GENERAL QUARTIC-CUBIC-QUADRATIC FUNCTIONAL EQUATION IN NON-ARCHIMEDEAN NORMED SPACES

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1 U.P.B. Sci. Bull., Series A, Vol. 72, Iss. 3, 200 ISSN GENERAL QUARTIC-CUBIC-QUADRATIC FUNCTIONAL EQUATION IN NON-ARCHIMEDEAN NORMED SPACES M. Eshaghi Gordji, H. Khodaei 2, R. Khodabakhsh 3 The aim of this paper is to find the general solution of a mixed type quartic, cubic and quadratic functional equation f(x + ky) + f(x ky) = k 2 f(x + y) + k 2 f(x y) + 2( k 2 )f(x) + k2 (k 2 ) (f(2y) + 2f( y) f(y)) (k Z {0, ±}) in the class of functions between real vector spaces and to obtain the generalized Hyers-Ulam stability problem for the equation in non-archimedean spaces. Keywords: Quartic, cubic and quadratic functions; Non-Archimedean spaces; p adic field; Stability MSC2000: 39B82, 39B52.. Introduction In 897, Hensel [9] has introduced a normed space which does not have the Archimedean property. It turned out that non-archimedean spaces have many nice applications [3, 20, 22, 23]. A non-archimedean field is a field K equipped with a function (valuation). from K into [0, ) such that r = 0 if and only if r = 0, rs = r s, and r + s max{ r, s } for all r, s K. Clearly = = and n for all n N. An example of a non-archimedean valuation is the function. taking everything but 0 into and 0 = 0. This valuation is called trivial. Definition.. Let X be a vector space over a scalar field K with a non Archimedean non-trivial valuation.. A function. : X R is a non Archimedean norm (valuation) if it satisfies the following conditions: (NA) x = 0 if and only if x = 0; (NA2) rx = r x for all r K and x X; (NA3) x + y max{ x, y } for all x, y X (the strong triangle inequality). Then (X,. ) is called a non Archimedean space. Associate Professor, Department of Mathematics, Semnan University, P. O. Box , Semnan, Iran, madjid.eshaghi@gmail.com 2 Department of Mathematics, Semnan University, P. O. Box , Semnan, Iran, khodaei.hamid.math@gmail.com 3 Department of Mathematics, Semnan University, P. O. Box , Semnan, Iran, raziehkhodabakhsh@gmail.com 9

2 70 M. Eshaghi Gordji, H. Khodaei, R. Khodabakhsh Remark.. Thanks to the inequality x m x l max{ x j+ x j : l j m } (m > l) a sequence {x m } is Cauchy if and only if {x m+ x m } converges to zero in a non Archimedean space. By a complete non Archimedean space we mean one in which every Cauchy sequence is convergent. The most important examples of non-archimedean spaces are p-adic numbers. A key property of p-adic numbers is that they do not satisfy the Archimedean axiom: for x, y > 0, there exists n N such that x < ny. Example.. Let p be a prime number. For any nonzero rational number x = a b pr such that a and b are coprime to the prime number p, define the p-adic absolute value x p := p r. Then. p is a non-archimedean norm on Q. The completion of Q with respect to. p is denoted by Q p and is called the p-adic number field. Note that if p > 3, then 2 n = in for each integer n. The concept of stability of a functional equation arises when one replaces a functional equation by an inequality which acts as a perturbation of the equation. In 940, Ulam [2] posed the first stability problem. In the next year, Hyers [0] gave an affirmative answer to the question of Ulam. Hyers theorem was generalized by Aoki [2] for additive mappings and by Rassias [] for linear mappings by considering an unbounded Cauchy difference. The concept of the Hyers-Ulam-Rassias stability originated from Rassias paper [] for the stability of functional equations (see also [8, 7, 8, 9]). The functional equation f(x + y) + f(x y) = 2f(x) + 2f(y) () is related to a symmetric bi additive function [, 2]. It is natural that this equation is called a quadratic functional equation. In particular, every solution of the quadratic equation () is said to be a quadratic function. It is well known that a function f between real vector spaces X and Y is quadratic if and only if there exists a unique symmetric bi-additive function B such that f(x) = B (x, x) for all x X. The bi-additive function B is given by B (x, y) = (f(x + y) f(x y)) 4 for all x, y X. In the paper [5], Czerwik proved the Hyers Ulam Rassias stability of the equation (). Jun and Kim [] introduced the following functional equation f(2x + y) + f(2x y) = 2f(x + y) + 2f(x y) + 2f(x) (2) and they established the general solution and the generalized Hyers Ulam Rassias stability for functional equation (2). They proved that a function f between real vector spaces X and Y is a solution of (2) if and only if there exits a unique function C : X X X Y such that f(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. The function C is given by C(x, y, z) = (f(x + y + z) + f(x y z) f(x + y z) f(x y + z)) 24

3 General quartic-cubic-quadratic functional equation... 7 for all x, y, z X. It is easy to see that the function f(x) = cx 3 satisfies functional equation (2), so it is natural to call (2) the cubic functional equation and every solution of the cubic functional equation (2) is said to be a cubic function. Lee et. al. [4] considered the following functional equation f(2x + y) + f(2x y) = 4f(x + y) + 4f(x y) + 24f(x) f(y) (3) In fact, they proved that a function f between two real vector spaces X and Y is a solution of (3) if and only if there exists a unique symmetric bi-quadratic function B 2 : X X Y such that f(x) = B 2 (x, x) for all x. The bi-quadratic function B 2 is given by B 2 (x, y) = (f(x + y) + f(x y) 2f(x) 2f(y)) 2 for all x, y X. It is easy to show that the function f(x) = dx 4 satisfies the functional equation (3), which is called the quartic functional equation (see also [4]). In 2007, Moslehian and Rassias [5] proved the generalized Hyers Ulam stability of the Cauchy functional equation and the quadratic functional equation in non Archimedean normed spaces. Eshaghi Gordji and Khodaei [], have obtained the generalized Hyers Ulam Rassias stability for a mixed type of cubic, quadratic and additive functional equation. In addition the generalized Hyers Ulam Rassias stability for cubic and quartic functional equation in non-archimedean space has been investigated by Eshaghi Gordji and Bavand Savadkouhi [7]. In this paper, we deal with the following mixed type quartic, cubic and quadratic functional equation for fixed integers k ±, f(x + ky) + f(x ky) = k 2 f(x + y) + k 2 f(x y) + 2( k 2 )f(x) + k2 (k 2 ) (f(2y) + 2f( y) f(y)) It is easy to see that the function f(x) = dx 4 +cx 3 +bx 2 is a solution of the functional equation (4). The main purpose of this paper is to establish the general solution of Eq. (4) and investigate the generalized Hyers-Ulam stability for Eq. (4) in non-archimedean spaces. 2. Functional equations deriving from quartic, cubic and quadratic functions We here present the general solutions of (4). Theorem 2.. Let both X and Y be real vector spaces. A function f : X Y satisfies (4) for all x, y X if and only if there exist a unique symmetric bi-quadratic function B 2 : X X Y, a unique function C : X X X Y and a unique symmetric bi-additive function B : X X Y such that f(x) = B 2 (x, x) + C(x, x, x) + B (x, x) for all x X, and that C is symmetric for each fixed one variable and is additive for fixed two variables. (4)

4 72 M. Eshaghi Gordji, H. Khodaei, R. Khodabakhsh Proof. Let f satisfies (4). setting We decompose f into the even part and odd part by f e (x) = (f(x) + f( x)), 2 f o(x) = (f(x) f( x)) 2 for all x X. It is clear that f(x) = f e (x) + f o (x) for all x X. It is easy to show that the functions f e and f o satisfy (4). Now, let g, h : X Y be functions defined by g(x) = f e (2x) f e (x), h(x) = f e (2x) 4f e (x) for all x X. We show that the functions g and h are quadratic and quartic, respectively. Interchanging x with y in (4) and then using f e ( x) = f e (x), we have f e (kx + y) + f e (kx y) = k 2 f e (x + y) + k 2 f e (x y) + k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )f e (y) for all x, y X. Letting x = y = 0 in (5), we have f e (0) = 0. Putting y = x + y in (5), gives (5) f e ((k + )x + y) + f e ((k )x y) = k 2 f e (2x + y) + k 2 f e ( y) + k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )f e (x + y) for all x, y X. Replacing y by y in (), we obtain f e ((k + )x y) + f e ((k )x + y) = k 2 f e (2x y) + k 2 f e (y) for all x, y X. Adding () to (7), we get + k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )f e (x y) f e ((k + )x + y) + f e ((k + )x y) + f e ((k )x + y) + f e ((k )x y) = k 2 (f e (2x + y) + f e (2x y)) + 2k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )(f e (x + y) + f e (x y)) + 2k 2 f e (y) () (7) (8) for all x, y X. From the substitution y = kx + y in (5), we have f e (2kx + y) + f e (y) = k 2 f e ((k + )x + y) + k 2 f e ((k )x + y) + k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )f e (kx + y) for all x, y X. Replacing y by y in (9), we get f e (2kx y) + f e ( y) = k 2 f e ((k + )x y) + k 2 f e ((k )x y) + k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )f e (kx y) (9) (0)

5 General quartic-cubic-quadratic functional equation for all x, y X. Adding (9) to (0), we obtain f e (2kx + y) + f e (2kx y) = k 2 (f e ((k + )x + y) + f e ((k + )x y) + f e ((k )x + y) + f e ((k )x y)) + 2( k 2 )(f e (kx + y) + f e (kx y)) + 2k2 (k 2 ) (f e (2x) 4f e (x)) 2f e (y) for all x, y X. It follows from () using (5) and (8) that f e (2kx+y) + f e (2kx y) = k 2 [k 2 (f e (2x + y) + f e (2x y)) + 2k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )(f e (x + y) + f e (x y)) + 2k 2 f e (y)] + 2( k 2 )[k 2 f e (x + y) + k 2 f e (x y) + k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )f e (y)] + 2k2 (k 2 ) (f e (2x) 4f e (x)) 2f e (y) for all x, y X. If we replace x by 2x in (5), we get that f e (2kx + y) + f e (2kx y) = k 2 f e (2x + y) + k 2 f e (2x y) + k2 (k 2 ) (f e (4x) 4f e (2x)) + 2( k 2 )f e (y) for all x, y X. It follows from (2) and (3) that k 2 [k 2 (f e (2x + y) + f e (2x y)) + 2k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )(f e (x + y) + f e (x y)) + 2k 2 f e (y)] + 2( k 2 )[k 2 f e (x + y) + k 2 f e (x y) + k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )f e (y)] + 2k2 (k 2 ) (f e (2x) 4f e (x)) 2f e (y) = k 2 f e (2x + y) + k 2 f e (2x y) + k2 (k 2 ) (f e (4x) 4f e (2x)) + 2( k 2 )f e (y) for all x, y X. Also, putting y = 0 in (5), we get f e (kx) = k 2 f e (x) + k2 (k 2 ) (f e (2x) 4f e (x)) (5) 2 for all x X. Setting y = x in (5), we get f e ((k + )x) + f e ((k )x) = k 2 f e (2x) + k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )f e (x) () (2) (3) (4) ()

6 74 M. Eshaghi Gordji, H. Khodaei, R. Khodabakhsh for all x X. Putting y = kx in (5), we obtain f e (2kx) =k 2 (f e ((k + )x)) + f e ((k )x)) + k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )f e (kx) for all x X. Letting y = 0 in (3), we have f e (2kx) = k 2 f e (2x) + k2 (k 2 ) (f e (4x) 4f e (2x)) (8) 2 for all x X. It follows from (7) and (8) that k 2 (k 2 ) (f e (4x) 4f e (2x)) = k 2 (f e ((k + )x)) + f e ((k )x)) 2 + k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )f e (kx) k 2 f e (2x) for all x X. Now, using (5), () and (9), we are lead to k 2 (k 2 ) (f e (4x) 4f e (2x)) 2 = k 2 [k 2 f e (2x) + k2 (k 2 ) (f e (2x) 4f e (x)) + 2( k 2 )f e (x)] + 2( k 2 )[k 2 f e (x) + k2 (k 2 ) (f e (2x) 4f e (x))] 2 + k2 (k 2 ) (f e (2x) 4f e (x)) k 2 f e (2x) for all x X. Finally, if we compare (4) with (20), then we conclude that f e (2x + y) + f e (2x y) = 4f e (x + y) + 4f e (x y) + 2(f e (2x) 4f e (x)) f e (y) (2) for all x, y X. Replacing y by 2y in (2), we get f e (2x + 2y) + f e (2x 2y) = 4f e (x + 2y) + 4f e (x 2y) + 2(f e (2x) 4f e (x)) f e (2y) for all x, y X. Interchanging x with y in (2), we obtain f e (2y + x) + f e (2y x) = 4f e (y + x) + 4f e (y x) + 2(f e (2y) 4f e (y)) f e (x) (23) for all x, y X, which implies that f e (x + 2y) + f e (x 2y) = 4f e (x + y) + 4f e (x y) + 2(f e (2y) 4f e (y)) f e (x) (24) for all x, y X. It follows from (22) and (24) that f e (2(x + y)) f e (x + y) + f e (2(x y)) f e (x y) for all x, y X. This means that = 2(f e (2x) f e (x)) + 2(f e (2y) f e (y)) g(x + y) + g(x y) = 2g(x) + 2g(y) for all x, y X. So the function g : X Y defined by g(x) = f e (2x) f e (x) is quadratic. (7) (9) (20) (22)

7 General quartic-cubic-quadratic functional equation To prove that h : X Y defined by h(x) = f e (2x) 4f e (x) is quartic, we have to show that h(2x + y) + h(2x y) = 4h(x + y) + 4h(x y) + 24h(x) h(y) (25) for all x, y X. Replacing x and y by 2x and 2y in (2), respectively, we obtain f e (2(2x + y)) + f e (2(2x y)) = 4f e (2(x + y)) + 4f e (2(x y)) + 2(f e (4x) 4f e (2x)) f e (2y) for all x, y X. But, since g(2x) = 4g(x) for all x X, with g : X Y is the quadratic function defined above, thus we see that (2) f e (4x) = 20f e (2x) 4f e (x) (27) for all x X. Hence, according to (2) and (27), we get f e (2(2x + y)) + f e (2(2x y)) = 4f e (2(x + y)) + 4f e (2(x y)) + 32(f e (2x) 4f e (x)) f e (2y) for all x, y X. By multiplying both sides of (2) by 4, we get that 4f e (2x + y) + 4f e (2x y) = f e (x + y) + f e (x y) + 8(f e (2x) 4f e (x)) 24f e (y) for all x, y X. If we subtract the last equation from (28), we arrive at f e (2(2x + y)) 4f e (2x + y) + f e (2(2x y)) 4f e (2x y) = 4(f e (2(x + y)) 4f e (x + y)) + 4(f e (2(x y)) 4f e (x y)) + 24(f e (2x) 4f e (x)) (f e (2y) 4f e (y)) for all x, y X. This means that h satisfies the equation (25), so the function h : X Y is quartic. But, since, f e (x) = 2 h(x) 2g(x) for all x X, there exist a unique symmetric bi-quadratic function B 2 : X X Y and a unique symmetric bi-additive function B : X X Y such that h(x) = 2B 2 (x, x) and g(x) = 2B (x, x) for all x X (see [, 4]). So (28) (29) f e (x) = B 2 (x, x) + B (x, x) (30) for all x X. On the other hand, we show that the function f o : X Y is cubic. It follows from (4) and f o ( x) = f o (x) that f o (x + ky) + f o (x ky) = k 2 f o (x + y) + k 2 f o (x y) + 2( k 2 )f o (x) + k2 (k 2 ) (3) (f o (2y) 8f o (y)) for all x, y X. Putting x = 0 in (3), to get f o (2y) = 8f o (y) for all y X, so we get from (3) that f o (x + ky) + f o (x ky) = k 2 f o (x + y) + k 2 f o (x y) + 2( k 2 )f o (x) (32) for all x, y X. Setting x = x y in (32), we have f o (x + (k )y) + f o (x (k + )y) = k 2 f o (x) + k 2 f o (x 2y) + 2( k 2 )f o (x y) (33)

8 7 M. Eshaghi Gordji, H. Khodaei, R. Khodabakhsh for all x, y X. Replacing y by y in (33), gives f o (x (k )y) + f o (x + (k + )y) = k 2 f o (x) + k 2 f o (x + 2y) + 2( k 2 )f o (x + y) for all x, y X. Adding (33) to (34), we obtain f o (x + (k )y) + f o (x (k )y) + f o (x + (k + )y) + f o (x (k + )y) = k 2 (f o (x + 2y) + f o (x 2y)) + 2k 2 f o (x) + 2( k 2 )(f o (x + y) + f o (x y)) for all x, y X. Setting x = x + ky in (32), we get f o (x + 2ky) + f o (x) = k 2 (f o (x + (k )y) + f o (x + (k + )y))) + 2( k 2 )f o (x + ky) for all x, y X. Replacing y by y in (3), we have f o (x 2ky) + f o (x) = k 2 (f o (x (k )y) + f o (x (k + )y))) + 2( k 2 )f o (x ky) for all x, y X. Adding (3) to (37), one gets f o (x + 2ky) + f o (x 2ky) = k 2 [f o (x + (k )y) + f o (x + (k + )y) + f o (x (k )y) + f o (x (k + )y)] + 2( k 2 )(f o (x + ky) + f o (x ky)) 2f o (x) for all x, y X. Using (32), (35) and (38), we lead to f o (x + 2ky) + f o (x 2ky) = 4k 2 ( k 2 )(f o (x + y) + f o (x y)) + (k 4 8k 2 + 2)f o (x) + k 4 (f o (x + 2y) + f o (x 2y)) for all x, y X. Replacing y by 2y in (32), we get (34) (35) (3) (37) (38) (39) f o (x + 2ky) + f o (x 2ky) = k 2 (f o (x + 2y) + f o (x 2y)) + 2( k 2 )f o (x) (40) for all x, y X. If we compare (39) with (40), then we conclude that f o (x + 2y) + f o (x 2y) = 4(f o (x + y) + f o (x y)) f o (x) (4) for all x, y X. Replacing x by 2x in (4), gives f o (2(x + y)) + f o (2(x y)) = 4(f o (2x + y) + f o (2x y)) f o (2x) (42) for all x, y X, which by considering f o (2x) = 8f o (x) and (42), gives f o (2x + y) + f o (2x y) = 2f o (x + y) + 2f o (x y)2f o (x) for all x, y X, this means that f o is cubic. So f o (x) = C(x, x, x) (43) for all x X, that C is symmetric for each fixed one variable and is additive for fixed two variables []. Hence, according to (30) and (43), we obtain that f(x) = f e (x) + f o (x) = B 2 (x, x) + C(x, x, x) + B (x, x)

9 General quartic-cubic-quadratic functional equation for all x X. The proof of the converse is trivial. 3. Generalized Hyers Ulam stability in non-archimedean spaces Throughout this section, assume that G is an additive group and X is a complete non-archimedean space. Before taking up the main subject, for f : G G X, we define the difference operator Df(x, y) =f(x + ky) + f(x ky) k 2 f(x + y) k 2 f(x y) + 2(k 2 )f(x) k2 (k 2 ) (f(2y) + 2f( y) f(y)) (k Z {0, ±}) for all x, y G. Theorem 3.. Let ϕ : G G [0, ) be a function such that for all x, y G, and exists for all x G, where ϕ(x) := ϕ(2 n x, 2 n y) lim n 2 2n = 0 = lim n 2 2n ϕ(2n x) (44) ϕ q (x) = lim n max{ 2 2j ϕ(2j x) : 0 j < n} (45) k 2 (k 2 ) max{ max{ 2k2 ϕ(x, x), 2(k 2 ) ϕ(0, x)}, max{ ϕ(0, 2x), 2 ϕ(kx, x)}} for all x G. Suppose that an even function f : G X with f(0) = 0 satisfies the inequality Df(x, y) ϕ(x, y) (47) for all x, y G. Then there exist a quadratic function Q : G X such that for all x G. If (4) f(2x) f(x) Q(x) 2 2 ϕ q(x) (48) lim lim max{ i n 2 2j ϕ(2j x) : i j < n + i} = 0 (49) then Q is the unique quadratic function satisfying (48). Proof. Interchanging x with y in (47) and then using the evenness of f, we obtain f(kx + y) + f(kx y) k 2 f(x + y) k 2 f(x y) + 2(k 2 )f(y) k2 (k 2 ) (f(2x) 4f(x)) ϕ(y, x) for all x G. Setting y = 0 in (50), we have (50) 2f(kx) 2k 2 f(x) k2 (k 2 ) (f(2x) 4f(x)) ϕ(0, x) (5)

10 78 M. Eshaghi Gordji, H. Khodaei, R. Khodabakhsh for all x G. Putting y = x in (50), we get f((k + )x) + f((k )x) k 2 f(2x) + 2(k 2 )f(x) k2 (k 2 ) (f(2x) 4f(x)) ϕ(x, x) for all x G. Replacing x and y by 2x and 0 in (50), respectively, we obtain (52) 2f(2kx) 2k 2 f(2x) k2 (k 2 ) (f(4x) 4f(2x)) ϕ(0, 2x) (53) for all x G. Setting y = kx in (50) and using the evenness of f, it follows that f(2kx) k 2 f((k + )x) k 2 f((k )x) + 2(k 2 )f(kx) k2 (k 2 ) (f(2x) 4f(x)) ϕ(kx, x) for all x G. It follows from (5) (54) that f(4x) 20f(2x) + 4f(x) k 2 (k 2 ) max{max{ 2k2 ϕ(x, x), 2(k 2 ) ϕ(0, x)}, max{ ϕ(0, 2x), 2 ϕ(kx, x)}} (54) (55) for all x G. According to (4) and (55), we obtain f(4x) 20f(2x) + 4f(x) ϕ(x) (5) for all x G. Let g : X Y be a function defined by g(x) := f(2x) f(x) for all x G. From (5), we conclude that for all x G. This implies that for all x G. Replacing x by 2 n x in (58), we have g(2x) 4g(x) ϕ(x) (57) g(x) g(2x) 2 2 ϕ(x) (58) 2 2 g(2n x) 2 2(n ) g(2n x) 2 2n 2 2n ϕ(2n x) (59) for all x G. It follows from (44) and (59) that the sequence { g(2n x) } is Cauchy. 2 2n Since X is complete, we conclude that { g(2n x) } is convergent. So one can define the 2 2n g(2 function Q : X Y by Q(x) := lim n x) n for all x G. It follows from (58) 2 2n and (59) by using induction that g(x) g(2n x) 2 2n 2 2 max{ 2 2j ϕ(2j x) : 0 j < n} (0)

11 General quartic-cubic-quadratic functional equation for all n N and all x G. Taking n to approach infinity in (0) and using (45) one gets (48). By (44) and (47), we obtain DQ(x, y) = lim n 2 2n Dg(2n x, 2 n y) = lim n 2 2n Df(2n+ x, 2 n+ y) Df(2 n x, 2 n y) lim n 2 2n max{ Df(2n+ x, 2 n+ y), Df(2 n x, 2 n y) } = 0 for all x, y G. Therefore the function Q : G X satisfies (4). Thus by Theorem 2., the function x Q(2x) Q(x) is quadratic. If Q is another quadratic function satisfying (48), then Q(x) Q (x) = lim i 2 2i Q(2 i x) Q (2 i x) for all x G. Therefore Q = Q. lim i 2 2i max{ Q(2 i x) f(2 i x), f(2 i x) Q (2 i x) } 2 2 lim i lim n max{ 2 2j ϕ(2j x) : i j < n + i} = 0 Theorem 3.2. Let ϕ : G G [0, ) be a function such that for all x, y G, and ϕ(2 n x, 2 n y) lim n 2 4n = 0 = lim n 2 4n ϕ(2n x) () ϕ v (x) = lim n max{ 2 4j ϕ(2j x) : 0 j < n} (2) exists for all x G, with ϕ(x) satisfies the equation (4) for all x G. Suppose that an even function f : G X with f(0) = 0 satisfies the inequality (47) for all x, y G. Then there exist a quartic function V : G X such that f(2x) 4f(x) V (x) 2 4 ϕ v(x) (3) for all x G. If lim lim max{ i n 2 4j ϕ(2j x) : i j < n + i} = 0 (4) then V is the unique quartic function satisfying (3). Proof. Similar to the proof Theorem 3., we have f(4x) 20f(2x) + 4f(x) ϕ(x) (5) for all x G. Let h : X Y be a function defined by h(x) := f(2x) 4f(x) for all x G. From (5), we conclude that for all x G. Which implies that h(2x) h(x) ϕ(x) () h(x) h(2x) 2 4 ϕ(x) (7) 2 4

12 80 M. Eshaghi Gordji, H. Khodaei, R. Khodabakhsh for all x G. Replacing x by 2 n x in (7), we have h(2n x) 2 4(n ) h(2n x) 2 4n 2 4n ϕ(2n x) (8) for all x G. It follows from () and (8) that the sequence { h(2n x) } is Cauchy. 2 4n Since X is complete, we conclude that { h(2n x) } is convergent. So one can define the 2 4n h(2 function V : X Y by V (x) := lim n x) n for all x G. It follows from (7) 2 4n and (8) by using induction that h(x) h(2n x) 2 4n 2 4 max{ 2 4j ϕ(2j x) : 0 j < n} (9) for all n N and all x G. Taking n to approach infinity in (9) and using (2) one gets (3). By () and (47), we obtain DV (x, y) = lim n 2 4n Dh(2n x, 2 n y) = lim n 2 4n Df(2n+ x, 2 n+ y) 4Df(2 n x, 2 n y) lim n 2 4n max{ Df(2n+ x, 2 n+ y), 4 Df(2 n x, 2 n y) } = 0 for all x, y G. Therefore the function V : G X satisfies (4). Thus by Theorem 2., the function x V (2x) 4V (x) is quartic. The rest of the proof is similar to the proof of Theorem 3.. Theorem 3.3. Let ϕ : G G [0, ) be a function satisfies (44) for all x, y G, and the limit lim max{ n 2 2j ϕ(2j x) : 0 j < n} exists for all x G, with ϕ(x) satisfies the equation (4) for all x G. Suppose that an even function f : G X with f(0) = 0 satisfies the inequality (47) for all x, y G. Then there exist a quadratic function Q : G X and a quartic function V : G X such that f(x) Q(x) V (x) 48 max{ ϕ q(x), 2 2 ϕ v(x)} (70) for all x G, where ϕ q (x) and ϕ v (x) are defined as in Theorems 3. and 3.2. Moreover, if (49) holds, then Q is the unique quadratic function and V is the unique quartic function satisfying (70). Proof. By Theorem 3. and 3.2, there exists a quadratic function Q 0 : G X and a quartic function V 0 : G X such that f(2x) f(x) Q 0 (x) 2 2 ϕ q(x) f(2x) 4f(x) V 0 (x) 2 4 ϕ v(x) for all x G. So we obtain (70) by letting Q(x) = /2Q 0 (x) and V (x) = /2V 0 (x) for all x G.

13 General quartic-cubic-quadratic functional equation... 8 To prove the uniqueness property of Q and V, let Q, V : X Y be another quadratic and quartic functions satisfying (70). Let Q = Q Q and V = V V. Hence Q(x) + V (x) max{ f(x) Q(x) V (x) + f(x) Q (x) V (x) } 48 max{ ϕ q(x), 2 2 ϕ v(x)} for all x G. Since lim lim max{ i n 2 2j ϕ(2j x) : i j < n + i} = 0 = lim i lim n max{ 2 4j ϕ(2j x) : i j < n + i} for all x G. So lim n 2 4n Q(2n x) + V (2 n x) = 0 for all x X. Therefore, we get V = 0 and then Q = 0, and the proof is complete. Theorem 3.4. Let ϕ : G G [0, ) be a function such that for all x, y G and ϕ(2 n x, 2 n y) lim n 2 3n = 0 (7) ϕ c (x) = lim n max{ 2 3j ϕ(0, 2j x) : 0 j < n} (72) for all x G, exists. Suppose that f : G X is an odd function satisfying (47) for all x, y G. Then there exist a cubic function C : G X such that 3 f(x) C(x) 4k 2 (k 2 ) ϕ c(x) (73) for all x G. If lim lim max{ i n 2 3j ϕ(0, 2j x) : i j < n + i} = 0 then C is the unique cubic function satisfying (73). Proof. Putting x = 0 in (47) and then using oddness of f, we get k2 (k 2 ) (f(2y) 8f(y)) ϕ(0, y) for all y G. Which implies that f(x) f(2x) k 2 (k 2 ϕ(0, x) (74) ) for all x G. Replacing x by 2 n x in (74), we have f(2n x) 2 3(n ) f(2n x) 2 3n 3 2 3n k 2 (k 2 ) ϕ(0, 2n x) (75) for all x G. It follows from (7) and (75) that the sequence { f(2n x) 2 3n } is Cauchy. Since X is complete, we conclude that { f(2n x) 2 3n } is convergent. So one can define the

14 82 M. Eshaghi Gordji, H. Khodaei, R. Khodabakhsh function C : X Y by C(x) := lim n f(2 n x) 2 3n for all x G. It follows from (74) and (75) by using induction that f(x) f(2n x) 2 3n 3 4k 2 (k 2 ) max{ 2 3i ϕ(0, 2i x) : 0 i < n} (7) for all n N and all x G. By taking n to approach infinity in (7) and using (72) one obtains (73). By (7) and (47), one gets DC(x, y) = lim n 2 3n Df(2n x, 2 n ϕ(2 n x, 2 n y) y) lim n 2 3n = 0 for all x, y G. Therefore the function C : G X satisfies (4). Thus by Theorem 2., the function x C(x) is cubic. The rest of the proof is similar to the proof of Theorem 3.. Now, we are ready to prove the main theorem concerning the stability problem for the equation (4). Theorem 3.5. Let ϕ : G G [0, ) be a function satisfies (44) for all x, y G, and the limits lim max{ n 2 2j ϕ(2j x) : 0 j < n}, lim max{ n 2 3j ϕ(0, 2j x) : 0 j < n} exists for all x G, with ϕ(x) satisfies the equation (4) for all x G. Suppose that a function f : G X with f(0) = 0 satisfies the inequality (47) for all x, y G. Then there exist a quadratic function Q : G X, a cubic function C : G X and a quartic function V : G X such that f(x) Q(x) C(x) V (x) 2 3 max{ 3 k 2 (k 2 ) max{ ϕ c(x), ϕ c ( x)}, 2 max{max{ ϕ q(x), 2 2 ϕ v(x)}, max{ ϕ q ( x), 2 2 ϕ v( x)}}} for all x G, where ϕ q (x), ϕ c (x) and ϕ v (x) are defined as in Theorems 3., 3.2 and 3.4. If lim lim max{ i n 2 2j ϕ(2j x) : 0 j < n} = 0 = lim i lim n max{ 2 3j ϕ(0, 2j x) : i j < n + i} then Q is the unique quadratic function, C is the unique cubic function and V is the unique quartic function satisfying (77). Proof. Let f o (x) = 2 (f(x) f( x)) for all x G. Then Df o (x, y) max{ϕ(x, y), ϕ( x, y)} 2 for all x, y G. From Theorem 3.4, it follows that there exists a unique cubic function C : G X satisfying 3 f o (x) C(x) 8k 2 (k 2 ) max{ ϕ c(x), ϕ c ( x)} (78) (77)

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