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

$f ( x + y ) + f ( x - y ) = 2 f ( x ) + 2 f ( y )$
(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

$B ( x , y ) = 1 4 ( f ( x + y ) - f ( x - y ) )$
(1.2)

A generalized Hyers-Ulam stability problem for the quadratic functional equation (1.1) was proved by Skof for mappings f : AB, where A is normed space and B is a Banach space [8] (see [912]).

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

$f ( 2 x + y ) + f ( 2 x - y ) = 2 f ( x + y ) + 2 f ( x - y ) + 1 2 f ( x )$
(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 × XY 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

$C ( x , y , z ) = 1 2 4 ( f ( x + y + z ) + f ( x - y - z ) - f ( x + y - z ) - f ( x - y + z ) )$
(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; [1421].

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,

$f ( x + k y ) + f ( x - k y ) = k 2 f ( x + y ) + k 2 f ( x - y ) + 2 ( 1 - k 2 ) f ( x )$
(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 [2843]).

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 × → [0, 1] is said to be a fuzzy norm on X if for all x, y X and all s, t ;

(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) $N ( c x , t ) =N x , t | c |$ if c ≠ 0;

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

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

(N6) N(x, ·) is left continuous on 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, 4547]).

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 limn→∞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 - limn→∞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 such that N(x m - x n , δ) > 1 - ϵ (m, nn0).

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 : XY 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 : XY is continuous at each x X, then f : XY 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 : XY,

$D f ( x , y ) = f ( x + k y ) + f ( x - k y ) - k 2 f ( x + y ) - k 2 f ( x - y ) - 2 ( 1 - k 2 ) f ( x )$

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 : V1V2 satisfies (1.5), then f(x) is quadratic.

Theorem 3.2. Let ℓ {-1, 1} be fixed and let φ q : X × XY be a mapping such that

$φ q ( k x , k y ) = α φ q ( x , y )$
(3.1)

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

$N ( D f ( x , y ) , t ) ≥ N ′ ( φ q ( x , y ) , t )$
(3.2)

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

$Q ( x ) = N − lim n → ∞ f ( k ℓ n x ) k 2 ℓ n )$

exists for all x X and Q : XY is a unique quadratic mapping satisfying

$N ( f ( x ) - Q ( x ) , t ) ≥ N ′ φ q ( 0 , x ) , ℓ ( k 2 - α ) t 2$
(3.3)

for all x X and all t > 0.

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

$N ( 2 f ( k y ) - 2 k 2 f ( y ) , t ) ≥ N ′ ( φ q ( 0 , y ) , t )$
(3.4)

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

$N f ( k x ) - k 2 f ( x ) , t 2 ≥ N ′ ( φ q ( 0 , x ) , t )$
(3.5)

for all x X. So

$N f ( k x ) k 2 - f ( x ) , t 2 k 2 ≥ N ′ ( φ q ( 0 , x ) , t )$
(3.6)

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

$N ′ ( φ q ( 0 , k x ) , t ) = N ′ φ q ( 0 , x ) , t α$
(3.7)

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

$N f ( k n + 1 x ) k 2 ( n + 1 ) - f ( k n x ) k 2 n , t k 2 ( k 2 n ) ≥ N ′ ( φ q ( 0 , k n x ) , t ) = N ′ φ q ( 0 , x ) , t α n$
(3.8)

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

$N f ( k n + 1 x ) k 2 ( n + 1 ) - f ( k n x ) k 2 n , α n t k 2 ( k 2 n ) ≥ N ′ ( φ q ( 0 , x ) , t )$
(3.9)

for all x X, t > 0 and n > 0. It follows from $f ( k n x ) k 2 n -f ( x ) = ∑ j = 0 n - 1 f ( k j + 1 x ) k 2 ( j + 1 ) - f ( k j x ) k 2 j$ and (3.9) that

$N f ( k n x ) k 2 n - f ( x ) , ∑ j = 0 n - 1 α j t k 2 ( k 2 ) j ≥ min ⋃ j = 0 n - 1 N f ( k j + 1 x ) k 2 ( j + 1 ) - f ( k j x ) k 2 j , α j t k 2 ( k 2 ) j ≥ N ′ ( φ q ( 0 , x ) , t )$
(3.10)

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

$N f ( k n + m x ) k 2 ( n + m ) - f ( k m x ) k 2 m , ∑ j = 0 n - 1 α j t k 2 ( k 2 ) j + m ≥ N ′ ( φ q ( 0 , k m x ) , t ) = N ′ φ ( 0 , x ) , t α m$

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

$N f ( k n + m x ) k 2 ( n + m ) - f ( k m x ) k 2 m , ∑ j = m n + m - 1 α j t k 2 ( k 2 ) j ≥ N ′ ( φ q ( 0 , x ) , t )$

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

$N f ( k n + m x ) k 2 ( n + m ) - f ( k m x ) k 2 m , t ≥ N ′ φ q ( 0 , x ) , t ∑ j = m n + m - 1 α j k 2 ( k 2 ) j$
(3.11)

for all x X, all t > 0 and all m ≥ 0, n > 0. Since 0 < α < k2 and $∑ j = 0 ∞ α k 2 j <∞$, the Cauchy criterion for convergence and (N5) imply that $f ( k n x ) k 2 n$ 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 : XY by

$Q ( x ) = N − lim n → ∞ f ( k n x ) k 2 n )$
(3.12)

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

$N f ( k n x ) k 2 n - f ( x ) , t ≥ N ′ φ q ( 0 , x ) , t ∑ j = 0 n - 1 α j k 2 ( k 2 ) j$

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

$N ( Q ( x ) - f ( x ) , t ) ≥ min N ( Q ( x ) - f ( k n x ) k 2 n , t 2 ) , N f ( k n x ) k 2 n - f ( x ) , t 2 ≥ N ′ φ q ( 0 , x ) , t ∑ j = 0 n - 1 2 α j k 2 ( k 2 ) j$
(3.13)

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

$N ( Q ( x ) - f ( x ) , t ) ≥ N ′ φ q ( 0 , x ) , ( k 2 - α ) t 2$
(3.14)

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

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

$Q ( k x ) = k 2 Q ( x )$
(3.15)

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

$N 1 k 2 n D f ( k n x , k n y ) , t ≥ N ′ ( φ q ( k n x , k n y ) , k 2 n t ) = N ′ φ q ( x , y ) , k 2 n t α n$

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': XY be another quadratic function satisfying (3.3). It follows from (3.3), (3.7) and (3.15) that

$N ( Q ( x ) - Q ′ ( x ) , t ) = N Q ( k n x ) k 2 n - Q ′ ( k n x ) k 2 n , t ≥ min N Q ( k n x ) k 2 n - f ( k n x ) k 2 n , t 2 , N f ( k n x ) k 2 n - Q ′ ( k n x ) k 2 n , t 2 ≥ N ′ φ q ( 0 , k n x ) , k 2 n ( k 2 - α ) t 4 = N ′ φ q ( 0 , x ) , k 2 n ( k 2 - α ) t 4 α n$

for all x X and all t > 0. Since α < k2, we obtain $lim n → ∞ N ′ φ q ( 0 , x ) , k 2 n ( k 2 - α ) t 4 α n =1$. Thus, Q(x) = Q'(x).

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

Replacing x by $x k$ in (3.5), we obtain

$N f ( x ) - k 2 f x k , t 2 ≥ N ′ φ q 0 , x k , t$
(3.16)

for all x X and all t > 0. Replacing x and t by $x k n$ and $t k 2 n$ in (3.16), respectively, we obtain

$N ( k 2 n f ( x k n ) − k 2 ( n + 1 ) f ( x k n + 1 ) , t 2 ) ≥ N ′ ( φ q ( 0, x k n + 1 ) , t k 2 n ) = N ′ ( φ q ( 0, x ) , ( α k 2 ) n α t )$

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

$N k 2 ( n + m ) f x k n + m - k 2 m f x k m , t ≥ N ′ φ q ( 0 , x ) , t ∑ j = m + 1 n + m 2 k 2 j k 2 α j$
(3.17)

for all x X, all t > 0 and all m ≥ 0, n ≥ 0. From which we conclude that $k 2 n f x k n$ is a Cauchy sequence in the fuzzy Banach space (Y, N). Therefore, there is a mapping Q : XY defined by $Q ( x ) :=N- lim n → ∞ k 2 n f x k n$. Employing (3.17) with m = 0, we obtain

$N ( Q ( x ) - f ( x ) , t ) ≥ N ′ φ q ( 0 , x ) , ( α - k 2 ) t 2$

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

$N ( Q ( x ) − f ( x ) , t + s ) ≥ min { N ( Q ( x ) − f ( k n x ) k 2 n , s ) , N ( f ( k n x ) k 2 n − f ( x ) , t ) } ≥ N ′ ( φ q ( 0, x ) , t ∑ j = 0 n − 1 α j k 2 ( k ) 2 j ) ) ≥ N ′ ( φ q ( 0, x ) , ( k 2 − α ) t ) .$

Tending s to zero we infer that

$N ( Q ( x ) - f ( x ) , t ) ≥ N ′ ( φ q ( 0 , x ) , ( k 2 - α ) t )$

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 : XY with f(0) = 0 satisfies

$∥ D f ( x , y ) ∥ ≤ ε ( ∥ x ∥ λ + ∥ y ∥ λ )$
(3.18)

for all x, y X. Then, the limit

$Q ( x ) = N − lim n → ∞ f ( k ℓ n x ) k 2 ℓ n )$

exists for all x X and Q : XY is a unique quadratic mapping satisfying

$| | f ( x ) - Q ( x ) | | ≤ 2 ε ∥ x ∥ λ ℓ ( k 2 - k λ )$
(3.19)

for all x X, where λℓ < 2.

Proof. Define the function N by

$N ( x , t ) = t t + ∥ x ∥ , t > 0 0 , t ≤ 0$

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, the function sending each (x, y) to ε(||x||λ + ||y||λ). By assumption

$N ( D f ( x , y ) , t ) ≥ N ′ ( φ q ( x , y ) , t )$

note that N': × → [0, 1] given by

$N ′ ( x , t ) = t t + | x | , t > 0 0 , t ≤ 0$

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

$t t + ∥ f ( x ) - Q ( x ) ∥ = N ( f ( x ) - Q ( x ) , t ) ≥ N ′ φ q ( 0 , x ) , ℓ ( k 2 - k λ ) t 2 = N ′ ε ∥ x ∥ λ , ℓ ( k 2 - k λ ) t 2 = ℓ ( k 2 - k λ ) t ℓ ( k 2 - k λ ) t + 2 ε ∥ x ∥ λ$

and thus

$t t + ∥ f ( x ) - Q ( x ) ∥ ≥ ℓ ( k 2 - k λ ) t ℓ ( k 2 - k λ ) t + 2 ε ∥ x ∥ λ$

which implies that, (k2 - kλ)||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 .

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 . Let for all x X, the functions r f(rx) (from (, N1) into (Y, N)) and r φ q (0, rx) (from (, 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 .

Proof. Case (1): = 1. Let {r k } be a sequence in that converge to some r , and let t > 0. Let ε > 0 be given, since 0 < α < k2, so $lim n → ∞ ( k 2 - α ) k 2 n t 1 2 α n =∞$, there is m such that

$N ′ φ q ( 0 , r x ) , ( k 2 - α ) k 2 m t 1 2 α m > 1 - ε$
(3.20)

It follows from (3.14) and (3.20) that

$N f ( k m r x ) k 2 m - Q ( k m r x ) k 2 m , t 3 > 1 - ε$
(3.21)

By the fuzzy continuity of functions r f(rx) and r φ q (0, rx), we can find some $J∈ℕ$ such that for any nj,

$N f ( k m r k x ) k 2 m - f ( k m r x ) k 2 m , t 3 > 1 - ε$
(3.22)

and

$N ′ ( φ q ( 0 , r k x ) - φ q ( 0 , r x ) , ( k 2 - α ) k 2 m t 1 2 α m ) > 1 - ε$
(3.23)

It follows from (3.20) and (3.23) that

$N ′ ( φ q ( 0 , r k x ) , ( k 2 - α ) k 2 m t 6 α m ) > 1 - ε$
(3.24)

On the other hand,

$N Q ( r k x ) - f ( k m r k x ) k 2 m , t k 2 m = N Q ( k m r k x ) k 2 m - f ( k m r k x ) k 2 m , t k 2 m ≥ N ′ φ q ( 0 , r k x ) , ( k 2 - α ) t 2 α m$
(3.25)

It follows from (3.24) and (3.25) that

$N Q ( r k x ) - f ( k m r k x ) k 2 m , t 3 > 1 - ε$
(3.26)

So, it follows from (3.21), (3.22) and (3.26) that for any nj,

$N ( Q ( r k x ) - Q ( r x ) , t ) > 1 - ε$

Therefore, for every choice x X, t > 0, and ε > 0, we can find some $J∈ℕ$ such that N(Q(r k x) - Q(rx), t) > 1 - ε for every $n≥J$. This shows that Q(r k x) → Q(rx). The proof for = -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 : XY satisfies Q(rx) = r2Q(x) for each 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 : V1V2 satisfies (1.5), then the mapping g : V1V2, defined by g(x) = f(2x) - 8f(x), is additive.

Theorem 4.2. Let ℓ {-1, 1} be fixed and let φ q : X × XZ be a function such that

$φ a ( 2 x , 2 y ) = α φ a ( x , y )$
(4.1)

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

$N ( D f ( x , y ) , t ) ≥ N ′ ( φ a ( x , y ) , t )$
(4.2)

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

$A ( x ) = N - lim n → ∞ 1 2 ℓ n ( f ( 2 ℓ n + 1 x ) - 8 f ( 2 ℓ n x ) )$

exists for all x X and A : XY is a unique additive mapping satisfying

$N ( f ( 2 x ) - 8 f ( x ) - A ( x ) , t ) ≥ M a x , ℓ ( 2 - α ) 2 t$
(4.3)

for all x X and all t > 0, where

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

$N ( f ( k y + x ) - f ( k y - x ) - k 2 f ( x + y ) - k 2 f ( x - y ) + 2 ( k 2 - 1 ) f ( x ) , t ) ≥ N ′ ( φ a ( x , y ) , t )$
(4.4)

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

$N ( f ( ( k + 1 ) x ) - f ( ( k - 1 ) x ) - k 2 f ( 2 x ) + 2 ( k 2 - 1 ) f ( x ) , t ) ≥ N ′ ( φ a ( x , x ) , t )$
(4.5)

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

$N ( f ( 2 ( k + 1 ) x ) - f ( 2 ( k - 1 ) x ) - k 2 f ( 4 x ) + 2 ( k 2 - 1 ) f ( 2 x ) , t ) ≥ N ′ ( φ a ( 2 x , 2 x ) , t )$
(4.6)

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

$N ( f ( ( k + 2 ) x ) - f ( ( k - 2 ) x ) - k 2 f ( 3 x ) - k 2 f ( x ) + 2 ( k 2 - 1 ) f ( 2 x ) , t ) ≥ N ′ ( φ a ( 2 x , x ) , t )$
(4.7)

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

$N ( f ( ( 2 k + 1 ) x ) - f ( ( 2 k - 1 ) x ) - k 2 f ( 3 x ) - k 2 f ( - x ) + 2 ( k 2 - 1 ) f ( x ) , t ) ≥ N ′ ( φ a ( x , 2 x ) , t )$
(4.8)

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

$N ( f ( ( 3 k + 1 ) x ) - f ( ( 3 k - 1 ) x ) - k 2 f ( 4 x ) - k 2 f ( - 2 x ) + 2 ( k 2 - 1 ) f ( x ) , t ) ≥ N ′ ( φ a ( x , 3 x ) , t )$
(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

$N ( f ( ( 2 k + 1 ) x ) - f ( - x ) - k 2 f ( ( k + 2 ) x ) - k 2 f ( k x ) + 2 ( k 2 - 1 ) f ( ( k + 1 ) x ) , t ) ≥ N ′ ( φ a ( ( k + 1 ) x , x ) , t )$
(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

$N ( f ( ( 2 k - 1 ) x ) - f ( x ) - k 2 f ( ( k - 2 ) x ) - k 2 f ( k x ) + 2 ( k 2 - 1 ) f ( ( k - 1 ) x ) , t ) ≥ N ′ ( φ a ( ( k - 1 ) x , x ) , t )$
(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

$N ( f ( ( 3 k + 1 ) x ) - f ( - ( k + 1 ) x ) - k 2 f ( 2 ( k + 1 ) x ) - k 2 f ( 2 k x ) + 2 ( k 2 - 1 ) f ( ( 2 k + 1 ) x ) , t ) ≥ N ′ ( φ a ( ( 2 k + 1 ) x , x ) , t )$
(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

$N ( f ( ( 3 k - 1 ) x ) - f ( - ( k - 1 ) x ) - k 2 f ( 2 ( k - 1 ) x ) - k 2 f ( 2 k x ) + 2 ( k 2 - 1 ) f ( ( 2 k - 1 ) x ) , t ) ≥ N ′ ( φ a ( ( 2 k - 1 ) x , x ) , t )$
(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

(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

$N f ( 4 x ) - 2 f ( 3 x ) - 2 f ( 2 x ) + 6 f ( x ) , 1 k 2 ( k 2 - 1 ) ( 2 ( k 2 - 1 ) + k 2 + 4 ) t ≥ min { N ′ ( φ a ( x , x ) , t ) , N ′ ( φ a ( 2 x , 2 x ) , t ) , N ′ ( φ a ( x , 2 x ) , t ) , N ′ ( φ a ( x , 3 x ) , t ) , N ′ ( φ a ( ( 2 k + 1 ) x , x ) , t ) , N ′ ( φ a ( ( 2 k - 1 ) x , x ) , t ) }$
(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

$N f ( 4 x ) - 1 0 f ( 2 x ) + 1 6 f ( x ) , 9 k 2 + 4 k 2 ( k 2 - 1 ) t ≥ min { N ′ ( φ a ( x , x ) , t ) , N ′ ( φ a ( 2 x , x ) , t ) , N ′ ( φ a ( x , 2 x ) , t ) , N ′ ( φ a ( ( k + 1 ) x , x ) , t ) , N ′ ( φ a ( ( k - 1 ) x , x ) , t ) ) , N ′ ( φ a ( 2 x , 2 x ) , t ) , N ′ ( φ a ( x , 3 x ) , t ) , N ′ ( φ a ( ( 2 k + 1 ) x , x ) , t ) , N ′ ( φ a ( ( 2 k - 1 ) x , x ) , t ) }$
(4.16)

for all x X and all t > 0, where

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

$N ( f ( 4 x ) - 1 0 f ( 2 x ) + 1 6 f ( x ) , t ) ≥ M a ( x , t )$
(4.17)

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

$N ( g ( 2 x ) - 2 g ( x ) , t ) ≥ M a ( x , t )$
(4.18)

for all x X and all t > 0. So

$N g ( 2 x ) 2 - g ( x ) , t 2 ≥ M a ( x , t )$
(4.19)

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

$M a ( 2 x , t ) = M a x , t α$
(4.20)

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

$N ( g ( 2 n + 1 x ) 2 n + 1 − g ( 2 n x ) 2 n , t 2 ( 2 n ) ) ≥ M a ( 2 n x , t ) = M a ( x , t α n )$
(4.21)

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

$N ( g ( 2 n + 1 x ) 2 n + 1 − g ( 2 n x ) 2 n , t α n 2 ( 2 n ) ) ≥ M a ( x , t )$
(4.22)

for all x X, t > 0 and n > 0. It follows from $g ( 2 n x ) 2 n − g ( x ) = ∑ j = 0 n − 1 ( g ( 2 j + 1 x ) 2 j + 1 − g ( 2 j x ) 2 j )$ and (4.22) that

$N ( g ( 2 n x ) 2 n − g ( x ) , ∑ j = 0 n − 1 α j t 2 ( 2 ) j ) ≥ min ∪ j = 0 n − 1 { N ( g ( 2 j + 1 x ) 2 j + 1 − g ( 2 j x ) 2 j , α j t 2 ( 2 ) j ) } ≥ M a ( x , t )$
(4.23)

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

$N ( g ( 2 n + m x ) 2 n + m − g ( 2 m x ) 2 m , ∑ j = 0 n − 1 α j t 2 ( 2 ) j + m ) ≥ M a ( 2 m x , t ) = M a ( x , t α m )$

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

$N ( g ( 2 n + m x ) 2 n + m − g ( 2 m x ) 2 m , ∑ j = m n + m − 1 α j t 2 ( 2 ) j ) ≥ M a ( x , t )$

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

$N ( g ( 2 n + m x ) 2 n + m − g ( 2 m x ) 2 m , t ) ≥ M a ( x , t ∑ j = m n + m − 1 α j 2 ( 2 ) j )$
(4.24)

for all x X, all t > 0 and all m ≥ 0, n > 0. Since 0 < α < 2 and $∑ n = 0 ∞ α 2 n <∞$, the Cauchy criterion for convergence and (N5) imply that ${ g ( 2 n x ) 2 n }$ is a Cauchy sequence in (Y, N) to some point A(x) Y. So one can define the mapping A : XY by

$A ( x ) = N − lim n → ∞ g ( 2 n x ) 2 n )$
(4.25)

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

$N ( g ( 2 n x ) 2 n − g ( x ) , t ) ≥ M a ( x , t ∑ j = 0 n − 1 α j 2 ( 2 ) j )$

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

$N ( A ( x ) − g ( x ) , t ) ≥ min { N ( A ( x ) − g ( 2 n x ) 2 n , t 2 ) , N ( g ( 2 n x ) 2 n − g ( x ) , t 2 ) } ≥ M a ( x , t ∑ j = 0 n − 1 α j 2 j )$
(4.26)

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

$N ( A ( x ) - g ( x ) , t ) ≥ M a x , t ( 2 - α ) 2$
(4.27)

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

$N ( A ( 2 x ) 2 − A ( x ) , t ) ≥ min { N ( A ( 2 x ) 2 − g ( 2 n + 1 x ) 2 n + 1 ) , t 3 ) , N ( g ( 2 n x ) 2 n − A ( x ) , t 3 ) , N ( g ( 2 n + 1 x ) 2 n + 1 − g ( 2 n x ) 2 n ) , t 3 ) = M a ( x , 2 ( 2 ) n t 3 α n )$

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

$A ( 2 x ) = 2 A ( x )$
(4.28)

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

$N ( 1 2 n D g ( 2 n x ,2 n y ) , t ) = N ( D f ( 2 n + 1 x ,2 n + 1 y ) − 8 D f ( 2 n x ,2 n y ) , 2 n t ) = min { N ( D f ( 2 n + 1 x ,2 n + 1 y ) , 2 n t 2 ) , N ( D f ( 2 n x ,2 n y ) , 2 n t 16 ) } ≥ m i n { N ′ ( φ a ( 2 n + 1 x ,2 n + 1 y ) , 2 n t 2 ) , N ′ ( φ a ( 2 n x ,2 n y ) , 2 n t 16 ) } = min { N ′ ( φ a ( x , y ) , 2 n t 2 α n + 1 ) , N ′ ( φ a ( x , y ) , 2 n t 16 α n ) }$

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 : XY satisfies the inequality (3.18) for all x, y X. Then, the limit

$A ( x ) = lim n → ∞ 1 2 ℓ n ( f ( 2 ℓ n + 1 x ) - 8 f ( 2 ℓ n x ) )$

exists for all x X and A : XY is a unique additive mapping satisfying

$f ( 2 x ) - 8 f ( x ) - A ( x ) ≤ ( | 2 k + ℓ | λ + 1 ) ( 9 k 2 + 4 ) ε ∥ x ∥ λ ℓ k 2 ( k 2 - 1 ) ( 1 - 2 λ - 1 )$
(4.29)

for all x X, where λℓ < .

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 : V1V2 satisfies (1.5), then the mapping h : V1V2 defined by h(x) = f(2x) - 2f(x) is cubic.

Theorem 4.7. Let ℓ {-1, 1} be fixed and let φ c : X × XZ be a mapping such that

$φ c ( 2 x , 2 y ) = α φ c ( x , y )$
(4.30)

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

$N ( D f ( x , y ) , t ) ≥ N ′ ( φ c ( x , y ) , t )$
(4.31)

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

$C ( x ) = N - lim n → ∞ 1 8 ℓ n ( f ( 2 ℓ n + 1 x ) - 2 f ( 2 ℓ n x ) )$

exists for all x X and C : XY is a unique cubic mapping satisfying

$N ( f ( 2 x ) - 2 f ( x ) - C ( x ) , t ) ≥ M c x , ℓ ( 8 - α ) 2 t$
(4.32)

for all x X and all t > 0, where

$M c ( x , t ) = min { N ′ ( φ c ( x , x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ c ( 2 x , x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ c ( x ,2 x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ c ( ( k + 1 ) x , x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ c ( ( k − 1 ) x , x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) ) , N ′ ( φ c ( 2 x ,2 x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ c ( x ,3 x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ c ( ( 2 k + 1 ) x , x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ c ( ( 2 k − 1 ) x , x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) } .$

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

$N ( f ( 4 x ) - 1 0 f ( 2 x ) + 1 6 f ( x ) , t ) ≥ M c ( x , t )$

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

$N ( h ( 2 x ) - 8 h ( x ) , t ) ≥ M c ( x , t )$
(4.33)

for all x X and all t > 0. So

$N h ( 2 x ) 8 - h ( x ) , t 8 ≥ M c ( x , t )$
(4.34)

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

$M c ( 2 x , t ) = M c x , t α$
(4.35)

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

$N ( h ( 2 n + 1 x ) 8 n + 1 − h ( 2 n x ) 8 n , t 8 ( 8 n ) ) ≥ M c ( 2 n x , t ) = M c ( x , t α n )$
(4.36)

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

$N ( h ( 2 n + 1 x ) 8 n + 1 − h ( 2 n x ) 8 n , t α n 8 ( 8 n ) ) ≥ M c ( x , t )$
(4.37)

for all x X, t > 0 and n > 0. It follows from $h ( 2 n x ) 8 n − h ( x ) = ∑ j = 0 n − 1 ( h ( 2 j + 1 x ) 8 j + 1 − h ( 2 j x ) 8 j )$ and (4.37) that

$N ( h ( 2 n x ) 8 n − h ( x ) , ∑ j = 0 n − 1 α j t 8 ( 8 ) j ) ≥ min ∪ j = 0 n − 1 { N ( h ( 2 j + 1 x ) 8 j + 1 − h ( 2 j x ) 8 j , α j t 8 ( 8 ) j ) } ≥ M c ( x , t )$
(4.38)

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

$N ( h ( 2 n + m x ) 8 n + m − h ( 2 m x ) 8 m , ∑ j = 0 n − 1 α j t 8 ( 8 ) j + m ) ≥ M c ( 2 m x , t ) = M c ( x , t α m )$

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

$N ( h ( 2 n + m x ) 8 n + m − h ( 2 m x ) 8 m , ∑ j = m n + m − 1 α j t 8 ( 8 ) j ) ≥ M c ( x , t )$

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

$N ( h ( 2 n + m x ) 8 n + m − h ( 2 m x ) 8 m , t ) ≥ M c ( x , t ∑ j = m n + m − 1 α j 8 ( 8 ) j ) )$
(4.39)

for all x X, all t > 0 and all m ≥ 0, n > 0. Since 0 < α < 8 and $∑ n = 0 ∞ α 8 n <∞$, the Cauchy criterion for convergence and (N5) imply that ${ h ( 2 n x ) 8 n }$ is a Cauchy sequence in (Y, N) to some point C(x) Y. So one can define the mapping C : XY by

$C ( x ) = N − lim n → ∞ h ( 2 n x ) 8 n )$
(4.40)

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

$N ( h ( 2 n x ) 8 n − h ( x ) , t ) ≥ M c ( x , t ∑ j = 0 n − 1 α j 8 ( 8 ) j ) )$

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

$N ( C ( x ) − h ( x ) , t ) ≥ min { N ( C ( x ) − h ( 2 n x ) 8 n , t 2 ) , N ( h ( 2 n x ) 8 n − h ( x ) , t 2 ) } ≥ M c ( x , t ∑ j = 0 n − 1 α j 4 ( 8 j ) ) )$
(4.41)

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

$N ( C ( x ) - h ( x ) , t ) ≥ M a x , t ( 8 - α ) 2$
(4.42)

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

$N ( C ( 2 x ) 8 − C ( x ) , t ) ≥ min { N ( C ( 2 x ) 8 − h ( 2 n + 1 x ) 8 n + 1 ) , t 3 ) , N ( h ( 2 n x ) 8 n − C ( x ) , t 3 ) , N ( h ( 2 n + 1 x ) 8 n + 1 − h ( 2 n x ) 8 n ) , t 3 ) = M c ( x , 8 ( 8 ) n t 3 α n )$

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

$C ( 2 x ) = 8 C ( x )$
(4.43)

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

$N ( 1 8 n D h ( 2 n x ,2 n y ) , t ) = N ( D f ( 2 n + 1 x ,2 n + 1 y ) − 2 D f ( 2 n x ,2 n y ) , 8 n t ) = min { N ( D f ( 2 n + 1 x ,2 n + 1 y ) , 8 n t 2 ) , N ( D f ( 2 n x ,2 n y ) , 8 n t 4 ) } ≥ min { N ′ ( φ c ( 2 n + 1 x ,2 n + 1 y ) , 8 n t 2 ) , N ′ ( φ c ( 2 n x ,2 n y ) , 8 n t 4 ) } = min { N ′ ( φ c ( x , y ) , 8 n t 2 α n + 1 ) , N ′ ( φ c ( x , y ) , 8 n t 4 α n ) }$

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 : XY satisfies the inequality (3.18) for all x, y X. Then, the limit

$C ( x ) = lim n → ∞ 1 8 ℓ n ( f ( 2 ℓ n + 1 x ) - 2 f ( 2 ℓ n x ) )$

exists for all x X and C : XY is a unique cubic mapping satisfying

$∥ f ( 2 x ) - 2 f ( x ) - C ( x ) | | ≤ ( | 2 k + ℓ | λ + 1 ) ( 9 k 2 + 4 ) ε ∥ x ∥ λ ℓ k 2 ( k 2 - 1 ) ( 4 - 2 λ - 1 )$
(4.44)

for all x X, where λℓ < 3.

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 × XZ be a mapping such that

$φ ( 2 x , 2 y ) = α φ ( x , y )$
(4.45)

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

$N ( D f ( x , y ) , t ) ≥ N ′ ( φ ( x , y ) , t )$
(4.46)

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

$N ( f ( x ) - A ( x ) - C ( x ) , t ) ≥ min M ( x , 3 t ( 2 - α ) 2 ) , M ( x , 3 t ( 8 - α ) 2 ) , 0 < α < 2 min M ( x , 3 t ( α - 2 ) 2 ) , M ( x , 3 t ( 8 - α ) 2 ) , 2 < α < 8 min M ( x , 3 t ( α - 2 ) 2 ) , M ( x , 3 t ( α - 8 ) 2 ) , α > 8$
(4.47)

for all x X and all t > 0, where

$M ( x , t ) = min { N ′ ( φ ( x , x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ ( 2 x , x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ ( x ,2 x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ ( ( k + 1 ) x , x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ ( ( k − 1 ) x , x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) ) , N ′ ( φ ( 2 x ,2 x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ ( x ,3 x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ ( ( 2 k + 1 ) x , x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4 t ) , N ′ ( φ ( ( 2 k − 1 ) x , x ) , k 2 ( k 2 − 1 ) 9 k 2 + 4