Bulletin of Mathematical Analysis and Applications ISSN: 181-191, URL: http://www.bmathaa.org Volume Issue 4(010, Pages 130-136. A QUADRATIC TYPE FUNCTIONAL EQUATION (DEDICATED IN OCCASION OF THE 70-YEARS OF PROFESSOR HARI M. SRIVASTAVA GHADIR SADEGHI Abstract. In this paper, the solution and the Hyers Ulam stability of the following quadratic type functional equation f (x 1 + ε j x i = (k 1f(x 1 + f(x i is investigated. ε j { 1,1} 1. Introduction and preliminaries A classical question in the theory of functional equations is the following: When is it true that a function which approximately satisfies a functional equation E must be close to an exact solution of E? If there exists an affirmative answer, we say that the equation E is stable [9]. During the last decades several stability problems for various functional equations have been investigated by numerous mathematicians. We refer the reader to the survey articles [10, 9, 1] and monographs [11, 1, 8] and references therein. Let X and Y be normed spaces. A function f : X Y satisfying the functional equation f(x + y + f(x y = f(x + f(y (x, y X (1.1 is called the quadratic functional equation. It is well known that a function f between real vector spaces is quadratic if and only if there exists a unique symmetric bi-additive function B such that f(x = B(x, x for all x X ; see [9]. The biadditive function B is given by B(x, x = 1 (f(x + y f(x y. 4 The Hyers Ulam stability of the quadratic equation (1.1 was proved by Skof []. Cholewa [6] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an abelian group. Furthermore, Czerwik [7] deal with stability problem of the quadratic functional equation (1.1 in the spirit of Hyers Ulam 000 Mathematics Subject Classification. 39B, 39B8. Key words and phrases. Hyers Ulam stability; quadratic equation; normed space. c 010 Universiteti i Prishtinës, Prishtinë, Kosovë. Submitted August 1, 010. Published November 15, 010. 130
A QUADRATIC TYPE FUNCTIONAL EQUATION 131 Rassias. Also, Jung [13] proved the stability of (1.1 on a restricted domain. For more information on the stability of the quadratic equation, we refer the reader to [, 3, 16, 4, 14]. Theorem 1.1. (Czerwik Let ε 0 be fixed. If a mapping f : X Y satisfies the inequality f(x + y + f(x y f(x f(y ε (x X (1. then there exists a unique quadratic mapping Q : X Y such that f(x Q(x 1 ε (x X. Moreover, if f is measurable or if f(tx is continuous in t for each fixed x X, then Q(tx = t Q(tx for all x X and t R. The Hyers Ulam stability of equation (1.1 on a certain restricted domain was investigated by Jung [13] in the following theorem, Theorem 1.. (Jung Let d > 0 and ε 0 be given. Assume that a mapping f : X Y satisfies the inequality (1. for all x, y X with x + y d. Then there exists a unique quadratic mapping Q : X Y such that f(x Q(x 7 ε (x X. (1.3 If, moreover, f is measurable or f(tx is continuous in t for each fixed x X then Q(tx = t Q(tx for all x X and t R. The quadratic functional equation was used to characterize the inner product spaces [1]. A square norm on an inner product space satisfies the important parallelogram equality x + y + x y = ( x + y. It was shown by Moslehian and Rassias [19] that a normed space (X,. is an inner product space if and only if for any finite set of vectors x 1, x,, x k X, x 1 + ε j x i = ( x 1 + ε j x i. (1.4 Motivated by (1.4, we introduce the following functional equation deriving from the quadratic function f (x 1 + ε j x i = (k 1f(x 1 + f(x i, (1.5 where k is a fixed integer. It is easy to see that the function f(x = x is a solution of functional equation (1.5.. Solution of the equation (1.5 Theorem.1. A mapping f : X Y satisfies the equation (1.5 for all x 1, x,, x k X if and only if f is quadratic.
13 GH. SADEGHI Proof. If we replace x 1, x,, x k in (1.5 by 0, then we get f(0 = 0. Putting x 3 = x 4 = = x k = 0 in the equation (1.5 we see that f(x 1 x + f(x 1 + x + (k f(x 1 = (k 1f(x 1 + f(x. Hence f(x 1 x + f(x 1 + x = f(x 1 + f(x. The converse is trivial. Remark. We can prove the theorem above on the punching space X {0}. If we consider x = x 3 = = x k, then we observe that whence f (x 1 + ε j x = (k 1f(x 1 + f(x, (k 1 (f(x 1 x + f(x 1 + x = (k 1f(x 1 + (k 1f(x. Hence f is quadratic. 3. Stability Results Throughout this section, let X and Y be normed and Banach spaces also, we prove the Hyers Ulam stability of equation (1.5. From now on, we use the following abbreviation Df(x 1, x,, x k = f (x 1 + ε j x i (k 1f(x 1 f(x i. (3.1 Theorem 3.1. Let ε 0 be fixed. If a mapping f : X Y with f(0 = 0 satisfies Df(x 1, x, x k ε (3. for all x 1, x, x k X, then there exists a unique quadratic mapping Q : X Y such that f(x Q(x 1 ε. Moreover, if f is measurable or if f(tx is continuous in t for each fixed x X, then Q(tx = t Q(tx for all x X and t R. Proof. It is enough to put x 3 = x 4 = x k = 0 in (3. and use Theorem 1.1. By using an idea from the paper [13], we will prove the Hyers Ulam stability of equation (1.5 on a restricted domain. Theorem 3.. Let d > 0 and ε 0 be given. Suppose that a mapping f : X Y satisfies the inequality (3. for all x 1, x, x k X with x 1 + x + + x k d. Then there exists a unique quadratic mapping Q : X Y such that f(x Q(x 3 + k ε (3.3 for all x X. Moreover, if f is measurable or if f(tx is continuous in t for each fixed x X, then Q(tx = t Q(tx for all x X and t R.
A QUADRATIC TYPE FUNCTIONAL EQUATION 133 Proof. Assume x 1 + x + + x k < d. If x 1 = x = = x k = 0, then we chose a t X with t = d. Otherwise, let t = (1 + d x i0 x i 0, where x i0 = max{ x j : 1 j k}. Clearly, we see that x 1 t + x + t + + x k + t d (3.4 x 1 + t + x + t + + x k + t d x 1 + x + t + + x k + t d x + t + x 3 + t + + x k + t + t d since x j + t d and x j + t d, for 1 j k. From (3. and (3.4 and the relations x 1 + t d, f(x 1 + x + f(x 1 x f(x 1 f(x = f(x 1 + x + f(x 1 x t f(x 1 t f(x + t + f(x 1 + x + t + f(x 1 x f(x 1 + t f(x + t f(x + t f(x + 4f(x + t + 4f(t f(x 1 + x + t f(x 1 x t + f(x 1 + f(x + t + f(x 1 + t + f(x 1 t 4f(x 1 4f(t we get Df(x 1, x,, x k f (α 1 + ε j α i (k 1f(α 1 f(α i + f (β 1 + ε j β i (k 1f(β 1 f(β i + f (γ 1 + ε j γ i (k 1f(γ 1 f(γ i + f (θ 1 + ε j θ i (k 1f(θ 1 f(θ i + (k 1 f (η 1 + ε j η i (k 1f(η 1 f(η i, where α 1 = x 1 t, α i = x i + t, i k β 1 = x 1 + t, β i = x i + t, i k γ 1 = t, γ i = x i + t, i k θ 1 = x 1, θ i = x i + t, i k η i = x 1, η i+1 = t, i k.
134 GH. SADEGHI Hence we have Df(x 1, x,, x k Df(α 1, α,, α k + Df(β 1, β,, β k + Df(γ 1, γ,, γ k + Df(θ 1, θ,, θ k + (k 1 Df(η 1, η,, η k (3 + kε. (3.5 Obviously, inequality (3. holds for all x, y X. According to (3.5 and Theorem 3.1, there exists a unique quadratic mapping Q : X Y which satisfies the inequality (3.3 for all x 1, x,, x k X. Now we study asymptotic behavior of function equation (1.5. Theorem 3.3. Suppose that f : X Y is a mapping. Then f is quadratic if and only if for k N (k as x 1 + x + x k. Df(x 1, x,, x k 0 (3.6 Proof. If f is quadratic then (3.6 evidently holds. Conversely, by using the limits (3.6 we can find for every n N a sequence ε n such that Df(x 1, x,, x k 1 n for all x 1, x,, x k X with x 1 + x + x k ε n. By Theorem 3. for every n N there exists a unique quadratic mapping Q n such that f(x Q n (x 3 + k (3.7 n for all x X. Since f(x Q 1 (x 3+k and f(x Q n (x 3+k n 3+k, by the uniqueness of Q 1 we conclude that Q n = Q 1 for all n N. Now, by tending n to the infinity in (3.7 we deduce that f = Q 1. Therefore f is quadratic. 4. stability on bounded domains Throughout this section, we denote by B r (0 the closed ball of radius r around the origin and B r = B r (0 {0}. In this section we used some ideas from the paper s Moslehian et al [18]. Theorem 4.1. Let X and Y be normed and Banach spaces p >, r > 0, φ : X k [0, (k be a function such that φ( x1, x,, x k 1 φ(x p 1, x,, x k for all x 1, x,, x k B r. Suppose that f : X Y is a mapping satisfying f(0 = 0 and Df(x 1, x,, x k φ(x 1, x,, x k (4.1 for all x 1, x,, x K B r with x i ± x j B r for 1 i, j k. Then there exists a unique quadratic mapping Q : X Y such that 1 f(x Q(x ( p φ(x, x,, x, (4. 4(k 1 where x B r. Proof. Let x 1, x,, x K B r. If we consider x = x 3 = = x k in (4.1, then we see that f(x 1 + x + f(x 1 x f(x 1 f(x 1 k 1 φ(x 1, x,, x. (4.3
A QUADRATIC TYPE FUNCTIONAL EQUATION 135 Replacing x 1, x in (4.3 by x, we get f(x 4f( x 1. (4.4 ( x k 1 φ, x,, x Replacing x by x B n r and multiplying with 4 n in (4.4, we obtain 4 n f( x n x ( 4n+1 4n x f( n+1 k 1 φ n+1, x n+1,, x n+1. (4.5 It follows from (4.5 that 4 n f( x n x 4n+m f( n+m 1 m ( x 4 n+i 1 φ k 1 n+i, x n+i,, x n+i i=1 (n 1 m pn (k 1 φ(x, x,, x i=1 1. (4.6 (p i It follows that {4 n f( x } is Cauchy and so is convergent. Therefore we see that a n mapping ˆQ(x := lim n 4n f( x n (x B r, satisfies f(x ˆQ(x 1 ( p φ(x, x,, x, 4(k 1 and ˆQ(0 = 0, when taking the limit m in (4.6 with n = 0. Next fix x B r. Because of x B r, we have 4 ˆQ( x = lim x n 4n+1 f( = lim n+1 n 4n f( x = ˆQ(x. n x Therefore 4 n+m ˆQ( = ˆQ(x and so the mapping Q : X Y given by Q(x := n+m 4 n x ˆQ(, where n is least non-negative integer such that x n B n r is well-defined. It is easy to see that Q(x = lim n 4 n f( x (x X and Q n Br(0 = ˆQ. Now let x, y X. There is a large enough n such that x, y n, x+y n, x y n B n r (0. Put x 1 = x and x n = y in (4.3 and multiplying both sides with 4 n to obtain n 4 n x + y f( + 4 n x y f( n n 4n f( x n 4n f( y n 4 n ( x k 1 φ y y,,, n n n 4 n np φ(x, y, y,, y. (k 1 whence, by taking the limit as n, we get Q(x+y+Q(x y = q(x+q(y. Hence Q is quadratic. Uniqueness of Q can be proved by using the strategy used in the proof of Theorem 3.. Corollary 4.. Let X and Y be normed and Banach spaces p >, r > 0, θ > 0. Suppose that f : X Y is a mapping satisfying f(0 = 0 and Df(x 1, x,, x k θ x 1 p k x p k xk p k (4.7 for all x 1, x,, x k B r with x i ± x j B r for i, j k. Then there exists a unique quadratic mapping Q : X Y such that θr p f(x Q(x ( p 4(k 1, (4.8 where x B r.
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