# Some Fixed Point Theorems in Complex Valued b-Metric Spaces

Authors: A. K. Dubey; Rita Shukla; R. P. Dubey
Journal: Journal of Complex Systems
 (2015)

---

## Abstract

Recently, Azam et al. introduced the notion of complex valued metric spaces and proved fixed point theorems under the contraction condition. Rao et al. introduced the notion of complex valued b-metric spaces. In this paper, we obtain some fixed point results for the mapping satisfying rational expressions in complex valued b-metric spaces. Also, an example is given to illustrate our obtained result.


---

## Body

## 1. Introduction

Banach contraction principle [1] is one of the most important results in fixed point theory. Later, a large number of articles have been devoted to the improvement and generalization of the Banach contraction principle by using different form of contraction condition in various spaces. Bakhtin [2] introduced the notion of b-metric space which is a generalized form of metric spaces. Azam et al. [3] introduced the notion of complex valued metric spaces which is more general than well-known metric spaces and also proved common fixed point theorems for mappings satisfying rational expression. Afterwards, the concept of complex valued b-metric spaces was introduced in 2013 by Rao et al. [4]. In the sequel, Mukheimer [5] proved some common fixed point theorems in complex valued b-metric spaces.The aim of this paper is to prove some fixed point theorems for map with different type of rational expressions in complex valued b-metric spaces. Our results unify, generalize, and complement the comparable results from the current literature.


## 2. Preliminaries

LetC be the set of complex numbers and z
1
,
z
2
∈
C. Define a partial order ≾ on C as follows.z
1
≾
z
2 if and only if Re
z
1
≤
Re
z
2, Im
⁡
(
z
1
)
≤
Im
⁡
(
z
2
). Thus z
1
≾
z
2 if one of the following holds:(1)
Re
z
1
=
Re
z
2 and Im
⁡
z
1
=
Im
⁡
z
2,
(2)
Re
z
1
<
Re
z
2 and Im
⁡
z
1
=
Im
⁡
z
2,
(3)
Re
z
1
=
Re
z
2 and Im
⁡
z
1
<
Im
⁡
z
2,
(4)
Re
z
1
<
Re
z
2 and Im
⁡
z
1
<
Im
⁡
z
2.We will write z
1
⋨
z
2 if z
1
≠
z
2 and one of (2), (3), and (4) is satisfied; also we will write z
1
≺
z
2 if only (4) is satisfied.It follows that(i)
0
≾
z
1
⋨
z
2 implies z
1
<
z
2,
(ii)
z
1
≾
z
2 and z
2
≺
z
3 imply z
1
≺
z
3,
(iii)
0
≾
z
1
≾
z
2 implies z
1
≤
z
2,
(iv)
a
,
b
∈
R and a
≤
b imply a
z
≾
b
z for all z
∈
C.The following definitions and results [4] will be needed in the sequel.Definition 1 (see [4]).
LetX be a nonempty set and let s
≥
1 be a given real number. A function d
:
X
×
X
→
C is called a complex valued b-metric on X if, for all x
,
y
,
z
∈
X, the following conditions are satisfied:(i)
0
≾
d
(
x
,
y
) and d
x
,
y
=
0 if and only if x
=
y;
(ii)
d
x
,
y
=
d
y
,
x;
(iii)
d
x
,
y
≾
s
d
x
,
z
+
d
z
,
y.
The pair(
X
,
d
) is called a complex valued b-metric space.Example 2 (see [4]).
LetX
=
[
0,1
]. Define the mapping d
:
X
×
X
→
C by d
x
,
y
=
x
-
y
2
+
i
x
-
y
2 for all x
,
y
∈
X. Then (
X
,
d
) is a complex valued b-metric space with s
=
2.Definition 3 (see [4]).
Let(
X
,
d
) be a complex valued b-metric space.(i)
A pointx
∈
X is called interior point of a set A
⊆
X whenever there exists 0
≺
r
∈
C such that B
x
,
r
=
y
∈
X
:
d
x
,
y
≺
r
⊆
A.
(ii)
A pointx
∈
X is called limit point of a set A whenever, for every 0
≺
r
∈
C, B
x
,
r
∩
A
-
x
≠
∅.
(iii)
A subsetA
⊆
X is called open whenever each element of A is an interior point of A.
(iv)
A subsetA
⊆
X is called closed whenever each element of A belongs to A.
(v)
A subbasis for Hausdorff topologyτ on X is a family F
=
{
B
x
,
r
:
x
∈
X
and
0
≺
r
}.Definition 4 (see [4]).
Let(
X
,
d
) be a complex valued b-metric space; let {
x
n
} be a sequence in X and x
∈
X.(i)
If, for everyc
∈
C, with 0
≺
r, there is N
∈
N such that, for all n
∈
N, d
x
n
,
x
≺
c, then {
x
n
} is said to be convergent, {
x
n
} converges to x, and x is the limit point of {
x
n
}. One denotes this by lim
⁡
n
→
∞
x
n
=
x or {
x
n
}
→
x as n
→
∞.
(ii)
If, for everyc
∈
C, with 0
≺
r, there is N
∈
N such that, for all n
>
N, d
x
n
,
x
n
+
m
≺
c, where m
∈
N, then {
x
n
} is said to be Cauchy sequence.
(iii)
If every Cauchy sequence inX is convergent, then (
X
,
d
) is said to be a complete complex valued b-metric space.Lemma 5 (see [4]).
Let(
X
,
d
) be a complex valued b-metric space and let {
x
n
} be a sequence in X. Then {
x
n
} converges to x if and only if d
x
n
,
x
→
0 as n
→
∞.Lemma 6 (see [4]).
Let(
X
,
d
) be a complex valued b-metric space and let {
x
n
} be a sequence in X. Then {
x
n
} is a Cauchy sequence if and only if d
x
n
,
x
n
+
m
→
0 as n
→
∞, where m
∈
N.


## 3. Main Results

Theorem 7.
Let(
X
,
d
) be a complete complex valued b-metric space with the coefficient s
≥
1 and let T
:
X
→
X be a mapping satisfying
(1)
d
T
x
,
T
y
≾
λ
d
2
x
,
y
1
+
d
x
,
y
+
μ
d
y
,
T
y
,
for all x
,
y
∈
X, where λ
,
μ are nonnegative reals with s
λ
+
μ
<
1. Then T has a unique fixed point in X.Proof.
For any arbitrary point,x
o
∈
X. Define sequence {
x
n
} in X such that
(2)
x
2
n
+
1
=
T
x
2
n
for
n
=
0,1
,
2,3
,
…
.
Now, we show that the sequence {
x
n
} is Cauchy:
(3)
d
x
2
n
+
1
,
x
2
n
+
2
=
d
T
x
2
n
,
T
x
2
n
+
1
≾
λ
d
2
x
2
n
,
x
2
n
+
1
1
+
d
x
2
n
,
x
2
n
+
1
+
μ
d
x
2
n
+
1
,
T
x
2
n
+
1
=
λ
d
2
(
x
2
n
,
x
2
n
+
1
)
1
+
d
(
x
2
n
,
x
2
n
+
1
)
+
μ
d
(
x
2
n
+
1
,
x
2
n
+
2
)
which implies that
(4)
d
x
2
n
+
1
,
x
2
n
+
2
≤
λ
d
x
2
n
,
x
2
n
+
1
1
+
d
x
2
n
,
x
2
n
+
1
d
x
2
n
,
x
2
n
+
1
+
μ
d
x
2
n
+
1
,
x
2
n
+
2
.
Since 1
+
d
x
2
n
,
x
2
n
+
1
>
d
x
2
n
,
x
2
n
+
1, we get
(5)
d
x
2
n
+
1
,
x
2
n
+
2
≤
λ
d
x
2
n
,
x
2
n
+
1
+
μ
d
x
2
n
+
1
,
x
2
n
+
2
and hence
(6)
d
x
2
n
+
1
,
x
2
n
+
2
≤
λ
1
-
μ
d
x
2
n
,
x
2
n
+
1
.
Similarly, we obtain
(7)
d
x
2
n
+
2
,
x
2
n
+
3
≤
λ
1
-
μ
d
x
2
n
+
1
,
x
2
n
+
2
.
Since s
λ
+
μ
<
1 and s
≥
1, we get λ
+
μ
<
1.
Therefore, withδ
=
λ
/
(
1
-
μ
)
<
1 and for all n
≥
0 and consequently, we have
(8)
d
x
2
n
+
1
,
x
2
n
+
2
≤
δ
d
x
2
n
,
x
2
n
+
1
≤
δ
2
d
x
2
n
-
1
,
x
2
n
≤
⋯
≤
δ
2
n
+
1
d
x
0
,
x
1
,
d
x
n
+
1
,
x
n
+
2
≤
δ
d
x
n
,
x
n
+
1
≤
δ
2
d
x
n
-
1
,
x
n
≤
⋯
≤
δ
n
+
1
d
x
0
,
x
1
.
Thus for any m
>
n, m
,
n
∈
N and since s
δ
=
s
λ
/
(
1
-
μ
)
<
1, we get
(9)
d
x
n
,
x
m
≤
s
d
x
n
,
x
n
+
1
+
s
d
x
n
+
1
,
x
m
≤
s
d
x
n
,
x
n
+
1
+
s
2
d
x
n
+
1
,
x
n
+
2
+
s
2
d
x
n
+
2
,
x
m
≤
s
d
x
n
,
x
n
+
1
+
s
2
d
x
n
+
1
,
x
n
+
2
+
s
3
d
x
n
+
2
,
x
n
+
3
+
s
3
d
x
n
+
3
,
x
m
⋮
d
x
n
,
x
m
≤
s
d
x
n
,
x
n
+
1
+
s
2
d
x
n
+
1
,
x
n
+
2
+
s
3
d
x
n
+
2
,
x
n
+
3
+
⋯
+
s
m
-
n
-
1
d
x
m
-
2
,
x
m
-
1
+
s
m
-
n
d
x
m
-
1
,
x
m
.
By using (8), we get
(10)
d
x
n
,
x
m
≤
s
δ
n
d
x
0
,
x
1
+
s
2
δ
n
+
1
d
x
0
,
x
1
+
s
3
δ
n
+
2
d
x
0
,
x
1
+
⋯
+
s
m
-
n
-
1
δ
m
-
2
d
x
0
,
x
1
+
s
m
-
n
δ
m
-
1
d
x
0
,
x
1
=
∑
i
=
1
m
-
n
s
i
δ
i
+
n
-
1
d
x
0
,
x
1
.
Therefore,
(11)
d
x
n
,
x
m
≤
∑
i
=
1
m
-
n
s
i
+
n
-
1
δ
i
+
n
-
1
d
x
0
,
x
1
=
∑
t
=
n
m
-
1
s
t
δ
t
d
x
0
,
x
1
≤
∑
t
=
n
∞
s
δ
t
d
x
0
,
x
1
=
s
δ
n
1
-
s
δ
d
x
0
,
x
1
and hence
(12)
d
x
n
,
x
m
≤
s
δ
n
1
-
s
δ
d
x
0
,
x
1
⟶
0
as
m
,
n
⟶
∞
.
Thus, {
x
n
} is a Cauchy sequence in X. Since X is complete, there exists some u
∈
X such that x
n
→
u as n
→
∞. Suppose this is not possible; then there exists z
∈
X such that
(13)
d
u
,
T
u
=
z
>
0
.
Now,
(14)
z
=
d
u
,
T
u
≾
s
d
u
,
x
2
n
+
2
+
s
d
x
2
n
+
2
,
T
u
=
s
d
u
,
x
2
n
+
2
+
s
d
T
x
2
n
+
1
,
T
u
≾
s
d
u
,
x
2
n
+
2
+
s
λ
d
2
x
2
n
+
1
,
u
1
+
d
(
x
2
n
+
1
,
u
)
+
s
μ
d
(
u
,
T
u
)
which implies that
(15)
z
=
d
u
,
T
u
≤
s
d
u
,
x
2
n
+
2
+
s
λ
d
2
x
2
n
+
1
,
u
1
+
d
x
2
n
+
1
,
u
+
s
μ
d
u
,
T
u
.
Taking the limit of (15) as n
→
∞, we obtain that z
=
d
u
,
T
u
≤
0, a contradiction with (13).
Soz
=
0. Hence T
u
=
u.
Now, we show thatT has a unique fixed point in X. To show this, assume that u
* is another fixed point of T. Then,
(16)
d
u
,
u
*
=
d
T
u
,
T
u
*
≾
λ
d
2
u
,
u
*
1
+
d
u
,
u
*
+
μ
d
u
*
,
T
u
*
.
So
(17)
d
u
,
u
*
≤
λ
d
u
,
u
*
1
+
d
u
,
u
*
d
u
,
u
*
+
μ
d
u
*
,
T
u
*
,
since
(18)
1
+
d
u
,
u
*
>
d
u
,
u
*
.
Therefore,
(19)
d
u
,
u
*
<
λ
d
u
,
u
*
+
μ
d
u
*
,
u
*
=
λ
d
u
,
u
*
,
a
contradiction
.
So u
=
u
*, which proves the uniqueness of fixed point in X. This completes the proof.Corollary 8.
LetX
,
d be a complete complex valued b-metric space with the coefficient s
≥
1 and let T
:
X
→
X be a mapping satisfying (for some fixed n)
(20)
d
T
n
x
,
T
n
y
≾
λ
d
2
x
,
y
1
+
d
x
,
y
+
μ
d
y
,
T
n
y
,
for all x
,
y
∈
X, where λ
,
μ are nonnegative reals with s
λ
+
μ
<
1. Then T has a unique fixed point in X.Proof.
From Theorem7, we obtain u
∈
X such that
(21)
T
n
u
=
u
.
The uniqueness follows from
(22)
d
T
u
,
u
=
d
T
T
n
u
,
T
n
u
=
d
T
n
T
u
,
T
n
u
≾
λ
d
2
T
u
,
u
1
+
d
T
u
,
u
+
μ
d
u
,
T
n
u
.
By taking modulus (22), we have
(23)
d
T
u
,
u
≤
λ
d
T
u
,
u
1
+
d
T
u
,
u
d
T
u
,
u
+
μ
d
u
,
u
,
since
(24)
1
+
d
T
u
,
u
>
d
T
u
,
u
.
Therefore,
(25)
d
T
u
,
u
<
λ
d
T
u
,
u
,
a
contradiction
.
So T
u
=
u. Hence T
u
=
T
n
u
=
u.
Therefore, the fixed point ofT is unique. This completes the proof.Theorem 9.
Let(
X
,
d
) be a complete complex valued b-metric space with the coefficient s
≥
1 and let T
:
X
→
X be a mapping satisfying
(26)
d
T
x
,
T
y
≾
λ
d
x
,
y
+
μ
d
x
,
T
x
d
y
,
T
y
d
x
,
T
y
+
d
y
,
T
x
+
d
x
,
y
,
for all x
,
y
∈
X such that x
≠
y, d
(
x
,
T
y
)
+
d
y
,
T
x
+
d
(
x
,
y
)
≠
0, where λ
,
μ are nonnegative reals with λ
+
s
μ
<
1 or d
T
x
,
T
y
=
0 if d
x
,
T
y
+
d
y
,
T
x
+
d
x
,
y
=
0. Then T has a unique fixed point in X.Proof.
For any arbitrary point,x
o
∈
X. Define sequence {
x
n
} in X such that
(27)
x
2
n
+
1
=
T
x
2
n
for
n
=
0,1
,
2,3
,
…
.
Now, we show that the sequence {
x
n
} is Cauchy:
(28)
d
x
2
n
+
1
,
x
2
n
+
2
=
d
T
x
2
n
,
T
x
2
n
+
1
≾
λ
d
x
2
n
,
x
2
n
+
1
+
μ
d
x
2
n
,
T
x
2
n
d
x
2
n
+
1
,
T
x
2
n
+
1
d
x
2
n
,
T
x
2
n
+
1
+
d
x
2
n
+
1
,
T
x
2
n
+
d
x
2
n
,
x
2
n
+
1
=
λ
d
x
2
n
,
x
2
n
+
1
+
μ
d
(
x
2
n
,
x
2
n
+
1
)
d
(
x
2
n
+
1
,
x
2
n
+
2
)
d
x
2
n
,
x
2
n
+
2
+
d
x
2
n
+
1
,
x
2
n
+
1
+
d
(
x
2
n
,
x
2
n
+
1
)
which implies that
(29)
d
x
2
n
+
1
,
x
2
n
+
2
≤
λ
d
x
2
n
,
x
2
n
+
1
+
μ
d
x
2
n
+
1
,
x
2
n
+
2
d
x
2
n
,
x
2
n
+
2
+
d
x
2
n
,
x
2
n
+
1
×
d
x
2
n
,
x
2
n
+
1
,
since
(30)
d
x
2
n
+
1
,
x
2
n
+
2
≤
d
x
2
n
+
1
,
x
2
n
+
d
x
2
n
,
x
2
n
+
2
.
Therefore,
(31)
d
x
2
n
+
1
,
x
2
n
+
2
≤
λ
d
x
2
n
,
x
2
n
+
1
+
μ
d
x
2
n
,
x
2
n
+
1
=
λ
+
μ
d
x
2
n
,
x
2
n
+
1
.
Similarly, we obtain
(32)
d
x
2
n
+
2
,
x
2
n
+
3
≤
λ
+
μ
d
x
2
n
+
1
,
x
2
n
+
2
.
Since λ
+
s
μ
<
1 and s
≥
1, we get λ
+
μ
<
1.
Therefore, withδ
=
λ
+
μ
<
1 and for all n
≥
0 and consequently, we have
(33)
d
x
2
n
+
1
,
x
2
n
+
2
≤
δ
d
x
2
n
,
x
2
n
+
1
≤
δ
2
d
x
2
n
-
1
,
x
2
n
≤
⋯
≤
δ
2
n
+
1
d
x
0
,
x
1
,
d
x
n
+
1
,
x
n
+
2
≤
δ
d
x
n
,
x
n
+
1
≤
δ
2
d
x
n
-
1
,
x
n
≤
⋯
≤
δ
n
+
1
d
x
0
,
x
1
.
Thus, for any m
>
n, m
,
n
∈
N, we have
(34)
d
x
n
,
x
m
≤
s
d
x
n
,
x
n
+
1
+
s
d
x
n
+
1
,
x
m
≤
s
d
x
n
,
x
n
+
1
+
s
2
d
x
n
+
1
,
x
n
+
2
+
s
2
d
x
n
+
2
,
x
m
≤
s
d
x
n
,
x
n
+
1
+
s
2
d
x
n
+
1
,
x
n
+
2
+
s
3
d
x
n
+
2
,
x
n
+
3
+
s
3
d
x
n
+
3
,
x
m
⋮
d
x
n
,
x
m
≤
s
d
x
n
,
x
n
+
1
+
s
2
d
x
n
+
1
,
x
n
+
2
+
s
3
d
x
n
+
2
,
x
n
+
3
+
⋯
+
s
m
-
n
-
1
d
x
m
-
2
,
x
m
-
1
+
s
m
-
n
d
x
m
-
1
,
x
m
.
By using (33), we get
(35)
d
x
n
,
x
m
≤
s
δ
n
d
x
0
,
x
1
+
s
2
δ
n
+
1
d
x
0
,
x
1
+
s
3
δ
n
+
2
d
x
0
,
x
1
+
⋯
+
s
m
-
n
-
1
δ
m
-
2
d
x
0
,
x
1
+
s
m
-
n
δ
m
-
1
d
x
0
,
x
1
=
∑
i
=
1
m
-
n
s
i
δ
i
+
n
-
1
d
x
0
,
x
1
.
Therefore,
(36)
d
x
n
,
x
m
≤
∑
i
=
1
m
-
n
s
i
+
n
-
1
δ
i
+
n
-
1
d
x
0
,
x
1
=
∑
t
=
n
m
-
1
s
t
δ
t
d
x
0
,
x
1
≤
∑
t
=
n
∞
s
δ
t
d
x
0
,
x
1
=
s
δ
n
1
-
s
δ
d
x
0
,
x
1
and hence
(37)
d
x
n
,
x
m
≤
s
δ
n
1
-
s
δ
d
x
0
,
x
1
⟶
0
as
m
,
n
⟶
∞
.
Thus, {
x
n
} is a Cauchy sequence in X. Since X is complete, there exists some u
∈
X such that x
n
→
u as n
→
∞. Suppose this is not possible; then there exists z
∈
X such that
(38)
d
u
,
T
u
=
z
>
0
.
Now,
(39)
z
=
d
u
,
T
u
≾
s
d
u
,
x
2
n
+
2
+
s
d
x
2
n
+
2
,
T
u
=
s
d
u
,
x
2
n
+
2
+
s
d
T
x
2
n
+
1
,
T
u
≾
s
d
u
,
x
2
n
+
2
+
s
λ
d
x
2
n
+
1
,
u
+
s
μ
d
x
2
n
+
1
,
T
x
2
n
+
1
d
u
,
T
u
d
x
2
n
+
1
,
T
u
+
d
u
,
T
x
2
n
+
1
+
d
x
2
n
+
1
,
u
=
s
d
u
,
x
2
n
+
2
+
s
λ
d
x
2
n
+
1
,
u
+
s
μ
d
x
2
n
+
1
,
x
2
n
+
2
d
u
,
T
u
d
x
2
n
+
1
,
T
u
+
d
u
,
x
2
n
+
2
+
d
x
2
n
+
1
,
u
which implies that
(40)
z
=
d
u
,
T
u
≤
s
d
u
,
x
2
n
+
2
+
s
λ
d
x
2
n
+
1
,
u
+
s
μ
d
x
2
n
+
1
,
x
2
n
+
2
d
x
2
n
+
1
,
T
u
+
d
u
,
x
2
n
+
2
+
d
x
2
n
+
1
,
u
×
d
u
,
T
u
.
Taking the limit of (40) as n
→
∞, we obtain that z
=
d
u
,
T
u
≤
0, a contradiction with (38).
Soz
=
0. Hence T
u
=
u.
Now, we show thatT has a unique fixed point in X. To show this, assume that u
* is another fixed point of T. Then,
(41)
d
u
,
u
*
=
d
T
u
,
T
u
*
≾
λ
d
u
,
u
*
+
μ
d
u
,
T
u
d
(
u
*
,
T
u
*
)
d
u
,
T
u
*
+
d
u
*
,
T
u
+
d
(
u
,
u
*
)
so that
(42)
d
u
,
u
*
≤
λ
d
u
,
u
*
+
μ
d
u
,
T
u
d
u
*
,
T
u
*
d
u
,
T
u
*
+
d
u
*
,
T
u
+
d
u
,
u
*
<
λ
d
u
,
u
*
,
a
contradiction
.
So u
=
u
*, which proves the uniqueness of fixed point in X.
Now, we consider the following case:d
x
2
n
,
T
x
2
n
+
1
+
d
x
2
n
+
1
,
T
x
2
n
+
d
x
2
n
,
x
2
n
+
1
=
0 (for any n) implies d
T
x
2
n
,
T
x
2
n
+
1
=
0, so that x
2
n
=
T
x
2
n
=
x
2
n
+
1
=
T
x
2
n
+
1
=
x
2
n
+
2. Thus we have x
2
n
+
1
=
T
x
2
n
=
x
2
n, so there exist K
1 and l
1 such that K
1
=
T
l
1
=
l
1. Using foregoing arguments, one can also show that there exist K
2 and l
2 such that K
2
=
T
l
2
=
l
2. As d
(
l
1
,
T
l
2
)
+
d
l
2
,
T
l
1
+
d
l
1
,
l
2
=
0 (due to definition) implies d
(
T
l
1
,
T
l
2
)
=
0, K
1
=
T
l
1
=
T
l
2
=
K
2, which in turn yields that K
1
=
T
l
1
=
T
K
1. Similarly, one can also have K
2
=
T
K
2. As K
1
=
K
2 implies T
K
1
=
K
1, therefore K
1
=
K
2 is fixed point of T.
We now prove thatT has unique fixed point. For this, assume that K
1
* in X is another fixed point of T. Then we have T
K
1
*
=
K
1
*. As d
K
1
,
T
K
1
*
+
d
K
1
*
,
T
K
1
+
d
K
1
,
K
1
*
=
0, therefore d
K
1
,
K
1
*
=
d
T
K
1
,
T
K
1
*
=
0.
This implies thatK
1
*
=
K
1. This completes the proof of the theorem.Corollary 10.
LetX
,
d be a complete complex valued b-metric space with the coefficient s
≥
1 and let T
:
X
→
X be a mapping satisfying (for some fixed n):
(43)
d
T
n
x
,
T
n
y
≾
λ
d
x
,
y
+
μ
d
x
,
T
n
x
d
y
,
T
n
y
d
x
,
T
n
y
+
d
y
,
T
n
x
+
d
x
,
y
,
for all x
,
y
∈
X such that x
≠
y, d
x
,
T
y
+
d
y
,
T
x
+
d
(
x
,
y
)
≠
0, where λ
,
μ are nonnegative reals with λ
+
s
μ
<
1 or d
T
n
x
,
T
n
y
=
0 if d
x
,
T
n
y
+
d
y
,
T
n
x
+
d
x
,
y
=
0. Then T has a unique fixed point in X.Proof.
From Theorem9, we obtain u
∈
X such that
(44)
T
n
u
=
u
.
The uniqueness follows from
(45)
d
T
u
,
u
=
d
T
T
n
u
,
T
n
u
=
d
T
n
T
u
,
T
n
u
≾
λ
d
T
u
,
u
+
μ
d
T
u
,
T
n
T
u
d
u
,
T
n
u
d
T
u
,
T
n
u
+
d
u
,
T
n
T
u
+
d
T
u
,
u
=
λ
d
T
u
,
u
.
By taking modulus (45), we get
(46)
d
T
u
,
u
<
λ
d
T
u
,
u
,
a
contradiction
.
So T
u
=
u. Hence T
u
=
T
n
u
=
u.
Therefore, the fixed point ofT is unique. This completes the proof.Example 11.
LetX
=
C be the set of complex numbers.
Defined
:
C
×
C
→
C as
(47)
d
z
1
,
z
2
=
x
1
-
x
2
2
+
i
y
1
-
y
2
2
,
where z
1
=
x
1
+
i
y
1 and z
2
=
x
2
+
i
y
2.
Obviously,(
X
,
d
) is complete complex valued b-metric space with s
=
2. Define T
:
C
→
C as
(48)
T
x
+
i
y
=
0
;
x
,
y
∈
Q
,
2
+
2
i
;
x
,
y
∈
Q
c
,
2
;
x
∈
Q
c
,
y
∈
Q
,
2
i
;
x
∈
Q
,
y
∈
Q
c
.
Now, for x
=
1
/
2 and y
=
0, we get
(49)
d
T
1
2
,
T
0
=
d
2,0
=
4
≾
λ
d
1
2
,
0
=
λ
1
2
.
Thus λ
≥
2 which is a contradiction as 0
≤
λ
<
1.
However, notice thatT
n
z
=
0 for n
>
1, so
(50)
d
T
n
x
,
T
n
y
=
0
≾
λ
d
x
,
y
+
μ
d
x
,
T
n
x
d
y
,
T
n
y
d
x
,
T
n
y
+
d
y
,
T
n
x
+
d
x
,
y
,
for all x
,
y
∈
X and λ
,
μ
≥
0 with λ
+
2
μ
<
1. So all conditions of Corollary 10 are satisfied to get a unique fixed point 0 of T.Example 12.
LetX
=
0,1. Consider a b-metric d
:
X
×
X
→
C defined as
(51)
d
x
,
y
=
2
3
x
-
y
2
+
i
x
-
y
2
,
∀
x
,
y
∈
X
.
To verify that X
,
d is a complete complex valued b-metric space with s
=
2, it is enough to verify the triangular inequality condition:
(52)
d
x
,
y
=
2
3
x
-
y
2
+
i
x
-
y
2
=
2
3
x
-
z
+
z
-
y
2
+
i
x
-
z
+
z
-
y
2
≾
2
3
x
-
z
2
+
z
-
y
2
+
2
x
-
z
z
-
y
+
i
x
-
z
2
+
z
-
y
2
+
2
x
-
z
z
-
y
≾
2
3
x
-
z
2
+
z
-
y
2
+
x
-
z
2
+
z
-
y
2
+
i
x
-
z
2
+
z
-
y
2
+
x
-
z
2
+
z
-
y
2
=
2
2
3
x
-
z
2
+
i
x
-
z
2
+
2
3
z
-
y
2
+
i
z
-
y
2
=
2
d
x
,
z
+
d
z
,
y
.
That is,
(53)
d
x
,
y
≾
s
d
x
,
z
+
d
z
,
y
.
Therefore, s
=
2. Define T
:
X
→
X as T
x
=
x
/
2, T
y
=
y
/
2 for all x
,
y
∈
X. Then
(54)
d
T
x
,
T
y
=
d
x
2
,
y
2
d
T
x
,
T
y
=
2
3
x
2
-
y
2
2
+
i
x
2
-
y
2
2
d
T
x
,
T
y
=
1
6
x
-
y
2
+
i
x
-
y
2
=
1
6
d
x
,
y
,
d
T
x
,
T
y
≾
3
8
d
x
,
y
+
(
1
/
5
)
d
x
,
x
/
2
d
y
,
y
/
2
d
x
,
y
/
2
+
d
y
,
x
/
2
+
d
x
,
y
.
Here
(55)
λ
+
s
μ
=
3
8
+
2
1
5
=
31
40
<
1
.
It is easily and clearly verified that the map T satisfies contractive condition (26) of Theorem 9 with the coefficients s
=
2, λ
=
3
/
8, and μ
=
1
/
5. Observe that the point 0
∈
X remains fixed under T and is indeed unique.



---

*Source: 832467-2015-01-26.xml*