On the stability of a cubic functional equation in random 2-normed spaces

Abstract

In this article, we propose to determine some stability results for the functional equation of cubic in random 2-normed spaces which seems to be a quite new and interesting idea. Also, we define the notion of continuity, approximately and conditional cubic mapping in random 2-normed spaces and prove some interesting results.

1 Introduction and preliminaries

In 1940, Ulam [1] proposed the following question concerning the stability of group homomorphisms:

Let G₁ be a group and let G₂ be a metric group with the metric d(., .). Given ϵ > 0, does there exist a δ > 0 such that if a function h : G₁ → G₂ satisfies the inequality d(h(xy), h(x)h(y)) < δ for all x, y ∈ G₁, then there exists a homomorphism H : G₁ → G₁ with d(h(x), H(x)) < ϵ for all x ∈ G₁?

In next year, Hyers [2] answers the problem of Ulam under the assumption that the groups are Banach spaces and then generalized by Aoki [3] and Rassias [4] for additive mappings and linear mappings, respectively. Since then several stability problems for various functional equations have been investigated in [5–12].

The stability problem for the cubic functional equation was proved by Jun and Kim [5] for mappings f: X → Y, where X is a real normed space and Y is a Banach space. Later on, the problem of stability of cubic functional equation were discussed by many mathematician.

An interesting and important generalization of the notion of a metric space was introduced by Menger [13] under the name of statistical metric space, which is now called a probabilistic metric space. An important family of probabilistic metric spaces is that of probabilistic normed spaces. The theory of probabilistic normed spaces is important as a generalization of deterministic results of linear normed spaces. The theory of probabilistic normed spaces was initiated and developed in [14, 15] and further it was extended to random 2-normed spaces by Goleţ [16] using the concept of 2-norm of Gahler [17]. For more details of probabilistic and random/fuzzy 2-normed space, we refer to [18–22] and references therein.

In this article, we establish Hyers-Ulam stability concerning the cubic functional equations in random 2-normed spaces which is quite a new and interesting idea to study with.

In this section, we recall some notations and basic definitions used in this article.

A distribution function is an element of Δ⁺, where Δ⁺ = {f : ℝ → [0, 1]; f is left-continuous, nondecreasing, f(0) = 0 and f(+∞) = 1} and the subset D⁺ ⊆ Δ⁺ is the set D⁺ = {f ∈ Δ⁺; l^-f(+∞) = 1}. Here l^-f(+∞) denotes the left limit of the function f at the point x. The space Δ⁺ is partially ordered by the usual point-wise ordering of functions, i.e., f ≤ g if and only if f(x) ≤ g(x) for all x ∈ ℝ. For any a ∈ ℝ, H _a is a distribution function defined by

$$H_a\left(x\right)=\left\{\begin{array}{c}0\mathsf{\text{if}}x\le a;\hfill \\ 1\mathsf{\text{if}}x>a.\hfill \end{array}\right.$$

The set Δ, as well as its subsets, can be partially ordered by the usual pointwise order: in this order, H₀ is the maximal element in Δ⁺.

A triangle function is a binary operation on Δ⁺, namely a function τ : Δ⁺ × Δ⁺ → Δ⁺ that is associative, commutative nondecreasing and which has ε₀ as unit, that is, for all f, g, h ∈ Δ⁺, we have:

• (i) τ(τ(f, g), h) = τ(f, τ(g, h)),

• (ii) τ(f, g) = τ(g, f),

• (iii) τ(f, g) = τ(g, f) whenever f ≤ g,

• (iv) τ(f, H ₀) = f.

A t-norm is a continuous mapping * : [0, 1] × [0, 1] → [0, 1] such that ([0, 1], *) is abelian monoid with unit one and c * d ≥ a * b if c ≥ a and d ≥ b for all a, b, c, d ∈ [0, 1].

The concept of 2-normed space was first introduced in [17] and further studied in [23–25].

Let X is a linear space of a dimension d, where 2 ≤ d < ∞. A 2-normed on X is a function ∥., .∥ : X × X → ℝ satisfying the following conditions, for every x, y ∈ X (i) ∥x, y∥ = 0 if and only if x and y are linearly dependent; (ii) ∥x, y∥ = ∥y, x∥; (iii) ∥αx, y∥ = |α|∥x, y∥, α ∈ ℝ; (iv) ∥x + y, z∥ ≤ ∥x, z∥ + ∥y, z∥. In this case (X, ∥., . ∥) is called a 2-norm space.

Example 1.1. Take X = ℝ² being equipped with the 2-norm ∥x, y∥ = the area of the parallelogram spanned by the vectors x and y, which may be given explicitly by the formula

$$∥x,y∥=\left|x_1{y}_2-x_2{y}_1\right|,\mathsf{\text{where}}x=\left(x_1,x_2\right),y=\left(y_1,y_2\right).$$

Recently, Goleţ [16] introduced the notion of random 2-normed space and further studied by Mursaleen [26].

Let X be a linear space of a dimension greater than one, τ is a triangle function, and $ℱ:X\times X\to {\Delta }^{+}$. Then ℱ is called a probabilistic 2-norm on X and $\left(X,ℱ,\tau \right)$ a probabilistic 2-normed space if the following conditions are satisfied:

• (i) ${ℱ}_{x,y}\left(t\right)=H_0\left(t\right)$ if x and y are linearly dependent, where ${ℱ}_{x,y}\left(t\right)$ denotes the value of ${ℱ}_{x,y}$ at t ∈ ℝ,

• (ii) ${ℱ}_{x,y}\ne H_0$ if x and y are linearly independent,

• (iii) ${ℱ}_{x,y}={ℱ}_{y,x}$ for every x, y in X,

• (iv) ${ℱ}_{\alpha x,y}\left(t\right)={ℱ}_{x,y}\left(\frac{t}{\left|\alpha \right|}\right)$ for every t > 0, α ≠ 0 and x, y ∈ X,

• (v) ${ℱ}_{x+y,z}\ge \tau \left({ℱ}_{x,z},{ℱ}_{y,z}\right)$ whenever x, y, z ∈ X.

If (v) is replaced by

(v') ${ℱ}_{x+y,z}\left(t_1+t_2\right)\ge {ℱ}_{x,z}\left(t_1\right)*{ℱ}_{y,z}\left(t_2\right)$, for all x, y, z ∈ X and $t_1,t_2\in {ℝ}_0^{+}$, then triple $\left(X,ℱ,*\right)$ is called a random 2-normed space (for short, RTN-space).

Example 1.2. Let (X, ∥., .∥) be a 2-normed space with ∥x, z∥ = ∥x₁z₂ - x₂z₁∥, x = (x₁, x₂), z = (z₁, z₂) and a * b = ab for a, b ∈ [0, 1]. For all x ∈ X, t > 0 and nonzero z ∈ X, consider

$${ℱ}_{x,z}\left(t\right)=\left\{\begin{array}{cc}\frac{t}{t+∥x,z∥}\hfill & \mathsf{\text{if}}t>0\hfill \\ 0\hfill & \mathsf{\text{if}}t\le 0;\hfill \end{array}\right.$$

Then $\left(X,ℱ,*\right)$ is a random 2-normed space.

Remark 1.3. Note that every 2-normed space (X, ∥., .∥) can be made a random 2-normed space in a natural way, by setting ${ℱ}_{x,y}\left(t\right)=H_0\left(t-∥x,y∥\right)$, for every x, y ∈ X, t > 0 and a * b = min{a, b}, a, b ∈ [0, 1].

2 Stability of cubic functional equation

In the present section, we define the notion of convergence, Cauchy sequence and completeness in RTN-space and determine some stability results of the cubic functional equation in RTN-space.

The functional equation

$$f\left(2x+y\right)+f\left(2x-y\right)=2f\left(x+y\right)+2f\left(x-y\right)+12f\left(x\right)$$

(1)

is called the cubic functional equation, since the function f(x) = cx³ is its solution. Every solution of the cubic functional equation is said to be a cubic mapping.

We shall assume throughout this article that X and Y are linear spaces; $\left(X,ℱ,*\right)$ and $\left(Z,{ℱ}^{\prime },*\right)$ are random 2-normed spaces; and $\left(Y,ℱ,*\right)$ is a random 2-Banach space.

Let φ be a function from X × X to Z. A mapping f : X → Y is said to be φ-approximately cubic function if

$${ℱ}_{E_{x,y},z}\left(t\right)\ge {ℱ}_{\phi \left(x,y\right),z}^{\prime }\left(t\right),$$

(2)

for all x, y ∈ X, t > 0 and nonzero z ∈ X, where

$$E_{x,y}=f\left(2x+y\right)+f\left(2x-y\right)-2f\left(x+y\right)-2f\left(x-y\right)-12f\left(x\right).$$

We define:

We say that a sequence x = (x _k ) is convergent in $\left(X,ℱ,*\right)$ or simply ℱ-convergent to ℓ if for every ϵ > 0 and θ ∈ (0, 1) there exists k₀ ∈ ℕ such that ${ℱ}_{x_k-\ell ,z}\left(\epsilon \right)>1-\theta$ whenever k ≥ k₀ and nonzero z ∈ X. In this case we write $ℱ-\underset{k\to \infty }{\text{lim}}{x}_k=\ell$ and ℓ is called the ℱ-limit of x = (x _k ).

A sequence x = (x _k ) is said to be Cauchy sequence in $\left(X,ℱ,*\right)$ or simply ℱ-Cauchy if for every ϵ > 0, θ > 0 and nonzero z ∈ X there exists a number N = N(ϵ, z) such that $\mathsf{\text{lim}}{ℱ}_{x_n-x_m,z}\left(\epsilon \right)>1-\theta$ for all n, m ≥ N. RTN-space $\left(X,ℱ,*\right)$ is said to be complete if every ℱ-Cauchy is ℱ-convergent. In this case $\left(X,ℱ,*\right)$ is called random 2-Banach space.

Theorem 2.1. Suppose that a function φ : X × X → Z satisfies φ(2x, 2y) = αφ(x, y) for all x, y ∈ X and α ≠ 0. Let f : X → Y be a φ-approximately cubic function. If for some 0 < α < 8,

$${ℱ}_{\phi \left(2x,2y\right),z}^{\prime }\left(t\right)\ge {ℱ}_{\alpha \phi \left(x,y\right),z}^{\prime }\left(t\right),$$

(3)

and $\underset{n\to \infty }{\text{lim}}{ℱ}_{\phi \left(2^{n}x,2^{n}y\right),z}^{\prime }\left(8^{n}t\right)=1$ for all x, y ∈ X, t > 0 and nonzero z ∈ X. Then there exists a unique cubic mapping C : X → Y such that

$${ℱ}_{C\left(x\right)-f\left(x\right),z}\left(t\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(\left(8-\alpha \right)t\right),$$

(4)

for all x ∈ X, t > 0 and nonzero z ∈ X.

Proof. For convenience, let us fix y = 0 in (2). Then for all x ∈ X, t > 0 and nonzero z ∈ X

$${ℱ}_{\frac{f\left(2x\right)}{8}-f\left(x\right),z}\left(\frac{t}{16}\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(t\right).$$

(5)

Replacing x by 2ⁿx in (5) and using (3), we obtain

$${ℱ}_{\frac{f\left(2^{n+1}x\right)}{8^{n+1}}-\frac{f\left(2^{n}x\right)}{8^n},z}\left(\frac{t}{16\left(8^n\right)}\right)\ge {ℱ}_{\phi \left(2^{n}x,0\right),z}^{\prime }\left(t\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(t/{\alpha }^n\right),$$

for all x ∈ X, t > 0 and nonzero z ∈ X; and for all n ≥ 0. By replacing t by αⁿt, we get

$${ℱ}_{\frac{f\left(2^{n+1}x\right)}{8^{n+1}}-\frac{f\left(2^{n}x\right)}{8^n},z}\left(\frac{{\alpha }^{n}t}{16\left(8^n\right)}\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(t\right).$$

(6)

It follows from $\frac{f\left(2^{n}x\right)}{8^n}-f\left(x\right)=\sum _{k=0}^{n=1}\left(\frac{f\left(2^{k+1}x\right)}{8^{k+1}}-\frac{f\left(2^{k}x\right)}{8^k}\right)$ and (6) that

$${ℱ}_{\frac{f\left(2^{n}x\right)}{8^n}-f\left(x\right),z}\left(\sum _{k=0}^{n-1}\frac{{\alpha }^{k}t}{16\left(8^k\right)}\right)\ge \prod _{k=0}^{n-1}{ℱ}_{\frac{f\left(2^{k+1}x\right)}{8^{k+1}}-\frac{f\left(2^{k}x\right)}{8^k},z}\left(\frac{{\alpha }^{k}t}{16\left(8^k\right)}\right)\ge {{ℱ}^{\prime }}_{\phi \left(x,0\right),z}\left(t\right),$$

(7)

for all x ∈ X, t > 0 and n > 0 where ${\prod }_{j=1}^n{a}_j=a_1*a_2*\cdots *a_n$. By replacing x with 2^mx in (7), we have

$${ℱ}_{\frac{f\left(2^{n+m}x\right)}{8^{n+m}}-\frac{f\left(2^{m}x\right)}{8^m},z}\left(\sum _{k=0}^{n-1}\frac{{\alpha }^{k}t}{16{\left(8\right)}^{k+m}}\right)\ge {ℱ}_{\phi \left(2^{m}x,0\right),z}^{\prime }\left(t\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(t/{\alpha }^m\right).$$

Thus

$${ℱ}_{\frac{f\left(2^{n+m}x\right)}{8^{n+m}}-\frac{f\left(2^{m}x\right)}{8^m},z}\left(\sum _{k=m}^{n+m-1}\frac{{\alpha }^{k}t}{16{\left(8\right)}^k}\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(t\right),$$

for all x ∈ X, t > 0, m > 0, n ≥ 0 and nonzero z ∈ X. Hence

$${ℱ}_{\frac{f\left(2^{n+m}x\right)}{8^{n+m}}-\frac{f\left(2^{m}x\right)}{8^m},z}\left(t\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(\frac{t}{{\sum }_{k=m}^{n+m-1}\frac{{\alpha }^k}{16{\left(8\right)}^k}}\right),$$

(8)

for all x ∈ X, t > 0 m ≥ 0, n ≥ 0 and nonzero z ∈ X. Since 0 < α < 8 and $\sum _{k=0}^{\infty }{\left(\frac{\alpha }{8}\right)}^k<\infty$, the Cauchy criterion for convergence shows that $\left(\frac{f\left(2^{n}x\right)}{8^n}\right)$ is a Cauchy sequence in $\left(Y,ℱ,*\right)$. Since $\left(Y,ℱ,*\right)$ is complete, this sequence converges to some point C(x) ∈ Y. Fix x ∈ X and put m = 0 in (8) to obtain

$${ℱ}_{\frac{f\left(2^{n}x\right)}{8^n}-f\left(x\right),z}\left(t\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(\frac{t}{{\sum }_{k=0}^{n-1}\frac{{\alpha }^k}{16{\left(8\right)}^k}}\right),$$

for all t > 0, n > 0 and nonzero z ∈ X. Thus we obtain

$${ℱ}_{C\left(x\right)-f\left(x\right),z}\left(t\right)\ge {ℱ}_{C\left(x\right)-\frac{f\left(2^{n}x\right)}{8^n},z}\left(t/2\right)*{ℱ}_{\frac{f\left(2^{n}x\right)}{8^n}-f\left(x\right),z}\left(t/2\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(\frac{t}{{\sum }_{k=0}^{n-1}\frac{a^k}{8{\left(8\right)}^k}}\right),$$

for large n. Taking the limit as n → ∞ and using the definition of RTN-space, we get

$${ℱ}_{C\left(x\right)-f\left(x\right),z}\left(t\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(\left(8-\alpha \right)t\right).$$

Replace x and y by 2ⁿx and 2ⁿy, respectively, in (2), we have

$${ℱ}_{\frac{E_2{n}_{x,2}{n}_y}{8^n},z}\left(t\right)\ge {ℱ}_{\phi \left(2^{n}x,2^{n}y\right),z}^{\prime }\left(8^{n}t\right),$$

for all x, y ∈ X, t > 0 and nonzero z ∈ X. Since

$$\underset{n\to \infty }{\text{lim}}{ℱ}_{\phi \left(2^{n}x,2^{n}y\right),z}^{\prime }\left(8^{n}t\right)=1,$$

we observe that C fulfills (1). To Prove the uniqueness of the cubic function C, assume that there exists a cubic function D : X → Y which satisfies (4). For fix x ∈ X, clearly C(2ⁿx) = 8ⁿC(x) and D(2ⁿx) = 8ⁿD(x) for all n ∈ ℕ. It follows from (4) that

$$\begin{array}{ll}\hfill {ℱ}_{C\left(x\right)-D\left(x\right),z}\left(t\right)& ={ℱ}_{\frac{C\left(2^{n}x\right)}{8^n}-\frac{D\left(2^{n}x\right)}{8^n},z}\left(t\right)\ge {ℱ}_{\frac{C\left(2^{n}x\right)}{8^n}-\frac{f\left(2^{n}x\right)}{8^n},z}\left(\frac{t}{2}\right)*{ℱ}_{\frac{f\left(2^{n}x\right)}{8^n}-\frac{D\left(2^{n}x\right)}{8^n},z}\left(\frac{t}{2}\right)\\ \ge {ℱ}_{\phi \left(2^{n}x,0\right),z}^{\prime }\left(\frac{8^n\left(8-\alpha \right)t}{2}\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(\frac{8^n\left(8-\alpha \right)t}{2{\alpha }^n}\right).\end{array}$$

Therefore

$${ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(\frac{8^n\left(8-\alpha \right)t}{2{\alpha }^n}\right)=1.$$

Thus ${ℱ}_{C\left(x\right)-D\left(x\right),z}\left(t\right)=1$ for all x ∈ X, t > 0 and nonzero z ∈ X. Hence C(x) = D(x).

Example 2.2. Let X be a Hilbert space and Z be a normed space. By ℱ and ${ℱ}^{\prime }$, we denote the random 2-norms given as in Example 1.1 on X and Z, respectively. Let φ : X × X → Z be defined by φ(x, y) = 8(∥x∥² + ∥y∥²)z_ο, where z_ο is a fixed unit vector in Z. Define f : X → X by f(x) = ∥x∥²x + ∥x∥²x_ο for some unit vector x_ο ∈ X. Then

$${ℱ}_{E_{x,y},z}\left(t\right)=\frac{t}{t+8{∥x,z∥}^2+2{∥y,z∥}^2}\ge \frac{t}{t+8{∥x,z∥}^2+8{∥y,z∥}^2}={ℱ}_{\phi \left(x,y\right),z}^{\prime }\left(t\right).$$

Also

$${ℱ}_{\phi \left(2x,0\right),z}^{\prime }\left(t\right)=\frac{t}{t+32{∥x,z∥}^2}={ℱ}_{4\phi \left(x,0\right),z}^{\prime }\left(t\right).$$

Thus,

$$\underset{n\to \infty }{\text{lim}}{ℱ}_{\phi \left(2^{n}x,2^{n}y\right),z}^{\prime }\left(8^{n}t\right)=\underset{n\to \infty }{\text{lim}}\frac{8^{n}t}{8^{n}t+8\left(4^n\right)\left({∥x,z∥}^2+{∥y,z∥}^2\right)}=1.$$

Hence, conditions of Theorem 2.1 for α = 4 are fulfilled. Therefore, there is a unique cubic mapping C : X → X such that ${ℱ}_{C\left(x\right)-f\left(x\right),z}\left(t\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(4t\right)$ for all x ∈ X, t > 0 and nonzero z ∈ X.

By a modification in the proof of Theorem 2.1, one can easily prove the following:

Theorem 2.3. Suppose that a function φ : X × X → Z satisfies $\phi \left(x/2,y/2\right)=\frac{1}{\alpha }\phi \left(x,y\right)$ for all x, y ∈ X and α ≠ 0. Let f : X → Y be a φ-approximately cubic function. If for some α > 8

$${ℱ}_{\phi \left(x/2,y/2\right),z}^{\prime }\left(t\right)\ge {ℱ}_{\phi \left(x,y\right),z}^{\prime }\left(\alpha t\right)$$

and $\underset{n\to \infty }{\text{lim}}{ℱ}_{8^n\phi \left(2^{-n}x,2^{-n}y\right),z}^{\prime }\left(t\right)=1$ for all x, y ∈ X, t > 0 and nonzero z ∈ X. Then there exists a unique cubic mapping C : X → Y such that

$${ℱ}_{C\left(x\right)-f\left(x\right),z}\left(t\right)\ge {ℱ}_{\phi \left(x,0\right),z}^{\prime }\left(\left(\alpha -8\right)t\right),$$

for all x ∈ X, t > 0 and nonzero z ∈ X.

3 Continuity in random 2-normed spaces

In this section, we establish some interesting results of continuous approximately cubic mappings.

Let f : ℝ → X be a function, where ℝ is endowed with the Euclidean topology and X is an random 2-normed space equipped with random 2-norm ℱ. Then, f is said to be random 2-continuous or simply ℱ-continuous at a point s_ο ∈ ℝ if for all ϵ > 0 and all 0 < α < 1 there exists δ > 0 such that

$${ℱ}_{f\left(sx\right)-f\left(s_{\circ }x\right),z}\left(\epsilon \right)\ge \alpha ,$$

for each s with 0 < |s - s_ο| < δ and nonzero z ∈ X.

A mapping f : X → Y is said to be (p, q)-approximately cubic function if, for some p, q and some z_ο ∈ Z,

$${ℱ}_{E_{x,y},z}\left(t\right)\ge {ℱ}_{\left({∥x∥}^p+{∥y∥}^q\right)z_{\circ },z}^{\prime }\left(t\right),$$

for all x, y ∈ X, t > 0 and nonzero z ∈ X.

Theorem 3.2. Let X be a normed space and let f : X → Y be a (p, q)-approximately cubic function. If p, q < 3, there exists a unique cubic mapping C : X → Y such that

$${ℱ}_{C\left(x\right)-f\left(x\right),z}\left(t\right)\ge {ℱ}_{{∥x∥}^p{z}_{\circ },z}^{\prime }\left(\left(8-2^p\right)t\right),$$

(9)

for all x ∈ X, t > 0 and nonzero z ∈ X. Furthermore, if for some x ∈ X and all n ∈ ℕ, the mapping g : ℝ → Y defined by g(s) = f(2ⁿsx) is ℱ-continuous. Then the mapping s ↦ C(sx) from ℝ to Y is ℱ-continuous; in this case, C(rx) = r³C(x) for all r ∈ ℝ.

Proof. Suppose that a function φ : X × X → Z satisfies φ(x, y) = (∥x∥^p+∥y∥^q)z_ο. Existence and uniqueness of the cubic mapping C satisfying (9) are deduced from Theorem 2.1. Note that for each x ∈ X, t ∈ ℝ and n ∈ ℕ, we have

$${ℱ}_{C\left(x\right)-\frac{f\left(2^{n}x\right)}{8^n},z}\left(t\right)={ℱ}_{C\left(2^{n}x\right)-f\left(2^{n}x\right),z}\left(8^{n}t\right)\ge {ℱ}_{2^{np}{∥x∥}^p{z}_{\circ },z}^{\prime }\left(8^n\left(8-2^p\right)t\right)={ℱ}_{{∥x∥}^p{z}_{\circ },z}^{\prime }\left(\frac{8^n\left(8-2^p\right)t}{2^{np}}\right).$$

(10)

Fix x ∈ X and s_ο ∈ ℝ. Given ϵ > 0 and 0 < α < 1. From (10) follows that

$${ℱ}_{C\left(sx\right)-\frac{f\left(2^{n}sx\right)}{8^n},z}\left(t\right)\ge {ℱ}_{{∥x∥}^p{z}_{\circ },z}^{\prime }\left(\frac{8^n\left(8-2^p\right)t}{{\left|s\right|}^p{2}^{np}}\right)\ge {ℱ}_{{∥x∥}^p{z}_{\circ },z}^{\prime }\left(\frac{8^n\left(8-2^p\right)t}{{\left(1+\left|s_{\circ }\right|\right)}^p{2}^{np}}\right),$$

for all |s - s_ο| < 1 and s ∈ ℝ. Since $\underset{n\to \infty }{\text{lim}}\frac{8^n\left(8-2^p\right)t}{{\left(1+\left|s_{\circ }\right|\right)}^p{2}^{np}}=\infty$, there exists n_ο ∈ ℕ such that

$${ℱ}_{C\left(sx\right)-\frac{f\left(2^{n_{\circ }}sx\right)}{8^{n_{\circ }}},z}\left(\frac{\epsilon }{3}\right)\ge \alpha ,$$

for all |s - s_ο| < 1 and s ∈ ℝ. By the ℱ-continuity of the mapping $t\to f\left(2^{n_{\circ }}tx\right)$, there exists δ < 1 such that for each s with 0 < |s - s_ο| < δ, we have

$${ℱ}_{\frac{f\left(2^{n_{\circ }}sx\right)}{8^{n_{\circ }}}-\frac{f\left(2^{n_{\circ }}{s}_{\circ }x\right)}{8^{n_{\circ }}},z}\left(\frac{\epsilon }{3}\right)\ge \alpha .$$

It follows that

$$\begin{array}{c}{ℱ}_{C\left(sx\right)-C\left(s_{\circ }x\right),z}\left(\epsilon \right)\\ \ge {ℱ}_{C\left(sx\right)-\frac{f\left(2^{n_{\circ }}sx\right)}{8^{n_{\circ }}},z}\left(\frac{\epsilon }{3}\right)*{ℱ}_{\frac{f\left(2^{n_{\circ }}sx\right)}{8^{n_{\circ }}}-\frac{f\left(2^{n_{\circ }}{s}_{\circ }x\right)}{8^{n_{\circ }}},z}\left(\frac{\epsilon }{3}\right)*{ℱ}_{C\left(s_{\circ }x\right)-\frac{f\left(2^{n_{\circ }}{s}_{\circ }x\right)}{8^{n_{\circ }}},z}\left(\frac{\epsilon }{3}\right)\ge \alpha ,\end{array}$$

for each s with 0 < |s - s_ο| < δ. Hence, the mapping s ↦ C(sx) is ℱ-continuous.

Now, we use the ℱ-continuity of s ↦ C(sx) to establish that $C\left(r_{\circ }x\right)=r_{\circ }^{3}C\left(x\right)$ for all r_ο ∈ ℝ. For each r, ℚ is a dense subset of ℝ, we have C(rx) = r³C(x). Fix r_ο ∈ ℝ and t > 0. Then, for 0 < α < 1 there exists δ > 0 such that

$${ℱ}_{C\left(rx\right)-C\left(r_{\circ }x\right),z}\left(t/3\right)\ge \alpha ,$$

for each r ∈ ℝ and 0 < |r - r_ο| < δ. Choose a rational number r with 0 < |r - r_ο| < δ and $\left|r^3-r_{\circ }^3\right|<1-\alpha$. Then

$$\begin{array}{c}{ℱ}_{C\left(r_{\circ }x\right)-r_{\circ }^{3}C\left(x\right),z}\left(t\right)\ge {ℱ}_{C\left(r_{\circ }x\right)-C\left(rx\right),z}\left(t/3\right)*{ℱ}_{C\left(rx\right)-r^{3}C\left(x\right),z}\left(t/3\right)*{ℱ}_{r^{3}C\left(x\right)-r_{\circ }^{3}C\left(x\right),z}\left(t/3\right)\\ \ge \alpha *1*{ℱ}_{C\left(x\right),z}\left(t/3\left(1-\alpha \right)\right).\end{array}$$

Thus ${ℱ}_{C\left(r_{\circ }x\right)-r_{\circ }^{3}C\left(x\right),z}\left(t\right)=1$. Hence, we conclude that $C\left(r_{\circ }x\right)=r_{\circ }^{3}C\left(x\right)$.

Remark 3.2. We can also prove Theorem 3.1 for the case when p, q > 3. In this case, there exists a unique cubic mapping C : X → Y such that ${ℱ}_{C\left(x\right)-f\left(x\right),z}\left(t\right)\ge {ℱ}_{{∥x∥}^p{z}_{\circ },z}^{\prime }\left(\left(2^p-8\right)t\right)$ for all x ∈ X, t > 0 and nonzero z ∈ X.

4 Approximately and conditional cubic mapping in random 2-normed spaces

In this section, we obtain completeness in RTN-space through the existence of some solution of a stability problem for cubic functional equation.

A mapping f : ℕ∪{0} → X is said to be approximately cubic if for each α ∈ (0, 1) there exists some n _α ∈ ℕ such that ${ℱ}_{E\left(n,m\right),z}\left(1\right)\ge \alpha$, for all n ≥ 2m ≥ n _α and nonzero z ∈ X.

By a conditional cubic mapping, we mean a mapping f : ℕ ∪ {0} → X such that (1) holds whenever x ≥ 2y.

It can be easily verified that for each conditional cubic mapping f : ℕ ∪ {0} → X, we have f(2ⁿ) = 2³ⁿf(1).

Theorem 4.1. Let $\left(X,ℱ,*\right)$ be a RTN-space such that for each approximately cubic mapping f : ℕ ∪ {0} → X, there exists a conditional cubic mapping C : ℕ ∪ {0} → X, such that

$$\underset{n\to \infty }{\text{lim}}{ℱ}_{C\left(n\right)-f\left(n\right),z}\left(1\right)=1,$$

for nonzero z ∈ X. Then $\left(X,ℱ,*\right)$ is a random 2-Banach space.

Proof. Let (x _n ) be a Cauchy sequence in a RTN-space. By induction on k, we can find a strictly increasing sequence (n _k ) of natural numbers such that

$${ℱ}_{x_n-x_m,z}\left(\frac{1}{{\left(10k\right)}^3}\right)\ge 1-\frac{1}{k},$$

for each n, m ≥ n _k and nonzero z ∈ X. Let $y_k=x_{n_k}$ and define f : ℕ ∪ {0} → X by f(k) = k³y _k . Let α ∈ (0, 1). and find some n_ο ∈ ℕ such that $1-\frac{1}{n_{\circ }}>\alpha$. One can easily verified that

$$\begin{array}{ll}\hfill {ℱ}_{E_{k,j},z}\left(1\right)& \ge {ℱ}_{y_{2k+j}-y_{k+j},z}\left(\frac{1}{20k^3}\right)*{ℱ}_{y_{2k+j}-y_{k-j},z}\left(\frac{1}{20k^3}\right)*{ℱ}_{y_{2k+j}-y_k,z}\left(\frac{1}{40k^3}\right)\\ *{ℱ}_{{y_2}_{k-j}-yk,z}\left(\frac{1}{80k^3}\right)*{ℱ}_{y_{2k+j}-y_{2k-j},z}\left(\frac{1}{120k^{2}j}\right)*{ℱ}_{y_{k-j}-y_{k+j},z}\left(\frac{1}{60k^{2}j}\right)\\ *{ℱ}_{y_{2k+j}-y_{k+j},z}\left(\frac{1}{60k{j}^2}\right)*{ℱ}_{y_{2k-j}-y_{k-j},z}\left(\frac{1}{60k{j}^2}\right)\\ *{ℱ}_{y_{2k+j}-y_{2k-j},z}\left(\frac{1}{10j^3}\right)*{ℱ}_{y_{k-j}-y_{k+j},z}\left(\frac{1}{20j^3}\right),\end{array}$$

for each k ≥ 2j, and nonzero z ∈ X. Then

$${ℱ}_{y_{2k+j}-y_{k+j},z}\left(\frac{1}{20k^3}\right)\ge {ℱ}_{y_{2k+j}-y_{k+j},z}\left(\frac{1}{20{\left(k+j\right)}^3}\right)\ge {ℱ}_{y_{2k+j}-y_{k+j},z}\left(\frac{1}{{10}^3{\left(k+j\right)}^3}\right)\ge \alpha ,$$

for j > n_ο and nonzero z ∈ X. Since $k-j\ge \frac{k}{2}$ and k - j ≥ j, we have

$$\begin{array}{ll}\hfill {ℱ}_{y_{2k+j}-y_{k-j},z}\left(\frac{1}{20k^3}\right)& \ge {ℱ}_{y_{2k+j}-y_{k-j},z}\left(\frac{1}{{10}^3{\left(k-j\right)}^3}\right)\ge \alpha ,\\ \hfill {ℱ}_{y_{k-j}-y_{k+j},z}\left(\frac{1}{60k^{2}j}\right)& \ge {ℱ}_{y_{k-j}-y_{k+j},z}\left(\frac{1}{{10}^3{\left(k-j\right)}^3}\right)\ge \alpha ,\\ \hfill {ℱ}_{y_{2k-j}-y_{k-j},z}\left(\frac{1}{60k{j}^2}\right)& \ge {ℱ}_{y_{2k-j}-y_{k-j},z}\left(\frac{1}{{10}^3{\left(k-j\right)}^3}\right)\ge \alpha ,\\ \hfill {ℱ}_{y_{k-j}-y_{k+j},z}\left(\frac{1}{20j^3}\right)& \ge {ℱ}_{y_{2k-j}-y_{k-j},z}\left(\frac{1}{{10}^3{\left(k-j\right)}^3}\right)\ge \alpha .\end{array}$$

Clearly

$$\begin{array}{c}{ℱ}_{y_{2k+j}-y_k,z}\left(\frac{1}{40k^3}\right)\ge {ℱ}_{y_{2k+j}-y_k,z}\left(\frac{1}{{10}^3{k}^3}\right)\ge \alpha ,{ℱ}_{y_{2k-j}-y_k,z}\left(\frac{1}{80k^3}\right)\ge {ℱ}_{y_{2k-j}-y_k,z}\left(\frac{1}{{10}^3{k}^3}\right)\ge \alpha ,\\ \mathsf{\text{and}}{ℱ}_{y_{2k+j}-y_{k+j},z}\left(\frac{1}{60k{j}^2}\right)\ge {ℱ}_{y_{2k+j}-y_{k+j},z}\left(\frac{1}{{10}^3{\left(k+j\right)}^3}\right)\ge \alpha .\end{array}$$

The inequalities 2k - j ≥ j and 2k - j > k imply

$$\begin{array}{c}{ℱ}_{y_{2k+j}-y_{2k-j},z}\left(\frac{1}{120k^{2}j}\right)\ge {ℱ}_{y_{2k+j}-y_{2k-j},z}\left(\frac{1}{120{\left(2k-j\right)}^3}\right)\ge {ℱ}_{y_{2k+j}-y_{2k-j},z}\left(\frac{1}{{10}^3{\left(2k-j\right)}^3}\right)\ge \alpha \\ {ℱ}_{y_{2k+j}-y_{2k-j},z}\left(\frac{1}{10j^3}\right)\ge {ℱ}_{y_{2k+j}-y_{2k-j},z}\left(\frac{1}{10{\left(2k-j\right)}^3}\right)\ge {ℱ}_{y_{2k+j}-y_{2k-j},z}\left(\frac{1}{{10}^3{\left(2k-j\right)}^3}\right)\ge \alpha .\end{array}$$

Therefore ${ℱ}_{E_{k,j},z}\left(1\right)\ge \alpha$. This shows that f is approximately cubic type mapping. By our assumption, there exists a conditional cubic mapping C : ℕ ∪ {0} → X, such that $\underset{k\to \infty }{\text{lim}}{ℱ}_{C\left(k\right)-f\left(k\right),z}\left(1\right)=1$. In particular, $\underset{k\to \infty }{\text{lim}}{ℱ}_{C\left(2^k\right)-f\left(2^k\right),z}\left(1\right)=1$. This means that

$$\underset{k\to \infty }{\text{lim}}{ℱ}_{C\left(1\right)-y_{2}k,z}\left(\frac{1}{2^{3k}}\right)=1$$

Hence the subsequence $\left(y_{2^k}\right)$ converges to y = C(1). Therefore, the Cauchy sequence (x _n ) also converges to y.