# Remarks of Fixed Point for (ψ,φ)-Weakly Contractive Mappings

Authors: Zhiqun Xue; Guiwen Lv
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we obtain a new fixed point theorem for weak contraction mappings on complete generalized metric spaces. Our result is different from that of (ψ,φ)-weakly contractive mappings which had been obtained by Lakzian–Samet’s [Appl. Math. Lett. 25 (2012) 902–906]. Moreover, an example is given to illustrate the usability of the obtained result.


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## Body

## 1. Introduction

Fixed point theory of metric spaces is a classical subject. It is playing an important role in the theory of function analysis by reason of its wide application. Over the past decades, there have been many authors who have studied heavily in the existence of fixed points of noncontractive mappings from metric spaces to generalized metric spaces (see [1–8]). Two very typical familiar and important generalizations of the Banach contraction principle obtained by Dutta and Choudhury [9] and Lakzian and Samet [4] are as follows.Theorem 1 (see [9]).
LetX,d be a complete metric space and let T:X⟶X be a self-mapping satisfying the inequality:(1)ψdTx,Ty≤ψdx,y−φdx,y,x,y∈X,where ψ,φ:0,+∞⟶0,+∞ are both continuous and monotone nondecreasing functions with ψt=φt=0 if and only if t=0. Then, T has a unique fixed point.Theorem 2 (see [4]).
LetX,d be a Hausdorff and complete generalized metric space and let T:X⟶X be a self-mapping satisfying (1), where (i) ψ is the same as Theorem 1 and (ii) φ is only continuous function with φt=0 if and only if t=0. Then, T has a unique fixed point.It is worth noting thatψ:0,+∞⟶0,+∞ is continuous monotone nondecreasing function in above two theorems, but we cannot obtain that t1≤t2 if ψt1≤ψt2. Indeed, the false assertion had been widely applied to the proving process of [4, 9].The aim of this paper is to point out problems of [4, 9] and further correct and improve them. Before presenting the main results, let us first recall some definitions of the past (see [1]).Definition 1.
LetX be a nonempty set and d:X×X⟶0,+∞ be a mapping such that, for all x,y,z∈X and for all distinct points u,v∈X, each of them different from x and y satisfy the following conditions:d1dx,y=0 if and only if x=y. d2dx,y=dy,x. d3dx,y≤dx,u+du,v+dv,y (the rectangular inequality).
Then,d is called a generalized metric and X,d is called a generalized metric space (or shortly g.m.s).
Every metric space is a generalized metric space, and the converse is not true in general.Definition 2.
LetX,d be a g.m.s, xn be a sequence in X and x∈X. We call that the following:(i)
xn is convergent to x if dxn,x⟶0 as n⟶∞, and we denote this by xn⟶x.(ii)
xn is a Cauchy sequence if for each ϵ>0, there exists a natural number N such that dxn,xm<ϵ for all m,n>N.(iii)
X is complete if every Cauchy sequence is convergent in X.


## 2. The Main Results

In the sequel, we denote byΨ the set of functions ψ:0,+∞⟶0,+∞ satisfying the following: a1ψ is monotone nondecreasing, a2limt⟶rψt>0 for all r>0, a3ψt=0 if and only if t=0.We denote byΦ the set of functions φ:0,+∞⟶0,+∞ satisfying the following:b1limt⟶rinfφt>0 for all r>0, b2φt=0 if and only if t=0.Theorem 3.
LetX,d be a complete g.m.s and T:X⟶X be a self-mapping satisfying the following:(2)ψdTx,Ty≤ψdx,y−φdx,y,for all x,y∈X, where ψ∈Ψ, φ∈Φ. Then, T has a unique fixed point.Proof.
First, we show that ifx∗ is the fixed point of T in X, then it is unique. Otherwise, there exists y∗∈X with Ty∗=y∗≠x∗=Tx∗, then(3)ψdx∗,y∗=ψdTx∗,Ty∗≤ψdx∗,y∗−φdx∗,y∗<ψdx∗,y∗,is a contradiction and so y∗=x∗.
Letx0∈X be an arbitrary point, and we construct a sequence xn in X by xn+1=Txn for any positive integers n.
Second, we show thatT has a fixed point in X.Case 1.
T has a periodic point.(i)
Ifxn+1=xn for some natural number n, then Txn=xn, i.e., xn is the fixed point of T, and our proof is complete. Next, we assume that xn+1≠xn for all n.(ii)
If there exists some natural numbern such that xn+2=xn, by (2), we have(4)ψdxn,Txn=ψdT2xn,T3xn≤ψdTxn,T2xn−φdTxn,T2xn<ψdTxn,xn,is a contradiction. Thus,xn+2≠xn for all n.(iii)
If there are two natural numbersm,n such that xm=xn with m−n>2, again using (2), we get(5)ψdxn,Txn=ψdTm−nxn,Tm−n+1xn≤ψdTm−n−1xn,Tm−nxn−φdTm−n−1xn,Tm−nxn<ψdTm−n−1xn,Tm−nxn≤ψdTm−n−2xn,Tm−n−1xn−φdTm−n−2xn,Tm−n−1xn≤ψdTm−n−2xn,Tm−n−1xn⋯≤ψdxn,Txn,is a contradiction. Hence, xm≠xn for all m≠n.Case 2.
T has no periodic point, i.e., xn≠xm for all n≠m.Step 1.
We prove thatlimn⟶∞dxn+1,xn=0. Take x=xn,y=xn−1 in (2), and we have(6)ψdxn+1,xn=ψdTxn,Txn−1≤ψdxn,xn−1−φdxn,xn−1<ψdxn,xn−1.It shows that the sequence ψdxn+1,xn is monotone decreasing, and there exists nonnegative number r such that(7)limn⟶∞ψdxn+1,xn=r.
Hence, from (6),(8)φdxn,xn−1≤ψdxn,xn−1−ψdxn+1,xn.
Taking the lower limit asn⟶∞ in the above inequality, we have(9)limn⟶∞infφdxn,xn−1=0.
On the other hand, (6) implies that(10)dxn+1,xn<dxn,xn−1,for all n≥0. It follows that dxn,xn−1 is monotone decreasing, and consequently, there exists s≥0 such that(11)limn⟶∞dxn,xn−1=s.
Then,s=0. If it is not true, then s>0. By b2, we obtain that(12)limn⟶∞infφdxn,xn−1>0is a contradiction with (9). Therefore, limn⟶∞dxn+1,xn=0.Step 2.
We prove thatlimn⟶∞dxn+2,xn=0. Again, take x=xn+1,y=xn−1 in (2), and then(13)ψdxn+2,xn=ψdTxn+1,Txn−1≤ψdxn+1,xn−1−φdxn+1,xn−1<ψdxn+1,xn−1.
It follows thatψdxn+2,xn is monotone decreasing and so there exists R≥0 such that(14)limn⟶∞ψdxn+2,xn=R.
Take lower limits asn⟶∞ on each side of above inequality (13):(15)limn⟶∞infφdxn+1,xn−1=0.
(13) implies that(16)dxn+2,xn<dxn+1,xn−1,and then dxn+1,xn−1 is a monotone decreasing sequence, and there exists S≥0 such that(17)limn⟶∞dxn+1,xn−1=S.
Suppose thatS>0, then we obtain from b2 that(18)limn⟶∞infφdxn+1,xn−1>0,is a contradiction with (15) and so limn⟶∞dxn+2,xn=0.Step 3.
We prove thatlimn⟶∞dxm,xn=0 for m>n with m−n>2.
Suppose thatxn is a sequence of distinct points, i.e., xm≠xn for all m≠n. We claim that xn is a Cauchy sequence in X. Suppose to the contrary, there exists ϵ>0 for which we can find subsequences xmk and xnk of xn such that(19)dxmk,xnk≥ϵ,for mk>nk>k with mk is the smallest index. So, we have(20)dxnk,xmk−1<ϵ,for any k≥1. Applying the rectangular inequality, we obtain that(21)ϵ≤dxnk,xmk≤dxnk,xmk−1+dxmk−1,xmk−2+dxmk−2,xmk<ϵ+dxmk−1,xmk−2+dxmk−2,xmk,and therefore, dxnk,xmk⟶ε as k⟶∞. Since(22)dxnk,xmk−dxnk,xnk−1−dxmk−1,xmk≤dxnk−1,xmk−1≤dxnk−1,xnk+dxnk,xmk+dxmk,xmk−1,then dxnk−1,xmk−1⟶ϵ as k⟶∞, which implies(23)∗limk⟶∞infφdxnk−1,xmk−1>0.
Again applying (2), (24)ψdxnk,xmk=ψdTxnk−1,Txmk−1≤ψdxnk−1,xmk−1−φdxnk−1,xmk−1.
Taking below limits ask⟶∞ on either side of the above inequality, (24) yields(25)limk⟶∞infφdxnk−1xmk−1=0,is a contradiction with ∗. Hence, xn is a Cauchy sequence. Since X,d is complete, there exists q∈X such that limn⟶∞xn=q.Step 4.
Let us show thatq is a fixed point of T. If it is not the case, q is not the fixed point of T, i.e., dq,Tq>0. Since(26)dq,Tq−dq,xn−dxn,xn+1≤dTq,Txn≤dTq,q+dq,xn+dxn,xn+1,then(27)limn⟶∞dTq,Txn=dTq,q.
In view of (2), we get(28)ψdTq,Txn≤ψdq,xn−φdq,xn<ψdq,xn.
It implies that(29)dTq,xn+1<dq,xn⟶0,as n⟶∞. This is a contradiction and so q=Tq.Example 1.
(see [5]). Let X=A∪B, where A=1/2,1/3,1/4,1/5 and B=1,2. Define the generalized metric on X as follows:(30)dx,y=dy,x,x,y∈X,dx,y=0,x,y∈Xwithx=y,x−y,x,y∈Borx∈A,y∈B.d12,13=d14,15=0.3,d12,15=0.25,d13,14=0.2,d12,14=d15,13=0.6.Then, X,d is a Branciari-type generalized metric space, but it is not a metric space. In fact,(31)0.6=d12,14>d12,13+d13,14=0.5,and let T:X⟶X be defined by(32)Tx=15,x∈1,2,14,x∈12,13,14,13,x=15.
Defineψt=t,φt=t/6,t∈0,+∞. Then, T satisfies(33)ψdTx,Ty≤ψdx,y−φdx,y,for all x,y∈X, and T has a unique fixed point x=1/4. Indeed, the following holds:(i)
Ifx,y∈A, then(34)ψT12,dT13=ψd14,14=0≤0.25=ψd12,13−φd12,13,ψdT12,T14=ψd14,14=0≤0.5=ψd12,14−φd12,14,ψdT12,T15=ψd14,13=0.2<1.256=ψd12,15−φd12,15,ψdT13,T14=ψd14,14=0<0.53=ψd13,14−φd13,14,ψdT13,T15=ψd14,13=0.2≤0.5=ψd13,15−φd13,15,ψdT14,T15=ψd14,13=0.2≤0.25=ψd14,15−φd14,15.(ii)
Ifx,y∈B, then(35)ψdTx,Ty=ψd15,15=0≤56x−y=ψdx,y−φdx,y.(iii)
Ifx∈A,y∈B, i.e., 1≤y≤2, then(36)ψdT12,Ty=ψd14,15=0.3≤56y−12=ψd12,y−φd12,y,ψdT13,Ty=ψd14,15=0.3≤56y−13=ψd13,y−φd13,y,ψdT14,Ty=ψd14,15=0.3≤56y−14=ψd14,y−φd14,y,ψdT15,Ty=ψd13,15=0.6≤56y−15=ψd15,y−φd15,y.
If we denote thatψ−φ=ϕ in Theorem 3, the result is as follows.Theorem 4.
LetX,d be a complete g.m.s and T:X⟶X be a self-mapping such that(37)ψdTx,Ty≤ϕdx,y,for all x,y∈X, where ψ∈Ψ, and ψ and ϕ satisfy the following: ϕ1ψt>ϕt for all t>0. ϕ2limt⟶rψt>limt⟶rinfϕt>0 for all r>0. ϕ3ϕt=0 if and only if t=0.
Then,T has a unique fixed point.Remark 1.
Theorem 3 extends and improves in the following senses:(1)
We point out the problems of theorems proving in references [4, 9].(2)
The assumptions that the continuity of the functionsψ and φ in [4] have been replaced by corresponding conditions limt⟶rψt>0 and limt⟶rinfφt>0 for each r>0, respectively.(3)
The condition of Hausdorff in [4] is unnecessary. It has been removed.(4)
Our results are new in Branciari-type generalized metric spaces, and the proof techniques are simple and also different from literature [4].(5)
The conditionlimt⟶0+ψt=0 in [8] is superfluous.



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