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On the Stability of a Cauchy Type Functional Equation

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On the Stability of a Cauchy Type Functional Equation Demonstr. Math. 2018; 51:323–331 Research Article Open Access Abbas Najati, Jung Rye Lee*, Choonkil Park, and Themistocles M. Rassias On the stability of a Cauchy type functional equation https://doi.org/10.1515/dema-2018-0026 Received May 31, 2018; accepted November 11, 2018 Abstract: In this work, the Hyers-Ulam type stability and the hyperstability of the functional equation (︁ x + y )︁ (︁ x − y )︁ f + xy + f − xy = f(x) 2 2 are proved. Keywords: Hyers-Ulam stability, additive mapping, hyperstability, topological vector space MSC: 39B82, 34K20, 26D10 1 Introduction The functional equation (ξ) is called stable if any function g satisfying the equation (ξ) approximately, is near to a true solution of (ξ). Ulam, in 1940 [1], introduced the stability of homomorphisms between two groups. More precisely, he proposed the following problem: given a group (G1,.), a metric group (G2, *, d) and a positive number ϵ, does there exist a δ > 0 such that if a function f : G1 ! G2 satisfies the inequal- ity d(f (x.y), f (x) * f(y)) < δ for all x, y 2 G1, then there exists a homomorphism T : G1 ! G2 such that d(f(x), T(x)) < ϵ for all x 2 G1? If this problem has a solution, we say that the homomorphisms from G1 to G2 are stable. In 1941, Hyers [2] gave a partial solution of Ulam’s problem for the case of approximate addi- tive mappings under the assumption that G1 and G2 are Banach spaces. Aoki [3] and Rassias [4] provided a generalization of the Hyers’ theorem for additive and linear mappings, respectively, by allowing the Cauchy difference to be unbounded. During the last decades, several stability problems of functional equations have been investigated by several mathematicians. A large list of references concerning the stability of functional equations can be found in [5–15]. In this paper, we deal with the functional equations x + y x − y f(︀ + xy)︀ + f(︀ − xy)︀ = f (x), (1.1) 2 2 x + y x − y f (︀ + xy)︀ + g(︀ − xy)︀ = h(x), (1.2) 2 2 x + y x − y f(︀ + αxy)︀ + f (︀ + βxy)︀ = f (x) + (α + β)f (xy). (1.3) 2 2 Abbas Najati: Department of Mathematics, Faculty of Sciences, University of Mohaghegh Ardabili, Ardabil, Iran; E-mail: [email protected] *Corresponding Author: Jung Rye Lee: Department of Mathematics, Daejin University, Kyunggi 11159, Republic of Korea; E-mail: [email protected] Choonkil Park: Research Institute for Natural Sciences, Hanyang University, Seoul 04763, Republic of Korea; E-mail: [email protected] Themistocles M. Rassias: Department of Mathematics, National Technical University of Athens, Zografou Campus, 15780 Athens, Greece; E-mail: [email protected] Open Access. © 2018 Abbas Najati et al., published by De Gruyter. This work is licensed under the Creative Commons Attribution- NonCommercial-NoDerivs 4.0 License. 324 Ë Abbas Najati, Jung Rye Lee, Choonkil Park, and Themistocles M. Rassias 2 Solutions of the functional equations (1.1), (1.2) and (1.3) Theorem 2.1. Let X be a vector space. A mapping f : R ! X satisfies equation (1.1) if and only if f is additive. Proof. Let f satisfy equation (1.1). Letting x = y = 0 in equation (1.1), we get f (0) = 0. Letting x = 0 in equation 1 1 (1.1), we obtain that f is odd. Setting x = 2 and replacing y by y + 4 in equation (1.1) and using the oddness of f, we obtain 1 1 f (︀y + )︀ = f (y) + f(︀ )︀, y 2 . (2.1) 2 2 R a−b Let a, b 2 R with 2a + 2b ≠ −1. Let x, y 2 R such that x = a + b and y = 2a+2b+1 . Since f satisfies equation (1.1), we get f(a + b) = f(a) + f(b) (2.2) for all a, b 2 R with 2a+2b ≠ −1. Since f is odd, it follows from equations (2.1) and (2.2) that f (x+y) = f(x)+f (y) for all x, y 2 R. Therefore f is additive. Conversely, if f is additive, it is easy to check that f satisfies equation (1.1). Theorem 2.2. Let X be a vector space. Suppose that mappings f , g, h : R ! X satisfy equation (1.2). Then (︀ 1 )︀ (︀ 1 )︀ (i) f (x) + g − x − 2 = h − 2 , x 2 R, (ii) f (x) + g(y) = h(x + y), x, y 2 R, (iii) f − f (0), g − g(0) and h − h(0) are additive. Proof. Letting x = 0 and replacing y by 2y in equation equation (1.2), we get f(y) + g(−y) = h(0), y 2 R. (2.3) Letting y = 0 in equation (1.2), we get x x f (︀ )︀ + g(︀ )︀ = h(x), x 2 . (2.4) 2 2 R It follows from equations (2.3) and (2.4) that h x x i h x x i h(x) + h(−x) = f(︀ )︀ + g(︀ )︀ + f (︀ − )︀ + g(︀ − )︀ 2 2 2 2 h x x i h x x i = f(︀ )︀ + g(︀ − )︀ + f(︀ − )︀ + g(︀ )︀ 2 2 2 2 = h(0) + h(0) = 2h(0), x 2 R. In particular, 1 1 h(︀ )︀ + h(︀ − )︀ = 2h(0). (2.5) 2 2 1 1 Setting x = 2 and replacing y by −y − 4 in equation (1.2), we obtain 1 1 f(−y) + g(︀y + )︀ = h(︀ )︀, y 2 . (2.6) 2 2 R It follows from equations (2.3), (2.5) and (2.6) that 1 1 f(︀ − y − )︀ + g(y) = h(︀ − )︀, y 2 . (2.7) 2 2 R 1 Letting x = −y − 2 in equation (2.7), we get 1 1 f(x) + g(︀ − x − )︀ = h(︀ − )︀, x 2 . (2.8) 2 2 R x−y Replacing x and y by x + y and 1+2x+2y in equation (1.2), respectively, we get f (x) + g(y) = h(x + y), x, y 2 R, 2x + 2y ≠ −1. (2.9) By equations (2.8) and (2.9), we obtain that f(x) + g(y) = h(x + y) for all x, y 2 R. Then h(x) = f (x) + g(0) = f(0) + g(x) for all x 2 R. This implies f − f(0), g − g(0) and h − h(0) are additive. On the stability of a Cauchy type functional equation Ë 325 We need the following theorem from [16] to find the general solution of equation (1.3). Theorem 2.3. [16] Let X be a vector space and α be a real number. If a function f : R ! X satisfies f(x + y − αxy) + f (αxy) = f (x) + f(y) and f(0) = 0, then f is additive. Theorem 2.4. Let X be a vector space and α, β be real numbers. If a mapping f : R ! X satisfies equation (1.3) with f(0) = 0 or α + β ≠ 1, then f is additive. Proof. Letting x = y = 0 in equation (1.3), we get (α + β)f (0) = 0. Therefore we may assume that f (0) = 0. Letting x = 0 in equation (1.3), we obtain that f is odd. Letting y = 0 and replacing x by 2x in equation (1.3), we get f(2x) = 2f(x) for all x 2 R. Replacing y by −y in equation (1.3), we obtain x − y x + y f(︀ − αxy)︀ + f (︀ − βxy)︀ = f (x) + (α + β)f (−xy) (2.10) 2 2 for all x, y 2 R. Adding the equations (1.3) and (2.10) and using the oddness of f, we have h x + y x − y i h x + y x − y i f (︀ + αxy)︀ + f(︀ − αxy)︀ + f (︀ − βxy)︀ + f(︀ + βxy)︀ = 2f(x) (2.11) 2 2 2 2 for all x, y 2 R. Interchanging x with y in equation (1.3), we obtain x + y y − x f(︀ + αxy)︀ + f (︀ + βxy)︀ = f (y) + (α + β)f (−xy) (2.12) 2 2 for all x, y 2 R. Replacing y by −y in equation (2.12), and then adding the resulting equation to equation (2.12), we have h x + y x − y i h x + y x − y i f (︀ + αxy)︀ + f (︀ − αxy)︀ = f(︀ + βxy)︀ + f(︀ − βxy)︀ (2.13) 2 2 2 2 for all x, y 2 R. Using equation (2.13), we rewrite equation (2.11) as h x + y x + y i h x − y x − y i f (︀ + βxy)︀ + f(︀ − βxy)︀ + f (︀ − βxy)︀ + f(︀ + βxy)︀ = 2f(x) (2.14) 2 2 2 2 for all x, y 2 R. Interchanging x with y in equation (2.14), and then adding the resulting equation to (2.14), we obtain x + y x + y f(︀ + βxy)︀ + f (︀ − βxy)︀ = f (x) + f(y) (2.15) 2 2 for all x, y 2 R. Replacing x and y by 2x and 2y in equation (2.15), respectively, and using f (2t) = 2f (t), we get f (x + y + 4βxy) + f (x + y − 4βxy) = 2f(x) + 2f (y), x, y 2 R. (2.16) Let β = 0. It follows from equation (2.16) that f is additive.
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