# Complex Valuedb-Metric Spaces and Common Fixed Point Theorems under Rational Contractions

Authors: Anil Kumar Dubey
Journal: Journal of Complex Analysis
 (2016)

---

## Abstract

The aim of this paper is to prove the existence and uniqueness of a common fixed point for a pair of mappings satisfying certain rational contraction conditions in complex valuedb-metric space. The obtained results generalize and extend some of the well-known results in the literature.


---

## Body

## 1. Introduction

Banach contraction principle in [1] gives appropriate and simple conditions to establish the existence and uniqueness of a solution of an operator equation T
x
=
x. Later, a number of papers were devoted to the improvement and generalization of that result. Most of these results deal with the generalizations of the different contractive conditions in metric spaces. There have been a number of generalizations of metric spaces such as vector valued metric spaces, G-metric spaces, pseudometric spaces, fuzzy metric spaces, D-metric spaces, cone metric spaces, and modular metric spaces. Bakhtin [2] introduced the notion of b-metric space which is a generalized form of metric spaces.In [3], Czerwik proved the contraction mapping principle in b-metric spaces. Subsequently, many authors obtained fixed point results for single valued and multivalued operators in b-metric spaces.A new space called the complex valued metric space which is more general than the well-known metric space has been introduced by Azam et al. [4]. They proved some fixed point results for a pair of mappings for contraction condition satisfying a rational expression. Azam et al. [4] improved the Banach contraction principle in the context of complex valued metric space involving rational inequality which could not be meaningful in cone metric spaces. Several authors studied many common fixed point theorems on complex valued metric spaces (see [5–9]).The concept of complex valuedb-metric spaces was introduced in 2013 by Rao et al. [10]. In sequel, Mukheimer [11] proved some common fixed point theorems in complex valued b-metric spaces.In this paper, we continue the study of fixed point theorems in complex valuedb-metric spaces. The obtained results are generalizations of recent results proved by Dubey et al. [12, 13], Nashine et al. [5, 6], and Rao et al. [10].


## 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 R
e
(
z
1
)
≤
R
e
(
z
2
), I
m
(
z
1
)
≤
I
m
(
z
2
).Thusz
1
≾
z
2 if one of the following holds:(1)
R
e
(
z
1
)
=
R
e
(
z
2
) and I
m
(
z
1
)
=
I
m
(
z
2
);
(2)
R
e
(
z
1
)
<
R
e
(
z
2
) and I
m
(
z
1
)
=
I
m
(
z
2
);
(3)
R
e
(
z
1
)
=
R
e
(
z
2
) and I
m
(
z
1
)
<
I
m
(
z
2
);
(4)
R
e
z
1
<
R
e
z
2 and I
m
(
z
1
)
<
I
m
(
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)
ifa
,
b
∈
R, 0
≤
a
≤
b and z
1
≾
z
2, then a
z
1
≾
b
z
2 for all z
1
,
z
2
∈
C.The following definition is recently introduced by Rao et al. [10].Definition 1.
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 [10]).
IfX
=
[
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 complex valued b-metric space with s
=
2.Definition 3 (see [10]).
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 a limit point of a set A whenever for every 0
≺
r
∈
C, B
(
x
,
r
)
∩
(
A
-
x
)
≠
ϕ.
(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 a Hausdorff topologyτ on X is a family F
=
{
B
(
x
,
r
)
:
x
∈
X
a
n
d
0
≺
r
}.Definition 4 (see [10]).
Let(
X
,
d
) be a complex valued b-metric space and 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. We denote this by l
i
m
n
→
∞
x
n
=
x or x
n
→
x
a
s
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 [10]).
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 [10]).
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 S
,
T
:
X
→
X be mappings satisfying(1)
d
S
x
,
T
y
≾
⁡
α
d
x
,
y
+
β
d
x
,
S
x
d
y
,
T
y
d
x
,
T
y
+
d
y
,
S
x
+
d
x
,
yfor all x
,
y
∈
X, such that x
≠
y, d
(
x
,
T
y
)
+
d
(
y
,
S
x
)
+
d
(
x
,
y
)
≠
0, where α, β are nonnegative reals with α
+
s
β
<
1 or d
(
S
x
,
T
y
)
=
0 if d
(
x
,
T
y
)
+
d
(
y
,
S
x
)
+
d
(
x
,
y
)
=
0.
ThenS and T have a unique common fixed point.Proof.
For any arbitrary pointx
0
∈
X, define sequence x
n in X such that(2)
x
2
n
+
1
=
S
x
2
n
,
x
2
n
+
2
=
T
x
2
n
+
1
f
o
r
n
=
0,1
,
2,3
,
…
.Now, we show that the sequence x
n is Cauchy.
Letx
=
x
2
n and y
=
x
2
n
+
1 in (1); we have(3)
d
x
2
n
+
1
,
x
2
n
+
2
=
d
S
x
2
n
,
T
x
2
n
+
1
≾
⁡
α
d
x
2
n
,
x
2
n
+
1
+
β
d
x
2
n
,
S
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
,
S
x
2
n
+
d
x
2
n
,
x
2
n
+
1
=
α
d
x
2
n
,
x
2
n
+
1
+
β
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
+
1so that(4)
d
x
2
n
+
1
,
x
2
n
+
2
≤
α
d
x
2
n
,
x
2
n
+
1
+
β
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
,
x
2
n
+
1
.As (owing to triangular inequality)(5)
d
x
2
n
+
1
,
x
2
n
+
2
≤
d
x
2
n
+
1
,
x
2
n
+
d
x
2
n
,
x
2
n
+
2therefore(6)
d
x
2
n
+
1
,
x
2
n
+
2
≤
α
d
x
2
n
,
x
2
n
+
1
+
β
d
x
2
n
,
x
2
n
+
1
=
α
+
β
d
x
2
n
,
x
2
n
+
1
d
x
2
n
+
1
,
x
2
n
+
2
≤
α
+
β
d
x
2
n
,
x
2
n
+
1
.Similarly, we obtain(7)
d
x
2
n
+
2
,
x
2
n
+
3
≤
α
+
β
d
x
2
n
+
1
,
x
2
n
+
2
.Since α
+
s
β
<
1 and s
≥
1, we get α
+
β
<
1. Therefore, with δ
=
α
+
β
<
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
.That is,(9)
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(10)
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
≤
⋯
≤
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
-
2
d
x
m
-
3
,
x
m
-
2
+
s
m
-
n
-
1
d
x
m
-
2
,
x
m
-
1
+
s
m
-
n
d
x
m
-
1
,
x
m
.By using (9), we get(11)
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
-
2
δ
m
-
3
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,(12)
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
1and hence(13)
d
x
n
,
x
m
≤
s
δ
n
1
-
s
δ
d
x
0
,
x
1
⟶
0
a
s
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
→
∞. Assume not, then there exists z
∈
X such that(14)
d
u
,
S
u
=
z
>
0
.So by using the triangular inequality and (1), we get(15)
z
=
d
u
,
S
u
≾
s
d
u
,
x
2
n
+
2
+
s
d
x
2
n
+
2
,
S
u
=
s
d
u
,
x
2
n
+
2
+
s
d
T
x
2
n
+
1
,
S
u
≾
s
d
u
,
x
2
n
+
2
+
s
α
⁡
d
u
,
x
2
n
+
1
+
s
β
d
u
,
S
u
d
x
2
n
+
1
,
T
x
2
n
+
1
d
u
,
T
x
2
n
+
1
+
d
x
2
n
+
1
,
S
u
+
d
u
,
x
2
n
+
1
=
s
d
u
,
x
2
n
+
2
+
s
α
⁡
d
u
,
x
2
n
+
1
+
s
β
d
u
,
S
u
d
x
2
n
+
1
,
x
2
n
+
2
d
u
,
x
2
n
+
2
+
d
x
2
n
+
1
,
S
u
+
d
u
,
x
2
n
+
1which implies that(16)
z
=
d
u
,
S
u
≤
s
d
u
,
x
2
n
+
2
+
s
α
⁡
d
u
,
x
2
n
+
1
+
s
β
d
u
,
S
u
d
x
2
n
+
1
,
x
2
n
+
2
d
u
,
x
2
n
+
2
+
d
x
2
n
+
1
,
S
u
+
d
u
,
x
2
n
+
1
.Taking the limit of (16) as n
→
∞, we obtain that z
=
d
u
,
S
u
≤
0, a contradiction with (14). So z
=
0
. Hence S
u
=
u. Similarly, we obtain T
u
=
u.
Now we show thatS and T have unique common fixed point of S and T. To prove this, assume that u
∗ is another common fixed point of S and T. Then(17)
d
u
,
u
∗
=
d
S
u
,
T
u
∗
≾
⁡
α
d
u
,
u
∗
+
β
d
u
,
S
u
d
u
∗
,
T
u
∗
d
u
,
T
u
∗
+
d
u
∗
,
S
u
+
d
u
,
u
∗so that(18)
d
u
,
u
∗
≤
α
d
u
,
u
∗
+
β
d
u
,
S
u
d
u
∗
,
T
u
∗
d
u
,
T
u
∗
+
d
u
∗
,
S
u
+
d
u
,
u
∗
≾
α
⁡
d
u
,
u
∗so that u
=
u
∗ which proves the uniqueness of common fixed point.
Now, we consider the second case:d
(
x
,
T
y
)
+
d
(
y
,
S
x
)
+
d
(
x
,
y
)
=
0. Put x
=
x
2
n and y
=
x
2
n
+
1 in this expression; we get d
(
x
2
n
,
T
x
2
n
+
1
)
+
d
(
x
2
n
+
1
,
S
x
2
n
)
+
d
(
x
2
n
,
x
2
n
+
1
)
=
0 (for any n) which implies d
(
S
x
2
n
,
T
x
2
n
+
1
)
=
0 so that x
2
n
=
S
x
2
n
=
x
2
n
+
1
=
T
x
2
n
+
1
=
x
2
n
+
2. Thus, we have x
2
n
+
1
=
S
x
2
n
=
x
2
n, so there exist K
1 and l
1 such that K
1
=
S
l
1
=
l
1, where K
1
=
x
2
n
+
1 and l
1
=
x
2
n. Using foregoing arguments, one can also show that there exist K
2 and l
2 such that K
2
=
T
l
2
=
l
2, where K
2
=
x
2
n
+
2 and l
2
=
x
2
n
+
1. As d
(
l
1
,
T
l
2
)
+
d
(
l
2
,
S
l
1
)
+
d
(
l
1
,
l
2
)
=
0 (due to definition) implies d
(
S
l
1
,
T
l
2
)
=
0, therfore K
1
=
S
l
1
=
T
l
2
=
K
2. Thus we obtain that K
1
=
S
l
1
=
S
K
1. Similarly, one can also have K
2
=
T
K
2. As K
1
=
K
2 implies S
K
1
=
T
K
1
=
K
1, therefore K
1
=
K
2 is common fixed point of S and T. For uniqueness of common fixed point, assume that K
1
∗ in X is another common fixed point of S and T. Then we have S
K
1
∗
=
T
K
1
∗
=
K
1
∗.
Asd
(
K
1
,
T
K
1
∗
)
+
d
(
K
1
∗
,
S
K
1
)
+
d
(
K
1
,
K
1
∗
)
=
0, therefore d
(
K
1
,
K
1
∗
)
=
d
(
S
K
1
,
T
K
1
∗
)
=
0.
This implies thatK
1
=
K
1
∗. This completes the proof of the theorem.Corollary 8 (see [12]).
Let(
X
,
d
) be a complete complex valued b-metric space with the coefficient s
≥
1. Let T
:
X
→
X be a mapping satisfying(19)
d
T
x
,
T
y
≾
⁡
α
d
x
,
y
+
β
d
x
,
T
x
d
y
,
T
y
d
x
,
T
y
+
d
y
,
T
x
+
d
x
,
yfor all x
,
y
∈
X such that x
≠
y, d
(
x
,
T
y
)
+
d
(
y
,
T
x
)
+
d
(
x
,
y
)
≠
0, where α, β are nonnegative reals with α
+
s
β
<
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.
We can prove this result by applying Theorem7 with S
=
T.Corollary 9 (see [12]).
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 (for some fixed n)(20)
d
T
n
x
,
T
n
y
≾
⁡
α
d
x
,
y
+
β
d
x
,
T
n
x
d
y
,
T
n
y
d
x
,
T
n
y
+
d
y
,
T
n
x
+
d
x
,
yfor all x
,
y
∈
X such that x
≠
y, d
(
x
,
T
y
)
+
d
(
y
,
T
x
)
+
d
(
x
,
y
)
≠
0, where α, β are nonnegative reals with α
+
s
β
<
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.Theorem 10.
Let(
X
,
d
) be a complete complex valued b-metric space with the coefficient s
≥
1 and let S
,
T
:
X
→
X be mappings satisfying(21)
d
S
x
,
T
y
≾
⁡
α
d
x
,
y
+
β
d
2
x
,
T
y
+
d
2
y
,
S
x
d
x
,
T
y
+
d
y
,
S
x
+
γ
d
x
,
S
x
+
d
y
,
T
yfor all x
,
y
∈
X such that x
≠
y, where α, β, and γ are nonnegative reals with α
+
2
s
β
+
2
γ
<
1 or d
(
S
x
,
T
y
)
=
0 if d
(
x
,
T
y
)
+
d
(
y
,
S
x
)
=
0. Then S and T have a unique common fixed point in X.Proof.
For any arbitrary pointx
0
∈
X, define sequence x
n in X such that(22)
x
2
n
+
1
=
S
x
2
n
,
x
2
n
+
2
=
T
x
2
n
+
1
,
f
o
r
n
=
0,1
,
2
,
…
.Now, we show that the sequence x
n is a Cauchy sequence. Let x
=
x
2
n and y
=
x
2
n
+
1 in (21); we have(23)
d
x
2
n
+
1
,
x
2
n
+
2
=
d
S
x
2
n
,
T
x
2
n
+
1
≾
α
d
x
2
n
,
x
2
n
+
1
+
β
d
2
x
2
n
,
T
x
2
n
+
1
+
d
2
x
2
n
+
1
,
S
x
2
n
d
x
2
n
,
T
x
2
n
+
1
+
d
x
2
n
+
1
,
S
x
2
n
+
γ
d
x
2
n
,
S
x
2
n
+
d
x
2
n
+
1
,
T
x
2
n
+
1
=
α
d
x
2
n
,
x
2
n
+
1
+
β
d
2
x
2
n
,
x
2
n
+
2
+
d
2
x
2
n
+
1
,
x
2
n
+
1
d
x
2
n
,
x
2
n
+
2
+
d
x
2
n
+
1
,
x
2
n
+
1
+
γ
d
x
2
n
,
x
2
n
+
1
+
d
x
2
n
+
1
,
x
2
n
+
2so that(24)
d
x
2
n
+
1
,
x
2
n
+
2
≤
⁡
α
d
x
2
n
,
x
2
n
+
1
+
β
d
2
x
2
n
,
x
2
n
+
2
d
x
2
n
,
x
2
n
+
2
+
γ
d
x
2
n
,
x
2
n
+
1
+
d
x
2
n
+
1
,
x
2
n
+
2
d
x
2
n
+
1
,
x
2
n
+
2
≤
⁡
α
d
x
2
n
,
x
2
n
+
1
+
β
d
x
2
n
,
x
2
n
+
2
+
γ
d
x
2
n
,
x
2
n
+
1
+
d
x
2
n
+
1
,
x
2
n
+
2
.As(25)
d
x
2
n
,
x
2
n
+
2
≤
s
d
x
2
n
,
x
2
n
+
1
+
d
x
2
n
+
1
,
x
2
n
+
2
,therefore(26)
d
x
2
n
+
1
,
x
2
n
+
2
≤
α
d
x
2
n
,
x
2
n
+
1
+
s
β
d
x
2
n
,
x
2
n
+
1
+
d
x
2
n
+
1
,
x
2
n
+
2
+
γ
d
x
2
n
,
x
2
n
+
1
+
d
x
2
n
+
1
,
x
2
n
+
2
d
x
2
n
+
1
,
x
2
n
+
2
≤
α
+
s
β
+
γ
1
-
s
β
-
γ
d
x
2
n
,
x
2
n
+
1
.Similarly, we obtain(27)
d
x
2
n
+
2
,
x
2
n
+
3
≤
α
+
s
β
+
γ
1
-
s
β
-
γ
d
x
2
n
+
1
,
x
2
n
+
2
.Since α
+
2
s
β
+
2
γ
<
1 and s
≥
1 we get α
+
2
β
+
2
γ
<
1.
Therefore, withδ
=
α
+
β
+
γ
/
1
-
β
-
γ
<
1, and for all n
≥
0, and consequently, we have(28)
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
.That is,(29)
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 get(30)
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
≤
⋯
≤
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
-
2
d
x
m
-
3
,
x
m
-
2
+
s
m
-
n
-
1
d
x
m
-
2
,
x
m
-
1
+
s
m
-
n
d
x
m
-
1
,
x
m
.By using (29) we get(31)
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
-
2
δ
m
-
3
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(32)
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
1and hence(33)
d
x
n
,
x
m
≤
s
δ
n
1
-
s
δ
d
x
0
,
x
1
⟶
0
a
s
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
→
∞. Assume not, then there exists z
∈
X such that(34)
d
u
,
S
u
=
z
>
0
.So by using the triangular inequality and (21), we get(35)
z
=
d
u
,
S
u
≾
s
d
u
,
x
2
n
+
2
+
s
d
x
2
n
+
2
,
S
u
=
s
d
u
,
x
2
n
+
2
+
s
d
S
u
,
T
x
2
n
+
1
≾
s
d
u
,
x
2
n
+
2
+
s
α
d
u
,
x
2
n
+
1
+
s
β
d
2
u
,
T
x
2
n
+
1
+
d
2
x
2
n
+
1
,
S
u
d
u
,
T
x
2
n
+
1
+
d
x
2
n
+
1
,
S
u
+
s
γ
d
u
,
S
u
+
d
x
2
n
+
1
,
T
x
2
n
+
1
=
s
d
u
,
x
2
n
+
2
+
s
α
⁡
d
u
,
x
2
n
+
1
+
s
β
d
2
u
,
x
2
n
+
2
+
d
2
x
2
n
+
1
,
S
u
d
u
,
x
2
n
+
2
+
d
x
2
n
+
1
,
S
u
+
s
γ
z
+
d
x
2
n
+
1
,
x
2
n
+
2which implies that(36)
z
=
d
u
,
S
u
≤
s
d
u
,
x
2
n
+
2
+
s
α
⁡
d
u
,
x
2
n
+
1
+
s
β
d
2
u
,
x
2
n
+
2
+
d
2
x
2
n
+
1
,
S
u
d
u
,
x
2
n
+
2
+
d
x
2
n
+
1
,
S
u
+
s
γ
z
+
d
x
2
n
+
1
,
x
2
n
+
2
.Taking the limit of (36) as n
→
∞, we obtain that z
=
d
u
,
S
u
≤
0, a contradiction with (34). So z
=
0; hence S
u
=
u.
Similarly, we obtain thatT
u
=
u. Now, we show that S and T have a unique common fixed point of S and T. To show this, assume that u
∗ is another common fixed point of S and T. Then(37)
d
u
,
u
∗
=
d
S
u
,
T
u
∗
≾
α
d
u
,
u
∗
+
β
d
2
u
,
T
u
∗
+
d
2
u
∗
,
S
u
d
u
,
T
u
∗
+
d
u
∗
,
S
u
+
γ
d
u
,
S
u
+
d
u
∗
,
T
u
∗
,
≺
d
(
u
,
u
∗
), a contradiction. So u
=
u
∗, which proves the uniqueness of common fixed point in X
. For the second case d
(
S
x
,
T
y
)
=
0 if d
(
x
,
T
y
)
+
d
(
y
,
S
x
)
=
0, the proof of unique common fixed point can be completed in the line of Theorem 7. This completes the proof of the theorem.Corollary 11.
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(38)
d
T
x
,
T
y
≾
α
d
x
,
y
+
β
d
2
x
,
T
y
+
d
2
y
,
T
x
d
x
,
T
y
+
d
y
,
T
x
+
γ
d
x
,
T
x
+
d
y
,
T
yfor all x
,
y
∈
X such that x
≠
y, where α, β, and γ are nonnegative reals with α
+
2
s
β
+
2
γ
<
1 or d
(
T
x
,
T
y
)
=
0 if d
(
x
,
T
y
)
+
d
(
y
,
T
x
)
=
0. Then T has a unique fixed point.Proof.
We can prove this result by applying Theorem10 with S
=
T.Corollary 12.
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 (for some fixed n)(39)
d
T
n
x
,
T
n
y
≾
⁡
α
d
x
,
y
+
β
d
2
x
,
T
n
y
+
d
2
y
,
T
n
x
d
x
,
T
n
y
+
d
y
,
T
n
x
+
γ
d
x
,
T
n
x
+
d
y
,
T
n
yfor all x
,
y
∈
X, such that x
≠
y, where α, β, and γ are nonnegative reals with α
+
2
s
β
+
2
γ
<
1 or d
(
T
n
x
,
T
n
y
)
=
0 if d
(
x
,
T
n
y
)
+
d
(
y
,
T
n
x
)
=
0. Then T has a unique fixed point in X.Proof.
From Corollary11, we obtain that u
∈
X such that T
n
u
=
u. The uniqueness follows from(40)
d
T
u
,
u
=
d
T
T
n
u
,
T
n
u
=
d
T
n
T
u
,
T
n
u
≾
α
d
T
u
,
u
+
β
d
2
T
u
,
T
n
u
+
d
2
u
,
T
n
T
u
d
T
u
,
T
n
u
+
d
u
,
T
n
T
u
+
γ
d
T
u
,
T
n
T
u
+
d
u
,
T
n
u
=
α
+
2
β
d
T
u
,
u
.By Taking modulus of (40) and since (
α
+
2
β
)
<
1, we obtain d
T
u
,
u
≤
α
+
2
β
d
T
u
,
u
<
d
T
u
,
u, a contradiction. So T
u
=
u. Hence T
u
=
T
n
u
=
u. Therefore, the fixed point of T is unique. This completes the proof.



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*Source: 9786063-2016-06-27.xml*