# Two New Geraghty Type Contractions inGb-Metric Spaces

Authors: Yixiao Wang; Chunfang Chen
Journal: Journal of Function Spaces
 (2019)

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## Abstract

In this work, two Geraghty type contractions are introduced inGb-metric spaces, and some fixed point theorems about the contractions are proved. At the end of this article, a theorem about unique solution of an integral function is proved.


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## Body

## 1. Introduction

It is well known that fixed point theorem is an important tool for solving many equations in the study of mathematics, such as integral equations [1] and differential equations [2]. It can be applied in several subjects as well, like game theory [3] and economics [4].The notion ofG-metric spaces was introduced by Mustafa and Sims [5] as a generalization of metric spaces. Thereafter, G-metric spaces have been studied and applied to obtain different kinds of fixed point theorems, see [6–12]. Aghajani et al. [13] introduced the notion of Gb-metric spaces based on G-metric spaces and b-metric spaces introduced by Bakhtin in [14]. Also, some further fixed point theorems were studied after Gp-metric spaces, which related to partial metric spaces and G-metric spaces, introduced by Zand and Nezhad in [15]. Ansari et al. [16] proved some common fixed point results in complete Gp-metric spaces with a new approach. More recently, some Geraghty type contraction results were studied in various metric spaces, see [17–19].In this work, we introduce two Geraghty type contractions inGb-metric spaces and investigate some fixed point theorems about such contractions. In [13], Gb-metric space was introduced as follows.Definition 1 (see [13]).
LetX be a nonempty set and s≥1 be a given real number. Suppose that Gb:X×X×X→[0,∞) is a function satisfying the following properties: (1)
Gbu,v,w=0 if u=v=w;(2)
0<Gbu,u,v for all u,v∈X with u≠v;(3)
Gb(u,u,v)≤Gb(u,v,w) for all u,v,w∈X with v≠w;(4)
Gb(u,v,w)=Gb(pu,v,w), where p is a permutation of u,v,w;(5)
Gb(u,v,w)≤s(Gb(u,c,c)+Gb(c,v,w)) for all u,v,w,c∈X.Then the function is called a generalizedb-metric or a Gb-metric on X. The pair (X,Gb) is called a Gb-metric space.It is obvious thatGb-metric space is effectively larger than that of G-metric space. Actually, each G-metric space is a Gb-metric space with s=1.Definition 2 (see [13]).
AGb-metric space is said to be symmetric if Gb(u,v,v)=Gb(v,u,u) for all u,v∈X.Proposition 3 (see [13]).
LetX be a Gb-metric space. Then for each u,v,w,c∈X, it satisfies the following properties: (1)
IfGb(u,v,w)=0, then u=v=w;(2)
Gb(u,v,w)≤s(Gb(u,u,v)+Gb(u,u,w));(3)
Gb(u,v,v)≤2sGb(v,u,u);(4)
Gb(u,v,w)≤s(Gb(u,c,w)+Gb(c,v,w)).In this paper, we denoteN as the set of all positive integers and R as the set of all real numbers.Definition 4 (see [13]).
Let(X,Gb) be a Gb-metric space and {xn} be a sequence in X such that limn,m→∞Gb(x,xn,xm)=0,x∈X, then {xn} is Gb-convergent; that is, for any ε>0, there exists N∈N such that Gb(x,xn,xm)<ε, for all n,m≥N.Proposition 5 (see [13]).
LetX be a Gb-metric space. The following statements are equivalent: (1)
{xn} is Gb-convergent to x;(2)
Gb(xn,xn,x)→0asn→+∞;(3)
Gb(xn,x,x)→0asn→+∞;(4)
Gb(xn,xm,x)→0 as n,m→+∞.Definition 6 (see [13]).
LetX be a Gb-metric space. A sequence {xn} is called a Gb-Cauchy sequence if for any ε>0, there is N∈N such that Gb(xn,xm,xl)<ε for all n,m,l≥N; that is, Gb(xn,xm,xl)→0 as n,m,l→∞.Definition 7 (see [13]).
AGb-metric space X is called Gb-complete if every Gb-Cauchy sequence is Gb-convergent in X.LetA be the family of all functions α:[0,∞)→[0,1) which satisfies the condition limn→∞α(pn)=1 implying limn→∞pn=0.Example 8.
Let(1)α1p=11+p,p>012,p=0,α2p=p2+13p2+1,p>013,p=0.Then α1,α2∈A.LetB be the family of all functions β:[0,∞)→[0,1/s) which satisfies the condition limn→∞β(qn)=1/s implying limn→∞qn=0.Example 9.
Let(2)βq=1se-q,q>01s+1,q=0.Then, β(q)∈B.In [20], Karapınar et al. proved the following result.Theorem 10.
Let(X,σ) be a complete metric-like space and T:X→X be a mapping. Suppose that there exists α∈A such that(3)σTx,Ty≤ασx,yσx,y,for all x,y∈X, then T has a unique fixed point u∈X with σ(u,u)=0.Recently, Aydi et al. [21] introduced a type of Geraghty contraction in metric-like spaces and proved a fixed point theorem about such contraction as follows:Theorem 11.
Let(X,σ) be a complete metric-like space and T:X→X be a mapping. Suppose that there exists α∈A such that(4)σTx,Ty≤αFx,yFx,yfor all x,y∈X, where(5)Fx,y=σx,y+σx,Tx-σy,Ty.Then T has a unique fixed point u∈X with σ(u,u)=0.In our work, enlightened by the preceding works, we introduce the Geraghty contraction to theGb-metric space and prove fixed point theorems for Geraghty type contractions. At the end, we give an application about a unique solution of an integral function.


## 2. Main Results

Theorem 12.
Let(X,Gb) be a complete Gb-metric space and T:X→X be a given mapping. Suppose there exists β∈B such that (6)GbTx,Ty,Ty≤βFx,yFx,y,for all x,y∈X where(7)Fx,y=Gbx,y,y+Gbx,Tx,Tx-Gby,Ty,Ty.Then T has a unique fixed point u∈X.Proof.
Letx0∈X. Define a sequence {xn} in X by xn+1=Txn=Tn+1x0 for all n∈N. Assume that Gb(xn0,xn0+1,xn0+1)=0 for some n0, then x0 is the fixed point of T; the proof is completed. Thus, we assume Gb(xn,xn+1,xn+1)≠0 for all n∈N. From (6), we have(8)0<Gbxn,xn+1,xn+1=GbTxn-1,Txn,Txn≤βFxn-1,xnFxn-1,xn,n≥1,where (9)Fxn-1,xn=Gbxn-1,xn,xn+Gbxn-1,Txn-1,Txn-1-Gbxn,Txn,Txn=Gbxn-1,xn,xn+Gbxn-1,xn,xn-Gbxn,xn+1,xn+1.Take(10)Gbn=Gbxn-1,xn,xn.Then, (8) becomes(11)0<Gbn+1≤βGbn+Gbn-Gbn+1Gbn+Gbn-Gbn+1.Suppose that there exists n0>0 such that Gb(n0)≤Gb(n0+1), then from (11), we have (12)Gbn0+1≤βGbn0+1Gbn0+1<1sGbn0+1,s≥1,which is a contradiction.
Thus, for alln>0, Gb(n)>Gb(n+1). From (11), we have(13)0<Gbn+1≤β2Gbn-Gbn+12Gbn-Gbn+1.The real sequence {Gb(n)} is decreasing; suppose there exists γ≥0 such that limn→∞Gb(n)=γ. Assume that γ>0. Taking n→∞ in (13), we get(14)γ=limn→∞Gbn+1≤limn→∞β2Gbn-Gbn+12Gbn-Gbn+1.There are two situations that need to be discussed.
(A) When s=1, then (14) becomes (15)γ=limn→∞Gbn+1≤limn→∞β2Gbn-Gbn+12Gbn-Gbn+1≤1slimn→∞2Gbn-Gbn+1=1sγ=γ.We can obtain(16)limn→∞β2Gbn-Gbn+1=1.Since β∈B, we get(17)γ=limn→∞2Gbn-Gbn+1=0,which is a contradiction.
(B) When s>1, according to (14) we have (18)γ=limn→∞Gbn+1≤limn→∞β2Gbn-Gbn+12Gbn-Gbn+1≤1slimn→∞2Gbn-Gbn+1=1sγ,s>1,which is a contradiction.
In conclusion of the above two conditions, we haveγ=0, that is(19)limn→∞Gbxn,xn+1,xn+1=0.Note that (20)Gbxn+1,xn,xn≤2sGbxn,xn+1,xn+1,thus, we get (21)limn→∞Gbxn+1,xn,xn=0.We shall prove that {xn} is a Cauchy sequence in (X,Gb). Equation(22)limn,m→∞Gbxn,xm,xm=0,m>n,will be proved.
Suppose (22) does not hold. Then there exists ε>0 for which we can find subsequences {xmi} and {xni} of {xn} with mi>ni>i such that for every i(23)Gbxni,xmi,xmi≥ε.Corresponding to ni, we can find mi with the smallest index and mi>ni and satisfying (23), then(24)Gbxni,xmi-1,xmi-1<ε.By (23) and (24), we have(25)ε≤Gbxni,xmi,xmi≤sGbxni,xni+1,xni+1+Gbxni+1,xmi,xmi.Taking i→∞ in (25) and using (19), we obtain(26)ε≤liminfi→∞Gbxni,xmi,xmi≤sliminfi→∞Gbxni+1,xmi,xmi.Back to (6), there is (27)Gbxni+1,xmi,xmi=GbTxni,Txmi-1,Txmi-1≤βFxni,xmi-1Fxni,xmi-1=βGbxni,xmi-1,xmi-1+Gbxni,xni+1,xni+1-Gbxmi-1,xmi,xmi·Gbxni,xmi-1,xmi-1+Gbxni,xni+1,xni+1-Gbxmi-1,xmi,xmi.Let (28)M=Gbxni,xmi-1,xmi-1+Gbxni,xni+1,xni+1-Gbxmi-1,xmi,xmi.Taking i→∞ in the above inequations and by (19), (24), we have (29)liminfi→∞Gbxni+1,xmi,xmi≤εliminfi→∞βM.Then we obtain(30)1εliminfi→∞Gbxni+1,xmi,xmi≤liminfi→∞βM.From (26) and (30), we get (31)1s=1ε·εs≤1εliminfi→∞Gbxni+1,xmi,xmi≤liminfi→∞βM≤limsupi→∞βM≤1s.We deduce that(32)limi→∞βM=1s.Since β∈B and by (19), then(33)limi→∞M=limi→∞Gbxni,xmi-1,xmi-1+Gbxni,xni+1,xni+1-Gbxmi-1,xmi,xmi=limi→∞Gbxni,xmi-1,xmi-1=0.From (19) and (33), we have (34)ε≤Gbxni,xmi,xmi≤sGbxni,xmi-1,xmi-1+Gbxmi-1,xmi,xmi→0asi→∞,which is a contradiction. Thus(35)limn,m→∞Gbxn,xm,xm=0,and {xn} is a Cauchy sequence in complete Gb-metric space. So there exists u∈X, such that(36)limn→∞Gbxn,u,u=0.By Proposition 3, we have Gb(u,xn,xn)≤2sGb(xn,u,u); therefore by (36) we get(37)limn→∞Gbu,xn,xn=0.We shall prove that u is a fixed point of T. Assume that u≠Tu, then Gb(u,Tu,Tu)>0. From (6) and (7), we have(38)Gbxn+1,Tu,Tu=GbTxn,Tu,Tu≤βFxn,uFxn,u<1sFxn,u,where (39)Fxn,u=Gbxn,u,u+Gbxn,xn+1,xn+1-Gbu,Tu,Tu.We also have (40)Gbu,Tu,Tu≤sGbu,Txn,Txn+GbTxn,Tu,Tu=sGbu,xn+1,xn+1+sGbxn+1,Tu,Tu≤sGbu,xn+1,xn+1+sβFxn,uFxn,u<sGbu,xn+1,xn+1+Fxn,u.Taking n→∞ in the above inequation and using (36), (37), and (38) obtains (41)Gbu,Tu,Tu≤slimn→∞Gbu,xn+1,xn+1+slimn→∞βFxn,uFxn,u≤slimn→∞Gbu,xn+1,xn+1+limn→∞Fxn,u≤0+limn→∞Gbxn,u,u+limn→∞Gbxn,xn+1,xn+1-limn→∞Gbu,Tu,Tu≤Gbu,Tu,Tu,which implies(42)limn→∞sβFxn,u=1.Since β∈B, then(43)limn→∞Fxn,u=0,which is a contradiction. Thus Gb(u,Tu,Tu)=0 and so u=Tu. Consequently, u is a fixed point of T.
We shall prove that suchu is the unique fixed point of T. We argue by contradiction. Assume there exists v,v≠u such that v=Tv. We have(44)Fu,v=Gbu,v,v+Gbu,u,u-Gbv,v,v=Gbu,v,v.From (6), we have (45)0<Gbu,v,v=GbTu,Tv,Tv≤βFu,vFu,v=βGbu,v,vGbu,v,v<1sGbu,v,v,which is a contradiction. Thus, T has a unique fixed point.We now present the following result.Theorem 13.
Let(X,Gb) be a complete Gb-metric space and T:X→X be a given mapping. Suppose there exists α∈A such that(46)GbTx,Ty,Ty≤αFx,yFx,yfor all x,y∈X where(47)Fx,y=1s2Gbx,y,y+Gbx,Tx,Tx-Gby,Ty,Ty.Then T has a unique fixed point u∗∈X.Proof.
Letx0∈X. Define a sequence {xn} in X by xn+1=Txn=Tn+1x0. Assume that Gb(xn,xn+1,xn+1)=0 for n=n0, that is, Gb(xn0,xn0+1,xn0+1)=0, then xn0=xn0+1, i.e., Txn0=xn0. Therefore xn0 is a fixed point of T.
SupposeGb(xn,xn+1,xn+1)≠0 for all n∈N. From (46) we have(48)0<Gbxn,xn+1,xn+1=GbTxn-1,Txn,Txn≤αFxn-1,xnFxn-1,xn,where (49)Fxn-1,xn=1s2Gbxn-1,xn,xn+Gbxn-1,Txn-1,Txn-1-Gbxn,Txn,Txn=1s2Gbxn-1,xn,xn+Gbxn-1,xn,xn-Gbxn,xn+1,xn+1.Take(50)Gbn∗=Gbxn-1,xn,xn.Then (48) becomes(51)Gbn+1∗≤α1s2Gbn∗+Gbn∗-Gbn+1∗·1s2Gbn∗+Gbn∗-Gbn+1∗.Suppose there exists n0>0 such that Gb(n0)∗≤Gb(n0+1)∗. From (51), we have (52)Gbn0+1∗≤α1s2Gbn0+1∗·1s2Gbn0+1∗<1s2Gbn0+1∗,which is a contradiction. Thus Gb(n)∗>Gb(n+1)∗ for all n>0, and the real sequence {Gb(n)∗} is decreasing.
Suppose there existsδ∈R such that limn→∞Gb(n)∗=δ. Now we shall prove that(53)δ=0.Applying (51), we get(54)Gbn+1∗≤α1s22Gbn∗-Gbn+1∗·1s22Gbn∗-Gbn+1∗<α1s22Gbn∗-Gbn+1∗·2Gbn∗-Gbn+1∗≤2Gbn∗-Gbn+1∗.Take n→∞ in (54) to write (55)δ=limn→∞Gbn+1∗≤limn→∞α1s22Gbn∗-Gbn+1∗·limn→∞2Gbn∗-Gbn+1∗≤limn→∞2Gbn∗-Gbn+1∗=δ.We obtain(56)limn→∞α1s22Gbn∗-Gbn+1∗=1.Since α∈A, then(57)limn→∞1s22Gbn∗-Gbn+1∗=1s2·δ=0.Thus(58)δ=0.That is(59)limn→∞Gbxn,xn+1,xn+1=0.Note that(60)Gbxn+1,xn,xn≤2sGbxn,xn+1,xn+1,then we can get(61)limn→∞Gbxn+1,xn,xn=0.Now we shall prove that {xn} is a Cauchy sequence in Gb-metric spaces. We will prove that(62)limn,m→∞Gbxn,xm,xm=0,m>n.Suppose (62) does not hold, then there exists ε>0 for which we can find subsequences {xmi} and {xni} of {xn} with mi>ni>i such that(63)Gbxni,xmi,xmi≥ε.Corresponding to ni, we can find mi with the smallest index satisfying (63) and mi>ni. That is(64)Gbxni,xmi-1,xmi-1<ε.From (46) and (63), we have(65)ε≤Gbxni,xmi,xmi=GbTxni-1,Txmi-1,Txmi-1≤αFxni-1,xmi-1·Fxni-1,xmi-1<Fxni-1,xmi-1,where(66)Fxni-1,xmi-1=1s2Gbxni-1,xmi-1,xmi-1+Gbxni-1,xni,xni-Gbxmi-1,xmi,xmi.From (65), (66), and applying (59), we have(67)ε≤liminfi→∞Fxni-1,xmi-1=1s2liminfi→∞Gbxni-1,xmi-1,xmi-1.On the other hand, applying (64) we have (68)Gbxni-1,xmi-1,xmi-1≤sGbxni-1,xni,xni+sGbxni,xmi-1,xmi-1<sGbxni-1,xni,xni+sε.Taking n→∞ in the above inequations, we have(69)limsupi→∞Gbxni-1,xmi-1,xmi-1≤sε.From (67) and (69) we obtain (70)ε≤1s2liminfi→∞Gbxni-1,xmi-1,xmi-1≤1s2limsupi→∞Gbxni-1,xmi-1,xmi-1≤1s2·sε=εs,which is a contradiction with s>1.
We shall consider the situation which withs=1. From (63) and (64) we have(71)ε≤Gbxni,xmi,xmi≤Gbxni,xmi-1,xmi-1+Gbxmi-1,xmi,xmi<ε+Gbxmi-1,xmi,xmi.Taking n→∞ in (71) and by (59) we have(72)limi→∞Gbxni,xmi,xmi=ε.In the meantime, (73)Gbxni-1,xmi-1,xmi-1-Gbxni,xmi,xmi≤Gbxni-1,xni,xni+Gbxni,xmi-1,xmi-1-Gbxni,xmi-1,xmi-1-Gbxmi-1,xmi,xmi=Gbxni-1,xni,xni+Gbxmi-1,xmi,xmi.By taking n→∞ and using (59) in the above inequation, we can deduce that(74)limi→∞Gbxni-1,xmi-1,xmi-1=ε.By (46), we have(75)ε≤Gbxni,xmi,xmi=GbTxni-1,Txmi-1,Txmi-1≤αFxni-1,xmi-1Fxni-1,xmi-1<Fxni-1,xmi-1,where (76)Fxni-1,xmi-1=Gbxni-1,xmi-1,xmi-1+Gbxni-1,xni,xni-Gbxmi-1,xmi,xmi.By using (59), we obtain(77)limi→∞Fxni-1,xmi-1=ε.Taking i→∞ in (75), we can deduce that(78)limi→∞αFxni-1,xmi-1=1.Since α∈A, we have(79)limi→∞Fxni-1,xmi-1=0,which is a contradiction. In conclusion of two situations, {xn} is a Cauchy sequence in complete Gb-metric spaces. So there exists u∗∈X such that(80)limn→∞Gbxn,u∗,u∗=0.We can confirm that Gb(u∗,Tu∗,Tu∗)=0. In fact, if Gb(u∗,Tu∗,Tu∗)≠0, from (46) and (47) we have (81)Gbxn+1,Tu∗,Tu∗=GbTxn,Tu∗,Tu∗≤αFxn,u∗·Fxn,u∗<Fxn,u∗,where (82)Fxn,u∗=1s2Gbxn,u∗,u∗+Gbxn,Txn,Txn-Gbu∗,Tu∗,Tu∗=1s2Gbxn,u∗,u∗+Gbxn,xn+1,xn+1-Gbu∗,Tu∗,Tu∗→1s2Gbu∗,Tu∗,Tu∗asn→∞.We also have (83)Gbu∗,Tu∗,Tu∗≤sGbu∗,Txn,Txn+sGbTxn,Tu∗,Tu∗≤sGbu∗,xn+1,xn+1+sαFxn,u∗Fxn,u∗<sGbu∗,xn+1,xn+1+sFxn,u∗.Taking n→∞ in the above inequation and using (80), we obtain(84)Gbu∗,Tu∗,Tu∗≤limn→∞Gbu∗,xn+1,xn+1+slimn→∞αFxn,u∗Fxn,u∗≤slimn→∞Fxn,u∗=1sGbu∗,Tu∗,Tu∗.When s=1, we have(85)limn→∞αFxn,u∗=1.Since α∈A, then(86)limn→∞Fxn,u∗=Gbu∗,Tu∗,Tu∗=0,which is a contradiction.
Whens>1, we get a contradiction from (84).
Thus, in conclusion of two situations,Gb(u∗,Tu∗,Tu∗)=0 and so Tu∗=u∗. u∗ is a fixed point of T. We shall prove that such u∗ is the unique fixed point of T. We argue by contradiction. Assume there exists v∗,v∗≠u∗ such that(87)v∗=Tv∗.We have(88)Fu∗,v∗=Gbu∗,v∗,v∗+Gbu∗,u∗,u∗-Gbv∗,v∗,v∗=Gbu∗,v∗,v∗.From (46), we have (89)0<Gbu∗,v∗,v∗=GbTu∗,Tv∗,Tv∗≤αFu∗,v∗Fu∗,v∗=α1s2Gbu∗,v∗,v∗1s2Gbu∗,v∗,v∗<1s2Gbu∗,v∗,v∗,which is a contradiction. Thus, there exists a unique fixed point u∗∈X such that u∗=Tu∗.


## 3. Application

LetX=C([0,1],R) be the set of real continuous functions defined on [0,1]. Take the Gb-metric Gb:X×X×X→[0,∞) given by (90)Gbx,y,z=supt∈0,1xt-yt+supt∈0,1xt-zt+supt∈0,1yt-zt2,for all x,y∈X. Then (X,Gb) is Gb-metric spaces with s≥1. Consider the following integral equation(91)xt=Pt+∫01St,ufu,xudu,t∈0,1,where f:[0,1]×R→R and P:[0,1]→R are two continuous functions and S:[0,1]×[0,1]→[0,∞) is a function such that S(t,.)∈L1([0,1]) for all t∈[0,1]. Consider the operator T:X→X defined by(92)Txt=Pt+∫01St,ufu,xudu,t∈0,1.Theorem 14.
Suppose that the following conditions are satisfied:(1)
there existsη:X×X→[0,∞) for all u∈[0,1](93)0≤fx,xu-fu,yu≤ηx,yxu-yu∀x,y∈X,(2)
there existsβ:[0,∞)→[0,1/s) such that (94)supt∈0,1∫01St,uηx,ydu≤β2supt∈0,1J2+2supt∈0,1K2-2supt∈0,1L2
where(95)J=x-y,K=x-Tx,L=y-Ty.Then the integral equation (91) has a unique solution in X.Proof.
It is clear that any fixed point of (92) is a solution of (91). By conditions (1) and (2), we get (96)GbTxt,Tyt,Tyt=2supt∈0,1Txt-Tyt2=2supt∈0,1∫01St,ufu,xudu-∫01St,ufu,yudu2=2supt∈0,1∫01St,ufu,xu-fu,yudu)2≤2supt∈0,1∫01St,uηx,yxu-yudu2=2supt∈0,1∫01St,uηx,y·122xu-yu21/2du2≤Fx,ysupt∈0,1∫01St,uηx,ydu2≤Fx,y·βFx,ywhere (97)Fx,y=2supt∈0,1J2+2supt∈0,1K2-2supt∈0,1L2,with (98)J=x-y,K=x-Tx,L=y-Ty.Then for all x,y∈X we obtain (99)GbTx,Ty,Ty≤βFx,yFx,y.This implies that Theorem 12 holds. Thus the operator T has a unique fixed point; that is, the integral function has a unique solution in X.


## 4. Conclusion

In this paper, we present some fixed point theorems about two new Geraghty type contractions in the setting ofGb-metric spaces.In the third section, we study an application about the unique solution of a class of integral functions to illustrate our fixed point theorems.



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