Stability of an additive-quadratic-quartic functional equation

In this paper, we investigate the stability of an additive-quadratic-quartic functional equation$$\begin{align*}f(x+2y)& +f(x-2y)-2f(x+y)-2f(-x- y)-2f(x-y)-2f(y-x)\nonumber \\ &+4f(-x)+ 2f(x)-f(2y)-f(-2y)+4f(y)+4f(-y)=0 \end{align*}$$by the direct method in the sense of Găvruta.


Introduction
A. K. Hassan et al. [1], M. Mohamadi et al. [2] and C. Park et al. [3] investigated the stability of the AQQ (additivequadratic-quartic) functional equation in various spaces. For the terminology "AQQ (additive-quadratic-quartic) functional equation", refer to the papers [1][2][3]. The second author [4] also studied different type of the additive-quadratic-quartic functional equation f (x + ky) + f (x − ky) − k 2 f (x + y) − k 2 f (x − y) + 2(k 2 − 1)f (x) + (k 2 + k)f (y) + (k 2 − k)f (−y) − 2f (ky) = 0, where k is a fixed real constant with k ≠ 0, ±1. In this paper, let V and W be real vector spaces and Y be a real Banach space. For a given mapping f : V → W, we use the following abbreviations for all x, y ∈ V. In this paper, we will prove the stability of the functional equation (1.1) in the sense of Găvruta [5] (See also [6,7]). In other words, from the given mapping f that approximately satisfies the functional equation (1.1), we will show that the mapping F, which is the solution of the functional equation (1.1), can be constructed using the formula )︁ )︂ and we will prove that the mapping F is the unique solution mapping of functional equation (1.1) near the mapping f .

Lemma 1.
If a mapping f : V → W satisfies Df (x, y) = 0 for all x, y ∈ V, then the equalities hold for all x ∈ V and all n ∈ N ∪ {0}.
Proof. If a mapping f : V → W satisfies Df (x, y) = 0 for all x, y ∈ V, then the equality (2.1) can be derived from the equalities for all x ∈ V and n ∈ N ∪ {0}. The equality (2.2) can be easily obtained in a similar way.
In the following theorem, we can prove the generalized Hyers-Ulam stability of the functional equation (1.1) in the sense of Găvruta.

. In particular, F is represented by
for all x ∈ V.
Proof. First, we define a set A : for all x ∈ V∖{0}. It follows from (2.7) that for all x ∈ V∖{0}. In view of (2.4) and (2.8), the sequence {Jn f (x)} is a Cauchy sequence for all x ∈ V∖{0}.
Since Y is complete and f (0) = 0, the sequence {Jn f (x)} converges for all x ∈ V. Hence, we can define a mapping F : V → Y by for all x ∈ V. Moreover, letting n = 0 and passing the limit m → ∞ in (2.8) we get the inequality (2.5). With the definition of F, we easily get the equality DF(x, y) = 0 from the relations for all x, y ∈ V∖{0}.
. By (2.1), the equality F ′ (x) = Jn F ′ (x) holds for all n ∈ N. Therefore, we have for all x ∈ V∖{0} and all n ∈ N. Taking the limit in the above inequality as n → ∞, we conclude that In the following corollary, we obtain the hyperstability of the functional equation (1.1).

Corollary 1. Let p < 0 be a real number and X be a real normed space. If f : X → Y is a mapping such that
‖Df (x, y)‖ ≤ θ(‖x‖ p + ‖y‖ p ) (2.9) for all x, y ∈ X∖{0} and f (0) = 0, then f : X → Y satisfies the equality Df (x, y) = 0 for all x, y ∈ X.
Theorem 2. Let f : V → Y be a mapping for which there exists a function φ : holds for all x, y ∈ V and let f (0) = 0. If φ has the property for all x, y ∈ V, then there exists a unique solution mapping F : V → Y of the functional equation (1.1) satisfying the inequality for all x ∈ V∖{0}. In particular, F is represented by for all x ∈ V.
In view of (2.11) and (2.15), the sequence {Jn f (x)} is a Cauchy sequence for all x ∈ V∖{0}. Since Y is complete and f (0) = 0, the sequence {Jn f (x)} converges for all x ∈ V. Hence, we can define a mapping for all x ∈ V. Moreover, letting n = 0 and passing the limit n → ∞ in (2.15) we get the inequality (2.12). From the definition of F, we easily get for all x, y ∈ V∖{0}, which means that DF(x, y) = 0 for all x, y ∈ V from the same reason in Theorem 1.
To prove the uniqueness of F, let F ′ : V → Y be another solution of the functional equation (1.1) satisfying (2.12). Instead of the condition (2.12), it is sufficient to show that there is a unique mapping satisfying the simpler condition for all x ∈ V. By (2.2), the equality F ′ (x) = Jn F ′ (x) holds for all x ∈ V and all n ∈ N. Therefore, we have for all x ∈ V and all n ∈ N. Taking the limit in the above inequality as n → ∞, we can conclude that F ′ (x) = limn→∞ Jn f (x) for all x ∈ V. This means that F(x) = F ′ (x) for all x ∈ V.