6_677_rassias.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 4, No. 1, 2011, 50-58 ISSN 1307-5543 – www.ejpam.com Generalised Hyers-Ulam Product-Sum Stability of a Cauchy Type Additive Functional Equation Matina J. Rassias Department of Statistics, University of Glasgow, Mathematics Building, Office No. 208, University Gardens, Glasgow G12 8QW, U.K. Abstract. In 1940 (and 1964) S.M. Ulam proposed the well-known Ulam stability problem. In 1941 D.H. Hyers solved the Hyers-Ulam problem for linear mappings. In 2008, J. M. Rassias introduced the generalised Hyers-Ulam “product-sum” stability. In this paper we introduce a Cauchy type ad- ditive functional equation and investigate the generalised Hyers-Ulam “product-sum” stability of this equation. 2000 Mathematics Subject Classifications: Primary 39B. Secondary 26D. Key Words and Phrases: Generalised “product-sum” Hyers-Ulam stability, Cauchy type additive func- tional equation. 1. Introduction and Preliminaries In 1940 (and 1964) Stanislaw M. Ulam [9] proposed the following stability problem, well-known as Ulam stability problem: “When is true that by slightly changing the hypotheses of a theorem one can still assert that the thesis of the theorem remains true or approximately true?” In particular he stated the stability question: “Let G1 be a group and G2 a metric group with the metric ρ(., .). Given a constant δ > 0, does there exist a constant c > 0 such that if a mapping f : G1 → G2 satisfies ρ( f (x y), f (x) f (y)) < c for all x , y ∈ G1, then a unique homomorphism h : G1→ G2 exists with ρ( f (x),h(x))< δ for all x ∈ G1 ?” In 1941 D.H. Hyers [2] solved this problem for linear mappings as follows: Email address: rassias.matina�gmail. om http://www.ejpam.com 50 c© 2010 EJPAM All rights reserved. M. Rassias / Eur. J. Pure Appl. Math, 4 (2011), 50-58 51 Theorem 1 (D.H. Hyers, 1941: [2]). If a mapping f : E → E′ satisfies the approximately additive inequality || f (x + y)− f (x)− f (y)|| ≤ ǫ, for some fixed ǫ > 0 and all x , y ∈ E, where E and E′ are Banach spaces, then there exists a unique additive mapping A : E→ E′, satisfying the formula A(x) = lim n→∞2−n f (2n x), and inequality || f (x)− A(x)|| ≤ ǫ for some fixed ǫ > 0 and all x ∈ E. No continuity conditions are required for this result. Theorem 2 (T. Aoki, 1950: [1]). Let f : E → E′ be a mapping from a normed vector space E into a Banach space E′ subject to the inequality || f (x + y)− f (x)− f (y)|| ≤ ǫ(||x ||p+ ||y||p), (1) for all x , y ∈ E, where ǫ > 0 and p < 1 constants. Then the limit A(x) = lim n→∞2−n f (2n x), exists for all x ∈ E and A : E→ E′ is the unique additive mapping which satisfies || f (x)− A(x)|| ≤ 2ǫ 2− 2p ||x ||p (2) for all x ∈ E. If p < 0 then the inequality (1) holds for x , y 6= 0 and (2) for x 6= 0. Theorem 3 (Th. M. Rassias, 1978: [6]). Let f : E → E′ be a mapping from a normed vector space E into a Banach space E′ subject to the inequality || f (x + y)− f (x)− f (y)|| ≤ ǫ(||x ||p+ ||y||p), (3) for all x , y ∈ E, where ǫ > 0 and p < 1 constants. Then the limit A(x) = lim n→∞2−n f (2n x), exists for all x ∈ E and A : E→ E′ is the unique additive mapping which satisfies || f (x)− A(x)|| ≤ 2ǫ 2− 2p ||x ||p (4) for all x ∈ E. If p < 0 then the inequality (3) holds for x , y 6= 0 and (4) for x 6= 0. If, moreover, f (t x) is continuous in t ∈ R for each fixed x ∈ E, then A(t x) = tA(x) for all x ∈ E and t ∈ R. A : E→ E′ is a unique linear additive mapping satisfying equation A(x + y) = A(x)+ A(y). M. Rassias / Eur. J. Pure Appl. Math, 4 (2011), 50-58 52 Theorem 4 (J. M. Rassias, 1982-1989: [3, 4, 5]). Let X be a real normed linear space and Y a real Banach space. Assume that f : X → Y is a mapping for which there exist constants θ ≥ 0 and p,q ∈ R such that r = p+ q 6= 1 and f satisfies the functional inequality || f (x + y)− f (x)− f (y)|| ≤ θ ||x ||p||y||q, for all x , y ∈ X . Then the limit A(x) = lim n→∞2−n f (2n x), exists for all x ∈ X and A : X → Y is the unique additive mapping which satisfies || f (x)− A(x)|| ≤ θ |2r − 2| ||x || r for all x ∈ X . If, moreover, f (t x) is continuous in t ∈ R for each fixed x ∈ X , then A(t x) = tA(x) for all x ∈ X and t ∈ R. A : X → Y is a unique linear additive mapping satisfying equation A(x + y) = A(x)+ A(y). For the theorem that follows, let (E,⊥) denote an orthogonality normed space with norm ||.||E and (F, ||.||F ) is a Banach space. Theorem 5 (Ravi, K., Arunkumar, M. and Rassias, J. M., 2008: [7]). Let f : E → F be a mapping which satisfies the inequality || f (mx + y) + f (mx − y)− 2 f (x + y)− 2 f (x − y)− 2(m2− 2) f (x)+ 2 f (y)||F ≤ ǫ�||x ||pE||y||pE + �||x ||2p E + ||y||2p E � (5) for all x , y ∈ E with x ⊥ y, where ǫ and p are constants with ǫ, p > 0 and either m > 1; p < 1 or m< 1; p > 1 with m 6= 0; m 6= ±1; m 6= ±p2 and −1 6= |m|p−1 < 1. Then the limit Q(x) = lim n→∞ f (mn x) m2n exists for all x ∈ E and Q : E→ F is the unique orthogonally Euler-Lagrange quadratic mapping such that || f (x)−Q(x)||F ≤ ǫ 2|m2 −m2p| ||x || 2p E for all x ∈ E. Note that the mixed type product-sum function (x , y)→ ǫ�||x ||pE||y||pE + �||x ||2p E + ||y||2p E �� was introduced by J. M. Rassias ([7, 8]). In this paper we introduce a Cauchy type additive functional equation and investigate the generalised Hyers-Ulam “product-sum” stability of this equation. M. Rassias / Eur. J. Pure Appl. Math, 4 (2011), 50-58 53 2. Cauchy Type Additive Functional Equation Let X be a real normed linear space and Y a real Banach space. Definition 1. A mapping f : X → Y is called approximately Cauchy type additive, if the approx- imately Cauchy additive functional inequality || f (x + y) + f (x − y) + f (y − x)− f (x)− f (y)|| ≤ ǫ�||x || α2 ||y|| α2 + ||x ||α+ ||y||α� (6) holds for every x , y ∈ X with ǫ ≥ 0 and α 6= 1. Lemma 1. Mapping A : X → Y satisfies the Cauchy-type additive equation A(x + y) + A(x − y) + A(y − x) = A(x)+ A(y) for all x , y ∈ X if and only if there exists a mapping C : X → Y satisfying the Cauchy additive equation C(x + y) = C(x)+ C(y) for all x , y ∈ X such that A(x) = C(x) for all x ∈ X . Proof. (⇒) Let mapping A : X → Y satisfy the Cauchy-type additive equation A(x + y) + A(x − y) + A(y − x) = A(x)+ A(y) (7) for all x , y ∈ X . Assume that there exists a mapping C : X → Y such that A(x) = C(x) for all x ∈ X . Observe that for x = y = 0 and x = x , y = x from (7) we obtain respectively C(0) = A(0) = 0 and C(−x) = A(−x) = −A(x) = −C(x), for x ∈ X . (8) From (7) and (8) it is obvious that C(x + y) + C(x − y) + C(y − x) = C(x)+ C(y), or C(x + y) + C(x − y) + C(−(x − y)) = C(x)+ C(y), or C(x + y) = C(x)+ C(y). Hence, C satisfies the Cauchy additive equation. (⇐) Let mapping C : X → Y satisfy the Cauchy additive equation C(x + y) = C(x)+ C(y) (9) for all x , y ∈ X . Assume that there exists a mapping A : X → Y such that A(x) = C(x) for all x ∈ X . Observe that for x = y = 0, from (9) we obtain A(0) = C(0) = 0. (10) M. Rassias / Eur. J. Pure Appl. Math, 4 (2011), 50-58 54 Thus, from (9) and (10) one gets A(x)+ A(y) = C(x)+ C(y) = C(x + y) = A(x + y) = A(x + y) + A(0) = A(x + y) + A((x − y) + (y − x)) = A(x + y) + A(x − y) + A(y − x). Hence, A satisfies the Cauchy type additive equation. Thus the proof of Lemma 1 is complete. Theorem 6. Assume that f : X → Y is an approximately Cauchy type additive mapping satisfy- ing (6). Then, there exists a unique Cauchy type additive mapping A : X → Y which satisfies the formula A(x) = lim n→∞ fn(x), where fn(x) = �2−n f (2n x), −∞<α<1 2n f (2−n x), α>1 for all x ∈ X and n ∈ N = {0,1,2, . . .}, which is the set of natural numbers and || f (x)− A(x)|| ≤ 3ǫ |2− 2α| ||x || α for some fixed ǫ > 0, α 6= 1 and all x ∈ X . If, moreover, f (t x) is continuous in t ∈ R for each fixed x ∈ X , then A(t x) = tA(x) for all t ∈ R and x ∈ X . A : X → Y is a unique linear Cauchy type additive mapping satisfying equation A(x + y) + A(x − y) + A(y − x) = A(x)+ A(y). (11) Proof. We start our proof considering: −∞< α < 1. Step 1 By substituting x = y = 0 and x = y in (6), respectively, we can observe that f (0) = 0 and || f (x)− 2−1 f (2x)|| ≤ 3 2 ǫ||x ||α. Hence, for n ∈ N − {0} || f (x)− 2−n f (2n x)|| ≤ || f (x)− 2−1 f (2x)||+ ||2−1 f (2x)− 2−2 f (22 x)||+ . . . + ||2−(n−1) f (2n−1 x)− 2−n f (2nx)|| ≤ 3 2 (1+ 2α−1+ ...+ 2(n−1)(α−1))ǫ||x ||α = 3 2− 2α (1− 2n(α−1))ǫ||x ||α. M. Rassias / Eur. J. Pure Appl. Math, 4 (2011), 50-58 55 Thus, || f (x)− 2−n f (2nx)|| ≤ 3 2− 2α (1− 2n(α−1))ǫ||x ||α, for n ∈ N − {0} and −∞< α < 1. Step 2 Following, we need to show that if there is a sequence { fn} : fn(x) = 2−n f (2nx), then { fn} converges. For every n> m> 0, we can obtain || fn(x)− fm(x)|| = ||2−n f (2nx)− 2−m f (2mx)|| = 2−m|| f (2mx)− 2−(n−m) f (2(n−m)2mx)|| ≤ 2−m 3ǫ 2− 2α (1− 2(n−m)(α−1))||x ||α < 2−m 3ǫ 2− 2α ||x ||α→ 0, for m → ∞, as α < 1. Therefore, { fn} is a Cauchy sequence. Since Y is complete we can conclude that { fn} is convergent. Thus, there is a well-defined A : X → Y such that A(x) = limn→∞ 2−n f (2nx), for −∞< α < 1. Step 3 Observe that || f (x)− fn(x)||= || f (x)− 2−n f (2nx)|| ≤ 3ǫ 2− 2α (1− 2n(α−1))||x ||α, from which by letting n→∞ we obtain || f (x)− A(x)|| ≤ 3ǫ 2− 2α ||x ||α. (12) Step 4 Claim that mapping A : X → Y satisfies (11). In fact, by letting x → 2nx and y → 2n y, from (6), we have: || f �2n(x + y) � + f � 2n(x − y) � + f � 2n(y − x) �− f � 2n x �− f � 2n y �|| ≤ ǫ�||2nx || α2 ||2n y|| α2 + ||2nx ||α+ ||2n y||α�. Next, by multiplying with 2−n we obtain 0 ≤ ||2−n f � 2n(x + y) � + 2−n f � 2n(x − y) � + 2−n f � 2n(y − x) �− 2−n f � 2n x �− 2−n f � 2n y �|| ≤ 2n(α−1)ǫ �||x || α2 ||y|| α2 + ||x ||α+ ||y||α� and by letting n→∞, for −∞ < α < 1 we can conclude that an A : X → Y truly exists such that: A(x) = limn→∞ 2−n f (2nx) satisfies the Cauchy-type additivity property A(x + y) +A(x − y) + A(y − x) = A(x)+ A(y). (13) Therefore, existence of Theorem holds. M. Rassias / Eur. J. Pure Appl. Math, 4 (2011), 50-58 56 Step 5 We need to prove that A is unique. Observe, from (13), that A(0) = 0 and A(2x) = 2A(x). Therefore, by induction we can show that A(2nx) = 2A(2n−1x) = 2nA(x) or equivalently A(x) = 2−nA(2nx). (14) Assume, now, the existence of another A′ : X → Y, such that A′(x) = 2−nA′(2n x). With the aid of the (12)-(14) and the triangular inequality, one gets 0≤ ||A(x)− A′(x)|| = ||2−nA(2nx)− 2−nA′(2nx)|| ≤ ||2−nA(2nx)− 2−n f (2n x)||+ ||2−n f (2nx)− 2−nA′(2nx)|| ≤ 2n(α−1) 3ǫ 2− 2α ||x ||α → 0, as n → ∞, (−∞ < α < 1). Thus, the uniqueness of A is proved and the stability of Cauchy-type additive mapping A : X → Y is established. Step 6 To complete the proof of Theorem 6, we only need to examine whether A : X → Y is a linear Cauchy-type mapping. To be more precise, we need to show that: (1) A(x + y) + A(x − y) + A(y − x) = A(x)+ A(y), and (2) A(r x) = rA(x), ∀r ∈ R. Recall that we have shown already that (1) holds. Therefore, we only need to show that (2) is valid ∀r ∈ R. For that we will study four cases. Case 1: Let r = k ∈ N = {0,1,2, . . .}. For k = 0, from (2), we have A(0) = 0. This is verified if we substitute x = y = 0 in (13). Assume, that A � (k− 1)x � = (k− 1)A(x) is true ∀k. Then, we need to prove that A(kx) = kA(x). Note that for x = x , and y = 0 from (13), we can easily obtain A(−x) = (−1)A(x). Let x = x and y = (k− 1)x in (13). Then, A(kx)+ A �− (k− 2)x � + A � (k− 2)x � = A(x)+ A � (k− 1)x � , or A(kx) = kA(x), ∀k ∈ N = {0,1,2, . . .}. M. Rassias / Eur. J. Pure Appl. Math, 4 (2011), 50-58 57 Case 2: Let r = k ∈ Z . We only need to observe that A is odd. Since, we have already proved that (2) is valid ∀k ∈ N = {0,1,2, . . .} we can then conclude that A(kx) = kA(x), ∀k ∈ Z . Case 3: Let r = k l ∈Q, for k ∈ Z , l ∈ Z − {0}. Then, A(x) = A � l 1 l x � = lA �1 l x � , for l ∈ Z − {0}. Hence, A �1 l x � = 1 l A(x). Besides, for k ∈ Z , A � k l x � = A � k 1 l x � = kA(1 l x), from Case 2. Thus, A � k l x � = k l A(x), or A(r x) = rA(x) for r ∈Q. Case 4: Let r ∈ R, where r = qn : rational numbers. Since R is a complete space, every sequence {qn} converges in R, i.e. limn→∞ qn = q ∈ R. Recall that A(x) = limn→∞ 2−n f (2n x) and f (t x) is continuous in t for each fixed x in X . Therefore, A(t x) is continuous in t for each fixed x in X . Besides, lim n→∞A(qnx) = A � lim n→∞qn x � = A(qx) (15) and lim n→∞A(qnx) = lim n→∞qnA(x) = qA(x). (16) From (15) and (16) Case 4. is now proved, which completes Step 6. and thus the proof of our Theorem 6 for the case of −∞ < α < 1. The proof for the case of α > 1 is similar to the proof for −∞< α < 1. In fact, we can find the general inequality || f (x)− 2n f (2−nx)|| ≤ 3ǫ 2α− 2 (1− 2n(1−α))||x ||α, (17) for all n ∈ N −{0}. Thus from this inequality (17) and the formula A(x) = lim n→∞2n f (2−n x), for n→∞, we get the inequality || f (x)−A(x)|| ≤ 3ǫ 2α− 2 ||x ||α, for α > 1. The rest of the proof for α > 1 is omitted as similar to the above mentioned proof for −∞ < α < 1. 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