Fuzzy stability for a class of cubic functional equations

Abstract

In this paper, we investigate the following typical form of a class of cubic functional equations: $\begin{matrix} f (x + 2 y) - 3 f (x + y) + 3 f (x) - f (x - y) - 6 f (y) \\ + k [f (mx + y) + f (mx - y) - m [f (x + y) + f (x - y)] - 2 (m^{3} - m) f (x)] = 0 \end{matrix}$ for some rational number m and some real number k. Furthermore, we provide a systematic program to prove the generalized Hyers-Ulam stability for the class of functional equations via the stability for the typical form in fuzzy normed spaces.

Keywords

Fuzzy Banach space fixed point theorem stability cubic mapping

1 Introduction

In 1940, Ulam proposed the following stability problem (cf. [22]):

“Let G₁ be a group and G₂ a metric group with the metric d. Given a constant δ > 0, does there exist a constant c > 0 such that if a mapping f : G₁ ⟶ G₂ satisfies d (f (xy) , f (x) f (y)) < c for all x, y ∈ G₁, then there exists a unique homomorphism h : G₁ ⟶ G₂ with d (f (x) , h (x)) < δ for all x ∈ G₁?”

In the next year, Hyers [9] gave a partial solution of Ulam^,s problem for the case of approximate additive mappings. Subsequently, his result was generalized by Aoki ([1]) for additive mappings and by Rassias [21] for linear mappings to consider the stability problem with unbounded Cauchy differences. During the last decades, the stability problem of functional equations have been extensively investigated by a number of mathematicians (see [4 –6], [8], and [19]).

In 2001, Rassias [20] introduced the following cubic functional equation

$\begin{matrix} f (x + 2 y) - 3 f (x + y) + 3 f (x) - f (x - y) \\ - 6 f (y) = 0 \end{matrix}$ (1.1) and every solution of the cubic functional equation is called a cubic mapping and the following cubic functional equations were investigated

$\begin{matrix} f (x + 2 y) + f (x - 2 y) + 6 f (x) = 4 f (x + y) \\ + 4 f (x - y), and \end{matrix}$ (1.2)

$\begin{matrix} f (mx + y) + f (mx - y) - m [f (x + y) + f (x - y)] \\ - 2 (m^{3} - m) f (x) = 0 \end{matrix}$ (1.3) for some rational number m(see [11 –13], and [20]).

In 1996, Isac and Rassias [10] were the first to provide applications of stability theory of functional equations for the proof of new fixed point theorems with applications.

Katsaras [15] defined a fuzzy norm on a vector space to construct a fuzzy vector topological structure on the space. Later, some mathematicians have defined fuzzy norms on a vector space in different points of view. In particular, Bag and Samanta [2], following Cheng and Mordeson [3], gave an idea of fuzzy norm in such a manner that the corresponding fuzzy metric is of Kramosil and Michalek type [14].

In this paper, we investigate linear sums of two functional equations (1.1) and (1.3) $\begin{matrix} f (x + 2 y) - 3 f (x + y) + 3 f (x) - f (x - y) \\ - 6 f (y) + k [f (mx + y) + f (mx - y) \\ - m [f (x + y) + f (x - y)] \\ - 2 (m^{3} - m) f (x)] = 0 \end{matrix}$ (1.4) for a fixed rational number m and a fixed real number k and prove the generalized Hyers-Ulam stability for (1.4) in fuzzy Banach spaces by fixed point methods.

2 Preliminaries

In this paper, we use the definition of fuzzy normed spaces given in [2], [17], [18].

Definition 2.1. Let X be a real vector space. A function $N : X \times ℝ ⟶ [0, 1]$ is called a fuzzy norm on X if for any x, y ∈ X and any $s, t \in ℝ$ ,

(N1) N (x, t) =0 for t ≤ 0;

(N2) x = 0 if and only if N (x, t) =1 for all t > 0;

(N3) $N (cx, t) = N (x, \frac{t}{| c |})$ if c ≠ 0;

(N4) N (x + y, s + t) ≥ min {N (x, s) , N (y, t)};

(N5) N (x, ·) is a nondecreasing function of $ℝ$ and $lim_{t \to \infty} N (x, t) = 1$ ;

(N6) for any x ≠ 0, N (x, ·) is continuous on $ℝ$ .

In this case, the pair (X, N) is called a fuzzy normed space.

Let (X, N) be a fuzzy normed space. (i) A sequence {x_n} in X is said to be convergent in (X, N) if there exists an x ∈ X such that $lim_{n \to \infty} N (x_{n} - x, t) = 1$ for all t > 0. In this case, x is called the limit of the sequence {x_n} in X and one denotes it by $N - lim_{n \to \infty} x_{n} = x$ . (ii) A sequence {x_n} in X is said to be Cauchy in (X, N) if for any ɛ > 0 and any t > 0, there exists an $m \in ℕ$ such that N (x_n+p - x_n, t) >1 - ɛ for all n ≥ m and all positive integer p.

For example, it is well known that for any normed space (X, || · ||), the mapping $N_{X} : X \times ℝ ⟶ [0, 1]$ , defined by $N_{X} (x, t) = {\begin{matrix} 0, & if t \leq 0 \\ \frac{t}{t + | | x | |}, & if t > 0 \end{matrix}$ is a fuzzy norm on X. In [17], [18] and [19], some examples are provided for the fuzzy norm N_X and other fuzzy norms.

It is well known that every convergent sequence in a fuzzy normed space is Cauchy. A fuzzy normed space is said to be complete if each Cauchy sequence in it is convergent and a complete fuzzy normed space is called a fuzzy Banach space.

Definition 2.2. Let X be a non-empty set. Then a mapping d : X² ⟶ [0, ∞] is called a generalized metric on X if d satisfies the following conditions:

(D1) d (x, y) =0 if and only if x = y,

(D2) d (x, y) = d (y, x), and

(D3) d (x, y) ≤ d (x, z) + d (z, y).

In case, (X, d) is called a generalized metric space.

Theorem 2.3. [7] Let (X, d) be a complete generalized metric space and let J : X ⟶ X be a strictly contractive mapping with some Lipschitz constant L with 0 < L < 1. Then for each given element x ∈ X, either d (Jⁿx, Jⁿ⁺¹x) =∞ for all nonnegative integer n or there exists a positive integer n₀ such that

(1) d (Jⁿx, Jⁿ⁺¹x)< ∞ for all n ≥ n₀ ;

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

(3) y^* is the unique fixed point of J in the set Y = {y ∈ X | d (J^{n
₀}x, y) < ∞} and

(4) $d (y, y^{*}) \leq \frac{1}{1 - L} d (y, Jy)$ for all y ∈ Y.

Throughout this paper, we assume that X is a linear space, (Y, N) is a fuzzy Banach space, and (Z, N′) is a fuzzy normed space.

3 Solutions of (1.4)

In this section, we investigate solutions of (1.4). In Theorem 3, it can be concluded that any solution of (1.4) is cubic. We start with the following lemma.

Lemma 3.1. Let f : X ⟶ Y be a mapping with f (0) =0 and m = 0. Then f satisfies (1.4) if and only if f is cubic.

Proof. Sincce m = 0, the functional equation (1.4) is equivalent to the following functional equation

$\begin{matrix} f (x + 2 y) - 3 f (x + y) + 3 f (x) - f (x - y) \\ - 6 f (y) + k [f (y) + f (- y)] = 0 \end{matrix}$ (3.1) for all x, y ∈ X. Letting x = 0 in (3.1), we have $f (2 y) - f (- y) - 9 f (y) + k [f (y) + f (- y)] = 0$ (3.2) for all y ∈ X. By (3.1) and (3.2), we have

$\begin{matrix} f (x + 2 y) - 3 f (x + y) + 3 f (x) - f (x - y) \\ - f (2 y) + f (- y) + 3 f (y) = 0 \end{matrix}$ (3.3) for all x, y ∈ X. Letting y = - x in (3.1), we have $f (2 x) = 3 f (x) - 5 f (- x) + k [f (x) + f (- x)]$ (3.4) for all x ∈ X. By (3.2) and (3.4), we have $f (2 x) = 6 f (x) - 2 f (- x)$ (3.5) for all x ∈ X. By (3.3) and (3.5), we have

$\begin{matrix} f (x + 2 y) - 3 f (x + y) + 3 f (x) - f (x - y) \\ - 3 f (y) + 3 f (- y) = 0 \end{matrix}$ (3.6) for all x, y ∈ X. Letting y = - y in (3.6), we have

$\begin{matrix} f (x - 2 y) - 3 f (x - y) + 3 f (x) - f (x + y) \\ - 3 f (- y) + 3 f (y) = 0 \end{matrix}$ (3.7) for all x, y ∈ X and by (3.6) and (3.7), we can show that f satisfies (1.2) and hence f is cubic. □

Lemma 3.2. Let f : X ⟶ Y be an even mapping with f (0) =0. Suppose that f satisfies (1.4) and m ≠ 0, 1. Then f (x) =0 for all x ∈ X.

Proof. Suppose that f satisfies (1.4). Suppose that k = 0. Then f is cubic and so f is odd. Hence f (x) =0 for all x ∈ X.

Suppose that k ≠ 0. Setting x = 0 in (1.4), since f is even, we have $f (2 y) = 10 f (y) + 2 k (m - 1) f (y)$ (3.8) for all y ∈ X. Letting y = - y in (1.4), we have $\begin{matrix} f (x - 2 y) - 3 f (x - y) + 3 f (x) - f (x + y) \\ - 6 f (y) + k [f (mx + y) + f (mx - y) \\ - m [f (x + y) + f (x - y)] \\ - 2 (m^{3} - m) f (x)] = 0 \end{matrix}$ (3.9) for all x, y ∈ X. By (1.4) and (3.9), we have $f (x - 2 y) - f (x + 2 y) + 2 [f (x + y) f (x - y)] = 0$ (3.10) for all x, y ∈ X.

case.1k (m - 1) = -3. By (3.8), we have $f (2 y) = 4 f (y)$ (3.11) for all y ∈ X and letting x = y in (3.10), by (3.11), we have f (3y) =9f (y) for all y ∈ X. Suppose that f (lx) = l²f (x) for all x ∈ X and all positive integer l with l ≤ n (n ≥ 3). Letting x = (n - 1) y in (3.10), by the above assumption, we have f ((n + 1) y) = (n + 1) ²f (y) for all y ∈ X and by induction, f (ny) = n²f (y) for all $n \in ℕ$ and for all $y \in ℝ$ . Hence we can show that f (ry) = r²f (y) for all rational numbers r and all y ∈ X. Since m is a rational number, f (mx) = m²f (x) for all x ∈ X. Since c ≠ 0, by (1.4), we have f (mx) = m³f (x) for all x ∈ X. Hence m = 0 or m = 1 which is a contradiction.

case.2k (m - 1) ≠ -3. Letting x = 2x in (3.10), by (3.8), we have

$\begin{matrix} [5 + k (m - 1)] [f (x - y) - f (x + y)] + f (2 x + y) \\ - f (2 x - y) = 0 \end{matrix}$ (3.12) for all x, y ∈ X. Interchanging x and y in (3.12), we have

$\begin{matrix} [5 + k (m - 1)] [f (x - y) - f (x + y)] + f (x + 2 y) \\ - f (x - 2 y) = 0 \end{matrix}$ (3.13) for all x, y ∈ X. Since k (m - 1) ≠ -3, by (3.10) and (3.13), we have $f (x + y) = f (x - y)$ (3.14) for all x, y ∈ X and letting $x = y = \frac{x}{2}$ in (3.14), we have f (x) =0 for all x ∈ X.□

Theorem 3.3. Let f : X ⟶ Y be an odd mapping with f (0) =0. Then f satisfies (1.4) if and only if f is cubic.

Proof. Suppose that f satisfies (1.4). If m = 0, k = 0 or m = 1, then f is cubic. Suppose that k ≠ 0 and m ≠ 0, 1. Setting y = 0 in (1.4), since c ≠ 0, we have $f (mx) = m^{3} f (x)$ (3.15) for all x ∈ X. Replacing y by my in (1.4), we have

$\begin{matrix} f (x + 2 my) - 3 f (x + my) + 3 f (x) \\ - f (x - my) - 6 m^{3} f (y) + {km}^{3} [f (x + y) \\ + f (x - y)] - km [f (x + my) + f (x - my)] \\ - 2 k (m^{3} - m) f (x) = 0 \end{matrix}$ (3.16) for all x, y ∈ X and setting x = 0 in (1.4), we have $f (2 x) = 8 f (x)$ (3.17) for all x ∈ X. Interchanging x and y in (3.16), since f is odd, we have

$\begin{matrix} f (2 mx + y) - 3 f (mx + y) + 3 f (y) + f (mx - y) \\ - 6 m^{3} f (x) + {km}^{3} [f (x + y) - f (x - y)] \\ - km [f (mx + y) - f (mx - y)] \\ - 2 k (m^{3} - m) f (y) = 0 \end{matrix}$ (3.18) for all x, y ∈ X. Relpacing y by 2y in (3.18), by (3.17), we have

$\begin{matrix} 8 f (mx + y) - 3 f (mx + 2 y) + 24 f (y) \\ + f (mx - 2 y) - 6 m^{3} f (x) + {km}^{3} [f (x + 2 y) \\ - f (x - 2 y)] - km [f (mx + 2 y) \\ - f (mx - 2 y)] - 16 k (m^{3} - m) f (y) = 0 \end{matrix}$ (3.19) for all x, y ∈ X. Letting y = - y in (3.19), we have

$\begin{matrix} 8 f (mx - y) - 3 f (mx - 2 y) - 24 f (y) \\ + f (mx + 2 y) - 6 m^{3} f (x) + {km}^{3} [f (x - 2 y) \\ - f (x + 2 y)] - km [f (mx - 2 y) \\ - f (mx + 2 y)] + 16 k (m^{3} - m) f (y) = 0 \end{matrix}$ (3.20) for all x, y ∈ X. By (3.19) and (3.20), we have

$\begin{matrix} 8 [f (mx + y) + f (mx - y)] - 2 [f (mx + 2 y) \\ + f (mx - 2 y)] - 12 m^{3} f (x) = 0 \end{matrix}$ (3.21) for all x, y ∈ X. Letting y = my in (3.21), by (3.15), we can show that f satisfies (1.2) and so f is cubic.□

Combining Lemma 3, Lemma 3 and Theorem 3, we can get the followings as the conclusion of this section.

Theorem 3.4. Let f : X ⟶ Y be a mapping with f (0) =0. Then f satisfies (1.4) if and only if f is cubic.

Corollary 3.5. Let f : X ⟶ Y be a mapping with f (0) =0 and s, t real numbers with st ≠ 0. Then f is a solution of the following functional equation $\begin{matrix} \begin{matrix} s (f (x + 2 y) - 3 f (x + y) + 3 f (x) - f (x - y) \\ - 6 f (y)) + t (f (mx + y) + f (mx - y) \\ - m [f (x + y) + f (x - y)] - 2 (m^{3} - m) f (x)) = 0 \end{matrix} \end{matrix}$ if and only if f is cubic.

4 The Generalized Hyers-Ulam stability for (1.4)

In this section, we prove the generalized Hyers-Ulam stability for the functional equation (1.4) in fuzzy Banach spaces. For any mapping f : X ⟶ Y, we define the difference operator Df : X² ⟶ Y by $\begin{matrix} \begin{matrix} Df (x, y) = f (x + 2 y) - 3 f (x + y) + 3 f (x) \\ - f (x - y) - 6 f (y) + k {f (mx + y) + f (mx - y) \\ - m [f (x + y) + f (x - y)] - 2 (m^{3} - m) f (x)} \end{matrix} \end{matrix}$ for some non-zero rational number m and some non-zero real number k.

Theorem 4.1. Assume that φ : X² ⟶ Z is a function such that

$N^{'} (φ (x, y), t) \geq N^{'} (\frac{L}{m^{3}} φ (mx, my), t)$ (4.1) for all x, y ∈ X, t > 0 and some L with 0 < L < 1. Let f : X ⟶ Y be a mapping such that f (0) =0 and $N (Df (x, y), t) \geq N^{'} (φ (x, y), t)$ (4.2) for all x, y ∈ X and all t > 0. Then there exists a unique cubic mapping F : X ⟶ Y such that $N (F (x) - f (x), \frac{L}{2 (1 - L) | {km}^{3} |} t) \geq N^{'} (φ (x, 0), t)$ (4.3)for all x ∈ X and all t > 0.

Proof. Consider the set S = {g | g : X ⟶ Y} and the generalized metric d on S defined by $\begin{matrix} d (g, h) = inf {c \in [0, \infty) | N (g (x) - h (x), ct) \\ \geq N^{'} (φ (x, 0), t), \forall x \in X, \forall t > 0} . \end{matrix}$

Then (S, d) is a generalized complete metric space([16]). Define a mapping J : S ⟶ S by $Jg (x) = m^{3} g (\frac{x}{m})$ for all g ∈ S and all x ∈ X. Let g, h ∈ S and d (g, h) ≤ c for some c ∈ [0, ∞). Then by (4.1), we have $\begin{matrix} N (Jg (x) - Jh (x), cLt) \geq N^{'} (m^{3} φ (\frac{x}{m}, 0), Lt) \\ \geq N^{'} (φ (x, 0), t) \end{matrix}$ for all x ∈ X and all t > 0. Hence we have d (Jg, Jh) ≤ Ld (g, h) for any g, h ∈ S and so J is a strictly contractive mapping.

Next, we claim that d (Jf, f)< ∞. Putting y = 0 in (4.2), we get $N (f (x) - Jf (x), \frac{L}{2 | {km}^{3} |} t) \geq N^{'} (φ (x, 0), t)$ for all x ∈ X and all t > 0. So we have $d (f, Jf) \leq \frac{L}{2 | {km}^{3} |} < \infty$ . By Theorem 2.3, there exists a mapping F : X ⟶ Y which is a fixed point of J such that d (Jⁿf, F) →0 as n→ ∞. That is, $lim_{n \to \infty} N (m^{3 n} f (\frac{x}{m^{n}}) - F (x), t) = 1$ for all x ∈ X and all t > 0. Moreover, by (4) of Theorem 2, we have (4.3).

Replacing x, y by $\frac{x}{m^{n}}$ , $\frac{x}{m^{n}}$ in (4.2), respectively, by (4.1), we have $\begin{matrix} N (m^{3 n} Df (\frac{x}{m^{n}}, \frac{y}{m^{n}}), t) \\ \geq N^{'} (m^{3 n} φ (\frac{x}{m^{n}}, \frac{y}{m^{n}}), t) \geq N^{'} (φ (x, y), \frac{1}{L^{n}} t) \end{matrix}$ for all x, y ∈ X, t > 0, and all $n \in ℕ$ . Since 0 < L < 1, letting n→ ∞ in the last inequality, F is a solution of (1.4). By Theorem 3, F is a cubic mapping.

Now, we will show the uniqueness of F. Let G be a cubic mapping with (4.3). Then clearly, G is a fixed point of J and by (4.3), we get

$\begin{matrix} d (Jf, G) = d (Jf, JG) \leq Ld (f, G) \\ \leq \frac{L^{2}}{2 | {km}^{3} | (1 - L)} < \infty \end{matrix}$ (4.4) and hence by (3) in Theorem 2, F = G.□

We can get an example for Theorem 4 as follows:

Corollary 4.2. Let θ be a nonnegative real number and p a positive real number with |m|³ < |m|^2p. Let f : X ⟶ Y be a mapping such that f (0) =0 and $N (Df (x, y), t) \geq \frac{t}{t + θ (∥ x ∥^{p} ∥ x ∥^{p} + ∥ x ∥^{2 p} + ∥ y ∥^{2 p})}$ (4.5) for all x, y ∈ X. Then there exists a unique cubic mapping F : X ⟶ Y such that $N (F (x) - f (x), t) \geq \frac{2 (| m |^{2 p} - | m |^{3}) | k | t}{2 (| m |^{2 p} - | m |^{3}) | k | t + θ ∥ x ∥^{2 p}}$ for all x ∈ X and all t > 0.

Related with Theorem 4, we can also have the following theorem. And the proof is similar to that of Theorem 4.

Theorem 4.3. Assume that φ : X² ⟶ Z is a function such that $N^{'} (φ (mx, my), t) \geq N^{'} (m^{3} L φ (x, y), t)$ (4.6) for all x, y ∈ X, t > 0 and some L with 0 < L < 1. Let f : X ⟶ Y be a mapping satisfying f (0) =0 and (4.2). Then there exists a unique quadratic-cubic mapping F : X ⟶ Y such that $N (F (x) - f (x), \frac{1}{2 | {km}^{3} | (1 - L)} t) \geq N^{'} (φ (x, 0), t)$ (4.7)for all x ∈ X and all t > 0.

We can get an example for Theorem 4 as follows:

Corollary 4.4. Let θ be a nonnegative real number and p a positive real number with |m|^2p < |m|³. Let f : X ⟶ Y be a mapping such that f (0) =0 and $N (Df (x, y), t) \geq \frac{t}{t + θ (∥ x ∥^{p} ∥ x ∥^{p} + ∥ x ∥^{2 p} + ∥ y ∥^{2 p})}$ for all x, y ∈ X. Then there exists a unique cubic mapping F : X ⟶ Y such that $N (F (x) - f (x), t) \geq \frac{2 (| m |^{3} - | m |^{2 p}) | k | t}{2 (| m |^{3} - | m |^{2 p}) | k | t + θ ∥ x ∥^{2 p}}$ for all x ∈ X and all t > 0.

We can use Corollary 4 and Corollary 4 to get a classical result in the framework of normed spaces by using the fuzzy norms N_X and N_Y.

Corollary 4.5. Let X be a normed space and Y a Banach space. Let θ be a nonnegative real number and p a positive real number with 2p ≠ 3. Let f : X ⟶ Y be a mapping such that f (0) =0 and $∥ Df (x, y) ∥ \leq θ (∥ x ∥^{p} ∥ x ∥^{p} + ∥ x ∥^{2 p} + ∥ y ∥^{2 p}) .$ for all x, y ∈ X. Then there exists a unique cubic mapping F : X ⟶ Y such that $∥ F (x) - f (x) ∥ \leq \frac{θ}{2 | | m |^{2 p} - | m |^{3} | | k |} ∥ x ∥^{2 p}$ for all x ∈ X and all t > 0.

5 Applications

In this section, we will give some applications which uses the cubic functional equation (1.4). First, from (1.4), we can get varius cubic functional equations. For an examlpe, putting m = 3 and k = -1 in (1.4), we get the following cubic functional equation.

$\begin{matrix} f (3 x + y) + f (3 x - y) = f (x + 2 y) + 2 f (x - y) \\ + 51 f (x) - 6 f (y) . \end{matrix}$

For any mapping f : X ⟶ Y, let $\begin{matrix} D_{3} f (x, y) = f (3 x + y) + f (3 x - y) - f (x + 2 y) \\ - 2 f (x - y) - 51 f (x) + 6 f (y) . \end{matrix}$

Using Theorem 4 and Theorem 4, we get the following propositions:

Proposition 5.1. Let φ : X² ⟶ Z be a mapping with (4.1) and f : X ⟶ Y a mapping such that f (0) =0 and $N (D_{3} f (x, y), t) \geq N^{'} (φ (x, y), t)$ (5.1) for all x, y ∈ X. Then there exists a unique cubic mapping F : X ⟶ Y such that $N (F (x) - f (x), \frac{L}{54 (1 - L)} t) \geq N^{'} (φ (x, 0), t)$ for all x ∈ X and all t > 0.

Proposition 5.2. Let φ : X² ⟶ Z be a mapping with (4.6) and f : X ⟶ Y a mapping with f (0) =0 and (5.1). Then there exists a unique cubic mapping F : X ⟶ Y such that $N (F (x) - f (x), \frac{1}{54 (1 - L)} t) \geq N^{'} (φ (x, 0), t)$ for all x ∈ X and all t > 0.

Using Corollary 4 and Corollary 4, we get the following corollaries:

Corollary 5.3. Let θ be a nonnegative real number and p a positive real number with 3 ≠ 2p. Let f : X ⟶ Y be a mapping such that $\begin{matrix} N (D_{3} f (x, y), t) \\ \geq \frac{t}{t + θ (∥ x ∥^{p} ∥ x ∥^{p} + ∥ x ∥^{2 p} + ∥ y ∥^{2 p})} \end{matrix}$ for all x, y ∈ X. Then there exists a unique cubic mapping F : X ⟶ Y such that $N (F (x) - f (x), t) \geq \frac{2 | 3^{2 p} - 3^{3} | t}{2 | 3^{2 p} - 3^{3} | t + θ ∥ x ∥^{2 p}}$ for all x ∈ X and all t > 0.

Proof. Letting x = y = 0 in (5.1), we have N (46f (0) , t) =1 for all t > 0 and hence by (N2), f (0) =0 . By Proposition 4-r1 and Proposition 4-r2, we have the results.□

Further, it turns out that some functional inequalities can be deformed into inequality (4.2). So we can regard inequality (4.2) as a typical form of a certain class of functional inequalities. In this point of view, we have a following systematic program to prove the generalized Hyers-Ulam stability of certain functional inequalities.

Step 1. Deform a given inequality into the type of (4.2) and get a modified bound function.

Step 2. Apply Theorem 4 and Theorem 4 for the modified bound function.

It should be remarked that if a functional inequality can be deformed into the type of (4.2), then a solution of the original functional equation is cubic. And, it can be easily checked that the resulting unique cubic mapping F in Step 2 also satisfies the original functional equation.

First, we consider the following functional equation:

$\begin{matrix} f (2 x + y) + af (my + x) + af (my - x) - 2 f (x + y) \\ - 6 f (x) + (2 m^{2} + 1) f (y) = 0 \end{matrix}$ (5.2) for some rational numbers a, m with am = -1. For any mapping f : X ⟶ Y, let

$\begin{matrix} D_{1} f (x, y) = f (2 x + y) + af (my + x) + af (my - x) \\ - 2 f (x + y) - 6 f (x) + (2 m^{2} + 1) f (y) . \end{matrix}$

By above theoremes, we get the following:

Theorem 5.4. Let φ : X² ⟶ Z be a function with (4.1) and f : X ⟶ Y a mapping such that f (0) =0 and $N (D_{1} f (x, y), t) \geq N^{'} (φ (x, y), t)$ (5.3) for all x, y ∈ X. Then there exists a unique cubic mapping F : X ⟶ Y such that $N (F (x) - f (x), \frac{L}{2 (1 - L) m^{2}} t) \geq N^{'} (φ (0, x), t)$ for all x ∈ X and all t > 0.

Proof. Interchang x and y in (5.3), we have $\begin{matrix} N (f (x + 2 y) - 3 f (x + y) + 3 f (x) - f (x - y) \\ - 6 f (y) + a {f (mx + y) + f (mx - y) \\ - m [f (x + y) + f (x - y)] - 2 (m^{3} - m) \\ f (x)}, t) \geq N^{'} (φ (y, x), t) \end{matrix}$ for all x, y ∈ X. By Theorem 4, we get the result.□

Similar to Theorem 5.4, we have the following theorem:

Theorem 5.5. Let φ : X² ⟶ Z be a function with (4.6) and f : X ⟶ Y a mapping with f (0) =0 and (5.3). Then there exists a unique cubic mapping F : X ⟶ Y such that $N (F (x) - f (x), \frac{1}{2 (1 - L) m^{2}} t) \geq N^{'} (φ (0, x), t)$ for all x ∈ X and all t > 0.

Corollary 5.6. Let θ be a nonnegative real number and p a positive real number with 3 ≠ 2p. Let f : X ⟶ Y be a mapping such that f (0) =0 and $\begin{matrix} N (D_{1} f (x, y), t) \\ \geq \frac{t}{t + θ (∥ x ∥^{p} ∥ x ∥^{p} + ∥ x ∥^{2 p} + ∥ y ∥^{2 p})} \end{matrix}$ for all x, y ∈ X. Then there exists a unique cubic mapping F : X ⟶ Y such that $N (F (x) - f (x), t) \geq \frac{2 | | m |^{2 p - 1} - m^{2} | t}{2 | | m |^{2 p - 1} - m^{2} | t + θ ∥ x ∥^{2 p}}$ for all x ∈ X and all t > 0.

Finally, we consider the following functional equation

$\begin{matrix} f (x + 3 y) - 6 f (x + y) - 3 f (x - y) + f (2 x) \\ - 24 f (y) = 0 . \end{matrix}$ (5.4)

For any mapping f : X ⟶ Y, let $\begin{matrix} D_{2} f (x, y) = f (x + 3 y) - 6 f (x + y) - 3 f (x - y) \\ + f (2 x) - 24 f (y) . \end{matrix}$

Theorem 5.7. Let θ be a nonnegative real number and p a positive real number with 3 < 2p. Let f : X ⟶ Y be a mapping such that

$\begin{matrix} N (D_{2} f (x, y), t) \\ \geq \frac{t}{t + θ (∥ x ∥^{p} ∥ x ∥^{p} + ∥ x ∥^{2 p} + ∥ y ∥^{2 p})} \end{matrix}$ (5.5) for all x, y ∈ X. Then there exists a unique cubic mapping F : X ⟶ Y such that

$\begin{matrix} N (F (x) - f (x), t) \\ \geq \frac{2 (2^{2 p} - 2^{3}) t}{2 (2^{2 p} - 2^{3}) t + 85 (2^{2 p} + 6^{2 p} + 12^{p}) θ ∥ x ∥^{2 p}} \end{matrix}$ (5.6) for all x ∈ X and all t > 0.

Proof. Let φ (x, y) = θ (∥ x ∥ ^p ∥ x ∥ ^p + ∥ x ∥ ^2p + ∥ y ∥ ^2p) and $N^{'} (φ (x, y), t) = \frac{t}{t + φ (x, y)}$ for all x, y ∈ X. and all t > 0. Letting x = y = 0 in (5.5), we have N (31f (0) , t) =1 for all t > 0 and hence by (N2), f (0) =0 . Letting y = 0 in (5.5), we have $N (f (2 x) - 8 f (x), t) \geq N^{'} (φ (x, 0), t)$ (5.7) for all x ∈ X and t > 0 and letting x = 3x and y = x in (5.5), we have

$\begin{matrix} N (2 f (6 x) - 6 f (4 x) - 3 f (2 x) - 24 f (x), t) \\ \geq N^{'} (φ (3 x, x), t) \end{matrix}$ (5.8) for all x ∈ X and t > 0. Since p is positive, by (5.7) and (5.8), we get

$\begin{matrix} N (f (3 x) - 27 f (x), t) = N (16 f (3 x) - 432 f (x), 16 t) \\ \geq min {N^{'} (φ (3 x, 0), \frac{4 t}{15}), N^{'} (φ (2 x, 0), \frac{4 t}{15}), \\ N^{'} (φ (x, 0), \frac{4 t}{15}), N^{'} (φ (3 x, x), \frac{4 t}{15})} \\ \geq N^{'} (φ (3 x, x), \frac{4 t}{15}) \end{matrix}$ (5.9) for all x ∈ X and t > 0. Letting x = 0 in (5.5), we have $N (f (3 y) - 3 f (- y) - 30 f (y), t) \geq N^{'} (φ (0, y), t)$ (5.10) for all y ∈ X and t > 0 and by (5.9) and (5.10), we have

$\begin{matrix} N (f (x) + f (- x), t) = N (3 f (x) + 3 f (- x), 3 t) \\ \geq min {N^{'} (φ (3 x, x), \frac{4 t}{15}), N^{'} (φ (0, x), 2 t)} \\ \geq N^{'} (φ (3 x, x), \frac{4 t}{15}) \end{matrix}$ (5.11) for all x ∈ X and t > 0. Letting x = 2x + y in (5.5), we have $\begin{matrix} N (f (2 x + 4 y) - 6 f (2 x + 2 y) - 3 f (2 x) \\ + f (4 x + 2 y) - 24 f (y), t) \\ \geq N^{'} (φ (2 x + y, y), t) \end{matrix}$ (5.12) for all x, y ∈ X and t > 0 and so we get

$\begin{matrix} N (f (x + 2 y) - 6 f (x + y) - 3 f (x) + f (2 x + y) \\ - 3 f (y), t) \geq min {N^{'} (φ (2 x + y, y), \frac{2 t}{3}), \\ N^{'} (φ (x + 2 y, 0), \frac{2 t}{3}), N^{'} (φ (x + y, 0), \frac{2 t}{3}), \\ N^{'} (φ (x, 0), \frac{2 t}{3}), N^{'} (φ (2 x + y, 0), \frac{2 t}{3})} \end{matrix}$ (5.13) for all x, y ∈ X and t > 0. Letting x = - x - y and y = x in (5.5), we have $\begin{matrix} N (f (2 x - y) - 6 f (- y) - 3 f (- 2 x - y) \\ + f (- 2 x - 2 y) - 24 f (x), t) \\ \geq N^{'} (φ (- x - y, x), t) \end{matrix}$ (5.14) for all x, y ∈ X and t > 0 and so we get

$\begin{matrix} N (f (2 x - y) + 6 f (y) + 3 f (2 x + y) - 8 f (x + y) \\ - 24 f (x), t) \geq min {N^{'} (φ (- x - y, x), \frac{2 t}{79}), \\ N^{'} (φ (3 y, y), \frac{2 t}{79}), N^{'} (φ (x + y, 0), \frac{2 t}{79}), \\ N^{'} (φ (6 x + 6 y, 2 x + 2 y), \frac{2 t}{79}), \\ N^{'} (φ (6 x + 3 y, 2 x + y), \frac{2 t}{79})} \end{matrix}$ (5.15) for all x, y ∈ X and t > 0. Since p is positive, by (5.13) and (5.15), we have $\begin{matrix} \begin{matrix} N (2 [f (x + 2 y) - 3 f (x + y) + 3 f (x) - f (x - y) \\ - 6 f (y)] - [f (2 x + y) + f (2 x - y) \\ - 2 f (x + y) - 2 f (x - y) - 12 f (x)], t) \\ \geq min {N^{'} (φ (2 x + y, y), \frac{2 t}{85}), \\ N^{'} (φ (x + 2 y, 0), \frac{2 t}{85}), N^{'} (φ (- x - y, x), \frac{2 t}{85}), \\ N^{'} (φ (3 y, y), \frac{2 t}{85}), N^{'} (φ (6 x + 3 y, 2 x + y), \frac{2 t}{85}), \\ N^{'} (φ (6 x + 6 y, 2 x + 2 y), \frac{2 t}{85})} \end{matrix} \end{matrix}$ for all x, y ∈ X and t > 0. Similar to the proof of Theorem 4 for $k = - \frac{1}{2}$ and m = 2 in (1.4), there exists a unique cubic mapping F : X ⟶ Y with (5.6).□

Corollary 5.8. Let f : X ⟶ Y be a mapping. Then f is a solution of (5.4) if and only if f is cubic.

Proof. Suppose that f is a solution of (5.4) and 3 < 2p. Let θ = 0. Then by Theorem 5, there is a unique cubic mapping F : X ⟶ Y with (5.6). By (5.6) and (N2), we get F = f.

For the converse, suppose that f is cubic. Letting x = x + y in (1.1), we get

$\begin{matrix} f (x + 3 y) - 3 f (x + 2 y) + 3 f (x + y) - f (x) \\ - 6 f (y) = 0 \end{matrix}$ (5.16) and since f (2x) =8f (x) for all x ∈ X, by (1.1) and (1.11), we can show that f is a solution of (5.4).□

Similar to Theorem 5, we have the following theorems:

Theorem 5.9. Let θ be a nonnegative real number and p a positive real number with 2p < 3. Let f : X ⟶ Y be a mapping with (5.5). Then there exists a unique cubic mapping F : X ⟶ Y such that $\begin{matrix} N (F (x) - f (x), t) \\ \geq \frac{2 (2^{3} - 2^{2 p}) t}{2 (2^{3} - 2^{2 p}) t + 85 (2^{2 p} + 6^{2 p} + 12^{p}) θ ∥ x ∥^{2 p}} \end{matrix}$ for all x ∈ X and all t > 0.

6 Conclusion

The linear combination (1.4) of cubic functional equations (1.1) and (1.3) is also cubic and the generalized Hyers-Ulam stability for (1.4) hold good in fuzzy normed spaces. From (1.4), we can have many cubic functional equations and we can prove the generalized Hyers-Ulam stability of some functional inequality, e.g. (5.8), which can be deformed into the typical form (4.2). In particular, it should be remarked that if a functional inequality can be deformed into (4.2), then the original functional equation is cubic(cf. Theorem 5.7 and Corollary 5.8). Moreover, it is an interesting question that every linear combination of cubic functional equations is also cubic.

References

Aoki

, On the stability of the linear transformation in Banach spaces, J Math Soc Japan2 (1950), 64–66.

Bag

and Samanta

S.K.

, Finite dimensional fuzzy normed linear spaces, J Fuzzy Math11 (2003), 687–705.

Cheng

S.C.

and Mordeson

J.N.

, Fuzzy linear operator and fuzzy normed linear spaces, Bull Calcutta Math Soc86 (1994), 429–436.

Cholewa

P.W.

, Remarkes on the stability of functional equations, Aequationes Math27 (1984), 76–86.

Cieplinski

, Applications of fixed point theorems to the hyers-ulam stability of functional equation-A survey, Ann Funct Anal3(1) (2012), 151–164.

Czerwik

, On the stability of the quadratic mapping in normed spaces, Bull Abh Math Sem Univ Hamburg62 (1992), 59–64.

Diaz

J.B.

and Margolis

, A fixed point theorem of the alternative for contractions on a generalized complete metric space, Bull Amer Math Soc74 (1968), 305–309.

Găvruta

, A generalization of the Hyer-Ulam-Rassias stability of approximately additive mappings, J Math Anal Appl184 (1994), 431–436.

Hyers

D.H.

, On the stability of the linear functional equation, Proc Nat Acad Sci27 (1941), 222–224.

10.

Isac

and Rassias

Th.M.

, Stability of -additive mappings, Appications to nonlinear analysis, Internat J Math and Math Sci19 (1996), 219–228.

11.

Jun

K.W.

and Kim

H.M.

, The generalized Hyers-Ulam-Rassias stability of a cubic functional equation, J Math Anal Appl274 (2002), 867–878.

12.

Jun

K.W.

and Kim

H.M.

, On the Hyers-Ulam-Rassias stability of a general cubic functional equation, Mathe-matical Inequalities and Applications6 (2003), 289–302.

13.

Jun

, Kim

and Chang

, On the Hyers-Ulam stability of an Euler-Lagrange type cubic functional equation, J Comput Anal Appl7 (2005), 21–33.

14.

Kramosil

and Michálek

, Fuzzy metric and statistical metric spaces, Kybernetica11 (1975), 336–344.

15.

Katsaras

A.K.

, Fuzzy topological vector spaces II, Fuzzy Sets Syst12 (1984), 143–154.

16.

Mihet

and Radu

, On the stability of the additive Cauchy functional equation in random normed spaces, J Math Anal Appl343 (2008), 567–572.

17.

Mirmostafaee

A.K.

, Mirzavaziri

and Moslehian

M.S.

, Fuzzy stability of the Jensen functional equation, Fuzzy Sets Syst159 (2008), 730–738.

18.

Mirmostafaee

A.K.

and Moslehian

M.S.

, Fuzzy versions of Hyers-Ulam-Rassias theorem, Fuzzy Sets Syst159 (2008), 720–729.

19.

Mirzavaziri

and Moslehian

M.S.

, A fixed point approach to stability of a quadratic equation, Bulletin of the Brazilian Mathematical Society37(3) (2006), 361–376.

20.

Rassias

J.M.

, Solution of the Ulam stability problem for cubic mappings, Glasnik Matematički36 (2001), 63–72.

21.

Rassias

Th.M.

, On the stability of the linear mapping in Banach sapces, Proc Amer Math Sco72 (1978), 297–300.

22.

Ulam

S.M.

, Problems in modern mathematics, Science Editions John Wiley and Sons, Inc., New York, 1964.