Bulletin of Mathematical Analysis and Applications ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 2 Issue 4(2010), 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 π ∑ ∑ π (π₯1 + ππ π₯π ) = 2(π − 1)π (π₯1 ) + 2 π=2 ππ ∈{−1,1} π ∑ π (π₯π ) π=2 is investigated. 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 satisο¬es a functional equation β° must be close to an exact solution of β°?” If there exists an aο¬rmative answer, we say that the equation β° 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, 21] and monographs [11, 12, 8] and references therein. Let π³ and π΄ be normed spaces. A function π : π³ → π΄ satisfying the functional equation π (π₯ + π¦) + π (π₯ − π¦) = 2π (π₯) + 2π (π¦) (π₯, π¦ ∈ π³ ) (1.1) is called the quadratic functional equation. It is well known that a function π between real vector spaces is quadratic if and only if there exists a unique symmetric bi-additive function π΅ such that π (π₯) = π΅(π₯, π₯) for all π₯ ∈ π³ ; see [9]. The biadditive function π΅ is given by 1 π΅(π₯, π₯) = (π (π₯ + π¦) − π (π₯ − π¦)) . 4 The Hyers–Ulam stability of the quadratic equation (1.1) was proved by Skof [22]. Cholewa [6] noticed that the theorem of Skof is still true if the relevant domain π³ 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– 2000 Mathematics Subject Classiο¬cation. 39B22, 39B82. Key words and phrases. Hyers–Ulam stability; quadratic equation; normed space. c β2010 Universiteti i PrishtineΜs, PrishtineΜ, KosoveΜ. Submitted August 1, 2010. Published November 15, 2010. 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 [2, 3, 16, 4, 14]. Theorem 1.1. (Czerwik) Let π ≥ 0 be ο¬xed. If a mapping π : π³ → π΄ satisο¬es the inequality β₯π (π₯ + π¦) + π (π₯ − π¦) − 2π (π₯) − 2π (π¦)β₯ ≤ π (π₯ ∈ π³ ) (1.2) then there exists a unique quadratic mapping π : π³ → π΄ such that β₯π (π₯) − π(π₯)β₯ ≤ 1 π 2 (π₯ ∈ π³ ). Moreover, if π is measurable or if π (π‘π₯) is continuous in π‘ for each ο¬xed π₯ ∈ π³ , then π(π‘π₯) = π‘2 π(π‘π₯) for all π₯ ∈ π³ and π‘ ∈ β. The Hyers–Ulam stability of equation (1.1) on a certain restricted domain was investigated by Jung [13] in the following theorem, Theorem 1.2. (Jung) Let π > 0 and π ≥ 0 be given. Assume that a mapping π : π³ → π΄ satisο¬es the inequality (1.2) for all π₯, π¦ ∈ π³ with β₯π₯β₯ + β₯π¦β₯ ≥ π. Then there exists a unique quadratic mapping π : π³ → π΄ such that β₯π (π₯) − π(π₯)β₯ ≤ 7 π 2 (π₯ ∈ π³ ). (1.3) If, moreover, π is measurable or π (π‘π₯) is continuous in π‘ for each ο¬xed π₯ ∈ π³ then π(π‘π₯) = π‘2 π(π‘π₯) for all π₯ ∈ π³ and π‘ ∈ β. The quadratic functional equation was used to characterize the inner product spaces [1]. A square norm on an inner product space satisο¬es the important parallelogram equality β₯π₯ + π¦β₯2 + β₯π₯ − π¦β₯2 = 2(β₯π₯β₯2 + β₯π¦β₯2 ). It was shown by Moslehian and Rassias [19] that a normed space (π³ , β₯.β₯) is an inner product space if and only if for any ο¬nite set of vectors π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π ∈ π³ , )2 ( 2 π π ∑ ∑ ∑ ∑ ππ β₯π₯π β₯ . (1.4) ππ π₯π = β₯π₯1 β₯ + π₯1 + π=2 ππ ∈{−1,1} ππ ∈{−1,1} π=2 Motivated by (1.4), we introduce the following functional equation deriving from the quadratic function π ∑ ∑ π=2 ππ ∈{−1,1} π (π₯1 + ππ π₯π ) = 2(π − 1)π (π₯1 ) + 2 π ∑ π (π₯π ) , (1.5) π=2 where π ≥ 2 is a ο¬xed integer. It is easy to see that the function π (π₯) = π₯2 is a solution of functional equation (1.5). 2. Solution of the equation (1.5) Theorem 2.1. A mapping π : π³ → π΄ satisο¬es the equation (1.5) for all π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π ∈ π³ if and only if π is quadratic. 132 GH. SADEGHI Proof. If we replace π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π in (1.5) by 0, then we get π (0) = 0. Putting π₯3 = π₯4 = ⋅ ⋅ ⋅ = π₯π = 0 in the equation (1.5) we see that π (π₯1 − π₯2 ) + π (π₯1 + π₯2 ) + 2(π − 2)π (π₯1 ) = 2(π − 1)π (π₯1 ) + 2π (π₯2 ) . Hence π (π₯1 − π₯2 ) + π (π₯1 + π₯2 ) = 2π (π₯1 ) + 2π (π₯2 ). The converse is trivial. β‘ Remark. We can prove the theorem above on the punching space π³ − {0}. If we consider π₯2 = π₯3 = ⋅ ⋅ ⋅ = π₯π , then we observe that π ∑ ∑ π (π₯1 + ππ π₯2 ) = 2(π − 1)π (π₯1 ) + 2 π=2 ππ ∈{−1,1} π ∑ π (π₯2 ) , π=2 whence (π − 1) (π (π₯1 − π₯2 ) + π (π₯1 + π₯2 )) = 2(π − 1)π (π₯1 ) + 2(π − 1)π (π₯2 ). Hence π is quadratic. 3. Stability Results Throughout this section, let π³ and π΄ be normed and Banach spaces also, we prove the Hyers–Ulam stability of equation (1.5). From now on, we use the following abbreviation ππ (π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π ) = π ∑ ∑ π (π₯1 + ππ π₯π )−2(π−1)π (π₯1 )−2 π ∑ π (π₯π ) . (3.1) π=2 π=2 ππ ∈{−1,1} Theorem 3.1. Let π ≥ 0 be ο¬xed. If a mapping π : π³ → π΄ with π (0) = 0 satisο¬es β₯ππ (π₯1 , π₯2 , ⋅ ⋅ ⋅ π₯π )β₯ ≤ π (3.2) for all π₯1 , π₯2 , ⋅ ⋅ ⋅ π₯π ∈ π³ , then there exists a unique quadratic mapping π : π³ → π΄ such that 1 β₯π (π₯) − π(π₯)β₯ ≤ π . 2 Moreover, if π is measurable or if π (π‘π₯) is continuous in π‘ for each ο¬xed π₯ ∈ π³ , then π(π‘π₯) = π‘2 π(π‘π₯) for all π₯ ∈ π³ and π‘ ∈ β. Proof. It is enough to put π₯3 = π₯4 = ⋅ ⋅ ⋅ π₯π = 0 in (3.2) and use Theorem 1.1. β‘ By using an idea from the paper [13], we will prove the Hyers–Ulam stability of equation (1.5) on a restricted domain. Theorem 3.2. Let π > 0 and π ≥ 0 be given. Suppose that a mapping π : π³ → π΄ satisο¬es the inequality (3.2) for all π₯1 , π₯2 , ⋅ ⋅ ⋅ π₯π ∈ π³ with β₯π₯1 β₯+β₯π₯2 β₯+⋅ ⋅ ⋅+β₯π₯π β₯ ≥ π. Then there exists a unique quadratic mapping π : π³ → π΄ such that β₯π (π₯) − π(π₯)β₯ ≤ 3 + 2π π 2 (3.3) for all π₯ ∈ π³ . Moreover, if π is measurable or if π (π‘π₯) is continuous in π‘ for each ο¬xed π₯ ∈ π³ , then π(π‘π₯) = π‘2 π(π‘π₯) for all π₯ ∈ π³ and π‘ ∈ β. A QUADRATIC TYPE FUNCTIONAL EQUATION 133 Proof. Assume β₯π₯1 β₯ + β₯π₯2 β₯ + ⋅ ⋅ ⋅ + β₯π₯π β₯ < π. If π₯1 = π₯2 = ⋅ ⋅ ⋅ = π₯π = 0, then we chose a π‘ ∈ π³ with β₯π‘β₯ = π. Otherwise, let π‘ = (1 + β₯π₯ππ β₯ )π₯π0 , where 0 β₯π₯π0 β₯ = max{β₯π₯π β₯ : 1 ≤ π ≤ π}. Clearly, we see that β₯π₯1 − π‘β₯ + β₯π₯2 + π‘β₯ + ⋅ ⋅ ⋅ + β₯π₯π + π‘β₯ ≥ π (3.4) β₯π₯1 + π‘β₯ + β₯π₯2 + π‘β₯ + ⋅ ⋅ ⋅ + β₯π₯π + π‘β₯ ≥ π β₯π₯1 β₯ + β₯π₯2 + 2π‘β₯ + ⋅ ⋅ ⋅ + β₯π₯π + 2π‘β₯ ≥ π β₯π₯2 + π‘β₯ + β₯π₯3 + π‘β₯ + ⋅ ⋅ ⋅ + β₯π₯π + π‘β₯ + β₯π‘β₯ ≥ π β₯π₯1 β₯ + β₯π‘β₯ ≥ π, since β₯π₯π + π‘β₯ ≥ π and β₯π₯π + 2π‘β₯ ≥ π, for 1 ≤ π ≤ π. From (3.2) and (3.4) and the relations π (π₯1 + π₯2 ) + π (π₯1 − π₯2 ) − 2π (π₯1 ) − 2π (π₯2 ) = π (π₯1 + π₯2 ) + π (π₯1 − π₯2 − 2π‘) − 2π (π₯1 − π‘) − 2π (π₯2 + π‘) + π (π₯1 + π₯2 + 2π‘) + π (π₯1 − π₯2 ) − 2π (π₯1 + π‘) − 2π (π₯2 + π‘) − 2π (π₯2 + 2π‘) − 2π (π₯2 ) + 4π (π₯2 + π‘) + 4π (π‘) − π (π₯1 + π₯2 + 2π‘) − π (π₯1 − π₯2 − 2π‘) + 2π (π₯1 ) + 2π (π₯2 + 2π‘) + 2π (π₯1 + π‘) + 2π (π₯1 − π‘) − 4π (π₯1 ) − 4π (π‘) we get β₯ππ (π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π )β₯ π ∑ ≤ ∑ π (πΌ1 + ππ πΌπ ) − 2(π − 1)π (πΌ1 ) − 2 π ∑ ∑ π (πΌπ ) π=2 π=2 ππ ∈{−1,1} + π ∑ π (π½1 + ππ π½π ) − 2(π − 1)π (π½1 ) − π (π½π ) π=2 ππ ∈{−1,1} + 2 π ∑ ∑ π (πΎ1 + ππ πΎπ ) − 2(π − 1)π (πΎ1 ) − 2 π ∑ ∑ π (π1 + ππ ππ ) − 2(π − 1)π (π1 ) − 2 π=2 ππ ∈{−1,1} + 2(π − 1) π ∑ π (πΎπ ) π=2 π=2 ππ ∈{−1,1} + π ∑ π ∑ π (ππ ) π=2 ∑ π (π1 + ππ ππ ) − 2(π − 1)π (π1 ) − 2 π=2 ππ ∈{−1,1} π=2 where πΌ1 = π₯1 − π‘ , πΌπ = π₯ π + π‘ , 2≤π≤π π½1 = π₯ 1 + π‘ , π½π = π₯ π + π‘ , 2≤π≤π 2≤π≤π πΎ1 = π‘ , πΎπ = π₯π + π‘ , π 1 = π₯1 , ππ = π₯π + 2π‘ , π π = π₯1 , ππ+1 = π‘ , π ∑ 2≤π≤π 2 ≤ π ≤ π. π (ππ ) , 134 GH. SADEGHI Hence we have β₯ππ (π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π )β₯ ≤ β₯ππ (πΌ1 , πΌ2 , ⋅ ⋅ ⋅ , πΌπ )β₯ + β₯ππ (π½1 , π½2 , ⋅ ⋅ ⋅ , π½π )β₯ + 2β₯ππ (πΎ1 , πΎ2 , ⋅ ⋅ ⋅ , πΎπ )β₯ + β₯ππ (π1 , π2 , ⋅ ⋅ ⋅ , ππ )β₯ + 2(π − 1)β₯ππ (π1 , π2 , ⋅ ⋅ ⋅ , ππ )β₯ ≤ (3 + 2π)π. (3.5) Obviously, inequality (3.2) holds for all π₯, π¦ ∈ π³ . According to (3.5) and Theorem 3.1, there exists a unique quadratic mapping π : π³ → π΄ which satisο¬es the inequality (3.3) for all π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π ∈ π³ . β‘ Now we study asymptotic behavior of function equation (1.5). Theorem 3.3. Suppose that π : π³ → π΄ is a mapping. Then π is quadratic if and only if for π ∈ β (π ≥ 2) β₯ππ (π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π )β₯ → 0 (3.6) as β₯π₯1 β₯ + β₯π₯2 β₯ + ⋅ ⋅ ⋅ β₯π₯π β₯ → ∞. Proof. If π is quadratic then (3.6) evidently holds. Conversely, by using the limits (3.6) we can ο¬nd for every π ∈ β a sequence ππ such that β₯ππ (π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π )β₯ ≤ π1 for all π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π ∈ π³ with β₯π₯1 β₯ + β₯π₯2 β₯ + ⋅ ⋅ ⋅ β₯π₯π β₯ ≥ ππ . By Theorem 3.2 for every π ∈ β there exists a unique quadratic mapping ππ such that 3 + 2π (3.7) β₯π (π₯) − ππ (π₯)β₯ ≤ 2π 3+2π for all π₯ ∈ π³ . Since β₯π (π₯) − π1 (π₯)β₯ ≤ 3+2π and β₯π (π₯) − ππ (π₯)β₯ ≤ 3+2π 2 2π ≤ 2 , by the uniqueness of π1 we conclude that ππ = π1 for all π ∈ π© . Now, by tending π to the inο¬nity in (3.7) we deduce that π = π1 . Therefore π is quadratic. β‘ 4. stability on bounded domains Throughout this section, we denote by π΅π (0) the closed ball of radius π around the origin and π΅π = π΅π (0) − {0}. In this section we used some ideas from the paper’s Moslehian et al [18]. Theorem 4.1. Let π³ and π΄ be normed and Banach spaces π > 2, π > 0, π : π π → [0, ∞)(π ≥ 2) be a function such that π( π₯21 , π₯22 , ⋅ ⋅ ⋅ , π₯2π ) ≤ 21π π(π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π ) for all π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π ∈ π΅π . Suppose that π : π³ → π΄ is a mapping satisfying π (0) = 0 and β₯ππ (π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π )β₯ ≤ π(π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π ) (4.1) for all π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯πΎ ∈ π΅π with π₯π ± π₯π ∈ π΅π for 1 ≤ π, π ≤ π. Then there exists a unique quadratic mapping π : π³ → π΄ such that 1 π(π₯, π₯, ⋅ ⋅ ⋅ , π₯), (4.2) β₯π (π₯) − π(π₯)β₯ ≤ π (2 − 4)(π − 1) where π₯ ∈ π΅π . Proof. Let π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯πΎ ∈ π΅π . If we consider π₯2 = π₯3 = ⋅ ⋅ ⋅ = π₯π in (4.1), then we see that 1 π(π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯2 ). (4.3) β₯π (π₯1 + π₯2 ) + π (π₯1 − π₯2 ) − 2π (π₯1 ) − 2π (π₯2 )β₯ ≤ π−1 A QUADRATIC TYPE FUNCTIONAL EQUATION Replacing π₯1 , π₯2 in (4.3) by π₯ 2, 135 we get (π₯ π₯ π₯ 1 π₯) β₯π (π₯) − 4π ( )β₯ ≤ . (4.4) π , ,⋅⋅⋅ , 2 π−1 2 2 2 Replacing π₯ by 2π₯π ∈ π΅π and multiplying with 4π in (4.4), we obtain ( π₯ 4π π₯ π₯ ) π₯ π₯ β₯4π π ( π ) − 4π+1 π ( π+1 )β₯ ≤ (4.5) π π+1 , π+1 , ⋅ ⋅ ⋅ , π+1 . 2 2 π−1 2 2 2 It follows from (4.5) that π 1 ∑ π+π−1 ( π₯ π₯ π₯ ) π₯ π₯ β₯4π π ( π ) − 4π+π π ( π+π )β₯ ≤ 4 π π+π , π+π , ⋅ ⋅ ⋅ , π+π 2 2 π − 1 π=1 2 2 2 ≤ π ∑ 22(π−1) 1 . π(π₯, π₯, ⋅ ⋅ ⋅ , π₯) (π−2)π 2ππ (π − 1) 2 π=1 (4.6) It follows that {4π π ( 2π₯π )} is Cauchy and so is convergent. Therefore we see that a mapping π₯ ˆ π(π₯) := lim 4π π ( π ) (π₯ ∈ π΅π ), π→∞ 2 satisο¬es 1 ˆ π(π₯, π₯, ⋅ ⋅ ⋅ , π₯), β₯π (π₯) − π(π₯)β₯ ≤ π (2 − 4)(π − 1) ˆ and π(0) = 0, when taking the limit π → ∞ in (4.6) with π = 0. Next ο¬x π₯ ∈ π΅π . Because of π₯2 ∈ π΅π , we have ˆ π₯ ) = lim 4π+1 π ( π₯ ) = lim 4π π ( π₯ ) = π(π₯). ˆ 4π( π→∞ π→∞ 2 2π+1 2π π₯ ˆ π+π ˆ ) = π(π₯) and so the mapping π : π³ → π΄ given by π(π₯) := Therefore 4π+π π( 2 πˆ π₯ 4 π( 2π ), where π is least non-negative integer such that 2π₯π ∈ π΅π is well-deο¬ned. ˆ It is easy to see that π(π₯) = limπ→∞ 4π π ( π₯π ) (π₯ ∈ π³ ) and πβ£π΅ (0) = π. 2 π π₯−π¦ Now let π₯, π¦ ∈ π³ . There is a large enough π such that 2π₯π , 2π¦π , π₯+π¦ 2π , 2π ∈ π΅π (0). π¦ π₯ π Put π₯1 = 2π and π₯2 = 2π in (4.3) and multiplying both sides with 4 to obtain (π₯ π¦ π₯+π¦ π₯−π¦ π₯ π¦ 4π π¦ ) β₯4π π ( π ) + 4π π ( π ) − 4π 2π ( π ) − 4π 2π ( π )β₯ ≤ π π, π,⋅⋅⋅ , π 2 2 2 π−1 2 2 2 4π ≤ π(π₯, π¦, π¦, ⋅ ⋅ ⋅ , π¦). 2ππ (π − 1) whence, by taking the limit as π → ∞, we get π(π₯ + π¦) + π(π₯ − π¦) = 2π(π₯) + 2π(π¦). Hence π is quadratic. Uniqueness of π can be proved by using the strategy used in the proof of Theorem 3.2. β‘ Corollary 4.2. Let π³ and π΄ be normed and Banach spaces π > 2, π > 0, π > 0. Suppose that π : π³ → π΄ is a mapping satisfying π (0) = 0 and π π π β₯ππ (π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π )β₯ ≤ πβ₯π₯1 β₯ π β₯π₯2 β₯ π ⋅ ⋅ ⋅ β₯π₯π β₯ π (4.7) for all π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π ∈ π΅π with π₯π ± π₯π ∈ π΅π for 2 ≤ π, π ≤ π. Then there exists a unique quadratic mapping π : π³ → π΄ such that πππ , (4.8) β₯π (π₯) − π(π₯)β₯ ≤ π (2 − 4)(π − 1) where π₯ ∈ π΅π . 136 GH. SADEGHI π π π Proof. Apply Theorem 4.1 with π(π₯1 , π₯2 , ⋅ ⋅ ⋅ , π₯π ) = πβ₯π₯1 β₯ π β₯π₯2 β₯ π ⋅ ⋅ ⋅ β₯π₯π β₯ π . β‘ References [1] D. Amir, Characterizations of inner product spaces, BirkhaΜuesr, Basel, 1986. [2] J.-H. Bae and I.-S. Chang, On the Ulam stability problem of a quadratic functional equation, Korean. J. Comput. Appl. Math. (Series A) 8 (2001), 561–567. [3] J.-H. Bae and Y.-S. Jung, THE Hyers–Ulam stability of the quadratic functional equations on abelian groups, Bull. Korean Math. Soc. 39 (2002), no.2, 199–209. [4] B. Belaid, E. 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