3.10)

for all x, y ∈ X. If f : X → Y is an even mapping with f(0) = 0 satisfying (2.5), then the limit

$$Q\left(x\right):=\underset{n\to \infty }{lim}16^{n}f\left(\frac{x}{2^n}\right)$$

exists for all x ∈ X and defines a unique quartic mapping Q : X → Y such that

$$\parallel f\left(x\right)-Q\left(x\right)\parallel \le \frac{L}{\left|16\right|-\left|16\right|L}max\left\{\frac{1}{\left|22\right|}\gamma \left(0,x\right),\left|\frac{6}{11}\right|\gamma \left(x,x\right)\right\}.$$

(3.11)

Corollary 3.2. Let θ ≥ 0 and r be a real number with 0 < r < 1. Let f : X → Y be an even mapping with f(0) = 0 satisfying (2.15). Then, the limit$Q\left(x\right)=\underset{n\to \infty }{lim}16^{n}f\left(\frac{x}{2^n}\right)$exists for all x ∈ X and Q : X → Y is a unique quartic mapping such that

$$\parallel f\left(x\right)-Q\left(x\right)\parallel \le \frac{\left|16\right|}{{\left|16\right|}^{r+1}-{\left|16\right|}^2}max\left\{\frac{\theta \parallel x{\parallel }^r}{\left|22\right|},2\left|\frac{6}{11}\right|\theta \parallel x{\parallel }^r\right\}$$

for all x ∈ X.

Proof. The proof follows from Theorem 3.2 if we take

$$\gamma \left(x,y\right)=\theta \left(\parallel x{\parallel }^r+\parallel y{\parallel }^r\right)$$

for all x, y ∈ X. In fact, if we choose L = |16|^1-r, then we get the desired result.   □

4. Non-Archimedean stability of Equation (1.1): a direct method-odd case

Throughout this section, using direct method, we prove the generalized Hyers-Ulam stability of the functional Equation (1.1) in non-Archimedean spaces for an odd case.

Theorem 4.1. Let G be an additive semigroup and X a complete non-Archimedean space. Assume that φ : G² → [0, +∞) is a function such that

$$\underset{n\to \infty }{lim}|8{|}^n\phi \left(\frac{x}{2^n},\frac{y}{2^n}\right)=0$$

(4.1)

for all x, y ∈ G. Let for all x ∈ G

$$\Phi \left(x\right)=\underset{n\to \infty }{lim}max\left\{{\left|8\right|}^{k+1}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^k},\frac{x}{2^{k+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{k+1}},0\right)\right\};0\le k<n\right\}$$

(4.2)

exist. Suppose that f : G → X is an odd mapping satisfying the inequality

$${∥{\Phi }_f\left(x,y\right)∥}_X\le \phi \left(x,y\right)$$

(4.3)

for all x, y ∈ G. Then the limit

$$C\left(x\right):=\underset{n\to \infty }{lim}{8}^n\left(f\left(\frac{x}{2^{n-1}}\right)-2f\left(\frac{x}{2^n}\right)\right)$$

exists for all x ∈ G and C : G → X is a cubic mapping satisfying

$$\left|\right|f\left(2x\right)-2f\left(x\right)-C\left(x\right)|{|}_X\le \frac{1}{\left|8\right|}\Phi \left(x\right)$$

(4.4)

for all x ∈ G. Moreover, if

$$\underset{j\to \infty }{lim}\underset{n\to \infty }{lim}max\left\{{\left|8\right|}^{k+1}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^k},\frac{x}{2^{k+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{k+1}},0\right)\right\};j\le k<n+j\right\}=0,$$

then C is the unique mapping satisfying (4.4).

Proof. Proceeding as in the proof of Theorem 2.1, we obtain

$${∥f\left(4y\right)-10f\left(2y\right)+16f\left(y\right)∥}_X\le max\left\{\frac{1}{\left|11\right|}\phi \left(2y,y\right),\left|\frac{14}{33}\right|\phi \left(y,0\right)\right\}$$

(4.5)

for all y ∈ X. Letting $y:=\frac{x}{2}$ and g(x) := f(2x) - 2f(x) for all x ∈ X, we get

$${∥g\left(x\right)-8g\left(\frac{x}{2}\right)∥}_X\le max\left\{\frac{1}{\left|11\right|}\phi \left(x,\frac{x}{2}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2},0\right)\right\}.$$

(4.6)

Replacing x by $\frac{x}{2^n}$ in (4.6), we get

$${∥8^{n}g\left(\frac{x}{2^n}\right)-8^{n+1}g\left(\frac{x}{2^{n+1}}\right)∥}_X\le {\left|8\right|}^{n}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^n},\frac{x}{2^{n+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{n+1}},0\right)\right\}.$$

(4.7)

It follows from (4.1) and (4.7) that the sequence ${\left\{8^{n}g\left(\frac{x}{2^n}\right)\right\}}_{n=1}^{\infty }$ is a Cauchy sequence. Since X is complete, so ${\left\{8^{n}g\left(\frac{x}{2^n}\right)\right\}}_{n=1}^{\infty }$ is convergent. Set

$$C\left(x\right):=\underset{n\to \infty }{lim}{8}^{n}g\left(\frac{x}{2^n}\right)=\underset{n\to \infty }{lim}{8}^n\left(f\left(\frac{x}{2^{n-1}}\right)-2f\left(\frac{x}{2^n}\right)\right).$$

Using induction, we see that

$$\begin{array}{c}{∥8^{n}g\left(\frac{x}{2^n}\right)-g\left(x\right)∥}_X\\ \le \frac{1}{\left|8\right|}max\left\{{\left|8\right|}^{k+1}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^k},\frac{x}{2^{k+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{k+1}},0\right)\right\};0\le k<n\right\}.\end{array}$$

(4.8)

By taking n to approach infinity in (4.8), one obtains (4.4). If L is another mapping satisfying (4.4), then, for x ∈ G, we get

$$\begin{array}{c}\parallel C\left(x\right)-L\left(x\right){\parallel }_X\\ =\underset{j\to \infty }{lim}{∥8^{j}L\left(\frac{x}{2^j}\right)-8^{j}C\left(\frac{x}{2^j}\right)∥}_X\\ =\underset{j\to \infty }{lim}{∥8^{j}L\left(\frac{x}{2^j}\right)\pm 8^{j}g\left(\frac{x}{2^j}\right)-8^{j}C\left(\frac{x}{2^j}\right)∥}_X\\ \le \underset{j\to \infty }{lim}max\left\{{∥8^j\left[L\left(\frac{x}{2^j}\right)-g\left(\frac{x}{2^j}\right)\right]∥}_X,{∥8^j\left[g\left(\frac{x}{2^j}\right)-C\left(\frac{x}{2^j}\right)\right]∥}_X\right\}\\ \le \frac{1}{\left|8\right|}\underset{j\to \infty }{lim}\underset{n\to \infty }{lim}max\left\{{\left|8\right|}^{k+1}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^k},\frac{x}{2^{k+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{k+1}},0\right)\right\};j\le k<n+j\right\}\\ =0.\end{array}$$

Therefore, L = C. This completes the proof.   □

Corollary 4.1. Let ξ : [0, ∞) → [0, ∞) be a function satisfying

$$\xi \left(\frac{t}{\left|2\right|}\right)\le \xi \left(\frac{1}{\left|2\right|}\right)\xi \left(t\right),\xi \left(\frac{1}{\left|2\right|}\right)<\frac{1}{\left|8\right|}$$

for all t ≥ 0. Let δ > 0 and f : G → X be an odd mapping satisfying the inequality

$${∥{\Phi }_f\left(x,y\right)∥}_X\le \delta \left(\xi \left(\left|x\right|\right)+\xi \left(\left|y\right|\right)\right)$$

(4.9)

for all x, y ∈ G. Then the limit$C\left(x\right)=\underset{n\to \infty }{lim}{8}^n\left(f\left(\frac{x}{2^{n-1}}\right)-2f\left(\frac{x}{2^n}\right)\right)$exists for all x ∈ G and C : G → X is a unique cubic mapping such that

$$\left|\right|f\left(2x\right)-2f\left(x\right)-C\left(x\right)|{|}_X\le max\left\{\frac{1}{\left|11\right|}\delta \xi \left(\left|x\right|\right)\left(1+\frac{1}{\left|8\right|}\right),\left|\frac{7}{132}\right|\xi \left(\left|x\right|\right)\right\}$$

for all x ∈ G.

Proof. Defining φ : G² → [0, ∞) by φ(x, y) := δ(ξ(|x|) + ξ(|y|)). Since $\left|8\right|\xi \left(\frac{1}{\left|2\right|}\right)<1$, we have

$$\underset{n\to \infty }{lim}|8{|}^n\phi \left(\frac{x}{2^n},\frac{y}{2^n}\right)\le \underset{n\to \infty }{lim}{\left[\left|8\right|\xi \left(\frac{1}{\left|2\right|}\right)\right]}^n\phi \left(x,y\right)=0$$

for all x, y ∈ G. Also for all x ∈ G

$$\begin{array}{ll}\hfill \Phi \left(x\right)& =\underset{n\to \infty }{lim}max\left\{{\left|8\right|}^{k+1}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^k},\frac{x}{2^{k+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{k+1}},0\right)\right\};0\le k<n\right\}\\ =\left|8\right|max\left\{\frac{1}{\left|11\right|}\phi \left(x,\frac{x}{2}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2},0\right)\right\}\\ =\left|8\right|max\left\{\frac{1}{\left|11\right|}\delta \xi \left(\left|x\right|\right)\left(1+\frac{1}{\left|8\right|}\right),\left|\frac{7}{132}\right|\xi \left(\left|x\right|\right)\right\}\end{array}$$

exists for all x ∈ G. On the other hand,

$$\begin{array}{c}\underset{j\to \infty }{lim}\underset{n\to \infty }{lim}max\left\{{\left|8\right|}^{k+1}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^k},\frac{x}{2^{k+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{k+1}},0\right)\right\};j\le k<n+j\right\}\\ \underset{j\to \infty }{lim}|8{|}^{j+1}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^j},\frac{x}{2^{j+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{j+1}},0\right)\right\}\\ =0.\end{array}$$

Applying Theorem 4.1, we get the desired result.   □

Theorem 4.2. Let G be an additive semigroup and X a complete non-Archimedean space. Assume that φ : G² → [0, +∞) is a function such that

$$\underset{n\to \infty }{\lim }\frac{\phi {(2}^{n}x{\mathrm{,2}}^{n}y)}{{\left|8\right|}^n}=0$$

(4.10)

for all x, y ∈ G. Let for each x ∈ G

$$\Phi (x)=\underset{n\to \infty }{\lim }\max \left\{\frac{1}{{\left|8\right|}^{k+1}}\max \left\{\frac{1}{\left|11\right|}\phi {(2}^{k+1}x{\mathrm{,2}}^{k}x)\mathrm{,}\left|\frac{14}{33}\right|\phi {(2}^{k}x\mathrm{,0})\right\};0\le k<n\right\}$$

(4.11)

exist. Suppose that f : G → X is an odd mapping satisfying the inequality (4.3). Then the limit

$$C(x):=\underset{n\to \infty }{\lim }\frac{f{(2}^{n+1}x)-2f{(2}^{n}x)}{8^n}$$

exists for all x ∈ G and C : G → X is a cubic mapping satisfying

$$\left|\right|f\left(2x\right)-2f\left(x\right)-C\left(x\right)|{|}_X\le \Phi \left(x\right)$$

(4.12)

for all x ∈ G. Moreover, if

$$\underset{j\to \infty }{\lim }\underset{n\to \infty }{\lim }\max \left\{\frac{1}{{\left|8\right|}^{k+1}}\max \left\{\frac{1}{\left|11\right|}\phi {(2}^{k+1}x{\mathrm{,2}}^{k}x)\mathrm{,}\left|\frac{14}{33}\right|\phi {(2}^{k}x\mathrm{,0})\right\};j\le k<n+j\right\}=\mathrm{0,}$$

then C is the unique mapping satisfying (4.12).

Proof. It follows from (4.5) that

$${∥\frac{g\left(2x\right)}{8}-g\left(x\right)∥}_X\le \frac{1}{\left|8\right|}max\left\{\frac{1}{\left|11\right|}\phi \left(2x,x\right),\left|\frac{14}{33}\right|\phi \left(x,0\right)\right\}$$

(4.13)

for all x ∈ G. Replacing x by 2 ⁿx in (4.13), we get

$${\Vert \frac{g{(2}^{n+1}x)}{8^{n+1}}-\frac{g{(2}^{n}x)}{8^n}\Vert }_X\le \frac{1}{{\left|8\right|}^{n+1}}\max \left\{\frac{1}{\left|11\right|}\phi {(2}^{n+1}x{\mathrm{,2}}^{n}x)\mathrm{,}\left|\frac{14}{33}\right|\phi {(2}^{n}x\mathrm{,0})\right\}\mathrm{.}$$

(4.14)

It follows from (4.10) and (4.14) that the sequence ${\left\{\frac{g{(2}^{n}x)}{8^n}\right\}}_{n=1}^{\infty }$ is a Cauchy sequence. Since X is complete, ${\left\{\frac{g{(2}^{n}x)}{8^n}\right\}}_{n=1}^{\infty }$ is convergent. It follows from (4.14) that

$$\begin{array}{c}{\Vert \frac{g{(2}^{p}x)}{8^p}-\frac{g{(2}^{q}x)}{8^q}\Vert }_X={\Vert {\displaystyle \sum _{k=p}^{q-1}\frac{g{(2}^{k+1}x)}{8^{k+1}}}-\frac{g{(2}^{k}x)}{8^k}\Vert }_X\\ \le \max \left\{{\Vert \frac{g{(2}^{k+1}x)}{8^{k+1}}-\frac{g{(2}^{k}x)}{8^k}\Vert }_X;p\le k<q-1\right\}\\ \le \max \left\{\frac{1}{{\left|8\right|}^{k+1}}\max \left\{\frac{1}{\left|11\right|}\phi {(2}^{k+1}x{\mathrm{,2}}^{k}x)\mathrm{,}\left|\frac{14}{33}\right|\phi {(2}^{k}x\mathrm{,0})\right\};p\le k<q-1\right\}\end{array}$$

(4.15)

for all x ∈ G and all non-negative integers q, p with q > p ≥ 0. Letting p = 0 and passing the limit q → ∞ in the last inequality, we obtain (4.12).

The rest of the proof is similar to the proof of Theorem 4.1.   □

Corollary 4.2. Let ξ : [0, ∞) → [0, ∞) be a function satisfying

$$\xi \left(\left|2\right|t\right)\le \xi \left(\left|2\right|\right)\xi \left(t\right),\xi \left(\left|2\right|\right)<\left|8\right|$$

for all t ≥ 0. Let δ > 0 and f : G → X be a mapping satisfying the inequality (4.9). Then the limit$C(x)={\lim }_{n\to \infty }\frac{f{(2}^{n+1}x)-2f{(2}^{n}x)}{8^n}$exists for all x ∈ G and C : G → X is a unique cubic mapping such that

$$\left|\right|f\left(2x\right)-2f\left(x\right)-C\left(x\right)|{|}_X\le \frac{1}{\left|8\right|}max\left\{\frac{1+\left|8\right|}{\left|11\right|}\delta \xi \left(\left|x\right|\right),\left|\frac{14}{33}\right|\delta \xi \left(\left|x\right|\right)\right\}$$

(4.16)

for all x ∈ G.

Proof. Define φ : G² → [0, ∞) by φ(x, y) := δ(ξ(|x|) + ξ(|y|)). Proceeding as in the proof of Corollary 4.1, we have

$$\underset{n\to \infty }{\lim }\frac{\phi {(2}^{n}x{\mathrm{,2}}^{n}y)}{{\left|8\right|}^n}=0$$

for all x, y ∈ G. Also

$$\begin{array}{c}\Phi (x)=li{m}_{n\to \infty }\max \left\{\frac{1}{{\left|8\right|}^{k+1}}\max \left\{\frac{1}{\left|11\right|}\phi {(2}^{k+1}x{\mathrm{,2}}^{k}x)\mathrm{,}\left|\frac{14}{33}\right|\phi {(2}^{k}x\mathrm{,0})\right\};0\le k<n\right\}\\ =\frac{1}{\left|8\right|}\max \left\{\frac{1}{\left|11\right|}\phi (2x\mathrm{,}x)\mathrm{,}\left|\frac{14}{33}\right|\phi (x\mathrm{,0})\right\}\\ \le \frac{1}{\left|8\right|}\max \left\{\frac{1+\left|8\right|}{\left|11\right|}\delta \xi (|x|)\mathrm{,}\left|\frac{14}{33}\right|\delta \xi (|x|)\right\}\end{array}$$

exists for all x ∈ G. Applying Theorem 4.2, we get the desired result.   □

Theorem 4.3. Let G be an additive semigroup and X a complete non-Archimedean space. Assume that φ : G² → [0, +∞) is a function such that

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

(4.17)

for all x, y ∈ G. Let for all x ∈ G

$$\Phi \left(x\right)=\underset{n\to \infty }{lim}max\left\{{\left|2\right|}^{k}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^k},\frac{x}{2^{k+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{k+1}},0\right)\right\};0\le k<n\right\}$$

(4.18)

exist. Suppose that f : G → X is an odd mapping satisfying the inequality (4.3). Then the limit

$$A\left(x\right):=\underset{n\to \infty }{lim}{2}^n\left(f\left(\frac{x}{2^{n-1}}\right)-8f\left(\frac{x}{2^n}\right)\right)$$

exists for all x ∈ G and A : G → X is an additive mapping satisfying

$$\left|\right|f\left(2x\right)-8f\left(x\right)-A\left(x\right)|{|}_X\le \Phi \left(x\right)$$

(4.19)

for all x ∈ G. Moreover, if

$$\underset{j\to \infty }{lim}\underset{n\to \infty }{lim}max\left\{{\left|2\right|}^{k}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^k},\frac{x}{2^{k+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{k+1}},0\right)\right\};j\le k<n+j\right\}=0,$$

then A is the unique mapping satisfying (4.19).

Proof. Letting $y:=\frac{x}{2}$ and h(x) := f(2x) - 8f(x) for all x ∈ G in (4.5), we get

$${∥h\left(x\right)-2h\left(\frac{x}{2}\right)∥}_X\le max\left\{\frac{1}{\left|11\right|}\phi \left(x,\frac{x}{2}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2},0\right)\right\}.$$

(4.20)

Replacing x by $\frac{x}{2^n}$ in (4.20), we obtain

$$\begin{array}{c}{∥2^{n}h\left(\frac{x}{2^n}\right)-2^{n+1}h\left(\frac{x}{2^{n+1}}\right)∥}_X\\ \le |2{|}^{n}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^n},\frac{x}{2^{n+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{n+1}},0\right)\right\}.\end{array}$$

(4.21)

Using induction, one can easily show that

$$\begin{array}{c}{∥2^{n}h\left(\frac{x}{2^n}\right)-h\left(x\right)∥}_X\\ \le max\left\{{\left|2\right|}^{k}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^k},\frac{x}{2^{k+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{k+1}},0\right)\right\};0\le k<n\right\}.\end{array}$$

(4.22)

The rest of the proof is similar to the proof of Theorem 4.1.   □

Corollary 4.3. Let ξ : [0, ∞) → [0, ∞) be a function satisfying

$$\xi \left(\frac{t}{\left|2\right|}\right)\le \xi \left(\frac{1}{\left|2\right|}\right)\xi \left(t\right),\xi \left(\frac{1}{\left|2\right|}\right)<\frac{1}{\left|2\right|}$$

for all t ≥ 0. Let δ > 0 and f : G → X be an odd mapping satisfying the inequality (4.9). Then the limit$A\left(x\right)=\underset{n\to \infty }{lim}{2}^n\left(f\left(\frac{x}{2^{n-1}}\right)-8f\left(\frac{x}{2^n}\right)\right)$ exists for all x ∈ G and A : G → X is a unique additive mapping such that

$$\left|\right|f\left(2x\right)-8f\left(x\right)-A\left(x\right)|{|}_X\le max\left\{\left(1+\frac{1}{\left|2\right|}\right)\frac{\delta \xi \left(\left|x\right|\right)}{\left|11\right|},\left|\frac{7}{33}\right|\xi \left(\left|x\right|\right)\right\}$$

for all x ∈ G.

Proof. Define φ : G² → [0, ∞) by φ(x, y) := δ((ξ(|x|) + ξ(|y|)). Also

$$\begin{array}{ll}\hfill \Phi \left(x\right)& =\underset{n\to \infty }{lim}max\left\{{\left|2\right|}^{k}max\left\{\frac{1}{\left|11\right|}\phi \left(\frac{x}{2^k},\frac{x}{2^{k+1}}\right),\left|\frac{14}{33}\right|\phi \left(\frac{x}{2^{k+1}},0\right)\right\};0\le k<n\right\}\\ =max\left\{\left(1+\frac{1}{\left|2\right|}\right)\frac{\delta \xi \left(\left|x\right|\right)}{\left|11\right|},\left|\frac{7}{33}\right|\xi \left(\left|x\right|\right)\right\}\end{array}$$

exists for all x ∈ G. Applying Theorem 4.3, we get the desired result.   □

Similarly, we can obtain the following. We will omit the proof.

Theorem 4.4. Let G be an additive semigroup and X a complete non-Archimedean space. Assume that φ : G² → [0, +∞) is a function such that

$$\underset{n\to \infty }{\lim }\frac{\phi {(2}^{n}x{\mathrm{,2}}^{n}y)}{{\left|2\right|}^n}=0$$

(4.23)

for all x, y ∈ G. Let for each x ∈ G

$$\Phi (x)=\underset{n\to \infty }{\lim }\max \left\{\frac{1}{{\left|2\right|}^k}\max \left\{\frac{1}{\left|11\right|}\phi {(2}^{k+1}x{\mathrm{,2}}^{k}x)\mathrm{,}\left|\frac{14}{33}\right|\phi {(2}^{k}x\mathrm{,0})\right\};0\le k<n\right\}$$

(4.24)

exist. Suppose that f : G → X be an odd mapping satisfying the inequality (4.3). Then the limit

$$A(x):=\underset{n\to \infty }{\lim }\frac{f{(2}^{n+1}x)-8f{(2}^{n}x)}{2^n}$$

exists for all x ∈ G and A : G → X is an additive mapping satisfying

$$\left|\right|f\left(2x\right)-8f\left(x\right)-A\left(x\right)|{|}_X\le \frac{1}{\left|2\right|}\Phi \left(x\right)$$

(4.25)

for all x ∈ G. Moreover, if

$$\underset{j\to \infty }{\lim }\underset{n\to \infty }{\lim }\max \left\{\frac{1}{{\left|2\right|}^k}\max \left\{\frac{1}{\left|11\right|}\phi {(2}^{k+1}x{\mathrm{,2}}^{k}x)\mathrm{,}\left|\frac{14}{33}\right|\phi {(2}^{k}x\mathrm{,0})\right\};j\le k<n+j\right\}=\mathrm{0,}$$

then A is the unique mapping satisfying (4.25).

5. Non-Archimedean stability of Equation (1.1): a direct method-even case

Theorem 5.1. Let G be an additive semigroup and X a complete non-Archimedean space. Assume that φ : G² → [0, +∞) is a function such that

$$\underset{n\to \infty }{\lim }\frac{\phi {(2}^{n}x{\mathrm{,2}}^{n}y)}{{\left|16\right|}^n}=0$$

(5.1)

for all x, y ∈ G. Let for all x ∈ G

$$\Phi (x)=\underset{n\to \infty }{\lim }\max \left\{\frac{1}{{\left|16\right|}^k}\max \left\{\frac{1}{\left|22\right|}\phi {(\mathrm{0,2}}^{k}x)\mathrm{,}\left|\frac{6}{11}\right|\phi {(2}^{k}x{\mathrm{,2}}^{k}x)\right\};0\le k<n\right\}$$

(5.2)

exist. Suppose that f : G → X is an even mapping with f(0) = 0 satisfying the inequality (4.3). Then the limit

$$Q(x):=\underset{n\to \infty }{\lim }\frac{f{(2}^{n}x)}{{16}^n}$$

exists for all x ∈ G and Q : G → X is a quartic mapping satisfying

$$\left|\right|f\left(x\right)-Q\left(x\right)|{|}_X\le \frac{1}{\left|16\right|}\Phi \left(x\right)$$

(5.3)

for all x ∈ G. Moreover, if

$$\underset{j\to \infty }{\lim }\underset{n\to \infty }{\lim }\max \left\{\frac{1}{{\left|16\right|}^k}\max \left\{\frac{1}{\left|22\right|}\phi {(\mathrm{0,2}}^{k}x)\mathrm{,}\left|\frac{6}{11}\right|\phi {(2}^{k}x{\mathrm{,2}}^{k}x)\right\};j\le k<n+j\right\}=\mathrm{0,}$$

then Q is the unique mapping satisfying (5.3).

Proof. Proceeding as in the proof of Theorem 3.1, we obtain

$${∥\frac{f\left(2x\right)}{16}-f\left(x\right)∥}_X\le \frac{1}{\left|16\right|}max\left\{\frac{1}{\left|22\right|}\phi \left(0,x\right),\left|\frac{6}{11}\right|\phi \left(x,x\right)\right\}.$$

One can easily show that

$$\begin{array}{l}{\Vert \frac{f{(2}^{n}x)}{{16}^n}-f(x)\Vert }_X\\ \le \frac{1}{\left|16\right|}\max \left\{\frac{1}{{\left|16\right|}^k}\max \left\{\frac{1}{\left|22\right|}\phi {(\mathrm{0,2}}^{k}x)\mathrm{,}\left|\frac{6}{11}\right|\phi {(2}^{k}x{\mathrm{,2}}^{k}x)\right\};0\le k<n\right\}\mathrm{.}\end{array}$$

The rest of the proof is similar to the proof of Theorem 4.1.   □

Corollary 5.1. Let ξ : [0, ∞) → [0, ∞) be a function satisfying

$$\xi \left(\left|2\right|t\right)\le \xi \left(\left|2\right|\right)\xi \left(t\right),\xi \left(\left|2\right|\right)<\left|16\right|$$

for all t ≥ 0. Let δ > 0 and f : G → X be an even mapping with f(0) = 0 satisfying the inequality (4.9). Then the limit$Q(x)={\lim }_{n\to \infty }\frac{f{(2}^{n}x)}{{16}^n}$exists for all x ∈ G and Q : G → X is a unique quartic mapping such that

$$\left|\right|f\left(x\right)-Q\left(x\right)|{|}_X\le \frac{1}{\left|16\right|}max\left\{\frac{1}{\left|22\right|}\delta \xi \left(\left|x\right|\right),2\left|\frac{6}{11}\right|\xi \left(\left|x\right|\right)\right\}$$

for all x ∈ G.

Proof. Define φ : G² → [0, ∞) by φ(x, y) := δ(ξ(|x|) + ξ(|y|)). Also

$$\Phi \left(x\right)=max\left\{\frac{1}{\left|22\right|}\delta \xi \left(\left|x\right|\right),2\left|\frac{6}{11}\right|\xi \left(\left|x\right|\right)\right\}$$

exists for all x ∈ G. Applying Theorem 5.1, we get the desired result.   □

Similarly, we can obtain the following. We will omit the proof.

Theorem 5.2. Let G be an additive semigroup and X a complete non-Archimedean space. Assume that φ : G² → [0, +∞) is a function such that

$$\underset{n\to \infty }{lim}16{|}^n\phi \left(\frac{x}{2^n},\frac{y}{2^n}\right)=0$$

for all x, y ∈ G. Let for all x ∈ G

$$\Phi \left(x\right)=\underset{n\to \infty }{lim}max\left\{{\left|16\right|}^{k}max\left\{\frac{1}{\left|22\right|}\phi \left(0,\frac{x}{2^{k+1}}\right),\left|\frac{6}{11}\right|\phi \left(\frac{x}{2^{k+1}},\frac{x}{2^{k+1}}\right)\right\};0\le k<n\right\}$$

exist. Suppose that f : G → X is an even mapping satisfying the inequality (4.3). Then the limit

$$Q\left(x\right):=\underset{n\to \infty }{lim}16^{n}f\left(\frac{x}{2^n}\right)$$

exists for all x ∈ G and Q : G → X is a quartic mapping satisfying

$$\left|\right|f\left(x\right)-Q\left(x\right)|{|}_X\le \Phi \left(x\right)$$

(5.4)

for all x ∈ G. Moreover, if

$$\underset{j\to \infty }{lim}\underset{n\to \infty }{lim}max\left\{{\left|16\right|}^{k}max\left\{\frac{1}{\left|22\right|}\phi \left(0,\frac{x}{2^{k+1}}\right),\left|\frac{6}{11}\right|\phi \left(\frac{x}{2^{k+1}},\frac{x}{2^{k+1}}\right)\right\};j\le k<n+j\right\}=0,$$

then Q is the unique mapping satisfying (5.4).

6. Conclusion

We linked here three different disciplines, namely, the non-Archimedean normed spaces, functional equations and fixed point theory. We established the generalized Hyers-Ulam stability of the functional Equation (1.1) in non-Archimedean normed spaces.