# Various Fixed Point Theorems in Complex Valuedb-Metric Spaces

Authors: Anil Kumar Dubey; Manjula Tripathi; Ravi Prakash Dubey
Journal: International Journal of Engineering Mathematics
 (2016)

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## Abstract

We prove some common fixed point results for a pair of mappings which satisfy generalized contractive conditions with rational expressions having point-dependent control functions as coefficients in complex valuedb-metric spaces. The results of this paper generalize and extend the several known results in complex valued b-metric spaces. Finally, examples are provided to verify the effectiveness and to usability of our main results.


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## Body

## 1. Introduction and Preliminaries

The concept of complex valued metric space was introduced by Azam et al. [1], proving some fixed point results for mappings satisfying a rational inequality in complex valued metric spaces. Since then, several papers have dealt with fixed point theory in complex valued metric spaces (see [2–10] and references therein). Rao et al. [11] initiated the studying of fixed point results on complex valued b-metric spaces, which was more general than the complex valued metric spaces [1]. Following this paper, a number of authors have proved several fixed point results for various mapping satisfying a rational inequalities in the context of complex valued b-metric spaces (see [12–15]) and the related references therein.Recently, Sintunavarat et al. [8, 9], Sitthikul and Saejung [10], and Singh et al. [7] obtained common fixed point results by replacing the constant of contractive condition to control functions in complex valued metric spaces. In a continuation of [7, 10, 14, 16], in this paper, we establish some common fixed point results for a pair of mappings satisfying more general contractive conditions involving rational expressions having point-dependent control functions as coefficients in complex valued b-metric spaces.Consistent with Rao et al. [11], the following definitions and results will be needed in the sequel.LetC be the set of complex numbers and z1,z2∈C. Define a partial order ≾ on C as follows:z 1 ≾ z 2 if and only if Re(z1)≤Re(z2),Im(z1)≤Im(z2).Consequently, one can infer thatz1≾z2 if one of the following conditions is satisfied:(i)
Rez1=Rez2,Imz1<Imz2.(ii)
Rez1<Rez2,Imz1=Imz2.(iii)
Rez1<Rez2,Imz1<Imz2.(iv)
Re(z1)=Re(z2),Im(z1)=Im(z2).In particular, we writez1⋨z2 if z1≠z2 and one of (i), (ii), and (iii) is satisfied and we write z1≺z2 if only (iii) is satisfied. Notice that(a)
if0≾z1⋨z2, then z1<z2;(b)
ifz1≾z2 and z2≺z3, then z1≺z3;(c)
ifa,b∈R and a≤b, then az≾bz for all z∈C+.The following definition is recently introduced by Rao et al. [11].Definition 1 (see [11]).
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)
dx,y=dy,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 [11]).
IfX=0,1, define a 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 [11]).
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 an open set whenever each element of A is an interior point of A.(iv)
A subsetA⊆X is called closed set whenever each limit point of A belongs to A.(v)
The familyF={B(x,r):x∈Xand0≺r} is a subbasis for a Hausdorff topology τ on X.Definition 4 (see [11]).
Let(X,d) be a complex valued b-metric space, and let xn be a sequence in X and x∈X.(i)
If for everyc∈C, with 0≺c, there is N∈N such that for all n>N,d(xn,x)≺c, then xn is said to be convergent and converges to x. We denote this by limn→∞xn=x or xn→xasn→∞.(ii)
If for everyc∈C, with 0≺c there is N∈N such that for all n>N,d(xn,xn+m)≺c, where m∈N, then xn is said to be a Cauchy sequence.(iii)
If every Cauchy sequence inX is convergent in X, then (X,d) is said to be a complete complex valued b-metric space.Lemma 5 (see [11]).
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 6 (see [11]).
Let(X,d) be a complex valued b-metric space and let xn be a sequence in X. Then, xn is Cauchy sequence if and only if |d(xn,xn+m)|→0 as n→∞, where m∈N.


## 2. Main Result

Throughout this paper, let(X,d) be a complete complex valued b-metric space and S,T:X→X be mappings. In our results, we will use the following family of functions.Let(X,d) be a complete complex valued b-metric space with the coefficient s≥1 and let S,T:X→X be mappings. Let Ψ be the family of all functions ψ:X×X×X→[0,1) such that for all x,y∈X and for fixed a∈X,(F1)
ψ(TSx,y,a)≤ψ(x,y,a);(F2)
ψ(x,STy,a)≤ψ(x,y,a).We start this section with the following observation.Proposition 7.
Let(X,d) be a complex valued b-metric space and let S,T:X→X be mappings. Let x0∈X and define the sequence {xn} by (1)x2n+1=Sx2n,x2n+2=Tx2n+1,∀n=0,1,2,….Assume that there exists a mapping α∈Ψ for all x,y∈X and for a fixed element a∈X and n=0,1,2,…. Then, α(x2n,y,a)≤α(x0,y,a) and α(x,x2n+1,a)≤α(x,x1,a).Proof.
Letx,y∈X and n=0,1,2,…. Then, we have (2)αx2n,y,a=αTx2n-1,y,a=αTSx2n-2,y,a≤αx2n-2,y,a=αTSx2n-4,y,a≤⋯≤αx0,y,a.Similarly, we have (3)αx,x2n+1,a=αx,Sx2n,a=αx,STx2n-1,a≤αx,x2n-1,a=αx,STx2n-3,a≤⋯≤αx,x1,a.Lemma 8 (see [10]).
Let{xn} be a sequence in X and h∈[0,1). If an=|d(xn,xn+1)| satisfies an≤han-1, for all n∈N, then {xn} is a Cauchy sequence.Now, we proceed to establish common fixed point theorems for the general contraction conditions in complex valuedb-metric space.Theorem 9.
Let(X,d) be a complete complex valued b-metric space with the coefficient s≥1 and let S,T:X→X be mappings. If there exist mappings α,β,γ,δ∈Ψ such that for all x,y∈X and for fixed a∈X, (4)aαx,y,a+βx,y,a+2γx,y,a+2sδx,y,a<1;(5)bdSx,Ty≾αx,y,adx,y+βx,y,ady,Tydx,Sx1+dx,y+γx,y,adx,Sx+dy,Ty+δx,y,adx,Ty+dy,Sx.Then, S and T have a unique common fixed point.Proof.
Letx,y∈X, from (5) we have (6)dSx,TSx≾αx,Sx,adx,Sx+βx,Sx,adSx,TSxdx,Sx1+dx,Sx+γx,Sx,adx,Sx+dSx,TSx+δx,Sx,adx,TSx+dSx,Sx≾αx,Sx,adx,Sx+βx,Sx,adSx,TSxdx,Sx1+dx,Sx+γx,Sx,adx,Sx+dSx,TSx+sδx,Sx,adx,Sx+dSx,TSxwhich implies that(7)dSx,TSx≤αx,Sx,adx,Sx+βx,Sx,adSx,TSxdx,Sx1+dx,Sx+γx,Sx,adx,Sx+dSx,TSx+sδx,Sx,adx,Sx+dSx,TSx.Since |1+d(x,Sx)|≥|d(x,Sx)|, (8)dSx,TSx≤αx,Sx,adx,Sx+βx,Sx,adSx,TSx+γx,Sx,adx,Sx+dSx,TSx+sδx,Sx,adx,Sx+dSx,TSx.In a similar way, by setting x=Ty in (5), we have(9)dSTy,Ty≤αTy,y,adTy,y+βTy,y,adTy,STy+γTy,y,ady,Ty+dTy,STy+sδTy,y,ady,Ty+dTy,STy.Let x0∈X and the sequence {xn} be defined by (1). We show that {xn} is a Cauchy sequence. From Proposition 7 and for all K=0,1,2,…, we obtain(10)dx2K+1,x2K=dSTx2K-1,Tx2K-1≤αx2K,x2K-1,adx2K,x2K-1+βx2K,x2K-1,adx2K+1,x2K+γx2K,x2K-1,adx2K+1,x2K+dx2K,x2K-1+sδx2K,x2K-1,adx2K+1,x2K+dx2K,x2K-1≤αx0,x1,adx2K-1,x2K+βx0,x1,adx2K,x2K+1+γx0,x1,adx2K-1,x2K+dx2K,x2K+1+sδx0,x1,adx2K-1,x2K+dx2K,x2K+1which yields that(11)dx2K+1,x2K≤αx0,x1,a+γx0,x1,a+sδx0,x1,a1-βx0,x1,a-γx0,x1,a-sδx0,x1,adx2K-1,x2K.Similarly, one can obtain(12)dx2K+2,x2K+1≤αx0,x1,a+γx0,x1,a+sδx0,x1,a1-βx0,x1,a-γx0,x1,a-sδx0,x1,adx2K,x2K+1.Let μ=α(x0,x1,a)+γ(x0,x1,a)+sδ(x0,x1,a)/1-β(x0,x1,a)-γ(x0,x1,a)-sδ(x0,x1,a)<1.
Sinceα(x0,x1,a)+β(x0,x1,a)+2γ(x0,x1,a)+2sδ(x0,x1,a)<1, thus we have |d(x2K+2,x2K+1)|≤μ|d(x2K,x2K+1)| and |d(x2K+1,x2K)|≤μ|d(x2K-1,x2K)|, or in fact(13)dxn+1,xn+2≤μdxn,xn+1.Thus, by Lemma 8 we get that this sequence is Cauchy sequence in (X,d). Since X is complete, there exists some u∈X such that xn→u as n→∞. Let, on contrary, u≠Su; then(14)du,Su>0.So by using the triangular inequality and (5), we get(15)du,Su≾sdu,x2n+2+sdx2n+2,Su=sdu,x2n+2+sdTx2n+1,Su≾sdu,x2n+2+sαu,x2n+1,adu,x2n+1+sβu,x2n+1,adx2n+1,Tx2n+1du,Su1+du,x2n+1+sγu,x2n+1,adu,Su+dx2n+1,Tx2n+1+sδu,x2n+1,adu,Tx2n+1+dx2n+1,Su≾sdu,x2n+2+sαu,x1,adu,x2n+1+sβu,x1,adx2n+1,x2n+2du,Su1+du,x2n+1+sγu,x1,adu,Su+dx2n+1,x2n+2+sδu,x1,adu,x2n+2+dx2n+1,Su.This implies that(16)du,Su≤sdu,x2n+2+sαu,x1,adu,x2n+1+sβu,x1,adx2n+1,x2n+2du,Su1+du,x2n+1+sγu,x1,adu,Su+dx2n+1,x2n+2+sδu,x1,adu,x2n+2+du,Su.Letting n→∞, it follows that(17)du,Su≤sγu,x1,adu,Su+sδu,x1,adu,Su=sγu,x1,a+δu,x1,adu,Su≤sαu,x1,a+βu,x1,a+2γu,x1,a+2δu,x1,adu,Su<du,Su,a contradiction, and so du,Su=0; that is, u=Su. It follows similarly that u=Tu. This implies that u is a common fixed point of S and T.
We now prove that thisu is unique:(18)du,u∗=dSu,Tu∗≾αu,u∗,adu,u∗+βu,u∗,adu∗,Tu∗du,Su1+du,u∗+γu,u∗adu,Su+du∗,Tu∗+δu,u∗,adu,Tu∗+du∗,Su≾αu,u∗,a+2δu,u∗adu,u∗.Therefore, we have(19)du,u∗≤αu,u∗,a+2δu,u∗,adu,u∗.Since α(u,u∗,a)+2δ(u,u∗,a)<1, we have |d(u,u∗)|=0.
Thus,u=u∗, which proves the uniqueness of common fixed point in X. This concludes the theorem.Remark 10.
If we replaceα,β:X×X×X→[0,1) by Λ,E:X→[0,1), with α(x,y,a)=Λ(x)andβ(x,y,a)=E(x) for all x,y∈X and so sΛ(x)+E(x)<1, then we get the result of Theorem 3.1 of Sintunavarat and Kumam [8] (complex valued b-metric space version).Remark 11.
If we set mappingsα,β:X×X×X→[0,1) as α(x,y,a)=α′ and β(x,y,a)=β′, where α′,β′∈[0,1) such that sα′+β′<1 and for all x,y∈X, we get Theorem 4 of Azam et al. [1] (complex valued b-metric space version).Next theorem is presented for single mapping satisfying slightly different conditions.Theorem 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. If there exist mappings α,β,γ,δ∈Ψ such that for all x,y∈X and for fixed a∈X,(20)aαx,y,a+βx,y,a+2γx,y,a+2sδx,y,a<1;(21)bdTx,Ty≾αx,y,adx,y+βx,y,a1+dx,Txdy,Ty1+dx,y+γx,y,adx,Tx+dy,Ty+δx,y,adx,Ty+dy,Tx,then T has a unique fixed point.Proof.
Letx0∈X and the sequence {xn} be defined by xn+1=Txn, where n=0,1,2,…. Now we show that {xn} is a Cauchy sequence. From condition (21), we have(22)dxn+1,xn+2=dTxn,Txn+1≾αxn,xn+1,adxn,xn+1+βxn,xn+1,a1+dxn,Txndxn+1,Txn+11+dxn,xn+1+γxn,xn+1,adxn,Txn+dxn+1,Txn+1+δxn,xn+1,adxn,Txn+1+dxn+1,Txn≾αx0,x0,adxn,xn+1+βx0,x0,adxn+1,xn+2+γx0,x0,adxn,xn+1+dxn+1,xn+2+sδx0,x0,adxn,xn+1+dxn+1,xn+2.Therefore,(23)dxn+1,xn+2≤αx0,x0,adxn,xn+1+βx0,x0,adxn+1,xn+2+γx0,x0,adxn,xn+1+dxn+1,xn+2+sδx0,x0,adxn,xn+1+dxn+1,xn+2≤αx0,x0,a+γx0,x0,a+sδx0,x0,a1-βx0,x0,a-γx0,x0,a-sδx0,x0,adxn,xn+1.Let μ=α(x0,x0,a)+γ(x0,x0,a)+sδ(x0,x0,a)/1-β(x0,x0,a)-γ(x0,x0,a)-sδ(x0,x0,a)<1; then,(24)dxn+1,xn+2≤μdxn,xn+1∀n=0,1,2,….By Lemma 8, this sequence is Cauchy sequence in (X,d). Since X is complete, there exists some u∈X such that xn→u as n→∞. Next, we show that u is a fixed point of T.
From (21), we have(25)du,Tu≾sdu,Txn+sdTxn,Tu≾sdu,Txn+sαxn,u,adxn,u+sβxn,u,a1+dxn,Txndu,Tu1+dxn,u+sγxn,u,adxn,Txn+du,Tu+sδxn,u,adxn,Tu+du,Txn≾sdu,xn+1+sαx0,u,adxn,u+sβx0,u,a1+dxn,xn+1du,Tu1+dxn,u+sγx0,u,adxn,xn+1+du,Tu+sδx0,u,adxn,Tu+du,xn+1.This implies that(26)du,Tu≤sdu,xn+1+sαx0,u,adxn,u+sβx0,u,a1+dxn,xn+1du,Tu1+dxn,u+sγx0,u,adxn,xn+1+du,Tu+sδx0,u,adxn,Tu+du,xn+1which on making n→∞ reduces to(27)du,Tu≤sβx0,u,adu,Tu+sγx0,u,adu,Tu+sδx0,u,adu,Tu=sβx0,u,a+sγx0,u,a+sδx0,u,adu,Tu≤sαx0,u,a+βx0,u,a+2γx0,u,a+2δx0,u,adu,Tu,a contradiction, and so du,Tu=0; that is, u=Tu. This implies that u is a fixed point of T.
Uniqueness of fixed point is an easy consequence of condition (22). This completes the proof.Corollary 13.
Let(X,d) be a complete complex valued b-metric space with the coefficient s≥1 and let T:X→X be a mapping. If there exist mappings α,β,γ,δ∈Ψ such that for all x,y∈X and for some fixed n,(28)aαx,y,a+βx,y,a+2γx,y,a+2sδx,y,a<1;(29)bdTnx,Tny≾αx,y,adx,y+βx,y,a1+dx,Tnxdy,Tny1+dx,y+γx,y,adx,Tnx+dy,Tny+δx,y,adx,Tny+dy,Tnx,then T has a unique fixed point.Proof.
By Theorem12, there exists v∈X such that Tnv=v. Then,(30)dTv,v=dTTnv,Tnv=dTnTv,Tnv≾αTv,v,adTv,v+βTv,v,a1+dx,TnTvdv,Tnv1+dTv·v+γTv,v,adTv,TnTv+dv,Tnv+δTv,v,adTv,Tnv+dv,TnTv≾αTv,v,adTv,v+βTv,v,a1+dx,TTnvdv,v1+dTv·v+γTv,v,adTv,TTnv+dv,v+δTv,v,adTv,v+dv,TTnv=αTv,v,adTv,v+2δTv,v,adTv,v=α+2δTv,v,adTv,vand so d(Tv,v)=0. So Tv=v. Therefore, the fixed point of T is unique.Example 14.
LetX=[0,1] and d:X×X→C be defined by d(x,y)=i|x-y|2 for all x,y∈X. Then, (X,d) is a complex valued b-metric space with the coefficient s=2. Now we define self-mappings S,T:X→X by S(x)=x/4 and T(y)=y/4. Further, for all x,y∈X and for fixed a=1/3∈X, we define the functions α,β,γ,δ:X×X×X→[0,1) by(31)αx,y,a=x4+y5+a,βx,y,a=xya10,γx,y,a=x2y2a210,δx,y,a=x3y3a310.Clearly α(x,y,a)+β(x,y,a)+2γ(x,y,a)+2sδ(x,y,a)<1 for all x,y∈X and for a fixed a=1/3∈X.Now consider(32)αTSx,y,a=αTx4,y,a=αx16,y,a=x64+y5+a≤x4+y5+a=αx,y,a.That is, α(TSx,y,a)≤α(x,y,a) for all x,y∈X and for a fixed a=1/3∈X. Now(33)αx,STy,a=αx,Sy4,a=αx,y16,a=x4+y80+a≤x4+y5+a=αx,y,a.That is, α(x,STy,a)≤α(x,y,a) for all x,y∈X and for a fixed a=1/3∈X. Similarly, we can show that(34)βTSx,y,a≤βx,y,a,βx,STy,a≤βx,y,a,γTSx,y,a≤γx,y,a,γx,STy,a≤γx,y,a,δTSx,y,a≤δx,y,a,δx,STy,a≤δx,y,a.Now for the verification of inequality (5), it is sufficient to show that (Sx,Ty)≾α(x,y,a)d(x,y).Consider(35)dSx,Ty=dx4,y4=x4-y42i=116x-y2i≾13x-y2i≾x4+y5+13x-y2i=αx,y,adx,y,∀x,y∈X,fora=13∈X.That is, d(Sx,Ty)≾α(x,y,a)d(x,y) for all x,y∈X and for fixed a=1/3∈X. Therefore, all the conditions of Theorem 9 are satisfied, also x=0 remains fixed under S and T and is indeed unique.



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