Stability of a functional equation deriving from cubic and quartic functions

In this paper, we obtain the general solution and the generalized Ulam-Hyers stability of the cubic and quartic functional equation&4(f(3x+y)+f(3x-y))=-12(f(x+y)+f(x-y))&+12(f(2x+y)+f(2x-y))-8f(y)-192f(x)+f(2y)+30f(2x).


Introduction
The stability problem of functional equations originated from a question of Ulam [34] in 1940, concerning the stability of group homomorphisms. Let (G1, .) be a group and let (G2, * ) be a metric group with the metric d(., .). Given ǫ > 0, does there exist a δ > 0, such that if a mapping h : G1 −→ G2 satisfies the inequality d(h(x.y), h(x) * h(y)) < δ for all x, y ∈ G1, then there exists a homomorphism H : G1 −→ G2 with d(h(x), H(x)) < ǫ for all x ∈ G1? In the other words, Under what condition does there exists a homomorphism near an approximate homomorphism? The concept of stability for functional equation arises when we replace the functional equation by an inequality which acts as a perturbation of the equation. In 1941, D. H. Hyers [9] gave the first affirmative answer to the question of Ulam for Banach spaces. Let f : E −→ E ′ be a mapping between Banach spaces such that for all x, y ∈ E, and for some δ > 0. Then there exists a unique additive mapping T : for all x ∈ E. Moreover if f (tx) is continuous in t ∈ R for each fixed x ∈ E, then T is linear. Finally in 1978, Th. M. Rassias [31] proved the following theorem. Theorem 1.1. Let f : E −→ E ′ be a mapping from a normed vector space E into a Banach space E ′ subject to the inequality for all x, y ∈ E, where ǫ and p are constants with ǫ > 0 and p < 1. Then there exists a unique additive mapping T : E −→ E ′ such that for all x ∈ E. If p < 0 then inequality (1.1) holds for all x, y = 0, and (1.2) for x = 0. Also, if the function t → f (tx) from R into E ′ is continuous in real t for each fixed x ∈ E, then T is linear.
The oldest cubic functional equation, and was introduced by J. M. Rassias (in -2001: [17][18], as follows: f (x + 2y) + 3f (x) = 3f (x + y) + f (x − y) + 6f (y). Jun and Kim [11] introduced the following cubic functional equation and they established the general solution and the generalized Hyers-Ulam-Rassias stability for the functional equation (1.3). The function f (x) = x 3 satisfies the functional equation (1.3), which is thus called a cubic functional equation. Every solution of the cubic functional equation is said to be a cubic function. Jun and Kim proved that a function f between real vector spaces X and Y is a solution of (1.3) if and only if there exists a unique function for all x ∈ X, and C is symmetric for each fixed one variable and is additive for fixed two variables. The oldest quartic functional equation, and was introduced by J. M. Rassias(in 1999: [16], [27], and then (in 2005) was employed by Won-Gil Park [15] and others, such that: In fact they proved that a function f between real vector spaces X and Y is a solution of (1.4) if and only if there exists a unique symmetric multi-additive function Q : for all x (see also [3,4], [12][13][14][15], [33]). It is easy to show that the function f (x) = x 4 satisfies the functional equation (1.4), which is called a quartic functional equation and every solution of the quartic functional equation is said to be a quartic function.
We deal with the following functional equation deriving from quartic and cubic functions: It is easy to see that the function f (x) = ax 4 + bx 3 is a solution of the functional equation (1.5). In the present paper we investigate the general solution and the generalized Hyers-Ulam-Rassias stability of the functional equation (1.5).

General solution
In this section we establish the general solution of functional equation (1.5). Theorem 2.1. Let X,Y be vector spaces, and let f : X −→ Y be a function. Then f satisfies (1.5) if and only if there exists a unique symmetric multi-additive function Q : X × X × X × X −→ Y and a unique function C : X × X × X → Y such that C is symmetric for each fixed one variable and is additive for fixed two variables, and that f ( Proof. Suppose there exists a symmetric multi-additive function Q : X × X × X × X −→ Y and a function C : X × X × X → Y such that C is symmetric for each fixed one variable and is additive for fixed two variables, and that f ( Then it is easy to see that f satisfies (1.5). For the converse let f satisfy (1.5). We decompose f into the even part and odd part by setting for all x ∈ X. By (1.5), we have for all x, y ∈ X. This means that fe satisfies (1.5), or Now putting x = y = 0 in (1.5(e)), we get fe(0) = 0. Setting x = 0 in (1.5(e)), by evenness of fe we obtain for all y ∈ X. Hence (1.5(e)) can be written as for all x, y ∈ X. With the substitution y := 2y in (2.2), we have Replacing y by x + 2y in (2.2), we obtain Interchanging x and y in (2.3), we get fe(2x + 3y) + fe(2x − 3y) + 3fe(2x + y) If we add (2.3) to (2.7), we have And by utilizing equations (2.4), (2.5) and (2.8), we arrive at Let us interchange x and y in (2.9). Then we see that If we compare (2.11) and (2.6), we conclude that This means that fe is quartic function. Thus there exists a unique symmetric multi-additive function Q : On the other hand we can show that fo satisfies (1.5), or Hence (1.5(o)) can be written as fo(3x + y) + fo(3x − y) + 3fo(x + y) for all x, y ∈ X. Replacing x by x + y, and y by x − y in (2.13) we have 8fo(2x + y) + 8fo(x + 2y) + 24fo(x) + 24fo(y) = 3fo(3x + y) + 3fo(x + 3y) + 12fo(x + y) (2.14) and interchanging x and y in (2.13) yields Let us interchange x and y in (2.19). Then we see that With the substitution y := x + y in (2.13), we have fo(4x + y) + fo(2x − y) + 3fo(2x + y) − 3fo(y) = 3fo(3x + y) + 3fo(x − y) + 12fo(x) (2.21) and replacing −y by y gives fo(4x − y) + fo(2x + y) + 3fo(2x − y) + 3fo(y) = 3fo(3x − y) + 3fo(x + y) + 12fo(x Let us interchange x and y in (2.25). Then we see that Which, by putting y := 2y in (2.14), leads to 64fo(x + y) + 8fo(x + 4y) + 24fo(x) + 192fo(y) = 3fo(3x + 2y) + 3fo(x + 6y) + 12fo(x + 2y). If we subtract (2.30) from (2.31), we obtain 8fo(x + 4y) − 8fo(x − 4y) = 3fo(3x + 2y) − 3fo(3x − 2y) Setting x instead of y and y instead of x in (2.24), we get Interchanging y with 2y in (2.29) gives the equation We obtain from (2.36) and (2.37) (2.40) If we compare (2.40) and (2.26), we conclude that 8fo(x + y) + 8fo(x − y) + 48fo(x) = 4fo(2x + y) + 4fo(2x − y). This means that fo is cubic function and that there exits a unique function C : X ×X ×X −→ Y such that fo(x) = C(x, x, x) for all x ∈ X, and C is symmetric for each fixed one variable and is additive for fixed two variables. Thus for all x ∈ X, we have This completes the proof of Theorem.
The following Corollary is an alternative result of above Theorem 2.1.
Corollary 2.2. Let X,Y be vector spaces, and let f : X −→ Y be a function satisfying (1.5). Then the following assertions hold.
a) If f is even function, then f is quartic. b) If f is odd function, then f is cubic.

Stability
We now investigate the generalized Hyers-Ulam-Rassias stability problem for functional equation (1.5). From now on, let X be a real vector space and let Y be a Banach space. Now before taking up the main subject, given f : X → Y , we define the difference operator for all x, y ∈ X. We consider the following functional inequality : for an upper bound φ : X × X → [0, ∞). for all x, y ∈ X. If the upper bound φ : X × X → [0, ∞) is a mapping such that the series P ∞ i=0 2 4si φ(0, x 2 si ) converges, and that limn→∞ 2 4sn φ( x 2 sn , y 2 sn ) = 0 for all x, y ∈ X, then the limit Q(x) = limn→∞ 2 4sn f ( x 2 sn ) exists for all x ∈ X, and Q : X → Y is a unique quartic function satisfying (1.5), and for all x ∈ X.
Interchanging x with x 2 in (3.4), and multiplying by 16 it follows that 16f ( Combining (3.4) and (3.5), we lead to From the inequality (3.4) we use iterative methods and induction on n to prove our next relation: (3.7) We multiply (3.7) by 16 m and replace x by x 2 m to obtain that This shows that {16 n f ( x 2 n )} is a Cauchy sequence in Y by taking the limit m → ∞. Since Y is a Banach space, it follows that the sequence {16 n f ( x 2 n )} converges. We define Q : for all x ∈ X, and it follows from (3.1) that y 2 n ) = 0 for all x, y ∈ X. Hence by Corollary 2.2, Q is quartic. It remains to show that Q is unique. Suppose that there exists another quartic function Q ′ : X → Y which satisfies (1.5) and (3.2). Since Q(2 n x) = 16 n Q(x), and Q ′ (2 n x) = 16 n Q ′ (x) for all x ∈ X, we conclude that for all x ∈ X. By letting n → ∞ in this inequality, it follows that Q(x) = Q ′ (x) for all x ∈ X, which gives the conclusion. For s = −1, we obtain from which one can prove the result by a similar technique. is a mapping such that P ∞ i=0 2 3si φ(0, x 2 si ) converges, and that limn→∞ 2 3si φ( x 2 si , y 2 si ) = 0 for all x, y ∈ X, then the limit C(x) = limn→∞ 2 3sn f ( x 2 sn ) exists for all x ∈ X, and C : X → Y is a unique cubic function satisfying (1.5), and for all x ∈ X.
Proof. Let s = 1. Set x = 0 in (3.8). We obtain An induction argument now implies (3.12) Multiply (3.12) by 8 m and replace x by x 2 m , we obtain that (3.13) The right hand side of the inequality (3.13) tends to 0 as m → ∞ because of by assumption, and thus the sequence {2 3n f ( x 2 n )} is Cauchy in Y, as desired. Therefore we may define a mapping C : X → Y as C(x) = limn→∞ 2 3n f ( x 2 n ). The rest of proof is similar to the proof of Theorem 3.1.

14)
and for all x, y ∈ X. Then there exists a unique quartic function Q : X → Y and a unique cubic function C : X → Y satisfying for all x ∈ X.
Proof. Let fe(x) = 1 2 (f (x) + f (−x)) for all x ∈ X. Then fe(0) = 0 and fe is even function satisfying for all x, y ∈ X. From Theorem 3.1, it follows that there exists a unique quartic function Q : X → Y satisfies Then fo is odd function satisfying for all x, y ∈ X. Hence in view of Theorem 3.2, it follows that there exists a unique cubic function C : X → Y such that {2 3s(i+1) φ(0, x 2 s(i+1) ) + 2 3s(i+1) φ(0, −x 2 s(i+1) )} (3.17) for all x ∈ X. On the other hand we have f (x) = fe(x) + fo(x) for all x ∈ X. Then by combining (3.16) and (3.17), it follows that for all x ∈ X, and the proof of Theorem is complete.
We are going to investigate the Hyers-Ulam -Rassias stability problem for functional equation (1.5). for all x, y ∈ X. Then there exists a unique quartic function Q : X → Y , and a unique cubic function C : X → Y satisfying Proof. Let s = 1 in Theorem 3.3. Then by taking φ(x, y) = θ( x p + y p ) for all x, y ∈ X, the relations (3.14) and (3.15) hold for p > 4. Then there exists a unique quartic function Q : X → Y and a unique cubic function C : X → Y satisfying for all x ∈ X. Let now s = −1 in Theorem 3.3 and put φ(x, y) = θ( x p + y p ) for all x, y ∈ X. Then the relations (3.14) and (3.15) hold for p < 3. Then there exists a unique quartic function Q : X → Y and a unique cubic function C : X → Y satisfying Similarly, we can prove the following Ulam stability problem for functional equation (1.5) controlled by the mixed type product-sum function (x, y) → θ( x u X y v X + x p + y p ) introduced by J. M. Rassias (see for example [32]).