Abstract

We consider the partial metric on a set 𝑋, 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.

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.

If 𝑝 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-metric 𝑑 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.

If 𝑑 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 𝑑𝑝𝑇𝑥0,𝑥0≤𝜖.(2.1)

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 ğ‘›â†’âˆž for some 𝑘>0, then 𝑇𝑘 has an 𝜖-fixed point.

Proof. Since 𝑑𝑝(𝑇𝑛(𝑥0),𝑇𝑛+𝑘(𝑥0))→0 as ğ‘›â†’âˆž, 𝜖>0, ∃𝑛0>0s.t.∀𝑛≥𝑛0𝑑𝑝𝑇𝑛𝑥0,𝑇𝑛+𝑘𝑥0<𝜖.(2.2) 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, 𝑑𝑝(𝑇𝑥,𝑇𝑦)≤𝑐𝑑𝑝(𝑥,𝑦)0<𝑐<1,(2.3) 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. Suppose 𝑥∈𝑋, then 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).(2.4) Therefore 𝑑𝑝(𝑇𝑛(𝑥),𝑇𝑛+1(𝑥))→0 as ğ‘›â†’âˆž. From Theorem 2.3, 𝑇 has an 𝜖-fixed point and Since 𝑑𝑝(𝑥,𝑦)≤𝑑𝑝(𝑥,𝑇𝑥)+𝑑𝑝(𝑇𝑥,𝑇𝑦)+𝑑𝑝(𝑦,𝑇𝑦)≤2𝜖+𝑐𝑑𝑝(𝑥,𝑦).(2.5) Then 𝑑𝑝(𝑥,𝑦)≤2𝜖/(1−𝑐).

Theorem 2.5. Let 𝑇 be a mapping of a dualistic partial metric space (𝑋,𝑝) into itself such that ||||||||𝑝(𝑇𝑥,𝑇𝑦)≤𝑐𝑝(𝑥,𝑦)0<𝑐<1(2.6) for all 𝑥,𝑦∈𝑋. Then 𝑇𝑘 has an 𝜖-fixed point, for all 𝑘.

Proof. Fix 𝑥∈𝑋, then it is clear that, for each 𝑥∈𝑁, ||𝑝(𝑇𝑛𝑥,𝑇𝑛||𝑥)≤𝑐𝑛||||,||𝑝𝑇𝑝(𝑥,𝑥)𝑛𝑥,𝑇𝑛+1𝑥||≤𝑐𝑛||||,𝑝(𝑥,𝑇𝑥)(2.7) also 𝑑𝑝𝑇𝑛𝑥,𝑇𝑛+1𝑥+𝑝(𝑇𝑛𝑥,𝑇𝑛𝑇𝑥)=𝑝𝑛𝑥,𝑇𝑛+1𝑥.(2.8) We deduce that 𝑑𝑝𝑇𝑛𝑥,𝑇𝑛+1𝑥∣+𝑝(𝑇𝑛𝑥,𝑇𝑛𝑥)≤𝑐𝑛||||𝑝(𝑥,𝑇𝑥).(2.9) Hence 𝑑𝑝𝑇𝑛(𝑥),𝑇𝑛+1(𝑥)=𝑐𝑛||||−||𝑝(𝑥,𝑇𝑥)𝑝(𝑇𝑛𝑥,𝑇𝑛||𝑥)≤𝑐𝑛||||+||𝑝(𝑥,𝑇𝑥)𝑝(𝑇𝑛𝑥,𝑇𝑛||𝑥)≤𝑐𝑛||||−||||.𝑝(𝑥,𝑇𝑥)𝑝(𝑥,𝑥)(2.10)
Therefore, for 𝑘,𝑛∈𝑁, 𝑑𝑝𝑇𝑛(𝑥),𝑇𝑛+𝑘(𝑥)≤𝑑𝑝𝑇𝑛(𝑥),𝑇𝑛+1(𝑥)+⋯+𝑑𝑝𝑇𝑛+𝑘−1(𝑥),𝑇𝑛+𝑘≤𝑐(𝑥)𝑛+⋯+𝑐𝑛+𝑘−1||||+||||≤𝑐𝑝(𝑥,𝑇𝑥)𝑝(𝑥,𝑥)𝑛||||+||||.1−𝑐𝑝(𝑥,𝑇𝑥)𝑝(𝑥,𝑥)(2.11)
Similarly, we obtain that 𝑑𝑝𝑇𝑛+𝑘(𝑥),𝑇𝑛≤𝑐(𝑥)𝑛||||+||||.1−𝑐𝑝(𝑥,𝑇𝑥)𝑝(𝑥,𝑥)(2.12) Then limğ‘›â†’âˆžğ‘‘ğ‘î€·ğ‘‡ğ‘›+𝑘(𝑥),𝑇𝑛(𝑥)=0.(2.13) Therefore, 𝑇𝑘 has an 𝜖-fixed point.

Example 2.6. Let 𝑋=(−∞,−2], and let p be the dualistic metric on 𝑋 given by 𝑝(𝑥,𝑦)=𝑥∨𝑦,(2.14) for all 𝑥,𝑦∈𝑋.

Let 𝑇 be the mapping from 𝑋 into itself defined by 𝑇(𝑥)=𝑥+1, for all 𝑋=(−∞,−2]. It is immediate to see that 1𝑝(𝑇(𝑥),𝑇(𝑦))≤2𝑝(𝑥,𝑦),(2.15) 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 𝑑𝑝𝑇𝑥0,𝑥0≤𝜖,𝜖>0,(2.16) since 𝑑𝑝𝑇𝑥0,𝑥0𝑇𝑥=𝑝0,𝑥0𝑇𝑥−𝑝0,𝑇𝑥0𝑥=𝑝0+1,𝑥0𝑥−𝑝0+1,𝑥0+1=𝑥0𝑥+1−0+1=0≤𝜖.(2.17)

Theorem 2.7. Let 𝑇 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.
If 𝑥0 is an 𝜖-fixed point for 𝑇, then 𝑇𝑥0 is an 𝜖-fixed point for 𝑇2.

Proof. We have 𝑑𝑝𝑇𝑥,𝑇2𝑥𝑑≤𝛽𝑝(𝑥,𝑇𝑥)+𝑑𝑝𝑇𝑥,𝑇2𝑥,(2.18) therefore 𝑑𝑝𝑇𝑥,𝑇2𝑥≤𝛽𝑑1−𝛽𝑝(𝑥,𝑇𝑥).(2.19) Since 2𝛽<1, 𝑑𝑝𝑇𝑥,𝑇2𝑥≤𝑑𝑝(𝑥,𝑇𝑥).(2.20) 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 𝑑𝑝𝑇𝑥,𝑇2𝑥≤𝛼𝑑𝑝(𝑥,𝑇𝑥)+𝛽𝑑𝑝𝑇𝑥,𝑇2𝑥.(2.21) Therefore 𝑑𝑝𝑇𝑥,𝑇2𝑥≤𝛼𝑑1−𝛽𝑝(𝑥,𝑇𝑥),(2.22) also 𝑑𝑝𝑇2𝑥,𝑇3𝑥≤𝛼𝑑𝑝𝑇𝑥,𝑇2𝑥+𝛽𝑑𝑝𝑇2𝑥,𝑇3𝑥,(2.23) so 𝑑𝑝𝑇2𝑥,𝑇3𝑥≤𝛼1−𝛽2𝑑𝑝𝑇𝑥,𝑇2𝑥(2.24) and for every 𝑛≥1, we have 𝑑𝑝𝑇𝑛𝑥,𝑇𝑛+1𝑥≤𝛼1−𝛽𝑛𝑑𝑝𝛼(𝑥,𝑇𝑥),1−𝛽<1.(2.25) Thus, since 𝛼/(1−𝛽)<1,   𝑑𝑝(𝑇𝑛𝑥,𝑇𝑛+1𝑥)→0 as ğ‘›â†’âˆž. Now, by Theorem 2.3, 𝑇 has an 𝜖-fixed point and since 𝑑𝑝(𝑥,𝑦)≤𝑑𝑝(𝑥,𝑇𝑥)+𝑑𝑝(𝑇𝑥,𝑇𝑦)+𝑑𝑝(𝑦,𝑇𝑦)≤(1+𝛼)𝑑𝑝(𝑥,𝑇𝑥)+(1+𝛽)𝑑𝑝(𝑦,𝑇𝑦).(2.26) 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.