2_440_gordji.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 2, No. 4, 2009, (494-507)

ISSN 1307-5543 – www.ejpam.com

Approximation of the Quadratic and Cubic Functional

Equations in RN–spaces

M. Eshaghi Gordji1∗, J. M. Rassias2 and M. Bavand Savadkouhi1

1 Department of Mathematics, University of Semnan, P. O. Box 35195-363, Semnan, Iran

2 Section of Mathematics and Informatics, Pedagogical Department National and Capodis-

trian University of Athens, 4, Agamemnonos St., Aghia Paraskevi, Athens 15342, Greece

Abstract. We prove a stability result for the quadratic and cubic functional equations in ran-

dom normed (RN) spaces (in the sense of Sherstnev) under arbitrary t–norms.

2000 Mathematics Subject Classifications: Primary 54E40; Secondary 39B82, 46S50, 46S40.

Key Words and Phrases: Stability; Quadratic functional equation; Cubic functional equation;

Random normed space.

1. Introduction

The study of stability problems for functional equations is related to a question of

Ulam [37], concerning the stability of group homomorphisms, affirmatively answered

∗Corresponding author.

Email addresses: madjid.eshaghi�gmail.
om (M. Gordji), jrassias�primedu.uoa.gr (J.

Rassias), bavand.m�gmail.
om (M. Savadkouhi)

http://www.ejpam.com 494 c© 2009 EJPAM All rights reserved.



M. Gordji, J. Rassias, and M. Savadkouhi / Eur. J. Pure Appl. Math, 2 (2009), (494-507) 495

for Banach spaces by Hyers [13]. Subsequently, the result of Hyers was generalized

by Aoki [2], Bourgin [4], Gǎvruta [8] and Rassias [24] (see also [9] and [31]). The

functional equation

f (x + y) + f (x − y) = 2 f (x) + 2 f (y) (1)

is related to a symmetric bi–additive function. It is natural that this equation is called

a quadratic functional equation. In particular, every solution of the quadratic equation

(1) is said to be a quadratic function. It is well known that a function f between real

vector spaces is quadratic if and only if there exists a unique symmetric bi–additive

function B such that f (x) = B(x , x) for all x (see [1, 16]). The bi–additive function

B is given by

B(x , y) =
1

4
[ f (x + y)− f (x − y)]. (2)

Hyers–Ulam–Rassias stability problem for the quadratic functional equation (1) was

proved by Skof for functions f : A→ B, where A is normed space and B is a Banach

space (see [36]). Cholewa [5] noticed that the theorem of Skof is still true if rele-

vant domain A is replaced by an abelian group. In the paper [7], Czerwik proved the

Hyers–Ulam–Rassias stability of the functional equation (1). Grabiec [10] has gener-

alized these result mentioned above. We only mention here the papers [14], [16],

[23], [32], [33] [27–30] concerning the stability of the quadratic functional equa-

tions.

The following cubic functional equation, which is the oldest cubic functional equation,

and was introduced by J. M. Rassias [25](in 2001):

f (x + 2y) + 3 f (x) = 3 f (x + y) + f (x − y) + 6 f (y).

Jun and Kim [15] introduced the following cubic functional equation

f (2x + y) + f (2x − y) = 2 f (x + y) + 2 f (x − y) + 12 f (x) (3)



M. Gordji, J. Rassias, and M. Savadkouhi / Eur. J. Pure Appl. Math, 2 (2009), (494-507) 496

and they established the general solution and the generalized Hyers–Ulam–Rassias

stability for the functional equation (3) (in this case we have a much better pos-

sible upper bound for (1.3) than the Hyers–Ulam–Rassias stability). The function

f (x) = x3 satisfies the functional equation (3), which is thus called a cubic functional

equation. Every solution of the cubic functional equation is said to be a cubic function.

There are many works in the very active area of the stability of functional equations.

We only mention here the papers [26] and [14] concerning the stability of the cubic

functional equation.

The generalized Hyers–Ulam–Rassias stability of different functional equations in ran-

dom normed and fuzzy normed spaces has been recently studied in [17]– [22].

In the sequel we adopt the usual terminology, notations and conventions of the

theory of random normed spaces, as in [3, 6, 17, 19, 34, 35]. Throughout this paper,

∆+ is the space of distribution functions, that is, the space of all mappings F : R ∪

{−∞,∞}→ [0, 1], such that F is left-continuous and non-decreasing on R, F(0) = 0

and F(+∞) = 1. D+ is a subset of ∆+ consisting of all functions F ∈ ∆+ for which

l−F(+∞) = 1, where l− f (x) denotes the left limit of the function f at the point x ,

that is, l− f (x) = limt→x− f (t). The space ∆+ is partially ordered by the usual point–

wise ordering of functions, i.e., F ≤ G if and only if F(t) ≤ G(t) for all t in R. The

maximal element for ∆+ in this order is the distribution function ǫ0 given by

ǫ0(t) =







0, if t ≤ 0,

1, if t > 0.

Definition 1. ( [34]). A mapping T : [0, 1]×[0, 1]→ [0, 1] is a continuous triangular

norm (briefly, a continuous t–norm) if T satisfies the following conditions:

(a) T is commutative and associative;

(b) T is continuous;

(c) T (a, 1) = a for all a ∈ [0, 1];



M. Gordji, J. Rassias, and M. Savadkouhi / Eur. J. Pure Appl. Math, 2 (2009), (494-507) 497

(d) T (a, b) ≤ T (c, d) whenever a ≤ c and b ≤ d for all a, b, c, d ∈ [0, 1].

Typical examples of continuous t–norms are TP(a, b) = ab, TM(a, b) = min(a, b)

and TL(a, b) = max(a + b − 1, 0) (the Lukasiewicz t-norm). Recall (see [11], [12])

that if T is a t-norm and {xn} is a given sequence of numbers in [0, 1], T n
i=1

x i is

defined recurrently by T 1
i=1

x i = x1 and T n
i=1

x i = T (T n−1
i=1

x i, xn) for n ≥ 2. T∞
i=n

x i is

defined as T∞
i=1

xn+i. It is known( [12]) that for the Lukasiewicz t–norm the following

implication holds:

lim
n→∞
(TL)

∞
i=1

xn+i = 1⇐⇒
∞
∑

n=1

(1− xn)<∞ (4)

for all t ≥ 0.

Definition 2. ( [35]). A random normed space (briefly, RN–space) is a triple (X ,µ, T ),

where X is a vector space, T is a continuous t-norm, and µ is a mapping from X into D+

such that, the following conditions hold:

(RN1) µx(t) = ǫ0(t) for all t > 0 if and only if x = 0;

(RN2) µαx(t) = µx(
t

|α|
) for all x ∈ X , α 6= 0 and all t ≥ 0;

(RN3) µx+y(t + s) ≥ T (µx(t),µy(s)) for all x , y ∈ X and t , s ≥ 0.

Every normed space (X ,‖.‖) defines a random normed space (X ,µ, TM ) where

µx(t) =
t

t + ‖x‖
,

for all t > 0, and TM is the minimum t-norm. This space is called the induced random

normed space.

Definition 3. Let (X ,µ, T ) be an RN–space.

(1) A sequence {xn} in X is said to be convergent to x in X if, for every ε > 0 and λ > 0,

there exists positive integer N such that µxn−x(ε)> 1−λ whenever n≥ N.

(2) A sequence {xn} in X is called Cauchy sequence if, for every ε > 0 and λ > 0, there



M. Gordji, J. Rassias, and M. Savadkouhi / Eur. J. Pure Appl. Math, 2 (2009), (494-507) 498

exists positive integer N such that µxn−xm
(ε)> 1−λ whenever n ≥ m ≥ N.

(3) An RN–space (X ,µ, T ) is said to be complete if and only if every Cauchy sequence in

X is convergent to a point in X .

Theorem 1. ( [34]). If (X ,µ, T ) is an RN–space and {xn} is a sequence such that

xn→ x, then limn→∞µxn
(t) = µx(t) almost everywhere.

The aim of this paper is to investigate the stability of the quadratic and cubic

functional equations in random normed spaces (in the sense of Sherstnev), under

arbitrary continuous t–norms.

2. Main Results

In this section we establish the stability of the quadratic and cubic functional equa-

tion

f (2x + y) + f (2x − y)− 2 f (x + y)− 2 f (x − y)− 12 f (x) = 0 (5)

in the setting of random normed spaces.

Theorem 2. Let X be a real linear space, (Y,µ, T ) be a complete RN-space and f : X → Y

be a mapping with f (0) = 0 for which there is ρ : X × X → D+ ( ρ(x , y) is denoted by

ρx ,y ) with the property:

µ f (x+y)+ f (x−y)−2 f (x)−2 f (y)(t)≥ ρx ,y(t) (6)

for all x , y ∈ X and all t > 0. If

lim
n→∞

T∞
i=1
(ρ2n+i−1 x ,2n+i−1 x(2

2n+2i t)) = 1 (7)

and

lim
n→∞
ρ2n x ,2n y(2

2n t) = 1 (8)



M. Gordji, J. Rassias, and M. Savadkouhi / Eur. J. Pure Appl. Math, 2 (2009), (494-507) 499

for all x , y ∈ X and all t > 0, then there exists a unique quadratic mapping Q : X → Y

such that

µQ(x)− f (x)(t)≥ T∞
i=1
(ρ2i−1 x ,2i−1 x(2

2i t)). (9)

for all x ∈ X and all t > 0.

Proof. Putting y = x in (6), we get

µ f (2x)

22 − f (x)
(t)≥ ρx ,x(2

2 t). (10)

Therefore,

µ f (2k+1 x)

22(k+1)
− f (2k x)

22k

(t)≥ ρ2k x ,2k x(2
2(k+1) t), (11)

for all k ∈ N and all t > 0. By the triangle inequality it follows that

µ f (2n x)

22n − f (x)
(t)≥ T n−1

k=0
(µ f (2k+1 x)

22(k+1) −
f (2k x)

22k

(t))≥ T n−1
k=0
(ρ2k x ,2k x(2

2(k+1) t))

= T n
i=1
(ρ2i−1 x ,2i−1 x(2

2i t)) (12)

for all x ∈ X and all t > 0. In order to prove the convergence of the sequence { f (2n x)

22n },

we replace x with 2mx in (12) to find that

µ f (2n+m x)

22(n+m)
− f (2m x)

22m

(t) = µ f (2n+m x)

22n − f (2m x)
(22m t)

≥ T n
i=1
(ρ2i+m−1 x ,2i+m−1 x(2

2i+2m t)). (13)

Since the right hand side of the inequality tends to 1 as m and n tend to infinity, the se-

quence { f (2n x)

22n } is a Cauchy sequence. Therefore, we may define Q(x) = limn→∞
f (2n x)

22n

for all x ∈ X . Now, we show that Q is a quadratic function. Replacing x , y with 2n x

and 2n y, respectively, in (6), it follows that

µ f (2n x+2n y)

22n +
f (2n x−2n y)

22n −2
f (2n x)

22n −2
f (2n y)

22n
(t)≥ ρ2n x ,2n y(2

2n t). (14)

Taking the limit as n→∞, we find that Q satisfies (5) for all x , y ∈ X .

To prove (9), take the limit as n → ∞ in (12). Finally, to prove the uniqueness of



M. Gordji, J. Rassias, and M. Savadkouhi / Eur. J. Pure Appl. Math, 2 (2009), (494-507) 500

the quadratic function Q subject to (9), let us assume that there exists a quadratic

function Q′ which satisfies (9) Since Q(2nx) = 22nQ(x) and Q′(2nx) = 22nQ′(x) for

all x ∈ X and all n ∈ N, from (9) it follows that

µQ(x)−Q′(x)(2t) = µQ(2n x)−Q′(2n x)(2
2n+1 t)

≥ T (µQ(2n x)− f (2n x)(2
2n t),µ f (2n x)−Q′(2n x)(2

2n t))

≥ T (T∞
i=1
(ρ2n+i−1 x ,2n+i−1 x(2

2n+2i t)), T∞
i=1
(ρ2n+i−1 x ,2n+i−1(22n+2i t))) (15)

for all x ∈ X and all t > 0. By letting n→∞ in (15), we find that Q =Q′.

Theorem 3. Let X be a real linear space, (Y,µ, T ) be a complete RN-space and f : X → Y

be a mapping which there is τ : X × X → D+ ( τ(x , y) is denoted by τx ,y ) with the

property:

µ f (2x+y)+ f (2x−y)−2 f (x+y)−2 f (x−y)−12 f (x)(t)≥ τx ,y(t) (16)

for all x , y ∈ X and all t > 0. If

lim
n→∞

T∞
i=1
(τ2n+i−1 x ,0(2

3n+2i t)) = 1 (17)

and

lim
n→∞
τ2n x ,2n y(2

3n t) = 1 (18)

for all x , y ∈ X and all t > 0, then there exists a unique cubic mapping C : X → Y such

that

µC(x)− f (x)(t)≥ T∞
i=1
(τ2i−1 x ,0(2

2i t)). (19)

for all x ∈ X and all t > 0.

Proof. Putting y = 0 in (16), we get

µ f (2x)

23 − f (x)
(t)≥ τx ,0(2

4 t)≥ τx ,0(2
3 t). (20)

Therefore,

µ f (2k+1 x)

23(k+1)
− f (2k x)

23k

(t)≥ τ2k x ,0(2
3(k+1) t), (21)



M. Gordji, J. Rassias, and M. Savadkouhi / Eur. J. Pure Appl. Math, 2 (2009), (494-507) 501

for every k ∈ N and t > 0. Thus we have

µ f (2k+1 x)

23(k+1) −
f (2k x)

23k

(
t

2k+1
)≥ τ2k x ,0(2

2(k+1) t), (22)

for every k ∈ N and t > 0. As 1> 1

2
+ 1

22 + ...+ 1

2n , by the triangle inequality it follows

µ f (2n x)

23n − f (x)
(t)≥ T n−1

k=0
(µ f (2k+1 x)

23(k+1)
− f (2k x)

23k

(
t

2k+1
))≥ T n−1

k=0
(τ2k x ,0(2

2(k+1) t))

= T n
i=1
(τ2i−1 x ,0(2

2i t)) (23)

for all x ∈ x and all t > 0. In order to prove the convergence of the sequence { f (2n x)

23n },

we replace x with 2mx in (23) to find that

µ f (2n+m x)

23(n+m)
− f (2m x)

23m

(t) = µ f (2n+m x)

23n − f (2m x)
(23m t)

≥ T n
i=1
(τ2i+m−1 x ,0(2

2i+3m t)). (24)

Since the right hand side of the inequality tends to 1 as m and n tend to infinity, the se-

quence { f (2n x)

23n } is a Cauchy sequence. Therefore, we may define C(x) = limn→∞
f (2n x)

23n

for all x ∈ X . Now, we show that C is a cubic function. Replacing x , y with 2n x and

2n y respectively in (16), it follows that

µ f (2n+1 x+2n y)

23n +
f (2n+1x−2n y)

23n −2
f (2n x+2n y)

23n −2
f (2n x−2n y)

23n −12
f (2n x)

23n

(t)≥ τ2n x ,2n y(2
3n t). (25)

Taking the limit as n→∞, we find that C satisfies (25) for all x , y ∈ X .

To prove (19), take the limit as n→∞ in (23). Finally, to prove the uniqueness of the

cubic function C subject to (19), let us assume that there exists a cubic function C ′

which satisfies (19). Since C(2n x) = 23nC(x) and C ′(2n x) = 23nC ′(x) for all x ∈ X

and n ∈ N, from (19) it follows that

µC(x)−C ′(x)(2t) = µC(2n x)−C ′(2n x)(2
3n+1 t)

≥ T (µC(2n x)− f (2n x)(2
3n t),µ f (2n x)−C ′(2n x)(2

3n t))

≥ T (T∞
i=1
(τ2n+i−1 x ,0(2

3n+2i t)), T∞
i=1
(τ2n+i−1 x ,0(2

3n+2i t))), (26)

for all x ∈ X and all t > 0. By letting n→∞ in (26), we find that C = C ′.



M. Gordji, J. Rassias, and M. Savadkouhi / Eur. J. Pure Appl. Math, 2 (2009), (494-507) 502

Example 1. Let (A,‖.‖) be a Banach algebra and

µx(t) =







max{1− ‖x‖
t

, 0}, if t > 0,

0, if t ≤ 0,

for every x , y ∈ A. Let

ρx ,y(t) =max{1−
32‖x‖+ 32‖y‖

t
, 0} (t > 0)

and ρx ,y(t) = 0 if t ≤ 0. We note that ρx ,y is a distribution function and

lim
n→∞
ρ2n x ,2n y(2

2n t) = 1

for all x , y ∈ A and all t > 0. It is straightforward to show that (A,µ, TL) is an RN-space.

Indeed,

(∀t > 0 ; µx(t) = 1) =⇒

(∀t > 0 ;
‖x‖

t
= 0) =⇒ x = 0

and

µλx(t) = 1−
‖λx‖

t
= 1−

|λ|‖x‖

t
= 1−

‖x‖
t

λ

= µx(
t

λ
) (27)

for all x ∈ A and all t > 0. Also, for every x , y ∈ A and t , s > 0 we have

µx+y(t + s) =max{1−
‖x + y‖

t + s
, 0}=max{1−‖

x + y

t + s
‖, 0}

=max{1−‖
x

t + s
+

y

t + s
‖, 0}

≥max{1−‖
x

t
+

y

s
‖, 0}

≥max{1−‖
x

t
‖− ‖

y

s
‖, 0}

= TL(µx(t),µy(s)).

It is also easy to see that (A,µ, TL) is complete, for

µx−y(t)≥ 1−
‖x − y‖

t
; (x , y ∈ A, t > 0) (28)



M. Gordji, J. Rassias, and M. Savadkouhi / Eur. J. Pure Appl. Math, 2 (2009), (494-507) 503

and (A,‖.‖) is complete. Define f : A→ A, f (x) = ‖x‖x0, where x0 is a unit vector in A.

A simple computation shows that

‖ f (x + y) + f (x − y)− 2 f (x)− 2 f (y)‖

= ‖x + y‖+ ‖x − y‖ − 2‖x‖− 2‖y‖

≤ 32‖x‖+ 32‖y‖

for all x , y ∈ A, hence

µ f (x+y)+ f (x−y)−2 f (x)−2 f (y)(t)≥ ρx ,y(t), (29)

for all x , y ∈ A and all t > 0. Fix x ∈ A and t > 0, then

(TL)
∞
i=1
(ρ2n+i−1 x ,2n+i−1 x(2

2n+2i t)) =max{
∞
∑

i=1

(ρ2n+i−1 x ,2n+i−1 x(2
2n+2i t)− 1) + 1, 0}

=max{1−
32‖x‖

2nt
, 0},

hence limn→∞(TL)
∞
i=1
(ρ2n+i−1 x ,2n+i−1 x(2

2n+2i t)) = 1. Hence, all the conditions of Theorem

3 hold. Since

(TL)
∞
i=1
(ρ2i−1 x ,2i−1 x(2

2i t)) =max{
∞
∑

i=1

(ρ2i−1 x ,2i−1 x(2
2i t)− 1) + 1, 0}

=max{1−
32‖x‖

t
, 0},

we obtain that there exists a unique quadratic mapping Q : A−→ A such that

µQ(x)− f (x)(t)≥max{1−
32‖x‖

t
, 0} (30)

for all x ∈ A and all t > 0.

Let

τx ,y(t) =max{1−
64‖x‖+ 64‖y‖

t
, 0} (t > 0) (31)

and τx ,y(t) = 0 if t ≤ 0. We note that τx ,y is a distribution function and

lim
n→∞
τ2n x ,2n y(2

3n t) = 1 (32)



M. Gordji, J. Rassias, and M. Savadkouhi / Eur. J. Pure Appl. Math, 2 (2009), (494-507) 504

for all x , y ∈ A and all t > 0. It is obviously that (A,µ, TL) is an RN–space. It is also

easy to see that (A,µ, TL) is complete, for

µx−y(t)≥ 1−
‖x − y‖

t
(x , y ∈ A, t > 0) (33)

and (A,‖.‖) is complete. Define g : A→ A, g(x) = x3+ ‖x‖x0, where x0 is a unit vector

in A. A simple computation shows that

‖g(2x + y) + g(2x − y)− 2g(x + y)− 2g(x − y)− 12g(x)‖

≤ 64‖x‖+ 64‖y‖

for all x , y ∈ A, hence

µg(2x+y)+g(2x−y)−2g(x−y)−2g(x−y)−12g(x)(t)≥ τx ,y(t),

for all x , y ∈ A and all t > 0. Fix x ∈ A and t > 0, then

(TL)
∞
i=1
(τ2n+i−1 x ,0(2

3n+2i t)) =max{
∞
∑

i=1

(τ2n+i−1 x ,0(2
3n+2i t)− 1) + 1, 0}

=max{1−
32‖x‖

22n t
, 0},

hence limn→∞(TL)
∞
i=1
(τ2n+i−1 x ,0(2

3n+2i t)) = 1.

Thus, all the conditions of Theorem 3 hold. Since

(TL)
∞
i=1
(τ2i−1 x ,0(2

2i t)) =max{
∞
∑

i=1

(τ2i−1 x ,0(2
2i t)− 1) + 1, 0}

=max{1−
32‖x‖

t
, 0},

we obtain that there exists a unique cubic mapping C : A−→ A such that

µc(x)−g(x)(t)≥max{1−
32‖x‖

t
, 0} (34)

for all x ∈ A and all t > 0.



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