IJMMS International Journal of Mathematics and Mathematical Sciences 1687-0425 0161-1712 Hindawi Publishing Corporation 597074 10.1155/2012/597074 597074 Research Article A Fixed Point Result for Boyd-Wong Cyclic Contractions in Partial Metric Spaces Aydi Hassen 1 Karapinar Erdal 2 Rhoades Billy 1 Institut Supérieur d'Informatique et des Technologies de Communication de Hammam Sousse Université de Sousse, Route GP1-4011, Hammam Sousse, 4002 Sousse Tunisia uc.rnu.tn 2 Department of Mathematics Atilim University 06836 İncek Turkey atilim.edu.tr 2012 12 7 2012 2012 31 03 2012 31 05 2012 2012 Copyright © 2012 Hassen Aydi and Erdal Karapinar. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

1. Introduction and Preliminaries

Partial metric spaces were introduced by Matthews  to the study of denotational semantics of data networks. In particular, he proved a partial metric version of the Banach contraction principle . Subsequently, many fixed points results in partial metric spaces appeared (see, e.g., [1, 319] for more details).

Throughout this paper, the letters and * 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 set X is a function p:X×X[0,) such that for all x,y,zX

x=yp(x,x)=p(x,y)=p(y,y),

p(x,x)p(x,y),

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

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.

If p 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., [<xref ref-type="bibr" rid="B22">1</xref>, <xref ref-type="bibr" rid="B2">3</xref>, <xref ref-type="bibr" rid="B15">11</xref>, <xref ref-type="bibr" rid="B16">12</xref>]).

Consider X=[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., [<xref ref-type="bibr" rid="B22">1</xref>]).

Let X={[a,b]:a,b,,ab}, 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., [<xref ref-type="bibr" rid="B22">1</xref>, <xref ref-type="bibr" rid="B13">20</xref>]).

Let X:=[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 metric p on X generates a T0 topology τp on X, which has as a base the family of open p-balls {Bp(x,ε),xX,ε>0}, where Bp(x,ε)={yX:p(x,y)<p(x,x)+ε} for all xX and ε>0.

Definition 1.5.

Let (X,p) be a partial metric space and {xn} a sequence in X. Then,

{xn} converges to a point xX if and only if p(x,x)=limn+p(x,xn),

{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 xX, such that p(x,x)=limn,m+p(xn,xm).

Lemma 1.7 (see, e.g., [<xref ref-type="bibr" rid="B2">3</xref>, <xref ref-type="bibr" rid="B15">11</xref>, <xref ref-type="bibr" rid="B16">12</xref>]).

Let (X,p) be a partial metric space. Then,

{xn} is a Cauchy sequence in (X,p) if and only if it is a Cauchy sequence in the metric space (X,dp),

(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)=limn+p(xn,x)=limn,m+p(xn,xm).

Lemma 1.8 (see, e.g., [<xref ref-type="bibr" rid="B2">3</xref>, <xref ref-type="bibr" rid="B15">11</xref>, <xref ref-type="bibr" rid="B16">12</xref>]).

Let (X,p) be a partial metric space. Then,

if p(x,y)=0, then x=y,

if xy, then p(x,y)>0.

Remark 1.9.

If x=y, p(x,y) may not be 0.

Lemma 1.10 (see, e.g., [<xref ref-type="bibr" rid="B2">3</xref>, <xref ref-type="bibr" rid="B15">11</xref>, <xref ref-type="bibr" rid="B16">12</xref>]).

Let xnz as n in a partial metric space (X,p) where p(z,z)=0. Then, limnp(xn,y)=p(z,y) for every yX.

Let Φ be the set of functions ϕ:[0,)[0,) such that

ϕ is upper semicontinuous (i.e., for any sequence {tn} in [0,) such that tnt as n, we have limsupnϕ(tn)ϕ(t)),

ϕ(t)<t for each t>0.

Recently, Romaguera  obtained the following fixed point theorem of Boyd-Wong type .

Theorem 1.11.

Let (X,p) be a complete partial metric space, and let T:XX be a map such that for all x,yX(1.4)p(Tx,Ty)ϕ(M(x,y)), where ϕΦ 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.  introduced the following definition.

Definition 1.12 (see [<xref ref-type="bibr" rid="B20">23</xref>]).

Let X be a nonempty set, m a positive integer, and T:XX a mapping. X=i=1mAi is said to be a cyclic representation of X with respect to T if

Ai,i=1,2,,m are nonempty closed sets,

T(A1)A2,,T(Am-1)Am,T(Am)A1.

Recently, fixed point theorems involving a cyclic representation of X with respect to a self-mapping T have appeared in many papers (see, e.g., ).

Very recently, Abbas et al.  extended Theorem 1.11 to a class of cyclic mappings and proved the following result, but with ϕΦ 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:YY be a mapping such that

Y=i=1mAi is a cyclic representation of Y with respect to T,

there exists ϕ:[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 xAi, yAi+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 zi=1mAi.

In the following example, ϕΦ, but it is not continuous.

Example 1.14.

Define ϕ:[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*. 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.

Following Example 1.14, the main aim of this paper is to present the analog of Theorem 1.13 for a weaker hypothesis on ϕ, that is, with ϕΦ. Our proof is simpler than that in . 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:YY be a mapping such that

Y=i=1mAi is a cyclic representation of Y with respect to T,

there exists ϕΦ such that (2.1)p(Tx,Ty)ϕ(M(x,y)), for any xAi, yAi+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 zi=1mAi.

Proof.

Let x0Y=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 that xnxn+1, for each n0. Having in mind that Y=i=1mAi, so for each n0, there exists in{1,2,,m} such that xnAin and xn+1=TxnT(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)}n0. If for some k, 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)n0. 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 γ0 such that limnp(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<γlimsupnϕ(p(xn+1,xn+2))ϕ(γ)<γ, which is a contradiction, so that γ=0, that is, (2.9)limnp(xn,xn+1)=0. By (1.1), we have dp(x,y)2p(x,y) for all x,yX, and then from (2.9) (2.10)limndp(xn,xn+1)=0. Also, by (p2), (2.11)limnp(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)εdp(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)limkdp(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)limkdp(xn(k)+1,xm(k)+1)=ε. Similarly, we have (2.18)limkdp(xn(k),xm(k)+1)=limk+dp(xm(k),xn(k)+1)=ε. Also, by (1.1), (2.11), and (2.15)–(2.18), we may find (2.19)limkp(xn(k),xm(k))=limkp(xn(k),xm(k)+1)=ϵ2,limkp(xn(k)+1,xm(k)+1)=limkp(xm(k),xn(k)+1)=ϵ2. On the other hand, for all k, there exists j(k), 0j(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)limkp(xn(k),xm(k)-j(k))=limkp(xn(k)+1,xm(k)-j(k)+1)=ϵ2,(2.23)limkp(xn(k),xm(k)-j(k)+1)=limkp(xn(k)+1,xm(k)-j(k))=ϵ2. By (2.22) and (2.23), we get that (2.24)limkM(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<ϵ2limsupkϕ(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=limnxn(Y,dp). Notice that the sequence {xn}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)ui=1mAi. Thus, i=1mAi.

For simplicity, set A=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:AA. 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.

Let X= and p(x,y)=max{|x|,|y|}. It is obvious that (X,p) is a complete partial metric space.

Set A1=[-8,0], A2=[0,8], and Y=A1A2=[-8,8]. Define T:TY by (3.1)Tx={-x4      if    x[-1,1],0      otherwise.

Notice that T([-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)={t3if  t[0,1),n2n2+1if  t[n,n+1),nN*. 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}=0ϕ(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}=y4ϕ(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 1/2ϕ(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|4ϕ(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 1/2ϕ(t)<1 for all t>1.

Case4. (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 Theorem 2.1 are also satisfied. The function T has 0 as a unique fixed point.

However, since ϕ is not a continuous function, we could not apply Theorem 1.13.

Example 3.2.

Let X=[0,1] and p(x,y)=max{x,y} for all x,yX. Then, (X,p) is a complete partial metric space. Take A1==Ap=X. Define T:XX by Tx=x/2. Consider ϕ:[0,)[0,) given by Example 1.14.

For all x,yX, 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, Theorem 1.13 is not applicable.

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