A New Fixed Point Theorem and a New Generalized Hyers-Ulam-Rassias Stability in Incomplete Normed Spaces

: In this study, our goal is to apply a new ﬁxed point method to prove the Hyers-Ulam-Rassias stability of a quadratic functional equation in normed spaces which are not necessarily Banach spaces. The results of the present paper improve and extend some previous results.


Introduction
The notion of the stability of functional equations was presented in 1940 by Ulam [1], "Under what conditions does there exist an additive mapping near an approximately additive mapping?" One year later, Hyers [2] found a partial answer to Ulam's question in a Banach space. Since then, the stability of such forms is known as Hyers-Ulam stability. In 1978, Rassias [3] proved the existence of unique linear mapping near approximate additive mapping, which provides a remarkable generalization of the Hyers-Ulam stability. Gavruta [4] investigated a different generalization of the Hyers-Ulam-Rassias theorem. For more details, see References [5][6][7][8][9][10][11]. Also, there are several applications of this concept in pure mathematics, sociology, financial and actuarial mathematics and psychology [12].
A Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [13] for mappings f : X → Y, where X is a normed space and Y is a Banach space. Cholewa [14] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group. The Hyers-Ulam-Rassias stability of the quadratic functional equation was proved in Reference [15]. Several functional equations have been presented in References [16,17].
There are many forms of the quadratic functional equation, one among them of great interest to us is the following: The fixed point method for studying the stability of functional equations was used for the first time in 1991 by Baker [18]. Yang [19] proved the Hyers-Ulam-Rassias stability of the quadratic functional Equation (1) in F-spaces.

A New Hyers-Ulam-Rassias Stability
In this section, we will assume that (X, . X ) and (Y, . Y ) are two normed spaces. We denote by d the induced metric by . Y and ⊥ is an orthogonal relation on Y which is R-preserving. Theorem 3. Let (Y, d, ⊥) be an SO-complete orthogonal metric space (not necessarily complete metric space). Assume that f : X → Y is a function such that and φ : for each x, y ∈ X. Suppose there exists a function α : [0, ∞) → [0, 1) satisfying the following statements: for all x ∈ X. Then there exists a quadratic function F : X → Y and a nonempty subset X * in X such that for some positive real number L < 1 we have for all x ∈ X * .
Proof. Consider S 0 := {g : X → Y | g(0) = 0} with the following generalized metric, for all h, g ∈ S 0 . Taking x = y = 0 in (A2), we see that φ(0, 0) = 0 and by using (3) we observe that f (0) = 0. Hence f ∈ S 0 and S 0 is a nonempty set. Let S = {g ∈ S 0 | D(g, f ) < ∞} and T : S → S 0 be a function given by for every x ∈ X. In order to show that T(S) ⊆ S, substitute y = 0 in (3) we have for all x ∈ X. Replacing x with x 2 in the above equation and employing (A2), we have for all x ∈ X. This implies that D(T f , f ) ≤ 1 8 . Now if g ∈ S, then the definition of D and the relation (A2) conclude that D(Tg, T f ) ≤ D(g, f ) and the triangle inequality results that So Tg ∈ S and hence T is self-adjoint mapping, that is T(S) ⊆ S. Consider for all x ∈ X and for each g, h ∈ S we define ⊥ S on S as follows: Clearly, (S, ⊥ S ) is an O-set. We now show that (S, d, ⊥ S ) is an SO-complete orthogonal metric space, first of all we need to prove that for each x ∈ X, the sequence {(T n f )(x)} is a Cauchy SO-sequence in Y. To see this, since the relation ⊥ is R-preserving, the definition of ⊥ S implies that T is ⊥ S -preserving. According to the assumptions (2) and R-preserving of ⊥, we obtain ∀x ∈ X, ∀n ∈ N, Replacing x by x 2 k and multiplying both sides of the above relations by 4 k , we obtain and multiplying both sides of the inequality (7) by 4 n and making use of (A2) and (A3), we get for all x ∈ X and n ∈ N.
for all x ∈ X and m, n ∈ N. Since L x < 1, taking the limit as m, n → ∞ in the above inequality, we deduce that the sequence {(T n f )(x)} is a Cauchy sequence for each x ∈ X. By SO-completeness of Y, we obtain that for every x ∈ X, there exists an element F(x) ∈ Y which is a limit point of {(T n f )(x)}. That is, F : X → Y is well-defined and is given by Let x 0 be an arbitrary point in X. We can see that the following cases can occur: On the other hand, since every Cauchy sequence with a convergent subsequence is convergent, the sequence {g n (x 0 )} is convergent.
Case 2. {g n (x 0 )} is an SO-sequence in Y. Let > 0 be given. Since {g n } is a Cauchy sequence in S, then there exists N ∈ N such that D(g n , g m ) < for every n, m ≥ N which implies the following inequality: for every n, m ≥ N and x ∈ X. This means that for every x ∈ X, {g n (x)} is a Cauchy sequence in Y.
The SO-completeness of Y implies that {g n (x 0 )} is a convergent sequence.
In the above two cases, there is a point g(x 0 ) ∈ Y such that lim n→∞ g n (x 0 ) = g(x 0 ). According to the choice of x 0 , we can see that g : X → Y is well-defined and also, g(x) = lim n→∞ g n (x) for each x ∈ X. If we take the limit as m → ∞ in the inequality (10), then for every n ≥ N and x ∈ X. From the definition of D, we gain D(g n , g) ≤ for all n ≥ N, that is, g ∈ S and {g n } is a convergent sequence. Therefore, (S, D, ⊥ S ) is an SO-complete orthogonal metric space.
On the other hand, since lim sup t→0 + α(t) < 1, then there exist r ∈ (0, ∞] and 0 < L < 1 such that Hence we see that D(Tg, Th) ≤ LD(g, h) for all g, h ∈ S. It follows from L < 1 that T is a contraction. Consequently, T is an SO-continuous mapping and is a contraction on ⊥ S -comparable elements with Lipschitz constant L. Since (S, D, ⊥ S ) is SO-complete and T is also ⊥ S -preserving, then from the fixed point Theorem 2, we conclude that T has a unique fixed point and T is a Picard operator. This means that the sequence {T n f } converges to the fixed point of T. It follows from (8) that F is a unique fixed point of T. Moreover, (7) ensures that the inequality (4) holds. Finally, we will show that F is a quadratic mapping. To this aim, fix x and y in X. Since {φ( x 2 n , y 2 n )} is a non-negative and decreasing sequence, then there is τ ≥ 0 for which φ( x 2 n , y 2 n ) → τ as n → ∞. Taking into account (A1), we have lim sup t→τ + α(t) < 1, so there exist δ > 0 and ν < 1 such that for all t ∈ [τ, τ + δ), α(t) < ν. Consider the positive integer N such that for all n ≥ N, φ( x 2 n , y 2 n ) ∈ [τ, τ + δ). By virtue of (3), we obtain This completes the proof.

Corollary 1.
Let Y be a Banach space and f : X → Y be a function such that there exists a function φ : X 2 → [0, ∞) satisfying (3). If there exists a positive real number L < 1 such that for all x, y ∈ X. Then there exists a unique quadratic mapping F : X → Y which satisfies the inequality for all x ∈ X.
Proof. For every y 1 , y 2 ∈ Y, we define y 1 ⊥y 2 if and only if y 1 Y ≤ y 2 Y . It is easy to see that (Y, ⊥) is an O-set. Moreover, since Y is a Banach space, then (Y, d, ⊥) is an SO-complete orthogonal metric space which d is the induced metric by norm. From the definition of ⊥, it follows that Setting α(t) = L for all t ∈ [0, ∞), from the proof of Theorem 3 we can see the result.
Assume that there exists a function φ : X 2 → [0, ∞) satisfying the Equation ( y) for all x, y ∈ X not both being zero; Then there exist a quadratic function F : X → Y and a nonempty subset X * of X such that for some positive real number L < 1 we have for all x ∈ X * .
Proof. By the same reasoning as in the proof of Theorem 3, there are λ ∈ (0, ∞] and 0 < L < 1, such that α(t) ≤ L for each 0 ≤ t < λ. Set X * := {x ∈ X| x = 0, [φ(x, 0)] −1 < λ} ∪ {0}. By the same argument of Theorem 3, one can show that the mapping T : S → S defined by Tg(x) = 1 4 g(2x) for all x ∈ X, is a ⊥ S -preserving mapping. Define F : X → Y by for all x ∈ X. Replacing X * by X in definition of S 0 we obtain that T is a contraction with Lipschitz constant L. Applying Theorem 2 we can see F is a unique fixed point of T. Dividing both sides of the inequality (6) by 4, we have for all x ∈ X. In fact, D( f , T f ) ≤ 1 8 . It follows that and consequently, .
To show that the function F is quadratic, let us consider x, y are elements in X which not both zero. Since [φ(2 n x, 2 n y)] −1 is a non-negative and decreasing sequence in R + , so the rest of the proof is similar to the proof of Theorem 3.

Corollary 2.
Let Y be a Banach space and f : X → Y be a mapping such that there exists a function φ : X 2 → [0, ∞) satisfying the condition (B1) and inequality (3) of Theorem 4. If there exists a positive real number L < 1 such that for all x, y ∈ X. Then there exists a unique quadratic mapping F : X → Y which satisfies the inequality for all x ∈ X.
Proof. Take the same metric d and the orthogonal relation of Corollary 1. By the same argument of Corollary 1, one can show that (Y, d, ⊥) is an SO-complete orthogonal metric space and the relation (12) holds. Putting α(t) = L for all t ∈ [0, ∞) and applying Theorem 4, we can easily obtain the result.

Corollary 3.
Suppose that Y is a Banach space and θ ≥ 0 and r = 2 are fixed. Assume that f : X → Y is a function which satisfies the functional inequality for all x, y ∈ X. Then there exists a unique quadratic mapping F : X → Y such that the inequality holds for all x ∈ X, where r > 2, or the inequality holds for all x ∈ X, where r < 2.
Proof. Take the same metric d and the orthogonal relation of Corollary 1. By the same argument of Corollary 1, one can show that (Y, d, ⊥) is an SO-complete orthogonal metric space. Moreover, the definition of ⊥ ensures that the relations (2) and (12) hold. Let φ(x, y) = θ( x r X + y r X ) for each x, y ∈ X. It follows that for all x, y ∈ X where r > 2. Set α(t) = 1 2 r−2 for all t ∈ [0, ∞). This ensures that X * = X and the relations (A1) and (A3) of Theorem 3 hold. Applying Theorem 3, we see that inequality (4) holds with L = 1 2 r−2 which yields the inequality (16). On the other hand, the function φ satisfies in the properties (B1), (B2) and also, for all x, y ∈ X, where r < 2. Putting α(t) = 1 2 2−r for every t ∈ [0, ∞), it is easily seen that X * = X and the conditions (A1) and (B3) hold. Employing Theorem 4, we see that the inequality (13) holds with L = 1 2 2−r . This implies the inequality (17).
The next example shows that Theorem 3 is a real extension of Corollary 1.

Example 1.
Let Y be a Banach space. Suppose that a function f : X → Y has the property for all x, y ∈ X, where φ : We define a function α : [0, ∞) → [0, 1) as (C3) For every positive real number and r, there exist a constant L ∈ (0, 1) and a quadratic mapping F : X → Y such that the inequality (4) holds for any x ∈ X with x X ≤ r.
Proof. Take the same metric d and the orthogonal relation of Corollary 1. By the same argument of Corollary 1, one can show that (Y, d, ⊥) is an SO-complete orthogonal metric space and the relation (2) holds. Let us take x, y ∈ X with and m be the smallest natural number such that Then From the inequality (18), we observe that This follows that there exists k 0 ∈ N for which Assume k is the smallest natural number satisfying the above condition. Clearly, k > m and Suppose that r is the smallest natural number such that k ( x X + y X ) ≤ r, then α(φ(x, y)) = r−1 r . Since x X + y X > 1, then k < r and we conclude that This implies that Therefore, the property (C2) holds. From the definition of the function α, it is easily seen that α is an increasing mapping. Finally, it follows from lim sup t→ 0 + α(t) = 0 that for every r > 0 there exists L < 1 such that α(φ(x, 0)) ≤ L for all x ∈ X with x X ≤ r. By the same proof of Theorem 3, we prove (C3).
Note that there is no L < 1 such that the inequality (11) holds and hence the stability of f does not imply by Corollary 1.
In the following example, we observe that our results go further than the stability on Banach spaces.

Example 2.
Assume that θ and r are two real numbers such that θ ≥ 0 and r = 2. Consider Y = {x = {x n } ⊂ R; ∃n 1 , n 2 , ..., n k ; ∀n = n 1 , n 2 , . . . , n k , x n = 0} with norm x Y = ∑ ∞ n=1 |x n | p 1 p where 1 < p < ∞. Suppose f : X → Y is a mapping satisfying the inequality (7) and the following condition Then there exists a unique quadratic mapping F : X → Y such that the inequality (8) holds for all x ∈ X, where r > 2, or the inequality (9) holds for all x ∈ X, where r < 2.
Proof. Let q be the conjugate of p; that is, 1 p + 1 q = 1. Note that (Y, . Y ) is not a Banach space because, A n = {1, 1 2 , ..., 1 2 n , 0, 0, 0, ...}, n ∈ N, is a sequence in Y where the limit of {A n } does not belong to Y. For all A = {x n } and B = {y n } in Y, define