# Further Remarks on Fixed-Point Theorems in the Context of Partial Metric Spaces

Authors: Mohamed Jleli; Erdal Karapınar; Bessem Samet
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

New fixed-point theorems on metric spaces are established, and analogous results on partial
metric spaces are deduced. This work can be considered as a continuation of the paper Samet et al. (2013).


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## Body

## 1. Introduction

In 1994, Matthews [1] introduced the concept of partial metric spaces as a part of the study of denotational semantics of dataflow networks and showed that the Banach’s contraction principle can be generalized to the partial metric context for applications in program verification. Later on, many authors studied fixed-point theorems on partial metric spaces (see, e.g., [2–9] and references therein).We start by recalling some basic definitions and properties of partial metric spaces (see [1, 5] for more details).Definition 1.
A partial metric on a nonempty setX is a function p:X×X→[0,+∞) such that for all x,y,z∈X, we have(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).
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, ifp(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 ([0,+∞),p), where p(x,y)=max{x,y} for all x,y∈[0,+∞). Other examples of partial metric spaces which are interesting from a computational point of view may be found in [1].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
(1)Bp(x,ε)={y∈X:p(x,y)<p(x,x)+ε},
for all x∈X and ε>0.Definition 2.
Let(X,p) be a partial metric space and {xn} a sequence in X. Then {xn} converges to a point x∈X if and only if p(x,x)=limn→+∞p(x,xn). We may write this as xn→x; {xn} is called a Cauchy sequence if limn,m→+∞p(xn,xm) exists and is finite; (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).
Ifp is a partial metric on X, then the function ps:X×X→[0,+∞) given by
(2)ps(x,y)=2p(x,y)-p(x,x)-p(y,y)
is a metric on X.Lemma 3.
Let(X,p) be a partial metric space. Then(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)
a partial metric space(X,p) is complete if and only if the metric space (X,ps) is complete; furthermore, limn→+∞ps(xn,x)=0 if and only if
(3)p(x,x)=limn→+∞p(xn,x)=limn,m→+∞p(xn,xm).In [1], Matthews extended the Banach contraction principle to the setting of partial metric spaces.Theorem 4 (see [1]).
Let(X,p) be a complete partial metric space and T:X→X a given mapping. Suppose that there exists a constant k∈(0,1) such that
(4)p(Tx,Ty)≤kp(x,y),∀x,y∈X.
Then T has a unique fixed point z∈X. Moreover, we have p(z,z)=0.Very recently, in [10, 11], the authors proved that a large class of fixed-point theorems in partial metric spaces, including Matthews result, are immediate consequences of fixed-point theorems in metric spaces. More precisely, in [11], the authors established the following result.Theorem 5 (see [11]).
Let(X,d) be a complete metric space, T:X→X, and φ:X→[0,+∞) a lower semicontinuous function. Suppose that there exists k∈(0,1) such that
(5)d(Tx,Ty)+φ(Tx)+φ(Ty)≤k(d(x,y)+φ(x)+φ(y)),
for all x,y∈X. Then T has a unique fixed point z∈X. Moreover, one has φ(z)=0.Now, takingφ(x)=p(x,x) in Theorem 4, we obtain that
(6)ps(Tx,Ty)+φ(Tx)+φ(Ty)≤k(ps(x,y)+φ(x)+φ(y)),
for all x,y∈X. Applying Theorem 5, we obtain immediately the result of Matthews. Another technique suggested by Haghi et al. [10] let D be a metric on X defined by
(7)D(x,y):={p(x,y)ifx≠y,0ifx=y.
It is not difficult to show that if (X,p) is complete, then (X,D) is complete. Now, Matthews’ contraction is equivalent to
(8)D(Tx,Ty)≤kD(x,y),
for all x,y∈X. So, the result of Matthews can be deduced immediately from the Banach contraction principle. Observe that in Theorem 5, if we consider the mapping Dφ:X×X→[0,+∞) defined by
(9)Dφ(x,y):={d(x,y)+φ(x)+φ(y)ifx≠y,0ifx=y,
then Dφ is a metric on X and (X,Dφ) is a complete metric space. Moreover, the contractive condition in Theorem 5 is equivalent to
(10)Dφ(Tx,Ty)≤kDφ(x,y),
for all x,y∈X. So, Theorem 5 can also be deduced from the Banach contraction principle.Define the functionF:[0,+∞)3→[0,+∞) by
(11)F(a,b,c):=a+b+c,∀a,b,c≥0.
Then, the contractive condition in Theorem 5 is equivalent to
(12)F(d(Tx,Ty),φ(Tx),φ(Ty))≤kF(d(x,y),φ(x),φ(y)),
for all x,y∈X. In this paper, we establish new fixed-point theorems in metric spaces involving a function F:[0,+∞)3→[0,+∞), where the mapping
(13)Dφ(x,y):={F(d(x,y),φ(x),φ(y))ifx≠y,0ifx=y
is not a metric on X. Some fixed-point theorems in partial metric spaces are deduced from our main results in metric spaces.


## 2. Main Result

We denote byℱ the set of functions F:[0,+∞)3→[0,+∞) satisfying the following conditions:(i)
max{a,b}≤F(a,b,c) for all a,b,c∈[0,+∞),(ii)
F(0,0,0)=0,(iii)
F is continuous.As examples, the following functions belong toℱ:(a)
F(a,b,c)=(a+b)(c+1),(b)
F(a,b,c)=a(ln(c+1)+1)+b(ec+1),(c)
F(a,b,c)=a+b+c,(d)
F(a,b,c)=max{a,b,c},(e)
F(a,b,c)=a+max{a,b,c}.We denote byΨ the set of functions ψ:[0,+∞)→[0,+∞) satisfying the following conditions:(j)
ψ is nondecreasing,(jj)
∑n≥1‍ψn(t)<∞ for each t>0, where ψn is the nth iterate of ψ.It is not difficult to show that ifψ∈Ψ, then ψ(t)<t for every t>0. These functions are known in the literature as (c)-comparison functions.Definition 6.
Let(X,d) be a metric space, T:X→X, φ:X→[0,+∞), and F∈ℱ. We say that the pair (T,φ) satisfies an F-generalized ψ-contraction if there exists ψ∈Ψ such that
(14)F(d(Tx,Ty),φ(Tx),φ(Ty))≤ψ(max{+F2-1(d(y,Ty),φ(Ty),φ(y)))F(d(x,y),φ(x),φ(y)),(F(d(x,Tx),φ(Tx),φ(x))+F(d(y,Ty),φ(Ty),φ(y)))×2-1}),
for every x,y∈X.Our main result is giving by the following theorem.Theorem 7.
Let(X,d) be a complete metric space, T:X→X, and φ:X→[0,+∞). Suppose that the following conditions hold: (1)
φ is lower semicontinuous,(2)
there existψ∈Ψ and F∈ℱ such that the pair (T,φ) satisfies an F-generalized ψ-contraction.
ThenT has a unique fixed point z∈X. Moreover, one has φ(z)=0.Proof.
Letx0∈X. Define the sequence {xn}⊂X by xn+1=Txn for all positive integer n. Let n be a positive integer such that n≥1. Using condition (2) with x=xn and y=xn-1, we obtain that
(15)F(d(xn+1,xn),φ(xn+1),φ(xn))≤ψ(max{F(d(xn,xn-1),φ(xn),φ(xn-1)),F(d(xn,xn+1),φ(xn+1),φ(xn))}).
Without restriction of the generality, we can suppose that xm+1≠xm for every m. Suppose that
(16)max{F(d(xn,xn-1),φ(xn),φ(xn-1)),F(d(xn,xn+1),φ(xn+1),φ(xn))}=F(d(xn,xn+1),φ(xn+1),φ(xn)).
In this case, we have
(17)F(d(xn+1,xn),φ(xn+1),φ(xn))≤ψ(F(d(xn,xn+1),φ(xn+1),φ(xn))).
Note that F(d(xn,xn+1),φ(xn+1),φ(xn))>0. Indeed, if F(d(xn,xn+1),φ(xn+1),φ(xn))=0, from condition (i), we have d(xn,xn+1)≤F(d(xn,xn+1),φ(xn+1),φ(xn))=0, which implies that xn=xn+1, that is a contradiction with the assumption xm+1≠xm for every m. Since ψ(t)<t for every t>0, we get that
(18)F(d(xn+1,xn),φ(xn+1),φ(xn))<F(d(xn+1,xn),φ(xn+1),φ(xn)),
which is a contradiction. Then, we deduce that
(19)F(d(xn+1,xn),φ(xn+1),φ(xn))≤ψ(F(d(xn,xn-1),φ(xn),φ(xn-1))),
for every n≥1. Using the property (i) of F and the monotony property of ψ, we obtain that
(20)max{φ(xn+1),d(xn+1,xn)}≤ψn(F(d(x1,x0),φ(x1),φ(x0))),
for every n≥1. Fix ε>0, and let h=h(ε) be a positive integer (given by (jj)) such that
(21)∑n≥h‍ψn(F(d(x1,x0),φ(x1),φ(x0)))<ε.
Let m>n>h; using the triangular inequality and (20), we obtain
(22)d(xn,xm)≤∑k=nm-1‍d(xk,xk+1)≤∑n≥h‍ψn(F(d(x1,x0),φ(x1),φ(x0)))<ε.
Thus we proved that {xn} is a Cauchy sequence in the metric space (X,d). Since (X,d) is complete, there exists some z∈X such that xn→z as n→∞. We shall prove that z is a fixed point of T. At first, observe that from condition (jj) and (20), we have
(23)limn→+∞φ(xn)=0,
which implies (since φ is lower semicontinuous) that
(24)φ(z)=0.
Using condition (2) with x=z and y=xn, we get that
(25)F(d(Tz,xn+1),φ(Tz),φ(xn+1))≤ψ(max{(d(xn,xn+1)(2)-1,φ(xn+1),φ(xn))F(d(z,xn),φ(z),φ(xn)),((d(xn,xn+1),φ(xn+1),φ(xn))F(d(z,Tz),φ(Tz),φ(z))+F(d(xn,xn+1),φ(xn+1),φ(xn)))×2-1}),
for every n. Suppose that z≠Tz. Using the continuity of F, (23), and (24), we have
(26)limn→+∞max{F2-1(d(xn,xn+1),φ(xn+1),φ(xn)))F(d(z,xn),φ(z),φ(xn)),((d(xn,xn+1),φ(xn+1),φ(xn))F(d(z,Tz),φ(Tz),φ(z))+F(d(xn,xn+1),φ(xn+1),φ(xn)))×2-1}=F(d(z,Tz),φ(Tz),0)2>0.
Then, there exists a positive integer N such that
(27)max{F2-1(d(xn,xn+1),φ(xn+1),φ(xn)))F(d(z,xn),φ(z),φ(xn)),((d(xn,xn+1),φ(xn+1),φ(xn))F(d(z,Tz),φ(Tz),φ(z))+F(d(xn,xn+1),φ(xn+1),φ(xn)))×2-1}>0,
for every n≥N. Thus, we have
(28)F(d(Tz,xn+1),φ(Tz),φ(xn+1))<max{F2-1(d(xn,xn+1),φ(xn+1),φ(xn)))F(d(z,xn),φ(z),φ(xn)),((d(xn,xn+1),φ(xn+1),φ(xn))F(d(z,Tz),φ(Tz),φ(z))+F(d(xn,xn+1),φ(xn+1),φ(xn)))×2-1}.
Letting n→+∞ in the above inequality, we get that
(29)F(d(Tz,z),φ(Tz),0)≤F(d(z,Tz),φ(Tz),0)2,
which is a contradiction. Thus, we proved that Tz=z. Now, suppose that z′∈X is another fixed point of T. We can observe that
(30)φ(z′)=0.
Indeed, applying (14) with x=y=z′, we get that
(31)F(0,φ(z′),φ(z′))≤ψ(F(0,φ(z′),φ(z′))),
which implies (30). Now, applying (14) with x=z and y=z′, we obtain
(32)F(d(z,z′),φ(z),φ(z′))≤ψ(max{+F(0,φ(z′),φ(z′))2F(d(z,z′),φ(z),φ(z′)),F(0,φ(z),φ(z))+F(0,φ(z′),φ(z′))2}).
Using (24) and (30), we obtain
(33)F(d(z,z′),0,0)≤ψ(F(d(z,z′),0,0)),
which implies that d(z,z′)=0; that is, z=z′. Thus, we proved that T has a unique fixed point z∈X with φ(z)=0.Example 8.
LetT:[0,1]→[0,1] be the mapping defined by
(34)Tx={x4if0≤x<1,12ifx=1.
We endow X=[0,1] with the standard metric d(x,y)=|x-y|. Let φ(x)=x for all x∈X, ψ(t)=(3/4)t for all t≥0, and F(a,b,c)=a+b+c for all a,b,c≥0. Clearly, φ is continuous, ψ∈Ψ, and F∈ℱ. Moreover, we have
(35)F(d(Tx,Ty),φ(Tx),φ(Ty))≤ψ(F(d(x,y),φ(x),φ(y))),
for all x,y∈X. By Theorem 7, T has a unique fixed point z∈X (z=0) with φ(z)=0. Note that in this example, the Banach contraction principle cannot be used since the mapping T is not continuous.


## 3. Particular Cases

In this section, new fixed point results are deduced from our main result given by Theorem7.Corollary 9.
Let(X,d) be a complete metric space, T:X→X, and φ:X→[0,+∞). Suppose that the following conditions hold: (1)
φ is lower semicontinuous,(2)
there existsψ∈Ψ such that
(36)d(Tx,Ty)≤(ψ((d(x,y)+φ(x))(φ(y)+1))-φ(Tx)(φ(Ty)+1)ψ((d(x,y)+φ(x))(φ(y)+1)))×(φ(Ty)+1)-1,for allx,y∈X.ThenT has a unique fixed point z∈X. Moreover, we have φ(z)=0.Proof.
It follows from Theorem7 with F(a,b,c)=(a+b)(c+1).Corollary 10.
Let(X,d) be a complete metric space, T:X→X, and φ:X→[0,+∞). Suppose that the following conditions hold:(1)
φ is lower semicontinuous,(2)
there existsψ∈Ψ such that
(37)d(Tx,Ty)≤(ψ((d(x,Tx)+φ(Tx))(φ(x)+1)2)-φ(Tx)(φ(Ty)+1)((d(x,Tx)+φ(Tx))(φ(x)+1)2))×(φ(Ty)+1)-1,for allx,y∈X.ThenT has a unique fixed point z∈X. Moreover, we have φ(z)=0.Proof.
It follows from Theorem7 with F(a,b,c)=(a+b)(c+1).Corollary 11.
Let(X,d) be a complete metric space, T:X→X, and φ:X→[0,+∞). Suppose that the following conditions hold:(1)
φ is lower semicontinuous,(2)
there existsψ∈Ψ such that
(38)d(Tx,Ty)≤(ψ((d(y,Ty)+φ(Ty))(φ(y)+1)2)-φ(Tx)(φ(Ty)+1)((d(y,Ty)+φ(Ty))(φ(y)+1)2))×(φ(Ty)+1)-1,for allx,y∈X.ThenT has a unique fixed point z∈X. Moreover, we have φ(z)=0.Proof.
It follows from Theorem7 with F(a,b,c)=(a+b)(c+1).Corollary 12.
Let(X,d) be a complete metric space, T:X→X, and φ:X→[0,+∞). Suppose that the following conditions hold:(1)
φ is lower semicontinuous,(2)
there existsψ∈Ψ such that
(39)d(Tx,Ty)≤(ψ(d(x,y)[ln(φ(y)+1)+1]+φ(x)(eφ(y)+1))-φ(Tx)(eφ(Ty)+1))×(ln(φ(Ty)+1)+1)-1,for allx,y∈X.ThenT has a unique fixed point z∈X. Moreover, we have φ(z)=0.Proof.
It follows from Theorem7 with F(a,b,c)=a(ln(c+1)+1)+b(ec+1).Corollary 13.
Let(X,d) be a complete metric space, T:X→X, and φ:X→[0,+∞). Suppose that the following conditions hold:(1)
φ is lower semicontinuous,(2)
there existsψ∈Ψ such that
(40)d(Tx,Ty)≤(ψ(d(x,Tx)[ln(φ(x)+1)+1]+φ(Tx)(eφ(x)+1)2)-φ(Tx)(eφ(Ty)+1)(d(x,Tx)[ln(φ(x)+1)+1]+φ(Tx)(eφ(x)+1)2))×(ln(φ(Ty)+1)+1)-1,for allx,y∈X.ThenT has a unique fixed point z∈X. Moreover, we have φ(z)=0.Proof.
It follows from Theorem7 with F(a,b,c)=a(ln(c+1)+1)+b(ec+1).Corollary 14.
Let(X,d) be a complete metric space, T:X→X, and φ:X→[0,+∞). Suppose that the following conditions hold: (1)
φ is lower semicontinuous,(2)
there existsψ∈Ψ such that
(41)d(Tx,Ty)≤(ψ(d(y,Ty)[ln(φ(y)+1)+1]+φ(Ty)(eφ(y)+1)2)-φ(Tx)(eφ(Ty)+1)(d(y,Ty)[ln(φ(y)+1)+1]+φ(Ty)(eφ(y)+1)2))×(ln(φ(Ty)+1)+1)-1,for allx,y∈X.ThenT has a unique fixed point z∈X. Moreover, we have φ(z)=0.Proof.
It follows from Theorem7 with F(a,b,c)=a(ln(c+1)+1)+b(ec+1).Many other results can be deduced from Theorem7 by considering different choices of F.


## 4. Applications to Partial Metric Spaces

We will show that the following fixed-point theorem in partial metric spaces can be deduced from Theorem7.Corollary 15.
Let(X,p) be a complete partial metric space, and let T:X→X. Suppose that there exists ψ∈Ψ such that
(42)p(Tx,Ty)≤ψ(max{p(x,y),p(x,Tx)+p(y,Ty)2}),
for all x,y∈X. Then T has a unique fixed point z∈X. Moreover, one has p(z,z)=0.Proof.
Letd(x,y):=ps(x,y)/2 for all x,y∈X and φ(x)=p(x,x)/2 for all x∈X. We have
(43)p(x,y)=F(d(x,y),φ(x),φ(y)),∀x,y∈X,
where
(44)F(a,b,c)=a+b+c,∀a,b,c≥0.
Now, inequality (42) is equivalent to
(45)F(d(Tx,Ty),φ(Tx),φ(Ty))≤ψ(max{F(d(y,Ty)×(2)-1,φ(Ty),φ(y)))F(d(x,y),φ(x),φ(y)),(F(d(x,Tx),φ(Tx),φ(x))+F(d(y,Ty),φ(Ty),φ(y)))×2-1}),
for every x,y∈X. Applying Theorem 7, we obtain that T has a unique fixed point z∈X with φ(z)=0; that is, p(z,z)=0.The following results follow immediately from Corollary15.Corollary 16.
Let(X,p) be a complete partial metric space, and let T:X→X. Suppose that there exists ψ∈Ψ such that
(46)p(Tx,Ty)≤ψ(p(x,y)),
for all x,y∈X. Then T has a unique fixed point z∈X. Moreover, one has p(z,z)=0.Corollary 17.
Let(X,p) be a complete partial metric space, and let T:X→X. Suppose that there exists ψ∈Ψ such that
(47)p(Tx,Ty)≤ψ(p(x,Tx)+p(y,Ty)2),
for all x,y∈X. Then T has a unique fixed point z∈X. Moreover, one has p(z,z)=0.Note that Matthews result (see Theorem4) follows from Corollary 16 by taking ψ(t)=kt, where k∈(0,1).



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