8_lee.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 5, No. 4, 2012, 540-553 ISSN 1307-5543 – www.ejpam.com Random Stability of a Functional Equation Related to an Inner Product Space Dong Yun Shin1, Jung Rye Lee2,∗, Choonkil Park3 1 Department of Mathematics, University of Seoul, Seoul 130-743, Korea 2 Department of Mathematics, Daejin University, Kyeonggi 487-711, Korea 3 Research Institute for Natural Sciences, Hanyang University, Seoul 133-791, Korea Abstract. In [14], Th.M. Rassias introduced the following equality n ∑ i, j=1 ‖x i − x j‖ 2 = 2n n ∑ i=1 ‖x i‖ 2, n ∑ i=1 x i = 0 for a fixed integer n ≥ 3. For a mapping f : X → Y , where X is a vector space and Y is a complete random normed space, we consider the following functional equation n ∑ i, j=1 f (x i − x j) = 2n n ∑ i=1 f (x i) (1) for all x1, . . . , xn ∈ X with ∑n i=1 x i = 0. In this paper, we prove the Hyers-Ulam stability of the func- tional equation (1) related to an inner product space. 2010 Mathematics Subject Classifications: 39B52, 46S50, 46C05, 47S50, 26E50. Key Words and Phrases: random normed space, Hyers-Ulam stability, quadratic functional equation, inner product space. 1. Introduction A square norm on an inner product space satisfies the parallelogram equality ‖x + y‖2 + ‖x − y‖2 = 2‖x‖2+ 2‖y‖2. From the above equation, we consider the following functional equation f (x + y) + f (x − y) = 2 f (x)+ 2 f (y) ∗Corresponding author. Email addresses: dyshin�uos.a .kr (D. Shin), jrlee�daejin.a .kr (J. Lee), baak�hanyang.a .kr (C. Park) http://www.ejpam.com 540 c© 2012 EJPAM All rights reserved. D. Shin, J. Lee, C. Park / Eur. J. Pure Appl. Math, 5 (2012), 540-553 541 related to an inner product space. The stability problem of functional equations originated from a question of S.M. Ulam [18] concerning the stability of group homomorphisms. D.H. Hyers [5] gave a first affirmative partial answer to the question of Ulam for Banach spaces and Hyers’ Theorem was generalized by Th.M. Rassias [13] for linear mappings by considering an unbounded Cauchy difference. Especially, the Hyers-Ulam stability of the above functional equation related to an inner product space has been studied [see 7, 17]. A square norm on an inner product space also satisfies 3 ∑ i, j=1 ‖x i − x j‖ 2 = 6 3 ∑ i=1 ‖x i‖ 2 for all x1, x2, x3 ∈ R with x1 + x2 + x3 = 0 [see 14]. From the above equality we can define the functional equation f (x − y) + f (2x + y) + f (x + 2y) = 3 f (x)+ 3 f (y) + 3 f (x + y), which is called a quadratic functional equation. In fact, f (x) = ax2 in R satisfies the quadratic functional equation. The aim of this paper is to investigate the Hyers-Ulam stability of additive-quadratic func- tional equation in a random normed space related to an inner product space. Throughout this paper, we use the definition of a random normed space as in [1, 10, 15, 16]. ∆+ is the space of distribution functions that is, the space of all mappings F : R ∪ {−∞,∞} → [0,1] which 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 . The space ∆+ is partially ordered by the usual point-wise ordering of functions. 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 ([15]). 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]. Recall that if T is a t-norm and {xn} is a sequence of numbers in [0,1], then 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 [see 3]. T∞ i=1 x i is defined as limm→∞ T m i=1 x i. D. Shin, J. Lee, C. Park / Eur. J. Pure Appl. Math, 5 (2012), 540-553 542 Definition 2 ([16]). 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+ satisfies the following conditions: (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; (RN3) µx+y(t + s)≥ T (µx(t),µy (s)) for all x , y ∈ X and t, s ≥ 0. A sequence {xn} in an RN-space (X ,µ, T ) is said to be convergent to x in X if, for every ε > 0 and λ > 0, there exists a positive integer N such that µxn−x(ε) > 1 − λ whenever n≥ N . 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 . The Hyers-Ulam stability of functional equations in random normed spaces and fuzzy normed spaces has been studied [see 3, 4, 6, 8, 9, 11, 12]. Let V,W be vector spaces. It is shown that if a mapping f : V → W satisfies the functional equation (1), then the mapping f is the sum of an additive mapping and a quadratic mapping [see 2]. In this paper, we investigate the Hyers-Ulam stability of the functional equation (1) in RN-spaces. Throughout this paper, assume that X is a vector space and that (Y,µ, T ) is a complete RN-space. 2. Hyers-Ulam Stability of the Functional Equation (1): An Odd Case We investigate the functional equation (1) for an odd mapping in RN-spaces. For a given mapping f : X → Y , we define D f (x1, . . . , xn) := n ∑ i, j=1 f (x i − x j)− 2n n ∑ i=1 f (x i) for all x1, . . . , xn ∈ X with ∑n i=1 x i = 0. For an odd mapping f : X → Y , we note that if f satisfies D f (x1, x2, . . . , xn) = 0 for all x1, . . . , xn ∈ X with ∑n i=1 x i = 0 then the mapping f is additive. We prove the Hyers-Ulam stability of the functional equation (1) of an odd mapping in RN-spaces. Theorem 1. Let f : X → Y be an odd mapping for which there is a ρ : X n→ D+ ( ρ(x1, x2, . . . , xn) is denoted by ρ(x1,x2,...,xn) ) such that µD f (x1,x2,...,xn) (t) ≥ ρ(x1,x2,...,xn) (t) (2) for all (x1, x2, . . . , xn) ∈ X n and all t > 0. If T∞k=1ρ � x 2k+l , x 2k+l ,− x 2k+l−1 ,0,...,0 � � nt 22k+l−2 � = 1 (3) D. Shin, J. Lee, C. Park / Eur. J. Pure Appl. Math, 5 (2012), 540-553 543 and lim m→∞ ρ� x 2m , y 2m ,− x+y 2m ,0,...,0 � � nt 2m−1 � = 1 (4) for all x , y ∈ X , all t > 0 and all l = 0,1,2, . . ., then there exists a unique additive mapping A : X → Y such that µ f (x)−A(x)(t)≥ T∞k=1ρ � x 2k , x 2k ,− x 2k−1 ,0,...,0 � � nt 22k−2 � (5) for all x ∈ X and all t > 0. Proof. Putting x1 = x2 = x 2 , x3 = −x , x4 = . . . = xn = 0 in (2), we get µ2n � f (x)−2 f � x 2 �� (t) ≥ ρ� x 2 , x 2 ,−x ,0,...,0 �(t) which is equivalent to µ f (x)−2 f � x 2 �(t) ≥ ρ� x 2 , x 2 ,−x ,0,...,0 �(2nt) for all x ∈ X and all t > 0. Replacing x and t by x 2k−1 and t 22k−1 , respectively in the above inequality, we get µ 2k−1 f � x 2k−1 � −2k f � x 2k � � t 2k � ≥ ρ� x 2k , x 2k ,− x 2k−1 ,0,...,0 � � nt 22k−2 � for all x ∈ X and all t > 0. Since µx(s)≤ µx(t) for all s and t with 0< s ≤ t, we obtain µ f (x)−2m f � x 2m �(t) =µ∑m k=1 � 2k−1 f � x 2k−1 � −2k f � x 2k ��(t) ≥µ∑m k=1 � 2k−1 f � x 2k−1 � −2k f � x 2k �� m ∑ k=1 t 2k ! ≥T m k=1 ρ� x 2k , x 2k ,− x 2k−1 ,0,...,0 � � nt 22k−2 � Replacing x by x 2l in the above inequality, we get µ f � x 2l � −2m f � x 2m+l �(t) ≥ T m k=1 ρ� x 2k+l , x 2k+l ,− x 2k+l−1 ,0,...,0 � � nt 22k−2 � which is equivalent to µ 2l f � x 2l � −2m+l f � x 2m+l �(t) ≥ T m k=1 ρ� x 2k+l , x 2k+l ,− x 2k+l−1 ,0,...,0 � � nt 22k+l−2 � (6) for all x ∈ X , all t > 0 and all l = 0,1,2, . . .. D. Shin, J. Lee, C. Park / Eur. J. Pure Appl. Math, 5 (2012), 540-553 544 Since the right hand side of the inequality (6) tends to 1 as m→∞ by (3), the sequence {2m f � x 2m � } is a Cauchy sequence. Thus we define A(x) := limm→∞ 2m f � x 2m � for all x ∈ X , which is an odd mapping. Now we show that A is an additive mapping. By (2), we get µ2m � f � x+y 2m � − f � x 2m � − f � y 2m �� (t) ≥ ρ� x 2m , y 2m ,− � x+y 2m � ,0,...,0 � � nt 2m−1 � . Taking the limit as m → ∞ in the above inequality, by (4), the mapping A is additive. By letting l = 0 and taking the limit as m→∞ in (6), we get (5). Finally, to prove the uniqueness of the additive mapping A subject to (5), let us assume that there exists another additive mapping B which satisfies (5). Since µA(x)−B(x)(2t) =µA(x)−2m f � x 2m � +2m f � x 2m � −B(x)(2t) ≥T � µA(x)−2m f � x 2m �(t),µ2m f � x 2m � −B(x)(t) � and lim m→∞ µA(x)−2m f � x 2m � = lim m→∞ µB(x)−2m f � x 2m � = 1 for all x ∈ X and all t > 0, we get lim m→∞ T � µA(x)−2m f � x 2m �(t),µ2m f � x 2m � −B(x)(t) � = 1. Thus we have A= B. Corollary 1. Let θ ≥ 0 and let p be a constant with p > 1. For a normed vector space X and complete RN-space Y , let f : X → Y be an odd mapping satisfying µD f (x1 ,x2,...,xn) (t) ≥ t t + θ ∑n i=1 ||x i||p for all (x1, x2, . . . , xn) ∈ X with ∑n i=1 x i = 0 and all t > 0. If T∞k=1 � 2(k+l)pnt 2(k+l)pnt + 22k+l−2(2+ 2p)θ ||x ||p � = 1 for all x ∈ X , all t > 0 and all l = 0,1,2, . . ., then there exists a unique additive mapping A : X → Y such that µ f (x)−A(x)(t) ≥ T∞k=1 � 2kpnt 2kpnt + 22k−2(2+ 2p)θ ||x ||p � for all x ∈ X and all t > 0. D. Shin, J. Lee, C. Park / Eur. J. Pure Appl. Math, 5 (2012), 540-553 545 Proof. If we define ρ(x1,x2,...,xn) (t) = t t + θ ∑n i=1 ||x i||p and apply Theorem 1, then we get the desired result. Theorem 2. Let f : X → Y be an odd mapping for which there is a ρ : X n→ D+ satisfying (2). If T∞k=1ρ(2k+l−2 x ,2k+l−2 x ,−2k+l−1 x ,0,...,0) � 2l+1nt � = 1 (7) and lim m→∞ ρ(2m x ,2m y,−2m(x+y),0,...,0) � 2m+1nt � = 1 (8) for all x , y ∈ X , all t > 0 and all l = 0,1,2, . . ., then there exists a unique additive mapping A : X → Y such that µ f (x)−A(x)(t) ≥ T∞k=1ρ(2k−2 x ,2k−2 x ,−2k−1 x ,0,...,0) (2nt) (9) for all x ∈ X and all t > 0. Proof. Putting x1 = x2 = x , x3 = −2x , x4 = . . . = xn = 0 in (2), we get µ2n( f (2x)−2 f (x)) (t) ≥ ρ(x ,x ,−2x ,0,...,0)(t) which is equivalent to µ f (x)− 1 2 f (2x)(t) ≥ ρ � x 2 , x 2 ,−x ,0,...,0 �(4nt) for all x ∈ X and all t > 0. Replacing x and t by 2k−1x and 2t, respectively, in the above inequality, we get µ 1 2k−1 f (2k−1 x)− 1 2k f (2k x) � t 2k � ≥ ρ(2k−2 x ,2k−2 x ,−2k−1 x ,0,...,0)(2nt) for all x ∈ X and all t > 0. Since µx(s)≤ µx(t) for all s and t with 0< s ≤ t, we obtain µ f (x)− 1 2m f (2m x)(t) =µ∑m k=1 � 1 2k−1 f (2k−1 x)− 1 2k f (2k x) �(t) ≥µ∑m k=1 � 1 2k−1 f (2k−1 x)− 1 2k f (2k x) � m ∑ k=1 t 2k ! ≥T m k=1 ρ(2k−2 x ,2k−2 x ,−2k−1 x ,0,...,0) (2nt) Replacing x by 2l x in the above inequality, we get µ f (2l x)− 1 2m f (2m+l x)(t) ≥ T m k=1ρ(2k+l−2 x ,2k+l−2 x ,−2k+l−1 x ,0,...,0) (2nt) D. Shin, J. Lee, C. Park / Eur. J. Pure Appl. Math, 5 (2012), 540-553 546 which is equivalent to µ 1 2l f (2l x)− 1 2m+l f (2m+l x)(t)≥ T m k=1ρ(2k+l−2 x ,2k+l−2 x ,−2k+l−1 x ,0,...,0) � 2l+1nt � (10) for all x ∈ X , all t > 0 and all l = 0,1,2, . . .. Since the right hand side of the inequality (10) tends to 1 as m→∞ by (7), the sequence { 1 2m f (2mx)} is a Cauchy sequence. Thus we define A(x) := limm→∞ 1 2m f (2mx) for all x ∈ X , which is an odd mapping. Now we show that A is an additive mapping. By (2), we get µ 1 2m ( f (2 m(x+y))− f (2m x)− f (2m y))(t) ≥ ρ(2m x ,2m y,−2m(x+y),0,...,0)(2 m+1nt). Taking the limit as m→∞ in the above inequality, by (8) the mapping A is additive. By letting l = 0 an taking the limit as m→∞ in (10), we get (9). The rest of the proof is the same as in the proof of Theorem 1. Corollary 2. Let θ ≥ 0 and let p be a constant with 0 < p < 1. For a normed vector space X and complete RN-space Y , let f : X → Y be an odd mapping satisfying µD f (x1 ,x2,...,xn) (t) ≥ t t + θ ∑n i=1 ||x i||p for all (x1, x2, . . . , xn) ∈ X with ∑n i=1 x i = 0 and all t > 0. If T∞k=1 � 2l+1nt 2l+1nt + 2(k+l−1)p(21−p + 1)θ ||x ||p � = 1 for all x ∈ X , all t > 0 and all l = 0,1,2, . . ., then there exists a unique additive mapping A : X → Y such that µ f (x)−A(x)(t) ≥ T∞k=1 � 2nt 2nt + 2(k−1)p(21−p + 1)θ ||x ||p � for all x ∈ X and all t > 0. Proof. If we define ρ(x1,x2,...,xn) (t) = t t + θ ∑n i=1 ||x i||p and apply Theorem 2, then we get the desired result. D. Shin, J. Lee, C. Park / Eur. J. Pure Appl. Math, 5 (2012), 540-553 547 3. Hyers-Ulam Stability of the Functional Equation (1): An Even Case We prove the Hyers-Ulam stability of the functional equation (1) of an even mapping in RN-spaces. For an even mapping f : X → Y with f (0) = 0, we note that if f satisfies D f (x1, x2, . . . , xn) = 0 for all x1, . . . , xn ∈ X with ∑n i=1 x i = 0 then the mapping f is quadratic. Theorem 3. Let f : X → Y be an even mapping with f (0) = 0 for which there is a ρ : X n→ D+ satisfying (2). If T∞k=1ρ � x 2k+l ,− x 2k+l ,0,...,0 � � t 23k+2l−3 � = 1 (11) and lim m→∞ ρ� x 2m , y 2m ,− x+y 2m ,0,...,0 � � t 22m−1 � = 1 (12) for all x , y ∈ X , all t > 0 and all l = 0,1,2, . . ., then there exists a unique quadratic mapping Q : X → Y such that µ f (x)−Q(x)(t) ≥ T∞k=1ρ � x 2k ,− x 2k ,0,...,0 � � t 23k−3 � (13) for all x ∈ X and all t > 0. Proof. Putting x1 = x , x2 = −x , x3 = . . . = xn = 0 in (2), we get µ2( f (2x)−4 f (x)) (t) ≥ ρ(x ,−x ,0,...,0)(t) which is equivalent to µ f (x)−4 f � x 2 � (t) ≥ ρ� x 2 ,− x 2 ,0,...,0 � (2t) for all x ∈ X and all t > 0. Replacing x and t by x 2k−1 and t 23k−2 , respectively in the above inequality, we get µ 4k−1 f � x 2k−1 � −4k f � x 2k � � t 2k � ≥ ρ� x 2k ,− x 2k 0,...,0 � � t 23k−3 � for all x ∈ X and all t > 0. Since µx(s)≤ µx(t) for all s and t with 0< s ≤ t, we obtain µ f (x)−4m f � x 2m �(t) =µ∑m k=1 � 4k−1 f � x 2k−1 � −4k f � x 2k ��(t) ≥µ∑m k=1 � 4k−1 f � x 2k−1 � −4k f � x 2k �� m ∑ k=1 t 2k ! ≥T m k=1ρ � x 2k ,− x 2k ,0,...,0 � � t 23k−3 � D. Shin, J. Lee, C. Park / Eur. J. Pure Appl. Math, 5 (2012), 540-553 548 Replacing x by x 2l in the above inequality, we get µ f � x 2l � −4m f � x 2m+l �(t) ≥ T m k=1ρ � x 2k+l ,− x 2k+l ,0,...,0 � � t 23k−3 � which is equivalent to µ 4l f � x 2l � −4m+l f � x 2m+l �(t) ≥ T m k=1 ρ� x 2k+l ,− x 2k+l ,0,...,0 � � t 23k+2l−3 � (14) for all x ∈ X , all t > 0 and all l = 0,1,2, . . .. Since the right hand side of the inequality (14) tends to 1 as m→∞ by (11), the sequence {4m f � x 2m � } is a Cauchy sequence. Thus we define Q(x) := limm→∞ 4m f � x 2m � for all x ∈ X , which is an even mapping. Now we show that Q is an quadratic mapping. By (2), we get µ 4m � f � x−y 2m � + f � 2x+y 2m � + f � x+2y 2m � −3 f � x+y 2m � −3 f � x 2m � −3 f � y 2m � � (t) ≥ρ� x 2m , y 2m ,− x+y 2m ,0,...,0 � � t 22m−1 � . Taking the limit as m → ∞ in the above inequality, by (12), the mapping Q is quadratic. Moreover, letting l = 0 and taking the limit as m→∞ in (14), we get (13). The rest of the proof is the same as in the proof of Theorem 1. Corollary 3. Let θ ≥ 0 and let p be a constant with p > 2. For a normed vector space X and complete RN-space Y , let f : X → Y be an even mapping satisfying µD f (x1 ,x2,...,xn) (t) ≥ t t + θ ∑n i=1 ||x i||p for all (x1, x2, . . . , xn) ∈ X with ∑n i=1 x i = 0 and all t > 0. If T∞k=1 � 2(k+l)p t 2(k+l)p t + 23k+2l−2θ ||x ||p � = 1 for all x ∈ X , all t > 0 and all l = 0,1,2, . . ., then there exists a unique quadratic mapping Q : X → Y such that µ f (x)−Q(x)(t) ≥ T∞k=1 � 2kp t 2kp t + 23k−2θ ||x ||p � for all x ∈ X and all t > 0. Proof. If we define ρ(x1,x2,...,xn) (t) = t t + θ ∑n i=1 ||x i||p and apply Theorem 3, then we get the desired result. D. Shin, J. Lee, C. Park / Eur. J. Pure Appl. Math, 5 (2012), 540-553 549 Theorem 4. Let f : X → Y be an even mapping with f (0) = 0 for which there is a ρ : X n→ D+ satisfying (2). If T∞k=1ρ(2k+l−1 x ,−2k+l−1 x ,0,...,0) � 2k+2l−1t � = 1 (15) and lim m→∞ ρ(2m x ,2m y,−2m(x+y),0,...,0) � 2m+1t � = 1 (16) for all x , y ∈ X , all t > 0 and all l = 0,1,2, . . ., then there exists a unique quadratic mapping Q : X → Y such that µ f (x)−Q(x)(t) ≥ T∞k=1ρ(2k x ,−2k x ,0,...,0) � 2k−1t � (17) for all x ∈ X and all t > 0. Proof. Letting x1 = x , x2 =−x , x3 = . . . = xn = 0 in (2), we get µ2( f (2x)−4 f (x)) (t) ≥ ρ(x ,−x ,0,...,0)(t) which is equivalent to µ f (x)− 1 4 f (2x) � t 4 � ≥ ρ(x ,−x ,0,...,0)(2t) for all x ∈ X and all t > 0. Replacing x and t by 2k−1 x and 2k−2t, respectively in the above inequality, we get µ 1 4k−1 f (2k−1 x)− 1 4k f (2k x) � t 2k � ≥ ρ(2k−1 x ,−2k−1 x ,0,...,0)(2 k−1t) for all x ∈ X and all t > 0. Since µx(s)≤ µx(t) for all s and t with 0< s ≤ t, we obtain µ f (x)− 1 4m f (2m x)(t) =µ∑m k=1 � 1 4k−1 f (2k−1 x)− 1 4k f (2k x) �(t) ≥µ∑m k=1 � 1 4k−1 f (2k−1 x)− 1 4k f (2k x) � m ∑ k=1 t 2k ! ≥T m k=1 ρ(2k−1 x ,−2k−1 x ,0,...,0) � 2k−1t � Replacing x by 2l x in the above inequality, we get µ f (2l x)− 1 4m f (2m+l x)(t) ≥ T m k=1ρ(2k+l−1 x ,−2k+l−1 x ,0,...,0) � 2k−1t � which is equivalent to µ 1 4l f (2l x)− 1 4m+l f (2m+l x) (t) ≥ T m k=1ρ(2k+l−1 x ,−2k+l−1 x ,0,...,0) � 2k+2l−1t � (18) for all x ∈ X , all t > 0 and all l = 0,1,2, . . .. D. Shin, J. Lee, C. Park / Eur. J. Pure Appl. Math, 5 (2012), 540-553 550 Since the right hand side of the inequality (18) tends to 1 as m→∞ by (15), the sequence { 1 4m f (2mx)} is a Cauchy sequence. Thus we define Q(x) := limm→∞ 1 4m f (2mx) for all x ∈ X , which is an even mapping. Now we show that Q is a quadratic mapping. By (2), we get µ 1 4m ( f (2 m(x−y))+ f (2m(2x+y))+ f (2m(x+2y))−3 f (2m(x+y))−3 f (2m x)−3 f (2m y))(t) ≥ρ(2m x ,2m y,−2m(x+y),0,...,0)(2 m+1t). Taking the limit as m → ∞ in the above inequality, by (16), the mapping Q is quadratic. Moreover, letting l = 0 and taking the limit as m→∞ in (18), we get (17). The rest of the proof is the same as in the proof of Theorem 3. Corollary 4. Let θ ≥ 0 and let p be a constant with 0 < p < 2. For a normed vector space X and complete RN-space Y , let f : X → Y be an even mapping satisfying µD f (x1 ,x2,...,xn) (t) ≥ t t + θ ∑n i=1 ||x i||p for all (x1, x2, . . . , xn) ∈ X with ∑n i=1 x i = 0 and all t > 0. If T∞k=1 � 2k+2l−2t 2k+2l−2t + 2(k+l)pθ ||x ||p � = 1 for all x ∈ X , all t > 0 and all l = 0,1,2, . . ., then there exists a unique quadratic mapping Q : X → Y such that µ f (x)−Q(x)(t)≥ lim m→∞ T m k=1 � 2k−2t 2k−2t + 2kpθ ||x ||p � for all x ∈ X and all t > 0. Proof. If we define ρ(x1,x2,...,xn) (t) = t t + θ ∑n i=1 ||x i||p and apply Theorem 4, then we get the desired result. 4. Hyers-Ulam Stability of the Functional Equation (1) We note that if a mapping f : X → Y satisfies the functional equation (1), then the mapping f is realized as the sum of an additive mapping and a quadratic mapping [see 2, Lemma 2.1]. Here, we let g(x) := 1 2 ( f (x)− f (−x)) and h(x) := 1 2 ( f (x) + f (−x)) for all x ∈ X . Then g(x) is an odd mapping and h(x) is an even mapping satisfying f (x) = g(x)+h(x). Moreover, we get the following: Dg(x1, x2, . . . , xn) = 1 2 {D f (x1, x2, . . . , xn)− D f (−x1,−x2, . . . ,−xn)} D. Shin, J. Lee, C. Park / Eur. J. Pure Appl. Math, 5 (2012), 540-553 551 Dh(x1, x2, . . . , xn) = 1 2 {D f (x1, x2, . . . , xn) + D f (−x1,−x2, . . . ,−xn)} for all x1, x2, . . . , xn ∈ X . Note that D f (x1, . . . , xn) = 0 implies that Dg(x1, . . . , xn) = 0 and Dh(x1, . . . , xn) = 0. Theorem 5. Let f : X → Y be a mapping with f (0) = 0 for which there is a ρ : X n→ D+ such that µD f (x1,x2,...,xn)+D f (−x1,−x2,...,−xn) (2t) ≥ ρ(x1,x2,...,xn) (t) (19) µD f (x1,x2,...,xn)−D f (−x1,−x2,...,−xn) (2t) ≥ ρ(x1,x2,...,xn) (t) (20) for all (x1, x2, . . . , xn) ∈ X n and all t > 0. If ρ satisfies (3), (11) and (12), then there exist an additive mapping A : X → Y and a quadratic mapping Q : X → Y such that µ f (x)−A(x)−Q(x)(2t) ≥T � T∞k=1ρ � x 2k , x 2k ,− x 2k−1 ,0,...,0 � � nt 22k−2 � , T∞k=1ρ � x 2k ,− x 2k ,0,...,0 � � t 23k−3 � � for all x ∈ X and all t > 0. Proof. Consider an odd mapping g(x) := 1 2 ( f (x)− f (−x)) and an even mapping h(x) := 1 2 ( f (x) + f (−x)) for all x ∈ X with f (x) = g(x) + h(x). By Theorem 1, there exists a unique additive mapping A : X → Y such that µg(x)−A(x)(t) ≥ T∞k=1ρ � x 2k , x 2k ,− x 2k−1 ,0,...,0 � � nt 22k−3 � for all x ∈ X and all t > 0. And by Theorem 3, there exists a unique quadratic mapping Q : X → Y such that µh(x)−Q(x)(t) ≥ T∞k=1ρ � x 2k ,− x 2k ,0,...,0 � � t 23k−3 � for all x ∈ X and all t > 0. Since f (x) = g(x)+ h(x), we obtain µ f (x)−A(x)−Q(x) (2t) = µg(x)−A(x)+h(x)−Q(x)(2t) ≥ T (µg(x)−A(x)(t),µh(x)−Q(x)(t)) ≥ T � T∞k=1ρ � x 2k , x 2k ,− x 2k−1 ,0,...,0 � � nt 22k−2 � , T∞k=1ρ � x 2k ,− x 2k ,0,...,0 � � t 23k−3 � � for all x ∈ X and all t > 0, as desired. Similarly, we can obtain the following. We will omit the proof. Theorem 6. Let f : X → Y be a mapping with f (0) = 0 for which there is a ρ : X n → D+ satisfying (19) and (20). If ρ satisfies (7), (15) and (16), then there exist an additive mapping A : X → Y and a quadratic mapping Q : X → Y such that µ f (x)−A(x)−Q(x)(2t) ≥T � T∞k=1ρ(2k−2 x ,2k−2 x ,−2k−1 x ,0,...,0)(2nt), T∞k=1ρ(2k x ,−2k x ,0,...,0) � 2k−1t � � for all x ∈ X and all t > 0. REFERENCES 552 ACKNOWLEDGEMENTS D. Y. 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