Fuzzy Stability of Generalized Mixed Type Cubic, Quadratic, and Additive Functional Equation

Abstract

In this paper, we prove the generalized Hyers-Ulam stability of generalized mixed type cubic, quadratic, and additive functional equation, in fuzzy Banach spaces.

2010 Mathematics Subject Classification: 39B82; 39B52.

1. Introduction

The stability problem of functional equations originated from a question of Ulam [1] concerning the stability of group homomorphisms. Hyers [2] gave a first affirmative answer to the question of Ulam for Banach spaces. Hyers' theorem was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference. The paper of Rassias [4] has provided a lot of influence in the development of what we now call generalized Hyers-Ulam stability of functional equations. In 1994, a generalization of the Rassias' theorem was obtained by Găvruta [5] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias' approach.

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 [6, 7]. It is natural that this equation is called a quadratic functional equation. In particular, every solution of the quadratic 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 exits a unique symmetric bi-additive mapping B such that f(x) = B(x, x) for all x (see [6, 7]). The bi-additive mapping B is given by

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

(1.2)

A generalized Hyers-Ulam stability problem for the quadratic functional equation (1.1) was proved by Skof for mappings f : A → B, where A is normed space and B is a Banach space [8] (see [9–12]).

Jun and Kim [13] introduced the following cubic functional equation

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

(1.3)

and they established the general solution and the generalized Hyers-Ulam stability for the functional equation (1.3). 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 mapping C : X × X × X → Y such that f (x) = C(x, x, x) for all x ∈ X, moreover, C is symmetric for each fixed one variable and is additive for fixed two variables. The mapping C is given by

$$C\left(x,y,z\right)=\frac{1}{24}\left(f\left(x+y+z\right)+f\left(x-y-z\right)-f\left(x+y-z\right)-f\left(x-y+z\right)\right)$$

(1.4)

for all x, y, z ∈ X. During the last decades, several stability problems for various functional equations have been investigated by many mathematicians; [14–21].

Eshaghi and Khodaei [22] have established the general solution and investigated the generalized Hyers-Ulam stability for a mixed type of cubic, quadratic, and additive functional equation with f (0) = 0,

$$f\left(x+ky\right)+f\left(x-ky\right)=k^{2}f\left(x+y\right)+k^{2}f\left(x-y\right)+2\left(1-k^2\right)f\left(x\right)$$

(1.5)

in quasi-Banach spaces, where k is nonzero integer numbers with k ≠ ± 1. Obviously, the function f (x) = ax + bx² + cx³ is a solution of the functional equation (1.5). Interesting new results concerning mixed functional equations has recently been obtained by Najati et. al. [23, 24], Jun and Kim [25, 26] as well as for the fuzzy stability of a mixed type of additive and quadratic functional equation by Park [27] (see also [28–43]).

This paper is organized as follows: In Section 3, we prove the generalized Hyers-Ulam stability of the functional equation (1.5) in fuzzy Banach spaces for an even case. In Section 4, we prove the generalized Hyers-Ulam stability of the functional equation (1.5) in fuzzy Banach spaces for an odd case. In Section 5, we prove the generalized Hyers-Ulam stability of generalized mixed cubic, quadratic, and additive functional equation (1.5) in fuzzy Banach spaces.

2. Preliminaries

We use the definition of fuzzy normed spaces given in [44] to investigate a fuzzy version of the generalized Hyers-Ulam stability for the functional equation (1.5) in the fuzzy normed space setting.

Definition 2.1. (Bag and Samanta [44], Mirmostafaee [45]). Let X be a real linear space. A function N : X × ℝ → [0, 1] is said to be a fuzzy norm on X if for all x, y ∈ X and all s, t ∈ ℝ ;

(N₁) N(x, t) = 0 for all t ≤ 0;

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

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

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

(N₅) N(x, ·) is non-decreasing function on ℝ and lim_t→∞N(x, t) = 1;

(N₆) N(x, ·) is left continuous on ℝ for every x ≠ 0.

The pair (X, N) is called a fuzzy normed linear space.

The properties of fuzzy normed vector spaces and examples of fuzzy norms are given in ([3, 45–47]).

Definition 2.2. (Bag and Samanta [44], Mirmostafaee [45]). Let (X, N) be a fuzzy normed linear space. A sequence {x _n } in X is said to be convergent if there exists x ∈ X such that lim_n→∞N(x _n - x, t) = 1 for all t > 0. In that case, x is called the limit of the sequence (x _n ) and we write N - lim_n→∞x _n = x.

Definition 2.3. (Bag and Samanta [44], Mirmostafaee [45]). Let (X, N) be a fuzzy normed linear space. A sequence {x _n } in X is called Cauchy if for each ϵ > 0 and each δ > 0, there exists n₀ ∈ ℕ such that N(x _m - x _n , δ) > 1 - ϵ (m, n ≥ n₀).

It is well known that every convergent sequence in a fuzzy normed vector space is Cauchy. If each Cauchy sequence is convergent, then the fuzzy norm is said to be complete and the fuzzy normed vector space is called a fuzzy Banach space.

We say that a function f : X → Y between fuzzy normed vector spaces X and Y is continuous at a point x₀ ∈ X if for each sequence {x _k } converging to x₀ in X, then the sequence {f(x _k )} converges to f(x₀). If f : X → Y is continuous at each x ∈ X, then f : X → Y is said to be continuous on X (see [48]).

In the rest of this paper, unless otherwise explicitly stated, we will assume that X is a vector space, (Z, N') is a fuzzy normed space, and (Y, N) is a fuzzy Banach space. For convenience, we use the following abbreviation for a given function f : X → Y,

$$D_f\left(x,y\right)=f\left(x+ky\right)+f\left(x-ky\right)-k^{2}f\left(x+y\right)-k^{2}f\left(x-y\right)-2\left(1-k^2\right)f\left(x\right)$$

for all x, y ∈ X, where k is nonzero integer numbers with k ≠ ± 1.

3. Fuzzy stability of the functional equation (1.5): an even case

In this section, we prove the generalized Hyers-Ulam stability of the functional equation (1.5) in fuzzy Banach spaces, for an even case. From now on, V₁ and V₂ will be real vector spaces.

Lemma 3.1. [22]. If an even mapping f : V₁ → V₂ satisfies (1.5), then f(x) is quadratic.

Theorem 3.2. Let ℓ ∈ {-1, 1} be fixed and let φ _q : X × X → Y be a mapping such that

$${\phi }_q\left(kx,ky\right)=\alpha {\phi }_q\left(x,y\right)$$

(3.1)

for all x, y ∈ X and for some positive real number α with αℓ < k²ℓ. Suppose that an even mapping f : X → Y with f(0) = 0 satisfies the inequality

$$N\left(D_f\left(x,y\right),t\right)\ge N^{\prime }\left({\phi }_q\left(x,y\right),t\right)$$

(3.2)

for all x, y ∈ X and all t > 0. Then, the limit

$$Q(x)=N-\underset{n\to \infty }{\lim }\frac{f(k^{\ell n}x)}{k^{2\ell n}})$$

exists for all x ∈ X and Q : X → Y is a unique quadratic mapping satisfying

$$N\left(f\left(x\right)-Q\left(x\right),t\right)\ge N^{\prime }\left({\phi }_q\left(0,x\right),\frac{\ell \left(k^2-\alpha \right)t}{2}\right)$$

(3.3)

for all x ∈ X and all t > 0.

Proof. Case (1): ℓ = 1. By putting x = 0 in (3.2) and then using evenness of f and f(0) = 0, we obtain

$$N\left(2f\left(ky\right)-2k^{2}f\left(y\right),t\right)\ge N^{\prime }\left({\phi }_q\left(0,y\right),t\right)$$

(3.4)

for all x, y ∈ X and all t > 0. If we replace y in (3.4) by x, we get

$$N\left(f\left(kx\right)-k^{2}f\left(x\right),\frac{t}{2}\right)\ge N^{\prime }\left({\phi }_q\left(0,x\right),t\right)$$

(3.5)

for all x ∈ X. So

$$N\left(\frac{f\left(kx\right)}{k^2}-f\left(x\right),\frac{t}{2k^2}\right)\ge N^{\prime }\left({\phi }_q\left(0,x\right),t\right)$$

(3.6)

for all x ∈ X and all t > 0. Then by our assumption

$$N^{\prime }\left({\phi }_q\left(0,kx\right),t\right)=N^{\prime }\left({\phi }_q\left(0,x\right),\frac{t}{\alpha }\right)$$

(3.7)

for all x ∈ X and all t > 0. Replacing x by kⁿx in (3.6) and using (3.7), we obtain

$$N\left(\frac{f\left(k^{n+1}x\right)}{k^{2\left(n+1\right)}}-\frac{f\left(k^{n}x\right)}{k^{2n}},\frac{t}{k^2\left(k^{2n}\right)}\right)\ge N^{\prime }\left({\phi }_q\left(0,k^{n}x\right),t\right)=N^{\prime }\left({\phi }_q\left(0,x\right),\frac{t}{{\alpha }^n}\right)$$

(3.8)

for all x ∈ X, t > 0 and n ≥ 0. Replacing t by αⁿt in (3.8), we see that

$$N\left(\frac{f\left(k^{n+1}x\right)}{k^{2\left(n+1\right)}}-\frac{f\left(k^{n}x\right)}{k^{2n}},\frac{{\alpha }^{n}t}{k^2\left(k^{2n}\right)}\right)\ge N^{\prime }\left({\phi }_q\left(0,x\right),t\right)$$

(3.9)

for all x ∈ X, t > 0 and n > 0. It follows from $\frac{f\left(k^{n}x\right)}{k^{2n}}-f\left(x\right)={\sum }_{j=0}^{n-1}\left(\frac{f\left(k^{j+1}x\right)}{k^{2\left(j+1\right)}}-\frac{f\left(k^{j}x\right)}{k^{2j}}\right)$ and (3.9) that

$$\begin{array}{ll}\hfill N\left(\frac{f\left(k^{n}x\right)}{k^{2n}}-f\left(x\right),\sum _{j=0}^{n-1}\frac{{\alpha }^{j}t}{k^2{\left(k^2\right)}^j}\right)& \ge \mathsf{\text{min}}\bigcup _{j=0}^{n-1}\left\{N\left(\frac{f\left(k^{j+1}x\right)}{k^{2\left(j+1\right)}}-\frac{f\left(k^{j}x\right)}{k^{2j}},\frac{{\alpha }^{j}t}{k^2{\left(k^2\right)}^j}\right)\right\}\\ \ge N^{\prime }\left({\phi }_q\left(0,x\right),t\right)\end{array}$$

(3.10)

for all x ∈ X, t > 0 and n > 0. Replacing x by k^mx in (3.10), we observe that

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

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

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

for all x ∈ X, all t > 0 and all m ≥ 0, n > 0. By last inequality, we obtain

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

(3.11)

for all x ∈ X, all t > 0 and all m ≥ 0, n > 0. Since 0 < α < k² and ${\sum }_{j=0}^{\infty }{\left(\frac{\alpha }{k^2}\right)}^j<\infty$, the Cauchy criterion for convergence and (N₅) imply that $\left\{\frac{f\left(k^{n}x\right)}{k^{2n}}\right\}$ is a Cauchy sequence in Y. Since Y is a fuzzy Banach space, this sequence converges to some point Q(x) ∈ Y. So one can define the function Q : X → Y by

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

(3.12)

for all x ∈ X. Fix x ∈ X and put m = 0 in (3.11) to obtain

$$N\left(\frac{f\left(k^{n}x\right)}{k^{2n}}-f\left(x\right),t\right)\ge N^{\prime }\left({\phi }_q\left(0,x\right),\frac{t}{\sum _{j=0}^{n-1}\frac{{\alpha }^j}{k^2{\left(k^2\right)}^j}}\right)$$

for all x ∈ X, all t > 0 and all n > 0. From which we obtain

$$\begin{array}{ll}\hfill N\left(Q\left(x\right)-f\left(x\right),t\right)& \ge min\left\{N\left(Q\left(x\right)-\frac{f\left(k^{n}x\right)}{k^{2n}},\frac{t}{2}\right),N\left(\frac{f\left(k^{n}x\right)}{k^{2n}}-f\left(x\right),\frac{t}{2}\right)\right\}\\ \ge N^{\prime }\left({\phi }_q\left(0,x\right),\frac{t}{\sum _{j=0}^{n-1}\frac{2{\alpha }^j}{k^2{\left(k^2\right)}^j}}\right)\end{array}$$

(3.13)

for n large enough. Taking the limit as n → ∞ in (3.13), we obtain

$$N\left(Q\left(x\right)-f\left(x\right),t\right)\ge N^{\prime }\left({\phi }_q\left(0,x\right),\frac{\left(k^2-\alpha \right)t}{2}\right)$$

(3.14)

for all x ∈ X and all t > 0. It follows from (3.8) and (3.12) that

$$\begin{array}{ll}\hfill N\left(\frac{Q\left(kx\right)}{k^2}-Q\left(x\right),t\right)& \ge \mathsf{\text{min}}\left\{N\left(\frac{Q\left(kx\right)}{k^2}-\frac{f\left(k^{n+1}x\right)}{k^{2\left(n+1\right)}},\frac{t}{3}\right),N\left(\frac{f\left(k^{n}x\right)}{k^{2n}}-Q\left(x\right),\frac{t}{3}\right),\right.\\ \left.N\left(\frac{f\left(k^{n+1}x\right)}{k^{2\left(n+1\right)}}-\frac{f\left(k^{n}x\right)}{k^{2n}},\frac{t}{3}\right)\right\}=N^{\prime }\left({\phi }_q\left(0,x\right),\frac{k^2\left(k^{2n}\right)t}{3{\alpha }^n}\right)\end{array}$$

for all x ∈ X and all t > 0. Therefore,

$$Q\left(kx\right)=k^{2}Q\left(x\right)$$

(3.15)

for all x ∈ X. Replacing x, y by kⁿx, kⁿy in (3.2), respectively, we obtain

$$N\left(\frac{1}{k^{2n}}{D}_f\left(k^{n}x,k^{n}y\right),t\right)\ge N^{\prime }\left({\phi }_q\left(k^{n}x,k^{n}y\right),k^{2n}t\right)=N^{\prime }\left({\phi }_q\left(x,y\right),\frac{k^{2n}t}{{\alpha }^n}\right)$$

which tends to 1 as n → ∞ for all x, y ∈ X and all t > 0. So, we see that Q satisfies (1.5). Thus, by Lemma 3.1, the function x ⇝ f(x) is quadratic. Therefore, (3.15) implies that the function Q is quadratic.

Now, to prove the uniqueness property of Q, let Q': X → Y be another quadratic function satisfying (3.3). It follows from (3.3), (3.7) and (3.15) that

$$\begin{array}{ll}\hfill N\left(Q\left(x\right)-Q^{\prime }\left(x\right),t\right)& =N\left(\frac{Q\left(k^{n}x\right)}{k^{2n}}-\frac{Q^{\prime }\left(k^{n}x\right)}{k^{2n}},t\right)\\ \ge \mathsf{\text{min}}\left\{N\left(\frac{Q\left(k^{n}x\right)}{k^{2n}}-\frac{f\left(k^{n}x\right)}{k^{2n}},\frac{t}{2}\right),N\left(\frac{f\left(k^{n}x\right)}{k^{2n}}-\frac{Q^{\prime }\left(k^{n}x\right)}{k^{2n}},\frac{t}{2}\right)\right\}\\ \ge N^{\prime }\left({\phi }_q\left(0,k^{n}x\right),\frac{k^{2n}\left(k^2-\alpha \right)t}{4}\right)=N^{\prime }\left({\phi }_q\left(0,x\right),\frac{k^{2n}\left(k^2-\alpha \right)t}{4{\alpha }^n}\right)\end{array}$$

for all x ∈ X and all t > 0. Since α < k², we obtain $\underset{n\to \infty }{lim}{N}^{\prime }\left({\phi }_q\left(0,x\right),\frac{k^{2n}\left(k^2-\alpha \right)t}{4{\alpha }^n}\right)=1$. Thus, Q(x) = Q'(x).

Case (2): ℓ = -1. We can state the proof in the same pattern as we did in the first case.

Replacing x by $\frac{x}{k}$ in (3.5), we obtain

$$N\left(f\left(x\right)-k^{2}f\left(\frac{x}{k}\right),\frac{t}{2}\right)\ge N^{\prime }\left({\phi }_q\left(0,\frac{x}{k}\right),t\right)$$

(3.16)

for all x ∈ X and all t > 0. Replacing x and t by $\frac{x}{k^n}$ and $\frac{t}{k^{2n}}$ in (3.16), respectively, we obtain

$$N\left(k^{2n}f\left(\frac{x}{k^n}\right)-k^{2(n+1)}f(\frac{x}{k^{n+1}})\mathrm{,}\frac{t}{2}\right)\ge N^{\prime }\left({\phi }_q(\mathrm{0,}\frac{x}{k^{n+1}})\mathrm{,}\frac{t}{k^{2n}}\right)=N^{\prime }\left({\phi }_q(\mathrm{0,}x)\mathrm{,}(\frac{\alpha }{k^2}{)}^n\alpha t\right)$$

for all x ∈ X, all t > 0 and all n > 0. One can deduce

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

(3.17)

for all x ∈ X, all t > 0 and all m ≥ 0, n ≥ 0. From which we conclude that $\left\{k^{2n}f\left(\frac{x}{k^n}\right)\right\}$ is a Cauchy sequence in the fuzzy Banach space (Y, N). Therefore, there is a mapping Q : X → Y defined by $Q\left(x\right):=N-\underset{n\to \infty }{lim}{k}^{2n}f\left(\frac{x}{k^n}\right)$. Employing (3.17) with m = 0, we obtain

$$N\left(Q\left(x\right)-f\left(x\right),t\right)\ge N^{\prime }\left({\phi }_q\left(0,x\right),\frac{\left(\alpha -k^2\right)t}{2}\right)$$

for all x ∈ X and all t > 0. The proof for uniqueness of Q for this case proceeds similarly to that in the previous case, hence it is omitted.   □

Remark 3.3. Let 0 < α < k². Suppose that the function t ↦ N(f(x) - Q(x), .) from (0, ∞) into [0, 1] is right continuous. Then, we obtain a better fuzzy (3.14) as follows.

We obtain

$$\begin{array}{c}N(Q(x)-f(x)\mathrm{,}t+s)\ge \min \left\{N\left(Q(x)-\frac{f(k^{n}x)}{k^{2n}}\mathrm{,}s\right)\mathrm{,}N\left(\frac{f(k^{n}x)}{k^{2n}}-f(x)\mathrm{,}t\right)\right\}\\ \ge N^{\prime }\left({\phi }_q(\mathrm{0,}x)\mathrm{,}\frac{t}{{\displaystyle \sum _{j=0}^{n-1}\frac{{\alpha }^j}{k^2{(k)}^{2j}}})}\right)\\ \ge N^{\prime }\left({\phi }_q(\mathrm{0,}x)\mathrm{,}(k^2-\alpha )t\right)\mathrm{.}\end{array}$$

Tending s to zero we infer that

$$N\left(Q\left(x\right)-f\left(x\right),t\right)\ge N^{\prime }\left({\phi }_q\left(0,x\right),\left(k^2-\alpha \right)t\right)$$

for all x ∈ X and all t > 0.

From Theorem 3.2, we obtain the following corollary concerning the generalized Hyers-Ulam stability [4] of quadratic mappings satisfying (1.5), in normed spaces.

Corollary 3.4. Let X be a normed space and Y be a Banach space. Let ε, λ be non-negative real numbers such that λ ≠ 2. Suppose that an even mapping f : X → Y with f(0) = 0 satisfies

$$\parallel D_f\left(x,y\right)\parallel \le \epsilon \left(\parallel x{\parallel }^{\lambda }+\parallel y{\parallel }^{\lambda }\right)$$

(3.18)

for all x, y ∈ X. Then, the limit

$$Q(x)=N-\underset{n\to \infty }{\lim }\frac{f(k^{\ell n}x)}{k^{2\ell n}})$$

exists for all x ∈ X and Q : X → Y is a unique quadratic mapping satisfying

$$\left|\right|f\left(x\right)-Q\left(x\right)\left|\right|\le \frac{2\epsilon \parallel x{\parallel }^{\lambda }}{\ell \left(k^2-k^{\lambda }\right)}$$

(3.19)

for all x ∈ X, where λℓ < 2ℓ.

Proof. Define the function N by

$$N\left(x,t\right)=\left\{\begin{array}{cc}\frac{t}{t+\parallel x\parallel },\hfill & t>0\hfill \\ 0,\hfill & t\le 0\hfill \end{array}\right.$$

It is easy to see that (X, N) is a fuzzy normed space and (Y, N) is a fuzzy Banach space. Denote φ _q : X × X → ℝ, the function sending each (x, y) to ε(||x||^λ + ||y||^λ). By assumption

$$N\left(D_f\left(x,y\right),t\right)\ge N^{\prime }\left({\phi }_q\left(x,y\right),t\right)$$

note that N': ℝ × ℝ → [0, 1] given by

$$N^{\prime }\left(x,t\right)=\left\{\begin{array}{cc}\frac{t}{t+\left|x\right|},\hfill & t>0\hfill \\ 0,\hfill & t\le 0\hfill \end{array}\right.$$

is a fuzzy norm on ℝ. By Theorem 3.2, there exists a unique quadratic mapping Q : X → Y satisfying the equation (1.5) and

$$\begin{array}{ll}\hfill \frac{t}{t+\parallel f\left(x\right)-Q\left(x\right)\parallel }& =N\left(f\left(x\right)-Q\left(x\right),t\right)\\ \ge N^{\prime }\left({\phi }_q\left(0,x\right),\frac{\ell \left(k^2-k^{\lambda }\right)t}{2}\right)\\ =N^{\prime }\left(\epsilon \parallel x{\parallel }^{\lambda },\frac{\ell \left(k^2-k^{\lambda }\right)t}{2}\right)=\frac{\ell \left(k^2-k^{\lambda }\right)t}{\ell \left(k^2-k^{\lambda }\right)t+2\epsilon \parallel x{\parallel }^{\lambda }}\end{array}$$

and thus

$$\frac{t}{t+\parallel f\left(x\right)-Q\left(x\right)\parallel }\ge \frac{\ell \left(k^2-k^{\lambda }\right)t}{\ell \left(k^2-k^{\lambda }\right)t+2\epsilon \parallel x{\parallel }^{\lambda }}$$

which implies that, ℓ(k² - k^λ)||f(x) - Q(x)|| ≤ 2ε||x||^λ for all x ∈ X.   □

In the following theorem, we will show that under some extra conditions on Theorem 3.2, the quadratic function r ↦ Q(rx) is fuzzy continuous. It follows that in such a case, Q(rx) = r²Q(x) for all x ∈ X and r ∈ ℝ.

In the following result, we will assume that all conditions of the theorem 3.2 hold.

Theorem 3.5. Denote N₁ the fuzzy norm obtained as Corollary 3.4 on ℝ . Let for all x ∈ X, the functions r ↦ f(rx) (from (ℝ, N₁) into (Y, N)) and r ↦ φ _q (0, rx) (from (ℝ, N₁) into (Z, N')) be fuzzy continuous. Then, for all x ∈ X, the function r ↦ Q(rx) is fuzzy continuous and Q(rx) = r²Q(x) for all r ∈ ℝ .

Proof. Case (1): ℓ = 1. Let {r _k } be a sequence in ℝ that converge to some r ∈ ℝ, and let t > 0. Let ε > 0 be given, since 0 < α < k², so $\underset{n\to \infty }{lim}\frac{\left(k^2-\alpha \right)k^{2n}t}{12{\alpha }^n}=\infty$, there is m ∈ ℕ such that

$$N^{\prime }\left({\phi }_q\left(0,rx\right),\frac{\left(k^2-\alpha \right)k^{2m}t}{12{\alpha }^m}\right)>1-\epsilon$$

(3.20)

It follows from (3.14) and (3.20) that

$$N\left(\frac{f\left(k^{m}rx\right)}{k^{2m}}-\frac{Q\left(k^{m}rx\right)}{k^{2m}},\frac{t}{3}\right)>1-\epsilon$$

(3.21)

By the fuzzy continuity of functions r ↦ f(rx) and r ↦ φ _q (0, rx), we can find some $\mathcal{J}\in \mathbb{N}$ such that for any n ≥ j,

$$N\left(\frac{f\left(k^m{r}_{k}x\right)}{k^{2m}}-\frac{f\left(k^{m}rx\right)}{k^{2m}},\frac{t}{3}\right)>1-\epsilon$$

(3.22)

and

$$N^{\prime }\left({\phi }_q\left(0,r_{k}x\right)-{\phi }_q\left(0,rx\right),\frac{\left(k^2-\alpha \right)k^{2m}t}{12{\alpha }^m}\right)>1-\epsilon$$

(3.23)

It follows from (3.20) and (3.23) that

$$N^{\prime }\left({\phi }_q\left(0,r_{k}x\right),\frac{\left(k^2-\alpha \right)k^{2m}t}{6{\alpha }^m}\right)>1-\epsilon$$

(3.24)

On the other hand,

$$\begin{array}{ll}\hfill N\left(Q\left(r_{k}x\right)-\frac{f\left(k^m{r}_{k}x\right)}{k^{2m}},\frac{t}{k^{2m}}\right)& =N\left(\frac{Q\left(k^m{r}_{k}x\right)}{k^{2m}}-\frac{f\left(k^m{r}_{k}x\right)}{k^{2m}},\frac{t}{k^{2m}}\right)\\ \ge N^{\prime }\left({\phi }_q\left(0,r_{k}x\right),\frac{\left(k^2-\alpha \right)t}{2{\alpha }^m}\right)\end{array}$$

(3.25)

It follows from (3.24) and (3.25) that

$$N\left(Q\left(r_{k}x\right)-\frac{f\left(k^m{r}_{k}x\right)}{k^{2m}},\frac{t}{3}\right)>1-\epsilon$$

(3.26)

So, it follows from (3.21), (3.22) and (3.26) that for any n ≥ j,

$$N\left(Q\left(r_{k}x\right)-Q\left(rx\right),t\right)>1-\epsilon$$

Therefore, for every choice x ∈ X, t > 0, and ε > 0, we can find some $\mathcal{J}\in \mathbb{N}$ such that N(Q(r _k x) - Q(rx), t) > 1 - ε for every $n\ge \mathcal{J}$. This shows that Q(r _k x) → Q(rx). The proof for ℓ = -1 proceeds similarly to that in the previous case.

It is not hard to see that Q(rx) = r²Q(x) for each rational number r. Since Q is a fuzzy continuous function, by the same reasoning as in the proof of [45], the quadratic mapping Q : X → Y satisfies Q(rx) = r²Q(x) for each r ∈ ℝ. □

4. Fuzzy stability of the functional equation (1.5): an odd case

In this section, we prove the generalized Hyers-Ulam stability of the functional equation (1.5) in fuzzy Banach spaces for an odd case.

Lemma 4.1. [22, 24]. If an odd mapping f : V₁ → V₂ satisfies (1.5), then the mapping g : V₁ → V₂, defined by g(x) = f(2x) - 8f(x), is additive.

Theorem 4.2. Let ℓ ∈ {-1, 1} be fixed and let φ _q : X × X → Z be a function such that

$${\phi }_a\left(2x,2y\right)=\alpha {\phi }_a\left(x,y\right)$$

(4.1)

for all x, y ∈ X and for some positive real number α with αℓ < 2ℓ. Suppose that an odd mapping f : X → Y satisfies the inequality

$$N\left(D_f\left(x,y\right),t\right)\ge N^{\prime }\left({\phi }_a\left(x,y\right),t\right)$$

(4.2)

for all x, y ∈ X and all t > 0. Then, the limit

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

exists for all x ∈ X and A : X → Y is a unique additive mapping satisfying

$$N\left(f\left(2x\right)-8f\left(x\right)-A\left(x\right),t\right)\ge M_a\left(x,\frac{\ell \left(2-\alpha \right)}{2}t\right)$$

(4.3)

for all x ∈ X and all t > 0, where

$$\begin{array}{ll}\hfill M_a\left(x,t\right)& =min\left\{N^{\prime }\left({\phi }_a\left(x,x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),N^{\prime }\left({\phi }_a\left(2x,x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),\right.\\ N^{\prime }\left({\phi }_a\left(x,2x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),N^{\prime }\left({\phi }_a\left(\left(k+1\right)x,x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),\\ N^{\prime }\left({\phi }_a\left(\left(k-1\right)x,x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right)),N^{\prime }\left({\phi }_a\left(2x,2x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),\\ N^{\prime }\left({\phi }_a\left(x,3x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),N^{\prime }\left({\phi }_a\left(\left(2k+1\right)x,x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),\\ \left.N^{\prime }\left({\phi }_a\left(\left(2k-1\right)x,x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right)\right\}.\end{array}$$

Proof. Case (1): ℓ = 1. It follows from (4.2) and using oddness of f that

$$\begin{array}{c}N\left(f\left(ky+x\right)-f\left(ky-x\right)-k^{2}f\left(x+y\right)-k^{2}f\left(x-y\right)+2\left(k^2-1\right)f\left(x\right),t\right)\\ \ge N^{\prime }\left({\phi }_a\left(x,y\right),t\right)\end{array}$$

(4.4)

for all x, y ∈ X and all t > 0. Putting y = x in (4.4), we have

$$N\left(f\left(\left(k+1\right)x\right)-f\left(\left(k-1\right)x\right)-k^{2}f\left(2x\right)+2\left(k^2-1\right)f\left(x\right),t\right)\ge N^{\prime }\left({\phi }_a\left(x,x\right),t\right)$$

(4.5)

for all x ∈ X and all t > 0. It follows from (4.5) that

$$N\left(f\left(2\left(k+1\right)x\right)-f\left(2\left(k-1\right)x\right)-k^{2}f\left(4x\right)+2\left(k^2-1\right)f\left(2x\right),t\right)\ge N^{\prime }\left({\phi }_a\left(2x,2x\right),t\right)$$

(4.6)

for all x ∈ X and all t > 0. Replacing x and y by 2x and x in (4.4), respectively, we get

$$\begin{array}{c}N\left(f\left(\left(k+2\right)x\right)-f\left(\left(k-2\right)x\right)-k^{2}f\left(3x\right)-k^{2}f\left(x\right)+2\left(k^2-1\right)f\left(2x\right),t\right)\\ \ge N^{\prime }\left({\phi }_a\left(2x,x\right),t\right)\end{array}$$

(4.7)

for all x ∈ X. Setting y = 2x in (4.4), we have

$$\begin{array}{c}N\left(f\left(\left(2k+1\right)x\right)-f\left(\left(2k-1\right)x\right)-k^{2}f\left(3x\right)-k^{2}f\left(-x\right)+2\left(k^2-1\right)f\left(x\right),t\right)\\ \ge N^{\prime }\left({\phi }_a\left(x,2x\right),t\right)\end{array}$$

(4.8)

for all x ∈ X and all t > 0. Putting y = 3x in (4.4), we obtain

$$\begin{array}{c}N\left(f\left(\left(3k+1\right)x\right)-f\left(\left(3k-1\right)x\right)-k^{2}f\left(4x\right)-k^{2}f\left(-2x\right)+2\left(k^2-1\right)f\left(x\right),t\right)\\ \ge N^{\prime }\left({\phi }_a\left(x,3x\right),t\right)\end{array}$$

(4.9)

for all x ∈ X and all t > 0. Replacing x and y by (k + 1)x and x in (4.4), respectively, we get

$$\begin{array}{c}N\left(f\left(\left(2k+1\right)x\right)-f\left(-x\right)-k^{2}f\left(\left(k+2\right)x\right)-k^{2}f\left(kx\right)+2\left(k^2-1\right)f\left(\left(k+1\right)x\right),t\right)\\ \ge N^{\prime }\left({\phi }_a\left(\left(k+1\right)x,x\right),t\right)\end{array}$$

(4.10)

for all x ∈ X and all t > 0. Replacing x and y by (k - 1)x and x in (4.4), respectively, one gets

$$\begin{array}{c}N\left(f\left(\left(2k-1\right)x\right)-f\left(x\right)-k^{2}f\left(\left(k-2\right)x\right)-k^{2}f\left(kx\right)+2\left(k^2-1\right)f\left(\left(k-1\right)x\right),t\right)\\ \ge N^{\prime }\left({\phi }_a\left(\left(k-1\right)x,x\right),t\right)\end{array}$$

(4.11)

for all x ∈ X and all t > 0. Replacing x and y by (2k + 1)x and x in (4.4), respectively, we obtain

$$\begin{array}{c}N(f\left(\left(3k+1\right)x\right)-f\left(-\left(k+1\right)x\right)-k^{2}f\left(2\left(k+1\right)x\right)-k^{2}f\left(2kx\right)\\ +2\left(k^2-1\right)f\left(\left(2k+1\right)x\right),t)\ge N^{\prime }\left({\phi }_a\left(\left(2k+1\right)x,x\right),t\right)\end{array}$$

(4.12)

for all x ∈ X and all t > 0. Replacing x and y by (2k - 1)x and x in (4.4), respectively, we have

$$\begin{array}{c}N(f\left(\left(3k-1\right)x\right)-f\left(-\left(k-1\right)x\right)-k^{2}f\left(2\left(k-1\right)x\right)-k^{2}f\left(2kx\right)\\ +2\left(k^2-1\right)f\left(\left(2k-1\right)x\right),t)\ge N^{\prime }\left({\phi }_a\left(\left(2k-1\right)x,x\right),t\right)\end{array}$$

(4.13)

for all x ∈ X and all t > 0. It follows from (4.5), (4.7), (4.8), (4.10) and (4.11) that

$$\begin{array}{c}N\left(2f\left(3x\right)-8f\left(2x\right)+10f\left(x\right),\frac{2}{k^2\left(k^2-1\right)}\left(2\left(k^2-1\right)+k^2+3\right)t\right)\\ \ge \mathsf{\text{min}}\left\{N^{\prime }\left({\phi }_a\left(x,x\right),t\right),N^{\prime }\left({\phi }_a\left(2x,x\right),t\right),N^{\prime }\left({\phi }_a\left(x,2x\right),t\right),\right.\\ \left.N^{\prime }\left({\phi }_a\left(\left(k+1\right)x,x\right),t\right),N^{\prime }\left({\phi }_a\left(\left(k-1\right)x,x\right),t\right)\right\}\end{array}$$

(4.14)

for all x ∈ X and all t > 0. And, from (4.5), (4.6), (4.8), (4.9), (4.12) and (4.14), we conclude that

$$\begin{array}{c}N\left(f\left(4x\right)-2f\left(3x\right)-2f\left(2x\right)+6f\left(x\right),\frac{1}{k^2\left(k^2-1\right)}\left(2\left(k^2-1\right)+k^2+4\right)t\right)\\ \ge \mathsf{\text{min}}\{N^{\prime }\left({\phi }_a\left(x,x\right),t\right),N^{\prime }\left({\phi }_a\left(2x,2x\right),t\right),N^{\prime }\left({\phi }_a\left(x,2x\right),t\right),N^{\prime }\left({\phi }_a\left(x,3x\right),t\right),\\ N^{\prime }\left({\phi }_a\left(\left(2k+1\right)x,x\right),t\right),N^{\prime }\left({\phi }_a\left(\left(2k-1\right)x,x\right),t\right)\}\end{array}$$

(4.15)

for all x ∈ X and all t > 0. Finally, by using (4.14) and (4.15), we obtain that Similar to the proof Theorem 3.2, we have

$$\begin{array}{c}N\left(f\left(4x\right)-10f\left(2x\right)+16f\left(x\right),\frac{9k^2+4}{k^2\left(k^2-1\right)}t\right)\ge \mathsf{\text{min}}\{N^{\prime }\left({\phi }_a\left(x,x\right),t\right),N^{\prime }\left({\phi }_a\left(2x,x\right),t\right),\\ N^{\prime }\left({\phi }_a\left(x,2x\right),t\right),N^{\prime }\left({\phi }_a\left(\left(k+1\right)x,x\right),t\right),N^{\prime }\left({\phi }_a\left(\left(k-1\right)x,x\right),t\right)),N^{\prime }\left({\phi }_a\left(2x,2x\right),t\right),\\ N^{\prime }\left({\phi }_a\left(x,3x\right),t\right),N^{\prime }\left({\phi }_a\left(\left(2k+1\right)x,x\right),t\right),N^{\prime }\left({\phi }_a\left(\left(2k-1\right)x,x\right),t\right)\}\end{array}$$

(4.16)

for all x ∈ X and all t > 0, where

$$\begin{array}{ll}\hfill M_a\left(x,t\right)& =\mathsf{\text{min}}\left\{N^{\prime }\left({\phi }_a\left(x,x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),N^{\prime }\left({\phi }_a\left(2x,x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),\right.\\ N^{\prime }\left({\phi }_a\left(x,2x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),N^{\prime }\left({\phi }_a\left(\left(k+1\right)x,x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),\\ N^{\prime }\left({\phi }_a\left(\left(k-1\right)x,x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right)),N^{\prime }\left({\phi }_a\left(2x,2x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),\\ N^{\prime }\left({\phi }_a\left(x,3x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),N^{\prime }\left({\phi }_a\left(\left(2k+1\right)x,x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right),\\ \left.N^{\prime }\left({\phi }_a\left(\left(2k-1\right)x,x\right),\frac{k^2\left(k^2-1\right)}{9k^2+4}t\right)\right\}\end{array}$$

for all x ∈ X and all t > 0. Thus, (4.16) means that

$$N\left(f\left(4x\right)-10f\left(2x\right)+16f\left(x\right),t\right)\ge M_a\left(x,t\right)$$

(4.17)

for all x ∈ X and all t > 0. Let g : X → Y be a mapping defined by g(x):= f(2x) - 8f(x) for all x ∈ X. From (4.17), we conclude that

$$N\left(g\left(2x\right)-2g\left(x\right),t\right)\ge M_a\left(x,t\right)$$

(4.18)

for all x ∈ X and all t > 0. So

$$N\left(\frac{g\left(2x\right)}{2}-g\left(x\right),\frac{t}{2}\right)\ge M_a\left(x,t\right)$$

(4.19)

for all x ∈ X and all t > 0. Then, by our assumption

$$M_a\left(2x,t\right)=M_a\left(x,\frac{t}{\alpha }\right)$$

(4.20)

for all x ∈ X and all t > 0. Replacing x by 2ⁿx in (4.19) and using (4.20), we obtain

$$N\left(\frac{g{(2}^{n+1}x)}{2^{n+1}}-\frac{g{(2}^{n}x)}{2^n}\mathrm{,}\frac{t}{{2(2}^n)}\right)\ge M_a{(2}^{n}x\mathrm{,}t)=M_a\left(x\mathrm{,}\frac{t}{{\alpha }^n}\right)$$

(4.21)

for all x ∈ X, t > 0 and n ≥ 0. Replacing t by αⁿt in (4.21), we see that

$$N\left(\frac{g{(2}^{n+1}x)}{2^{n+1}}-\frac{g{(2}^{n}x)}{2^n}\mathrm{,}\frac{t{\alpha }^n}{{2(2}^n)}\right)\ge M_a(x\mathrm{,}t)$$

(4.22)

for all x ∈ X, t > 0 and n > 0. It follows from $\frac{g{(2}^{n}x)}{2^n}-g(x)={\displaystyle {\sum }_{j=0}^{n-1}\left(\frac{g{(2}^{j+1}x)}{2^{j+1}}-\frac{g{(2}^{j}x)}{2^j}\right)}$ and (4.22) that

$$\begin{array}{c}N\left(\frac{g{(2}^{n}x)}{2^n}-g(x)\mathrm{,}{\displaystyle \sum _{j=0}^{n-1}\frac{{\alpha }^{j}t}{{2(2)}^j}}\right)\ge \text{min}{\displaystyle \underset{j=0}{\overset{n-1}{\cup }}\left\{N\left(\frac{g{(2}^{j+1}x)}{2^{j+1}}-\frac{g{(2}^{j}x)}{2^j}\mathrm{,}\frac{{\alpha }^{j}t}{{2(2)}^j}\right)\right\}}\\ \ge M_a(x\mathrm{,}t)\end{array}$$

(4.23)

for all x ∈ X, t > 0 and n > 0. Replacing x by 2^mx in (4.23), we observe that

$$N\left(\frac{g{(2}^{n+m}x)}{2^{n+m}}-\frac{g{(2}^{m}x)}{2^m}\mathrm{,}{\displaystyle \sum _{j=0}^{n-1}\frac{{\alpha }^{j}t}{{2(2)}^{j+m}}}\right)\ge M_a{(2}^{m}x\mathrm{,}t)=M_a\left(x\mathrm{,}\frac{t}{{\alpha }^m}\right)$$

for all x ∈ X, all t > 0 and all m ≥ 0, n > 0. So

$$N\left(\frac{g{(2}^{n+m}x)}{2^{n+m}}-\frac{g{(2}^{m}x)}{2^m}\mathrm{,}{\displaystyle \sum _{j=m}^{n+m-1}\frac{{\alpha }^{j}t}{{2(2)}^j}}\right)\ge M_a(x\mathrm{,}t)$$

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

$$N\left(\frac{g{(2}^{n+m}x)}{2^{n+m}}-\frac{g{(2}^{m}x)}{2^m}\mathrm{,}t\right)\ge M_a\left(x\mathrm{,}\frac{t}{{\displaystyle \sum _{j=m}^{n+m-1}\frac{{\alpha }^j}{{2(2)}^j}}}\right)$$

(4.24)

for all x ∈ X, all t > 0 and all m ≥ 0, n > 0. Since 0 < α < 2 and ${\sum }_{n=0}^{\infty }{\left(\frac{\alpha }{2}\right)}^n<\infty$, the Cauchy criterion for convergence and (N₅) imply that $\left\{\frac{g{(2}^{n}x)}{2^n}\right\}$ is a Cauchy sequence in (Y, N) to some point A(x) ∈ Y. So one can define the mapping A : X → Y by

$$A(x)=N-\underset{n\to \infty }{\lim }\frac{g{(2}^{n}x)}{2^n})$$

(4.25)

for all x ∈ X. Fix x ∈ X and put m = 0 in (4.24) to obtain

$$N\left(\frac{g{(2}^{n}x)}{2^n}-g(x)\mathrm{,}t\right)\ge M_a\left(x\mathrm{,}\frac{t}{{\displaystyle \sum _{j=0}^{n-1}}\frac{{\alpha }^j}{{2(2)}^j}}\right)$$

for all x ∈ X, t > 0 and n > 0. From which we obtain

$$\begin{array}{c}N(A(x)-g(x)\mathrm{,}t)\ge \text{min}\left\{N\left(A(x)-\frac{g{(2}^{n}x)}{2^n}\mathrm{,}\frac{t}{2}\right)\mathrm{,}N\left(\frac{g{(2}^{n}x)}{2^n}-g(x)\mathrm{,}\frac{t}{2}\right)\right\}\\ \ge M_a\left(x\mathrm{,}\frac{t}{{\displaystyle \sum _{j=0}^{n-1}\frac{{\alpha }^j}{2^j}}}\right)\end{array}$$

(4.26)

for n large enough. Taking the limit as n → ∞ in (4.26), we obtain

$$N\left(A\left(x\right)-g\left(x\right),t\right)\ge M_a\left(x,\frac{t\left(2-\alpha \right)}{2}\right)$$

(4.27)

for all x ∈ X and all t > 0. It follows from (4.21) and (4.25) that

$$\begin{array}{l}N\left(\frac{A(2x)}{2}-A(x)\mathrm{,}t\right)\ge \text{min}\{N\left(\frac{A(2x)}{2}-\frac{g{(2}^{n+1}x)}{2^{n+1}}\right)\mathrm{,}\frac{t}{3})\mathrm{,}N\left(\frac{g{(2}^{n}x)}{2^n}-A(x)\mathrm{,}\frac{t}{3}\right)\\ \mathrm{,}N\left(\frac{g{(2}^{n+1}x)}{2^{n+1}}-\frac{g{(2}^{n}x)}{2^n}\right)\mathrm{,}\frac{t}{3})=M_a\left(x\mathrm{,}\frac{{2(2)}^{n}t}{3{\alpha }^n}\right)\end{array}$$

for all x ∈ X and all t > 0. Therefore,

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

(4.28)

for all x ∈ X. Replacing x, y by 2ⁿx, 2ⁿy in (4.2), respectively, we obtain

$$\begin{array}{c}N\left(\frac{1}{2^n}{D}_g{(2}^{n}x{\mathrm{,2}}^{n}y)\mathrm{,}t\right)=N(D_f(2^{n+1}x{\mathrm{,2}}^{n+1}y)-8D_f(2^{n}x{\mathrm{,2}}^{n}y),2^{n}t)\\ =\text{min}\left\{N(D_f{(2}^{n+1}x{\mathrm{,2}}^{n+1}y)\mathrm{,}\frac{2^{n}t}{2})\mathrm{,}N(D_f{(2}^{n}x{\mathrm{,2}}^{n}y)\mathrm{,}\frac{2^{n}t}{16})\right\}\\ \ge \text{m}in\left\{N^{\prime }\left({\phi }_a{(2}^{n+1}x{\mathrm{,2}}^{n+1}y)\mathrm{,}\frac{2^{n}t}{2}\right)\mathrm{,}{N}^{\prime }\left({\phi }_a{(2}^{n}x{\mathrm{,2}}^{n}y)\mathrm{,}\frac{2^{n}t}{16}\right)\right\}\\ =\text{min}\left\{N^{\prime }\left({\phi }_a(x\mathrm{,}y)\mathrm{,}\frac{2^{n}t}{2{\alpha }^{n+1}}\right)\mathrm{,}{N}^{\prime }\left({\phi }_a(x\mathrm{,}y)\mathrm{,}\frac{2^{n}t}{16{\alpha }^n}\right)\right\}\end{array}$$

which tends to 1 as n → ∞ for all x, y ∈ X and all t > 0. So we see that A satisfies (1.5). Thus, by Lemma 4.1, the mapping x ⇝ A(2x) - 8A(x) is additive. So (4.28) implies that the mapping A is additive.

The rest of the proof is similar to the proof of Theorem 3.2 and we omit the details.   □

Remark 4.3. Let 0 < α < 2. Suppose that the function t ↦ N(f(2x) - 8f(x) - A(x), .) from (0, ∞) into [0, 1] is right continuous. Then, we obtain a better fuzzy approximation than (4.27).

Corollary 4.4. Let X be a normed space and Y be a Banach space. Let ε, λ be non-negative real numbers such that λ ≠ 1. Suppose that an odd mapping f : X → Y satisfies the inequality (3.18) for all x, y ∈ X. Then, the limit

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

exists for all x ∈ X and A : X → Y is a unique additive mapping satisfying

$$∥f\left(2x\right)-8f\left(x\right)-A\left(x\right)∥\le \frac{\left(|2k+\ell {|}^{\lambda }+1\right)\left(9k^2+4\right)\epsilon \parallel x{\parallel }^{\lambda }}{\ell k^2\left(k^2-1\right)\left(1-2^{\lambda -1}\right)}$$

(4.29)

for all x ∈ X, where λℓ < ℓ.

Proof. The proof is similar to the proof of Corollary 3.4 and the result follows from Theorem 4.2.   □

Theorem 4.5. Denote N₁ the fuzzy norm obtained as Corollary 3.4 on R. Let for all x ∈ X, the functions r ↦ f(2rx) - 8f(rx) (from (R, N₁) into (Y, N)) and r ↦ φ _a (ι₁rx, ι₂ry) (from (R, N₁) into (Z, N')) be fuzzy continuous, where ι₁ ∈ {1, 2, (k + 1), (k - 1), (2k + 1), (2k - 1)} and ι₂ ∈ {1, 2, 3}. Then, for all x ∈ X, the function r ↦ A(rx) is fuzzy continuous and A(rx) = rA(x) for all r ∈ R.

Proof. The proof is similar to the proof of Theorem 3.5 and the result follows from Theorem 4.2.   □

Lemma 4.6. [22, 24]. If an odd mapping f : V₁ → V₂ satisfies (1.5), then the mapping h : V₁ → V₂ defined by h(x) = f(2x) - 2f(x) is cubic.

Theorem 4.7. Let ℓ ∈ {-1, 1} be fixed and let φ _c : X × X → Z be a mapping such that

$${\phi }_c\left(2x,2y\right)=\alpha {\phi }_c\left(x,y\right)$$

(4.30)

for all x, y ∈ X and for some positive real number α with αℓ < 8ℓ. Suppose that an odd mapping f : X → Y satisfies the inequality

$$N\left(D_f\left(x,y\right),t\right)\ge N^{\prime }\left({\phi }_c\left(x,y\right),t\right)$$

(4.31)

for all x, y ∈ X and all t > 0. Then, the limit

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

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

$$N\left(f\left(2x\right)-2f\left(x\right)-C\left(x\right),t\right)\ge M_c\left(x,\frac{\ell \left(8-\alpha \right)}{2}t\right)$$

(4.32)

for all x ∈ X and all t > 0, where

$$\begin{array}{c}{M}_c(x\mathrm{,}t)=\text{min}\{N^{\prime }\left({\phi }_c(x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}{N}^{\prime }\left({\phi }_c(2x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}\\ N^{\prime }\left({\phi }_c(x\mathrm{,2}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}{N}^{\prime }\left({\phi }_c((k+1)x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}\\ N^{\prime }\left({\phi }_c((k-1)x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right))\mathrm{,}{N}^{\prime }\left({\phi }_c(2x\mathrm{,2}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}\\ N^{\prime }\left({\phi }_c(x\mathrm{,3}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}{N}^{\prime }\left({\phi }_c((2k+1)x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}\\ N^{\prime }\left({\phi }_c((2k-1)x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\}\mathrm{.}\end{array}$$

Proof. Case (1): ℓ = 1. Similar to the proof of Theorem 4.2, we have

$$N\left(f\left(4x\right)-10f\left(2x\right)+16f\left(x\right),t\right)\ge M_c\left(x,t\right)$$

for for all x ∈ X and all t > 0, where M _c (x, t) is defined as in above. Letting h : X → Y be a mapping defined by h(x):= f(2x) - 2f(x). Then, we conclude that

$$N\left(h\left(2x\right)-8h\left(x\right),t\right)\ge M_c\left(x,t\right)$$

(4.33)

for all x ∈ X and all t > 0. So

$$N\left(\frac{h\left(2x\right)}{8}-h\left(x\right),\frac{t}{8}\right)\ge M_c\left(x,t\right)$$

(4.34)

for all x ∈ X and all t > 0. Then, by our assumption

$$M_c\left(2x,t\right)=M_c\left(x,\frac{t}{\alpha }\right)$$

(4.35)

for all x ∈ X and all t > 0. Replacing x by 2ⁿx in (4.34) and using (4.35), we obtain

$$N\left(\frac{h{(2}^{n+1}x)}{8^{n+1}}-\frac{h{(2}^{n}x)}{8^n}\mathrm{,}\frac{t}{{8(8}^n)}\right)\ge M_c{(2}^{n}x\mathrm{,}t)=M_c\left(x\mathrm{,}\frac{t}{{\alpha }^n}\right)$$

(4.36)

for all x ∈ X, t > 0 and n ≥ 0. Replacing t by αⁿt in (4.36), we see that

$$N\left(\frac{h{(2}^{n+1}x)}{8^{n+1}}-\frac{h{(2}^{n}x)}{8^n}\mathrm{,}\frac{t{\alpha }^n}{{8(8}^n)}\right)\ge M_c(x\mathrm{,}t)$$

(4.37)

for all x ∈ X, t > 0 and n > 0. It follows from $\frac{h{(2}^{n}x)}{8^n}-h(x)={\displaystyle {\sum }_{j=0}^{n-1}\left(\frac{h{(2}^{j+1}x)}{8^{j+1}}-\frac{h{(2}^{j}x)}{8^j}\right)}$ and (4.37) that

$$\begin{array}{r}N\left(\frac{h{(2}^{n}x)}{8^n}-h(x)\mathrm{,}{\displaystyle \sum _{j=0}^{n-1}\frac{{\alpha }^{j}t}{{8(8)}^j}}\right)\ge \text{min}{\displaystyle \underset{j=0}{\overset{n-1}{\cup }}\left\{N\left(\frac{h{(2}^{j+1}x)}{8^{j+1}}-\frac{h{(2}^{j}x)}{8^j}\mathrm{,}\frac{{\alpha }^{j}t}{{8(8)}^j}\right)\right\}}\\ \ge M_c(x\mathrm{,}t)\end{array}$$

(4.38)

for all x ∈ X, t > 0 and n > 0. Replacing x by 2^mx in (4.38), we observe that

$$N\left(\frac{h{(2}^{n+m}x)}{8^{n+m}}-\frac{h{(2}^{m}x)}{8^m}\mathrm{,}{\displaystyle \sum _{j=0}^{n-1}\frac{{\alpha }^{j}t}{{8(8)}^{j+m}}}\right)\ge M_c{(2}^{m}x\mathrm{,}t)=M_c\left(x\mathrm{,}\frac{t}{{\alpha }^m}\right)$$

for all x ∈ X, all t > 0 and all m ≥ 0, n > 0. So

$$N\left(\frac{h{(2}^{n+m}x)}{8^{n+m}}-\frac{h{(2}^{m}x)}{8^m}\mathrm{,}{\displaystyle \sum _{j=m}^{n+m-1}\frac{{\alpha }^{j}t}{{8(8)}^j}}\right)\ge M_c(x\mathrm{,}t)$$

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

$$N\left(\frac{h{(2}^{n+m}x)}{8^{n+m}}-\frac{h{(2}^{m}x)}{8^m}\mathrm{,}t\right)\ge M_c\left(x\mathrm{,}\frac{t}{{\displaystyle \sum _{j=m}^{n+m-1}\frac{{\alpha }^j}{{8(8)}^j}})}\right)$$

(4.39)

for all x ∈ X, all t > 0 and all m ≥ 0, n > 0. Since 0 < α < 8 and ${\sum }_{n=0}^{\infty }{\left(\frac{\alpha }{8}\right)}^n<\infty$, the Cauchy criterion for convergence and (N₅) imply that $\left\{\frac{h{(2}^{n}x)}{8^n}\right\}$ is a Cauchy sequence in (Y, N) to some point C(x) ∈ Y. So one can define the mapping C : X → Y by

$$C(x)=N-\underset{n\to \infty }{\lim }\frac{h{(2}^{n}x)}{8^n})$$

(4.40)

for all x ∈ X. Fix x ∈ X and put m = 0 in (4.39) to obtain

$$N\left(\frac{h{(2}^{n}x)}{8^n}-h(x)\mathrm{,}t\right)\ge M_c\left(x\mathrm{,}\frac{t}{{\displaystyle \sum _{j=0}^{n-1}}\frac{{\alpha }^j}{{8(8)}^j})}\right)$$

for all x ∈ X, t > 0 and n > 0. From which we obtain

$$\begin{array}{c}N(C(x)-h(x)\mathrm{,}t)\ge \text{min}\left\{N(C(x)-\frac{h{(2}^{n}x)}{8^n}\mathrm{,}\frac{t}{2})\mathrm{,}N(\frac{h{(2}^{n}x)}{8^n}-h(x)\mathrm{,}\frac{t}{2})\right\}\\ \ge M_c\left(x\mathrm{,}\frac{t}{{\displaystyle \sum _{j=0}^{n-1}\frac{{\alpha }^j}{{4(8}^j)}})}\right)\end{array}$$

(4.41)

for n large enough. Taking the limit as n → ∞ in (4.41), we obtain

$$N\left(C\left(x\right)-h\left(x\right),t\right)\ge M_a\left(x,\frac{t\left(8-\alpha \right)}{2}\right)$$

(4.42)

for all x ∈ X and all t > 0. It follows from (4.36) and (4.40) that

$$\begin{array}{l}N\left(\frac{C(2x)}{8}-C(x)\mathrm{,}t\right)\ge \text{min}\{N\left(\frac{C(2x)}{8}-\frac{h{(2}^{n+1}x)}{8^{n+1}}\right)\mathrm{,}\frac{t}{3})\mathrm{,}N\left(\frac{h{(2}^{n}x)}{8^n}-C(x)\mathrm{,}\frac{t}{3}\right)\mathrm{,}\\ N\left(\frac{h{(2}^{n+1}x)}{8^{n+1}}-\frac{h{(2}^{n}x)}{8^n}\right)\mathrm{,}\frac{t}{3})=M_c\left(x\mathrm{,}\frac{{8(8)}^{n}t}{3{\alpha }^n}\right)\end{array}$$

for all x ∈ X and all t > 0. Therefore,

$$C\left(2x\right)=8C\left(x\right)$$

(4.43)

for all x ∈ X. Replacing x, y by 2ⁿx, 2ⁿy in (4.31), respectively, we obtain

$$\begin{array}{c}N\left(\frac{1}{8^n}{D}_h{(2}^{n}x{\mathrm{,2}}^{n}y)\mathrm{,}t\right)=N(D_f(2^{n+1}x{\mathrm{,2}}^{n+1}y)-2D_f(2^{n}x{\mathrm{,2}}^{n}y),8^{n}t)\\ =\text{min}\left\{N\left(D_f{(2}^{n+1}x{\mathrm{,2}}^{n+1}y)\mathrm{,}\frac{8^{n}t}{2}\right)\mathrm{,}N\left(D_f{(2}^{n}x{\mathrm{,2}}^{n}y)\mathrm{,}\frac{8^{n}t}{4}\right)\right\}\\ \ge \text{min}\left\{N^{\prime }\left({\phi }_c{(2}^{n+1}x{\mathrm{,2}}^{n+1}y)\mathrm{,}\frac{8^{n}t}{2}\right)\mathrm{,}{N}^{\prime }\left({\phi }_c{(2}^{n}x{\mathrm{,2}}^{n}y)\mathrm{,}\frac{8^{n}t}{4}\right)\right\}\\ =\text{min}\left\{N^{\prime }\left({\phi }_c(x\mathrm{,}y)\mathrm{,}\frac{8^{n}t}{2{\alpha }^{n+1}}\right)\mathrm{,}{N}^{\prime }({\phi }_c(x\mathrm{,}y)\mathrm{,}\frac{8^{n}t}{4{\alpha }^n})\right\}\end{array}$$

which tends to 1 as n → ∞ for all x, y ∈ X and all t > 0. So we see that C, satisfies (1.5). Thus, by Lemma 4.6, the mapping x ⇝ C(2x) - 2C(x) is cubic. So (4.43) implies that the function C is cubic. The rest of the proof is similar to the proof of Theorem 3.2 and we omit the details.   □

Remark 4.8. Let 0 < α < 8. Suppose that the function t ↦ N(f(2x) - 2f(x) - C(x), .) from (0, ∞) into [0, 1] is right continuous. Then, we obtain a better fuzzy approximation than(4.42).

Corollary 4.9. Let X be a normed space and Y be a Banach space. Let ε, λ be non-negative real numbers such that λ ≠ 3. Suppose that an odd mapping f : X → Y satisfies the inequality (3.18) for all x, y ∈ X. Then, the limit

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

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

$$\parallel f\left(2x\right)-2f\left(x\right)-C\left(x\right)\left|\right|\le \frac{\left(|2k+\ell {|}^{\lambda }+1\right)\left(9k^2+4\right)\epsilon \parallel x{\parallel }^{\lambda }}{\ell k^2\left(k^2-1\right)\left(4-2^{\lambda -1}\right)}$$

(4.44)

for all x ∈ X, where λℓ < 3ℓ.

Theorem 4.10. Denote N₁ the fuzzy norm obtained as Corollary 3.4 on R. Let for all x ∈ X, the functions r ↦ f(2rx) - 2f(rx) (from (R, N₁) into (Y, N)) and r ↦ φ _q (ι₁rx, ι₂ry) (from (R, N₁) into (Z, N')) be fuzzy continuous, where ι₁ ∈ {1, 2, (k + 1), (k - 1), (2k + 1), (2k - 1)} and ι₂ ∈ {1, 2, 3}. Then, for all x ∈ X, the function r ↦ C(rx) is fuzzy continuous and C(rx) = r³C(x) for all r ∈ R.

Proof. The proof is similar to the proof of Theorem 3.5 and the result follows from Theorem 4.7.   □

Theorem 4.11. Let φ : X × X → Z be a mapping such that

$$\phi \left(2x,2y\right)=\alpha \phi \left(x,y\right)$$

(4.45)

for all x, y ∈ X and for some positive real number α. Suppose that an odd mapping f : X → Y satisfies the inequality

$$N\left(D_f\left(x,y\right),t\right)\ge N^{\prime }\left(\phi \left(x,y\right),t\right)$$

(4.46)

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

$$N\left(f\left(x\right)-A\left(x\right)-C\left(x\right),t\right)\ge \left\{\begin{array}{c}\hfill \mathsf{\text{min}}\left\{M\left(x,\frac{3t\left(2-\alpha \right)}{2}\right),M\left(x,\frac{3t\left(8-\alpha \right)}{2}\right)\right\},0<\alpha <2\hfill \\ \hfill \mathsf{\text{min}}\left\{M\left(x,\frac{3t\left(\alpha -2\right)}{2}\right),M\left(x,\frac{3t\left(8-\alpha \right)}{2}\right)\right\},2<\alpha <8\hfill \\ \hfill \mathsf{\text{min}}\left\{M\left(x,\frac{3t\left(\alpha -2\right)}{2}\right),M\left(x,\frac{3t\left(\alpha -8\right)}{2}\right)\right\},\alpha >8\hfill \end{array}\right.$$

(4.47)

for all x ∈ X and all t > 0, where

$$\begin{array}{c}M(x\mathrm{,}t)=\text{min}\{N^{\prime }\left(\phi (x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}{N}^{\prime }\left(\phi (2x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}\\ N^{\prime }\left(\phi (x\mathrm{,2}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}{N}^{\prime }\left(\phi ((k+1)x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}\\ N^{\prime }\left(\phi ((k-1)x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t)\right)\mathrm{,}{N}^{\prime }\left(\phi (2x\mathrm{,2}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}\\ N^{\prime }\left(\phi (x\mathrm{,3}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}{N}^{\prime }\left(\phi ((2k+1)x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}\\ N^{\prime }\left(\phi ((2k-1)x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\}\mathrm{.}\end{array}$$

Proof. Case (1): 0 < α < 2. By Theorems 4.2 and 4.7, there exist an additive mapping A₀ : X → Y and a cubic mapping C₀ : X → Y such that

$$N\left(f\left(2x\right)-8f\left(x\right)-A_0\left(x\right),t\right)\ge M\left(x,\frac{t\left(2-\alpha \right)}{2}\right)$$

(4.48)

and

$$N\left(f\left(2x\right)-2f\left(x\right)-C_0\left(x\right),t\right)\ge M\left(x,\frac{t\left(8-\alpha \right)}{2}\right)$$

(4.49)

for all x ∈ X and all t > 0. It follows from (4.48) and (4.49) that

$$N\left(f\left(x\right)+\frac{1}{6}{A}_0\left(x\right)-\frac{1}{6}{C}_0\left(x\right),t\right)\ge \mathsf{\text{min}}\left\{M\left(x,\frac{3t\left(2-\alpha \right)}{2}\right),M\left(x,\frac{3t\left(8-\alpha \right)}{2}\right)\right\}$$

(4.50)

for all x ∈ X and all t > 0. Letting $A\left(x\right)=-\frac{1}{6}{A}_0\left(x\right)$ and $C\left(x\right)=\frac{1}{6}{C}_0\left(x\right)$ in (4.50), we obtain

$$N\left(f\left(x\right)-A\left(x\right)-C\left(x\right),t\right)\ge \mathsf{\text{min}}\left\{M\left(x,\frac{3t\left(2-\alpha \right)}{2}\right),M\left(x,\frac{3t\left(8-\alpha \right)}{2}\right)\right\}$$

(4.51)

for all x ∈ X and all t > 0. To prove the uniqueness of A and C, let A', C': X → Y be another additive and cubic mappings satisfying (4.51). Let $Ā=A-A^{\prime }$ and $\bar{C}=C-C^{\prime }$. So

$$\begin{array}{ll}\hfill N\left(Ā\left(x\right)+\bar{C}\left(x\right),t\right)& \ge \mathsf{\text{min}}\left\{N\left(f\left(x\right)-A\left(x\right)-C\left(x\right),\frac{t}{2}\right),N\left(f\left(x\right)-A^{\prime }\left(x\right)-C^{\prime }\left(x\right),\frac{t}{2}\right)\right\}\\ \ge \mathsf{\text{min}}\left\{M\left(x,\frac{3t\left(2-\alpha \right)}{4}\right),M\left(x,\frac{3t\left(8-\alpha \right)}{4}\right)\right\}\end{array}$$

for all x ∈ X and all t > 0. Therefore, it follows from the last inequalities that

$$\begin{array}{c}N(\overline{A}(2^{n}x)+\overline{C}(2^{n}x),8^{n}t)\ge \text{min}\left\{M\left(2^{n}x\mathrm{,}\frac{{3(8}^n)t(2-\alpha )}{4}\right)\mathrm{,}M\left(2^{n}x\mathrm{,}\frac{{3(8}^n)t(8-\alpha )}{4}\right)\right\}\\ =\text{min}\left\{M\left(x\mathrm{,}\frac{{3(8}^n)t(2-\alpha )}{4{\alpha }^n}\right)\mathrm{,}M\left(x\mathrm{,}\frac{{3(8}^n)t(8-\alpha )}{4{\alpha }^n}\right)\right\}\end{array}$$

for all x ∈ X and all t > 0. So, ${\lim }_{n\to \infty }N\left(\frac{1}{8^n}(\overline{A}{(2}^{n}x)+\overline{C}{(2}^{n}x))\mathrm{,}t\right)=1$, hence $\bar{C}=0$ and then $Ā=0$. The rest of the proof, proceeds similarly to that in the previous case.   □

Remark 4.12. Let 0 < α < 2. Suppose that the function t ↦ N(f(x) - A(x) - C(x), .) from (0, ∞) into [0, 1] is right continuous. Then, we obtain a better fuzzy approximation than (4.51).

Corollary 4.13. Let X be a normed space and Y be a Banach space. Let ε, λ be non-negative real numbers. Suppose that an odd mapping f : X → Y satisfies the inequality (3.18) for all x, y ∈ X. Then there exist a unique additive mapping A : X → Y and a unique cubic mapping C : X → Y such that

$$\begin{array}{l}\parallel f(x)-A(x)-C(x)\parallel \\ \le (\begin{array}{ll}\frac{1}{6}\frac{(|2k+\ell {|}^{\lambda }+1)(9k^2+4)\epsilon }{k^2(k^2-1)}\left(\frac{1}{(1-2^{\lambda -1})}+\frac{1}{(4-2^{\lambda -1})}\right)\parallel x{\parallel }^{\lambda }\mathrm{,}\hfill & \lambda <1\hfill \\ \frac{1}{6}\frac{(|2k+\ell {|}^{\lambda }+1)(9k^2+4)\epsilon }{k^2(k^2-1)}\left(\frac{1}{{(2}^{\lambda -1}-1)}+\frac{1}{(4-2^{\lambda -1})}\right)\parallel x{\parallel }^{\lambda }\mathrm{,}\hfill & 1<\lambda <3\hfill \\ \frac{1}{6}\frac{(|2k+\ell {|}^{\lambda }+1)(9k^2+4)\epsilon }{k^2(k^2-1)}\left(\frac{1}{{(2}^{\lambda -1}-1)}+\frac{1}{{(2}^{\lambda -1}-4)}\right)\parallel x{\parallel }^{\lambda }\mathrm{,}\hfill & \lambda >3\hfill \end{array}\end{array}$$

(4.52)

for all x ∈ X.

Proof. The result follows by Corollaries 4.4 and 4.9.   □

Theorem 4.14. Denote N₁ the fuzzy norm obtained as Corollary 3.4 on R. Let for all x ∈ X, the functions r ↦ f(rx) (from (R, N₁) into (Y, N)) and r ↦ φ(ι₁rx, ι₂ry) (from (R, N₁) into (Z, N')) be fuzzy continuous, where ι₁ ∈ {1, 2, (k + 1), (k - 1), (2k + 1), (2k - 1)} and ι₂ ∈ {1, 2, 3}. Then, for all x ∈ X, the function r ↦ A(rx) + C(rx) is fuzzy continuous and A(rx) + C(rx) = rA(x) + r³C(x) for all r ∈ R.

Proof. The result follows by Theorems 4.5 and 4.10.   □

5. Fuzzy stability of the functional equation (1.5)

In this section, we prove the generalized Hyers-Ulam stability of a mixed cubic, quadratic, and additive functional equation (1.5) in fuzzy Banach spaces.

Theorem 5.1. Let φ : X × X → Z be a function which satisfies (3.1) and (4.45) for all x, y ∈ X and for some positive real number α. Suppose that a mapping f : X → Y satisfies the inequality

$$N\left(D_f\left(x,y\right),t\right)\ge N^{\prime }\left(\phi \left(x,y\right),t\right)$$

(5.1)

for all x, y ∈ X and all t > 0. Furthermore, assume that f(0) = 0 in (5.1) for the case f is even. If |k| = 2, then there exist a unique cubic mapping C : X → Y, a unique quadratic mapping Q : X → Y and a unique additive mapping A : X → Y such that

$$N\left(f\left(x\right)-C\left(x\right)-Q\left(x\right)-A\left(x\right),t\right)\ge \left\{\begin{array}{cc}\hfill \mathsf{\text{min}}\left\{{\tilde{M}}_1\left(x,t\right),{\tilde{M}}_1\left(-x,t\right)\right\},\hfill & \hfill 0<\alpha <2\hfill \\ \hfill \mathsf{\text{min}}\left\{{\tilde{M}}_2\left(x,t\right),{\tilde{M}}_2\left(-x,t\right)\right\},\hfill & \hfill 2<\alpha <k^2\hfill \\ \hfill \mathsf{\text{min}}\left\{{\tilde{M}}_3\left(x,t\right),{\tilde{M}}_3\left(-x,t\right)\right\},\hfill & \hfill k^2<\alpha <8\hfill \\ \hfill \mathsf{\text{min}}\left\{{\tilde{M}}_4\left(x,t\right),{\tilde{M}}_4\left(-x,t\right)\right\},\hfill & \hfill \alpha >8\hfill \end{array}\right.$$

(5.2)

for all x ∈ X and all t > 0, otherwise

$$N\left(f\left(x\right)-C\left(x\right)-Q\left(x\right)-A\left(x\right),t\right)\ge \left\{\begin{array}{cc}\hfill \mathsf{\text{min}}\left\{{\tilde{M}}_1\left(x,t\right),{\tilde{M}}_1\left(-x,t\right)\right\},\hfill & \hfill 0<\alpha <2\hfill \\ \hfill \mathsf{\text{min}}\left\{{\tilde{M}}_2\left(x,t\right),{\tilde{M}}_2\left(-x,t\right)\right\},\hfill & \hfill 2<\alpha <8\hfill \\ \hfill \mathsf{\text{min}}\left\{{\tilde{M}}_5\left(x,t\right),{\tilde{M}}_5\left(-x,t\right)\right\},\hfill & \hfill 8<\alpha <k^2\hfill \\ \hfill \mathsf{\text{min}}\left\{{\tilde{M}}_4\left(x,t\right),{\tilde{M}}_4\left(-x,t\right)\right\},\hfill & \hfill \alpha >k^2\hfill \end{array}\right.$$

(5.3)

for all x ∈ X and all t > 0, where

$$\begin{array}{l}{\tilde{M}}_1(x\mathrm{,}t)=\text{min}\left\{N^{\prime }\left(\phi (\mathrm{0,}x)\mathrm{,}\frac{(k^2-\alpha )}{4}t\right)\mathrm{,}M\left(x\mathrm{,}\frac{3t(2-\alpha )}{4}\right)\mathrm{,}M(x\mathrm{,}\frac{3t(8-\alpha )}{4}\right\}\mathrm{,}\\ {\tilde{M}}_2(x\mathrm{,}t)=\text{min}\{N^{\prime }\left(\phi (\mathrm{0,}x)\mathrm{,}\frac{(k^2-\alpha )}{4}t\right)\mathrm{,}M\left(x\mathrm{,}\frac{3t(\alpha -2)}{4}\right)\mathrm{,}M(x\mathrm{,}\frac{3t(8-\alpha )}{4}\}\mathrm{,}\\ {\tilde{M}}_3(x\mathrm{,}t)=\text{min}\left\{N^{\prime }\left(\phi (\mathrm{0,}x)\mathrm{,}\frac{(\alpha -k^2)}{4}t\right)\mathrm{,}M\left(x\mathrm{,}\frac{3t(\alpha -2)}{4}\right)\mathrm{,}M(x\mathrm{,}\frac{3t(8-\alpha )}{4}\right\}\mathrm{,}\\ {\tilde{M}}_4(x\mathrm{,}t)=\text{min}\left\{N^{\prime }\left(\phi (\mathrm{0,}x)\mathrm{,}\frac{(\alpha -k^2)}{4}t\right)\mathrm{,}M\left(x\mathrm{,}\frac{3t(\alpha -2)}{4}\right)\mathrm{,}M(x\mathrm{,}\frac{3t(\alpha -8)}{4}\right\}\mathrm{,}\\ {\tilde{M}}_5(x\mathrm{,}t)=\text{min}\left\{N^{\prime }\left(\phi (\mathrm{0,}x)\mathrm{,}\frac{(k^2-\alpha )}{4}t\right)\mathrm{,}M\left(x\mathrm{,}\frac{3t(\alpha -2)}{4}\right)\mathrm{,}M(x\mathrm{,}\frac{3t(\alpha -8)}{4}\right\}\end{array}$$

and

$$\begin{array}{c}M(x\mathrm{,}t)=\text{min}\{N^{\prime }\left(\phi (x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}{N}^{\prime }\left(\phi (2x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}\\ N^{\prime }\left(\phi (x\mathrm{,2}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}{N}^{\prime }\left(\phi ((k+1)x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}\\ N^{\prime }\left(\phi ((k-1)x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t)\right)\mathrm{,}{N}^{\prime }\left(\phi (2x\mathrm{,2}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}\\ N^{\prime }\left(\phi (x\mathrm{,3}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}{N}^{\prime }\left(\phi ((2k+1)x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\mathrm{,}\\ N^{\prime }\left(\phi ((2k-1)x\mathrm{,}x)\mathrm{,}\frac{k^2(k^2-1)}{9k^2+4}t\right)\}\mathrm{.}\end{array}$$

Proof. Case (1): 0 < α < 2. Assume that φ : X × X → Z satisfies (1.6) for all x, y ∈ X. Let $f_e\left(x\right)=\frac{1}{2}\left(f\left(x\right)+f\left(-x\right)\right)$ for all x ∈ X, then f _e (0) = 0, f _e (-x) = f _e (x), and

$$N\left(D_{f_e}\left(x,y\right),t\right)\ge \mathsf{\text{min}}\left\{N^{\prime }\left(\phi \left(x,y\right),t\right),N^{\prime }\left(\phi \left(-x,-y\right),t\right)\right\}$$

for all x, y ∈ X and all t > 0. By Theorem 3.2 for all x, y ∈ X, there exist a unique quadratic mapping Q : X → Y such that

$$N\left(f_e\left(x\right)-Q\left(x\right),t\right)\ge \mathsf{\text{min}}\left\{N^{\prime }\left(\phi \left(0,x\right),\frac{\left(k^2-\alpha \right)}{2}t\right),N^{\prime }\left(\phi \left(0,-x\right),\frac{\left(k^2-\alpha \right)}{2}t\right)\right\}$$

(5.4)

for all x ∈ X and all t > 0. Now, if φ : X × X → Z satisfies (4.45) for all x, y ∈ X, and let $f_o\left(x\right)=\frac{1}{2}\left(f\left(x\right)-f\left(-x\right)\right)$ for all x ∈ X, then

$$N\left(D_{f_o}\left(x,y\right),t\right)\ge \mathsf{\text{min}}\left\{N^{\prime }\left(\phi \left(x,y\right),t\right),N^{\prime }\left(\phi \left(-x,-y\right),t\right)\right\}$$

for all x, y ∈ X and all t > 0. By Theorem 4.11, it follows that there exist a unique cubic mapping C : X → Y and a unique additive mapping A ; X → Y such that

$$\begin{array}{ll}\hfill N\left(f_o\left(x\right)-C\left(x\right)-A\left(x\right),t\right)& \ge \mathsf{\text{min}}\left\{M\left(x,\frac{3t\left(2-\alpha \right)}{2}\right),M\left(x,\frac{3t\left(8-\alpha \right)}{2}\right)\right.\\ \left.,M\left(-x,\frac{3t\left(2-\alpha \right)}{2}\right),M\left(-x,\frac{3t\left(8-\alpha \right)}{2}\right)\right\}\end{array}$$

(5.5)

for all x ∈ X and all t > 0. It follows from (5.4) and (5.5) that

$$\begin{array}{l}N(f(x)-C(x)-Q(x)-A(x)\mathrm{,}t)\\ \ge \text{min}\{N^{\prime }\left(\phi (\mathrm{0,}x)\mathrm{,}\frac{(k^2-\alpha )}{4}t\right)\mathrm{,}M\left(x\mathrm{,}\frac{3t(2-\alpha )}{4}\right)\mathrm{,}M\left(x\mathrm{,}\frac{3t(8-\alpha )}{4}\right)\mathrm{,}\\ N^{\prime }\left(\phi (\mathrm{0,}-x)\mathrm{,}\frac{(k^2-\alpha )}{4}t\right)\mathrm{,}M\left(-x\mathrm{,}\frac{3t(2-\alpha )}{4}\right)\mathrm{,}M(-x\mathrm{,}\frac{3t(8-\alpha )}{4}\}\\ =\text{min}\left\{{\tilde{M}}_1(x\mathrm{,}t)\mathrm{,}{\tilde{M}}_1(-x\mathrm{,}t)\right\}\end{array}$$

(5.6)

The rest of the proof proceeds similarly to that in the previous case.   □

Remark 5.2. Let 0 < α < 2. Suppose that the function t ↦ N(f(x) - C(x) - Q(x) - A(x), .) from (0, ∞) into [0, 1] is right continuous. Then, we obtain a better fuzzy approximation than (5.2) or (5.3).

Corollary 5.3. Let X be a normed space and Y be a Banach space. Let ε, λ be non-negative real numbers. Suppose that f(0) = 0 in (3.18) for the case f : X → Y is even. Then, there exist a unique cubic mapping C : X → Y, a unique quadratic mapping Q : X → Y and a unique additive mapping A : X → Y such that

$$\begin{array}{l}\parallel f(x)-C(x)-Q(x)-A(x)\parallel \\ \le (\begin{array}{ll}\frac{1}{6}\frac{(|2k+\ell {|}^{\lambda }+1)(9k^2+4)\epsilon }{k^2(k^2-1)}\left(\frac{1}{(1-2^{\lambda -1})}+\frac{1}{(4-2^{\lambda -1})}\right)\parallel x{\parallel }^{\lambda }+\frac{2\epsilon }{k^2-k^{\lambda }}\parallel x{\parallel }^{\lambda }\mathrm{,}\hfill & \lambda <1\hfill \\ \frac{1}{6}\frac{(|2k+\ell {|}^{\lambda }+1)(9k^2+4)\epsilon }{k^2(k^2-1)}\left(\frac{1}{{(2}^{\lambda -1}-1)}+\frac{1}{(4-2^{\lambda -1})}\right)\parallel x{\parallel }^{\lambda }+\frac{2\epsilon }{k^2-k^{\lambda }}\parallel x{\parallel }^{\lambda }\mathrm{,}\hfill & 1<\lambda <2\hfill \\ \frac{1}{6}\frac{(|2k+\ell {|}^{\lambda }+1)(9k^2+4)\epsilon }{k^2(k^2-1)}\left(\frac{1}{{(2}^{\lambda -1}-1)}+\frac{1}{(4-2^{\lambda -1})}\right)\parallel x{\parallel }^{\lambda }+\frac{2\epsilon }{k^{\lambda }-k^2}\parallel x{\parallel }^{\lambda }\mathrm{,}\hfill & 2<\lambda <3\hfill \\ \frac{1}{6}\frac{(|2k+\ell {|}^{\lambda }+1)(9k^2+4)\epsilon }{k^2(k^2-1)}\left(\frac{1}{{(2}^{\lambda -1}-1)}+\frac{1}{{(2}^{\lambda -1}-4)}\right)\parallel x{\parallel }^{\lambda }+\frac{2\epsilon }{k^{\lambda }-k^2}\parallel x{\parallel }^{\lambda }\mathrm{,}\hfill & \lambda >3\hfill \end{array}\end{array}$$

(5.7)

for all x ∈ X.

Proof. The result follows by Corollaries 3.4 and 4.13.   □

Theorem 5.4. Denote N₁ the fuzzy norm obtained as Corollary 3.4 on R. Let for all x ∈ X, the functions r ↦ f(rx) (from (R, N₁) into (Y, N)) and r ↦ φ(ι₁rx, ι₂ry) (from (R, N₁) into (Z, N')) be fuzzy continuous, where ι₁ ∈ {0, ± 1, ± 2, ± (k + 1), ± (k - 1), ± (2k + 1), ± (2k - 1)} and ι₂ ∈ { ± 1, ± 2, ± 3}. Then, for all x ∈ X, the function r ↦ C(rx) + Q(rx) + A(rx) is fuzzy continuous and C(rx) + Q(rx) + A(rx) = r³C(x) + r²Q(x) + rA(x) for all r ∈ R.

Proof. The result follows by Theorems 3.5 and 4.14.   □