# General Quadratic-Additive Type Functional Equation and Its Stability

Authors: Yang-Hi Lee; Soon-Mo Jung
Journal: International Journal of Mathematics and Mathematical Sciences
 (2016)

---

## Abstract

We investigate the general functional equation of the formf
(
a
x
+
b
y
+
c
z
)
-
a
b
f
(
x
+
y
)
-
b
c
f
(
y
+
z
) − a
c
f
(
x
+
z
)
-
a
(
(
a
+
1
)
/
2
-
b
-
c
)
f
(
x
)
-
b
(
(
b
+
1
)
/
2
-
a
-
c
)
f
(
y
)
-
c
(
(
c
+
1
)
/
2
-
a
-
b
)
f
(
z
)
-
(
a
(
a
-
1
)
/
2
)
f
(
-
x
) − (
b
(
b
-
1
)
/
2
)
f
(
-
y
)
-
(
c
(
c
-
1
)
/
2
)
f
(
-
z
)
=
0, whose solutions are quadratic-additive mappings in connection with stability problems.


---

## Body

## 1. Introduction

In 1940, Ulam [1] posed an important problem concerning the stability of group homomorphisms. In the following year, Hyers [2] solved the problem for the case of Cauchy additive functional equation. After a period longer than two decades, Rassias [3] generalized Hyers’ result and then Găvruta [4] extended Rassias’ result by allowing unbounded control functions. The concept of stability introduced by Rassias and Găvruta is known today with the term “generalized Hyers-Ulam stability” of functional equations.A solution to the functional equation(1)
f
x
+
y
-
f
x
-
f
y
=
0is called an additive mapping and a solution to the functional equation(2)
f
x
+
y
+
f
x
-
y
-
2
f
x
-
2
f
y
=
0is called a quadratic mapping. If a mapping can be expressed by the sum of an additive mapping and a quadratic mapping, then we call the mapping a quadratic-additive mapping. Now, we consider the general quadratic-additive type functional equation(3)
f
a
x
+
b
y
+
c
z
-
a
b
f
x
+
y
-
b
c
f
y
+
z
-
a
c
f
x
+
z
-
a
a
+
1
2
-
b
-
c
f
x
-
b
b
+
1
2
-
a
-
c
f
y
-
c
c
+
1
2
-
a
-
b
f
z
-
a
a
-
1
2
f
-
x
-
b
b
-
1
2
f
-
y
-
c
c
-
1
2
f
-
z
=
0with nonzero real constants a, b, and c. The mapping f
(
x
)
=
d
x
2
+
e
x is a solution to this functional equation, where d, e are real constants. For the case a
=
b
=
c, the stability of the functional equation (3) was investigated by some mathematicians (see [5] for a
=
b
=
c
=
1
/
3).In this paper, we will prove that ifa, b, and c are nonzero real constants, then every solution to the functional equation (3) is a quadratic-additive mapping and, conversely, we will also prove that every quadratic-additive mapping is a solution to (3) provided a, b, and c are rational constants. Moreover, we will prove the generalized Hyers-Ulam stability of the functional equation (3).We remark here that (3) is a special form of the general linear equation, whose stability was investigated by Bahyrycz and Olko [6] via a different method from that we apply in this paper. In particular, the main result of this paper is a generalization of [6].


## 2. Preliminaries

Throughout this paper, letV and W be vector spaces over R, let X be a real normed space, let Y be a real Banach space, and let a, b, and c be positive real constants.For a given mappingf
:
V
→
W, we use the following abbreviations: (4)
f
o
x
≔
f
x
-
f
-
x
2
,
f
e
x
≔
f
x
+
f
-
x
2
,
A
f
x
,
y
≔
f
x
+
y
-
f
x
-
f
y
,
Q
f
x
,
y
≔
f
x
+
y
+
f
x
-
y
-
2
f
x
-
2
f
y
,
D
a
,
b
f
x
,
y
≔
f
a
x
+
b
y
-
a
b
f
x
+
y
-
a
a
+
1
2
-
b
f
x
-
b
b
+
1
2
-
a
f
y
-
a
a
-
1
2
f
-
x
-
b
b
-
1
2
f
-
y
,
C
a
,
b
f
x
,
y
≔
f
a
x
+
b
y
+
a
b
f
x
-
y
-
a
2
b
+
a
+
1
2
f
x
-
b
b
+
1
2
f
y
-
a
a
-
1
2
f
-
x
-
b
2
a
+
b
-
1
2
f
-
y
,
D
a
,
b
,
c
f
x
,
y
,
z
≔
f
a
x
+
b
y
+
c
z
-
a
b
f
x
+
y
-
b
c
f
y
+
z
-
a
c
f
x
+
z
-
a
a
+
1
2
-
b
-
c
f
x
-
b
b
+
1
2
-
a
-
c
f
y
-
c
c
+
1
2
-
a
-
b
f
z
-
a
a
-
1
2
f
-
x
-
b
b
-
1
2
f
-
y
-
c
c
-
1
2
f
-
z for all x
,
y
,
z
∈
V.The following lemmas were proved in [7].Lemma 1.
Leta, b be real numbers with (
a
b
+
1
)
(
a
+
b
)
a
b
(
a
2
+
b
2
+
a
b
-
1
)
≠
0. If a mapping f
:
V
→
W satisfies the functional equation C
a
,
b
f
(
x
,
y
)
=
0 for all x
,
y
∈
V, then f
e is quadratic and f
o is additive.Lemma 2.
Leta, b be rational constants with (
a
b
+
1
)
(
a
+
b
)
a
b
(
a
2
+
b
2
+
a
b
-
1
)
≠
0. A mapping f
:
V
→
W satisfies C
a
,
b
f
(
x
,
y
)
=
0 for all x
,
y
∈
V if and only if f
e is quadratic and f
o is additive.Since the equality(5)
C
a
,
b
f
x
,
y
=
D
a
,
-
b
f
x
,
-
y holds for all x
,
y
∈
V, the next couple of lemmas are direct consequences of Lemmas 1 and 2.Lemma 3.
Leta, b be real numbers with (
a
b
-
1
)
(
a
-
b
)
a
b
(
a
2
+
b
2
-
a
b
-
1
)
≠
0. If a mapping f
:
V
→
W satisfies the functional equation D
a
,
b
f
(
x
,
y
)
=
0 for all x
,
y
∈
V, then f
e is quadratic and f
o is additive.Lemma 4.
Leta, b be rational constants with (
a
b
-
1
)
(
a
-
b
)
a
b
(
a
2
+
b
2
-
a
b
-
1
)
≠
0. A mapping f
:
V
→
W satisfies D
a
,
b
f
(
x
,
y
)
=
0 for all x
,
y
∈
V if and only if f
e is quadratic and f
o is additive.Lemma 5.
Leta, b be rational constants. If f
:
V
→
W is a quadratic-additive mapping, then f is a solution to the functional equation D
a
,
b
f
(
x
,
y
)
=
0 for all x
,
y
∈
V.Proof.
Letf
:
V
→
W be a quadratic-additive mapping. Then, f
e is a quadratic mapping and f
o is an additive mapping. Hence, we have f
e
(
r
x
)
=
r
2
f
e
(
x
), f
o
(
r
x
)
=
r
f
o
(
x
), and f
o
(
x
+
y
)
=
f
o
(
x
)
+
f
o
(
y
), where r is an arbitrary rational number.
First, we will prove that the mappingf
e satisfies D
a
,
b
f
e
(
x
,
y
)
=
0 for arbitrary rational numbers a and b. If a
=
0 or b
=
0, then the equality (6)
D
a
,
b
f
e
x
,
y
=
0 holds for all x
,
y
∈
V. When a
=
b, the equality D
a
,
a
f
e
(
x
,
y
)
=
0 follows from the equality (7)
D
a
,
a
f
e
x
,
y
=
f
e
a
x
+
y
-
a
2
f
e
x
+
y for all x
,
y
∈
V. If a and b are nonzero rational constants with a
≠
b, then there exists an integer k satisfying (
a
k
-
b
k
)
(
a
k
b
k
-
1
)
(
(
a
k
)
2
+
(
b
k
)
2
-
a
k
b
k
-
1
)
≠
0. According to Lemma 4, the equality(8)
D
a
,
b
f
e
x
,
y
=
D
a
k
,
b
k
f
e
x
k
,
y
k implies that f
e satisfies D
a
,
b
f
e
(
x
,
y
)
=
0 for all x
,
y
∈
V. Altogether, the mapping f
e satisfies D
a
,
b
f
e
(
x
,
y
)
=
0 for arbitrary rational numbers a and b.
On the other hand, the equality(9)
D
a
,
b
f
o
x
,
y
=
A
f
o
a
x
,
b
y
-
a
b
A
f
o
x
,
y
+
f
o
a
x
-
a
f
o
x
+
f
o
b
y
-
b
f
o
y implies that f
o satisfies D
a
,
b
f
o
(
x
,
y
)
=
0 for arbitrary rational numbers a and b. Since the equality D
a
,
b
f
(
x
,
y
)
=
D
a
,
b
f
e
(
x
,
y
)
+
D
a
,
b
f
o
(
x
,
y
) holds for all x
,
y
∈
V, we conclude that the mapping f satisfies D
a
,
b
f
(
x
,
y
)
=
0 for all x
,
y
∈
V.Lemma 6.
If real numbersa, b, and c satisfy (
a
-
b
)
(
b
-
c
)
(
c
-
a
)
≠
0, then either a
2
+
b
2
-
a
b
-
1
≠
0, or a
2
+
c
2
-
a
c
-
1
≠
0, or c
2
+
b
2
-
c
b
-
1
≠
0.Proof.
First, assume thata
=
0 and a, b, and c are real constants satisfying (
a
-
b
)
(
b
-
c
)
(
c
-
a
)
≠
0, a
2
+
b
2
-
a
b
-
1
=
0, a
2
+
c
2
-
a
c
-
1
=
0, and c
2
+
b
2
-
c
b
-
1
=
0. We then obtain the equalities b
2
=
1, c
2
=
1, and c
b
=
1, which contradict the fact b
≠
c.
Second, assume thata, b, and c are nonzero real constants satisfying conditions (
a
-
b
)
(
b
-
c
)
(
c
-
a
)
≠
0, a
2
+
b
2
-
a
b
-
1
=
0, a
2
+
c
2
-
a
c
-
1
=
0, and c
2
+
b
2
-
c
b
-
1
=
0. We then obtain the equalities a
+
b
=
c, a
+
c
=
b, and a
=
0, which contradict the fact a
≠
0.We will now prove thatf is a quadratic-additive mapping provided f is a solution to the functional equation D
a
,
b
,
c
f
(
x
,
y
,
z
)
=
0 for all x
,
y
,
z
∈
V.Remark 7.
We remark that ifσ
:
{
a
,
b
,
c
}
→
{
a
,
b
,
c
} and τ
:
{
x
,
y
,
z
}
→
{
x
,
y
,
z
} are permutations, then we have(10)
D
a
,
b
,
c
f
x
,
y
,
z
=
0
∀
x
,
y
,
z
∈
V
iff
D
σ
a
,
σ
b
,
σ
c
f
τ
x
,
τ
y
,
τ
z
=
0
∀
x
,
y
,
z
∈
V
.Theorem 8.
Leta, b, and c be nonzero real numbers. If a mapping f
:
V
→
W satisfies the functional equation D
a
,
b
,
c
f
(
x
,
y
,
z
)
=
0 (with f
(
0
)
=
0, when a
2
+
b
2
+
c
2
-
a
b
-
b
c
-
a
c
=
1), then f is a quadratic-additive mapping.Proof.
In view of Remark7, it is enough to check the following three cases: (1), (2), and (3). Notice that f
(
0
)
=
D
a
,
b
,
c
f
(
0,0
,
0
)
/
a
2
+
b
2
+
c
2
-
a
b
-
b
c
-
a
c
-
1
=
0, when a
2
+
b
2
+
c
2
-
a
b
-
b
c
-
a
c
-
1
≠
0.
(1) Ifa, b, and c are real constants with (
a
-
b
)
(
a
-
c
)
(
b
-
c
)
=
0, then we can assume that a
=
b without loss of generality by Remark 7. In view of Lemma 1, the equality -
a
c
C
1,1
f
(
x
,
y
)
=
D
a
,
a
,
c
f
(
y
,
-
y
,
x
)
-
D
a
,
a
,
c
f
(
0,0
,
x
) implies that f is a quadratic-additive mapping. In particular, we know that every solution to D
a
,
a
,
c
f
(
x
,
y
,
z
)
=
0 is a quadratic-additive mapping.
(2) Ifa, b, and c are real constants with (
a
b
-
1
)
(
a
c
-
1
)
(
c
b
-
1
)
=
0, then we can assume that a
b
=
1 without loss of generality due to Remark 7. We can easily show the validity of the following equalities:(11)
f
a
x
-
a
2
+
a
2
f
x
-
a
2
-
a
2
f
-
x
=
D
a
,
1
/
a
,
c
f
x
,
0,0
,
f
y
a
-
1
+
a
2
a
2
f
y
-
1
-
a
2
a
2
f
-
y
=
D
a
,
1
/
a
,
c
f
0
,
y
,
0
,
f
z
c
-
1
+
c
2
c
2
f
c
-
1
-
c
2
c
2
f
-
z
=
D
a
,
1
/
a
,
c
f
0,0
,
zfor all x
,
y
,
z
∈
V. Since every solution to D
a
,
a
,
c
f
(
x
,
y
,
z
)
=
0 is a quadratic-additive mapping, the equality(12)
D
1,1
,
1
f
x
,
y
,
z
=
D
a
,
1
/
a
,
c
f
x
a
,
a
y
,
z
c
-
D
a
,
1
/
a
,
c
f
x
a
,
a
y
,
0
-
D
a
,
1
/
a
,
c
f
x
a
,
0
,
z
c
-
D
a
,
1
/
a
,
c
f
0
,
a
y
,
z
c
+
D
a
,
1
/
a
,
c
f
x
a
,
0,0
+
D
a
,
1
/
a
,
c
f
0,0
,
z
c
+
D
a
,
1
/
a
,
c
f
0
,
a
y
,
0 implies that f is a quadratic-additive mapping.
(3) By Lemma6, we conclude that if a, b, and c are real constants with (
a
-
b
)
(
b
-
c
)
(
c
-
a
)
(
a
b
-
1
)
(
a
c
-
1
)
(
c
b
-
1
)
≠
0, then either (
a
-
b
)
(
a
b
-
1
)
(
a
2
+
b
2
-
a
b
-
1
)
≠
0, or (
a
-
c
)
(
a
c
-
1
)
(
a
2
+
c
2
-
a
c
-
1
)
≠
0, or (
b
-
c
)
(
b
c
-
1
)
(
c
2
+
b
2
-
c
b
-
1
)
≠
0. By Remark 7, we can assume that (
a
-
b
)
(
a
b
-
1
)
(
a
2
+
b
2
-
a
b
-
1
)
≠
0 without loss of generality. On account of Lemma 3, the equality D
a
,
b
f
(
x
,
y
)
=
D
a
,
b
,
c
f
(
x
,
y
,
0
)
=
0 implies that f is a quadratic-additive mapping.Theorem 9.
Leta, b, and c be rational numbers. If f
:
V
→
W is a quadratic-additive mapping, then f satisfies the functional equation D
a
,
b
,
c
f
(
x
,
y
,
z
)
=
0.Proof.
Letf
:
V
→
W be a quadratic-additive mapping. Then we easily show that f
e is a quadratic mapping and f
o is an additive mapping. Hence, we have f
e
(
r
x
)
=
r
2
f
e
(
x
), f
o
(
r
x
)
=
r
f
o
(
x
), and f
o
(
x
+
y
)
=
f
o
(
x
)
+
f
o
(
y
), where r is an arbitrary rational number. In view of Lemma 5, we see that the mappings f
e and f
o satisfy the equalities D
a
,
b
f
e
(
x
,
y
)
=
0 and D
a
,
b
f
o
(
x
,
y
)
=
0. Therefore, the equalities(13)
D
a
,
b
,
c
f
e
x
,
y
,
z
=
Q
f
e
a
x
+
c
z
2
,
b
y
+
c
z
2
-
Q
f
e
a
x
+
c
z
2
,
c
z
2
-
Q
f
e
b
y
+
c
z
2
,
c
z
2
-
Q
f
e
a
x
,
b
y
+
D
a
,
b
f
e
x
,
y
+
D
a
,
c
f
e
x
,
z
+
D
b
,
c
f
e
y
,
z
-
f
e
a
x
+
a
2
f
e
x
-
f
e
b
y
+
b
2
f
e
y
-
4
f
e
c
z
2
+
c
2
f
e
z
,
D
a
,
b
,
c
f
o
x
,
y
,
z
=
A
f
o
a
x
+
b
y
,
c
z
+
A
f
o
a
x
,
b
y
-
a
b
A
f
o
x
,
y
-
b
c
A
f
o
y
,
z
-
a
c
A
f
o
x
,
z
+
f
o
a
x
-
a
f
o
x
+
f
o
b
y
-
b
f
o
y
+
f
o
c
z
-
c
f
o
z imply that the equalities D
a
,
b
,
c
f
e
(
x
,
y
,
z
)
=
0 and D
a
,
b
,
c
f
o
(
x
,
y
,
z
)
=
0 hold for all x
,
y
,
z
∈
V. From the equality D
a
,
b
,
c
f
(
x
,
y
,
z
)
=
D
a
,
b
,
c
f
e
(
x
,
y
,
z
)
+
D
a
,
b
,
c
f
o
(
x
,
y
,
z
) for all x
,
y
,
z
∈
V, we get the equality D
a
,
b
,
c
f
(
x
,
y
,
z
)
=
0, as desired.The next theorem is a direct consequence of Theorems8 and 9.Theorem 10.
Leta, b, and c be nonzero rational constants. A mapping f
:
V
→
W satisfies D
a
,
b
,
c
f
(
x
,
y
,
z
)
=
0 for all x
,
y
,
z
∈
V if and only if f is a quadratic-additive mapping.


## 3. Main Results

In the following theorems, Theorems3.1
–
3.3 of [8] can be slightly modified for the case when n
=
3 and D
f
(
x
,
y
,
z
)
=
D
a
,
b
,
c
f
(
x
,
y
,
z
) without altering their proofs.Theorem 11.
Given a real numberk with k
≠
1, let μ
:
V
∖
{
0
}
→
[
0
,
∞
) be a function satisfying the condition(14)
∑
i
=
0
∞
μ
k
i
x
k
i
<
∞
w
h
e
n
k
>
1
,
∑
i
=
0
∞
μ
k
i
x
k
2
i
<
∞
w
h
e
n
k
<
1for all x
∈
V
∖
{
0
} and let φ
:
(
V
∖
{
0
}
)
3
→
[
0
,
∞
) be a function satisfying the condition(15)
∑
i
=
0
∞
φ
k
i
x
,
k
i
y
,
k
i
z
k
i
<
∞
w
h
e
n
k
>
1
,
∑
i
=
0
∞
φ
k
i
x
,
k
i
y
,
k
i
z
k
2
i
<
∞
w
h
e
n
k
<
1for all x
,
y
,
z
∈
V
∖
{
0
}. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and(16)
f
k
x
-
k
2
+
k
2
f
x
-
k
2
-
k
2
f
-
x
≤
μ
xfor all x
∈
V
∖
{
0
} and if f satisfies(17)
D
a
,
b
,
c
f
x
,
y
,
z
≤
φ
x
,
y
,
zfor all x
,
y
,
z
∈
V
∖
{
0
}, then there exists a unique mapping F
:
V
→
Y such that(18)
D
a
,
b
,
c
F
x
,
y
,
z
=
0holds for all x
,
y
,
z
∈
V
∖
{
0
},(19)
F
e
k
x
=
k
2
F
e
x
,
F
o
k
x
=
k
F
o
xhold for all x
∈
V, and (20)
f
x
-
F
x
≤
∑
i
=
0
∞
k
i
+
1
+
1
μ
k
i
x
+
k
i
+
1
-
1
μ
-
k
i
x
2
k
2
i
+
2holds for all x
∈
V
∖
{
0
}.Theorem 12.
Given a real constantk with k
≠
1, let μ
:
V
∖
{
0
}
→
[
0
,
∞
) be a function satisfying the condition(21)
∑
i
=
0
∞
k
2
i
μ
x
k
i
<
∞
w
h
e
n
k
>
1
,
∑
i
=
0
∞
k
i
μ
x
k
i
<
∞
w
h
e
n
k
<
1for all x
∈
V
∖
{
0
} and let φ
:
(
V
∖
{
0
}
)
3
→
[
0
,
∞
) be a function satisfying the condition(22)
∑
i
=
0
∞
k
2
i
φ
x
k
i
,
y
k
i
,
z
k
i
<
∞
w
h
e
n
k
>
1
,
∑
i
=
0
∞
k
i
φ
x
k
i
,
y
k
i
,
z
k
i
<
∞
w
h
e
n
k
<
1for all x
,
y
,
z
∈
V
∖
{
0
}. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and (16) for all x
∈
V
∖
{
0
} and if f satisfies (17) for all x
,
y
,
z
∈
V
∖
{
0
}, then there exists a unique mapping F
:
V
→
Y satisfying (18) for all x
,
y
,
z
∈
V
∖
{
0
} and (19) for all x
∈
V and satisfying(23)
f
x
-
F
x
≤
∑
i
=
0
∞
k
2
i
+
k
i
2
μ
x
k
i
+
1
+
k
2
i
-
k
i
2
μ
-
x
k
i
+
1for all x
∈
V
∖
{
0
}.Theorem 13.
Given a real constantk with k
≠
1, let μ
:
V
∖
{
0
}
→
[
0
,
∞
) be a function satisfying the conditions(24)
∑
i
=
0
∞
μ
k
i
x
k
2
i
<
∞
,
∑
i
=
0
∞
k
i
μ
x
k
i
<
∞
w
h
e
n
k
>
1
,
∑
i
=
0
∞
μ
k
i
x
k
i
<
∞
,
∑
i
=
0
∞
k
2
i
μ
x
k
i
<
∞
w
h
e
n
k
<
1for all x
∈
V
∖
{
0
} and let φ
:
(
V
∖
{
0
}
)
3
→
[
0
,
∞
) be a function satisfying the conditions(25)
∑
i
=
0
∞
φ
k
i
x
,
k
i
y
,
k
i
z
k
2
i
<
∞
,
∑
i
=
0
∞
k
i
φ
x
k
i
,
y
k
i
,
z
k
i
<
∞
w
h
e
n
k
>
1
,
∑
i
=
0
∞
φ
k
i
x
,
k
i
y
,
k
i
z
k
i
<
∞
,
∑
i
=
0
∞
k
2
i
φ
x
k
i
,
y
k
i
,
z
k
i
<
∞
w
h
e
n
k
<
1for all x
,
y
,
z
∈
V
∖
{
0
}. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and (16) for all x
∈
V
∖
{
0
} and if f satisfies (17) for all x
,
y
,
z
∈
V
∖
{
0
}, then there exists a unique mapping F
:
V
→
Y satisfying (18) for all x
,
y
,
z
∈
V
∖
{
0
} and the equalities in (19) for all x
∈
V, such that(26)
f
x
-
F
x
≤
∑
i
=
0
∞
μ
k
i
x
+
μ
-
k
i
x
2
k
2
i
+
2
+
k
i
2
μ
x
k
i
+
1
+
μ
-
x
k
i
+
1
w
h
e
n
k
>
1
,
∑
i
=
0
∞
k
2
i
2
μ
x
k
i
+
1
+
μ
-
x
k
i
+
1
+
μ
k
i
x
+
μ
-
k
i
x
2
k
i
+
1
w
h
e
n
k
<
1holds for all x
∈
V
∖
{
0
}.It is to be noted that we can replaceV
∖
{
0
} with V in Theorems 11–13.Lemma 14.
Leta, b, and c be real constants. If a mapping f
:
V
→
W satisfies f
(
0
)
=
0 and D
a
,
b
,
c
f
(
x
,
y
,
z
)
=
0 for all x
,
y
,
z
∈
V, then(i)
f
e
(
a
x
)
=
a
2
f
e
(
x
),
(ii)
f
e
(
b
x
)
=
b
2
f
e
(
x
),
(iii)
f
e
(
c
x
)
=
c
2
f
e
(
x
),
(iv)
f
e
(
(
a
-
b
)
x
)
=
(
a
-
b
)
2
f
e
(
x
),
(v)
f
e
(
(
a
-
c
)
x
)
=
(
a
-
c
)
2
f
e
(
x
),
(vi)
f
e
(
(
b
-
c
)
x
)
=
(
b
-
c
)
2
f
e
(
x
),
(vii)
f
o
(
a
x
)
=
a
f
o
(
x
),
(viii)
f
o
(
b
x
)
=
b
f
o
(
x
),
(ix)
f
o
(
c
x
)
=
c
f
o
(
x
),
(x)
f
e
(
(
a
-
b
)
x
)
=
(
a
-
b
)
2
f
e
(
x
),
(xi)
f
e
(
(
a
-
c
)
x
)
=
(
a
-
c
)
2
f
e
(
x
),
(xii)
f
e
(
(
b
-
c
)
x
)
=
(
b
-
c
)
2
f
e
(
x
) for all x
∈
V.Proof.
Using the equalities(27)
f
e
a
x
-
a
2
f
e
x
=
D
a
,
b
,
c
f
e
x
,
0,0
=
0
,
f
o
a
x
-
a
f
o
x
=
D
a
,
b
,
c
f
o
x
,
0,0
=
0
,
f
e
a
-
b
x
-
a
-
b
2
f
e
x
=
D
a
,
b
,
c
f
e
x
,
-
x
,
0
=
0
,
f
o
a
-
b
x
-
a
-
b
f
o
x
=
D
a
,
b
,
c
f
o
x
,
-
x
,
0
=
0 for all x
∈
V and by Remark 7, we can easily prove the assertions.Since every solution to the functional equation (3) is a quadratic-additive mapping according to Theorem 8 provided a, b, and c are nonzero real constants, we can prove the following set of theorems by using Theorems 11–13.Theorem 15.
For nonzero real constantsa, b, and c with a
-
b
∉
{
-
1,0
,
1
}, let φ
:
V
3
→
[
0
,
∞
) be a function satisfying the condition(28)
∑
i
=
0
∞
φ
a
-
b
i
x
,
a
-
b
i
y
,
a
-
b
i
z
a
-
b
i
<
∞
w
h
e
n
a
-
b
>
1
,
∑
i
=
0
∞
φ
a
-
b
i
x
,
a
-
b
i
y
,
a
-
b
i
z
a
-
b
2
i
<
∞
w
h
e
n
a
-
b
<
1for all x
,
y
,
z
∈
V. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and(29)
D
a
,
b
,
c
f
x
,
y
,
z
≤
φ
x
,
y
,
zfor all x
,
y
,
z
∈
V, then there exists a unique quadratic-additive mapping F
:
V
→
Y such that (18) holds for all x
,
y
,
z
∈
V and(30)
f
x
-
F
x
≤
∑
i
=
0
∞
a
-
b
i
+
1
+
1
φ
a
-
b
i
x
,
-
a
-
b
i
x
,
0
2
a
-
b
2
i
+
2
+
∑
i
=
0
∞
a
-
b
i
+
1
-
1
φ
-
a
-
b
i
x
,
a
-
b
i
x
,
0
2
a
-
b
2
i
+
2holds for all x
∈
V.Proof.
It follows from (29) that(31)
f
a
-
b
x
-
a
-
b
2
+
a
-
b
2
f
x
-
a
-
b
2
-
a
-
b
2
f
-
x
=
D
a
,
b
,
c
f
x
,
-
x
,
0
≤
φ
x
,
-
x
,
0 for all x
∈
V. If we put μ
(
x
)
=
φ
(
x
,
-
x
,
0
) and k
=
a
-
b, then μ
(
x
) satisfies condition (14) and inequality (16) for all x
∈
V. In view of Theorem 11, there exists a unique mapping F
:
V
→
Y satisfying equality (18) for all x
,
y
,
z
∈
V and the equalities in (19) for all x
∈
V, such that inequality (30) holds for all x
∈
V.
In view of Lemma14, the equalities in (19) follow from equality (18). Hence, there exists a unique mapping F
:
V
→
Y satisfying equality (18) for all x
,
y
,
z
∈
V and inequality (30) for all x
∈
V.By the same way as in the proof of Theorem15, we can prove the following couple of theorems. Hence, we omit their proofs.Theorem 16.
Given nonzero real constantsa, b, and c with a
-
b
∉
{
-
1,0
,
1
}, let φ
:
V
3
→
[
0
,
∞
) be a function satisfying the condition(32)
∑
i
=
0
∞
a
-
b
i
φ
x
a
-
b
i
,
y
a
-
b
i
,
z
a
-
b
i
<
∞
w
h
e
n
a
-
b
<
1
,
∑
i
=
0
∞
a
-
b
2
i
φ
x
a
-
b
i
,
y
a
-
b
i
,
z
a
-
b
i
<
∞
w
h
e
n
a
-
b
>
1for all x
,
y
,
z
∈
V. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and inequality (29) for all x
,
y
,
z
∈
V, then there exists a unique quadratic-additive mapping F
:
V
→
Y satisfying (18) for all x
,
y
,
z
∈
V and(33)
f
x
-
F
x
≤
∑
i
=
0
∞
a
-
b
2
i
+
a
-
b
i
2
φ
x
a
-
b
i
+
1
,
-
x
a
-
b
i
+
1
,
0
+
a
-
b
2
i
-
a
-
b
i
2
φ
-
x
a
-
b
i
+
1
,
x
a
-
b
i
+
1
,
0for all x
∈
V.Theorem 17.
Given nonzero real constantsa, b, and c satisfying a
-
b
∉
{
-
1,0
,
1
}, let φ
:
V
3
→
[
0
,
∞
) be a function satisfying the conditions(34)
∑
i
=
0
∞
φ
a
-
b
i
x
,
a
-
b
i
y
,
a
-
b
i
z
a
-
b
2
i
<
∞
,
∑
i
=
0
∞
a
-
b
i
φ
x
a
-
b
i
,
y
a
-
b
i
,
z
a
-
b
i
<
∞
w
h
e
n
a
-
b
>
1
,
∑
i
=
0
∞
φ
a
-
b
i
x
,
a
-
b
i
y
,
a
-
b
i
z
a
-
b
i
<
∞
,
∑
i
=
0
∞
a
-
b
2
i
φ
x
a
-
b
i
,
y
a
-
b
i
,
z
a
-
b
i
<
∞
w
h
e
n
a
-
b
<
1for all x
,
y
,
z
∈
V. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and equality (29) for all x
,
y
,
z
∈
V, then there exists a unique quadratic-additive mapping F
:
V
→
Y satisfying equality (18) for all x
,
y
,
z
∈
V and(35)
f
x
-
F
x
≤
∑
i
=
0
∞
φ
a
-
b
i
x
,
-
a
-
b
i
x
,
0
+
φ
-
a
-
b
i
x
,
a
-
b
i
x
,
0
2
a
-
b
2
i
+
2
+
a
-
b
i
2
φ
x
a
-
b
i
+
1
,
-
x
a
-
b
i
+
1
,
0
+
φ
-
x
a
-
b
i
+
1
,
x
a
-
b
i
+
1
,
0
w
h
e
n
a
-
b
>
1
,
∑
i
=
0
∞
φ
a
-
b
i
x
,
-
a
-
b
i
x
,
0
+
φ
-
a
-
b
i
x
,
a
-
b
i
x
,
0
2
a
-
b
i
+
1
+
a
-
b
2
i
2
φ
x
a
-
b
i
+
1
,
-
x
a
-
b
i
+
1
,
0
+
φ
-
x
a
-
b
i
+
1
,
x
a
-
b
i
+
1
,
0
w
h
e
n
a
-
b
<
1for all x
∈
V.By the same way as in the proofs of Theorems18–23, we can use Lemma 14 to prove Theorems 15–17.Theorem 18.
For nonzero real constantsa, b, and c with a
∉
{
-
1,0
,
1
}, let φ
:
V
3
→
[
0
,
∞
) be a function satisfying the condition(36)
∑
i
=
0
∞
φ
a
i
x
,
a
i
y
,
a
i
z
a
i
<
∞
w
h
e
n
a
>
1
,
∑
i
=
0
∞
φ
a
i
x
,
a
i
y
,
a
i
z
a
2
i
<
∞
w
h
e
n
a
<
1for all x
,
y
,
z
∈
V. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and (29) for all x
,
y
,
z
∈
V, then there exists a unique quadratic-additive mapping F
:
V
→
Y satisfying (18) for all x
,
y
,
z
∈
V and(37)
f
x
-
F
x
≤
∑
i
=
0
∞
a
i
+
1
+
1
φ
a
i
x
,
0,0
2
a
2
i
+
2
+
∑
i
=
0
∞
a
i
+
1
-
1
φ
-
a
i
x
,
0,0
2
a
2
i
+
2for all x
∈
V.Proof.
If we putμ
(
x
)
=
φ
(
x
,
0,0
) and k
=
a, then it follows from (29) that(38)
f
a
x
-
a
2
+
a
2
f
x
-
a
2
-
a
2
f
-
x
=
D
a
,
b
,
c
f
x
,
0,0
≤
φ
x
,
0,0for all x
∈
V. For this case, conditions (14)–(17) in Theorem 11 are satisfied. Hence, there exists a unique mapping F
:
V
→
Y satisfying equality (18) for all x
,
y
,
z
∈
V and the equalities in (19) for all x
∈
V and satisfying inequality (37) for all x
∈
V. Since equality (18) implies equalities (19) in Lemma 14, there exists a unique mapping F
:
V
→
Y satisfying equality (18) for all x
,
y
,
z
∈
V as well as inequality (37) for all x
∈
V.Theorem 19.
Given nonzero real constantsa, b, and c with a
∉
{
-
1,0
,
1
}, assume that φ
:
V
3
→
[
0
,
∞
) is a function satisfying the condition(39)
∑
i
=
0
∞
a
i
φ
x
a
i
,
y
a
i
,
z
a
i
<
∞
w
h
e
n
a
<
1
,
∑
i
=
0
∞
a
2
i
φ
x
a
i
,
y
a
i
,
z
a
i
<
∞
w
h
e
n
a
>
1for all x
,
y
,
z
∈
V. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and equality (29) for all x
,
y
,
z
∈
V, then there exists a unique quadratic-additive mapping F
:
V
→
Y satisfying (18) for all x
,
y
,
z
∈
V and(40)
f
x
-
F
x
≤
∑
i
=
0
∞
a
2
i
+
a
i
2
φ
x
a
i
+
1
,
0,0
+
a
2
i
-
a
i
2
φ
-
x
a
i
+
1
,
0,0for all x
∈
V.Theorem 20.
Leta, b, and c be nonzero real constants with a
∉
{
-
1,0
,
1
}. Assume that φ
:
V
3
→
[
0
,
∞
) is a function satisfying the conditions(41)
∑
i
=
0
∞
φ
a
i
x
,
a
i
y
,
a
i
z
a
2
i
<
∞
,
∑
i
=
0
∞
a
i
φ
x
a
i
,
y
a
i
,
z
a
i
<
∞
w
h
e
n
a
>
1
,
∑
i
=
0
∞
φ
a
i
x
,
a
i
y
,
a
i
z
a
i
<
∞
,
∑
i
=
0
∞
a
2
i
φ
x
a
i
,
y
a
i
,
z
a
i
<
∞
w
h
e
n
a
<
1for all x
,
y
,
z
∈
V. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and equality (29) for all x
,
y
,
z
∈
V, then there exists a unique quadratic-additive mapping F
:
V
→
Y satisfying (18) for all x
,
y
,
z
∈
V and(42)
f
x
-
F
x
≤
∑
i
=
0
∞
φ
a
i
x
,
0,0
+
φ
-
a
i
x
,
0,0
2
a
2
i
+
2
+
a
i
2
φ
x
a
i
+
1
,
0,0
+
φ
-
x
a
i
+
1
,
0,0
w
h
e
n
a
>
1
,
∑
i
=
0
∞
φ
a
i
x
,
0,0
+
φ
-
a
i
x
,
0,0
2
a
i
+
1
+
a
2
i
2
φ
x
a
i
+
1
,
0,0
+
φ
-
x
a
i
+
1
,
0,0
w
h
e
n
a
<
1for all x
∈
V.Since every quadratic-additive mapping is a solution to the functional equation (3) and every solution to the functional equation (3) is a quadratic-additive mapping provided a, b, and c are nonzero rational numbers by Theorem 9, the following set of theorems are direct consequences of Theorems 15–20.Theorem 21.
For nonzero rational constantsa, b, and c, let φ
:
V
3
→
[
0
,
∞
) be a function satisfying condition (28) for all x
,
y
,
z
∈
V. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and (29) for all x
,
y
,
z
∈
V, then there exists a unique quadratic-additive mapping F
:
V
→
Y satisfying (30) for all x
∈
V.Theorem 22.
For nonzero rational constantsa, b, and c, let φ
:
V
3
→
[
0
,
∞
) be a function satisfying condition (32) for all x
,
y
,
z
∈
V. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and inequality (29) for all x
,
y
,
z
∈
V, then there exists a unique quadratic-additive mapping F
:
V
→
Y satisfying (33) for all x
∈
V.Theorem 23.
Given nonzero rational constantsa, b, and c, assume that φ
:
V
3
→
[
0
,
∞
) is a function satisfying the conditions in (34) for all x
,
y
,
z
∈
V. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and equality (29) for all x
,
y
,
z
∈
V, then there exists a unique quadratic-additive mapping F
:
V
→
Y satisfying (35) for all x
∈
V.Theorem 24.
Leta, b, and c be nonzero rational constants and let φ
:
V
3
→
[
0
,
∞
) be a function satisfying condition (36) for all x
,
y
,
z
∈
V. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and (29) for all x
,
y
,
z
∈
V, then there exists a unique quadratic-additive mapping F
:
V
→
Y satisfying (37) for all x
∈
V.Theorem 25.
For nonzero rational constantsa, b, and c, let φ
:
V
3
→
[
0
,
∞
) be a function satisfying condition (39) for all x
,
y
,
z
∈
V. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and equality (29) for all x
,
y
,
z
∈
V, then there exists a unique quadratic-additive mapping F
:
V
→
Y satisfying (40) for all x
∈
V.Theorem 26.
Given nonzero rational constantsa, b, and c, suppose φ
:
V
3
→
[
0
,
∞
) is a function satisfying the conditions in (41) for all x
,
y
,
z
∈
V. If a mapping f
:
V
→
Y satisfies f
(
0
)
=
0 and equality (29) for all x
,
y
,
z
∈
V, then there exists a unique quadratic-additive mapping F
:
V
→
Y satisfying (42) for all x
∈
V.We remember thatX is a real normed space and Y is a real Banach space.Corollary 27.
Assume thata, b, and c are nonzero real constants with a
-
b
∉
{
-
1,0
,
1
}, p is a real number with p
∉
{
1,2
}, and θ is a positive real number. If a mapping f
:
X
→
Y satisfies f
(
0
)
=
0 and the inequality(43)
D
a
,
b
,
c
f
x
,
y
,
z
≤
θ
x
p
+
y
p
+
z
pfor all x
,
y
,
z
∈
X, then there exists a unique quadratic-additive mapping F
:
X
→
Y such that D
a
,
b
,
c
F
(
x
,
y
,
z
)
=
0 holds for all x
,
y
,
z
∈
X and(44)
f
x
-
F
x
≤
2
θ
x
p
a
-
b
-
a
-
b
p
w
h
e
n
p
<
1
,
2
θ
x
p
a
-
b
-
a
-
b
p
+
2
x
p
a
-
b
2
-
a
-
b
p
w
h
e
n
1
<
p
<
2
,
2
θ
x
p
a
-
b
2
-
a
-
b
p
w
h
e
n
p
>
2holds for all x
∈
X.Proof.
If we putφ
(
x
,
y
,
z
)
≔
θ
(
x
p
+
y
p
+
z
p
) for all x
,
y
,
z
∈
X, then φ satisfies either the first inequality in (28) when p
<
1 and a
-
b
>
1, or the second inequality in (28) when p
>
2 and a
-
b
<
1, or the first inequality in (32) when p
>
2 and a
-
b
>
1, or the second inequality in (32) when p
<
1 and a
-
b
<
1, or the first inequality in (34) when 1
<
p
<
2 and a
-
b
>
1, or the second inequality in (34) when 1
<
p
<
2 and a
-
b
<
1 for all x
,
y
,
z
∈
X. Therefore, by Theorems 15–17, we obtain the desired inequality (44).The following corollary follows from Theorems18–20.Corollary 28.
Assume thata, b, and c are nonzero real constants with a
∉
{
-
1,0
,
1
}, p is a real number with p
∉
{
1,2
}, and θ is a positive real number. If a mapping f
:
X
→
Y satisfies f
(
0
)
=
0 and inequality (43) for all x
,
y
,
z
∈
X, then there exists a unique quadratic-additive mapping F
:
X
→
Y such that D
a
,
b
,
z
F
(
x
,
y
,
z
)
=
0 holds for all x
,
y
,
z
∈
X and(45)
f
x
-
F
x
≤
θ
x
p
a
-
a
p
w
h
e
n
p
<
1
,
θ
x
p
a
-
a
p
+
x
p
a
2
-
a
p
w
h
e
n
1
<
p
<
2
,
θ
x
p
a
2
-
a
p
w
h
e
n
p
>
2holds for all x
∈
X.Proof.
If we putφ
(
x
,
y
,
z
)
≔
θ
(
x
p
+
y
p
+
z
p
) for all x
,
y
,
z
∈
X, then φ satisfies either the first inequality in (36) when p
<
1 and a
>
1, or the second inequality in (36) when p
>
2 and a
<
1, or the first inequality in (39) when p
>
2 and a
>
1, or the second inequality in (39) when p
<
1 and a
<
1, or the first inequality in (41) when 1
<
p
<
2 and a
>
1, or the second inequality in (41) when 1
<
p
<
2 and a
<
1 for all x
,
y
∈
X. Therefore, by Theorems 18–20, we obtain the desired inequality (45).The following corollary follows from Theorems21–23.Corollary 29.
Leta, b, and c be nonzero rational constants with a
-
b
∉
{
-
1,0
,
1
} and let θ
,
p be nonnegative real numbers with p
∉
{
1,2
}. If a mapping f
:
X
→
Y satisfies f
(
0
)
=
0 and inequality (43) for all x
,
y
,
z
∈
X, then there exists a unique quadratic-additive mapping F
:
X
→
Y such that inequality (44) holds for all x
∈
X.The following corollary follows from Theorems24–26.Corollary 30.
Leta, b, and c be nonzero rational constants with a
∉
{
-
1,0
,
1
} and let θ
,
p be nonnegative real numbers with p
∉
{
1,2
}. If a mapping f
:
X
→
Y satisfies f
(
0
)
=
0 and inequality (43) for all x
,
y
,
z
∈
X, then there exists a unique quadratic-additive mapping F
:
X
→
Y such that inequality (45) holds for all x
∈
X.



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*Source: 1793065-2016-05-22.xml*