Advances in Difference Equations
Volume 2009 (2009), Article ID 826130, 17 pages
doi:10.1155/2009/826130
Research Article

Solution and Stability of a Mixed Type Additive, Quadratic, and Cubic Functional Equation

1Department of Mathematics, Semnan University, P.O. Box 35195-363, Semnan, Iran
2Department of Mathematics, Payame Noor University of Mashhad, Mashhad, Iran
3Section of Mathematics and Informatics, Pedagogical Department, National and Capodistrian University of Athens, 4 Agamemnonos St., Aghia Paraskevi, Athens 15342, Greece

Received 24 January 2009; Revised 13 April 2009; Accepted 26 June 2009

Academic Editor: Patricia J. Y. Wong

Copyright © 2009 M. Eshaghi Gordji et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We obtain the general solution and the generalized Hyers-Ulam-Rassias stability of the mixed type additive, quadratic, and cubic functional equation 𝑓 ( 𝑥 + 2 𝑦 ) 𝑓 ( 𝑥 2 𝑦 ) = 2 ( 𝑓 ( 𝑥 + 𝑦 ) 𝑓 ( 𝑥 𝑦 ) ) + 2 𝑓 ( 3 𝑦 ) 6 𝑓 ( 2 𝑦 ) + 6 𝑓 ( 𝑦 ) .

1. Introduction

The stability problem of functional equations originated from a question of Ulam [1] in 1940, concerning the stability of group homomorphisms. Let ( 𝐺 1 , ) be a group, and let ( 𝐺 2 , ) be a metric group with the metric 𝑑 ( , ) . Given 𝜖 > 0 , does there exist a 𝛿 > 0 , such that if a mapping 𝐺 1 𝐺 2 satisfies the inequality 𝑑 ( ( 𝑥 𝑦 ) , ( 𝑥 ) ( 𝑦 ) ) < 𝛿 for all 𝑥 , 𝑦 𝐺 1 , then there exists a homomorphism 𝐻 𝐺 1 𝐺 2 with 𝑑 ( ( 𝑥 ) , 𝐻 ( 𝑥 ) ) < 𝜖 for all 𝑥 𝐺 1 ? In other words, under what condition does there exist a homomorphism near an approximate homomorphism?

In 1941, Hyers [2] gave a first affirmative answer to the question of Ulam for Banach spaces. Let 𝑓 𝐸 𝐸 be a mapping between Banach spaces such that 𝑓 ( 𝑥 + 𝑦 ) 𝑓 ( 𝑥 ) 𝑓 ( 𝑦 ) 𝛿 , ( 1 . 1 ) for all 𝑥 , 𝑦 𝐸 and for some 𝛿 > 0 . Then there exists a unique additive mapping 𝑇 𝐸 𝐸 such that 𝑓 ( 𝑥 ) 𝑇 ( 𝑥 ) 𝛿 , ( 1 . 2 ) for all 𝑥 𝐸 . Moreover if 𝑓 ( 𝑡 𝑥 ) is continuous in 𝑡 for each fixed 𝑥 𝐸 , then 𝑇 is linear (see also [3]). In 1950, Aoki [4] generalized Hyers' theorem for approximately additive mappings. In 1978, Th. M. Rassias [5] provided a generalization of Hyers' theorem which allows the Cauchy difference to be unbounded. This new concept is known as Hyers-Ulam-Rassias stability of functional equations (see [224]).

The functional equation 𝑓 ( 𝑥 + 𝑦 ) + 𝑓 ( 𝑥 𝑦 ) = 2 𝑓 ( 𝑥 ) + 2 𝑓 ( 𝑦 ) ( 1 . 3 ) is related to symmetric biadditive function. In the real case it has 𝑓 ( 𝑥 ) = 𝑥 2 among its solutions. Thus, it has been called quadratic functional equation, and each of its solutions is said to be a quadratic function. Hyers-Ulam-Rassias stability for the quadratic functional equation (1.3) was proved by Skof for functions 𝑓 𝐴 𝐵 , where 𝐴 is normed space and 𝐵 Banach space (see [2528]).

The following cubic functional equation was introduced by the third author of this paper, J. M. Rassias [29, 30] (in 2000-2001): 𝑓 ( 𝑥 + 2 𝑦 ) + 3 𝑓 ( 𝑥 ) = 3 𝑓 ( 𝑥 + 𝑦 ) + 𝑓 ( 𝑥 𝑦 ) + 6 𝑓 ( 𝑦 ) . ( 1 . 4 )

Jun and Kim [13] introduced the following cubic functional equation: 𝑓 ( 2 𝑥 + 𝑦 ) + 𝑓 ( 2 𝑥 𝑦 ) = 2 𝑓 ( 𝑥 + 𝑦 ) + 2 𝑓 ( 𝑥 𝑦 ) + 1 2 𝑓 ( 𝑥 ) , ( 1 . 5 ) and they established the general solution and the generalized Hyers-Ulam-Rassias stability for the functional equation (1.5).

The function 𝑓 ( 𝑥 ) = 𝑥 3 satisfies the functional equation (1.5), which explains why it is called cubic functional equation.

Jun and Kim proved that a function 𝑓 between real vector spaces 𝑋 and 𝑌 is a solution of (1.5) if and only if there exists 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 (see also [3133]).

We deal with the following functional equation deriving from additive, cubic and quadratic functions: 𝑓 ( 𝑥 + 2 𝑦 ) 𝑓 ( 𝑥 2 𝑦 ) = 2 ( 𝑓 ( 𝑥 + 𝑦 ) 𝑓 ( 𝑥 𝑦 ) ) + 2 𝑓 ( 3 𝑦 ) 6 𝑓 ( 2 𝑦 ) + 6 𝑓 ( 𝑦 ) . ( 1 . 6 )

It is easy to see that the function 𝑓 ( 𝑥 ) = 𝑎 𝑥 3 + 𝑏 𝑥 2 + 𝑐 𝑥 is a solution of the functional equation (1.6). In the present paper we investigate the general solution and the generalized Hyers-Ulam-Rassias stability of the functional equation (1.6).

2. General Solution

In this section we establish the general solution of functional equation (1.6).

Theorem 2.1. Let 𝑋 , 𝑌 be vector spaces, and let 𝑓 𝑋 𝑌 be a function. Then 𝑓 satisfies (1.6) if and only if there exists a unique additive function 𝐴 𝑋 𝑌 , a unique symmetric and biadditive function 𝑄 𝑋 × 𝑋 𝑌 , and a unique symmetric and 3-additive function 𝐶 𝑋 × 𝑋 × 𝑋 𝑌 such that 𝑓 ( 𝑥 ) = 𝐴 ( 𝑥 ) + 𝑄 ( 𝑥 , 𝑥 ) + 𝐶 ( 𝑥 , 𝑥 , 𝑥 ) for all 𝑥 𝑋 .

Proof. Suppose that 𝑓 ( 𝑥 ) = 𝐴 ( 𝑥 ) + 𝑄 ( 𝑥 , 𝑥 ) + 𝐶 ( 𝑥 , 𝑥 , 𝑥 ) for all 𝑥 𝑋 , where 𝐴 𝑋 𝑌 is additive, 𝑄 𝑋 × 𝑋 𝑌 is symmetric and biadditive, and 𝐶 𝑋 × 𝑋 × 𝑋 𝑌 is symmetric and 3-additive. Then it is easy to see that 𝑓 satisfies (1.6). For the converse let 𝑓 satisfy (1.6). We decompose 𝑓 into the even part and odd part by setting 𝑓 𝑒 1 ( 𝑥 ) = 2 ( 𝑓 ( 𝑥 ) + 𝑓 ( 𝑥 ) ) , 𝑓 𝑜 1 ( 𝑥 ) = 2 ( 𝑓 ( 𝑥 ) 𝑓 ( 𝑥 ) ) , ( 2 . 1 ) for all 𝑥 𝑋 . By (1.6), we have 𝑓 𝑒 ( 𝑥 + 2 𝑦 ) 𝑓 𝑒 = 1 ( 𝑥 2 𝑦 ) 2 [ ] = 1 𝑓 ( 𝑥 + 2 𝑦 ) + 𝑓 ( 𝑥 2 𝑦 ) 𝑓 ( 𝑥 2 𝑦 ) 𝑓 ( 𝑥 + 2 𝑦 ) 2 [ 𝑓 ] + 1 ( 𝑥 + 2 𝑦 ) 𝑓 ( 𝑥 2 𝑦 ) 2 [ 𝑓 ] = 1 ( ( 𝑥 ) + ( 2 𝑦 ) ) 𝑓 ( ( 𝑥 ) ( 2 𝑦 ) ) 2 [ ] + 1 2 𝑓 ( 𝑥 + 𝑦 ) 2 𝑓 ( 𝑥 𝑦 ) + 2 𝑓 ( 3 𝑦 ) 6 𝑓 ( 2 𝑦 ) + 6 𝑓 ( 𝑦 ) 2 [ ] 1 2 𝑓 ( 𝑥 𝑦 ) 2 𝑓 ( 𝑥 + 𝑦 ) + 2 𝑓 ( 3 𝑦 ) 6 𝑓 ( 2 𝑦 ) + 6 𝑓 ( 𝑦 ) = 2 2 1 ( 𝑓 ( 𝑥 + 𝑦 ) + 𝑓 ( 𝑥 𝑦 ) ) 2 2 1 ( 𝑓 ( 𝑥 𝑦 ) + 𝑓 ( 𝑥 + 𝑦 ) ) + 2 2 1 ( 𝑓 ( 3 𝑦 ) + 𝑓 ( 3 𝑦 ) ) 6 2 1 ( 𝑓 ( 2 𝑦 ) + 𝑓 ( 2 𝑦 ) ) + 6 2 𝑓 ( 𝑓 ( 𝑦 ) + 𝑓 ( 𝑦 ) ) = 2 𝑒 ( 𝑥 + 𝑦 ) 𝑓 𝑒 ( 𝑥 𝑦 ) + 2 𝑓 𝑒 ( 3 𝑦 ) 6 𝑓 𝑒 ( 2 𝑦 ) + 6 𝑓 𝑒 ( 𝑦 ) , ( 2 . 2 ) for all 𝑥 , 𝑦 𝑋 . This means that 𝑓 𝑒 satisfies (1.6), that is, 𝑓 𝑒 ( 𝑥 + 2 𝑦 ) 𝑓 𝑒 𝑓 ( 𝑥 2 𝑦 ) = 2 𝑒 ( 𝑥 + 𝑦 ) 𝑓 𝑒 ( 𝑥 𝑦 ) + 2 𝑓 𝑒 ( 3 𝑦 ) 6 𝑓 𝑒 ( 2 𝑦 ) + 6 𝑓 𝑒 ( 𝑦 ) . ( 2 . 3 ) Now putting 𝑥 = 𝑦 = 0 in (2.3), we get 𝑓 𝑒 ( 0 ) = 0 . Setting 𝑥 = 0 in (2.3), by evenness of 𝑓 𝑒 we obtain 3 𝑓 𝑒 ( 2 𝑦 ) = 𝑓 𝑒 ( 3 𝑦 ) + 3 𝑓 𝑒 ( 𝑦 ) . ( 2 . 4 ) Replacing 𝑥 by 𝑦 in (2.3), we obtain 4 𝑓 𝑒 ( 2 𝑦 ) = 𝑓 𝑒 ( 3 𝑦 ) + 7 𝑓 𝑒 ( 𝑦 ) . ( 2 . 5 ) Comparing (2.4) with (2.5), we get 𝑓 𝑒 ( 3 𝑦 ) = 9 𝑓 𝑒 ( 𝑦 ) . ( 2 . 6 ) By utilizing (2.5) with (2.6), we obtain 𝑓 𝑒 ( 2 𝑦 ) = 4 𝑓 𝑒 ( 𝑦 ) . ( 2 . 7 ) Hence, according to (2.6) and (2.7), (2.3) can be written as 𝑓 𝑒 ( 𝑥 + 2 𝑦 ) 𝑓 𝑒 ( 𝑥 2 𝑦 ) = 2 𝑓 𝑒 ( 𝑥 + 𝑦 ) 2 𝑓 𝑒 ( 𝑥 𝑦 ) . ( 2 . 8 ) With the substitution 𝑥 = 𝑥 + 𝑦 , 𝑦 = 𝑥 𝑦 in (2.8), we have 𝑓 𝑒 ( 3 𝑥 𝑦 ) 𝑓 𝑒 ( 𝑥 3 𝑦 ) = 8 𝑓 𝑒 ( 𝑥 ) 8 𝑓 𝑒 ( 𝑦 ) . ( 2 . 9 ) Replacing 𝑦 by 𝑦 in above relation, we obtain 𝑓 𝑒 ( 3 𝑥 + 𝑦 ) 𝑓 𝑒 ( 𝑥 + 3 𝑦 ) = 8 𝑓 𝑒 ( 𝑥 ) 8 𝑓 𝑒 ( 𝑦 ) . ( 2 . 1 0 ) Setting 𝑥 + 𝑦 instead of 𝑥 in (2.8), we get 𝑓 𝑒 ( 𝑥 + 3 𝑦 ) 𝑓 𝑒 ( 𝑥 𝑦 ) = 2 𝑓 𝑒 ( 𝑥 + 2 𝑦 ) 2 𝑓 𝑒 ( 𝑥 ) . ( 2 . 1 1 ) Interchanging 𝑥 and 𝑦 in (2.11), we get 𝑓 𝑒 ( 3 𝑥 + 𝑦 ) 𝑓 𝑒 ( 𝑥 𝑦 ) = 2 𝑓 𝑒 ( 2 𝑥 + 𝑦 ) 2 𝑓 𝑒 ( 𝑦 ) . ( 2 . 1 2 ) If we subtract (2.12) from (2.11) and use (2.10), we obtain 𝑓 𝑒 ( 𝑥 + 2 𝑦 ) 𝑓 𝑒 ( 2 𝑥 + 𝑦 ) = 3 𝑓 𝑒 ( 𝑦 ) 3 𝑓 𝑒 ( 𝑥 ) , ( 2 . 1 3 ) which, by putting 𝑦 = 2 𝑦 and using (2.7), leads to 𝑓 𝑒 ( 𝑥 + 4 𝑦 ) 4 𝑓 𝑒 ( 𝑥 + 𝑦 ) = 1 2 𝑓 𝑒 ( 𝑦 ) 3 𝑓 𝑒 ( 𝑥 ) . ( 2 . 1 4 ) Let us interchange 𝑥 and 𝑦 in (2.14). Then we see that 𝑓 𝑒 ( 4 𝑥 + 𝑦 ) 4 𝑓 𝑒 ( 𝑥 + 𝑦 ) = 1 2 𝑓 𝑒 ( 𝑥 ) 3 𝑓 𝑒 ( 𝑦 ) , ( 2 . 1 5 ) and by adding (2.14) and (2.15), we arrive at 𝑓 𝑒 ( 𝑥 + 4 𝑦 ) + 𝑓 𝑒 ( 4 𝑥 + 𝑦 ) = 8 𝑓 𝑒 ( 𝑥 + 𝑦 ) + 9 𝑓 𝑒 ( 𝑥 ) + 9 𝑓 𝑒 ( 𝑦 ) . ( 2 . 1 6 ) Replacing 𝑦 by 𝑥 + 𝑦 in (2.8), we obtain 𝑓 𝑒 ( 3 𝑥 + 2 𝑦 ) 𝑓 𝑒 ( 𝑥 + 2 𝑦 ) = 2 𝑓 𝑒 ( 2 𝑥 + 𝑦 ) 2 𝑓 𝑒 ( 𝑦 ) . ( 2 . 1 7 ) Let us Interchange 𝑥 and 𝑦 in (2.17). Then we see that 𝑓 𝑒 ( 2 𝑥 + 3 𝑦 ) 𝑓 𝑒 ( 2 𝑥 + 𝑦 ) = 2 𝑓 𝑒 ( 𝑥 + 2 𝑦 ) 2 𝑓 𝑒 ( 𝑥 ) . ( 2 . 1 8 ) Thus by adding (2.17) and (2.18), we have 𝑓 𝑒 ( 2 𝑥 + 3 𝑦 ) + 𝑓 𝑒 ( 3 𝑥 + 2 𝑦 ) = 3 𝑓 𝑒 ( 𝑥 + 2 𝑦 ) + 3 𝑓 𝑒 ( 2 𝑥 + 𝑦 ) 2 𝑓 𝑒 ( 𝑥 ) 2 𝑓 𝑒 ( 𝑦 ) . ( 2 . 1 9 ) Replacing 𝑥 by 2 𝑥 in (2.11) and using (2.7) we have 𝑓 𝑒 ( 2 𝑥 + 3 𝑦 ) 𝑓 𝑒 ( 2 𝑥 𝑦 ) = 8 𝑓 𝑒 ( 𝑥 + 𝑦 ) 8 𝑓 𝑒 ( 𝑥 ) , ( 2 . 2 0 ) and interchanging 𝑥 and 𝑦 in (2.20) yields 𝑓 𝑒 ( 3 𝑥 + 2 𝑦 ) 𝑓 𝑒 ( 𝑥 2 𝑦 ) = 8 𝑓 𝑒 ( 𝑥 + 𝑦 ) 8 𝑓 𝑒 ( 𝑦 ) . ( 2 . 2 1 ) If we add (2.20) to (2.21), we have 𝑓 𝑒 ( 2 𝑥 + 3 𝑦 ) + 𝑓 𝑒 ( 3 𝑥 + 2 𝑦 ) = 𝑓 𝑒 ( 2 𝑥 𝑦 ) + 𝑓 𝑒 ( 𝑥 2 𝑦 ) + 1 6 𝑓 𝑒 ( 𝑥 + 𝑦 ) 8 𝑓 𝑒 ( 𝑥 ) 8 𝑓 𝑒 ( 𝑦 ) . ( 2 . 2 2 ) Interchanging 𝑥 and 𝑦 in (2.8), we get 𝑓 𝑒 ( 2 𝑥 + 𝑦 ) 𝑓 𝑒 ( 2 𝑥 𝑦 ) = 2 𝑓 𝑒 ( 𝑥 + 𝑦 ) 2 𝑓 𝑒 ( 𝑥 𝑦 ) , ( 2 . 2 3 ) and by adding the last equation and (2.8) with (2.19), we get 𝑓 𝑒 ( 2 𝑥 + 3 𝑦 ) + 𝑓 𝑒 ( 3 𝑥 + 2 𝑦 ) 𝑓 𝑒 ( 2 𝑥 𝑦 ) 𝑓 𝑒 ( 𝑥 2 𝑦 ) = 2 𝑓 𝑒 ( 𝑥 + 2 𝑦 ) + 2 𝑓 𝑒 ( 2 𝑥 + 𝑦 ) + 4 𝑓 𝑒 ( 𝑥 + 𝑦 ) 4 𝑓 𝑒 ( 𝑥 𝑦 ) 2 𝑓 𝑒 ( 𝑥 ) 2 𝑓 𝑒 ( 𝑦 ) . ( 2 . 2 4 ) Now according to (2.22) and (2.24), it follows that 𝑓 𝑒 ( 𝑥 + 2 𝑦 ) + 𝑓 𝑒 ( 2 𝑥 + 𝑦 ) = 6 𝑓 𝑒 ( 𝑥 + 𝑦 ) + 2 𝑓 𝑒 ( 𝑥 𝑦 ) 3 𝑓 𝑒 ( 𝑥 ) 3 𝑓 𝑒 ( 𝑦 ) . ( 2 . 2 5 ) From the substitution 𝑦 = 𝑦 in (2.25) it follows that 𝑓 𝑒 ( 𝑥 2 𝑦 ) + 𝑓 𝑒 ( 2 𝑥 𝑦 ) = 6 𝑓 𝑒 ( 𝑥 𝑦 ) + 2 𝑓 𝑒 ( 𝑥 + 𝑦 ) 3 𝑓 𝑒 ( 𝑥 ) 3 𝑓 𝑒 ( 𝑦 ) . ( 2 . 2 6 ) Replacing 𝑦 by 2 𝑦 in (2.25) we have 𝑓 𝑒 ( 𝑥 + 4 𝑦 ) + 4 𝑓 𝑒 ( 𝑥 + 𝑦 ) = 6 𝑓 𝑒 ( 𝑥 + 2 𝑦 ) + 2 𝑓 𝑒 ( 𝑥 2 𝑦 ) 3 𝑓 𝑒 ( 𝑥 ) 1 2 𝑓 𝑒 ( 𝑦 ) , ( 2 . 2 7 ) and interchanging 𝑥 and 𝑦 yields 𝑓 𝑒 ( 4 𝑥 + 𝑦 ) + 4 𝑓 𝑒 ( 𝑥 + 𝑦 ) = 6 𝑓 𝑒 ( 2 𝑥 + 𝑦 ) + 2 𝑓 𝑒 ( 2 𝑥 𝑦 ) 1 2 𝑓 𝑒 ( 𝑥 ) 3 𝑓 𝑒 ( 𝑦 ) . ( 2 . 2 8 ) By adding (2.27) and (2.28) and then using (2.25) and (2.26), we lead to 𝑓 𝑒 ( 𝑥 + 4 𝑦 ) + 𝑓 𝑒 ( 4 𝑥 + 𝑦 ) = 3 2 𝑓 𝑒 ( 𝑥 + 𝑦 ) + 2 4 𝑓 𝑒 ( 𝑥 𝑦 ) 3 9 𝑓 𝑒 ( 𝑥 ) 3 9 𝑓 𝑒 ( 𝑦 ) . ( 2 . 2 9 ) If we compare (2.16) and (2.29), we conclude that 𝑓 𝑒 ( 𝑥 + 𝑦 ) + 𝑓 𝑒 ( 𝑥 𝑦 ) = 2 𝑓 𝑒 ( 𝑥 ) + 2 𝑓 𝑒 ( 𝑦 ) . ( 2 . 3 0 ) This means that 𝑓 𝑒 is quadratic. Thus there exists a unique quadratic function 𝑄 𝑋 × 𝑋 𝑌 such that 𝑓 𝑒 ( 𝑥 ) = 𝑄 ( 𝑥 , 𝑥 ) , for all 𝑥 𝑋 . On the other hand we can show that 𝑓 𝑜 satisfies (1.6), that is, 𝑓 𝑜 ( 𝑥 + 2 𝑦 ) 𝑓 𝑜 𝑓 ( 𝑥 2 𝑦 ) = 2 𝑜 ( 𝑥 + 𝑦 ) 𝑓 𝑜 ( 𝑥 𝑦 ) + 2 𝑓 𝑜 ( 3 𝑦 ) 6 𝑓 𝑜 ( 2 𝑦 ) + 6 𝑓 𝑜 ( 𝑦 ) . ( 2 . 3 1 ) Now we show that the mapping 𝑔 𝑋 𝑌 defined by 𝑔 ( 𝑥 ) = 𝑓 𝑜 ( 2 𝑥 ) 8 𝑓 𝑜 ( 𝑥 ) is additive and the mapping 𝑋 𝑌 defined by ( 𝑥 ) = 𝑓 𝑜 ( 2 𝑥 ) 2 𝑓 𝑜 ( 𝑥 ) is cubic. Putting 𝑥 = 0 in (2.31), then by oddness of 𝑓 𝑜 , we have 4 𝑓 𝑜 ( 2 𝑦 ) = 5 𝑓 𝑜 ( 𝑦 ) + 𝑓 𝑜 ( 3 𝑦 ) . ( 2 . 3 2 ) Hence (2.31) can be written as 𝑓 𝑜 ( 𝑥 + 2 𝑦 ) 𝑓 𝑜 ( 𝑥 2 𝑦 ) = 2 𝑓 𝑜 ( 𝑥 + 𝑦 ) 2 𝑓 𝑜 ( 𝑥 𝑦 ) + 2 𝑓 𝑜 ( 2 𝑦 ) 4 𝑓 𝑜 ( 𝑦 ) . ( 2 . 3 3 ) From the substitution 𝑦 = 𝑦 in (2.33) it follows that 𝑓 𝑜 ( 𝑥 2 𝑦 ) 𝑓 𝑜 ( 𝑥 + 2 𝑦 ) = 2 𝑓 𝑜 ( 𝑥 𝑦 ) 2 𝑓 𝑜 ( 𝑥 + 𝑦 ) 2 𝑓 𝑜 ( 2 𝑦 ) + 4 𝑓 𝑜 ( 𝑦 ) . ( 2 . 3 4 ) Interchange 𝑥 and 𝑦 in (2.33), and it follows that 𝑓 𝑜 ( 2 𝑥 + 𝑦 ) + 𝑓 𝑜 ( 2 𝑥 𝑦 ) = 2 𝑓 𝑜 ( 𝑥 + 𝑦 ) + 2 𝑓 𝑜 ( 𝑥 𝑦 ) + 2 𝑓 𝑜 ( 2 𝑥 ) 4 𝑓 𝑜 ( 𝑥 ) . ( 2 . 3 5 ) With the substitutions 𝑥 = 𝑥 𝑦 and 𝑦 = 𝑥 + 𝑦 in (2.35), we have 𝑓 𝑜 ( 3 𝑥 𝑦 ) + 𝑓 𝑜 ( 𝑥 3 𝑦 ) = 2 𝑓 𝑜 ( 2 𝑥 2 𝑦 ) 4 𝑓 𝑜 ( 𝑥 𝑦 ) + 2 𝑓 𝑜 ( 2 𝑥 ) 2 𝑓 𝑜 ( 2 𝑦 ) . ( 2 . 3 6 ) Replace 𝑥 by 𝑥 𝑦 in (2.34). Then we have 𝑓 𝑜 ( 𝑥 3 𝑦 ) 𝑓 𝑜 ( 𝑥 + 𝑦 ) = 2 𝑓 𝑜 ( 𝑥 2 𝑦 ) 2 𝑓 𝑜 ( 𝑥 ) 2 𝑓 𝑜 ( 2 𝑦 ) + 4 𝑓 𝑜 ( 𝑦 ) . ( 2 . 3 7 ) Replacing 𝑦 by 𝑦 in (2.37) gives 𝑓 𝑜 ( 𝑥 + 3 𝑦 ) 𝑓 𝑜 ( 𝑥 𝑦 ) = 2 𝑓 𝑜 ( 𝑥 + 2 𝑦 ) 2 𝑓 𝑜 ( 𝑥 ) + 2 𝑓 𝑜 ( 2 𝑦 ) 4 𝑓 𝑜 ( 𝑦 ) . ( 2 . 3 8 ) Interchanging 𝑥 and 𝑦 in (2.38), we get 𝑓 𝑜 ( 3 𝑥 + 𝑦 ) + 𝑓 𝑜 ( 𝑥 𝑦 ) = 2 𝑓 𝑜 ( 2 𝑥 + 𝑦 ) 2 𝑓 𝑜 ( 𝑦 ) + 2 𝑓 𝑜 ( 2 𝑥 ) 4 𝑓 𝑜 ( 𝑥 ) . ( 2 . 3 9 ) If we add (2.38) to (2.39), we have 𝑓 𝑜 ( 𝑥 + 3 𝑦 ) + 𝑓 𝑜 ( 3 𝑥 + 𝑦 ) = 2 𝑓 𝑜 ( 𝑥 + 2 𝑦 ) + 2 𝑓 𝑜 ( 2 𝑥 + 𝑦 ) + 2 𝑓 𝑜 ( 2 𝑥 ) + 2 𝑓 𝑜 ( 2 𝑦 ) 6 𝑓 𝑜 ( 𝑥 ) 6 𝑓 𝑜 ( 𝑦 ) . ( 2 . 4 0 ) Replacing 𝑦 by 𝑦 in (2.36) gives 𝑓 𝑜 ( 𝑥 + 3 𝑦 ) + 𝑓 𝑜 ( 3 𝑥 + 𝑦 ) = 2 𝑓 𝑜 ( 2 𝑥 + 2 𝑦 ) 4 𝑓 𝑜 ( 𝑥 + 𝑦 ) + 2 𝑓 𝑜 ( 2 𝑥 ) + 2 𝑓 𝑜 ( 2 𝑦 ) . ( 2 . 4 1 ) By comparing (2.40) with (2.41), we arrive at 𝑓 𝑜 ( 𝑥 + 2 𝑦 ) + 𝑓 𝑜 ( 2 𝑥 + 𝑦 ) = 𝑓 𝑜 ( 2 𝑥 + 2 𝑦 ) 2 𝑓 𝑜 ( 𝑥 + 𝑦 ) + 3 𝑓 𝑜 ( 𝑥 ) + 3 𝑓 𝑜 ( 𝑦 ) . ( 2 . 4 2 ) Replacing 𝑦 by 𝑦 in (2.42) gives 𝑓 𝑜 ( 𝑥 2 𝑦 ) + 𝑓 𝑜 ( 2 𝑥 𝑦 ) = 𝑓 𝑜 ( 2 𝑥 2 𝑦 ) 2 𝑓 𝑜 ( 𝑥 𝑦 ) + 3 𝑓 𝑜 ( 𝑥 ) 3 𝑓 𝑜 ( 𝑦 ) . ( 2 . 4 3 ) With the substitution 𝑦 = 𝑥 + 𝑦 in (2.43), we have 𝑓 𝑜 ( 𝑥 𝑦 ) 𝑓 𝑜 ( 𝑥 + 2 𝑦 ) = 𝑓 𝑜 ( 2 𝑦 ) 3 𝑓 𝑜 ( 𝑥 + 𝑦 ) + 3 𝑓 𝑜 ( 𝑥 ) + 2 𝑓 𝑜 ( 𝑦 ) , ( 2 . 4 4 ) and replacing 𝑦 by 𝑦 gives 𝑓 𝑜 ( 𝑥 + 𝑦 ) 𝑓 𝑜 ( 𝑥 2 𝑦 ) = 𝑓 𝑜 ( 2 𝑦 ) 3 𝑓 𝑜 ( 𝑥 𝑦 ) + 3 𝑓 𝑜 ( 𝑥 ) 2 𝑓 𝑜 ( 𝑦 ) . ( 2 . 4 5 ) Let us interchange 𝑥 and 𝑦 in (2.45). Then we see that 𝑓 𝑜 ( 𝑥 + 𝑦 ) + 𝑓 𝑜 ( 2 𝑥 𝑦 ) = 𝑓 𝑜 ( 2 𝑥 ) + 3 𝑓 𝑜 ( 𝑥 𝑦 ) 2 𝑓 𝑜 ( 𝑥 ) + 3 𝑓 𝑜 ( 𝑦 ) . ( 2 . 4 6 ) If we add (2.45) to (2.46), we have 𝑓 𝑜 ( 2 𝑥 𝑦 ) 𝑓 𝑜 ( 𝑥 2 𝑦 ) = 𝑓 𝑜 ( 2 𝑥 ) 2 𝑓 𝑜 ( 𝑥 + 𝑦 ) + 𝑓 𝑜 ( 𝑥 ) + 𝑓 𝑜 ( 2 𝑦 ) + 𝑓 𝑜 ( 𝑦 ) . ( 2 . 4 7 ) Adding (2.42) to (2.47) and using (2.33) and (2.35), we obtain 𝑓 𝑜 ( 2 ( 𝑥 + 𝑦 ) ) 8 𝑓 𝑜 𝑓 ( 𝑥 + 𝑦 ) = 𝑜 ( 2 𝑥 ) 8 𝑓 𝑜 + 𝑓 ( 𝑥 ) 𝑜 ( 2 𝑦 ) 8 𝑓 𝑜 ( 𝑦 ) , ( 2 . 4 8 ) for all 𝑥 , 𝑦 𝑋 . The last equality means that 𝑔 ( 𝑥 + 𝑦 ) = 𝑔 ( 𝑥 ) + 𝑔 ( 𝑦 ) , ( 2 . 4 9 ) for all 𝑥 , 𝑦 𝑋 . Therefore the mapping 𝑔 𝑋 𝑌 is additive. With the substitutions 𝑥 = 2 𝑥 and 𝑦 = 2 𝑦 in (2.35), we have 𝑓 𝑜 ( 4 𝑥 + 2 𝑦 ) + 𝑓 𝑜 ( 4 𝑥 2 𝑦 ) = 2 𝑓 𝑜 ( 2 𝑥 + 2 𝑦 ) + 2 𝑓 𝑜 ( 2 𝑥 2 𝑦 ) + 2 𝑓 𝑜 ( 4 𝑥 ) 4 𝑓 𝑜 ( 2 𝑥 ) . ( 2 . 5 0 ) Let 𝑔 𝑋 𝑌 be the additive mapping defined above. It is easy to show that 𝑓 𝑜 is cubic-additive function. Then there exists a unique function 𝐶 𝑋 × 𝑋 × 𝑋 𝑌 and a unique additive function 𝐴 𝑋 𝑌 such that 𝑓 𝑜 ( 𝑥 ) = 𝐶 ( 𝑥 , 𝑥 , 𝑥 ) + 𝐴 ( 𝑥 ) , for all 𝑥 𝑋 , and 𝐶 is symmetric and 3-additive. Thus for all 𝑥 𝑋 , we have 𝑓 ( 𝑥 ) = 𝑓 𝑒 ( 𝑥 ) + 𝑓 𝑜 ( 𝑥 ) = 𝑄 ( 𝑥 , 𝑥 ) + 𝐶 ( 𝑥 , 𝑥 , 𝑥 ) + 𝐴 ( 𝑥 ) . ( 2 . 5 1 ) This completes the proof of theorem.

The following corollary is an alternative result of Theorem 2.1.

Corollary 2.2. Let 𝑋 , 𝑌 be vector spaces, and let 𝑓 𝑋 𝑌 be a function satisfying (1.6). Then the following assertions hold. (a)If 𝑓 is even function, then 𝑓 is quadratic.(b)If 𝑓 is odd function, then 𝑓 is cubic-additive.

3. Stability

We now investigate the generalized Hyers-Ulam-Rassias stability problem for functional equation (1.6). 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 𝐷 𝑓 𝑋 × 𝑋 𝑌 by 𝐷 𝑓 [ ] ( 𝑥 , 𝑦 ) = 𝑓 ( 𝑥 + 2 𝑦 ) 𝑓 ( 𝑥 2 𝑦 ) 2 𝑓 ( 𝑥 + 𝑦 ) 𝑓 ( 𝑥 𝑦 ) 2 𝑓 ( 3 𝑦 ) + 6 𝑓 ( 2 𝑦 ) 6 𝑓 ( 𝑦 ) , ( 3 . 1 ) for all 𝑥 , 𝑦 𝑋 . We consider the following functional inequality: 𝐷 𝑓 ( 𝑥 , 𝑦 ) 𝜙 ( 𝑥 , 𝑦 ) , ( 3 . 2 ) for an upper bound 𝜙 𝑋 × 𝑋 [ 0 , ) .

Theorem 3.1. Let 𝑠 { 1 , 1 } be fixed. Suppose that an even mapping 𝑓 𝑋 𝑌 satisfies 𝑓 ( 0 ) = 0 , and 𝐷 𝑓 ( 𝑥 , 𝑦 ) 𝜙 ( 𝑥 , 𝑦 ) , ( 3 . 3 ) for all 𝑥 , 𝑦 𝑋 . If the upper bound 𝜙 𝑋 × 𝑋 [ 0 , ) is a mapping such that 𝑖 = 0 4 𝑠 𝑖 𝜙 2 𝑠 𝑖 𝑥 , 2 𝑠 𝑖 𝑥 + 1 2 𝜙 0 , 2 𝑠 𝑖 𝑥 < ( 3 . 4 ) and that l i m 𝑛 4 𝑠 𝑛 𝜙 ( 2 𝑠 𝑛 𝑥 , 2 𝑠 𝑛 𝑦 ) = 0 , ( 3 . 5 ) for all 𝑥 , 𝑦 𝑋 , then the limit 𝑄 ( 𝑥 ) = l i m 𝑛 4 𝑠 𝑛 𝑓 ( 2 𝑠 𝑛 𝑥 ) ( 3 . 6 ) exists for all 𝑥 𝑋 , and 𝑄 𝑋 𝑌 is a unique quadratic function satisfying (1.6), and ( 1 𝑓 𝑥 ) 𝑄 ( 𝑥 ) 8 𝑖 = ( 𝑠 + 1 ) / 2 4 𝑠 𝑖 𝜙 2 𝑠 𝑖 𝑥 , 2 𝑠 𝑖 𝑥 + 1 2 𝜙 0 , 2 𝑠 𝑖 𝑥 , ( 3 . 7 ) for all 𝑥 𝑋 .

Proof. Let 𝑠 = 1 . Putting 𝑥 = 0 in (3.3), we get [ ] 2 𝑓 ( 3 𝑦 ) 3 𝑓 ( 2 𝑦 ) + 3 𝑓 ( 𝑦 ) 𝜙 ( 0 , 𝑦 ) , ( 3 . 8 ) for all 𝑦 𝑋 . On the other hand by replacing 𝑦 by 𝑥 in (3.3), it follows that 𝑓 ( 3 𝑦 ) + 4 𝑓 ( 2 𝑦 ) 7 𝑓 ( 𝑦 ) 𝜙 ( 𝑦 , 𝑦 ) , ( 3 . 9 ) for all 𝑦 𝑋 . Combining (3.8) and (3.9), we lead to 2 𝑓 ( 2 𝑦 ) 8 𝑓 ( 𝑦 ) 2 𝜙 ( 𝑦 , 𝑦 ) + 𝜙 ( 0 , 𝑦 ) , ( 3 . 1 0 ) for all 𝑦 𝑋 . With the substitution 𝑦 = 𝑥 / 2 in (3.10) and then dividing both sides of inequality by 2, we get 𝑥 𝑓 ( 𝑥 ) 4 𝑓 2 1 2 𝑥 2 𝜙 2 , 𝑥 2 𝑥 + 𝜙 0 , 2 . ( 3 . 1 1 )
Now, using methods similar as in [8, 34, 35], we can easily show that the function 𝑄 𝑋 𝑌 defined by 𝑄 ( 𝑥 ) = l i m 𝑛 4 𝑛 𝑓 ( 𝑥 / 2 𝑛 ) for all 𝑥 𝑋 is unique quadratic function satisfying (1.6) and (3.7). Let 𝑠 = 1 . Then by (3.10) we have 𝑓 ( 2 𝑥 ) 4 1 𝑓 ( 𝑥 ) 8 ( 2 𝜙 ( 𝑥 , 𝑥 ) + 𝜙 ( 0 , 𝑥 ) ) , ( 3 . 1 2 ) for all 𝑥 𝑋 . And analogously, as in the case 𝑠 = 1 , we can show that the function 𝑄 𝑋 𝑌 defined by 𝑄 ( 𝑥 ) = l i m 𝑛 4 𝑛 𝑓 ( 2 𝑛 𝑥 ) is unique quadratic function satisfying (1.6) and (3.7).

Theorem 3.2. Let 𝑠 { 1 , 1 } be fixed. Let 𝜙 𝑋 × 𝑋 [ 0 , ) is a mapping such that 𝑖 = 1 2 𝑠 𝑖 𝜙 𝑥 2 𝑠 𝑖 , 𝑥 2 𝑠 𝑖 + 1 𝑥 + 𝜙 0 , 2 𝑠 𝑖 + 1 < ( 3 . 1 3 ) and that l i m 𝑛 2 𝑠 𝑛 𝜙 𝑥 2 𝑠 𝑛 , 𝑦 2 𝑠 𝑛 = 0 , ( 3 . 1 4 ) for all 𝑥 , 𝑦 𝑋 .
Suppose that an odd mapping 𝑓 𝑋 𝑌 satisfies 𝐷 𝑓 ( 𝑥 , 𝑦 ) 𝜙 ( 𝑥 , 𝑦 ) , ( 3 . 1 5 ) for all 𝑥 , 𝑦 𝑋 .
Then the limit 𝐴 ( 𝑥 ) = l i m 𝑛 2 𝑠 𝑛 𝑓 𝑥 2 𝑠 𝑛 1 𝑥 8 𝑓 2 𝑠 𝑛 ( 3 . 1 6 ) exists, for all 𝑥 𝑋 , and 𝐴 𝑋 𝑌 is a unique additive function satisfying (1.6), and ( 𝑓 2 𝑥 ) 8 𝑓 ( 𝑥 ) 𝐴 ( 𝑥 ) | | | | 𝑖 = 𝑠 1 / 2 2 𝑠 𝑖 𝜙 𝑥 2 𝑠 𝑖 , 𝑥 2 𝑠 𝑖 + 1 + 2 | | | | 𝑖 = 𝑠 1 / 2 2 𝑠 𝑖 𝜙 𝑥 0 , 2 𝑠 𝑖 + 1 , ( 3 . 1 7 ) for all 𝑥 𝑋 .

Proof. Let 𝑠 = 1 . set 𝑥 = 0 in (3.15). Then by oddness of 𝑓 we have 2 𝑓 ( 3 𝑦 ) 8 𝑓 ( 2 𝑦 ) + 1 6 𝑓 ( 𝑦 ) 𝜙 ( 0 , 𝑦 ) , ( 3 . 1 8 ) for all 𝑦 𝑋 . Replacing 𝑥 by 2 𝑦 in (3.15) we get 𝑓 ( 4 𝑦 ) 4 𝑓 ( 3 𝑦 ) + 6 𝑓 ( 2 𝑦 ) 4 𝑓 ( 𝑦 ) 𝜙 ( 2 𝑦 , 𝑦 ) . ( 3 . 1 9 ) Combining (3.18) and (3.19), we lead to 𝑓 ( 4 𝑦 ) 1 0 𝑓 ( 2 𝑦 ) + 1 6 𝑓 ( 𝑦 ) 𝜙 ( 2 𝑦 , 𝑦 ) + 2 𝜙 ( 0 , 𝑦 ) , ( 3 . 2 0 ) for all 𝑦 𝑋 . Putting 𝑦 = 𝑥 / 2