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Fuzzy Hyers-Ulam approximation of a mixed AQ mapping

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.

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(xp+ yp), (ε > 0, p [0,1)). Furthermore, in 1994, a generalization of Rassias' theorem was obtained by Găvruta [4] by replacing the bound ϵ(xp+ yp) 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 : XY, 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 [917]).

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

r 2 f x + y + z r + r 2 f x - y + z r + r 2 f x + y - z r + r 2 f - x + y + z r = 4 f ( x ) + 4 f ( y ) + 4 f ( z )
(1.1)

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

Lee and Jun [22] proved that an even (odd) mapping f : XY satisfies the functional Equation (1.1) if and only if the mapping f : XY 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 [2333]).

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

(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 , 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 and limt→∞N(x, t) = 1;

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

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 ) = α t α t + β x t > 0 , x X 0 t 0 , x X

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 limt→∞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 - limt→∞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 such that for all nn0 and all p > 0, we have N(xn+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 × → [0,1] be a fuzzy norm on defined by

N ( x , t ) = t t + x t > 0 0 t 0 .

Then (, N) is a fuzzy Banach space. Let {x n } be a Cauchy sequence in , δ > 0 andε= δ 1 + δ . Then there exist m such that for all nm and all p > 0, we have

1 1 + x n + p - x n 1 - ε .

So |xn+p-x n | < δ for all nm and all p > 0. Therefore {x n } is a Cauchy sequence in (, |·|). Let x n x0 as n → ∞. Then limn→∞N (x n - x0, t) = 1 for all t > 0.

We say that a mapping f : XY 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: XY is continuous at each x X, then f : XY 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. (1)

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

  2. (2)

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

  3. (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 : XX be a strictly contractive mapping with Lipschitz constant L < 1. Then, for all x X, either

d ( J n x , J n + 1 x ) =

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

  1. (1)

    d(J n x, J n+ 1 x) < ∞ for all n 0n 0;

  2. (2)

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

  3. (3)

    y* is the unique fixed point of J in the set Y= { y X : d ( J n 0 x , y ) < } ;

  4. (4)

    d ( y , y * ) 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 ) 2 α φ ( x , y , z )
(2.1)

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

N r 2 f x + y + z r + r 2 f x - y + z r + r 2 f x + y - z r + r 2 f - x + y + z r - 4 f ( x ) - 4 f ( y ) - 4 f ( z ) , t 1 t + φ ( x , y , z )
(2.2)

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

A ( x ) : = N - lim n f ( 2 n x ) 2 n

exists for each x X and defines a unique additive mapping A:XY such that

N ( f ( x ) - A ( x ) , t ) ( 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 2 x r - 4 f ( 2 x ) , t 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 2 x r - 8 f ( x ) , t 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

N f ( 2 x ) 2 - f ( x ) , t 4 = N 2 r 2 f 2 x r - 8 f ( x ) - 2 r 2 f 2 x r + 4 f ( 2 x ) , 2 t min N 2 r 2 f 2 x r - 4 f ( 2 x ) , t , N 2 r 2 f 2 x r - 8 f ( x ) , t t t + φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 ) .
(2.6)

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

d ( g , h ) = inf μ ( 0 , + ) N ( g ( x ) - h ( x ) , μ t ) t t + φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 ) , x 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 ) : = 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 ) t t + φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 )

for all x X and all t > 0. Hence

N ( J g ( x ) - J h ( x ) , α β t ) = N 1 2 g ( 2 x ) - 1 2 h ( 2 x ) , α β t = N ( g ( 2 x ) - h ( 2 x ) , 2 α β t ) 2 α t 2 α t + φ ( 4 x , 0 , 0 ) + φ ( 2 x , 2 x , 0 ) 2 α t 2 α t + 2 α ( φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 ) ) = t t + φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 )

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 ) 1 4 .

By Theorem 1.7, there exists a mapping A : XY satisfying the following:

  1. (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 ) t t + φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 )

for all x X;

  1. (2)

    d(J n f, A) → 0 as n → ∞. This implies the equality lim n f ( 2 n x ) 2 n =A ( x ) for all x X;

  2. (3)

    d ( f , A ) 1 1 - α d ( f , J f ) , which implies the inequality d ( f , A ) 1 4 - 4 α . This implies that the inequalities (2.3) holds.

It follows from (2.1) and (2.2) that

N 1 2 n r 2 f 2 n ( x + y + z ) r + r 2 f 2 n ( x - y + z ) r + r 2 f 2 n ( x + y - z ) r + r 2 f 2 n ( - x + y + z ) r - 4 f ( 2 n x ) - 4 f ( 2 n y ) - 4 f ( 2 n z ) , 1 2 n t t + φ ( 2 n x , 2 n y , 2 n z )

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

N 1 2 n r 2 f 2 n ( x + y + z ) r + r 2 f 2 n ( x - y + z ) r + r 2 f 2 n ( x + y - z ) r + r 2 f 2 n ( - x + y + z ) r - 4 f ( 2 n x ) - 4 f ( 2 n y ) - 4 f ( 2 n z ) , t 2 n t 2 n t + ( 2 α ) n φ ( x , y , z )

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

lim n 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

N r 2 A x + y + z r + 2 r A x - y + z r + r 2 A x + y - z r + r 2 A - x + y + z s - 4 A ( x ) - 4 A ( y ) - 4 A ( z ) , t = 1

for all x, y, z X and all t > 0. Hence the mapping A : XY 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 : XY be an odd mapping satisfying

N r 2 f x + y + z r + r 2 f x - y + z r + r 2 f x + y - z r + r 2 f - x + y + z r - 4 f ( x ) - 4 f ( y ) - 4 f ( z ) , t t t + θ x r + y r + z r
(2.8)

for all x, y, z X and all t > 0. Then the limitA ( x ) :=N- lim n f ( 2 n x ) 2 n exists for each x X and defines a unique additive mapping A: XY such that

N ( f ( x ) - A ( x ) , t ) 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(xr+ yr+ zr) for all x, y, z X. Then we can choose α = 2r-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 x 2 , y 2 , z 2 α 2 φ ( x , y , z )
(2.9)

for all x, y, z X. Let f : XY be an odd mapping satisfying (2.2). Then the limitA ( x ) :=N- lim n 2 n f x 2 n exists for each x X and defines a unique additive mapping A : XY such that

N f ( x ) - A ( x ) , t ( 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 : SS such that

J g ( x ) : = 2 g x 2 .

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

N ( g ( x ) - h ( x ) , β t ) t t + φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 )

for all x X and all t > 0. Hence

N ( J g ( x ) - J h ( x ) , α β t ) = N 2 g x 2 - 2 h x 2 , α β t = N g x 2 - g x 2 , α β t 2 α 2 t α 2 t + φ ( x , 0 , 0 ) + φ x 2 , x 2 , 0 t t + φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 )

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 x 2 - f ( x ) , t 2 t t + φ ( x , 0 , 0 ) + φ x 2 , x 2 , 0 t t + α 2 ( φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 ) )

for all x X and t > 0. Therefore

N 2 f x 2 - f ( x ) , α t 4 t t + φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 ) .
(2.11)

So d ( f , J f ) α 4 . By Theorem 1.7, there exists a mapping A : XY satisfying the following:

  1. (1)

    A is a fixed point of J, that is,

    A x 2 = 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 ) t t + φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 )

for all x X and t > 0.

  1. (2)

    d(J n f, A) → 0 as n → ∞. This implies the equality N- lim n 2 n f x 2 n =A ( x ) for all x X.

  2. (3)

    d ( f , A ) d ( f , J f ) 1 - L with f Ω, which implies the inequality d ( f , A ) 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 : XY be an odd mapping satisfying (2.8). Then, the limitA ( x ) :=N- lim n 2 n f x 2 n exists for each x X and defines a unique additive mapping A : XY such that

N ( f ( x ) - A ( x ) , t ) 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(xr+ yr+ zr) for all x, y, z X. Then we can choose α = 21-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 ) 4 α φ ( x , y , z )
(2.13)

for all x, y, z X. Let f : XY be an even mapping with f(0) = 0 and satisfying (2.2). ThenQ ( x ) :=N- lim n f ( 2 n x ) 4 n exists for each x X and defines a unique quadratic mapping Q : XY such that

N ( f ( x ) - Q ( x ) , t ) ( 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 2 x r - 8 f ( x ) , t 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 2 x r - 4 f ( 2 x ) , t 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

N f ( 2 x ) 4 - f ( x ) , t 16 = N 2 2 r 2 f 2 x r - 8 f ( x ) - 4 r 2 f 2 x r - 4 f ( 2 x ) , t min N 2 r 2 f 2 x r - 8 f ( x ) , t 4 , N 4 r 2 f 2 x r - 4 f ( 2 x ) , t 2

for all x X and all t > 0. So

N f ( 2 x ) 4 - f ( x ) , 3 t 8 min N 2 r 2 f 2 x r - 8 f ( x ) , t , N 4 r 2 f 2 x r - 4 f ( x ) , t t t + φ ( x , x , 0 ) + φ ( 2 x , 0 , 0 ) .
(2.17)

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

d ( g , h ) = inf μ ( 0 , + ) N ( g ( x ) - h ( x ) , μ t ) t t + φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 ) , x 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 Jg ( x ) := 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 ) 3 8 .

By Theorem 1.7, there exists a mapping Q : XY satisfying the following:

  1. (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 ) t t + φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 ) for all x X;

  1. (2)

    d(J n f, Q) → 0 as n → ∞. This implies the equality lim n f ( 2 n x ) 4 n =Q ( x ) for all x X;

  2. (3)

    d ( f , Q ) 1 1 - α d ( f , J f ) , which implies the inequality d ( f , Q ) 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 : XY be an even mapping satisfying (2.8) and f(0) = 0. Then the limitQ ( x ) :=N- lim n f ( 2 n x ) 4 n exists for each x X and defines a unique quadratic mapping Q : XY such that

N ( f ( x ) - Q ( x ) , t ) 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(xr+ yr+ zr) for all x, y, z X. Then we can choose α = 4r-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 x 2 , y 2 , z 2 α 4 φ ( x , y , z )

for all x, y, z X. Let f : XY be an even mapping satisfying (2.2) and f(0) = 0. Then the limitQ ( x ) :=N- lim n 4 n f x 2 n exists for each x X and defines a unique quadratic mapping Q : XY such that

N f ( x ) - Q ( x ) , t ( 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 x 2 , 3 t 2 t t + φ ( x , 0 , 0 ) + φ x 2 , x 2 , 0 t t + α 4 ( φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 ) )

for all x X and t > 0. So

N f ( x ) - 4 f x 2 , 3 α t 8 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 : XY be an even mapping satisfying f(0) = 0 and (2.8). ThenQ ( x ) :=N- lim n 4 n f x 2 n exists for each x X and defines a unique quadratic mapping Q : XY such that

N ( f ( x ) - Q ( x ) , t ) 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(xr+ yr+ zr) for all x, y, z X. Then we can choose α = 41-rand we get the desired result.

Let f : XY be a mapping satisfying f(0) = 0 and (1.1). Let f e ( x ) := f ( x ) + f ( - x ) 2 and f o ( x ) = f ( x ) - f ( - x ) 2 . Then f e is an even mapping satisfying (1.1) and f o is 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 x 2 , y 2 , z 2 α 4 φ ( x , y , z )

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

N ( f ( x ) - A ( x ) - Q ( x ) , 2 t ) = N f ( x ) - f ( - x ) 2 + f ( x ) + f ( - x ) 2 - A ( x ) - Q ( x ) , 2 t min N f ( x ) - f ( - x ) 2 - A ( x ) , t , N f ( x ) + f ( - x ) 2 - Q ( x ) , t min ( 8 - 8 α ) t ( 8 - 8 α ) t + α ( φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 ) ) , ( 8 - 8 α ) t ( 8 - 8 α ) t + 3 α ( φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 ) ) = ( 8 - 8 α ) t ( 8 - 8 α ) t + 3 α ( φ ( 2 x , 0 , 0 ) + φ ( x , x , 0 ) ) .

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 : XY be a mapping with f(0) = 0 and satisfying (2.8). Then there exist a unique additive mapping A : XY and a unique quadratic mapping Q : XY such that

N ( f ( x ) - A ( x ) - Q ( x ) , t ) 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(xr+ yr+ zr) for all x, y, z X. Then we can choose α = 21-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 .

Theorem 3.1. Assume that a mapping f : XY is an odd mapping satisfying the inequality

N r 2 f x + y + z r + r 2 f x - y + z r + r 2 f x + y - z r + r 2 f - x + y + z r - 4 f ( x ) - 4 f ( y ) - 4 f ( z ) , t N ( φ ( x , y , z ) , t )
(3.1)

for all x, y, z X, t > 0 and that φ : X3Z is a mapping for which there is a constant r satisfying0< r < 1 2 such that

N φ x 2 , y 2 , z 2 , t N φ ( x , y , z ) , t r
(3.2)

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

N ( f ( x ) - A ( x ) , t ) min N ( φ ( x , 0 , 0 ) , 2 ( 1 - 2 r ) t ) , N φ ( x , x , 0 ) , 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 x 2 j , y 2 j , z 2 j , t N φ ( x , y , z ) , 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 2 x r - 4 f ( 2 x ) , t 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 2 x r - 8 f ( x ) , t 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

N f ( 2 x ) 2 - f ( x ) , t 4 = N 2 r 2 f 2 x r - 8 f ( x ) - 2 r 2 f 2 x r + 4 f ( 2 x ) , 2 t min N 2 r 2 f 2 x r - 4 f ( 2 x ) , t , N 2 r 2 f 2 x r - 8 f ( x ) , t min N ( φ ( 2 x , 0 , 0 ) , t ) , N ( φ ( x , x , 0 ) , t )
(3.6)

for all x X and all t > 0. Replacing x by x 2 in (3.6), we have

N f ( x ) - 2 f x 2 , t 2 min N ( φ ( x , 0 , 0 ) , t ) , N φ x 2 , x 2 , 0 , t .
(3.7)

Replacing x by x 2 j in (3.7), we have

N 2 j + 1 f x 2 j + 1 - 2 j f x 2 j , 2 j - 1 t min N φ x 2 j , 0 , 0 , t , N φ x 2 j + 1 , x 2 j + 1 , 0 , t min N φ ( x , 0 , 0 ) , t r j , N φ ( x , x , 0 ) , t r j + 1
(3.8)

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

N f ( x ) - 2 n f x 2 n , j = 0 n - 1 2 j - 1 r j t = N j = 0 n - 1 2 j + 1 f x 2 j + 1 - 2 j f x 2 j , j = 0 n - 1 2 j - 1 r j t min 0 j n - 1 N 2 j + 1 f x 2 j + 1 - 2 j f x 2 j , 2 j - 1 r j t min N ( φ ( x , 0 , 0 ) , t ) , N φ ( x , x , 0 ) , t r

which yields

N 2 n + p f x 2 n + p - 2 p f x 2 p , j = 0 n - 1 2 j + p - 1 r j min N φ x 2 p , 0 , 0 , t , N φ x 2 p , x 2 p , 0 , t r min N φ ( x , 0 , 0 ) , t r p , N φ ( x , x , 0 ) , t r p + 1
(3.9)

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

N 2 n + p f x 2 n + p - 2 p f x 2 p , j = 0 n - 1 2 j + p - 1 r j + p t min N ( φ ( x , 0 , 0 ) , t ) , N φ ( x , x , 0 ) , t r

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

N 2 n + p f x 2 n + p - 2 p f x 2 p , t min N φ ( x , 0 , 0 ) , t j = 0 n - 1 2 j + p - 1 r j + p , N φ ( x , x , 0 ) , t j = 0 n - 1 2 j + p - 1 r j + p + 1
(3.10)

for all x X, t > 0 and any integers n > 0, p ≥ 0. Since the series j = 0 + 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 x 2 n is a Cauchy sequence in the fuzzy Banach space (Y, N) and so it converges in Y. Therefore a mapping A : XY defined by A ( x ) :=N- lim n 2 n f x 2 n is well-defined for all x X. It means that

lim n N A ( x ) - 2 n f 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 x 2 n , t min N φ ( x , 0 , 0 ) , 2 t j = 0 n - 1 2 j r j , N φ ( x , x , 0 ) , 2 t j = 0 n - 1 2 j r j + 1

for all x X and all t > 0. So

N ( f ( x ) - A ( x ) , t ) min N f ( x ) - 2 n f x 2 n , ( 1 - ε ) t , N A ( x ) - 2 n f x 2 n , ε t min N φ ( x , 0 , 0 ) , 2 ε t j = 0 n - 1 2 j r j , N φ ( x , x , 0 ) , 2 ε t j = 0 n - 1 2 j r j + 1 min N ( φ ( x , 0 ) , 2 ( 1 - 2 ) r ε t ) , N φ ( x , x , 0 ) , 2 ( 1 - 2 r ) ε t r

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 ) min N ( φ ( x , 0 , 0 ) , 2 ( 1 - 2 r ) t ) , N φ ( x , x , 0 ) , 2 ( 1 - 2 r t ) r

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

N 2 n r 2 f x + y + z r . 2 n + 2 n . r 2 f x - y + z r . 2 n + 2 n . r 2 f x + y - z r . 2 n + 2 n . r 2 f - x + y + z r . 2 n - 2 n + 2 f x 2 n - 2 n + 2 f y 2 n - 2 n + 2 f z 2 n , t N φ x 2 n , y 2 n , z 2 n , t 2 n N φ ( x , y , z ) , t 2 n r n 1 as n +

for all x, y, z X, t > 0. So

N 2 n r 2 f x + y + z r . 2 n + 2 n . r 2 f x - y + z r . 2 n + 2 n . r 2 f x + y - z r . 2 n + 2 n . r 2 f - x + y + z r . 2 n - 2 n + 2 f x 2 n - 2 n + 2 f y 2 n - 2 n + 2 f z 2 n , t 1

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

N r 2 A x + y + z r + r 2 A x - y + z r + r 2 A x + y - z r + r 2 A - x + y + z r - 4 A ( x ) - 4 A ( y ) - 4 A ( z ) , t min N r 2 A x + y + z r + r 2 A x - y + z r + r 2 A x + y - z r + r 2 A - x + y + z r - 4 A ( x ) - 4 A ( y ) - 4 A ( z ) - 2 n r 2 f x + y + z r . 2 n + 2 n . r 2 f x - y + z r . 2 n + 2 n . r 2 f x + y - z r . 2 n + 2 n . r 2 f - x + y + z r . 2 n - 2 n + 2 f x 2 n - 2 n + 2 f y 2 n - 2 n + 2 f z 2 n , t 2 , N 2 n r 2 f x + y + z r . 2 n + 2 n . r 2 f x - y + z r . 2 n + 2 n . r 2 f x + y - z r . 2 n + 2 n . r 2 f - x + y + z r . 2 n - 2 n + 2 f x 2 n - 2 n + 2 f y 2 n - 2 n + 2 f z 2 n , t 2 = N 2 n r 2 f x + y + z r . 2 n + 2 n . r 2 f x - y + z r . 2 n + 2 n . r 2 f x + y - z r . 2 n + 2 n . r 2 f - x + y + z r . 2 n - 2 n + 2 f x 2 n - 2 n + 2 f y 2 n - 2 n + 2 f z 2 n , t 2 ( for sufficiently large n ) N φ ( x , y , z ) , t 2 n + 1 r n 1 as n

which implies

r 2 A x + y + z r + r 2 A x - y + z r + r 2 A x + y - z r + r 2 A - x + y + z r = 4 A ( x ) + 4 A ( y ) + 4 A ( z )

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

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

N ( A ( x ) - L ( x ) , t ) = N 2 n A x 2 n - 2 n L x 2 n , t min N 2 n A x 2 n - 2 n f x 2 n , t 2 , N 2 n f x 2 n - 2 n L x 2 n , t 2 min N φ x 2 n + 1 , 0 , 0 , ( 1 - 2 r ) t 2 n + 1 , N φ x 2 n , x 2 n , 0 , ( 1 - 2 r ) t 2 n + 1 r min N φ ( x , 0 , 0 ) , ( 1 - 2 r ) t r n 2 n + 1 , N φ ( x , x , 0 ) , ( 1 - 2 r ) t 2 n + 1 r n + 1 1 as n

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 (, N') a fuzzy Banach space. Assume that there exist real numbers θ ≥ 0 and p > 1 such that an odd mapping f : XY satisfies the inequality

N r 2 f x + y + z r + r 2 f x - y + z r + r 2 f x + y - z r + r 2 f - x + y + z r - 4 f ( x ) - 4 f ( y ) - 4 f ( z ) , t N θ x p + y p + z p , t
(3.12)

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

N ( f ( x ) - A ( x ) , t ) min N θ x p , ( 2 p + 1 - 4 ) t 2 p , N θ x p , ( 2 p - 2 ) t

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

Theorem 3.3. Assume that an odd mapping f : XY satisfies the inequality (3.1) and that φ:X3Z is a mapping for which there is a constant r satisfying 0 < |r| < 2 such that

N ( φ ( x , y , z ) , r t ) N φ x 2 , y 2 , z 2 , t
(3.13)

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

N ( f ( x ) - A ( x ) , t ) min N φ ( x , 0 , 0 ) , 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 f ( 2 x ) 2 - f ( x ) , t 4 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

N f ( 2 n + 1 x ) 2 n + 1 - f ( 2 n x ) 2 n , t 2 n + 2 min { N ( φ ( 2 n + 1 x , 0 , 0 ) , t ) , N ( φ ( 2 n x , 2 n x , 0 ) , t ) } min N φ ( x , 0 , 0 ) , t r n + 1 , N φ ( x , x , 0 ) , t r n .
(3.16)

So

N f ( 2 n + 1 x ) 2 n + 1 - f ( 2 n x ) 2 n , r n t 2 n + 2 min N φ ( x , 0 , 0 ) , 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 ) - f ( 2 n x ) 2 n , j = 0 n - 1 r j t 2 j + 2 min N φ ( x , 0 , 0 ) , t r , N ( φ ( x , x , 0 ) , t )

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

N f ( x ) - f ( 2 n x ) 2 n , t min N φ ( x , 0 , 0 ) , t r j = 0 n - 1 r j 2 j + 2 , N φ ( x , x , 0 ) , t j = 0 n - 1 r j 2 j + 2 min N φ ( x , 0 , 0 ) , 2 ( 2 - r t ) r , N ( φ ( x , x , 0 ) , 2 ( 2 - r ) t ) .

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 (, N') a fuzzy Banach space. Assume that there exist real numbers θ ≥ 0 and 0 < p < 1 such that an odd mapping f : XY satisfies (3.12). Then there is a unique additive mapping A : XY satisfying (1.1) and the inequality

N ( f ( x ) - A ( x ) , t ) min N θ x p , 2 ( 2 - 2 p ) 2 p , N θ x p , ( 2 - 2 p ) t

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

Theorem 3.5. Assume that a mapping f : XY is an even mapping satisfies (3.1) and f(0) = 0 and that φ : X3Z is a mapping for which there is a constant r satisfying 0 < |r| < 4 such that

N ( φ ( x , y , z ) , r t ) N φ x 2 , y 2 , z 2 , t
(3.18)

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

N ( f ( x ) - Q ( x ) , t ) min N φ ( x , 0 , 0 ) , 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 2 x r - 8 f ( x ) , t 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 2 x r - 4 f ( 2 x ) , t 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

N f ( 2 x ) 4 - f ( x ) , t 16 = N 2 2 r 2 f 2 x r - 8 f ( x ) - 4 r 2 f 2 x r - 4 f ( 2 x ) , t min N 2 r 2 f 2 x r - 8 f ( x ) , t 4 , N 4 r 2 f 2 x r - 4 f ( 2 x ) , t 2 min N φ ( x , x , 0 ) , t 4 , N φ ( 2 x , 0 , 0 ) , t 2
(3.22)

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

N f ( 2 n + 1 ) x 4 n + 1 - f ( 2 n x ) 4 n , t 4 n + 2 min N φ ( 2 n x , 2 n x , 0 ) , t 4 , N φ ( 2 n + 1 x , 0 , 0 ) , t 2 min N φ ( x , x , 0 ) , t 4 r n , N φ ( x , 0 , 0 ) , t 2 r n + 1

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

N f ( x ) - f ( 2 n x ) 4 n , j = 0 n - 1 r j t 4 j + 2 min N φ ( x , x , 0 ) , t 4 , N φ ( x , 0 , 0 ) , t 2 r

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

N f ( x ) - f ( 2 n x ) 4 n , t min N φ ( x , x , 0 ) , 4 t j = 0 n - 1 r j 4 j , N φ ( x , 0 , 0 ) , 8 t r j = 0 n - 1 r j 4 j min N φ ( x , 0 , 0 ) , 2 ( 4 - r t ) r , N ( φ ( x , x , 0 ) , ( 4 - r ) t ) .

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

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

N ( f ( x ) - Q ( x ) , t ) min N θ x p , 2 ( 4 - 4 p ) t 4 p , N θ x p , ( 4 - 4 p ) t 2

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

Theorem 3.7. Assume that an even mapping f : XY satisfies the inequality (3.1) and f(0) = 0 and that φ : X3Z is a mapping for which there is a constant r satisfying0< r < 1 4 such that

N φ x 2 , y 2 , z 2 , t N φ ( x , y , z ) , t r
(3.23)

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

N ( f ( x ) - Q ( x ) , t ) min N φ ( x , x , 0 ) , ( 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 x 2 , t 4 min N φ x 2 , x 2 , 0 , t 4 , N φ ( x , 0 , 0 ) , t 2
(3.25)

for all x X and all t > 0. Replacing x by x 2 j in (3.25), we have

N 4 j + 1 f x 2 j + 1 - 4 j f x 2 j , 4 j t min N φ ( x , x , 0 ) , t r j + 1 , N φ ( x , 0 , 0 ) , 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

N f ( x ) - 4 n f x 2 n , t min N φ ( x , x , 0 ) , t j = 0 n - 1 4 j r j + 1 , N φ ( x , 0 , 0 ) , 2 t j = 0 n - 1 4 j r j min N φ ( x , x , 0 ) , ( 1 - 4 r ) t r , N ( φ ( x , 0 , 0 ) , ( 2 - 8 r ) t )

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

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

N ( f ( x ) - Q ( x ) , t ) min N θ x p , ( 4 p - 4 t ) 2 , N θ x p , 2 - 8 4 p t

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

Let f : XY be a mapping satisfying f(0) = 0 and (1.1). Let f e ( x ) := f ( x ) + f ( - x ) 2 and f o ( x ) = f ( x ) - f ( - x ) 2 . Then f e is an even mapping satisfying (1.1) and f o is 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 : XY is a mapping satisfying (3.1) and f(0) = 0 and that φ : X3Z is a mapping for which there is a constant r satisfying0< r < 1 4 such that

N φ x 2 , y 2 , z 2 , t N φ ( x , y , z ) , t r

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

N ( f ( x ) - A ( x ) - Q ( x ) , 2 t ) min min N φ ( x , 0 , 0 ) , 2 ( 1 - 2 r t ) , N φ ( x , x , 0 ) , 2 ( 1 - 2 ) r t r , min N φ ( x , 0 , 0 ) , ( 1 - 4 r ) t r , N ( φ ( x , 0 , 0 ) , ( 2 - 8 r ) t )

for all x X and all t > 0.

Proof. It follows from Theorems 3.1 and 3.7 that

N ( f ( x ) - A ( x ) - Q ( x ) , 2 t ) = N f ( x ) + f ( - x ) 2 + f ( x ) - f ( - x ) 2 - A ( x ) - Q ( x ) , 2 t min N f ( x ) + f ( - x ) 2 - Q ( x ) , t , N f ( x ) - f ( - x ) 2 - A ( x ) , t min min N ( φ ( x , 0 , 0 ) , 2 ( 1 - 2 r ) t ) , N φ ( x , x , 0 ) , 2 ( 1 - 2 r ) t r , min N φ ( x , 0 , 0 ) , ( 1 - 4 r ) t r , N ( φ ( x , 0 , 0 ) , ( 2 - 8 r ) t ) .

Corollary 3.10. Let X be a normed spaces and that (, N') a fuzzy Banach space. Assume that there exists real number θ ≥ 0 and 0 < p < 1 such that a mapping f : XY satisfies (3.12) and f(0) = 0. Then there are unique quadratic and additive mappings Q : XY and A : XY (respectively) satisfying (1.1) and the inequality

N ( f ( x ) - - A ( x ) - Q ( x ) , t ) min N θ x p , ( 1 - 4 p + 1 ) t 4 p , N θ x p , ( 2 - 2 . 4 p + 1 ) t

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

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