# A Fixed Point Result for Boyd-Wong Cyclic Contractions in Partial Metric Spaces

Authors: Hassen Aydi; Erdal Karapinar
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

A fixed point theorem involving Boyd-Wong-type cyclic contractions in partial metric spaces is proved. We also provide examples to support the concepts and results presented herein.


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## Body

## 1. Introduction and Preliminaries

Partial metric spaces were introduced by Matthews [1] to the study of denotational semantics of data networks. In particular, he proved a partial metric version of the Banach contraction principle [2]. Subsequently, many fixed points results in partial metric spaces appeared (see, e.g., [1, 3–19] for more details).Throughout this paper, the lettersℝ and ℕ* will denote the sets of all real numbers and positive integers, respectively. We recall some basic definitions and fixed point results of partial metric spaces.Definition 1.1.
A partial metric on a nonempty setX is a function p:X×X→[0,∞) 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,y)≤p(x,z)+p(z,y)-p(z,z). A partial metric space is a pair (X,p) such that X is a nonempty set and p is a partial metric on X.Ifp is a partial metric on X, then the function dp:X×X→[0,∞) given by
(1.1)dp(x,y)=2p(x,y)-p(x,x)-p(y,y)
is a metric on X.Example 1.2 (see, e.g., [1, 3, 11, 12]).
ConsiderX=[0,∞) with p(x,y)=max{x,y}. Then, (X,p) is a partial metric space. It is clear that p is not a (usual) metric. Note that in this case dp(x,y)=|x-y|.Example 1.3 (see, e.g., [1]).
LetX={[a,b]:a,b,∈ℝ,a≤b}, and define p([a,b],[c,d])=max{b,d}-min{a,c}. Then, (X,p) is a partial metric space.Example 1.4 (see, e.g., [1, 20]).
LetX:=[0,1]∪[2,3], and define p:X×X→[0,∞) by
(1.2)p(x,y)={max⁡{x,y}if{x,y}∩[2,3]≠∅,|x-y|if{x,y}⊂[0,1].
Then, (X,p) is a complete partial metric space.Each partial metricp on X generates a T0 topology τp on X, which has as a base 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 1.5.
Let(X,p) be a partial metric space and {xn} a sequence in X. Then, (i)
{xn} converges to a point x∈X if and only if p(x,x)=limn→+∞p(x,xn),(ii)
{xn} is called a Cauchy sequence if limn,m→+∞p(xn,xm) exists and is finite.Definition 1.6.
A partial metric space(X,p) is said to be complete if every Cauchy sequence {xn} in X converges, with respect to τp, to a point x∈X, such that p(x,x)=limn,m→+∞p(xn,xm).Lemma 1.7 (see, e.g., [3, 11, 12]).
Let(X,p) be a partial metric space. Then,(a)
{xn} is a Cauchy sequence in (X,p) if and only if it is a Cauchy sequence in the metric space (X,dp),(b)
(X,p) is complete if and only if the metric space (X,dp) is complete. Furthermore, limn→+∞dp(xn,x)=0 if and only if
(1.3)p(x,x)=lim⁡n→+∞p(xn,x)=lim⁡n,m→+∞p(xn,xm).Lemma 1.8 (see, e.g., [3, 11, 12]).
Let(X,p) be a partial metric space. Then,  (a)
ifp(x,y)=0, then x=y,(b)
ifx≠y, then p(x,y)>0.Remark 1.9.
Ifx=y, p(x,y) may not be 0.Lemma 1.10 (see, e.g., [3, 11, 12]).
Letxn→z as n→∞ in a partial metric space (X,p) where p(z,z)=0. Then, limn→∞p(xn,y)=p(z,y) for every y∈X.LetΦ be the set of functions ϕ:[0,∞)→[0,∞) such that(i)
ϕ is upper semicontinuous (i.e., for any sequence {tn} in [0,∞) such that tn→t as n→∞, we have limsupn→∞ϕ(tn)≤ϕ(t)),(ii)
ϕ(t)<t for each t>0.Recently, Romaguera [21] obtained the following fixed point theorem of Boyd-Wong type [22].Theorem 1.11.
Let(X,p) be a complete partial metric space, and let T:X→X be a map such that for all x,y∈X(1.4)p(Tx,Ty)≤ϕ(M(x,y)),
where ϕ∈Φ and
(1.5)M(x,y)=max⁡{p(x,y),p(x,Tx),p(y,Ty),12[p(x,Ty)+p(y,Tx)]}.
Then, T has a unique fixed point.In 2003, Kirk et al. [23] introduced the following definition.Definition 1.12 (see [23]).
LetX be a nonempty set, m a positive integer, and T:X→X a mapping. X=⋃i=1mAi is said to be a cyclic representation of X with respect to T if  (i)
Ai,i=1,2,…,m are nonempty closed sets,(ii)
T(A1)⊂A2,…,T(Am-1)⊂Am,T(Am)⊂A1.Recently, fixed point theorems involving a cyclic representation ofX with respect to a self-mapping T have appeared in many papers (see, e.g., [24–28]).Very recently, Abbas et al. [24] extended Theorem 1.11 to a class of cyclic mappings and proved the following result, but with ϕ∈Φ being a continuous map.Theorem 1.13.
Let(X,p) be a complete partial metric space. Let m be a positive integer, A1,A2,…,Am nonempty closed subsets of (X,dp), and Y=⋃i=1mAi. Let T:Y→Y be a mapping such that(i)
Y=⋃i=1mAi is a cyclic representation of Y with respect to T,(ii)
there existsϕ:[0,∞)→[0,∞) such that ϕ is continuous and ϕ(t)<t for each t>0, satisfying(1.6)p(Tx,Ty)≤ϕ(M(x,y)), for any x∈Ai, y∈Ai+1, i=1,2,…,m, where Am+1=A1 and M(x,y) is defined by (1.5).Then, T has a unique fixed point z∈⋂i=1mAi.In the following example,ϕ∈Φ, but it is not continuous.Example 1.14.
Defineϕ:[0,∞)→[0,∞) by ϕ(t)=t/2 for all t∈[0,1) and ϕ(t)=n(n+1)/(n+2) for t∈[n,n+1), n∈ℕ*. Then, ϕ is upper semicontinuous on [0,∞) with ϕ(t)<t for all t>0. However, it is not continuous at t=n for all n∈ℕ.Following Example1.14, the main aim of this paper is to present the analog of Theorem 1.13 for a weaker hypothesis on ϕ, that is, with ϕ∈Φ. Our proof is simpler than that in [24]. Also, some examples are given.


## 2. Main Results

Our main result is the following.Theorem 2.1.
Let(X,p) be a complete partial metric space. Let m be a positive integer, A1,A2,…,Am nonempty closed subsets of (X,dp), and Y=⋃i=1mAi. Let T:Y→Y be a mapping such that(1)
Y=⋃i=1mAi is a cyclic representation of Y with respect to T,(2)
there existsϕ∈Φ such that
(2.1)p(Tx,Ty)≤ϕ(M(x,y)),
for any x∈Ai, y∈Ai+1, i=1,2,…,m, where Am+1=A1 and M(x,y) is defined by (1.5).Then, T has a unique fixed point z∈⋂i=1mAi.Proof.
Letx0∈Y=⋃i=1mAi. Consider the Picard iteration {xn} given by Txn=xn+1 for n=0,1,2,…. If there exists n0 such that xn0+1=xn0, then xn0+1=Txn0=xn0 and the existence of the fixed point is proved.
Assume thatxn≠xn+1, for each n≥0. Having in mind that Y=⋃i=1mAi, so for each n≥0, there exists in∈{1,2,…,m} such that xn∈Ain and xn+1=Txn∈T(Ain)⊆Ain+1. Then, by (2.1)
(2.2)p(xn+1,xn+2)=p(Txn,Txn+1)≤ϕ(M(xn,xn+1)),
where
(2.3)M(xn,xn+1)=max⁡{p(xn,Txn+1)+p(xn+1,Txn)2p(xn,xn+1),p(xn,Txn),p(xn+1,Txn+1),p(xn,Txn+1)+p(xn+1,Txn)2}=max⁡{p(xn,xn+1),p(xn+1,xn+2),p(xn,xn+2)+p(xn+1,xn+1)2}=max⁡{p(xn,xn+1),p(xn+1,xn+2),p(xn,xn+1)+p(xn+1,xn+2)2}=max⁡{p(xn,xn+1),p(xn+1,xn+2)}.
Therefore,
(2.4)M(xn,xn+1)=max⁡{p(xn,xn+1),p(xn+1,xn+2)}∀n≥0.
If for some k∈ℕ, we have M(xk,xk+1)=p(xk+1,xk+2), so by (2.2)
(2.5)0<p(xk+1,xk+2)≤ϕ(p(xk+1,xk+2))<p(xk+1,xk+2),
which is a contradiction. It follows that
(2.6)M(xn,xn+1)=p(xn,xn+1)∀n≥0.
Thus, from (2.2), we get that
(2.7)0<p(xn+1,xn+2)≤ϕ(p(xn+1,xn+2))<p(xn+1,xn+2).
Hence, {p(xn,xn+1)} is a decreasing sequence of positive real numbers. Consequently, there exists γ≥0 such that limn→∞p(xn,xn+1)=γ. Assume that γ>0. Letting n→∞ in the above inequality, we get using the upper semicontinuity of ϕ(2.8)0<γ≤limsup⁡n→∞ϕ(p(xn+1,xn+2))≤ϕ(γ)<γ,
which is a contradiction, so that γ=0, that is,
(2.9)lim⁡n→∞p(xn,xn+1)=0.
By (1.1), we have dp(x,y)≤2p(x,y) for all x,y∈X, and then from (2.9)
(2.10)lim⁡n→∞dp(xn,xn+1)=0.
Also, by (p2),
(2.11)lim⁡n→∞p(xn,xn)=0.
In the sequel, we will prove that {xn} is a Cauchy sequence in the partial metric space (Y=⋃i=1mAi,p). By Lemma 1.7, it suffices to prove that {xn} is Cauchy sequence in the metric space (Y,dp). We argue by contradiction. Assume that {xn} is not a Cauchy sequence in (Y,dp). Then, there exists ε>0 for which we can find subsequences {xm(k)} and {xn(k)} of {xn} with n(k)>m(k)≥k such that
(2.12)dp(xn(k),xm(k))≥ε.
Further, corresponding to m(k), we can choose n(k) in such a way that it is the smallest integer with n(k)>m(k) and satisfying (2.12). Then,
(2.13)dp(xn(k)-1,xm(k))<ε.We use (2.13) and the triangular inequality
(2.14)ε≤dp(xn(k),xm(k))≤dp(xn(k),xn(k)-1)+dp(xn(k)-1,xm(k))<ε+dp(xn(k),xn(k)-1).
Letting k→∞ in (2.14) and using (2.10), we find
(2.15)lim⁡k→∞dp(xn(k),xm(k))=ε.
On the other hand
(2.16)dp(xn(k),xm(k))≤dp(xn(k),xn(k)+1)+dp(xn(k)+1,xm(k)+1)+dp(xm(k)+1,xm(k)),dp(xn(k)+1,xm(k)+1)≤dp(xn(k)+1,xn(k))+dp(xn(k),xm(k))+dp(xm(k),xm(k)+1).
Letting k→+∞ in the two above inequalities and using (2.10) and (2.15),
(2.17)lim⁡k→∞dp(xn(k)+1,xm(k)+1)=ε.
Similarly, we have
(2.18)lim⁡k→∞dp(xn(k),xm(k)+1)=lim⁡k→+∞dp(xm(k),xn(k)+1)=ε.
Also, by (1.1), (2.11), and (2.15)–(2.18), we may find
(2.19)lim⁡k→∞p(xn(k),xm(k))=lim⁡k→∞p(xn(k),xm(k)+1)=ϵ2,lim⁡k→∞p(xn(k)+1,xm(k)+1)=lim⁡k→∞p(xm(k),xn(k)+1)=ϵ2.
On the other hand, for all k, there exists j(k), 0≤j(k)≤p, such that n(k)-m(k)+j(k)≡1(p). Then, xm(k)-j(k) (for k large enough, m(k)>j(k)) and xn(k) lie in different adjacently labeled sets Ai and Ai+1 for certain i=1,…,p. Using the contractive condition (2.1), we get
(2.20)p(xn(k)+1,xm(k)-j(k)+1)=p(Txn(k),Txm(k)-j(k))≤ϕ(M(xn(k),xm(k)-j(k)),
where
(2.21)M(xn(k),xm(k)-j(k))=max⁡{p(xn(k),Txm(k)-j(k))+p(xm(k)-j(k),Txn(k))2p(xn(k),xm(k)-j(k)),p(xn(k),Txn(k)),p(xm(k)-j(k),Txm(k)-j(k)),p(xn(k),Txm(k)-j(k))+p(xm(k)-j(k),Txn(k))2}=max⁡{p(xn(k),xm(k)-j(k)+1)+p(xm(k)-j(k),xn(k)+1)2p(xn(k),xm(k)-j(k)),p(xn(k),xn(k)+1),p(xm(k)-j(k),xm(k)-j(k)+1),p(xn(k),xm(k)-j(k)+1)+p(xm(k)-j(k),xn(k)+1)2}.
As (2.19), using (2.9), we may get
(2.22)lim⁡k→∞p(xn(k),xm(k)-j(k))=lim⁡k→∞p(xn(k)+1,xm(k)-j(k)+1)=ϵ2,(2.23)lim⁡k→∞p(xn(k),xm(k)-j(k)+1)=lim⁡k→∞p(xn(k)+1,xm(k)-j(k))=ϵ2.
By (2.22) and (2.23), we get that
(2.24)lim⁡k→∞M(xn(k),xm(k)-j(k))=ϵ2.
Letting n→∞ in (2.20), we get using (2.22), (2.24), and the upper semicontinuity of ϕ(2.25)0<ϵ2≤limsup⁡k→∞ϕ(M(xn(k),xm(k)-j(k)))≤ϕ(ϵ2)<ϵ2,
which is a contradiction.
This shows that{xn} is a Cauchy sequence in the complete subspace Y=⋃i=1mAi equipped with the metric dp. Thus, there exists u=limn→∞xn∈(Y,dp). Notice that the sequence {xn}n∈ℕ has an infinite number of terms in each Ai, i=1,…,m, so since (Y,dp) is complete, from each Ai, i=1,…,m one can extract a subsequence of {xn} that converges to u. Because Ai, i=1,…,m are closed in (Y,dp), it follows that
(2.26)u∈⋂i=1mAi.
Thus, ⋂i=1mAi≠∅.
For simplicity, setA=⋂i=1mAi. Clearly, A is also closed in (Y,dp), so it is a complete subspace of (Y,dp) and then (A,p) is a complete partial metric space. Consider the restriction of T on A, that is, T/A:A→A. Then, T/A satisfies the assumptions of Theorem 1.11, and thus T/A has a unique fixed point in Z.


## 3. Examples

We give some examples illustrating our results.Example 3.1.
LetX=ℝ and p(x,y)=max{|x|,|y|}. It is obvious that (X,p) is a complete partial metric space.
SetA1=[-8,0], A2=[0,8], and Y=A1∪A2=[-8,8]. Define T:T→Y by
(3.1)Tx={-x4ifx∈[-1,1],0otherwise.
Notice thatT([-8,-1))=0 and T([-1,0])=[0,1/4], and hence T(A1)⊆A2. Analogously, T((1,8])=0 and T([0,1])=[-1/4,0], and hence T(A2)⊆A1.Take(3.2)ϕ(t)={t3ift∈[0,1),n2n2+1ift∈[n,n+1),n∈N*.
Clearly, T satisfies condition (2.1). Indeed, we have the following cases.Case  1.
(
x
∈
[
-
8
,
-
1
) and y∈(1,8]). Inequality (2.1) turns into
(3.3)p(Tx,Ty)=max⁡{0,0}=0≤ϕ(M(x,y)),
which is necessarily true.Case   2.
(
x
∈
[
-
8
,
-
1
) and y∈[0,1]). Inequality (2.1) becomes
(3.4)p(Tx,Ty)=max⁡{0,|y|4}=y4≤ϕ(M(x,y))=ϕ(max⁡{p(x,y),p(x,Tx),p(Ty,y),12[p(x,Ty)+p(Tx,y)]})=ϕ(max⁡{|x|,|x|,|y|,12[|x|+|y|]})=ϕ(|x|).
It is clear that 1/2≤ϕ(t)<1 for all t>1. Hence, (3.4) holds.Case   3.
(
x
∈
[
-
1,0
] and y∈(1,8]). Inequality (2.1) turns into
(3.5)p(Tx,Ty)=max⁡{|x|4,0}=|x|4≤ϕ(M(x,y))=ϕ(max⁡{p(x,y),p(x,Tx),p(Ty,y),12[p(x,Ty)+p(Tx,y)]})=ϕ(max⁡{|y|,|x|,|y|,12[|x|+|y|]})=ϕ(|y|)=ϕ(y),
which is true again by the fact that 1/2≤ϕ(t)<1 for all t>1.Case 4.(x∈[-1,0] and y∈[0,1]). Inequality (2.1) becomes
(3.6)p(Tx,Ty)=max⁡{|x|4,|y|4}≤ϕ(M(x,y))=ϕ(max⁡{p(x,y),p(x,Tx),p(Ty,y),12[p(x,Ty)+p(Tx,y)]})=ϕ(max⁡{max⁡{|x|,|y|},|x|,|y|,12[max⁡{|x|4,|y|}+max⁡{|x|,|y|4}]}).
Let use examine all possibilities:
(3.7)p(Tx,Ty)={|x|4if|x|≥|y|,|y|4if|y|4≤|x|≤|y|,|y|4if|x|≤|y|4,M(x,y)≤{|x|if|x|≥|y|,|y|if|y|4≤|x|≤|y|,|y|if|x|≤|y|4.
Thus, (2.1) holds for ϕ(t)=t/3.The rest of the assumptions of Theorem2.1 are also satisfied. The function T has 0 as a unique fixed point.However, sinceϕ is not a continuous function, we could not apply Theorem 1.13.Example 3.2.
LetX=[0,1] and p(x,y)=max{x,y} for all x,y∈X. Then, (X,p) is a complete partial metric space. Take A1=⋯=Ap=X. Define T:X→X by Tx=x/2. Consider ϕ:[0,∞)→[0,∞) given by Example 1.14.For allx,y∈X, we have
(3.8)p(Tx,Ty)=max⁡{x2,y2}=ϕ(p(x,y)≤ϕ(M(x,y)).
Then all the assumptions of Theorem 2.1 are satisfied. The function T has 0 as a unique fixed point.Similarly, Theorem1.13 is not applicable.



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