# Some Common Coupled Fixed Point Results for Generalized Contraction in Complex-Valued Metric Spaces

Authors: Marwan Amin Kutbi; Akbar Azam; Jamshaid Ahmad; Cristina Di Bari
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We introduce and study the notion of common coupled fixed points for a pair of mappings in complex valued metric space and demonstrate the existence and uniqueness of the common coupled fixed points in a complete complex-valued metric space in view of diverse contractive conditions. In addition, our investigations
are well supported by nontrivial examples.


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## Body

## 1. Introduction

Azam et al. [1] introduced the concept of complex-valued metric spaces and obtained sufficient conditions for the existence of common fixed points of a pair of contractive type mappings involving rational expressions. Subsequently, several authors have studied the existence and uniqueness of the fixed points and common fixed points of self-mappings in view of contrasting contractive conditions. Some of these investigations are noted in [2–26].In [27], Bhaskar and Lakshmikantham introduced the concept of coupled fixed points for a given partially ordered set X. Recently Samet et al. [28, 29] proved that most of the coupled fixed point theorems (on ordered metric spaces) are in fact immediate consequences of well-known fixed point theorems in the literature. In this paper, we deal with the corresponding definition of coupled fixed point for mappings on a complex-valued metric space along with generalized contraction involving rational expressions. Our results extend and improve several fixed point theorems in the literature.


## 2. Preliminaries

Letℂ be the set of complex numbers and z1,z2∈ℂ. Define a partial order  ⪯ on ℂ as follows:
(1)z1⪯z2iffRe(z1)≤Re(z2),Im(z1)≤Im(z2).
Note that 0⪯z1,z2 and z1≠z2,  z1⪯z2 implies |z1|<|z2|.Definition 1.
LetX be a nonempty set. Suppose that the self-mapping d:X×X→ℂ satisfies the following:(1)
0⪯d(x,y),for all x,y∈X and d(x,y)=0 if and only if x=y;(2)
d(x,y)=d(y,x) for all x,y∈X;(3)
d(x,y)⪯d(x,z)+d(z,y), for all x,y,z∈X.
Thend is called a complex valued metric on X, and (X,d) is known as a complex valued metric space. A point x∈X is called interior point of a set A⊆X whenever, there exists 0≺r∈ℂ such that
(2)B(x,r)={y∈X:d(x,y)≺r}⊆A.A pointx∈X is a limit point of A whenever, for every 0≺r∈ℂ,
(3)B(x,r)∩(A∖{x})≠∅.A is called open whenever each element of A is an interior point of A. Moreover, a subset B⊆X is called closed whenever each limit point of B belongs to B. The family
(4)F={B(x,r):x∈X,0≺r∈ℂ}
is a subbasis for a Hausdorff topology τ on X.Let{xn} be a sequence in X and x∈X. If for every c∈ℂ with 0≺c there is n0∈ℕ such that, for all n>n0,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 limn→+∞xn=x, or xn→x, as n→+∞. If for every c∈ℂ with 0≺c there is n0∈ℕ such that, for all n>n0,d(xn,xn+m)≺c, then {xn} is called a Cauchy sequence in (X,d). If every Cauchy sequence is convergent in (X,d), then (X,d) is called a complete complex valued metric space. We require the following lemmas.Lemma 2 (see [1]).
Let(X,d) be a complex valued metric space, and let {xn} be a sequence in X. Then {xn} converges to x if and only if  |d(xn,x)|→0asn→+∞.Lemma 3 (see [1]).
Let(X,d) be a complex valued metric space, and let {xn} be a sequence in X. Then {xn} is a Cauchy sequence if and only if|d(xn,xn+m)|→0asn→+∞.Definition 4 (see [27]).
An element(x,y)∈X×X is called a coupled fixed point of T:X×X→X if
(5)x=T(x,y),y=T(y,x).Definition 5.
An element(x,y)∈X×X is called a coupled coincidence point of S,T:X×X→X if
(6)S(x,y)=T(x,y),S(y,x)=T(y,x).Example 6.
LetX=ℝ and S,T:X×X→X defined as S(x,y)=x2y2 and T(x,y)=(4/3)(x+y) for all x,y∈X. Then (0,0), (1,2), and (2,1) are coupled coincidence points of S and T.Example 7.
LetX=ℝ and S,T:X×X→X defined as S(x,y)=x+y+sin(x+y) and T(x,y)=x+y+xy+cos(x+y) for all x,y∈X. Then (0,π/4) and (π/4,0) are coupled coincidence points of S and T.Definition 8.
An element(x,y)∈X×X is called a common coupled fixed point of S,T:X×X→X if
(7)x=S(x,y)=T(x,y),y=S(y,x)=T(y,x).Example 9.
LetX=ℝ and S,T:X×X→X defined as S(x,y)=x((x+(y-1)2)/2) and T(x,y)=x(x2+y2+4-2) for all x,y∈X. Then (0,0), (1,2), and (2,1) are common coupled fixed points of S and T.In the following, we provide common coupled fixed point theorem for a pair of mappings satisfying a rational inequality in complex valued metric spaces.Theorem 10.
Let(X,d) be a complete complex-valued metric space, and let the mappings S,T:X×X→Xsatisfy
(8)d(S(x,y),T(u,v))⪯α(d(x,u)+d(y,v))2+(βd(x,S(x,y))d(u,T(u,v))(x,u)+γd(u,S(x,y))d(x,T(u,v)))×(1+d(x,u)+d(y,v))-1
for all x,y,u,v∈X and α,β, and γ are nonnegative reals with α+β+γ<1. Then S and T have a unique common coupled fixed point.Proof.
Letx0 and y0 be arbitrary points in X. Define x2k+1=S(x2k,y2k), y2k+1=S(y2k,x2k) and x2k+2=T(x2k+1,y2k+1), y2k+2=T(y2k+1,x2k+1), for k=0,1,…
Then,(9)d(x2k+1,x2k+2)=d(S(x2k,y2k),T(x2k+1,y2k+1))⪯α(d(x2k,x2k+1)+d(y2k,y2k+1))2+(βd(x2k,S(x2k,y2k))(zS(y))×d(x2k+1,T(x2k+1,y2k+1)))×(1+d(x2k,x2k+1)+d(y2k,y2k+1))-1+(γd(x2k+1,S(x2k,y2k))(zS(y))×d(x2k,T(x2k+1,y2k+1)))×(1+d(x2k,x2k+1)+d(y2k,y2k+1))-1⪯α(d(x2k,x2k+1)+d(y2k,y2k+1))2+βd(x2k,x2k+1)d(x2k+1,x2k+2)1+d(x2k,x2k+1)+d(y2k,y2k+1)+γd(x2k+1,x2k+1)d(x2k,x2k+2)1+d(x2k,x2k+1)+d(y2k,y2k+1)⪯α(d(x2k,x2k+1)+d(y2k,y2k+1))2+βd(x2k,x2k+1)d(x2k+1,x2k+2)1+d(x2k,x2k+1)+d(y2k,y2k+1),
which implies that
(10)|d(x2k+1,x2k+2)|≤α|d(x2k,x2k+1)+d(y2k,y2k+1)|2+β|d(x2k,x2k+1)d(x2k+1,x2k+2)||1+d(x2k,x2k+1)+d(y2k,y2k+1)|.
Since |1+d(x2k,x2k+1)+d(y2k,y2k+1)|>|d(x2k,x2k+1)|, so we get
(11)|d(x2k+1,x2k+2)|≤α|d(x2k,x2k+1)|+α|d(y2k,y2k+1)|2+β|d(x2k+1,x2k+2)|,
and hence
(12)|d(x2k+1,x2k+2)|≤12(α1-β)|d(x2k,x2k+1)|+12(α1-β)|d(y2k,y2k+1)|.
Similarly, one can show that
(13)|d(y2k+1,y2k+2)|≤12(α1-β)|d(y2k,y2k+1)|+12(α1-β)|d(x2k,x2k+1)|.
Also,
(14)d(x2k+2,x2k+3)=d(T(x2k+1,y2k+1),S(x2k+2,y2k+2))⪯α(d(x2k+1,x2k+2)+d(y2k+1,y2k+2))2+(βd(x2k+1,T(x2k+1,y2k+1))(x2k+2,)×d(x2k+2,S(x2k+2,y2k+2)))×(1+d(x2k+1,x2k+2)+d(y2k+1,y2k+2))-1+(γd(x2k+2,T(x2k+1,y2k+1))(x2k+2,)×d(x2k+1,S(x2k+2,y2k+2)))×(1+d(x2k+1,x2k+2)+d(y2k+1,y2k+2))-1⪯α(d(x2k+1,x2k+2)+d(y2k+1,y2k+2))2+βd(x2k+1,x2k+2)d(x2k+2,x2k+3)1+d(x2k+1,x2k+2)+d(y2k+1,y2k+2)+γd(x2k+2,x2k+2)d(x2k+1,x2k+3)1+d(x2k+1,x2k+2)+d(y2k+1,y2k+2)⪯α(d(x2k+1,x2k+2)+d(y2k+1,y2k+2))2+βd(x2k+1,x2k+2)d(x2k+2,x2k+3)1+d(x2k+1,x2k+2)+d(y2k+1,y2k+2),
so that
(15)|d(x2k+2,x2k+3)|≤α|d(x2k+1,x2k+2)+d(y2k+1,y2k+2)|2+β|d(x2k+1,x2k+2)||d(x2k+2,x2k+3)||1+d(x2k+1,x2k+2)+d(y2k+1,y2k+2)|.
As |1+d(x2k+1,x2k+2)+d(y2k+1,y2k+2)|>|d(x2k+1,x2k+2)|, therefore
(16)|d(x2k+2,x2k+3)|≤12(α1-β)|d(x2k+1,x2k+2)|+12(α1-β)|d(y2k+1,y2k+2)|.
Similarly, one can show that
(17)|d(y2k+2,y2k+3)|≤α1-β|d(y2k+1,y2k+2)|+12(α1-β)|d(x2k+1,x2k+2)|.
Adding (12)–(17), we get
(18)|d(x2k+1,x2k+2)|+|d(y2k+1,y2k+2)|≤α1-β|d(x2k,x2k+1)|+α1-β|d(y2k,y2k+1)||d(x2k+2,x2k+3)|+|d(y2k+2,y2k+3)|≤α1-β|d(x2k+1,x2k+2)|+α1-β|d(y2k+1,y2k+2)|.
If h=α/(1-β)<1, then from (18), we get
(19)|d(xn,xn+1)|+|d(yn,yn+1)|≤h(|d(xn-1,xn)|+|d(yn-1,yn)|)≤⋯≤hn(|d(x0,x1)|+|d(y0,y1)|).
Now if |d(xn,xn+1)|+|d(yn,yn+1)|=δn, then
(20)δn≤hδn-1≤⋯≤hnδ0.
Without loss of generality, we take m>n. Since 0≤h<1, so we get
(21)|d(xn,xm)|+|d(yn,ym)|≤|d(xn,xn+1)|+|d(yn,yn+1)|+⋯+|d(xm-1,xm)|+|d(ym-1,ym)|≤[hnδ0+hn+1δ0+⋯+hm-1δ0]≤∑‍i=nm-1hiδ0→0,asm,n→+∞.
This implies that {xn} and {yn} are Cauchy sequences in X. Since X is complete, there exists x,y∈X such that xn→x and yn→y as n→+∞. We now show that x=S(x,y) and y=S(y,x). We suppose on the contrary that x≠S(x,y) and y≠S(y,x) so that 0≺d(x,S(x,y))=l1 and 0≺d(y,S(y,x))=l2; we would then have
(22)l1=d(x,S(x,y))⪯d(x,x2k+2)+d(x2k+2,S(x,y))⪯d(x,x2k+2)+d(T(x2k+1,y2k+1),S(x,y))⪯d(x,x2k+2)+α(d(x2k+1,x)+d(y2k+1,y))2+βd(x2k+1,T(x2k+1,y2k+1))d(x,S(x,y))1+d(x2k+1,x)+d(y2k+1,y)+γd(x,T(x2k+1,y2k+1))d(x2k+1,S(x,y))1+d(x2k+1,x)+d(y2k+1,y)=d(x,x2k+2)+α(d(x2k+1,x)+d(y2k+1,y))2+βd(x2k+1,x2k+2)d(x,S(x,y))1+d(x2k+1,x)+d(y2k+1,y)+γd(x,x2k+2)d(x2k+1,S(x,y))1+d(x2k+1,x)+d(y2k+1,y)
so that
(23)|l1|≤|d(x,x2k+2)|+α|d(x2k+1,x)+d(y2k+1,y)|2+β|d(x2k+1,x2k+2)||d(x,S(x,y))||1+d(x2k+1,x)+d(y2k+1,y)|+γ|d(x,x2k+2)||d(x2k+1,S(x,y))||1+d(x2k+1,x)+d(y2k+1,y)|.
By taking k→+∞, we get |d(x,S(x,y))|=0 which is a contradiction so that x=S(x,y). Similarly, one can prove that y=S(y,x). It follows similarly that x=T(x,y) and y=T(y,x). So we have proved that (x,y) is a common coupled fixed point of S and T. We now show that S and T have a unique common coupled fixed point. For this, assume that (x*,y*)∈X is a second common coupled fixed point of S and T. Then
(24)d(x,x*)=d(S(x,y),T(x*,y*))⪯α(d(x,x*)+d(y,y*))2+βd(x,S(x,y))d(x*,T(x*,y*))1+d(x,x*)+d(y,y*)+γd(x,T(x*,y*))d(x*,S(x,y))1+d(x,x*)+d(y,y*)⪯α(d(x,x*)+d(y,y*))2+βd(x,x)d(x*,x*)1+d(x,x*)+d(y,y*)+γd(x,x*)d(x*,x)1+d(x,x*)+d(y,y*)
so that
(25)|d(x,x*)|≤|α(d(x,x*)+d(y,y*))2|+γ|d(x,x*)||d(x*,x)||1+d(x,x*)+d(y,y*)|.
Since |1+d(x,x*)+d(y,y*)|>|d(x,x*)|, so we get
(26)|d(x,x*)|≤|α(d(x,x*)+d(y,y*))2|+γ|d(x,x*)|=(α2-α-2γ)|d(y,y*)|.
Similarly, one can easily prove that
(27)|d(y,y*)|≤(α2-α-2γ)|d(x,x*)|.
If we add (26) and (27), we get
(28)|d(x,x*)|+|d(y,y*)|≤(α2-α-2γ)(|d(x,x*)|+|d(y,y*)|),
which is a contradiction because α+β+γ<1. Thus, we get x*=x and y*=y, which proves the uniqueness of common coupled fixed point of S and T.By settingS=T in Theorem 10, one deduces the following.Corollary 11.
Let(X,d) be a complete complex-valued metric space, and let the mapping T:X×X→X satisfy
(29)d(T(x,y),T(u,v))⪯α(d(x,u)+d(y,v))2+(βd(x,T(x,y))d(u,T(u,v))T(x,y),+γd(u,T(x,y))d(x,T(u,v)))×(1+d(x,u)+d(y,v))-1
for all x,y,u,v∈X, where α,β, and γ are nonnegative reals with α+β+γ<1. Then T has a unique coupled fixed point.Corollary 12.
Let(X,d) be a complete complex-valued metric space, and let the mapping T:X×X→X satisfy
(30)d(Tn(x,y),Tn(u,v))⪯α(d(x,u)+d(y,v))2+(βd(x,Tn(x,y))d(u,Tn(u,v))T(x,y),+γd(u,Tn(x,y))d(x,Tn(u,v)))×(1+d(x,u)+d(y,v))-1
for all x,y,u,v∈X, where α,β, and γ are nonnegative reals with α+β+γ<1. Then, T has a unique coupled fixed point.Theorem 13.
Let(X,d) be a complete complex-valued metric space, and let the mappings S,T:X×X→X satisfy
(31)d(S(x,y),T(u,v))⪯{α(d(x,u)+d(y,v))2+βd(x,S(x,y))d(u,T(u,v))d(x,T(u,v))+d(u,S(x,y))+d(x,u)+d(y,v),d(x,T(u,v))+d(u,S(x,y))d(x,u)d(y,v)ifD≠00,(x,T(u,v))+d(u,S(x,y))d(xu)d(y,v)ifD=0
for all x,y,u,v∈X, where D=d(x,T(u,y))+d(u,S(x,y))+d(x,u)+d(y,v) and α,β are nonnegative reals with α+β<1. Then S and T have a unique common coupled fixed point.Proof.
Letx0 and y0 be arbitrary points in X. Define x2k+1=S(x2k,y2k), y2k+1=S(y2k,x2k) and x2k+2=T(x2k+1,y2k+1), y2k+2=T(y2k+1,x2k+1), for k=0,1,….
Now, we assume that(32)DS(x2k,y2k)=d(x2k,T(x2k+1,y2k+1))+d(x2k+1,S(x2k,y2k))+d(x2k,x2k+1)+d(y2k,y2k+1)=d(x2k,x2k+2)+d(x2k,x2k+1)+d(y2k,y2k+1)≠0,DS(y2k,x2k)=d(y2k,T(y2k+1,x2k+1))+d(y2k+1,S(y2k,x2k))+d(x2k,x2k+1)+d(y2k,y2k+1)=d(y2k,y2k+2)+d(x2k,x2k+1)+d(y2k,y2k+1)≠0.
Then,(33)d(x2k+1,x2k+2)=d(S(x2k,y2k),T(x2k+1,y2k+1))⪯α(d(x2k,x2k+1)+d(y2k,y2k+1))2+βd(x2k,S(x2k,y2k))d(x2k+1,T(x2k+1,y2k+1))DS(x2k,y2k)=α(d(x2k,x2k+1)+d(y2k,y2k+1))2+(βd(x2k,x2k+1)d(x2k+1,x2k+2))×(d(x2k,x2k+2)+d(x2k,x2k+1)x2kx2k1+d(y2k,y2k+1))-1
which implies that
(34)|d(x2k+1,x2k+2)|≤α|d(x2k,x2k+1)+d(y2k,y2k+1)|2+(β|d(x2k,x2k+1)||d(x2k+1,x2k+2)|)×(|d(x2k,x2k+2)+d(x2k,x2k+1)x2k1x2k2+d(y2k,y2k+1)|)-1≤α|d(x2k,x2k+1)+d(y2k,y2k+1)|2+β|d(x2k,x2k+1)|
as
(35)|d(x2k+1,x2k+2)|≤|d(x2k+1,x2k)+d(x2k,x2k+2)+d(y2k,y2k+1)|.
Therefore,
(36)|d(x2k+1,x2k+2)|≤(α+2β)2|d(x2k,x2k+1)|+α2|d(y2k,y2k+1)|.
Similarly, one can easily prove that
(37)|d(y2k+1,y2k+2)|≤(α+2β)2|d(y2k,y2k+1)|+α2|d(x2k,x2k+1)|.
Now, if(38)DT(x2k+1,y2k+1)=d(x2k+2,T(x2k+1,y2k+1))+d(x2k+1,S(x2k+2,y2k+2))+d(x2k+2,x2k+1)+d(y2k+2,y2k+1)=d(x2k+1,x2k+3)+d(x2k+2,x2k+1)+d(y2k+2,y2k+1)≠0,
we get
(39)d(x2k+2,x2k+3)=d(T(x2k+1,y2k+1),S(x2k+2,y2k+2))⪯α(d(x2k+2,x2k+1)+d(y2k+2,y2k+1))2+(βd(x2k+2,S(x2k+2,y2k+2))x2k+2x3×d(x2k+1,T(x2k+1,y2k+1)))×(DT(x2k+1,y2k+1))-1=α(d(x2k+2,x2k+1)+d(y2k+2,y2k+1))2+(βd(x2k+2,x2k+3)d(x2k+1,x2k+2))×(d(x2k+1,x2k+3)+d(x2k+2,x2k+1)x2k+2,x+d(y2k+2,y2k+1))-1,
which implies that
(40)|d(x2k+2,x2k+3)|≤α|α(d(x2k+2,x2k+1)+d(y2k+2,y2k+1))2|+(β|d(x2k+2,x2k+3)||d(x2k+1,x2k+2)|)×(|d(x2k+1,x2k+3)+d(x2k+2,x2k+1)x2k1x2k3+d(y2k+2,y2k+1)|)-1≤α|α(d(x2k+2,x2k+1)+d(y2k+2,y2k+1))2|+β|d(x2k+1,x2k+2)|
as
(41)|d(x2k+2,x2k+3)|≤|d(x2k+2,x2k+1)+d(x2k+1,x2k+3)x2k2x1+d(y2k+2,y2k+1)|.
Therefore,
(42)|d(x2k+2,x2k+3)|≤α|d(x2k+2,x2k+1)|2+α|d(y2k+2,y2k+1)|2+β|d(x2k+1,x2k+2)|=(α+2β)2|d(x2k+1,x2k+2)|+α2|d(y2k+1,y2k+2)|.
Similarly, ifDT(y2k+1,x2k+1)≠0, one can easily prove that
(43)|d(y2k+2,y2k+3)|≤(α+2β)2|d(y2k+1,y2k+2)|+α2|d(x2k+1,x2k+2)|.
Adding the inequalities (36)–(43), we get
(44)|d(x2k+1,x2k+2)|+|d(y2k+1,y2k+2)|≤(α+β)(|d(x2k,x2k+1)|+|d(y2k,y2k+1)|).|d(x2k+2,x2k+3)|+|d(y2k+2,y2k+3)|≤(α+β)(|d(x2k+1,x2k+2)|+|d(y2k+1,y2k+2)|).
If h=(α+β)<1, then, from (44), we get
(45)|d(xn,xn+1)|+|d(yn,yn+1)|≤h(|d(xn-1,xn)|+|d(yn-1,yn)|)≤⋯≤hn(|d(x0,x1)|+|d(y0,y1)|).
Now if |d(xn,xn+1)|+|d(yn,yn+1)|=δn, then
(46)δn≤hδn-1≤⋯≤hnδ0.
Without loss of generality, we take m>n. Since 0≤h<1, so we get
(47)|d(xn,xm)|+|d(yn,ym)|≤|d(xn,xn+1)|+|d(yn,yn+1)|+⋯+|d(xm-1,xm)|+|d(ym-1,ym)|≤[hnδ0+hn+1δ0+⋯+hm-1δ0]≤∑‍i=nm-1hiδ0→0,asm,n→+∞.
This implies that {xn} and {yn} are Cauchy sequences in X. Since X is complete, so there exists x,y∈X such that xn→x and yn→y as n→+∞. We now show that x=S(x,y) and y=S(y,x). We suppose on the contrary that x≠S(x,y) and y≠S(y,x) so that 0≺d(x,S(x,y))=l1 and 0≺d(y,S(y,x))=l2; we would then have
(48)l1=d(x,S(x,y))⪯d(x,x2k+2)+d(x2k+2,S(x,y))⪯d(x,x2k+2)+d(T(x2k+1,y2k+1),S(x,y))⪯d(x,x2k+2)+α(d(x2k+1,x)+d(y2k+1,y))2+(βd(x2k+1,T(x2k+1,y2k+1))d(x,S(x,y)))×(d(x2k+1,S(x,y))+d(x,T(x2k+1,y2k+1))xk1yk+d(x2k+1,x)+d(y2k+1,y))-1⪯d(x,x2k+2)+α(d(x2k+1,x)+d(y2k+1,y))2+(βl1d(x2k+1,x2k+2))×(d(x2k+1,S(x,y))+d(x,x2k+2)xk1yk+d(x2k+1,x)+d(y2k+1,y)(d(x2k+1,S(x,y))+d(x,x2k+2)))-1,
so that
(49)|l1|≤|d(x,x2k+2)|+α2|d(x2k+1,x)+d(y2k+1,y)|+(β|l1||d(x2k+1,x2k+2)|)×(|d(x2k+1,S(x,y))+d(x,x2k+2)x,x2k2+d(x2k+1,x)+d(y2k+1,y)(|d(x2k+1,S(x,y))+d(x,x2k+2))|)-1.
By taking k→+∞, we get |d(x,S(x,y))|=0 which is a contradiction so that x=S(x,y). Now
(50)l2=d(y,S(y,x))⪯d(y,y2k+2)+d(y2k+2,S(y,x))⪯d(y,y2k+2)+d(T(y2k+1,x2k+1),S(y,x))⪯d(y,y2k+2)+α(d(y2k+1,y)+d(x2k+1,x))2+(βd(y2k+1,T(y2k+1,x2k+1))d(y,S(y,x)))×(d(y2k+1,S(y,x))+d(y,T(y2k+1,x2k+1))(y+1)+d(y2k+1,y)+d(x2k+1,x))-1⪯d(y,y2k+2)+α(d(y2k+1,y)+d(x2k+1,x))2+(βl2d(y2k+1,y2k+2))×(d(y2k+1,S(y,x))+d(y,y2k+2)d(y,)+d(y2k+1,y)+d(x2k+1,x)(d(y2k+1,S(y,x))+d(y,y2k+2)))-1,
which implies that
(51)|l2|≤|d(y,y2k+2)|+α2|d(y2k+1,y)d(x2k+1,x)|+(β|l2||d(y2k+1,y2k+2)|)×(|d(y2k+1,S(y,x))+d(y,y2k+2)(y2k)+d(y2k+1,y)+d(x2k+1,x)|d(y2k+1,S(y,x))+d(y,y2k+2))|)-1,
Which, on making k→+∞, gives us |d(y,S(y,x))|=0 which is a contradiction so that y=S(y,x). It follows similarly that x=T(x,y) and y=T(y,x). So we have proved that (x,y) is a common coupled fixed point of S and T. As in Theorem 10, the uniqueness of common coupled fixed point remains a consequence of contraction condition (31).
We have obtained the existence and uniqueness of a unique common coupled fixed point if(52)DS(x2k,y2k),DS(y2k,x2k),DT(x2k+1,y2k+1),DT(y2k+1,x2k+1)≠0
for all k∈ℕ. Now, assume that DS(x2k,y2k)=0 for some k∈ℕ. From
(53)d(x2k,x2k+2)+d(x2k,x2k+1)+d(y2k,y2k+1)=0,
we obtain that x2k=x2k+1=x2k+2 and y2k=y2k+1. If DS(y2k,x2k)≠0, using (8), we deduce
(54)d(y2k+1,y2k+2)=d(S(y2k,x2k),T(y2k+1,x2k+1))=0.
That is, y2k+1=y2k+2 (this equality holds also if DS(y2k,x2k)=0). The equalities
(55)x2k=x2k+1=x2k+2,y2k=y2k+1=y2k+2,
ensure that (x2k+1,y2k+1) is a unique common coupled fixed point of S and T. The same holds if either DS(y2k,x2k)=0, DT(x2k+1,y2k+1)=0, or DT(y2k+1,x2k+1)=0.From Theorem13, if we assume α=0, we obtain the following corollary.Corollary 14.
Let(X,d) be a complete complex-valued metric space, and let the self-mappings S,T:X×X→X satisfy
(56)d(S(x,y),T(u,v))⪯{βd(x,S(x,y))d(u,T(u,v))d(x,T(u,v))+d(u,S(x,y))+d(x,u)+d(y,v),d(x,T(u,v))+d(u,S(x,y))(x,u)(y,v)ifD≠00,d(x,T(u,v))+(u,S(x,y))(x,u)(y,v)ifD≠0
for all x,y,u,v∈X, where D=d(x,T(u,y))+d(u,S(x,y))+d(x,y)+d(y,v) and β is a nonnegative real such that 0<β<1. Then S and T have a unique common coupled fixed point.Corollary 15.
Let(X,d) be a complete complex-valued metric space, and let the mapping T:X×X→X satisfy
(57)d(T(x,y),T(u,v))⪯{α(d(x,u)+d(y,v))2+βd(x,T(x,y))d(u,T(u,v))d(x,T(u,v))+d(u,T(x,y))+d(x,u)+d(y,v),d(x,T(u,v))+d(u,T(x,y))d(x,u)d(y,v)ifD≠00,d(x,T(u,v))(u,T(x,y))cd(x,u)d(y,v)ifD=0
for all x,y,u,v∈X, where D=d(x,T(u,y))+d(u,T(x,y))+d(x,u)+d(y,v) and α,β are nonnegative reals with α+β<1. Then T has a unique coupled fixed point.Corollary 16.
Let(X,d) be a complete complex-valued metric space, and let the mapping T:X×X→X satisfy
(58)d(Tn(x,y),Tn(u,v))⪯{α(d(x,u)+d(y,v))2+βd(x,Tn(x,y))d(u,Tn(u,v))d(x,Tn(u,v))+d(u,Tn(x,y))+d(x,u)+d(y,v),d(x,Tn(u,v))+d(u,Tn(x,y))+(x,u)d(y,v)ifD≠00,d(x,Tn(u,v))(u,Tn(x,y))ds(x,u)d(y,v)ifD=0
for all x,y,u,v∈X, where D=d(x,Tn(u,y))+d(u,Tn(x,y))+d(x,u)+d(y,v) and α,β are nonnegative reals with α+β<1. Then T has a unique coupled fixed point.Now, we furnished a nontrivial example to support our main result (Theorem10).Example 17.
Let(59)X1={z∈ℂ:Re(z)≥0,Im(z)=0},X2={z∈ℂ:Im(z)≥0,Re(z)=0},
and let X=X1∪X2. Consider a complex valued metric d:X×X→ℂ as follows:
(60)d(z1,z2)={23|x1-x2|+i2|x1-x2|,23|x1-x2|ifz1,z2∈X112|y1-y2|+i3|y1-y2|,23|x1-x2|ifz1,z2∈X229(x1+y2)+i6(x1+y2),23|xx2|ifz1∈X1,z2∈X2i3(x2+y1)+2i9(x2+y1),23|xx2|ifz1∈X2,z2∈X1,
with z1=x1+iy1 and z2=x2+iy2. Then (X,d) is a complex valued metric space. Define S,T:X×X→X as follows:
(61)S(z1,z2)={0+x1x24iifz1,z2∈X1y1y25+0iifz1,z2∈X20+x1y28iifz1∈X1andz2∈X2y1x29+0iifz1∈X2andz2∈X1,T(z1,z2)={0+x1x26iifz1,z2∈X1y1y27+0iifz1,z2∈X20+x1y210iifz1∈X1andz2∈X2y1x211+0iifz1∈X2andz2∈X1.
By a routine calculation, one can easily verify that the maps S and T satisfy the contraction condition (8) with α=3/4, β=1/15, and γ=2/15. Notice that the point (0,0) remains fixed under S and T and is indeed unique common coupled fixed point.



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*Source: 352927-2013-06-20.xml*