extensions of the zamfirescu theorem to partial metric spaces

Mathematical and Computer Modelling 55 (2012) 801–809

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Mathematical and Computer Modelling

journal homepage: www.elsevier.com/locate/mcm

Extensions of the Zamfirescu theorem to partial metric spaces✩

Dejan Ilić, Vladimir Pavlović ∗, Vladimir RakočevićUniversity of Niš, Faculty of Sciences and Mathematics, Department of Mathematics, Višegradska 33, 18000 Niš, Serbia

a r t i c l e i n f o

Article history:Received 22 June 2011Received in revised form 5 September 2011Accepted 6 September 2011

Keywords:Fixed pointBanach contractionPartial metric spaceZamfirescu theoremZamfirescu operator

a b s t r a c t

Zamfirescu [T. ZamfirescuFix point theorems inmetric spaces Arch.Math. (Basel) 23 (1972)292-298], obtained a very interesting fixed point theorem on complete metric spaces bycombining the results of S. Banach, R. Kannan and S.K. Chatterjea. In [S.G. Matthews, Partialmetric topology, in: Proc. 8th Summer Conference on General Topology and Applications,in: Ann. New York Acad. Sci., vol. 728, 1994, pp. 183-197], the author introduced andstudied the concept of partial metric spaces, and obtained a Banach type fixed pointtheorem on complete partial metric spaces. In this paper, we study new extensions ofthe Zamfirescu theorem to the context of partial metric spaces, and among other things,we give some generalized versions of the fixed point theorem of Matthews. The theory isillustrated by some examples.

© 2011 Elsevier Ltd. All rights reserved.

1. Introduction

Zamfirescu [1] obtained a very interesting fixed point theorem on complete metric spaces by combining the results ofBanach [2], Kannan [3] and Chatterjea [4].

There exist many generalizations of the concept of metric spaces in the literature. In particular, Matthews [5] introducedthe notion of a partial metric space as a part of the study of denotational semantics of dataflow networks, showing thatthe Banach contraction mapping theorem can be generalized to the partial metric context for applications in programverification. After that, fixed point results in partial metric spaces have been studied bymany other authors. Refs. [6–16], aresome works in this line of research. The existence of several connections between partial metrics and topological aspects ofdomain theory has been pointed in, e,g. [10,17,18,14,19,20].

In this paper, we study new extensions of the Zamfirescu theorem to the context of partial metric spaces, and amongother things, we give some generalized versions of the fixed point theorem of Matthews. The theory is illustrated by someexamples.

2. Preliminaries

In this section, we present some basic results and definitions concerning the Zamfirescu theorem [1] (see also [21–26])and partial metric spaces.

Theorem 2.1 (Zamfirescu [1]). Let (X, d) be a completemetric space and T : X → X amap for which there exist the real numbersa, b and c satisfying 0 ≤ a < 1, 0 ≤ b, c < 1/2 such that for each pair x, y ∈ X, at least one of the following is true:(1) d(Tx, Ty) ≤ ad(x, y);(2) d(Tx, Ty) ≤ b[d(x, Tx) + d(y, Ty)];(3) d(Tx, Ty) ≤ c[d(x, Ty) + d(y, Tx)].

✩ Supported by Grant No. 174025 of the Ministry of Science, Technology and Development, Republic of Serbia.∗ Corresponding author.

E-mail addresses: [email protected] (D. Ilić), [email protected] (V. Pavlović), [email protected] (V. Rakočević).

0895-7177/$ – see front matter© 2011 Elsevier Ltd. All rights reserved.doi:10.1016/j.mcm.2011.09.005

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802 D. Ilić et al. / Mathematical and Computer Modelling 55 (2012) 801–809

Then T has a unique fixed point p and the Picard iteration {xn}∞n=0 defined by xn+1 = Txn, n = 0, 1, 2, . . . converges to p, for anyx0 ∈ X.

An operator T which satisfies the contractive conditions in Theorem 2.1 is called a Zamfirescu operator. The class ofZamfirescu operators is one of the most studied classes of quasicontractive type operators. In this class, all important fixedpoint iteration procedures, i.e., the Picard, Mann and Ishikawa iterations, are known to converge to the unique fixed point ofT . The class of Zamfirescu operators is independent (see [25]) of the class of strictly (strongly) pseudocontractive operators,extensively studied by several authors in the past years. For a recent survey and a comprehensive bibliography, we refer thereader to the recent Berinde’s monograph [23].

Throughout this paper the letters R, R+, Q, N will denote the set of real numbers, of nonnegative real numbers, rationalnumbers and natural numbers, respectively.

Let us recall [5] that a mapping p : X × X → R+, where X is a nonempty set, is said to be a partial metric on X if for anyx, y, z ∈ X the following four conditions hold true:

(P1) x = y if and only if p(x, x) = p(y, y) = p(x, 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).

The pair (X, p) is then called a partial metric space. It is clear that, if p(x, y) = 0, then from (P1) to (P2) x = y. But if x = y,p(x, y) may not be 0.

A sequence {xm}∞

m=0 of elements of X is called p-Cauchy if the limit limm,n p(xn, xm) exists and is finite. The partial metricspace (X, p) is called complete if for each p-Cauchy sequence {xm}

∞

m=0 there is some z ∈ X such that

p(z, z) = limn

p(z, xn) = limn,m

p(xn, xm). (2.1)

If (X, p) is a partial metric space, then ps(x, y) = 2p(x, y)− p(x, x)− p(y, y), x, y ∈ X , is a metric on X , {xn}n≥1 convergesto z ∈ X with respect to ps if and only if (2.1) holds, and (X, p) is a complete partial metric space if and only if (X, ps) is acomplete metric space (see [5,13]).

A sequence xn in a partial metric space (X, p), is called 0-Cauchy (see [11] e.g.) if limm,n p(xn, xm) = 0. We say that (X, p)is 0-complete if every 0-Cauchy sequence in X converges, with respect to p, to a point x ∈ X such that p(x, x) = 0. Notethat every 0-Cauchy sequence in (X, p) is Cauchy in (X, ps), and that every complete partial metric space is 0-complete. Aparadigm for partial metric spaces is the pair (X, p) where X = Q ∩ [0, +∞) and p(x, y) = max{x, y} for x, y ≥ 0, whichprovides an example of a 0-complete partial metric space which is not complete.

For a partial metric space (X, p) and T : X → X we introduce the following notation:

XT := {x ∈ X |p(x, x) = p(x, Tx)}, ρT := inf{p(x, x)|x ∈ XT }, and rx := infip(T ix, T i+1x)

where we take inf∅ = 0. Notice that if p is a metric then XT is exactly the set of fixed points of T .

3. Auxiliary results

In this section, we define and study two extensions of the notion of Zamfirescu operators to the context of partial metricspaces.

Definition 3.1. Let (X, p) be a partial metric space, α, γ ∈ [0, 1/2), λ ∈ [0, 1) and T : X → X .(i) We say that T is Z1-operator on X if

p(Tx, Ty) ≤ maxλp(x, y), α [p(x, Ty) + p(Tx, y)] , γ [p(x, Tx) + p(y, Ty)] ,

p(x, x) + p(y, y)2

(3.1)

for each x, y ∈ X .(ii) We say that T is Z2-operator on X if

p(Tx, Ty) ≤ max {λp(x, y), α [p(x, Ty) + p(Tx, y)] , γ [p(x, Tx) + p(y, Ty)]} , (3.2)

for each x, y ∈ X .

The following auxiliary results will be used in our main section (Section 4).

Lemma 3.1. Let T be a Z1-operator on a partial metric space (X, p) and x ∈ X. Let i ≥ 1, β := max{λ, α/(1 − α), γ /(1 − γ )} ∈ [0, 1), and

Bx :=i ≥ 1|p(T ix, T i+1x) ≤ βp(T i−1x, T ix)

.

D. Ilić et al. / Mathematical and Computer Modelling 55 (2012) 801–809 803

Then the following hold:

(1) For each integer i ≥ 1 we have

p(T ix, T i+1x) ≤ maxβp(T i−1x, T ix),

p(T ix, T ix) + p(T i−1x, T i−1x)2

. (3.3)

(2) If for some integer i ≥ 1

p(T i−1x, T i−1x) < p(T ix, T ix), then p(T ix, T i+1x) ≤ βp(T i−1x, T ix). (3.4)

(3) If for some integer i ≥ 1

p(T ix, T i+1x) > βp(T i−1x, T ix), then p(T ix, T i+1x) ≤ p(T i−1x, T i−1x). (3.5)

(4) {p(T ix, T i+1x)}i≥0 is a nonincreasing sequence.(5)

If Bx is infinite, then rx = 0. (3.6)

(6)

rx = limn,m

p(T nx, Tmx). (3.7)

Proof. (1) Note that if i ∈ N is such that

p(T ix, T i+1x) ≤ maxα

p(T i−1x, T i+1x) + p(T ix, T ix)

, γ

p(T i−1x, T ix) + p(T ix, T i+1x)

holds, then using

p(T i−1x, T i+1x) ≤ p(T i−1x, T ix) + p(T ix, T i+1x) − p(T ix, T ix)

it can easily be seen that we must have

p(T ix, T i+1x) ≤ max

α

1 − α,

γ

1 − γ

p(T i−1x, T ix).

From this implication and the contractive condition (3.1) one readily sees that (3.3) must be satisfied.(2) Observe that under the assumption p(T i−1x, T i−1x) < p(T ix, T ix), the condition

p(T ix, T i+1x) ≤p(T ix, T ix) + p(T i−1x, T i−1x)

2

implies p(T ix, T i+1x) < p(T ix, T ix), which is not possible.(3) Assume p(T ix, T i+1x) > βp(T i−1x, T ix). Then (3.4) entails p(T ix, T ix) ≤ p(T i−1x, T i−1x), so (3.3) now gives

p(T ix, T i+1x) ≤ p(T i−1x, T i−1x).(4) This follows from (1) to (3).(5) We only need to check that limk p(T nkx, T nk+1x) = 0 where {nk|k ≥ 1} = Bx is a strictly increasing enumeration of

Bx. But this follows directly from

p(T nk+1x, T nk+1+1x) ≤ βp(T nk+1−1x, T nk+1x) ≤ βp(T nkx, T nk+1x) ≤ · · · ≤ βkp(T n1x, T n1+1x),

where we used that facts that the sequence {p(T ix, T i+1x)}i≥0 is nonincreasing and that nk ≤ nk+1 − 1.(6) Note that (4) implies

rx = limi

p(T ix, T i+1x). (3.8)

Case 1. Bx is infinite.Suppose that if n ≥ 1 and i ≥ 2. Because T is Z1-operator

p(T nx, T n+ix) ≤ max

λp(T n−1x, T n+i−1x), α

p(T n−1x, T n+ix) + p(T nx, T n+i−1x)

,

γp(T n−1x, T nx) + p(T n+i−1x, T n+ix)

,p(T n−1x, T n−1x) + p(T n+i−1x, T n+i−1x)

2

,

for each x ∈ X . Now, using

p(T n−1x, T n+i−1x) ≤p(T n−1x, T nx) + p(T nx, T n+ix) + p(T n+ix, T n+i−1x)

,

p(T n−1x, T n+ix) ≤ p(T n−1x, T nx) + p(T nx, T n+ix)


and

p(T nx, T n+i−1x) ≤ p(T nx, T n+ix) + p(T n+ix, T n+i−1x),

one can easily deduce the inequality

p(T nx, T n+ix) ≤ Cp(T n−1x, T nx) + p(T n+i−1x, T n+ix)

,

where C := max {α/(1 − 2α), γ , λ/(1 − λ), 1/2}. Now (3.7) follows by (3.1), (3.6) and (3.8).Case 2. Bx is finite or empty.Then there is a positive integer n1 such that p(T ix, T i+1x) > βp(T i−1x, T ix), for all i ≥ n1. Thus by (3.5)

p(T ix, T ix) ≤ p(T ix, T i+1x) ≤ p(T i−1x, T i−1x) for all i ≥ n1. (3.9)

This implies

rx = infi≥n1

p(T ix, T ix) = limi

p(T ix, T ix). (3.10)

Now, using (3.8) and (3.10), for a given ε > 0 we can find some n2 ≥ n1 such thatp(T nx, T n+1x), p(T nx, T nx)

⊆

rx, rx +

ε

2

, n ≥ n2.

Let n ≥ n2 and i ≥ 2 be arbitrary. From (P4) and (3.9) we have

rx ≤ p(T nx, T nx) ≤ p(T nx, T n+ix)

≤

i−1−k=0

p(T n+kx, T n+k+1x) −

i−1−k=1

p(T n+kx, T n+kx)

≤ p(T nx, T n+1x) + p(T n+1x, T n+2x) − p(T n+i−1x, T n+i−1x) +

i−1−k=2

[p(T n+kx, T n+k+1x) − p(T n+k−1x, T n+k−1x)]

≤ p(T nx, T n+1x) + [p(T n+1x, T n+2x) − p(T n+i−1x, T n+i−1)] < rx +ε

2+

ε

2= rx + ε.

Now, by (3.8) we get (3.7). �

Throughout this paper we keep the notations of Lemma 3.1.

Remark. A few comments are perhaps in order concerning the preceding lemma.(a) In general for Z1-operators, there may even be no i ∈ N such that

p(T ix, T i+1x) ≤ maxα


, γ


(3.11)

is true (this is an inequality occurring in the proof of (1)). Indeed let (R+, p) be the partial metric space with p(x, y) =

max{x, y}. Set α = γ = λ = 0. Define the mapping T : R+→ R+ by Tx =

x2 . Then T is a Z1-operator since

p(Tx, Ty) = max x2,y2

≤

x + y2

=p(x, x) + p(y, y)

2for all x, y ∈ R+. Nevertheless there is no x ∈ R+ and i ∈ N such that (3.11) holds since

x2i

= p(T ix, T i+1x) > 0

for all x ∈ R+. Note that here we have

maxα


, γ


= 0.

(b) The set Bx may be empty for all x ∈ X as is witnessed by the same space and mapping that were considered in (a)above. Indeed we have x

2i= p(T ix, T i+1x) > βp(T i−1x, T i−1x) = 0 for all x > 0 since here β = 0.

(c) Here is an example of a Z1-operator such that Bx is infinite for all x ∈ X . Let (R+, p) be as in (a) above, set α = λ = 0and choose any γ ∈ (0, 1/2). Define T : R+

→ R+ by Tx = γ x for x ∈ R+. T is a Z1-operator since

max{γ x, γ y} = p(Tx, Ty) ≤ γ [p(x, Tx) + p(y, Ty)] = γ x + γ y.

From γ ≤γ

1−γ= β it follows

p(T ix, T i+1x) = γ ix ≤γ

1 − γγ i−1x = βp(T i−1x, T ix)

for all x > 0 and i ∈ N. So Bx = N for all x ∈ R+. �


Lemma 3.2. Let T be a Z1-operator on a partial metric space (X, p). If x, y ∈ X and

p(y, y) = limn

p(y, T nx) = limn,m

p(T nx, Tmx), (3.12)

then p(y, y) = p(y, Ty).

Proof. Set K := max {1/(1 − α), 1/(1 − γ )}, θn := p(y, T nx) − p(T nx, T nx) and µn := p(y, T n−1x) − p(y, y). From

p(y, Ty) ≤ θn + p(Ty, T nx)

≤ θn + maxλ p(y, T n−1x), α

p(y, T nx) + p(Ty, T n−1x)

,

γp(y, Ty) + p(T n−1x, T nx)

,p(y, y) + p(T n−1x, T n−1x)

2

,

using p(Ty, T n−1x) ≤ p(Ty, y) + µn, we obtain

p(y, Ty) ≤ Kθn + maxλ p(y, T n−1x),

α

1 − α

p(y, T nx) + µn

,

γ

1 − γp(T n−1x, T nx),

p(y, y) + p(T n−1x, T n−1x)2

.

Now p(y, Ty) ≤ p(y, y) follows from (3.12) after taking the limit above as n → ∞. �

Lemma 3.3. Let T be a Z1-operator on a complete partial metric space (X, p). Then

for each ε > 0 there exists v ∈ XT such that p(v, v) < ρT + ε and p(Tv, Tv) > ρT − ε. (3.13)

Proof. By (3.7) and the completeness of (X, p), for any x ∈ X there is some x ∈ X such that

p(x, x) = limn

p(x, T nx) = limn,m

p(T nx, Tmx) = rx,

implying in particular, by the previous lemma, that XT = ∅.If ρT = 0 then (3.13) clearly holds. Thus assume ρT > 0 and suppose (3.13) were not true. Then we would have some

ε > 0 such that all v ∈ XT for which p(v, v) < ρT + ε would have to satisfy p(Tv, Tv) ≤ ρT − ε. Take some v ∈ XT suchthat p(v, v) < ρT + min {ε, (1/β − 1)ρT }. By Lemma 3.2 we have v ∈ XT and so ρT ≤ p(v, v). Therefore

ρT ≤ p(v, v) = rv ≤ p(Tv, T 2v) ≤ maxβp(v, Tv),

p(v, v) + p(Tv, Tv)

2

. (3.14)

If

p(Tv, T 2v) ≤p(v, v) + p(Tv, Tv)

2,

then (3.14) yields

ρT ≤p(v, v) + ρT − ε

2,

i.e. ρT + ε ≤ p(v, v) which is not possible.If p(Tv, T 2v) ≤ βp(v, Tv), then from (3.14) it follows

ρT ≤ βp(v, Tv) = βp(v, v) < β

ρT +

(1 − β)ρT

β

= ρT ,

again a contradiction. �

4. Main results

In this section, we prove the main results in the paper, i.e., the fixed point theorems for Z1 and Z2-operators on completepartial metric spaces. Among other things, as corollaries we give some generalized versions of the fixed point theorem ofMatthews. The theory is illustrated by some examples.

Theorem 4.1. Let T be a Z1-operator on a complete partial metric space (X, p). Then:(1) there is a unique z ∈ X such that Tz = z.(2) p(z, z) = min{p(x, x)|x ∈ XT } = ρT ,(3) if x ∈ X and p(x, x) = p(x, Tx) = p(z, z), then x = z.


Proof. By Lemma 3.3, there exists vn ∈ XT such that

p(vn, vn) < ρT +1n, and ρT −

1n

< p(Tvn, Tvn), n = 1, 2, . . . . (4.1)

Because of p(Tvn, Tvn) ≤ p(vn, Tvn) = p(vn, vn) we have

0 ≤ p(vn, vn) − p(Tvn, Tvn) <2n.

Also observe that p(y, Tw) ≤ p(y, w) + p(w, Tw) − p(w, w) implies

p(y, Tw) ≤ p(y, w) for all y ∈ X and all w ∈ XT . (4.2)

Let n,m ≥ k ≥ 1 be arbitrary and set L := max {5, 4/(1 − λ), 4/(1 − 2α)}. We have

ρT ≤ p(vn, vn) ≤ p(vn, vm)

≤ p(Tvn, Tvm) + [p(vn, Tvn) − p(Tvn, Tvn)] + [p(vm, Tvm) − p(Tvm, Tvm)].

This further yields

ρT ≤ p(vn, vm) ≤ p(Tvn, Tvm) +4k

≤4k

+ maxλp(vn, vm), α [p(vn, Tvm) + p(Tvn, vm)] , γ [p(vn, Tvn) + p(vm, Tvm)] ,

p(vn, vn) + p(vm, vm)

2

.

Thus, using (3.1), (4.1) and (4.2), we obtain ρT ≤ p(vn, vm) ≤ ρT + L/k.This proves that limn,m p(vn, vm) = ρT . Now, by completeness, there exists z ∈ X such that

p(z, z) = limn

p(z, vn) = limn,m

p(vn, vm) = ρT . (4.3)

Let us show z ∈ XT . Set δn := p(z, vn) − p(Tvn, Tvn). Thus, we have

p(z, Tz) ≤ [p(z, vn) − p(Tvn, Tvn)] + [p(vn, Tvn) − p(vn, vn)] + p(Tvn, Tz) = δn + p(Tvn, Tz), i.e.p(z, Tz) ≤ δn + p(Tvn, Tz). (4.4)

From

ρT −1n

< p(Tvn, Tvn) ≤ p(vn, Tvn) = p(vn, vn) ≤ p(z, vn)

and (4.3) it follows that δn → 0.By (4.3), we obtain

p(z, Tz) ≤ δn + maxλp(vn, z), α [p(vn, Tz) + p(Tvn, z)] , γ [p(vn, Tvn) + p(z, Tz)] ,

p(vn, vn) + p(z, z)2

.

Set N := max {1/(1 − α), 1/(1 − γ )}. Hence, by p(vn, Tz) ≤ p(vn, z) + p(z, Tz) − p(z, z) and p(Tvn, z) ≤ p(vn, z), weobtain

p(z, Tz) ≤ Nδn + maxλp(vn, z),

α

1 − α[2p(vn, z) − p(z, z)] ,

γ

1 − γp(vn, vn),

p(vn, vn) + p(z, z)2

.

Now, in view of (4.3), the inequality above for n ∈ N proves p(z, Tz) = p(z, z). i.e. z ∈ XT . Hence p(z, z) = min{p(x, x)|x ∈

XT } = ρT .Next we show Tz = z. More generally, let us prove that p(x, x) = p(x, Tx) = ρT implies Tx = x.Assume ρT > 0, otherwise there is nothing to prove. Since

0 < ρT ≤ infn

p(T n−1x, T nx) ≤ p(Tx, T 2x) ≤ p(x, Tx) = ρT

we see that p(Tx, T 2x) = p(x, Tx) = ρT . As p(Tx, T 2x) = 0, from (3.3) it follows

ρT = p(Tx, T 2x) ≤p(x, x) + p(Tx, Tx)

2i.e. ρT ≤ p(Tx, Tx). So x = Tx.

Suppose h, g ∈ X are such that Th = h, Tg = g . By (3.1) we have

p(g, h) = p(Tg, Th) ≤ maxλp(g, h), 2αp(g, h),

p(g, g) + p(h, h)2

.

Thus, p(g, h) ≤ 0, i.e., p(g, h) = 0, or p(g, h) ≤p(g,g)+p(h,h)

2 , i.e. ps(g, h) = 0. Thus g = h. �


Theorem 4.2. Let T be a Z2-operator on a 0-complete partial metric space (X, p). Then there is a unique z ∈ X such that Tz = z.Also p(z, z) = 0 and for each x ∈ X the sequence {T nx}n≥1 converges to z with respect to ps.

Proof. For given x ∈ X , observe that (3.2) implies

p(T ix, T i+1x) ≤ βp(T i−1x, T ix), for each i ≥ 1.

Therefore rx = 0, by (3.6). So by (3.7) it follows that {T nx}n≥1 is a 0-Cauchy sequence. Hence for some y ∈ X there holds

p(y, y) = limn

p(y, T nx) = limn,m

p(T nx, Tmx) = 0.

By (3.12) of Lemma 3.2 we now have p(y, Ty) = 0, i.e. Ty = y. The uniqueness of the fixed point follows from the proofof Theorem 4.1. �

Now we give corollaries of Theorem 4.1.

Corollary 4.1 ([11]). Let (X, p) be a complete partial metric space, λ ∈ [0, 1) and T : X → X a given mapping. Suppose foreach x, y ∈ X the following condition holds

p(Tx, Ty) ≤ maxλp(x, y),

p(x, x) + p(y, y)2

.

Then we have the conclusions (1)–(3) of Theorem 4.1.

Corollary 4.2. Let (X, p) be a complete partial metric space, α ∈ [0, 1/2) and T : X → X a given mapping. Suppose for eachx, y ∈ X the following condition holds

p(Tx, Ty) ≤ maxα [p(x, Ty) + p(Tx, y)] ,

p(x, x) + p(y, y)2

.


Corollary 4.3. Let (X, p) be a complete partial metric space, γ ∈ [0, 1/2) and T : X → X a given mapping. Suppose for eachx, y ∈ X the following condition holds

p(Tx, Ty) ≤ maxγ [p(x, Tx) + p(y, Ty)] ,

p(x, x) + p(y, y)2

.


Now we give corollaries of Theorem 4.2.

Corollary 4.4 ([5]). Let (X, p) be a 0-complete partial metric space, λ ∈ [0, 1) and T : X → X a given mapping. Suppose foreach x, y ∈ X the following condition holds

p(Tx, Ty) ≤ λ p(x, y). (4.5)

Then there is a unique z ∈ X such that Tz = z. Also p(z, z) = 0 and for each x ∈ X the sequence {T nx}n≥1 converges with respectto the metric ps to z.

Corollary 4.5. Let (X, p) be a 0-complete partial metric space, α ∈ [0, 1/2) and T : X → X a given mapping. Suppose for eachx, y ∈ X the following condition holds

p(Tx, Ty) ≤ α [p(x, Ty) + p(Tx, y)] . (4.6)


Corollary 4.6. Let (X, p) be a 0-complete partial metric space, γ ∈ [0, 1/2) and T : X → X a given mapping. Suppose for eachx, y ∈ X the following condition holds

p(Tx, Ty) ≤ γ [p(x, Tx) + p(y, Ty)] . (4.7)


It is clear that Z2-operator T is Z1-operator. The following example illustrates that there exists Z1-operator T which is nota Z2-operator.


Example 4.1. Let X := [0, 1] ∪ [2, 3] and define p : X2→ R by

p(x, y) =

max{x, y}, {x, y} ∩ [2, 3] = ∅,|x − y|, {x, y} ⊆ [0, 1].

Then (X, p) is a complete partial metric space. Define T : X → X by

Tx =

x + 12

, 0 ≤ x ≤ 1,1, x = 2,2 + x2

, 2 < x ≤ 3.

Given any λ ∈ [0, 1) observe that

p(Tx, Ty) > λp(x, y)

holds if λ ≤ 1/2 and x, y ∈ (2, 3], or if λ ∈ (1/2, 1) and 2 < y ≤ x < 2/(2λ − 1).Also, given any α, γ ∈ [0, 1/2) we have that

p(Tx, Ty) > max {α [p(x, Ty) + p(Tx, y)] , γ [p(x, Tx) + p(y, Ty)]}

holds if θ ≤ 1/2 and 2 < x = y ≤ 3, or if 1/2 < θ and 2 < x = y < 2/(2θ − 1), where we set θ := max{α, γ }.On the other hand, we have that

p(Tx, Ty) ≤12p(x, y), {x, y} ⊆ [0, 1],

and

p(Tx, Ty) ≤p(x, x) + p(y, y)

2, {x, y} ∩ [2, 3] = ∅.

Thus T is a Z1-operator on X which is not a Z2-operator. By Theorem 4.1 there is a unique fixed point z = 1. AlsoXT = {1} ∪ (2, 3] and we have p(1, 1) = 0 = min{p(x, x)| x ∈ XT }.

In the next example, we show that the conclusion (3) of Theorem 4.1 does not hold, in general.

Example 4.2. Let (X, p0) be a partial metric space, a > 0 and f : X → [0, a) an arbitrary mapping. If x, y ∈ X are suchthat x = y define p(x, y) = p0(x, y) + a and p(x, x) = f (x). Then (X, p) is a partial metric space, as is easily verified. LetT = I : X → X be the identity on X , i.e., I(x) = x, x ∈ X .

If b := sup f [X] < a then, given a sequence {xn}n≥1, we have lim sup p(xn, xn) ≤ b < a and p(xn, xm) ≥ a wheneverxn = xm. Thus there are no nonstationary p-Cauchy sequences. Hence (X, p) is complete. Now, if inf f [X] ∈ f [X], there is noz ∈ X such that p(z, z) = inf{p(x, x)|x ∈ X}.

Acknowledgment

The authors would like to thank the referee for useful comments which subsequently lead to a more complete finalversion of this paper and a clearer exposition.

References

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extensions of the zamfirescu theorem to partial metric spaces

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