A generalized mixed type of quartic, cubic, quadratic and additive functional equation
We determine the general solution of the functional equation $f(x + ky) + f(x — ky) = g(x + y) + g(x — y) + h(x) + \tilde{h}(y)$ forfixed integers $k$ with $k \neq 0, \pm 1$ without assuming any regularity condition on the unknown functions $f, g, h, \tilde{h}$. The method used for solving these f...
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| Дата: | 2011 |
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Institute of Mathematics, NAS of Ukraine
2011
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860508683227627520 |
|---|---|
| author | Rassias, J. M. Xu, T. Z. Xu, W. X. Расіас, Дж. М. Ху, Т. З. Ху, В. Х. |
| author_facet | Rassias, J. M. Xu, T. Z. Xu, W. X. Расіас, Дж. М. Ху, Т. З. Ху, В. Х. |
| author_sort | Rassias, J. M. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2020-03-18T19:34:39Z |
| description | We determine the general solution of the functional equation $f(x + ky) + f(x — ky) = g(x + y) + g(x — y) + h(x) + \tilde{h}(y)$
forfixed integers $k$ with $k \neq 0, \pm 1$ without assuming any regularity condition on the unknown functions $f, g, h, \tilde{h}$.
The method used for solving these functional equations is elementary but exploits an important result due to Hosszii.
The solution of this functional equation can also be determined in certain type of groups using two important results due to Szekelyhidi. |
| first_indexed | 2026-03-24T02:29:06Z |
| format | Article |
| fulltext |
UDC 517.9
T. Z. Xu* (School Sci., Beijing Inst. Technology, China),
J. M. Rassias (Nat. Capodistrian Univ. Athens, Greece),
W. X. Xu (Univ. Electron. Sci. and Technology, Chengdu, China)
A GENERALIZED MIXED TYPE OF QUARTIC, CUBIC,
QUADRATIC AND ADDITIVE FUNCTIONAL EQUATION
УЗАГАЛЬНЕНИЙ МIШАНИЙ ТИП КВАРТИЧНОГО,
КУБIЧНОГО, КВАДРАТИЧНОГО ТА ДОДАТКОВОГО
ФУНКЦIОНАЛЬНОГО РIВНЯННЯ
We determine the general solution of the functional equation f(x+ky)+f(x−ky) = g(x+y)+g(x−y)+
+h(x)+ h̃(y) for fixed integers k with k 6= 0,±1 without assuming any regularity condition on the unknown
functions f, g, h, h̃. The method used for solving these functional equations is elementary but exploits an
important result due to Hosszú. The solution of this functional equation can also be determined in certain type
of groups using two important results due to Székelyhidi.
Визначено загальний розв’язок функцiонального рiвняння f(x + ky) + f(x − ky) = g(x + y) +
+ g(x− y) + h(x) + h̃(y) для фiксованих цiлих k при k 6= 0,±1 без припущення наявностi будь-якої
умови регулярностi для невiдомих функцiй f, g, h, h̃. Метод, що використано для розв’язку цих
функцiональних рiвнянь, елементарний, але базується на важливому результатi Хозу. Розв’язок цього
функцiонального рiвняння може бути визначений у певному типi груп з використанням двох важливих
результатiв Чекелiхiдi.
1. Introduction and preliminaries. J. M. Rassias [11] (in 2001) introduced the first
cubic functional equation
f(x+ 2y)− 3f(x+ y) + 3f(x)− f(x− y) = 6f(y) (1.1)
and established the solution of the Ulam – Hyers stability problem for this cubic functi-
onal equation. Since the function f(x) = x3 satisfies the functional equation (1.1),
this equation is called cubic functional equation. Every solution of the cubic functional
equation is said to be a cubic function.
J. M. Rassias [12] (in 1999) introduced the first quartic functional equation
f(x+ 2y) + f(x− 2y) + 6f(x) = 4[f(x+ y)− f(x− y) + 6f(y)]. (1.2)
and established the solution of the Ulam – Hyers stability problem for the quartic functi-
onal equation. Since the function f(x) = x4 satisfies the functional equation (1.2),
this equation is called quartic functional equation. J. K. Chung and P. K. Sahoo [2]
determined the general solution of the functional equation (1.2).
M. Eshaghi Gordji and H. Khodaei [5] (in 2009) introduced the following generalized
mixed type of cubic, quadratic and additive functional equation
f(x+ ky) + f(x− ky) = k2f(x+ y) + k2f(x− y) + 2(1− k2)f(y) (1.3)
and established the general solution and the generalized Ulam – Hyers stability for the
functional equation (1.3). They proved that a function f with f(0) = 0 between real
*The first author was supported by the National Natural Science Foundation of China (10671013,
60972089).
c© T. Z. XU, J. M. RASSIAS, W. X. XU, 2011
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 3 399
400 T. Z. XU, J. M. RASSIAS, W. X. XU
vector spacesX and Y is a solution of (1.3) if and only if there exist functions C : X3 →
→ Y and B : X2 → Y and A : X → Y, such that f(x) = C(x, x, x) +B(x, x) +A(x)
for all x ∈ X, where the function C is symmetric for each fixed variable and is additive
for fixed two variables and B is symmetric bi-additive and A is additive. In this paper,
we determine the general solution of the functional equation (1.3) using an elementary
technique but without assuming f(0) = 0.
M. Eshaghi Gordji, S. Kaboli-Gharetapeh, C. Park, and S. Zolfaghari [7] introduced
an additive-cubic-quartic functional equation
11[f(x+ 2y) + f(x− 2y)] =
= 44[f(x+ y) + f(x− y)] + 12f(3y)− 48f(2y) + 60f(y)− 66f(x) (1.4)
and established the general solution and the generalized Ulam – Hyers stability for the
functional equation (1.4). In [8], M. Eshaghi Gordji, H. Khodaei and Th. M. Rassias
introduced a generalized mixed type of quartic, cubic, quadratic and additive functional
equation
f(x+ ky) + f(x− ky) = k2f(x+ y) + k2f(x− y)+
+2(1− k2)f(x) + k2(k2 − 1)
12
[f̃(2y)− 4f̃(y)], (1.5)
where f̃(y) = f(y) + f(−y) for all y ∈ X . They proved that a function f between real
vector spaces X and Y is a solution of (1.5) if and only if there exist a symmetric multi-
additive functionM : X4 → Y, a function C : X3 → Y, a symmetric bi-additive function
B : X2 → Y and an additive function A : X → Y, such that f(x) = M(x, x, x, x) +
+C(x, x, x)+B(x, x)+A(x) for all x ∈ X, where the function C is symmetric for each
fixed variable and is additive for two fixed variables. In this paper, we also determine
the general solution of the functional equation (1.5) using an elementary technique.
Let k be a fixed integer with k 6= 0,±1, X and Y are real vector spaces. Equati-
ons (1.2) – (1.5) can be generalized to
f(x+ ky) + f(x− ky) = g(x+ y) + g(x− y) + h(x) + h̃(y) (1.6)
for all x, y ∈ X, where f, g, h, h̃ : 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 [2, 14, 15, 22, 23] 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 polynomi-
als. The interested reader should refer to [4] and [13] and references therein (see also
[17 – 21]).
A function A : X → Y is said to be additive if A(x + y) = A(x) + A(y) for all
x, y ∈ X . It is easy to see that A(rx) = rA(x) for all x ∈ X and all r ∈ Q (the set of
rational numbers).
Let n ∈ N (the set of natural numbers). A function An : X
n → Y is called n-
additive if it is additive in each of its variables. A function An is called symmetric if
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 3
A GENERALIZED MIXED TYPE OF QUARTIC, CUBIC, QUADRATIC AND ADDITIVE . . . 401
An(x1, x2, . . . , xn) = An(xπ(1), xπ(2), . . . , xπ(n)) for every permutation {π(1),
π(2), . . . , π(n)} of {1, 2, . . . , n}. If An(x1, x2, . . . , xn) is an n-additive symmetric
map, then An(x) will denote the diagonal An(x, x, . . . , x) for x ∈ X and note that
An(rx) = rnAn(x) whenever x ∈ X and r ∈ Q. Such a function An(x) will be called
a monomial function of degree n (assuming An 6≡ 0). Furthermore the resulting function
after substitution x1 = x2 = . . . = xl = x and xl+1 = xl+2 = . . . = xn = y in
An(x1, x2, . . . , xn) will be denoted by Al,n−l(x, y).
A function p : X → Y is called a generalized polynomial (GP) function of degree
n ∈ N provided that there exist A0(x) = A0 ∈ Y and i-additive symmetric functions
Ai : X
i → Y (for 1 ≤ i ≤ n) such that
p(x) =
n∑
i=0
Ai(x), for all x ∈ X
and An 6≡ 0.
For f : X → Y, let 4h be the difference operator defined as follows:
4hf(x) = f(x+ h)− f(x)
for h ∈ X . Furthermore, let 40
hf(x) = f(x), 41
hf(x) = 4hf(x) and 4h ◦4nhf(x) =
= 4n+1
h f(x) for all n ∈ N and all h ∈ X . Here4h ◦4nh denotes the composition of the
operators 4h and 4nh . For any given n ∈ N, the functional equation 4n+1
h f(x) = 0 for
all x, h ∈ X is well studied. In explicit form the last functional equation can be written
as
4n+1
h f(x) =
n+1∑
j=0
(−1)n+1−j
(
n+ 1
j
)
f(x+ jh) = 0.
The following theorem was proved by Mazur and Orlicz, and in greater generality
by Djoković (see [3]).
Theorem 1.1. Let X and Y be real vector spaces, n ∈ N and f : X → Y, then
the following are equivalent:
(1) 4n+1
h f(x) = 0 for all x, h ∈ X .
(2) 4x1,...,xn+1f(x0) = 0 for all x0, 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 of Y and Ai(x), i = 1, 2, . . . , n, is the diagonal of
an i-additive symmetric function Ai : Xi → Y .
2. Solution of equation (1.6) on real vector spaces. In this section, we determine
the general solution of the functional equation (1.6) and some other related equations
without assuming any regularity condition on the unknown functions.
Theorem 2.1. LetX and Y be real vector spaces. If the functions f, g, h, h̃ : X →
→ Y satisfy the functional equation
f(x+ ky) + f(x− ky) = g(x+ y) + g(x− y) + h(x) + h̃(y), for all x, y ∈ X
(2.1)
for fixed integers k with k 6= 0,±1, then f is a solution of the Fréchet functional
equation 4x1,x2,x3,x4,x5f(x0) = 0 for all x0, x1, x2, x3, x4, x5 ∈ X .
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 3
402 T. Z. XU, J. M. RASSIAS, W. X. XU
Proof. Replacing x + ky by x0 and x − ky by y1
(
that is, x =
1
2
x0 +
1
2
y1 and
y =
1
2k
x0 −
1
2k
y1
)
in (2.1), respectively, we get
f(x0) + f(y1) = g
(
k + 1
2k
x0 +
k − 1
2k
y1
)
+
+g
(
k − 1
2k
x0 +
k + 1
2k
y1
)
+ h
(
1
2
x0 +
1
2
y1
)
+ h̃
(
1
2k
x0 −
1
2k
y1
)
. (2.2)
Replacing x0 by x0 + x1 in (2.2), we have
f(x0 + x1) + f(y1) = g
(
k + 1
2k
(x0 + x1) +
k − 1
2k
y1
)
+
+g
(
k − 1
2k
(x0 + x1) +
k + 1
2k
y1
)
+ h
(
1
2
(x0 + x1) +
1
2
y1
)
+
+h̃
(
1
2k
(x0 + x1)−
1
2k
y1
)
. (2.3)
Subtracting (2.2) from (2.3), we get
f(x0 + x1)− f(x0) = g
(
k + 1
2k
(x0 + x1) +
k − 1
2k
y1
)
+
+g
(
k − 1
2k
(x0 + x1) +
k + 1
2k
y1
)
− g
(
k + 1
2k
x0 +
k − 1
2k
y1
)
−
−g
(
k − 1
2k
x0 +
k + 1
2k
y1
)
+ h
(
1
2
(x0 + x1) +
1
2
y1
)
− h
(
1
2
x0 +
1
2
y1
)
+
+h̃
(
1
2k
(x0 + x1)−
1
2k
y1
)
− h̃
(
1
2k
x0 −
1
2k
y1
)
. (2.4)
Letting y2 =
1
2
x0 +
1
2
y1 (that is, y1 = 2y2 − x0) in (2.4), we have
f(x0 + x1)− f(x0) = g
(
1
k
x0 +
k + 1
2k
x1 +
k − 1
k
y2
)
+
+g
(
−1
k
x0 +
k − 1
2k
x1 +
k + 1
k
y2
)
− g
(
1
k
x0 +
k − 1
k
y2
)
−
−g
(
−1
k
x0 +
k + 1
k
y2
)
+ h
(
1
2
x1 + y2
)
−
−h(y2) + h̃
(
1
k
x0 +
1
2k
x1 −
1
k
y2
)
− h̃
(
1
k
x0 −
1
k
y2
)
. (2.5)
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 3
A GENERALIZED MIXED TYPE OF QUARTIC, CUBIC, QUADRATIC AND ADDITIVE . . . 403
Replacing x0 by x0 + x2 in (2.5), we get
f(x0 + x1)− f(x0) = g
(
1
k
(x0 + x2) +
k + 1
2k
x1 +
k − 1
k
y2
)
+
+g
(
−1
k
(x0 + x2) +
k − 1
2k
x1 +
k + 1
k
y2
)
−
−g
(
1
k
(x0 + x2) +
k − 1
k
y2
)
− g
(
−1
k
(x0 + x2) +
k + 1
k
y2
)
+ h
(
1
2
x1 + y2
)
−
−h(y2) + h̃
(
1
k
(x0 + x2) +
1
2k
x1 −
1
k
y2
)
− h̃
(
1
k
(x0 + x2)−
1
k
y2
)
. (2.6)
Subtracting (2.5) from (2.6), we get
f(x0 + x1 + x2)− f(x0 + x1)− f(x0 + x2) + f(x0) =
= g
(
1
k
x0 +
k + 1
2k
x1 +
1
k
x2 +
k − 1
k
y2
)
+
+g
(
−1
k
x0 +
k − 1
2k
x1 −
1
k
x2 +
k + 1
k
y2
)
−
−g
(
1
k
x0 +
1
k
x2 +
k − 1
k
y2
)
− g
(
−1
k
x0 −
1
k
x2 +
k + 1
k
y2
)
−
−g
(
1
k
x0 +
k + 1
2k
x1 +
k − 1
k
y2
)
− g
(
−1
k
x0 +
k − 1
2k
x1 +
k + 1
k
y2
)
+
+g
(
1
k
x0 +
k − 1
k
y2
)
+ g
(
−1
k
x0 +
k + 1
k
y2
)
+
+h̃
(
1
k
x0 +
1
2k
x1 +
1
k
x2 −
1
k
y2
)
− h̃
(
1
k
x0 +
1
k
x2 −
1
k
y2
)
−
−h̃
(
1
k
x0 +
1
2k
x1 −
1
k
y2
)
+ h̃
(
1
k
x0 −
1
k
y2
)
. (2.7)
Letting y3 =
1
k
x0 −
1
k
y2 (that is, y2 = x0 − ky3) in (2.7), we have
f(x0 + x1 + x2)− f(x0 + x1)− f(x0 + x2) + f(x0) =
= g
(
x0 +
k + 1
2k
x1 +
1
k
x2 − (k − 1)y3
)
+
+g
(
x0 +
k − 1
2k
x1 −
1
k
x2 − (k + 1)y3
)
− g
(
x0 +
1
k
x2 − (k − 1)y3
)
−
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 3
404 T. Z. XU, J. M. RASSIAS, W. X. XU
−g
(
x0 −
1
k
x2 − (k + 1)y3
)
− g
(
x0 +
k + 1
2k
x1 − (k − 1)y3
)
−
−g
(
x0 +
k − 1
2k
x1 − (k + 1)y3
)
+ g(x0 − (k − 1)y3) + g(x0 − (k + 1)y3)+
+h̃
(
1
2k
x1 +
1
k
x2 + y3
)
− h̃
(
1
k
x2 + y3
)
− h̃
(
1
2k
x1 + y3
)
+ h̃(y3). (2.8)
Again replacing x0 by x0+x3 in (2.8) and subtracting (2.8) 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
(
x0 +
k + 1
2k
x1 +
1
k
x2 + x3 − (k − 1)y3
)
+
+g
(
x0 +
k − 1
2k
x1 −
1
k
x2 + x3 − (k + 1)y3
)
−
−g
(
x0 +
1
k
x2 + x3 − (k − 1)y3
)
− g
(
x0 −
1
k
x2 + x3 − (k + 1)y3
)
−
−g
(
x0 +
k + 1
2k
x1 + x3 − (k − 1)y3
)
− g
(
x0 +
k − 1
2k
x1 + x3 − (k + 1)y3
)
+
+g(x0 + x3 − (k − 1)y3) + g(x0 + x3 − (k + 1)y3)−
−g
(
x0 +
k + 1
2k
x1 +
1
k
x2 − (k − 1)y3
)
− g
(
x0 +
k − 1
2k
x1 −
1
k
x2 − (k + 1)y3
)
+
+g
(
x0 +
1
k
x2 − (k − 1)y3
)
− g
(
x0 −
1
k
x2 − (k + 1)y3
)
+
+g
(
x0 +
k + 1
2k
x1 − (k − 1)y3
)
− g
(
x0 +
k − 1
2k
x1 − (k + 1)y3
)
−
−g(x0 − (k − 1)y3) + g(x0 − (k + 1)y3). (2.9)
Putting y4 = x0 − (k − 1)y3
(
that is, y3 =
1
k − 1
x0 −
1
k − 1
y4
)
in (2.9), 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) =
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 3
A GENERALIZED MIXED TYPE OF QUARTIC, CUBIC, QUADRATIC AND ADDITIVE . . . 405
= g
(
k + 1
2k
x1 +
1
k
x2 + x3 + y4
)
+
+g
(
−2
k − 1
x0 +
k − 1
2k
x1 −
1
k
x2 + x3 +
k + 1
k − 1
y4
)
−
−g
(
1
k
x2 + x3 + y4
)
− g
(
−2
k − 1
x0 −
1
k
x2 + x3 +
k + 1
k − 1
y4
)
−
−g
(
k + 1
2k
x1 + x3 + y4
)
− g
(
−2
k − 1
x0 +
k − 1
2k
x1 + x3 +
k + 1
k − 1
y4
)
+
+g(x3 + y4) + g
(
−2
k − 1
x0 + x3 +
k + 1
k − 1
y4
)
−
−g
(
k + 1
2k
x1 +
1
k
x2 + y4
)
− g
(
−2
k − 1
x0 +
k − 1
2k
x1 −
1
k
x2 +
k + 1
k − 1
y4
)
+
+g
(
1
k
x2 + y4
)
− g
(
−2
k − 1
x0 −
1
k
x2 +
k + 1
k − 1
y4
)
+ g
(
k + 1
2k
x1 + y4
)
−
−g
(
−2
k − 1
x0 +
k − 1
2k
x1 +
k + 1
k − 1
y4
)
− g(y4) + g
(
−2
k − 1
x0 +
k + 1
k − 1
y4
)
. (2.10)
Replacing x0 by x0 + x4 in (2.10) to get
f(x0 + x1 + x2 + x3 + x4)− f(x0 + x1 + x2 + x4)−
−f(x0 + x1 + x3 + x4)− f(x0 + x4)− f(x0 + x2 + x3 + x4)+
+f(x0 + x3 + x4) + f(x0 + x1 + x4) + f(x0 + x2 + x4) =
= g
(
k + 1
2k
x1 +
1
k
x2 + x3 + y4
)
+
+g
(
−2
k − 1
(x0 + x4) +
k − 1
2k
x1 −
1
k
x2 + x3 +
k + 1
k − 1
y4
)
−
−g
(
1
k
x2 + x3 + y4
)
− g
(
−2
k − 1
(x0 + x4)−
1
k
x2 + x3 +
k + 1
k − 1
y4
)
−
−g
(
k + 1
2k
x1 + x3 + y4
)
− g
(
−2
k − 1
(x0 + x4) +
k − 1
2k
x1 + x3 +
k + 1
k − 1
y4
)
+
+g(x3 + y4) + g
(
−2
k − 1
(x0 + x4) + x3 +
k + 1
k − 1
y4
)
− g
(
k + 1
2k
x1 +
1
k
x2 + y4
)
−
−g
(
−2
k − 1
(x0 + x4) +
k − 1
2k
x1 −
1
k
x2 +
k + 1
k − 1
y4
)
+
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406 T. Z. XU, J. M. RASSIAS, W. X. XU
+g
(
1
k
x2 + y4
)
− g
(
−2
k − 1
(x0 + x4)−
1
k
x2 +
k + 1
k − 1
y4
)
+
+g
(
k + 1
2k
x1 + y4
)
− g
(
−2
k − 1
(x0 + x4) +
k − 1
2k
x1 +
k + 1
k − 1
y4
)
−
−g(y4) + g
(
−2
k − 1
(x0 + x4) +
k + 1
k − 1
y4
)
. (2.11)
Subtract (2.10) from (2.11), 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
(
−2
k − 1
(x0 + x4) +
k − 1
2k
x1 −
1
k
x2 + x3 +
k + 1
k − 1
y4
)
−
−g
(
−2
k − 1
(x0 + x4)−
1
k
x2 + x3 +
k + 1
k − 1
y4
)
−
−g
(
−2
k − 1
(x0 + x4) +
k − 1
2k
x1 + x3 +
k + 1
k − 1
y4
)
+
+g
(
−2
k − 1
(x0 + x4) + x3 +
k + 1
k − 1
y4
)
−
−g
(
−2
k − 1
(x0 + x4) +
k − 1
2k
x1 −
1
k
x2 +
k + 1
k − 1
y4
)
−
−g
(
−2
k − 1
(x0 + x4)−
1
k
x2 +
k + 1
k − 1
y4
)
−
−g
(
−2
k − 1
(x0 + x4) +
k − 1
2k
x1 +
k + 1
k − 1
y4
)
+ g
(
−2
k − 1
(x0 + x4) +
k + 1
k − 1
y4
)
−
−g
(
−2
k − 1
x0 +
k − 1
2k
x1 −
1
k
x2 + x3 +
k + 1
k − 1
y4
)
+
+g
(
−2
k − 1
x0 −
1
k
x2 + x3 +
k + 1
k − 1
y4
)
+
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 3
A GENERALIZED MIXED TYPE OF QUARTIC, CUBIC, QUADRATIC AND ADDITIVE . . . 407
+g
(
−2
k − 1
x0 +
k − 1
2k
x1 + x3 +
k + 1
k − 1
y4
)
− g
(
−2
k − 1
x0 + x3 +
k + 1
k − 1
y4
)
+
+g
(
−2
k − 1
x0 +
k − 1
2k
x1 −
1
k
x2 +
k + 1
k − 1
y4
)
+ g
(
−2
k − 1
x0 −
1
k
x2 +
k + 1
k − 1
y4
)
+
+g
(
−2
k − 1
x0 +
k − 1
2k
x1 +
k + 1
k − 1
y4
)
− g
(
−2
k − 1
x0 +
k + 1
k − 1
y4
)
. (2.12)
Setting y5 =
−2
k − 1
x0+
k + 1
k − 1
y4
(
that is, y4 =
2
k + 1
x0+
k − 1
k + 1
y5
)
in (2.12), 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
(
−2
k − 1
x4 +
k − 1
2k
x1 −
1
k
x2 + x3 + y5
)
− g
(
−2
k − 1
x4 −
1
k
x2 + x3 + y5
)
−
−g
(
−2
k − 1
x4 +
k − 1
2k
x1 + x3 + y5
)
+ g
(
−2
k − 1
x4 + x3 + y5
)
−
−g
(
−2
k − 1
x4 +
k − 1
2k
x1 −
1
k
x2 + y5
)
− g
(
−2
k − 1
x4 −
1
k
x2 + y5
)
−
−g
(
−2
k − 1
x4 +
k − 1
2k
x1 + y5
)
+ g
(
−2
k − 1
x4 + y5
)
+ g
(
−1
k
x2 + x3 + y5
)
−
−g
(
k − 1
2k
x1 −
1
k
x2 + x3 + y5
)
+ g
(
k − 1
2k
x1 + x3 + y5)− g(x3 + y5
)
+
+g
(
k − 1
2k
x1 −
1
k
x2 + y5
)
+ g
(
−1
k
x2 + y5
)
+ g
(
k − 1
2k
x1 + y5
)
− g(y5).
(2.13)
Replacing x0 by x0 + x5 in (2.13) 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)+
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408 T. Z. XU, J. M. RASSIAS, W. X. XU
+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
(
−2
k − 1
x4 +
k − 1
2k
x1 −
1
k
x2 + x3 + y5
)
− g
(
−2
k − 1
x4 −
1
k
x2 + x3 + y5
)
−
−g
(
−2
k − 1
x4 +
k − 1
2k
x1 + x3 + y5
)
+ g
(
−2
k − 1
x4 + x3 + y5
)
−
−g
(
−2
k − 1
x4 +
k − 1
2k
x1 −
1
k
x2 + y5
)
− g
(
−2
k − 1
x4 −
1
k
x2 + y5
)
−
−g
(
−2
k − 1
x4 +
k − 1
2k
x1 + y5
)
+ g
(
−2
k − 1
x4 + y5
)
+ g
(
−1
k
x2 + x3 + y5
)
−
−g
(
k − 1
2k
x1 −
1
k
x2 + x3 + y5
)
+ g
(
k − 1
2k
x1 + x3 + y5
)
− g(x3 + y5)+
+g
(
k − 1
2k
x1 −
1
k
x2 + y5
)
+ g
(
−1
k
x2 + y5
)
+ g
(
k − 1
2k
x1 + y5
)
− g(y5).
(2.14)
Subtract (2.13) from (2.14), 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 is 4x1,x2,x3,x4,x5
f(x0) = 0 for all x0, x1, x2, x3, x4, x5 ∈ X .
Theorem 2.1 is proved.
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A GENERALIZED MIXED TYPE OF QUARTIC, CUBIC, QUADRATIC AND ADDITIVE . . . 409
As an application of Theorem 2.1, we can get the following theorem which is a
further improvement of the Theorem 2.3 in [5].
Theorem 2.2. Let X and Y be real vector spaces, then the function f : X → Y
satisfies the functional equation (1.3) for all x, y ∈ X if and only if f is of the form
f(x) = A3(x)+A2(x)+A1(x)+A0(x) (∀x ∈ X), where A0(x) = A0 is an arbitrary
element of Y, Ai(x) is the diagonal of the i-additive symmetric map Ai : Xi → Y for
i = 1, 2, 3.
Proof. By Theorems 2.1 and 1.1, we have
f(x) = A4(x) +A3(x) +A2(x) +A1(x) +A0(x), x ∈ X, (2.15)
where A0(x) = A0 is an arbitrary element of Y, and Ai(x) is the diagonal of the i-
additive symmetric map Ai : Xi → Y for i = 1, 2, 3, 4. Putting (2.15) into (1.3), and
noting that
A4(x+ y) +A4(x− y) = 2A4(x) + 2A4(y) + 12A2,2(x, y),
A3(x+ y) +A3(x− y) = 2A3(x) + 6A1,2(x, y),
A2(x+ y) +A2(x− y) = 2A2(x) + 2A2(y),
andA2,2(x, ky) = k2A2,2(x, y), A1,2(x, ky) = k2A1,2(x, y), we conclude thatA4(y) =
= 0 for all y ∈ X . Therefore
f(x) = A3(x) +A2(x) +A1(x) +A0(x)
for all x ∈ X . The converse is easily verified.
Theorem 2.2 is proved.
Using the techniques are similar to that of Theorem 2.2, we have the following results
(Theorems 2.3 – 2.6).
Theorem 2.3 ([6], Theorem 2.1). Let X and Y be real vector spaces, then the
function f : X → Y satisfies the functional equation
3[f(x+ 2y) + f(x− 2y)] =
= 12[f(x+ y) + f(x− y)] + 4f(3y)− 18f(2y) + 36f(y)− 18f(x) (2.16)
for all x, y ∈ X if and only if f is of the form f(x) = A4(x) + A3(x) + A2(x)
(∀x ∈ X), where Ai(x) is the diagonal of the i-additive symmetric map Ai : Xi → Y
for i = 2, 3, 4.
Theorem 2.4 ([7], Theorem 2.4). Let X and Y be real vector spaces, then the
function f : X → Y satisfies the functional equation (1.4) for all x, y ∈ X if and only if
f is of the form f(x) = A4(x)+A3(x)+A1(x) (∀x ∈ X), where Ai(x) is the diagonal
of the i-additive symmetric map Ai : Xi → Y for i = 1, 3, 4.
Theorem 2.5 ([2], Theorem 3.1). Let X and Y be real vector spaces, then the
function f : X → Y satisfies the functional equation (1.2) for all x, y ∈ X if and only if
f is of the form f(x) = A4(x) (∀x ∈ X), where A4(x) is the diagonal of the 4-additive
symmetric map A4 : X4 → Y .
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410 T. Z. XU, J. M. RASSIAS, W. X. XU
Theorem 2.6 ([8], Theorem 2.3). Let X and Y be real vector spaces, then the
function f : X → Y satisfies the functional equation (1.5) for all y ∈ X if and only if f
is of the form f(x) = A4(x) + A3(x) + A2(x) + A1(x) (∀x ∈ X), where Ai(x) is the
diagonal of the i-additive symmetric map Ai : Xi → Y for i = 1, 2, 3, 4.
Theorem 2.7. Let X and Y be real vector spaces, then the functions f, g, h, h̃ :
X → Y satisfy the functional equation (1.6) for all x, y ∈ X if and only if
f(x) = A4(x) +A3(x) +A2(x) +A1(x) +A0(x),
g(x) = k2A4(x) + k2A3(x) +B2(x) +B0(x) + C1(x) +D0(x),
h(x) = (2− 2k2)A4(x) + (2− 2k2)A3(x) + 2A2(x) + 2A1(x) + 2A0(x)− (2.17)
−2B2(x)− 2C1(x)− 2B0(x),
h̃(x) = (2k4 − 2k2)A4(x) + 2k2A2(x)− 2B2(x)− 2D0(x),
where A0(x) = A0, B0(x) = B0 and D0(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, for i = 1, 2, 3, 4.
Proof. Assume that f, g, h, h̃ satisfy the functional equation (1.6). By Theorem 2.1
we see that f is a solution of the Fréchet functional equation 4x1,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 all x ∈ X, (2.18)
where A0(x) = A0 is an arbitrary element of Y, and Ai(x) is the diagonal of the i-
additive symmetric map Ai : Xi → Y for i = 1, 2, 3, 4. Putting (2.18) into (1.6), and
noting that
A4(x+ y) +A4(x− y) = 2A4(x) + 2A4(y) + 12A2,2(x, y),
A3(x+ y) +A3(x− y) = 2A3(x) + 6A1,2(x, y),
A2(x+ y) +A2(x− y) = 2A2(x) + 2A2(y),
and A2,2(x, ky) = k2A2,2(x, y), A1,2(x, ky) = k2A1,2(x, y), we conclude that
g(x+ y) + g(x− y) + h(x) + h̃(y) =
= 2A4(x) + 2k4A4(y) + 12k2A2,2(x, y) + 2A3(x) + 6k2A1,2(x, y)+
+2A2(x) + 2k2A2(y) + 2A1(x) + 2A0.
Therefore
g(x+ y) + g(x− y) + h(x) + h̃(y) =
= k2A4(x+ y) + k2A4(x− y) + k2A3(x+ y) + k2A3(x− y)+
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A GENERALIZED MIXED TYPE OF QUARTIC, CUBIC, QUADRATIC AND ADDITIVE . . . 411
+(2− 2k2)A4(x) + (2− 2k2)A3(x)+
+2A2(x) + 2A1(x) + 2A0 + (2k4 − 2k2)A4(y) + 2k2A2(y). (2.19)
Letting
G(x) = g(x)− k2A4(x)− k2A3(x),
H̃(x) = −h̃(x) + (2k4 − 2k2)A4(x) + 2k2A2(x)
(2.20)
and
H(x) = −h(x) + (2− 2k2)A4(x) + (2− 2k2)A3(x) + 2A2(x) + 2A1(x) + 2A0.
(2.21)
From (2.19) we have
G(x+ y) +G(x− y) = H(x) + H̃(y). (2.22)
Let G satisfies (2.22). We decompose G into 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 all x ∈ X . It is clear that G(x) = Ge(x) +Go(x) for all x ∈ X . Similarly, we have
H(x) = He(x) +Ho(x) and H̃(x) = H̃e(x) + H̃o(x). Thus
Ge(x+ y) +Ge(x− y) = He(x) + H̃e(y), (2.23)
and
Go(x+ y) +Go(x− y) = Ho(x) + H̃o(y). (2.24)
Letting y = 0 in (2.23), we have He(x) = 2Ge(x)− H̃e(0). Setting x = 0 in (2.23) to
get H̃e(y) = 2Ge(y)−He(0). Hence
Ge(x+ y) +Ge(x− y) = 2Ge(x) + 2Ge(y)− 2Ge(0) (2.25)
for all x, y ∈ X . Setting M(x) = Ge(x)−Ge(0), we get
M(x+ y) +M(x− y) = 2M(x) + 2M(y) (2.26)
which is the quadratic functional equation and its solution is given byM(x) = B2(x) for
all x ∈ X, where B2(x) is the diagonal of the 2-additive symmetric map B2 : X
2 → Y .
In this case, we obtain
Ge(x) = B2(x) +G(0), He(x) = 2B2(x) +He(0),
H̃e(x) = 2B2(x) + H̃e(0).
(2.27)
Similarly, letting y = 0 in (2.24), we have Ho(x) = 2Go(x). Setting x = 0 in (2.24) to
get H̃o(y) = 0. Then from (2.24) we have
Go(x+ y) +Go(x− y) = 2Go(x), (2.28)
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412 T. Z. XU, J. M. RASSIAS, W. X. XU
which is the Jensen functional equation and its solution is given by Go(x) = C1(x),
where C1 : 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), (2.29)
H̃(x) = H̃e(x) + H̃o(x) = 2B2(x) + 2D0(x),
where B0(x) = B0 (that is, He(0) = 2B0(x) = 2B0) and D0(x) = D0 (that is,
H̃e(0) = 2D0(x) = 2D0) are arbitrary elements of Y . Therefore from (2.20), (2.21),
(2.29), we obtain the asserted solution (2.17). The converse is easily verified.
Theorem 2.7 is proved.
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 group G is said to be divisible if for every element b ∈ G and every n ∈ N, there
exists an element a ∈ G such that na = b. If this element a is unique, thenG is said to be
uniquely divisible. In a uniquely divisible group, this unique element a is denoted by
b
n
.
That the equation na = b has a solution is equivalent to saying that the multiplication by
n is surjective. Similarly, that the equation na = b has a unique solution is equivalent to
saying that the multiplication by n is bijective. Hence the notions of n-divisibility and n-
unique divisibility refer, respectively, to surjectivity and bijectivity of the multiplication
by n.
Lemma 3.1 (Hosszú [9]). Let n ≥ 0 be an integer, G and S be abelian groups.
Furthermore let S be uniquely divisible. The map F fromG into S satisfies the functional
equation 4x1,...,xn+1
F (x0) = 0 for all x0, x1, . . . , xn+1 ∈ G if and only if F is given
by F (x) = An(x) + . . . + A1(x) + A0(x) for all x ∈ G, where A0(x) = A0 is an
arbitrary element of S and An(x) is the diagonal of an n-additive symmetric function
An : Gn → S.
Using Lemma 3.1, one can prove the similar results (Theorems 2.2 – 2.7) for unknown
functions map a commutative uniquely divisible group into another one. Also, Theorems
2.2 – 2.7 can be further strengthened using two important results due to Székelyhidi
[16]. By the use of the two important results, the proofs become even shorter but not
so elementary any more. The results needed for this improvement are the following
(see [16]).
Theorem 3.1. Let G be a commutative semigroup with identity, S a commutative
group and n a nonnegative integer. Let the multiplication by n! be bijective in S. The
function f : G→ S is a solution of Fréchet functional equation
4x1,...,xn+1
f(x0) = 0 (3.1)
for all x0, x1, . . . , xn+1 ∈ G if and only if f is a polynomial of degree at most n.
Theorem 3.2. Let G and S be commutative groups, n a nonnegative integer, ϕi,
ψi additive functions from G into G and ϕi(G) ⊆ ψi(G), i = 1, 2, . . . , n + 1. If the
functions f, fi : G→ S, i = 1, 2, . . . , n+ 1, satisfy
f(x) +
n+1∑
i=1
fi(ϕi(x) + ψi(y)) = 0, (3.2)
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 3
A GENERALIZED MIXED TYPE OF QUARTIC, CUBIC, QUADRATIC AND ADDITIVE . . . 413
then f satisfies Fréchet functional equation 4x1,...,xn+1
f(x0) = 0.
Using these two theorems, Theorems 2.2 – 2.7 can be further improved.
Theorem 3.3. Let G and S be commutative groups. Let the multiplication by 6
and 2(k2 − 1) be bijective in S, respectively. Then the function f : G → S satisfies
the functional equation (1.3) for all x, y ∈ G if and only if f is of the form f(x) =
= A3(x) + A2(x) + A1(x) + A0(x) for all x ∈ G, where A0(x) = A0 is an arbitrary
element of S, Ai(x) is the diagonal of the i-additive symmetric map Gi : Xi → S for
i = 1, 2, 3.
Proof. Assume that f satisfies the functional equation (1.3). Using the unique
divisibility of S by 2(k2 − 1), we can rewrite the functional equation (1.3) in the form
f(x) +
1
2(k2 − 1)
f(kx+ y) +
1
2(k2 − 1)
f(−kx− y)−
− k2
2(k2 − 1)
f(x+ y)− k2
2(k2 − 1)
f(−x+ y) = 0.
Thus by Theorem 3.2, f satisfies the Fréchet functional equation (3.1). By Theorem 3.1,
f is a generalized polynomial function of degree at most 3, that is f is of the form
f(x) = A3(x) + A2(x) + A1(x) + A0(x), where A0(x) = A0 is an arbitrary element
of S, and Ai(x) is the diagonal of the i-additive symmetric map Ai : G
i → S for
i = 1, 2, 3. The remaining assertion goes through by the similar way to corresponding
part of Theorem 2.2.
Theorem 3.3 is proved.
Using the techniques are similar to that of Theorem 3.3, we have the following
results.
Theorem 3.4. Let G and S be commutative groups. Let the multiplication by 2 be
surjective in G and let the multiplication by 24 and 18 be bijective in S, respectively.
Then the function f : G → S satisfies the functional equation (2.16) for all x, y ∈ G if
and only if f is of the form f(x) = A4(x)+A3(x)+A2(x) for all x ∈ G, where Ai(x)
is the diagonal of the i-additive symmetric map Ai : Gi → S for i = 2, 3, 4.
Theorem 3.5. Let G and S be commutative groups. Let the multiplication by 2 be
surjective in G and let the multiplication by 24 and 66 be bijective in S, respectively.
Then the function f : G → S satisfies the functional equation (1.4) for all x, y ∈ G if
and only if f is of the form f(x) = A4(x)+A3(x)+A1(x) for all x ∈ G, where Ai(x)
is the diagonal of the i-additive symmetric map Ai : Gi → S for i = 1, 3, 4.
Theorem 3.6. Let G and S be commutative groups. Let the multiplication by 2
be surjective in G and let the multiplication by 24 be bijective in S. Then the function
f : G → S satisfies the functional equation (1.2) for all x, y ∈ G if and only if f is of
the form f(x) = A4(x) for all x ∈ G, where A4(x) is the diagonal of the 4-additive
symmetric map A4 : G4 → S.
Theorem 3.7. Let G and S be commutative groups. Let the multiplication by k
be surjective in G and let the multiplication by 24 and 12(k2 − 1) be bijective in S,
respectively. Then the function f : G → S satisfies the functional equation (1.5) for all
x, y ∈ G if and only if f is of the form f(x) = A4(x) +A3(x) +A2(x) +A1(x) for all
x ∈ G, where Ai(x) is the diagonal of the i-additive symmetric map Ai : Gi → S for
i = 1, 2, 3, 4.
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 3
414 T. Z. XU, J. M. RASSIAS, W. X. XU
Theorem 3.8. Let G and S be commutative groups. Let the multiplication by
2k(k2−1) be surjective in G and let the multiplication by 24 be bijective in S. Then the
function f : G→ S satisfies the functional equation (1.6) for all x, y ∈ G if and only if
f(x) = A4(x) +A3(x) +A2(x) +A1(x) +A0(x),
g(x) = k2A4(x) + k2A3(x) +B2(x) +B0(x) + C1(x) +D0(x),
h(x) = (2− 2k2)A4(x) + (2− 2k2)A3(x) + 2A2(x) + 2A1(x) + 2A0(x)−
−2B2(x)− 2C1(x)− 2B0(x),
h̃(x) = (2k4 − 2k2)A4(x) + 2k2A2(x)− 2B2(x)− 2D0(x),
where A0(x) = A0, B0(x) = B0 and D0(x) = D0 are arbitrary elements of S, and
Ai(x), Bi(x), Ci(x) are the diagonal of the i-additive symmetric mapsAi, Bi, Ci : Gi →
→ S, respectively, for i = 1, 2, 3, 4.
Proof. Assume that f satisfies the functional equation (1.6). Using the multiplication
by 2k(k2 − 1) be surjective in G, we can rewrite the functional equation (1.6) in the
form
f(x) + f(x− 2ky)− f(x+ (1− k)y)− f(x− (1 + k)y)− h(x− ky)− h̃(y) = 0.
Thus by Theorem 3.2, f satisfies the Fréchet functional equation (3.1). By Theorem 3.1,
f is a generalized polynomial function of degree at most 4, that is f is of the form
f(x) = A4(x) + A3(x) + A2(x) + A1(x) + A0(x), where A0(x) = A0 is an arbitrary
element of S, and Ai(x) is the diagonal of the i-additive symmetric map Ai : Gi → S for
i = 1, 2, 3, 4. The remaining assertion goes through by the similar way to corresponding
part of Theorem 2.7.
Theorem 3.8 is proved.
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A GENERALIZED MIXED TYPE OF QUARTIC, CUBIC, QUADRATIC AND ADDITIVE . . . 415
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Received 13.04.10,
after revision — 28.10.10
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 3
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| id | umjimathkievua-article-2725 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:29:06Z |
| publishDate | 2011 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/7c/c007c53bd0b5a80d98cf928cedfca97c.pdf |
| spelling | umjimathkievua-article-27252020-03-18T19:34:39Z A generalized mixed type of quartic, cubic, quadratic and additive functional equation Узагальнений мiшаний тип квартичного, кубiчного, квадратичного та додаткового функцiонального рiвняння Rassias, J. M. Xu, T. Z. Xu, W. X. Расіас, Дж. М. Ху, Т. З. Ху, В. Х. We determine the general solution of the functional equation $f(x + ky) + f(x — ky) = g(x + y) + g(x — y) + h(x) + \tilde{h}(y)$ forfixed integers $k$ with $k \neq 0, \pm 1$ without assuming any regularity condition on the unknown functions $f, g, h, \tilde{h}$. The method used for solving these functional equations is elementary but exploits an important result due to Hosszii. The solution of this functional equation can also be determined in certain type of groups using two important results due to Szekelyhidi. Визначено загальний розв’язок функцiонального рiвняння $f(x + ky) + f(x — ky) = g(x + y) + g(x — y) + h(x) + \tilde{h}(y)$ для фiксованих цiлих $k$ при $k \neq 0, \pm 1$ без припущення наявностi будь-якої умови регулярностi для невiдомих функцiй $f, g, h, \tilde{h}$. Метод, що використано для розв’язку цих функцiональних рiвнянь, елементарний, але базується на важливому результатi Хозу. Розв’язок цього функцiонального рiвняння може бути визначений у певному типi груп з використанням двох важливих результатiв Чекелiхiдi. Institute of Mathematics, NAS of Ukraine 2011-03-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2725 Ukrains’kyi Matematychnyi Zhurnal; Vol. 63 No. 3 (2011); 399-415 Український математичний журнал; Том 63 № 3 (2011); 399-415 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2725/2206 https://umj.imath.kiev.ua/index.php/umj/article/view/2725/2207 Copyright (c) 2011 Rassias J. M.; Xu T. Z.; Xu W. X. |
| spellingShingle | Rassias, J. M. Xu, T. Z. Xu, W. X. Расіас, Дж. М. Ху, Т. З. Ху, В. Х. A generalized mixed type of quartic, cubic, quadratic and additive functional equation |
| title | A generalized mixed type of quartic, cubic, quadratic and additive functional equation |
| title_alt | Узагальнений мiшаний тип квартичного, кубiчного, квадратичного та додаткового функцiонального рiвняння |
| title_full | A generalized mixed type of quartic, cubic, quadratic and additive functional equation |
| title_fullStr | A generalized mixed type of quartic, cubic, quadratic and additive functional equation |
| title_full_unstemmed | A generalized mixed type of quartic, cubic, quadratic and additive functional equation |
| title_short | A generalized mixed type of quartic, cubic, quadratic and additive functional equation |
| title_sort | generalized mixed type of quartic, cubic, quadratic and additive functional equation |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2725 |
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