Available online through
ISSN 2229 – 5046
COMMON FIXED POINTS OF THREE SELF MAPPINGS
IN COMPLEX VALUED METRIC SPACES
Sushanta Kumar Mohanta* & Rima Maitra
Department of Mathematics, West Bengal State University, Barasat, 24 Parganas(North), Kolkata-700126, West Bengal, India
(Received on: 25-07-12; Accepted on: 18-08-12)
ABSTRACT
W
e prove a common fixed point theorem for three self mappings in complex valued metric spaces. Our result generalizes some recent results in the literature due to Azam et. al.[1] and Sintunavarat et. al.[14]. Also, an example is given to illustrate our obtained result.Keywords and phrases: Complex valued metric space, point of coincidence, weakly compatible mappings, common fixed point.
2010 Mathematics Subject Classification: 54H25, 47H10.
1. INTRODUCTION
Banach’s fixed point theorem plays a major role in fixed point theory. It has applications in many branches of mathematics. Because of its usefulness, a lot of articles have been dedicated to the improvement and generalization of that result. Most of these generalizations have been made by considering different contractive type conditions in different spaces. In 2011, Azam et. al.[1] made one such generalization by introducing a complex valued metric space. In fact, they obtained a sufficient condition for the existence of common fixed points of a pair of mappings satisfying some contractive type conditions in this setting. Very recently, Sintunavarat et. al. [14] generalized this result by replacing the constants of contraction by some control functions. The purpose of this work is to obtain a common fixed point result for three self mappings in complex valued metric spaces which generalizes the results of [1] and [14].
2. PRELIMINARIES
Let
C
/
be the set of complex numbers andz
1,
z
2∈
C
/
. We can define a partial ordering
onC
/
as follows:z
1
z
2if and only if
Re(
z
1)
≤
Re(
z
2)
and
Im(
z
1)
≤
Im(
z
2)
.Thus,
z
1
z
2 if one of the following conditions is satisfied: (i)Re(
z
1)
=
Re(
z
2)
and
Im(
z
1)
=
Im(
z
2)
; (ii)Re(
z
1)
<
Re(
z
2)
and
Im(
z
1)
=
Im(
z
2)
; (iii)Re(
z
1)
=
Re(
z
2)
and
Im(
z
1)
<
Im(
z
2)
; (iv)Re(
z
1)
<
Re(
z
2)
and
Im(
z
1)
<
Im(
z
2)
.In particular, we will write
z
1
≠z
2 ifz
1≠
z
2 and one of (ii), (iii), and (iv) is satisfied and we will writez
1
z
2 ifonly (iv) is satisfied. It follows that (i)
0
z
1
z
2⇒
z
1≤
z
2 ; (ii)0
z
1≠
z
2⇒
z
1<
z
2 ;(iii)
z
1
z
2 andz
2
z
3⇒
z
1
z
3;(iv)
a
,
b
∈
R
,
0
≤
a
≤
b
and
z
1
z
2⇒
az
1
bz
2.Definition 2.1. ([1]) Let
X
be a nonempty set. Suppose that the mappingd
:
X
×
X
→
C
/
satisfies the following conditions:(i)
0
d
(
x
,
y
),
for
all
x
,
y
∈
X
and
d
(
x
,
y
)
=
0
if
and
only
if
x
=
y
;
(ii)d
(
x
,
y
)
=
d
(
y
,
x
),
for
all
x
,
y
∈
X
;
(iii)
d
(
x
,
y
)
d
(
x
,
z
)
+
d
(
z
,
y
),
for
all
x
,
y
,
z
∈
X
.Then
d
is called a complex valued metric onX
and(
X
,
d
)
is called a complex valued metric space. Note that)
,
(
x
y
d
1
+
d
(
x
,
y
)
and so,1
)
,
(
1
)
,
(
≤
+
d
x
y
y
x
d
.
Example 2.2. ( [14] ) Let
X
=
C
/
. Define the mappingd
:
X
×
X
→
C
/
byd
(
z
1,
z
2)
=
e
ikz
1−
z
2 , wherek
∈
R
. Then(
X
,
d
)
is a complex valued metric space.Definition 2.3. ( [1] ) Let
(
X
,
d
)
be a complex valued metric space,(
x
n)
be a sequence inX
andx
∈
X
.(i) If for every
c
∈
C
/
, with0
c
there isn
0∈
N
such that for alln
>
n
0,
d
(
x
n,
x
)
c
, then(
x
n)
is said to be convergent,(
x
n)
converges tox
andx
is the limit point of(
x
n)
. We denote this byx
nx
n→∞
=
lim
or∞
→
→
x
as
n
x
n .(ii) If for every
c
∈
C
/
, with0
c
there isn
0∈
N
such that for alln
>
n
0,
d
(
x
n,
x
n+m)
c
, wherem
∈
N
, then(
x
n)
is said to be Cauchy sequence.(iii) If every Cauchy sequence in
X
is convergent, then(
X
,
d
)
is said to be a complete complex valued metric space.Lemma 2.4. ([1]) Let
(
X
,
d
)
be a complex valued metric space and let(
x
n)
be a sequence inX
. Then(
x
n)
converges to
x
if and only ifd
(
x
n,
x
)
→
0
as
n
→
∞
.Lemma 2.5. ([1])Let
(
X
,
d
)
be a complex valued metric space and let(
x
n)
be a sequence inX
. Then(
x
n)
is a Cauchy sequence if and only ifd
(
x
n,
x
n+m)
→
0
as
n
→
∞
, wherem
∈
N
.Definition 2.6. ([4]) Let
T
andS
be self mappings of a setX
. Ifw
=
Tx
=
Sx
for somex
inX
, thenx
is called a coincidence point ofT
andS
andw
is called a point of coincidence ofT
andS
.Definition 2.7. ([7]) Let
T
andS
be self mappings of a nonempty setX
. The mappingsT
andS
are weakly compatible ifTSx
=
STx
wheneverTx
=
Sx
.Definition 2.8. A mapping
T
:
X
→
X
in a complex valued metric space(
X
,
d
)
is said to be expansive if there is a real constantc
>
1
satisfying
c
d
(
x
,
y
)
d
(
Tx
,
Ty
)
for allx
,
y
∈
X
.3. MAIN RESULTS
In this section, we always suppose that
C
/
is the set of complex numbers and
is a partial ordering onC
/
. Throughoutthe paper we denote by
N
the set of all positive integers.Theorem 3.2. Let
(
X
,
d
)
be a complex valued metric space andf
,
S
,
T
:
X
→
X
. Suppose there exist mappings)
1
,
0
[
:
,
21
∧
→
∧
X
such that for allx
,
y
∈
X
:
(i)
∧
i(
Sx
)
≤
∧
i(
fx
)
and
∧
i(
Tx
)
≤
∧
i(
fx
)
for
i
=
1
,
2
;
(ii)∧
1(
fx
)
+
∧
2(
fx
)
<
1
;
(iii)
d
(
Sx
,
Ty
)
)
,
(
1
)
,
(
)
,
(
)
(
)
,
(
)
(
2 1fy
fx
d
Ty
fy
d
Sx
fx
d
fx
fy
fx
d
fx
+
∧
+
∧
.If
S
(
X
)
∪
T
(
X
)
⊆
f
(
X
)
andf
(
X
)
is complete, thenf
,
S
andT
have a unique point of coincidence. Moreover, if(
S
,
f
)
and(
T
,
f
)
are weakly compatible , thenf
,
S
andT
have a unique common fixed point inX
.Proof. Let
x
0∈
X
be arbitrary. Choose a pointx
1∈
X
such thatfx
1=
Sx
0 which is possible since)
(
)
(
X
f
X
S
⊆
. Also, we may choose a pointx
2∈
X
satisfyingfx
2=
Tx
1 sinceT
(
X
)
⊆
f
(
X
)
. Continuing in this way, we can construct a sequence(
fx
n)
inf
(
X
)
such thatodd
is
n
if
Sx
fx
n=
n−1,
=
Tx
n−1,
if
n
is
even
.If
n
∈
N
is odd, then by using hypothesis we obtain)
,
(
)
,
(
fx
nfx
n 1d
Sx
n 1Tx
nd
+=
−
)
,
(
1
)
,
(
)
,
(
)
(
)
,
(
)
(
1 1 1 1 2 1 1 1 n n n n n n n n n nfx
fx
d
Tx
fx
d
Sx
fx
d
fx
fx
fx
d
fx
− − − − − −+
∧
+
∧
)
,
(
1
)
,
(
)
,
(
)
(
)
,
(
)
(
1 1 1 1 2 1 1 1 n n n n n n n n n nfx
fx
d
fx
fx
d
fx
fx
d
fx
fx
fx
d
fx
− + − − − −+
∧
+
∧
=
. Therefore,)
,
(
)
(
)
,
(
fx
nfx
n 1 1fx
n 1d
fx
n 1fx
nd
+≤
∧
− −)
,
(
1
)
,
(
)
,
(
)
(
1 1 1 1 2 n n n n n n nfx
fx
d
fx
fx
d
fx
fx
d
fx
− − + −+
∧
+
≤
∧
1(
fx
n−1)
d
(
fx
n−1,
fx
n)
+
∧
2(
fx
n−1)
d
(
fx
n,
fx
n+1)
=
∧
1(
Tx
n−2)
d
(
fx
n−1,
fx
n)
+
∧
2(
Tx
n−2)
d
(
fx
n,
fx
n+1)
≤
∧
1(
fx
n−2)
d
(
fx
n−1,
fx
n)
+
∧
2(
fx
n−2)
d
(
fx
n,
fx
n+1)
=
∧
1(
Sx
n−3)
d
(
fx
n−1,
fx
n)
+
∧
2(
Sx
n−3)
d
(
fx
n,
fx
n+1)
≤
∧
1(
fx
n−3)
d
(
fx
n−1,
fx
n)
+
∧
2(
fx
n−3)
d
(
fx
n,
fx
n+1)
≤
∧
1(
fx
0)
d
(
fx
n−1,
fx
n)
+
∧
2(
fx
0)
d
(
fx
n,
fx
n+1)
which implies that
)
,
(
)
(
1
)
(
)
,
(
1 0 2 0 11 n n
n
n
d
fx
fx
fx
fx
fx
fx
d
+ −∧
−
∧
≤
.If
n
∈
N
is even, then)
,
(
)
,
(
)
,
(
fx
nfx
n+1=
d
Tx
n−1Sx
n=
d
Sx
nTx
n−1
)
,
(
1
)
,
(
)
,
(
)
(
)
,
(
)
(
1 1 1 2 1 1 − − + −+
∧
+
∧
=
n n n n n n n n n nfx
fx
d
fx
fx
d
fx
fx
d
fx
fx
fx
d
fx
. Therefore,)
,
(
)
(
)
,
(
fx
nfx
n+1≤
∧
1fx
nd
fx
nfx
n−1d
)
,
(
1
)
,
(
)
,
(
)
(
1 1 1 2 − − ++
∧
+
n n n n n n nfx
fx
d
fx
fx
d
fx
fx
d
fx
≤
∧
1(
fx
n)
d
(
fx
n,
fx
n−1)
+
∧
2(
fx
n)
d
(
fx
n,
fx
n+1)
=
∧
1(
Tx
n−1)
d
(
fx
n,
fx
n−1)
+
∧
2(
Tx
n−1)
d
(
fx
n,
fx
n+1)
≤
∧
1(
fx
n−1)
d
(
fx
n,
fx
n−1)
+
∧
2(
fx
n−1)
d
(
fx
n,
fx
n+1)
=
∧
1(
Sx
n−2)
d
(
fx
n,
fx
n−1)
+
∧
2(
Sx
n−2)
d
(
fx
n,
fx
n+1)
≤
∧
1(
fx
0)
d
(
fx
n,
fx
n−1)
+
∧
2(
fx
0)
d
(
fx
n,
fx
n+1)
which gives that
)
,
(
)
(
1
)
(
)
,
(
1 0 2 0 1 1 − +−
∧
∧
≤
n nn
n
d
fx
fx
fx
fx
fx
fx
d
.Thus for any positive integer
n
, it must be the case that)
,
(
)
(
1
)
(
)
,
(
1 0 2 0 11 n n
n
n
d
fx
fx
fx
fx
fx
fx
d
+ −∧
−
∧
≤
. (3.1)If we let
)
(
1
)
(
:
0 2 0 1fx
fx
∧
−
∧
=
α
, then by repeated application of (3.1))
,
(
)
,
(
fx
nfx
n 1d
fx
n 1fx
nd
+≤
α
−
.
)
,
(
)
,
(
1 0 1 2 2fx
fx
d
fx
fx
d
n n nα
α
≤
≤
− −
Now, for all
m
,
n
∈
N
,
m
>
n
, we have)
,
(
fx
nfx
md
d
(
fx
n,
fx
n+1)
+
d
(
fx
n+1,
fx
n+2)
+
+
d
(
fx
m−1,
fx
m)
.Therefore,
)
,
(
fx
nfx
md
≤
d
(
fx
n,
fx
n+1)
+
d
(
fx
n+1,
fx
n+2)
+
+
d
(
fx
m−1,
fx
m)
(
)
.
)
,
(
1
)
,
(
1 0 1 0 1 1fx
fx
d
fx
fx
d
n m n nα
α
α
α
α
−
≤
+
+
+
≤
+
−Since
α
∈
[
0
,
1
)
, taking limit asm
,
n
→
∞
, we haved
(
fx
n,
fx
m)
→
0
which implies that(
fx
n)
is a Cauchysequence in
f
(
X
)
. By completeness off
(
X
)
, there existu
,
v
∈
X
such thatfx
n→
v
=
fu
.Now,
)
,
(
fu
Tu
d
d
(
fu
,
fx
2n+1)
+
d
(
fx
2n+1,
Tu
)
=
d
(
fu
,
fx
2n+1)
+
d
(
Sx
2n,
Tu
)
d
(
fu
,
fx
2n+1)
+
∧
1(
fx
2n)
d
(
fx
2n,
fu
)
which implies that
)
,
(
)
(
)
,
(
)
,
(
fu
Tu
d
fu
fx
2 1 1fx
2d
fx
2fu
d
≤
n++
∧
n n)
,
(
1
)
,
(
)
,
(
)
(
2 2 2 2 2fu
fx
d
Tu
fu
d
Sx
fx
d
fx
n n n n+
∧
+
≤
d
(
fu
,
fx
2n+1)
+
∧
1(
fx
2n)
d
(
fx
2n,
fu
)
+
∧
2(
fx
2n)
d
(
fx
2n,
Sx
2n)
d
(
fu
,
Tu
)
,
since 1
1
+
d
(
fx
2n,
fu
)
.≤
d
(
fu
,
fx
2n+1)
+
∧
1(
fx
0)
d
(
fx
2n,
fu
)
+
∧
2(
fx
0)
d
(
fx
2n,
fx
2n+1)
d
(
fu
,
Tu
)
.Taking
n
→
∞
, it follows thatd
(
fu
,
Tu
)
=
0
and henced
(
fu
,
Tu
)
=
0
. Therefore,fu
=
Tu
=
v
. Similarly, we can show thatfu
=
Su
=
v
.Thus,
fu
=
Su
=
Tu
=
v
and sov
becomes a common point of coincidence off
,
S
andT
.For uniqueness, let there exists another point
w
(
≠
v
)
∈
X
such thatfx
=
Sx
=
Tx
=
w
for somex
∈
X
. Thus,)
,
(
)
,
(
v
w
d
Su
Tx
d
=
)
,
(
1
)
,
(
)
,
(
)
(
)
,
(
)
(
2 1fx
fu
d
Tx
fx
d
Su
fu
d
fu
fx
fu
d
fu
+
∧
+
∧
)
,
(
)
(
)
,
(
1
)
,
(
)
,
(
)
(
)
,
(
)
(
1 2 1w
v
d
v
w
v
d
w
w
d
v
v
d
v
w
v
d
v
∧
=
+
∧
+
∧
=
which implies that
)
,
(
)
(
)
,
(
v
w
1v
d
v
w
d
≤
∧
.Since
0
≤
∧
1(
v
)
<
1
, it follows thatd
(
v
,
w
)
=
0
and sov
=
w
. If(
S
,
f
)
and(
T
,
f
)
are weakly compatible, then by Lemma 3.1,f
,
S
andT
have a unique common fixed point inX
.As an application of Theorem 3.2, we have the following results.
Corollary 3.3. [[14], Theorem 3.1] Let
(
X
,
d
)
be a complete complex valued metric space andS
,
T
:
X
→
X
. Suppose there exist mappings∧
1,
∧
2:
X
→
[
0
,
1
)
such that for allx
,
y
∈
X
:
(i)
∧
i(
Sx
)
≤
∧
i(
x
)
and
∧
i(
Tx
)
≤
∧
i(
x
)
for
i
=
1
,
2
;
(ii)∧
1(
x
)
+
∧
2(
x
)
<
1
;
(iii)
d
(
Sx
,
Ty
)
)
,
(
1
)
,
(
)
,
(
)
(
)
,
(
)
(
2 1y
x
d
Ty
y
d
Sx
x
d
x
y
x
d
x
+
∧
+
∧
.Then
S
andT
have a unique common fixed point inX
.Proof. The result follows from Theorem 3.2 by taking
f
=
I
, the identity mapping.Corollary 3.4. [[1], Theorem 4] Let
(
X
,
d
)
be a complete complex valued metric space andS
,
T
:
X
→
X
. IfS
andT
satisfy)
,
(
Sx
Ty
d
)
,
(
1
)
,
(
)
,
(
)
,
(
y
x
d
Ty
y
d
Sx
x
d
y
x
d
+
+
µ
λ
Proof. The desired result can be obtained from Theorem 3.2 by setting
∧
1(
x
)
=
λ
,
∧
2(
x
)
=
µ
and
f
=
I
.Corollary 3.5. [[14], Theorem 3.7] Let
(
X
,
d
)
be a complex valued metric space,f
,
T
:
X
→
X
be such that)
(
)
(
X
f
X
T
⊆
andf
(
X
)
is complete. Suppose there exist mappings∧
1,
∧
2:
X
→
[
0
,
1
)
such that for all:
,
y
X
x
∈
(i)
∧
i(
Tx
)
≤
∧
i(
fx
)
for
i
=
1
,
2
;
(ii)∧
1(
fx
)
+
∧
2(
fx
)
<
1
;
(iii)
d
(
Tx
,
Ty
)
)
,
(
1
)
,
(
)
,
(
)
(
)
,
(
)
(
21
fy
fx
d
Ty
fy
d
Tx
fx
d
fx
fy
fx
d
fx
+
∧
+
∧
.Then
f
andT
have a unique point of coincidence. Moreover, iff
andT
are weakly compatible, thenf
andT
have a unique common fixed point inX
.Proof. The conclusion of the Corollary follows from Theorem 3.2 by considering
S
=
T
.Corollary 3.6. Let
(
X
,
d
)
be a complex valued metric space and letf
,
T
:
X
→
X
satisfy)
,
(
Tx
Ty
d
)
,
(
1
)
,
(
)
,
(
)
,
(
fy
fx
d
Ty
fy
d
Tx
fx
d
fy
fx
d
+
+
µ
λ
for all
x
,
y
∈
X
, whereλ
,
µ
are nonnegative reals withλ
+
µ
<
1
. IfT
(
X
)
⊆
f
(
X
)
andf
(
X
)
is complete, thenf
andT
have a unique point of coincidence. Moreover, iff
andT
are weakly compatible, thenf
andT
have a unique common fixed point inX
.Proof. Putting
S
=
T
,
∧
1(
x
)
=
λ
,
∧
2(
x
)
=
µ
in Theorem 3.2, we can prove this result.Corollary 3.7. Let
(
X
,
d
)
be a complete complex valued metric space andT
:
X
→
X
. Suppose there exist mappings∧
1,
∧
2:
X
→
[
0
,
1
)
such that for allx
,
y
∈
X
:
(i)
∧
i(
Tx
)
≤
∧
i(
x
)
for
i
=
1
,
2
;
(ii)∧
1(
x
)
+
∧
2(
x
)
<
1
;
(iii)
d
(
Tx
,
Ty
)
)
,
(
1
)
,
(
)
,
(
)
(
)
,
(
)
(
21
y
x
d
Ty
y
d
Tx
x
d
x
y
x
d
x
+
∧
+
∧
.Then
T
has a unique fixed point inX
.Proof. The conclusion of the Corollary follows from Theorem 3.2 by considering
S
=
T
andf
=
I
.Theorem 3.8. Let
(
X
,
d
)
be a complete complex valued metric space and letf
:
X
→
X
be an onto expansive mapping i.e.,f
(
X
)
=
X
and there exists a real constantc
>
1
such that)
,
(
x
y
d
c
d
(
fx
,
fy
)
for all
x
,
y
∈
X
. Thenf
has a unique fixed point inX
.Proof. We can prove this result by applying Corollary 3.6 with
T
=
I
,
and
µ
=
0
. We conclude with an example.Example 3.9. Let
X
=
[
1
,
∞
)
. DefineT
,
f
:
X
→
X
byTx
=
2
x
−
1
andfx
=
5
x
−
4
. Ifd
uis the usual metric onX
, thenT
andf
are not contraction mappings as for allx
,
y
∈
X
y
x
Ty
Tx
and
y
x
fy
fx
d
u(
,
)
=
5
−
.So, we can not apply Banach contraction theorem to find the unique fixed point 1 of
T
andf
.We consider a complex valued metric
d
:
X
×
X
→
C
/
byX
y
x
all
for
y
x
i
y
x
y
x
d
(
,
)
=
−
+
−
,
,
∈
.Then
(
X
,
d
)
is a complete complex valued metric space.Now,
[
x
y
i
x
y
]
Ty
Tx
d
(
,
)
=
2
−
+
−
(
,
)
5
2
fy
fx
d
=
(
,
)
2
1
fy
fx
d
.Since
T
(
X
)
=
f
(
X
)
=
X
, we have all the conditions of Corollary 3.6 with,
0
2
1
=
=
µ
λ
. So, applying Corollary3.6 we can obtain a unique common fixed point
1
ofT
andf
inX
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ztu
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