## 1. Introduction

The study of stability problems for functional equations is related to a question of Ulam [1] concerning the stability of group homomorphisms and affirmatively answered for Banach spaces by Hyers [2]. Subsequently, this result of Hyers was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference. The article of Rassias [4] has provided a lot of influence in the development of what we now call generalized Ulam-Hyers stability of functional equations. We refer the interested readers for more information on such problems to the article [517].

Recently, Alsina [18], Chang, et al. [19], Mirmostafaee et al. [20], [21], Miheţ and Radu [22], Miheţ et al. [23], [24], [25], [26], Baktash et al. [27], Eshaghi et al. [28], Saadati et al. [29], [30] investigated the stability in the settings of fuzzy, probabilistic, and random normed spaces.

In this article, we study the stability of the following functional equation

(1.1)

in the various random normed spaces via different methods. Since ax4 is a solution of above functional equation, we say it quartic functional equation.

## 2. Preliminaries

In this section, we recall some definitions and results which will be used later on in the article.

A triangular norm (shorter t-norm) is a binary operation on the unit interval [0, 1], i.e., a function T : [0, 1] × [0, 1] → [0, 1] such that for all a, b, c ∈ [0, 1] the following four axioms satisfied:

1. (i)

T(a, b) = T(b, a) (commutativity);

2. (ii)

T(a, (T(b, c))) = T(T(a, b), c) (associativity);

3. (iii)

T(a, 1) = a (boundary condition);

4. (iv)

T(a, b) ≤ T(a, c) whenever bc (monotonicity).

Basic examples are the Lukasiewicz t-norm T L , T L (a, b) = max (a + b - 1, 0) ∀a, b ∈ [0, 1] and the t-norms T P , T M , T D , where T P (a, b) := ab, T M (a, b) := min {a, b},

${T}_{D}\left(a,b\right):=\left\{\begin{array}{cc}\hfill min\left(a,b\right),\hfill & \hfill \mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{max}}\left(a,b\right)=1;\hfill \\ \hfill 0,\hfill & \hfill \mathsf{\text{otherwise}}.\hfill \end{array}\right\$

If T is a t-norm then ${x}_{T}^{\left(n\right)}$ is defined for every x ∈ [0, 1] and nN ∪ {0} by 1, if n = 0 and $T\left({x}_{T}^{\left(n-1\right)},x\right)$, if n ≥ 1. A t-norm T is said to be of Hadžić-type (we denote by $T\in \mathcal{H}$) if the family ${\left({x}_{T}^{\left(n\right)}\right)}_{n\in N}$ is equicontinuous at x = 1 (cf. [31]).

Other important triangular norms are (see [32]):

• the Sugeno-Weber family ${\left\{{T}_{\lambda }^{SW}\right\}}_{\lambda \in \left[-1,\infty \right]}$ is defined by ${T}_{-1}^{SW}={T}_{D}$, ${T}_{\infty }^{SW}={T}_{P}$ and

${T}_{\lambda }^{SW}\left(x,y\right)=max\left(0,\frac{x+y-1+\lambda xy}{1+\lambda }\right)$

if λ ∈ (-1, ∞).

• the Domby family ${\left\{{T}_{\lambda }^{D}\right\}}_{\lambda \in \left[0,\infty \right]}$, defined by T D , if λ = 0, T M , if λ = ∞ and

${T}_{\lambda }^{D}\left(x,y\right)=\frac{1}{1+{\left(\left(\frac{1-x}{x}{\right)}^{\lambda }+{\left(\frac{1-y}{y}\right)}^{\lambda }\right)}^{1/\lambda }}$

if λ ∈ (0, ∞).

• the Aczel-Alsina family ${\left\{{T}_{\lambda }^{AA}\right\}}_{\lambda \in \left[0,\infty \right]}$, defined by T D , if λ = 0, T M , if λ = ∞ and

${T}_{\lambda }^{AA}\left(x,y\right)={e}^{-{\left(|logx{|}^{\lambda }+|logy{|}^{\lambda }\right)}^{1∕\lambda }}$

if λ ∈ (0, ∞).

A t-norm T can be extended (by associativity) in a unique way to an n-array operation taking for (x1, ..., x n ) ∈ [0, 1]n the value T (x1, ..., x n ) defined by

${\mathsf{\text{T}}}_{i=1}^{0}{x}_{i}=1,{\mathsf{\text{T}}}_{i=1}^{n}{x}_{i}=T\left({\mathsf{\text{T}}}_{i=1}^{n-1}{x}_{i},{x}_{n}\right)=T\left({x}_{1},\dots ,{x}_{n}\right).$

T can also be extended to a countable operation taking for any sequence (x n )nNin [0, 1] the value

${\mathsf{\text{T}}}_{i=1}^{\infty }{x}_{i}=\underset{n\to \infty }{lim}{\mathsf{\text{T}}}_{i=1}^{n}{x}_{i}.$
(2.1)

The limit on the right side of (2.1) exists since the sequence ${\left\{{\mathsf{\text{T}}}_{i=1}^{n}{x}_{i}\right\}}_{n\in ℕ}$ is non-increasing and bounded from below.

Proposition 2.1. [32] (i) For TT L the following implication holds:

$\underset{n\to \infty }{lim}{\mathsf{\text{T}}}_{i=1}^{\infty }{x}_{n+i}=1⇔\sum _{n=1}^{\infty }\left(1-{x}_{n}\right)<\infty .$

(ii) If T is of Hadžić-type then

$\underset{n\to \infty }{lim}{\mathsf{\text{T}}}_{i=1}^{\infty }{x}_{n+i}=1$

for every sequence {x n }nNin [0, 1] such that limn→∞x n = 1.

(iii) If $T\in {\left\{{T}_{\lambda }^{AA}\right\}}_{\lambda \in \left(0,\infty \right)}\cup {\left\{{T}_{\lambda }^{D}\right\}}_{\lambda \in \left(0,\infty \right)}$, then

$\underset{n\to \infty }{lim}{\mathsf{\text{T}}}_{i=1}^{\infty }{x}_{n+i}=1⇔\sum _{n=1}^{\infty }{\left(1-{x}_{n}\right)}^{\alpha }<\infty .$

(iv) If $T\in {\left\{{T}_{\lambda }^{SW}\right\}}_{\lambda \in \left[-1,\infty \right)}$, then

$\underset{n\to \infty }{lim}{\mathsf{\text{T}}}_{i=1}^{\infty }{x}_{n+i}=1⇔\sum _{n=1}^{\infty }\left(1-{x}_{n}\right)<\infty .$

Definition 2.2. [33] A random normed space (briefly, RN-space) is a triple (X, μ, T), where X is a vector space, T is a continuous t-norm, and μ is a mapping from X into D+ such that, the following conditions hold:

(RN1) μ x (t) = ε0(t) for all t > 0 if and only if x = 0;

(RN2) ${\mu }_{\alpha x}\left(t\right)={\mu }_{x}\left(\frac{t}{|\alpha |}\right)$ for all xX, α ≠ 0;

(RN3) μx+y(t + s) ≥ T (μ x (t), μ y (s)) for all x, y, zX and t, s ≥ 0.

Definition 2.3. Let (X, μ, T) be an RN-space.

1. (1)

A sequence {x n } in X is said to be convergent to x in X if, for every ε > 0 and λ > 0, there exists a positive integer N such that ${\mu }_{{x}_{n}-x}\left(\epsilon \right)>1-\lambda$ whenever nN.

2. (2)

A sequence {x n } in X is called Cauchy if, for every ε > 0 and λ > 0, there exists a positive integer N such that ${\mu }_{{x}_{n}-{x}_{m}}\left(\epsilon \right)>1-\lambda$ whenever nmN.

3. (3)

An RN-space (X, μ, T) is said to be complete if every Cauchy sequence in X is convergent to a point in X.

Theorem 2.4. [34] If (X, μ, T) is an RN-space and {x n } is a sequence such that x n x, then ${lim}_{n\to \infty }{\mu }_{{x}_{n}}\left(t\right)={\mu }_{x}\left(t\right)$ almost everywhere.

## 3. Non-Archimedean random normed space

By a non-Archimedean field we mean a field $\mathcal{K}$ equipped with a function (valuation) | · | from K into [0, ∞] such that |r| = 0 if and only if r = 0, |rs| = |r| |s|, and |r + s| ≤ max{|r|, |s|} for all $r,s\in \mathcal{K}$. Clearly |1| = | -1| = 1 and |n| ≤ 1 for all n ∈ ℕ. By the trivial valuation we mean the mapping | · | taking everything but 0 into 1 and |0| = 0. Let $\mathcal{X}$ be a vector space over a field $\mathcal{K}$ with a non-Archimedean non-trivial valuation | · |. A function $||\cdot ||:\mathcal{X}\to \left[0,\infty \right]$ is called a non-Archimedean norm if it satisfies the following conditions:

1. (i)

||x|| = 0 if and only if x = 0;

2. (ii)

for any $r\in \mathcal{K}$, $x\in \mathcal{X}$, ||rx|| = ||r|||x||;

3. (iii)

the strong triangle inequality (ultrametric); namely,

$||x+y||\le max\left\{||x||,||y||\right\}\phantom{\rule{1em}{0ex}}\left(x,y\in \mathcal{X}\right).$

Then $\left(\mathcal{X},||\cdot ||\right)$ is called a non-Archimedean normed space. Due to the fact that

$||{x}_{n}-{x}_{m}||\le max\left\{||{x}_{j+1}-{x}_{j}||:m\le j\le n-1\right\}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(n>m\right),$

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

In 1897, Hensel [35] discovered the p-adic numbers as a number theoretical analogue of power series in complex analysis. Fix a prime number p. For any non-zero rational number x, there exists a unique integer n x ∈ 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 Q. The completion of Q with respect to the metric d(x, y) = |x - y| p is denoted by Q p , which is called the p-adic number field.

Throughout the article, we assume that $\mathcal{X}$ is a vector space and $\mathcal{Y}$ is a complete non-Archimedean normed space.

Definition 3.1. A non-Archimedean random normed space (briefly, non-Archimedean RN-space) is a triple $\left(\mathcal{X},\mu ,T\right)$, where X is a linear space over a non-Archimedean field $\mathcal{K}$, T is a continuous t-norm, and μ is a mapping from X into D+ such that the following conditions hold:

(NA-RN1) μ x (t) = ε0(t) for all t > 0 if and only if x = 0;

(NA-RN2) ${\mu }_{\alpha x}\left(t\right)={\mu }_{x}\left(\frac{t}{|\alpha |}\right)$ for all $x\in \mathcal{X}$, t > 0, α ≠ 0;

(NA-RN3) μx+y(max{t, s}) ≥ T (μ x (t), μ y (s)) for all $x,y,z\in \mathcal{X}$ and t, s ≥ 0.

It is easy to see that if (NA-RN3) holds then so is

(RN3) μx+y(t + s) ≥ T (μ x (t), μ y (s)).

As a classical example, if $\left(\mathcal{X},||.||\right)$ is a non-Archimedean normed linear space, then the triple $\left(\mathcal{X},\mu ,{T}_{M}\right)$, where

${\mu }_{x}\left(t\right)=\left\{\begin{array}{cc}\hfill 0\hfill & \hfill t\le ||x||\hfill \\ \hfill 1\hfill & \hfill t>||x||\hfill \end{array}\right\$

is a non-Archimedean RN-space.

Example 3.2. Let $\left(\mathcal{X},||.||\right)$ be is a non-Archimedean normed linear space. Define

${\mu }_{x}\left(t\right)=\frac{t}{t+||x||},\phantom{\rule{1em}{0ex}}\forall x\in \mathcal{X}\phantom{\rule{1em}{0ex}}t>0.$

Then $\left(\mathcal{X},\mu ,{T}_{M}\right)$ is a non-Archimedean RN-space.

Definition 3.3. Let $\left(\mathcal{X},\mu ,T\right)$ be a non-Archimedean RN-space. Let {x n } be a sequence in $\mathcal{X}$. Then {x n } is said to be convergent if there exists $x\in \mathcal{X}$ such that

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

for all t > 0. In that case, x is called the limit of the sequence {x n }.

A sequence {x n } in $\mathcal{X}$ is called Cauchy if for each ε > 0 and each t > 0 there exists n0 such that for all nn0 and all p > 0 we have ${\mu }_{{x}_{n+p}-{x}_{n}}\left(t\right)>1-\epsilon$.

If each Cauchy sequence is convergent, then the random norm is said to be complete and the non-Archimedean RN-space is called a non-Archimedean random Banach space.

Remark 3.4. [36] Let $\left(\mathcal{X},\mu ,{T}_{M}\right)$ be a non-Archimedean RN-space, then

${\mu }_{{x}_{n+p}-{x}_{n}}\left(t\right)\ge min\left\{{\mu }_{{x}_{n+j+1}-{x}_{n+j}}\left(t\right):j=0,1,2,\dots ,p-1\right\}$

So, the sequence {x n } is Cauchy if for each ε > 0 and t > 0 there exists n0 such that for all nn0 we have

${\mu }_{{x}_{n+1}-{x}_{n}}\left(t\right)>1-\epsilon .$

## 4. Generalized Ulam-Hyers stability for a quartic functional equation in non-Archimedean RN-spaces

Let $\mathcal{K}$ be a non-Archimedean field, $\mathcal{X}$ a vector space over $\mathcal{K}$ and let $\left(\mathcal{Y},\mu ,T\right)$ be a non-Archimedean random Banach space over $\mathcal{K}$.

We investigate the stability of the quartic functional equation

where f is a mapping from $\mathcal{X}$ to $\mathcal{Y}$ and f(0) = 0.

Next, we define a random approximately quartic mapping. Let Ψ be a distribution function on $\mathcal{X}×\mathcal{X}×\left[0,\infty \right]$ such that Ψ (x, y, ·) is symmetric, nondecreasing and

$\Psi \left(cx,cx,t\right)\ge \Psi \left(x,x,\frac{t}{|c|}\right)\phantom{\rule{1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{1em}{0ex}}c\ne 0\right).$

Definition 4.1. A mapping $f:\mathcal{X}\to \mathcal{Y}$ is said to be Ψ-approximately quartic if

$\begin{array}{c}{\mu }_{16f\left(x+4y\right)+f\left(4x-y\right)-306\left[9f\left(x+\frac{y}{3}\right)+f\left(x+2y\right)\right]-136f\left(x-y\right)+1394f\left(x+y\right)-425f\left(y\right)+1530f\left(x\right)}\left(t\right)\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\ge \Psi \left(x,y,t\right)\phantom{\rule{1em}{0ex}}\left(x,y\in \mathcal{X},\phantom{\rule{1em}{0ex}}t>0\right).\end{array}$
(4.1)

In this section, we assume that 4 ≠ 0 in $\mathcal{K}$ (i.e., characteristic of $\mathcal{K}$ is not 4). Our main result, in this section, is the following:

Theorem 4.2. Let $\mathcal{K}$ be a non-Archimedean field, $\mathcal{X}$ a vector space over $\mathcal{K}$ and let $\left(\mathcal{Y},\mu ,T\right)$ be a non-Archimedean random Banach space over $\mathcal{K}$. Let $f:\mathcal{X}\to \mathcal{Y}$ be a Ψ-approximately quartic mapping. If for some α ∈ ℝ, α > 0, and some integer k, k > 3 with |4k| < α,

$\Psi \left({\mathsf{\text{4}}}^{-k}x,{\mathsf{\text{4}}}^{-k}y,t\right)\ge \Psi \left(x,y,\alpha t\right)\phantom{\rule{1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{1em}{0ex}}t>0\right)$
(4.2)

and

$\underset{n\to \infty }{lim}{\mathsf{\text{T}}}_{j=n}^{\infty }M\left(x,\frac{{\alpha }^{j}t}{{|4|}^{kj}}\right)=1\phantom{\rule{1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{1em}{0ex}}t>0\right),$
(4.3)

then there exists a unique quartic mapping $Q:\mathcal{X}\to \mathcal{Y}$ such that

${\mu }_{f\left(x\right)-Q\left(x\right)}\left(t\right)\ge {\mathsf{\text{T}}}_{i=1}^{\infty }M\left(x,\frac{{\alpha }^{i+1}t}{{|4|}^{ki}}\right)$
(4.4)

for all xX and t > 0, where

$M\left(x,t\right):=T\left(\Psi \left(x,0,t\right),\Psi \left(\mathsf{\text{4}}x,0,t\right),\cdots \phantom{\rule{0.3em}{0ex}},\Psi \left({\mathsf{\text{4}}}^{k-1}x,0,t\right)\right)\phantom{\rule{1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{1em}{0ex}}t>0\right).$

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

${\mu }_{f\left({\mathsf{\text{4}}}^{j}x\right)-\mathsf{\text{2}}{56}^{j}f\left(x\right)}\left(t\right)\ge {M}_{j}\left(x,t\right):=T\left(\Psi \left(x,0,t\right),\cdots \phantom{\rule{0.3em}{0ex}},\Psi \left({\mathsf{\text{4}}}^{j-1}x,0,t\right)\right).$
(4.5)

Putting y = 0 in (4.1), we obtain

${\mu }_{f\left(\mathsf{\text{4}}x\right)-\mathsf{\text{2}}56f\left(x\right)}\left(t\right)\ge \Psi \left(x,0,t\right)\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{1em}{0ex}}t>0\right).$

This proves (4.5) for j = 1. Assume that (4.5) holds for some j ≥ 1. Replacing y by 0 and x by 4jx in (4.1), we get

${\mu }_{f\left({\text{4}}^{j+1}x\right)-256f{\left(4}^{j}x\right)}\left(t\right)\ge \Psi \left({\text{4}}^{j}x,0,t\right)\phantom{\rule{0.1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{0.1em}{0ex}}t>0\right).$

Since |256| ≤ 1,

$\begin{array}{c}{\mu }_{f{\left(4}^{j+1}x\right)-{256}^{j+1}f\left(x\right)}\left(t\right)\ge T\left({\mu }_{f{\left(4}^{j+1}x\right)-256f{\left(4}^{j}x\right)}\left(t\right),{\mu }_{256f{\left(4}^{j}x\right)-{256}^{j+1}f\left(x\right)}\left(t\right)\right)\\ =T\left({\mu }_{f{\left(4}^{j+1}x\right)-256f{\left(4}^{j}x\right)}\left(t\right),{\mu }_{f{\left(4}^{j}x\right)-{256}^{j}f\left(x\right)}\left(\frac{t}{|256|}\right)\right)\\ \ge T\left({\mu }_{f{\left(4}^{j+1}x\right)-256f{\left(4}^{j}x\right)}\left(t\right),{\mu }_{f{\left(4}^{j}x\right)-{256}^{j}f\left(x\right)}\left(t\right)\right)\\ \ge T\left(\Psi \left({\text{4}}^{j}x,0,t\right),{M}_{j}\left(x,t\right)\right)\\ ={M}_{j+1}\left(x,t\right)\end{array}$

for all $x\in \mathcal{X}$. Thus (4.5) holds for all j ≥ 1. In particular

${\mu }_{f{\left(4}^{k}x\right)-{256}^{k}f\left(x\right)}\left(t\right)\ge M\left(x,t\right)\phantom{\rule{0.1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{0.1em}{0ex}}t>0\right).$
(4.6)

Replacing x by 4-(kn+k)x in (4.6) and using inequality (4.2), we obtain

(4.7)

Then

${\mu }_{{{\left(4}^{4k}\right)}^{n}f\left(\frac{x}{{{\left(4}^{k}\right)}^{n}}\right)-{{\left(4}^{4k}\right)}^{n+1}f\left(\frac{x}{{{\left(4}^{k}\right)}^{n+1}}\right)}\left(t\right)\ge M\left(x,\frac{{\alpha }^{n+1}}{{|\left(4}^{4k}{\right)}^{n}|}t\right)\phantom{\rule{0.1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{0.1em}{0ex}}t>0,\phantom{\rule{0.1em}{0ex}}n=0,1,2,\dots \right).$

Hence,

$\begin{array}{l}{\mu }_{{{\left(4}^{4k}\right)}^{n}f\left(\frac{x}{{{\left(4}^{k}\right)}^{n}}\right)-{{\left(4}^{4k}\right)}^{n+p}f\left(\frac{x}{{{\left(4}^{k}\right)}^{n+p}}\right)}\left(t\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\ge {T}_{j=n}^{n+p}\left({\mu }_{{{\left(4}^{4k}\right)}^{j}f\left(\frac{x}{{{\left(4}^{k}\right)}^{j}}\right)-{{\left(4}^{4k}\right)}^{j+p}f\left(\frac{x}{{{\left(4}^{k}\right)}^{j+p}}\right)}\left(t\right)\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\ge {T}_{j=n}^{n+p}M\left(x,\frac{{\alpha }^{j+1}}{{|\left(4}^{4k}{\right)}^{j}|}t\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\ge {T}_{j=n}^{n+p}M\left(x,\frac{{\alpha }^{j+1}}{{|\left(4}^{k}{\right)}^{j}|}t\right)\phantom{\rule{0.1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{0.1em}{0ex}}t>0,\phantom{\rule{0.1em}{0ex}}n=0,1,2,\dots \right).\end{array}$

Since ${\mathrm{lim}}_{n\to \infty }{T}_{j=n}^{\infty }M\left(x,\frac{{\alpha }^{j+1}}{{|\left(4}^{k}{\right)}^{j}|}t\right)=1\phantom{\rule{0.1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{0.1em}{0ex}}t>0\right),\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}{\left\{{{\left(4}^{4k}\right)}^{n}f\left(\frac{x}{{{\left(4}^{k}\right)}^{n}}\right)\right\}}_{n\in N}$, is a Cauchy sequence in the non-Archimedean random Banach space $\left(\mathcal{Y},\mu ,T\right)$. Hence, we can define a mapping $Q:\mathcal{X}\to \mathcal{Y}$ such that

$\underset{n\to \infty }{\mathrm{lim}}{\mu }_{{{\left(4}^{4k}\right)}^{n}f\left(\frac{x}{{{\left(4}^{k}\right)}^{n}}\right)-Q\left(x\right)}\left(t\right)=1\phantom{\rule{0.1em}{0ex}}\left(x\in X,\phantom{\rule{0.1em}{0ex}}t>0\right).$
(4.8)

Next, for each n ≥ 1, $x\in \mathcal{X}$ and t > 0,

$\begin{array}{c}{\mu }_{f\left(x\right)-{{\left(4}^{4k}\right)}^{n}f\left(\frac{x}{{{\left(4}^{k}\right)}^{n}}\right)}\left(t\right)={\mu }_{{\sum }_{i=0}^{n-1}{{\left(4}^{4k}\right)}^{i}f\left(\frac{x}{{{\left(4}^{k}\right)}^{i}}\right)-{{\left(4}^{4k}\right)}^{i+1}f\left(\frac{x}{{{\left(4}^{k}\right)}^{i+1}}\right)}\left(t\right)\\ \ge {T}_{i=0}^{n-1}\left({\mu }_{{{\left(4}^{4k}\right)}^{i}f\left(\frac{x}{{{\left(4}^{k}\right)}^{i}}\right)-{{\left(4}^{4k}\right)}^{i+1}f\left(\frac{x}{{{\left(4}^{k}\right)}^{i+1}}\right)}\left(t\right)\right)\\ \ge {T}_{i=0}^{n-1}M\left(x,\frac{{\alpha }^{i+1}t}{{|4}^{4k}{|}^{i}}\right).\end{array}$

Therefore,

$\begin{array}{c}{\mu }_{f\left(x\right)-Q\left(x\right)}\left(t\right)\ge T\left({\mu }_{f\left(x\right)-{{\left(4}^{4k}\right)}^{n}f\left(\frac{x}{{{\left(4}^{k}\right)}^{n}}\right)}\left(t\right),{\mu }_{{{\left(4}^{4k}\right)}^{n}f\left(\frac{x}{{{\left(4}^{k}\right)}^{n}}\right)-Q\left(x\right)}\left(t\right)\right)\\ \ge T\left({T}_{i=0}^{n-1}M\left(x,\frac{{\alpha }^{i+1}t}{{|4}^{4k}{|}^{i}}\right),{\mu }_{{{\left(4}^{4k}\right)}^{n}f\left(\frac{x}{{{\left(4}^{k}\right)}^{n}}\right)-Q\left(x\right)}\left(t\right)\right).\end{array}$

By letting n → ∞, we obtain

${\mu }_{f\left(x\right)-Q\left(x\right)}\left(t\right)\ge {T}_{i=1}^{\infty }M\left(x,\frac{{\alpha }^{i+1}t}{{|4}^{k}{|}^{i}}\right).$

This proves (4.4).

As T is continuous, from a well-known result in probabilistic metric space (see e.g., [[34], Chapter 12]), it follows that

$\begin{array}{l}\underset{n\to \infty }{\mathrm{lim}}{\mu }_{{{\left(4}^{k}\right)}^{n}\cdot 16f{\left(4}^{-kn}\left(x+4y\right)\right)+{{\left(4}^{k}\right)}^{n}f{\left(4}^{-kn}\left(4x-y\right)\right)-306\left[{{\left(4}^{k}\right)}^{n}\cdot 9f{\left(4}^{-kn}\left(x+\frac{y}{3}\right)\right)+{{\left(4}^{k}\right)}^{n}f{\left(4}^{-kn}\left(x+2y\right)\right)\right]}\\ \phantom{\rule{0.1em}{0ex}}{\phantom{\rule{0.1em}{0ex}}}_{-{136\left(4}^{k}{\right)}^{n}f{\left(4}^{-kn}\left(x-y\right)\right)+{1394\left(4}^{k}{\right)}^{n}f{\left(4}^{-kn}\left(x+y\right)\right)-{425\left(4}^{k}{\right)}^{n}f{\left(4}^{-kn}y\right)+{1530\left(4}^{k}{\right)}^{n}f{\left(4}^{-kn}x\right)}\left(t\right)\\ ={\mu }_{16Q\left(x+4y\right)+Q\left(4x-y\right)-306\left[9Q\left(x+\frac{y}{3}\right)+Q\left(x+2y\right)\right]-136Q\left(x-y\right)+1394Q\left(x+y\right)-425Q\left(y\right)+1530Q\left(x\right)}\left(t\right)\end{array}$

for almost all t > 0.

On the other hand, replacing x, y by 4-knx, 4-kny, respectively, in (4.1) and using (NA-RN2) and (4.2), we get

$\begin{array}{l}{\mu }_{{{\left(4}^{k}\right)}^{n}\cdot 16f{\left(4}^{-kn}\left(x+4y\right)\right)+{{\left(4}^{k}\right)}^{n}f{\left(4}^{-kn}\left(4x-y\right)\right)-306\left[{{\left(4}^{k}\right)}^{n}\cdot 9f{\left(4}^{-kn}\left(x+\frac{y}{3}\right)\right)+{{\left(4}^{k}\right)}^{n}f{\left(4}^{-kn}\left(x+2y\right)\right)\right]}\\ {\phantom{\rule{0.1em}{0ex}}}_{-{136\left(4}^{k}{\right)}^{n}f{\left(4}^{-kn}\left(x-y\right)\right)+{1394\left(4}^{k}{\right)}^{n}f{\left(4}^{-kn}\left(x+y\right)\right)-{425\left(4}^{k}{\right)}^{n}f{\left(4}^{-kn}y\right)+{1530\left(4}^{k}{\right)}^{n}f{\left(4}^{-kn}x\right)}\left(t\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\ge \Psi \left({4}^{-kn}x{,4}^{-kn}y,\frac{t}{{|4}^{k}{|}^{n}}\right)\ge \Psi \left(x,y,\frac{{\alpha }^{n}t}{{|4}^{k}{|}^{n}}\right)\end{array}$

for all $x,y\in \mathcal{X}$ and all t > 0. Since ${lim}_{n\to \infty }\Psi \left(x,y,\frac{{\alpha }^{n}t}{{|4}^{k}{|}^{n}}\right)=1$, we infer that Q is a quartic mapping.

If ${Q}^{\prime }:\mathcal{X}\to \mathcal{Y}$ is another quartic mapping such that μQ'(x)-f(x)(t) ≥ M(x, t) for all $x\in \mathcal{X}$ and t > 0, then for each nN, $x\in \mathcal{X}$ and t > 0,

${\mu }_{Q\left(x\right)-{Q}^{\prime }\left(x\right)}\left(t\right)\ge T\left({\mu }_{Q\left(x\right)-{{\left(4}^{4k}\right)}^{n}f\left(\frac{x}{{{\left(4}^{k}\right)}^{n}}\right)}\left(t\right),{\mu }_{{{\left(4}^{4k}\right)}^{n}f\left(\frac{x}{{{\left(4}^{k}\right)}^{n}}\right)-{Q}^{\prime }\left(x\right)}\left(t\right),t\right)\right).$

Thanks to (4.8), we conclude that Q = Q'. □

Corollary 4.3. Let $\mathcal{K}$ be a non-Archimedean field, $\mathcal{X}$ a vector space over $\mathcal{K}$ and let $\left(\mathcal{Y},\mu ,T\right)$ be a non-Archimedean random Banach space over $\mathcal{K}$ under a t-norm $T\in \mathcal{H}$. Let $f:\mathcal{X}\to \mathcal{Y}$ be a Ψ-approximately quartic mapping. If, for some α ∈ ℝ, α > 0, and some integer k, k > 3, with |4k| < α,

$\Psi \left({\mathsf{\text{4}}}^{-k}x,{\mathsf{\text{4}}}^{-k}y,t\right)\ge \Psi \left(x,y,\alpha t\right)\phantom{\rule{1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{1em}{0ex}}t>0\right),$

then there exists a unique quartic mapping $Q:\mathcal{X}\to \mathcal{Y}$ such that

${\mu }_{f\left(x\right)-Q\left(x\right)}\left(t\right)\ge {T}_{i=1}^{\infty }M\left(x,\frac{{\alpha }^{i+1}t}{|\mathsf{\text{4}}{|}^{ki}}\right)$

for all $x\in \mathcal{X}$ and all t > 0, where

$M\left(x,t\right):=T\left(\Psi \left(x,0,t\right),\Psi \left(\mathsf{\text{4}}x,0,t\right),\cdots \phantom{\rule{0.3em}{0ex}},\Psi \left({\mathsf{\text{4}}}^{k-1}x,0,t\right)\right)\phantom{\rule{1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{1em}{0ex}}t>0\right).$

Proof. Since

$\underset{n\to \infty }{lim}M\left(x,\frac{{\alpha }^{j}t}{{|4|}^{kj}}\right)=1\phantom{\rule{1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{1em}{0ex}}t>0\right)$

and T is of Hadžić type, from Proposition 2.1, it follows that

$\underset{n\to \infty }{lim}{T}_{j=n}^{\infty }M\left(x,\frac{{\alpha }^{j}t}{{|4|}^{kj}}\right)=1\phantom{\rule{1em}{0ex}}\left(x\in \mathcal{X},\phantom{\rule{1em}{0ex}}t>0\right).$

Now we can apply Theorem 4.2 to obtain the result. □

Example 4.4. Let $\left(\mathcal{X},\mu ,{T}_{M}\right)$ non-Archimedean random normed space in which

${\mu }_{x}\left(t\right)=\frac{t}{t+||x||},\phantom{\rule{1em}{0ex}}\forall x\in \mathcal{X},\phantom{\rule{1em}{0ex}}t>0,$

and $\left(\mathcal{Y},\mu ,{T}_{M}\right)$ a complete non-Archimedean random normed space (see Example 3.2). Define

$\Psi \left(x,y,t\right)=\frac{t}{1+t}.$

It is easy to see that (4.2) holds for α = 1. Also, since

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

we have

$\begin{array}{lll}\hfill \underset{n\to \infty }{lim}{T}_{M,j=n}^{\infty }M\left(x,\frac{{\alpha }^{j}t}{{|4|}^{kj}}\right)& =\underset{n\to \infty }{lim}\left(\underset{m\to \infty }{lim}{T}_{M,j=n}^{m}M\left(x,\frac{t}{{|4|}^{kj}}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\underset{n\to \infty }{lim}\underset{m\to \infty }{lim}\left(\frac{t}{t+{|4}^{k}{|}^{n}}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ =1,\phantom{\rule{1em}{0ex}}\forall x\in \mathcal{X},\phantom{\rule{1em}{0ex}}t>0.\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$

Let $f:\mathcal{X}\to \mathcal{Y}$ be a Ψ-approximately quartic mapping. Thus all the conditions of Theorem 4.2 hold and so there exists a unique quartic mapping $Q:\mathcal{X}\to \mathcal{Y}$ such that

${\mu }_{f\left(x\right)-Q\left(x\right)}\left(t\right)\ge \frac{t}{t+{|4}^{k}|}.$

## 5. Fixed point method for random stability of the quartic functional equation

In this section, we apply a fixed point method for achieving random stability of the quartic functional equation. The notion of generalized metric space has been introduced by Luxemburg [37], by allowing the value +∞ for the distance mapping. The following lemma (Luxemburg-Jung theorem) will be used in the proof of Theorem 5.3.

Lemma 5.1. [38]. Let (X, d) be a complete generalized metric space and let A : XX be a strict contraction with the Lipschitz constant k such that d(x0, A(x0)) < +∞ for some x0X. Then A has a unique fixed point in the set Y := {yX, d(x0, y) < ∞} and the sequence (An(x))nNconverges to the fixed point x* for every xY. Moreover, d(x0, A(x0)) ≤ δ implies $d\left({x}^{*},{x}_{0}\right)\le \frac{\delta }{1-k}$.

Let X be a linear space, (Y, ν, T M ) a complete RN-space and let G be a mapping from X × R into [0, 1], such that G(x, .) ∈ D+ for all x. Consider the set E := {g : XY, g(0) = 0} and the mapping d G defined on E × E by

where, as usual, inf ∅ = +∞. The following lemma can be proved as in [22]:

Lemma 5.2. cf. [22, 39] d G is a complete generalized metric on E.

Theorem 5.3. Let X be a real linear space, t f a mapping from X into a complete RN-space (Y, μ , T M ) with f(0) = 0 and let Φ : X2D+ be a mapping with the property

$\exists \alpha \in \left(0,256\right):{\Phi }_{4x,4y}\left(\alpha t\right)\ge {\Phi }_{x,y}\left(t\right),\phantom{\rule{2.77695pt}{0ex}}\forall x,y\in X,\phantom{\rule{1em}{0ex}}\forall t>0.$
(5.1)

If

$\begin{array}{c}{\mu }_{16f\left(x+4y\right)+f\left(4x-y\right)-306\left[9f\left(x+\frac{y}{3}\right)+f\left(x+2y\right)\right]\phantom{\rule{0.3em}{0ex}}-136f\left(x-y\right)+1394f\left(x+y\right)-425f\left(y\right)+1530f\left(x\right)}\left(t\right)\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\ge {\Phi }_{x,y}\left(t\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.3em}{0ex}}\forall x,y\in X,\end{array}$
(5.2)

then there exists a unique quartic mapping g : XY such that

${\mu }_{g\left(x\right)-f\left(x\right)}\left(t\right)\ge {\Phi }_{x,0}\left(Mt\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.3em}{0ex}}\forall x\in X,\phantom{\rule{1em}{0ex}}\forall t>0,$
(5.3)

where

$M=\left(256-\alpha \right).$

Moreover,

$g\left(x\right)=\underset{n\to \infty }{\mathrm{lim}}\frac{f{\left(4}^{n}x\right)}{{4}^{4n}}.$

Proof. By setting y = 0 in (5.2), we obtain

${\mu }_{f\left(4x\right)-256f\left(x\right)}\left(t\right)\ge {\Phi }_{x,0}\left(t\right)$

for all xX, whence

$\begin{array}{lll}\hfill {\mu }_{\frac{1}{256}f\left(4x\right)-f\left(x\right)}\left(t\right)& ={\mu }_{\frac{1}{256}\left(f\left(4x\right)-256f\left(x\right)\right)}\left(t\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ ={\mu }_{f\left(4x\right)-256f\left(x\right)}\left(256t\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \ge {\Phi }_{x,0}\left(256t\right),\phantom{\rule{1em}{0ex}}\forall x\in X,\phantom{\rule{1em}{0ex}}\forall t>0.\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$

Let

$G\left(x,t\right):={\Phi }_{x,0}\left(256t\right).$

Consider the set

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

and the mapping d G defined on E × E by

By Lemma 5.2, (E, d G ) is a complete generalized metric space. Now, let us consider the linear mapping J : EE,

$Jg\left(x\right):=\frac{1}{256}g\left(4x\right).$

We show that J is a strictly contractive self-mapping of E with the Lipschitz constant k = α/256.

Indeed, let g, hE be mappings such that d G (g, h) < ε. Then

${\mu }_{g\left(x\right)-h\left(x\right)}\left(\epsilon t\right)\ge G\left(x,t\right),\phantom{\rule{0.3em}{0ex}}\forall x\in X,\phantom{\rule{1em}{0ex}}\forall t>0,$

whence

$\begin{array}{lll}\hfill {\mu }_{Jg\left(x\right)-Jh\left(x\right)}\left(\frac{\alpha }{256}\epsilon t\right)& ={\mu }_{\frac{1}{256}\left(g\left(4x\right)-h\left(4x\right)\right)}\left(\frac{\alpha }{256}\epsilon t\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ ={\mu }_{g\left(4x\right)-h\left(4x\right)}\left(\alpha \epsilon t\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \ge G\left(4x,\alpha t\right)\phantom{\rule{1em}{0ex}}\left(x\in X,\phantom{\rule{1em}{0ex}}t>0\right).\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$

Since G(4x, αt) ≥ G(x, t), ${\mu }_{Jg\left(x\right)-Jh\left(x\right)}\left(\frac{\alpha }{256}\epsilon t\right)\ge G\left(x,t\right)$, that is,

${d}_{G}\left(g,h\right)<\epsilon ⇒{d}_{G}\left(Jg,Jh\right)\le \frac{\alpha }{256}\epsilon .$

This means that

${d}_{G}\left(Jg,Jh\right)\le \frac{\alpha }{256}{d}_{G}\left(g,h\right)$

for all g, h in E.

Next, from

${\mu }_{f\left(x\right)-\frac{1}{256}f\left(4x\right)}\left(t\right)\ge G\left(x,t\right)$

it follows that d G (f, Jf ) ≤ 1. Using the Luxemburg-Jung theorem, we deduce the existence of a fixed point of J, that is, the existence of a mapping g : XY such that g(4x) = 256g(x) for all xX.

Since, for any xX and t > 0,

${d}_{G}\left(u,v\right)<\epsilon ⇒{\mu }_{u\left(x\right)-v\left(x\right)}\left(t\right)\ge G\left(x,\frac{t}{\epsilon }\right),$

from d G (Jnf, g) → 0, it follows that ${lim}_{n\to \infty }\frac{f\left({4}^{n}x\right)}{{4}^{4n}}=g\left(x\right)$ for any xX.

Also, ${d}_{G}\left(f,g\right)\le \frac{1}{1-L}d\left(f,Jf\right)$ implies the inequality ${d}_{G}\left(f,g\right)\le \frac{1}{1-\frac{\alpha }{256}}$ from which it immediately follows ${\nu }_{g\left(x\right)-f\left(x\right)}\left(\frac{256}{256-\alpha }t\right)\ge G\left(x,t\right)$ for all t > 0 and all xX. This means that

${\mu }_{g\left(x\right)-f\left(x\right)}\left(t\right)\ge G\left(x,\frac{256-\alpha }{256}t\right),\phantom{\rule{1em}{0ex}}\forall x\in X,\phantom{\rule{1em}{0ex}}\forall t>0.$

It follows that

${\mu }_{g\left(x\right)-f\left(x\right)}\left(t\right)\ge {\Phi }_{x,0}\left(\left(256-\alpha \right)t\right)\phantom{\rule{1em}{0ex}}\forall x\in X,\phantom{\rule{1em}{0ex}}\forall t>0.$

The uniqueness of g follows from the fact that g is the unique fixed point of J with the property: there is C ∈ (0, ∞) such that μg(x)-f(x)(Ct) ≥ G(x, t) for all xX and all t > 0, as desired. □

## 6. Intuitionistic random normed spaces

Recently, the notation of intuitionistic random normed space introduced by Chang et al. [19]. In this section, we shall adopt the usual terminology, notations, and conventions of the theory of intuitionistic random normed spaces as in [22], [31], [33], [34], [40], [41], [42].

Definition 6.1. A measure distribution function is a function μ : R → [0, 1] which is left continuous, non-decreasing on R, inftRμ(t) = 0 and suptRμ(t) = 1.

We will denote by D the family of all measure distribution functions and by H a special element of D defined by

$H\left(t\right)=\left\{\begin{array}{cc}\hfill 0,\hfill & \hfill \mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}t\le 0,\hfill \\ \hfill 1,\hfill & \hfill \mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}t>0.\hfill \end{array}\right\$

If X is a nonempty set, then μ : XD is called a probabilistic measure on X and μ (x) is

denoted by μ x .

Definition 6.2. A non-measure distribution function is a function ν : R → [0, 1] which is right continuous, non-increasing on R, inftRν(t) = 0 and suptRν(t) = 1.

We will denote by B the family of all non-measure distribution functions and by G a special element of B defined by

$G\left(t\right)=\left(\begin{array}{cc}\hfill 1,\hfill & \hfill \mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}t\le 0,\hfill \\ \hfill 0,\hfill & \hfill \mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}t>0.\hfill \end{array}\right$

If X is a nonempty set, then ν : XB is called a probabilistic non-measure on X and ν (x) is denoted by ν x .

Lemma 6.3. [43], [44] Consider the set L* and operation ${\le }_{{L}^{*}}$ defined by:

$\begin{array}{c}{L}^{*}=\left\{\left({x}_{1},{x}_{2}\right):\left({x}_{1},{x}_{2}\right)\in {\left[0,1\right]}^{2}\phantom{\rule{0.3em}{0ex}}and\phantom{\rule{0.3em}{0ex}}{x}_{1}+{x}_{2}\le 1\right\},\\ \left({x}_{1},{x}_{2}\right){\le }_{L*}\left({y}_{1},{y}_{2}\right)⇔{x}_{1}\le {y}_{1},{x}_{2}\ge {y}_{2},\phantom{\rule{1em}{0ex}}\forall \left({x}_{1},{x}_{2}\right),\left({y}_{1},{y}_{2}\right)\in {L}^{*}.\end{array}$

Then $\left({L}^{*},{\le }_{{L}^{*}}\right)$ is a complete lattice.

We denote its units by ${0}_{{L}^{*}}=\left(0,1\right)$ and ${1}_{{L}^{*}}=\left(1,0\right)$. In Section 2, we presented classical t-norm. Using the lattice $\left({L}^{*},{\le }_{{L}^{*}}\right)$, these definitions can be straightforwardly extended.

Definition 6.4. [44] A triangular norm (t-norm) on L* is a mapping $\mathcal{T}:{\left({L}^{*}\right)}^{2}\to {L}^{*}$ satisfying the following conditions:

1. (a)

$\left(\forall x\in {L}^{*}\right)\left(\mathcal{T}\left(x,{1}_{{L}^{*}}\right)=x\right)$ (boundary condition);

2. (b)

$\left(\forall \left(x,y\right)\in {\left({L}^{*}\right)}^{2}\right)\left(\mathcal{T}\left(x,y\right)=\mathcal{T}\left(y,x\right)\right)$ (commutativity);

3. (c)

$\left(\forall \left(x,y,z\right)\in {\left({L}^{*}\right)}^{3}\right)\left(\mathcal{T}\left(x,\mathcal{T}\left(y,z\right)\right)=\mathcal{T}\left(\mathcal{T}\left(x,y\right),z\right)\right)$ (associativity);

4. (d)

(monotonicity).

If $\left({L}^{*},{\le }_{{L}^{*}},\mathcal{T}\right)$ is an Abelian topological monoid with unit ${1}_{{L}^{*}}$, then $\mathcal{T}$ is said to be a continuous t-norm.

Definition 6.5. [44] A continuous t-norm $\mathcal{T}$ on L* is said to be continuous t-representable if there exist a continuous t-norm * and a continuous t-conorm ◇ on [0, 1] such that, for all x = (x1, x2), y = (y1, y2) ∈ L*,

$\mathcal{T}\left(x,y\right)=\left({x}_{1}*{y}_{1},{x}_{2}♢{y}_{2}\right).$

For example,

$\mathcal{T}\left(a,b\right)=\left({a}_{1}{b}_{1},min\left\{{a}_{2}+{b}_{2},1\right\}\right)$

and

$M\left(a,b\right)=\left(min\left\{{a}_{1},{b}_{1}\right\},max\left\{{a}_{2},{b}_{2}\right\}\right)$

are continuous t-representable for all a = (a1, a2), b = (b1, b2) ∈ L*.

Now, we define a sequence ${\mathcal{T}}^{n}$ recursively by ${\mathcal{T}}^{1}=\mathcal{T}$ and

${\mathcal{T}}^{n}\left({x}^{\left(1\right)},\dots ,{x}^{\left(n+1\right)}\right)=\mathcal{T}\left({\mathcal{T}}^{n-1}\left({x}^{\left(1\right)},\dots ,{x}^{\left(n\right)}\right),{x}^{\left(n+1\right)}\right),\phantom{\rule{1em}{0ex}}\forall n\ge 2,\phantom{\rule{1em}{0ex}}{x}^{\left(i\right)}\in {L}^{*}.$

Definition 6.6. A negator on L* is any decreasing mapping $\mathcal{N}:{L}^{*}\to {L}^{*}$ satisfying $\mathcal{N}\left({0}_{{L}^{*}}\right)={1}_{{L}^{*}}$and $\mathcal{N}\left({1}_{{L}^{*}}\right)={0}_{{L}^{*}}$. If $\mathcal{N}\left(\mathcal{N}\left(x\right)\right)=x$ for all xL*, then $\mathcal{N}$ is called an involutive negator. A negator on [0, 1] is a decreasing function N : [0, 1] → [0, 1] satisfying N(0) = 1 and N(1) = 0. N s denotes the standard negator on [0, 1] defined by

${N}_{s}\left(x\right)=1-x,\phantom{\rule{1em}{0ex}}\forall x\in \left[0,1\right].$

Definition 6.7. Let μ and ν be measure and non-measure distribution functions from X × (0, +∞) to [0, 1] such that μ x (t) + ν x (t) ≤ 1 for all xX and t > 0. The triple $\left(X,{\mathcal{P}}_{\mu ,\nu },\mathcal{T}\right)$ is said to be an intuitionistic random normed space (briefly IRN-space) if X is a vector space, $\mathcal{T}$ is continuous t-representable and ${\mathcal{P}}_{\mu ,\nu }$ is a mapping X × (0, +∞) → L* satisfying the following conditions: for all x, yX and t, s > 0,

1. (a)

${\mathcal{P}}_{\mu ,\nu }\left(x,0\right)={0}_{{L}^{*}}$;

2. (b)

${\mathcal{P}}_{\mu ,\nu }\left(x,t\right)={1}_{{L}^{*}}$ if and only if x = 0;

3. (c)

${\mathcal{P}}_{\mu ,\nu }\left(\alpha x,t\right)={\mathcal{P}}_{\mu ,\nu }\left(x,\frac{t}{|\alpha |}\right)$ for all α ≠ 0;

4. (d)

${\mathcal{P}}_{\mu ,\nu }\left(x+y,t+s\right){\ge }_{{L}^{*}}\phantom{\rule{0.3em}{0ex}}\mathcal{T}\left({\mathcal{P}}_{\mu ,\nu }\left(x,t\right),{\mathcal{P}}_{\mu ,\nu }\left(y,s\right)\right)$.

In this case, ${\mathcal{P}}_{\mu ,\nu }$ is called an intuitionistic random norm. Here,

${\mathcal{P}}_{\mu ,\nu }\left(x,t\right)=\left({\mu }_{x}\left(t\right),{\nu }_{x}\left(t\right)\right).$

Example 6.8. Let (X, || · ||) be a normed space. Let $\mathcal{T}\left(a,b\right)=\left({a}_{1}{b}_{1},min\left({a}_{2}+{b}_{2},1\right)\right)$ for all a = (a1, a2), b = (b1, b2) ∈ L* and let μ, ν be measure and non-measure distribution functions defined by

${\mathcal{P}}_{\mu ,\nu }\left(x,t\right)=\left({\mu }_{x}\left(t\right),{\nu }_{x}\left(t\right)\right)=\left(\frac{t}{t+||x||},\frac{||x||}{t+||x||}\right),\phantom{\rule{1em}{0ex}}\forall t\in {R}^{+}.$

Then $\left(X,{\mathcal{P}}_{\mu ,\nu },\mathcal{T}\right)$ is an IRN-space.

Definition 6.9. (1) A sequence {x n } in an IRN-space $\left(X,{\mathcal{P}}_{\mu ,\nu },\mathcal{T}\right)$ is called a Cauchy sequence if, for any ε > 0 and t > 0, there exists an n0 ∈ ℕ such that

${\mathcal{P}}_{\mu ,\nu }\left({x}_{n}-{x}_{m},t\right){>}_{{L}^{*}}\left({N}_{s}\left(\epsilon \right),\epsilon \right),\phantom{\rule{1em}{0ex}}\forall n,m\ge {n}_{0},$

where N s is the standard negator.

1. (2)

The sequence {x n } is said to be convergent to a point xX (denoted by${x}_{n}\stackrel{{\mathcal{P}}_{\mu ,\nu }}{\to }x$) if ${\mathcal{P}}_{\mu ,\nu }\left({x}_{n}-x,t\right)\to {1}_{{L}^{*}}$ as n → ∞ for every t > 0.

2. (3)

An IRN-space $\left(X,{\mathcal{P}}_{\mu ,\nu },\mathcal{T}\right)$ is said to be complete if every Cauchy sequence in X is convergent to a point xX.

## 7. Stability results in intuitionistic random normed spaces

In this section, we prove the generalized Ulam-Hyers stability of the quartic functional equation in intuitionistic random normed spaces.

Theorem 7.1. Let X be a linear space and let $\left(X,{\mathcal{P}}_{\mu ,\nu },\mathcal{T}\right)$ be a complete IRN-space. Let f : XY be a mapping with f(0) = 0 for which there are ξ, ζ : X2D+, where ξ (x, y) is denoted by ξx,yand ζ(x, y)is denoted by ζx,y, further, (ξx,y(t), ζx,y(t)) is denoted by Qξ,ζ(x, y, t), with the property:

$\begin{array}{l}{\mathcal{P}}_{\mu ,\nu }\left(16f\left(x+4y\right)+f\left(4x-y\right)-306\left[9f\left(x+\frac{y}{3}\right)+f\left(x+2y\right)\right]\\ \phantom{\rule{0.1em}{0ex}}-136f\left(x-y\right)+1394f\left(x+y\right)-425f\left(y\right)+1530f\left(x\right),t\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}{\ge }_{{L}^{*}}{Q}_{\xi ,\zeta }\left(x,y,t\right).\end{array}$
(7.1)

If

${\mathcal{T}}_{i=1}^{\infty }\left({Q}_{\xi ,\zeta }\left({4}^{n+i-1}x,0,{4}^{4n+3i+3}t\right)\right)={1}_{{L}^{*}}$
(7.2)

and

$\underset{n\to \infty }{lim}{Q}_{\xi ,\zeta }\left({4}^{n}x,{4}^{n}y,{4}^{4n}t\right)={1}_{{L}^{*}}$
(7.3)

for all x, yX and all t > 0, then there exists a unique quartic mapping Q : XY such that

${\mathcal{P}}_{\mu ,\nu }\left(f\left(x\right)-Q\left(x\right),t\right){\ge }_{{L}^{*}}\phantom{\rule{0.3em}{0ex}}{\mathcal{T}}_{i=1}^{\infty }\left({Q}_{\xi ,\zeta }\left({4}^{i-1}x,0,{4}^{3i+3}t\right)\right).$
(7.4)

Proof. Putting y = 0 in (7.1), we have

${\mathcal{P}}_{\mu ,\nu }\left(\frac{f\left(4x\right)}{256}-f\left(x\right),t\right){\ge }_{{L}^{*}}\phantom{\rule{0.3em}{0ex}}{Q}_{\xi ,\zeta }\left(x,0,{4}^{4}t\right).$
(7.5)

Therefore, it follows that

${\mathcal{P}}_{\mu ,\nu }\left(\frac{f{\left(4}^{k+1}x\right)}{{4}^{4\left(k+1\right)}}-\frac{f{\left(4}^{k}x\right)}{{4}^{4k}},\frac{t}{{4}^{4k}}\right){\ge }_{{L}^{*}}\phantom{\rule{0.1em}{0ex}}{Q}_{\xi ,\zeta }{\left(4}^{k}x{,0,4}^{4}t\right),$
(7.6)

which implies that

${\mathcal{A}}_{\mu ,\nu }\left(\frac{f{\left(4}^{k+1}x\right)}{{4}^{4\left(k+1\right)}}-\frac{f{\left(4}^{k}x\right)}{{4}^{4k}},t\right){\ge }_{{L}^{*}}\phantom{\rule{0.1em}{0ex}}{Q}_{\xi ,\zeta }{\left(4}^{k}x{,0,4}^{4\left(k+1\right)}t\right),$
(7.7)

that is,

${\mathcal{P}}_{\mu ,\nu }\left(\frac{f\left({4}^{k+1}x\right)}{{4}^{4\left(k+1\right)}}-\frac{f\left({4}^{k}x\right)}{{4}^{4k}},\frac{t}{{4}^{k+1}}\right){\ge }_{L*}{Q}_{\xi ,\zeta }\left({4}^{k}x,0,{4}^{4\left(k+1\right)}t\right)$
(7.8)

for all kN and all t > 0. As 1 > 1/4 + ⋯ + 1/4n, from the triangle inequality, it follows

(7.9)

In order to prove convergence of the sequence $\left\{\frac{f\left({4}^{n}x\right)}{{256}^{n}}\right\}$, replacing x with 4mx in (7.9), we get that for m, n > 0

${\mathcal{P}}_{\mu ,\nu }\left(\frac{f\left({4}^{n+m}x\right)}{{256}^{\left(n+m\right)}}-\frac{f\left({4}^{m}x\right)}{{256}^{m}},t\right){\ge }_{{L}^{*}}\phantom{\rule{0.3em}{0ex}}{\mathcal{T}}_{i=1}^{n}\left({Q}_{\xi ,\zeta }\left({4}^{i+m-1}x,0,{4}^{3i+4m+3}t\right)\right).$
(7.10)

Since the right-hand side of the inequality tends 1L*as m tends to infinity, the sequence $\left\{\frac{f\left({4}^{n}x\right)}{{4}^{4n}}\right\}$ is a Cauchy sequence. So we may define $Q\left(x\right)={lim}_{n\to \infty }\frac{f\left({4}^{n}x\right)}{{4}^{4n}}$ for all xX.

Now, we show that Q is a quartic mapping. Replacing x, y with 4nx and 4ny, respectively, in (7.1), we obtain

$\begin{array}{l}{\mathcal{P}}_{\mu ,\nu }\left(\frac{f{\left(4}^{n}\left(x+4y\right)\right)}{{256}^{n}}+\frac{f{\left(4}^{n}\left(4x-y\right)\right)}{{256}^{n}}-\frac{306\left[9f{\left(4}^{n}\left(x+\frac{y}{3}\right)\right)+f{\left(4}^{n}\left(x+2y\right)\right)}{{256}^{n}}\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}-\frac{136f{\left(4}^{n}\left(x-y\right)\right)}{{256}^{n}}+\frac{1394f{\left(4}^{n}\left(x+y\right)\right)}{{256}^{n}}-\frac{425f{\left(4}^{n}\left(y\right)\right)}{{256}^{n}}+\frac{1530f{\left(4}^{n}\left(x\right)\right)}{{256}^{n}},t\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}{\ge }_{L*}{Q}_{\xi ,\zeta }{\left(4}^{n}x{,4}^{n}y{,4}^{4n}t\right).\end{array}$
(7.11)

Taking the limit as n → ∞, we find that Q satisfies (1.1) for all x, yX.

Taking the limit as n → ∞ in (7.9), we obtain (7.4).

To prove the uniqueness of the quartic mapping Q subject to (7.4), let us assume that there exists another quartic mapping Q' which satisfies (7.4). Obviously, we have xX and all n ∈ ℕ. Hence it follows from (7.4) that

$\begin{array}{l}{\mathcal{P}}_{\mu ,\nu }\left(Q\left(x\right)-{Q}^{\prime }\left(x\right),t\right)\\ \phantom{\rule{0.1em}{0ex}}{\ge }_{{L}^{*}}{\mathcal{P}}_{\mu ,\nu }\left(Q{\left(4}^{n}x\right)-{Q}^{\prime }{\left(4}^{n}x{\right),4}^{4n}t\right)\\ \phantom{\rule{0.1em}{0ex}}{\ge }_{{L}^{*}}\mathcal{T}\left({\mathcal{P}}_{\mu ,\nu }\left(Q{\left(4}^{n}x\right)-f{\left(4}^{n}x{\right),4}^{4n-1}t\right),{\mathcal{P}}_{\mu ,\nu }\left(f{\left(4}^{n}x\right)-{Q}^{\prime }{\left(4}^{n}x{\right),4}^{4n-1}t\right)\right)\\ \phantom{\rule{0.1em}{0ex}}{\ge }_{{L}^{*}}\mathcal{T}\left({\mathcal{T}}_{i=1}^{\infty }\left({Q}_{\xi ,\zeta }{\left(4}^{n+i-1}x{,0,4}^{4n+3i+3}t\right)\right),{\mathcal{T}}_{i=1}^{\infty }\left({Q}_{\xi ,\zeta }{\left(4}^{n+i-1}x{,0,4}^{4n+3i+3}t\right)\right)\end{array}$

for all xX. By letting n → ∞ in (7.4), we prove the uniqueness of Q. This completes the proof of the uniqueness, as desired. □

Corollary 7.2. Let $\left(X,{{\mathcal{P}}^{\prime }}_{{\mu }^{\prime },{\nu }^{\prime }},\mathcal{T}\right)$ be an IRN-space and let $\left(Y,{\mathcal{P}}_{\mu ,\nu },\mathcal{T}\right)$ be a complete IRN-space. Let f : XY be a mapping such that

$\begin{array}{l}{\mathcal{P}}_{\mu ,\nu }\left(16f\left(x+4y\right)+f\left(4x-y\right)-306\left[9f\left(x+\frac{y}{3}\right)+f\left(x+2y\right)\right]\\ \phantom{\rule{0.1em}{0ex}}-136f\left(x-y\right)+1394f\left(x+y\right)-425f\left(y\right)+1530f\left(x\right),t\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}{\ge }_{L*}{{\mathcal{P}}^{\prime }}_{{\mu }^{\prime },{\nu }^{\prime }}\left(x+y,t\right)\end{array}$

for all t > 0 in which

$\underset{n\to \infty }{lim}{\mathcal{T}}_{i=1}^{\infty }\left({\mathcal{P}}_{{\mu }^{\prime },{\nu }^{\prime }}^{\prime }\left(x,{4}^{4n+3i+3}t\right)\right)={1}_{{L}^{*}}$

for all x, yX. Then there exists a unique quartic mapping Q : XY such that

${\mathcal{P}}_{\mu ,\nu }\left(f\left(x\right)-Q\left(x\right),t\right){\ge }_{L*}{\mathcal{T}}_{i=1}^{\infty }\left({\mathcal{P}}_{{\mu }^{\prime },{\nu }^{\prime }}^{\prime }\left(x,{4}^{3i+3}t\right)\right).$

Now, we give an example to illustrate the main result of Theorem 7.1 as follows.

Example 7.3. Let (X, ||.||) be a Banach algebra, $\left(X,{\mathcal{P}}_{\mu ,\nu },M\right)$ an IRN-space in which

${\mathcal{P}}_{\mu ,\nu }\left(x,t\right)=\left(\frac{t}{t+||x||},\frac{||x||}{t+||x||}\right)$

and let $\left(Y,{\mathcal{P}}_{\mu ,\nu },M\right)$ be a complete IRN-space for all xX. Define f : XX by f (x) = x4 + x0, where x0 is a unit vector in X. A straightforward computation shows that

$\begin{array}{l}{\mathcal{P}}_{\mu ,\nu }\left(16f\left(x+4y\right)+f\left(4x-y\right)-306\left[9f\left(x+\frac{y}{3}\right)+f\left(x+2y\right)\right]\\ \phantom{\rule{0.1em}{0ex}}-136f\left(x-y\right)+1394f\left(x+y\right)-425f\left(y\right)+1530f\left(x\right),t\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}{\ge }_{L*}{\mathcal{P}}_{\mu ,\nu }\left(x+y,t\right),\phantom{\rule{0.1em}{0ex}}\forall t>0.\end{array}$

Also

Therefore, all the conditions of 7.1 hold and so there exists a unique quartic mapping Q : XY such that

${\mathcal{P}}_{\mu ,\nu }\left(f\left(x\right)-Q\left(x\right),t\right){\ge }_{{L}^{*}}\phantom{\rule{0.3em}{0ex}}{\mathcal{P}}_{\mu ,\nu }\left(x,{4}^{6}t\right).$