Fuzzy Hyers-Ulam approximation of a mixed AQ mapping

Hassan Azadi Kenary; Saedeh Talebzadeh; Hamid Rezaei; Jung Rye Lee; Dong Yun Shin

doi:10.1186/1029-242X-2012-64

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Fuzzy Hyers-Ulam approximation of a mixed AQ mapping

Hassan Azadi Kenary¹,
Saedeh Talebzadeh²,
Hamid Rezaei¹,
Jung Rye Lee³ and
Dong Yun Shin⁴Email author

Journal of Inequalities and Applications20122012:64

https://doi.org/10.1186/1029-242X-2012-64

Received: 19 November 2011
Accepted: 14 March 2012
Published: 14 March 2012

Abstract

In this article, we adopt the fixed point and direct methods to prove the Hyers-Ulam stability for an additive-quadratic functional equation in fuzzy Banach spaces.

Mathematics Subject Classification 2010: 39B52, 46S40; 26E50.

Keywords

fuzzy Banach space
functional equation
fixed point
Hyers-Ulam stability

1. Introduction and preliminaries

A classical question in the theory of functional equations is the following: "When is it true that a function which approximately satisfies a functional equation must be close to an exact solution of the equation?". If the problem accepts a solution, we say that the equation is stable. The first stability problem concerning group homomorphisms was raised by Ulam [1]. In the next year, Hyers [2] gave a positive answer to the above question for additive groups under the assumption that the groups are Banach spaces. In 1978, Rassias [3] extended the theorem of Hyers by considering the unbounded Cauchy difference ∥f(x + y)- f(x) - f(y)∥ ≤ ε(∥x∥^p+ ∥y∥^p), (ε > 0, p ∈[0,1)). Furthermore, in 1994, a generalization of Rassias' theorem was obtained by Găvruta [4] by replacing the bound ϵ(∥x∥^p+ ∥y∥^p) by a general control function φ(x, y).

In 1983, a Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [5] for mappings f : X → Y, where X is a normed space and Y is a Banach space. In 1984, Cholewa [6] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group and, in 2002, Czerwik [7] proved the Hyers-Ulam stability of the quadratic functional equation.

Katsaras [8] defined a fuzzy norm on a vector space to construct a fuzzy vector topological structure on the space. Some mathematicians have defined fuzzy norms of a vector space from various points of view (see [9–17]).

In particular, Bag and Samanta [18], following Cheng and Mordeson [19], gave an idea of fuzzy norm in such a manner that the corresponding fuzzy metric is of Karmosil and Michalek type [20]. They established a decomposition theorem of a fuzzy norm into a family of crisp norms and investigated some properties of fuzzy normed spaces [21].

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

\begin{align} r^{2} f (\frac{x + y + z}{r}) & + r^{2} f (\frac{x - y + z}{r}) + r^{2} f (\frac{x + y - z}{r}) + r^{2} f (\frac{- x + y + z}{r}) \\ = 4 f (x) + 4 f (y) + 4 f (z) \end{align}

(1.1)

where r is a positive real number, in fuzzy Banach spaces.

Lee and Jun [22] proved that an even (odd) mapping f : X → Y satisfies the functional Equation (1.1) if and only if the mapping f : X → Y is quadratic (additive). Moreover, they proved the Hyers-Ulam stability of the functional Equation (1.1) in non-Archimedean normed spaces. The stability problems of several functional equations have been extensively investigated by a number of authors, and there are many interesting results concerning this problem (see [23–33]).

Definition 1.1. [18] Let X be a real vector space. A function N : X × ℝ → [0,1] is called a fuzzy norm on X if for all x, y ∈ X and all s,t ∈ ℝ,

(N 1) N(x, t) = 0 for t ≤ 0;

(N 2) x = 0 if and only if N(x, t) = 1 for all t > 0;

(N 3) $N (c x, t) = N (x, \frac{t}{|c|})$ if c ≠ 0;

(N 4) N(x + y, c + t) ≥ min{N(x, s), N(y, t)};

(N 5) N(x,.) is a non-decreasing function of ℝ and lim_t→∞N(x, t) = 1;

(N 6) for x ≠ 0, N(x,.) is continuous on ℝ.

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

Example 1.2. Let (X, ∥⋅∥) be a normed linear space and α, β > 0. Then

N (x, t) = \{\begin{matrix} \frac{α t}{α t + β ∥x∥} & t > 0, & x \in X \\ 0 & t \leq 0, & x \in X \end{matrix}

is a fuzzy norm on X.

Definition 1.3. [18] Let (X, N) be a fuzzy normed vector space. A sequence {x_n} in X is said to be convergent or converge if there exists an x ∈ X such that lim_t→∞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 we denote it by N - lim_t→∞x_n= x.

Definition 1.4. [18] Let (X, N) be a fuzzy normed vector space. A sequence {x_n} in X is called Cauchy if for each ϵ > 0 and each t > 0 there exists an n₀ ∈ ℕ such that for all n ≥ n₀ and all p > 0, we have N(x_n+p- x_n, t) > 1 - ϵ.

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.

Example 1.5. Let N : X × ℝ → [0,1] be a fuzzy norm on ℝ defined by

N (x, t) = \{\begin{matrix} \frac{t}{t + |x|} & t > 0 \\ 0 & t \leq 0 \end{matrix} .

Then (ℝ, N) is a fuzzy Banach space. Let {x_n} be a Cauchy sequence in ℝ, δ > 0 and

ε = \frac{δ}{1 + δ}

. Then there exist m ∈ ℕ such that for all n ≥ m and all p > 0, we have

\frac{1}{1 + |x_{n + p} - x_{n}|} \geq 1 - ε .

So |x_n+p-x_n| < δ for all n ≥ m and all p > 0. Therefore {x_n} is a Cauchy sequence in (ℝ, |·|). Let x_n→ x₀ ∈ ℝ as n → ∞. Then lim_n→∞N (x_n- x₀, t) = 1 for all t > 0.

We say that a mapping f : X → Y between fuzzy normed vector spaces X and Y is continuous at a point x ∈ X if for each sequence {x_n} converging to x₀ ∈ X, then the sequence {f(x_n)} 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 ([21]).

Definition 1.6. Let X be a set. A function d : X × X → [0, ∞] is called a generalized metric on X if d satisfies the following conditions:

(1)

d(x, y) = 0 if and only if x = y for all x, y ∈ X;
(2)

d(x, y) = d(y, x) for all x, y ∈ X;
(3)

d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X.

Theorem 1.7. Let (X, d) be a complete generalized metric space and J : X → X be a strictly contractive mapping with Lipschitz constant L < 1. Then, for all x ∈ X, either

d (J^{n} x, J^{n + 1} x) = \infty

for all nonnegative integers n or there exists a positive integer n ₀ such that

(1)

d(J ⁿ x, J ^{n+ 1} x) < ∞ for all n ₀ ≥ n ₀;
(2)

the sequence {J ⁿ x} converges to a fixed point y* of J;
(3)

y* is the unique fixed point of J in the set $Y = {y \in X : d (J^{n_{0}} x, y) < \infty}$ ;
(4)

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

2. Fuzzy stability of the functional equation (1.1): fixed point method

In this section, using the fixed point method, we prove the Hyers-Ulam stability of the additive-quadratic functional equation (1.1) in fuzzy Banach spaces.

Remark 2.1. If f be an odd mapping satisfying (1.1) and r = 2, then an additive mapping f is nonzero in general. But if r is a rational number and r ≠ 2 in (1.1), then f ≡ 0. So, to avoid the trivial case, let r = 2 or r is an positive irrational real number.

Theorem 2.2. Let φ: X³ → [0, ∞) be a function such that there exists an α < 1 with

φ (2 x, 2 y, 2 z) \leq 2 α φ (x, y, z)

(2.1)

for all x, y, z ∈ X. Let f : X → Y be an odd mapping satisfying

\begin{gathered} N (r^{2} f (\frac{x + y + z}{r}) + r^{2} f (\frac{x - y + z}{r}) + r^{2} f (\frac{x + y - z}{r}) + r^{2} f (\frac{- x + y + z}{r}) \\ - 4 f (x) - 4 f (y) - 4 f (z), t) \geq \frac{1}{t + φ (x, y, z)} \end{gathered}

(2.2)

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

A (x) : = N - lim_{n \to \infty} \frac{f (2^{n} x)}{2^{n}}

exists for each x ∈ X and defines a unique additive mapping A:X → Y such that

N (f (x) - A (x), t) \geq \frac{(4 - 4 α) t}{(4 - 4 α) t + φ (2 x, 0, 0) + φ (x, x, 0)}

(2.3)

Proof. Note that f (0) = 0 and f (-x) = -f(x) for all x ∈ X since f is an odd function. Putting y = z = 0 in (2.2) and replacing x by 2x, we get

N (2 r^{2} f (\frac{2 x}{r}) - 4 f (2 x), t) \geq \frac{t}{t + φ (2 x, 0, 0)}

(2.4)

for all x ∈ X and all t > 0. Putting y = x and z = 0 in (2.2), we have

N (2 r^{2} f (\frac{2 x}{r}) - 8 f (x), t) \geq \frac{t}{t + φ (x, x, 0)}

(2.5)

for all x ∈ X and all t > 0. By (2.4), (2.5), (N 3) and (N 4), we get

\begin{align} N (\frac{f (2 x)}{2} - f (x), \frac{t}{4}) & = N (2 r^{2} f (\frac{2 x}{r}) - 8 f (x) - 2 r^{2} f (\frac{2 x}{r}) + 4 f (2 x), 2 t) \\ \geq min \{N (2 r^{2} f (\frac{2 x}{r}) - 4 f (2 x), t), N (2 r^{2} f (\frac{2 x}{r}) - 8 f (x), t)\} \\ \geq \frac{t}{t + φ (2 x, 0, 0) + φ (x, x, 0)} . \end{align}

(2.6)

for all x ∈ X and all t > 0. Consider the set S := {h : X → Y} and introduce the generalized metric on S:

d (g, h) = inf_{μ \in (0, + \infty)} \{N (g (x) - h (x), μ t) \geq \frac{t}{t + φ (2 x, 0, 0) + φ (x, x, 0)}, \forall x \in X\},

where, as usual, inf ϕ = + ∞. It is easy to show that (S, d) is complete (see [11]). Now we consider the linear mapping J : (S, d) → (S, d) such that

J g (x) : = \frac{1}{2} g (2 x)

for all x ∈ X.

Let g, h ∈ S be given such that d(g, h) = β. Then

N (g (x) - h (x), β t) \geq \frac{t}{t + φ (2 x, 0, 0) + φ (x, x, 0)}

for all x ∈ X and all t > 0. Hence

\begin{align} N (J g (x) - J h (x), α β t) = N (\frac{1}{2} g (2 x) - \frac{1}{2} h (2 x), α β t) & = N (g (2 x) - h (2 x), 2 α β t) \\ \geq \frac{2 α t}{2 α t + φ (4 x, 0, 0) + φ (2 x, 2 x, 0)} \\ \geq \frac{2 α t}{2 α t + 2 α (φ (2 x, 0, 0) + φ (x, x, 0))} \\ = \frac{t}{t + φ (2 x, 0, 0) + φ (x, x, 0)} \end{align}

for all x ∈ X and all t > 0. So d (g, h) = β implies that d (Jg, Jh) ≤ αβ. This means that d (Jg, Jh) ≤ αd(g, h) for all g, h ∈ S. It follows from (2.6) that

d (f, J f) \leq \frac{1}{4} .

By Theorem 1.7, there exists a mapping A : X → Y satisfying the following:

(1)

A is a fixed point of J, i.e.,
$2 A (x) = A (2 x)$

(2.7)

for all x ∈ X. The mapping A is a unique fixed point of J in the set M = {g ∈ S : d (h, g) < ∞}. This implies that A is a unique mapping satisfying (2.7) such that there exists a μ ∈ (0, ∞) satisfying

N (f (x) - A (x), μ t) \geq \frac{t}{t + φ (2 x, 0, 0) + φ (x, x, 0)}

for all x ∈ X;

(2)

d(J ⁿ f, A) → 0 as n → ∞. This implies the equality $lim_{n \to \infty} \frac{f (2^{n} x)}{2^{n}} = A (x)$ for all x ∈ X;
(3)

$d (f, A) \leq \frac{1}{1 - α} d (f, J f)$ , which implies the inequality $d (f, A) \leq \frac{1}{4 - 4 α}$ . This implies that the inequalities (2.3) holds.

It follows from (2.1) and (2.2) that

\begin{align} N (\frac{1}{2^{n}} [r^{2} f (\frac{2^{n} (x + y + z)}{r}) + r^{2} f (\frac{2^{n} (x - y + z)}{r}) + r^{2} f (\frac{2 n (x + y - z)}{r}) \\ + r^{2} f (\frac{2^{n} (- x + y + z)}{r}) - 4 f (2^{n} x) - 4 f (2^{n} y) - 4 f (2^{n} z)], \frac{1}{2^{n}}) \\ \geq \frac{t}{t + φ (2^{n} x, 2^{n} y, 2^{n} z)} \end{align}

for all x, y, z ∈ X, all t > 0 and all n ∈ ℕ. So

\begin{align} N (\frac{1}{2^{n}} [r^{2} f (\frac{2^{n} (x + y + z)}{r}) + r^{2} f (\frac{2^{n} (x - y + z)}{r}) + r^{2} f (\frac{2^{n} (x + y - z)}{r}) \\ + r^{2} f (\frac{2^{n} (- x + y + z)}{r}) - 4 f (2^{n} x) - 4 f (2^{n} y) - 4 f (2^{n} z)], t) \\ \geq \frac{2^{n} t}{2^{n} t + {(2 α)}^{n} φ (x, y, z)} \end{align}

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

lim_{n \to \infty} \frac{2^{n} t}{2^{n} t + {(2 α)}^{n} φ (x, y, z)} = 1

for all x, y, z ∈ X and all t > 0, we obtain that

\begin{align} N (r^{2} A (\frac{x + y + z}{r}) + 2^{r} A (\frac{x - y + z}{r}) \\ + r^{2} A (\frac{x + y - z}{r}) + r^{2} A (\frac{- x + y + z}{s}) - 4 A (x) - 4 A (y) - 4 A (z), t) = 1 \end{align}

for all x, y, z ∈ X and all t > 0. Hence the mapping A : X → Y is additive, as desired.

Corollary 2.3. Let θ ≥ 0 and r be a real positive number with r < 1. Let X be a normed vector space with norm ∥·∥. Let f : X → Y be an odd mapping satisfying

\begin{align} N (r^{2} f (\frac{x + y + z}{r}) + r^{2} f (\frac{x - y + z}{r}) + r^{2} f (\frac{x + y - z}{r}) + r^{2} f (\frac{- x + y + z}{r}) \\ - 4 f (x) - 4 f (y) - 4 f (z), t) \geq \frac{t}{t + θ ({∥x∥}^{r} + {∥y∥}^{r} + {∥z∥}^{r})} \end{align}

(2.8)

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

A (x) : = N - lim_{n \to \infty} \frac{f (2^{n} x)}{2^{n}}

exists for each x ∈ X and defines a unique additive mapping A: X → Y such that

N (f (x) - A (x), t) \geq \frac{4 (1 - 2^{r - 1}) t}{4 (1 - 2^{r - 1}) t + (2^{r} + 2) θ {∥x∥}^{r}} .

for all x ∈ X and all t > 0.

Proof. The proof follows from Theorem 2.2 by taking φ(x, y, z) := θ(∥x∥^r+ ∥y∥^r+ ∥z∥^r) for all x, y, z ∈ X. Then we can choose α = 2^r-1and we get the desired result.

Theorem 2.4. Let φ : X³ → [0, ∞) be a function such that there exists an α < 1 with

φ (\frac{x}{2}, \frac{y}{2}, \frac{z}{2}) \leq \frac{α}{2} φ (x, y, z)

(2.9)

for all x, y, z ∈ X. Let f : X → Y be an odd mapping satisfying (2.2). Then the limit

A (x) : = N - lim_{n \to \infty} 2^{n} f (\frac{x}{2^{n}})

exists for each x ∈ X and defines a unique additive mapping A : X → Y such that

N (f (x) - A (x), t) \geq \frac{(4 - 4 α) t}{(4 - 4 α) t + α (φ (2 x, 0, 0) + φ (x, x, 0))}

(2.10)

Proof. Let (S, d) be the generalized metric space defined as in the proof of Theorem 2.2. Consider the linear mapping J : S → S such that

J g (x) : = 2 g (\frac{x}{2}) .

Let g,h ∈ S be given such that d (g, h) = β. Then

N (g (x) - h (x), β t) \geq \frac{t}{t + φ (2 x, 0, 0) + φ (x, x, 0)}

for all x ∈ X and all t > 0. Hence

\begin{align} N (J g (x) - J h (x), α β t) = N (2 g (\frac{x}{2}) - 2 h (\frac{x}{2}), α β t) & = N (g (\frac{x}{2}) - g (\frac{x}{2}), \frac{α β t}{2}) \\ \geq \frac{\frac{α}{2} t}{\frac{α}{2} t + φ (x, 0, 0) + φ (\frac{x}{2}, \frac{x}{2}, 0)} \\ \geq \frac{t}{t + φ (2 x, 0, 0) + φ (x, x, 0)} \end{align}

for all x ∈ X and all t > 0. So d (g, h) = β implies that d (Jg, Jh) ≤ αβ. This means that d(Jg, Jh) ≤ αd (g, h) for all g, h ∈ S. It follows from (2.6) that

N (2 f (\frac{x}{2}) - f (x), \frac{t}{2}) \geq \frac{t}{t + φ (x, 0, 0) + φ (\frac{x}{2}, \frac{x}{2}, 0)} \geq \frac{t}{t + \frac{α}{2} (φ (2 x, 0, 0) + φ (x, x, 0))}

for all x ∈ X and t > 0. Therefore

N (2 f (\frac{x}{2}) - f (x), \frac{α t}{4}) \geq \frac{t}{t + φ (2 x, 0, 0) + φ (x, x, 0)} .

(2.11)

So

d (f, J f) \leq \frac{α}{4}

. By Theorem 1.7, there exists a mapping A : X → Y satisfying the following:

(1)

A is a fixed point of J, that is,
$A (\frac{x}{2}) = \frac{1}{2} A (x)$

(2.12)

for all x ∈ X. The mapping A is a unique fixed point of J in the set Ω = {h ∈ S : d(g, h) < ∞}. This implies that A is a unique mapping satisfying (2.12) such that there exists μ ∈ (0, ∞) satisfying

N (f (x) - A (x), μ t) \geq \frac{t}{t + φ (2 x, 0, 0) + φ (x, x, 0)}

for all x ∈ X and t > 0.

(2)

d(J ⁿ f, A) → 0 as n → ∞. This implies the equality $N - lim_{n \to \infty} 2^{n} f (\frac{x}{2^{n}}) = A (x)$ for all x ∈ X.
(3)

$d (f, A) \leq \frac{d (f, J f)}{1 - L}$ with f ∈ Ω, which implies the inequality $d (f, A) \leq \frac{α}{4 - 4 α}$ . This implies that the inequality (2.10) holds.

The rest of the proof is similar to that of the proof of Theorem 2.2.

Corollary 2.5. Let θ ≥ 0 and let r be a real number with r > 1. Let X be a normed vector space with norm ∥·∥. Let f : X → Y be an odd mapping satisfying (2.8). Then, the limit

A (x) : = N - lim_{n \to \infty} 2^{n} f (\frac{x}{2^{n}})

exists for each x ∈ X and defines a unique additive mapping A : X → Y such that

N (f (x) - A (x), t) \geq \frac{4 (2^{r} - 2) t}{4 (2^{r} - 2) t + (2^{r + 1} + 4) θ {∥x∥}^{r}} .

for all x ∈ X and all t > 0.

Proof. The proof follows from Theorem 2.4 by taking φ(x, y, z) := θ(∥x∥^r+ ∥y∥^r+ ∥z∥^r) for all x, y, z ∈ X. Then we can choose α = 2^1-rand we get the desired result.

Theorem 2.6. Let φ : X³ → [0, ∞) be a function such that there exists an α < 1 with

φ (2 x, 2 y, 2 z) \leq 4 α φ (x, y, z)

(2.13)

for all x, y, z ∈ X. Let f : X → Y be an even mapping with f(0) = 0 and satisfying (2.2). Then

Q (x) : = N - lim_{n \to \infty} \frac{f (2^{n} x)}{4^{n}}

exists for each x ∈ X and defines a unique quadratic mapping Q : X → Y such that

N (f (x) - Q (x), t) \geq \frac{(8 - 8 α) t}{(8 - 8 α) t + 3 (φ (2 x, 0, 0) + φ (x, x, 0))}

(2.14)

Proof. Putting y = x and z = 0 in (2.2), we get

N (2 r^{2} f (\frac{2 x}{r}) - 8 f (x), t) \geq \frac{t}{t + φ (x, x, 0)}

(2.15)

for all x ∈ X and all t > 0. Putting y = z = 0 and replacing x by 2x in (2.2), we have

N (4 r^{2} f (\frac{2 x}{r}) - 4 f (2 x), t) \geq \frac{t}{t + φ (2 x, 0, 0)}

(2.16)

for all x ∈ X and all t > 0. By (2.15), (2.16), (N 3) and (N 4), we get

\begin{align} N (\frac{f (2 x)}{4} - f (x), \frac{t}{16}) & = N (2 (2 r^{2} f (\frac{2 x}{r}) - 8 f (x)) - (4 r^{2} f (\frac{2 x}{r}) - 4 f (2 x)), t) \\ \geq min \{N (2 r^{2} f (\frac{2 x}{r}) - 8 f (x), \frac{t}{4}), N (4 r^{2} f (\frac{2 x}{r}) - 4 f (2 x), \frac{t}{2})\} \end{align}

for all x ∈ X and all t > 0. So

\begin{align} N (\frac{f (2 x)}{4} - f (x), \frac{3 t}{8}) & \geq min \{N (2 r^{2} f (\frac{2 x}{r}) - 8 f (x), t), N (4 r^{2} f (\frac{2 x}{r}) - 4 f (x), t)\} \\ \geq \frac{t}{t + φ (x, x, 0) + φ (2 x, 0, 0)} . \end{align}

(2.17)

Consider the set S* := {h : X → Y; h(0) = 0} and introduce the generalized metric on S*:

d (g, h) = inf_{μ \in (0, + \infty)} \{N (g (x) - h (x), μ t) \geq \frac{t}{t + φ (2 x, 0, 0) + φ (x, x, 0)}, \forall x \in X\},

where, as usual, inf ϕ = + ∞. It is easy to show that (S*, d) is complete (see [11]). Now we consider the linear mapping J : (S*, d) → (S*, d) such that

J g (x) : = \frac{1}{4} g (2 x)

for all x ∈ X. Proceeding as in the proof of Theorem 2.2, we obtain that d(g, h) = β implies that d(Jg, Jh) ≤ αβ. This means that d(Jg, Jh) ≤ αd (g, h) for all g, h ∈ S. It follows from (2.17) that

d (f, J f) \leq \frac{3}{8} .

By Theorem 1.7, there exists a mapping Q : X → Y satisfying the following:

(1)

Q is a fixed point of J, i.e.,
$4 Q (x) = Q (2 x)$

(2.18)

for all x ∈ X. The mapping Q is a unique fixed point of J in the set M = {g ∈ S* : d(h, g) < ∞}. This implies that Q is a unique mapping satisfying (2.18) such that there exists a μ ∈ (0, ∞) satisfying

N (f (x) - Q (x), μ t) \geq \frac{t}{t + φ (2 x, 0, 0) + φ (x, x, 0)}

for all x ∈ X;

(2)

d(J ⁿ f, Q) → 0 as n → ∞. This implies the equality $lim_{n \to \infty} \frac{f (2^{n} x)}{4^{n}} = Q (x)$ for all x ∈ X;
(3)

$d (f, Q) \leq \frac{1}{1 - α} d (f, J f)$ , which implies the inequality $d (f, Q) \leq \frac{3}{8 - 8 α}$ . This implies that the inequalities (2.14) holds.

The rest of the proof is similar to the proof of Theorem 2.2.

Corollary 2.7. Let θ ≥ 0 and r be a real positive number with r < 1. Let X be a normed vector space with norm ∥·∥. Let f : X → Y be an even mapping satisfying (2.8) and f(0) = 0. Then the limit

Q (x) : = N - lim_{n \to \infty} \frac{f (2^{n} x)}{4^{n}}

exists for each x ∈ X and defines a unique quadratic mapping Q : X → Y such that

N (f (x) - Q (x), t) \geq \frac{2 (4 - 4^{r}) t}{2 (4 - 4^{r}) t + 3 (2^{r} + 2) θ {∥x∥}^{r}} .

for all x ∈ X and all t > 0.

Proof. The proof follows from Theorem 2.6 by taking φ(x, y, z) := θ(∥x∥^r+ ∥y∥^r+ ∥z∥^r) for all x, y, z ∈ X. Then we can choose α = 4^r-1and we get the desired result.

Theorem 2.8. Let φ : X³ → [0, ∞) be a function such that there exists an α < 1 with

φ (\frac{x}{2}, \frac{y}{2}, \frac{z}{2}) \leq \frac{α}{4} φ (x, y, z)

for all x, y, z ∈ X. Let f : X → Y be an even mapping satisfying (2.2) and f(0) = 0. Then the limit

Q (x) : = N - lim_{n \to \infty} 4^{n} f (\frac{x}{2^{n}})

exists for each x ∈ X and defines a unique quadratic mapping Q : X → Y such that

N (f (x) - Q (x), t) \geq \frac{(8 - 8 α) t}{(8 - 8 α) t + 3 α (φ (2 x, 0, 0) + φ (x, x, 0))}

Proof. Let (S*, d) be the generalized metric space defined as in the proof of Theorem 2.6.

It follows from (2.17) that

N (f (x) - 4 f (\frac{x}{2}), \frac{3 t}{2}) \geq \frac{t}{t + φ (x, 0, 0) + φ (\frac{x}{2}, \frac{x}{2}, 0)} \geq \frac{t}{t + \frac{α}{4} (φ (2 x, 0, 0) + φ (x, x, 0))}

for all x ∈ X and t > 0. So

N (f (x) - 4 f (\frac{x}{2}), \frac{3 α t}{8}) \geq \frac{t}{t + φ (2 x, 0, 0) + φ (x, x, 0)} .

(2.19)

The rest of the proof is similar to the proof of Theorems 2.2 and 2.4.

Corollary 2.9. Let θ ≥ 0 and r be a real number with r > 1. Let X be a normed vector space with norm ∥·∥. Let f : X → Y be an even mapping satisfying f(0) = 0 and (2.8). Then

Q (x) : = N - lim_{n \to \infty} 4^{n} f (\frac{x}{2^{n}})

exists for each x ∈ X and defines a unique quadratic mapping Q : X → Y such that

N (f (x) - Q (x), t) \geq \frac{2 (4^{r + 1} - 16) t}{2 (4^{r + 1} - 16) t + 3 (2^{r + 2} + 8) θ ∥x∥ r} .

for all x ∈ X and all t > 0.

Proof. The proof follows from Theorem 2.8 by taking φ(x, y, z) := θ(∥x∥^r+ ∥y∥^r+ ∥z∥^r) for all x, y, z ∈ X. Then we can choose α = 4^1-rand we get the desired result.

Let f : X → Y be a mapping satisfying f(0) = 0 and (1.1). Let $f_{e} (x) : = \frac{f (x) + f (- x)}{2}$ and $f_{o} (x) = \frac{f (x) - f (- x)}{2}$ . Then f_eis an even mapping satisfying (1.1) and f_ois an odd mapping satisfying (1.1) such that f(x) = f_e(x) + f_o(x). So we obtain the following.

Theorem 2.10. Let φ : X³ → [0, ∞) be a function such that there exists an α < 1 with

φ (\frac{x}{2}, \frac{y}{2}, \frac{z}{2}) \leq \frac{α}{4} φ (x, y, z)

for all x, y, z ∈ X. Let f : X → Y be a mapping satisfying (2.2) and f(0) = 0. Then there exist a unique additive mapping A: X → Y and a unique quadratic mapping Q : X → Y such that

\begin{align} N (f (x) - A (x) - Q (x), 2 t) \\ = N (\frac{f (x) - f (- x)}{2} + \frac{f (x) + f (- x)}{2} - A (x) - Q (x), 2 t) \\ \geq min \{N (\frac{f (x) - f (- x)}{2} - A (x), t), N (\frac{f (x) + f (- x)}{2} - Q (x), t)\} \\ \geq min \{\frac{(8 - 8 α) t}{(8 - 8 α) t + α (φ (2 x, 0, 0) + φ (x, x, 0))}, \frac{(8 - 8 α) t}{(8 - 8 α) t + 3 α (φ (2 x, 0, 0) + φ (x, x, 0))}\} \\ = \frac{(8 - 8 α) t}{(8 - 8 α) t + 3 α (φ (2 x, 0, 0) + φ (x, x, 0))} . \end{align}

Corollary 2.11. Let θ ≥ 0 and let r be a real number with r > 1. Let X be a normed vector space with norm ∥·∥. Let f : X → Y be a mapping with f(0) = 0 and satisfying (2.8). Then there exist a unique additive mapping A : X → Y and a unique quadratic mapping Q : X → Y such that

N (f (x) - A (x) - Q (x), t) \geq \frac{2 (2^{r} - 2) t}{2 (2^{r} - 2) t + 3 (2^{r} + 2) θ {∥x∥}^{r}} .

for all x ∈ X and all t > 0.

Proof. The proof follows from Theorem 2.10 by taking φ(x, y,z) := θ(∥x∥^r+ ∥y∥^r+ ∥z∥^r) for all x, y, z ∈ X. Then we can choose α = 2^1-rand we get the desired result.

3. Fuzzy stability of the functional equation (1.1): a direct method

In this section, using direct method, we prove the Hyers-Ulam stability of functional equation (1.1) in fuzzy Banach spaces. Throughout this section, we assume that X is a linear space, (Y, N) is a fuzzy Banach space and (Z, N') is a fuzzy normed spaces. Moreover, we assume that N(x,.) is a left continuous function on ℝ.

Theorem 3.1. Assume that a mapping f : X → Y is an odd mapping satisfying the inequality

\begin{align} N (r^{2} f (\frac{x + y + z}{r}) + r^{2} f (\frac{x - y + z}{r}) + r^{2} f (\frac{x + y - z}{r}) + r^{2} f (\frac{- x + y + z}{r}) \\ - 4 f (x) - 4 f (y) - 4 f (z), t) \geq N^{'} (φ (x, y, z), t) \end{align}

(3.1)

for all x, y, z ∈ X, t > 0 and that φ : X³ → Z is a mapping for which there is a constant r ∈ ℝ satisfying

0 < |r| < \frac{1}{2}

such that

N^{'} (φ (\frac{x}{2}, \frac{y}{2}, \frac{z}{2}), t) \geq N^{'} (φ (x, y, z), \frac{t}{|r|})

(3.2)

for all x, y, z ∈ X and all t > 0. Then there exists a unique additive mapping A : X → Y satisfying (1.1) and the inequality

N (f (x) - A (x), t) \geq min \{N^{'} (φ (x, 0, 0), 2 (1 - 2 |r|) t), N^{'} (φ (x, x, 0), \frac{2 (1 - 2 |r| t)}{|r|})\}

(3.3)

for all x ∈ X and all t > 0.

Proof. It follows from (3.2) that

N^{'} (φ (\frac{x}{2^{j}}, \frac{y}{2^{j}}, \frac{z}{2^{j}}), t) \geq N^{'} (φ (x, y, z), \frac{t}{{|r|}^{j}})

for all x, y, z ∈ X and all t > 0. Putting y = z = 0 in (3.1) and replacing x by 2x, we get

N (2 r^{2} f (\frac{2 x}{r}) - 4 f (2 x), t) \geq N^{'} (φ (2 x, 0, 0), t)

(3.4)

for all x ∈ X and all t > 0. Putting y = x and z = 0 in (3.1), we have

N (2 r^{2} f (\frac{2 x}{r}) - 8 f (x), t) \geq N^{'} (φ (x, x, 0), t)

(3.5)

for all x ∈ X and all t > 0. By (3.4), (3.5), (N 3) and (N 4), we get

\begin{align} N (\frac{f (2 x)}{2} - f (x), \frac{t}{4}) & = N (2 r^{2} f (\frac{2 x}{r}) - 8 f (x) - 2 r^{2} f (\frac{2 x}{r}) + 4 f (2 x), 2 t) \\ \geq min \{N (2 r^{2} f (\frac{2 x}{r}) - 4 f (2 x), t), N (2 r^{2} f (\frac{2 x}{r}) - 8 f (x), t)\} \\ \geq min \{N^{'} (φ (2 x, 0, 0), t), N^{'} (φ (x, x, 0), t)\} \end{align}

(3.6)

for all x ∈ X and all t > 0. Replacing x by

\frac{x}{2}

in (3.6), we have

N (f (x) - 2 f (\frac{x}{2}), \frac{t}{2}) \geq min \{N^{'} (φ (x, 0, 0), t), N^{'} (φ (\frac{x}{2}, \frac{x}{2}, 0), t)\} .

(3.7)

Replacing x by

\frac{x}{2^{j}}

in (3.7), we have

\begin{align} N (2^{j + 1} f (\frac{x}{2^{j + 1}}) - 2^{j} f (\frac{x}{2^{j}}), 2^{j - 1} t) \\ \geq min \{N^{'} (φ (\frac{x}{2^{j}}, 0, 0), t), N^{'} (φ (\frac{x}{2^{j + 1}}, \frac{x}{2^{j + 1}}, 0), t)\} \\ \geq min \{N^{'} (φ (x, 0, 0), \frac{t}{{|r|}^{j}}), N^{'} (φ (x, x, 0), \frac{t}{{|r|}^{j + 1}})\} \end{align}

(3.8)

for all x ∈ X, all t > 0 and any integer j ≥ 0. So

\begin{align} N (f (x) - 2^{n} f (\frac{x}{2^{n}}), \sum_{j = 0}^{n - 1} 2^{j - 1} {|r|}^{j} t) & = N (\sum_{j = 0}^{n - 1} 2^{j + 1} f (\frac{x}{2^{j + 1}}) - 2^{j} f (\frac{x}{2^{j}}), \sum_{j = 0}^{n - 1} 2^{j - 1} {|r|}^{j} t) \\ \geq min_{0 \leq j \leq n - 1} \{N (2^{j + 1} f (\frac{x}{2^{j + 1}}) - 2^{j} f (\frac{x}{2^{j}}), 2^{j - 1} {|r|}^{j} t)\} \\ \geq min \{N^{'} (φ (x, 0, 0), t), N^{'} (φ (x, x, 0), \frac{t}{|r|})\} \end{align}

which yields

\begin{align} N (2^{n + p} f (\frac{x}{2^{n + p}}) - 2^{p} f (\frac{x}{2^{p}}), \sum_{j = 0}^{n - 1} 2^{j + p - 1} {|r|}^{j}) \\ \geq min \{N^{'} (φ (\frac{x}{2^{p}}, 0, 0), t), N^{'} (φ (\frac{x}{2^{p}}, \frac{x}{2^{p}}, 0), \frac{t}{|r|})\} \\ \geq min \{N^{'} (φ (x, 0, 0), \frac{t}{{|r|}^{p}}), N^{'} (φ (x, x, 0), \frac{t}{{|r|}^{p + 1}})\} \end{align}

(3.9)

for all x ∈ X, t > 0 and any integers n > 0, p ≥ 0. So

N (2^{n + p} f (\frac{x}{2^{n + p}}) - 2^{p} f (\frac{x}{2^{p}}), \sum_{j = 0}^{n - 1} 2^{j + p - 1} {|r|}^{j + p} t) \geq min \{N^{'} (φ (x, 0, 0), t), N^{'} (φ (x, x, 0), \frac{t}{|r|})\}

for all x ∈ X, t > 0 and any integers n > 0, p ≥ 0. Hence one obtains

\begin{align} N (2^{n + p} f (\frac{x}{2^{n + p}}) - 2^{p} f (\frac{x}{2^{p}}), t) \\ \geq min \{N^{'} (φ (x, 0, 0), \frac{t}{\sum_{j = 0}^{n - 1} 2^{j + p - 1} {|r|}^{j + p}}), N^{'} (φ (x, x, 0), \frac{t}{\sum_{j = 0}^{n - 1} 2^{j + p - 1} {|r|}^{j + p + 1}})\} \end{align}

(3.10)

for all x ∈ X, t > 0 and any integers n > 0, p ≥ 0. Since the series

\sum_{j = 0}^{+ \infty} 2^{j} {|r|}^{j}

is a convergent series, we see by taking the limit p → ∞ in the last inequality that the sequence

\{2^{n} f (\frac{x}{2^{n}})\}

is a Cauchy sequence in the fuzzy Banach space (Y, N) and so it converges in Y. Therefore a mapping A : X → Y defined by

A (x) : = N - lim_{n \to \infty} 2^{n} f (\frac{x}{2^{n}})

is well-defined for all x ∈ X. It means that

lim_{n \to \infty} N (A (x) - 2^{n} f (\frac{x}{2^{n}}), t) = 1

(3.11)

for all x ∈ X and all t > 0. In addition, it follows from (3.10) that

N (f (x) - 2^{n} f (\frac{x}{2^{n}}), t) \geq min \{N^{'} (φ (x, 0, 0), \frac{2 t}{\sum_{j = 0}^{n - 1} 2^{j} {|r|}^{j}}), N^{'} (φ (x, x, 0), \frac{2 t}{\sum_{j = 0}^{n - 1} 2^{j} {|r|}^{j + 1}})\}

for all x ∈ X and all t > 0. So

\begin{align} N (f (x) - A (x), t) & \geq min \{N (f (x) - 2^{n} f (\frac{x}{2^{n}}), (1 - ε) t), N (A (x) - 2^{n} f (\frac{x}{2^{n}}), ε t)\} \\ \geq min \{N^{'} (φ (x, 0, 0), \frac{2 ε t}{\sum_{j = 0}^{n - 1} 2^{j} {|r|}^{j}}), N^{'} (φ (x, x, 0), \frac{2 ε t}{\sum_{j = 0}^{n - 1} 2^{j} {|r|}^{j + 1}})\} \\ \geq min \{N^{'} (φ (x, 0), 2 (1 - 2) |r| ε t), N^{'} (φ (x, x, 0), \frac{2 (1 - 2 |r|) ε t}{|r|})\} \end{align}

for sufficiently large n and for all x ∈ X, t > 0 and ϵ with 0 < ϵ < 1. Since ϵ is arbitrary and N' is left continuous, we obtain

N (f (x) - A (x), t) \geq min \{N^{'} (φ (x, 0, 0), 2 (1 - 2 |r|) t), N^{'} (φ (x, x, 0), \frac{2 (1 - 2 |r| t)}{|r|})\}

for all x ∈ X and t > 0. It follows from (3.1) that

\begin{gathered} N (2^{n} r^{2} f (\frac{x + y + z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{x - y + z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{x + y - z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{- x + y + z}{r {. 2}^{n}}) \\ - 2^{n + 2} f (\frac{x}{2^{n}}) - 2^{n + 2} f (\frac{y}{2^{n}}) - 2^{n + 2} f (\frac{z}{2^{n}}), t) \\ \geq N^{'} (φ (\frac{x}{2^{n}}, \frac{y}{2^{n}}, \frac{z}{2^{n}}), \frac{t}{2^{n}}) \geq N^{'} (φ (x, y, z), \frac{t}{2^{n} {|r|}^{n}}) \to 1 as n \to + \infty \end{gathered}

for all x, y, z ∈ X, t > 0. So

\begin{gathered} N (2^{n} r^{2} f (\frac{x + y + z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{x - y + z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{x + y - z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{- x + y + z}{r {. 2}^{n}}) \\ - 2^{n + 2} f (\frac{x}{2^{n}}) - 2^{n + 2} f (\frac{y}{2^{n}}) - 2^{n + 2} f (\frac{z}{2^{n}}), t) \to 1 \end{gathered}

for all x, y, z ∈ X and all t > 0. Therefore, we obtain in view of (3.11)

\begin{gathered} N (r^{2} A (\frac{x + y + z}{r}) + r^{2} A (\frac{x - y + z}{r}) + r^{2} A (\frac{x + y - z}{r}) + r^{2} A (\frac{- x + y + z}{r}) \\ - 4 A (x) - 4 A (y) - 4 A (z), t) \\ \geq min \{N (r^{2} A (\frac{x + y + z}{r}) + r^{2} A (\frac{x - y + z}{r}) + r^{2} A (\frac{x + y - z}{r}) + r^{2} A (\frac{- x + y + z}{r}) \\ - 4 A (x) - 4 A (y) - 4 A (z) - 2^{n} r^{2} f (\frac{x + y + z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{x - y + z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{x + y - z}{r {. 2}^{n}}) \\ + 2^{n} . r^{2} f (\frac{- x + y + z}{r {. 2}^{n}}) - 2^{n + 2} f (\frac{x}{2^{n}}) - 2^{n + 2} f (\frac{y}{2^{n}}) - 2^{n + 2} f (\frac{z}{2^{n}}), \frac{t}{2}) \\ , N (2^{n} r^{2} f (\frac{x + y + z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{x - y + z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{x + y - z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{- x + y + z}{r {. 2}^{n}}) \\ - 2^{n + 2} f (\frac{x}{2^{n}}) - 2^{n + 2} f (\frac{y}{2^{n}}) - 2^{n + 2} f (\frac{z}{2^{n}}), \frac{t}{2})\} \\ = N (2^{n} r^{2} f (\frac{x + y + z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{x - y + z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{x + y - z}{r {. 2}^{n}}) + 2^{n} . r^{2} f (\frac{- x + y + z}{r {. 2}^{n}}) \\ - 2^{n + 2} f (\frac{x}{2^{n}}) - 2^{n + 2} f (\frac{y}{2^{n}}) - 2^{n + 2} f (\frac{z}{2^{n}}), \frac{t}{2}) (for sufficiently large n) \\ \geq N^{'} (φ (x, y, z), \frac{t}{2^{n + 1} {|r|}^{n}}) \to 1 as n \to \infty \end{gathered}

which implies

r^{2} A (\frac{x + y + z}{r}) + r^{2} A (\frac{x - y + z}{r}) + r^{2} A (\frac{x + y - z}{r}) + r^{2} A (\frac{- x + y + z}{r}) = 4 A (x) + 4 A (y) + 4 A (z)

for all x, y, z ∈ X. Thus A : X → Y is a mapping satisfying the Equation (1.1) and the inequality (3.3). Thus the mapping A : X → Y is additive, as desired.

To prove the uniqueness, assume that there is another mapping L : X → Y which satisfies the inequality (3.3). Since L(2ⁿx) = 2ⁿL(x) for all x ∈ X, we have

\begin{align} N (A (x) - L (x), t) & = N (2^{n} A (\frac{x}{2^{n}}) - 2^{n} L (\frac{x}{2^{n}}), t) \\ \geq min \{N (2^{n} A (\frac{x}{2^{n}}) - 2^{n} f (\frac{x}{2^{n}}), \frac{t}{2}), N (2^{n} f (\frac{x}{2^{n}}) - 2^{n} L (\frac{x}{2^{n}}), \frac{t}{2})\} \\ \geq min \{N^{'} (φ (\frac{x}{2^{n + 1}}, 0, 0), \frac{(1 - 2 |r|) t}{2^{n + 1}}), N^{'} (φ (\frac{x}{2^{n}}, \frac{x}{2^{n}}, 0), \frac{(1 - 2 |r|) t}{2^{n + 1} |r|})\} \\ \geq min \{N^{'} (φ (x, 0, 0), \frac{(1 - 2 |r|) t}{{|r|}^{n} 2^{n + 1}}), N^{'} (φ (x, x, 0), \frac{(1 - 2 |r|) t}{2^{n + 1} {|r|}^{n + 1}})\} \to 1 as n \to \infty \end{align}

for all t > 0. Therefore, A(x) = L(x) for all x ∈ X. This completes the proof.

Corollary 3.2. Let X be a normed spaces and (ℝ, N') a fuzzy Banach space. Assume that there exist real numbers θ ≥ 0 and p > 1 such that an odd mapping f : X → Y satisfies the inequality

\begin{gathered} N (r^{2} f (\frac{x + y + z}{r}) + r^{2} f (\frac{x - y + z}{r}) + r^{2} f (\frac{x + y - z}{r}) + r^{2} f (\frac{- x + y + z}{r}) \\ - 4 f (x) - 4 f (y) - 4 f (z), t) \geq N^{'} (θ ({∥x∥}^{p} + {∥y∥}^{p} + {∥z∥}^{p}), t) \end{gathered}

(3.12)

for all x, y, z ∈ X and t > 0. Then there is a unique additive mapping A : X → Y satisfying (1.1) and the inequality

N (f (x) - A (x), t) \geq min \{N^{'} (θ {∥x∥}^{p}, \frac{(2^{p + 1} - 4) t}{2^{p}}), N^{'} (θ {∥x∥}^{p}, (2^{p} - 2) t)\}

Proof. Let φ(x, y, z) := θ (∥x∥^p+ ∥y∥^p+ ∥z∥^p) and |r| = 2^-p. Applying Theorem 3.1, we get the desired result.

Theorem 3.3. Assume that an odd mapping f : X → Y satisfies the inequality (3.1) and that φ:X³ → Z is a mapping for which there is a constant r ∈ ℝ satisfying 0 < |r| < 2 such that

N^{'} (φ (x, y, z), |r| t) \geq N^{'} (φ (\frac{x}{2}, \frac{y}{2}, \frac{z}{2}), t)

(3.13)

for all x, y, z ∈ X and all t > 0. Then there exists a unique additive mapping A : X → Y satisfying (1.1) and the inequality

N (f (x) - A (x), t) \geq min \{N^{'} (φ (x, 0, 0), \frac{2 (2 - |r| t)}{|r|}), N^{'} (φ (x, x, 0), 2 (2 - |r|) t)\}

(3.14)

for all x ∈ X and all t > 0.

Proof. It follows from (3.6) that

N (\frac{f (2 x)}{2} - f (x), \frac{t}{4}) \geq min {N^{'} (φ (2 x, 0, 0), t), N^{'} (φ (x, x, 0), t)}

(3.15)

for all x ∈ X and all t > 0. Replacing x by 2ⁿx in (3.15), we obtain

\begin{align} N (\frac{f (2^{n + 1} x)}{2^{n + 1}} - \frac{f (2^{n} x)}{2^{n}}, \frac{t}{2^{n + 2}}) & \geq min {N^{'} (φ (2^{n + 1} x, 0, 0), t), N^{'} (φ (2^{n} x, 2^{n} x, 0), t)} \\ \geq min \{N^{'} (φ (x, 0, 0), \frac{t}{{|r|}^{n + 1}}), N^{'} (φ (x, x, 0), \frac{t}{{|r|}^{n}})\} . \end{align}

(3.16)

So

N (\frac{f (2^{n + 1} x)}{2^{n + 1}} - \frac{f (2^{n} x)}{2^{n}}, \frac{{|r|}^{n} t}{2^{n + 2}}) \geq min \{N^{'} (φ (x, 0, 0), \frac{t}{|r|}), N^{'} (φ (x, x, 0), t)\}

(3.17)

for all x ∈ X and all t > 0. Proceeding as in the proof of Theorem 3.1, we obtain that

N (f (x) - \frac{f (2^{n} x)}{2^{n}}, \sum_{j = 0}^{n - 1} \frac{{|r|}^{j} t}{2^{j + 2}}) \geq min \{N^{'} (φ (x, 0, 0), \frac{t}{|r|}), N^{'} (φ (x, x, 0), t)\}

for all x ∈ X, all t > 0 and any integer n > 0. So

\begin{align} N (f (x) - \frac{f (2^{n} x)}{2^{n}}, t) & \geq min \{N^{'} (φ (x, 0, 0), \frac{t}{|r| \sum_{j = 0}^{n - 1} \frac{{|r|}^{j}}{2^{j + 2}}}), N^{'} (φ (x, x, 0), \frac{t}{\sum_{j = 0}^{n - 1} \frac{{|r|}^{j}}{2^{j + 2}}})\} \\ \geq min \{N^{'} (φ (x, 0, 0), \frac{2 (2 - |r| t)}{|r|}), N^{'} (φ (x, x, 0), 2 (2 - |r|) t)\} . \end{align}

The rest of the proof is similar to the proof of Theorem 3.1.

Corollary 3.4. Let X be a normed spaces and that (ℝ, N') a fuzzy Banach space. Assume that there exist real numbers θ ≥ 0 and 0 < p < 1 such that an odd mapping f : X → Y satisfies (3.12). Then there is a unique additive mapping A : X → Y satisfying (1.1) and the inequality

N (f (x) - A (x), t) \geq min \{N^{'} (θ {∥x∥}^{p}, \frac{2 (2 - 2^{p})}{2^{p}}), N^{'} (θ {∥x∥}^{p}, (2 - 2^{p}) t)\}

Proof. Let φ(x, y, z) := θ(∥x∥^p+ ∥y∥^p+ ∥z∥^p) and |r| = 4^p. Applying Theorem 3.3, we get the desired results.

Theorem 3.5. Assume that a mapping f : X → Y is an even mapping satisfies (3.1) and f(0) = 0 and that φ : X³ → Z is a mapping for which there is a constant r ∈ ℝ satisfying 0 < |r| < 4 such that

N^{'} (φ (x, y, z), |r| t) \geq N^{'} (φ (\frac{x}{2}, \frac{y}{2}, \frac{z}{2}), t)

(3.18)

for all x, y, z ∈ X and all t > 0. Then there exists a unique quadratic mapping Q : X → Y satisfying (1.1) and the inequality

N (f (x) - Q (x), t) \geq min \{N^{'} (φ (x, 0, 0), \frac{2 (4 - |r|) t}{|r|}), N^{'} (φ (x, x, 0), (4 - |r|) t)\}

(3.19)

for all x ∈ X and all t > 0.

Proof. Putting y = x and z = 0 in (3.1), we get

N (2 r^{2} f (\frac{2 x}{r}) - 8 f (x), t) \geq N^{'} (φ (x, x, 0), t)

(3.20)

for all x ∈ X and all t > 0. Putting y = z = 0 and replacing x by 2 x in (3.1), we have

N (4 r^{2} f (\frac{2 x}{r}) - 4 f (2 x), t) \geq N^{'} (φ (2 x, 0, 0), t)

(3.21)

for all x ∈ X and all t > 0. By (3.20), (3.21), (N 3) and (N 4), we get

\begin{align} N (\frac{f (2 x)}{4} - f (x), \frac{t}{16}) & = N (2 (2 r^{2} f (\frac{2 x}{r}) - 8 f (x)) - (4 r^{2} f (\frac{2 x}{r}) - 4 f (2 x)), t) \\ \geq min \{N (2 r^{2} f (\frac{2 x}{r}) - 8 f (x), \frac{t}{4}), N (4 r^{2} f (\frac{2 x}{r}) - 4 f (2 x), \frac{t}{2})\} \\ \geq min \{N^{'} (φ (x, x, 0), \frac{t}{4}), N^{'} (φ (2 x, 0, 0), \frac{t}{2})\} \end{align}

(3.22)

for all x ∈ X and all t > 0. Replacing x by 2ⁿx in (3.22), we obtain

\begin{align} N (\frac{f (2^{n + 1}) x}{4^{n + 1}} - \frac{f (2^{n} x)}{4^{n}}, \frac{t}{4^{n + 2}}) & \geq min \{N^{'} (φ (2^{n} x, 2^{n} x, 0), \frac{t}{4}), N^{'} (φ (2^{n + 1} x, 0, 0), \frac{t}{2})\} \\ \geq min \{N^{'} (φ (x, x, 0), \frac{t}{4 {|r|}^{n}}), N^{'} (φ (x, 0, 0), \frac{t}{2 {|r|}^{n + 1}})\} \end{align}

for all x ∈ X and all t > 0. Proceeding as in the proof of Theorem 3.1, we obtain that

N (f (x) - \frac{f (2^{n} x)}{4^{n}}, \sum_{j = 0}^{n - 1} \frac{{|r|}^{j} t}{4^{j + 2}}) \geq min \{N^{'} (φ (x, x, 0), \frac{t}{4}), N^{'} (φ (x, 0, 0), \frac{t}{2 |r|})\}

for all x ∈ X, all t > 0 and any integer n > 0. So

\begin{align} N (f (x) - \frac{f (2^{n} x)}{4^{n}}, t) & \geq min \{N^{'} (φ (x, x, 0), \frac{4 t}{\sum_{j = 0}^{n - 1} \frac{{|r|}^{j}}{4^{j}}}), N^{'} (φ (x, 0, 0), \frac{8 t}{|r| \sum_{j = 0}^{n - 1} \frac{{|r|}^{j}}{4^{j}}})\} \\ \geq min \{N^{'} (φ (x, 0, 0), \frac{2 (4 - |r| t)}{|r|}), N^{'} (φ (x, x, 0), (4 - |r|) t)\} . \end{align}

The rest of the proof is similar to the proof of Theorem 3.5.

Corollary 3.6. Let X be a normed spaces and (ℝ, N') a fuzzy Banach space. Assume that there exist real numbers θ ≥ 0 and 0 < p < 1 such that an even mapping f : X → Y satisfies (3.12). Then there is a unique quadratic mapping Q: X → Y satisfying (1.1) and the inequality

N (f (x) - Q (x), t) \geq min \{N^{'} (θ {∥x∥}^{p}, \frac{2 (4 - 4^{p}) t}{4^{p}}), N^{'} (θ {∥x∥}^{p}, \frac{(4 - 4^{p}) t}{2})\}

Proof. Let φ(x, y, z) := θ(∥x∥^p+ ∥y∥^p+ ∥z∥^p) and |r| = 4^p. Applying Theorem 3.1, we get the desired result.

Theorem 3.7. Assume that an even mapping f : X → Y satisfies the inequality (3.1) and f(0) = 0 and that φ : X³ → Z is a mapping for which there is a constant r ∈ ℝ satisfying

0 < |r| < \frac{1}{4}

such that

N^{'} (φ (\frac{x}{2}, \frac{y}{2}, \frac{z}{2}), t) \geq N^{'} (φ (x, y, z), \frac{t}{|r|})

(3.23)

for all x, y, z ∈ X and all t > 0. Then there exists a unique quadratic mapping Q : X → Y satisfying (1.1) and the inequality

N (f (x) - Q (x), t) \geq min \{N^{'} (φ (x, x, 0), \frac{(1 - 4 |r| t)}{|r|}), N^{'} (φ (x, 0, 0), (2 - 8 |r|) t)\}

(3.24)

for all x ∈ X and all t > 0.

Proof. It follows from (3.22) that

N (f (x) - 4 f (\frac{x}{2}), \frac{t}{4}) \geq min \{N^{'} (φ (\frac{x}{2}, \frac{x}{2}, 0), \frac{t}{4}), N^{'} (φ (x, 0, 0), \frac{t}{2})\}

(3.25)

for all x ∈ X and all t > 0. Replacing x by

\frac{x}{2^{j}}

in (3.25), we have

N (4^{j + 1} f (\frac{x}{2^{j + 1}}) - 4^{j} f (\frac{x}{2^{j}}), 4^{j} t) \geq min \{N^{'} (φ (x, x, 0), \frac{t}{{|r|}^{j + 1}}), N^{'} (φ (x, 0, 0), \frac{2 t}{{|r|}^{j}})\}

for all x ∈ X, all t > 0 and any integer j ≥ 0. By the same technique as in Theorem 3.1, we find that

\begin{align} N (f (x) - 4^{n} f (\frac{x}{2^{n}}), t) & \geq min \{N^{'} (φ (x, x, 0), \frac{t}{\sum_{j = 0}^{n - 1} 4^{j} {|r|}^{j + 1}}), N^{'} (φ (x, 0, 0), \frac{2 t}{\sum_{j = 0}^{n - 1} 4^{j} {|r|}^{j}})\} \\ \geq min \{N^{'} (φ (x, x, 0), \frac{(1 - 4 |r|) t}{|r|}), N^{'} (φ (x, 0, 0), (2 - 8 |r|) t)\} \end{align}

The rest of the proof is similar to the proof of Theorem 3.1.

Corollary 3.8. Let X be a normed spaces and (ℝ, N') a fuzzy Banach space. Assume that there exists real numbers θ ≥ 0 and p > 1 such that an even mapping f : X → Y satisfies (3.12) and f(0) = 0. Then there is a unique quadratic mapping Q : X → Y satisfying (1.1) and the inequality

N (f (x) - Q (x), t) \geq min \{N^{'} (θ {∥x∥}^{p}, \frac{(4^{p} - 4 t)}{2}), N^{'} (θ {∥x∥}^{p}, (2 - \frac{8}{4^{p}}) t)\}

Proof. Let φ(x, y, z) := θ (∥x∥^p+ ∥y∥^p+ ∥z∥^p) and |r| = 4^-p. Applying Theorem 3.7, we get the desired results.

Let f : X → Y be a mapping satisfying f(0) = 0 and (1.1). Let $f_{e} (x) : = \frac{f (x) + f (- x)}{2}$ and $f_{o} (x) = \frac{f (x) - f (- x)}{2}$ . Then f_eis an even mapping satisfying (1.1) and f_ois an odd mapping satisfying (1.1) such that f(x) = f_e(x) + f_o(x). So we obtain the following.

Theorem 3.9. Assume that a mapping f : X → Y is a mapping satisfying (3.1) and f(0) = 0 and that φ : X³ → Z is a mapping for which there is a constant r ∈ ℝ satisfying

0 < |r| < \frac{1}{4}

such that

N^{'} (φ (\frac{x}{2}, \frac{y}{2}, \frac{z}{2}), t) \geq N^{'} (φ (x, y, z), \frac{t}{|r|})

for all x, y, z ∈ X and all t > 0. Then there exist a unique additive mapping A : X → Y and a unique quadratic mapping Q : X → Y satisfying (1.1) and the inequality

\begin{gathered} N (f (x) - A (x) - Q (x), 2 t) \\ \geq min \{min \{N^{'} φ (x, 0, 0), 2 (1 - 2 |r| t), N^{'} (φ (x, x, 0), \frac{2 (1 - 2) |r| t}{|r|})\}, \\ min \{N^{'} (φ (x, 0, 0), \frac{(1 - 4 |r|) t}{|r|}), N^{'} (φ (x, 0, 0), (2 - 8 |r|) t)\}\} \end{gathered}

for all x ∈ X and all t > 0.

Proof. It follows from Theorems 3.1 and 3.7 that

\begin{gathered} N (f (x) - A (x) - Q (x), 2 t) \\ = N (\frac{f (x) + f (- x)}{2} + \frac{f (x) - f (- x)}{2} - A (x) - Q (x), 2 t) \\ \geq min \{N (\frac{f (x) + f (- x)}{2} - Q (x), t), N (\frac{f (x) - f (- x)}{2} - A (x), t)\} \end{gathered}