ULAM STABILITY OF A MIXED TYPE ADDITIVE-QUARTIC FUNCTIONAL EQUATION

In this paper, we obtain the generalized Hyers-Ulam stability of a the functional equation f(2x + y) + f(x + 2y) − 9f(x + y) + f(x − y)


Introduction
In 1940, S.M. Ulam [23] raised the following question concerning the stability of group homomorphisms: "Let G be a group and H be a metric group with metric d(., .).Given ǫ > 0, does there exist a δ > 0 such that if a function f : G → H satisfies d(f (xy), f (x)f (y)) < δ for all x, y ∈ G, then there exists a homomorphism a : G → H with d(f (x), a(x)) < ǫ for all x ∈ G?" In 1941, D.H. Hyers [10] gave an answer to the Ulam's stability problem.He proved the following celebrated theorem.Theorem 1. (D.H. Hyers) Let X, Y be Banach spaces and let f : X → Y be a mapping satisfying for all x, y ∈ X.Then the limit exists for all x ∈ X and a : X → Y is the unique additive mapping satisfying for all x ∈ X.
In 1950, Aoki [2] generalized the Hyers theorem for additive mappings.In 1978, Th.M. Rassias [22] provided a generalized version of the theorem of Hyers which permitted the cauchy difference to become unbounded.Th.M. Rassias proved the following theorem for sum of powers of norms.
Theorem 2. (Th.M. Rassias) Let X and Y be two Banach spaces.Let θ ∈ [0, ∞) and let p ∈ [0, 1).If a function f : X → Y satisfies the inequality for all x, y ∈ X.Then there exists a unique linear mapping T : for all x ∈ X.Moreover, if f (tx) is continuous in t for each fixed x ∈ X, then the function T is linear.
The theorem of Th.M. Rassias was later extended for all p = 1.The stability phenomenon that was presented by Th.M. Rassias is called the generalized Hyers-Ulam stability.
In the year 2010, M.E.Gordji [8] obtained the general solution and generalized Hyers-Ulam stability of the mixed type quartic-additive functional equation In this paper, we consider the following functional equation deriving from additive and quartic functions It is easy to see that the function f (x) = ax + bx 4 is a solution of the functional equation (6).In this paper, we obtain the general solution and the generalized Hyers-Ulam stability of the functional equation ( 6).

General Solution of Equation (6)
In this section, we obtain the general solution of functional equation (6).
Theorem 3. Let X, Y be vector spaces.An odd function f : X → Y satisfies (6) if and only if f is additive.
Proof.Since f is an odd function, equation ( 6) can be written as (7) for all x, y ∈ X.Now, replacing (x, y) by (x, x) in (7), we get for all x ∈ X.Now, plugging (x, y) by (0, 2x) in (7), we obtain for all x ∈ X.Now, substituting x by x 2 in (9), we have for all x ∈ X.Now, setting (x, y) = (x, x + 2y) in (7) and using (10) in the resulting equation, we deduce that 2f (x + 2y) = 2f (2x + y) + 2f (x − y) for all x, y ∈ X. Replacing (x, y) by (x − y, x + y) in (7), we get for all x, y ∈ X. Switching (x, y) to (x + y, x − y) in (12), we obtain for all x, y ∈ X.By the virtue of equation ( 11) and after simplification, equation ( 13) can be written as for all x, y ∈ X.Now, replacing (x, y) by (x, y − x) in ( 14), we obtain for all x, y ∈ X.This means that f is an additive function.
Conversely, suppose f is an additive function.Then f satisfies (15).Replacing y by −y in (15), we get for all x, y ∈ X.Now, multiplying −9 in (15), we obtain for all x, y ∈ X. Summing the equation ( 16) with (17), we get for all x, y ∈ X. Plugging x into 2x in (15), we obtain for all x, y ∈ X.Now, switching y to 2y in (15), we get for all x, y ∈ X. Adding the equations ( 18), ( 19) and ( 20), we get equation ( 6).
Theorem 4. Let X, Y be vector spaces.An even function f : X → Y satisfies (6) if and only if f is quartic.
Proof.Since f is an even function, equation ( 6) can be written as for all x, y ∈ X. Plugging (x, y) = (0, 0) in ( 21), we get f (0) = 0. Now, replacing (x, y) by (0, x) in ( 21), we obtain for all x ∈ X.Now, subtituting (x, y) = (x − y, x + y) in ( 21), we get for all x, y ∈ X. Replacing y by −y in (23), we obtain for all x, y ∈ X. Setting (x, y) = (x + y, −y) in ( 21) and further simplification yields, for all x, y ∈ X.Using ( 21) in (37), we have for all x, y ∈ X.Using (38) in (36), we obtain for all x, y ∈ X.Now, interchanging x and y and simplifying further, we get for all x ∈ X.Using equations ( 22) and ( 28) in (39) and further simplification gives, for all x, y ∈ X, which is a quartic functional equation dealt by M.E.Gordji in his paper [5] with n = 3.
Conversely, suppose f is quartic.Then f satisfies the standard quartic functional equation of the form for fixed integer n = 0, ±1, for all x, y ∈ X.Now, when n = 2, the equation (30) becomes for all x, y ∈ X. Replacing (x, y) by 2x−y 2 , y in (31) and simplifying further, we get for all x, y ∈ X. Plugging (x, y) into (x − y, y) in (31), we obtain for all x, y ∈ X.Using (33) in (32), we arrive at for all x, y ∈ X.Now, replacing y by −y in (34) and simplifying further, we get for all x, y ∈ X.Using (31) in (35) and using f is quartic, we arrive at (6).This completes the proof of the theorem.
Theorem 5. Let X, Y be vector spaces, and let f : X → Y be a function.Then f satisfies (1) if and only if there exists a unique additive function A : X → Y and a unique symmetric multi-additive function B : for all x ∈ X.Then we have A(0) = 0 and and for all x, y ∈ X.Now, substituting x = 0 and y = x in (36), we get for all x ∈ X.Now, putting x = 0 and y = x in (37), we obtain for all x ∈ X. Substituting (38) in (36), we get for all x ∈ X. Theorem 3 guarantees that A is additive.
Using (39) in (37), we obtain for all x ∈ X. Theorem 4 guarantees that Q is quartic.Hence there exists a unique multi-additive function for all x ∈ X.
Conversely, suppose if f (x) = A(x) + B(x, x, x, x), then we have A(2x) = 2A(x) and Q(2x) = 16Q(x), for all x ∈ X. Hence it is easy to show that f satisfies (1), which completes the proof of the theorem.

Generalized Hyers-Ulam Stability of Equation (6)
Throughout this section, X and Y will be a real normed space and a real Banach space, respectively.Let f : X → Y be a function.Then we define D for all x, y ∈ X.Let N denote the set of natural numbers.
for all x, y ∈ X.If f : X → Y is an odd function such that f (0) = 0, and that for all x, y ∈ X, then there exists a unique additive function A : X → Y satisfying (1) and for all x ∈ X.
Proof.Replacing (x, y) by (0, x) in (40) and dividing by 2, we get for all x ∈ X.Now, plugging x by 2x in (41), dividing by 2 and then adding the resulting inequality with (41), we obtain for all x ∈ X. Proceeding further and using induction on a positive integer n, we get for all x ∈ X.Now, replacing x by 2 m x in (43) and dividing by 2 m , we obtain Since Y is a Banach space, then the sequence f (2 n x) for all x ∈ X.Since f is odd function, then A is odd.On the other hand, we have for all x, y ∈ X. Hence by Theorem 3, A is an additive function.To show that A is unique, suppose that there exists another additive function A ′ : X → Y which satisfies ( 1) and (41).We have for all x ∈ X.By taking n → ∞ in this inequality, we have A ′ (x) = A(x).
for all x, y ∈ X. Suppose that an odd function f : X → Y satisfies f (0) = 0, and (40) for all x, y ∈ X, then there exists a unique additive function A : X → Y satisfying (1) and for all x ∈ X.
Proof.Replacing x by x 2 in (42) and multiplying by 2, we get for all x ∈ X. Again replacing x by x 2 in (45) and summing the resulting inequality with (45), we get for all x ∈ X.By induction on n ∈ N , we have for all x ∈ X. Replacing x by 2 −m x in (46) and multiplying by 2 m , we obtain The rest of proof is similar to that of Theorem 6.
for all x, y ∈ X.If f : X → Y is an even function such that f (0) = 0, and that for all x, y ∈ X, then there exists a unique additive function Q : X → Y satisfying (1) and for all x ∈ X.
Proof.Replacing (x, y) by (0, x) in (47) and dividing by 16, we get for all x ∈ X.Now, substituting x by 2x in (49), dividing by 16 and then adding the resulting inequality with (49), we obtain for all x ∈ X. Proceeding further and using induction on a positive integer n, we get 1 for all x ∈ X.Now, replacing x by 2 m x in (50) and dividing by 16 m , we obtain Since Y is a Banach space, then the sequence f (2 n x) 16 n converges.We define 16 n for all x ∈ X.Since f is even function, then Q is even.On the other hand, we have 16 n = 0 for all x, y ∈ X. Hence by Theorem 3, Q is a quartic function.To show that Q is unique, suppose that there exists another additive function Q ′ : X → Y which satisfies (1) and (47).We have for all x ∈ X.By taking n → ∞ in this inequality, we have for all x ∈ X with the condition lim n→∞ for all x, y ∈ X. Suppose that an even function f : X → Y satisfies f (0) = 0, and (47) for all x, y ∈ X, then there exists a unique quartic function Q : X → Y satisfying (1) and for all x ∈ X.
Proof.Replacing x by x 2 in (47), we get for all x ∈ X. Again replacing x by x 2 in (52) and summing the resulting inequality with (52), we get for all x ∈ X.By induction on n ∈ N , we have for all x ∈ X. Replacing x by 2 −m x in (46) and multiplying by 16 m , we obtain , for all x ∈ X.The rest of proof is similar to that of Theorem 8.
for all x ∈ X with the condition for all x, y ∈ X. Suppose that a function f : X → Y satisfies the inequality for all x, y ∈ X, and f (0) = 0. Then there exist a unique additive function A : X → Y and a unique quartic function Q : X → Y satisfying (1) and for all x ∈ X.
Proof.We have for all x, y ∈ X.Since f e (0) = 0 and f e is an even function, then by Theorem 6, there exists a unique additive function A : for all x ∈ X.On the other hand f 0 is odd function and for all x, y ∈ X.Then by Theorem 8, there exists a unique quartic function for all x ∈ X. Combining (55) and (56) to obtain (54).This completes the proof of theorem.
Using Theorem 10, we solve the Hyers-Ulam stability problem for the functional equation (6) in the following corollary.x p for all x ∈ X.
Proof.Taking φ(x, y) = θ ( x p + y p ), for all x, y ∈ X in Theorem 10, we arrive at x p for all x ∈ X.