# Suzuki-Type Fixed Point Results in Metric-Like Spaces

Authors: Nabiollah Shobkolaei; Shaban Sedghi; Jamal Rezaei Roshan; Nawab Hussain
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We demonstrate a fundamental lemma for the convergence of sequences in metric-like spaces, and by using it we prove some Suzuki-typefixed point results in the setup of metric-like spaces. As an immediate consequence of our results we obtain certain recent results in partial metric spaces as corollaries. Finally, three examples are presented to verify the effectiveness and applicability of our main results.


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## Body

## 1. Introduction

There are a lot of generalizations of Banach fixed-point principle in the literature. So far several authors have studied the problem of existence and uniqueness of a fixed point for mappings satisfying different contractive conditions (e.g., [1–20]). In 2008, Suzuki introduced an interesting generalization of Banach fixed-point principle. This interesting fixed-point result is as follows.Theorem 1 (see [19]).
Let(X,d) be a complete metric space, and let T be a mapping on X. Define a nonincreasing function θ from [0,1] into [1/2,1] by
(1)θ(r)={1,0≤r≤5-121-rr2,5-12≤r≤1211+r,12≤r<1.Assume that there existsr∈[0,1], such that
(2)θ(r)d(x,Tx)≤d(x,y)⟹d(Tx,Ty)≤rd(x,y),
for all x,y∈X, then there exists a unique fixed-point z of T. Moreover, limn→∞Tnx=z for all x∈X.Suzuki proved also the following version of Edelstein's fixed point theorem.Theorem 2.
Let(X,d) be a compact metric space. Let T:X→X be a self-map, satisfying for all x,y∈X, x≠y the condition
(3)12d(x,Tx)≤d(x,y)⟹d(Tx,Ty)<d(x,y).
ThenT has a unique fixed point in X.This theorem was generalized in [3].In addition to the above results, Kikkawa and Suzuki [8] provided a Kannan type version of the theorems mentioned before. In [14], Chatterjea type version is provided. Popescu [15] presented a Cirić type version. Recently, Kikkawa and Suzuki also provided multivalued versions which can be found in [9, 10].Very recently Hussain et al. [4] have extended Suzuki's Theorems 1 and 2, as well as Popescu's results from [15] to the case of metric type spaces and cone metric type spaces (see also [5–7, 11]).The aim of this paper is to generalize the above-mentioned results. Indeed we prove a fixed point theorem in the set up of metric-like spaces and derive certain new results as corollaries. Finally, three examples are presented to verify the effectiveness and applicability of our main results.In the rest of this section, we recall some definitions and facts which will be used throughout the paper. First, we present some known definitions and propositions in partial metric and metric-like spaces.A partial metric on a nonempty setX is a mapping p:X×X→ℝ+ such that for all x,y,z∈X,(p1)
x=y if and only if p(x,x)=p(x,y)=p(y,y),(p2)
p(x,x)≤p(x,y),(p3)
p(x,y)=p(y,x),(p4)
p(x,y)≤p(x,z)+p(z,y)-p(z,z).A partial metric space is a pair(X,p) such that X is a nonempty set and p is a partial metric on X. It is clear that if p(x,y)=0, then from (p1) and (p2) x=y. But if x=y, p(x,y) may not be 0. A basic example of a partial metric space is the pair (ℝ+,p), where p(x,y)=max{x,y} for all x,y∈ℝ+.Lemma 3 (see [17]).
Let(X,d) and (X,p) be a metric space and partial metric space, respectively. Then(i)
the functionρ:X×X→ℝ+ defined by ρ(x,y)=d(x,y)+p(x,y) is a partial metric;(ii)
letρ:X×X→ℝ+ be defined by ρ(x,y)=d(x,y)+max{ω(x),ω(y)}; then ρ is a partial metric on X, where ω:X→ℝ+ is an arbitrary function;(iii)
Letρ:ℝ×ℝ→ℝ be defined by ρ(x,y)=max{2x,2y}; then ρ is a partial metric on ℝ;(iv)
Letρ:X×X→ℝ+ be defined by ρ(x,y)=d(x,y)+a; then ρ is a partial metric on X, where a≥0.Other examples of the partial metric spaces which are interesting from a computational point of view may be found in [7, 11, 12, 18].Each partial metricp on X generates a T0 topology τp on X which has as a base the family of open p-balls {Bp(x,ε):x∈X,ε>0}, where Bp(x,ε)={y∈X:p(x,y)<p(x,x)+ε} for all x∈X and ε>0.Let(X,p) be a partial metric.A sequence{xn} in a partial metric space (X,p) converges to a point x∈X if and only if p(x,x)=limn→∞p(x,xn).A sequence{xn} in a partial metric space (X,p) is called a Cauchy sequence if there exists (and is finite) limn,m→∞p(xn,xm).A partial metric space(X,p) is said to be complete if every Cauchy sequence {xn} in X converges, with respect to τp, to a point x∈X such that p(x,x)=limn,m→∞p(xn,xm).Suppose that{xn} is a sequence in partial metric space (X,p); then we define L(xn)={x∣xn→x}.The following example shows that every convergent sequence{xn} in a partial metric space (X,p) may not be a Cauchy sequence. In particular, it shows that the limit is not unique.Example 4 (see [17]).
LetX=[0,∞) and p(x,y)=max{x,y}. Let
(4)xn={0,n=2k,1,n=2k+1.
Then clearly it is a convergent sequence, and for every x≥1 we have limn→∞p(xn,x)=p(x,x), hence L(xn)=[1,∞). But limn,m→∞p(xn,xm) does not exist; that is, it is not a Cauchy sequence.Definition 5 (see [2]).
A metric-like on a nonempty setX is a mapping σ:X×X→ℝ+ such that for all x,y,z∈X,(σ1)
σ(x,y)=0⇒x=y,(σ2)
σ(x,y)=σ(y,x),(σ3)
σ(x,y)≤σ(x,z)+σ(z,y).The pair(X,σ) is called a metric-like space. Then a metric-like on X satisfies all of the conditions of a metric except that σ(x,x) may be positive for x∈X. Each metric-like σ on X generates a topology τσ on X whose base is the family of open σ-ball, {Bσ(x,ε):x∈X,ε>0}, where Bσ(x,ε)={y∈X:|σ(x,y)-σ(x,x)|<ε} for all x∈X and ε>0.A sequence{xn} in a metric-like space (X,σ) converges to a point x∈X if and only if limn→∞σ(x,xn)=σ(x,x).A sequence{xn} in a metric-like space (X,σ) is called a σ-Cauchy sequence if there exists (and is finite) limn,m→∞σ(xn,xm).A metric-like space(X,σ) is said to be complete if every σ-Cauchy sequence {xn} in X converges, with respect to τσ, to a point x∈X such that
(5)limn→∞σ(xn,x)=σ(x,x)=limn,m→∞σ(xn,xm).Every partial metric space is a metric-like space. Below we give some examples of a metric-like space.Example 6.
LetX=[0,1]; then  mapping σ1:X×X→ℝ+ defined by σ1(x,y)=x+y-xy is a metric-like on X.Example 7.
LetX=ℝ; then  mappings σi:X×X→ℝ+(i∈{2,3,4}) defined by
(6)σ2(x,y)=|x|+|y|+a,σ3(x,y)=|x-b|+|y-b|,σ4(x,y)=x2+y2
are metric-like space on X, where a≥0 and b∈ℝ.


## 2. Main Results

We start our work by proving the following crucial lemma.Lemma 8.
Let(X,σ) be a metric-like space, and suppose that {xn} is σ-convergent to x. Then for every y∈X, one has
(7)σ(x,y)-σ(x,x)≤liminfn→∞σ(xn,y)≤limsupn→∞σ(xn,y)≤σ(x,y)+σ(x,x).
In particular, if σ(x,x)=0, then one has limn→∞σ(xn,y)=σ(x,y).Proof.
Using the triangle inequality in a metric-like space, it is easy to see that(8)σ(xn,y)≤σ(xn,x)+σ(x,y),σ(x,y)≤σ(x,xn)+σ(xn,y).
Taking the upper limit asn→∞ in the first inequality and the lower limit as n→∞ in the second inequality, we obtain the desired result.Theorem 9.
Let(X,σ) be a complete metric-like space. Let T:X→X be a self-map, and let θ=:[0,1)→(1/2,1] be defined by
(9)θ(r)={1,0≤r≤5-12,1-rr2,5-12≤r≤12,11+r,12≤r<1.
If there existsr∈[0,1) such that for each x,y∈X(10)θ(r)σ(x,Tx)≤σ(x,y)⟹σ(Tx,Ty)≤rσ(x,y).
Then T has a unique fixed point z∈X, and for each x∈X, the sequence {Tnx} converges to z.Proof.
Puttingy=Tx in (10), hence from
(11)θ(r)σ(x,Tx)≤σ(x,Tx),
it follows
(12)σ(Tx,T2x)≤rσ(x,Tx),
for every x∈X. Let x0∈X be arbitrary and form the sequence {xn} by x1=Tx0 and xn+1=Txn for n∈ℕ∪{0}. By (12), we have
(13)σ(xn,xn+1)=σ(Txn-1,T2xn-1)≤rσ(xn-1,Txn-1)=rσ(xn-1,xn)⋮≤rnσ(x0,x1).
Also, by the condition σ3 of the definition of metric-like space, for all m≥n, we have
(14)σ(xn,xm)≤σ(xn,xn+1)+σ(xn+1,xn+2)+⋯+σ(xm-1,xm)≤rnσ(x0,x1)+rn+1σ(x0,x1)+⋯+rm-1σ(x0,x1)=rn-rm1-rσ(x0,x1)<rn1-rσ(x0,x1)⟶0asn⟶∞.
Hence,{xn} is a σ-Cauchy sequence.
SinceX is σ-complete, there exists z∈X such that
(15)limn→∞σ(xn,z)=σ(z,z)=limm,n→∞σ(xm,xn)=0.
That is, limn→∞xn+1=limn→∞Txn=z. We prove that Tz=z. Putting x=Tn-1z in (12), we get that
(16)σ(Tnz,Tn+1z)≤rσ(Tn-1z,Tnz)
holds for each n∈ℕ (where T0z=z). It follows by induction that
(17)σ(Tnz,Tn+1z)≤rnσ(z,Tz).
Let us prove now that(18)σ(z,Tx)≤rσ(z,x)
holds for each x≠z. Since σ(xn,Txn)→0 and by Lemma 8σ(xn,x)→σ(z,x)≠0, it follows that there exists n0∈ℕ such that
(19)θ(r)σ(xn,Txn)≤σ(xn,x)
holds for every n≥n0. Assumption (10) implies that for such nσ(Txn,Tx)≤rσ(xn,x), thus as n→∞, we get that
(20)σ(z,Tx)≤rσ(z,x).
We will prove that
(21)σ(Tnz,z)≤σ(Tz,z),
for each n∈ℕ. For n=1, this relation is obvious. Suppose that it holds for some m∈ℕ. If Tmz=z, then Tm+1z=Tz and σ(Tm+1z,z)=σ(Tz,z)≤σ(Tz,z). If Tmz≠z, then applying (18) and the induction hypothesis; we get that
(22)σ(Tm+1z,z)≤rσ(Tmz,z)≤rσ(Tz,z)≤σ(Tz,z),
and (21) is proved by induction.
In order to prove thatTz=z, we consider two possible cases.
Case I. 0≤r<1/2 (and hence θ(r)≤(1-r)/r2). We will prove first that
(23)σ(Tnz,Tz)≤rσ(Tz,z)
for n≥2. For n=2, it follows from (16). Suppose that (23) holds for some n>2. Then
(24)σ(Tz,z)≤σ(z,Tnz)+σ(Tnz,Tz)≤σ(z,Tnz)+rσ(z,Tz),
which implies (1-r)σ(z,Tz)≤σ(z,Tnz). Using (17) we obtain
(25)θ(r)σ(Tnz,Tn+1z)≤1-rrnσ(Tnz,Tn+1z)≤1-rrn·rnσ(z,Tz)=(1-r)σ(z,Tz)≤σ(z,Tnz).
Assumption (10) and relation (21) imply that
(26)σ(Tz,Tn+1z)≤rσ(z,Tnz)≤rσ(z,Tz).
So relation (23) is proved by induction.
NowTz≠z and (23) implies  that Tnz≠z for each n∈ℕ. Hence, (18) imply that
(27)σ(z,Tn+1z)≤rσ(z,Tnz)≤r2σ(z,Tn-1z)≤rnσ(z,Tz).
Hence limn→∞σ(z,Tn+1z)=0=σ(z,z), thus Tnz→z and; using Lemma 8 in (23), we have σ(z,Tz)≤rσ(Tz,z) as n→∞ which implies that σ(z,Tz)=0, a contradiction.
Case II. 1/2≤r<1 (and so θ(r)=1/(1+r)). We will prove that there exists a subsequence {xnk} of {xn} such that
(28)θ(r)σ(xnk,Txnk)=θ(r)σ(xnk,xnk+1)≤σ(xnk,z)
holds for each k∈ℕ. From (12) we know that σ(xn,xn+1)≤rσ(xn-1,xn) holds for each n∈ℕ. Suppose that
(29)11+rσ(xn-1,xn)>σ(xn-1,z),11+rσ(xn,xn+1)>σ(xn,z)
hold for  some n∈ℕ. Then
(30)σ(xn-1,xn)≤σ(xn-1,z)+σ(z,xn)<11+rσ(xn-1,xn)+σ(xn,z)<11+rσ(xn-1,xn)+11+rσ(xn,xn+1)≤11+rσ(xn-1,xn)+r1+rσ(xn-1,xn)=σ(xn-1,xn),
which is impossible. Hence one of the following holds for each n:
(31)θ(r)σ(xn-1,xn)≤σ(xn-1,z),
or
(32)θ(r)σ(xn,xn+1)≤σ(xn,z).
In particular,
(33)θ(r)σ(x2n-1,x2n)≤σ(x2n-1,z),
or
(34)θ(r)σ(x2n,x2n+1)≤σ(x2n,z).
In other words, there is a subsequence {xnk} of {xn} such that (28) holds for each k∈ℕ. But then assumption (10) implies that
(35)σ(Txnk,Tz)≤rσ(xnk,z),
or
(36)σ(Txnk-1,Tz)≤rσ(xnk-1,z).
Passing to the limit when k→∞, we get that σ(z,Tz)≤0, which is possible only if Tz=z.
Thus, we have proved thatz is a fixed point of T. The uniqueness of the fixed point follows easily from (10). Indeed, if y and z are two fixed points of T such that y≠z, then from (18) we have
(37)σ(y,z)=σ(y,Tz)≤rσ(y,z),
which is a contradiction.According to Theorem9, we get the following result.Corollary 10 (see [19]).
Let(X,d) be a complete metric space, and let T be a mapping on X. Define a nonincreasing function θ from [0,1] into [1/2,1] by
(38)θ(r)={1,0≤r≤5-12,1-rr2,5-12≤r≤12,11+r,12≤r<1.
Assume that there existsr∈[0,1], such that
(39)θ(r)d(x,Tx)≤d(x,y)⟹d(Tx,Ty)≤rd(x,y),
for all x,y∈X; then there exists a unique fixed-point z of T. Moreover, limn→∞Tnx=z for all x∈X.Proof.
Using a similar argument given in Theorem9 for σ(x,y)=d(x,y), the desired result is obtained.Now, in order to support the useability of our results, let us introduce the following example.Example 11.
LetX=[0,∞). Define σ:X×X→ℝ+ by
(40)σ(x,y)=x+y
for all x,y∈X. Then (X,σ) is a complete metric-like space. Define a map T:X→X by
(41)T(x)=ln(1+12x)forx∈X. Then for each x,y∈X,  we have
(42)11+1/2σ(x,Tx)=22+1(x+ln(1+12x))≤22+1(x+12x)=x≤x+y=σ(x,y).On the other hand, we have(43)σ(Tx,Ty)=ln(1+12x)+ln(1+12y)≤12x+12y=12σ(x,y).
Thus T  satisfies all the hypotheses of Theorem 9, and hence T  has a unique fixed point. Indeed, r=1/2, θ(r)=1/(1+r), and 0 is the unique fixed point of T.Theorem 12.
Let(X,σ) be a complete metric-like space. Let S, T:X→X be two self-mappings. Suppose that there exists r∈[0,1) such that
(44)max{σ(S(x),TS(x)),σ(T(x),ST(x))}≤rmin{σ(x,S(x)),σ(x,T(x))}
for every x∈X and that
(45)α(y)=inf{σ(x,y)+min{σ(x,S(x)),σ(x,T(x))}:x∈X{{σ(x,S(x)),σ(x,T(x))}}}>0
for every y∈X with y that  is not a common fixed point of S and T. Then there exists z∈X such that z=S(z)=T(z). Moreover, if v=S(v)=T(v), then σ(v,v)=0.Proof.
Letx0∈X be arbitrary, and define a sequence {xn} by
(46)xn=S(xn-1),ifnisodd=T(xn-1),ifniseven.
Then if n∈ℕ is odd, we have
(47)σ(xn,xn+1)=σ(S(xn-1),T(xn))=σ(S(xn-1),TS(xn-1))≤max{σ(S(xn-1),TS(xn-1)),σ(T(xn-1),ST(xn-1))}≤rmin{σ(xn-1,S(xn-1)),σ(xn-1,T(xn-1))}≤rσ(xn-1,S(xn-1))=rσ(xn-1,xn).
Ifn is even, then by (44), we have
(48)σ(xn,xn+1)=σ(T(xn-1),S(xn))=σ(T(xn-1),ST(xn-1))≤max{σ(T(xn-1),ST(xn-1)),σ(S(xn-1),TS(xn-1))}≤rmin{σ(xn-1,T(xn-1)),σ(xn-1,S(xn-1))}≤rσ(xn-1,T(xn-1))=rσ(xn-1,xn).
Thus for any positive integer n, it must be the case that
(49)σ(xn,xn+1)≤rσ(xn-1,xn).
Repeating (49), we obtain
(50)σ(xn,xn+1)≤rnσ(x0,x1).
So, if m>n, then
(51)σ(xn,xm)≤σ(xn,xn+1)+σ(xn+1,xn+2)+⋯+σ(xm-1,xm)≤[rn+rn+1+⋯+rm-1]σ(x0,x1)≤rn1-rσ(x0,x1).
Thuslimn,m→∞σ(xn,xm)=0.
That is,{xn} is a σ-Cauchy sequence in the metric-like space (X,σ). Since (X,σ) is σ-complete, there exist z∈X such that
(52)σ(z,z)=limn→∞σ(xn,z)=limn,m→∞σ(xn,xm)=0.
Assume thatz is not a common fixed point of S and T. Then by hypothesis
(53)0<inf{σ(x,z)+min{σ(x,S(x)),σ(x,T(x))}:x∈X}≤inf{σ(xn,z)+min{σ(xn,S(xn)),σ(xn,T(xn))}:n∈ℕ{(xn,T(xn))}}≤inf{rn1-rσ(x0,x1)+σ(xn,xn+1):n∈ℕ}≤inf{rn1-rσ(x0,x1)+rnσ(x0,x1):n∈ℕ}=0
which is a contradiction. Therefore, z=S(z)=T(z).
Ifv=S(v)=T(v) for some v∈X, then
(54)σ(v,v)=max{σ(S(v),TS(v)),σ(T(v),ST(v))}≤rmin{σ(v,S(v)),σ(v,T(v))}=rmin{σ(v,v),σ(v,v)}=rσ(v,v)
which gives that σ(v,v)=0.Example 13.
Let(X,σ) be a metric-like space where X={1/n}n=1∞∪{0} and σ(x,y)=x+y. Define S:X→X by S(0)=0, S(1/2n)=1/(4n+3), and S(1/(2n-1))=0 and T(0)=0, T(1/(2n-1))=1/(4n+4), and T(1/2n)=0. Then for x=1/2n, we have
(55)max{σ(S(x),TS(x)),σ(T(x),ST(x))}=max{σ(S(12n),T(S(12n))),σ(T(12n),S(T(12n)))}=max{14n+3+18n+12,0}=14n+3+18n+12≤rmin{σ(x,S(x)),σ(x,T(x))}=rmin{12n+14n+3,12n+0}=r12n.
It is easy to see that the above inequality is true for x=1/(2n-1) and for 3/4≤r<1. Also,
(56)α(y)=inf{σ(x,y)+min{σ(x,S(x)),σ(x,T(x))}:x∈X{{σ(x,S(x)),σ(x,T(x))}}}>0
for every y∈X with y is not a common fixed point of S and T. This shows that all conditions of Theorem 12 are satisfied and 0 is a common fixed point for S and T.Corollary 14.
Let(X,σ) be a complete metric-like space, and let T:X→X be a mapping. Suppose that there exists r∈[0,1) such that
(57)σ(T(x),T2(x))≤rσ(x,T(x))
for every x∈X and that
(58)α(y)=inf{σ(x,y)+σ(x,T(x)):x∈X}>0
for every y∈X with y≠T(y). Then there exists z∈X such that z=T(z). Moreover, if v=T(v), then σ(v,v)=0.Proof.
TakingS=T in Theorem 12, the conclusion of the corollary follows.Theorem 15.
Let(X,σ) be a complete metric-like space. Let S, T be mappings from X onto itself. Suppose that there exists r>1 such that
(59)min{σ(TS(x),S(x)),σ(ST(x),T(x))}≥rmax{σ(Sx,x),σ(Tx,x)}
for every x∈X and that
(60)α(y)=inf{σ(x,y)+min{σ(x,S(x)),σ(x,T(x))}:x∈X{{σ(x,S(x)),σ(x,T(x))}}}>0
for every y∈X with y that is not a common fixed point of S and T. Then there exists z∈X such that z=S(z)=T(z). Moreover, if v=S(v)=T(v), then σ(v,v)=0.Proof.
Letx0∈X be arbitrary. Since S is onto, there is an element x1 satisfying x1∈S-1(x0). Since T is also onto, there is an element x2 satisfying x2∈T-1(x1). Proceeding in the same way, we can find that x2n+1∈S-1(x2n) and x2n+2∈T-1(x2n+1) for n=1,2,3,…. Therefore, x2n=Sx2n+1 and x2n+1=Tx2n+2 for n=0,1,2,…. If n=2m, then using (59)
(61)σ(xn-1,xn)=σ(x2m-1,x2m)=σ(Tx2m,Sx2m+1)=σ(TSx2m+1,Sx2m+1)≥min{σ(TS(x2m+1),S(x2m+1)),σ(ST(x2m+1),T(x2m+1))}≥rmax{σ(Sx2m+1,x2m+1),σ(Tx2m+1,x2m+1)}≥rσ(Sx2m+1,x2m+1)=rσ(x2m,x2m+1)=rσ(xn,xn+1).
If n=2m+1, then using (59)
(62)σ(xn-1,xn)=σ(x2m,x2m+1)=σ(Sx2m+1,Tx2m+2)=σ(STx2m+2,Tx2m+2)≥min{σ(TS(x2m+2),S(x2m+2)),σ(ST(x2m+2),T(x2m+2))}≥rmax{σ(Sx2m+2,x2m+2),σ(Tx2m+2,x2m+2)}≥rσ(Tx2m+2,x2m+2)=rσ(x2m+1,x2m+2)=rσ(xn,xn+1).
Thus for any positive integer n, it must be the case that
(63)σ(xn-1,xn)≥rσ(xn,xn+1)
which implies that
(64)σ(xn,xn+1)≤1rσ(xn-1,xn)≤⋯≤(1r)nσ(x0,x1).
Let α=1/r; then 0<α<1 since r>1.
Now, (64) becomes
(65)σ(xn,xn+1)≤αnσ(x0,x1).
So, if m>n, then
(66)σ(xn,xm)≤σ(xn,xn+1)+σ(xn+1,xn+2)+⋯+σ(xm-1,xm)≤[αn+αn+1+⋯+αm-1]σ(x0,x1)≤αn1-ασ(x0,x1).
Thuslimn,m→∞σ(xn,xm)=0. That is, {xn} is a σ-Cauchy sequence in the metric-like space (X,σ). Since (X,σ) is σ-complete, there exists  z∈X such that
(67)σ(z,z)=limn→∞σ(xn,z)=limn,m→∞σ(xn,xm)=0.
Assume that z is not a common fixed point of S and T. Then by hypothesis
(68)0<inf{σ(x,z)+min{σ(x,S(x)),σ(x,T(x))}:x∈X}≤inf{σ(xn,z)+min{σ(xn,S(xn)),σ(xn,T(xn))}:n∈ℕ{(xn,S(xn))}}≤inf{αn1-ασ(x0,x1)+σ(xn-1,xn):n∈ℕ}≤inf{αn1-ασ(x0,x1)+αn-1σ(x0,x1):n∈ℕ}=0
which is a contradiction. Therefore, z=S(z)=T(z).
Ifv=S(v)=T(v) for some v∈X, then
(69)σ(v,v)=min{σ(TS(v),S(v)),σ(ST(v),T(v))}≥rmax{σ(S(v),v),σ(T(v),v)}=rmax{σ(v,v),σ(v,v)}=rσ(v,v)
which gives that σ(v,v)=0.Corollary 16.
Let(X,σ) be a complete metric-like space, and let T:X→X be an onto mapping. Suppose that there exists r∈[0,1) such that
(70)σ(T2(x),T(x))≥rσ(T(x),x)
for every x∈X and that
(71)α(y)=inf{σ(x,y)+σ(T(x),x):x∈X}>0
for every y∈X with y≠T(y). Then there exists z∈X such that z=T(z). Moreover, if v=T(v), then σ(v,v)=0.Proof.
TakingS=T in Theorem 15, we have the desired result.Definition 17.
Let(X,σ) and (Y,τ) be metric-like spaces. Then f:X→Y is said to be a continuous mapping, if limn→∞xn=x implies that limn→∞f(xn)=f(x).Corollary 18.
Let(X,σ) be a complete metric-like space, and let T be a mapping of X into itself. If there is a real number r with r>1 satisfying
(72)σ(T2(x),T(x))≥rσ(T(x),x)
for every x∈X and T is onto and continuous, then T has a fixed point.Proof.
Assume that there existsy∈X with y≠T(y) and
(73)inf{σ(x,y)+σ(T(x),x):x∈X}=0.
Then there exists a sequence {xn} such that
(74)limn→∞{σ(xn,y)+σ(T(xn),xn)}=0.
So, we have σ(xn,y)→0 and σ(T(xn),xn)→0 as n→∞. Since, σ(y,y)≤σ(y,xn)+σ(xn,y), hence σ(y,y)→0 as n→∞. Now,
(75)σ(T(xn),y)≤σ(T(xn),xn)+σ(xn,y)⟶0asn⟶∞.
Since T is continuous, we have
(76)T(y)=T(limn→∞xn)=limn→∞T(xn)=y.
This is a contradiction. Hence if y≠T(y), then
(77)inf{σ(x,y)+σ(T(x),x):x∈X}>0,
which is condition (71) of Corollary 16. By Corollary 16, there exists z∈X such that z=T(z).Now we give an example to support our result.Example 19.
LetX=[0,∞) and σ(x,y)=x+y. Define T:X→X by T(x)=2x.
ObviouslyT is onto and continuous. Also for each x,y∈X, we have
(78)σ(T2x,Tx)=4x+2x=6x≥r3x=rσ(Tx,x),
where r=2. Thus T  satisfies the conditions given in Corollary 18, and 0 is the fixed point of T.Corollary 20.
Let(X,σ) be a complete metric-like space, and T be a mapping of X into itself. If there is a real number r with r>1 satisfying
(79)σ(T(x),T(y))≥rmin{σ(x,T(x)),σ(T(y),y),σ(x,y)}
for every x,y∈X and T is onto and continuous, then T has a fixed point.Proof.
Replacingy by T(x) in (79), we obtain
(80)σ(T(x),T2(x))≥rmin{σ(x,T(x)),σ(T2(x),T(x)),σ(x,T(x))}
for all x∈X.
Without loss of generality, we may assume thatT(x)≠T2(x). Otherwise T has a fixed point. Since r>1, it follows from (80) that
(81)σ(T2(x),T(x))≥rσ(T(x),x)
for every x∈X. By the argument similar to that used in Corollary 18, we can prove that if y≠T(y), then
(82)inf{σ(x,y)+σ(T(x),x):x∈X}>0,
which is condition (71) of Corollary 16. So, Corollary 16 applies to obtain a fixed point of T.According to Theorem12, we get the following result.Corollary 21 (see [17, Theorem 1]).
Let(X,p) be a complete partial metric space. Let S, T:X→X be two self-mappings. Suppose that there exists r∈[0,1) such that
(83)max{p(S(x),TS(x)),p(T(x),ST(x))}≤rmin{p(x,S(x)),p(x,T(x))}
for every x∈X and that
(84)α(y)=inf{p(x,y)+min{p(x,S(x)),p(x,T(x))}:x∈X{{p(x,S(x)),p(x,T(x))}}}>0
for every y∈X with y that is not a common fixed point of S and T. Then there exists z∈X such that z=S(z)=T(z). Moreover, if v=S(v)=T(v), then p(v,v)=0.Proof.
Using a similar argument given in the Theorem12 for σ(x,y)=p(x,y), the desired result is obtained, where p is a partial metric on X.Also, according to Theorem15, we get Theorem 2 from [17].



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*Source: 143686-2013-08-27.xml*