# Nonlinear Contractions in 0-Complete Partial Metric Spaces

Authors: Abd Ghafur Bin Ahmad; Zaid Mohammed Fadail; Vesna Ćojbašić Rajić; Stojan Radenović
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

Using the setting of 0-complete partial metric spaces, some common
fixed point results of maps that satisfy the nonlinear contractive conditions are obtained. These results generalize and improve the existing fixed point results in the literature in the sense that weaker conditions are used. An example shows how our result can be used when the corresponding result in standard metric cannot.


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## Body

## 1. Introduction and Preliminaries

Matthews [1] generalized the concept of a metric space by introducing partial metric spaces. Based on the notion of partial metric spaces, Matthews [1, 2], Oltra and Valero [3], Ilić et al. [4, 5] obtained some fixed point theorems for mappings satisfying different contractive conditions. Recently, Abdeljawad et al. [6] proved one fixed point result for generalized contraction principle with control functions on partial metric spaces. For some new results on partial metric and cone metric spaces, see [1–27].The aim of this paper is to continue the study of common fixed points of mappings but now in 0-complete partial metric spaces, under nonlinear generalized contractive conditions.Consistent with Matthews [1, 2], O'Neill [21, 22], and Oltra et al. [23], the following definitions and results will be needed throughout this paper.Definition 1.1.
A partial metric on a nonempty setX 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,z)≤p(x,y)+p(y,z)-p(y,y).
A partial metric space is a pair(X,p) such that X is a nonempty set and p is a partial metric on X.
For a partial metricp on X, the function ps:X×X→R+ given by
(1.1)ps(x,y)=2p(x,y)-p(x,x)-p(y,y)
is a (usual) metric on X. Each partial metric p on X generates a T0 topology τp on X with a base of 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.Definition 1.2 (see [1, 20]).
(i) A sequence{xn} in a partial metric space (X,p) converges to x∈X if and only if p(x,x)=limn→+∞p(xn,x).(ii)
a sequence{xn} in a partial metric space (X,p) is called 0-Cauchy if limn,m→+∞p(xn,xm)=0.(iii)
a partial metric space(X,p) is said to be 0-complete if every 0-Cauchy sequence {xn} in X converges, with respect to τp, to a point x∈X such that p(x,x)=0. In this case, p is a 0-complete partial metric on X.Remark 1.3.
(1) (see [20]) Clearly, a limit of a sequence in a partial metric space does not need to be unique. Moreover, the function p(·,·) does not need to be continuous in the sense that xn→x and yn→y implies p(xn,yn)→p(x,y). For example, if X=[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→∞.(2)
(see [6]) However, if p(xn,x)→p(x,x)=0, then p(xn,y)→p(x,y) for all y∈X.
Assertions similar to the following lemma were used (and proved) in the course of proofs of several fixed point results in various papers [8, 9, 20, 28].Lemma 1.4.
Let(X,p) be a partial metric space and let {yn} be a sequence in X such that p(yn,yn+1) is nonincreasing and that
(1.2)limn→+∞p(yn,yn+1)=0.
If {y2n} is not a 0-Cauchy sequence in (X,p), then there exist ε>0 and two sequences {mk} and {nk} of positive integers such that mk>nk>k and the following four sequences tend to ε+ when k→+∞:(1.3)p(y2mk,y2nk),p(y2mk,y2nk+1),p(y2mk-1,y2nk),p(y2mk-1,y2nk+1).Definition 1.5 (see [29]).
Letf and g be self-maps of a set X. If w=fx=gx for some x∈X, then x is called a coincidence point of f and g, and w is called a point of coincidence of f and g. The pair f,g of self-maps is weakly compatible if they commute at their coincidence points.The following lemma is Proposition 1.4 of [29].Lemma 1.6.
Letf and g be weakly compatible self-maps of a set X. If f and g have a unique point of coincidence w=fx=gx, then w is the unique common fixed point of f and g.Definition 1.7 (see [30]).
The following two classes of mappings are defined asΨ={ψ|ψ:[0,∞)→[0,∞) is a nondecreasing right continuous function such that limn→+∞ψn(t)=0 for all t>0, and Φ={ψ|ψ:[0,∞)→[0,∞) is a nondecreasing function such that limn→+∞ψn(t)=0 for all t>0.
It is clear for the function(1.4)ψ(t)={13t,0≤t<112,t=123t,t>1,
that ψ∈Φ but ψ∉Ψ.Lemma 1.8 (see [19]).
Ifψ∈Φ, then ψ(t)<t for all t>0 and ψ(0)=0.


## 2. Common Fixed Point Results

In this section, we obtain some common fixed point results defined on 0-complete partial metric spaces.Theorem 2.1.
Let(X,p) be a partial metric space and T,f be mappings on X with TX⊂fX. Assume that
(2.1)p(Tx,Ty)≤ψ(max{p(fx,fy),p(fx,Tx),p(fy,Ty),p(fx,Ty)+p(fy,Tx)2})
for all x,y∈X, where ψ∈Ψ. If TX or fX is a 0-complete subspace of X, then T and f have a unique point of coincidence. Moreover, if T and f are weakly compatible, then T and f have a unique common fixed point.Proof.
First, we prove thatT and f have a unique point of coincidence (if it exists). If y∈X with y=Tu=fu and z∈X with z=Ts=fs, we assume z≠y. Applying (2.1) we have
(2.2)p(y,z)=p(Tu,Ts)≤ψ(max{p(fu,fs),p(fu,Tu),p(fs,Ts),p(fu,Ts)+p(fs,Tu)2})=ψ(max{p(y,z),p(y,y),p(z,z),p(y,z)+p(z,y)2})=ψ(p(y,z))(by(p2))<p(y,z).
This implies a contradiction, so we conclude thatz=y.
Now, consider the sequence{xn}⊂X defined by x0∈X and Txn=fxn+1. Consider the two possible cases: (i)
p(Txn+1,Txn)=0 for some n∈ℕ.In this caseTxn=fxn(=y a point of coincidence). Because of Lemma 1.6 (unique point of coincidence and weakly compatible mappings) we have a unique common fixed point.(iii)
p(Txn+1,Txn)>0 for every n∈ℕ.
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.Ifp(Txn+1,Txn) is a maximum, then we have
(2.4)p(Txn+1,Txn)≤ψ(p(Txn,Txn+1))<p(Txn,Txn+1),
which is a contradiction. So, we conclude that the maximum is p(Txn,Txn-1). Now, we have the following:
(2.5)p(Txn+1,Txn)≤ψ(p(Txn,Txn-1))<p(Txn,Txn-1).
It follows from (2.5) that the sequence p(Txn+1,Txn) is monotone decreasing. According to the properties of function ψ, it follows
(2.6)p(Txn+1,Txn)≤ψ(p(Txn,Txn-1))≤ψ2(p(Txn-1,Txn-2))≤⋯≤ψn(p(Tx1,Tx0))→0,
when n→+∞. Therefore, limn→+∞p(Txn+1,Txn)=0.Next, we prove that{Txn} is a 0-Cauchy sequence in the space (X,p). It is sufficient to show that {Tx2n} is a 0-Cauchy sequence. Assume the opposite. Then using Lemma 1.4 we get that there exist ε>0 and two sequences {m(k)} and {n(k)} of positive integers and sequences
(2.7)p(y2mk,y2nk),p(y2mk,y2nk+1),p(y2mk-1,y2nk),p(y2mk-1,y2nk+1),
all tend to ε+ when k→+∞. Applying condition (2.1) to elements x=x2m(k) and y=x2n(k)+1 and putting yn=Txn=fxn+1 for each n≥0, we get that
(2.8)p(Tx2m(k),Tx2n(k)+1)=p(y2m(k),y2n(k)+1)≤ψ({p(fx2m(k),Tx2n(k)+1)+p(fx2n(k)+1,Tx2m(k))2}max{{p(fx2m(k),Tx2n(k)+1)+p(fx2n(k)+1,Tx2m(k))2}p(fx2m(k),fx2n(k)+1),p(fx2m(k),Tx2m(k)),p(fx2n(k)+1,Tx2n(k)+1),p(fx2m(k),Tx2n(k)+1)+p(fx2n(k)+1,Tx2m(k))2})=ψ({p(y2m(k)-1,y2n(k)+1)+p(y2n(k),y2m(k))2}max{{p(y2m(k)-1,y2n(k)+1)+p(y2n(k),y2m(k))2}p(y2m(k)-1,y2n(k)),p(y2m(k)-1,y2m(k)),p(y2n(k),y2n(k)+1),p(y2m(k)-1,y2n(k)+1)+p(y2n(k),y2m(k))2}).Whenk→+∞, we have
(2.9)max{{p(y2m(k)-1,y2n(k)+1)+p(y2n(k),y2m(k))2}p(y2m(k)-1,y2n(k)),p(y2m(k)-1,y2m(k)),p(y2n(k),y2n(k)+1),maxp(y2m(k)-1,y2n(k)+1)+p(y2n(k),y2m(k))2}→max{ε+,0,0,12(ε++ε+)}=ε+.
Using properties of ψ, we obtain a contradiction ε≤ψ(ε)<ε, since ε>0.This shows that{Tx2n} is a 0-Cauchy sequence in the space (X,p) and hence {Txn} is a 0-Cauchy sequence in (X,p). If we suppose that TX is a 0-complete subspace of (X,p), then there exists y∈TX⊂fX such that
(2.10)p(y,y)=limn→+∞p(Txn,y)=limn→+∞p(fxn,y)=limn,m→+∞p(Txn,Txm)=0.
If fX is a 0-complete subspace of (X,p) with y∈fX, (2.10) also holds.Letu∈X and y=fu. We show that y is a point of coincidence of T and f. Suppose that p(Tu,fu)>0. We have
(2.11)p(Txn,Tu)≤ψ(max{p(fxn,fu),p(fxn,Txn),p(fu,Tu),p(fxn,Tu)+p(fu,Txn)2}).Asn→+∞, we have p(fxn,fu)→0,p(fxn,Txn)→0 and by Remark 1.3 (2) (p(fxn,Tu)+p(fu,Txn))/2→p(fu,Tu)/2. Since p(fu,Tu)>0, there exists n0∈ℕ such that for every n>n0 we have p(fxn,fu)<(1/2)p(fu,Tu) and p(fxn,Txn)<(1/2)p(fu,Tu). Now, for n>n0, we obtain
(2.12)p(fu,Tu)≤p(fu,fxn+1)+p(Txn,Tu)≤p(fu,fxn+1)+ψ(p(fu,Tu))
or
(2.13)p(fu,Tu)≤p(fu,fxn+1)+ψ(p(fu,Tu)).
As n→+∞ we have a contradiction:
(2.14)p(fu,Tu)≤ψ(p(fu,Tu))<p(fu,Tu).
This implies p(Tu,fu)=0, that is, Tu=fu. Hence, T and f have a (unique) point of coincidence. From Lemma 1.6 follows that this is a unique common fixed point of T and f.Remark 2.2.
Assumption about right continuity of the function is only used in the proof that{Txn} is a 0-Cauchy sequence.
In the following theorem we consider a weaker assumption for the functionψ. As a compensation we assume a bit stronger contractive condition.Theorem 2.3.
Let(X,p) be a partial metric space and T,f be mappings on X with TX⊂fX. Assume that
(2.15)p(Tx,Ty)≤ψ(max{p(fx,fy),p(fx,Tx),p(fy,Ty),p(fx,Ty)2})
for all x,y∈X, where ψ∈Φ. If TX or fX is a 0-complete subspace of X, then T and f have a unique point of coincidence. Moreover, if T and f are weakly compatible, then T and f have a unique common fixed point.Proof.
We can prove thatT and f have a unique point of coincidence in a similar way like in Theorem 2.1. If we consider the sequence {xn}⊂X defined by x0∈X and Txn=fxn+1, we used Theorem 2.1 to show limn→+∞p(Txn+1,Txn)=0.
Here, we only prove that{Txn} is a 0-Cauchy sequence in the space (X,p) by using induction. Let us denote yn=Txn=fxn+1 for each n≥0, We have p(yn,yn+1)→0, as n→+∞. For ε>0 there exists n(ε) such that for every n>n(ε) we have p(yn,yn+1)<ε-ψ(ε). Assuming that for some n>n(ε) and k∈ℕ it holds that p(yn,yn+k)<ε and we need to prove that p(yn,yn+k+1)<ε.
We have(2.16)p(yn,yn+k+1)≤p(yn,yn+1)+p(yn+1,yn+k+1),p(yn+1,yn+k+1)=p(Txn+1,Txn+k+1)≤ψ(max{p(fxn+1,fxn+k+1),p(fxn+1,Txn+1),p(fxn+k+1,Txn+k+1),p(fxn+1,Txn+k+1)2})=ψ(max{p(yn,yn+k),p(yn,yn+1),p(yn+k,yn+k+1),p(yn,yn+k+1)2})≤ψ(max{p(yn,yn+k),p(yn,yn+1),p(yn+k,yn+k+1),p(yn,yn+k)+p(yn+k,yn+k+1)2})≤ψ(max{ε,ε-ψ(ε),ε-ψ(ε),ε+ε-ψ(ε)2})=ψ(ε).
Then (2.16) becomes p(yn,yn+k+1)≤p(yn,yn+1)+p(yn+1,yn+k+1)<ε-ψ(ε)+ψ(ε)=ε. This shows that {Txn} is a 0-Cauchy sequence in (X,p). Further, proof is similar as in Theorem 2.1, so we omit it.Corollary 2.4.
Let(X,p) be a partial metric space and let T:X→X be a map such that
(2.17)p(Tx,Ty)≤ψ(max{p(x,y),p(x,Tx),p(y,Ty),p(x,Ty)+p(y,Tx)2})
for all x,y∈X, where ψ∈Ψ. If TX is a 0-complete subspace of X, then T has a unique fixed point.Proof.
It follows from Theorem2.1 by taking f=iX (the identity map).Corollary 2.5.
Let(X,p) be a partial metric space and let T:X→X be a map such that
(2.18)p(Tx,Ty)≤ψ(max{p(x,y),p(x,Tx),p(y,Ty),p(x,Ty)2})
for all x,y∈X, where ψ∈Φ. If TX is a 0-complete subspace of X, then T has a unique fixed point.Proof.
It follows from Theorem2.3 by taking f=iX (the identity map).Corollary 2.6.
Let(X,p) be a partial metric space and let T,f be mappings on X with TX⊂fX. Assume that
(2.19)p(Tx,Ty)≤λmax{p(fx,fy),p(fx,Tx),p(fy,Ty),p(fx,Ty)+p(fy,Tx)2}
for all x,y∈X, where λ∈[0,1). If TX or fX is a 0-complete subspace of X, then T and f have a unique point of coincidence. Moreover, if T and f are weakly compatible, then T and f have a unique fixed point.Proof.
It follows from Theorem2.1 by taking ψ(t)=λt, where λ∈[0,1).Corollary 2.7.
Let(X,p) be a partial metric space and let T,f be mappings on X with TX⊂fX. Assume that
(2.20)p(Tx,Ty)≤a1p(fx,fy)+a2p(fx,Tx)+a3p(fy,Ty)+a4(p(fx,Ty)+p(fy,Tx))
for all x,y∈X, where a1,a2,a3,a4≥0 and a1+a2+a3+2a4∈[0,1). If TX or fX is a 0-complete subspace of X, then T and f have a unique point of coincidence. Moreover, if T and f are weakly compatible, then T and f have a unique fixed point.Proof.
It follows from Corollary2.6 by noting that
(2.21)a1p(fx,fy)+a2p(fx,Tx)+a3p(fy,Ty)+a4(p(fx,Ty)+p(fy,Tx))≤(a1+a2+a3+2a4)max{p(fx,fy),p(fx,Tx),p(fy,Ty),p(fx,Ty)+p(fy,Tx)2}Corollary 2.8.
Let(X,p) be a partial metric space and let T,f be mappings on X with TX⊂fX. Assume that
(2.22)p(Tx,Ty)≤λp(fx,fy),
for all x,y∈X, where λ∈[0,1). If TX or fX is a 0-complete subspace of X, then T and f have a unique point of coincidence. Moreover, if T and f are weakly compatible, then T and f have a unique fixed point.Proof.
It follows from Corollary2.7.From Corollary2.8 follows the theorem proved by Jungck [31], where we consider a 0-complete partial metric space instead of a complete metric space, under the assumption that T and f are commuting mappings.Corollary 2.9.
Let(X,p) be a 0-complete partial metric space and T:X→X be a map such that
(2.23)p(Tx,Ty)≤ψ(p(x,y))
for all x,y∈X, where ψ∈Φ. If TX is a closed subspace of X, then T has a unique fixed point.Proof.
It follows from Theorem2.3 and Corollary 2.5. by taking f=iX (the identity map).Under a weaker assumption for the functionψ, from Corollary 2.9 follows the theorem proved by Boyd and Wong [32] where we consider a partial metric instead of a standard metric.From Corollary2.7 we deduce the following corollaries, which are extensions of some well-known theorems.Corollary 2.10.
Let(X,p) be a 0-complete partial metric space and let T:X→X be a map such that
(2.24)p(Tx,Ty)≤λp(x,y)
for all x,y∈X, where λ∈[0,1). Then T has a unique fixed point.This corollary is an extension of Banach contraction theorem on a 0-complete partial metric space. This corollary is already mentioned in Matthews [2] for a complete partial metric space, but is true for a 0-complete partial metric spaces.Corollary 2.11.
Let(X,p) be a 0-complete partial metric space and let T:X→X be a map such that
(2.25)p(Tx,Ty)≤λ(p(x,Tx)+p(y,Ty))
for all x,y∈X, where λ∈[0,(1/2)). Then T has a unique fixed point.This corollary is extension of Kannan theorem [33] (4) on a 0-complete partial metric spaces.Corollary 2.12.
Let(X,p) be a 0-complete partial metric space and let T:X→X be a map such that
(2.26)p(Tx,Ty)≤a1p(x,y)+a2p(x,Tx)+a3p(y,Ty)
for all x,y∈X, where a1,a2,a3≥ and a1+a2+a2∈[0,1). Then T has a unique fixed point.This corollary is extension of Reich theorem [33] (8) on a 0-complete partial metric spaces.Corollary 2.13.
Let(X,p) be a 0-complete partial metric space and let T:X→X be a map such that
(2.27)p(Tx,Ty)≤λ(p(x,Ty)+p(y,Tx))
for all x,y∈X, where λ∈[0,(1/2)). Then T has a unique fixed point.This corollary is extension of Chatterjea theorem [33] (11) on a 0-complete partial metric spaces.Corollary 2.14.
Let(X,p) be a 0-complete partial metric space and let T:X→X be a map such that at least one of the following is true
(2.28)p(Tx,Ty)≤a1p(x,Tx)p(Tx,Ty)≤a2(p(x,Tx)+p(y,Ty))p(Tx,Ty)≤a3(p(x,Ty)+p(y,Tx))
for all x,y∈X, where a1∈[0,1),a2,a3∈[0,(1/2)). Then T has a unique fixed point.This corollary is extension of Zamfirescu theorem [33] (19)  on a 0-complete partial metric spaces.We demonstrate the use of Theorem2.1 with the help of the following example.Example 2.15.
Let(X,p)=(ℚ∩[0,1),p), where ℚ denotes the set of rational numbers and p is given by p(x,y)=max{x,y}. Then (X,p) is a 0-complete partial metric space which is not complete. Suppose that T,f:X→X are such that Tx=x2/(1+x),fx=x for all x∈X and
(2.29)ψ(t)={t2,t>1515t,t≤15.
Then ψ∈Ψ. Without loss of generality assume that x<y. Then x2/(1+x)<y/(1+y),x2/(1+x)≤x and (y2/(1+y))≤y. From (2.1) follows
(2.30)L=max{x21+x,y21+y}=y21+y,(2.31)R=ψ(max{p(x,y),p(x,x21+x),p(y,y21+y),p(x,y2/(1+y))+p(y,x2/(1+x))2})=ψ(max{y,x,y,max(x,y2/(1+y))+max(y,x2/(1+x))2})≤ψ(y)
and so y2/(1+y)≤ψ(y). If y≤1/5, then we have y2/(1+y)≤y/5 (this is true because y2/(1+y)≤y2≤y/5). If y>1/5 then we have y2/(1+y)≤y2, which is true. It follows that T has a unique fixed point.
On the other hand, consider the same problem in the standard metricd(x,y)=|x-y| and take x=0.10,y=0.30. Then, from (2.1) follows L=0.009-0.069=0.06 and
(2.32)R=ψ(max{0.10-0.30,0.10-0.009,0.30-0.069,0.10-0.069+0.30-0.0092})=ψ(0.231)=0.053.
Hence L≤R does not hold and the existence of a unique fixed point cannot be obtained.Remark 2.16.
Note that Theorem2.1 improves [12, Theorem 1 and Corollary 1], [25, Theorem 3, Corollaries 1 and 2 and Theorem 4], and [13, Corollary 2.3] since our assumptions are weaker than the assumptions from [12, 13, 25] in several places.Finally, it is worth to notice that all new results in recently papers [7, 10, 15, 20, 27] are true if partial metric space (X,p) is 0-complete instead complete.



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*Source: 451239-2012-08-22.xml*