# Comparison Functions and Fixed Point Results in Partial Metric Spaces

Authors: Nawab Hussain; Zoran Kadelburg; Stojan Radenović; Falleh Al-Solamy
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

The aim of this article is to clearly formulate various possible assumptions for a comparison function in contractive conditions and to deduce respective (common) fixed point results in partial metric spaces. Since standard metric spaces are special cases, these results also apply for them. We will show by examples that there exist situations when a partial metric result can be applied, while the standard metric one cannot.


---

## Body

## 1.  Introduction

In recent years many authors have worked on domain theory in order to equip semantics domain with a notion of distance. In particular, Matthews [1] introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks and obtained, among other results, a nice relationship between partial metric spaces and so-called weightable quasimetric spaces. He showed that the Banach contraction mapping theorem can be generalized to the partial metric context for applications in program verification. Subsequently, several authors (see, e.g., [2–11]) proved fixed point theorems in partial metric spaces.Contractive conditions with the so-called comparison functionφ of the form
(1.1)d(Tx,Ty)≤φ(M(x,y))
have been used for obtaining (common) fixed point results of mappings in metric spaces since the celebrated result of Boyd and Wong [12]. In various articles, different assumptions for function φ were made. Sometimes these assumptions were not clearly stated, and sometimes the assumptions were stronger than needed. This includes some recent fixed point results in partial metric spaces.The aim of this paper is to clearly formulate various possible conditions for a comparison function and to deduce respective (common) fixed point results in partial metric spaces. Since standard metric spaces are special cases, these results also apply for them. We will show by examples that there exist situations when a partial metric result can be applied, while the standard metric one cannot.


## 2.  Preliminaries

The following definitions and details can be seen, for example, in [1, 3, 6, 9, 11].Definition 2.1.
A partial metric space is a pair(X,p) of a nonempty set X and a partial metric p on X,that is, a function p:X×X→ℝ+ such that for all x,y,z∈X:(p1)
x=y⇔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).It is clear that ifp(x,y)=0, then, from (p1) and (p2), it follows that x=y. But p(x,x) may not be 0.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. A sequence {xn} in (X,p) converges to a point x∈X, with respect to τp, if limn→∞p(x,xn)=p(x,x). This will be denoted as xn→x, n→∞ or limn→∞xn=x. If T:X→X is continuous at x0∈X (in τp), then, for each sequence {xn} in X,
(2.1)xn⟶x0⟹Txn⟶Tx0.Remark 2.2.
Clearly, a limit of a sequence in a partial metric space needs not to be unique. For example, ifX=[0,+∞) and p(x,y)=max{x,y}for x,y∈X, then for {xn}={1},p(xn,x)=x=p(x,x) for each x≥1 and so, for example, xn→2 and xn→3 when n→∞.Moreover, the function p(·,·) needs not to be continuous in the sense that xn→x and yn→y imply p(xn,yn)→p(x,y).Definition 2.3.
Let(X,p)be a partial metric space. Then we have the following.(1)
A sequence{xn} in (X,p) is called a Cauchy sequence if limn,m→∞p(xn,xm) exists (and is finite).(2)
The 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).It is easy to see that every closed subset of a complete partial metric space is complete.Ifp is a partial metric on X, then the function ps:X×X→ℝ+ given by
(2.2)ps(x,y)=2p(x,y)-p(x,x)-p(y,y)
is a metric on X. Furthermore, limn→∞ps(xn,x)=0 if and only if
(2.3)p(x,x)=lim⁡n→∞p(xn,x)=lim⁡n,m→∞p(xn,xm).Lemma 2.4.
Let(X,p) be a partial metric space: (a)
{xn} is a Cauchy sequence in (X,p) if and only if it is a Cauchy sequence in the metric space (X,ps),(b)
the space(X,p) is complete if and only if the metric space (X,ps) is complete.


## 3.  Auxiliary Results

We will consider the following properties of functionsφ:[0,+∞)→[0,+∞). φn will denote the nth iteration of φ:(I)
φ(t)<t for each t>0 and φn(t)→0, n→∞ for each t≥0,(II)
φ is nondecreasing and φn(t)→0, n→∞ for each t≥0,(III)
φ is right-continuous, and φ(t)<t for each t>0,(IV)
φ is nondecreasing and ∑n≥1φn(t)<+∞ for each t≥0.Lemma 3.1.
(1) (II)⇒ (I).
(2) (III) +φ is nondecreasing ⇒ (II).
(3) (IV)⇒ (II).
(4) (III) and (IV) are not comparable (even ifφ is nondecreasing).Proof.
(1) (see [13]). Suppose that (II) holds and that there is some t0∈(0,+∞) such that φ(t0)≥t0. Then monotonicity of φ implies that φ2(t0)≥φ(t0)≥t0. Continuing by induction we get that φn(t0)≥t0 and so φn(t0)→0, n→∞ is impossible.
(2) Let (III) hold and letφ be nondecreasing. Monotonicity of φ implies that, for each fixed t≥0, the sequence {φn(t)} is nonincreasing (and nonnegative); hence, there exists limn→∞φn(t)=α≥0. Suppose that α>0. Then it follows by (III) that
(3.1)0<α≤lim⁡n→∞φn+1(t)=lim⁡φn(t)→α+φ(φn(t))=φ(lim⁡φn(t)→α+φn(t))=φ(α),
which is a contradiction with φ(t)<t.
(3) Obvious.
(4) It is demonstrated in the following example.Example 3.2.
(1) The function(3.2)φ(t)={13t,0≤t<1,12,t=1,23t,t>1,
satisfies (IV) but not (III).
(2) The functionφ(t)=t/(1+t) is nondecreasing and satisfies (III) but not (IV), since φn(t)=t/(1+nt).Assertions similar to the following lemma (see, e.g., [14]) were used (and proved) in the course of proofs of several fixed point results in various papers.Lemma 3.3.
Let(X,d) be a metric space and let  {yn} be a sequence in  X such that  {d(yn+1,yn)} is nonincreasing and
(3.3)lim⁡n→∞d(yn+1,yn)=0.
If {y2n} is not a Cauchy sequence, then there exist ε>0 and two subsequences {mk} and {nk} of positive integers such that the following four sequences tend to ε+0 when k→∞:
(3.4)d(y2mk,y2nk),d(y2mk,y2nk+1),d(y2mk-1,y2nk),d(y2mk-1,y2nk+1).As a consequence we obtain the following.Lemma 3.4.
Let(X,p) be a partial metric space and let  {yn} be a sequence in  X such that  {p(yn+1,yn)} is nonincreasing and
(3.5)lim⁡n→∞p(yn+1,yn)=0.
If {y2n} is not a Cauchy sequence in (X,p), then there exist ε>0 and two subsequences {mk} and {nk} of positive integers such that the following four sequences tend to ε+0 when k→∞:
(3.6)p(y2mk,y2nk),p(y2mk,y2nk+1),p(y2mk-1,y2nk),p(y2mk-1,y2nk+1).Proof.
Suppose that{yn} is a sequence in (X,p) satisfying (3.5) such that {y2n} is not Cauchy. According to Lemma 2.4, it is not a Cauchy sequence in the metric space (X,ps), either. Applying Lemma 3.3 we get the sequences
(3.7)ps(y2mk,y2nk),ps(y2mk,y2nk+1),ps(y2mk-1,y2nk),ps(y2mk-1,y2nk+1)
tending (from above) to some 2ε>0 when k→∞. Using definition (2.2) of the associated metric and (3.5) (which implies that also limn→∞p(yn,yn)=0), we get that the sequences (3.6) tend to ε+0 when k→∞.


## 4.  Common Fixed Point Results for Four Mappings

In this section we prove two common fixed point results for four mappings in partial metric spaces, using two distinct properties of a comparison function mentioned in Section3.Theorem 4.1.
Let(X,p) be a complete partial metric space, and let  A,B,S,T:X→X. Suppose that  AX⊂TX,  BX⊂SX and one of these four subsets of  X is closed. Let further
(4.1)p(Ax,By)≤φ(M(x,y))
hold for all x,y∈X, where
(4.2)M(x,y)=max⁡{p(Sx,Ty),p(Ax,Sx),p(By,Ty),12(p(Sx,By)+p(Ax,Ty))},
and φ:[0,+∞)→[0,+∞) satisfies property (III), that is, φ is right-continuous, and φ(t)<t for each t>0. Then A and S, as well as B and T, have a coincidence point. If, moreover, pairs (A,S) and (B,T) are weakly compatible, then A,B,S,T have a unique common fixed point.Recall that a pointy∈X is called a point of coincidence for mappings f,g:X→X if there exists x∈X such that f(x)=g(x)=y. Then, x is called a coincidence point. Mappings f and g are called weakly compatible if they commute at their coincidence points.Proof.
Starting from arbitraryx0∈X and using that AX⊂TX, BX⊂SX construct a Jungck sequence {yn} by
(4.3)y2n=Ax2n=Tx2n+1,y2n+1=Bx2n+1=Sx2n+2,n=0,1,2,….
Consider two possible cases.
(1)p(yn,yn+1)=0 (and so yn=yn+1) for some n∈ℕ. Let, for example, n=2k-1, k∈ℕ. Then
(4.4)p(y2k,y2k+1)=p(Ax2k,Bx2k+1)≤φ(M(x2k,x2k+1)),
where
(4.5)M(x2k,x2k+1)=max⁡{p(y2k-1,y2k),p(y2k,y2k-1),p(y2k+1,y2k),12(p(y2k-1,y2k+1)+p(y2k,y2k))}=p(y2k,y2k+1)(by(p4)).
It follows that p(y2k,y2k+1)≤φ(p(y2k,y2k+1))<p(y2k,y2k+1), which is impossible, unless p(y2k,y2k+1)=0 and y2k=y2k+1. In a similar way, if n=2k, k∈ℕ, it follows that also y2k+1=y2k+2. Hence, in both cases we obtain that the sequence {yn} is eventually constant, and so a Cauchy one.
(2) Suppose thatp(yn,yn+1)>0 for all n∈ℕ. Then, as above,
(4.6)p(y2n,y2n+1)=p(Ax2n,Bx2n+1)≤φ(M(x2n,x2n+1)),
where
(4.7)M(x2n,x2n+1)=max⁡{p(y2n-1,y2n),p(y2n,y2n-1),p(y2n+1,y2n),12(p(y2n-1,y2n+1)+p(y2n,y2n))}=max⁡{p(y2n-1,y2n),p(y2n,y2n+1)}(by(p4)).
Similarly, p(y2n+1,y2n+2)≤φ(M(x2n+1,x2n+2)),M(x2n+1,x2n+2)=max{p(y2n,y2n+1),p(y2n+1,y2n+2)}, that is
(4.8)p(yn,yn+1)≤φ(max⁡{p(yn-1,yn),p(yn,yn+1)}),n=1,2,….
If p(yn,yn+1)≥p(yn-1,yn), then p(yn,yn+1)≤φ(p(yn,yn+1))<p(yn,yn+1), a contradiction. It follows that(4.9)p(yn,yn+1)<p(yn-1,yn),n=1,2,….
Thus, in this case {p(yn,yn+1)} is a decreasing sequence of positive numbers. Denote limn→∞p(yn,yn+1)=r≥0. In fact, r=0. Indeed, if r>0, then passing to the limit when n→∞ in p(yn,yn+1)≤φ(p(yn-1,yn)), and using properties (III) of φ, we get that
(4.10)r≤φ(r)<r,
a contradiction.
We have proved thatlimn→∞p(yn,yn+1)=0. Suppose that {y2n} is not a Cauchy sequence. Then, Lemma 3.4 implies that there exist ε>0 and two subsequences {mk} and {nk} of positive integers such that the sequences (3.6) tend to ε (from above) when k→∞. Now, using (4.1) with x=x2mk and y=x2nk+1, we get that(4.11)p(y2mk,y2nk+1)≤φ(M(x2mk,x2nk+1)),
where
(4.12)M(x2mk,x2nk+1)=max⁡{p(y2mk-1,y2nk),p(y2mk,y2mk-1),p(y2nk+1,y2nk),12(p(y2mk-1,y2nk+1)+p(y2mk,y2nk))}⟶ε+0,k⟶∞.
Using properties (III) of φ, we obtain a contradiction ε≤φ(ε)<ε, since ε>0.
Thus{y2n} and so also {yn} is a Cauchy sequence, both in (X,p) and in (X,ps). Suppose that, for example, SX is closed in X, and hence complete. It follows that sequence {yn} converges in the metric space (X,ps), say limn→∞ps(yn,z)=0, where z=Su for some u∈X. Again from Lemma 2.4, we have
(4.13)p(z,z)=lim⁡n→∞p(yn,z)=lim⁡n,m→∞p(yn,ym).
Moreover, since{yn} is a Cauchy sequence in the metric space (X,ps), we have limn,m→∞ps(yn,ym)=0 and so, by the definition of ps, we have limn,m→∞p(yn,ym)=0. Then (4.13) implies that p(z,z)=0 and
(4.14)lim⁡n→∞p(yn,z)=p(z,z)=0.
We will prove that z is a point of coincidence of A and S, that is, Au=z.
Suppose thatp(Au,z)>0. Then, using (p4),(4.15)p(Au,z)≤p(Au,Bx2n+1)+p(Bx2n+1,z)-p(Bx2n+1,Bx2n+1)≤φ(max⁡{p(z,y2n),p(Au,z),p(y2n+1,y2n),12(p(z,y2n+1)+p(Au,y2n))})+p(y2n+1,z)≤φ(max⁡{p(z,y2n),p(Au,z),p(y2n+1,y2n),12(p(z,y2n+1)+p(Au,z)+p(z,y2n))})+p(y2n+1,z).
All terms in the previous set {⋯}, which depend on n, tend to 0 when n→∞ and they are smaller, for n large enough, than, say, (1/2)p(Au,z). It follows that
(4.16)p(Au,z)≤φ(p(Au,z))+p(y2n+1,z).
Letting n→∞, and using that φ(t)<t, we get that p(Au,z)≤φ(p(Au,z))<p(Au,z), a contradiction. Hence, p(Au,z)=0 and Au=z.
NowAX⊂TX implies that z=Au∈TX and so there exists v∈X such that Tv=z. If p(z,Bv)>0, then
(4.17)p(z,Bv)=p(Au,Bv)≤φ(max⁡{p(Su,Tv),p(Au,Su),p(Bv,Tv),12(p(Su,Bv)+p(Au,Tv))})≤φ(p(z,Bv))<p(z,Bv),
a contradiction. Hence,p(z,Bv)=0 and Bv=z, so z is a point of coincidence, both for (A,S) and for (B,T). It order to show that this point of coincidence is unique, one has only to use that φ(t)<t for t>0 and property (p2) of partial metric. Hence, if these pairs of mappings are weakly compatible, it follows by a well-known result that z is a unique common fixed point of A,B,S,T. The theorem is proved.Theorem 4.2.
Let(X,p) be a complete partial metric space, and let  A,B,S,T:X→X. Suppose that  AX⊂TX,  BX⊂SX and one of these four subsets of  X is closed. Let further
(4.18)p(Ax,By)≤φ(M(x,y))
hold for all x,y∈X, where
(4.19)M(x,y)=max⁡{p(Sx,Ty),p(Ax,Sx),p(By,Ty),12(p(Sx,By)+p(Ax,Ty))},
and φ:[0,+∞)→[0,+∞) satisfies property (IV), that is, φ is nondecreasing and ∑n≥1φn(t)<+∞ for each t≥0. Then A and S, as well as B and T, have a coincidence point. If, moreover, pairs (A,S) and (B,T) are weakly compatible, then A,B,S,T have a unique common fixed point.Proof.
Construct a Jungck sequence{yn} as in the proof of Theorem 4.1. By Lemma 3.1, conditions (IV) imply that φn(t)→0, n→∞ for each t∈[0,+∞). Since φ is nondecreasing, it further implies that φ(t)<t for each t>0. Condition (4.9) follows as in the proof of Theorem 4.1. It further implies that
(4.20)p(yn,yn+1)<φn(p(y0,y1)),n∈N.
By the definition of associated metric (2.2), we get that
(4.21)ps(yn,yn+1)<4φn(p(y0,y1)),n∈N.
Now, for arbitrary k∈ℕ,
(4.22)ps(yn,yn+k)≤ps(yn,yn+1)+ps(yn+1,yn+2)+⋯+ps(yn+k-1,yn+k)≤4∑i=nn+k-1φi(p(y0,y1))≤4∑i=n∞φi(p(y0,y1))⟶0
as n→∞, by conditions (IV). Hence {yn} is a Cauchy sequence in X.
The rest of the proof is the same as in Theorem4.1, since it uses only the contractive condition and that φ(t)<t for t>0.Remark 4.3.
(1) It follows from the proofs of previous theorems that, under the respective hypotheses, an arbitrary Jungck sequence{yn} converges to the (unique) common fixed point z of A,B,S,T.
(2) Takingφ(t)=kt, k∈[0,1) in Theorems 4.1 and 4.2, we obtain an extension of Fisher’s theorem for four mappings [15] to the setting of partial metric spaces.
(3) TakingA=B=S=T, Theorems 3 and 4 from [8] are obtained, and so also extensions of [3, Theorem 1] and [2, Theorem 1]. For [4, Theorem 2.1], precise assumptions for the function φ are stated.
(4) Our results are far extensions of the classical Boyd-Wong result [12] to the setting of partial metric spaces.
(5) A related result for so-called Geraghty-type mappings was recently obtained in [5].We illustrate the results of this section with an example.Example 4.4.
LetX=[0,1] be equipped with the partial metric p(x,y)=max{x,y} and let φ(t)=(t/(t+1))e-t for t≥0 (φ satisfies condition (III)). The mappings
(4.23)Ax=16x2,Bx=18x2,Tx=12x2,Sx=x2
are such that AX⊂TX, BX⊂SX, and (A,S) as well as (B,T) are weakly compatible. In order to check condition (4.1), take arbitrary x,y∈X with, first, x≥y. Then,
(4.24)p(Ax,By)=p(16x2,18y2)=16x2,M(x,y)=max⁡{p(x2,12y2),p(16x2,x2),p(18y2,12y2),12[p(x2,18y2)+p(16x2,12y2)]}=max⁡{x2,x2,12x2,12[x2+max⁡{16x2,12y2}]}=x2,φ(M(x,y))=x2x2+1e-x2≥16x2,
since e-x2/(x2+1)≥1/2e>1/6 for x∈[0,1]. In the case x<y, the same can be checked after careful calculations. Hence, A,B,S,T satisfy all the conditions of Theorem 4.1 and they have a unique common fixed point (z=0).


## 5.  Common Fixed Points for Two Mappings under Weaker Condition for the Comparison Function

In the next theorem we consider weaker condition (II) for the comparison functionφ. As a compensation, we assume a bit stronger contractive condition.Theorem 5.1.
Let(X,p) be a complete partial metric space and let  A,T:X→X be two self-maps. Let  AX⊂TX and at least one of these subsets of  X is closed. Suppose that there is a function  φ:[0,+∞)→[0,+∞) satisfying property  (II), such that
(5.1)p(Ax,Ay)≤φ(max⁡{p(Tx,Ty),p(Ax,Tx),p(Ay,Ty),12p(Tx,Ay)})
holds for all x,y∈X. Then A and T have a unique point of coincidence. If A and T are weakly compatible, then they have a unique common fixed point u∈X.Proof.
We will construct a Jungck sequence in the usual way. Take arbitraryx0∈X and using that AX⊂TX, choose xn∈X such that yn=Axn=Txn+1, n=0,1,2,…. First, suppose that there exists an n such that p(yn,yn-1)=0. Then the sequence {yn} is eventually constant. Indeed, from (5.1) it follows that
(5.2)p(yn+1,yn)=p(Axn+1,Axn)≤φ(max⁡{p(Txn+1,Txn),p(Axn+1,Txn+1),p(Axn,Txn),12p(Txn+1,Axn)})=φ(max⁡{p(yn,yn-1),p(yn,yn+1),p(yn-1,yn),12p(yn,yn)})=φ(p(yn+1,yn)),
since (1/2)p(yn,yn)≤p(yn,yn)≤p(yn+1,yn) by (p2). This is a contradiction with φ(t)<t (which follows from assumption (I)), unless p(yn+1,yn)=0. Hence, yn+1=yn=yn-1. Continuing this process, we obtain that {yn} is an eventually constant sequence and hence a Cauchy one. Moreover, from yn=Axn=Txn+1=yn+1=Axn+1=Txn+1 it follows that Axn+1=Txn+1 and A and T have a coincidence point.
Suppose now thatp(yn,yn-1)>0 for all n∈ℕ. Then, similarly as above, we get
(5.3)p(yn+1,yn)≤φ(max⁡{p(yn,yn-1),p(yn+1,yn)}).
Using property (I) of functionφ it follows that
(5.4)p(yn+1,yn)≤φ(p(yn,yn-1))<p(yn,yn-1).
Hence, p(yn+1,yn)<φn(p(y0,y1))→0 when n→∞ and {p(yn+1,yn)} is a decreasing sequence, tending to 0.
Now, using mathematical induction, we prove that{yn} is a Cauchy sequence. Since p(yn,yn+1)→0, for each ε>0, there exists n(ε) such that p(yn,yn+1)<ε-φ(ε) for n>n(ε). Let, for some k∈ℕ, p(yn,yn+k)<ε. Then we have
(5.5)p(yn,yn+k+1)≤p(yn,yn+1)+p(yn+1,yn+k+1),
where(5.6)p(yn+1,yn+k+1)=p(Axn+1,Axn+k+1)≤φ(max⁡{p(Txn+1,Txn+k+1),p(Axn+1,Txn+1),p(Axn+k+1,Txn+k+1),12p(Txn+1,Axn+k+1)})=φ(max⁡{p(yn,yn+k),p(yn+1,yn),p(yn+k+1,yn+k),12p(yn,yn+k+1)}).
The first three members of the last set are smaller than ε. Concerning the fourth one, we have that
(5.7)12p(yn,yn+k+1)≤12p(yn,yn+k)+12p(yn+k,yn+k+1)<12ε+12(ε-φ(ε))<ε,
and it follows that
(5.8)p(yn+1,yn+k+1)<φ(ε).
Hence,
(5.9)p(yn,yn+k+1)<ε-φ(ε)+φ(ε)=ε,
and the inductive proof is over.
Using the definition of the associated metricps, we get that
(5.10)ps(yn,yn+k)≤4p(yn,yn+k)<4ε,
for eachn>n(ε) and k∈ℕ. Thus, {yn} is also a ps-Cauchy sequence, and so there exists y∈X such that
(5.11)p(y,y)=lim⁡n→∞p(yn,y)=lim⁡m,n→∞p(yn,ym).
Suppose that, for example, TX is closed. Then y=Tu for some u∈X. We will prove that also Au=y. Suppose that p(Au,y)>0. Then, using (5.1), we get
(5.12)p(y,Au)≤p(y,Axn)+p(Axn,Au)≤p(y,Axn)+φ(max⁡{p(Txn,Tu),p(Axn,Txn),p(Au,Tu),12p(Tu,Axn)})=p(y,Axn)+φ(max⁡{p(yn-1,Tu),p(yn,yn-1),p(Au,y),12p(Tu,yn)}).
Since yn→y=Tu, there exists n0 such that p(yn-1,Tu)<p(Au,y) and p(yn,yn-1)<p(Au,y) whenever n>n0. It follows that
(5.13)p(y,Au)≤p(y,Axn)+φ(p(Au,y))
and, passing to the limit when n→∞, we obtain that
(5.14)p(y,Au)≤0+φ(p(Au,y))<p(Au,y),
which is a contradiction. Hence, p(Au,y)=0 and Au=y=Tu is a point of coincidence of A and T.
Note that continuity ofφ was not needed in the previous conclusions.
In order to prove that the point of coincidence is unique, suppose that there existsy1≠y (and so p(y1,y)>0) such that Au1=Tu1=y1 for some u1∈X. Then (5.1) implies that
(5.15)p(y1,y)=p(Au1,Au)≤φ(max⁡{p(Tu1,Tu),p(Au1,Tu1),p(Au,Tu),12p(Tu1,Au)})=φ(max⁡{p(y1,y),p(y1,y1),p(y,y),12p(y1,y)})≤φ(p(y1,y))<p(y1,y),since p(y,y)≤p(y,y1) and p(y1,y1)≤p(y,y1). This contradiction shows that p(y1,y)=0 and y1=y.
The last assertion of the theorem follows easily from the definition of weak compatibility and the uniqueness of the coincidence point.We present an example where the existence of a common fixed point can be proved using Theorem5.1 (and conditions formulated in terms of a partial metric) but cannot be obtained using respective conditions in the associated (standard) metric.Example 5.2.
Consider onX={0,1,2} both the partial metric p(x,y)=max{x,y} and the associated metric d(x,y)=|x-y|. Let A:X→X be given by A0=A1=0, A2=1, let T=iX (identity map), and let φ(t)=(3/4)t, t∈[0,+∞). Then
(5.16)p(A0,A1)=p(0,0)=0,p(A0,A2)=p(0,1)=1<3/2=φ(2)=φ(max⁡{p(0,2),p(0,0),p(2,1),1/2p(0,1)}),p(A1,A2)=p(0,1)=1<3/2=φ(2)=φ(max⁡{p(1,2),p(1,0),p(2,1),1/2p(1,1)}),
and condition (5.1) is satisfied. The respective condition in the standard metric does not hold since
(5.17)d(A1,A2)=d(0,1)=1>3/4=φ(1)=φ(max⁡{d(1,2),d(1,0),d(2,1),1/2d(1,1)}).


## 6.  Meir Keeler-Type Result in Partial Metric Spaces

It is well-known that the celebrated Meir-Keeler fixed point result [16] can also be formulated in a form with a comparison function. We present a partial metric version of this result. The function φ:[0,+∞)→[0,+∞) will be called a Meir-Keeler function if
(6.1)∀ε>0∃δ>0∀tε≤t<ε+δ⟹φ(t)<ε.Theorem 6.1.
Let(X,p) be a complete partial metric space and let a mapping  f:X→X satisfies the following condition
(6.2)∀x,y∈Xp(fx,fy)≤φ(p(x,y)),
for a Meir-Keeler function φ. Then f has a unique fixed point, say u, and for each x∈X, the Picard sequence {fnx} converges to u, satisfying p(u,u)=0.Proof.
We prove first the uniqueness. Letu,v be two fixed points of f and let p(u,v)>0. Then, taking ε=p(u,v) in (9), it follows that there exists δ>0 such that p(u,v)≤t<p(u,v)+δ implies that φ(t)<p(u,v). In particular, φ(p(u,v))<p(u,v). But then, by (6.2), p(u,v)=p(fu,fv)≤φ(p(u,v))<p(u,v), a contradiction. Hence, p(u,v)=0 and u=v.
Take now arbitraryx∈X and form the Picard sequence {fnx}. If, for some n, p(fnx,fn+1x)=0, then ffnx=fnx and fnx is a (unique) fixed point of f. Suppose that p(fnx,fn+1x)>0 for each n=0,1,2,…. Taking (for fixed n) ε=p(fnx,fn+1x) in (6.1), there exists δ>0 such that
(6.3)p(fnx,fn+1x)≤t<p(fnx,fn+1x)+δ⟹φ(t)<p(fnx,fn+1x).
Then, by (6.2),
(6.4)p(fn+1x,fn+2x)=p(ffnx,ffn+1x)≤φ(p(fnx,fn+1x))<p(fnx,fn+1x).
It follows that {p(fnx,fn+1x)} is a decreasing sequence of positive numbers, tending to some r≥0. If r>0 then, again by (6.1) and (6.2), we can find δ such that
(6.5)r≤p(fnx,fn+1x)<r+δ⟹p(fn+1x,fn+2x)<r.
This is a contradiction since p(fn+1x,fn+2x)≥r. We conclude that p(fnx,fn+1x)→0 when n→∞.
In order to prove that{fnx} is a Cauchy sequence, we use again Lemma 3.4. If we suppose the contrary, then there exist ε>0 and two sequences {mk} and {nk} of positive integers such that p(fnkx,fmkx)→ε+ and also p(fnk+1x,fmk+1x)→ε+. But, (6.1) and (6.2) imply that there exists δ>0 such that ε≤p(fnkx,fmkx)<ε+δ implies p(fnk+1x,fmk+1x)<ε. This is a contradiction since p(fnk+1x,fmk+1x)≥ε for large k.
Thus,{fnx} is a Cauchy sequence, both in (X,p) and in (X,ps) (Lemma 2.4) and there exists an element u in the complete (partial) metric space X such that limn→∞ps(fnx,u)=0, wherefrom
(6.6)p(u,u)=lim⁡n→∞p(fnx,u)=lim⁡m,n→∞p(fnx,fmx).
By the definition (2.2) of metric ps, and
(6.7)lim⁡n→∞p(fnx,fnx)=lim⁡m→∞p(fmx,fmx)=0,
it follows that
(6.8)p(u,u)=lim⁡n→∞p(fnx,u)=lim⁡m,n→∞p(fnx,fmx)=0.
We have proved that
(6.9)0=p(u,u)=lim⁡n→∞p(fnx,u).
It remains to show that fu=u. Take fnx instead of x and u instead of y in (6.2). Then, using (6.1) we get that
(6.10)p(fn+1x,fu)<p(fnx,u),p(u,fu)≤p(u,fn+1x)+p(fn+1x,fu)-p(fn+1x,fn+1x)≤p(u,fn+1x)+p(fn+1x,fu)<p(u,fn+1x)+p(fnx,u)⟶p(u,u)+p(u,u)=0asn⟶∞.
If follows that p(u,fu)=0 and fu=u.Example 6.2.
LetX, p and d be as in Example 5.2. Consider mapping f:X→X and function φ:[0,+∞)→[0,+∞) given by
(6.11)f=(012001),φ(t)=34t.
Then φ is a Meir-Keeler function. Indeed, for arbitrary ε>0 choose δ=(1/3)ε and ε≤t<ε+δ=(4/3)ε implies that φ(t)<ε. We will check that f satisfies condition (6.2) of Theorem 6.1.
In the casesx=y=0; x=y=1; and x=0,y=1, the left-hand side of (6.2) is equal to zero. In all other cases (x=y=2; x=0,y=2; and x=1,y=2), it is p(fx,fy)=1 and φ(p(x,y))=φ(2)=3/2. Hence, condition (6.2) always holds true, and mapping f has a unique fixed point (u=0).
Note again that in the case when standard metricd is used instead of partial metric p, this conclusion cannot be obtained. Indeed, for x=1, y=2 we have that
(6.12)d(f1,f2)=d(0,1)=1>34=φ(1)=φ(d(1,2)).



---

*Source: 605781-2012-06-13.xml*