# Ulam-Hyers Stability for MKC Mappings via Fixed Point Theory

Authors: Anisa Mukhtar Hassan; Erdal Karapınar; Hamed H. Alsulami
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

We consider some extension of MKC mappings in the framework of complete dislocated metric spaces. Besides the theoretical results, we also consider some illustrative examples. Further, we state and prove that our main results improved the related results in the frame of generalized Ulam-Hyers stability theory.


---

## Body

## 1. Introduction and Preliminaries

At the beginning of the nineteenth century, Fréchet [1] explicated the concept of “distance” formally by introducing the notion of metric. This conception has been generalized, extended, and improved in several ways, such as fuzzy metric, quasi-metric, partial metric, G-metric, D-metric, b-metric, 2-metric, ultrametric, and modular metric. Among them, the concept of dislocated metric, defined by Hitzler [2], deserves a special interest due to its application potential in several quantitative sciences, in particular, computer science. The concept of dislocated metric spaces was rediscovered by Amini-Harandi [3] as a “metric-like space.”In this paper, we continue to use dislocated metric space instead of metric-like space.Definition 1 (see [2]).
For a nonempty setM, a dislocated metric is a function ρ
:
M
×
M
→
R
0
+ such that, for all ξ
,
η
,
ζ
∈
M,ρ
1 if ρ
ξ
,
η
=
0, then ξ
=
η;
ρ
2
ρ
ξ
,
η
=
ρ
η
,
ξ;
ρ
3
ρ
ξ
,
η
≤
ρ
ξ
,
ζ
+
ρ
ζ
,
η.
Moreover, the pair(
M
,
ρ
) is called dislocated metric space (DMS).The topology of dislocated spaces and basic topological properties (convergence, completeness, etc.) was also considered in [2]. Among them, let us recall the essential notions. Each dislocated metric ρ on a nonempty set M has a topology τ
ρ that was generated by the family of open balls(1)
O
ε
ξ
=
η
∈
M
:
ρ
ξ
,
η
-
ρ
ξ
,
ξ
<
ε
∀
ξ
∈
M
,
ε
>
0
.In context of the DMS(
M
,
ρ
), a sequence {
ξ
n
} converges to a point ξ
∈
M if the following limit exists (and is finite):(2)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
=
ρ
ξ
,
ξ
.Moreover, a sequence{
ξ
n
} is said to be Cauchy if the limit(3)
L
1
=
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
mexists and is finite. Furthermore, if L
1
=
0 in (3), then we say that {
ξ
n
} is a 0-Cauchy sequence.As it is expected, a pair(
M
,
ρ
) is called complete DMS if, for each Cauchy sequence {
ξ
n
}, there is some ξ
∈
M such that(4)
L
2
=
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
=
ρ
ξ
,
ξ
=
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
m
.Analogously, a pair(
M
,
ρ
) is called 0-complete DMS if, for each 0-Cauchy sequence {
ξ
n
}, it converges to a point ξ
∈
M such that L
2
=
0 in (4).Let(
M
,
ρ
) and (
K
,
σ
) be DMSs. A mapping T
:
M
→
K is called continuous if (5)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
=
ρ
ξ
,
ξ
=
l
i
m
n
,
m
→
∞
ρ
ξ
n
,
ξ
m
; then, we have (6)
l
i
m
n
→
∞
ρ
T
ξ
n
,
T
ξ
=
ρ
T
ξ
,
T
ξ
=
l
i
m
n
,
m
→
∞
ρ
T
ξ
n
,
T
ξ
m
.Remark 2.
Each partial metric space is a DMS. Notice also that every metric is necessarily partial metric.Definition 3 (see [3]).
Let(
M
,
ρ
) be a DMS and let S be a subset of M. One says that S is an open subset of M, if for all ξ
∈
M there exists r
>
0 such that O
r
(
ξ
)
⊆
S
. Also, F
⊆
X is a closed subset of M if (
M
∖
F
) is an open subset of M.Lemma 4 (see [4]).
For a DMS(
M
,
ρ
), one has the following observations:(A)
Ifρ
(
ξ
,
η
)
=
0, then ρ
(
ξ
,
ξ
)
=
ρ
(
η
,
η
)
=
0.
(B)
For a sequence{
ξ
n
} with l
i
m
n
→
∞
ρ
(
ξ
n
,
ξ
n
+
1
)
=
0, one has(7)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
=
l
i
m
n
→
∞
ρ
ξ
n
+
1
,
ξ
n
+
1
=
0
.
(C)
Ifξ
≠
η, then ρ
(
ξ
,
η
)
>
0
.
(D)
ρ
(
ξ
,
ξ
)
≤
2
/
n
∑
i
=
1
i
=
n
‍
ρ
(
ξ
,
ξ
i
) holds for all ξ
i
,
ξ
∈
M, where 1
≤
i
≤
n.
(E)
LetV be a closed subset of M and let {
ξ
n
} be a sequence in V. If ξ
n
→
ξ as n
→
∞, then ξ
∈
V.
(F)
For a sequence{
ξ
n
} in M such that ξ
n
→
ξ as n
→
∞ with ρ
(
ξ
,
ξ
)
=
0, then l
i
m
n
→
∞
ρ
(
ξ
n
,
η
)
=
ρ
(
ξ
,
η
) for all η
∈
M.Definition 5 (see [5]).
LetM be a nonempty set and let α
:
M
×
M
→
[
0
,
∞
) be a mapping. A self-mapping T
:
M
→
M is called α-admissible if the following implications hold:(8)
α
ξ
,
η
≥
1
⟹
α
T
ξ
,
T
η
≥
1
∀
ξ
,
η
∈
M
.
Anα-admissible T is said to be triangular α-admissible if(9)
α
ξ
,
η
≥
1
,
α
η
,
ζ
≥
1
⟹
α
ξ
,
ζ
≥
1
,
∀
ξ
,
η
,
ζ
∈
M
.The letterN represents the set of positive integers and N
0
=
N
∪
{
0
}. Further, the nonnegative real numbers will be denoted by R
0
+
=
[
0
,
∞
).


### 1.1. (c)-Comparison Functions

LetΨ be the family of nondecreasing functions ψ
:
R
0
+
→
R
0
+ satisfying the following conditions:(
Ψ
1
)
∑
n
=
1
+
∞
‍
ψ
n
(
t
)
<
∞ for all t
>
0, where ψ
n is the nth iterate of ψ.
(
Ψ
2
)  ψ
-
1
(
0
)
=
0.It is well known that ifψ is a (c)-comparison function, then ψ
(
t
)
<
t for any t
>
0 and ψ is continuous at 0 (see, e.g., [6]).


## 1.1. (c)-Comparison Functions

LetΨ be the family of nondecreasing functions ψ
:
R
0
+
→
R
0
+ satisfying the following conditions:(
Ψ
1
)
∑
n
=
1
+
∞
‍
ψ
n
(
t
)
<
∞ for all t
>
0, where ψ
n is the nth iterate of ψ.
(
Ψ
2
)  ψ
-
1
(
0
)
=
0.It is well known that ifψ is a (c)-comparison function, then ψ
(
t
)
<
t for any t
>
0 and ψ is continuous at 0 (see, e.g., [6]).


## 2. Fixed Point Theorems for(
α
,
ψ
)-Meir-Keeler Contractions on Dislocated Metric Spaces

### 2.1.(
α
,
ψ
)-Meir-Keeler Contractive Mapping

We introduce the following notion which is an improved version of Meir-Keeler contractive mapping.Definition 6.
Let(
M
,
ρ
) be a DMS. A self-mapping T
:
M
→
M is called (
α
,
ψ
)-Meir-Keeler contractive (in brief, (
α
,
ψ
)-MKC) if there exist two functions ψ
∈
Ψ and α
:
M
×
M
→
R
0
+ satisfying the following condition.
For eachε
>
0, there exists δ
>
0 such that (10)
ε
≤
ψ
ρ
ξ
,
η
<
ε
+
δ
i
m
p
l
i
e
s
α
ξ
,
η
ρ
T
ξ
,
T
η
<
ε
.Notice that, for(
α
,
ψ
)-MKC mapping T
:
M
→
M, we have (11)
α
ξ
,
η
ρ
T
ξ
,
T
η
≤
ψ
ρ
ξ
,
η
,
f
o
r
a
n
y
ξ
,
η
∈
M
.In what follows, we will state and prove the first main result of this section.Theorem 7.
Suppose that(
M
,
ρ
) is a complete DMS and a self-mapping T
:
M
→
M is (
α
,
ψ
)-MKC. Assume also that(i)
T is α-admissible;
(ii)
there existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1;
(iii)
T is continuous.
Then, there existsθ
∈
M such that T
θ
=
θ.Proof.
Due to (ii), there existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1. We define an iterative sequence {
ξ
n
} in M as(12)
ξ
n
+
1
=
T
ξ
n
,
∀
n
∈
N
0
.Notice that if there exists some k
∈
N such that ξ
k
=
ξ
k
+
1, then the proof is completed since θ
=
ξ
k
=
ξ
k
+
1
=
T
ξ
k
=
T
θ.
So, we assume thatξ
n
≠
ξ
n
+
1, for all n. Due to Lemma 4 part (C), for ξ
n
≠
ξ
n
+
1, we have(13)
ρ
ξ
n
,
ξ
n
+
1
>
0
,
∀
n
∈
N
0
.Since T is α-admissible, again by (ii), we get(14)
α
ξ
0
,
ξ
1
=
α
ξ
0
,
T
ξ
0
≥
1
⟹
α
T
ξ
0
,
T
ξ
1
=
α
ξ
1
,
ξ
2
≥
1
.By repeating this process, we derive that(15)
α
ξ
n
,
ξ
n
+
1
≥
1
,
∀
n
∈
N
0
.Regarding (12) and (15) together with the assumption of the theorem that T is an (
α
,
ψ
)-MKC mapping, for each n
∈
N, we find(16)
ρ
ξ
n
,
ξ
n
+
1
=
ρ
T
ξ
n
-
1
,
T
ξ
n
≤
α
ξ
n
-
1
,
ξ
n
ρ
T
ξ
n
-
1
,
T
ξ
n
≤
ψ
ρ
ξ
n
-
1
,
ξ
n
.Taking (13) and the property of ψ into account, the above inequalities yield that(17)
ρ
ξ
n
,
ξ
n
+
1
≤
ψ
ρ
ξ
n
-
1
,
ξ
n
<
ρ
ξ
n
-
1
,
ξ
n
.Eventually, we conclude that {
ρ
(
ξ
n
-
1
,
ξ
n
)
} is a nonincreasing and bounded sequence. Hence, there exists t
∈
[
0
,
∞
) such that(18)
l
i
m
n
→
∞
ρ
ξ
n
-
1
,
ξ
n
=
t
.In what follows, we will prove that t
=
0
. Suppose, on the contrary, that t
>
0. Since T is an (
α
,
ψ
)-MKC mapping, for ε
=
ψ
(
t
)
>
0, there exist δ
>
0 and a natural number m such that(19)
ε
≤
ψ
ρ
ξ
m
-
1
,
ξ
m
<
ε
+
δ
i
m
p
l
i
e
s
α
ξ
m
-
1
,
ξ
m
ρ
T
ξ
m
-
1
,
T
ξ
m
<
ε
.By taking (15) into account, we get that(20)
ρ
ξ
m
,
ξ
m
+
1
=
ρ
T
ξ
m
-
1
,
T
ξ
m
≤
α
ξ
m
-
1
,
ξ
m
ρ
T
ξ
m
-
1
,
T
ξ
m
<
ε
=
ψ
t
<
t
,which is a contradiction since t
=
inf
⁡
{
ρ
(
ξ
n
,
ξ
n
+
1
)
:
n
∈
N
}.
Consequently, we have(21)
l
i
m
n
→
∞
ρ
ξ
n
-
1
,
ξ
n
=
0
.
Now, we will prove that{
ξ
n
} is a Cauchy sequence. That is,(22)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
k
=
0
,
∀
k
∈
N
.First, we observe from (17) that(23)
ρ
ξ
n
,
ξ
n
+
1
≤
ψ
ρ
ξ
n
-
1
,
ξ
n
∀
n
∈
N
.Since ψ is nondecreasing, by iteration, we conclude that(24)
ρ
ξ
n
,
ξ
n
+
1
≤
ψ
n
ρ
ξ
0
,
ξ
1
∀
n
∈
N
.
Now, by using (ρ
3), (12), (15), and (24), we have the following:(25)
ρ
ξ
n
,
ξ
n
+
k
≤
ρ
ξ
n
,
ξ
n
+
1
+
⋯
+
ρ
ξ
n
+
k
-
1
,
ξ
n
+
k
=
∑
p
=
n
n
+
k
-
1
ρ
ξ
p
,
ξ
p
+
1
=
∑
p
=
n
n
+
k
-
1
ρ
T
ξ
p
-
1
,
T
ξ
p
≤
∑
p
=
n
n
+
k
-
1
α
ξ
p
-
1
,
ξ
p
ρ
T
ξ
p
-
1
,
T
ξ
p
≤
∑
p
=
n
n
+
k
-
1
ψ
ρ
ξ
p
-
1
,
ξ
p
=
∑
p
=
n
n
+
k
-
1
ψ
p
-
1
ρ
ξ
0
,
ξ
1
≤
∑
p
=
n
∞
ψ
p
-
1
ρ
ξ
0
,
ξ
1
.
Lettingn
→
∞ in the above inequality, we derive expression (22). That is, the iterative sequence {
ξ
n
} is Cauchy in the DMS (
M
,
ρ
).
Since(
M
,
ρ
) is a complete DMS, then there exists θ
∈
M such that(26)
l
i
m
n
→
∞
ρ
ξ
n
,
θ
=
ρ
θ
,
θ
=
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
k
=
0
.
AsT is continuous, then we deduce that(27)
l
i
m
n
→
∞
ρ
ξ
n
+
1
,
T
θ
=
l
i
m
n
→
∞
ρ
T
ξ
n
,
T
θ
=
ρ
T
θ
,
T
θ
=
l
i
m
n
→
∞
ρ
T
ξ
n
,
T
ξ
n
+
k
=
0
.
Sinceξ
n
+
1
→
u as n
→
∞ and ρ
(
θ
,
θ
)
=
0, then, by using Lemma 4, we get(28)
l
i
m
n
→
∞
ρ
ξ
n
+
1
,
T
θ
=
ρ
θ
,
T
θ
.From (27) and (28), we derive that ρ
(
θ
,
T
θ
)
=
0. By (ρ
1), we conclude that T
θ
=
θ.Theorem 8.
Suppose that(
M
,
ρ
) is a complete DMS and a self-mapping T
:
M
→
M is (
α
,
ψ
)-MKC. Assume also that(i)
T is α-admissible;
(ii)
there existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1;
(iii)
if{
ξ
n
} is a sequence in M such that α
(
ξ
n
,
ξ
n
+
1
)
≥
1 for all n and ξ
n
→
ξ
∈
M as n
→
∞, then α
(
ξ
n
,
ξ
)
≥
1 for all n.
Then, there existsθ
∈
M such that T
θ
=
θ.Proof.
By following the lines in the proof of Theorem7, we deduce that there exists a Cauchy sequence {
ξ
n
}. Since M is complete, it converges to some θ
∈
M. On the other hand, from (15) and (iii), we have(29)
α
ξ
n
,
θ
≥
1
,
∀
n
.By using (ρ
3) and (29) with the assumption of the theorem that T is an (α
-
ψ)-MKC mapping, we obtain(30)
ρ
T
θ
,
θ
≤
ρ
T
θ
,
T
ξ
n
+
ρ
ξ
n
+
1
,
θ
≤
α
ξ
n
,
θ
ρ
T
ξ
n
,
T
θ
+
ρ
ξ
n
+
1
,
θ
≤
ψ
ρ
ξ
n
,
θ
+
ρ
ξ
n
+
1
,
θ
.
Sinceψ is continuous at t
=
0, by letting n
→
∞ in the inequality above, we find (31)
ρ
T
θ
,
θ
=
0
.
It yieldsT
θ
=
θ due to (ρ
1).Example 9.
LetM
=
[
0
,
∞
) endowed with the dislocated metric ρ
(
ξ
,
η
)
=
max
⁡
{
x
,
y
} for all ξ
,
η
∈
[
0
,
∞
). Define T
:
M
→
M and α
:
M
×
M
→
R
0
+ by(32)
T
ξ
=
ξ
2
,
α
ξ
,
η
=
1
,
i
f
ξ
=
η
=
0
;
0
,
o
t
h
e
r
w
i
s
e
.We can prove easily thatT is an (
α
-
ψ
)-MKC mapping and it is α-admissible.Moreover, there existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1. In fact, for ξ
0
=
0, we have(33)
α
0
,
T
0
=
1
.Now, we show thatT is continuous. Let(34)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
=
ρ
ξ
,
ξ
=
l
i
m
n
,
m
→
∞
ρ
ξ
n
,
ξ
m
. That is,(35)
l
i
m
n
→
∞
ξ
n
=
ξ
. Now, we have that T is continuous since(36)
l
i
m
n
→
∞
ρ
T
ξ
n
,
T
ξ
=
l
i
m
n
→
∞
max
⁡
T
ξ
n
,
T
ξ
=
T
ξ
=
ρ
T
ξ
,
T
ξ
=
l
i
m
n
,
m
→
∞
ρ
T
ξ
n
,
T
ξ
m
.So all hypotheses of Theorem7 are satisfied. Consequently, T has a fixed point. Notice that ξ
=
0 is a fixed point of T.In the following example, a self-mappingT is not continuous.Example 10.
LetM
=
[
0
,
∞
) endowed with the dislocated metric ρ
(
ξ
,
η
)
=
m
a
x
{
x
,
y
} for all ξ
,
η
∈
[
0
,
∞
). Define T
:
M
→
M and α
:
M
×
M
→
R
0
+ by(37)
T
ξ
=
1
2
ξ
-
1
2
,
ξ
>
1
,
0
,
0
≤
ξ
≤
1
,
α
ξ
,
η
=
1
,
i
f
ξ
,
η
∈
0,1
,
0
,
o
t
h
e
r
w
i
s
e
.Clearly,T is not continuous at 1 which shows that Theorem 7 is not applicable in this case.We will prove that a self-mappingT is (
α
-
ψ
)-MKC. Let ε
>
0 be given. Take δ
>
0 and suppose that ε
≤
ψ
(
ρ
(
ξ
,
η
)
)
<
ε
+
δ; we want to show that (38)
α
ξ
,
η
ρ
T
ξ
,
T
η
<
ε
.Suppose thatα
(
ξ
,
η
)
=
1; then ξ
,
η
∈
[
0,1
] and so T
ξ
=
0
,
T
η
=
0. Hence, (39)
ρ
T
ξ
,
T
η
=
ρ
0,0
=
max
⁡
0,0
=
0
<
ε
.Also,T is α-admissible. To see this, let α
(
ξ
,
η
)
≥
1; then, both ξ
,
η
∈
[
0,1
]. Due to the definition of T, we have T
ξ
=
0
∈
[
0,1
] and T
η
=
0
∈
[
0,1
]
. Thus, we get α
(
T
ξ
,
T
η
)
≥
1.Moreover, there existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1. Indeed, for ξ
0
=
1
/
2, we have(40)
α
1
2
,
T
1
2
=
α
1
2
,
0
=
1
.Finally, let{
ξ
n
} be a sequence in M such that α
(
ξ
n
,
ξ
n
+
1
)
≥
1 for all n and ξ
n
→
ξ
∈
M as n
→
∞. Since α
(
ξ
n
,
ξ
n
+
1
)
≥
1 for all n, by the definition of α, we have ξ
n
∈
[
0,1
] for all n and ξ
∈
[
0,1
]; then, α
(
ξ
n
,
ξ
)
=
1.So, we conclude that all hypotheses of Theorem8 are fulfilled. Hence, we proved that T has a fixed point.


### 2.2. Generalized-(
α
,
ψ
)-Meir-Keeler Contractive Mapping

Definition 11.
Suppose that(
M
,
ρ
) is a DMS. A self-mapping T
:
M
→
M is said to be generalized-(
α
,
ψ
)-Meir-Keeler contractive (in brief, generalized-(
α
,
ψ
)-MKC) if there exist ψ
∈
Ψ and α
:
M
×
M
→
R
0
+ such that, for each ε
>
0, there exists δ
>
0 such that (41)
ε
≤
ψ
m
a
x
T
ξ
,
η
<
ε
+
δ
i
m
p
l
i
e
s
α
ξ
,
η
ρ
T
ξ
,
T
η
<
ε
,where(42)
m
a
x
T
ξ
,
η
=
max
⁡
ρ
ξ
,
η
,
ρ
ξ
,
T
ξ
,
ρ
η
,
T
η
,
1
4
ρ
ξ
,
T
η
+
ρ
η
,
T
ξ
.If a self-mappingT
:
M
→
M is a generalized-(
α
,
ψ
)-MKC, then we have (43)
α
ξ
,
η
ρ
T
ξ
,
T
η
≤
ψ
m
a
x
T
ξ
,
η
,
f
o
r
a
n
y
ξ
,
η
∈
M
.Very recently, Popescu [7] improved the triangular α-admissible notion as follows.Definition 12 (see [7]).
Suppose thatT
:
M
→
M is a self-mapping and α
:
M
×
M
→
R
0
+ is a function. A self-mapping T is called α-orbital admissible if (44)
T
3
α
ξ
,
T
ξ
≥
1
⟹
α
T
ξ
,
T
2
ξ
≥
1
.
A self-mappingT is called triangular α-orbital admissible if T is α-orbital admissible and (45)
T
4
α
ξ
,
η
≥
1
,
α
η
,
T
η
≥
1
⟹
α
ξ
,
T
η
≥
1
.As it was mentioned in [7], every (triangular) α-admissible function is (triangular) α-orbital admissible function. The converse is false; see, for example, [7, Example 7].Lemma 13 (see [7]).
Suppose thatT
:
M
→
M is a triangular α-orbital admissible mapping. Suppose also that there exists ξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1. Define an iterative sequence {
ξ
n
} by ξ
n
+
1
=
T
ξ
n for each n
∈
N
0. Then, one has α
(
ξ
n
,
ξ
m
)
≥
1 for all m
,
n
∈
N with n
<
m.The following is the first main result of this section.Theorem 14.
Suppose that(
M
,
ρ
) is a complete DMS, a self-mapping T
:
M
→
M is a generalized-(
α
,
ψ
)-MKC, and the following conditions are fulfilled:(i)
T is triangular α-orbital admissible mapping.
(ii)
There existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1.
(iii)
T is continuous.
Then,T has a fixed point; that is, there exists θ
∈
M such that T
θ
=
θ.Proof.
Takeξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1. As in the proof of Theorem 7, we define an iterative sequence {
ξ
n
} in M as(46)
ξ
n
+
1
=
T
ξ
n
,
f
o
r
e
a
c
h
n
∈
N
0
.Without loss of generality, we assume that ξ
n
≠
ξ
n
+
1, for all n; then,(47)
ρ
ξ
n
,
ξ
n
+
1
>
0
,
∀
n
∈
N
.
Indeed, if there exists somen
0
∈
N such that ξ
n
0
=
ξ
n
0
+
1, then the proof is completed since u
=
ξ
n
0
=
ξ
n
0
+
1
=
T
ξ
n
0
=
T
u.
SinceT is triangular α-orbital admissible mapping, by using Lemma 13, we derive that(48)
α
ξ
n
,
ξ
m
≥
1
,
∀
m
,
n
∈
N
w
i
t
h
n
<
m
.
Step 1. We will prove that(49)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
1
=
0
.Taking (47) and (48) into account together with the fact that T is a generalized-(
α
,
ψ
)-MKC mapping, for each n
∈
N, we derive(50)
ρ
ξ
n
,
ξ
n
+
1
=
ρ
T
ξ
n
-
1
,
T
ξ
n
≤
α
ξ
n
-
1
,
ξ
n
ρ
T
ξ
n
-
1
,
T
ξ
n
≤
ψ
m
a
x
T
ξ
n
-
1
,
ξ
n
<
m
a
x
T
ξ
n
-
1
,
ξ
n
,where(51)
m
a
x
T
ξ
n
-
1
,
ξ
n
=
max
⁡
ρ
ξ
n
-
1
,
ξ
n
,
ρ
ξ
n
-
1
,
T
ξ
n
-
1
,
ρ
ξ
n
,
T
ξ
n
,
1
4
ρ
ξ
n
-
1
,
T
ξ
n
+
ρ
ξ
n
,
T
ξ
n
-
1
=
max
⁡
ρ
ξ
n
-
1
,
ξ
n
,
ρ
ξ
n
,
ξ
n
+
1
,
1
4
ρ
ξ
n
-
1
,
ξ
n
+
1
+
ρ
ξ
n
,
ξ
n
.Regarding (
ρ
3
), we estimate the last term in the expression m
a
x
T
(
ξ
n
-
1
,
ξ
n
) as follows:(52)
1
4
ρ
ξ
n
-
1
,
ξ
n
+
1
+
ρ
ξ
n
,
ξ
n
≤
max
⁡
ρ
ξ
n
-
1
,
ξ
n
,
ρ
ξ
n
,
ξ
n
+
1
.Consequently, we get(53)
m
a
x
T
ξ
n
-
1
,
ξ
n
=
max
⁡
ρ
ξ
n
-
1
,
ξ
n
,
ρ
ξ
n
,
ξ
n
+
1
.
Let us examine two cases. Letmax
⁡
{
ρ
(
ξ
n
-
1
,
ξ
n
)
,
ρ
(
ξ
n
,
ξ
n
+
1
)
}
=
ρ
(
ξ
n
,
ξ
n
+
1
). Since ψ is nondecreasing, from (50), we get(54)
ρ
ξ
n
,
ξ
n
+
1
≤
ψ
ρ
ξ
n
,
ξ
n
+
1
<
ρ
ξ
n
,
ξ
n
+
1
,which is a contradiction. Thus, m
a
x
{
ρ
(
ξ
n
-
1
,
ξ
n
)
,
ρ
(
ξ
n
,
ξ
n
+
1
)
}
=
ρ
(
ξ
n
-
1
,
ξ
n
) and again, by (50), we find(55)
ρ
ξ
n
,
ξ
n
+
1
≤
ψ
ρ
ξ
n
-
1
,
ξ
n
<
ρ
ξ
n
-
1
,
ξ
n
,
∀
n
∈
N
.
So, we deduce that the sequence{
ρ
(
ξ
n
,
ξ
n
+
1
)
} is nonincreasing and bounded below by zero.
Hence, there existst
∈
[
0
,
∞
) such that(56)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
1
=
t
.
Recursively, we derive from (55) that(57)
ρ
ξ
n
,
ξ
n
+
1
≤
ψ
n
ρ
ξ
0
,
ξ
1
,
∀
n
,by keeping in mind that ψ is nondecreasing.
On account of (57) and (Ψ
1), we obtain(58)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
1
=
0
.
It is obvious from the triangle inequality that(59)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
≤
2
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
1
=
0
.
Step 2. We will prove that {
ξ
n
} is a Cauchy sequence.
Suppose, on the contrary, that there existε
>
0 and a subsequence {
ξ
n
(
i
)
} of {
ξ
n
} such that(60)
ρ
ξ
n
i
,
ξ
n
i
+
1
>
2
ε
.
First, we will show the existence ofk
∈
N such that n
(
i
)
<
k
≤
n
(
i
+
1
). Later, we will prove that, for given ε
>
0 above, there exists δ
>
0 such that(61)
ε
≤
ψ
m
a
x
T
ξ
n
i
,
ξ
k
<
ε
+
δ
b
u
t
α
ξ
n
i
,
ξ
k
ρ
T
ξ
n
i
,
T
ξ
k
≥
ε
,which contradicts (60), where(62)
m
a
x
T
ξ
n
i
,
ξ
k
=
m
a
x
ρ
ξ
n
i
,
ξ
k
,
ρ
ξ
n
i
,
ξ
n
i
+
1
,
ρ
ξ
k
,
ξ
k
+
1
,
1
4
ρ
ξ
n
i
,
ξ
k
+
1
+
ρ
ξ
k
,
ξ
n
i
+
1
.
Letr
=
m
i
n
{
ε
,
δ
}. On account of Step 1, we will choose n
0
∈
N in a way that(63)
d
n
=
ρ
ξ
n
,
ξ
n
+
1
<
r
4
,
(64)
ρ
ξ
n
,
ξ
n
<
r
2
,for all n
≥
n
0. Due to our construction, clearly, we have n
(
i
)
≤
n
(
i
+
1
)
-
1. If ρ
(
ξ
n
(
i
)
,
ξ
n
(
i
+
1
)
-
1
)
≤
ε
+
r
/
2, then, by using (ρ
3), we have(65)
ρ
ξ
n
i
,
ξ
n
i
+
1
≤
ρ
ξ
n
i
,
ξ
n
i
+
1
-
1
+
ρ
ξ
n
i
+
1
-
1
,
ξ
n
i
+
1
≤
ε
+
r
2
+
d
n
i
+
1
-
1
<
ε
+
r
2
+
r
4
=
ε
+
3
r
4
<
2
ε
,which is a contradiction due to (60). Consequently, there are values of k
∈
N such that n
(
i
)
≤
k
≤
n
(
i
+
1
) and ρ
(
ξ
n
(
i
)
,
ξ
k
)
>
ε
+
r
/
2. Indeed, if(66)
ρ
ξ
n
i
,
ξ
n
i
+
1
≥
ε
+
r
2
,then we obtain(67)
d
n
i
=
ρ
ξ
n
i
,
ξ
n
i
+
1
≥
ε
+
r
2
≥
r
+
r
2
>
r
4which contradicts (63).
Moreover, the case(68)
ρ
ξ
n
i
,
ξ
n
i
≥
ε
+
r
2is impossible due to (64).
Hence, we choose the smallest integerk with k
>
n
(
i
) in a way that(69)
ρ
ξ
n
i
,
ξ
k
≥
ε
+
r
2
.So, necessarily, we have ρ
(
ξ
n
(
i
)
,
ξ
k
-
1
)
<
ε
+
r
/
2. Hence, we find that (70)
ρ
ξ
n
i
,
ξ
k
≤
ρ
ξ
n
i
,
ξ
k
-
1
+
ρ
ξ
k
-
1
,
ξ
k
<
ε
+
r
2
+
r
4
=
ε
+
3
r
4
,and, hence, we get the following approximation:(71)
ε
+
r
2
≤
ρ
ξ
n
i
,
ξ
k
<
ε
+
3
r
4
,for a natural number k satisfying n
(
i
)
<
k
≤
n
(
i
+
1
).
Thus, the first three terms of (62) are bounded above by ε
+
r; that is,(72)
ρ
ξ
n
i
,
ξ
k
<
ε
+
3
r
4
<
ε
+
r
,
(73)
ρ
ξ
n
i
,
ξ
n
i
+
1
=
d
n
i
<
r
4
<
ε
+
r
,
(74)
ρ
ξ
k
,
ξ
k
+
1
=
d
k
<
r
4
<
ε
+
r
.
Eventually, the last terms of (62) can be estimated from above as follows:(75)
1
4
ρ
ξ
n
i
,
ξ
k
+
1
+
ρ
ξ
n
i
+
1
,
ξ
k
≤
1
4
ρ
ξ
n
i
,
ξ
k
+
ρ
ξ
k
,
ξ
k
+
1
+
ρ
ξ
n
i
+
1
,
ξ
n
i
+
ρ
ξ
n
i
,
ξ
k
≤
1
2
ρ
ξ
n
i
,
ξ
k
+
1
4
d
k
+
d
n
i
≤
ρ
ξ
n
i
,
ξ
k
+
1
4
d
k
+
d
n
i
<
ε
+
3
r
4
+
1
4
2
r
4
=
ε
+
7
r
8
<
ε
+
r
.
Combining estimations (72), (73), (74), and (75), we conclude that(76)
ψ
m
a
x
T
ξ
n
i
,
ξ
k
≤
m
a
x
T
ξ
n
i
,
ξ
k
<
ε
+
r
≤
ε
+
δ
.
SinceT is generalized-(
α
-
ψ
)-MKC mapping and since it is triangular α-orbital admissible mapping, we have(77)
ρ
ξ
n
i
+
1
,
ξ
k
+
1
≤
α
ξ
n
i
,
ξ
k
ρ
ξ
n
i
+
1
,
ξ
k
+
1
<
ε
.
On the other hand, by (ρ
3), we have (78)
ρ
ξ
n
i
,
ξ
k
≤
ρ
ξ
n
i
,
ξ
n
i
+
1
+
ρ
ξ
n
i
+
1
,
ξ
k
≤
ρ
ξ
n
i
,
ξ
n
i
+
1
+
ρ
ξ
n
i
+
1
,
ξ
k
+
1
+
ρ
ξ
k
+
1
,
ξ
k
<
r
4
+
ε
+
r
4
=
ε
+
r
2
,which is a contradiction with (71).
Thus, claim (60) is false and the constructive sequence {
ξ
n
} is Cauchy; that is,(79)
l
i
m
n
,
m
→
∞
ρ
ξ
n
,
ξ
m
=
0
.
Since(
M
,
ρ
) is a complete DMS, then there exists θ
∈
M such that(80)
l
i
m
n
→
∞
ρ
ξ
n
,
θ
=
ρ
θ
,
θ
=
l
i
m
n
,
m
→
∞
ρ
ξ
n
,
ξ
m
=
0
.
On account of the continuity of the self-mappingT, we deduce that(81)
l
i
m
n
→
∞
ρ
ξ
n
+
1
,
T
θ
=
l
i
m
n
→
∞
ρ
T
ξ
n
,
T
θ
=
ρ
T
θ
,
T
θ
=
l
i
m
n
,
m
→
∞
ρ
T
ξ
n
,
T
ξ
m
=
0
.
Sinceξ
n
+
1
→
θ as n
→
∞ and ρ
(
θ
,
θ
)
=
0, then, by using Lemma 4, we get(82)
l
i
m
n
→
∞
ρ
ξ
n
+
1
,
T
θ
=
ρ
θ
,
T
θ
.
Regarding (81) and (82), we get ρ
(
θ
,
T
θ
)
=
0. Thus, by (ρ
1), we conclude that T
θ
=
θ.Theorem 15.
Suppose that(
M
,
ρ
) is a complete DMS, a self-mapping T
:
M
→
M is a generalized-(
α
,
ψ
)-MKC, and the following conditions are fulfilled:(i)
T is triangular α-orbital admissible mapping.
(ii)
There existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1.
(iii)
If{
ξ
n
} is a sequence in M such that α
(
ξ
n
,
ξ
n
+
1
)
≥
1 for all n and ξ
n
→
ξ
∈
M as n
→
∞, then α
(
ξ
n
,
ξ
)
≥
1 for all n.
Then,T has a fixed point; that is, there exists θ
∈
M such that T
θ
=
θ.Proof.
Define an iterative sequence{
ξ
n
} in M as in Theorem 14. Suppose that there is k
0
∈
N such that ξ
k
0
=
ξ
k
0
+
1; then, the proof is completed since u
=
ξ
k
0
=
ξ
k
0
+
1
=
T
ξ
k
0
=
T
u. So, it is interesting to assume that ξ
n
≠
ξ
n
+
1, for all n
∈
N
0. Consequently, we have (83)
ρ
ξ
n
,
ξ
n
+
1
>
0
,
∀
n
∈
N
0
.
From (48), we find that(84)
α
ξ
n
,
ξ
n
+
1
≥
1
,
∀
n
∈
N
0
.
Following the lines for the proofs of Steps1 and 2 in Theorem 14, we derive that {
ξ
n
} is a Cauchy sequence and(85)
l
i
m
n
,
m
→
∞
ρ
ξ
n
,
ξ
m
=
0
.Since (
M
,
ρ
) is a complete DMS, then there exists θ
∈
M such that(86)
l
i
m
n
→
∞
ρ
ξ
n
,
θ
=
ρ
θ
,
θ
=
l
i
m
n
,
m
→
∞
ρ
ξ
n
,
ξ
m
=
0
.
We will prove thatθ
=
T
θ. Suppose, on the contrary, that ρ
(
θ
,
T
θ
)
>
0.
Notice from (84), (86), and (iii) that(87)
α
ξ
n
,
θ
≥
1
,
∀
n
.
By using (ρ
3) and (87) together with the assumption of the theorem that T is a generalized-(α
-
ψ)-MKC mapping, we obtain(88)
ρ
T
θ
,
θ
≤
ρ
T
θ
,
T
ξ
n
+
ρ
ξ
n
+
1
,
θ
≤
α
ξ
n
,
θ
ρ
T
ξ
n
,
T
θ
+
ρ
ξ
n
+
1
,
θ
≤
ψ
m
a
x
T
ξ
n
,
θ
+
ρ
ξ
n
+
1
,
θ
,where(89)
m
a
x
T
ξ
n
,
θ
=
m
a
x
ρ
ξ
n
,
θ
,
ρ
ξ
n
,
ξ
n
+
1
,
ρ
θ
,
T
θ
,
1
4
ρ
ξ
n
,
T
θ
+
ρ
ξ
n
+
1
,
θ
.
Notice that asρ
(
θ
,
T
θ
)
>
0, then we must have m
a
x
T
(
ξ
n
,
θ
)
>
0.
Regarding (ρ
3), we have (90)
1
4
ρ
ξ
n
,
T
θ
+
ρ
ξ
n
+
1
,
θ
≤
max
⁡
ρ
ξ
n
,
θ
,
ρ
ξ
n
,
ξ
n
+
1
,
ρ
θ
,
T
θ
.
By the above inequality, we have(91)
m
a
x
T
ξ
n
,
θ
=
max
⁡
ρ
ξ
n
,
θ
,
ρ
ξ
n
,
ξ
n
+
1
,
ρ
θ
,
T
θ
.
Suppose thatm
a
x
T
(
ξ
n
,
θ
)
=
ρ
(
ξ
n
,
θ
); then, from (88), we get(92)
ρ
T
θ
,
θ
≤
ψ
ρ
ξ
n
,
θ
+
ρ
ξ
n
+
1
,
θ
<
ρ
ξ
n
,
θ
+
ρ
ξ
n
+
1
,
θ
.
Takingn
→
∞ in (92), we have(93)
ρ
T
θ
,
θ
<
ρ
θ
,
θ
+
ρ
θ
,
θ
=
0
,which is a contradiction.
Now, we suppose thatm
a
x
T
(
ξ
n
,
θ
)
=
ρ
(
ξ
n
,
ξ
n
+
1
); then, by (88), we find that (94)
ρ
T
θ
,
θ
≤
ψ
ρ
ξ
n
,
ξ
n
+
1
+
ρ
ξ
n
+
1
,
θ
<
ρ
ξ
n
,
ξ
n
+
1
+
ρ
ξ
n
+
1
,
θ
.
Lettingn
→
∞, this implies that (95)
ρ
T
θ
,
θ
<
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
1
+
ρ
ξ
n
+
1
,
θ
=
0
,which is again a contradiction.
Finally, we suppose thatm
a
x
T
(
ξ
n
,
θ
)
=
ρ
(
θ
,
T
θ
); then, again from (88), we have (96)
ρ
T
θ
,
θ
≤
ψ
ρ
θ
,
T
θ
+
ρ
ξ
n
+
1
,
θ
.
Lettingn
→
∞, in the above inequality, we get(97)
ρ
T
θ
,
θ
≤
ψ
ρ
θ
,
T
θ
+
ρ
θ
,
θ
<
ρ
θ
,
T
θ
+
ρ
θ
,
θ
=
ρ
T
θ
,
θ
,so we also have a contradiction. Thus, we have ρ
(
T
θ
,
θ
)
=
0 and, by (ρ
1), we have T
θ
=
θ.


### 2.3. The Uniqueness of the Fixed Point

We propose the following conditions for the uniqueness of the fixed points of the mappings discussed in Sections2.1 and 2.2. Let F
i
x
(
T
) denote the set of fixed points of the mapping T.We, first, recollect the following interesting condition for uniqueness of a fixed point of an(
α
-
ψ
)-MKC mapping.(
H
) For all ξ
,
η
∈
F
i
x
(
T
), there exists ζ
∈
M such that α
(
ξ
,
ζ
)
≥
1 and α
(
η
,
ζ
)
≥
1, where we have the following.Theorem 16.
Putting condition (H) to the statements of Theorem 7 (resp., Theorem 8), one obtains that θ is the unique fixed point of T.Proof.
Letθ and ν be two distinct fixed points of T. From (
H
), there exists ζ
∈
M such that (98)
α
θ
,
ζ
≥
1
,
α
ν
,
ζ
≥
1
. Due to the fact that T is α-admissible, we have (99)
α
θ
,
T
ζ
≥
1
,
α
ν
,
T
ζ
≥
1
.
Inductively, we obtain(100)
α
θ
,
T
n
ζ
≥
1
,
α
ν
,
T
n
ζ
≥
1
,
∀
n
∈
N
.
From the above relation and sinceT is an (
α
,
ψ
)-MKC mapping, we have (101)
ρ
θ
,
T
n
ζ
≤
α
θ
,
T
n
-
1
ζ
ρ
T
θ
,
T
n
ζ
≤
ψ
ρ
θ
,
T
n
-
1
ζ
.
Iteratively, we get(102)
ρ
θ
,
T
n
ζ
≤
ψ
n
ρ
θ
,
ζ
.
Lettingn
→
∞, we obtain(103)
l
i
m
n
→
∞
ρ
θ
,
T
n
ζ
=
0
.Similarly, we can prove that(104)
l
i
m
n
→
∞
ρ
ν
,
T
n
ζ
=
0
.Using (ρ
3), we have (105)
ρ
θ
,
ν
≤
ρ
θ
,
T
n
ζ
+
ρ
T
n
ζ
,
ν
.
Takingn
→
∞, we find that(106)
ρ
θ
,
ν
=
0
, and so, by (ρ
1), θ
=
ν.As an alternative uniqueness condition for fixed points of(
α
-
ψ
)-MKC mapping, we suggest the following hypothesis:(U)
For allξ
,
η
∈
F
i
x
(
T
), then α
(
ξ
,
η
)
≥
1.Theorem 17.
Putting condition (U) to the statements of Theorem7 (resp., Theorem 8), one finds that θ is the unique fixed point of T.Proof.
Letθ
,
ν be two distinct fixed points of T. Then, by Lemma 4 part (C), we have(107)
ρ
θ
,
ν
>
0
.Due to the property of ψ (Ψ
2), we get(108)
ψ
ρ
θ
,
ν
>
0
.
Letε
=
ψ
(
ρ
(
θ
,
ν
)
); then, for any δ
>
0, we find that (109)
ε
=
ψ
ρ
θ
,
ν
<
ε
+
δ
.
Regarding (U) and the assumption of the theorem thatT is an (
α
,
ψ
)-MKC mapping, we obtain(110)
ρ
θ
,
ν
≤
α
θ
,
ν
ρ
T
θ
,
T
ν
<
ψ
ρ
θ
,
ν
<
ρ
θ
,
ν
,which is a contradiction. Thus, θ
=
ν.In what follows, we propose the conditions for the uniqueness of a fixed point of a generalized-(
α
-
ψ
)-MKC mapping:(H1)
For allξ
,
η
∈
F
i
x
(
T
), there exists ζ
∈
M such that α
(
ξ
,
ζ
)
≥
1, α
(
η
,
ζ
)
≥
1, and α
(
ζ
,
T
ζ
)
≥
1.
(H2)
Letξ
,
η
∈
F
i
x
(
T
). If there exists a sequence {
ζ
n
} in M such that α
(
ξ
,
ζ
n
)
≥
1, α
(
η
,
ζ
n
)
≥
1, and α
(
ζ
n
,
ζ
n
+
1
)
≥
1, then (111)
ρ
ζ
n
,
ζ
n
+
1
≤
inf
⁡
ρ
ξ
,
ζ
n
,
ρ
η
,
ζ
n
.
(H3)
For anyξ
∈
F
i
x
(
T
), then α
(
ξ
,
ξ
)
≥
1.Theorem 18.
Putting conditions (H1), (H2), and (H3) to the statements of Theorem14 (resp., Theorem 15), one has that θ is the unique fixed point of T.Proof.
Letθ
,
ν be two distant fixed points of T. From (H1), there exists ζ
∈
M such that(112)
α
θ
,
ζ
≥
1
,
α
ν
,
ζ
≥
1
,
α
ζ
,
T
ζ
≥
1
.
Owing to the fact thatT is triangular α-orbital admissible and α
(
ζ
,
T
ζ
)
≥
1, by (T
3), we have(113)
α
T
ζ
,
T
2
ζ
≥
1
.
Inductively, we find(114)
α
T
n
-
1
ζ
,
T
n
ζ
≥
1
,
∀
n
∈
N
.
Now, sinceα
(
θ
,
ζ
)
≥
1 and α
(
ζ
,
T
ζ
)
≥
1, then, by (T
4), we deduce(115)
α
θ
,
T
z
≥
1
.
Again, by (T
4), since α
(
θ
,
T
ζ
)
≥
1 and α
(
T
ζ
,
T
2
ζ
)
≥
1, we derive(116)
α
θ
,
T
2
ζ
≥
1
.
Inductively, we get(117)
α
θ
,
T
n
ζ
≥
1
,
∀
n
∈
N
.
In an analogous way, we will prove that(118)
α
ν
,
T
n
ζ
≥
1
,
∀
n
∈
N
.
Define an iterative sequence{
ζ
n
} by ζ
n
+
1
=
T
ζ
n, for all n
≥
0 and ζ
0
=
ζ
.
Step 1. We will prove that(119)
l
i
m
n
→
∞
ρ
θ
,
ζ
n
=
0
.
By (117) and the statement of the theorem that T is generalized-(
α
,
ψ
)-MKC mapping, we have (120)
ρ
θ
,
ζ
n
+
1
≤
α
θ
,
ζ
n
ρ
T
θ
,
T
ζ
n
≤
ψ
m
a
x
T
θ
,
ζ
n
.
Ifψ
(
m
a
x
T
(
θ
,
ζ
n
)
)
=
0, then(121)
l
i
m
n
→
∞
ρ
θ
,
ζ
n
+
1
=
0
. Now, suppose that ψ
(
m
a
x
T
(
θ
,
ζ
n
)
)
>
0; then, m
a
x
T
(
θ
,
ζ
n
)
>
0. As T is a generalized-(
α
,
ψ
)-MKC mapping, we get (122)
ρ
θ
,
ζ
n
+
1
≤
α
θ
,
ζ
n
ρ
T
θ
,
T
ζ
n
≤
ψ
m
a
x
T
θ
,
ζ
n
<
m
a
x
T
θ
,
ζ
n
,where(123)
m
a
x
T
ζ
n
,
θ
=
m
a
x
ρ
ζ
n
,
θ
,
ρ
ζ
n
,
ζ
n
+
1
,
ρ
θ
,
T
θ
,
1
4
ρ
ζ
n
,
T
θ
+
ρ
ζ
n
+
1
,
θ
.Regarding (H2) and (ρ
3), we have(124)
m
a
x
T
ζ
n
,
θ
=
ρ
ζ
n
,
θ
.Thus, we have(125)
ρ
θ
,
ζ
n
+
1
<
ρ
θ
,
ζ
n
.
Lettingn
→
∞ in the inequality above, we obtain(126)
l
i
m
n
→
∞
ρ
θ
,
ζ
n
+
1
<
l
i
m
n
→
∞
ρ
θ
,
ζ
n
,which is a contradiction. Then,(127)
m
a
x
T
ζ
n
,
θ
=
ρ
ζ
n
,
θ
=
0
.
Hence, we get(128)
l
i
m
n
→
∞
ρ
θ
,
ζ
n
=
0
.
Step 2. We will prove that (129)
l
i
m
n
→
∞
ρ
ν
,
ζ
n
=
0
.
By (H3) and the assumption of the theorem thatT is a generalized-(
α
,
ψ
)-MKC mapping, we find that(130)
ρ
ν
,
ν
≤
α
ν
,
ν
ρ
T
v
,
T
ν
≤
ψ
m
a
x
T
ν
,
ν
=
ψ
ρ
ν
,
ν
.Suppose, on the contrary, that ρ
(
ν
,
ν
)
>
0. Then, from the above inequality, we obtain(131)
ρ
ν
,
ν
≤
ψ
ρ
ν
,
ν
<
ρ
ν
,
ν
,which is a contradiction. Thus, ρ
(
ν
,
ν
)
=
ρ
(
ν
,
T
ν
)
=
0.
In an analogous way of Step1, we can complete the proof of(132)
l
i
m
n
→
∞
ρ
ν
,
ζ
n
=
0
.
By (ρ
3), we have(133)
ρ
θ
,
ν
≤
ρ
θ
,
T
n
-
1
ζ
+
ρ
T
n
-
1
ζ
,
ν
=
ρ
θ
,
ζ
n
+
ρ
ζ
n
,
ν
.
Lettingn
→
∞ in the above inequality, we find that(134)
ρ
θ
,
ν
=
0
;by (ρ
1), we have θ
=
ν.


## 2.1.(
α
,
ψ
)-Meir-Keeler Contractive Mapping

We introduce the following notion which is an improved version of Meir-Keeler contractive mapping.Definition 6.
Let(
M
,
ρ
) be a DMS. A self-mapping T
:
M
→
M is called (
α
,
ψ
)-Meir-Keeler contractive (in brief, (
α
,
ψ
)-MKC) if there exist two functions ψ
∈
Ψ and α
:
M
×
M
→
R
0
+ satisfying the following condition.
For eachε
>
0, there exists δ
>
0 such that (10)
ε
≤
ψ
ρ
ξ
,
η
<
ε
+
δ
i
m
p
l
i
e
s
α
ξ
,
η
ρ
T
ξ
,
T
η
<
ε
.Notice that, for(
α
,
ψ
)-MKC mapping T
:
M
→
M, we have (11)
α
ξ
,
η
ρ
T
ξ
,
T
η
≤
ψ
ρ
ξ
,
η
,
f
o
r
a
n
y
ξ
,
η
∈
M
.In what follows, we will state and prove the first main result of this section.Theorem 7.
Suppose that(
M
,
ρ
) is a complete DMS and a self-mapping T
:
M
→
M is (
α
,
ψ
)-MKC. Assume also that(i)
T is α-admissible;
(ii)
there existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1;
(iii)
T is continuous.
Then, there existsθ
∈
M such that T
θ
=
θ.Proof.
Due to (ii), there existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1. We define an iterative sequence {
ξ
n
} in M as(12)
ξ
n
+
1
=
T
ξ
n
,
∀
n
∈
N
0
.Notice that if there exists some k
∈
N such that ξ
k
=
ξ
k
+
1, then the proof is completed since θ
=
ξ
k
=
ξ
k
+
1
=
T
ξ
k
=
T
θ.
So, we assume thatξ
n
≠
ξ
n
+
1, for all n. Due to Lemma 4 part (C), for ξ
n
≠
ξ
n
+
1, we have(13)
ρ
ξ
n
,
ξ
n
+
1
>
0
,
∀
n
∈
N
0
.Since T is α-admissible, again by (ii), we get(14)
α
ξ
0
,
ξ
1
=
α
ξ
0
,
T
ξ
0
≥
1
⟹
α
T
ξ
0
,
T
ξ
1
=
α
ξ
1
,
ξ
2
≥
1
.By repeating this process, we derive that(15)
α
ξ
n
,
ξ
n
+
1
≥
1
,
∀
n
∈
N
0
.Regarding (12) and (15) together with the assumption of the theorem that T is an (
α
,
ψ
)-MKC mapping, for each n
∈
N, we find(16)
ρ
ξ
n
,
ξ
n
+
1
=
ρ
T
ξ
n
-
1
,
T
ξ
n
≤
α
ξ
n
-
1
,
ξ
n
ρ
T
ξ
n
-
1
,
T
ξ
n
≤
ψ
ρ
ξ
n
-
1
,
ξ
n
.Taking (13) and the property of ψ into account, the above inequalities yield that(17)
ρ
ξ
n
,
ξ
n
+
1
≤
ψ
ρ
ξ
n
-
1
,
ξ
n
<
ρ
ξ
n
-
1
,
ξ
n
.Eventually, we conclude that {
ρ
(
ξ
n
-
1
,
ξ
n
)
} is a nonincreasing and bounded sequence. Hence, there exists t
∈
[
0
,
∞
) such that(18)
l
i
m
n
→
∞
ρ
ξ
n
-
1
,
ξ
n
=
t
.In what follows, we will prove that t
=
0
. Suppose, on the contrary, that t
>
0. Since T is an (
α
,
ψ
)-MKC mapping, for ε
=
ψ
(
t
)
>
0, there exist δ
>
0 and a natural number m such that(19)
ε
≤
ψ
ρ
ξ
m
-
1
,
ξ
m
<
ε
+
δ
i
m
p
l
i
e
s
α
ξ
m
-
1
,
ξ
m
ρ
T
ξ
m
-
1
,
T
ξ
m
<
ε
.By taking (15) into account, we get that(20)
ρ
ξ
m
,
ξ
m
+
1
=
ρ
T
ξ
m
-
1
,
T
ξ
m
≤
α
ξ
m
-
1
,
ξ
m
ρ
T
ξ
m
-
1
,
T
ξ
m
<
ε
=
ψ
t
<
t
,which is a contradiction since t
=
inf
⁡
{
ρ
(
ξ
n
,
ξ
n
+
1
)
:
n
∈
N
}.
Consequently, we have(21)
l
i
m
n
→
∞
ρ
ξ
n
-
1
,
ξ
n
=
0
.
Now, we will prove that{
ξ
n
} is a Cauchy sequence. That is,(22)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
k
=
0
,
∀
k
∈
N
.First, we observe from (17) that(23)
ρ
ξ
n
,
ξ
n
+
1
≤
ψ
ρ
ξ
n
-
1
,
ξ
n
∀
n
∈
N
.Since ψ is nondecreasing, by iteration, we conclude that(24)
ρ
ξ
n
,
ξ
n
+
1
≤
ψ
n
ρ
ξ
0
,
ξ
1
∀
n
∈
N
.
Now, by using (ρ
3), (12), (15), and (24), we have the following:(25)
ρ
ξ
n
,
ξ
n
+
k
≤
ρ
ξ
n
,
ξ
n
+
1
+
⋯
+
ρ
ξ
n
+
k
-
1
,
ξ
n
+
k
=
∑
p
=
n
n
+
k
-
1
ρ
ξ
p
,
ξ
p
+
1
=
∑
p
=
n
n
+
k
-
1
ρ
T
ξ
p
-
1
,
T
ξ
p
≤
∑
p
=
n
n
+
k
-
1
α
ξ
p
-
1
,
ξ
p
ρ
T
ξ
p
-
1
,
T
ξ
p
≤
∑
p
=
n
n
+
k
-
1
ψ
ρ
ξ
p
-
1
,
ξ
p
=
∑
p
=
n
n
+
k
-
1
ψ
p
-
1
ρ
ξ
0
,
ξ
1
≤
∑
p
=
n
∞
ψ
p
-
1
ρ
ξ
0
,
ξ
1
.
Lettingn
→
∞ in the above inequality, we derive expression (22). That is, the iterative sequence {
ξ
n
} is Cauchy in the DMS (
M
,
ρ
).
Since(
M
,
ρ
) is a complete DMS, then there exists θ
∈
M such that(26)
l
i
m
n
→
∞
ρ
ξ
n
,
θ
=
ρ
θ
,
θ
=
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
k
=
0
.
AsT is continuous, then we deduce that(27)
l
i
m
n
→
∞
ρ
ξ
n
+
1
,
T
θ
=
l
i
m
n
→
∞
ρ
T
ξ
n
,
T
θ
=
ρ
T
θ
,
T
θ
=
l
i
m
n
→
∞
ρ
T
ξ
n
,
T
ξ
n
+
k
=
0
.
Sinceξ
n
+
1
→
u as n
→
∞ and ρ
(
θ
,
θ
)
=
0, then, by using Lemma 4, we get(28)
l
i
m
n
→
∞
ρ
ξ
n
+
1
,
T
θ
=
ρ
θ
,
T
θ
.From (27) and (28), we derive that ρ
(
θ
,
T
θ
)
=
0. By (ρ
1), we conclude that T
θ
=
θ.Theorem 8.
Suppose that(
M
,
ρ
) is a complete DMS and a self-mapping T
:
M
→
M is (
α
,
ψ
)-MKC. Assume also that(i)
T is α-admissible;
(ii)
there existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1;
(iii)
if{
ξ
n
} is a sequence in M such that α
(
ξ
n
,
ξ
n
+
1
)
≥
1 for all n and ξ
n
→
ξ
∈
M as n
→
∞, then α
(
ξ
n
,
ξ
)
≥
1 for all n.
Then, there existsθ
∈
M such that T
θ
=
θ.Proof.
By following the lines in the proof of Theorem7, we deduce that there exists a Cauchy sequence {
ξ
n
}. Since M is complete, it converges to some θ
∈
M. On the other hand, from (15) and (iii), we have(29)
α
ξ
n
,
θ
≥
1
,
∀
n
.By using (ρ
3) and (29) with the assumption of the theorem that T is an (α
-
ψ)-MKC mapping, we obtain(30)
ρ
T
θ
,
θ
≤
ρ
T
θ
,
T
ξ
n
+
ρ
ξ
n
+
1
,
θ
≤
α
ξ
n
,
θ
ρ
T
ξ
n
,
T
θ
+
ρ
ξ
n
+
1
,
θ
≤
ψ
ρ
ξ
n
,
θ
+
ρ
ξ
n
+
1
,
θ
.
Sinceψ is continuous at t
=
0, by letting n
→
∞ in the inequality above, we find (31)
ρ
T
θ
,
θ
=
0
.
It yieldsT
θ
=
θ due to (ρ
1).Example 9.
LetM
=
[
0
,
∞
) endowed with the dislocated metric ρ
(
ξ
,
η
)
=
max
⁡
{
x
,
y
} for all ξ
,
η
∈
[
0
,
∞
). Define T
:
M
→
M and α
:
M
×
M
→
R
0
+ by(32)
T
ξ
=
ξ
2
,
α
ξ
,
η
=
1
,
i
f
ξ
=
η
=
0
;
0
,
o
t
h
e
r
w
i
s
e
.We can prove easily thatT is an (
α
-
ψ
)-MKC mapping and it is α-admissible.Moreover, there existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1. In fact, for ξ
0
=
0, we have(33)
α
0
,
T
0
=
1
.Now, we show thatT is continuous. Let(34)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
=
ρ
ξ
,
ξ
=
l
i
m
n
,
m
→
∞
ρ
ξ
n
,
ξ
m
. That is,(35)
l
i
m
n
→
∞
ξ
n
=
ξ
. Now, we have that T is continuous since(36)
l
i
m
n
→
∞
ρ
T
ξ
n
,
T
ξ
=
l
i
m
n
→
∞
max
⁡
T
ξ
n
,
T
ξ
=
T
ξ
=
ρ
T
ξ
,
T
ξ
=
l
i
m
n
,
m
→
∞
ρ
T
ξ
n
,
T
ξ
m
.So all hypotheses of Theorem7 are satisfied. Consequently, T has a fixed point. Notice that ξ
=
0 is a fixed point of T.In the following example, a self-mappingT is not continuous.Example 10.
LetM
=
[
0
,
∞
) endowed with the dislocated metric ρ
(
ξ
,
η
)
=
m
a
x
{
x
,
y
} for all ξ
,
η
∈
[
0
,
∞
). Define T
:
M
→
M and α
:
M
×
M
→
R
0
+ by(37)
T
ξ
=
1
2
ξ
-
1
2
,
ξ
>
1
,
0
,
0
≤
ξ
≤
1
,
α
ξ
,
η
=
1
,
i
f
ξ
,
η
∈
0,1
,
0
,
o
t
h
e
r
w
i
s
e
.Clearly,T is not continuous at 1 which shows that Theorem 7 is not applicable in this case.We will prove that a self-mappingT is (
α
-
ψ
)-MKC. Let ε
>
0 be given. Take δ
>
0 and suppose that ε
≤
ψ
(
ρ
(
ξ
,
η
)
)
<
ε
+
δ; we want to show that (38)
α
ξ
,
η
ρ
T
ξ
,
T
η
<
ε
.Suppose thatα
(
ξ
,
η
)
=
1; then ξ
,
η
∈
[
0,1
] and so T
ξ
=
0
,
T
η
=
0. Hence, (39)
ρ
T
ξ
,
T
η
=
ρ
0,0
=
max
⁡
0,0
=
0
<
ε
.Also,T is α-admissible. To see this, let α
(
ξ
,
η
)
≥
1; then, both ξ
,
η
∈
[
0,1
]. Due to the definition of T, we have T
ξ
=
0
∈
[
0,1
] and T
η
=
0
∈
[
0,1
]
. Thus, we get α
(
T
ξ
,
T
η
)
≥
1.Moreover, there existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1. Indeed, for ξ
0
=
1
/
2, we have(40)
α
1
2
,
T
1
2
=
α
1
2
,
0
=
1
.Finally, let{
ξ
n
} be a sequence in M such that α
(
ξ
n
,
ξ
n
+
1
)
≥
1 for all n and ξ
n
→
ξ
∈
M as n
→
∞. Since α
(
ξ
n
,
ξ
n
+
1
)
≥
1 for all n, by the definition of α, we have ξ
n
∈
[
0,1
] for all n and ξ
∈
[
0,1
]; then, α
(
ξ
n
,
ξ
)
=
1.So, we conclude that all hypotheses of Theorem8 are fulfilled. Hence, we proved that T has a fixed point.


## 2.2. Generalized-(
α
,
ψ
)-Meir-Keeler Contractive Mapping

Definition 11.
Suppose that(
M
,
ρ
) is a DMS. A self-mapping T
:
M
→
M is said to be generalized-(
α
,
ψ
)-Meir-Keeler contractive (in brief, generalized-(
α
,
ψ
)-MKC) if there exist ψ
∈
Ψ and α
:
M
×
M
→
R
0
+ such that, for each ε
>
0, there exists δ
>
0 such that (41)
ε
≤
ψ
m
a
x
T
ξ
,
η
<
ε
+
δ
i
m
p
l
i
e
s
α
ξ
,
η
ρ
T
ξ
,
T
η
<
ε
,where(42)
m
a
x
T
ξ
,
η
=
max
⁡
ρ
ξ
,
η
,
ρ
ξ
,
T
ξ
,
ρ
η
,
T
η
,
1
4
ρ
ξ
,
T
η
+
ρ
η
,
T
ξ
.If a self-mappingT
:
M
→
M is a generalized-(
α
,
ψ
)-MKC, then we have (43)
α
ξ
,
η
ρ
T
ξ
,
T
η
≤
ψ
m
a
x
T
ξ
,
η
,
f
o
r
a
n
y
ξ
,
η
∈
M
.Very recently, Popescu [7] improved the triangular α-admissible notion as follows.Definition 12 (see [7]).
Suppose thatT
:
M
→
M is a self-mapping and α
:
M
×
M
→
R
0
+ is a function. A self-mapping T is called α-orbital admissible if (44)
T
3
α
ξ
,
T
ξ
≥
1
⟹
α
T
ξ
,
T
2
ξ
≥
1
.
A self-mappingT is called triangular α-orbital admissible if T is α-orbital admissible and (45)
T
4
α
ξ
,
η
≥
1
,
α
η
,
T
η
≥
1
⟹
α
ξ
,
T
η
≥
1
.As it was mentioned in [7], every (triangular) α-admissible function is (triangular) α-orbital admissible function. The converse is false; see, for example, [7, Example 7].Lemma 13 (see [7]).
Suppose thatT
:
M
→
M is a triangular α-orbital admissible mapping. Suppose also that there exists ξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1. Define an iterative sequence {
ξ
n
} by ξ
n
+
1
=
T
ξ
n for each n
∈
N
0. Then, one has α
(
ξ
n
,
ξ
m
)
≥
1 for all m
,
n
∈
N with n
<
m.The following is the first main result of this section.Theorem 14.
Suppose that(
M
,
ρ
) is a complete DMS, a self-mapping T
:
M
→
M is a generalized-(
α
,
ψ
)-MKC, and the following conditions are fulfilled:(i)
T is triangular α-orbital admissible mapping.
(ii)
There existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1.
(iii)
T is continuous.
Then,T has a fixed point; that is, there exists θ
∈
M such that T
θ
=
θ.Proof.
Takeξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1. As in the proof of Theorem 7, we define an iterative sequence {
ξ
n
} in M as(46)
ξ
n
+
1
=
T
ξ
n
,
f
o
r
e
a
c
h
n
∈
N
0
.Without loss of generality, we assume that ξ
n
≠
ξ
n
+
1, for all n; then,(47)
ρ
ξ
n
,
ξ
n
+
1
>
0
,
∀
n
∈
N
.
Indeed, if there exists somen
0
∈
N such that ξ
n
0
=
ξ
n
0
+
1, then the proof is completed since u
=
ξ
n
0
=
ξ
n
0
+
1
=
T
ξ
n
0
=
T
u.
SinceT is triangular α-orbital admissible mapping, by using Lemma 13, we derive that(48)
α
ξ
n
,
ξ
m
≥
1
,
∀
m
,
n
∈
N
w
i
t
h
n
<
m
.
Step 1. We will prove that(49)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
1
=
0
.Taking (47) and (48) into account together with the fact that T is a generalized-(
α
,
ψ
)-MKC mapping, for each n
∈
N, we derive(50)
ρ
ξ
n
,
ξ
n
+
1
=
ρ
T
ξ
n
-
1
,
T
ξ
n
≤
α
ξ
n
-
1
,
ξ
n
ρ
T
ξ
n
-
1
,
T
ξ
n
≤
ψ
m
a
x
T
ξ
n
-
1
,
ξ
n
<
m
a
x
T
ξ
n
-
1
,
ξ
n
,where(51)
m
a
x
T
ξ
n
-
1
,
ξ
n
=
max
⁡
ρ
ξ
n
-
1
,
ξ
n
,
ρ
ξ
n
-
1
,
T
ξ
n
-
1
,
ρ
ξ
n
,
T
ξ
n
,
1
4
ρ
ξ
n
-
1
,
T
ξ
n
+
ρ
ξ
n
,
T
ξ
n
-
1
=
max
⁡
ρ
ξ
n
-
1
,
ξ
n
,
ρ
ξ
n
,
ξ
n
+
1
,
1
4
ρ
ξ
n
-
1
,
ξ
n
+
1
+
ρ
ξ
n
,
ξ
n
.Regarding (
ρ
3
), we estimate the last term in the expression m
a
x
T
(
ξ
n
-
1
,
ξ
n
) as follows:(52)
1
4
ρ
ξ
n
-
1
,
ξ
n
+
1
+
ρ
ξ
n
,
ξ
n
≤
max
⁡
ρ
ξ
n
-
1
,
ξ
n
,
ρ
ξ
n
,
ξ
n
+
1
.Consequently, we get(53)
m
a
x
T
ξ
n
-
1
,
ξ
n
=
max
⁡
ρ
ξ
n
-
1
,
ξ
n
,
ρ
ξ
n
,
ξ
n
+
1
.
Let us examine two cases. Letmax
⁡
{
ρ
(
ξ
n
-
1
,
ξ
n
)
,
ρ
(
ξ
n
,
ξ
n
+
1
)
}
=
ρ
(
ξ
n
,
ξ
n
+
1
). Since ψ is nondecreasing, from (50), we get(54)
ρ
ξ
n
,
ξ
n
+
1
≤
ψ
ρ
ξ
n
,
ξ
n
+
1
<
ρ
ξ
n
,
ξ
n
+
1
,which is a contradiction. Thus, m
a
x
{
ρ
(
ξ
n
-
1
,
ξ
n
)
,
ρ
(
ξ
n
,
ξ
n
+
1
)
}
=
ρ
(
ξ
n
-
1
,
ξ
n
) and again, by (50), we find(55)
ρ
ξ
n
,
ξ
n
+
1
≤
ψ
ρ
ξ
n
-
1
,
ξ
n
<
ρ
ξ
n
-
1
,
ξ
n
,
∀
n
∈
N
.
So, we deduce that the sequence{
ρ
(
ξ
n
,
ξ
n
+
1
)
} is nonincreasing and bounded below by zero.
Hence, there existst
∈
[
0
,
∞
) such that(56)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
1
=
t
.
Recursively, we derive from (55) that(57)
ρ
ξ
n
,
ξ
n
+
1
≤
ψ
n
ρ
ξ
0
,
ξ
1
,
∀
n
,by keeping in mind that ψ is nondecreasing.
On account of (57) and (Ψ
1), we obtain(58)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
1
=
0
.
It is obvious from the triangle inequality that(59)
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
≤
2
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
1
=
0
.
Step 2. We will prove that {
ξ
n
} is a Cauchy sequence.
Suppose, on the contrary, that there existε
>
0 and a subsequence {
ξ
n
(
i
)
} of {
ξ
n
} such that(60)
ρ
ξ
n
i
,
ξ
n
i
+
1
>
2
ε
.
First, we will show the existence ofk
∈
N such that n
(
i
)
<
k
≤
n
(
i
+
1
). Later, we will prove that, for given ε
>
0 above, there exists δ
>
0 such that(61)
ε
≤
ψ
m
a
x
T
ξ
n
i
,
ξ
k
<
ε
+
δ
b
u
t
α
ξ
n
i
,
ξ
k
ρ
T
ξ
n
i
,
T
ξ
k
≥
ε
,which contradicts (60), where(62)
m
a
x
T
ξ
n
i
,
ξ
k
=
m
a
x
ρ
ξ
n
i
,
ξ
k
,
ρ
ξ
n
i
,
ξ
n
i
+
1
,
ρ
ξ
k
,
ξ
k
+
1
,
1
4
ρ
ξ
n
i
,
ξ
k
+
1
+
ρ
ξ
k
,
ξ
n
i
+
1
.
Letr
=
m
i
n
{
ε
,
δ
}. On account of Step 1, we will choose n
0
∈
N in a way that(63)
d
n
=
ρ
ξ
n
,
ξ
n
+
1
<
r
4
,
(64)
ρ
ξ
n
,
ξ
n
<
r
2
,for all n
≥
n
0. Due to our construction, clearly, we have n
(
i
)
≤
n
(
i
+
1
)
-
1. If ρ
(
ξ
n
(
i
)
,
ξ
n
(
i
+
1
)
-
1
)
≤
ε
+
r
/
2, then, by using (ρ
3), we have(65)
ρ
ξ
n
i
,
ξ
n
i
+
1
≤
ρ
ξ
n
i
,
ξ
n
i
+
1
-
1
+
ρ
ξ
n
i
+
1
-
1
,
ξ
n
i
+
1
≤
ε
+
r
2
+
d
n
i
+
1
-
1
<
ε
+
r
2
+
r
4
=
ε
+
3
r
4
<
2
ε
,which is a contradiction due to (60). Consequently, there are values of k
∈
N such that n
(
i
)
≤
k
≤
n
(
i
+
1
) and ρ
(
ξ
n
(
i
)
,
ξ
k
)
>
ε
+
r
/
2. Indeed, if(66)
ρ
ξ
n
i
,
ξ
n
i
+
1
≥
ε
+
r
2
,then we obtain(67)
d
n
i
=
ρ
ξ
n
i
,
ξ
n
i
+
1
≥
ε
+
r
2
≥
r
+
r
2
>
r
4which contradicts (63).
Moreover, the case(68)
ρ
ξ
n
i
,
ξ
n
i
≥
ε
+
r
2is impossible due to (64).
Hence, we choose the smallest integerk with k
>
n
(
i
) in a way that(69)
ρ
ξ
n
i
,
ξ
k
≥
ε
+
r
2
.So, necessarily, we have ρ
(
ξ
n
(
i
)
,
ξ
k
-
1
)
<
ε
+
r
/
2. Hence, we find that (70)
ρ
ξ
n
i
,
ξ
k
≤
ρ
ξ
n
i
,
ξ
k
-
1
+
ρ
ξ
k
-
1
,
ξ
k
<
ε
+
r
2
+
r
4
=
ε
+
3
r
4
,and, hence, we get the following approximation:(71)
ε
+
r
2
≤
ρ
ξ
n
i
,
ξ
k
<
ε
+
3
r
4
,for a natural number k satisfying n
(
i
)
<
k
≤
n
(
i
+
1
).
Thus, the first three terms of (62) are bounded above by ε
+
r; that is,(72)
ρ
ξ
n
i
,
ξ
k
<
ε
+
3
r
4
<
ε
+
r
,
(73)
ρ
ξ
n
i
,
ξ
n
i
+
1
=
d
n
i
<
r
4
<
ε
+
r
,
(74)
ρ
ξ
k
,
ξ
k
+
1
=
d
k
<
r
4
<
ε
+
r
.
Eventually, the last terms of (62) can be estimated from above as follows:(75)
1
4
ρ
ξ
n
i
,
ξ
k
+
1
+
ρ
ξ
n
i
+
1
,
ξ
k
≤
1
4
ρ
ξ
n
i
,
ξ
k
+
ρ
ξ
k
,
ξ
k
+
1
+
ρ
ξ
n
i
+
1
,
ξ
n
i
+
ρ
ξ
n
i
,
ξ
k
≤
1
2
ρ
ξ
n
i
,
ξ
k
+
1
4
d
k
+
d
n
i
≤
ρ
ξ
n
i
,
ξ
k
+
1
4
d
k
+
d
n
i
<
ε
+
3
r
4
+
1
4
2
r
4
=
ε
+
7
r
8
<
ε
+
r
.
Combining estimations (72), (73), (74), and (75), we conclude that(76)
ψ
m
a
x
T
ξ
n
i
,
ξ
k
≤
m
a
x
T
ξ
n
i
,
ξ
k
<
ε
+
r
≤
ε
+
δ
.
SinceT is generalized-(
α
-
ψ
)-MKC mapping and since it is triangular α-orbital admissible mapping, we have(77)
ρ
ξ
n
i
+
1
,
ξ
k
+
1
≤
α
ξ
n
i
,
ξ
k
ρ
ξ
n
i
+
1
,
ξ
k
+
1
<
ε
.
On the other hand, by (ρ
3), we have (78)
ρ
ξ
n
i
,
ξ
k
≤
ρ
ξ
n
i
,
ξ
n
i
+
1
+
ρ
ξ
n
i
+
1
,
ξ
k
≤
ρ
ξ
n
i
,
ξ
n
i
+
1
+
ρ
ξ
n
i
+
1
,
ξ
k
+
1
+
ρ
ξ
k
+
1
,
ξ
k
<
r
4
+
ε
+
r
4
=
ε
+
r
2
,which is a contradiction with (71).
Thus, claim (60) is false and the constructive sequence {
ξ
n
} is Cauchy; that is,(79)
l
i
m
n
,
m
→
∞
ρ
ξ
n
,
ξ
m
=
0
.
Since(
M
,
ρ
) is a complete DMS, then there exists θ
∈
M such that(80)
l
i
m
n
→
∞
ρ
ξ
n
,
θ
=
ρ
θ
,
θ
=
l
i
m
n
,
m
→
∞
ρ
ξ
n
,
ξ
m
=
0
.
On account of the continuity of the self-mappingT, we deduce that(81)
l
i
m
n
→
∞
ρ
ξ
n
+
1
,
T
θ
=
l
i
m
n
→
∞
ρ
T
ξ
n
,
T
θ
=
ρ
T
θ
,
T
θ
=
l
i
m
n
,
m
→
∞
ρ
T
ξ
n
,
T
ξ
m
=
0
.
Sinceξ
n
+
1
→
θ as n
→
∞ and ρ
(
θ
,
θ
)
=
0, then, by using Lemma 4, we get(82)
l
i
m
n
→
∞
ρ
ξ
n
+
1
,
T
θ
=
ρ
θ
,
T
θ
.
Regarding (81) and (82), we get ρ
(
θ
,
T
θ
)
=
0. Thus, by (ρ
1), we conclude that T
θ
=
θ.Theorem 15.
Suppose that(
M
,
ρ
) is a complete DMS, a self-mapping T
:
M
→
M is a generalized-(
α
,
ψ
)-MKC, and the following conditions are fulfilled:(i)
T is triangular α-orbital admissible mapping.
(ii)
There existsξ
0
∈
M such that α
(
ξ
0
,
T
ξ
0
)
≥
1.
(iii)
If{
ξ
n
} is a sequence in M such that α
(
ξ
n
,
ξ
n
+
1
)
≥
1 for all n and ξ
n
→
ξ
∈
M as n
→
∞, then α
(
ξ
n
,
ξ
)
≥
1 for all n.
Then,T has a fixed point; that is, there exists θ
∈
M such that T
θ
=
θ.Proof.
Define an iterative sequence{
ξ
n
} in M as in Theorem 14. Suppose that there is k
0
∈
N such that ξ
k
0
=
ξ
k
0
+
1; then, the proof is completed since u
=
ξ
k
0
=
ξ
k
0
+
1
=
T
ξ
k
0
=
T
u. So, it is interesting to assume that ξ
n
≠
ξ
n
+
1, for all n
∈
N
0. Consequently, we have (83)
ρ
ξ
n
,
ξ
n
+
1
>
0
,
∀
n
∈
N
0
.
From (48), we find that(84)
α
ξ
n
,
ξ
n
+
1
≥
1
,
∀
n
∈
N
0
.
Following the lines for the proofs of Steps1 and 2 in Theorem 14, we derive that {
ξ
n
} is a Cauchy sequence and(85)
l
i
m
n
,
m
→
∞
ρ
ξ
n
,
ξ
m
=
0
.Since (
M
,
ρ
) is a complete DMS, then there exists θ
∈
M such that(86)
l
i
m
n
→
∞
ρ
ξ
n
,
θ
=
ρ
θ
,
θ
=
l
i
m
n
,
m
→
∞
ρ
ξ
n
,
ξ
m
=
0
.
We will prove thatθ
=
T
θ. Suppose, on the contrary, that ρ
(
θ
,
T
θ
)
>
0.
Notice from (84), (86), and (iii) that(87)
α
ξ
n
,
θ
≥
1
,
∀
n
.
By using (ρ
3) and (87) together with the assumption of the theorem that T is a generalized-(α
-
ψ)-MKC mapping, we obtain(88)
ρ
T
θ
,
θ
≤
ρ
T
θ
,
T
ξ
n
+
ρ
ξ
n
+
1
,
θ
≤
α
ξ
n
,
θ
ρ
T
ξ
n
,
T
θ
+
ρ
ξ
n
+
1
,
θ
≤
ψ
m
a
x
T
ξ
n
,
θ
+
ρ
ξ
n
+
1
,
θ
,where(89)
m
a
x
T
ξ
n
,
θ
=
m
a
x
ρ
ξ
n
,
θ
,
ρ
ξ
n
,
ξ
n
+
1
,
ρ
θ
,
T
θ
,
1
4
ρ
ξ
n
,
T
θ
+
ρ
ξ
n
+
1
,
θ
.
Notice that asρ
(
θ
,
T
θ
)
>
0, then we must have m
a
x
T
(
ξ
n
,
θ
)
>
0.
Regarding (ρ
3), we have (90)
1
4
ρ
ξ
n
,
T
θ
+
ρ
ξ
n
+
1
,
θ
≤
max
⁡
ρ
ξ
n
,
θ
,
ρ
ξ
n
,
ξ
n
+
1
,
ρ
θ
,
T
θ
.
By the above inequality, we have(91)
m
a
x
T
ξ
n
,
θ
=
max
⁡
ρ
ξ
n
,
θ
,
ρ
ξ
n
,
ξ
n
+
1
,
ρ
θ
,
T
θ
.
Suppose thatm
a
x
T
(
ξ
n
,
θ
)
=
ρ
(
ξ
n
,
θ
); then, from (88), we get(92)
ρ
T
θ
,
θ
≤
ψ
ρ
ξ
n
,
θ
+
ρ
ξ
n
+
1
,
θ
<
ρ
ξ
n
,
θ
+
ρ
ξ
n
+
1
,
θ
.
Takingn
→
∞ in (92), we have(93)
ρ
T
θ
,
θ
<
ρ
θ
,
θ
+
ρ
θ
,
θ
=
0
,which is a contradiction.
Now, we suppose thatm
a
x
T
(
ξ
n
,
θ
)
=
ρ
(
ξ
n
,
ξ
n
+
1
); then, by (88), we find that (94)
ρ
T
θ
,
θ
≤
ψ
ρ
ξ
n
,
ξ
n
+
1
+
ρ
ξ
n
+
1
,
θ
<
ρ
ξ
n
,
ξ
n
+
1
+
ρ
ξ
n
+
1
,
θ
.
Lettingn
→
∞, this implies that (95)
ρ
T
θ
,
θ
<
l
i
m
n
→
∞
ρ
ξ
n
,
ξ
n
+
1
+
ρ
ξ
n
+
1
,
θ
=
0
,which is again a contradiction.
Finally, we suppose thatm
a
x
T
(
ξ
n
,
θ
)
=
ρ
(
θ
,
T
θ
); then, again from (88), we have (96)
ρ
T
θ
,
θ
≤
ψ
ρ
θ
,
T
θ
+
ρ
ξ
n
+
1
,
θ
.
Lettingn
→
∞, in the above inequality, we get(97)
ρ
T
θ
,
θ
≤
ψ
ρ
θ
,
T
θ
+
ρ
θ
,
θ
<
ρ
θ
,
T
θ
+
ρ
θ
,
θ
=
ρ
T
θ
,
θ
,so we also have a contradiction. Thus, we have ρ
(
T
θ
,
θ
)
=
0 and, by (ρ
1), we have T
θ
=
θ.


## 2.3. The Uniqueness of the Fixed Point

We propose the following conditions for the uniqueness of the fixed points of the mappings discussed in Sections2.1 and 2.2. Let F
i
x
(
T
) denote the set of fixed points of the mapping T.We, first, recollect the following interesting condition for uniqueness of a fixed point of an(
α
-
ψ
)-MKC mapping.(
H
) For all ξ
,
η
∈
F
i
x
(
T
), there exists ζ
∈
M such that α
(
ξ
,
ζ
)
≥
1 and α
(
η
,
ζ
)
≥
1, where we have the following.Theorem 16.
Putting condition (H) to the statements of Theorem 7 (resp., Theorem 8), one obtains that θ is the unique fixed point of T.Proof.
Letθ and ν be two distinct fixed points of T. From (
H
), there exists ζ
∈
M such that (98)
α
θ
,
ζ
≥
1
,
α
ν
,
ζ
≥
1
. Due to the fact that T is α-admissible, we have (99)
α
θ
,
T
ζ
≥
1
,
α
ν
,
T
ζ
≥
1
.
Inductively, we obtain(100)
α
θ
,
T
n
ζ
≥
1
,
α
ν
,
T
n
ζ
≥
1
,
∀
n
∈
N
.
From the above relation and sinceT is an (
α
,
ψ
)-MKC mapping, we have (101)
ρ
θ
,
T
n
ζ
≤
α
θ
,
T
n
-
1
ζ
ρ
T
θ
,
T
n
ζ
≤
ψ
ρ
θ
,
T
n
-
1
ζ
.
Iteratively, we get(102)
ρ
θ
,
T
n
ζ
≤
ψ
n
ρ
θ
,
ζ
.
Lettingn
→
∞, we obtain(103)
l
i
m
n
→
∞
ρ
θ
,
T
n
ζ
=
0
.Similarly, we can prove that(104)
l
i
m
n
→
∞
ρ
ν
,
T
n
ζ
=
0
.Using (ρ
3), we have (105)
ρ
θ
,
ν
≤
ρ
θ
,
T
n
ζ
+
ρ
T
n
ζ
,
ν
.
Takingn
→
∞, we find that(106)
ρ
θ
,
ν
=
0
, and so, by (ρ
1), θ
=
ν.As an alternative uniqueness condition for fixed points of(
α
-
ψ
)-MKC mapping, we suggest the following hypothesis:(U)
For allξ
,
η
∈
F
i
x
(
T
), then α
(
ξ
,
η
)
≥
1.Theorem 17.
Putting condition (U) to the statements of Theorem7 (resp., Theorem 8), one finds that θ is the unique fixed point of T.Proof.
Letθ
,
ν be two distinct fixed points of T. Then, by Lemma 4 part (C), we have(107)
ρ
θ
,
ν
>
0
.Due to the property of ψ (Ψ
2), we get(108)
ψ
ρ
θ
,
ν
>
0
.
Letε
=
ψ
(
ρ
(
θ
,
ν
)
); then, for any δ
>
0, we find that (109)
ε
=
ψ
ρ
θ
,
ν
<
ε
+
δ
.
Regarding (U) and the assumption of the theorem thatT is an (
α
,
ψ
)-MKC mapping, we obtain(110)
ρ
θ
,
ν
≤
α
θ
,
ν
ρ
T
θ
,
T
ν
<
ψ
ρ
θ
,
ν
<
ρ
θ
,
ν
,which is a contradiction. Thus, θ
=
ν.In what follows, we propose the conditions for the uniqueness of a fixed point of a generalized-(
α
-
ψ
)-MKC mapping:(H1)
For allξ
,
η
∈
F
i
x
(
T
), there exists ζ
∈
M such that α
(
ξ
,
ζ
)
≥
1, α
(
η
,
ζ
)
≥
1, and α
(
ζ
,
T
ζ
)
≥
1.
(H2)
Letξ
,
η
∈
F
i
x
(
T
). If there exists a sequence {
ζ
n
} in M such that α
(
ξ
,
ζ
n
)
≥
1, α
(
η
,
ζ
n
)
≥
1, and α
(
ζ
n
,
ζ
n
+
1
)
≥
1, then (111)
ρ
ζ
n
,
ζ
n
+
1
≤
inf
⁡
ρ
ξ
,
ζ
n
,
ρ
η
,
ζ
n
.
(H3)
For anyξ
∈
F
i
x
(
T
), then α
(
ξ
,
ξ
)
≥
1.Theorem 18.
Putting conditions (H1), (H2), and (H3) to the statements of Theorem14 (resp., Theorem 15), one has that θ is the unique fixed point of T.Proof.
Letθ
,
ν be two distant fixed points of T. From (H1), there exists ζ
∈
M such that(112)
α
θ
,
ζ
≥
1
,
α
ν
,
ζ
≥
1
,
α
ζ
,
T
ζ
≥
1
.
Owing to the fact thatT is triangular α-orbital admissible and α
(
ζ
,
T
ζ
)
≥
1, by (T
3), we have(113)
α
T
ζ
,
T
2
ζ
≥
1
.
Inductively, we find(114)
α
T
n
-
1
ζ
,
T
n
ζ
≥
1
,
∀
n
∈
N
.
Now, sinceα
(
θ
,
ζ
)
≥
1 and α
(
ζ
,
T
ζ
)
≥
1, then, by (T
4), we deduce(115)
α
θ
,
T
z
≥
1
.
Again, by (T
4), since α
(
θ
,
T
ζ
)
≥
1 and α
(
T
ζ
,
T
2
ζ
)
≥
1, we derive(116)
α
θ
,
T
2
ζ
≥
1
.
Inductively, we get(117)
α
θ
,
T
n
ζ
≥
1
,
∀
n
∈
N
.
In an analogous way, we will prove that(118)
α
ν
,
T
n
ζ
≥
1
,
∀
n
∈
N
.
Define an iterative sequence{
ζ
n
} by ζ
n
+
1
=
T
ζ
n, for all n
≥
0 and ζ
0
=
ζ
.
Step 1. We will prove that(119)
l
i
m
n
→
∞
ρ
θ
,
ζ
n
=
0
.
By (117) and the statement of the theorem that T is generalized-(
α
,
ψ
)-MKC mapping, we have (120)
ρ
θ
,
ζ
n
+
1
≤
α
θ
,
ζ
n
ρ
T
θ
,
T
ζ
n
≤
ψ
m
a
x
T
θ
,
ζ
n
.
Ifψ
(
m
a
x
T
(
θ
,
ζ
n
)
)
=
0, then(121)
l
i
m
n
→
∞
ρ
θ
,
ζ
n
+
1
=
0
. Now, suppose that ψ
(
m
a
x
T
(
θ
,
ζ
n
)
)
>
0; then, m
a
x
T
(
θ
,
ζ
n
)
>
0. As T is a generalized-(
α
,
ψ
)-MKC mapping, we get (122)
ρ
θ
,
ζ
n
+
1
≤
α
θ
,
ζ
n
ρ
T
θ
,
T
ζ
n
≤
ψ
m
a
x
T
θ
,
ζ
n
<
m
a
x
T
θ
,
ζ
n
,where(123)
m
a
x
T
ζ
n
,
θ
=
m
a
x
ρ
ζ
n
,
θ
,
ρ
ζ
n
,
ζ
n
+
1
,
ρ
θ
,
T
θ
,
1
4
ρ
ζ
n
,
T
θ
+
ρ
ζ
n
+
1
,
θ
.Regarding (H2) and (ρ
3), we have(124)
m
a
x
T
ζ
n
,
θ
=
ρ
ζ
n
,
θ
.Thus, we have(125)
ρ
θ
,
ζ
n
+
1
<
ρ
θ
,
ζ
n
.
Lettingn
→
∞ in the inequality above, we obtain(126)
l
i
m
n
→
∞
ρ
θ
,
ζ
n
+
1
<
l
i
m
n
→
∞
ρ
θ
,
ζ
n
,which is a contradiction. Then,(127)
m
a
x
T
ζ
n
,
θ
=
ρ
ζ
n
,
θ
=
0
.
Hence, we get(128)
l
i
m
n
→
∞
ρ
θ
,
ζ
n
=
0
.
Step 2. We will prove that (129)
l
i
m
n
→
∞
ρ
ν
,
ζ
n
=
0
.
By (H3) and the assumption of the theorem thatT is a generalized-(
α
,
ψ
)-MKC mapping, we find that(130)
ρ
ν
,
ν
≤
α
ν
,
ν
ρ
T
v
,
T
ν
≤
ψ
m
a
x
T
ν
,
ν
=
ψ
ρ
ν
,
ν
.Suppose, on the contrary, that ρ
(
ν
,
ν
)
>
0. Then, from the above inequality, we obtain(131)
ρ
ν
,
ν
≤
ψ
ρ
ν
,
ν
<
ρ
ν
,
ν
,which is a contradiction. Thus, ρ
(
ν
,
ν
)
=
ρ
(
ν
,
T
ν
)
=
0.
In an analogous way of Step1, we can complete the proof of(132)
l
i
m
n
→
∞
ρ
ν
,
ζ
n
=
0
.
By (ρ
3), we have(133)
ρ
θ
,
ν
≤
ρ
θ
,
T
n
-
1
ζ
+
ρ
T
n
-
1
ζ
,
ν
=
ρ
θ
,
ζ
n
+
ρ
ζ
n
,
ν
.
Lettingn
→
∞ in the above inequality, we find that(134)
ρ
θ
,
ν
=
0
;by (ρ
1), we have θ
=
ν.


## 3. Ulam-Hyers Stability

Definition 19 (see [8]).
Let(
M
,
ρ
) be a complete DMS and let T
:
M
→
M be a self-mapping. The fixed point inclusion(135)
ξ
=
T
ξ
,
ξ
∈
M
,is called generalized Ulam-Hyers stable if and only if ς
:
[
0
,
∞
)
→
[
0
,
∞
) is increasing and continuous in 0 and ς
(
0
)
=
0, such that, for each ε
>
0 and for each solution η
∗
∈
M of the equation(136)
ρ
y
,
T
η
≤
ε
,there exists a solution ξ
∗ of the fixed point inclusion (135) such that (137)
ρ
η
∗
,
ξ
∗
≤
ς
ε
.
If there existsc
>
0 such that ς
(
t
)
:
=
c
·
t, for each t
∈
[
0
,
∞
), then the fixed point (135) is said to be Ulam-Hyers stable.Theorem 20.
Let(
M
,
ρ
) be a complete DMS and let T
:
M
→
M be a self-mapping. Suppose that all the hypotheses of Theorem 18 hold. Additionally, one supposes that (i)
ψ
(
t
)
<
t and the function β
:
R
0
+
→
R
0
+, β
(
r
)
:
=
r
-
ψ
(
r
) is strictly increasing and onto;
(ii)
for any solutionη
∗
∈
M of (136) one has α
ξ
∗
,
η
∗
≥
1, where ξ
∗
∈
F
i
x
T
.
In these conditions, the fixed point inclusion (135) is generalized Ulam-Hyers stable.Proof.
We are in the conditions of Theorem18; hence, there exists ξ
∗
∈
F
i
x
T such that ρ
(
ξ
∗
,
ξ
∗
)
=
0. Let ε
>
0 and η
∗ be a solution of (136).
From (ii), we have thatα
ξ
∗
,
η
∗
≥
1, and because T is α-admissible we will obtain that α
(
T
ξ
∗
,
T
η
∗
)
=
α
(
ξ
∗
,
T
η
∗
)
≥
1.
We have(138)
ρ
ξ
∗
,
η
∗
=
ρ
T
ξ
∗
,
η
∗
≤
ρ
T
ξ
∗
,
T
η
∗
+
ρ
η
∗
,
T
η
∗
≤
ρ
T
ξ
∗
,
T
η
∗
+
ρ
η
∗
,
T
η
∗
≤
α
T
ξ
∗
,
T
η
∗
ρ
T
ξ
∗
,
T
η
∗
+
ρ
η
∗
,
T
η
∗
≤
ψ
m
a
x
T
ξ
∗
,
η
∗
+
ε
,where(139)
m
a
x
T
ξ
∗
,
η
∗
=
m
a
x
ρ
ξ
∗
,
η
∗
,
ρ
ξ
∗
,
T
ξ
∗
,
ρ
η
∗
,
T
η
∗
,
ρ
ξ
∗
,
T
η
∗
+
ρ
η
∗
,
T
ξ
∗
4
.We have(140)
ρ
ξ
∗
,
T
η
∗
+
ρ
η
∗
,
T
ξ
∗
4
≤
1
4
ρ
ξ
∗
,
η
∗
+
ρ
η
∗
,
T
η
∗
+
ρ
η
∗
,
ξ
∗
=
1
2
ρ
ξ
∗
,
η
∗
+
1
4
ρ
η
∗
,
T
η
∗
≤
1
2
ρ
ξ
∗
,
η
∗
+
1
4
ε
≤
m
a
x
ρ
ξ
∗
,
η
∗
,
ε
.
From here, we have(141)
m
a
x
T
ξ
∗
,
η
∗
≤
m
a
x
ρ
ξ
∗
,
η
∗
,
ε
.
It is clear that ifρ
(
ξ
∗
,
η
∗
)
≤
ε the proof is completed.
Suppose thatm
a
x
ρ
(
ξ
∗
,
η
∗
)
,
ε
=
ρ
(
ξ
∗
,
η
∗
). Consequently, we have m
a
x
T
(
ξ
∗
,
η
∗
)
≤
ρ
(
ξ
∗
,
η
∗
). On account of (138), we derive that (142)
ρ
ξ
∗
,
η
∗
≤
ψ
ρ
ξ
∗
,
y
∗
+
ε
,
ρ
ξ
∗
,
η
∗
-
ψ
ρ
ξ
∗
,
η
∗
≤
ε
.
From condition (i), we get that(143)
β
ρ
ξ
∗
,
η
∗
≤
ε
,and hence (144)
ρ
ξ
∗
,
η
∗
≤
β
-
1
ε
.Hence, (135) is generalized Ulam-Hyers stable.Corollary 21.
Suppose that(
M
,
ρ
) is a complete DMS, T
:
M
→
M is a self-mapping, and all the hypotheses of Theorem 16 hold. One supposes also that(i)
ψ
(
t
)
<
t and the function β
:
R
0
+
→
R
0
+, β
(
r
)
:
=
r
-
ψ
(
r
) is strictly increasing and onto;
(ii)
for any solutionη
∗
∈
M of (136) one has α
ξ
∗
,
η
∗
≥
1, where ξ
∗
∈
F
i
x
T.
In these conditions, the fixed point inclusion (135) is generalized Ulam-Hyers stable.The proof is an analog of the proof of Theorem20.


## 4. Conclusion

It is possible to list several existing results as a consequence of our main results by choosing both auxiliary functionsα and ψ in a proper way like in the literature [5, 6, 9–11] and so on.



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*Source: 9623597-2016-05-25.xml*