Solution and Stability of a Mixed Type Cubic and Quartic Functional Equation in Quasi-Banach Spaces

and Applied Analysis 3 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 20 . The quartic functional equation 1.6 was introduced by Rassias 21 in 2000 and then in 2005 was employed by Park and Bae 22 and others, such that: f ( x 2y ) f ( x − 2y) 4[f(x y) f(x − y)] 24f(y) − 6f x . 1.6 In fact they proved that a function f between real vector spaces X and Y is a solution of 1.6 if and only if there exists a unique symmetric multiadditive function Q : X ×X ×X ×X → Y such that f x Q x, x, x, x for all x see also 21–29 . It is easy to show that the function f x x4 satisfies the functional equation 1.6 , which is called a quartic functional equation and every solution of the quartic functional equation is said to be a quartic function. In this paper we deal with the following functional equation: f ( x 2y ) f ( x − 2y) 4(f(x y) f(x − y)) − 24f(y) − 6f x 3f(2y) 1.7 in quasiBanach spaces. It is easy to see that the function f x ax3 bx4 is a solution of the functional equation 1.7 . In the present paper we investigate the general solution of functional equation 1.7 when f is a mapping between vector spaces, and we establish the generalized Hyers-Ulam-Rassias stability of the functional equation 1.7 whenever f is a mapping between two quasiBanach spaces. We only mention here the papers 30, 31 concerning the stability of the mixed type functional equations. 2. General Solution Throughout this section, X and Y will be real vector spaces. Before proceeding to the proof of Theorem 2.3 which is the main result in this section, we shall need the following two lemmas. Lemma 2.1. If an even function f : X → Y satisfies 1.7 , then f is quartic. Proof. Putting x y 0 in 1.7 , we get f 0 0. Setting x 0 in 1.7 , by evenness of f we obtain f ( 2y ) 16f ( y ) 2.1 for all y ∈ X. Hence 1.7 can be written as f ( x 2y ) f ( x − 2y) 4(f(x y) f(x − y)) 24f(y) − 6f x . 2.2 This means that f is quartic function,which completes the proof of the lemma. 4 Abstract and Applied Analysis Lemma 2.2. If an odd function f : X → Y satisfies 1.7 , then f is a cubic function. Proof. Setting x y 0 in 1.7 gives f 0 0. Putting x 0 in 1.7 , then by oddness of f , we have f ( 2y ) 8f ( y ) . 2.3 Hence 1.7 can be written as f ( x 2y ) f ( x − 2y) 4f(x y) 4f(x − y) − 6f x . 2.4 Replacing x by x y in 2.4 ,we obtain f ( x 3y ) f ( x − y) 4f(x 2y) − 6f(x y) 4f x . 2.5 Substituting −y for y in 2.5 gives f ( x − 3y) f(x y) 4f(x − 2y) − 6f(x − y) 4f x . 2.6 If we subtract 2.5 from 2.6 ,we obtain f ( x 3y ) − f(x − 3y) 4f(x 2y) − 4f(x − 2y) − 5f(x y) 5f(x − y). 2.7 Let us interchange x and y in 2.7 . Then we see that f ( 3x y ) f ( 3x − y) 4f(2x y) 4f(2x − y) − 5f(x y) − 5f(x − y). 2.8 With the substitution y : x y in 2.4 , we have f ( 3x 2y ) − f(x 2y) 4f(2x y) − 4f(y) − 6f x . 2.9 From the substitution y : −y in 2.9 it follows that f ( 3x − 2y) − f(x − 2y) 4f(2x − y) 4f(y) − 6f x . 2.10 If we add 2.9 to 2.10 ,we have f ( 3x 2y ) f ( 3x − 2y) 4f(2x y) 4f(2x − y) f(x 2y) f(x − 2y) − 12f x . 2.11 Replacing x by 2x in 2.7 and using 2.3 , we obtain f ( 2x 3y ) − f(2x − 3y) 32f(x y) − 32f(x − y) − 5f(2x y) 5f(2x − y). 2.12 Abstract and Applied Analysis 5 Interchanging x with y in 2.12 gives the equationand Applied Analysis 5 Interchanging x with y in 2.12 gives the equation f ( 3x 2y ) f ( 3x − 2y) 32f(x y) 32f(x − y) − 5f(x 2y) − 5f(x − 2y). 2.13 If we compare 2.11 and 2.13 and employ 2.4 , we conclude that f ( 2x y ) f ( 2x − y) 2f(x y) 2f(x − y) 12f x . 2.14 This means that f is cubic function. This completes the proof of Lemma. Theorem 2.3. A function f : X → Y satisfies 1.7 for all x, y ∈ X if and only if there exists a unique functionC : X×X×X → Y and a unique symmetric multiadditive functionQ : X×X×X×X → Y such that f x C x, x, x Q x, x, x, x for all x ∈ X, and that C is symmetric for each fixed one variable and is additive for fixed two variables. Proof. Let f satisfy 1.7 . We decompose f into the even part and odd part by setting fe x 1 2 ( f x f −x ), fo x 12 ( f x − f −x ) 2.15 for all x ∈ X. By 1.7 , we have fe ( x 2y ) fe ( x − 2y) 1 2 [ f ( x 2y ) f (−x − 2y) f(x − 2y) f(−x 2y)] 4 ( fe ( x y ) fe ( x − y)) − 24fe ( y ) − 6fe x 3fe ( 2y ) 2.16 for all x, y ∈ X. This means that fe satisfies in 1.7 . Similarly we can show that fo satisfies 1.7 . By Lemmas 2.1 and 2.2, fe and fo are quartic and cubic, respectively. Thus there exists a unique function C : X × X × X → Y and a unique symmetric multiadditive function Q : X×X×X×X → Y such that fe x Q x, x, x, x and 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 f x C x, x, x Q x, x, x, x for all x ∈ X. The proof of the converse is trivial. 3. Stability Throughout this section, X and Y will be a uniquely two-divisible abelian group and a quasiBanach spaces respectively, and p will be a fixed real number in 0, 1 . We need the following lemma in the main theorems. Now before taking up the main subject, given f : X → Y , we define the difference operator Df : X ×X → Y by Df ( x, y ) f ( x 2y ) f ( x − 2y) − 4[f(x y) f(x − y)] − 3f(2y) 24f(y) 6f x 3.1 6 Abstract and Applied Analysis for all x, y ∈ X.We consider the following functional inequality: ∥ Df ( x, y )∥ ∥ ≤ φ(x, y) 3.2 for an upper bound φ : X ×X → 0,∞ . Lemma 3.1. Let x1, x2, . . . , xn be nonnegative real numbers. Then