# Caristi’s Fixed Point Theorem and Subrahmanyam’s Fixed Point Theorem inν-Generalized Metric Spaces

Authors: Badriah Alamri; Tomonari Suzuki; Liaqat Ali Khan
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We discuss the completeness ofν-generalized metric spaces
in the sense of Branciari.
We also prove generalizations
of Subrahmanyam’s and Caristi’s fixed point theorem.


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## Body

## 1. Introduction

In 2000, Branciari introduced the following very interesting concept.Definition 1 (Branciari [1]).
LetX be a set, let d be a function from X×X into [0,∞), and let ν∈N. Then, (X,d) is said to be a ν-generalized metric space if the following hold.(N1)
d(x,y)=0 iff x=y for any x,y∈X.(N2)
d(x,y)=d(y,x) for any x,y∈X.(N3)
d(x,y)≤d(x,u1)+d(u1,u2)+⋯+d(uν-1,uν)+d(uν,y) for any x,u1,u2,…,uν,y∈X such that x,u1,u2,…,uν,y are all different.Remark 2.
Every metric space(X,d) is a 1-generalized metric space.Branciari in [1] also proved a generalization of the Banach contraction principle; however, the proof is not correct; see [2–4]. When proving theorems on ν-generalized metric spaces X, we may have to be careful because X does not necessarily have the compatible topology; see [4]. See also [5, 6].In this paper, we first discuss the completeness ofν-generalized metric spaces. We next prove generalizations of Subrahmanyam’s and Caristi’s fixed point theorem.


## 2. Preliminaries

Throughout this paper we denote byN the set of all positive integers and by R the set of all real numbers.In this section, we give some preliminaries.As we mentioned in Section1, ν-generalized metric spaces do not necessarily have the compatible topology. So we have to define something connected with convergence.Definition 3.
Let(X,d) be a ν-generalized metric space. (i)
A sequence{xn} in X is said toconverge to x iff limnd(x,xn)=0.(ii)
A sequence{xn} in X is said toconverge only to x iff limnd(x,xn)=0 holds and limnd(y,xn)=0 does not hold for y∈X∖{x}.(iii)
A mappingT on X is said to besequentially continuous iff {Txn} converges to Tx whenever {xn} converges to x.(iv)
A functionf from X into (-∞,∞] is said to besequentially lower semicontinuous iff f(x)≤liminf⁡nf(xn) whenever {xn} converges to x.We introduce new concepts.Definition 4.
Let(X,d) be a ν-generalized metric space and let κ∈N. (i)
A sequence{xn} in X is said to be κ-Cauchy iff (1)limn→∞sup⁡{d(xn,xn+1+jκ):j=0,1,2,…}=0holds.(ii)
X is κ-complete iff every κ-Cauchy sequence converges.Remark 5.
We sometimes write “Cauchy” instead of “1-Cauchy” and “complete” instead of “1-complete.”The following is obvious.Proposition 6.
Let(X,d) be a ν-generalized metric space and let κ,λ∈N such that λ is divisible by κ. Then, the following hold.(i)
Everyκ-Cauchy sequence is λ-Cauchy.(ii)
IfX is λ-complete, then X is κ-complete.The following are partially converse to Proposition6(i).Proposition 7.
Let(X,d) be a ν-generalized metric space where ν is odd. Let {xn} be a ν-Cauchy sequence such that xn are all different. Then, {xn} is Cauchy.Proof.
In the case whereν=1, the conclusion clearly holds. So we assume ν≥3. Fix ε>0. Then, there exists l∈N such that (2)dxn,xn+1+jν<εfor any j∈N∪{0} and n∈N with n≥l. Fix j∈N∪{0} and n∈N with n≥l. We first show(3)dxn,xn+1+jν+2k<kν+1εfor k=0,1,…,(ν-1)/2. It is obvious that (3) holds when k=0. We assume that (3) holds for some k with 0≤k<(ν-1)/2. Then, we have by (N3) (4)dxn,xn+1+jν+2k+1≤dxn,xn+1+jν+2k+dxn+1+jν+2k,xn+2+j+1ν+2k+∑i=n+1+jν+2(k+1)n+1+(j+1)ν+2k‍dxi,xi+1≤(kν+1)ε+ε+(ν-1)ε=((k+1)ν+1)ε.Hence, (3) holds when k≔k+1. Therefore, (3) holds for every k, which implies (5)dxn,xn+1+jν+2k<ν22-ν2+1εfor any j∈N∪{0}, k=0,1,…,(ν-1)/2, and n∈N with n≥l. Using this, we have (6)dxn,xn+1+jν+2k+1≤dxn,xn+1+jν+2k+dxn+1+jν+2k,xn+2+jν+2k+ν-1+∑i=n+2+jν+2kn+(j+1)ν+2k‍d(xi,xi+1)<2ν22-ν2+1ε+(ν-1)ε=ν2+1ε,for any j∈N∪{0}, k=0,1,…,(ν-3)/2, and n∈N with n≥l. So {xn} is Cauchy.Proposition 8.
Let(X,d) be a ν-generalized metric space where ν is even. Let {xn} be a ν-Cauchy sequence such that xn are all different. Then, {xn} is 2-Cauchy.Proof.
Fixε>0. Then, there exists l∈N such that (7)dxn,xn+1+jν<εfor any j∈N∪{0} and n∈N with n≥l. Fix j∈N∪{0} and n∈N with n≥l. Then, as in the proof of Proposition 7, by induction, we can show (8)dxn,xn+1+jν+2k<kν+1εfor k=0,1,…,ν/2-1. Therefore, we obtain (9)dxn,xn+1+2j<ν22-ν+1εfor any j∈N∪{0}. So {xn} is 2-Cauchy.Lemma 9.
Let(X,d) be a ν-generalized metric space and let {xn} be a sequence in X such that xn are all different and ∑n=1∞‍d(xn,xn+1)<∞. Then, {xn} is ν-Cauchy.Remark 10.
Example 1 in [7] tells that there exists some sequence {xn} in a 2-generalized metric space such that ∑n=1∞‍d(xn,xn+1)<∞ and {xn} is not Cauchy.Proof.
Fixε>0. Then, there exists l∈N such that ∑i=l∞‍d(xi,xi+1)<ε. Fix n∈N with n≥l. We will show(10)dxn,xn+1+jν≤∑i=nn+jν‍d(xi,xi+1)by induction. It is obvious that (10) holds for j=0. We assume that (10) holds for some j∈N∪{0}. Then, by (N3), we have (11)dxn,xn+1+(j+1)ν≤dxn,xn+1+jν+∑i=n+1+jνn+(j+1)ν‍dxi,xi+1≤∑i=nn+(j+1)ν‍dxi,xi+1.Therefore, (10) holds for j≔j+1. By induction, (10) holds for any j∈N∪{0}. Hence, (12)dxn,xn+1+jν≤∑i=nn+jν‍dxi,xi+1<∑i=l∞‍dxi,xi+1<εholds. Therefore, {xn} is ν-Cauchy.Lemma 11.
Let(X,d) be a ν-generalized metric space and let {xn} be a sequence in X such that xn are all different, limnd(xn,xn+1)=0, and {xn} converges to some z∈X. Then, {xn} converges only to z∈X.Proof.
Arguing by contradiction, we assume that{xn} converges to w which differs from z. Since xn are all different, xn≠z and xn≠w for sufficiently large n∈N. By (N3), we have (13)d(w,z)≤limn→∞dw,xn+∑i=nn+ν-2‍dxi,xi+1+dxn+ν-1,z=0,where we define ∑i=nn-1‍d(xi,xi+1)=0. By (N1), we obtain w=z. This is a contradiction.Lemma 12.
Let(X,d) be a ν-generalized metric space satisfying either of the following: (i)
ν is odd and X is complete;(ii)
ν is even and X is 2-complete.Let {xn} be a sequence in X such that xn are all different and ∑n=1∞‍d(xn,xn+1)<∞. Then, there exists z∈X such that {xn} converges only to z.Proof.
By Lemma9, {xn} is ν-Cauchy. By Propositions 7 and 8, the following hold. (i)
Ifν is odd, then {xn} is Cauchy.(ii)
Ifν is even, then {xn} is 2-Cauchy.From the assumption on the completeness of x, {xn} converges to some point z∈X. By Lemma 11, {xn} converges only to z.


## 3. Fixed Point Theorems

The following is a generalization of Subrahmanyam’s fixed point theorem [8]; see [9–11].Theorem 13.
Let(X,d) be as in Lemma 12. Let T be a sequentially continuous mapping on X. Assume that there exists r∈[0,1) satisfying (14)dTx,T2x≤rd(x,Tx)for all x∈X. Then, for any x∈X, {Tnx} converges only to a fixed point of T.Proof.
Define a sequence{uj} in X by uj=Tjx for j∈N. We prove the conclusion, dividing the following three cases. (i)
There existsn∈N such that un+1=un.(ii)
uj+1≠uj for all j∈N and there exist m,n∈N such that m+2≤n and um=un.(iii)
u1,u2,… are all different.In the first case, un is a fixed point of T. By (N1), {uj} converges only to un. In the second case, since um+1=un+1, we have (15)dum,um+1=dun,un+1≤rd(un-1,un)≤⋯≤rn-mdum,um+1,which implies d(um,um+1)=0. This is a contradiction. Thus, the second case cannot be possible. In the third case, we have (16)∑j=1∞‍duj,uj+1≤∑j=1∞‍rj-1d(u1,u2)=d(u1,u2)1-r<∞.So by Lemma 12, there exists z∈X such that {uj} converges only to z. We note that {Tuj}={uj+1} also converges only to z. Since T is sequentially continuous, we obtain Tz=z.A functionf from X into (-∞,+∞] isproper iff {x∈X:f(x)∈R} is nonempty.The following is a generalization of Caristi’s fixed point theorem [12, 13].Theorem 14.
Let(X,d) be as in Lemma 12. Let T be a mapping on X. Let f be a proper, sequentially lower semicontinuous function from X into (-∞,+∞] bounded from below. Assume that(17)fTx+dx,Tx≤fxfor all x∈X. Then, T has a fixed point.Remark 15.
This theorem is connected with Theorem 2 in [14]. See Section 4.Proof.
In the case whereν=1, Theorem 14 becomes the original Caristi fixed point theorem. So we assume ν≥2. Arguing by contradiction, we assume that T does not have a fixed point. Then, we note f(Tx)<f(x) for every x∈X with f(x)<∞. By induction, we define a sequence {uj} in X satisfying the following:(18)uj+1≠ujforanyj∈N,fuj+1+duj,uj+1≤fuj<∞foranyj∈N,fuj+1<inf⁡fx:fx+duj,x≤fuj+1jforanyj∈N.Fix u1∈X with f(u1)<∞. We assume that uj is defined for some j∈N. Then, we put (19)Sj=x∈X:fx+duj,x≤fuj.Since Tuj∈Sj, Sj is nonempty. Since Tuj≠uj, we can define uj+1 satisfying (18). By induction, we have defined {uj}. We note that uj are all different because f(uj+1)<f(uj) for any j∈N. Since f is bounded from below, {f(uj)} converges. We have (20)∑j=1∞‍duj,uj+1≤∑j=1∞‍fuj-fuj+1=fu1-limj→∞fuj≤fu1-inf⁡fx:x∈X<∞.So by Lemma 12, there exists z∈X such that {uj} converges only to z. Since f is sequentially lower semicontinuous, we have (21)fz≤liminf⁡j→∞fujandhencefz<fujforanyj∈N.We note that z≠uj for any j∈N. Since f(Tz)<f(z)<f(uj), Tz≠uj for any j∈N. Fix m∈N with 1/m<f(z)-f(Tz). Then, for any k∈N, we have (22)f(Tz)+d(um,Tz)≤f(Tz)+∑j=mm+ν-2‍d(uj,uj+1)+d(um+ν-1,z)hhhhhhhhhhhhhhh+d(z,Tz)≤f(z)+∑j=mm+ν-2‍d(uj,uj+1)+d(um+ν-1,z)≤f(z)+∑j=mm+2ν-2‍d(uj,uj+1)+dum+2ν-1,z≤⋯≤fz+∑j=mm+kν-2‍duj,uj+1+dum+kν-1,z<f(um+kν-1)+∑j=mm+kν-2‍fuj-fuj+1hhhhhhhhhhhhhhhhhhh+dum+kν-1,z=f(um)+d(um+kν-1,z).As k tends to infinity, we obtain f(Tz)+d(um,Tz)≤f(um) and hence Tz∈Sm. Then, we have (23)fum+1<inf⁡{f(x):x∈Sm}+1m<f(Tz)+f(z)-f(Tz)=f(z).This is a contradiction.


## 4. Counterexample

Kirk and Shahzad in [7] gave a counterexample to Theorem 2 in [14]. In this section, we give another example.Lemma 16 (see [4]).
Let(X,ρ) be a bounded metric space and let M be a real number satisfying (24)sup⁡{ρ(x,y):x,y∈X}≤M.Let A and B be two subsets of X with A∩B=⌀. Define a function d from X×X into [0,∞) by (25)dx,x=0dx,y=d(y,x)=ρx,yifx∈A,y∈Bdx,y=Motherwise.Then, (X,d) is a 2-generalized metric space.Remark 17.
We assumeX=A∪B in Lemma 4 in [4]. However, we do not use this assumption in the proof.Example 18.
LetX={0}∪{1/n:n∈N} and define a metric ρ on X as usual. Define two subsets A and B of X by (26)A=1n:n∈N,nisodd,B=1n:n∈N,niseven.Define a function d from X×X into [0,1] as in Lemma 16 with M=1. Define a mapping T on X by (27)Tx=1ifx=01n+1ifx=1nforsomen∈N.And define a function f from X into [0,2] by (28)f(x)=2ifx=0xotherwise.Then, (X,d) is a complete, 2-generalized metric space, f is sequentially continuous with respect to d, and (17) holds. However, T does not have a fixed point.Proof.
By Lemma16, (X,d) is a 2-generalized metric space. We will show that X is complete. Let {xj} be a Cauchy sequence in X. We consider the following three cases. (i)
#{j:xj=0}=∞, where #{j:xj=0} is the number of the elements of {j:xj=0}.(ii)
#{j:xj=1/n}=∞ for some n∈N.(iii)
#{j:xj=0}<∞ and #{j:xj=1/n}<∞ for any n∈N. In the first case, since 0∉A∪B, xj=0 holds for sufficiently large j∈N. Thus, {xj} converges to 0. In the second case, since (29)inf⁡dx,1n:x∈X∖1n=1n-1n+1>0,xj=1/n holds for sufficiently large j∈N. Thus, {xj} converges to 1/n. We consider the third case. Since {xj} is Cauchy, there exists l1∈N such that (30)sup⁡dxj,xk:k>j<1for any j∈N with j≥l1. We note that xl1∈A∪B because of the definition of d. Without loss of generality, we may assume that xl1∈A. There exists l2∈N such that l2>l1 and xj≠xl1 for j∈N with j≥l2. Then, xl2∈B clearly holds. Also, there exists l3∈N such that l3>l2 and xl3≠xl2. Then, xl3∈A and xl3≠xl1 clearly hold. We obtain d(xl1,xl3)<1. This is a contradiction. Therefore, the third case cannot be possible. We have shown that (X,d) is complete. We next show that f is sequentially continuous. Let a sequence {yj} in X converge to some y. Then, from the definition of d, there exists l∈N such that yj=y for j∈N with j≥l. This fact implies that f is sequentially continuous. For any x∈X, f(Tx)+d(x,Tx)=f(x) holds. So (17) is satisfied. However, it is clear that T does not have a fixed point.



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*Source: 709391-2015-05-18.xml*