# Common Fixed Point Theorems on Generalized Ratio Contraction Mapping in Extended Rectangularb-Metric Spaces

Authors: Budi Nurwahyu
Journal: International Journal of Mathematics and Mathematical Sciences
 (2019)

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## Abstract

In this paper, we propose and prove the common fixed point theorems on generalized contraction mappings in extended rectangularb-metric spaces by utilizing the weakly compatible function property.


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## Body

## 1. Introduction

The notion ofb-metric space was introduced by Czerwik in 1993 [1], as a generalization of a metric space, by modifying the third condition in the metric space. Mohanta [2] and Husain et al. [3] utilized that space to prove existence and uniqueness of common fixed points. Recently, Kamran et al. [4] in 2017 generalized b-metric to become extended b-metric for utilizing in fixed point results and Alqahtani et al. [5] in 2018 utilized extended b-metric to prove the common fixed point. In 2015, George et al. [6] introduced the notion of rectangular b-metric space as a generalization of rectangular metric space. Kadelburg and Radenović [7] and Mitrovic [8] established and proved the common fixed point in such space. Very recently, Asim et al. [9] in 2019 generalized the rectangular b-metric and termed it the extended rectangular b-metric. By utilizing this space, Mustafa et al. [10] proved some fixed point results for contracting mappings.Inspired of some fixed point results in extended rectangularb-metric, in the main results, we establish and prove existence and uniqueness of the common fixed point in the extended rectangular b-metric and furnish examples to clarify the results.


## 2. Preliminaries

In the following section, we need some definitions to govern our theorems.Definition 1 (see [1]).
LetX be a nonempty set. A mapping db:X×X⟶0,∞ is called a b-metric, if there exists b≥1 such that db satisfies the following conditions:(1)
dbx,y=0, if and only if x=y(2)
dbx,y=dby,x(3)
dbx,y≤bdbx,s+dbs,yfor all x,y,s∈X.
The pairX,db is called a b-metric space.Definition 2 (see [4]).
LetX be a nonempty set. A mapping db:X×X⟶0,∞ is called an extended b-metric, if there exists a function b:X×X⟶1,∞ such that db satisfies the following conditions:(1)
dbx,y=0, if and only if x=y(2)
dbx,y=dby,x(3)
dbx,y≤bx,ydbx,s+dbs,yfor all x,y,s∈X.
The pairX,db is called an extended b-metric space.Definition 3 (see [6]).
LetX be a nonempty set. A mapping db:X×X⟶0,∞ is called rectangular b-metric, if there exists b≥1 such that db satisfies the following conditions:(1)
dbx,y=0, if and only if x=y(2)
dbx,y=dby,x(3)
dbx,y≤bdbx,s+dbs,t+dbt,yfor all x,y∈X and s,t∈X\x,y.
The pairX,db is called a rectangular b-metric space.Definition 4 (see [9]).
LetX be a nonempty set. A mapping db:X×X⟶0,∞ is called an extended rectangular b-metric, if there exists a function b:X×X⟶1,∞ such that db satisfies the following conditions:(1)A1dbx,y=0,if and only ifx=y,A2dbx,y=dby,x,A3dbx,y≤bx,ydbx,s+dbs,t+dbt,y.for all x,y∈X and s,t∈X\x,y.
The pairX,db is called an extended rectangular b-metric space.Example 1.
LetX=0,∞ and dbx,y=x−yp, and p>1 with bx,y=x+y+3p−1.
It is obvious for condition A1 and A2. For condition A3, we consider from the Jensen inequality.
Thus, we havedbx,y=x−y2≤3p−1x−s2+s−t2+t−y2≤x+y+3p−1x−s2+s−t2+t−y2=bx,ydbx,s+dbs,t+dbt,y for all x,y,s,t∈X.
So, this shows thatdbx,y=x−yp is an extended rectangular b-metric with bx,y=x+y+3p−1. However, in general dbx,y is not continuous.Example 2 (see [7]).
LetX=0,2∪1/nn∈N, and choose a function bx,y=x+y+2, for x,y∈X. Define db:X×X⟶0,∞ a function as follows:(2)dbx,y=0,x=y,1,x≠y,x,y∈0,2orx,y∈1nn∈N,y,x∈0,2,y∈1nn∈N,x,x∈1nn∈N,y∈0,2.
It is easy to show thatdb is the complete extended rectangular b-metric on X with bx,y=x+y+2.
If we choose a sequence1/nn∈N, it is easy to show that sequence 1/nn∈N is convergent to 0 and 2, so it is not a Cauchy sequence. Moreover, db is not continuous, since limn⟶∞1/n=0 and 1=limn⟶∞db1/n,1/3≠db0,1/3=1/3. Thus, X,db is not metrizable because there exists a sequence in X,db that is not convergent to unique point in X.
Our proofs in the main results need uniqueness of the limit point of sequence inX,db as given in the following lemma.Lemma 1 (see [9, 11]).
LetX,db be a complete extended rectangular b-metric space and yn is a Cauchy sequence. If yn≠ym for all n≠m, then limit point of yn is unique.Definition 5 (see [12]).
LetX be a nonempty set and T1,T2:X⟶X be self-mapping. If T1x=T2x=y for some x∈X, then y is called a point of coincidence of T1 and T2 and x is called a coincidence point of T1 and T2.Definition 6 (see [13]).
LetX be a nonempty set and T1,T2:X⟶X be self-mappings. T1,T2 is called weakly compatible. For every x∈X, if T1x=T2x, then T2T1x=T1T2x.Definition 7 (see [9]).
LetX,db be an extended rectangular b-metric space, then sequence(a)
xn is called convergent if and only if there exists x∈X such that dbxn,x⟶0, as n⟶∞.(b)
xn is called Cauchy if and only if dbxn,xm⟶0, as n,m⟶∞.
In addition,X is called complete if every Cauchy sequence in X is convergent in X.


## 3. Main Results

In this section, we establish and prove the common fixed point theorems in complete extended rectangularb-metric space and furnish some examples to clarify the theorem.Lemma 2.
LetX,db be an extended rectangular b-metric space and yn be a sequence in X.
Ifdbyn,yn+1≤Kdbyn−1,yn, 0<K<1, and limn,m⟶∞byn,ym<1/K, then yn is a Cauchy.Proof.
Letm∈N, m>n.Case 1.
Form−n is odd, let m−n=2k+1.Consider that from rectangular condition in (1) we have(3)dbyn,ym≤byn,ymdbyn,yn+1+dbyn+1,yn+2+dbyn+2,ym≤byn,ymdbyn,yn+1+dbyn+1,yn+2+byn,ymdbyn+2,ym≤byn,ymKndby0,y1+Kn+1dby0,y1+byn,ymdbyn+2,ym≤Kn+Kn+1dby0,y1byn,ym+Kn+2+Kn+3dby0,y1byn,ymbyn+2,ym+byn,ymbyn+2,ymdbyn+4,ym≤Kn+Kn+1dby0,y1byn,ym+Kn+2+Kn+3dby0,y1byn,ymbyn+2,ym+⋯+Kn+2k−2+Kn+2k−1dby0,y1byn,ymbyn+2,ym⋯byn+2k−2,ym+byn,ymbyn+2,ym⋯byn+2k−2,ymdbyn+2k,ym+Kn+2k−2+Kn+2k−1dby0,y1byn,ymbyn+2,ym⋯byn+2k−2,ym+byn,ymbyn+2,ym⋯byn+2k−2,ymdbyn+2k,ym≤Kn+Kn+1dby0,y1byn,ym+Kn+2+Kn+3dby0,y1byn,ymbyn+2,ym+⋯+Kn+2k−2+Kn+2k−1dby0,y1byn,ymbyn+2,ym⋯byn+2k−2,ym+⋯+byn,ymbyn+2,ym⋯byn+2k−2,ymKn+2kdby0,y1=1+Kdby0,y1Kn∑j=0k−1K2j∏i=0jbyn+2i,ym+byn,ymbyn+2,ym⋯byn+2k−2,ymKn+2kdby0,y1.Sincebyi,yj≥1, using (3) we have that(4)dbyn,ym≤1+Kdby0,y1Kn∑j=0k−1K2j∏i=0jbyn+2i,ym+byn,ymbyn+2,ym⋯byn+2k−2,ymK2k≤1+Kdby0,y1Kn∑j=nm−1Kj−n∏i=1jbyi,ym=1+Kdby0,y1∑j=nm−1Kj∏i=1jbyi,ym=1+Kdby0,y1Sm−1−Sn−1,where Sn=∑j=0nKj∏i=1jbyi,ym.Letaj=Kj∏i=1jbyi,ym, so by applying ratio criteria and as known that limn,m⟶∞byn,ym<1/K, then we have limj⟶∞aj+1/aj=limj⟶∞Kbyj,ym<1 for each m∈N.This implies thatSn=∑j=1nKj∏i=1jbyi,ym is convergent for each m∈N.Hence, forn,m⟶∞ in (4) we get dbyn,ym⟶0; thus, yn is the Cauchy sequence in X.Case 2.
Form−n is even, let m−n=2k.(5)dbyn,ym≤byn,ymdbyn,yn+1+dbyn+1,yn+2+dbyn+2,ym≤byn,ymdbyn,yn+1+dbyn+1,yn+2+byn,ymdbyn+2,ym≤byn,ymKndby0,y1+Kn+1dby0,y1+byn,ymdbyn+2,ym≤Kn+Kn+1dby0,y1byn,ym+byn,ymdbyn+2,ym≤Kn+Kn+1dby0,y1byn,ym+byn,ymbyn+2,ymdbyn+2,yn+3+dbyn+3,yn+4+byn,yn+2kbyn+2,ymdbyn+4,ym≤Kn+Kn+1dby0,y1byn,ym+byn,ymbyn+2,ymKn+2dby0,y1+Kn+3dby0,y1+byn,ymbyn+2,ymdbyn+4,ym≤Kn+Kn+1dbdby0,y1byn,ym+Kn+2+Kn+3dby0,y1byn,ymbyn+2,ym+⋯+Kn+2k−2+Kn+2k−1dby0,y1byn,ymbyn+2,ym⋯bym−2,ym+Kn+2k−2dby0,y2byn,ymbyn+2,ym⋯bym−2,ym=1+Kdby0,y1Kn∑j=1k−1K2j∏i=0jbyn+2i,ym+Km−2dby0,y2∏i=0k−1byn+2i,ym.
LetL=maxdby0,y1,dby0,y2, so by using (5) we have(6)dbyn,ym≤1+KLKn∑j=1k−1K2j∏i=0jbyn+2i,ym+Km−2∏i=0k−1byn+2i,ym=1+KLKn∑j=1k−1K2j∏i=0jbyn+2i,ym+Kn+2k−2∏i=0k−1byn+2i,ym=1+KLKn∑j=1k−1K2j∏i=0jbyn+2i,ym+K2k−2∏i=0k−1byn+2i,ym≤1+KLKn∑j=nm−1Kj−n∏i=1jbyi,ym=1+KL∑j=nm−1Kj∏i=1jbyi,ym=1+KL∑j=0m−1Kj∏i=1jbyi,ym−∑j=0n−1Kj∏i=1jbyi,ym=1+KLSm−1−Sn−1.
This impliesdbyn,ym⟶0 as m,n⟶∞, thus yn is Cauchy in X.Theorem 1.
LetX,db be a complete extended rectangular b-metric space and let f,g:X⟶X be self-mappings such that fX⊆gX and gXclosed. If(7)bx,ydbfx,fy≤pdbgx,gy+qdbgx,fx+rgy,fy+sβdbfx,gydbfy,gxλbfx,gx+βdbfx,gy+dbfy,gx,holds for all x,y∈X, β>0, 0<s<1, 0<p+q+r<λ,p/1−s<λ, and limn,m⟶∞byn,ym<λ−r/p+q, then f and g have a unique coincidence in X.Moreover, iff and g are weakly compatible, then f and g have a unique common fixed point.Proof.
Letx0∈X as an initial point, define sequence xn and yn in X such that yn=fxn=gxn+1, n=0,1,2,.…. Suppose that yn≠yn+1, n=0,1,2,…, because if yn=yn+1 for some n, then gxn+1=yn=yn+1=fxn+1, so f and g have a coincidence point.
Considering (7), we have(8)bxn,xn+1dbyn,yn+1=bxn,xn+1dbfxn,fxn+1≤pdbgxn,gxn+1+qdbgxn,fxn+rdbgxn+1,fxn+1+sβdbfxn,gxn+1dbfxn+1,gxnλbfxn,gxn+βdbfxn,gxn+1+dbfxn+1,gxn=pdbyn−1,yn+qdbyn−1,yn+rdbyn,yn+1+sβdbyn,yndbyn+1,ynλbyn,yn−1+βdbyn,yn+dbyn+1,yn=p+qdbyn−1,yn+rdbyn,yn+1λbyn,yn−1+βdbyn+1,yn≤p+qdbyn−1,yn+rdbyn,yn+1λbyn,yn−1.
Sincexn,xn+1≥1, then we have(9)dbyn,yn+1≤p+qdbyn−1,yn+rdbyn,yn+1λbxn,xn+1byn,yn−1≤p+qdbyn−1,yn+rdbyn,yn+1λ,
Thus, we get(10)dbyn,yn+1≤δdbyn−1,yn,where δ=p+q/λ−r. Since p+q+r<λ, we have 0<δ<1, and as known that limn,m⟶∞byn,ym<1/δ, then by using Lemma 2, we get yn is a Cauchy sequence in gX.
SinceX is complete, then there exists u∗∈X such that(11)dbyn,u∗⟶0,and since yn=gxn+1⊆X and gX is a closed set, then u∗∈gX and u∗=gx∗ for some x∗∈X.
Furthermore, we will show thatf and g have a point coincidence as follows.
From (1) and (7) we have(12)dbfx∗,gx∗≤bfx∗,gx∗dbfx∗,yn+dbyn,yn+1+dbyn+1,gx∗=bfx∗,gx∗dbfx∗,fxn+dbyn,yn+1+dbyn+1,u∗≤bfx∗,gx∗1bx∗,xnpdbgx∗,gxn+qdbgx∗,fx∗+rdbgxn,fxn+sβdbfx∗,gxndbfxn,gx∗λbfx∗,gx∗+βdbfx∗,gxn+dbfxn,gx∗+dbyn,yn+1+dbyn+1,u∗≤bfx∗,gx∗pdbu∗,yn−1+qdbu∗,fx∗+rdbyn−1,yn+sβdbfx∗,yn−1dbyn,u∗λbfx∗,u∗+βdbfx∗,yn−1+dbyn,u∗+dbyn,yn+1+dbyn+1,u∗≤bfx∗,gx∗pdbu∗,yn−1+qdbu∗,fx∗+rdbyn−1,ynλbfx∗,u∗+βdbfx∗,yn−1+dbyn,u∗+sdbyn,u∗+dbyn,yn+1+dbyn+1,u∗≤bfx∗,gx∗pdbu∗,yn−1+qdbu∗,fx∗+rdbyn−1,ynλbfx∗,u∗+sdbyn,u∗+dbyn,yn+1+dbyn+1,u∗.
It follows from (11) to (12) that if n⟶∞, then we get(13)dbfx∗,gx∗≤bfx∗,gx∗dbu∗,fx∗λbfx∗,gx∗≤dbgx∗,fx∗λ.
This implies thatdbfx∗,gx∗=0, so then we get fx∗=gx∗=u∗. Hence, u∗ is a coincidence point of f and g.
To show uniqueness of the coincidence point off and g, we suppose there exists w∗∈X such that fy∗=gy∗=w∗.
From (1) and (7), we have(14)dbu∗,w∗=dbfx∗,fy∗≤1bx∗,y∗pdbu∗,w∗+qdbu∗,u∗+rdbw∗,w∗+sβdbu∗,w∗dbw∗,u∗λbu∗,u∗+βdbu∗,w∗+dbw∗,u∗≤pdbu∗,w∗+sβdbu∗,w∗dbw∗,u∗λbu∗,u∗+βdbu∗,w∗+dbw∗,u∗=pdbu∗,w∗+sβdbu∗,w∗dbw∗,u∗λbu∗,u∗+βdbu∗,w∗+dbw∗,u∗=pdbu∗,w∗λ+sdbw∗,u∗.
Thus, we get(15)1−pλ−sdbw∗,u∗≤0.
Sincep/1−s<λ, we have 1−p/λ−s>0 and thus we obtain dbu∗,w∗=0; hence, u∗=w∗.
Furthermore, we have to show thatu∗ is a common unique fixed point of f and g.
From (1) and (7), we have(16)dbu∗,fu∗=dbfx∗,fu∗≤≤1bx∗,u∗pdbgx∗,gu∗+qdbgx∗,fx∗+rdbgu∗,fu∗+sβdbfx∗,gu∗dbfu∗,gx∗λbfx∗,gx∗+βdbfx∗,gu∗+dbfu∗,gx∗≤pdbgx∗,gu∗+qdbgx∗,fx∗+rdbgu∗,fu∗+sβdbfx∗,gu∗dbfu∗,gx∗λbfx∗,gx∗+βdbfx∗,gu∗+dbfu∗,gx∗.
Sincef and g are weakly compatible and fx∗=gx∗=u∗, so we have gfx∗=fgx∗=gu∗=fu∗. Thus, from (16) we have(17)dbu∗,fu∗=dbfx∗,fu∗≤pdbu∗,gu∗+qdbu∗,u∗+rdbgu∗,fu∗+sβdbu∗,gu∗dbfu∗,u∗λbu∗,u∗+βdbu∗,gu∗+dbfu∗,u∗=pdbu∗,fu∗+qdbu∗,u∗+rdbfu∗,fu∗+sβdbu∗,fu∗dbfu∗,u∗λbu∗,u∗+βdbu∗,gu∗+dbfu∗,u∗=pdbu∗,fu∗+sβdbu∗,fu∗dbfu∗,u∗λbu∗,u∗+βdbu∗,gu∗+dbfu∗,u∗≤pdbu∗,fu∗λbu∗,u∗+sdbu∗,fu∗≤pdbu∗,fu∗λ+sdbu∗,fu∗.
Thus, we get(18)1−pλ−sdbu∗,fu∗≤0.
Sincep/1−s<λ, we have 1−p/λ−s>0 and thus we obtain dbu∗,fu∗=0; hence, fu∗=u∗. This implies that u∗=fu∗=gu∗.
Hence,u∗ is a common fixed point of f and g.
Uniqueness of the common fixed point off and g are shown as follows.
Supposew∗ is another common fixed point of f and g,(19)dbu∗,w∗=dbfu∗,fw∗≤1bu∗,w∗pdbgu∗,gw∗+bqgu∗,fu∗+rdbgw∗,fw∗+sβdbfu∗,gw∗dbfw∗,gu∗λbfu∗,gu∗+βdbfu∗,gw∗+dbfw∗,gu∗≤pdbu∗,w∗+bqu∗,u∗+rdbw∗,w∗+sβdbu∗,w∗dbw∗,u∗λbu∗,u∗+βdbu∗,w∗+dbw∗,u∗=pdbu∗,w∗+sβdbu∗,w∗dbw∗,u∗λbu∗,u∗+βdbu∗,w∗+dbw∗,u∗≤pdbu∗,w∗λ+sdbu∗,w∗.
Thus, we get(20)1−pλ−sdbu∗,w∗≤0.
Sincep/1−s<λ, we have 1−p/λ−s>0 and thus we obtain dbu∗,w∗=0; hence, u∗=w∗.
Hence,f and g have a unique common fixed point in X.
By takinggx=x for all x∈X in Theorem 1, we obtain the following.Corollary 1.
LetX,db be a complete extended rectangular b-metric space, and let f:X⟶X be self-mappings such that(21)bx,ydbfx,fy≤pdbgx,y+qdbx,fx+ry,fy+sβdbfx,ydbfy,xλbfx,x+βdbfx,y+dbfy,x,holds for all x,y∈X, β>0, 0<s<1, 0<p+q+r<λ,p/1−s<λ, and limn,m⟶∞byn,ym<λ−r/p+q, then f has a unique fixed point in X.Example 3.
LetA=2/3,1 and B=0,1/2. Let X=A∪B, and define bx,y=2x+y+1 for all x,y∈X,λ=.9, and define db:X×X⟶0,∞ as follows:(22)dbx,y=0,ifx=y,12,ifx,y∈A,1,ifx,y∈B,x−y2,otherwise.
It is easy thatX,db is an extended rectangular space with bx,y=0.2x+y+1.
Letfx=x4/4 and gx=x2/2, for all x∈X. It is clear that fX⊆gX.
By usingyn=fxn=gxn+1, fx=x4/24, and gx=x2/22, we have(23)y0=fx0=x424=gx1thusx1=x22,y1=fx1=fx22=x828=gx2thusx2=x423,y2=fx2=fx423=x16216=gx2thusx3=x827.
Generally, we have(24)yn=x2n+22n+2,xn=x2n22n−1,limn,m⟶∞byn,ym=limn,m⟶∞2+yn+ym=limn,m⟶∞1+x2n+22n+2+x2m+22m+2=1<λ−rp+q=0.9−0.40.1+0.3=0.50.4=1.25.
Sincex∈0,1, the sequence yn=x2n+2/2n+2 is convergent to 0. Indeed, limn⟶∞dbxn,0=0.(1)
Forx,y∈A=2/3,1 or x,y∈B=0,1/2, it is clear that bx,ydbfx,fy=0(2)
Forx∈A=2/3,1 and y∈B=0,1/2(25)Forx∈A=23,1thenfx=x44∈16324,14⊆B=0,12,Forx∈A=23,1thengx=x22∈418,12⊆B=0,12,Fory∈B=0,12thenfy=y44∈0,132⊆B=0,12,Fory∈B=0,12thengy=y22∈0,14⊆B=0,12,(26)pdbgx,gy+qdbgx,fx+rgy,fy+sβdbfx,gydbfy,gxλbfx,gx+βdbfx,gy+dbfy,gx=0.1dbx2/2,y2/2+0.3dbx2/2,x4/4+0.4y2/2,y4/4+0.50.1dbx4/4,y2/2dby4/4,x2/20.9bx4/4,x2/2+0.1dbx4/4,y2/2+dby4/4,x2/2=0.1+0.3+0.4+0.50.10.90.2x4/4,x2/2+1+0.11+1=0.850.18x4/4,x2/2+1+0.2,bx,ydbfx,fy=0.2x4/4,x2/2+1dbx4/4,y4/4=0.2x4/4,x2/2+1≤0.850.18x4/4,x2/2+1+0.1.
The largest value ofbx,ydbfx,fy occurs if x=1 and y=1/2. So, we have(27)bx,ydbfx,fy=0.2x44+x22+1=0.214+18+1=1.275≤pdbgx,gy+qdbgx,fx+rgy,fy+sβdbfx,gydbfy,gxλbfx,gx+βdbfx,gy+dbfy,gx≤0.850.1811/8+0.2=1.899.(3)
Forx∈B=0,1/2 and y∈A=2/3,1,(28)Forx∈B=0,12thenfx=x44∈0,132⊆B=0,12,Forx∈B=0,12thengx=x22∈0,18⊆B=0,12,Fory∈A=23,1thenfy=y44∈16324,14⊆B=0,12,Fory∈A=23,1thengy=y22∈418,12⊆B=0,12,(29)pdbgx,gy+qdbgx,fx+rgy,fy+sβdbfx,gydbfy,gxλbfx,gx+βdbfx,gy+dbfy,gx=0.1dbx2/2,y2/2+0.3dbx2/2,x4/4+0.4y2/2,y4/4+0.50.1dbx4/4,y2/2dby4/4,x2/20.9bx4/4,x2/2+0.1dbx4/4,y2/2+dby4/4,x2/2=0.1+0.3+0.4+0.50.10.90.2x4/4+x2/2+1+0.11+1=0.850.18x4/4+x2/2+1+0.2,bx,ydbfx,fy=0.2x44+x22+1dbx44,y44=0.2x44+x22+1≤0.850.18x4/4+x2/2+1+0.2.
The largest value ofbx,ydbfx,fy occurs if x=1/2 and y=1. So, we have(30)bx,ydbfx,fy=0.2x44+x22+1dbx44,y44=0.214+18+1=1.275≤pdbgx,gy+qdbgx,fx+rgy,fy+sβdbfx,gydbfy,gxλbfx,gx+βdbfx,gy+dbfy,gx≤0.850.1811/8+0.2=1.899.
Hence, we get that condition(31)bx,ydbfx,fy≤pdbgx,gy+qdbgx,fx+rgy,fy+sβdbfx,gydbfy,gxλbfx,gx+βdbfx,gy+dbfy,gx,is hold.
Hence, from Theorem1, we have x=0 is a unique common fixed point of f and g.Theorem 2.
LetX,db be a complete extended rectangular b-metric space, and let f,g:X⟶X be self-mappings such that fX⊆gX and gXclosed. If the conditions(32)bx,ydbfx,fy≤pdbgx,gy+qbfx,gxdbgx,fx+rdbgy,fy+sdbfx,gydbfy,gx1+dbfx,gy+dbfy,gx,hold for all x,y∈X, where 0<q,r<1,0<p+s<1, q,r<1,0<p+s<1, 0<p+q/1−r<1, and limn,m⟶∞bxn,ym<1−r/p+q, and then f and g have a unique coincidence in X.
Moreover, iff and g have weakly compatible property, then f and g have a unique fixed point.Proof.
Let takingx0∈X as initial point define sequence xn and yn in X such that yn=fxn=gxn+1, n=0,1,2,…. Suppose that yn≠yn+1, n=0,1,2,…, because if yn=yn+1 for some n, then gxn+1=yn=yn+1=fxn+1, so f and g have a coincidence point.
Consider from (1) and (32), we have(33)bxn,xn+1dbyn,yn+1=bxn,xn+1dbfxn,fxn+1≤pdbgxn,gxn+1+qbfxn,gxndbgxn,fxn+rdbgxn+1,fxn+1+sdbfxn,gxn+1dbfxn+1,gxn1+dbfxn,gxn+1+dbfxn+1,gxn=pdbyn−1,yn+qbyn,yn−1dbyn−1,yn+rdbyn,yn+1+sdbyn,yndbyn+,yn−11+dbyn,yn+dbyn+,yn−1≤p+qdbyn−1,yn+rdbyn,yn+1.
Sincebxn,xn+1≥1, we get(34)dbyn,yn+1≤p+qbxn,xn+1−rdbyn−1,yn≤p+q1−rdbyn−1,yn.
So, we get(35)dbyn,yn+1≤βdbyn−1,yn.where β=p+q/1−r. Since as known 0<β=p+q/1−r<1 and limn,m⟶∞byn,ym<1/β, then by using Lemma 2, we get yn is a Cauchy sequence in gX.
SinceX is complete, then there exists u∗∈X such that(36)dbyn,u∗⟶0.
Sinceyn=gxn+1⊆X and gX closed, then u∗∈gX and u∗=gx∗ for some x∗∈X.
Hence, forn,m⟶∞, in (16) we get dbyn,ym⟶0; thus, yn is the Cauchy sequence in gX.
Furthermore, we will show thatf and g have a point coincidence, as follows.
From (1) and (32), we have(37)dbfx∗,gx∗≤bfx∗,gx∗dbfx∗,yn+dbyn,yn+1+dbyn+1,gx∗=bx∗,x∗dbfx∗,fxn+dbyn,yn+1+dbyn+1,u∗≤bfx∗,gx∗1bx∗,xnpdbgx∗,gxn+qbfx∗,gx∗dbgx∗,fx∗+rdbgxn,fxn+sdbfx∗,gxndbfxn,gx∗1+dbfx∗,gxn+dbfxn,gx∗+dbyn,yn+1+dbyn+1,u∗≤bfx∗,gx∗1bx∗,xnpdbu∗,yn−1+qbfx∗,gx∗dbu∗,fx∗+rdbyn−1,yn+sdbfx∗,yn−1dbyn,u∗1+dbfx∗,yn−1+dbyn,u∗+dbyn,yn+1+dbyn+1,u∗≤bfx∗,gx∗1bx∗,xnpdbu∗,yn−1+qbfx∗,gx∗dbu∗,fx∗+rdbyn−1,yn+sdbyn,u∗+dbyn,yn+1+dbyn+1,u∗.
By using (36) and (37), if n⟶∞, then we get(38)dbfx∗,gx∗≤qdbu∗,fx∗bx∗,xn=qdbgx∗,fx∗bx∗,xn≤qdbgx∗,fx∗.
Since0<q<1, this implies that dbfx∗,gx∗=0, so then we get fx∗=gx∗=u∗. Hence, u∗ is a coincidence point of f and g.
To show uniqueness of the coincidence point off and g, we suppose there exists w∗∈X such that fy∗=gy∗=w∗:(39)dbu∗,w∗=dbfx∗,fy∗≤1bx∗,y∗pdbgx∗,gy∗+qbfx∗,gx∗dbgx∗,fx∗+rdbgy∗,fy∗+sdbfx∗,gy∗dbfy∗,gx∗1+dbfx∗,gy∗+dbfy∗,gx∗≤pdbu∗,w∗+qbu∗,u∗dbu∗,u∗+rdbw∗,w∗+sdbu∗,w∗dbw∗,u∗1+dbu∗,w∗+dbw∗,gu∗≤pdbu∗,w∗+qbu∗,u∗dbu∗,u∗+rdbw∗,w∗+sdbu∗,w∗≤pdbu∗,w∗+sdbu∗,w∗=p+sdbu∗,w∗.
Thus, we get(40)1−p−sdbu∗,w∗≤0.
Since0<p+s<1, then we get dbu∗,w∗=0; hence, u∗=w∗.
Next, we have to show thatu∗ is a common fixed point of fandg. Consider that(41)dbu∗,fu∗=dbfx∗,fu∗≤1bx∗,u∗pdbgx∗,gu∗+qbfx∗,gx∗dbgx∗,fx∗+rdbgu∗,fu∗+sdbfx∗,gu∗dbfu∗,gx∗1+dbfx∗,gu∗+dbfu∗,gx∗.
Sincef and g are weakly compatible and fx∗=gx∗=u∗, we have gfx∗=fgx∗=gu∗=fu∗. Thus, from (41) we get(42)dbu∗,fu∗=dbfx∗,fu∗≤1bx∗,u∗pdbgx∗,gu∗+qbfx∗,gx∗dbgx∗,fx∗+rdbgu∗,fu∗+sdbfx∗,gu∗dbfu∗,gx∗1+dbfx∗,gu∗+dbfu∗,gx∗≤1bx∗,u∗pdbu∗,fu∗+qbu∗,u∗dbu∗,u∗+rdbfu∗,fu∗+sdbu∗,fu∗dbfu∗,u∗1+dbu∗,fu∗+dbfu∗,u∗≤pbx∗,u∗dbu∗,fu∗+sdbu∗,fu∗≤pdbu∗,fu∗+sdbu∗,fu∗≤p+sdbu∗,fu∗.
Thus, we obtain1−p−sdbu∗,fu∗≤0.
Sincep+s<1, we obtain that dbu∗,fu∗=0. This implies that u∗=fu∗=gu∗.
Hence,u∗ is a common fixed point of f and g.
Uniqueness of the common fixed point off and g are shown as follows.
Supposew∗ is another common fixed point of f and g, that is, w∗=fw∗=gw∗.(43)dbu∗,w∗=dbfu∗,fw∗≤1bu∗,w∗pdbgu∗,gw∗+qbfu∗,gu∗dbgu∗,fu∗+rdbgw∗,fw∗+sdbfu∗,gw∗dbfw∗,fu∗1+dbfu∗,gw∗+dbfw∗,fu∗≤1bu∗,w∗pdbu∗,w∗+qbu∗,u∗dbu∗,u∗+rdbw∗,w∗+sdbu∗,w∗dbw∗,u∗1+dbu∗,w∗+dbw∗,u∗=pdbu∗,w∗bu∗,w∗+sdbu∗,w∗≤p+sdbu∗,w∗.
Since0<p+s<1, we obtain that dbu∗,w∗=0, and it implies that u∗=w∗.
Hence,f and g have a unique common fixed point in X.
By takinggx=x in Theorem 2 we obtainCorollary 2.
LetX,db be a complete extended rectangular b-metric space, and let f:X⟶X be a self-mapping such that(44)bx,ydbfx,fy≤pdbx,y+qbfx,xdbx,fx+rdby,fy+sdbfx,ydbfy,x1+dbfx,y+dbfy,x,holds for all x,y∈X, where 0<q,r<1,0<p+s<1, q,r<1,0<p+s<1, 0<p+q/1−r<1 and limn,m⟶∞bxn,ym<1−r/p+q, and then f has a unique fixed point in X.Example 4.
We usefx=x4/4,gx=x2/2 and dbx,y in Example 3, and taking bx,y=0.1x+y+1, p=0.2,q=0.3,r=0.45,ands=0.7. These parameters satisfy for conditions(45)0<r<1,0<p+s<1,0<p+q1−r<1,limn,m⟶∞bxn,ym<1−rp+q.
From Example 3 we have(46)yn=x2n+22n+2,xn=x2n22n−1.
Thus, we have(47)limn,m⟶∞byn,ym=limn,m⟶∞0.1yn+ym+1=limn,m⟶∞0.1x2n+22n+2+x2m+22m+2+1=1<1−rp+q=1−0.40.2+0.3=0.60.5=1.2.
Since0≤x≤1, the sequence yn=x2n+2/2n+2 is convergent to 0. Indeed, limn⟶∞dbyn,0=0.(1)
Forx,y∈A=2/3,1 or x,y∈B=0,1/2, it is clear that bx,ydbfx,fy=0(2)
Forx∈A=2/3,1 and y∈B=0,1/2(48)Forx∈A=23,1thenfx=x44∈16324,14⊆B=0,12,Forx∈A=23,1thengx=x22∈418,12⊆B=0,12,Fory∈B=0,12thenfy=y44∈0,132⊆B=0,12,Fory∈B=0,12thengy=y22∈0,14⊆B=0,12.(49)bx,ydbfx,fy≤0.2dbgx,gy+0.3bfx,gxdbgx,fx+0.45dbgy,fy+0.7dbfx,gydbfy,gx1+dbfx,gy+dbfy,gx=0.2dbx22,y44+0.3bx4/4,x2/2dbx2/2,x4/4+0.45dby22,y44+0.7dby4/4,y2/2dby4/4,x2/21+dby4/4,y2/2+dby4/4,x2/2=0.2+0.30.2x4/4+x2/2+1+0.45+0.71+1+1=0.65+0.30.2x4/4+x2/2+1+0.73bx,ydbfx,fy=0.2x+y+1dbx44,y44=0.2x+y+1.
The largest value ofbx,ydbfx,fy occurs if x=1 and y=1/2. So, we have(50)bx,ydbfx,fy=0.1x+y+1dbx44,y44=0.1x+y+1=0.11+12+1=1.1.
And we have(51)2dbgx,gy+0.3bfx,gxdbgx,fx+0.45dbgy,fy+0.7dbfx,gydbfy,gx1+dbfx,gy+dbfy,gx=0.65+0.30.1x4/4+x2/2+1+0.73==0.65+0.30.11/4+1/2+1+0.73=1.17.
So, we get that(52)bx,ydbfx,fy≤2dbgx,gy+0.3bfx,gxdbgx,fx+0.45dbgy,fy+0.7dbfx,gydbfy,gx1+dbfx,gy+dbfy,gx.(3)
Forx∈B=0,1/2 and y∈A=2/3,1
Similarly, we also obtain that(53)bx,ydbfx,fy≤2dbgx,gy+0.3bfx,gxdbgx,fx+0.45dbgy,fy+0.7dbfx,gydbfy,gx1+dbfx,gy+dbfy,gx.
Hence, we get that the condition(54)bx,ydbfx,fy≤pdbx,y+qbfx,xdbx,fx+rdby,fy,+sdbfx,ydbfy,x1+dbfx,y+dbfy,x.for all x,y∈X holds.
Hence, from Theorem2, we can conclude that x=0 is a unique common fixed point of f and g.



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*Source: 2756870-2019-12-01.xml*