# Fixed Point Theorems forL-Contractions in Generalized Metric Spaces

Authors: Seong-Hoon Cho
Journal: Abstract and Applied Analysis
 (2018)

---

## Abstract

In this paper, the notion ofL-contractions is introduced and a new fixed point theorem for such contractions is established.


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## Body

## 1. Introduction and Preliminaries

Branciari [1] introduced the notion of generalized metric spaces and obtained a generalization of the Banach contraction principle, whereafter many authors proved various fixed point results in such spaces, for example, [2–8] and references therein. Also, Suzukiet al. [9] and Abtahiet al. [10] studied ν-generalized metric spaces and proved the Banach and Kannan contraction principles in such spaces, and Mitrovićet al. [11] introduced the notion of b
ν
(
s
)-generalized metric spaces and Banach and Reich contraction principles in such spaces.In particular, Jleli and Samet [12] introduced the notion of θ-contractions and gave a generalization of the Banach contraction principle in generalized metric spaces, where θ
:
(
0
,
∞
)
→
(
1
,
∞
) is a function satisfying the following conditions:θ
1
θ is nondecreasing;
θ
2
∀
{
t
n
}
⊂
(
0
,
∞
), (1)
lim
n
→
∞
⁡
θ
t
n
=
1
⇔
lim
n
→
∞
⁡
t
n
=
0
+
;
θ
3
∃
r
∈
(
0,1
)
∧
l
∈
(
0
,
∞
): (2)
lim
t
→
0
+
⁡
θ
t
-
1
t
r
=
l
.Also, Ahmadet al. [13] extended the result of Jleli and Samet [12] to metric spaces by applying the following simple condition (θ4) instead of (θ3).θ
4
θ is continuous on (
0
,
∞
).Recently, Khojastehet al. [14] introduced the notion of Z-contractions by defining the concept of simulation functions. They unified some existing metric fixed point results. Afterward, many authors ([15–19] and references therein) obtained generalizations of the result of [14].In the paper, we introduce the concept of a new type of contraction maps, and we establish a new fixed point theorem for such contraction maps in the setting of generalized metric spaces.LetL be the family of all mappings ξ
:
[
1
,
∞
)
×
[
1
,
∞
)
→
R such thatξ
1
ξ
(
1,1
)
=
1;
ξ
2
ξ
(
t
,
s
)
<
s
/
t
∀
s
,
t
>
1;
ξ
3
for any sequence{
t
n
}
,
{
s
n
}
⊂
(
1
,
∞
) with t
n
≤
s
n
∀
n
=
1,2
,
3
,
⋯
(3)
lim
n
→
∞
⁡
t
n
=
lim
n
→
∞
⁡
s
n
>
1
⇒
lim
n
→
∞
⁡
sup
⁡
ξ
t
n
,
s
n
<
1
.We say thatξ
∈
L is a L-simulation function.Note thatξ
(
t
,
t
)
<
1
∀
t
>
1.Example 1.
Letξ
b
,
ξ
w
,
ξ
:
[
1
,
∞
)
×
[
1
,
∞
)
→
R be functions defined as follows, respectively:(1)
ξ
b
(
t
,
s
)
=
s
k
/
t
∀
t
,
s
≥
1, where k
∈
(
0,1
);
(2)
ξ
w
(
t
,
s
)
=
s
/
t
ϕ
(
s
)
∀
t
,
s
≥
1, where ϕ
:
[
1
,
∞
)
→
[
1
,
∞
) is nondecreasing and lower semicontinuous such that ϕ
-
1
(
{
1
}
)
=
1;(4)
ξ
t
,
s
=
1
i
f
s
,
t
=
1,1
,
s
2
t
i
f
s
<
t
,
s
λ
t
o
t
h
e
r
w
i
s
e
,
∀
s
,
t
≥
1, where λ
∈
(
0,1
).
Thenξ
b
,
ξ
w
,
ξ
∈
L.We recall the following definitions which are in [1].LetX be a nonempty set, and let d
:
X
×
X
→
[
0
,
∞
) be a map such that for all x
,
y
∈
X and for all distinct points u
,
v
∈
X, each of them is different from x and y:(d1)
d
(
x
,
y
)
=
0 if and only if x
=
y;
(d2)
d
(
x
,
y
)
=
d
(
y
,
x
);
(d3)
d
(
x
,
y
)
≤
d
(
x
,
u
)
+
d
(
u
,
v
)
+
d
(
v
,
y
).Thend is called a generalized metric on X and (
X
,
d
) is called a generalized metric space.Let(
X
,
d
) be a generalized metric space, let {
x
n
}
⊂
X be a sequence, and x
∈
X.Then we say that(1)
{
x
n
} is convergent to x (denoted by l
i
m
n
→
∞
x
n
=
x) if and only if l
i
m
n
→
∞
d
(
x
n
,
x
)
=
0;
(2)
{
x
n
} is Cauchy if and only if l
i
m
n
,
m
→
∞
d
(
x
n
,
x
m
)
=
0;
(3)
(
X
,
d
) is complete if and only if every Cauchy sequence in X is convergent to some point in X.Let(
X
,
d
) be a generalized metric space.A mapT
:
X
→
X is calledcontinuous at x
∈
X if, for any V
∈
τ containing T
x, there exists U
∈
τ containing x such that T
U
⊂
V, where τ is the topology on X induced by the generalized metric d. That is,(5)
τ
=
U
⊂
X
:
∀
x
∈
U
,
∃
B
∈
β
,
x
∈
B
⊂
U
,
β
=
B
x
,
r
:
x
∈
X
,
∀
r
>
0
,
B
x
,
r
=
y
∈
X
:
d
x
,
y
<
r
.IfT is continuous at each point x
∈
X, then it is calledcontinuous.Note thatT is continuous if and only if it is sequentially continuous, i.e., lim
n
→
∞
⁡
d
(
T
x
n
,
T
x
)
=
0 for any sequence {
x
n
}
⊂
X with lim
n
→
∞
⁡
d
(
x
n
,
x
)
=
0.Remark 2 (see [6]).
Ifd is a generalized metric on X, then it is not continuous in each coordinate.Lemma 3 (see [20]).
Let(
X
,
d
) be a generalized metric space, let {
x
n
}
⊂
X be a Cauchy sequence, and x
,
y
∈
X. If there exists a positive integer N such that (1)
x
n
≠
x
m
∀
n
,
m
>
N;
(2)
x
n
≠
x
∀
n
>
N;
(3)
x
n
≠
y
∀
n
>
N;
(4)
lim
n
→
∞
⁡
d
(
x
n
,
x
)
=
lim
n
→
∞
⁡
d
(
x
n
,
y
),
thenx
=
y.


## 2. Fixed Point Theorems

We denote byΘ the class of all functions θ
:
(
0
,
∞
)
→
(
1
,
∞
) such that conditions (θ1) and (θ2) hold.A mappingT
:
X
→
X is called L-contraction with respect to ξ if there exist θ
∈
Θ and ξ
∈
L such that, for all x
,
y
∈
X with d
(
T
x
,
T
y
)
>
0, (6)
ξ
θ
d
T
x
,
T
y
,
θ
d
x
,
y
≥
1
.Note that ifT is L-contraction with respect to ξ, then it is continuous. In fact, let x
∈
X be a point and let {
x
n
}
⊂
X be any sequence such that (7)
lim
n
→
∞
⁡
d
x
n
,
x
=
0
+
,
d
T
x
n
,
T
x
>
0
∀
n
=
1,2
,
3
,
⋯
.Then from (θ
2) lim
n
→
∞
⁡
θ
(
d
(
x
n
,
x
)
)
=
1.It follows from (6) and (ξ
2) that (8)
1
≤
ξ
θ
d
T
x
n
,
T
x
,
θ
d
x
n
,
x
<
θ
d
x
n
,
x
θ
d
T
x
n
,
T
x
, which implies (9)
θ
d
T
x
n
,
T
x
<
θ
d
x
n
,
x
.Sinceθ is nondecreasing, we have (10)
d
T
x
n
,
T
x
<
d
x
n
,
x
,and so(11)
lim
n
→
∞
⁡
d
T
x
n
,
T
x
=
0
.Hence T is continuous.Now, we prove our main result.Theorem 4.
Let(
X
,
d
) be a complete generalized metric space, and let T
:
X
→
X be a L-contraction with respect to ξ.
ThenT has a unique fixed point, and for every initial point x
0
∈
X, the Picard sequence {
T
n
x
0
} converges to the fixed point.Proof.
Firstly, we show uniqueness of fixed point whenever it exists.
Assume thatw and u are fixed points of T.
Ifu
≠
z, then d
(
w
,
u
)
>
0, and so it follows from (6) that(12)
1
≤
ξ
θ
d
T
w
,
T
u
,
θ
d
w
,
u
=
ξ
θ
d
w
,
u
,
θ
d
w
,
u
<
θ
d
w
,
u
θ
d
w
,
u
. Hence (13)
θ
d
w
,
u
<
θ
d
w
,
u which is a contradiction.
Hencew
=
u, and fixed point of T is unique.
Secondly, we prove existence of fixed point.
Letx
0
∈
X be a point. Define a sequence {
x
n
}
⊂
X by x
n
=
T
x
n
-
1
=
T
n
x
0
∀
n
=
1,2
,
3
⋯.
Ifx
n
0
=
x
n
0
+
1 for some n
0
∈
N, then x
n
0 is a fixed point of T, and the proof is finished.
Assume that(14)
x
n
-
1
≠
x
n
∀
n
=
1,2
,
3
⋯
.
It follows from (6) and (14) that ∀
n
=
1,2
,
3
,
⋯
(15)
1
≤
ξ
θ
d
T
x
n
-
1
,
T
x
n
,
θ
d
x
n
-
1
,
x
n
=
ξ
θ
d
x
n
,
x
n
+
1
,
θ
d
x
n
-
1
,
x
n
<
θ
d
x
n
-
1
,
x
n
θ
d
x
n
,
x
n
+
1
.
Consequently, we obtain that(16)
θ
d
x
n
,
x
n
+
1
<
θ
d
x
n
-
1
,
x
n
∀
n
=
1,2
,
3
,
⋯ which implies (17)
d
x
n
,
x
n
+
1
<
d
x
n
-
1
,
x
n
∀
n
=
1,2
,
3
,
⋯
.
Hence{
d
(
x
n
-
1
,
x
n
)
} is a decreasing sequence, and so there exists r
≥
0 such that (18)
lim
n
→
∞
⁡
d
x
n
-
1
,
x
n
=
r
.
We now show thatr
=
0.
Assume thatr
≠
0.
Then it follows from(
θ
2
)that(19)
lim
n
→
∞
⁡
θ
d
x
n
-
1
,
x
n
≠
1
,and so(20)
lim
n
→
∞
⁡
θ
d
x
n
-
1
,
x
n
>
1
.
Lets
n
=
θ
(
d
(
x
n
-
1
,
x
n
)
) and t
n
=
θ
(
d
(
x
n
,
x
n
+
1
)
)
∀
n
=
1,2
,
3
,
⋯
.
From(
ξ
3
) we obtain (21)
1
≤
lim
n
→
∞
⁡
sup
⁡
ξ
t
n
,
s
n
<
1 which is a contradiction.
Thus we have(22)
lim
n
→
∞
⁡
d
x
n
-
1
,
x
n
=
0and so (23)
lim
n
→
∞
⁡
θ
d
x
n
-
1
,
x
n
=
1
.
We show that(24)
lim
n
→
∞
⁡
d
x
n
-
1
,
x
n
+
1
=
0
.
We consider three cases.
Case
1. x
n
≠
x
n
+
2
∀
n
=
1,2
,
3
,
⋯
.
From (6) and (14) we obtain that ∀
n
=
1,2
,
3
,
⋯
(25)
1
≤
ξ
θ
d
T
x
n
-
1
,
T
x
n
+
1
,
θ
d
x
n
-
1
,
x
n
+
1
=
ξ
θ
d
x
n
,
x
n
+
2
,
θ
d
x
n
-
1
,
x
n
+
1
<
θ
d
x
n
-
1
,
x
n
+
1
θ
d
x
n
,
x
n
+
2
, and so (26)
θ
d
x
n
,
x
n
+
2
<
θ
d
x
n
-
1
,
x
n
+
1
∀
n
=
1,2
,
3
,
⋯ which implies (27)
d
x
n
,
x
n
+
2
<
d
x
n
-
1
,
x
n
+
1
∀
n
=
1,2
,
3
,
⋯
. Hence {
d
(
x
n
-
1
,
x
n
+
1
)
} is decreasing.
In a manner similar to that which proved (22), we have (28)
lim
n
→
∞
⁡
d
x
n
-
1
,
x
n
+
1
=
0
.
Case
2. There exists n
0
≥
1 such that x
n
0
=
x
n
0
+
2.
From the first term to then
0 th term shall be removed, and let x
n
=
x
n
0
+
n
∀
n
=
1,2
,
3
,
⋯
.
Thenx
n
≠
x
n
+
2
∀
n
=
1,2
,
3
,
⋯. By Case 1, we have (29)
lim
n
→
∞
⁡
d
x
n
-
1
,
x
n
+
1
=
0
.
Case
3. x
n
=
x
n
+
2
∀
n
=
0,1
,
2
,
⋯.
We have(30)
d
x
n
-
1
,
x
n
+
1
=
0
∀
n
=
1,2
,
3
,
⋯
.
Hence(31)
lim
n
→
∞
⁡
d
x
n
-
1
,
x
n
+
1
=
0
.
In all cases, (24) is satisfied.
Now, we show that{
x
n
} is bounded.
If{
x
n
} is not bounded, then there exists a subsequence {
x
n
(
k
)
} of {
x
n
} such that n
1
=
1 and ∀
k
=
1,2
,
3
,
…
,
n
(
k
+
1
) is the minimum integer greater than n
(
k
) with (32)
d
x
n
k
+
1
,
x
n
k
>
1
,
d
x
l
,
x
n
k
≤
1for n
(
k
)
≤
l
≤
n
(
k
+
1
)
-
1
.
Then we have(33)
1
<
d
x
n
k
+
1
,
x
n
k
≤
d
x
n
k
+
1
,
x
n
k
+
1
-
2
+
d
x
n
k
+
1
-
2
,
x
n
k
+
1
-
1
+
d
x
n
k
+
1
-
1
,
x
n
k
≤
d
x
n
k
+
1
,
x
n
k
+
1
-
2
+
d
x
n
k
+
1
-
2
,
x
n
k
+
1
-
1
+
1
.
By lettingk
→
∞ in the above, we obtain (34)
lim
k
→
∞
⁡
d
x
n
k
+
1
,
x
n
k
=
1
.
By using (22), (34), and condition (d3), we deduce that (35)
lim
k
→
∞
⁡
d
x
n
k
+
1
-
1
,
x
n
k
-
1
=
1
.
It follows from(
θ
2
), (34), and (35) that (36)
lim
k
→
∞
⁡
θ
d
x
n
k
+
1
,
x
n
k
>
1
,
(37)
lim
n
→
∞
⁡
θ
d
x
n
k
+
1
-
1
,
x
n
k
-
1
>
1
.
From (6) and (32) we infer that(38)
1
≤
ξ
θ
d
T
x
n
k
+
1
-
1
,
T
x
n
k
-
1
,
θ
d
x
n
k
+
1
-
1
,
x
n
k
-
1
=
ξ
θ
d
x
n
k
+
1
,
x
n
k
,
θ
d
x
n
k
+
1
-
1
,
x
n
k
-
1
<
θ
d
x
n
k
+
1
-
1
,
x
n
k
-
1
θ
d
x
n
k
+
1
,
x
n
kwhich implies (39)
θ
d
x
n
k
+
1
,
x
n
k
<
θ
d
x
n
k
+
1
-
1
,
x
n
k
-
1
.
Let(40)
s
n
=
θ
(
d
x
n
k
+
1
-
1
,
x
n
k
-
1
,
t
n
=
θ
d
x
n
k
+
1
,
x
n
k
∀
n
=
1,2
,
3
,
⋯
.
Thent
k
<
s
k
∀
n
=
1,2
,
3
,
⋯ and lim
k
→
∞
⁡
t
k
=
lim
k
→
∞
⁡
s
k
>
1.
It follows from(
ξ3) that (41)
1
≤
lim
k
→
∞
⁡
sup
⁡
ξ
t
k
,
s
k
<
1 which is a contradiction.
Thus{
x
n
} is bounded.
Now, we show that{
x
n
} is a Cauchy sequence.
Let(42)
M
n
=
sup
⁡
d
x
i
,
x
j
:
i
,
j
≥
n
.
Clearly,(43)
0
≤
M
n
+
1
≤
M
n
<
∞
∀
n
=
1,2
,
3
,
⋯and so there exists M
≥
0 such that (44)
lim
n
→
∞
⁡
M
n
=
M
.
Assume thatM
>
0.
It follows from (42) that ∀
k
=
1,2
,
3
,
⋯ there exist n
(
k
)
,
m
(
k
)
≥
k with (45)
M
k
-
1
k
<
d
x
m
k
,
x
n
k
≤
M
k
.
So(46)
lim
k
→
∞
⁡
d
x
m
k
,
x
n
k
=
lim
k
→
∞
⁡
M
k
=
M
.
It follows from (6) and (14) that(47)
ξ
θ
d
T
x
m
k
-
1
,
T
x
n
k
-
1
,
θ
d
x
m
k
-
1
,
x
n
k
-
1
=
ξ
θ
d
x
m
k
,
x
n
k
,
θ
d
x
m
k
-
1
,
x
n
k
-
1
<
θ
d
x
m
k
-
1
,
x
n
k
-
1
θ
d
x
m
k
,
x
n
k which implies (48)
θ
d
x
m
k
,
x
n
k
<
θ
d
x
m
k
-
1
,
x
n
k
-
1
. Hence we have(49)
d
x
m
k
,
x
n
k
<
d
x
m
k
-
1
,
x
n
k
-
1
≤
d
x
m
k
-
1
,
x
m
k
+
d
x
m
k
,
x
n
k
+
d
x
n
k
,
x
n
k
-
1
.
Lettingk
→
∞ in the above inequality, we obtain (50)
lim
n
→
∞
⁡
d
x
m
k
-
1
x
n
k
-
1
=
M
.
Let(51)
s
k
=
θ
d
x
m
k
-
1
,
x
n
k
-
1
,
t
k
=
θ
d
x
m
k
,
x
n
k
∀
k
=
1,2
,
3
,
⋯
.
Thent
k
<
s
k
∀
k
=
1,2
,
3
,
⋯
. Since M
>
0, (52)
lim
k
→
∞
⁡
t
k
=
lim
k
→
∞
⁡
s
k
>
1
.
Thus we have(53)
1
≤
lim
k
→
∞
⁡
sup
⁡
ξ
t
k
,
s
k
<
1 which is a contradiction.
HenceM
=
0, and hence {
x
n
} is a Cauchy sequence.
SinceX is complete, there exists z
∈
X such that (54)
lim
n
→
∞
⁡
d
x
n
,
z
=
0
.
BecauseT is continuous, (55)
lim
n
→
∞
⁡
d
x
n
,
T
z
=
lim
n
→
∞
⁡
d
T
x
n
-
1
,
T
z
=
0
.
By Lemma3, z
=
T
z
.We give an example to illustrate Theorem4.Example 5.
LetX
=
{
1,2
,
3,4
} and define d
:
X
×
X
→
[
0
,
∞
) as follows:(56)
d
1,2
=
d
2,1
=
3
,
d
2,3
=
d
3,2
=
d
1,3
=
d
3,1
=
1
,
d
1,4
=
d
4,1
=
d
2,4
=
d
4,2
=
d
3,4
=
d
4,3
=
4
,
d
x
,
x
=
0
∀
x
∈
X
.
Then(
X
,
d
) is a complete generalized metric space, but not a metric space (see [21]).
Define a mapT
:
X
→
X by (57)
T
x
=
3
x
≠
4
,
1
x
=
4
. And define a function θ
:
(
0
,
∞
)
→
(
1
,
∞
) by (58)
θ
t
=
e
t
.
We now show thatT is a L-contraction with respect to ξ
b, where ξ
b
(
t
,
s
)
=
s
k
/
t
∀
t
,
s
≥
1, k
=
1
/
2.
We have(59)
d
T
x
,
T
y
=
d
1,3
=
1
x
=
4
,
y
≠
4
,
d
1,1
=
0
x
=
4
,
y
=
4
,
d
3,3
=
0
x
≠
4
,
y
≠
4 so (60)
d
T
x
,
T
y
>
0
⇔
x
=
4
,
y
≠
4
.
We have, forx
=
4 and y
≠
4, (61)
d
x
,
y
=
4
,
d
T
x
,
T
y
=
1
.
We deduce that, for allx
,
y
∈
X with d
(
T
x
,
T
y
)
>
0,(62)
ξ
b
θ
d
T
x
,
T
y
,
θ
d
x
,
y
=
θ
d
x
,
y
k
θ
d
T
x
,
T
y
=
e
4
1
/
2
e
1
=
e
>
1
.
Thus all hypotheses of Theorem4 are satisfied, and T has a fixed point x
∗
=
3.Note that Banach’s contraction principle is not satisfied with the usual metricρ
x
,
y
=
x
-
y
∀
x
,
y
∈
X. In fact, if x
=
2, y
=
4, then (63)
ρ
T
2
,
T
4
≤
k
ρ
2,4
,
k
∈
0,1 which implies (64)
k
≥
1
.Also, note that theθ-contraction condition [13] does not hold.Letθ
(
t
)
=
e
t
,
∀
t
>
0.Thenθ
(
t
) satisfies conditions (θ
1
), (
θ2), and (
θ
4
).If(65)
θ
ρ
T
2
,
T
4
≤
θ
ρ
2,4
k
,
w
h
e
r
e
k
∈
0,1 then (66)
e
2
≤
e
2
k and so k
≥
1. Hence T is not θ-contraction map.By takingξ
=
ξ
b in Theorem 4, we obtain Corollary 6.Corollary 6.
Let(
X
,
d
) be a complete generalized metric space, and let T
:
X
→
X be a mapping such that for all x
,
y
∈
X with d
(
T
x
,
T
y
)
>
0
(67)
θ
d
T
x
,
T
y
≤
θ
d
x
,
y
k where θ
∈
Θ and k
∈
(
0,1
).
ThenT has a unique fixed point.Remark 7.
Corollary6 is a generalization of Theorem 2.1 of [12] without condition (θ
3) and Theorem 2.2 of [13] without condition (θ
4).By takingξ
=
ξ
w in Theorem 4, we obtain Corollary 8.Corollary 8.
Let(
X
,
d
) be a complete generalized metric space, and let T
:
X
→
X be a mapping such that for all x
,
y
∈
X with d
(
T
x
,
T
y
)
>
0
(68)
θ
d
T
x
,
T
y
≤
θ
d
x
,
y
ϕ
θ
d
x
,
y where θ
∈
Θ and ϕ
:
[
1
,
∞
)
→
[
1
,
∞
) is nondecreasing and lower semicontinuous such that ϕ
-
1
(
{
1
}
)
=
1.
ThenT has a unique fixed point.Corollary 9.
Let(
X
,
d
) be a complete generalized metric space, and let T
:
X
→
X be a mapping such that for all x
,
y
∈
X with d
(
T
x
,
T
y
)
>
0
(69)
d
T
x
,
T
y
≤
d
x
,
y
-
φ
d
x
,
ywhere φ
:
[
0
,
∞
)
→
[
0
,
∞
) is nondecreasing and lower semicontinuous such that φ
-
1
(
{
0
}
)
=
0.
ThenT has a unique fixed point.Proof.
Condition (69) implies T is continuous.
Letθ
(
t
)
=
e
t
,
∀
t
>
0
.
From (69) we have that, for all x
,
y
∈
X with d
(
T
x
,
T
y
)
>
0,(70)
θ
d
T
x
,
T
y
=
e
d
T
x
,
T
y
≤
e
d
x
,
y
-
φ
d
x
,
y
=
e
d
x
,
y
e
φ
d
x
,
y
.
Letφ
(
t
)
=
ln
⁡
(
ϕ
(
θ
(
t
)
)
)
,
∀
t
≥
0, where ϕ
:
[
1
,
∞
)
→
[
1
,
∞
) is nondecreasing and lower semicontinuous such that ϕ
-
1
(
{
1
}
)
=
1.
Thenφ is nondecreasing and lower semicontinuous, and (71)
φ
t
=
0
⇔
ϕ
θ
t
=
1
⇔
θ
t
=
e
t
=
1
⇔
t
=
0
.
It follows from (70) that, for all x
,
y
∈
X with d
(
T
x
,
T
y
)
>
0, (72)
θ
d
T
x
,
T
y
≤
θ
d
x
,
y
e
ln
⁡
ϕ
θ
d
x
,
y
=
θ
d
x
,
y
ϕ
θ
d
x
,
y
.
By Corollary8, T has a unique fixed point.By takingθ
(
t
)
=
2
-
2
/
π
arctan
⁡
(
1
/
t
α
), where α
∈
(
0,1
)
,
t
>
0 in Corollary 8, we obtain the following result.Corollary 10.
Let(
X
,
d
) be a complete generalized metric space, and let T
:
X
→
X be a mapping such that for all x
,
y
∈
X with d
(
T
x
,
T
y
)
>
0
(73)
2
-
2
π
arctan
⁡
1
d
T
x
,
T
y
α
≤
2
-
2
/
π
arctan
⁡
1
/
d
x
,
y
α
ϕ
2
-
2
/
π
arctan
⁡
1
/
d
x
,
y
α where α
∈
(
0,1
) and ϕ
:
[
1
,
∞
)
→
[
1
,
∞
) is nondecreasing and lower semicontinuous such that ϕ
-
1
(
{
1
}
)
=
1.
ThenT has a unique fixed point.



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*Source: 1327691-2018-12-02.xml*