# Fixed Point Theorems for Cyclic Weakly Contraction Mappings in Dislocated Quasi Extendedb-Metric Space

Authors: Budi Nurwahyu
Journal: Journal of Function Spaces
 (2019)

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## Abstract

In this paper, we establish and prove some theorems about existence and uniqueness of fixed point for cyclic weakly contraction mappings in dislocated quasi extendedb-metric space.


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## Body

## 1. Introduction

One of the famous generalizations of metric space which was introduced by Bakhtin in 1989 [1] isb-metric space. Many authors utilized the space for fixed point results on contraction mapping or weakly contraction mapping, such as Saluja et al. [2], Mostefaoui et al. [3], Chaudhury et al. [4] and Ansari et al. [5]. In 2012, Shah et al. [6] introduced quasib-metric space which removed symmetric conditions inb-metric and for utilizing in common fixed point results on contraction mapping. Some authors such as Zhu et al. [7] and Cvetkovic et al. [8] gave some results in that space. In 2013, Hussain et al. [9] introduced dislocatedb-metric which weakened first condition inb-metric for fixed point results, and Rasham et al. [10] utilized the space for multivalued fixed point results. In 2016, Rahman et al. [11] generalized the dislocatedb-metric to be dislocated quasib-metric. Several papers has published in dislocated quasib-metric for containing fixed point results on generalized Banach contraction mappings, such as Klin-eam et al. [12], Suanom et al. [13], and Tiwari et al. [14]. Recently, in 2017, Kamran et al. [15] generalized triangular inequality condition onb-metric such that to be extendedb-metric and utilized the space for fixed point results. Samreen et al. [16] yielded some theorems for fixed point results on nonlinear contraction mappings in the space and Alqahtani et al. [17, 18] utilized the space for common fixed point results on two self-mappings and on K-contraction mapping.Inspired by the extendedb-metric space of Samreen et al. [16]. In this work, we introduced a concept of dislocated quasi extendedb-metric space as a generalization of dislocated quasib-metric space [11]. We establish and prove some fixed point theorems in the dislocated quasi extendedb-metric space, by utilizing weakly contraction mapping which was introduced by Rhoades [19] and cyclic contraction which was introduced by Zoto et al. [20]. In addition, we also provide some examples to clarify the theorems.


## 2. Preliminaries

In the following section, we need some definitions to govern and prove our theorems.Definition 1 (see [1]).
LetXbe a non-empty set and a real number k≥1. Let d:X×X→[0,∞) be a function. The pair X,d is calledb-metric space if the following conditions are satisfied: (1)
dx,y=0 if and only if x=y,(2)
dx,y=dy,x,(3)
dx,y≤kdx,z+dz,y,
for allx,y,z∈X.Example 2 (see [15]).
LetX=lp(R) with 0<p<1, where lpR=ak⊆R∣∑k=1∞ak<∞. Let d:X×X→[0,∞) be a function, which is defined as dx,y=∑k=1∞ak-bk1/p, where x={ak} and y={bk}. Then d is ab-metric with parameter b=21/p.Definition 3 (see [11]).
LetXbe a nonempty set and a real number k≥1. Let d:X×X→[0,∞) be a function. The pair X,d is called adislocated quasi b-metric space (in short dqb- metric space) if the following conditions are satisfied: (1)
dx,y=0 then x=y,(2)
ddx,y≤kdx,z+dz,y,
for allx,y,z∈X.Example 4 (see [11]).
LetX=R and define dx,y=2x-y2+2x+y2. It is easy to show that (X,d) is a dislocated quasib-metric space with k=2.Definition 5 (see [15]).
LetXbe a non-empty set and k:X×X→[1,∞) be a function. Let d:X×X→[0,∞) be a function. The pair X,d is called anextended b-metric space if the following conditions are satisfied: (1)
dx,y=0 if and only if x=y,(2)
dx,y=dy,x,(3)
dx,y≤kx,ydx,z+dz,y,
for allx,y,z∈X.Example 6 (see [16]).
LetX=1,2,3,…. Define k:X×X→[1,∞) and d:X×X→[0,∞) as follows:(1)kx,y=x-y2ifx≠y1ifx=yanddx,y=x-y4.It is easy to show that (X,d) is a dislocated extendedb-metric space.Definition 7.
LetXbe a non-empty set and k:X×X→1,∞. Let dk:X×X→[0,∞) be a function. The pair X,dk is called aquasi extended b-metric space(in short qeb- metric space) if the following conditions are satisfied: (2)1dkx,y=0ifandonlyifx=y,2dkx,y≤kx,ydkx,z+dkz,y,for all x,y,z∈X.Example 8.
LetX=0,1 and dx,y=2x-y-1 for x,y∈0,1. Let kx,y=21-x+y/2 for x,y∈0,1.
It is obvious that for first condition anddx,y is not symmetric. For second condition, consider that (3)21-x+y/2dx,z+dz,y=21-x+y/22x-z-1+2z-y-1Since minz∈[0,1]⁡2x-z-1+2z-y-1=2x-x+y/2-1+2x+y/2-y-1, we get(4)21-x+y/22x-z-1+2z-y-1≥21-x+y/22x-x+y/2-1+2x+y/2-y-1=21-x+y/22x-y/2-1+2x-y/2-1=22-x+y/22x-y/2-1.If x≤y then we have x+y/2≥x, and 22-x+y/2≤22-x. Therefore, we get (5)21-x+y/22x-z-1+2z-y-1≥22-y-22-x+y/2≥22-y-22-x=222-y-2-xSince, for x∈0,1, 2-x≥1. Thus we get(6)21-x+y/22x-z-1+2z-y-1≥222-x2x-y-1=22-x2x-y-1≥2x-y-1=dx,yIf x≥y then we have 22-y≥22-x,x+y/2≤y, and 22-x+y/2≤22-y. Thus we get (7)21-x+y/22x-z-1+2z-y-1≥22-y-22-x+y/2≥22-x-22-y=222-x-2-y=222-y-2-x=222-x2x-y-1=22-x2x-y-1≥2x-y-1=dx,y.Hence, we have (8)dx,y≤21-x+y/2dx,z+dz,y.Thus d is a quasib-metric in X=[0,1].Definition 9.
LetXbe a non-empty set and k:X×X→1,∞ and let dk:X×X→[0,∞) be a function. The pair X,dk is called adislocated quasi extended b-metric space (in short dqeb- metric space) if the following conditions are satisfied: (9)1dkx,y=0thenx=y,2dkx,y≤kx,ydkx,z+dkz,y,for all x,y,z∈X.Remark 10.
Ifkx,y=k≥1, thendqeb isdqb.Example 11.
LetX=-1,1 and dkx,y=x+y+x2/m+y2/n,m≠n, for x,y∈-1,1.
Letkx,y=2+xy/2 for x∈-1,1.
In fact, it is clear that ifdx,y=0, then x=y=0, which is satisfied for first condition. For second condition, we consider, (10)kx,ydkx,z+dkz,y=2+xy2x+z+x2m+z2n+z+y+z2m+y2n≥2+xy2x+y+x2m+z2n+z2m+y2n≥x+y+x2m+y2n=dkx,y.Thus (X,dk) is a dislocated quasi extendedb-metric space, with kx,y=2+xy/2≥1.Definition 12 (see [12, 13]).
Let(X,dk) be a dislocated quasi extendedb-metric space and let {xn} be a sequence in X.(i)
{xn} convergent sequence to x∈X, if limn→∞⁡dkxn,x=limn→∞⁡dkx,xn=0.(ii)
{xn} is called Cauchy in X, if limn,m→∞⁡dkxn,xm=limn,m→∞⁡dkxm,xn=0.(iii)
(X,dk) is called complete if every Cauchy sequence in X is convergent in X.Definition 13 (see [6]).
Let X be nonempty set,G and H are subsets of X. A function T:G∪H→G∪H is called acyclic mapif TG⊆H and T(H)⊆G.Definition 14 (see [17]).
LetX,dk be a dislocated quasi extendedb-metric space, G and H be subsets of X. A function T:G∪H→G∪H is calleddqeb-cyclic weakly contraction if there exists continuous and non-decreasing function φ:[0,∞)→[0,∞) such that for every x∈G, y∈H,(11)kx,ydkTx,Ty≤dkx,y-φdkx,y,where φt=0 if and only if t=0.Definition 15 (see [20]).
LetX,dk be a dislocated quasi extendedb-metric space, G and H be subsets of X. A function T:G∪H→G∪H is called acyclicφcontractionif T is a cyclic and there exists a continuous and non-decreasing function φ:[0,∞)→[0,∞) such that for every x∈G, y∈H(12)dkTx,Ty≤φdkx,y.


## 3. Main Results

In this section, we show some theorems and examples of the existence and uniqueness of fixed point for generalizeddqeb-cyclic weakly contraction mapping in complete dislocated quasi extendedb-metric space.Theorem 16.
LetX,dk be a complete dislocated quasi extended b-metric space, G and H be closed subsets of X. If T:G∪H→G∪H is a cyclic map that satisfies the condition of dqeb-cyclic weakly contraction and limn,m→∞⁡k(xn,xm)=L>0, then T has a unique fixed point in G∩H.Proof.
SinceT is a cyclic map, if taking x0∈G, then Tx0∈H and T2x0∈G. Define a sequence {xn}, where xn=Txn-1=Tnx0. So we have x2n∈G and x2n-1∈H for n=1,2,3…
Sincekx,y≥1 for all x,y∈X, then for all n∈N we have(13)dkxn+1,xn≤kxn+1,xndkxn+1,xn=kxn+1,xndkTxn,Txn-1≤dkxn,xn-1-φdkxn,xn-1≤dkxn,xn-1.Thus we have {dkxn+1,xn} is a nonincreasing sequence of non-negative real numbers.
Claim thatlimn→∞⁡dkxn+1,xn=0. Suppose limn→∞⁡dkxn+1,xn=β.
Sinceφ is nondecreasing and kxn+1,xn≥1, we have (14)dkxn+1,xn≤kxn+1,xndkxn+1,xn≤dkxn,xn-1-φdkxn,xn-1.Since φ is continuous then for n→∞, we have β≤β-φ(β). Since φ≥0, thus we get φβ=0. Hence we have β=0. Similarly we have limn→∞⁡dkxn,xn+1=0.
Now, we have to prove that{xn} is a Cauchy sequence in X.
Suppose{xn} is not a Cauchy sequence. Then there exists ε>0 such that every n, there exists nk, mk>n such that dknk,mk≥ε and dknk-1,mk<ε.
From (11) we have (15)knk-1,mk-1dknk,mk=knk-1,mk-1dkTnk-1,Tmk-1≤dknk-1,mk-1-φdknk-1,mk-1≤dknk-1,mk-1.This implies (16)dknk-1,mk-1≥εknk-1,mk-1.We also have that (17)dknk-1,mk-1≤knk-1,mk-1dknk-1,mk+dkmk,mk-1<εknk-1,mk-1+knk-1,mk-1dkmk,mk-1.It implies limk→∞⁡dknk-1,mk-1≤εlimk→∞knk-1,mk-1. Therefore we have εlimk→∞knk-1,mk-1≤limk→∞⁡dknk-1,mk-1≤εlimk→∞knk-1,mk-1, so we have εL≤limk→∞⁡dknk-1,mk-1≤εL, thus we obtain limk→∞⁡dknk-1,mk-1=εL. From (15) and (16) we have (18)knk-1,mk-1dknk,mk≤dknk-1,mk-1-φdknk-1,mk-1≤dknk-1,mk-1εknk-1,mk-1≤dknk-1,mk-1-φdknk-1,mk-1≤dknk-1,mk-1.For k→∞ and using continuity of φ, we get (19)εL≤εL-φεL≤εL.Since kx,y≥1 and limn,m→∞⁡k(xn,xm)=L, we have L≥1. Thus we have φεL=0, this implies εL=0. Since ε>0 then we obtain L=0, which is a contradiction.
Hencexn is a Cauchy sequence in X. Since X complete, there exists x∗∈X such that dkxn,x∗→0 for n→∞. Similarly we can have dkxn,x∗→0.
Since the sequence{x2n}∈G, {x2n-1}∈H and G,H be closed, then we have x∗∈G∩H.
Now we prove thatx∗ is a fixed point of T. Using (2) and (11) we have(20)dkTx∗,x∗≤kTx∗,x∗dTx∗,Txn-1+dTxn-1,x∗≤kTx∗,x∗dx∗,xn-1-φdx∗,xn-1+dxn,x∗.Using continuity of φ and for n→∞, we have dkTx∗,x∗≤-kTx∗,x∗φ0≤0.
ThusdkTx∗,x∗=0, hence Tx∗=x∗.
Now we have to show thatT has unique fixed point in X. Suppose that u is an another fixed point of T,(21)dkx∗,u=dkTx∗,Tu≤kx∗,udkTx∗,Tu≤dkx∗,u-φdkx∗,u.Thus we get φ(dkx∗,u)≤0. Since φ≥0, we have φdkx∗,u=0. Which implies that dkx∗,u=0, so we have x∗=u.Example 17.
LetX=-1,1 and X,dk be a dislocated quasi extendedb-metric space which in Example 8. Let T:G∪H→G∪H be a function defined by Tx=-x/2, where G=-1,0,H=0,1. Let φ:0,∞→0,∞ be a function which is defined by φt=t/4
In fact, It is clear thatT is a cyclic map, indeed T(G)⊆H and T(H)⊆G.
Now, we have to show that(22)kx,ydkTx,Ty≤dkx,y-φdkx,ykx,ydkTx,Ty=2+xy2dk-x2,-y2=2+xy2-x2+-y2+-x/225+-y/226=2+xy4x+y+x210+y212=142x+y+x210+y212+xyx+y+x210+y212≤142x+y+x210+y212+x+y+x210+y212≤143x+y+x25+y26≤34x+y+x210+y212≤34x+y+x25+y26=x+y+x25+y26-14x+y+x25+y26=dkx,y-φdkx,y.Hence, T has adeqb-weak contraction property of Theorem 16 and x=0 is the unique fixed point of T.Theorem 18.
LetX,dk be a complete dislocated quasi extended b-metric space, G and H be closed subsets of X. If T:G∪H→G∪H is a cyclic map, continuous mapping and limn,m→∞⁡k(xn,xm)=L>0, such that (23)kx,ydkTx,Ty≤dkx,y-φdkTx,Ty,where φ:0,∞→0,∞ is a nondecreasing, continuous mapping and φt=0ifft=0.
ThenT has a unique fixed point in G∩H.Proof.
SinceT is a cyclic map, if taking x0∈G, then Tx0∈H and T2x0∈G. Define a sequence {xn}, where xn=Txn-1=Tnx0. So we have x2n∈G and x2n-1∈H for n=1,2,3…
By using (23) and for all n∈N, we have (24)dkxn+1,xn≤kxn+1,xndkxn+1,xn=kxn+1,xndkTxn,Txn-1≤dkxn,xn-1-φdkTxn,Txn-1≤dkxn,xn-1.Thus we have {dkxn+1,xn} is a nonincreasing sequence of non-negative real numbers. Claim that limn→∞⁡dkxn+1,xn=0. Suppose limn→∞⁡dkxn+1,xn=β.
Sinceφ is non-decreasing and kxn+1,xn≥1, then we have (25)dkxn+1,xn≤kxn+1,xndkxn+1,xn≤dkxn,xn-1-φdkTxn,Txn-1≤dkxn,xn-1-φdkTxn,Txn-1=dkxn,xn-1-φdkxn+1,xn.Since φ is a continuous mapping then for n→∞, we have β≤β-φ(β). Since φ≥0, thus we get φβ=0. Hence we have β=0. Similarly we have limn→∞⁡dkxn,xn+1=0.Now, we have to prove that{xn} is a Cauchy sequence in X.Suppose{xn} is not a Cauchy, then there exists ε>0 such that every n, there exists nk, mk>n such that(26)dknk,mk≥εanddknk-1,mk<ε.From (23) we have (27)knk-1,mk-1dknk,mk=knk-1,mk-1dkTnk-1,Tmk-1≤dknk-1,mk-1-φdkTnk-1,Tmk-1≤dknk-1,mk-1.By using (26) and (27), then we have (28)dknk-1,mk-1≥εknk-1,mk-1.By using (2) and (26) we also have that (29)dknk-1,mk-1≤knk-1,mk-1dknk-1,mk+dkmk,mk-1<εknk-1,mk-1+knk-1,mk-1dkmk,mk-1.It implies limk→∞⁡dknk-1,mk-1≤εlimk→∞knk-1,mk-1. By using (28) and (29), we have εlimk→∞knk-1,mk-1≤limk→∞⁡dknk-1,mk-1≤εlimk→∞knk-1,mk-1, so we get εL≤limk→∞⁡dknk-1,mk-1≤εL, thus we obtain limk→∞⁡dknk-1,mk-1=εL.From (23) and (28) we have (30)knk-1,mk-1dknk,mk≤dknk-1,mk-1-φdknk,mk≤dknk-1,mk-1.Since φ is a non-decreasing and φ≥0, we have(31)εknk-1,mk-1≤dknk-1,mk-1-φdknk,mk≤dknk-1,mk-1-φεknk-1,mk-1≤dknk-1,mk-1.For k→∞ and using continuity of φ, we get (32)εL≤εL-φεL≤εL.Since kx,y≥1 and limn,m→∞⁡k(xn,xm)=L, thus we have L≥1. However, from (32) we have φεL=0, this implies εL=0. Since ε>0 then we obtain L=0 which is a contradiction.Hence{xn} is a Cauchy sequence in X.SinceX complete, there exists x∗∈X such that dkxn,x∗→0 and dkx∗,xn→0 for n→∞.Since the sequence{x2n}∈G,{x2n-1}∈H and G,H closed, we have x∗∈G∩H.Now we have to prove thatx∗ is a fixed point of T. By using (2) and (23), we have(33)dkTx∗,x∗≤kTx∗,x∗dTx∗,Txn-1+dTxn-1,x∗=kTx∗,x∗dTx∗,Txn-1+dxn,x∗≤kTx∗,x∗kx∗,xn-1dTx∗,Txn-1+dxn,x∗≤kTx∗,x∗dx∗,xn-1-φdTx∗,Txn-1+dxn,x∗≤kTx∗,x∗dx∗,xn-1+dxn,x∗.Thus for n→∞, we have dkTx∗,x∗=0, hence Tx∗=x∗.Now we have to show thatT has unique fixed point in X. Suppose that u is an another fixed point T,(34)dkx∗,u=dkTx∗,Tu≤kx∗,udkTx∗,Tu≤dkx∗,u-φdkTx∗,Tu=dkx∗,u-φdkx∗,u.Thus we get φ(dkx∗,u)≤0. Since φ≥0, we have φdkx∗,u=0. Which implies that dkx∗,u=0, so we have x∗=u.Example 19.
LetX=-1,1 and X,dk be a dislocated quasi extendedb-metric space which in Example 8. Let T:G∪H→G∪H be a function defined by Tx=-x/2, where G=-1,0,H=0,1. Let φ:0,∞→0,∞ be a function and defined as, φt=t/8.
In fact, it is clear thatT is cyclic map, indeed T(G)⊆H and T(H)⊆G.
Now, for allx,y∈X we have to show that (35)
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LetX,dk be a complete dislocated quasi extended b-metric space, G and H be closed subsets of X. If T:G∪H→G∪H is a cyclic, continuous mapping and limn,m→∞⁡k(xn,xm)=L>0, such that (37)kTx,TydkTx,Ty≤dkx,y-φdkx,y,where φ:0,∞→0,∞ be a nondecreasing, continuous function and φt=0ifft=0.
ThenT has a unique fixed point in G∩H.Proof.
SinceT is a cyclic map, taking x0∈G, then Tx0∈H and T2x0∈G. Define a sequence {xn}, where xn=Txn-1=Tnx0. So we have x2n∈G and x2n-1∈H for n=1,2,3…
From (37), then for all n∈N we have (38)dkxn+1,xn≤kxn+1,xndkxn+1,xn=kTxn,Txn-1dkTxn,Txn-1≤dkxn,xn-1-φdkxn,xn-1≤dkxn,xn-1.Thus we have {dkxn+1,xn} be a nonincreasing sequence of non-negative real numbers. Claim that limn→∞⁡dkxn+1,xn=0. Suppose limn→∞⁡dkxn+1,xn=β.
Sinceφ is a nondecreasing and kxn+1,xn≥1, then we have (39)dkxn+1,xn≤kxn+1,xndkxn+1,xn≤dkxn,xn-1-φdkxn,xn-1.Since φ is continuous then for →∞, we have β≤β-φ(β). Since φ≥0, thus we get φβ=0. Hence we have β=0. Similarly we have limn→∞⁡dkxn,xn+1=0.
Now, we have to prove that{xn} is a Cauchy sequence in X.
Suppose{xn} is not a Cauchy, then there exists ε>0 such that every n, there exists nk, mk>n such that dknk,mk≥ε and dknk-1,mk<ε.
By using (37) we have (40)knk,mkdknk,mk=kTnk-1,Tmk-1dkTnk-1,Tmk-1≤dknk-1,mk-1-φdknk-1,mk-1≤dknk-1,mk-1.Since dknk,mk≥ε, we get (41)dknk-1,mk-1≥εknk,mk.By using (2), we also have that (42)dknk-1,mk-1≤knk-1,mk-1dknk-1,mk+dkmk,mk-1≤εknk-1,mk-1+knk-1,mk-1dkmk,mk-1.It implies that limk→∞⁡dknk-1,mk-1≤εlimk→∞knk-1,mk-1.
Therefore, from (40) and (41), we have(43)εlimk→∞knk,mk≤limk→∞⁡dknk-1,mk-1≤εlimk→∞knk-1,mk-1.Thus we have εL≤limk→∞⁡dknk-1,mk-1≤εL, and we obtain(44)limk→∞⁡dknk-1,mk-1=εL.From (40) and (44) and dknk,mk≥ε, we have (45)εknk,mk≤knk,mkdknk,mk≤dknk-1,mk-1-φdknk-1,mk-1≤dknk-1,mk-1.By using (45) and continuity of φ, then for k→∞ we get (46)εL≤εL-φεL≤εL.Since kx,y≥1 and limn,m→∞⁡k(xn,xm)=L, thus we have L≥1. However, from (46) we have φεL=0, this implies εL=0 and since ε>0 then we obtain L=0 which is a contradiction.
Hence{xn} is a Cauchy sequence in X.
SinceX complete, there exists x∗∈X such that dkxn,x∗→0 for n→∞. Since the sequence {x2n}∈G, {x2n-1}∈H and G,H closed, it implies that x∗∈G∩H.
Now, we have to prove thatx∗ is a fixed point of T.(47)dkTx∗,x∗≤kTx∗,x∗dTx∗,Txn-1+dTxn-1,x∗=kTx∗,x∗dTx∗,Txn-1+dxn,x∗≤kTx∗,x∗kTx∗,Txn-1dTx∗,Txn-1+dxn,x∗≤kTx∗,x∗dx∗,xn-1-φdx∗,xn-1+dxn,x∗≤kTx∗,x∗dx∗,xn-1+dxn,x∗.Thus for n→∞, we have dkTx∗,x∗=0, hence Tx∗=x∗.
Now we have to show thatT has unique fixed point in X. Suppose u is an another fixed point of T,(48)dkx∗,u=dkTx∗,Tu≤kTx∗,TudkTx∗,Tu≤dkx∗,u-φdkx∗,u.Thus we get φ(dkx∗,u)≤0. Since φ≥0, we have φdkx∗,u=0. Which implies that dkx∗,u=0, so we have x∗=u.Example 21.
LetX=-1,1 and X,dk be a dislocated quasi extendedb-metric space which in Example 8. Let T:G∪H→G∪H be a function defined by Tx=-x/2, where G=-1,0, H=0,1, and let φ:0,∞→0,∞ be a function and defined as, φt=7/16t.
In fact it clearT is cyclic, since T(G)⊆H and T(H)⊆G.
Now, we have to show that(49)kTx,TydkTx,Ty≤dkx,y-φdkx,y.kTx,TydkTx,Ty=k-x2,-y2dk-x2,-y2=8+xy8-x2+-y2+-x/225+-y/226=8+xy16x+y+x210+y212=1168x+y+x210+y212+xyx+y+x210+y212≤1168x+y+x210+y212+x+y+x210+y212=1169x+y+x25+y26≤916x+y+x25+y26=x+y+x25+y26-716x+y+x25+y26=dkx,y-φdkx,y.Hence, T has adeqb- weak contraction property of Theorem 20 and x=0 is the unique fixed point of T.Theorem 22.
LetX,dk be a complete dislocated quasi extended b-metric space, G and H be closed subsets of X and let 0<λ<1. If T:G∪H→G∪H is a cyclic, continuous function which satisfy the conditions (50)kTx,TydkTx,Ty≤λφdkx,y,where φ:0,∞→0,∞ be a φ nondecreasing and continuous function, φt=0if only ift=0 and φλt≤λφt, φn+1t≤φnt, φn+1t=φφnt, for n=1,2,3,…, and limn,m→∞⁡kxn,xm<1/λ.
ThenT has unique fixed point in G∩H.Proof.
SinceT is a cyclic map, for x0∈G, then Tx0∈H and T2x0∈G. Define a sequence {xn}, where xn=Txn-1=Tnx0. So we have x2n∈G and x2n-1∈H for n=1,2,3…
Sincekx,y≥1 and 0<λ<1 then for all n∈N, we have (51)kxn,xn+1dkxn,xn+1=kTxn-1,TxndkTxn-1,Txn≤λφdkxn-1,xn≤λφλφdkxn-2,xn-1=λ2φ2dkxn-2,xn-1≤λnφndkx0,x1.We have (52)dkxn,xn+1≤kxn,xn+1dkxn,xn+1≤λnφnt0,where t0=dkx0,x1.
By using (2) and (52), we have(53)dkxn,xm≤kxn,xmdkxn,xn+1+dkxn+1,xm≤kxn,xmdkxn,xn+1+dkxn+1,xm≤kxn,xmλnφnt0+dkxn+1,xm≤kxn,xmλnφnt0+kxn+1,xmdkxn+1,xn+2+dkxn+2,xm≤kxn,xmλnφnt0+kxn+1,xmdkxn+1,xn+2+dkxn+2,xm≤kxn,xmλnφnt0+kxn+1,xmλn+1φn+1t0+dkxn+2,xm≤kxn,xmλnφnt0+kxn+1,xmλn+1φn+1t0+kxn+1,xmdkxn+2,xm≤kxn,xmλnφnt0+kxn+1,xmλn+1φn+1t0+kxn+1,xmkxn+2,xmdkxn+2,xn+3+dkxn+3,xm≤kxn,xmλnφnt0+kxn+1,xmλn+1φn+1t0+kxn+1,xmkxn+2,xmλn+2φn+2t0+dkxn+3,xm≤kxn,xmλnφnt0+kxn+1,xmλn+1φn+1t0+kxn+1,xmkxn+2,xmλn+1φn+2t0+⋯+kxn+1,xmkxn+2,xm⋯kxm-1,xmλm-1φm-1t0=∑i=0m-n-1λn+iφn+it0∏j=0ikxn+j,xm=∑i=nm-1λiφit0∏j=0ikxn+j,xm.We have (54)dkxn,xm≤∑i=nm-1λiφit0∏j=0ikxn+j,xm.Let ai=λiφit0∏j=0ikxn+j,xm.
Sinceφn+1(t)≤φn(t) we have(55)ai+1ai=φn+i+1t0φn+it0λkxn+i+1,xm≤λkxn+i+1,xm.limi→∞⁡ai+1ai=limi→∞⁡φn+i+1t0φn+it0λkxn+i+1,xm≤limi→∞⁡λkxn+i+1,xm<1.By using the ratio test criteria, we get ∑i=0∞λn+iφn+it0∏j=0ikxn+j,xm=∑i=0∞ai convergence.
LetSp=∑i=0pai, then from (54), we get(56)dkxn,xm≤∑i=0m-n-1λn+iφn+it0∏j=0ikxn+j,xm=∑i=nm-1ai=Sm-1-Sn-1≤Sm-1-Sn-1.Thus for n,m→∞ we get dkxn,xm→0. Hence {xn} is a Cauchy sequence in X.
SinceX complete, there exists x∗∈X such that dkxn,x∗→0 for n→∞.
Similarly, we can havedkx∗,xn→0.
Since the sequence{x2n}∈G, {x2n-1}∈H and G,H closed, thus we have x∗∈G∩H.
Now we prove thatx∗ is a fixed point of T. Using (2) and (11) we have(57)dkTx∗,x∗≤kTx∗,x∗dTx∗,Txn-1+dTxn-1,x∗≤kTx∗,x∗kTx∗,Txn-1dTx∗,Txn-1+dTxn-1,x∗≤kTx∗,x∗λφdx∗,xn-1+dxn,x∗.Using continuity of φ, and φ0=0, then for n→∞, we have dkTx∗,x∗≤λkTx∗,x∗φ0≤0.
ThusdkTx∗,x∗=0, hence Tx∗=x∗.
Now we have to show thatT has unique fixed point in X. Suppose u is an another fixed point of T,(58)dkx∗,u=dkTx∗,Tu≤kTx∗,TudkTx∗,Tu≤λφdkx∗,u.We have (59)1-λφdkx∗,u≤0.Since 1-λ>0 thus we get φ(dkx∗,u)≤0. Since φ≥0, then φdkx∗,u=0. Which implies that dkx∗,u=0, so we have x∗=u.Example 23.
LetX=[-1,1] and X,dk be a dislocated quasi extended b-metric space which in Example 8. Let T:G∪H→G∪H be a function defined by Tx=-x3/8, where G=-1,0, H=0,1. Let φ:0,∞→0,∞ be a function and defined as, φt=3/4t2 and λ=1/4.
In fact, it clear thatφλt≤λφt, φn+1t≤φnt and T is cyclic, since T(G)⊆H and TH⊆G.
Sincexn,xm∈X=-1,1 and kxn,xm=2+xnxm/2, it is easy to show that limn,m→∞⁡k(xn,xm)<1/λ.
Now, we have to show that(60)kTx,TydkTx,Ty≤λφdkx,y.kTx,TydkTx,Ty=k-x38,-y38dk-x38,-y38=2+-x3/8-y3/82dk-x38,-y38≤2+x3y32dk-x38,-y38≤2+x3y32-x38+-y38+-x3/825+-y3/826=2+x3y316x3+y3+x3240+y3248=316x3+y3+x3240+y3248≤316x2+y2+x2225+y2236≤316x+y+x25+y262=14φdkx,y.Hence, T has adeqb-weak contraction property of Theorem 22 and x=0 is the unique fixed point of T.


## 4. Conclusion

In this article, we considered and proved the fixed point theorems for cyclic weakly contraction mapping in complete dislocated quasi extendedb-metric space. These results generalize the recent results of Samreen [14] and Rahman [9], which was in our results more general in the sense by utilizing dislocated quasi extendedb-metric and cyclic weakly contraction. Furthermore, In Theorems 16, 18, 20, and 22 one can derive several consequences in dislocated quasib-metric by letting kx,y=K≥1 and in dislocated quasi metric by letting kx,y=1.



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