# Fixed Point for Partial Metric Spaces

Authors: S. A. M. Mohsenalhosseini; H. Mazaheri; M. A. Dehghan; A. Zareh
Journal: ISRN Applied Mathematics
 (2011)

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## Abstract

We consider the partial metric on a setX, define ϵ-fixed point for maps, and obtain some sufficient and necessary conditions on that, also
we obtain some sufficient and necessary theorems on ϵ-fixed point.


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## Body

## 1. Introduction

The partial metric spaces were introduced in [1] as a part of the study of denotational semantics of dataflow networks. In particular, he established the precise rela- tionship between partial metric spaces and the so-called weightable quasimetric spaces and proved a partial metric generalization of Banach contraction mapping theorem.A partial metric [1] on a set X is a function p:X×X→[0,∞) such that, for all x,y,z∈X(1)
x=y⇔p(x,x)=p(x,y)=p(y,y);(2)
p(x,x)≤p(x,y);(3)
p(x,y)≤p(y,x);(4)
p(x,z)≤p(x,y)+p(y,z)-p(y,y).A partial metric space is a pair(X,p), where p is a partial metric on X. Each partial metric p on X induces a T0 topology τp on X which has as a base the family of open 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.Ifp is a partial metric on X, then the function ps:X×X→[0,∞) given by ps(x,y)=2p(x,y)-p(x,x)-p(y,y) is a metric on X.A sequence{xn}n∈N in a partial metric space (X,p) is called a Cauchy sequence if there exists (and is finite) limn,mp(xn,xm) [1, Definition 5.2].Note that{xn}n∈N is a Cauchy sequence in (X,p) if and only if it is a Cauchy sequence in the metric space (X,ps) [1, page 194].A partial metric space(X,p) is said to be complete if every Cauchy sequence {xn}n∈N in X converges, with respect to τp, to a point x∈X such that p(x,x)=limn,mp(xn,xm) [1, Definition 5.3].It is well known and easy to see that a partial metric space(X,p) is complete if and only if the metric space (X,ps) is complete.In [2], S. J. O’Neill  proposed one significant change to Matthews definition of the partial metrics, and that was to extend their range from R+ to R. In the following, partial metrics in the O’Neill sense will be called dualistic partial metrics and a pair (X,p) such that X is a nonempty set and p is a dualistic partial metric on X will be called a dualistic partial metric space.


## 2.ϵ-Fixed Point

Our basic references for quasi-metric spaces are [3, 4]. In our context, by a quasi-metric on a set X we mean a nonnegative real-valued function d on X×X such that, for all x,y,z∈X:(i)
d(x,y)=d(y,x)=0⇔x=y,(ii)
d(x,y)≤d(x,z)+d(z,y).A quasi-metric space is a pair(X,d) such that X is a (nonempty) set and d is a quasi-metric on X.Each quasi-metricd on X generates a T0-topology T(d) on X which has as a base the family of open d-balls Bd(x,ϵ)={x∈X:ϵ>0}, where Bd(x,ϵ)={y∈X:d(x,y)<ϵ>0} for all x∈X and ϵ>0.Ifd is a quasi-metric on X, then the function ds defined on X×X by ds(x,y)=max{d(x,y),d(y,x)} is a metric on X.Lemma 2.1.
If(X,p) is a dualistic partial metric space,  then the function dp:X×X→R+ defined by dp(x,y)=p(x,y)-p(x,x), is a quasi-metric on X such that τ(p)=τ(dp).Definition 2.2.
Let(X,p)be a dualistic partial metric space and let T:X→,Xbe a map. Then x0∈X is ϵ-fixed point for T if
(2.1)dp(Tx0,x0)≤ϵ.Theorem 2.3.
Let(X,p) be a dualistic partial metric space and let  T:X→X be a map,  x0∈X, and  ϵ>0. If  dp(Tn(x0),Tn+k(x0))→0 as  n→∞ for some  k>0, then  Tk has an  ϵ-fixed point.Proof.
Sincedp(Tn(x0),Tn+k(x0))→0 as n→∞, ϵ>0,
(2.2)∃n0>0s.t.∀n≥n0dp(Tn(x0),Tn+k(x0))<ϵ.
Then dp(Tn0(x0),Tk(Tn0(x0)))<ϵ. Therefore Tn0(x0) is an ϵ-fixed point of Tk.Theorem 2.4.
Let(X,p) be a dualistic partial metric space and let  T:X→X be a map also for all  x,y∈X,
(2.3)dp(Tx,Ty)≤cdp(x,y)0<c<1,
then T has an ϵ-fixed point in partial metric. Moreover, if x,y∈X are ϵ-fixed points of T, then dp(x,y)≤2ϵ/(1-c).Proof.
Supposex∈X, then
(2.4)dp(Tn(x),Tn+1(x))=dp(T(Tn-1(x)),T(Tn(x)))≤cdp(Tn-1(x),Tn(x))≤⋯≤cn-1dp(T(x),T2(x))≤cndp(x,Tx).
Therefore dp(Tn(x),Tn+1(x))→0 as n→∞. From Theorem 2.3, T has an ϵ-fixed point and Since
(2.5)dp(x,y)≤dp(x,Tx)+dp(Tx,Ty)+dp(y,Ty)≤2ϵ+cdp(x,y).
Then dp(x,y)≤2ϵ/(1-c).Theorem 2.5.
LetT be a mapping of a dualistic partial metric space  (X,p) into itself such that
(2.6)|p(Tx,Ty)|≤c|p(x,y)|0<c<1
for all x,y∈X. Then Tk has an ϵ-fixed point, for all k.Proof.
Fixx∈X, then it is clear that, for each x∈N,
(2.7)|p(Tnx,Tnx)|≤cn|p(x,x)|,|p(Tnx,Tn+1x)|≤cn|p(x,Tx)|,
also
(2.8)dp(Tnx,Tn+1x)+p(Tnx,Tnx)=p(Tnx,Tn+1x).
We deduce that
(2.9)dp(Tnx,Tn+1x)∣+p(Tnx,Tnx)≤cn|p(x,Tx)|.
Hence
(2.10)dp(Tn(x),Tn+1(x))=cn|p(x,Tx)|-|p(Tnx,Tnx)|≤cn|p(x,Tx)|+|p(Tnx,Tnx)|≤cn(|p(x,Tx)|-|p(x,x)|).
Therefore, fork,n∈N,
(2.11)dp(Tn(x),Tn+k(x))≤dp(Tn(x),Tn+1(x))+⋯+dp(Tn+k-1(x),Tn+k(x))≤(cn+⋯+cn+k-1)(|p(x,Tx)|+|p(x,x)|)≤cn1-c(|p(x,Tx)|+|p(x,x)|).
Similarly, we obtain that(2.12)dp(Tn+k(x),Tn(x))≤cn1-c(|p(x,Tx)|+|p(x,x)|).
Then
(2.13)lim⁡n→∞dp(Tn+k(x),Tn(x))=0.
Therefore, Tk has an ϵ-fixed point.Example 2.6.
LetX=(-∞,-2],and let p be the dualistic metric onXgiven by(2.14)p(x,y)=x∨y,
for all x,y∈X.LetT be the mapping from X into itself defined by T(x)=x+1, for all X=(-∞,-2]. It is immediate to see that (2.15)p(T(x),T(y))≤12p(x,y),
for all x·y∈X. However, T has no fixed point, of course. But by the Theorem 2.5, for every ϵ>0, T has an ϵ-fixed point. That is, there exists xo∈X such that (2.16)dp(T(x0),x0)≤ϵ,ϵ>0,
since (2.17)dp(T(x0),x0)=p(T(x0),x0)-p(T(x0),Tx0)=p(x0+1,x0)-p(x0+1,x0+1)=x0+1-(x0+1)=0≤ϵ.Theorem 2.7.
LetT be a mapping of a dualistic partial metric space  (X,p) into itself such that  dp(Tx,Ty)∣≤β(dp(x,Tx)+dp(y,Ty)) where  2β<1.
Ifx0 is an ϵ-fixed point for T, then Tx0 is an ϵ-fixed point for T2.Proof.
We have(2.18)dp(Tx,T2x)≤β(dp(x,Tx)+dp(Tx,T2x)),
therefore
(2.19)dp(Tx,T2x)≤β1-βdp(x,Tx).
Since 2β<1,
(2.20)dp(Tx,T2x)≤dp(x,Tx).
Since x0 is an ϵ-fixed point for T, then dp(Tx0,T2x0)≤ϵ. So Tx0 is an ϵ-fixed point for T2.Theorem 2.8.
Let(X,p) be a dualistic partial metric, and let   T:X→X be a mapping and  ϵ>0. If  dp(Tx,Ty)≤αdp(x,Tx)+βdp(y,Ty) and  α+β<1. then  T has  ϵ-fixed point. Moreover, if  x,y∈X are  ϵ-fixed points of  T, then  dp(x,y)≤(2+α+β)ϵ.Proof.
We have(2.21)dp(Tx,T2x)≤αdp(x,Tx)+βdp(Tx,T2x).
Therefore
(2.22)dp(Tx,T2x)≤α1-βdp(x,Tx),
also
(2.23)dp(T2x,T3x)≤αdp(Tx,T2x)+βdp(T2x,T3x),
so
(2.24)dp(T2x,T3x)≤(α1-β)2dp(Tx,T2x)
and for every n≥1, we have
(2.25)dp(Tnx,Tn+1x)≤(α1-β)ndp(x,Tx),α1-β<1.
Thus, since α/(1-β)<1,   dp(Tnx,Tn+1x)→0 as n→∞. Now, by Theorem 2.3, T has an ϵ-fixed point and since
(2.26)dp(x,y)≤dp(x,Tx)+dp(Tx,Ty)+dp(y,Ty)≤(1+α)dp(x,Tx)+(1+β)dp(y,Ty).
Then dp(x,y)≤(2+α+β)ϵ.Corollary 2.9.
Let(X,p) be a dualistic partial metric, and let   T:X→X be a mapping and  ϵ>0. If  dp(Tx,Ty)∣≤β(dp(x,Tx)+dp(y,Ty)) and  2β<1, then  T has an  ϵ-fixed point.



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*Source: 657868-2011-07-06.xml*