# A Generalized Meir-Keeler-Type Contraction on Partial Metric Spaces

Authors: Hassen Aydi; Erdal Karapınar; Shahram Rezapour
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We introduce a generalization of the Meir-Keeler-type contractions, referred to as generalized Meir-Keeler-type contractions, over partial metric spaces. Moreover, we
show that every orbitally continuous generalized Meir-Keeler-type contraction has a fixed point on a 0-complete partial metric space.


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## Body

## 1. Introduction

In 1992, Matthews introduced the notion of a partial metric space which is a generalization of usual metric space [1]. The main motivation behind the idea of a partial metric space is to transfer mathematical techniques into computer science. This is mostly apparent in the research areas of computer domains and semantics, which have many applications (see, e.g., [2–10]). Following this initial work, Matthews generalized the Banach contraction principle in the context of complete partial metric spaces. He proved that a self-mapping T on a complete partial metric space (X,p) has a unique fixed point if there exists 0≤k<1 such that p(Tx,Ty)≤kp(x,y) for all x,y∈X. After Matthews’ innovative approach, many authors conducted further studies on partial metric spaces and their topological properties (see, e.g., [2–4, 6, 11–41]).A partial metric is a functionp:X×X→[0,∞) satisfying the following conditions:(P1)p(x,y)=p(y,x),(P2) ifp(x,x)=p(x,y)=p(y,y), then x=y,(P3)p(x,x)≤p(x,y),(P4)p(x,z)+p(y,y)≤p(x,y)+p(y,z),for allx,y,z∈X. Then (X,p) is called a partial metric space.Example 1.1 (see [42]).
Let(X,d) and (X,p) be a metric space and partial metric space, respectively. Mappings ρi:X×X→ℝ+ (i∈{1,2,3}) defined by
(1.1)ρ1(x,y)=d(x,y)+p(x,y),ρ2(x,y)=d(x,y)+max⁡{ω(x),ω(y)},ρ3(x,y)=d(x,y)+a
induce partial metrics on X, where ω:X→ℝ+ is an arbitrary function and a≥0.Each partial metricp on X generates a T0 topology τp on X with the family of open p-balls {Bp(x,ε):x∈X,ε>0} as a base, where Bp(x,ε)={y∈X:p(x,y)<p(x,x)+ε} for all x∈X. Similarly, a closed p-ball is defined as Bp[x,ε]={y∈X:p(x,y)≤p(x,x)+ε}.In [1, page 187], Matthews gave the characterization of convergence in partial metric space as follows: a sequence {xn} in a partial metric space (X,p) converges to x∈X with respect to τp if and only if limn→∞p(x,xn)=p(x,x).Now we recall some basic concepts and useful facts on completeness of partial metric spaces. A sequence{xn} in a partial metric space (X,p) is called Cauchy whenever limn,m→∞p(xn,xm) exists (and is finite) [1, Definition  5.2].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 limn,m→∞p(xn,xm)=p(x,x) [1, Definition  5.3].In [35], Romaguera introduced the concepts 0-Cauchy sequence in a partial metric space and 0-complete partial metric space as follows.Definition 1.2.
A sequence{xn} in a partial metric space (X,p) is called 0-Cauchy if limn,m→∞p(xn,xm)=0. A partial metric space (X,p) is said to be 0-complete if every 0-Cauchy sequence in X converges, with respect to τp, to a point x∈X such that p(x,x)=0. In this case, p is said to be a 0-complete partial metric on X.Notice that each 0-Cauchy sequence is also a Cauchy sequence in a partial metric space. In particular, each complete partial metric is a 0-complete partial metric onX. But the converse is not true. The following example shows that there exists a 0-complete partial metric which is not complete.Example 1.3 (see [35, 39]).
Let(ℚ∩[0,∞),p) be the partial metric space, where ℚ and p(x,y) represent the set of rational numbers and the partial metric max{x,y}, respectively.A self-mappingF on a partial metric space (X,p) is continuous at x∈X if and only if for every ε>0, there exists δ>0 such that F(Bp(x,δ))⊆Bp(Fx,ε) (see, e.g., [15]).It is quite natural to consider characterizations of continuity of mappings in partial metric spaces. For example, Samet et al. [43] proved the following.Lemma 1.4.
Let(X,p) be a partial metric space. F:X→X is continuous if given a sequence {xn}∈ℕ and x∈X such that p(x,x)=limn→+∞p(x,xn); then, p(Fx,Fx)=limn→+∞p(Fx,Fxn).Very recently, Samet et al. [43] also observed the relationship between the continuity of a mapping in a partial metric space and in a metric space.Lemma 1.5.
ConsiderX=[0,∞) endowed with the partial metric p:X×X→[0,∞) defined by p(x,y)=max{x,y} for all x,y≥0. Let F:X→X be a nondecreasing function. If F is continuous with respect to the standard metric d(x,y)=|x-y| for all x,y≥0, then F is continuous with respect to the partial metric p.In 1971, Ćirić [44] introduced orbitally continuous maps on metric spaces as follows.Definition 1.6.
Let(X,d) be a metric space. A mapping T on X is orbitally continuous if limi→∞Tnix=u implies limi→∞TTnix=Tu for each x∈X.Recently, Karapınar and Erhan [28] renovated the definition above in the context of partial metric spaces in the following way.Definition 1.7.
Let(X,p) be a partial metric space, and let  T:X→X be a self-map. One says that T is orbitally continuous whenever limi→∞p(Tnix,z)=p(z,z) implies that limi→∞p(TTnix,Tz)=p(Tz,Tz) for each x∈X.It is clear that continuous mappings are orbitally continuous.We would like to point out the close relationship between metrics and partial metrics. In fact, ifp is a partial metric on X, then the function dp:X×X→[0,∞) given by(1.2)dp(x,y)=2p(x,y)-p(x,x)-p(y,y)
is a metric on X. Moreover,(1.3)lim⁡n→∞dp(x,xn)=0⟺lim⁡n→∞p(x,xn)=lim⁡n,m→∞p(xn,xm)=p(x,x).Lemma 1.8 (see, e.g., [1, 15]).
Let(X,p) be a partial metric space.
(a) A sequence{xn} is Cauchy if and only if {xn} is a Cauchy sequence in the metric space (X,dp);
(b)(X,p) is complete if and only if the metric space (X,dp) is complete.In 1969, Meir and Keeler [45] published their celebrated paper in which an interesting and general contraction condition for self-maps in metric spaces was considered.Definition 1.9.
Let(X,d) be a metric space, and let  T be a self-map on X. Then T is called a Meir-Keeler-type contraction whenever for each ε>0 there exists δ>0 such that
(1.4)d(x,y)<ε+δ⟹d(Tx,Ty)<ε.Many authors have discussed several variations, generalizations, and modifications of that condition both in metric spaces and other related structures (see, e.g., [46–49]). Following this trend, we introduce a generalized Meir-Keeler-type contraction on partial metric spaces. In this paper, we show an orbitally continuous self-mapping T on a 0-complete partial metric spaces satisfying that generalized Meir-Keeler-type contraction has a unique fixed point.


## 2. Main Results

We start this section by recalling the following two lemmas ([13]), which will be frequently used in the proofs of the main results.Lemma 2.1.
Let(X,p) be a partial metric space. Then
(a) ifp(x,y)=0, then x=y,
(b) ifx≠y, then p(x,y)>0,
(c) ifxn→z with p(z,z)=0, then limn→∞p(xn,y)=p(z,y) for all y∈X.We introduce the definition of a generalized Meir-Keeler-type contraction.Definition 2.2.
Let(X,p) be a partial metric space and T a self-map on X. Then T is called a generalized Meir-Keeler-type contraction whenever for each ε>0 there exists δ>0 such that
(2.1)ε≤M(x,y)<ε+δ⟹p(Tx,Ty)<ε,
where M(x,y)=max{p(x,y),p(Tx,x),p(Ty,y),(1/2)[p(Tx,y)+p(x,Ty)]}.Remark 2.3.
Note that ifT is a generalized Meir-Keeler-type contraction, we have
(2.2)p(Tx,Ty)≤M(x,y)∀x,y∈X.
If M(x,y)=0, it follows from (2.2) that p(Tx,Ty)=0. On the other hand, if M(x,y)>0, we get the strict inequality p(Tx,Ty)<M(x,y) by (2.1).Now, we are ready to state and prove our main results.Proposition 2.4.
Let(X,p) be a partial metric space and T:X→X a generalized Meir-Keeler-type contraction. Then, limn→∞p(Tn+1x,Tnx)=0 for all x∈X.Proof..
Takex∈X, and set x0=x. Define xn+1=Txn=Tn+1x0 for all n≥0. If p(xn0+1,xn0)=0 for some n0≥0, then Txn0=xn0+1=xn0 by Lemma 2.1. Then, p(xk+1,xk)=0 for all k≥n0. In this case, the proposition follows. In the rest of the proof, we assume that p(xn+1,xn)≠0 for every n≥0. As a consequence, we have M(xn+1,xn)>0 for every n≥0. By Remark 2.3,
(2.3)p(xn+2,xn+1)=p(Txn+1,Txn)≤M(xn+1,xn)=max⁡{p(xn+1,xn),p(Txn+1,xn+1),p(Txn,xn),12[p(Txn+1,xn)+p(xn+1,Txn)]}≤max⁡{p(xn+1,xn),p(xn+2,xn+1)}.
SinceM(xn+1,xn) is strictly positive for each n, we find that
(2.4)p(xn+2,xn+1)<M(xn+1,xn)≤max⁡{p(xn+1,xn),p(xn+2,xn+1)}
by the use of Remark 2.3 again. Notice that the case where
(2.5)max⁡{p(xn+1,xn),p(xn+2,xn+1)}=p(xn+2,xn+1)
is not possible. Hence we derive that
(2.6)p(xn+2,xn+1)<M(xn+1,xn)≤p(xn+1,xn)
for every n. Thus, {p(xn+1,xn)}n=0∞ is a decreasing sequence which is bounded below by 0. Hence, it converges to some ε∈[0,∞), that is,
(2.7)lim⁡n→∞p(xn+1,xn)=ε.
In particular, we have
(2.8)lim⁡n→∞M(xn+1,xn)=ε.
Notice that ε=inf{p(xn,xn+1):n∈ℕ}.
We claim thatε=0. Suppose, to the contrary, that ε>0. Regarding (2.8) together with the assumption that T is generalized Meir-Keeler-type contraction, for this ε, there exists δ>0 and a natural number m such that
(2.9)ε≤M(xm+1,xm)<ε+δimpliesthatp(Txm+1,Txm)=p(xm+2,xm+1)<ε.
This is a contradiction since ε=inf{p(xn,xn+1):n∈ℕ}.Theorem 2.5.
Let(X,p) be a 0-complete partial metric space, and let  T:X→X be an orbitally continuous generalized Meir-Keeler-type contraction. Then, T has a unique fixed point, say z∈X. Moreover, limn→∞p(Tnx,z)=p(z,z) for all x∈X and p(z,z)=0.Proof..
Takex∈X, and set x0=x. Define xn+1=Txn=Tn+1x0 for all n≥0. We claim that limm,n→∞p(xn,xm)=0. If this is not the case, then there exist a ε>0 and a subsequence {xn(i)} of {xn} such that
(2.10)p(xn(i),xn(i+1))>2ε.
For the sameε>0 above, there exists δ>0 such that ε≤M(x,y)<ε+δ which implies that p(Tx,Ty)<ε. Set r=min{ε,δ} and dn=p(xn,xn+1) for all n≥1. By Proposition 2.4, one can choose a natural number n0 such that
(2.11)dn=p(xn,xn+1)<r4
for all n≥n0. Let n(i)>n0. We have n(i)≤n(i+1)-1. If p(xn(i),xn(i+1)-1)≤ε+(r/2), then by using (P4) we derive
(2.12)p(xn(i),xn(i+1))≤p(xn(i),xn(i+1)-1)+p(xn(i+1)-1,xn(i+1))-p(xn(i+1)-1,xn(i+1)-1)≤p(xn(i),xn(i+1)-1)+p(xn(i+1)-1,xn(i+1))<ε+r2+dn(i+1)-1<ε+3r4<2ε,
which contradicts with assumption (2.10). Therefore, there are values of k such that n(i)≤k≤n(i+1) and p(xn(i),xk)>ε+(r/2). Now if p(xn(i),xn(i)+1)≥ε+(r/2), then
(2.13)dn(i)=p(xn(i),xn(i)+1)≥ε+r2>r+r2>r4.
This is a contradiction because of (2.11). Hence, there are values of k with n(i)≤k≤n(i+1) such that p(xn(i),xk)<ε+(r/2). Choose the smallest integer k with k≥n(i) such that p(xn(i),xk)≥ε+(r/2). Thus, we find p(xn(i),xk-1)<ε+(r/2). So we see that
(2.14)p(xn(i),xk)≤p(xn(i),xk-1)+p(xk-1,xk)-p(xk-1,xk-1)≤p(xn(i),xk-1)+p(xk-1,xk)<ε+r2+r4=ε+3r4.
Now, we can choose a natural number k satisfying n(i)≤k≤n(i+1) such that
(2.15)ε+r2≤p(xn(i),xk)<ε+3r4.
Therefore, we obtain the inequalities
(2.16)p(xn(i),xk)<ε+3r4<ε+r,(2.17)p(xn(i),xn(i)+1)=dn(i)<r4<ε+r,p(xk,xk+1)=dk<r4<ε+r.
Thus, we have
(2.18)12[p(xn(i),xk+1)+p(xn(i)+1,xk)]≤12[p(xn(i),xk)+p(xk,xk+1)-p(xk,xk)+p(xn(i)+1,xn(i))+p(xn(i),xk)-p(xn(i),xn(i))]≤12[p(xn(i),xk)+p(xk,xk+1)+p(xn(i)+1,xn(i))+p(xn(i),xk)]=p(xn(i),xk)+12[dk+dn(i)]<ε+3r4+12[r4+r4]=ε+r.
Now, inequalities (2.16)–(2.18) imply that M(xn(i),xk)<ε+r≤ε+δ. Hence, the fact that T is a generalized Meir-Keeler-type contraction yields p(xn(i)+1,xk+1)<ε. By using (P4), we obtain
(2.19)p(Tn(i)x0,Tkx0)≤p(Tn(i)x0,Tn(i)+1x0)+p(Tn(i)+1x0,Tkx0)-p(Tn(i)+1x0,Tn(i)+1x0)≤p(Tn(i)x0,Tn(i)+1x0)+p(Tn(i)+1x0,Tkx0)≤p(Tn(i)x0,Tn(i)+1x0)+p(Tn(i)+1x0,Tk+1x0)+p(Tk+1x0,Tkx0).
We combine the inequality above with (2.15) and (2.17) to conclude
(2.20)p(xn(i)+1,xk+1)≥p(xn(i),xk)-p(xn(i),xn(i)+1)-p(xk,xk+1)>ε+r2-r4-r4=ε,
which is a contradiction. Therefore, our claim is proved. So {xn}={Tnx0} is a 0-Cauchy sequence. Since (X,p) is 0-complete, then by Definition 1.2, the sequence {xn} converges with respect to τp to some z∈X such that p(z,z)=0. Thus
(2.21)lim⁡n→∞p(Tnx0,z)=p(z,z)=0.
Now, we will show that z is a fixed point of T.
SinceT is orbitally continuous and limn→∞p(Tnx0,z)=p(z,z), we get that
(2.22)lim⁡n→∞p(TTnx0,Tz)=p(Tz,Tz).
On the other hand, from Lemma 2.1, we have
(2.23)lim⁡n→∞p(TTnx0,Tz)=lim⁡n→∞p(xn+1,Tz)=p(z,Tz)
which follows from the fact that {xn+1} converges to z in (X,p) with p(z,z)=0, where xn+1=TTnx0=Tn+1x0. Combining this with (2.22), we get that p(z,Tz)=p(Tz,Tz).
We aim to show thatp(z,Tz)=0. Assume that p(z,Tz)>0. Then, we obtain M(z,z)≥p(z,Tz)>0. By (2.2), we have
(2.24)p(Tz,Tz)<M(z,z)=max⁡{p(z,z)=0,p(z,Tz)}=p(z,Tz)=p(Tz,Tz),
a contradiction. This implies Tz=z by Lemma 2.1.
Finally, we show thatT has a unique fixed point. If there exists w∈X such that Tw=w and p(z,w)≠0, then we get M(z,w)≥p(z,w)>0. Since T is a generalized Meir-Keeler-type contraction, we derive
(2.25)0<p(z,w)=p(Tz,Tw)<M(z,w)=max⁡{p(z,w),p(Tz,z),p(Tw,w),12[p(Tz,w)+p(z,Tw)]}=max⁡{p(z,w),p(w,w)}=p(w,z),
which is a contradiction. Thus, we find that p(z,w)=0. So by Lemma 2.1 we conclude that z=w. In particular, T has a unique fixed point.We state two examples to illustrate our results.Example 2.6.
Let(X,p) be the set [0,∞) equipped with the partial metric p(x,y)=max{x,y}. Clearly, (X,p) is a 0-complete partial metric space. Consider T:X→X defined by Tx=x/3(1+x). Given ε>0, we will show that there exists δ=δ(ε)≥0 such that (2.1) holds for all x,y∈X. Without loss of generality, take x≤y. Then, it is easy to show that
(2.26)p(Tx,Ty)=y3(1+y)M(x,y)=max⁡{p(x,y),p(Tx,x),p(Ty,y),12[p(Tx,y)+p(x,Ty)]}=y.
Thus, taking δ(ε)=2ε, we get that (2.1) holds. Also, by Lemma 1.5, the mapping T is continuous, and hence it is orbitally continuous. All hypotheses of Theorem 2.5 are satisfied and z=0 is the unique fixed point of T.Example 2.7.
Let(X,p) be the interval [0,2] equipped with the partial metric p(x,y)=max{x,y}. Consider T:X→X defined by
(2.27)Tx={x2if0≤x<1,12if1≤x≤2.
Take x≤y. Given ε>0, we have the two following cases.Case 1 . (0≤x≤y<1).
We have(2.28)p(Tx,Ty)=y2,M(x,y)=y.Case 2 . ((0≤x<1 and 1≤y<2) or (1≤x≤y≤2)).
We have(2.29)p(Tx,Ty)=12,M(x,y)=y.
In each case, it suffices to take δ=ε in order that (2.1) holds. Again, by Lemma 1.5, the mapping T is continuous, and hence it is orbitally continuous. All hypotheses of Theorem 2.5 are satisfied and z=0 is the unique fixed point of T.



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