On the Generalized Hyers-Ulam-Rassias Stability of Quadratic Functional Equations

and Applied Analysis 3 2. Solution of 1.5 , 1.6 Let X and Y be real vector spaces. We here present the general solution of 1.5 , 1.6 . Theorem 2.1. A function f : X → Y satisfies the functional equation 1.3 if and only if f : X → Y satisfies the functional equation 1.5 . Therefore, every solution of functional equation 1.5 is also a quadratic function. Proof. Let f satisfy the functional equation 1.3 . Putting x y 0 in 1.3 , we get f 0 0. Set x 0 in 1.3 to get f −y f y . Letting y x and y 2x in 1.3 , respectively, we obtain that f 2x 4f x and f 3x 9f x for all x ∈ X. By induction, we lead to f kx k2f x for all positive integers k. Replacing x and y by 2x y and 2x − y in 1.3 , respectively, gives f ( 2x y ) f ( 2x − y) 8f x 2f(y) 2.1 for all x, y ∈ X. Using 1.3 and 2.1 , we lead to f ( 2x y ) f ( 2x − y) 2f(x y) 2f(x − y) 4f x − 2f(y) 2.2 for all x, y ∈ X. Suppose that k / 0 is a fixed integer by using 1.3 , we get kf ( x y ) kf ( x − y) − 2kf x − 2kf(y) 0 2.3 for all x, y ∈ X. Using 2.2 and 2.3 , we obtain f ( 2x y ) f ( 2x − y) 2 k f(x y) 2 k f(x − y) 2 2 − k f x − 2 1 k f(y) 2.4 for all x, y ∈ X. Replacing x and y by 3x y and 3x − y in 1.3 , respectively, then using 1.3 and 2.3 , we have f ( 3x y ) f ( 3x − y) 3 k f(x y) 3 k f(x − y) 2 6 − k f x − 2 2 k f(y) 2.5 for all x, y ∈ X. By using the above method, by induction, we infer that f ( ax y ) f ( ax − y) a k f(x y) a k f(x − y) 2 ( a2 − a − k ) f x − 2 a k − 1 f(y) 2.6 4 Abstract and Applied Analysis for all x, y ∈ X and each positive integer a ≥ 1. For a negative integer a ≤ −1, replacing a by −a one can easily prove the validity of 2.6 . Therefore 1.3 implies 2.6 for any integer a/ 0. First, it is noted that 2.6 also implies the following equation f ( bx y ) f ( bx − y) b k f(x y) b k f(x − y) 2 ( b2 − b − k ) f x − 2 b k − 1 f(y) 2.7 for all integers b / 0. Setting y 0 in 2.7 gives f bx b2f x . Substituting y with by into 2.7 , one gets b k f ( x by ) b k f ( x − by) b2f(x y) b2f(x − y) − 2 ( b2 − b − k ) f x 2b2 b k − 1 f(y) 2.8 for all x, y ∈ X. Replacing y by by in 2.6 , we observe that f ( ax by ) f ( ax − by) a k f(x by) a k f(x − by) 2 ( a2 − a − k ) f x − 2 a k − 1 f(by) 2.9 for all x, y ∈ X.Hence, according to 2.8 and 2.9 , we get b k f ( ax by ) b k f ( ax − by) b2 a k f(x y) b2 a k f(x − y) 2 ( a2 b k − b2 a k ) f x − 2b2 a − b f(y) 2.10 for all x, y ∈ X. In particular, if we substitute k : b in 2.10 and dividing it by 2b, we conclude that f satisfies 1.5 . Let f satisfy the functional equation 1.5 , for nonzero fixed integers a,b with b / ± a,−3a. Putting x y 0 in 1.5 , we get ( 2a2 − ba b2 − 2 ) f 0 0, 2.11