# Approximate Generalized Cubic*-Derivations

Authors: Heejeong Koh; Dongseung Kang
Journal: Journal of Function Spaces
 (2014)

---

## Abstract

We will show the general solution of the functional equationf
(
a
x
+
b
y
)
+
f
(
a
x
-
b
y
) = a
b
2
[
f
(
x
+
y
)
+
f
(
x
-
y
)
]
+
2
a
(
a
2
-
b
2
)
f
(
x
) and investigate the stability of cubic *-derivations associated with the given functional equation on Banach *-algebras.


---

## Body

## 1. Introduction

Ulam [1] proposed the famous problem concerning the stability of group homomorphisms. We say that a functional equation is said to be stable if any approximate solution to the functional equation is near a true solution of that functional equation. The question of Ulam was affirmatively answered for Banach spaces by Hyers [2]. The result of Hyers was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference. This result of Rassias leads mathematicians working in stability of functional equations to establish what is known today as Hyers-Ulam-Rassias stability and to introduce new definitions of stability concepts. During the last three decades, several stability problems of a large variety of functional equations have been extensively studied and generalized by a number of authors; see [5]. By using fixed point methods the stability problems of several functional equations have been extensively investigated by a number of authors; see [6–9]. Also, Jang and Park [10] investigated the stability of *-derivations and of quadratic *-derivations with Cauchy functional equation and the Jensen functional equation on Banach *-algebra. In particular, the stability of *-derivations on Banach *-algebra by using fixed point alternative was proved by Park and Bodaghi; see [11].Jun and Kim [12] introduced the following functional equation:
(1)
f
(
2
x
+
y
)
+
f
(
2
x
-
y
)
=
2
f
x
+
y
+
2
f
x
-
y
+
12
f
x
,
and they established a general solution. Najati [13] investigated the cubic functional equation
(2)
f
(
a
x
+
y
)
+
f
(
a
x
-
y
)
=
a
f
(
x
+
y
)
+
a
f
(
x
-
y
)
+
2
(
a
3
-
a
)
f
(
x
)
for a fixed positive integer a
≥
2.In this paper, we deal with the functional equation(3)
f
(
a
x
+
b
y
)
+
f
(
a
x
-
b
y
)
=
a
b
2
[
f
(
x
+
y
)
+
f
(
x
-
y
)
]
+
2
a
(
a
2
-
b
2
)
f
(
x
)
for all x
,
y
∈
X and fixed a
,
b
∈
Z
(
a
≥
2
,
b
≥
1
,
a
≠
b
). We will provide the general solution of the functional equation (3) and investigate the stability of cubic *-derivations associated with the given functional equation on Banach *-algebras.


## 2. Cubic Functional Equations

In this section letX and Y be real vector spaces and we investigate the general solution of the functional equation (3). Before we proceed, we would like to introduce some basic definitions concerning n-additive symmetric mappings and key concepts which are found in [14, 15]. A function A
:
X
→
Y is said to beadditive if A
(
x
+
y
)
=
A
(
x
)
+
A
(
y
) for all x
,
y
∈
X. Let n be a positive integer. A function A
n
:
X
n
→
Y is called n
-additive if it is additive in each of its variables. A function A
n is said to besymmetric if A
n
(
x
1
,
…
,
x
n
)
=
A
n
(
x
σ
(
1
)
,
…
,
x
σ
(
n
)
) for every permutation {
σ
(
1
)
,
…
,
σ
(
n
)
} of {
1,2
,
…
,
n
}. If A
n
(
x
1
,
x
2
,
…
,
x
n
) is an n-additive symmetric map, then A
n
(
x
) will denote the diagonal A
n
(
x
,
x
,
…
,
x
) and A
n
(
r
x
)
=
r
n
A
n
(
x
) for all x
∈
X and all r
∈
Q. Such a function A
n
(
x
) will be called amonomial function of degree n (assuming A
n
≢
0). Furthermore the resulting function after substitution x
1
=
x
2
=
⋯
=
x
s
=
x and x
s
+
1
=
x
s
+
2
=
⋯
=
x
n
=
y in A
n
(
x
1
,
x
2
,
…
,
x
n
) will be denoted by A
s
,
n
-
s
(
x
,
y
).Theorem 1.
A functionf
:
X
→
Y is a solution of the functional equation (3) if and only if f is of the form f
(
x
)
=
A
3
(
x
) for all x
∈
X, where A
3
(
x
) is the diagonal of the 3-additive symmetric map A
3
:
X
3
→
Y.Proof.
Assume thatf satisfies the functional equation (3). Let x
=
y
=
0 in (3). We have
(4)
f
0
=
a
3
f
0
;
that is, f
(
0
)
=
0. It is easy to show that f is f
(
a
x
)
=
a
3
f
(
x
) and an odd mapping for all x
∈
X. We can rewrite the functional equation (3) in the form
(5)
f
(
x
)
-
1
2
a
(
a
2
-
b
2
)
f
(
a
x
+
b
y
)
-
1
2
a
(
a
2
-
b
2
)
f
(
a
x
-
b
y
)
+
b
2
2
(
a
2
-
b
2
)
f
(
x
+
y
)
+
b
2
2
(
a
2
-
b
2
)
f
(
x
-
y
)
=
0
for all x
,
y
∈
X. By Theorems 3.5 and 3.6 in [15], f is a generalized polynomial function of degree at most 3; that is, f is of the form
(6)
f
(
x
)
=
A
3
(
x
)
+
A
2
(
x
)
+
A
1
(
x
)
+
A
0
(
x
)
for all x
∈
X, where A
0
(
x
)
=
A
0 is an arbitrary element of Y and A
i
(
x
) is the diagonal of the i-additive symmetric map A
i
:
X
i
→
Y for i
=
1,2
,
3. By f
(
0
)
=
0 and f
(
-
x
)
=
-
f
(
x
) for all x
∈
X, we get A
0
(
x
)
=
A
0
=
0. Substituting (6) into (3) we have
(7)
A
3
(
a
x
+
b
y
)
+
A
2
(
a
x
+
b
y
)
+
A
1
(
a
x
+
b
y
)
+
A
3
(
a
x
-
b
y
)
+
A
2
(
a
x
-
b
y
)
+
A
1
(
a
x
-
b
y
)
=
a
b
2
A
3
(
x
+
y
)
+
A
2
(
x
+
y
)
+
A
1
(
x
+
y
)
+
A
3
(
x
-
y
)
+
A
2
(
x
-
y
)
+
A
1
(
x
-
y
)
+
2
a
(
a
2
-
b
2
)
[
A
3
(
x
)
+
A
2
(
x
)
+
A
1
(
x
)
]
for all x
,
y
∈
X. Note that
(8)
A
3
(
a
x
+
b
y
)
+
A
3
(
a
x
-
b
y
)
=
2
a
3
A
3
(
x
)
+
6
a
b
2
A
1,2
(
x
,
y
)
,
A
3
(
x
+
y
)
+
A
3
(
x
-
y
)
=
2
A
3
(
x
)
+
6
A
1,2
(
x
,
y
)
,
A
2
(
a
x
+
b
y
)
+
A
2
(
a
x
-
b
y
)
=
2
a
2
A
2
(
x
)
+
2
b
2
A
2
(
y
)
,
A
1
a
x
+
b
y
+
A
1
a
x
-
b
y
=
a
b
A
1
x
+
y
+
a
b
A
1
x
-
y
.
Then we have
(9)
2
a
2
(
a
-
1
)
A
2
(
x
)
+
2
b
2
(
a
-
1
)
A
2
(
y
)
+
2
a
(
a
2
-
b
2
)
A
1
(
x
)
=
0
for all x
,
y
∈
X. Letting x
=
0 in (9), we get A
2
(
y
)
=
0 for all y
∈
X. Hence we have A
2
(
x
)
=
A
1
(
x
)
=
0 for all x
∈
X. Thus
(10)
f
(
x
)
=
A
3
(
x
)
+
A
2
(
x
)
+
A
1
(
x
)
+
A
0
(
x
)
=
A
3
(
x
)
for all x
∈
X.
Conversely, assume thatf
(
x
)
=
A
3
(
x
) for all x
∈
X, where A
3
(
x
) is the diagonal of a 3-additive symmetric map A
3
:
X
3
→
Y. Note that
(11)
A
3
(
a
x
+
b
y
)
=
a
3
A
3
(
x
)
+
3
a
2
b
A
2,1
(
x
,
y
)
+
3
a
b
2
A
1,2
x
,
y
+
b
3
A
3
y
,
c
s
A
s
,
t
(
x
,
y
)
=
A
s
,
t
(
c
x
,
y
)
,
c
t
A
s
,
t
x
,
y
=
A
s
,
t
x
,
c
y
,
where 1
≤
s, t
≤
2, and c
∈
Q. Thus we may conclude that f satisfies (3).For this reason, we call a functionf ageneralized cubic functional equation if it satisfies (3).


## 3. Generalized Cubic*-Derivations

Throughout this section, we assume thatA is a Banach *-algebra. A mapping D
:
A
→
A is a cubic derivation if D is a cubic homogeneous mapping; that is, D is cubic, D
(
μ
a
)
=
μ
3
D
(
a
), and D
(
a
b
)
=
a
3
D
(
b
)
+
D
(
a
)
b
3 for all a
∈
A and μ
∈
C. In addition, if D satisfies D
(
a
*
)
=
D
(
a
)
* for all a
∈
A, then it is called a cubic *-derivation. For a given mapping f
:
A
→
A, we consider
(12)
D
μ
f
a
,
b
≔
f
s
μ
a
+
t
μ
b
+
f
s
μ
a
-
t
μ
b
-
s
t
2
μ
3
[
f
(
a
+
b
)
+
f
(
a
-
b
)
]
-
2
s
s
2
-
t
2
μ
3
f
a
,
D
f
(
a
,
b
)
:
=
f
(
a
b
)
-
f
(
a
)
b
3
-
a
3
f
(
b
)
for all μ
∈
C, a
,
b
∈
A, and s
,
t
∈
Z  (
s
≥
2
,
t
≥
1, and  s
≠
t
).Theorem 2.
Suppose thatf
:
A
→
A is a mapping with f
(
0
)
=
0 for which there exists a function ϕ
:
A
5
→
[
0
,
∞
) such that
(13)
ϕ
~
a
,
b
,
x
,
y
,
z
:
=
∑
j
=
0
∞
‍
1
s
3
j
ϕ
s
j
a
,
s
j
b
,
s
j
x
,
s
j
y
,
s
j
z
<
∞
,
(14)
D
μ
f
(
a
,
b
)
≤
ϕ
a
,
b
,
0,0
,
0
,
(15)
D
f
(
x
,
y
)
+
f
(
z
*
)
-
f
z
*
≤
ϕ
(
0,0
,
x
,
y
,
z
)
for all μ
∈
T
1
/
n
0
1
=
{
e
i
θ
∣
0
≤
θ
≤
2
π
/
n
0
} and all a
,
b
,
x
,
y
,
z
∈
A in which n
0
∈
N. Also assume that for each fixed a
∈
A the mapping t
↦
f
(
t
a
) from R to A is continuous; then there exists a unique generalized cubic *-derivation D on A satisfying
(16)
f
(
a
)
-
D
(
a
)
≤
1
2
s
3
ϕ
~
(
a
,
0,0
,
0,0
)
.Proof.
Lettingb
=
0 and μ
=
1 in inequality (14), we have
(17)
1
s
3
f
(
s
a
)
-
f
(
a
)
≤
1
2
s
3
ϕ
(
a
,
0,0
,
0,0
)
for all a
∈
A. Using induction, it can be shown that
(18)
1
s
3
k
f
(
s
k
a
)
-
1
s
3
l
f
(
s
l
a
)
≤
1
2
s
3
∑
j
=
l
k
-
1
‍
ϕ
(
s
j
a
,
0,0
,
0,0
)
s
3
j
for k
>
l
≥
0 and a
∈
A. Inequalities (13) and (18) imply that the sequence {
(
1
/
s
3
k
)
f
(
s
k
a
)
} is a Cauchy sequence. Since A is a Banach algebra, the sequence is convergent. Hence we can define a mapping D
:
A
→
A as
(19)
lim
⁡
k
→
∞
1
s
3
k
f
(
s
k
a
)
=
D
(
a
)
for a
∈
A. Also, letting l
=
0 in inequality (18), we have
(20)
1
s
3
k
f
(
s
k
a
)
-
f
(
a
)
≤
1
2
s
3
∑
j
=
0
k
-
1
‍
ϕ
(
s
j
a
,
0,0
,
0,0
)
s
3
j
for k
>
0 and a
∈
A. Inequalities (13) and (20) imply that inequality (16) holds.
In the following, we will show that the mappingD is a unique generalized cubic *-derivation such that inequality (16) holds for all a
∈
A. We note that
(21)
D
μ
D
(
a
,
b
)
=
lim
⁡
k
→
∞
1
s
3
k
D
μ
f
(
s
k
a
,
s
k
b
)
≤
lim
⁡
k
→
∞
ϕ
(
s
k
a
,
s
k
b
,
0,0
,
0
)
s
3
k
=
0
,
for all a
,
b
∈
A and μ
∈
T
1
/
n
0
1. By taking μ
=
1 in inequality (21), it follows that the mapping D is a generalized cubic mapping. Also, inequality (21) implies that D
μ
D
(
a
,
0
)
=
0. Hence
(22)
D
(
μ
a
)
=
μ
3
D
(
a
)
for all a
∈
A and μ
∈
T
1
/
n
0
1. Let μ
∈
T
1
=
{
λ
∈
C
∣
|
λ
|
=
1
}. Then μ
=
e
i
θ, where 0
≤
θ
≤
2
π. Let μ
1
=
μ
1
/
n
0
=
e
i
θ
/
n
0. Hence we have μ
1
∈
T
1
/
n
0
1. Then
(23)
D
(
μ
a
)
=
D
(
μ
1
n
0
a
)
=
μ
1
3
n
0
D
(
a
)
=
μ
3
D
(
a
)
for all μ
∈
T
1 and a
∈
A. Suppose that ρ is any continuous linear functional on A and a is a fixed element in A. Then we can define a function g
:
R
→
R by
(24)
g
(
r
)
=
ρ
(
D
(
r
a
)
)
for all r
∈
R. It is easy to check that g is cubic. Let
(25)
g
k
(
r
)
=
ρ
f
s
k
r
a
s
3
k
for all k
∈
N and r
∈
R.
Note thatg as the pointwise limit of the sequence of measurable functions g
k is measurable. Hence g as a measurable cubic function is continuous (see [16, 17]) and
(26)
g
(
r
)
=
r
3
g
(
1
)
for all r
∈
R. Thus
(27)
ρ
(
D
(
r
a
)
)
=
g
(
r
)
=
r
3
g
(
1
)
=
r
3
ρ
(
D
(
a
)
)
=
ρ
(
r
3
D
(
a
)
)
for all r
∈
R. Since ρ was an arbitrary continuous linear functional on A we may conclude that
(28)
D
(
r
a
)
=
r
3
D
(
a
)
for all r
∈
R. Let μ
∈
C  (
μ
≠
0
). Then μ
/
|
μ
|
∈
T
1. Hence
(29)
D
μ
a
=
D
(
μ
μ
μ
a
)
=
μ
μ
3
D
(
μ
a
)
=
μ
μ
3
μ
3
D
(
a
)
=
μ
3
D
(
a
)
for all a
∈
A and μ
∈
C  (
μ
≠
0
). Since a was an arbitrary element in A, we may conclude that D is cubic homogeneous.
Next, replacingx
,
y by s
k
x
,
s
k
y, respectively, and z
=
0 in inequality (15), we have
(30)
1
s
6
k
D
f
(
s
k
x
,
s
k
y
)
≤
1
s
6
k
ϕ
(
0,0
,
s
k
x
,
s
k
y
,
0
)
≤
1
s
3
k
ϕ
(
0,0
,
s
k
x
,
s
k
y
,
0
)
for all x
,
y
∈
A. Taking the limit as k tends to infinity, we have D
f
(
x
,
y
)
=
0 for all x
,
y
∈
A. Letting x
=
y
=
0 and replacing z by s
k
z in inequality (15), we get
(31)
f
(
s
k
z
*
)
s
3
k
-
f
s
k
z
*
s
3
k
≤
ϕ
(
0,0
,
0,0
,
s
k
z
)
s
3
k
for all z
∈
A. As k
→
∞ in inequality (31), we have
(32)
D
(
z
*
)
=
D
z
*
for all z
∈
A. This means that D is a cubic *-derivation. We now show that D is unique. Assume D
~
:
A
→
A is another generalized cubic *-derivation satisfying inequality (16). Then
(33)
D
(
a
)
-
D
~
(
a
)
=
1
s
3
k
D
(
s
k
a
)
-
D
~
(
s
k
a
)
≤
1
s
3
k
D
s
k
a
-
f
s
k
a
+
f
(
s
k
a
)
-
D
~
(
s
k
a
)
≤
1
2
s
3
k
+
1
∑
j
=
0
∞
‍
1
s
3
j
ϕ
s
j
+
k
a
,
0,0
,
0,0
=
1
2
s
3
∑
j
=
k
∞
‍
1
s
3
j
ϕ
s
j
a
,
0,0
,
0,0
,
which tends to zero as k
→
∞, for all a
∈
A. Thus D
(
a
)
=
D
~
(
a
) for all a
∈
A.Corollary 3.
Letθ
,
r be positive real numbers with r
<
3 and let f
:
A
→
A be a mapping with f
(
0
)
=
0 such that
(34)
D
μ
f
(
a
,
b
)
≤
θ
a
r
+
b
r
,
D
f
x
,
y
+
f
z
*
-
f
z
*
≤
θ
(
x
r
+
y
r
+
z
r
)
for all μ
∈
T
1
/
n
0
1 and a
,
b
,
x
,
y
,
z
∈
A. Then there exists a unique generalized cubic *-derivation D on A satisfying
(35)
f
(
a
)
-
D
(
a
)
≤
θ
a
r
2
(
s
3
-
s
r
)
for all a
∈
A.Proof.
The proof follows from Theorem2 by taking ϕ
(
a
,
b
,
x
,
y
,
z
)  =  θ
(
∥
a
∥
r  +  ∥
b
∥
r  +  ∥
x
∥
r
+
∥
y
∥
r
+
∥
z
∥
r
) for all a
,
b
,
x
,
y
,
z
∈
A.In the following corollaries, we show the hyperstability for the generalized cubic*-derivations.Corollary 4.
Letr be positive real numbers with r
<
3 and let f
:
A
→
A be a mapping with f
(
0
)
=
0 such that
(36)
D
μ
f
(
a
,
b
)
≤
a
r
b
r
,
D
f
(
x
,
y
)
+
f
(
z
*
)
-
f
z
*
≤
x
r
(
y
r
+
z
r
)
for all μ
∈
T
1
/
n
0
1 and a
,
b
,
x
,
y
,
z
∈
A. Then f is a generalized cubic *-derivation on A.Proof.
By takingϕ
(
a
,
b
,
x
,
y
,
z
)
=
(
∥
a
∥
r
+
∥
x
∥
r
)
(
∥
b
∥
r
+
∥
y
∥
r
+
∥
z
∥
r
) in Theorem 2 for all a
,
b
,
x
,
y
,
z
∈
A, we have ϕ
~
(
a
,
0,0
,
0,0
)
=
0. Equation (16) implies that f
=
D; that is, f is the generalized cubic *-derivation on A.Corollary 5.
Letr be positive real numbers with r
<
3 and let f
:
A
→
A be a mapping with f
(
0
)
=
0 such that
(37)
D
μ
f
(
a
,
b
)
≤
a
r
b
r
,
D
f
(
x
,
y
)
+
f
(
z
*
)
-
f
z
*
≤
x
r
y
r
z
r
for all μ
∈
T
1
/
n
0
1 and a
,
b
,
x
,
y
,
z
∈
A. Then f is a generalized cubic *-derivation on A.Proof.
By takingϕ
(
a
,
b
,
x
,
y
,
z
)
=
(
∥
a
∥
r
+
∥
x
∥
r
)
(
∥
b
∥
r
+
∥
y
∥
r
∥
z
∥
r
) in Theorem 2 for all a
,
b
,
x
,
y
,
z
∈
A, we have ϕ
~
(
a
,
0,0
,
0,0
)
=
0. Equation (16) implies that f
=
D; that is, f is the generalized cubic *-derivation on A.Now, we will investigate the stability of the given functional equation (12) using the alternative fixed point method. Before proceeding with the proof, we will state the theorem, the alternative of fixed point; see [18, 19].Definition 6.
LetX be a set. A function d
:
X
×
X
→
[
0
,
∞
] is called ageneralized metric on X if d satisfies the following:(1)
d
(
x
,
y
)
=
0 if and only if x
=
y;
(2)
d
(
x
,
y
)
=
d
(
y
,
x
) for all x
,
y
∈
X;
(3)
d
(
x
,
z
)
≤
d
(
x
,
y
)
+
d
(
y
,
z
) for all x
,
y
,
z
∈
X.Theorem 7 (the alternative of fixed point [18, 19]).
Suppose that one is given a complete generalized metric space(
Ω
,
d
) and a strictly contractive mapping T
:
Ω
→
Ω with Lipschitz constant L. Then, for each given x
∈
Ω, either
(38)
d
T
n
x
,
T
n
+
1
x
=
∞
∀
n
≥
0
or there exists a natural number n
0 such that (1)
d
(
T
n
x
,
T
n
+
1
x
)
<
∞ for all n
≥
n
0;
(2)
the sequence(
T
n
x
) is convergent to a fixed point y
* of T;
(3)
y
* is the unique fixed point of T in the set
(39)
Δ
=
{
y
∈
Ω
∣
d
(
T
n
0
x
,
y
)
<
∞
}
;
(4)
d
(
y
,
y
*
)
≤
(
1
/
1
-
L
)
d
(
y
,
T
y
) for all y
∈
Δ.Theorem 8.
Letf
:
A
→
A be a continuous mapping with f
(
0
)
=
0 and let ϕ
:
A
5
→
[
0
,
∞
) be a continuous mapping such that
(40)
D
μ
f
(
a
,
b
)
≤
ϕ
a
,
b
,
0,0
,
0
,
(41)
D
f
(
x
,
y
)
+
f
(
z
*
)
-
f
z
*
≤
ϕ
(
0,0
,
x
,
y
,
z
)
for all μ
∈
T
1
/
n
0
1 and a
,
b
,
x
,
y
,
z
∈
A. If there exists a constant l
∈
(
0,1
) such that
(42)
ϕ
(
s
a
,
s
b
,
s
x
,
s
y
,
s
z
)
≤
s
3
l
ϕ
(
a
,
b
,
x
,
y
,
z
)
for all a
,
b
,
x
,
y
,
z
∈
A, then there exists a unique generalized cubic *-derivation D on A satisfying
(43)
f
(
a
)
-
D
(
a
)
≤
1
2
s
3
(
1
-
l
)
ϕ
(
a
,
0,0
,
0,0
)
for all a
∈
A.Proof.
Consider the set(44)
Ω
=
{
g
∣
g
:
A
⟶
A
,
g
(
0
)
=
0
}
and introduce the generalized metric on Ω,
(45)
d
g
,
h
=
inf
⁡
∥
g
(
a
)
-
h
(
a
)
∥
≤
c
ϕ
(
a
,
0,0
,
0,0
)
,
∀
a
∈
A
.
c
∈
(
0
,
∞
)
∣
∥
g
(
a
)
-
h
(
a
)
∥
≤
c
ϕ
(
a
,
0,0
,
0,0
)
,
∀
a
∈
A
∥
g
(
a
)
-
h
(
a
)
∥
≤
c
ϕ
(
a
,
0,0
,
0,0
)
,
∀
a
∈
A
.
It is easy to show that (
Ω
,
d
) is complete. Now we define a function T
:
Ω
→
Ω by
(46)
T
(
g
)
(
a
)
=
1
s
3
g
(
s
a
)
for all a
∈
A. Note that, for all g
,
h
∈
Ω, one lets c
∈
(
0
,
∞
) be an arbitrary constant with d
(
g
,
h
)
≤
c. Then
(47)
g
(
a
)
-
h
(
a
)
≤
c
ϕ
(
a
,
0,0
,
0,0
)
for all a
∈
A. Letting a
=
s
a in inequality (47) and using (42) and (46), we have
(48)
T
g
a
-
T
h
a
=
1
s
3
g
(
s
a
)
-
h
(
s
a
)
≤
1
s
3
c
ϕ
s
a
,
0,0
,
0,0
≤
c
l
ϕ
a
,
0,0
,
0,0
;
that is,
(49)
d
(
T
g
,
T
h
)
≤
c
l
.
Hence we have that
(50)
d
(
T
g
,
T
h
)
≤
l
d
(
g
,
h
)
,
for all g
,
h
∈
Ω; that is, T is a strict self-mapping of Ω with the Lipschitz constant l. Letting μ
=
1, b
=
0 in inequality (40), we get
(51)
1
s
3
f
(
s
a
)
-
f
(
a
)
≤
1
2
s
3
ϕ
(
a
,
0,0
,
0,0
)
for all a
∈
A. This means that
(52)
d
(
T
f
,
f
)
≤
1
2
s
3
.
We can apply the alternative of fixed point and, since lim
⁡
k
→
∞
d
(
T
k
f
,
D
)
=
0, there exists a fixed point D of T in Ω such that
(53)
D
(
a
)
=
lim
⁡
k
→
∞
f
(
s
k
a
)
s
3
k
,
for all a
∈
A. Hence
(54)
d
(
f
,
D
)
≤
1
1
-
l
d
(
T
f
,
f
)
≤
1
2
s
3
1
1
-
l
.
This implies that inequality (43) holds for all a
∈
A. Since l
∈
(
0,1
), inequality (42) shows that
(55)
lim
⁡
k
→
∞
ϕ
(
s
k
a
,
s
k
b
,
s
k
x
,
s
k
y
,
s
k
z
)
s
3
k
=
0
.
Replacing a
,
b by s
k
a,  s
k
b, respectively, in inequality (40), we have
(56)
1
s
3
k
D
μ
f
(
s
k
a
,
s
k
b
)
≤
ϕ
(
s
k
a
,
s
k
b
,
0,0
,
0
)
s
3
k
.
Taking the limit as k tends to infinity, we have D
μ
f
(
a
,
b
)
=
0 for all a
,
b
∈
A and all μ
∈
T
1
/
n
0
1. The remains are similar to the proof of Theorem 2.Corollary 9.
Letθ
,
r be positive real numbers with r
<
3 and let f
:
A
→
A be a mapping with f
(
0
)
=
0 such that
(57)
D
μ
f
(
a
,
b
)
≤
θ
a
r
+
b
r
,
D
f
(
x
,
y
)
+
f
(
z
*
)
-
f
z
*
≤
θ
(
x
r
+
y
r
+
z
r
)
for all μ
∈
T
1
/
n
0
1 and a
,
b
,
x
,
y
,
z
∈
A. Then there exists a unique generalized cubic *-derivation D on A satisfying
(58)
f
(
a
)
-
D
(
a
)
≤
θ
a
r
2
s
3
(
1
-
l
)
for all a
∈
A.Proof.
The proof follows from Theorem8 by taking ϕ
(
a
,
b
,
x
,
y
,
z
)  =  θ
(
∥
a
∥
r  +  ∥
b
∥
r  +  ∥
x
∥
r
+
∥
y
∥
r
+
∥
z
∥
r
) for all a
,
b
,
x
,
y
,
z
∈
A.In the following corollary, we show the hyperstability for the generalized cubic*-derivations.Corollary 10.
Letr be positive real numbers with 2
r
<
3 and let f
:
A
→
A be a mapping with f
(
0
)
=
0 such that
(59)
D
μ
f
(
a
,
b
)
≤
a
r
b
r
,
D
f
(
x
,
y
)
+
f
(
z
*
)
-
f
z
*
≤
x
r
(
y
r
+
z
r
)
for all μ
∈
T
1
/
n
0
1 and a
,
b
,
x
,
y
,
z
∈
A. Then f is a generalized cubic *-derivation on A.Proof.
By takingϕ
(
a
,
b
,
x
,
y
,
z
)
=
(
∥
a
∥
r
+
∥
x
∥
r
)
(
∥
b
∥
r
+
∥
y
∥
r
+
∥
z
∥
r
) in Theorem 8 for all a
,
b
,
x
,
y
,
z
∈
A, we have ϕ
(
a
,
0,0
,
0,0
)
=
0. Equation (43) implies that f
=
D; that is, f is the generalized cubic *-derivation on A.



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*Source: 757956-2014-11-20.xml*