# A Fixed Point Approach to the Stability of an Additive-Quadratic-Cubic-Quartic Type Functional Equation

Authors: Yang-Hi Lee; Soon-Mo Jung
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

By applying a fixed point approach, we investigate the stability problems for an AQCQ functional equation of the formf
x
+
2
y
+
f
x
-
2
y
-
4
f
x
+
y
-
4
f
x
-
y
+
6
f
x
-
f
2
y
-
f
-
2
y
+
4
f
y
+
4
f
-
y
=
0
.


---

## Body

## 1. Introduction

In 1940, Ulam [1] posed the problem concerning the stability of group homomorphisms. In the following year, Hyers [2] gave an affirmative answer to Ulam’s problem for additive mappings between Banach spaces. Thereafter, many mathematicians came to deal with this subject (cf. [3, 4]).We now consider the following functional equation:(1)
f
x
+
2
y
+
f
x
-
2
y
-
4
f
x
+
y
-
4
f
x
-
y
+
6
f
x
-
f
2
y
-
f
-
2
y
+
4
f
y
+
4
f
-
y
=
0
.The mappings f
(
x
)
=
a
x
4
+
b
x
3
+
c
x
2
+
d
x are solutions of this functional equation, where a
,
b
,
c
,
d are real constants.The stability problems of (1) were investigated by Lee et al. [5]. Indeed, they proved the stability of the equation by dividing the relevant mapping into the additive, quadratic, cubic, and quartic parts and applying the fixed point method to each part separately. Unfortunately, they could not prove the uniqueness of the exact solution of (1) which approximates the (given) mapping that is a solution of the perturbed equation of (1).In this paper, we will show that every solution of functional equation (1) is an AQCQ mapping (see below for the definition of AQCQ mapping) and we introduce a strictly contractive mapping which allows us to use the fixed point theory in the sense of Cădariu and Radu (see [6–8]). In a different way from Lee et al. [5], we do not split the (given) mapping which satisfies the perturbed equation of (1) and we apply the fixed point method to construct the explicit form of the exact solution of (1) near the (given) mapping by (2)
F
x
=
lim
n
→
∞
⁡
∑
i
=
0
n
n
C
i
-
1
n
-
i
1
0
i
1
6
n
f
o
2
2
n
-
i
x
+
-
1
n
-
i
2
0
i
6
4
n
f
e
2
2
n
-
i
x or (3)
F
x
=
lim
n
→
∞
⁡
∑
i
=
0
n
n
C
i
1
0
i
-
16
n
-
i
f
o
x
2
2
n
-
i
+
2
0
i
-
64
n
-
i
f
e
x
2
2
n
-
i
, where we denote by f
o and f
e the odd and even parts of f, respectively.Using this method, we can prove the uniqueness of the exact solution of (1) near the (given) mapping.Throughout this paper, letN denote the set of all positive integers and let N
0 denote the set of all nonnegative integers.


## 2. Preliminaries

We recall the fixed point theorem of Diaz and Margolis [9].Theorem 1.
Assume that(
X
,
d
) is a complete generalized metric space and J
:
X
→
X is a strictly contractive mapping with the Lipschitz constant 0
<
L
<
1. It then holds that, for each x
∈
X, either d
(
J
n
x
,
J
n
+
1
x
)
=
∞ for all n
∈
N
0 or there exists k
∈
N
0 such that (i)
d
(
J
n
x
,
J
n
+
1
x
)
<
∞ for all n
≥
k;
(ii)
the sequence{
J
n
x
} is convergent to a fixed point y
∗ of J;
(iii)
y
∗ is the unique fixed point of J in Y
=
{
y
∈
X
:
d
(
J
k
x
,
y
)
<
∞
};
(iv)
d
(
y
,
y
∗
)
≤
(
1
/
(
1
-
L
)
)
d
(
y
,
J
y
) for all y
∈
Y.Throughout this paper, letV and W be real vector spaces, let X be a real normed space, and let Y be a real Banach space.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
,
C
f
x
,
y
≔
f
x
+
2
y
-
3
f
x
+
y
+
3
f
x
-
f
x
-
y
-
6
f
y
,
Q
′
f
x
,
y
≔
f
x
+
2
y
-
4
f
x
+
y
+
6
f
x
-
4
f
x
-
y
+
f
x
-
2
y
-
24
f
y
,
E
f
x
,
y
≔
f
x
+
2
y
-
4
f
x
+
y
+
6
f
x
-
4
f
x
-
y
+
f
x
-
2
y
,
D
f
x
,
y
≔
f
x
+
2
y
+
f
x
-
2
y
-
4
f
x
+
y
-
4
f
x
-
y
+
6
f
x
-
f
2
y
-
f
-
2
y
+
4
f
y
+
4
f
-
y
=
E
f
x
,
y
-
f
2
y
-
f
-
2
y
+
4
f
y
+
4
f
-
y
=
Q
′
x
,
y
-
f
2
y
-
f
-
2
y
+
28
f
y
+
4
f
-
y for all x
,
y
∈
V. Each solution of A
f
(
x
,
y
)
=
0, Q
f
(
x
,
y
)
=
0, C
f
(
x
,
y
)
=
0, and Q
′
f
(
x
,
y
)
=
0 is called an additive mapping, a quadratic mapping, a cubic mapping, and a quartic mapping, respectively. Every mapping f
:
V
→
W is called an AQCQ mapping (additive-quadratic-cubic-quartic mapping) if f can be expressed by the sum of an additive mapping, a quadratic mapping, a cubic mapping, and a quartic mapping.The following lemmas have been proved in [10].Lemma 2 ([10], Theorem 2.3).
Iff
:
V
→
W is an odd mapping satisfying E
f
(
x
,
y
)
=
0 for all x
,
y
∈
V, then there exists a cubic mapping C
:
V
→
W and an additive mapping A
:
V
→
W such that f
(
x
)
=
C
(
x
)
+
A
(
x
) for all x
∈
V.Lemma 3 ([10], Theorem 2.4).
Iff
:
V
→
W is an even mapping satisfying the equalities E
f
(
x
,
y
)
=
0 for all x
,
y
∈
V and f
(
2
x
)
=
4
f
(
x
) for all x
∈
V, then f is a quadratic mapping.In the following theorem, we investigate the general solution of the functional equationD
f
(
x
,
y
)
=
0.Theorem 4.
A mappingf
:
V
→
W satisfies D
f
(
x
,
y
)
=
0 for all x
,
y
∈
V if and only if f is an AQCQ mapping.Proof.
First, we assume that a mappingf
:
V
→
W satisfies D
f
(
x
,
y
)
=
0 for all x
,
y
∈
V. Then, the odd part f
o satisfies the equalities (5)
E
f
o
x
,
y
=
D
f
o
x
,
y
=
1
2
D
f
x
,
y
-
1
2
D
f
-
x
,
-
y
=
0 for all x
,
y
∈
V. It follows from Lemma 2 that f
o is a cubic-additive mapping.
Letf
1
(
x
)
≔
f
e
(
2
x
)
-
16
f
e
(
x
) and f
2
(
x
)
≔
f
e
(
2
x
)
-
4
f
e
(
x
) for all x
∈
V. By a direct calculation and using the assumption that D
f
(
x
,
y
)
=
0 for all x
,
y
∈
V, we obtain f
(
0
)
=
(
1
/
6
)
D
f
(
0,0
)
=
0 and(6)
f
1
2
x
-
4
f
1
x
=
f
e
4
x
-
20
f
e
2
x
+
64
f
e
x
=
f
2
2
x
-
16
f
2
x
=
D
f
e
2
x
,
x
+
4
D
f
e
x
,
x
=
0
;that is,(7)
f
1
2
x
=
4
f
1
x
,
f
2
2
x
=
16
f
2
xfor all x
∈
V. Using these equalities, we further get the equalities E
f
1
(
x
,
y
)
=
D
f
1
(
x
,
y
) and Q
′
f
2
(
x
,
y
)
=
D
f
2
(
x
,
y
) which imply the equalities E
f
1
(
x
,
y
)
=
0 and Q
′
f
2
(
x
,
y
)
=
0 for all x
,
y
∈
V. Hence, by (7) and the fact that E
f
1
(
x
,
y
)
=
0, Lemma 3 implies that f
1 is quadratic. Moreover, f
2 is quartic because of Q
′
f
2
(
x
,
y
)
=
0 for all x
,
y
∈
V. Since (8)
f
x
=
f
o
x
+
f
e
x
=
f
o
x
-
1
12
f
1
x
+
1
12
f
2
x for all x
∈
V, f is an AQCQ mapping.
Conversely, assume thatf
1
,
f
2
,
f
3
,
f
4 are mappings such that the equalities f
(
x
)
≔
f
1
(
x
)
+
f
2
(
x
)
+
f
3
(
x
)
+
f
4
(
x
), A
f
1
(
x
,
y
)
=
0, Q
f
2
(
x
,
y
)
=
0, C
f
3
(
x
,
y
)
=
0, and Q
′
f
4
(
x
,
y
)
=
0 hold for all x
,
y
∈
V. Then the equalities f
1
(
x
)
=
-
f
1
(
-
x
), f
2
(
x
)
=
f
2
(
-
x
), f
3
(
x
)
=
-
f
3
(
-
x
), f
4
(
x
)
=
f
4
(
-
x
), f
1
(
2
x
)
=
2
f
1
(
x
), f
2
(
2
x
)
=
4
f
2
(
x
), f
3
(
2
x
)
=
8
f
3
(
x
), and f
4
(
2
x
)
=
16
f
4
(
x
) hold for all x
∈
V. From the above equalities, we obtain the equalities (9)
D
f
1
x
,
y
=
4
A
f
1
x
+
y
,
x
-
y
-
A
f
1
x
+
2
y
,
x
-
2
y
,
D
f
2
x
,
y
=
Q
f
2
x
,
2
y
-
4
Q
f
2
x
,
y
,
D
f
3
x
,
y
=
C
f
3
x
,
y
-
C
f
3
x
-
y
,
y
,
D
f
4
x
,
y
=
Q
′
f
4
x
,
y for all x
,
y
∈
V, which imply that (10)
D
f
x
,
y
=
D
f
1
x
,
y
+
D
f
2
x
,
y
+
D
f
3
x
,
y
+
D
f
4
x
,
y
=
0for all x
,
y
∈
V.


## 3. Main Results

Throughout this section, letV be a real vector space and let Y be a real Banach space.In the following theorems, we can prove the generalized Hyers-Ulam stability of the functional equation (1) by using a fixed point approach.Theorem 5.
Letf
:
V
→
Y be a mapping, for which there exists a mapping φ
:
V
2
→
[
0
,
∞
) such that the inequality(11)
D
f
x
,
y
≤
φ
x
,
yholds for all x, y
∈
V, and let f
(
0
)
=
0. If there exists a constant 0
<
L
<
1 such that φ satisfies the condition(12)
φ
2
x
,
2
y
≤
41
-
5
L
φ
x
,
yfor all x, y
∈
V, then there exists a unique solution F
:
V
→
Y of (1) satisfying the inequality(13)
f
x
-
F
x
≤
5
128
1
-
L
Φ
xfor all x
∈
V, where Φ
(
x
)
≔
φ
(
2
x
,
x
)
+
4
φ
(
x
,
x
)
+
φ
(
-
2
x
,
-
x
)
+
4
φ
(
-
x
,
-
x
). In particular, F is represented by(14)
F
x
=
lim
n
→
∞
⁡
∑
i
=
0
n
n
C
i
-
1
n
-
i
1
0
i
1
6
n
f
o
2
2
n
-
i
x
+
-
1
n
-
i
2
0
i
6
4
n
f
e
2
2
n
-
i
xfor all x
∈
V.Proof.
LetS be the set of all functions g
:
V
→
Y with g
(
0
)
=
0. We introduce a generalized metric on S by (15)
d
g
,
h
=
inf
⁡
K
∈
0
,
∞
:
g
x
-
h
x
≤
K
Φ
x
∀
x
∈
V
. It is not difficult to see that (
S
,
d
) is a generalized complete metric space (see [11, 12]).
We now consider the mappingJ
:
S
→
S, which is defined by (16)
J
g
x
≔
-
5
128
g
4
x
+
3
128
g
-
4
x
+
60
128
g
2
x
-
20
128
g
-
2
x for all x
∈
V. We assert that the equality(17)
J
n
g
x
=
∑
i
=
0
n
n
C
i
-
1
n
-
i
1
0
i
1
6
n
g
o
2
2
n
-
i
x
+
∑
i
=
0
n
n
C
i
-
1
n
-
i
2
0
i
6
4
n
g
e
2
2
n
-
i
xholds for all n
∈
N
0 and x
∈
V. Equality (17) is true if n
=
0. We assume that (17) is true for some n
∈
N
0. Then we have(18)
J
n
+
1
g
x
=
J
n
J
g
x
=
∑
i
=
0
n
n
C
i
-
1
n
-
i
1
0
i
1
6
n
J
g
o
2
2
n
-
i
x
+
∑
i
=
0
n
n
C
i
-
1
n
-
i
2
0
i
6
4
n
J
g
e
2
2
n
-
i
x
=
∑
i
=
0
n
n
C
i
-
1
n
-
i
1
0
i
1
6
n
-
5
128
g
o
2
2
n
+
2
-
i
x
+
∑
i
=
0
n
n
C
i
-
1
n
-
i
1
0
i
1
6
n
3
128
g
o
-
2
2
n
+
2
-
i
x
+
∑
i
=
0
n
n
C
i
-
1
n
-
i
1
0
i
1
6
n
60
128
g
o
2
2
n
+
1
-
i
x
+
∑
i
=
0
n
n
C
i
-
1
n
-
i
1
0
i
1
6
n
-
20
128
g
o
-
2
2
n
+
1
-
i
x
+
∑
i
=
0
n
n
C
i
-
1
n
-
i
2
0
i
6
4
n
-
5
128
g
e
2
2
n
+
2
-
i
x
+
∑
i
=
0
n
n
C
i
-
1
n
-
i
2
0
i
6
4
n
3
128
g
e
-
2
2
n
+
2
-
i
x
+
∑
i
=
0
n
n
C
i
-
1
n
-
i
2
0
i
6
4
n
60
128
g
e
2
2
n
+
1
-
i
x
+
∑
i
=
0
n
n
C
i
-
1
n
-
i
2
0
i
6
4
n
-
20
128
g
e
-
2
2
n
+
1
-
i
x
=
∑
i
=
0
n
n
C
i
-
1
n
+
1
-
i
1
0
i
1
6
n
+
1
g
o
2
2
n
+
2
-
i
x
+
∑
i
=
0
n
n
C
i
-
1
n
-
i
1
0
i
+
1
1
6
n
+
1
g
o
2
2
n
+
1
-
i
x
+
∑
i
=
0
n
n
C
i
-
1
n
+
1
-
i
2
0
i
6
4
n
+
1
g
e
2
2
n
+
2
-
i
x
+
∑
i
=
0
n
n
C
i
-
1
n
-
i
2
0
i
+
1
6
4
n
+
1
g
e
2
2
n
+
1
-
i
x
=
∑
i
=
0
n
+
1
n
+
1
C
i
-
1
n
+
1
-
i
1
0
i
1
6
n
+
1
g
o
2
2
n
+
2
-
i
x
+
∑
i
=
0
n
+
1
n
+
1
C
i
-
1
n
+
1
-
i
2
0
i
6
4
n
+
1
g
e
2
2
n
+
2
-
i
xfor all n
∈
N
0 and x
∈
V, which we can obtain from (17) provided we replace n with n
+
1.
Letg
,
h
∈
S and let K
∈
[
0
,
∞
] be an arbitrary constant with d
(
g
,
h
)
≤
K. From the definition of d, we have(19)
J
g
x
-
J
h
x
≤
5
128
g
4
x
-
h
4
x
+
3
128
g
-
4
x
-
h
-
4
x
+
60
128
g
2
x
-
h
2
x
+
20
128
g
-
2
x
-
h
-
2
x
≤
K
5
128
Φ
4
x
+
3
128
Φ
-
4
x
+
60
128
Φ
2
x
+
20
128
Φ
-
2
x
≤
K
8
128
Φ
4
x
+
80
128
Φ
2
x
≤
K
41
-
5
16
L
Φ
2
x
+
10
16
Φ
2
x
≤
K
41
-
5
2
16
L
2
Φ
x
+
10
41
-
5
16
L
Φ
x
≤
K
41
-
5
2
+
10
41
-
5
16
L
Φ
x
≤
L
K
Φ
xfor all x
∈
V, which implies that (20)
d
J
g
,
J
h
≤
L
d
g
,
h for any g
,
h
∈
S. That is, J is a strictly contractive self-mapping of S with the Lipschitz constant L.
Moreover, by a long calculation and (11), we see that(21)
f
x
-
J
f
x
=
5
D
f
2
x
,
x
+
20
D
f
x
,
x
-
3
D
f
-
2
x
,
-
x
-
12
D
f
-
x
,
-
x
128
≤
5
φ
2
x
,
x
+
20
φ
x
,
x
+
3
φ
-
2
x
,
-
x
+
12
φ
-
x
,
-
x
128
≤
5
128
Φ
xfor all x
∈
V. It means that d
(
f
,
J
f
)
≤
5
/
128
<
∞ by the definition of d. Therefore, according to Theorem 1
(
i
i
) and (
i
i
i
), the sequence {
J
n
f
} converges to the unique fixed point F
:
V
→
Y of J in the set T
=
{
g
∈
S
:
d
(
f
,
g
)
<
∞
}, which is represented by (14) for all x
∈
V.
Notice that(22)
d
f
,
F
≤
1
1
-
L
d
f
,
J
f
≤
5
128
1
-
L
, which implies (13).
By the definition ofF, together with (11) and (12), we have(23)
D
F
x
,
y
=
lim
n
→
∞
⁡
D
J
n
f
x
,
y
=
lim
n
→
∞
⁡
∑
i
=
0
n
n
C
i
-
1
n
-
i
1
0
i
1
6
n
D
f
o
2
2
n
-
i
x
,
2
2
n
-
i
y
+
∑
i
=
0
n
n
C
i
-
1
n
-
i
2
0
i
6
4
n
D
f
e
2
2
n
-
i
x
,
2
2
n
-
i
y
≤
lim
n
→
∞
⁡
∑
i
=
0
n
n
C
i
2
1
0
i
1
6
n
+
2
0
i
6
4
n
×
φ
2
2
n
-
i
x
,
2
2
n
-
i
y
+
φ
-
2
2
n
-
i
x
,
-
2
2
n
-
i
y
≤
lim
n
→
∞
⁡
∑
i
=
0
n
n
C
i
1
0
i
1
6
n
×
41
-
5
n
-
i
L
n
-
i
φ
2
n
x
,
2
n
y
+
φ
-
2
n
x
,
-
2
n
y
≤
lim
n
→
∞
⁡
∑
i
=
0
n
n
C
i
41
-
5
n
-
i
1
0
i
1
6
n
φ
2
n
x
,
2
n
y
+
φ
-
2
n
x
,
-
2
n
y
≤
lim
n
→
∞
⁡
41
+
5
n
1
6
n
φ
2
n
x
,
2
n
y
+
φ
-
2
n
x
,
-
2
n
y
≤
lim
n
→
∞
⁡
41
+
5
n
41
-
5
n
1
6
n
L
n
φ
x
,
y
+
φ
-
x
,
-
y
≤
lim
n
→
∞
⁡
L
n
φ
x
,
y
+
φ
-
x
,
-
y
=
0for all x
,
y
∈
V; that is, F is a solution of the functional equation (1). Notice that if F is a solution of the functional equation (1), then the equality(24)
F
x
-
J
F
x
=
5
128
D
F
2
x
,
x
+
20
128
D
F
x
,
x
-
3
128
D
F
-
2
x
,
-
x
-
12
128
D
F
-
x
,
-
x implies that F is a fixed point of J.In the following theorem, we replace condition (12) for φ with another (25) and prove the generalized Hyers-Ulam stability of AQCQ equation (1) by applying a fixed point approach.Theorem 6.
Letf
:
V
→
Y be a mapping, for which there exists a mapping φ
:
V
2
→
[
0
,
∞
) such that inequality (11) holds for all x
,
y
∈
V, and let f
(
0
)
=
0. If there exists a constant 0
<
L
<
1 such that φ satisfies the condition(25)
L
φ
2
x
,
2
y
≥
32
41
-
5
φ
x
,
yfor all x
,
y
∈
V, then there exists a unique solution F
:
V
→
Y of (1) satisfying the inequality(26)
f
x
-
F
x
≤
41
-
5
2
L
2
3
2
2
1
-
L
Φ
xfor all x
∈
V. In particular, F is represented by(27)
F
x
=
lim
n
→
∞
⁡
∑
i
=
0
n
n
C
i
1
0
i
-
16
n
-
i
f
o
x
2
2
n
-
i
+
2
0
i
-
64
n
-
i
f
e
x
2
2
n
-
ifor all x
∈
V.Proof.
Let(
S
,
d
) be the set defined in the proof of Theorem 5. We now consider the mapping J
:
S
→
S defined by (28)
J
g
x
≔
15
g
x
2
+
5
g
-
x
2
-
40
g
x
4
-
24
g
-
x
4 for all x
∈
V. We assert that the equality(29)
J
n
g
x
=
∑
i
=
0
n
n
C
i
1
0
i
-
16
n
-
i
g
o
x
2
2
n
-
i
+
∑
i
=
0
n
n
C
i
2
0
i
-
64
n
-
i
g
e
x
2
2
n
-
iholds for all n
∈
N and x
∈
V. Equality (29) is obviously true when n
=
0. We assume that (29) holds for some n
∈
N
0. Then we have(30)
J
n
+
1
g
x
=
J
n
J
g
x
=
∑
i
=
0
n
n
C
i
1
0
i
-
16
n
-
i
J
g
o
x
2
2
n
-
i
+
∑
i
=
0
n
n
C
i
2
0
i
-
64
n
-
i
J
g
e
x
2
2
n
-
i
=
∑
i
=
0
n
n
C
i
1
0
i
-
16
n
-
i
15
g
o
x
2
2
n
-
1
-
i
+
∑
i
=
0
n
n
C
i
1
0
i
-
16
n
-
i
5
g
o
-
x
2
2
n
-
1
-
i
-
∑
i
=
0
n
n
C
i
1
0
i
-
16
n
-
i
40
g
o
x
2
2
n
-
2
-
i
-
∑
i
=
0
n
n
C
i
1
0
i
-
16
n
-
i
24
g
o
-
x
2
2
n
-
2
-
i
+
∑
i
=
0
n
n
C
i
2
0
i
-
64
n
-
i
15
g
e
x
2
2
n
-
1
-
i
+
∑
i
=
0
n
n
C
i
2
0
i
-
64
n
-
i
5
g
e
-
x
2
2
n
-
1
-
i
-
∑
i
=
0
n
n
C
i
2
0
i
-
64
n
-
i
40
g
e
x
2
2
n
-
2
-
i
-
∑
i
=
0
n
n
C
i
2
0
i
-
64
n
-
i
24
g
e
-
x
2
2
n
-
2
-
i
=
∑
i
=
0
n
n
C
i
1
0
i
+
1
-
16
n
-
i
g
o
x
2
2
n
-
1
-
i
+
∑
i
=
0
n
n
C
i
1
0
i
-
16
n
+
1
-
i
g
o
x
2
2
n
-
2
-
i
+
∑
i
=
0
n
n
C
i
2
0
i
+
1
-
64
n
-
i
g
e
x
2
2
n
-
1
-
i
+
∑
i
=
0
n
n
C
i
2
0
i
-
64
n
+
1
-
i
g
e
x
2
2
n
-
2
-
i
=
∑
i
=
0
n
+
1
n
+
1
C
i
1
0
i
-
16
n
+
1
-
i
g
o
x
2
2
n
+
2
-
i
+
∑
i
=
0
n
+
1
n
+
1
C
i
2
0
i
-
64
n
+
1
-
i
g
e
x
2
2
n
+
2
-
ifor all n
∈
N
0 and x
∈
V, which we get from (29) when we replace n by n
+
1.
Letg
,
h
∈
S and let K
∈
[
0
,
∞
] be an arbitrary constant with d
(
g
,
h
)
≤
K. From the definition of d, we have(31)
J
g
x
-
J
h
x
≤
15
g
x
2
-
h
x
2
+
5
g
-
x
2
-
h
-
x
2
+
40
g
x
4
-
h
x
4
+
24
g
-
x
4
-
h
-
x
4
≤
64
K
Φ
x
4
+
20
K
Φ
x
2
≤
L
2
2
41
-
5
2
32
K
Φ
x
+
20
41
-
5
32
L
K
Φ
x
≤
L
K
Φ
x for all x
∈
V, which implies that (32)
d
J
g
,
J
h
≤
L
d
g
,
h for any g
,
h
∈
S. That is, J is a strictly contractive self-mapping of S with the Lipschitz constant L.
Moreover, by (11), we see that(33)
f
x
-
J
f
x
=
D
f
x
2
,
x
4
+
4
D
f
x
4
,
x
4
≤
φ
x
2
,
x
4
+
4
φ
x
4
,
x
4
≤
41
-
5
2
L
2
3
2
2
φ
2
x
,
x
+
4
φ
x
,
x
≤
41
-
5
2
L
2
3
2
2
Φ
xfor all x
∈
V, which implies that (34)
d
f
,
J
f
≤
41
-
5
2
L
2
3
2
2
<
∞ by the definition of d.
Therefore, according to Theorem1, the sequence {
J
n
f
} converges to the unique fixed point F
:
V
→
Y of J in the set T
=
{
g
∈
S
:
d
(
f
,
g
)
<
∞
}, which is represented by (27) for all x
∈
V. Notice that (35)
d
f
,
F
≤
1
1
-
L
d
f
,
J
f
≤
41
-
5
2
L
2
3
2
2
1
-
L
, which implies (26).
By definition ofF, together with (11) and (25), we have(36)
D
F
x
,
y
=
lim
n
→
∞
⁡
D
J
n
f
x
,
y
=
lim
n
→
∞
⁡
∑
i
=
0
n
n
C
i
1
0
i
-
16
n
-
i
f
o
x
2
2
n
-
i
,
y
2
2
n
-
i
+
∑
i
=
0
n
n
C
i
2
0
i
-
64
n
-
i
f
e
x
2
2
n
-
i
,
y
2
2
n
-
i
≤
lim
n
→
∞
⁡
∑
i
=
0
n
n
C
i
2
1
0
i
1
6
n
-
i
+
2
0
i
6
4
n
-
i
×
φ
x
2
2
n
-
i
,
y
2
2
n
-
i
+
φ
-
x
2
2
n
-
i
,
-
y
2
2
n
-
i
≤
lim
n
→
∞
⁡
∑
i
=
0
n
n
C
i
2
0
i
6
4
n
-
i
41
-
5
n
-
i
L
n
-
i
3
2
n
-
i
×
φ
x
2
n
,
y
2
n
+
φ
-
x
2
n
,
-
y
2
n
≤
lim
n
→
∞
⁡
∑
i
=
0
n
n
C
i
2
0
i
2
41
-
5
n
-
i
φ
x
2
n
,
y
2
n
+
φ
-
x
2
n
,
-
y
2
n
≤
lim
n
→
∞
⁡
20
+
2
41
-
5
n
φ
x
2
n
,
y
2
n
+
φ
-
x
2
n
,
-
y
2
n
≤
lim
n
→
∞
⁡
2
n
41
+
5
n
41
-
5
n
L
n
3
2
n
φ
x
,
y
+
φ
-
x
,
-
y
≤
lim
n
→
∞
⁡
L
n
φ
x
,
y
+
φ
-
x
,
-
y
=
0for all x
,
y
∈
V; that is, F is a solution of the functional equation (1). Notice that if F is a solution of the functional equation (1), then the equality (37)
F
x
-
J
F
x
=
D
F
x
2
,
x
4
+
4
D
F
x
4
,
x
4 implies that F is a fixed point of J.



---

*Source: 8746728-2016-04-06.xml*