# Hyers-Ulam Stability of Iterative Equation in the Class of Lipschitz Functions

Authors: Chao Xia; Wei Song
Journal: Discrete Dynamics in Nature and Society
 (2014)

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## Abstract

Hyers-Ulam stability is a basic sense of stability for functional equations. In the present paper we discuss the Hyers-Ulam stability of a kind of iterative equations in the class of Lipschitz functions. By the construction of a uniformly convergent sequence of functions we prove that, for every approximate solution of such an equation, there exists an exact solution near it.


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## Body

## 1. Introduction

It is pointed out that the theory of Ulam’s type stability (also quite often connected, e.g., with the names of Bourgin, Ga
˘vruţa, Aoki, Hyers, and Rassias) is a very popular subject of investigations at the moment in [1]. For more details of Ulam’s type stability, we refer the audience to [1–3]. In [1], the authors present a survey of some selected recent developments (results and methods) in the theory of Ulam’s type stability, such as the Forti method and the methods of fixed points and stability in non-Archimedean spaces. These results and methods have not been treated at all or have been treated only marginally. The very book of Jung [2] covers and offers almost all classic results on the Hyers-Ulam-Rassias stability in an integrated and self-contained fashion. And the authors of [3] discussed Hyers-Ulam stability for functional equations in single variable, including the forms of linear functional equations, nonlinear functional equations, and iterative equations. And they also clarified the relation between Hyers-Ulam stability and other senses of stability which are used for functional equations.LetC
(
X
,
X
) be the set of all continuous self-mappings on a topological spaceX. For any f
∈
C
(
X
,
X
), let f
m denote themth iteration of f; that is, f
m
=
f
∘
f
m
-
1, f
0
=
id, m
=
1,2
,
…. Equations having iterations as their main operation, that is, including iterations of the unknown mapping, are called iterative equations. It is one of the most interesting classes of functional equations [4–7]. As a natural generalization of the problem of iterative roots, iterative equations of the following form:
(1)
λ
1
f
(
x
)
+
λ
2
f
2
(
x
)
+
⋯
+
λ
n
f
n
(
x
)
=
F
(
x
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
x
∈
I
=
[
a
,
b
]
,
are known as polynomial-like iterative equations. Here n
≥
2 is an integer, λ
i
∈
R
(
i
=
1,2
,
…
,
n
), F
:
I
→
R is a given mapping, and f
:
I
→
I is unknown. As mentioned in [8, 9], polynomial-like iterative equations are important not only in the study of functional equations but also in the study of dynamical systems.In 1986, Zhang [10] constructed an interesting operator called “structural operator” for (1) and used the fixed point theory in Banach space to get the solutions of (1). From then on (1) and other types of equations were discussed extensively by employing this idea (see [8, 9, 11–16] and references therein). In 2002, by means of a modification of Zhang’s method applied in [10], Kulczycki and Tabor [11] investigated the existence of Lipschitzian solutions of the iterative equation
(2)
∑
i
=
1
∞
‍
a
i
f
i
(
x
)
=
F
(
x
)
,
x
∈
B
,
where B is a compact convex subset of R
N and F
:
B
→
B is a given Lipschitz function. They generalized the results of Zhang [10] to a more general case.In 2002, Xu and Zhang [17] discussed the Hyers-Ulam stability for a nonlinear iterative equation which includes the polynomial-like iterative equation. In 2003, Agarwal et al. [3] investigated the Hyers-Ulam stability for a general form of iterative equations which include the polynomial-like iterative equation with variable coefficients. Motivated by the above results, in this paper, we will discuss the Hyers-Ulam stability of (2). As in [11], we give a result on the Hyers-Ulam stability of a functional equation with a more general form firstly. And, by means of this result, the Hyers-Ulam stability of (2) is investigated. In fact, we want to generalize the result of Xu and Zhang [17] to high-dimensional case.


## 2. Basic Assumptions, Definitions, and Lemmas

LetB be a compact convex subset of R
N, N
∈
Z
+, with nonempty interior. Let
(3)
C
(
B
,
R
N
)
=
{
f
:
B
⟶
R
N
∣
f
is
continuous
}
.
In C
(
B
,
R
N
), we use the supremum norm
(4)
∥
f
∥
B
=
sup
⁡
x
∈
B
∥
f
(
x
)
∥
,
for
∀
f
∈
C
(
B
,
R
N
)
,
where ∥
·
∥ denotes the usual metric of R
N. Obviously, C
(
B
,
R
N
) is a complete metric space.Definition 1 (see [11]).
LetB be a convex compact subset of R
N with nonempty interior. For m
∈
[
0,1
], and M
∈
[
1
,
∞
], one defines
(5)
Li
p
B
(
m
,
M
)
≔
{
f
:
B
⟶
B
∣
f
is
continuous
,
f
|
∂
B
=
id
|
∂
B
,
m
∥
x
-
y
∥
≤
∥
f
(
x
)
-
f
(
y
)
∥
≤
M
∥
x
-
y
∥
nnnnn
∀
x
,
y
∈
B
,
n
n
n
n
n
m
∥
x
-
y
∥
≤
∥
f
(
x
)
-
f
(
y
)
∥
≤
M
∥
x
-
y
∥
}
,
where ∂
B denotes the boundary of B.For a given functionf
∈
C
(
B
,
R
N
), we define its Lipschitz constant by
(6)
lip
(
f
)
≔
inf
⁡
{
M
≥
0
∣
∥
f
(
x
)
-
f
(
y
)
∥
≤
M
∥
x
-
y
∥
∀
x
,
y
∈
B
}
.Lemma 2 (see [11]).
Letk
∈
[
0,1
], K
∈
[
1
,
∞
], and f
∈
Lip
B
(
k
,
K
) be arbitrary. One assumes additionally that f is invertible. Then
(7)
f
-
1
∈
Lip
B
(
1
K
,
1
k
)
.Lemma 3 (see [11]).
For everym
∈
(
0,1
], the mapping
(8)
I
:
Lip
B
(
m
,
∞
)
∋
f
⟶
f
-
1
∈
Lip
B
(
0
,
1
m
)
is well defined with Lipschitz constant1
/
m.Lemma 4 (see [11]).
ForK, M
<
∞, F
∈
Lip
B
(
0
,
K
), the mapping defined by
(9)
S
F
:
Lip
B
(
0
,
M
)
∋
f
⟶
f
∘
F
∈
Lip
B
(
0
,
K
M
)
is Lipschitz with constant 1.Lemma 5.
Letm
∈
[
0,1
], and M
∈
[
1
,
∞
] be given. For any f
∈
Lip
B
(
m
,
M
), f
:
B
→
B is surjective.Proof.
Note thatB is homeomorphic toD
N, where D
N
=
{
x
∈
R
N
:
∥
x
∥
≤
1
} is the unit ball of R
N. Without any loss of generality, let B
=
D
N. Since f
|
∂
B
=
id
|
∂
B, set f
|
∂
B
=
f
0; we have deg
⁡
(
f
0
)
=
1, where deg
⁡
(
f
0
) denotes the degree of f
0. Suppose that f is not surjective. Let x
0
∈
B
∖
f
(
B
). Since f
|
∂
B
=
id
|
∂
B, we know that x
0
∉
∂
B; then there exists a retraction mapping r
:
B
∖
{
x
0
}
→
∂
B. Thus r
∘
f
:
B
→
∂
B is a continuous mapping and f
0
=
(
r
∘
f
)
|
∂
B. This means that deg
⁡
(
f
0
)
=
0, a contradiction. So f is surjective.Lemma 6.
Iff, g are homeomorphisms of B with Lipschitz constant M, then
(10)
∥
f
-
g
∥
B
≤
M
∥
f
-
1
-
g
-
1
∥
B
.Proof.
Note that(11)
0
=
∥
f
-
1
∘
f
-
g
-
1
∘
g
∥
B
=
∥
f
-
1
∘
f
-
g
-
1
∘
f
+
g
-
1
∘
f
-
g
-
1
∘
g
∥
B
.
So we have
(12)
f
-
1
∘
f
-
g
-
1
∘
f
=
-
g
-
1
∘
f
+
g
-
1
∘
g
.
Furthermore
(13)
∥
f
-
1
-
g
-
1
∥
B
=
∥
f
-
1
∘
f
-
g
-
1
∘
f
∥
B
=
∥
g
-
1
∘
f
-
g
-
1
∘
g
∥
B
≥
1
M
∥
f
-
g
∥
B
.
This means that ∥
f
-
g
∥
B
≤
M
∥
f
-
1
-
g
-
1
∥
B.Lemma 7.
Suppose thatf, g
:
B
→
B are continuous mappings and lip
(
f
)
=
M, whereMis a positive constant. Then
(14)
∥
f
k
-
g
k
∥
B
≤
∑
j
=
0
k
-
1
‍
M
j
∥
f
-
g
∥
B
,
∀
k
=
1,2
,
…
.Proof.
We use induction to prove this lemma. Fork
=
1, ∥
f
-
g
∥
B
≤
∥
f
-
g
∥
B. Suppose that this is true, for k; then
(15)
∥
f
k
+
1
-
g
k
+
1
∥
B
≤
∥
f
∘
f
k
-
f
∘
g
k
∥
B
+
∥
f
∘
g
k
-
g
∘
g
k
∥
B
≤
M
∥
f
k
-
g
k
∥
B
+
∥
f
-
g
∥
B
≤
M
∑
j
=
0
k
-
1
‍
M
j
∥
f
-
g
∥
B
+
∥
f
-
g
∥
B
=
∑
j
=
0
k
‍
M
j
∥
f
-
g
∥
B
.LetA be a subset of Li
p
B
(
0
,
∞
) and satisfy(A1):
for allf, g
∈
A, f
∘
g
∈
A;
(A2):
iff
∈
A is invertible, then f
-
1
∈
A;
(A3):
A is closed and convex.And let(16)
A
=
{
A
⊂
Li
p
B
(
0
,
∞
)
∣
(A1),
(
A2
)
,
and (A3) hold
}
.
It is easy to see that Li
p
B
(
0
,
∞
)
∈
A.LetA
∈
A and let P
:
A
→
A be a map. For any F
∈
A we will consider a functional equation of the following form:
(17)
P
(
f
)
∘
f
=
F
,
where f
∈
A is unknown.Definition 8.
If, for everyg
∈
Li
p
B
(
0
,
∞
) such that ∥
F
-
P
(
g
)
∘
g
∥
B
≤
δ, there exists a solution f of (17) such that ∥
f
-
g
∥
B
≤
K
δ, where K
>
0 is a constant and K does not depend on the choice of g, then (17) is said to have Hyers-Ulam stability.


## 3. Main Results

Lemma 9.
LetB be a convex compact subset of R
N with nonempty interior. Let A
∈
A and F
∈
A with lip
(
F
)
≤
M
0. Let P
:
A
→
A be given. If there exists a decreasing function α
:
[
1
,
∞
]
→
[
0,1
] and a constant M
∈
[
1
,
∞
] such that (P1):
for anyf
∈
A, P
(
f
)
∈
Lip
B
(
α
(
lip
(
f
)
)
,
∞
)
;
(P2):
M
·
α
(
M
)
≥
M
0.
Then the following two facts hold:(c1):
for anyf
∈
A
∩
Lip
B
(
0
,
M
), P
(
f
) is a homeomorphism with lip
(
(
P
(
f
)
)
-
1
)
≤
1
/
α
(
M
);
(c2):
for any givenf
0
∈
A
∩
Lip
B
(
0
,
M
), the sequence
(18)
f
k
=
S
F
∘
I
∘
P
(
f
k
-
1
)
,
k
=
1,2
,
…
,
is well defined and eachf
k
∈
A
∩
Lip
B
(
0
,
M
) is surjective.Proof.
(c1): since F
|
∂
B
=
id
|
∂
B, we get that M
0
≥
1. By inequality (P2) and the range of the function α, we know that both α
(
M
)
∈
(
0,1
] and M
≥
M
0 hold. Therefore F
∈
A
∩
Li
p
B
(
0
,
M
), which implies that A
∩
Li
p
B
(
0
,
M
) is nonempty. For any x
1
≠
x
2
∈
B and any f
∈
A
∩
Li
p
B
(
0
,
M
), since lip
(
f
)
≤
M and α is decreasing, we have
(19)
∥
P
(
f
)
(
x
1
)
-
P
(
f
)
(
x
2
)
∥
≥
α
(
lip
(
f
)
)
∥
x
1
-
x
2
∥
≥
α
(
M
)
∥
x
1
-
x
2
∥
>
0
.
This means that P
(
f
) is injective. Since P
(
f
)
∈
A, by Lemma 5, we obtain that P
(
f
) is surjective. Thus P
(
f
) is a homeomorphism. And lip
(
(
P
(
f
)
)
-
1
)
≤
1
/
α
(
M
) follows from Lemma 2.
(c2): by Lemma 3, the mapping
(20)
I
:
Li
p
B
(
α
(
M
)
,
∞
)
∋
g
⟶
g
-
1
∈
Li
p
B
(
0
,
1
α
(
M
)
)
is continuous. By Lemma 4 and assumptions of F, the mapping
(21)
S
F
:
Li
p
B
(
0
,
1
α
(
M
)
)
∋
g
⟶
g
∘
F
∈
Li
p
B
(
0
,
M
0
α
(
M
)
)
is well defined and continuous.
Now, by the assumption (P1) and the fact P
:
A
→
A, it is easy to see that, for every g
∈
Li
p
B
(
0
,
M
)
∩
A,
(22)
P
(
g
)
∈
Li
p
B
(
α
(
M
)
,
∞
)
∩
A
.
Since (A2) holds and by Lemma 2, we know that all elements of Li
p
B
(
α
(
M
)
,
∞
) are invertible. Moreover we can get that
(23)
I
∘
P
(
g
)
∈
Li
p
B
(
0
,
1
α
(
M
)
)
∩
A
,
and by means of (A1) we have
(24)
S
F
∘
I
∘
P
(
g
)
∈
Li
p
B
(
0
,
M
0
α
(
M
)
)
∩
A
.
Finally, by inequality (P2), we obtain that the mapping
(25)
S
F
∘
I
∘
P
:
Li
p
B
(
0
,
M
)
∩
A
⟶
Li
p
B
(
0
,
M
)
∩
A
is well defined.
Take anyf
0
∈
A
∩
Li
p
B
(
0
,
M
). By means of the mapping (25), it is easy to see that the sequence
(26)
f
k
=
S
F
∘
I
∘
P
(
f
k
-
1
)
,
k
=
1,2
,
…
,
is well defined and eachf
k
∈
A
∩
Li
p
B
(
0
,
M
). By virtue of Lemma 5, it is easy to see that all f
k are surjective.Theorem 10.
LetB be a convex compact subset of R
N with nonempty interior. Let A
∈
A and F
∈
A with lip
(
F
)
≤
M
0. Let P
:
A
→
A be given and let there exist a decreasing function α
:
[
1
,
∞
]
→
[
0,1
] and a constant M
∈
[
1
,
∞
] such that (P1) and (P2) of Lemma 9 hold. Moreover, if there is a constant 0
<
η
<
α
(
M
) such that
(27)
∥
P
(
g
1
)
-
P
(
g
2
)
∥
B
≤
η
∥
g
1
-
g
2
∥
B
,
∀
g
1
,
g
2
∈
A
∩
Lip
B
(
0
,
M
)
,
then, for any g
∈
A
∩
Lip
B
(
0
,
M
) with
(28)
∥
F
-
P
(
g
)
∘
g
∥
B
≤
δ
,
where δ
>
0 is a constant, there exists a unique continuous solution f
∈
A
∩
Lip
B
(
0
,
M
) of (17) such that
(29)
∥
f
-
g
∥
B
≤
K
δ
,
where
(30)
K
=
1
α
(
M
)
-
η
.Proof.
Since all the assumptions of Lemma9 hold, so we can construct a sequence {
f
k
} of functions as follows. We take
(31)
f
0
=
g
first and then by means of equality (18) repeatedly, define
(32)
f
k
=
S
F
∘
I
∘
P
(
f
k
)
,
k
=
1,2
,
…
.
By the conclusions (c1) and (c2) of Lemma 9, both P
(
f
k
-
1
) and f
k are well defined for all k
>
1. And each P
(
f
k
) is a homeomorphism of B onto itself with lip
(
(
P
(
f
k
)
)
-
1
)
≤
1
/
α
(
M
).
Now we claim that both(33)
∥
f
k
-
f
k
-
1
∥
B
≤
1
α
(
M
)
(
η
α
(
M
)
)
k
-
1
δ
,
∥
F
-
P
(
f
k
)
∘
f
k
∥
B
≤
(
η
α
(
M
)
)
k
δ
hold for k
=
1,2
,
…. We prove these two inequalities inductively.
Fork
=
1, by virtue of the definition of f
1, we have
(34)
∥
f
1
-
f
0
∥
B
=
∥
S
F
∘
I
∘
P
(
f
0
)
-
f
0
∥
B
=
∥
(
P
(
f
0
)
)
-
1
∘
F
-
(
P
(
f
0
)
)
-
1
∘
(
P
(
f
0
)
)
∘
f
0
∥
B
≤
1
α
(
M
)
∥
F
-
(
P
(
f
0
)
)
∘
f
0
∥
B
=
1
α
(
M
)
δ
.
And noting that F
=
P
(
f
k
-
1
)
∘
f
k and f
k
:
B
→
B is surjective for k
=
1,2
,
…, we have
(35)
∥
F
-
P
(
f
1
)
∘
f
1
∥
B
=
∥
P
(
f
0
)
∘
f
1
-
P
(
f
1
)
∘
f
1
∥
B
=
∥
P
(
f
0
)
-
P
(
f
1
)
∥
B
≤
η
∥
f
1
-
f
0
∥
B
=
η
α
(
M
)
δ
.
Assume that they are true for integerk. Then, by the same arguments, we can get that
(36)
∥
f
k
+
1
-
f
k
∥
B
=
∥
(
P
(
f
k
)
)
-
1
∘
F
-
(
P
(
f
k
)
)
-
1
∘
(
P
(
f
k
)
)
∘
f
k
∥
B
≤
1
α
(
M
)
∥
F
-
(
P
(
f
k
)
)
∘
f
k
∥
B
≤
1
α
(
M
)
(
η
α
(
M
)
)
k
δ
,
∥
F
-
P
(
f
k
+
1
)
∘
f
k
+
1
∥
B
=
∥
P
(
f
k
)
∘
f
k
+
1
-
P
(
f
k
+
1
)
∘
f
k
+
1
∥
B
=
∥
P
(
f
k
)
-
P
(
f
k
+
1
)
∥
B
≤
η
∥
f
k
+
1
-
f
k
∥
B
=
(
η
α
(
M
)
)
k
+
1
δ
.
Thus (33) is proved by induction.
For any positive integersk and s with k
>
s, we have
(37)
∥
f
k
-
f
s
∥
B
≤
∥
f
k
-
f
k
-
1
∥
B
+
∥
f
k
-
1
-
f
k
-
2
∥
B
+
⋯
+
∥
f
s
+
1
-
f
s
∥
B
≤
1
α
(
M
)
(
η
α
(
M
)
)
k
-
1
δ
+
1
α
(
M
)
(
η
α
(
M
)
)
k
-
2
δ
+
⋯
+
1
α
(
M
)
(
η
α
(
M
)
)
s
δ
=
δ
α
(
M
)
·
(
η
/
α
(
M
)
)
s
-
(
η
/
α
(
M
)
)
k
1
-
η
/
α
(
M
)
.
Since α
(
M
)
>
η
>
0, it follows from inequality (37) that
(38)
∥
f
k
-
f
s
∥
B
⟶
0
,
as
k
>
s
⟶
∞
.
As a Cauchy sequence, {
f
k
}
k
=
0
∞ converges uniformly in the Banach space C
(
B
,
R
N
). Let
(39)
lim
⁡
k
→
∞
f
k
=
f
.
Clearly, f
∈
A
∩
Li
p
B
(
0
,
M
). From
(40)
∥
F
-
P
(
f
)
∘
f
∥
B
=
lim
⁡
k
→
∞
∥
F
-
P
(
f
k
)
∘
f
k
∥
B
=
lim
⁡
k
→
∞
(
η
α
(
M
)
)
k
δ
=
0
,
we know that f is a solution of (17). Furthermore,
(41)
∥
f
-
g
∥
B
=
lim
⁡
k
→
∞
∥
f
k
-
f
0
∥
B
≤
lim
⁡
k
→
∞
{
∥
f
k
-
f
k
-
1
∥
B
+
⋯
+
∥
f
1
-
f
0
∥
B
}
≤
lim
⁡
k
→
∞
{
1
α
(
M
)
(
η
α
(
M
)
)
k
-
1
δ
+
1
α
(
M
)
(
η
α
(
M
)
)
k
-
2
δ
n
n
n
n
n
n
n
n
n
n
(
η
α
(
M
)
)
k
-
2
+
⋯
+
1
α
(
M
)
δ
}
=
1
α
(
M
)
-
η
δ
.
This proves (29).
Concerning uniqueness, we assume that there is another continuous solutionϕ
∈
A
∩
Li
p
B
(
0
,
M
) for (17), such that
(42)
∥
ϕ
-
g
∥
B
≤
ɛ
,
where ɛ
>
0 only depends on δ. By Lemma 5, we know that F
:
B
→
B is also surjective. By means of Lemma 6 and the assumption that P is Lipschitzian, it follows that
(43)
∥
f
-
ϕ
∥
B
=
∥
(
P
(
f
)
)
-
1
∘
F
-
(
P
(
ϕ
)
)
-
1
∘
F
∥
B
=
∥
(
P
(
f
)
)
-
1
-
(
P
(
ϕ
)
)
-
1
∥
B
≤
1
α
(
M
)
∥
P
(
f
)
-
P
(
ϕ
)
∥
B
≤
η
α
(
M
)
∥
f
-
ϕ
∥
B
.
This implies that
(44)
(
1
-
η
α
(
M
)
)
∥
f
-
ϕ
∥
B
≤
0
.
However, α
(
M
)
>
η
>
0; this implies that ∥
f
-
ϕ
∥
B
=
0; that is, f
=
ϕ.


## 4. Iterative Equations inR
N

Theorem 11.
LetB be a convex compact subset of R
N with nonempty interior. Let F
∈
Lip
B
(
0
,
∞
) be arbitrary with lip
(
F
)
≤
M
0. Assume that a
k
∈
[
0,1
]
,
k
=
0,1
,
2
,
…, and there exists a constant p
∈
[
1
,
∞
] such that(a1):
∑
k
=
0
∞
a
k
=
1
;
(a2):
a
0
p
>
a
0
p
-
∑
k
=
1
∞
a
k
p
k
+
1
≥
M
0
;
(a3):
∑
k
=
1
∞
a
k
∑
j
=
0
k
-
1
‍
p
j
+
∑
k
=
1
∞
‍
a
k
p
k
<
a
0hold. Then the equation
(45)
∑
k
=
0
∞
‍
a
k
f
k
+
1
(
x
)
=
F
(
x
)
,
x
∈
B
,
has Hyers-Ulam stability.Proof.
LetA
=
Li
p
B
(
0
,
∞
). For ∀
f
∈
A, define
(46)
P
(
f
)
=
∑
k
=
0
∞
‍
a
k
f
k
.
We claim that, for ∀
f
∈
A, P
(
f
)
∈
A. For ∀
x
∈
B, by the compactness and convexity of B and the assumption (a1), we have
(47)
∑
k
=
0
∞
‍
a
k
f
k
(
x
)
∈
co
⁡
{
x
,
f
(
x
)
,
f
2
(
x
)
,
…
}
¯
⊂
B
,
where co
⁡
{
x
,
f
(
x
)
,
f
2
(
x
)
,
…
} is the convex hull of the set {
x
,
f
(
x
)
,
f
2
(
x
)
,
…
}
.Moreover, by means of the assumption (a1) and the definition of the mapping P and the set A, it is easy to see that P
(
f
)
|
∂
B
=
id
|
∂
B. Now we prove the continuity of P
(
f
). By the compactness of B, we first denote
(48)
L
=
max
⁡
x
∈
B
{
∥
x
∥
}
.
By the assumption (a1), for any ϵ
>
0, there exists a positive integer l
0 such that
(49)
∑
k
=
l
0
+
1
∞
‍
a
k
≤
ϵ
4
L
.
Note that each f
k, k
=
0,1
,
2
,
…, l
0, is continuous. Then, for any x
∈
B, there is a δ
>
0 such that when ∥
y
-
x
∥
<
δ the fact
(50)
∥
f
k
(
y
)
-
f
k
(
x
)
∥
≤
ϵ
2
(
l
0
+
1
)
,
k
=
0,1
,
2
,
…
,
l
0
,
holds. Thus we can get that
(51)
∥
P
f
(
y
)
-
P
f
(
x
)
∥
≤
∑
k
=
0
l
0
‍
a
k
∥
f
k
(
y
)
-
f
k
(
x
)
∥
+
∑
k
=
l
0
+
1
∞
‍
a
k
∥
f
k
(
y
)
-
f
k
(
x
)
∥
≤
∑
k
=
0
l
0
‍
ϵ
2
(
l
0
+
1
)
+
2
L
∑
k
=
l
0
+
1
∞
‍
a
k
=
ϵ
.
This means that the fact P
:
A
→
A holds.
By virtue of the mappingP, (45) can be rewritten as
(52)
P
(
f
)
∘
f
=
F
.
SinceF
|
∂
B
=
id
|
∂
B, this implies that M
0
≥
1
.And by the assumption (a2), we see that a
0
>
0. For x
∈
[
1
,
∞
), define
(53)
α
(
x
)
=
max
⁡
{
a
0
-
∑
k
=
1
∞
‍
a
k
x
k
,
0
}
.
Since α
(
x
)
≤
a
0
≤
1 we obtain that α
:
[
1
,
∞
]
→
[
0,1
] and it is easy to see that α is decreasing.
For anyx
∈
[
1
,
∞
) and any g
∈
Li
p
B
(
0
,
∞
) with lip
(
g
)
=
x, we claim that P
(
g
)
∈
Li
p
B
(
α
(
x
)
,
∞
). It is enough to show that, for ∀
b
,
c
∈
B,
(54)
α
(
x
)
∥
b
-
c
∥
≤
∥
P
(
g
)
(
b
)
-
P
(
g
)
(
c
)
∥
.
But
(55)
∥
P
(
g
)
(
b
)
-
P
(
g
)
(
c
)
∥
=
∥
∑
k
=
0
∞
‍
a
k
(
g
k
(
b
)
-
g
k
(
c
)
)
∥
≥
a
0
∥
b
-
c
∥
-
∑
k
=
1
∞
‍
a
k
∥
g
k
(
b
)
-
g
k
(
c
)
∥
≥
a
0
∥
b
-
c
∥
-
∑
k
=
1
∞
‍
a
k
x
k
∥
b
-
c
∥
=
(
a
0
-
∑
k
=
1
∞
‍
a
k
x
k
)
∥
b
-
c
∥
=
α
(
x
)
∥
b
-
c
∥
.
It follows from the assumption(a2) that
(56)
p
·
α
(
p
)
=
a
0
p
-
∑
k
=
1
∞
‍
a
k
p
k
+
1
≥
M
0
.
Thus (P1) and (P2) of Lemma 9 also hold. Moreover, we can get that
(57)
∥
P
(
f
)
-
P
(
g
)
∥
B
=
∥
∑
k
=
0
∞
‍
a
k
(
f
k
-
g
k
)
∥
B
≤
∑
k
=
0
∞
‍
a
k
∥
f
k
-
g
k
∥
B
≤
∑
k
=
1
∞
‍
a
k
∑
j
=
0
k
-
1
‍
p
j
∥
f
-
g
∥
B
hold. Let η
=
∑
k
=
1
∞
‍
a
k
∑
j
=
0
k
-
1
‍
p
j. By the assumption (a2), we know that there must be a positive integer k
0 such that a
k
0
≠
0. So we know that η
>
0. Furthermore, by the assumption (a3), we have 0
<
η
<
α
(
p
). Then, by Theorem 10, we get the conclusion.Example 12.
ForF
(
x
)
=
(
x
1
,
x
2
,
x
3
,
…
,
x
N
) + dist
⁡
(
x
,
S
N
-
1
)
(
1,0
,
0
,
…
,
0
) = (
1
+
x
1
-
x
1
2
+
x
2
2
+
⋯
+
x
N
2
,
x
2
,
…
,
x
N
), where x
=
(
x
1
,
x
2
,
…
,
x
N
)
∈
D
N and dist
⁡
(
x
,
S
N
-
1
) denotes the distance of the point x from S
N
-
1, obviously, F
|
S
N
-
1
=
id
|
S
N
-
1. By simple calculation, we get that, for any x, y
∈
D
N, the following two inequalities:
(58)
∥
F
(
x
)
-
F
(
y
)
∥
≤
2
∥
x
-
y
∥
,
∥
F
(
x
)
∥
≤
∥
x
∥
+
dist
⁡
(
x
,
S
N
-
1
)
=
1
hold. This implies that F
∈
Li
p
D
N
(
0,2
). Then we consider the equation
(59)
254
255
f
(
x
)
+
∑
i
=
1
∞
‍
(
1
256
)
i
f
i
+
1
(
x
)
=
F
(
x
)
,
x
∈
D
N
,
F
:
D
N
⟶
D
N
,
F
|
S
N
-
1
=
id
|
S
N
-
1
.
Denote a
0
=
254
/
255, a
k
=
(
1
/
256
)
i, k
=
1,2
,
…. Define α
(
x
)
=
max
⁡
⁡
{
(
254
/
255
) − ∑
i
=
1
∞
(
1
/
256
)
i
x
i
,
0
}. Let M
0
=
2 and p
=
3. Then we have
(60)
∑
k
=
0
∞
‍
a
k
=
1
,
254
255
×
3
>
3
α
(
3
)
=
190491
64515
>
2
≥
lip
(
F
)
,
254
255
>
∑
k
=
1
∞
‍
(
1
256
)
k
∑
j
=
0
k
-
1
‍
3
j
+
∑
i
=
1
∞
‍
(
3
256
)
i
=
1021
64515
.
By Theorem 11, (59) has Hyers-Ulam stability.



---

*Source: 454569-2014-06-01.xml*