On the stability of an AQCQ-functional equation in random normed spaces

Abstract

In this paper, we prove the Hyers-Ulam stability of the following additive-quadratic-cubic-quartic functional equation

$$\begin{array}{lll}\hfill f\left(x+2y\right)+f\left(x-2y\right)& =4f\left(x+y\right)+4f\left(x-y\right)-6f\left(x\right)& \hfill \text{(1)}\\ +f\left(2y\right)+f\left(-2y\right)-4f\left(y\right)-4f\left(-y\right)& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

in random normed spaces.

2010 Mathematics Subject Classification: 46S40; 39B52; 54E70

1. Introduction

The stability problem of functional equations originated from a question of Ulam [1] in 1940, concerning the stability of group homomorphisms. Let (G₁, ·) be a group and let (G₂, *, d) be a metric group with the metric d(· , ·). Given ε > 0, does there exist a δ > 0 such that if a mapping h : G₁ → G₂ satisfies the inequality d(h(x·y), h(x) * h(y)) < δ for all x, y ∈ G₁, then there exists a homomorphism H : G₁ → G₂ with d(h(x), H(x)) < ε for all x ∈ G₁? In the other words, under what condition does there exists a homomorphism near an approximate homomorphism? The concept of stability for functional equation arises when we replace the functional equation by an inequality which acts as a perturbation of the equation. Hyers [2] gave a first affirmative answer to the question of Ulam for Banach spaces. Let f : E → E' be a mapping between Banach spaces such that

$$\parallel f\left(x+y\right)-f\left(x\right)-f\left(y\right)\parallel \le \delta$$

for all x, y ∈ E and some δ > 0. Then, there exists a unique additive mapping T : E → E' such that

$$\left|\right|f\left(x\right)-T\left(x\right)\left|\right|\le \delta$$

for all x ∈ E. Moreover, if f(tx) is continuous in t ∈ ℝ for each fixed x ∈ E, then T is ℝ-linear. In 1978, Th.M. Rassias [3] provided a generalization of the Hyers' theorem that allows the Cauchy difference to be unbounded. In 1991, Gajda [4] answered the question for the case p > 1, which was raised by Th.M. Rassias (see [5–11]).

On the other hand, in 1982-1998, J.M. Rassias generalized the Hyers' stability result by presenting a weaker condition controlled by a product of different powers of norms.

Theorem 1.1. ([12–18]). Assume that there exist constants Θ ≥ 0 and p₁, p₂ ∈ ℝ such that p = p₁ + p₂≠ 1, and f : E → E' is a mapping from a normed space E into a Banach space E' such that the inequality

$$\left|\right|f\left(x+y\right)-f\left(x\right)-f\left(y\right)\left|\right|\le \epsilon \left|\right|x\left|{|}^{p_1}\right|\left|y\right|{|}^{p_2}$$

for all x, y ∈ E. Then, there exists a unique additive mapping T : E → E' such that

$$\left|\right|f\left(x\right)-L\left(x\right)\left|\right|\le \frac{\Theta }{2-2^p}\left|\right|x|{|}^p$$

for all × ∈ E.

The control function ||x||^p· ||y|| ^q + ||x||^p+q+ ||y||^p+qwas introduced by Rassias [19] and was used in several papers (see [20–25]).

The functional equation

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

(1.1)

is related to a symmetric bi-additive mapping. It is natural that this equation is called a quadratic functional equation. In particular, every solution of the quadratic functional equation (1.1) is said to be a quadratic mapping. It is well known that a mapping f between real vector spaces is quadratic if and only if there exists a unique symmetric bi-additive mapping B such that f(x) = B(x, x) for all x (see [5, 26]). The bi-additive mapping B is given by

$$B\left(x,y\right)=\frac{1}{4}\left(f\left(x+y\right)-f\left(x-y\right)\right).$$

The Hyers-Ulam stability problem for the quadratic functional equation (1.1) was proved by Skof for mappings f : A → B, where A is a normed space and B is a Banach space (see [27]). Cholewa [28] noticed that the theorem of Skof is still true if relevant domain A is replaced by an abelian group. In [29], Czerwik proved the Hyers-Ulam stability of the functional equation (1.1). Grabiec [30] has generalized these results mentioned above.

In [31], Jun and Kim considered the following cubic functional equation:

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

(1.2)

It is easy to show that the function f(x) = x³ satisfies the functional equation (1.2), which is called a cubic functional equation and every solution of the cubic functional equation is said to be a cubic mapping.

In [32], Park and Bae considered the following quartic functional equation

$$f\left(x+2y\right)+f\left(x-2y\right)=4\left[f\left(x+y\right)+f\left(x-y\right)+6f\left(y\right)\right]-6f\left(x\right).$$

(1.3)

In fact, they proved that a mapping f between two real vector spaces X and Y is a solution of (1:3) if and only if there exists a unique symmetric multi-additive mapping M : X⁴ → Y such that f(x) = M(x, x, x, x) for all x. It is easy to show that the function f(x) = x⁴ satisfies the functional equation (1.3), which is called a quartic functional equation (see also [33]). In addition, Kim [34] has obtained the Hyers-Ulam stability for a mixed type of quartic and quadratic functional equation.

It should be noticed that in all these papers, the triangle inequality is expressed by using the strongest triangular norm T _M .

The aim of this paper is to investigate the Hyers-Ulam stability of the additive-quadratic-cubic-quartic functional equation

$$\begin{array}{ll}f(x+2y)+f(x-2y)\hfill & =4f(x+y)+4f(x-y)-6f(x)\hfill \\ +f(2y)+f(-2y)-4f(y)-4f(-y)\hfill \end{array}$$

(1.4)

in random normed spaces in the sense of Sherstnev under arbitrary continuous t-norms.

In the sequel, we adopt the usual terminology, notations and conventions of the theory of random normed spaces, as in [35–37]. Throughout this paper, Δ⁺ is the space of distribution functions, that is, the space of all mappings F : ℝ ∪ {-∞, ∞} → [0, 1] such that F is left-continuous and non-decreasing on ℝ, F(0) = 0 and F(+ ∞) = 1. D⁺ is a subset of Δ⁺ consisting of all functions F ∈ Δ⁺ for which l^- F(+ ∞) = 1, where l^- f (x) denotes the left limit of the function f at the point x, that is, $l^{-}f\left(x\right)=li{m_t}_{\to x^{-}}f\left(t\right)$. The space Δ⁺ is partially ordered by the usual point-wise ordering of functions, i.e., F ≤ G if and only if F(t) ≤ G(t) for all t in ℝ. The maximal element for Δ⁺ in this order is the distribution function ε₀ given by

$${\epsilon }_0\left(t\right)=\left\{\begin{array}{cc}\hfill 0,\hfill & \hfill ift\le 0,\hfill \\ \hfill 1,\hfill & \hfill ift>0.\hfill \end{array}\right.$$

Definition 1.2. [36]A mapping T : [0, 1] × [0, 1] → [0, 1] is a continuous triangular norm (briefly, a continuous t-norm) if T satisfies the following conditions:

(a) T is commutative and associative;

(b) T is continuous;

(c) T(a, 1) = a for all a ∈ [0, 1];

(d) T(a, b) ≤ T(c, d) whenever a ≤ c and b ≤ d for all a, b, c, d ∈ [0, 1].

Typical examples of continuous t-norms are T _P (a, b) = ab, T _M (a, b) = min(a, b) and T _L (a, b) = max(a+b - 1, 0) (the Lukasiewicz t-norm). Recall (see [38, 39]) that if T is a t-norm and {x _n } is a given sequence of numbers in [0, 1], then $T_{i=1}^n{x}_i$ is defined recurrently by $T_{i=1}^1{x}_i=x_1$ and $T_{i=1}^n{x}_i=T\left(T_{i=1}^{n-1}{x}_i,x_n\right)$ for n ≥ 2. $T_{i=n}^{\infty }{x}_i$ is defined as $T_{i=1}^{\infty }{x}_{n+i-1}$. It is known [39] that for the Lukasiewicz t-norm, the following implication holds:

$${\text{lim}}_{n\to \infty }{(T_L)}_{i=1}^{\infty }{x}_{n+i-1}=1\iff {\displaystyle \sum _{n=1}^{\infty }(1-x_n)}<\infty$$

Definition 1.3. [37]A random normed space (briefly, RN-space) is a triple (X, μ, T), where × is a vector space, T is a continuous t-norm, and μ is a mapping from × into D⁺such that the following conditions hold:

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

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

(RN₃) μ_x+y(t + s) ≥ T (μ _x (t), μ _y (s)) for all x, y ∈ X and all t, s ≥ 0.

Every normed space (X, ||·||) defines a random normed space (X, μ, T _M ),

where

$${\mu }_x\left(t\right)=\frac{t}{t+\left|\right|x\left|\right|}$$

for all t > 0, and T _M is the minimum t-norm. This space is called the induced random normed space.

Definition 1.4. Let (X, μ, T) be an RN-space.

• (1) A sequence {x _n } in × is said to be convergent to × in × 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 n ≥ N.

• (2) A sequence {x _n } in × is called a Cauchy sequence 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 n ≥ m ≥ N.

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

Theorem 1.5. [36]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.

Recently, Eshaghi Gordji et al. establish the stability of cubic, quadratic and additive-quadratic functional equations in RN-spaces (see [40–42]).

One can easily show that an odd mapping f : X → Y satisfies (1.4) if and only if the odd mapping f : X → Y is an additive-cubic mapping, i.e.,

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

It was shown in [[43], Lemma 2.2] that g(x) := f (2x) - 8f (x) and h(x) := f (2x) - 2f (x) are additive and cubic, respectively, and that $f\left(x\right)=\frac{1}{6}h\left(x\right)-\frac{1}{6}g\left(x\right)$.

One can easily show that an even mapping f : X → Y satisfies (1.4) if and only if the even mapping f : X → Y is a quadratic-quartic mapping, i.e.,

$$f\left(x+2y\right)+f\left(x-2y\right)=4f\left(x+y\right)+4f\left(x-y\right)-6f\left(x\right)+2f\left(2y\right)-8f\left(y\right).$$

It was shown in [[44], Lemma 2.1] that g (x) := f (2x) -16f (x) and h (x) := f (2x) -4f (x) are quadratic and quartic, respectively, and that $f\left(x\right)=\frac{1}{12}h\left(x\right)-\frac{1}{12}g\left(x\right)$

Lemma 1.6. Each mapping f : X → Y satisfying (1.4) can be realized as the sum of an additive mapping, a quadratic mapping, a cubic mapping and a quartic mapping.

This paper is organized as follows: In Section 2, we prove the Hyers-Ulam stability of the additive-quadratic-cubic-quartic functional equation (1.4) in RN-spaces for an odd case. In Section 3, we prove the Hyers-Ulam stability of the additive-quadratic-cubic-quartic functional equation (1.4) in RN-spaces for an even case.

Throughout this paper, assume that X is a real vector space and that (X, μ, T) is a complete RN-space.

2.Hyers-Ulam stability of the functional equation (1.4): an odd mapping Case

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

$$\begin{array}{ll}Df(x,y):=f(x+2y)\hfill & +f(x-2y)-4f(x+y)-4f(x-y)+6f(x)\hfill \\ -f(2y)-f(-2y)+4f(y)+4f(-y)\hfill \end{array}$$

for all x, y ∈ X.

In this section, we prove the Hyers-Ulam stability of the functional equation Df (x, y) = 0 in complete RN-spaces: an odd mapping case.

Theorem 2.1. Let f : X → Y be an odd mapping for which there is a ρ : X² → D⁺ (ρ (x, y) is denoted by ρ_{x, y}) such that

$${\mu }_{Df\left(x,y\right)}\left(t\right)\ge {\rho }_{x,y}\left(t\right)$$

(2.1)

for all x, y ∈ X and all t > 0. If

$${\text{lim}}_{n\to \infty }{T}_{k=1}^{\infty }(T({\rho }_{2^{k+n-1}x,2^{k+n-1}x}(2^{n-3}t),{\rho }_{2^{k+n}x,2^{k+n-1}x}(2^{n-1}t)))=1$$

(2.2)

and

$${\text{lim}}_{n\to \infty }{\rho }_{2^{n}x,2^{n}y}(2^{n}t)=1$$

(2.3)

for all x, y ∈ X and all t > 0, then there exist a unique additive mapping A : X → Y and a unique cubic mapping C : X → Y such that

$$\begin{array}{c}{\mu }_{f\left(2x\right)-8f\left(x\right)-A\left(x\right)}\left(t\right)\\ \ge T_{k=1}^{\infty }\left(T\left({\rho }_{2^{k-1}x{,2}^{k-1}x}\left(\frac{t}{8}\right),{\rho }_{2^{k}x{,2}^{k-1}x}\left(\frac{t}{2}\right)\right)\right),\end{array}$$

(2.4)

$$\begin{array}{c}{\mu }_{f\left(2x\right)-2f\left(x\right)-C\left(x\right)}\left(t\right)\\ \ge T_{k=1}^{\infty }\left(T\left({\rho }_{2^{k-1}x{,2}^{k-1}x}\left(\frac{t}{8}\right),{\rho }_{2^{k}x{,2}^{k-1}x}\left(\frac{t}{2}\right)\right)\right)\end{array}$$

(2.5)

for all × ∈ X and all t > 0.

Proof. Putting x = y in (2.1), we get

$${\mu }_{f\left(3y\right)-4f\left(2y\right)+5f\left(y\right)}\left(t\right)\ge {\rho }_{y,y}\left(t\right)$$

(2.6)

for all y ∈ X and all t > 0. Replacing x by 2y in (2.1), we get

$${\mu }_{f\left(4y\right)-4f\left(3y\right)+6f\left(2y\right)-4f\left(y\right)}\left(t\right)\ge {\rho }_{2y,y}\left(t\right)$$

(2.7)

for all y ∈ X and all t > 0. It follows from (2.6) and (2.7) that

$$\begin{array}{c}{\mu }_{f\left(4x\right)-10f\left(2x\right)+16f\left(x\right)}\left(t\right)\\ ={\mu }_{\left(4f\left(3x\right)-16f\left(2x\right)+20f\left(x\right)\right)+\left(f\left(4x\right)-4f\left(3x\right)+6f\left(2x\right)-4f\left(x\right)\right)}\left(t\right)\\ \ge T\left({\mu }_{4f\left(3x\right)-16f\left(2x\right)+20f\left(x\right)}\left(\frac{t}{2}\right),{\mu }_{f\left(4x\right)-4f\left(3x\right)+6f\left(2x\right)-4f\left(x\right)}\left(\frac{t}{2}\right)\right)\\ \ge T\left({\rho }_{x,x}\left(\frac{t}{8}\right),{\rho }_{2x,x}\left(\frac{t}{2}\right)\right)\end{array}$$

(2.8)

for all x ∈ X and all t > 0. Let g : X → Y be a mapping defined by g(x) := f (2x) - 8f (x). Then we conclude that

$${\mu }_{g\left(2x\right)-2g\left(x\right)}\left(t\right)\ge T\left({\rho }_{x,x}\left(\frac{t}{8}\right),{\rho }_{2x,x}\left(\frac{t}{2}\right)\right)$$

for all x ∈ X and all t > 0. Thus, we have

$${\mu }_{\frac{g\left(2x\right)}{2}-g\left(x\right)}\left(t\right)\ge T\left({\rho }_{x,x}\left(\frac{t}{4}\right),{\rho }_{2x,x}\left(t\right)\right)$$

for all x ∈ X and all t > 0. Hence,

$${\mu }_{\frac{g(2^{k+1}x)}{2^{k+1}}-\frac{g(2^{k}x)}{2^k}}(t)\ge T({\rho }_{2^{k}x,2^{k}x}(2^{k-2}t),{\rho }_{2^{k+1}x,2^{k}x}(2^{k}t))$$

for all x ∈ X, all t > 0 and all k ∈ ℕ: From $1>\frac{1}{2}+\frac{1}{2^2}+\cdot \cdot \cdot +\frac{1}{2^n}$, it follows that

$$\begin{array}{ll}{\mu }_{\frac{g(2^{n}x)}{2^n}-g(x)}(t)\hfill & \ge T_{k=1}^n\left({\mu }_{\frac{g(2^{k}x)}{2^k}-\frac{g(2^{k-1}x)}{2^{k-1}}}\left(\frac{t}{2^k}\right)\right)\hfill \\ \ge T_{k=1}^n\left(T\left({\rho }_{2^{k-1}x,2^{k-1}x}\left(\frac{t}{8}\right),{\rho }_{2^{k}x,2^{k-1}x}\left(\frac{t}{2}\right)\right)\right)\hfill \end{array}$$

(2.9)

for all x ∈ X and all t > 0. In order to prove the convergence of the sequence $\left\{\frac{g(2^{n}x)}{2^n}\right\}$, replacing x with 2 ^mx in (2.9), we obtain that

$$\begin{array}{l}{\mu }_{\frac{g(2^{n+m}x)}{2^{n+m}}-\frac{g(2^{m}x)}{2^m}}(t)\\ \ge T_{k=1}^n(T({\rho }_{2^{k+m-1}x,2^{k+m-1}x}(2^{m-3}t),{\rho }_{2^{k+m}x,2^{k+m-1}x}(2^{m-1}t))).\end{array}$$

(2.10)

Since the right-hand side of the inequality (2.10) tends to 1 as m and n tend to infinity, the sequence $\left\{\frac{g(2^{n}x)}{2^n}\right\}$ is a Cauchy sequence. Thus, we may define $A(x)={\text{lim}}_{n\to \infty }\frac{g(2^{n}x)}{2^n}$ for all x ∈ X.

Now, we show that A is an additive mapping. Replacing x and y with 2 ⁿx and 2 ⁿy in (2.1), respectively, we get

$${\mu }_{\frac{Df(2^{n}x,2^{n}y)}{2^n}}(t)\ge {\rho }_{2^{n}x,2^{n}y}(2^{n}t).$$

Taking the limit as n → ∞, we find that A : X → Y satisfies (1.4) for all x, y ∈ X. Since f : X → Y is odd, A : X → Y is odd. By [[43], Lemma 2.2], the mapping A : X → Y is additive. Letting the limit as n → ∞ in (2.9), we get (2.4).

Next, we prove the uniqueness of the additive mapping A : X → Y subject to (2.4). Let us assume that there exists another additive mapping L : X → Y which satisfies (2.4). Since A(2 ⁿx) = 2 ⁿA(x), L(2 ⁿx) = 2 ⁿL(x) for all x ∈ X and all n ∈ ℕ, from (2.4), it follows that

$$\begin{array}{l}{\mu }_{A(x)-L(x)}(2t)={\mu }_{A(2^{n}x)-L(2^{n}x)}(2^{n+1}t)\\ \ge T({\mu }_{A(2^{n}x)-g(2^{n}x)}(2^{n}t),{\mu }_{g(2^{n}x)-L(2^{n}x)}(2^{n}t))\\ \ge T(T_{k=1}^{\infty }(T({\rho }_{2^{n+k-1}x,2^{n+k-1}x}(2^{n-3}t),{\rho }_{2^{n+k}x,2^{n+k-1}x}(2^{n-1}t))),\\ T_{k=1}^{\infty }(T({\rho }_{2^{n+k-1}x,2^{n+k-1}x}(2^{n-3}t),{\rho }_{2^{n+k}x,2^{n+k-1}x}(2^{n-1}t)))\end{array}$$

(2.11)

for all x ∈ X and all t > 0. Letting n → ∞ in (2.11), we conclude that A = L.

Let h : X → Y be a mapping defined by h(x) := f (2x) -2f (x). Then, we conclude that

$${\mu }_{h\left(2x\right)-8h\left(x\right)}\left(t\right)\ge T\left({\rho }_{x,x}\left(\frac{t}{8}\right),{\rho }_{2x,x}\left(\frac{t}{2}\right)\right)$$

for all x ∈ X and all t > 0. Thus, we have

$${\mu }_{\frac{h\left(2x\right)}{8}-h\left(x\right)}\left(t\right)\ge T\left({\rho }_{x,x}\left(t\right),{\rho }_{2x,x}\left(4t\right)\right)$$

for all x ∈ X and all t > 0. Hence,

$${\mu }_{\frac{h(2^{k+1}x)}{8^{k+1}}-\frac{h(2^{k}x)}{8^k}}(t)\ge T({\rho }_{2^{k}x,2^{k}x}(8^{k}t),{\rho }_{2^{k+1}x,2^{k}x}(4·8^{k}t))$$

for all x ∈ X, all t > 0 and all k ∈ ℕ: From $1>\frac{1}{8}+\frac{1}{8^2}+\cdots +\frac{1}{8^n}$, it follows that

$$\begin{array}{ll}{\mu }_{\frac{h(2^{n}x)}{8^n}-h(x)}(t)\hfill & \ge T_{k=1}^n\left({\mu }_{\frac{h(2^{k}x)}{8^k}-\frac{h(2^{k-1}x)}{8^{k-1}}}\left(\frac{t}{8^k}\right)\right)\hfill \\ \ge T_{k=1}^n\left(T\left({\rho }_{2^{k-1}x,2^{k-1}x}\left(\frac{t}{8}\right),{\rho }_{2^{k}x,2^{k-1}x}\left(\frac{t}{2}\right)\right)\right)\hfill \end{array}$$

(2.12)

for all x ∈ X and all t > 0. In order to prove the convergence of the sequence $\left\{\frac{h(2^{n}x)}{8^n}\right\}$, replacing x with 2 ^mx in (2.12), we obtain that

$$\begin{array}{l}{\mu }_{\frac{h(2^{n+m}x)}{8^{n+m}}-\frac{h(2^{m}x)}{8^m}}(t)\\ \ge T_{k=1}^n(T({\rho }_{2^{k+m-1}x,2^{k+m-1}x}(8^{m-1}t),{\rho }_{2^{k+m}x,2^{k+m-1}x}(4·8^{m-1}t))).\end{array}$$

(2.13)

Since the right-hand side of the inequality (2.13) tends to 1 as m and n tend to infinity, the sequence $\left\{\frac{h(2^{n}x)}{8^n}\right\}$ is a Cauchy sequence. Thus, we may define $C(x)={\text{lim}}_{n\to \infty }\frac{h(2^{n}x)}{8^n}$ for all x ∈ X.

Now, we show that C is a cubic mapping. Replacing x and y with 2 ⁿx and 2 ⁿy in (2.1), respectively, we get

$${\mu }_{\frac{Df(2^{n}x,2^{n}y)}{8^n}}(t)\ge {\rho }_{2^{n}x,2^{n}y}(8^{n}t)\ge {\rho }_{2^{n}x,2^{n}y}(2^{n}t).$$

Taking the limit as n → ∞, we find that C : X → Y satisfies (1.4) for all x, y ∈ X. Since f : X → Y is odd, C : X → Y is odd. By [[43], Lemma 2.2], the mapping C : X → Y is cubic. Letting the limit as n → ∞ in (2.12), we get (2.5).

Finally, we prove the uniqueness of the cubic mapping C : X → Y subject to (2.5). Let us assume that there exists another cubic mapping L : X → Y which satisfies (2.5). Since C(2 ⁿx) = 8 ⁿC(x), L(2 ⁿx) = 8 ⁿL(x) for all x ∈ X and all n ∈ ℕ, from (2.5), it follows that

$$\begin{array}{l}{\mu }_{C(x)-L(x)}(2t)\\ ={\mu }_{C(2^{n}x)-L(2^{n}x)}(2·8^{n}t)\\ \ge T({\mu }_{C(2^{n}x)-h(2^{n}x)}(8^{n}t),{\mu }_{h(2^{n}x)-L(2^{n}x)}(8^{n}t))\\ \ge T(T_{k=1}^{\infty }(T({\rho }_{2^{n+k-1}x,2^{n+k-1}x}(8^{n-1}t),{\rho }_{2^{n+k}x,2^{n+k-1}x}(4·8^{n-1}t))),\\ T_{k=1}^{\infty }(T({\rho }_{2^{n+k-1}x,2^{n+k-1}x}(8^{n-1}t),{\rho }_{2^{n+k}x,2^{n+k-1}x}(4·8^{n-1}t)))\\ \ge T(T_{k=1}^{\infty }(T({\rho }_{2^{n+k-1}x,2^{n+k-1}x}(2^{n-3}t),{\rho }_{2^{n+k}x,2^{n+k-1}x}))),\\ T_{k=1}^{\infty }(T({\rho }_{2^{n+k-1}x,2^{n+k-1}x}(2^{n-3}t),{\rho }_{2^{n+k}x,2^{n+k-1}x}(2^{n-1}t)))\end{array}$$

(2.14)

for all x ∈ X and all t > 0. Letting n → ∞ in (2.14), we conclude that C = L, as desired. □

Similarly, one can obtain the following result.

Theorem 2.2. Let f : X → Y be an odd mapping for which there is a ρ : X² → D⁺(ρ(x, y) is denoted by ρ _{x, y} ) satisfying (2.1). If

$${\text{lim}}_{n\to \infty }{T}_{k=1}^{\infty }\left(T\left({\rho }_{\frac{x}{2^{k+n}},\frac{x}{2^{k+n}}}\left(\frac{t}{8^{n+2k}}\right),{\rho }_{\frac{x}{2^{k+n-1}},\frac{x}{2^{k+n}}}\left(\frac{4t}{8^{n+2k}}\right)\right)\right)=1$$

and

$${\text{lim}}_{n\to \infty }{\rho }_{\frac{x}{2^n},\frac{y}{2^n}}\left(\frac{t}{8^n}\right)=1$$

for all x, y ∈ X and all t > 0, then there exist a unique additive mapping A : X → Y and a unique cubic mapping C : X → Y such that

$$\begin{array}{c}{\mu }_{f\left(2x\right)-8f\left(x\right)-A\left(x\right)}\left(t\right)\ge T_{k=1}^{\infty }\left(T\left({\rho }_{\frac{x}{2^k},\frac{x}{2^k}}\left(\frac{t}{2^{2k+1}}\right),{\rho }_{\frac{x}{2^{k-1}},\frac{x}{2^k}}\left(\frac{t}{2^{2k-1}}\right)\right)\right),\\ {\mu }_{f\left(2x\right)-2f\left(x\right)-C\left(x\right)}\left(t\right)\ge T_{k=1}^{\infty }\left(T\left({\rho }_{\frac{x}{2^k},\frac{x}{2^k}}\left(\frac{t}{8^{2k}}\right),{\rho }_{\frac{x}{2^{k-1}},\frac{x}{2^k}}\left(\frac{4t}{8^{2k}}\right)\right)\right)\end{array}$$

for all × ∈ X and all t > 0.

3. Hyers-ulam stability of the functional equation (1.4): an even mapping case

In this section, we prove the Hyers-Ulam stability of the functional equation D f (x, y) = 0 in complete RN-spaces: an even mapping case.

Theorem 3.1. Let f : X → Y be an even mapping for which there is a ρ : X² → D⁺ (ρ (x, y) is denoted by ρ_{x, y}) satisfying f (0) = 0 and (2.1). If

$${\text{lim}}_{n\to \infty }{T}_{k=1}^{\infty }(T({\rho }_{2^{k+n-1}x,2^{k+n-1}x}(2·4^{n-2}t),{\rho }_{2^{k+n}x,2^{k+n-1}x}(2·4^{n-1}t)))=1$$

(3.1)

and

$${\text{lim}}_{n\to \infty }{\rho }_{2^{n}x,2^{n}y}(4^{n}t)=1$$

(3.2)

for all x, y ∈ X and all t > 0, then there exist a unique quadratic mapping P : X → Y and a unique quartic mapping Q : X → Y such that

$$\begin{array}{c}{\mu }_{f\left(2x\right)-16f\left(x\right)-P\left(x\right)}\left(t\right)\\ \ge T_{k=1}^{\infty }\left(T\left({\rho }_{2^{k-1}x{,2}^{k-1}x}\left(\frac{t}{8}\right),{\rho }_{2^{k}x{,2}^{k-1}x}\left(\frac{t}{2}\right)\right)\right),\end{array}$$

(3.3)

$$\begin{array}{c}{\mu }_{f\left(2x\right)-4f\left(x\right)-Q\left(x\right)}\left(t\right)\\ \ge T_{k=1}^{\infty }\left(T\left({\rho }_{2^{k-1}x{,2}^{k-1}x}\left(\frac{t}{8}\right),{\rho }_{2^{k}x{,2}^{k-1}x}\left(\frac{t}{2}\right)\right)\right)\end{array}$$

(3.4)

for all × ∈ X and all t > 0.

Proof. Putting x = y in (2.1), we get

$${\mu }_{f\left(3y\right)-6f\left(2y\right)+15f\left(y\right)}\left(t\right)\ge {\rho }_{y,y}\left(t\right)$$

(3.5)

for all y ∈ X and all t > 0. Replacing x by 2y in (2.1), we get

$${\mu }_{f\left(4y\right)-4f\left(3y\right)+4f\left(2y\right)+4f\left(y\right)}\left(t\right)\ge {\rho }_{2y,y}\left(t\right)$$

(3.6)

for all y ∈ X and all t > 0. It follows from (3.5) and (3.6) that

$$\begin{array}{c}{\mu }_{f\left(4x\right)-20f\left(2x\right)+64f\left(x\right)}\left(t\right)\\ ={\mu }_{\left(4f\left(3x\right)-24f\left(2x\right)+60f\left(x\right)\right)+\left(f\left(4x\right)-4f\left(3x\right)+4f\left(2x\right)+4f\left(x\right)\right)}\left(t\right)\\ \ge T\left({\mu }_{4f\left(3x\right)-24f\left(2x\right)+60f\left(x\right)}\left(\frac{t}{2}\right),{\mu }_{f\left(4x\right)-4f\left(3x\right)+4f\left(2x\right)+4f\left(x\right)}\left(\frac{t}{2}\right)\right)\\ \ge T\left({\rho }_{x,x}\left(\frac{t}{8}\right),{\rho }_{2x,x}\left(\frac{t}{2}\right)\right)\end{array}$$

(3.7)

for all x ∈ X and all t > 0. Let g : X → Y be a mapping defined by g(x) := f (2x) - 16 f (x). Then we conclude that

$${\mu }_{g\left(2x\right)-4g\left(x\right)}\left(t\right)\ge T\left({\rho }_{x,x}\left(\frac{t}{8}\right),{\rho }_{2x,x}\left(\frac{t}{2}\right)\right)$$

for all x ∈ X and all t > 0. Thus, we have

$${\mu }_{\frac{g\left(2x\right)}{4}-g\left(x\right)}\left(t\right)\ge T\left({\rho }_{x,x}\left(\frac{t}{2}\right),{\rho }_{2x,x}\left(2t\right)\right)$$

for all x ∈ X and all t > 0. Hence,

$${\mu }_{\frac{g(2^{k+1}x)}{4^{k+1}}-\frac{g(2^{k}x)}{4^k}}(t)\ge T({\rho }_{2^{k}x,2^{k}x}(2·4^{k-1}t),{\rho }_{2^{k+1}x,2^{k}x}(2·4^{k}t))$$

for all x ∈ X, all t > 0 and all k ∈ ℕ. From $1>\frac{1}{4}+\frac{1}{4^2}+\cdots +\frac{1}{4^n}$, it follows that

$$\begin{array}{ll}{\mu }_{\frac{g(2^{n}x)}{4^n}-g(x)}(t)\hfill & \ge T_{k=1}^n\left({\mu }_{\frac{g(2^{k}x)}{4^k}-\frac{g(2^{k-1}x)}{4^{k-1}}}\left(\frac{t}{4^k}\right)\right)\hfill \\ \ge T_{k=1}^n\left(T\left({\rho }_{2^{k-1}x,2^{k-1}x}\left(\frac{t}{8}\right),{\rho }_{2^{k}x,2^{k-1}x}\left(\frac{t}{2}\right)\right)\right)\hfill \end{array}$$

(3.8)

for all x ∈ X and all t > 0. In order to prove the convergence of the sequence $\left\{\frac{g(2^{n}x)}{4^n}\right\}$, replacing x with 2 ^mx in (3.8), we obtain that

$$\begin{array}{l}{\mu }_{\frac{g(2^{n+m}x)}{4^{n+m}}-\frac{g(2^{m}x)}{4^m}}(t)\\ \ge T_{k=1}^n(T({\rho }_{2^{k+m-1}x,2^{k+m-1}x}(2·4^{m-2}t),{\rho }_{2^{k+m}x,2^{k+m-1}x}(2·4^{m-1}t))).\end{array}$$

(3.9)

Since the right-hand side of the inequality (3.9) tends to 1 as m and n tend to infinity, the sequence $\left\{\frac{g(2^{n}x)}{4^n}\right\}$ is a Cauchy sequence. Thus, we may define $P(x)={\text{lim}}_{n\to \infty }\frac{g(2^{n}x)}{4^n}$ for all x ∈ X.

Now, we show that P is a quadratic mapping. Replacing x and y with 2 ⁿx and 2 ⁿy in (2.1), respectively, we get

$${\mu }_{\frac{Df(2^{n}x,2^{n}y)}{4^n}}(t)\ge {\rho }_{2^{n}x,2^{n}y}(4^{n}t).$$

Taking the limit as n → ∞, we find that P : X → Y satisfies (1.4) for all x, y ∈ X. Since f : X → Y is even, P : X → Y is even. By [[44], Lemma 2.1], the mapping P : X → Y is quadratic. Letting the limit as n → ∞ in (3.8), we get (3.3).

Next, we prove the uniqueness of the quadratic mapping P : X → Y subject to (3.3). Let us assume that there exists another quadratic mapping L : X → Y, which satisfies (3.3). Since P(2 ⁿx) = 4 ⁿP(x), L(2 ⁿx) = 4 ⁿL(x) for all x ∈ X and all n ∈ ℕ, from (3.3), it follows that

$$\begin{array}{l}{\mu }_{P(x)-L(x)}(2t)={\mu }_{P(2^{n}x)-L(2^{n}x)}(2·4^{n}t)\\ \ge T({\mu }_{P(2^{n}x)-g(2^{n}x)}(4^{n}t),{\mu }_{g(2^{n}x)-L(2^{n}x)}(4^{n}t))\\ \ge T(T_{k=1}^{\infty }(T({\rho }_{2^{n+k-1}x,2^{n+k-1}x}(2·4^{n-2}t),{\rho }_{2^{n+k}x,2^{n+k-1}x}(2·4^{n-1}t))),\\ T_{k=1}^{\infty }(T({\rho }_{2^{n+k-1}x,2^{n+k-1}x}(2·4^{n-2}t),{\rho }_{2^{n+k}x,2^{n+k-1}x}(2·4^{n-1}t))))\end{array}$$

(3.10)

for all x ∈ X and all t > 0. Letting n → ∞ in (3.10), we conclude that P = L.

Let h : X → Y be a mapping defined by h(x) := f (2x) -4f (x). Then, we conclude that

$${\mu }_{h\left(2x\right)-16h\left(x\right)}\left(t\right)\ge T\left({\rho }_{x,x}\left(\frac{t}{8}\right),{\rho }_{2x,x}\left(\frac{t}{2}\right)\right)$$

for all x ∈ X and all t > 0. Thus, we have

$${\mu }_{\frac{h\left(2x\right)}{16}-h\left(x\right)}\left(t\right)\ge T\left({\rho }_{x,x}\left(2t\right),{\rho }_{2x,x}\left(8t\right)\right)$$

for all x ∈ X and all t > 0. Hence,

$${\mu }_{\frac{h(2^{k+1}x)}{{16}^{k+1}}-\frac{h(2^{k}x)}{{16}^k}}(t)\ge T({\rho }_{2^{k}x,2^{k}x}(2·{16}^{k}t),{\rho }_{2^{k+1}x,2^{k}x}(8·{16}^{k}t))$$

for all x ∈ X, all t > 0 and all k ∈ ℕ. From $1>\frac{1}{16}+\frac{1}{{16}^2}+\cdot \cdot \cdot +\frac{1}{{16}^n}$, it follows that

$$\begin{array}{ll}{\mu }_{\frac{h(2^{n}x)}{{16}^n}-h(x)}(t)\hfill & \ge T_{k=1}^n\left({\mu }_{\frac{h(2^{k}x)}{{16}^k}-\frac{h(2^{k-1}x)}{{16}^{k-1}}}\left(\frac{t}{{16}^k}\right)\right)\hfill \\ \ge T_{k=1}^n\left(T\left({\rho }_{2^{k-1}x,2^{k-1}x}\left(\frac{t}{8}\right),{\rho }_{2^{k}x,2^{k-1}x}\left(\frac{t}{2}\right)\right)\right)\hfill \end{array}$$

(3.11)

for all x ∈ X and all t > 0. In order to prove the convergence of the sequence $\left\{\frac{h(2^{n}x)}{{16}^n}\right\}$, replacing x with 2 ^mx in (3.11), we obtain that

$$\begin{array}{l}{\mu }_{\frac{h(2^{n+m}x)}{{16}^{n+m}}-\frac{h(2^{m}x)}{{16}^m}}(t)\\ \ge T_{k=1}^n(T({\rho }_{2^{k+m-1}x,2^{k+m-1}x}(2·{16}^{m-1}t),{\rho }_{2^{k+m}x,2^{k+m-1}x}(8·{16}^{m-1}t))).\end{array}$$

(3.12)

Since the right-hand side of the inequality (3.12) tends to 1 as m and n tend to infinity, the sequence $\left\{\frac{h(2^{n}x)}{{16}^n}\right\}$ is a Cauchy sequence. Thus, we may define $Q(x)={\text{lim}}_{n\to \infty }\frac{h(2^{n}x)}{{16}^n}$x ∈ X.

Now, we show that Q is a quartic mapping. Replacing x and y with 2 ⁿx and 2 ⁿy in (2.1), respectively, we get

$${\mu }_{\frac{Df(2^{n}x,2^{n}y)}{{16}^n}}(t)\ge {\rho }_{2^{n}x,2^{n}y}({16}^{n}t)\ge {\rho }_{2^{n}x,2^{n}y}(4^{n}t).$$

Taking the limit as n → ∞, we find that Q : X → Y satisfies (1.4) for all x, y ∈ X. Since f : X → Y is even, Q : X → Y is even. By [[44], Lemma 2.1], the mapping Q : X → Y is quartic. Letting the limit as n → ∞ in (3.11), we get (3.4).

Finally, we prove the uniqueness of the quartic mapping Q : X → Y subject to (3.4). Let us assume that there exists another quartic mapping L : X → Y , which satisfies (3.4). Since Q(2 ⁿx) = 16 ⁿQ(x), L(2 ⁿx) = 16 ⁿL(x) for all x ∈ X and all n ∈ ℕ, from (3.4), it follows that

$$\begin{array}{l}{\mu }_{Q(x)-L(x)}(2t)={\mu }_{Q(2^{n}x)-L(2^{n}x)}(2·{16}^{n}t)\\ \ge T({\mu }_{Q(2^{n}x)-h(2^{n}x)}({16}^{n}t),{\mu }_{h(2^{n}x)-L(2^{n}x)}({16}^{n}t))\\ \ge T(T_{k=1}^{\infty }(T({\rho }_{2^{n+k-1}x,2^{n+k-1}x}(2·{16}^{n-1}t),{\rho }_{2^{n+k}x,2^{n+k-1}x}(8·{16}^{n-1}t))),\\ T_{k=1}^{\infty }(T({\rho }_{2^{n+k-1}x,2^{n+k-1}x}(2·{16}^{n-1}t),{\rho }_{2^{n+k}x,2^{n+k-1}x}(8·{16}^{n-1}t)))\\ \ge T(T_{k=1}^{\infty }(T({\rho }_{2^{n+k-1}x,2^{n+k-1}x}(2·4^{n-2}t),{\rho }_{2^{n+k}x,2^{n+k-1}x}(2·4^{n-1}t))),\\ T_{k=1}^{\infty }(T({\rho }_{2^{n+k-1}x,2^{n+k-1}x}(2·4^{n-2}t),{\rho }_{2^{n+k}x,2^{n+k-1}x}(2·4^{n-1}t))))\end{array}$$

(3.13)

for all x ∈ X and all t > 0. Letting n → ∞ in (3.13), we conclude that Q = L, as desired. □

Similarly, one can obtain the following result.

Theorem 3.2. Let f : X → Y be an even mapping for which there is a ρ : X² → D⁺(ρ (x, y) is denoted by ρ x, y) satisfying f (0) = 0 and (2.1). If

$${\text{lim}}_{n\to \infty }{T}_{k=1}^{\infty }\left(T\left({\rho }_{\frac{x}{2^{k+n}},\frac{x}{2^{k+n}}}\left(\frac{2t}{{16}^{n+2k}}\right),{\rho }_{\frac{x}{2^{k+n-1}},\frac{x}{2^{k+n}}}\left(\frac{8t}{{16}^{n+2k}}\right)\right)\right)=1$$

and

$${\text{lim}}_{n\to \infty }{\rho }_{\frac{x}{2^n},\frac{y}{2^n}}(\frac{t}{{16}^n})=1$$

for all x, y ∈ X and all t > 0, then there exist a unique quadratic mapping P : X → Y and a unique quartic mapping Q : X → Y such that

$$\begin{array}{c}{\mu }_{f\left(2x\right)-16f\left(x\right)-P\left(x\right)}\left(t\right)\ge T_{k=1}^{\infty }\left(T\left({\rho }_{\frac{x}{2^k},\frac{x}{2^k}}\left(\frac{2t}{4^{2k+1}}\right),{\rho }_{\frac{x}{2^{k-1}},\frac{x}{2^k}}\left(\frac{2t}{4^{2k}}\right)\right)\right),\\ {\mu }_{f\left(2x\right)-4f\left(x\right)-Q\left(x\right)}\left(t\right)\ge T_{k=1}^{\infty }\left(T\left({\rho }_{\frac{x}{2^k},\frac{x}{2^k}}\left(\frac{2t}{{16}^{2k}}\right),{\rho }_{\frac{x}{2^{k-1}},\frac{x}{2^k}}\left(\frac{8t}{{16}^{2k}}\right)\right)\right)\end{array}$$

for all × ∈ X and all t > 0.