On the stability of an AQCQ-functional equation in random normed spaces

Abstract

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

[Equation not available: see fulltext.]

in random normed spaces.

2010 Mathematics Subject Classification: 46S40; 39B52; 54E70 [PUBLICATION ABSTRACT]

Park et al. Journal of Inequalities and Applications 2011, 2011:34 http://www.journalofinequalitiesandapplications.com/content/2011/1/34

* Correspondence: mailto:[email protected]

Web End [email protected]. mailto:[email protected]

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4Department of Mathematics, University of Seoul, Seoul 130-743, Republic of KoreaFull list of author information is available at the end of the article

RESEARCH Open Access

On the stability of an AQCQ-functional equation in random normed spaces

Choonkil Park1, Sun Young Jang2, Jung Rye Lee3 and Dong Yun Shin4*

AbstractIn this paper, we prove the Hyers-Ulam stability of the following additive-quadratic-cubic-quartic functional equationf (x + 2y) + f (x 2y) = 4f (x + y) + 4f (x y) 6f (x)

+ f (2y) + f (2y) 4f (y) 4f (y) in random normed spaces.

2010 Mathematics Subject Classification: 46S40; 39B52; 54E70

Keywords: random normed space, additive-quadratic-cubic-quartic functional equation, Hyers-Ulam stability

1. Introduction

The stability problem of functional equations originated from a question of Ulam [1] in 1940, concerning the stability of group homomorphisms. Let (G1, ) be a group and let (G2, *, d) be a metric group with the metric d( , ). Given > 0, does there exist a > 0 such that if a mapping h : G1 G2 satisfies the inequality d(h(xy), h(x) * h(y)) < for all x, y G1, then there exists a homomorphism H : G1 G2 with d(h(x), H(x)) < for all x G1? In the other words, under what condition does there exists a homomorphism near an approximate homomorphism? The concept of stability for functional equation arises when we replace the functional equation by an inequality which acts as a perturbation of the equation. Hyers [2] gave a first affirmative answer to the question of Ulam for Banach spaces. Let f : E E be a mapping between Banach spaces such that

f (x + y) f (x) f (y) for all x, y E and some > 0. Then, there exists a unique additive mapping T : E E such that

||f (x) T(x)|| for all x E. Moreover, if f(tx) is continuous in t for each fixed x E, then T is -linear. In 1978, Th.M. Rassias [3] provided a generalization of the Hyers theorem that allows the Cauchy difference to be unbounded. In 1991, Gajda [4] answered the question for the case p > 1, which was raised by Th.M. Rassias (see [5-11]).

2011 Park et al; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0

Web End =http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Park et al. Journal of Inequalities and Applications 2011, 2011:34 http://www.journalofinequalitiesandapplications.com/content/2011/1/34

On the other hand, in 1982-1998, J.M. Rassias generalized the Hyers stability result by presenting a weaker condition controlled by a product of different powers of norms.

Theorem 1.1. ([12-18]). Assume that there exist constants 0 and p1, p2 such

that p = p1 + p2 1, and f : E E is a mapping from a normed space E into a Banach space E such that the inequality

||f (x + y) f (x) f (y)|| ||x||p1||y||p2for all x, y E. Then, there exists a unique additive mapping T : E E such that

||f (x) L(x)||

for all E.

The control function ||x||p ||y||q + ||x||p+q + ||y||p+q was introduced by Rassias [19] and was used in several papers (see [20-25]).

The functional equation

f (x + y) + f (x y) = 2f (x) + 2f (y) (1:1) is related to a symmetric bi-additive mapping. It is natural that this equation is called a quadratic functional equation. In particular, every solution of the quadratic functional 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 exists a unique symmetric bi-additive mapping B such that f(x) = B(x, x) for all x (see [5,26]). The bi-additive mapping B is given by

B(x, y) = 14(f (x + y) f (x y)).

The Hyers-Ulam stability problem for the quadratic functional equation (1.1) was proved by Skof for mappings f : A B, where A is a normed space and B is a Banach space (see [27]). Cholewa [28] noticed that the theorem of Skof is still true if relevant domain A is replaced by an abelian group. In [29], Czerwik proved the Hyers-Ulam stability of the functional equation (1.1). Grabiec [30] has generalized these results mentioned above.

In [31], Jun and Kim considered the following cubic functional equation:

f (2x + y) + f (2x y) = 2f (x + y) + 2f (x y) + 12f (x). (1:2) It is easy to show that the function f(x) = x3 satisfies the functional equation (1.2), which is called a cubic functional equation and every solution of the cubic functional equation is said to be a cubic mapping.

In [32], Park and Bae considered the following quartic functional equation

f (x + 2y) + f (x 2y) = 4[f (x + y) + f (x y) + 6f (y)] 6f (x). (1:3) In fact, 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 symmetric multi-additive mapping M : X4 Y such that f(x) = M(x, x, x, x) for all x. It is easy to show that the function f(x) = x4 satisfies the functional equation (1.3), which is called a quartic functional equation (see also [33]). In addition, Kim [34] has obtained the Hyers-Ulam stability for a mixed type of quartic and quadratic functional equation.

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2 2p ||

x||p

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It should be noticed that in all these papers, the triangle inequality is expressed by using the strongest triangular norm TM .

The aim of this paper is to investigate the Hyers-Ulam stability of the additive-quadratic-cubic-quartic functional equation

f (x + 2y) + f (x 2y) = 4f (x + y) + 4f (x y) 6f (x)

+ f (2y) + f (2y) 4f (y) 4f (y)

(1:4)

in random normed spaces in the sense of Sherstnev under arbitrary continuous t-norms.

In the sequel, we adopt the usual terminology, notations and conventions of the theory of random normed spaces, as in [35-37]. Throughout this paper, + is the space of distribution functions, that is, the space of all mappings F : {-, } [0, 1] such that F is left-continuous and non-decreasing on , 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, lf (x) = limtx 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 . The maximal element for + in this order is the distribution function 0 given by

0(t) =

0, if t 0, 1, if t > 0.

Definition 1.2. [36]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];(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 [38,39]) that if T is a t-norm and {xn} is a given sequence of numbers in [0, 1], then Tni=1xi is defined recurrently by T1i=1xi = x1 and Tni=1xi = T(Tn1i=1xi, xn) for n 2. Ti=nxi is defined as Ti=1xn+i1. It is known [39] that for the Lukasiewicz t-norm, the following implication holds:

lim

n

(TL)i=1xn+i1 = 1

n=1(1 xn) <

Definition 1.3. [37]A random normed space (briefly, RN-space) is a triple (X, , T), where is a vector space, T is a continuous t-norm, and is a mapping from into D

+ such that the following conditions hold:

(RN1) x (t) = 0(t) for all t > 0 if and only if = 0; (RN2) x(t) = x( t||)for all X, a 0;

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

Every normed space (X, ||||) defines a random normed space (X, , TM),

where

x(t) = t

t + ||x||

Park et al. Journal of Inequalities and Applications 2011, 2011:34 http://www.journalofinequalitiesandapplications.com/content/2011/1/34

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

Definition 1.4. Let (X, , T) be an RN-space.(1) A sequence {xn} in is said to be convergent to in if, for every > 0 and l > 0, there exists a positive integer N such that xnx() > 1 whenever n N.

(2) A sequence {xn} in is called a Cauchy sequence if, for every > 0 and l > 0, there exists a positive integer N such that xnxm () > 1 whenever n m N.

(3) An RN-space (X, , T) is said to be complete if and only if every Cauchy sequence in is convergent to a point in X.

Theorem 1.5. [36]If (X, , T) is an RN-space and {xn} is a sequence such that xn x, then limn xn(t) = x(t)almost everywhere.

Recently, Eshaghi Gordji et al. establish the stability of cubic, quadratic and additive-quadratic functional equations in RN-spaces (see [40-42]).

One can easily show that an odd mapping f : X Y satisfies (1.4) if and only if the odd mapping f : X Y is an additive-cubic mapping, i.e.,

f (x + 2y) + f (x 2y) = 4f (x + y) + 4f (x y) 6f (x).

It was shown in [[43], Lemma 2.2] that g(x) := f (2x) - 8f (x) and h(x) := f (2x) - 2f (x) are additive and cubic, respectively, and that f (x) = 16h(x) 16g(x).

One can easily show that an even mapping f : X Y satisfies (1.4) if and only if the even mapping f : X Y is a quadratic-quartic mapping, i.e.,

f (x + 2y) + f (x 2y) = 4f (x + y) + 4f (x y) 6f (x) + 2f (2y) 8f (y).

It was shown in [[44], Lemma 2.1] that g (x) := f (2x) -16f (x) and h (x) := f (2x) -4f (x) are quadratic and quartic, respectively, and that f (x) = 1

12 h(x)

Lemma 1.6. Each mapping f : X Y satisfying (1.4) can be realized as the sum of an additive mapping, a quadratic mapping, a cubic mapping and a quartic mapping.

This paper is organized as follows: In Section 2, we prove the Hyers-Ulam stability of the additive-quadratic-cubic-quartic functional equation (1.4) in RN-spaces for an odd case. In Section 3, we prove the Hyers-Ulam stability of the additive-quadratic-cubic-quartic functional equation (1.4) in RN-spaces for an even case.

Throughout this paper, assume that X is a real vector space and that (X, , T) is a complete RN-space.

2.Hyers-Ulam stability of the functional equation (1.4): an odd mapping Case For a given mapping f : X Y , we define

Df (x, y) : = f (x + 2y) + f (x 2y) 4f (x + y) 4f (x y) + 6f (x)

f (2y) f (2y) + 4f (y) + 4f (y)for all x, y X.

In this section, we prove the Hyers-Ulam stability of the functional equation Df (x, y) = 0 in complete RN-spaces: an odd mapping case.

Theorem 2.1. Let f : X Y be an odd mapping for which there is a r : X 2 D+ (r

(x, y) is denoted by r x, y) such that

Df(x,y)(t) x,y(t) (2:1)

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112 g(x)

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for all x, y X and all t > 0. If

lim

n

Tk=1(T(2k+n1 x,2k+n1x(2n3t), 2k+n x,2k+n1x(2n1t))) = 1 (2:2)

and

lim

n

2nx,2ny(2nt) = 1 (2:3)

for all x, y X and all t > 0, then there exist a unique additive mapping A : X Y and a unique cubic mapping C : X Y such that

f(2x)8f(x)A(x)(t)

Tk=1 [parenleftbigg]

T

2k1x,2k1x[parenleftbigg]t 8

, 2k x,2k1x[parenleftbigg]t2[parenrightbigg][parenrightbigg][parenrightbigg], (2:4)

f(2x)2f(x)C(x)(t)

Tk=1 [parenleftbigg]

T

2k1x,2k1x[parenleftbigg]t 8

, 2k x,2k1x[parenleftbigg]t2 [parenrightbigg][parenrightbigg][parenrightbigg]

(2:5)

for all X and all t > 0.

Proof. Putting x = y in (2.1), we get

f(3y)4f(2y)+5f(y)(t) y,y(t) (2:6)

for all y X and all t > 0. Replacing x by 2y in (2.1), we get

f(4y)4f(3y)+6f(2y)4f(y)(t) 2y,y(t) (2:7)

for all y X and all t > 0. It follows from (2.6) and (2.7) that

f(4x)10f(2x)+16f(x)(t)= (4f(3x)16f(2x)+20f(x))+(f(4x)4f(3x)+6f(2x)4f(x))(t)

T

4f(3x)16f(2x)+20f(x)

t 2

, f(4x)4f(3x)+6f(2x)4f(x)

t 2

[parenrightbigg][parenrightbigg]

(2:8)

T

x,x[parenleftbigg]t 8

, 2x,x[parenleftbigg]t2 [parenrightbigg][parenrightbigg]

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

g(2x)2g(x)(t) T

x,x[parenleftbigg]t 8

, 2x,x[parenleftbigg]t2 [parenrightbigg][parenrightbigg]

for all x X and all t > 0. Thus, we have

g(2x)

2 g(x)

(t) T

x,x

t 4

, 2x,x (t)

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

g(2k+1x)

2k+1

g(2kx)

2k

(t) T(2

kx,2kx(2k2t), 2k+1 x,2kx(2kt))

Park et al. Journal of Inequalities and Applications 2011, 2011:34 http://www.journalofinequalitiesandapplications.com/content/2011/1/34

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for all x X, all t > 0 and all k N: From 1 > 12 +

1

22 + +

1

2n, it follows that

g(2nx)

2n g(x)

(t) Tnk=1 [parenleftBigg]

g(2kx)

2k

g(2k1x)

2k1

[parenleftbigg]

t 2k

[parenrightbigg][parenrightBigg]

(2:9)

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

nx) 2n

Tnk=1 [parenleftbigg]

T

2k1x,2k1x[parenleftbigg]t 8

, 2k x,2k1x[parenleftbigg]t2 [parenrightbigg][parenrightbigg][parenrightbigg]

},

replacing x with 2mx in (2.9), we obtain that

g(2n+mx) 2n+m

g(2mx)

2m

(t)

(2:10)

Since the right-hand side of the inequality (2.10) tends to 1 as m and n tend to infinity, the sequence {g(2

nx) 2n

Tnk=1(T(2

k+m1 x,2k+m1x(2m3t), 2k+m x,2k+m1x(2m1t))).

} is a Cauchy sequence. Thus, we may define

A(x) = limn g(2

nx)2n for all x X.

Now, we show that A is an additive mapping. Replacing x and y with 2nx and 2ny in(2.1), respectively, we get

Df (2nx,2ny)

2n

(t) 2

nx,2ny(2nt).

Taking the limit as n , we find that A : X Y satisfies (1.4) for all x, y X. Since f : X Y is odd, A : X Y is odd. By [[43], Lemma 2.2], the mapping A : X Y is additive. Letting the limit as n in (2.9), we get (2.4).

Next, we prove the uniqueness of the additive mapping A : X Y subject to (2.4). Let us assume that there exists another additive mapping L : X Y which satisfies(2.4). Since A(2nx) = 2nA(x), L(2nx) = 2nL(x) for all x X and all n N, from (2.4), it follows that

A(x)L(x)(2t) = A(2

nx)L(2nx)(2n+1t)

T(A(2

nx)g(2nx)(2nt), g(2

nx)L(2nx)(2nt))

T(Tk=1(T(2

n+k1 x,2n+k1x(2n3t), 2n+k x,2n+k1x(2n1t))), Tk=1(T(2n+k1 x,2n+k1x(2n3t), 2n+k x,2n+k1x(2n1t)))

(2:11)

for all x X and all t > 0. Letting n in (2.11), we conclude that A = L.

Let h : X Y be a mapping defined by h(x) := f (2x) -2f (x). Then, we conclude that

h(2x)8h(x)(t) T

x,x[parenleftbigg]t 8

, 2x,x[parenleftbigg]t2 [parenrightbigg][parenrightbigg]

for all x X and all t > 0. Thus, we have

h(2x)

8 h(x)

(t) T(x,x(t), 2x,x(4t))

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

h(2k+1x)

8k+1

h(2kx)

8k

(t) T(2

kx,2kx(8kt), 2k+1x,2kx(4 8kt))

Park et al. Journal of Inequalities and Applications 2011, 2011:34 http://www.journalofinequalitiesandapplications.com/content/2011/1/34

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for all x X, all t > 0 and all k N: From 1 > 18 + 182 + + 18n, it follows that

h(2nx)8n h(x)

(t) Tnk=1 [parenleftBigg]

h(2kx)

8k

h(2k1x)

8k1

[parenleftbigg]

t 8k

[parenrightbigg][parenrightBigg]

(2:12)

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

nx) 8n

Tnk=1 [parenleftbigg]

T

2k1x,2k1x[parenleftbigg]t 8

, 2kx,2k1x[parenleftbigg]t2 [parenrightbigg][parenrightbigg][parenrightbigg]

},

replacing x with 2mx in (2.12), we obtain that

h(2n+mx) 8n+m

h(2mx)

8m

(t)

(2:13)

Since the right-hand side of the inequality (2.13) tends to 1 as m and n tend to infinity, the sequence {h(2

nx) 8n

Tnk=1(T(2

k+m1 x,2k+m1x(8m1t), 2k+m x,2k+m1x(4 8m1t))).

} is a Cauchy sequence. Thus, we may define

C(x) = limn h(2

nx)8n for all x X.

Now, we show that C is a cubic mapping. Replacing x and y with 2nx and 2ny in(2.1), respectively, we get

Df (2nx,2ny)

8n

(t) 2

nx,2ny(8nt) 2

nx,2ny(2nt).

Taking the limit as n , we find that C : X Y satisfies (1.4) for all x, y X. Since f : X Y is odd, C : X Y is odd. By [[43], Lemma 2.2], the mapping C : X Y is cubic. Letting the limit as n in (2.12), we get (2.5).

Finally, we prove the uniqueness of the cubic mapping C : X Y subject to (2.5). Let us assume that there exists another cubic mapping L : X Y which satisfies (2.5). Since C(2nx) = 8nC(x), L(2nx) = 8nL(x) for all x X and all n N, from (2.5), it follows that

C(x)L(x)(2t)= C(2nx)L(2nx)(2 8nt)

T(C(2

nx)h(2nx)(8nt), h(2

nx)L(2nx)(8nt))

T(Tk=1(T(2

n+k1 x,2n+k1x(8n1t), 2n+k x,2n+k1x(4 8n1t))),

Tk=1(T(2n+k1 x,2n+k1x(8n1t), 2n+kx,2n+k1x(4 8n1t))) T(Tk=1(T(2

n+k1 x,2n+k1x(2n3t), 2n+k x,2n+k1x))), Tk=1(T(2n+k1 x,2n+k1x(2n3t), 2n+kx,2n+k1x(2n1t)))

(2:14)

for all x X and all t > 0. Letting n in (2.14), we conclude that C = L, as desired.

Similarly, one can obtain the following result.

Theorem 2.2. Let f : X Y be an odd mapping for which there is a r : X2 D+ (r

(x, y) is denoted by rx, y) satisfying (2.1). If

lim

n

Tk=1

T

x2k+n ,x2k+n[parenleftbigg]t 8n+2k

, x2k+n1 ,x2k+n[parenleftbigg]4t8n+2k[parenrightbigg][parenrightbigg][parenrightbigg]= 1

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and

lim

n

x

2n ,

y 2n

[parenleftbigg]

t 8n

[parenrightbigg]

= 1

for all x, y X and all t > 0, then there exist a unique additive mapping A : X Y and a unique cubic mapping C : X Y such that

f(2x)8f(x)A(x)(t) Tk=1 [parenleftbigg]

T

x2k ,x2k[parenleftbigg]t 22k+1

, x2k1 ,x2k[parenleftbigg]t22k1[parenrightbigg][parenrightbigg][parenrightbigg],

f(2x)2f(x)C(x)(t) Tk=1 [parenleftbigg]

, x2k1 ,x2k[parenleftbigg]4t82k [parenrightbigg][parenrightbigg][parenrightbigg]

T

x2k ,x2k[parenleftbigg]t 82k

for all X and all t > 0.

3. Hyers-ulam stability of the functional equation (1.4): an even mapping case

In this section, we prove the Hyers-Ulam stability of the functional equation D f (x, y) = 0 in complete RN-spaces: an even mapping case.

Theorem 3.1. Let f : X Y be an even mapping for which there is a r : X2 D+ (r

(x, y) is denoted by rx, y) satisfying f (0) = 0 and (2.1). If

lim

n

Tk=1(T(2k+n1x,2k+n1x(2 4n2t), 2

k+n x,2k+n1x(2 4n1t))) = 1 (3:1)

and

lim

n

2nx,2ny(4nt) = 1 (3:2)

for all x, y X and all t > 0, then there exist a unique quadratic mapping P : X Y and a unique quartic mapping Q : X Y such that

f(2x)16f(x)P(x)(t)

Tk=1 [parenleftbigg]

T

2k1x,2k1x[parenleftbigg]t 8

, 2k x,2k1x[parenleftbigg]t2[parenrightbigg][parenrightbigg][parenrightbigg], (3:3)

f(2x)4f(x)Q(x)(t)

Tk=1 [parenleftbigg]

T

2k1x,2k1x[parenleftbigg]t 8

, 2k x,2k1x[parenleftbigg]t2 [parenrightbigg][parenrightbigg][parenrightbigg]

(3:4)

for all X and all t > 0.

Proof. Putting x = y in (2.1), we get

f(3y)6f(2y)+15f(y)(t) y,y(t) (3:5)

for all y X and all t > 0. Replacing x by 2y in (2.1), we get

f(4y)4f(3y)+4f(2y)+4f(y)(t) 2y,y(t) (3:6)

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for all y X and all t > 0. It follows from (3.5) and (3.6) that

f(4x)20f(2x)+64f(x)(t)= (4f(3x)24f(2x)+60f(x))+(f(4x)4f(3x)+4f(2x)+4f(x))(t)

T

4f(3x)24f(2x)+60f(x)

t 2

, f(4x)4f(3x)+4f(2x)+4f(x)

t 2

[parenrightbigg][parenrightbigg]

(3:7)

T

x,x[parenleftbigg]t 8

, 2x,x[parenleftbigg]t2 [parenrightbigg][parenrightbigg]

for all x X and all t > 0. Let g : X Y be a mapping defined by g(x) := f (2x) - 16 f (x). Then we conclude that

g(2x)4g(x)(t) T

x,x[parenleftbigg]t 8

, 2x,x[parenleftbigg]t2 [parenrightbigg][parenrightbigg]

for all x X and all t > 0. Thus, we have

g(2x)4 g(x)

(t) T

x,x

t 2

[parenrightbigg]

, 2x,x (2t)

[parenrightbigg]

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

g(2k+1x) 4k+1

g(2kx)

4k

(t) T(2

kx,2kx(2 4k1t), 2

k+1 x,2kx(2 4kt))

for all x X, all t > 0 and all k N. From 1 > 14 +

142 + +

14n, it follows that

g(2nx)

4n g(x)

(t) Tnk=1 [parenleftBigg]

g(2kx)

4k

g(2k1x)

4k1

[parenleftbigg]

t 4k

[parenrightbigg][parenrightBigg]

(3:8)

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

nx) 4n

Tnk=1 [parenleftbigg]

T

2k1x,2k1x[parenleftbigg]t 8

, 2k x,2k1x[parenleftbigg]t2 [parenrightbigg][parenrightbigg][parenrightbigg]

},

replacing x with 2mx in (3.8), we obtain that

g(2n+mx)

4n+m

g(2mx)

4m

(t)

(3:9)

Since the right-hand side of the inequality (3.9) tends to 1 as m and n tend to infinity, the sequence {g(2

nx) 4n

Tnk=1(T(2

k+m1 x,2k+m1x(2 4m2t), 2

k+m x,2k+m1x(2 4m1t))).

} is a Cauchy sequence. Thus, we may define

P(x) = limn g(2

nx)4n for all x X.

Now, we show that P is a quadratic mapping. Replacing x and y with 2nx and 2ny in(2.1), respectively, we get

Df (2nx,2ny)

4n

(t) 2

nx,2ny(4nt).

Taking the limit as n , we find that P : X Y satisfies (1.4) for all x, y X. Since f : X Y is even, P : X Y is even. By [[44], Lemma 2.1], the mapping P : X Y is quadratic. Letting the limit as n in (3.8), we get (3.3).

Next, we prove the uniqueness of the quadratic mapping P : X Y subject to (3.3). Let us assume that there exists another quadratic mapping L : X Y, which satisfies

Park et al. Journal of Inequalities and Applications 2011, 2011:34 http://www.journalofinequalitiesandapplications.com/content/2011/1/34

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(3.3). Since P(2nx) = 4nP(x), L(2nx) = 4nL(x) for all x X and all n N, from (3.3), it follows that

P(x)L(x)(2t) = P(2

nx)L(2nx)(2 4nt)

T(P(2

nx)g(2nx)(4nt), g(2

nx)L(2nx)(4nt))

T(Tk=1(T(2

n+k1 x,2n+k1x(2 4n2t), 2

n+k x,2n+k1x(2 4n1t))), Tk=1(T(2n+k1x,2n+k1x(2 4n2t), 2

n+k x,2n+k1x(2 4n1t))))

(3:10)

for all x X and all t > 0. Letting n in (3.10), we conclude that P = L.

Let h : X Y be a mapping defined by h(x) := f (2x) -4f (x). Then, we conclude that

h(2x)16h(x)(t) T

x,x[parenleftbigg]t 8

, 2x,x[parenleftbigg]t2 [parenrightbigg][parenrightbigg]

for all x X and all t > 0. Thus, we have

h(2x)16 h(x)

(t) T(x,x(2t), 2x,x(8t))

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

h(2k+1x)

16k+1

h(2kx)

16k

(t) T(2

kx,2kx(2 16kt), 2

k+1 x,2kx(8 16kt))

for all x X, all t > 0 and all k N. From 1 > 1

16 +

1162 + +

116n, it follows that

h(2nx)

16n h(x)

(t) Tnk=1 [parenleftBigg]

h(2kx)

16k

h(2k1x)

16k1

[parenleftbigg]

t 16k

[parenrightbigg][parenrightBigg]

(3:11)

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

nx) 16n

Tnk=1 [parenleftbigg]

T

2k1x,2k1x[parenleftbigg]t 8

, 2kx,2k1x[parenleftbigg]t2 [parenrightbigg][parenrightbigg][parenrightbigg]

},

replacing x with 2mx in (3.11), we obtain that

h(2n+mx)

16n+m

h(2mx)

16m

(t)

(3:12)

Since the right-hand side of the inequality (3.12) tends to 1 as m and n tend to infinity, the sequence {h(2

nx) 16n

Tnk=1(T(2

k+m1x,2k+m1x(2 16m1t), 2

k+m x,2k+m1x(8 16m1t))).

} is a Cauchy sequence. Thus, we may define

Q(x) = limn h(2

nx)16n x X.

Now, we show that Q is a quartic mapping. Replacing x and y with 2nx and 2ny in(2.1), respectively, we get

Df (2nx,2ny)

16n

(t) 2

nx,2ny(16nt) 2

nx,2ny(4nt).

Taking the limit as n , we find that Q : X Y satisfies (1.4) for all x, y X. Since f : X Y is even, Q : X Y is even. By [[44], Lemma 2.1], the mapping Q : X Y is quartic. Letting the limit as n in (3.11), we get (3.4).

Finally, we prove the uniqueness of the quartic mapping Q : X Y subject to (3.4). Let us assume that there exists another quartic mapping L : X Y , which satisfies(3.4). Since Q(2nx) = 16nQ(x), L(2nx) = 16nL(x) for all x X and all n N, from (3.4),

Park et al. Journal of Inequalities and Applications 2011, 2011:34 http://www.journalofinequalitiesandapplications.com/content/2011/1/34

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it follows that

Q(x)L(x)(2t) = Q(2

nx)L(2nx)(2 16nt)

T(Q(2

nx)h(2nx)(16nt), h(2

nx)L(2nx)(16nt))

T(Tk=1(T(2

n+k1 x,2n+k1x(2 16n1t), 2

n+k x,2n+k1x(8 16n1t))),

Tk=1(T(2n+k1x,2n+k1x(2 16n1t), 2

n+k x,2n+k1x(8 16n1t)))

T(Tk=1(T(2

n+k1 x,2n+k1x(2 4n2t), 2

n+k x,2n+k1x(2 4n1t))), Tk=1(T(2n+k1x,2n+k1x(2 4n2t), 2

n+k x,2n+k1x(2 4n1t))))

(3:13)

for all x X and all t > 0. Letting n in (3.13), we conclude that Q = L, as desired.

Similarly, one can obtain the following result.

Theorem 3.2. Let f : X Y be an even mapping for which there is a r : X2 D+ (r

(x, y) is denoted by r x, y) satisfying f (0) = 0 and (2.1). If

lim

n

Tk=1

T

x2k+n ,x2k+n[parenleftbigg]2t 16n+2k

, x2k+n1 ,x2k+n[parenleftbigg]8t16n+2k[parenrightbigg][parenrightbigg][parenrightbigg]= 1

and

t16n ) = 1

for all x, y X and all t > 0, then there exist a unique quadratic mapping P : X Y and a unique quartic mapping Q : X Y such that

f(2x)16f(x)P(x)(t) Tk=1 [parenleftbigg]

T

lim

n

x

2n ,

y 2n (

x2k ,x2k[parenleftbigg]2t 42k+1

, x2k1 ,x2k[parenleftbigg]2t42k[parenrightbigg][parenrightbigg][parenrightbigg],

f(2x)4f(x)Q(x)(t) Tk=1 [parenleftbigg]

T

x2k ,x2k[parenleftbigg]2t162k[parenrightbigg], x2k1 ,x2k[parenleftbigg]8t162k [parenrightbigg][parenrightbigg][parenrightbigg]

for all X and all t > 0.

AcknowledgementsChoonkil Park, Jung Rye Lee and Dong Yun Shin were supported by Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technology (NRF-2009-0070788), (NRF-2010-0009232) and (NRF-2010-0021792), respectively. Sun Young Jang was supported by NRF Research Fund 2010-0013211 and has written during visiting the research Institute of Mathematics, Seoul National University.

Author details

1Department of Mathematics, Hanyang University, Seoul 133-791, Republic of Korea 2Department of Mathematics, University of Ulsan, Ulsan 680-749, Republic of Korea 3Department of Mathematics, Daejin University, Kyeonggi 487-711, Republic of Korea 4Department of Mathematics, University of Seoul, Seoul 130-743, Republic of Korea

Authors contributionsAll authors conceived of the study, participated in its design and coordination, drafted the manuscript, participated in the sequence alignment, and read and approved the final manuscript.

Competing interestsThe authors declare that they have no competing interests.

Received: 18 March 2011 Accepted: 18 August 2011 Published: 18 August 2011

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doi:10.1186/1029-242X-2011-34Cite this article as: Park et al.: On the stability of an AQCQ-functional equation in random normed spaces. Journal of Inequalities and Applications 2011 2011:34.

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