Fixed points of mappings with a contractive iterate at a point in partial metric spaces

R E S E A R C H

Open Access

Fixed points of mappings with a contractive

iterate at a point in partial metric spaces

Dejan Ili´c, Vladimir Pavlovi´c

_{and Vladimir Rakoˇcevi´c}

*_{Correspondence:}

vlada@pmf.ni.ac.rs Department of Mathematics, Faculty of Sciences and Mathematics, University of Niš, Višegradska 33, Niš, 18000, Serbia

Abstract

In 1994, Matthews introduced and studied the concept of partial metric space and obtained a Banach-type fixed point theorem on complete partial metric spaces. In this paper we study fixed point results of new mappings with a contractive iterate at a point in partial metric spaces. Our results generalize and unify some results of Sehgal, Guseman and ´Ciri´c for mappings with a generalized contractive iterate at a point to partial metric spaces. We give some generalized versions of the fixed point theorem of Matthews. The theory is illustrated by some examples.

Keywords: fixed point; partial metric space; contractive iterate at a point

1 Introduction

In , Banach proved the following famous fixed point theorem []. Let (X,d) be a com-plete metric space. LetT be a contractive mapping onX, that is, one for which exists q∈[, ) satisfying

d(Tx,Ty)≤q·d(x,y) (.)

for allx,y∈X. Then there exists a unique fixed pointx∈XofT. This theorem, called the Banach contraction principle, is a forceful tool in nonlinear analysis. This principle has many applications and has been extended by a great number of authors. For the con-venience of the reader, let us recall the following results [–].

In , Sehgal [] proved the following interesting generalization of the contraction mapping principle.

Theorem .([]) Let(X,d)be a complete metric space,q∈[, )and T:X→X be a continuous mapping.If for each x∈X there exists a positive integer n=n(x)such that

dTnx,Tny≤q·d(x,y) (.)

for all y∈X,then T has a unique fixed point u∈X.Moreover,for any x∈X,u=limmTmx. In , Guseman [] (see also []) generalized the result of Sehgal to mappings which are both necessarily continuous and which have a contractive iterate at each point in a (possibly proper) subset of the space.

In , Ćirić [], among other things, proved the following interesting generalization of Sehgal’s result.

Theorem .([]) Let(X,d)be a complete metric space,q∈[, )and T:X→X.If for each x∈X there exists a positive integer n=n(x)such that

dTnx,Tny≤q·maxd(x,y),d(x,Ty), . . . ,dx,Tny,dx,Tnx (.) holds for all y∈X,then T has a unique fixed point u∈X.Moreover,for every x∈X,u=

limmTmx.

Partial metric spaces were introduced in [] by Matthews as part of the study of deno-tational semantics of dataflow networks and served as a device to solve some difficulties in domain theory of computer science (see also [, ]), in particular the ones which arose in the modeling of a parallel computation program given in []. The concept has since proved extremely useful in domain theory (see,e.g., [–]) and in constructing models in the theory of computation (see,e.g., [–]).

On the other hand, fixed point theory of mappings defined on partial metric spaces since the first results obtained in [] has flourished in the meantime, a fact evidenced by quite a number of papers dedicated to this subject (see,e.g., [–]). The task seems to have been set forth of determining what known fixed point results from the usual metric setting remain valid - after adequate modifications that should as much as possible reflect the nature of the concept of partial metric - when formulated in the partial metric setting. The potentially nonzero self-distance, built into Matthew’s definition of partial metrics, was taken into account in [] in an essential way by a rather mild variation of the classi-cal Banach contractive condition, and in [] further considerations in this direction were carried out which were in turn generalized by Chiet al.in []. This paper represents a continuation of the previous work by the authors. Now we study fixed point results of new mappings with a contractive iterate at a point in partial metric spaces. Our results general-ize and unify some results of Sehgal, Guseman and Ćirić for mappings with a generalgeneral-ized contractive iterate at a point to partial metric spaces. We give some generalized versions of the fixed point theorem of Matthews. The theory is illustrated by some examples.

2 Preliminaries

Throughout this paper the lettersRandNwill denote the set of real numbers and positive integers, respectively.

Let us recall [] that a nonnegative mappingp:X×X→R, whereXis a nonempty set, is said to be apartial metric on Xif for anyx,y,z∈Xthe following four conditions hold true:

(P) p(x,y) =p(y,x), (P) p(x,x)≤p(x,y),

(P) ifp(x,x) =p(y,y) =p(x,y), thenx=y, (P) p(x,z)≤p(x,y) +p(y,z) –p(y,y).

The pair (X,p) is then called apartial metric space. A sequence{xm}∞_m=of elements ofX is calledp-Cauchyif the limitlim_m,np(xn,xm) exists and is finite. The partial metric space (X,p) is calledcompleteif for eachp-Cauchy sequence{xm}∞_m=there is somez∈Xsuch that

p(z,z) =lim

Observe that condition (P) is a strengthening of the triangle inequality and thatp(x,y) =  impliesx=yas in the case of an ordinary metric.

It can be shown that if (X,p) is a partial metric space, then byps_{(x,y) = p(x,y) –p(x,x) –} p(y,y), forx,y∈X, a metricps_{is defined on the setXsuch that{x}

n}n≥converges toz∈X with respect topsif and only if (.) holds. Also (X,p) is a complete partial metric space if and only if (X,ps) is a complete metric space. For proofs of these facts, see [, ].

A paradigm for partial metric spaces is the pair (X,p) whereX= [, +∞) andp(x,y) =

max{x,y}forx,y≥. Below we give two more examples of partial metrics both of which are taken from [].

Example . IfX:={[a,b]|a,b∈R,a≤b}, thenp([a,b], [c,d]) =max{b,d}–min{a,b} defines a partial metricponX.

Example . LetX:=RN∪_n≥R{,,...,n–}_{, whereN}

is the set of nonnegative integers. ByL(x) denote the set{, , . . . ,n}if x∈R{,,...,n–} _{for somen∈N, and the setN}

 if x∈RN. Then a partial metric is defined onXby

p(x,y) =inf–i|i∈L(x)∩L(y) and∀j∈N

j<i⇒x(j) =y(j).

For applications of partial metrics to problems in theoretical computer science, the reader is referred to [, , , ].

In [] Matthews proved the following extension of the Banach contraction principle to the setting of partial metric spaces.

Theorem . Let(X,p)be a complete partial metric space,α∈[, )and T:X→X be a given mapping.Suppose that for each x,y∈X the following condition holds

p(Tx,Ty)≤αp(x,y). (.)

Then there is a unique z∈X such that Tz=z.Also p(z,z) = and for each x∈X the se-quence{Tn_x}

n≥converges with respect to the metric psto z.

A variant of the result above concerning the so-called dualistic partial metric spaces was later given in []. Altunet al.[] further generalized the result of Matthews as well as extended to partial metric spaces several other well-known results about fixed points of mappings on metric spaces.

Taking a different approach to the way in which contractive condition (.) can be gen-eralized for partial metrics, we [, ] have obtained other extensions of Theorem .. To state one of them, we will use the following notation. Given a partial metric space (X,p) set rp:=inf{p(x,y) :x,y∈X}=inf{p(x,x) :x∈X}andRp:={x∈X:p(x,x) =rp}. Notice thatRp may be empty and that ifpis a metric, then, clearly,rp=  andRp=X.

Theorem .(Theorem . of []) Let(X,p)be a complete partial metric space,α∈[, ) and T:X→X be a given mapping.Suppose that for each x,y∈X the following condition holds

Then the set Rpis nonempty.There is a unique u∈Rpsuch that Tu=u.For each x∈Rp, the sequence{Tn_x}

n≥converges with respect to the metric psto u.

Remark . Although Theorem . does not imply uniqueness of the fixed point, it is easy to see that, under the assumptions made, ifuandvare both fixed points satisfying p(u,u) =p(v,v), thenu=v.

Remark . Completeness of a partial metric does not necessarily entail that Rp is nonempty. A (class of ) counterexample(s) is easily constructed as follows.

Let (X,d) be a partial metric space,a>  andf :X→[,a) be an arbitrary mapping. If x,y∈Xare such thatx=y, definep(x,y) =d(x,y) +aandp(x,x) =f(x). Then (X,p) is a partial metric space, as is easily verified.

Now ifb:=supf[X] <a, then, given a sequence{xn}n≥, we havelim sup_np(xn,xn)≤b<a andp(xn,xm)≥awheneverxn=xm. Thus there are no nonstationaryp-Cauchy sequences. Hence (X,p) is complete. ButRp=∅wheneverinff[X] /∈f[X].

If condition (.) is replaced by the somewhat stronger condition below, then the uniqueness of the fixed point is guaranteed.

Theorem .(Theorem . of []) Let(X,p)be a complete partial metric space,α∈[, ) and T:X→X be a given mapping.Suppose that for each x,y∈X the following condition holds

p(Tx,Ty)≤max

αp(x,y),p(x,x) +p(y,y) 

. (.)

Then there is a unique z∈X such that Tz=z.Furthermore,z∈Rpand for each x∈Rpthe sequence{Tn_x}n≥

converges with respect to the metric psto z.

3 Auxiliary results

We now introduce the two types of contractive conditions that we shall be considering in this paper. Let us remark that ifT:X→X, then we writeT_{=Ifor the identity mapping} I:X→X,i.e.,I(x) =x,x∈X.

Definition . Let (X,p) be a partial metric space,α∈(, ) andT:X→X.

(i) We say thatTis aC-operator onXif for eachx∈Xthere is somen(x)∈Nsuch

that for eachy∈Xthere holds

pTn(x)x,Tn(x)y

≤maxαpx,Tjy,αpx,Tn(x)x,p(x,x),pTn(x)–y,Tn(x)–y (.)

for somej∈ {, , . . . ,n(x)}.

(ii) We say thatTis aC-operator onXif for eachx∈Xthere is somen(x)∈Nsuch that for eachy∈Xthere holds

pTn(x)x,Tn(x)y≤αmaxpx,Tjy,px,Tn(x)x (.)

For a C-operatorT on a partial metric space (X,p) andx∈X define the supporting sequence at the pointxas the sequence{sk}k≥, wheres=  andsk+=sk+n(Tskx). Clearly, this is a strictly increasing sequence. Also setRT(X) :={x∈X|Tmx=Tm+xfor somem∈

N}.

Lemma . Let T be aC-operator on a partial metric space(X,p),x∈X\RT(X),{sk}k≥

be the supporting sequence at x and k≥and i≥skbe given integers.Then we must have

pTskx,Tix≤maxαpTsk–x,Tjx,pTsk–x,Tsk–x

for some j≥sk–. (.)

Proof Case. Supposei=sk+ . By (.) we know that if

pTskx,Tsk+x>maxαpTsk–x,Tjx,pTsk–x,Tsk–x

for allj∈ {sk–, . . . ,sk+ }, (.)

then

pTsk_x,Tsk+_x≤_pTsk+_x,Tsk+_{x. (.)}

Likewise, if

pTsk_x,Tsk+_x>maxαpTsk–_x,Tj_x,pTsk–_x,Tsk–_x

for allj∈ {sk–, . . . ,sk+ }, (.)

then

pTsk_x,Tsk+_x≤_pTsk_x,Tsk_{x. (.)}

Now if (.) were to hold, then (.) and (P) would imply that (.) is true as well. So (.) also holds and thus

pTskx,Tsk+_x≤pTsk+_x,Tsk+_x≤pTsk_x,Tsk+_x

≤pTskx,Tskx≤pTskx,Tsk+x

meaning (by (P)) thatTsk_x=Tsk+_{x. But this contradicts the assumptionx∈/ R} T(X), so (.) must be true.

Case. Supposei=sk. Sincep(Tsk_x,Tsk_x)≤p(Tsk_x,Tsk+_{x), the assertion here follows} from the previous case.

Case. Suppose nowi=sk+ . Assume that

since otherwise there is nothing to prove. Then by (.) we must havep(Tsk_x,Tsk+_x)≤ p(Tsk_x,Tsk_{x). But then, by the previous case, there is somej≥s}

k–such that

pTsk_x,Tsk+_x≤_pTsk_x,Tsk_x≤_maxαpTsk–_x,Tj_x,pTsk–_x,Tsk–_x

and we are done.

Fori>sk+ , the argument carries on by induction. Suppose that (.) holds for some i≥sk. Ifp(Tsk_x,Ti+_{x) >{αp(T}sk–_x,Tj_x),p(Tsk–_x,Tsk–_{x)}for allj∈ {sk–, . . . ,i+ }, then} we must havep(Tsk_x,Ti+_x)≤_p(Ti_x,Ti_{x). By the induction hypothesis, there is somej}≥ sk–such that

pTskx,Tix≤αpTsk–x,Tjx,pTsk–x,Tsk–x.

But sincep(Tsk_x,Ti+_x)≤p(Ti_x,Ti_x)≤p(Tsk_x,Ti_{x), the last inequality clashes with our}

assumption.

To shorten the foregoing considerations, we introduce some auxiliary notions as follows. Fixx∈X\RT(X). For integersk≥ andi≥sk, use Lemma . repeatedly to fix integers lj≥sj, ≤j<kandt, . . . ,tk∈ {, }such that, puttinglk:=i, there holds

pTsjx,Tljx≤αtjpTsj–x,Tlj–x

for all ≤j≤k, where

tj=  ifsj–<lj–,  ifsj–=lj–.

We shall refer to (l, . . . ,lk–) and (t, . . . ,tk) as the (k,i)-descent and the (k,i)-signature at x, respectively. SetS_kx_,i:={j∈ {, . . . ,k} |tj= }. We shall say thatxis of type  if there are sequences of positive integers{km}m≥and{im}m≥, the first one strictly increasing, such that for allm≥ we haveim≥skmandcard(S_kmx _,im) <card(Sx_{km+,im+}); here and henceforth, for a finite setP, we denote bycard(P) the number of its elements. We shall say thatxis of type  ifxis not of type ,i.e., if there arek,D∈Nsuch that for allk≥kand alli≥sk there holdscard(Sx

k,i) <D.

To make the proof of our main result more transparent, we have extracted from it several parts and presented them first in form of the next five lemmas.

Lemma . Let T be aC-operator on a partial metric space(X,p),x∈/ RT(X),and let

{sk}k≥be the supporting sequence at x.Then:

(a) If(l, . . . ,lk–)is the(k,i)-descent atx,then

pTsk_x,Ti_x≤_αcard(S x

k,i_)_px,Tl_{x and}

pTskx,Tix≤pTsjx,Tljx for all≤j≤k,

(b) IfP⊆ {, . . . ,k– }is such thatcard(Sx_k,i) <card(P),then for somej∈Pthere must hold

pTsk_x,Ti_x≤_pTsjx,Tsjx.

Proof Regarding Lemma ., we getp(Tsk_x,Ti_x)≤αki=j+ti_p(Tsj_x,Tlj_{x) for alljsuch that} k>j≥, recursively. Now (a) follows directly.

To prove (b) simply observe thatcard(Sx_k,i) <card(P) implies that the set{j+ |j∈P}is a subset of{, . . . ,k}withcard(P) >card(Sx

k,i) elements so that there must be somej∈P withtj+= . Hence

pTskx,Tix≤pTsj+_x,Tlj+_x≤_αtj+_pTsj_x,Tlj_x=pTsj_x,Tsj_x,

where we used (a) and the fact thatlj=sj.

Lemma . If T is aC-operator on a partial metric space(X,p)and x∈X,then there is some Mx> such that for all i≥we must have p(x,Ti_x)≤Mx.

Proof Ifx∈RT(X), then this is obvious. Thus supposex∈/RT(X) and setMx:=_–_αmax{p(x, Tj_{x)| <j≤n(x)}+p(x,x). Ifk=n(x), then it is certainly true thatp(x,T}i_{x)≤Mxfor all} ≤i≤k. Now suppose that the same is valid for somek≥n(x).

Ifp(x,Tk+_x)≤p(x,Ti_{x) for some ≤i≤k, then by the induction hypothesis there holds} p(x,Tk+_x)≤p(x,Ti_x)≤M

x. Otherwise, we must have

px,Tk+x>maxpx,Tix|≤i≤k. (.)

Now using (.)

px,Tk+x≤px,Tn(x)x+pTn(x)x,Tk+x

≤px,Tn(x)_x+maxαpx,Tj_x,p(x,x),pTk_x,Tk_x,αpx,Tn(x)_x

for somek+  –n(x)≤j≤k+ . Hence we either have

px,Tk+x≤px,Tn(x)x+p(x,x)≤Mx

or, using (.) and the fact thatp(Tk_x,Tk_x)≤p(x,Tk_{x), it must be that}

px,Tk+x≤px,Tn(x)x+αpx,Tk+x, i.e.,

px,Tk+x≤   –αp

x,Tn(x)x≤Mx.

By induction the desired conclusion follows.

Proof Fixm≥. If (l, . . . ,lkm–) is the (skm,im)-descent, then by (a) of Lemma . we have

pTskm_x,Tim_x≤_αcard(Sxkm,im)_px,Tl_x≤αcard(Skmx ,im)_Mx.

Sincelimmcard(Sx_km,im) =∞, this implieslimmp(Tskmx,Timx) =limmp(Tskmx,Tskmx) = . Now givenε>  choosem≥ such thatαmMx<εand such that for allm≥mit holdsp(Tskm_x,Tskm_{x) <}ε. Leti≥skm_be arbitrary.

Suppose first thatcard(Sx_km

,i)≥m. Then

pTskm_x,Ti_x≤_α card(Sx_k

m,i)_Mx≤_αm_Mx<ε.

Suppose now thatcard(Sx_km

,i) <m. ForP:={km, . . . ,km–} ⊆ {, , . . . ,km– }, we havecard(P) >card(Sx

km_,i), so by (b) of Lemma . there must be somem≤j≤m–  such thatp(Tskm_x,Ti_x)≤_p(Tskj_x,Tskj_{x) <ε.}

We have thus shown that p(Tskm_x,Ti_{x) <ε must hold for alli≥s}

km_. Therefore if i,j≥skm_, then

pTix,Tjx≤pTskm_x,Ti_x+pTskm_x,Tj_{x< ε.}

The previous analysis proveslimi,jp(Tix,Tjx) = .

Lemma . Let T be aC-operator on a partial metric space(X,p)and x∈X\RT(X).If x is of type,then the sequence{Ti_x}

i≥is p-Cauchy.

Proof Let{sk}k≥be the supporting sequence atx. We first showlim inf_mp(Tsm_x,Tsm_{x) =lim sup}

mp(Tsmx,Tsmx). Suppose that this is not true and pick a realθ with lim infmp(Tsmx,Tsmx) <θ <lim supmp(Tsmx,Tsmx). Letk< k<· · ·<kD<kD+andi>kD+be positive integers, wherekD+≥k, such that

pTskj_x,Tskj_{x<θ for all ≤j≤D and pT}si_x,Tsi_x>θ.

The fact that si >skD+ implies that Sx

kD+,si is defined and since kD+ ≥k, we have

card(Sx

kD+,si) < D. For P := {k, . . . ,kD} ⊆ {, , . . . ,kD+ – }, we have card(P) = D >

card(Sx

kD+,si)|so, by (b) of Lemma ., there is somej∈ {, . . . ,D}such that

θ<pTsix,Tsix≤pTskD+_x,Tsi_x≤pTskj_x,Tskj_x<θ,

a contradiction.

By the preceding part and since ≤p(Tsm_x,Tsm_{x)≤Mx, withMxas in Lemma ., we} haverx:=limmp(Tsmx,Tsmx)∈R.

Let us prove

∀ε> ∃m∀m≥m∀i≥sm p

Tsmx,Tix∈(rx–ε,rx+ε). (.)

card(P) =D>card(S_mx_,i), so for somem≤j≤m+D–  it must be rx–ε<p

Tsmx,Tsmx≤pTsmx,Tix≤pTsjx,Tsjx<rx+ε

and we are done.

From (.) it now immediately follows that

∀ε> ∃k∀i,j≥k pTix,Tjx<rx+ε. (.)

Indeed, givenε> , considermas in (.) and leti,j≥smbe arbitrary. Then

pTix,Tjx≤pTsm_x,Ti_x+pTsm_x,Tj_x–pTsm_x,Tsm_x

<rx+ε+ ε=rx+ ε.

To prove

lim

i,j p

Tix,Tjx=rx, (.)

we now only need to show that

∀ε> ∃k∀i,j≥k rx–ε<p

Tix,Tjx. (.)

Let ε∈(,rx_+(–_αα)) be arbitrary and let k∈Nbe as in (.). We claim thatrx–ε< p(Ti_x,Ti_{x) holds for alli≥k. This would prove (.) sincep(T}i_x,Ti_x)≤p(Ti_x,Tj_x).

Suppose to the contrary that there is some i≥k with p(Tix,Tix)≤rx–ε. Put z:=Tix. x∈/ RT(X) impliesz∈/ RT(X). If z is of type , then by Lemma . we have  =limi,jp(Tiz,Tjz) =limi,jp(Tix,Tjx), so{Tix}i≥isp-Cauchy and we are finished. Sup-pose now that z is of type , so that by what we have proved thus far we know that rz=limmp(Tsmz,Tsmz)∈Rand also that (.) holds withztaken instead ofx. It cannot bep(Tn(z)_z,Tn(z)_{z) >r}

x–εbecause this would mean thatp(z,z) <p(Tn(z)z,Tn(z)z), so using Lemma ., it would followrx–ε<p(Tn(z)_z,Tn(z)_z)≤αp(z,Tj_{z) for somej∈N,i.e.,rx–ε<}

αp(Ti_x,Ti+j_{x)≤α(rx+ε), giving}rx(–α)

+α <ε, a contradiction. Sop(Tn(z)z,Tn(z)z)≤rx–ε.

The argument actually shows that p(Tqm_z,Tqm_{z)≤rx–εholds for everym≥, where}

{qm}m≥is the supporting sequence at the pointz. Sorz =limmp(Tqmz,Tqmz)≤rx–ε. Now use the fact thatrz<rz+_rx and (.) (withztaken instead ofxandrx–_rz instead ofε

of course) to findj∈Nsuch that

pTjz,Tjz<rz+rx

 for allj≥j. (.)

Aslimmp(Tsmx,Tsmx) =rxandrz+_rx<rx, there is somem≥i+jwithp(Tsmx,Tsmx) > rz+rx

 . Now, usingsm–i≥m–i≥j, we obtain

rz+rx  <p

Tsmx,Tsmx=pTsm–iz,Tsm–iz<rz+rx

 ,

Lemma . Let T:X→X and let p:X×X→Rbe any mapping satisfying(P).Suppose that x∈X is such that Tk_{x=x holds for some positive integer k,and that there exists y∈X} such that

p(y,y) =lim

i p

y,Tix=lim

i,j p

Tix,Tjx. (.)

Then Tx=x.

Proof FromTki_{x=x,i≥, we have}

p(y,y) =lim

i p

y,Tkix=p(y,x) and p(y,y) =lim

i p

Tkix,Tkix=p(x,x),

hencey=x. But (.) now gives

p(x,x) =lim

i p

x,Tki+x=p(x,Tx) and p(x,x) =lim

i p

Tki+x,Tki+x=p(Tx,Tx)

soTx=x.

4 Main results

Having made the necessary preparations, we are now able to prove fixed point results for C-operators on complete partial metric spaces. But first we prove a proposition giving some insight into the structure of this type of mappings.

Proposition . If T is aC-operator on a complete partial metric space(X,p),then

() for eachx∈X,the sequence{Tix}i≥ps-converges to somevx∈X;

() for allx,y∈X,there holdsp(vx,vy) =max{p(vx,vx),p(vy,vy)}.

Proof The existence of such pointsvxis assured by Lemmas . and . and completeness ifx∈X\RT(X), and is self-evident ifx∈RT(X).

To prove (), letx,y∈Xbe arbitrary and suppose thatp(vx,vx)≥p(vy,vy). Ifp(vx,vy) = , thenvx=vy(by (P) and (P)) and we are done. Thus assume thatp(vx,vy) >  and letε>  be arbitrary such that ε_–(+_αα) <p(vx,vy). There is somem∈Nsuch that for alli,j≥m there holds

maxpTiy,Tjy–p(vy,vy),pvy,Tjy–p(vy,vy)<ε,

maxpTix,Tjx–p(vx,vx),pvx,Tjx–p(vx,vx)<ε.

Fori,j≥mwe have

pTiy,Tjx≤pTiy,vy–p(vy,vy) +p(vy,vx) –p(vx,vx) +pvx,Tjx< ε+p(vy,vx)

and, similarly,

p(vy,vx)≤pvy,Tiy–pTiy,Tiy+pTiy,Tjx–pTjx,Tjx+pTjx,vx

Fix anyi≥mand seti:=n(Tix). By (.) there is somej∈ {i, . . . ,i+i}such that

p(vy,vx) – ε

<pTi+ix,Ti+iy

≤maxαpTix,Tjy,αpTix,Ti+ix,pTix,Tix,pTi+i–y,Ti+i–y

≤maxαε+p(vy,vx)

,αp(vx,vx) +ε

,p(vx,vx) +ε,p(vy,vy) +ε

=maxαε+p(vy,vx)

,p(vx,vx) +ε

.

Nowp(vy,vx) – ε<α[ε+p(vy,vx)] is just ε_–(+_αα) >p(vx,vy), which is false by our choice ofε. This leaves us with the only other possibility:p(vy,vx) – ε<p(vx,vx) +ε,i.e.,p(vy,vx) < p(vx,vx) + ε.

From the preceding analysis it follows thatp(vy,vx)≤p(vx,vx), which by (P) actually means thatp(vy,vx) =p(vx,vx) =max{p(vx,vx),p(vy,vy)}.

Theorem . If T is aC-operator on a complete partial metric space(X,p),then there is a fixed point z∈X of T such that p(z,z) =inf_x∈Xp(vx,vx),where vxare as in Proposition.. Proof Forx∈Xputrx:=p(vx,vx) (this is consistent with the notation of Lemma .). Set I:=infx∈Xrx. Form≥ pickxm∈Xsuch that for alli,j≥ it holds

pTixm,Tjxm

∈(I– /m,I+ /m). (.)

(You can first pickx_m∈Xsuch thatlimi,jp(Tixm ,Tjxm) =vxm∈[I,I+ 

m), then choose k(m)∈Nsuch that for alli,j≥k(m) there holdsI– 

m <p(T i_x

m,Tjxm) <I+m and finally putxm:=Tk(m)x_m.)

Notice that ifi(m) andj(m) are nonnegative integers form∈N, then we have

lim

m p

Ti(m)xm,Tj(m)xm= .

First we prove thatlim_m,kp(xm,xk) =I.

Form,k≥ letCm,k>  be such thatp(Tixm,Tjxk) <Cm,kholds for alli,j≥.

Fixm,k≥ and let{sq}q∈N be the supporting sequence atxm. Letl≥ be any integer such thatαl_Ck< 

k+m. We have

p(xm,xk)≤pxm,Tslxm–pTslxm,Tslxm

+pTslxm,Tslxk+pxk,Tslxk–pTslxk,Tslxk.

Now

δm,k:=p

xm,Tslxm

–pTsl_x m,Tslxm

< /m

and

μm,k:=p

First suppose that p(Tsl_xm,Tsl_{xk) >p(T}i_xk,Ti_{xk) for all i∈ {, . . . ,sl}, and p(T}sl_xm, Tsl_{xk) >p(T}i_xm,Tj_{xm) for alli,j∈ {, . . . ,sl}. Then, by repeated use of (.), we obtain}

pTsl_x m,Tslxk

≤αpTsl–_x m,Tixk

for somei≥sl–,

pTsl_x m,Tslxk

≤αpTsl–_x m,Tixk

for somei≥sl–,

and continuing in this manner finally

pTslxm,Tslxk≤αlpxm,Tilxk for someil≥.

Thusp(Tsl_xm,Tsl_xk)≤αl_Cm ,k<_k+_m.

On the other hand, ifp(Tsl_xm,Tsl_xk)≤p(Ti_xk,Ti_{xk) for somei∈ {, . . . ,sl}, orp(T}sl_xm, Tsl_x

k)≤p(Tixm,Tjxm) for some i,j∈ {, . . . ,sl}, then by (.) we must have p(Tslxm, Tsl_{xk) <I+max}{

m,  k}. Therefore

p(xm,xk)≤δm,k+μm,k+p

Tslxm,Tslxk

<   m+  k

+I+max

 m,  k .

From the above considerations and fromI– /m<p(xm,xm)≤p(xm,xk), it is now clear thatlimm,kp(xm,xk) =I.

So by completeness there is someu∈Xsuch that

I=lim

m,kp(xm,xk) =limk p(u,xk) =p(u,u). (.)

Let{sm}m≥be the supporting sequence atu.

Let us show by induction onkthat iff :N→Nis such thatf(m)≥sk, for allm∈N, then

I=lim

m p

Tsku,Tf(m)xm=p(u,u). (.)

Suppose first thatk= .

p(u,u)≤pu,Tf(m)xm

≤p(u,xm) +p

xm,Tf(m)xm

–p(xm,xm)

< p(u,xm) +  m,

so the desired conclusion immediately follows. Now suppose that the assertion is true for somek≥, take anyf :N→Nsuch thatf(m)≥sk+,m≥, and proceed as follows.

We have

pTsk_u,Tsk_u≤_pTsk_u,Tf(m)_x m

–pTf(m)xm,Tf(m)xm

≤pTsk_u,Tf(m)_x m

–I+  m

Now for eachm∈N, sincef(m)–(sk+–sk)≥sk, there must be someh(m)∈ {sk, . . . ,f(m)} such that

pTsk+_u,Tf(m)_x m

≤maxαpTsk_u,Th(m)_x m

,pTsk_u,Tsk_u,

pTf(m)–xm,Tf(m)–xm,αpTsku,Tsk+u.

Usingp(Tsk_u,Tsk+_u)≤p(Tsk_u,Tf(m)_{xm) –p(T}f(m)_xm,Tf(m)_{xm) +p(T}sk+_u,Tf(m)_{xm), we} proceed to obtain

pTf(m)xm,Tf(m)xm≤pTsk+_u,Tf(m)_xm≤_max

αpTsk_u,Th(m)_xm,I,I+  m,

α

 –α

pTsku,Tf(m)xm–pTf(m)xm,Tf(m)xm.

Now we have I– /m<p(Tf(m)xm,Tf(m)xm) <I + /m and also h(m)≥sk andf(m)≥ sk+>sk for all m≥. Hence, in view of the induction hypothesis, we finally arrive at

lim_mp(Tsk+_u,Tf(m)_x m) =I.

Using (.) it is straightforward to see that for allk,k≥ there holdsp(Tsku,Tsku)≤ I=p(u,u): indeed this follows by lettingm∈Ntend to infinity in

pTsku,Tsku≤pTsku,Tskxm+pTsku,Tskxm–pTskxm,Tskxm.

Thusru≤I. But by definition ofIwe must actually haveI=ru. We now claim that there are positive integersk<ksuch that

pTsk_u,Tsk_u=pTsk_u,Tsk_u=I.

Assume this is not the case. Thenp(Tsk_u,Tsk_{u) =Ican hold for at most onek∈N. As} we have ≤p(Tsk_u,Tsk_u)≤_{I for all k}∈N_{, our assumption implies in particular that} ru=I> . Thus we can take someε>  such thatru–ε>α(ru+ε). The assumption also allows us to find somem∈Nsuch that for allkwithsk≥mwe havep(Tsku,Tsku) <I, and such that for alli,j≥m it holdsp(Tiu,Tju)∈(ru–ε,ru+ε) (remember thatru= p(vu,vu) =limi,jp(Tiu,Tju)).

Take anykwithsk≥m. ThenL:=max{p(Tsk_u,Tsk_u),p(Tsk+_u,Tsk+_{u)}<I=ry. There is} some positiveε<εsuch thatL<ru–ε. Letibe the smallest integer withi>sk+such that p(Ti_u,Ti_{u) >ru–ε}

, and letm∈Nbe the greatest integer such thatsm≤i. Som≥k+ . By (.) there is somej≥sm–such that

pTiu,Tiu≤pTsmu,Tiu

≤maxαpTsm–u,Tju,pTsm–u,Tsm–u,pTi–u,Ti–u.

Clearly, we have sm– ≥sk ≥ m and i– ≥sk+ ≥m. The minimality of i and m and the fact that L <ru –ε can now easily be used to deduce that p(Tiu,Tiu) >

max{p(Tsm–_u,Tsm–_u),p(Ti–_u,Ti–_{u)}. Therefore r}

u – ε < ru – ε < p(Tiu,Tiu) ≤

So we have proved that there are positive integers k <k such thatp(Tsku,Tsku) = p(Tsk_u,Tsk_{u) =I. Sincep(T}sk_u,Tsk_u)≤_{I, we must haveT}sk_u=Tsk_{u (by (P) and} (P)), i.e., Tsk–skz=zforz:=Tsku. Asp(v_z,v_z) =lim_ip(v_z,Tiz) =lim_i,jp(Tiz,Tjz) and sk–sk∈N, by Lemma . it followsTz=z. Of courselimi,jp(Tiz,Tjz) =limi,jp(Tiu,Tju) =

ru=I=inf_x∈Xp(vx,vx).

Remark . To ensure uniqueness of the fixed point, we can strengthen condition (.) as follows. Given a partial metric space (X,p), callT:X→Xa C-operator if for eachx∈X there is somen(x)∈Nsuch that for eachy∈Xthere holds

pTn(x)x,Tn(x)y

≤max

αpx,Tjy,αpx,Tn(x)x,p(x,x) +p(T

n(x)–_y,Tn(x)–_y)



(.)

for somej∈ {, , . . . ,n(x)}. Evidently, each C-operator is a C-operator as well so that if (X,p) is complete, the conclusion of Theorem . holds. But now, in addition, ifTa=aand Tb=b, then

p(a,b) =pTn(a)a,Tn(b)b≤max

αp(a,b),αp(a,a),p(a,a) +p(b,b) 

so that either ( –α)p(a,b)≤ orps_{(a,b) = p(a,b) –p(a,a) –p(b,b) = , meaning that in} any case we must havea=b.

Recall that a sequencexnin a partial metric space (X,p) is called -Cauchy with re-spect top(see,e.g., []) iflimm,np(xn,xm) = . We say that (X,p) is -complete if every -Cauchy sequence inX ps-converges to somex∈X(for which we then necessarily must havep(x,x) = ). Note that every -Cauchy sequence in (X,p) is Cauchy in (X,ps_{), and that} every complete partial metric space is -complete.

Remark . Recently a very interesting paper by Haghi, Rezapour and Shahzad [] showed up in which the authors associated to each partial metric space (X,p) a metric space (X,d) by settingd(x,x) =  andd(x,y) =p(x,y) if x=y, and proved that (X,p) is -complete if and only if (X,d) is complete. They then proceeded to demonstrate how using the associated metricdsome of the fixed point results in partial metric spaces can easily be deduced from the corresponding known results in metric spaces.

Let us point out that these considerations cannot apply to C-operators since the terms p(x,x) andp(Tn(x)–_y,Tn(x)–_{y) on the right-hand side of (.) are not multiplied byα. Thus} our Theorem . cannot follow from the result of Ćirić it generalizes.

If we completely neglect the role of self-distances in (.), we can easily verify that the statement of Theorem . remains valid upon substituting the words ‘partial metric’ for ‘metric’ and ‘-complete’ for ‘complete’. We will prove this using the approach of Haghi, Rezapour and Shahzad [] that will allow us to deduce Theorem . directly from Ćirić’s result (Theorem .).

Theorem . If T is aC-operator on a-complete partial metric space(X,p),then there is a unique fixed point z of T.Furthermore,we have p(z,z) = and for each x∈X the sequence

{Ti_x}

Proof Letdbe defined as in Remark .. So (X,d) is a complete metric space (see Proposi-tion . of []). Observe that we haved(x,y)≤p(x,y) for allx,y∈X. Forx∈Xletn(x)∈N be as in (.), and forx,y∈Xset

S(x,y) =y,Ty,Ty, . . . ,Tn(x)y,Tn(x)x

and Mp=max{p(x,z)|z∈S(x,y)}, Md=max{d(x,z)|z∈S(x,y)}. We thus have that pTn(x)x,Tn(x)y≤αMp(x,y)

for allx,y∈X. We check that for allx,y∈Xit holds thatd(Tn(x)_x,Tn(x)_y)≤αM

d(x,y), so that Theorem . can immediately be applied.

Case. There is somez∈S(x,y) such thatx=zand Mp(x,y) =p(x,z). Here we have

dTn(x)x,Tn(x)y≤pTn(x)x,Tn(x)y≤αMp(x,y) =αp(x,z) =αd(x,z)≤αMd(x,y).

Case . For all z ∈ S(x,y) we have that Mp(x,y) =p(x,z)⇒ x =z. So it must be Mp(x,y) =p(x,x), in particular, and hencep(x,Tn(x)x)≤Mp(x,y) =p(x,x)≤p(x,Tn(x)x), i.e., Mp(x,y) =p(x,Tn(x)x). But by our assumption it now follows thatx=Tn(x)x. Similarly, from p(x,Tn(x)y)≤Mp(x,y) =p(x,x)≤p(x,Tn(x)y), we obtain Mp(x,y) =p(x,Tn(x)y) and consequentlyx=Tn(x)_{y. Nowd(T}n(x)_x,Tn(x)_{y) =d(x,x) = ≤αM}

d(x,y).

Remark . It should be pointed out, however, that even though the results of Haghiet al.can deduce the same fixed point as the corresponding partial metric fixed point result, using the partial metric version computers evaluate faster since many nonsense terms are omitted. This is very important from the aspect of computer science due to its cost and explains the vast body of partial metric fixed point results found in literature.

Given a C-operator and a pointx, one may ask what the minimal value ofn(x) is for which inequality (.) holds true. In the following example, for an arbitrary positive integer m, we construct a C-operator on a complete partial metric space (X,p) such that for some x∈Xit must ben(x) >m.

Example . Denote byX∞the set of all sequencesx:N→Nand forn∈NbyXnthe set of alln-tuplesx:{, . . . ,n} →Nof positive integers. PutX:=X∞∪n∈NXn. Forx,y∈X set

I(x,y) =i∈N∪ {} |j∈dom(x)∩dom(y)∧j≤i⇒x(j) =y(j) and definep(x,y) :=inf{

i|i∈I(x,y)}(thus ifx()=y(), thenI(x,y) ={}andp(x,y) = ). Here ‘dom(x)’ stands for the domain of the functionx. Then (X,p) is a partial metric space (see []) and a complete one as can easily be verified.

Fixl∈Nand defineT:X→Xas follows. Forx∈XletIx={i∈N|x(i)=i}. IfIx=∅, then setTx=x. IfIx=∅, then defineTx=ybyx∈Xn_⇔y∈Xn_,x∈X

∞⇔y∈ X∞and the following two conditions:

– ifIxis finite and has at mostlelements, theny(i) =iifi=maxIx, andy(i) =x(i)else;

Let us show that T is a C-operator withn(x) =l for allx∈N. Before we proceed, observe that ifkis a nonnegative integer such thatk+ ∈dom(x) andx(i) =ifor ≤i≤k, then fory=Tl_{xwe must havey(i) =ifor all ≤i≤k+ .}

Case. There is a nonnegative integeriwithi+ ∈dom(x)∩dom(y) such thatx(i+ )= i+ ∨y(i+ )=i+ . Denote bykthe least such nonnegative integer.

Ifx(k+ )=y(k+ ), thenp(Tlx,Tly)≤_k+ =_k =_p(x,y).

Ifx(k+ ) =y(k+ ), then sincex(k+ )=k+ ∨y(i+ )=i+  we must actually have x(k+ ) =y(k+ )=k+  and thusp(x,Tl_{x) =} 

k. Hencep(Tlx,Tly)≤_k+ =_p(x,Tlx). Case.x= (, , . . . ,k) for somek∈Nandx⊆y. Herep(Tl_x,Tl_{y) =} 

k =p(x,x). Case . y= (, , . . . ,k) for somek∈Nandy⊆x. Here p(Tl_x,Tl_{y) =} 

k =p(y,y) = p(Tl–_y,Tl–_y).

Condition (.) fails because the fixed point is not unique. SoT is not a C-operator, hence not a C-operator either.

Now suppose thatl<lis an arbitrary positive integer and takex,y∈Nl+such that x(i) =  for all ≤i≤l+ , andy() = ,y(i) =  for all ≤i≤l+ .

We havep(Tl_x,Tl_{y) =  =p(x,T}j_{y), for ≤j≤l}

,p(x,Tlx) =_ andp(Tl–y,Tl–y) = p(x,x) = 

l+. So we see that for this particular choice ofxandy, substitutinglforn(x) in (.) makes the inequality false.

Let us use this very example to illustrate Proposition .. Lett∈X∞be defined byt(i) =i for alli∈N. Forn∈Nlettn∈Xnbe defined bytn(i) =ifori= ,n.

If x∈X∞, we clearly have vx=t. Similarly, ifx∈Xn, then vx=tn. Sop(t,tn) = _n =

max{p(tn,tn),p(t,t)}becausep(tn,tn) =_n andp(t,t) = . Also

p(tm,tn) = 

n=p(tn,tn) ifn≤m, 

m=p(tm,tm) ifm≤n,

thusp(tm,tn) =max{p(tn,tn),p(tm,tm)}.

Remark . It should be noted that if in Theorem . we requiren(x) to be equal to  for allx∈X, then Theorem . is obtained as a corollary. On the other hand, as already pointed out, if in Theorem .pis a complete (ordinary) metric onX, then the result of Ćirić (Theorem .) is recovered.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors read and approved the final manuscript.

Acknowledgements

This work was supported by Grant No. 174025 of the Ministry of Science, Technology and Development, Republic of Serbia.

Received: 30 September 2013 Accepted: 20 November 2013 Published:13 Dec 2013 References

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