# Some Common Fixed Point Theorems in Complex Valuedb-Metric Spaces

Authors: Aiman A. Mukheimer
Journal: The Scientific World Journal
 (2014)

---

## Abstract

Azam et al. (2011), introduce the notion of complex valued metric spaces and obtained
common fixed point result for mappings in the context of complex valued metric spaces.
Rao et al. (2013) introduce the notion of complex valuedb-metric spaces. In this paper, we
generalize the results of Azam et al. (2011), and Bhatt et al. (2011), by improving the conditions
of contraction to establish the existence and uniqueness of common fixed point for two
self-mappings on complex valued b-metric spaces. Some examples are given to illustrate
the main results.


---

## Body

## 1. Introduction

Banach contraction principle in [1] was the starting point for many researchers during last decades in the field of nonlinear analysis. In 1989, Bakhtin [2] introduced the concept of b-metric space as a generalization of metric spaces. The concept of complex valued b-metric spaces was introduced in 2013 by Rao et al. [3], which was more general than the well-known complex valued metric spaces that were introduced in 2011 by Azam et al. [4]. The main purpose of this paper is to present common fixed point results of two self-mappings satisfying a rational inequality on complex valued b-metric spaces. The results presented in this paper are generalization of work done by Azam et al. in [4] and Bhatt et al in [5].Definition 1 (see [6]).
LetX be a nonempty set and let s≥1 be a given real number. A function d:X×X→[0,∞) is called a b-metric if for all x,y,z∈X the following conditions are satisfied: (i)
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 b-metric space. The number s≥1 is called the coefficient of (X,d).Example 2 (see [7]).
Let(X,d) be a metric space and ρ(x,y)=(d(x,y))p, where p>1 is a real number. Then (X,ρ) is a b-metric space with s=2p-1.LetC be the set of complex numbers and z1,z2∈C. Define a partial order ≾ on C as follows:
(1)z1≾z2iffRe(z1)≤Re(z1),Im⁡(z1)≤Im⁡(z2).
Thus z1≾z2 if one of the following holds:(1)
Re(z1)=Re(z2) and Im⁡(z1)=Im⁡(z2),(2)
Re(z1)<Re(z2) and Im⁡(z1)=Im⁡(z2),(3)
Re(z1)=Re(z2) and Im⁡(z1)<Im⁡(z2),(4)
Re(z1)<Re(z2) and Im⁡(z1)<Im⁡(z2).We will write z1⋨z2 if z1≠z2 and one of (2), (3), and (4) is satisfied; also we will write z1≺z2 if only (4) is satisfied.Remark 3.
We can easily check that the following statements are held:(i)
ifa,b∈R and a≤b, then az≾bz for all z∈C;(ii)
if0≾z1⋨z2, then |z1|<|z2|;(iii)
ifz1≾z2 and z2≺z3, then z1≺z3.Definition 4 (see [4]).
LetX be a nonempty set. A function d:X×X→C is called a complex valued 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)≾d(x,z)+d(z,y).The pair (X,d) is called a complex valued metric space.Example 5 (see [8]).
LetX=C. Define the mapping d:X×X→C by
(2)d(x,y)=i|x-y|,∀x,y∈X.
Then (X,d) is a complex valued metric space.Example 6 (see [9]).
LetX=C. Define the mapping d:X×X→C by
(3)d(x,y)=eik|x-y|,wherek∈R,∀x,y∈X.
Then (X,d) is a complex valued metric space.Definition 7 (see [3]).
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 8 (see [3]).
LetX=[0,1]. Define the mapping d:X×X→C by
(4)d(x,y)=|x-y|2+i|x-y|2,∀x,y∈X.
Then (X,d) is a complex valued b-metric space with s=2.Definition 9 (see [3]).
Let(X,d) be a complex valued b-metric space. Consider the following.(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)≠∅.(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 and 0≺r}.Definition 10 (see [3]).
Let(X,d) be a complex valued b-metric space and {xn} a sequence in X and x∈X. Consider the following.(i)
If for everyc∈C, with 0≺r, there is N∈N such that, for all n>N,  d(xn,x)≺c, then {xn} is said to be convergent, {xn} converges to x, and x is the limit point of {xn}. We denote this by lim⁡n→∞xn=x or {xn}→x  as  n→∞.(ii)
If for everyc∈C, with 0≺r, there is N∈N such that, for all n>N,  d(xn,xn+m)≺c, where m∈N, then {xn} 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 11 (see [3]).
Let(X,d) be a complex valued b-metric space and let {xn} be a sequence in X. Then {xn} converges to x if and only if |d(xn,x)|→0  as  n→∞.Lemma 12 (see [3]).
Let(X,d) be a complex valued b-metric space and let {xn} be a sequence in X. Then {xn} is a Cauchy sequence if and only if |d(xn,xn+m)|→0  as  n→∞, where m∈N.Theorem 13 (see [4]).
Let(X,d) be a complete complex valued metric space and let λ,μ be nonnegative real numbers such that λ+μ<1. Suppose that S,T:X→X are mappings satisfying
(5)d(Sx,Ty)≾λd(x,y)+μd(x,Sx)d(y,Ty)1+d(x,y),
for all x,y∈X. Then S,T have a unique common fixed point in X.Theorem 14 (see [5]).
Let(X,d) be a complete complex valued metric space and let S,T:X→X be mappings satisfying
(6)d(Sx,Ty)≾a[d(x,Sx)d(x,Ty)+d(y,Ty)d(y,Sx)]d(x,Ty)+d(y,Sx),
for all x,y∈X, where a∈[0,1). Then S,T have a unique common fixed point in X.


## 2. Main Result

Our next theorem is a generalization of Theorem13 in complex valued b-metric spaces.Theorem 15.
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
(7)d(Sx,Ty)≾λd(x,y)+μd(x,Sx)d(y,Ty)1+d(x,y),
for all x,y∈X, where λ,μ are nonnegative reals with sλ+μ<1. Then S,T have a unique common fixed point in X.Proof.
For any arbitrary point,x0∈X. Define sequence {xn} in X such that
(8)x2n+1=Sx2n,x2n+2=Tx2n+1,forn=0,1,2,3,….
Now, we show that the sequence {xn} is Cauchy. Let x=x2n and y=x2n+1 in (7); we have
(9)d(x2n+1,x2n+2)=d(Sx2n,Tx2n+1)≾λd(x2n,x2n+1)+μd(x2n,Sx2n)d(x2n+1,Tx2n+1)1+d(x2n,x2n+1)=λd(x2n,x2n+1)+μd(x2n,x2n+1)d(x2n+1,x2n+2)1+d(x2n,x2n+1),
which implies that
(10)|d(x2n+1,x2n+2)|≤λ|d(x2n,x2n+1)|+μ|d(x2n,x2n+1)||d(x2n+1,x2n+2)||1+d(x2n,x2n+1)|.
Since |1+d(x2n,x2n+1)|>|d(x2n,x2n+1)|, we get
(11)|d(x2n+1,x2n+2)|≤λ|d(x2n,x2n+1)|+μ|d(x2n+1,x2n+2)|,
and hence
(12)|d(x2n+1,x2n+2)|≤λ1-μ|d(x2n,x2n+1)|.
Similarly, we obtain
(13)|d(x2n+2,x2n+3)|≤λ1-μ|d(x2n+1,x2n+2)|.
Since sλ+μ<1 and s≥1, we get λ+μ<1.
Therefore, withδ=λ/(1-μ)<1, and for all n≥0, consequently, we have
(14)|d(x2n+1,x2n+2)|≤δ|d(x2n,x2n+1)|≤δ2|d(x2n-1,x2n)|≤⋯≤δ2n+1|d(x0,x1)|.
Thus for any m>n,  m,  n∈N, and since sδ=sλ/(1-μ)<1, we get
(15)|d(xn,xm)|≤s|d(xn,xn+1)|+s|d(xn+1,xm)|≤s|d(xn,xn+1)|+s2|d(xn+1,xn+2)|+s2|d(xn+2,xm)|≤s|d(xn,xn+1)|+s2|d(xn+1,xn+2)|+s3|d(xn+2,xm)|+s3|d(xn+2,xm)|≤s|d(xn,xn+1)|+s2|d(xn+1,xn+2)|+s3|d(xn+2,xm)|+⋯+sm-n-2|d(xm-3,xm-2)|+sm-n-1|d(xm-2,xm-1)|+sm-n-1|d(xm-1,xm)|≤s|d(xn,xn+1)|+s2|d(xn+1,xn+2)|+s3|d(xn+2,xm)|+⋯+sm-n-2|d(xm-3,xm-2)|+sm-n-1|d(xm-2,xm-1)|+sm-n|d(xm-1,xm)|.
By using (14) we get
(16)|d(xn,xm)|≤sδn|d(x0,x1)|+s2δn+1|d(x0,x1)|+s3δn+2|d(x0,x1)|+⋯+sm-n-2δm-3|d(x0,x1)|+sm-n-1δm-2|d(x0,x1)|+sm-nδm-1|d(x0,x1)|=∑i=1m-nsiδi+n-1|d(x0,x1)|.
Therefore,
(17)|d(xn,xm)|≤∑i=1m-nsi+n-1δi+n-1|d(x0,x1)|=∑t=nm-1stδt|d(x0,x1)|≤∑t=n∞(sδ)t|d(x0,x1)|=(sδ)n1-sδ|d(x0,x1)|,
and hence
(18)|d(xn,xm)|≤(sδ)n1-sδ|d(x0,x1)|⟶0asm,n⟶∞.
Thus, {xn} is a Cauchy sequence in X.
SinceX is complete, there exists some u∈X such that xn→u as n→∞. Assuming not, then there exist z∈X such that
(19)|d(u,Su)|=|z|>0.
So by using the triangular inequality and (7), we get
(20)z=d(u,Su)≾sd(u,x2n+2)+sd(x2n+2,Su)=sd(u,x2n+2)+sd(Tx2n+1,Su)≾sd(u,x2n+2)+sλd(u,x2n+2)+sμd(u,Su)d(x2n+1,Tx2n+1)1+d(u,x2n+2)=sd(u,x2n+2)+sλd(u,x2n+2)+sμd(u,Su)d(x2n+1,x2n+2)1+d(u,x2n+2),
which implies that
(21)|z|=|d(u,Su)|≤s|d(u,x2n+2)|+sμ|d(u,Su)||d(x2n+1,x2n+2)||1+d(u,x2n+2)|.
Taking the limit of (21) as n→∞, we obtain that |z|=|d(u,Su)|≤0, a contradiction with (19). So |z|=0. Hence Su=u. Similarly, we obtain Tu=u.
Now we show thatS and T have unique common fixed point of S and T. To show this, assume that u* is another common fixed point of S and T. Then
(22)d(u,u*)=d(Su,Tu*)≾λd(u,u*)+μd(u,Su)d(u*,Tu*)1+d(u,u*)≺d(u,u*).
This implies that |d(x,x*)|<|d(u,u*)|, a contradiction. So u=u* which proves the uniqueness of common fixed point in X. This completes the proof.Corollary 16.
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
(23)d(Tx,Ty)≾λd(x,y)+μd(x,Tx)d(y,Ty)1+d(x,y),
for all x,y∈X, where λ,μ are nonnegative reals with sλ+μ<1. Then T has a unique fixed point in X.Proof.
We can prove this result by applying Theorem15 with S=T.Corollary 17.
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
(24)d(Tnx,Tny)≾λd(x,y)+μd(x,Tnx)d(y,Tny)1+d(x,y),
for all x,y∈X, where λ,μ are nonnegative reals with sλ+μ<1. Then T has a unique fixed point in X.Proof.
From Corollary20, we obtain u∈X such that
(25)Tnu=u.
The uniqueness follows from
(26)d(Tu,u)=d(TTnu,Tnu)=d(TnTu,Tnu)≾λd(Tu,u)+μd(Tu,TnTu)d(u,Tnu)1+d(Tu,u)≾λd(Tu,u)+μd(Tu,TnTu)d(u,u)1+d(Tu,u)=λd(Tu,u).
By taking modulus of (26) and since λ<1, we obtain |d(Tu,u)|≤λ|d(Tu,u)|<|d(Tu,u)|, a contradiction. So, Tu=u. Hence
(27)Tu=Tnu=u.
Therefore, the fixed point of T is unique. This completes the proof.Example 18.
LetX=C. Define a function d:X×X→C such that
(28)d(z1,z2)=|x1-x2|2+i|y1-y2|2,
where z1=x1+iy1 and z2=x2+iy2.
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.
Letz1,z2, and z3∈X; then,
(29)d(z1,z2)=|x1-x2|2+i|y1-y2|2=|x1-x3+x3-x2|2+i|y1-y3+y3-y2|2≾|x1-x3|2+|x3-x2|2+2|x1-x3||x3-x2|+i[|y1-y3|2+|y3-y2|2+2|y1-y3||y3-y2|]≾|x1-x3|2+|x3-x2|2+|x1-x3|2+|x3-x2|2+i[|y1-y3|2+|y3-y2|2+|y1-y3|2+|y3-y2|2]=2{|x1-x3|2+|x3-x2|2+i[|y1-y3|2+|y3-y2|2]}=2[d(z1,z3)+d(z3,z2)].
Therefore, s=2.
Now, define two self-mappingsS,T:X→X as follows:
(30)Tz=T(x+iy)={0ifx,y∈Q2ifx∈Qc,y∈Q2iifx∈Qc,y∈Qc2+2iifx∈Q,y∈Qc
such that S=T and z=x+iy. Let x=1/π and y=0, and since λ∈[0,1), we have
(31)d(Tx,Ty)=d(T1π,T0)=d(2,0)=4≻λ1π2=λd(1π,0)+μd(1/π,T(1/π))d(0,T0)1+d(1/π,0).
Note that Tnz=0 for n>1, so
(32)d(Tnx,Tny)=0≾λd(x,y)+μd(x,Tnx)d(y,Tny)1+d(x,y),
for all x,y∈X and λ,μ≥0 with 2λ+μ<1. So all conditions of Corollary 21 are satisfied to get a unique fixed point 0 of T.Our next theorem is a generalization of Theorem14 in complex valued b-metric spaces.Theorem 19.
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
(33)d(Sx,Ty)≾a[d(x,Sx)d(x,Ty)+d(y,Ty)d(y,Sx)]d(x,Ty)+d(y,Sx),
for all x,y∈X, where sa∈[0,1). Then S,T have a unique common fixed point in X.Proof.
For any arbitrary point,x0∈X. Define sequence {xn} in X such that
(34)x2n+1=Sx2n,x2n+2=Tx2n+1,fork=0,1,2,3,….
Now, we show that the sequence {xn} is Cauchy. Let x=x2n and y=x2n+1 in (33); we have
(35)d(x2n+1,x2n+2)=d(Sx2n,Tx2n+1)≾a[d(x2n,Sx2n)d(x2n,Tx2n+1)+d(x2n+1,Tx2n+1)d(x2n+1,Sx2n]×(d(x2n,Tx2n+1)+d(x2n+1,Sx2n))-1=a[d(x2n,x2n+1)d(x2n,x2n+2)+d(x2n+1,x2n+2)d(x2n+1,x2n+1]×(d(x2n,x2n+2)+d(x2n+1,x2n+1))-1,
which implies that
(36)|d(x2n+1,x2n+2)|≤a[|d(x2n,x2n+1)||d(x2n,x2n+2)|+|d(x2n+1,x2n+2)||d(x2n+1,x2n+1)|]×(|d(x2n,x2n+2)+d(x2n+1,x2n+1)|)-1=a|d(x2n,x2n+1)||d(x2n,x2n+2)||d(x2n,x2n+2)|=a|d(x2n,x2n+1)|,
and hence
(37)|d(x2n+1,x2n+2)|≤a|d(x2n,x2n+1)|.
Similarly, we can see that
(38)|d(x2n+2,x2n+3)|≤a|d(x2n+1,x2n+2)|.
Since sa<1 and s≥1, we get a<1.
Therefore, for alln≥0, consequently, we have
(39)|d(x2n+1,x2n+2)|≤a|d(x2n,x2n+1)|≤a2|d(x2n-1,x2n)|≤⋯≤a2n+1|d(x0,x1)|.
Thus for any m>n,  m,  n∈N, we have
(40)|d(xn,xm)|≤s|d(xn,xn+1)|+s|d(xn+1,xm)|≤s|d(xn,xn+1)|+s2|d(xn+1,xn+2)|+s2|d(xn+2,xm)|≤s|d(xn,xn+1)|+s2|d(xn+1,xn+2)|+s3|d(xn+2,xm)|+s3|d(xn+2,xm)|≤s|d(xn,xn+1)|+s2|d(xn+1,xn+2)|+s3|d(xn+2,xm)|+⋯+sm-n-2|d(xm-3,xm-2)|+sm-n-1|d(xm-2,xm-1)|+sm-n-1|d(xm-1,xm)|≤s|d(xn,xn+1)|+s2|d(xn+1,xn+2)|+s3|d(xn+2,xm)|+⋯+sm-n-2|d(xm-3,xm-2)|+sm-n-1|d(xm-2,xm-1)|+sm-n|d(xm-1,xm)|.
By using (39), we get
(41)|d(xn,xm)|≤san|d(x0,x1)|+s2an+1|d(x0,x1)|+s3an+2|d(x0,x1)|+⋯+sm-n-2am-3|d(x0,x1)|+sm-n-1am-2|d(x0,x1)|+sm-nam-1|d(x0,x1)|=∑i=1m-n‍siai+n-1|d(x0,x1)|.
Therefore,
(42)|d(xn,xm)|≤∑i=1m-n‍si+n-1ai+n-1|d(x0,x1)|=∑t=nm-1‍stat|d(x0,x1)|≤∑t=n∞‍(sa)t|d(x0,x1)|=(sa)n1-sa|d(x0,x1)|.
Now, since sa<1, we deduce
(43)|d(xn,xm)|≤(sa)n1-sa|d(x0,x1)|⟶0asm,n⟶∞.
Thus, {xn} is a Cauchy sequence in X.
SinceX is complete, there exists some u∈X such that xn→u as n→∞. Assuming not, then there exist z∈X such that
(44)|d(u,Su)|=|z|>0.
So by using the triangular inequality and (33), we get
(45)z=d(u,Su)≾sd(u,x2n+2)+sd(x2n+2,Su)=sd(u,x2n+2)+sd(Tx2n+1,Su)≾sd(u,x2n+2)+(sad(u,Su)d(u,Tx2n+1)+sad(x2n+1,Tx2n+1)d(x2n+1,Su))×(d(u,Tx2n+1)+d(x2n+1,Su))-1=sd(u,x2n+2)+(sad(u,Su)d(u,x2n+2)+sad(x2n+1,x2n+2)d(x2n+1,Su))×(d(u,x2n+2)+d(x2n+1,Su))-1,
which implies that
(46)|z|=|d(u,Su)|≤s|d(u,x2n+2)|+(sa|d(u,Su)||d(u,x2n+2)|+sa|d(x2n+1,x2n+2)||d(x2n+1,Su)|)×(|d(u,x2n+2)+d(x2n+1,Su)|)-1.
Taking the limit of (48) as n→∞, we obtain that |z|=|d(u,Su)|≤0, a contradiction with (44). So |z|=0. Hence Su=u. Similarly, we obtain Tu=u.
Now we show thatS and T have unique common fixed point of S and T. To show this, assume that u* is another common fixed point of S and T. Then
(47)d(u,u*)=d(Su,Tu*)≾a[d(u,Su)d(u,Tu*)+d(u*,Tu*)d(u*,Su)]d(u,Tu*)+d(u*,Su)≺d(u,u*).
This implies that |d(x,x*)|≤0, and then u=u* which proves the uniqueness of common fixed point in X. This completes the proof.Corollary 20.
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
(48)d(Tx,Ty)≾a[d(x,Tx)d(x,Ty)+d(y,Ty)d(y,Tx)]d(x,Ty)+d(y,Tx),
for all x,y∈X, where sa∈[0,1). Then T has a unique fixed point in X.Corollary 21.
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
(49)d(Tnx,Tny)≾a[d(x,Tnx)d(x,Tny)+d(y,Tny)d(y,Tnx)]d(x,Tny)+d(y,Tnx),
for all x,y∈X, where sa∈[0,1) and n∈N. Then T has a unique fixed point in X.



---

*Source: 587825-2014-05-25.xml*