Stability of the Drygas Functional Equation on Restricted Domain

We study the stability of the Drygas functional equation on a restricted domain. The main tool used in the proofs is the fixed point theorem for functional spaces. Mathematics Subject Classification. 39B82, 47H10, 47J20.

The above equation was introduced in [3] in order to obtain a characterization of the quasi-inner-product spaces. Ebanks, Kannappan and Sahoo in [4] have obtained the general solution of the Eq. (1) as where a : R → R is an additive function and q : R → R is a quadratic function, i.e. q satisfies the quadratic functional equation q(x + y) + q(x − y) = 2q(x) + 2q(y), x,y∈ R.
The stability in the Hyers-Ulam sense of the Drygas equation has been investigated by Jung and Sahoo in [6]. They have proved that if a function f : X → Y , where X is a real vector space and Y is a Banach space satisfies the inequality Their result was improved first by Yang in [9] and later by Sikorska in [7]. In the case when X is an Abelian group they obtained sharper bounds: 3 2 ε and ε respectively instead of 25 3 ε (cf. Proposition 1 in [9] and Theorem 3.2 in [7]). The stability and solution of the Drygas equation under some additional conditions was also studied by Forti and Sikorska in [5] in the case when X and Y are amenable groups.
In the paper we present the stability results for the Drygas equation on restricted domain. Let X be a nonempty subset of a normed space and Y be a normed space. We say that a function f : One of the method of the proof is based on a fixed point result that can be derived from [1] (Theorem 1). To present it we need the following three hypothesis: Now we are in a position to present the above mentioned fixed point theorem.

Theorem 1.
Let hypotheses (H1)-(H3) be valid and functions ε : X → R + and ϕ : X → Y fulfil the following two conditions Then there exists a unique fixed point ψ of T with Throughout the paper N 0 denotes the set of all non-negative integers.

Stability Results
Theorem 2. Let X be a subset with 0 of a normed space, Y be a Banach space and c ≥ 0. Assume that p > 0 and a function f : X → Y satisfies for all x, y ∈ X such that x + y, x − y ∈ X.
(1) If p > 2 and −x, x 2 ∈ X for all x ∈ X, then there exists a function g : X → Y satisfying the Drygas equation on X such that (2) If 0 < p < 1 and −x, 2x ∈ X for all x ∈ X, then there exists a function g : X → Y satisfying the Drygas equation on X such that Moreover, g is the unique solution of the Eq.
(1) Replacing x and y by x 2 in (4) we obtain Consider functions T : Y X → Y X and ε : X → R + given as follows The inequality (5) now becomes For every ξ, μ ∈ Y X and x ∈ X In particular Since Λ is linear, we can prove by induction By Theorem 1, there exists a function g : X → Y such Next we prove that g satisfies the Drygas equation. Observe first that if a function h : X → Y satisfies the inequality for all x, y ∈ X such that x + y, x − y ∈ X and some M > 0, then for x, y ∈ X satisfying x + y, x − y ∈ X. Indeed, fix h : X → Y and assume (6). Then . Consequently, proceeding by induction we get that if a function h : X → Y satisfies the inequality (6), then for all n ∈ N 0 and x, y ∈ X, x + y, x − y ∈ X. On account of the above observation and (4) for every n ∈ N 0 and x, y ∈ X such that x + y, x − y ∈ X. Letting n → ∞ we get (2) The idea of the proof is the same as before so we only give a sketch. Replacing y by x in (4) we obtain Let functions T : Y X → Y X and ε : X → R + be define by formulas The inequality (7) takes now the form Obviously T satisfies the inequality (H2) with The operator Λ : R X + → R X + is given by Proceeding by induction, we obtain By Theorem 1, there exists a function g : X → Y such A trivial verification shows that for every n ∈ N 0 and x, y ∈ X satisfying x + y, x − y ∈ X. Hence, letting n → ∞ we obtain g(x + y) + g(x − y) = 2g(x) + g(y) + g(−y), x,y∈ X, x + y, x − y ∈ X.
(3) In this case let f e : X → Y and f o : X → Y be the even and the odd part of the function f , respectively. That means and analogously for every x, y ∈ X such that x + y, x − y ∈ X. Hence f e , f o satisfy the inequality (4).
Replace y by x in (4). By the evenness of f e , and ε(x) = 1 2 c x p , x ∈ X. By Theorem 1, there exists a function g e : X → Y such that Moreover, for every n ∈ N 0 and x, y ∈ X satisfying x + y, x − y ∈ X. Hence g e satisfies the Drygas equation.
In the same way, replacing y by x in (4) and using the oddness of f o , we obtain and ε(x) = 2 2 p c x p , x ∈ X. By Theorem 1, there exists a function g o : X → Y such that By g o satisfies the Drygas equation. Thus g = g e + g o also satisfies the Drygas equation and

Nonstability Results
In this section we show that for p ∈ {1, 2} the Drygas equation is not stable. The idea of the construction of the examples comes from the paper [2].
where α > 0. The function f : R → R given by but there exist no pair (g, k) of a function g : R → R satisfying the Drygas equation and a constant k ≥ 0 such that Proof. We observe that f is odd and bounded by 2α. Now, we show that (10) holds. For x = y = 0 and x, y ∈ R such that |x|+|y| ≥ 1 it is obvious. Consider the case 0 < |x| + |y| < 1. There exists N ∈ N such that Then |2 N −1 x| < 1, |2 N −1 y| < 1, |2 N −1 (x + y)| < 1, |2 N −1 (x − y)| < 1. Hence 2 n x, 2 n y, 2 n (x + y), 2 n (x − y) ∈ (−1, 1) for n = 0, 1, . . . , N − 1. By the definition of f , Assume that there exist a function g : R → R satisfying the Drygas equation and a constant k ≥ 0 such that Since g fulfills the Drygas equation, there exist an additive function h : R → R and a quadratic function q : Whence, as f is bounded by 2α, we have In particular, The function q satisfies the quadratic functional equation, which implies Hence q(x) = 0, x ∈ R and It follows that, the additive function h is bounded in the neighborhood of 0, and consequently h(x) = ax, x ∈ R for some constant a ∈ R. Thus Let N be such that Nα > k + |a| and take an x ∈ (0, 1 2 N −1 ). Then 2 n x ∈ (0, 1) for n = 0, 1, . . . , N − 1 and which is contrary to (11). For p = 2 we have the same example like in the case of the quadratic equation (see [2]). Open Access. This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited.