China E-mail addresses: [email protected] Abstract In this paper, we determine the general solution of the functional equation f(kx+y) +f(kx−y

Tian Zhou Xu*

Department of Mathematics, School of Science, Beijing Institute of Technology, Beijing100081, P. R. China

E-mail addresses: [email protected],[email protected] John Michael Rassias

Pedagogical Department E.E., Section of Mathematics and Informatics,

National and Capodistrian University of Athens, 4, Agamemnonos Str., Aghia Paraskevi, Athens 15342, Greece.

E-mail addresses: [email protected], [email protected], [email protected] URL: http://www.primedu.uoa.gr/∼jrassias/.

Wan Xin Xu

School of Communication and Information Engineering, University of Electronic Science and Technology of China, Chengdu 611731, P. R. China

E-mail addresses: [email protected]

Abstract In this paper, we determine the general solution of the functional equation f(kx+y) +f(kx−y) = g(x+y) +g(x−y) +h(x) + ˜h(y) for fixed integersk withk6= 0,±1 without assuming any regularity condition on the unknown functionsf, g, h,˜h. The method used for solving these functional equations is elementary but exploits an important result due to Hossz´u. The solution of this functional equation can also be determined in certain type of groups using two important results due to Sz´ekelyhidi. The results improve and extend some recent results.

Keywords Quadratic function, Quartic function, Difference operator, Fr´echet functional equation, Mixed quadratic- quartic functional equation.

MR(2000) Subject Classification 39B22, 39B82

1. Introduction and preliminaries

J. M. Rassias [12](in 2001) introduced the cubic functional equation:

f(x+ 2y)−3f(x+y) + 3f(x)−f(x−y) = 6f(y) (1.0) and established the solution of the Ulam-Hyers stability problem for this cubic functional equation. Since the functionf(x) =x³ satisfies the functional equation (1.0), this equation is called cubic functional equation.

Every solution of the cubic functional equation is said to be a cubic function.

K. W. Jun and H. M. Kim [7](in 2002) introduced the following cubic functional equation

f(2x+y) +f(2x−y) = 2f(x+y) + 2f(x−y) + 12f(x) (1.1) and established the general solution and the generalized Ulam-Hyers stability for the functional equation (1.1). They proved that a function f between real vector spacesX and Y is a solution of (1.1) 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. Recently, several further interesting discussions, modifications, extensions, and generalizations of the original problem of Ulam have been proposed (see, e.g., [5, 18-23] and the references therein). This cubic equation (1.1) can be generalized to

f(2x+y) +f(2x−y) = 2f(x+y) + 2f(x−y) + 2f(2x)−4f(x). (1.2)

*Corresponding author.

The first author was supported by the National Natural Science Foundation of China(10671013)

It is easy to see that the function f(x) = ax³+bxis a solution of the functional equation (1.2), which is called a general mixed additive-cubic functional equation. A. Najati and G. Z. Eskandani [11] established the general solution of the functional equation (1.2) and investigated the Ulam-Hyers stability of this equation in quasi-Banach spaces. In [18], we determined the general solution of a general mixed additive-cubic functional equation

f(kx+y) +f(kx−y) =kf(x+y) +kf(x−y) + 2f(kx)−2kf(x), (1.3) for a fixed integerkwithk6= 0,±1.

J. M. Rassias [13] (in 1999) introduced the first quartic functional equation:

f(x+ 2y)−4f(x+y) + 6f(x)−4f(x−y) +f(x−2y) = 24f(y) (1.4.0) and established the solution of the Ulam-Hyers stability problem for the quartic functional equation. Since the function f(x) = x⁴ satisfies the functional equation (1.4.0), this equation is called quartic functional equation.

S. H. Lee, S. M. Im and I. S. Hwang [9] determined the general solution of the quartic functional equation f(2x+y) +f(2x−y) = 4f(x+y) + 4f(x−y) + 24f(x)−6f(y). (1.4) In [5], M. Eshaghi Gordji, S. Abbaszadeh and C. Park established the general solution of a generalized quadratic and quartic type functional equation

f(kx+y) +f(kx−y) =k²f(x+y) +k²f(x−y) + 2(f(kx)−k²f(x))−2(k²−1)f(y), (1.5) in quasi-Banach spaces for a fixed integerkwithk6= 0,±1.

Let k be a fixed integer with k 6= 0,±1, X and Y are real vector spaces. The functional equations (1.2)-(1.5) can be generalized to

f(kx+y) +f(kx−y) =g(x+y) +g(x−y) +h(x) + ˜h(y). (1.6) for allx, y∈X, wheref, g, h,h˜:X→Y are unknown functions to be determined. In this paper, we determine the general solution of the functional equation (1.6) and some other related functional equations. We will first solve these functional equations using an elementary technique (see [2], [15]-[16], [20]) but without using any regularity condition on the unknown functions. The motivation for studying these functional equations came from the fact that recently polynomial equations have found applications in approximate checking, self- testing, and self-correcting of computer programs that compute polynomials. The interested reader should refer to [4] and [14] and references therein.

LetX andY be real vector spaces. A functionA:X→Y is said to be additive ifA(x+y) =A(x) +A(y) for allx, y∈X. It is easy to see thatA(rx) =rA(x) for allx∈X and allr∈Q(the set of rational numbers).

Letn∈N(the set of natural numbers). A functionA_n :Xⁿ →Y is calledn-additive if it is additive in each of its variables. A functionA_nis called symmetric ifA_n(x₁, x₂,· · · , x_n) =A_n(x_π(1), x_π(2),· · ·, x_π(n)) for every permutation{π(1), π(2),· · · , π(n)} of{1,2,· · · , n}. If A_n(x₁, x₂,· · ·, x_n) is an n-additive symmetric map, then Aⁿ(x) will denote the diagonal A_n(x, x,· · · , x) for x∈X and note thatAⁿ(rx) =rⁿAⁿ(x) whenever x∈X andr∈Q. Such a functionAⁿ(x) will be called a monomial function of degreen(assumingAⁿ 6≡0).

Furthermore the resulting function after substitutionx1=x2=· · ·=xl=xandxl+1=xl+2=· · ·=xn=y inAn(x1, x2,· · · , xn) will be denoted byA^l,n−l(x, y).

A function p:X →Y is called a generalized polynomial (GP) function of degreen ∈Nprovided that there existA⁰(x) =A⁰∈Y andi-additive symmetric functionsAi:Xⁱ→Y(for 1≤i≤n) such that

p(x) = Xn

i=0

Aⁱ(x), for allx∈X andAⁿ 6≡0.

Forf :X →Y, let4h be the difference operator defined as follows:

4hf(x) =f(x+h)−f(x)

forh∈X. Furthermore, let4⁰_hf(x) =f(x),4¹_hf(x) =4hf(x) and4h◦ 4ⁿ_hf(x) =4ⁿ⁺¹_h f(x) for alln∈N and all h∈X. Here 4h◦ 4ⁿ_h denotes the composition of the operators 4h and4ⁿ_h. For any givenn∈N, the functional equation 4ⁿ⁺¹_h f(x) = 0 for allx, h∈ X is well studied. In explicit form the last functional equation can be written as

4ⁿ⁺¹_h f(x) =

n+1X

j=0

(−1)^n+1−j

à n+ 1 j

!

f(x+jh) = 0.

The following theorem was proved by Mazur and Orlicz, and in greater generality by Djokovi´c (see [3]).

Theorem 1.1. Let X andY be real vector spaces, n∈Nandf :X→Y, then the following are equivalent.

(1)4ⁿ⁺¹_h f(x) = 0 for allx, h∈X.

(2)4x1,...,xn+1f(x0) = 0for allx0, x1, . . . , xn+1∈X.

(3) f(x) =Aⁿ(x) +Aⁿ⁻¹(x) +A²(x) +A¹(x) +A⁰(x) for all x∈X, where A⁰(x) =A⁰ is an arbitrary element ofY andAⁱ(x)(i= 1,2, . . . , n)is the diagonal of an i-additive symmetric functionAi:Xⁱ→Y. 2. Solution of equation (1.6) on real vector spaces

In this section, we determine the general solution of the functional equations (1.5) and (1.6) and some other related equations without assuming any regularity condition on the unknown functions.

Theorem 2.1. Let X andY be real vector spaces. If the functionsf, g, h,h˜ :X →Y satisfy the functional equation

f(kx+y) +f(kx−y) =g(x+y) +g(x−y) +h(x) + ˜h(y) (2.1.1) for all x, y ∈ X, then f is a solution of the Fr´echet functional equation 4x1,x2,x3,x4,x5f(x0) = 0 for all x0, x1, x2, x3, x4, x5∈X.

Proof. Replacingkx+y by x0 and kx−y byy1(that is, x= _2k¹x0+_2k¹y1 andy = ¹₂x0−¹₂y1) in (2.1.1), respectively, we get

f(x0) +f(y1) =g(k+ 1

2k x0−k−1

2k y1) +g(1−k

2k x0+k+ 1

2k y1) +h(1

2kx0+ 1

2ky1) + ˜h(1 2x0−1

2y1). (2.1.2) Replacingx0byx0+x1 in (2.1.2), we have

f(x0+x1) +f(y1) = g(^k+1_2k (x0+x1)−^k−1_2k y1) +g(^1−k_2k (x0+x1) +^k+1_2k y1)

+h(_2k¹(x0+x1) +_2k¹y1) + ˜h(¹₂(x0+x1)−¹₂y1). (2.1.3) Subtracting (2.1.2) from (2.1.3), we get

f(x0+x1)−f(x0) = g(^k+1_2k (x0+x1)−^k−1_2k y1) +g(^1−k_2k (x0+x1) +^k+1_2k y1)

−g(^k+1_2k x0−^k−1_2k y1)−g(^1−k_2k x0+^k+1_2k y1) +h(_2k¹(x0+x1) +_2k¹y1)−h(_2k¹x0+_2k¹y1) +˜h(¹₂(x0+x1)−¹₂y1)−˜h(¹₂x0−¹₂y1).

(2.1.4)

Lettingy2=_2k¹x0+_2k¹y1(that is, y1= 2ky2−x0) in (2.1.4), we have

f(x0+x1)−f(x0) = g(x0+^k+1_2k x1−(k−1)y2) +g(−x0+^1−k_2k x1+ (k+ 1)y2)

−g(x0−(k−1)y2)−g(−x0+ (k+ 1)y2)

+h(_2k¹x1+y2)−h(y2) + ˜h(x0+¹₂x1−ky2)−˜h(x0−ky2).

(2.1.5)

Replacingx0byx0+x2 in (2.1.5), we get

f(x0+x1+x2)−f(x0+x2) = g(x0+x2+^k+1_2k x1−(k−1)y2) +g(−x0−x2+^1−k_2k x1+ (k+ 1)y2)

−g(x0+x2−(k−1)y2)−g(−x0−x2+ (k+ 1)y2)

+h(_2k¹x₁+y₂)−h(y₂) + ˜h(x₀+x₂+¹₂x₁−ky₂)−˜h(x₀+x₂−ky₂).

(2.1.6) Subtracting (2.1.5) from (2.1.6), we get

f(x0+x1+x2)−f(x0+x1)−f(x0+x2) +f(x0)

=g(x0+x2+^k+1_2k x1−(k−1)y2) +g(−x0−x2+^1−k_2k x1+ (k+ 1)y2)

−g(x0+x2−(k−1)y2)−g(−x0−x2+ (k+ 1)y2) +g(x0−(k−1)y2)

−g(x0+^k+1_2k x1−(k−1)y2)−g(−x0+^1−k_2k x1+ (k+ 1)y2)−g(−x0+ (k+ 1)y2) +˜h(x0+x2+¹₂x1−ky2)−˜h(x0+x2−ky2)−˜h(x0+¹₂x1−ky2) + ˜h(x0−ky2).

(2.1.7)

Lettingy3=x0−ky2(that is, y2=_k¹(x0−y3) in (2.1.7), we have f(x0+x1+x2)−f(x0+x1)−f(x0+x2) +f(x0)

=g(¹_kx0+^k+1_2k x1+x2+^k−1_k y3) +g(¹_kx0+^1−k_2k x1−x2−^k+1_k y3)

−g(¹_kx0+x2+^k−1_k y3)−g(¹_kx0+^k+1_2k x1+^k−1_k y3)−g(_k¹x0+^1−k_2k x1−^k+1_k y3) +g(¹_kx0+^k−1_k y3)−g(_k¹x0−x2−^k+1_k y3) +g(_k¹x0−^k+1_k y3)

+˜h(¹₂x1+x2+y3)−h(x˜ 2+y3)−h(˜ ¹₂x1+y3) + ˜h(y3).

(2.1.8)

Again replacingx0 byx0+x3in (2.1.8) and subtracting (2.1.8) from the resulting expression, we get f(x₀+x₁+x₂+x₃)−f(x₀+x₁+x₂)−f(x₀+x₁+x₃)−f(x₀+x₂+x₃)

+f(x0+x3) +f(x0+x1) +f(x0+x2)−f(x0)

=g(¹_kx0+^k+1_2k x1+x2+¹_kx3+^k−1_k y3) +g(¹_kx0+^1−k_2k x1−x2−^k+1_k y3)

−g(¹_kx0+x2+¹_kx3+^k−1_k y3)−g(¹_kx0+^k+1_2k x1+¹_kx3+^k−1_k y3)

−g(¹_kx0+^1−k_2k x1+_k¹x3−^k+1_k y3) +g(¹_kx0+¹_kx3+^k−1_k y3)

−g(¹_kx0−x2+¹_kx3−^k+1_k y3) +g(_k¹x0+¹_kx3−^k+1_k y3)

−g(¹_kx0+^k+1_2k x1+x2+^k−1_k y3)−g(¹_kx0+^1−k_2k x1−x2−^k+1_k y3)

+g(¹_kx0+x2+^k−1_k y3) +g(¹_kx0+^k+1_2k x1+^k−1_k y3) +g(_k¹x0+^1−k_2k x1−^k+1_k y3)

−g(¹_kx0+^k−1_k y3) +g(_k¹x0−x2−^k+1_k y3)−g(_k¹x0−^k+1_k y3).

(2.1.9)

Puttingy4= ¹_kx0+^k−1_k y3 (that is, y3=_k−1^k y4−_k−1¹ x0) in (2.1.9), we get

f(x0+x1+x2+x3)−f(x0+x1+x2)−f(x0+x1+x3)−f(x0+x2+x3) +f(x0+x3) +f(x0+x1) +f(x0+x2)−f(x0)

=g(^k+1_2k x1+x2+¹_kx3+y4) +g(_k−1² x0+^1−k_2k x1−x2−^k+1_k−1y4)

−g(x2+_k¹x3+y4)−g(^k+1_2k x1+¹_kx3+y4)

−g(_k−1² x0+^1−k_2k x1+¹_kx3−^k+1_k−1y4) +g(¹_kx3+y4)

−g(_k−1² x0−x2+_k¹x3−^k+1_k−1y4) +g(_k−1² x0+_k¹x3−^k+1_k−1y3)

−g(^k+1_2k x1+x2+y4)−g(_k−1² x0+^1−k_2k x1−x2−^k+1_k−1y4) +g(x2+y4) +g(^k+1_2k x1+y4) +g(_k−1² x0+^1−k_2k x1−_k−1^k+1y4)

−g(y4) +g(_k−1² x0−x2−^k+1_k−1y4)−g(_k−1² x0−^k+1_k−1y4).

(2.1.10)

Replacingx0byx0+x4 in (2.1.10) to get

f(x0+x1+x2+x3+x4)−f(x0+x1+x2+x4)−f(x0+x1+x3+x4)−f(x0+x2+x3+x4) +f(x0+x3+x4) +f(x0+x1+x4) +f(x0+x2+x4)−f(x0+x4)

=g(^k+1_2k x1+x2+_k¹x3+y4) +g(_k−1² x0+^1−k_2k x1−x2+_k−1² x4−^k+1_k−1y4)

−g(x2+_k¹x3+y4)−g(^k+1_2k x1+_k¹x3+y4)

−g(_k−1² x0+^1−k_2k x1+¹_kx3+_k−1² x4−^k+1_k−1y4) +g(¹_kx3+y4)

−g(_k−1² x0−x2+_k¹x3+_k−1² x4−^k+1_k−1y4) +g(_k−1² x0+¹_kx3+_k−1² x4−^k+1_k−1y4)

−g(^k+1_2k x1+x2+y4)−g(_k−1² x0+^1−k_2k x1−x2+_k−1² x4−^k+1_k−1y4) +g(x2+y4) +g(^k+1_2k x1+y4) +g(_k−1² x0+^1−k_2k x1+_k−1² x4−^k+1_k−1y4)

−g(y4) +g(_k−1² x0−x2+_k−1² x4−^k+1_k−1y4)−g(_k−1² x0+_k−1² x4−^k+1_k−1y4).

(2.1.11) Subtract (2.1.10) from (2.1.11), we get

f(x0+x1+x2+x3+x4)−f(x0+x1+x2+x3)−f(x0+x1+x2+x4)

−f(x0+x1+x3+x4)−f(x0+x2+x3+x4) +f(x0+x1+x2) +f(x0+x1+x3) +f(x0+x2+x3) +f(x0+x3+x4) +f(x0+x1+x4) +f(x0+x2+x4)

−f(x0+x1)−f(x0+x2)−f(x0+x3)−f(x0+x4) +f(x0)

=g(_k−1² x0+^1−k_2k x1−x2+_k−1² x4−^k+1_k−1y4)−g(_k−1² x0+^1−k_2k x1+¹_kx3+_k−1² x4−^k+1_k−1y4)

−g(_k−1² x0−x2+¹_kx3+_k−1² x4−^k+1_k−1y4) +g(_k−1² x0+¹_kx3+_k−1² x4−^k+1_k−1y4)

−g(_k−1² x0+^1−k_2k x1−x2+_k−1² x4−^k+1_k−1y4) +g(_k−1² x0+^1−k_2k x1+_k−1² x4−^k+1_k−1y4) +g(_k−1² x0−x2+_k−1² x4−^k+1_k−1y4)−g(_k−1² x0+_k−1² x4−^k+1_k−1y4)

−g(_k−1² x0+^1−k_2k x1−x2−^k+1_k−1y4) +g(_k−1² x0+^1−k_2k x1+¹_kx3−^k+1_k−1y4) +g(_k−1² x0−x2+¹_kx3−^k+1_k−1y4)−g(_k−1² x0+¹_kx3−^k+1_k−1y4)

+g(_k−1² x0+^1−k_2k x1−x2−^k+1_k−1y4) +g(_k−1² x0+^1−k_2k x1−^k+1_k−1y4)

−g(_k−1² x0−x2−^k+1_k−1y4) +g(_k−1² x0−^k+1_k−1y4).

(2.1.12)

Settingy5=_k−1² x0−^k+1_k−1y4 (that is,y4= _k+1² x0−^k−1_k+1y5) in (2.1.12), we have

f(x0+x1+x2+x3+x4)−f(x0+x1+x2+x3)−f(x0+x1+x2+x4)

−f(x0+x1+x3+x4)−f(x0+x2+x3+x4) +f(x0+x1+x2) +f(x0+x1+x3) +f(x0+x2+x3) +f(x0+x3+x4) +f(x0+x1+x4) +f(x0+x2+x4)

−f(x0+x1)−f(x0+x2)−f(x0+x3)−f(x0+x4) +f(x0)

=g(^1−k_2k x1−x2+_k−1² x4+y5)−g(^1−k_2k x1+¹_kx3+_k−1² x4+y5)

−g(−x2+_k¹x3+_k−1² x4+y5) +g(¹_kx3+_k−1² x4+y5)−g(^1−k_2k x1−x2+_k−1² x4+y5) +g(^1−k_2k x1+_k−1² x4+y5) +g(−x2+_k−1² x4+y5)−g(_k−1² x4+y5)

−g(^1−k_2k x1−x2+y5) +g(^1−k_2k x1+_k¹x3+y5) +g(−x2+¹_kx3+y5)

−g(_k¹x3+y5) +g(^1−k_2k x1−x2+y5) +g(^1−k_2k x1+y5)−g(−x2+y5) +g(y5).

(2.1.13)

Replacingx0byx0+x5 in (2.1.13) to get

f(x0+x1+x2+x3+x4+x5)−f(x0+x1+x2+x3+x5)−f(x0+x1+x2+x4+x5)

−f(x0+x1+x3+x4+x5)−f(x0+x2+x3+x4+x5) +f(x0+x1+x2+x5) +f(x0+x1+x3+x5) +f(x0+x2+x3+x5) +f(x0+x3+x4+x5)

+f(x0+x1+x4+x5) +f(x0+x2+x4+x5)−f(x0+x1+x5)

−f(x0+x2+x5)−f(x0+x3+x5)−f(x0+x4+x5) +f(x0+x5)

=g(^1−k_2k x1−x2+_k−1² x4+y5)−g(^1−k_2k x1+_k¹x3+_k−1² x4+y5)

−g(−x2+¹_kx3+_k−1² x4+y5) +g(_k¹x3+_k−1² x4+y5)−g(^1−k_2k x1−x2+_k−1² x4+y5) +g(^1−k_2k x₁+_k−1² x₄+y₅) +g(−x₂+_k−1² x₄+y₅)−g(_k−1² x₄+y₅)

−g(^1−k_2k x1−x2+y5) +g(^1−k_2k x1+¹_kx3+y5) +g(−x2+_k¹x3+y5)

−g(¹_kx3+y5) +g(^1−k_2k x1−x2+y5) +g(^1−k_2k x1+y5)−g(−x2+y5) +g(y5).

(2.1.14)

Subtract (2.1.13) from (2.1.14), we get

f(x0+x1+x2+x3+x4+x5)−f(x0+x1+x2+x3+x4)−f(x0+x1+x2+x3+x5)

−f(x0+x1+x2+x4+x5)−f(x0+x1+x3+x4+x5)−f(x0+x2+x3+x4+x5) +f(x0+x1+x2+x3) +f(x0+x1+x2+x4) +f(x0+x1+x3+x4)

+f(x0+x2+x3+x4) +f(x0+x1+x2+x5) +f(x0+x1+x3+x5) +f(x0+x2+x3+x5) +f(x0+x3+x4+x5) +f(x0+x1+x4+x5)

+f(x0+x2+x4+x5)−f(x0+x1+x5)−f(x0+x2+x5)−f(x0+x3+x5)

−f(x0+x4+x5)−f(x0+x1+x2)−f(x0+x1+x3)−f(x0+x2+x3)

−f(x0+x3+x4)−f(x0+x1+x4)−f(x0+x2+x4) +f(x0+x5) +f(x0+x1) +f(x0+x2) +f(x0+x3) +f(x0+x4)−f(x0) = 0 which is4x1,x2,x3,x4,x5f(x0) = 0 for allx0, x1, x2, x3, x4, x5∈X. ¤

As an application of Theorem 2.1, we can get the following theorem which is proved in [5, Theorem 2.2].

Theorem 2.2. If X and Y are real vector spaces, then the function f : X → Y satisfies the functional equation

f(kx+y) +f(kx−y) =k²f(x+y) +k²f(x−y) + 2(f(kx)−k²f(x))−2(k²−1)f(y), (2.2.1) for allx, y∈X if and only if f is of the form

f(x) =A⁴(x) +A²(x), for allx∈X,

whereAⁱ(x)is the diagonal of the i-additive symmetric mapAi:Xⁱ→Y fori= 2,4.

Proof. Assume thatf satisfies the functional equation (2.2.1). By Theorem 2.1 we see thatf is a solution of the Fr´echet functional equation4x1,x2,x3,x4,x5f(x0) = 0 for allx0, x1, x2, x3, x4∈X. Thus from Theorem 1.1 we have

f(x) =A⁴(x) +A³(x) +A²(x) +A¹(x) +A⁰(x), for allx∈X, (2.2.2) where A⁰(x) =A⁰ is an arbitrary element ofY, andAⁱ(x) is the diagonal of thei-additive symmetric map Ai:Xⁱ→Y fori= 1,2,3,4.

By lettingx=y= 0 in (2.2.1), we getf(0) = 0. HenceA⁰(x) =A⁰= 0. Let us setx= 0 in (2.2.1) to get f(y) =f(−y) for ally ∈X. So the functionf is even. Thus we haveA³(x)≡0 andA¹(x)≡0. Therefore we havef(x) =A⁴(x) +A²(x). The proof of the converse can be easily checked. ¤

Theorem 2.3. If X and Y are real vector spaces, then the function f : X → Y satisfies the functional equation

f(kx+y) +f(kx−y) =k²f(x+y) +k²f(x−y) + 2k²(k²−1)f(x)−2(k²−1)f(y), (2.3.1) if and only if f is of the form

f(x) =A⁴(x), for allx∈X, whereA⁴(x) is the diagonal of the4-additive symmetric mapA4:X⁴→Y.

Proof. Assume that f satisfies the functional equation (2.3.1). Putting x = y = 0 in (2.3.1), we have f(0) = 0. Puttingx= 0 in (2.3.1), we have f(−y) =f(y) for ally∈X. By Theorem 2.1 we see thatf is a solution of the Fr´echet functional equation4x1,x2,x3,x4,x5f(x0) = 0 for all x0, x1, x2, x3, x4 ∈X. Thus from Theorem 1.1 we have

f(x) =A⁴(x) +A³(x) +A²(x) +A¹(x) +A⁰(x), for allx∈X, (2.3.2) where A⁰(x) =A⁰ is an arbitrary element ofY, andAⁱ(x) is the diagonal of thei-additive symmetric map Ai:Xⁱ→Y fori= 1,2,3,4. Hencef(x) =A⁴(x) +A²(x) for allx∈X. Puttingf(x) =A⁴(x) +A²(x) into (2.3.1), and noting that

( A⁴(x+y) +A⁴(x−y) = 2A⁴(x) + 2A⁴(y) + 12A^2,2(x, y), A²(x+y) +A²(x−y) = 2A²(x) + 2A²(y),

andA^2,2(kx, y) =k²A^2,2(x, y), we conclude thatA²(x) = 0 for allx∈X. Hencef(x) =A⁴(x) for allx∈X. The proof of the converse can be easily checked. ¤

Remark. We observe that in casek= 2, Eq.(2.3.1) yields the quartic functional equation (1.4). Therefore, Theorem 2.3 is a generalized version of the theorem for a solution of quartic functional equations (see [9, Theorem 2.1]).

Theorem 2.4. IfX andY are real vector spaces, then the functions f, g, h,˜h:X →Y satisfy the functional equation

f(kx+y) +f(kx−y) =g(x+y) +g(x−y) +h(x) + ˜h(y) (2.4.1) for allx, y∈X if and only if

















f(x) = A⁴(x) +A³(x) +A²(x) +A¹(x) +A⁰(x),

g(x) = k²A⁴(x) +kA³(x) +B²(x) +B⁰(x) +C¹(x) +D⁰(x),

h(x) = (2k⁴−2k²)A⁴(x) + (2k³−2k)A³(x) + 2k²A²(x) + 2kA¹(x) + 2A⁰(x)

−2B²(x)−2C¹(x)−2B⁰(x),

˜h(x) = (2−2k²)A⁴(x) + 2A²(x)−2B²(x)−2D⁰(x),

(2.4.2)

where A⁰(x) =A⁰, B⁰(x) =B⁰ andD⁰(x) =D⁰ are arbitrary elements of Y, and Aⁱ(x), Bⁱ(x), Cⁱ(x) are the diagonal of the i-additive symmetric mapsAi, Bi, Ci:Xⁱ→Y, respectively, fori= 1,2,3,4.

Proof. Assume that f, g, h,˜h satisfy the functional equation (2.4.1). By Theorem 2.1 we see that f is a solution of the Fr´echet functional equation4x1,x2,x3,x4,x5f(x0) = 0 for all x0, x1, x2, x3, x4, x5 ∈X. Hence from Theorem 1.1 we have

f(x) =A⁴(x) +A³(x) +A²(x) +A¹(x) +A⁰(x), for allx∈X, (2.4.3)

where A⁰(x) =A⁰ is an arbitrary element ofY, andAⁱ(x) is the diagonal of thei-additive symmetric map Ai:Xⁱ→Y fori= 1,2,3,4. Putting (2.4.3) into (2.4.1), and noting that







A⁴(x+y) +A⁴(x−y) = 2A⁴(x) + 2A⁴(y) + 12A^2,2(x, y), A³(x+y) +A³(x−y) = 2A³(x) + 6A^1,2(x, y),

A²(x+y) +A²(x−y) = 2A²(x) + 2A²(y), andA^2,2(kx, y) =k²A^2,2(x, y),A^1,2(kx, y) =kA^1,2(x, y), we conclude that

g(x+y) +g(x−y) +h(x) + ˜h(y) = 2k⁴A⁴(x) + 2A⁴(y) + 12k²A^2,2(x, y) + 2k³A³(x) + 6kA^1,2(x, y) +2k²A²(x) + 2A²(y) + 2kA¹(x) + 2A⁰(x).

Therefore

g(x+y) +g(x−y) +h(x) + ˜h(y)

=k²A⁴(x+y) +k²A⁴(x−y) +kA³(x+y) +kA³(x−y)

+(2k⁴−2k²)A⁴(x) + (2k³−2k)A³(x) + 2k²A²(x) + 2kA¹(x) + 2A⁰(x) +(2−2k²)A⁴(y) + 2A²(y).

(2.4.4)

Letting

G(x) =g(x)−k²A⁴(x)−kA³(x), H˜(x) =−˜h(x) + (2−2k²)A⁴(x) + 2A²(x) (2.4.5) and

H(x) =−h(x) + (2k⁴−2k²)A⁴(x) + (2k³−2k)A³(x) + 2k²A²(x) + 2kA¹(x) + 2A⁰(x). (2.4.6) Then from (2.4.4) we have

G(x+y) +G(x−y) =H(x) + ˜H(y). (2.4.7) LetGsatisfies (2.4.7). We decomposeGinto the even part and odd part by putting

Ge(x) =1

2(G(x) +G(−x)), Go(x) = 1

2(G(x)−G(−x))

for allx∈X. It is clear thatG(x) =Ge(x) +Go(x) for allx∈X. Similarly, we haveH(x) =He(x) +Ho(x) and ˜H(x) = ˜He(x) + ˜Ho(x). Thus

Ge(x+y) +Ge(x−y) =He(x) + ˜He(y), (2.4.8) and

Go(x+y) +Go(x−y) =Ho(x) + ˜Ho(y). (2.4.9) Letting y = 0 in (2.4.8), we have H_e(x) = 2G_e(x)−H˜_e(0). Setting x = 0 in (2.4.8) to get ˜H_e(y) = 2G_e(y)−H_e(0). Hence

Ge(x+y) +Ge(x−y) = 2Ge(x) + 2Ge(y)−2Ge(0). (2.4.10) for allx, y∈X. Setting M(x) =Ge(x)−Ge(0), we get

M(x+y) +M(x−y) = 2M(x) + 2M(y) (2.4.11) which is the quadratic functional equation and its solution is given by

M(x) =B²(x) for allx∈X,

whereB²(x) is the diagonal of the 2-additive symmetric mapB2:X²→Y. In this case, we obtain

Ge(x) =B²(x) +G(0), He(x) = 2B²(x) +He(0), H˜e(x) = 2B²(x) + ˜He(0). (2.4.12)

Similarly, lettingy= 0 in (2.4.9), we haveHo(x) = 2Go(x). Settingx= 0 in (2.4.9) to get ˜Ho(y) = 0. Then from (2.4.9) we have

G_o(x+y) +G_o(x−y) = 2G_o(x), (2.4.13) which is the Jensen functional equation and its solution is given by

Go(x) =C¹(x), (2.4.14)

whereC¹:X →Y is an additive function. Thus







G(x) =Ge(x) +Go(x) =B²(x) +B⁰(x) +D⁰(x) +C¹(x), H(x) =He(x) +Ho(x) = 2B²(x) + 2C¹(x) + 2B⁰(x), H(x) = ˜˜ He(x) + ˜Ho(x) = 2B²(x) + 2D⁰(x).

(2.4.15)

whereB⁰(x) =B⁰andD⁰(x) =D⁰ are arbitrary elements ofY. Therefore from (2.4.5), (2.4.6), (2.4.15), we obtain the asserted solution (2.4.2). The proof of the converse can be easily checked. ¤

3. Solution of equation (1.6) on commutative groups

In this section, we solve the functional equation (1.6) on commutative groups with some additional requirements.

A groupGis said to be divisible if for every elementb∈Gand everyn∈N, there exists an elementa∈G such thatna=b. If this elementais unique, thenGis said to be uniquely divisible. In a uniquely divisible group, this unique element a is denoted by _n^b. That the equation na = b has a solution is equivalent to saying that the multiplication bynis surjective. Similarly, that the equationna=bhas a unique solution is equivalent to saying that the multiplication bynis bijective. Hence the notions ofn-divisibility andn-unique divisibility refer, respectively, to surjectivity and bijectivity of the multiplication byn.

Lemma 3.1.(Hossz´u [6])Letn≥0 be an integer,GandS be abelian groups. Furthermore let S be uniquely divisible. The map F fromGintoS satisfies the functional equation

4x1,...,xn+1F(x0) = 0 for allx0, x1, . . . , xn+1∈Gif and only ifF is given by

F(x) =Aⁿ(x) +· · ·+A¹(x) +A⁰(x), for allx∈G,

whereA⁰(x) =A⁰ is an arbitrary element ofSandAⁿ(x)is the diagonal of ann-additive symmetric function An:Gⁿ→S.

The solution of the functional equation (2.2.1) can be determined in certain type of groups by using Lemma 3.1. As the proof is identical with the proof of Theorem 2.2 we simply state the theorem without a proof.

Theorem 3.2. Let GandS be uniquely divisible abelian groups. Then the functionf :G→S satisfies the functional equation (2.2.1) for all x, y∈G, if and only iff is of the form

f(x) =A⁴(x) +A²(x), for allx∈G,

whereAⁱ(x)is the diagonal of the i-additive symmetric mapAi:Gⁱ→S fori= 2,4.

Theorem 3.3. Let GandS be uniquely divisible abelian groups. Then the functionf :G→S satisfies the functional equation (2.3.1), if and only if f is of the form

f(x) =A⁴(x), for allx∈X, whereA⁴(x) is the diagonal of the4-additive symmetric mapA4:G⁴→S.