ANALYSIS International Journal of Analysis 2314-4998 2314-498X Hindawi Publishing Corporation 726387 10.1155/2013/726387 726387 Research Article Cyclic Contractions and Fixed Point Results via Control Functions on Partial Metric Spaces Nashine Hemant Kumar 1 Kadelburg Zoran 2 Sivasundaram Seenith 1 Department of Mathematics Disha Institute of Management and Technology Satya Vihar Vidhansabha-Chandrakhuri Marg, Mandir Hasaud Raipur, Chhattisgarh 492101 India dimatindia.com 2 Faculty of Mathematics University of Belgrade Studentski trg 16 11000 Beograd Serbia bg.ac.rs 2013 4 2 2013 2013 23 08 2012 31 10 2012 14 11 2012 2013 Copyright © 2013 Hemant Kumar Nashine and Zoran Kadelburg. 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.

Cyclic weaker-type contraction conditions involving a generalized control function (with two variables) are used for mappings on 0-complete partial metric spaces to obtain fixed point results, thus generalizing several known results. Various examples are presented showing how the obtained theorems can be used and that they are proper extensions of the known ones.

1. Introduction

The celebrated Banach contraction principle has been generalized in several directions and widely used to obtain various fixed point results, with applications in many branches of mathematics.

Cyclic representations and cyclic contractions were introduced by Kirk et al.  and further used by several authors to obtain various fixed point results. See, for example, papers . Note that while a classical contraction has to be continuous, cyclic contractions might not be.

On the other hand, Matthews  introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks. In partial metric spaces, self-distance of an arbitrary point need not be equal to zero. Several authors obtained many useful fixed point results in these spaces–-we just mention . Several results in ordered partial metric spaces have been obtained as well . Some results for cyclic contractions in partial metric spaces have been very recently obtained in .

Khan et al.  addressed a new category of fixed point problems for a single self-map with the help of a control function which they called an altering distance function. This idea was further used in many papers, such as Choudhury  where generalized control functions were used. This approach has been very recently used in [44, 45] to obtain fixed point results in partial metric spaces.

In this paper, we extend these results further, considering cyclic weaker-type contraction conditions involving a generalized control function (with two variables) for mappings on 0-complete partial metric spaces (Romaguera ). We obtain fixed point theorems for such mappings, thus generalizing several known results. Various examples are presented showing how the obtained results can be used and that they are proper extensions of the known ones.

2. Preliminaries

In 2003, Kirk et al. introduced the following notion of cyclic representation.

Definition 1 (see [<xref ref-type="bibr" rid="B1">1</xref>]).

Let X be a nonempty set, m, and let f:XX be a self-mapping. Then X=i=1mAi is a cyclic representation of X with respect to f if

Ai,i=1,,m are nonempty subsets of X;

f(A1)A2, f(A2)A3,,f(Am-1)Am, f(Am)A1.

They proved the following fixed point result.

Theorem 2 (see [<xref ref-type="bibr" rid="B1">1</xref>]).

Let (X,d) be a complete metric space, f:XX, and let X=i=1mAi be a cyclic representation of X with respect to f. Suppose that f satisfies the following condition: (1)d(fx,fy)ψ(d(x,y)),xAi,yAi+1,i{1,2,,m}, where Am+1=A1 and ψ:[0,1)[0,1) is a function, upper semicontinuous from the right and 0ψ(t)<t for t>0. Then, f has a fixed point zi=1mAi.

In 2010, Păcurar and Rus introduced the following notion of cyclic weaker φ-contraction.

Definition 3 (see [<xref ref-type="bibr" rid="B2">2</xref>]).

Let (X,d) be a metric space, mN, and let A1,A2,,Am be closed nonempty subsets of X and X=i=1mAi. An operator f:XX is called a cyclic weaker φ-contraction if

X=i=1mAi is a cyclic representation of X with respect to f;

there exists a continuous, nondecreasing function φ:[0,1)[0,1) with φ(t)>0 for t(0,1) and φ(0)=0 such that (2)d(fx,fy)d(x,y)-φ(d(x,y)),

for any xAi, yAi+1,i=1,2,,m, where Am+1=A1.

They proved the following result.

Theorem 4 (see [<xref ref-type="bibr" rid="B2">2</xref>]).

Suppose that f is a cyclic weaker φ-contraction on a complete metric space (X,d). Then, f has a fixed point zi=1mAi.

This was generalized by Karapınar in .

Khan et al. introduced the following notion.

Definition 5 (see [<xref ref-type="bibr" rid="B42">42</xref>]).

A function φ:[0,+)[0,+) is called an altering distance function if the following properties are satisfied:

φ is continuous and nondecreasing,

φ(t)=0t=0.

Choudhury introduced a generalization of Chatterjea type contraction as follows.

Definition 6 (see [<xref ref-type="bibr" rid="B43">43</xref>]).

A self-mapping T:XX, on a metric space (X,d), is said to be a weakly C-contractive (or a weak Chatterjea type contraction) if for all x,yX, (3)d(Tx,Ty)12[d(x,Ty)+d(y,Tx)]-ψ(d(x,Ty),d(y,Tx)), where ψ:[0,+)2[0,+) is a continuous function such that (4)ψ(x,y)=0iffx=y=0.

In , the author proved that every weak Chatterjea type contraction on a complete metric space has a unique fixed point.

The following definitions and details can be seen, for example, in [10, 12, 13, 15, 16].

Definition 7.

A partial metric on a nonempty set X is a function p:X×X+ such that for all x,y,zX:

(p1) x=yp(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).

The pair (X,p) is called a partial metric space.

It is clear that, if p(x,y)=0, then from (p1) and (p2)x=y. But if x=y, p(x,y) may not be 0.

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.

A sequence {xn} in (X,p) converges to a point xX (in the sense of τp) if limnp(x,xn)=p(x,x). This will be denoted as xnx (n) or limnxn=x. Clearly, a limit of a sequence in a partial metric space need not be unique. Moreover, the function p(·,·) need not be continuous in the sense that xnx and yny imply p(xn,yn)p(x,y).

Example 8 (see [<xref ref-type="bibr" rid="B10">10</xref>]).

(1) A paradigmatic example of a partial metric space is the pair (+,p), where p(x,y)=max{x,y} for all x,y+.

(2) Let X={[a,b]:a,b,ab} and let p([a,b],[c,d])=max{b,d}-min{a,c}. Then (X,p) is a partial metric space.

Definition 9.

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

A sequence {xn} in (X,p) is called a Cauchy sequence if limn,mp(xn,xm) exists (and is finite). The 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,mp(xn,xm).

(see ) A sequence {xn} in (X,p) is called 0-Cauchy if limn,mp(xn,xm)=0. The space (X,p) is said to be 0-complete if every 0-Cauchy sequence in X converges (in τp) to a point xX such that p(x,x)=0.

Lemma 10.

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

(a) (see [46, 47]) If p(xn,z)p(z,z)=0 as n, then p(xn,y)p(z,y) as n for each yX.

(b) (see ) If (X,p) is complete, then it is 0-complete.

The converse assertion of (b) does not hold as the following easy example shows.

Example 11 (see [<xref ref-type="bibr" rid="B16">16</xref>]).

The space X=[0,+) with the partial metric p(x,y)=max{x,y} is 0-complete, but is not complete. Moreover, the sequence {xn} with xn=1 for each n is a Cauchy sequence in (X,p), but it is not a 0-Cauchy sequence.

It is easy to see that every closed subset of a 0-complete partial metric space is 0-complete.

3. Main Results

In this section, we will prove some fixed point theorems for self-mappings defined on a 0-complete partial metric space and satisfying certain cyclic weak contractive condition involving a generalized control function. To achieve our goal, we introduce the new notion of a cyclic contraction.

Definition 12.

Let (X,p) be a partial metric space, q, and let A1,A2,,Aq be nonempty subsets of X and Y=i=1qAi. An operator T:YY is called a cyclic contraction under weak contractive condition if:

Y=i=1qAi is a cyclic representation of Y with respect to T;

for any (x,y)Ai×Ai+1,i=1,2,,q (with Aq+1=A1), (5)p(Tx,Ty)M(x,y)-ψ(p(x,y),p(x,Tx)),

where (6)M(x,y)=max{{12}p(x,y),p(x,Tx),p(y,Ty),12[p(x,Ty)+p(Tx,y)]}, and ψ:[0,)2[0,) is a lower semicontinuous mapping such that ψ(s,t)=0 if and only if s=t=0.

Our main result is the following.

Theorem 13.

Let (X,p) be a 0-complete partial metric space, let q,A1,A2,,Aq be nonempty closed subsets of (X,p), and let Y=i=1qAi. Suppose that T:YY is a cyclic contraction as defined in Definition 12. Then, T has a unique fixed point zY such that p(z,z)=0. Moreover, zi=1qAi. Each Picard sequence xn=Tnx0, x0Y converges to z in topology τp.

Proof.

Let x0 be an arbitrary point of Y. Then there exists some i0 such that x0Ai0. Now x1=Tx0Ai0+1 and, similarly, xn:=Txn-1=Tnx0Ai0+n for n, where Aq+k=Ak. In the case p(xn0,xn0+1)=0 for some n00, it is clear that xn0 is a fixed point of T.

Without loss of the generality, we may assume that (7)p(xn,xn+1)>0,n. From the condition (NZ1), we observe that for all n, there exists i=i(n){1,2,,q} such that (xn,xn+1)Ai×Ai+1.

Putting x=xn and y=xn+1 in (NZ2) condition, we have (8)p(xn+1,xn+2)=p(Txn,Txn+1)M(xn,xn+1)-ψ(p(xn,xn+1),p(xn,Txn))=max{{12}p(xn,xn+1),p(xn+1,xn+2),12[p(xn,xn+2)+p(xn+1,xn+1)]}-ψ(p(xn,xn+1),p(xn,xn+1)). By (p4), we have (9)p(xn,xn+2)+p(xn+1,xn+1)p(xn,xn+1)+p(xn+1,xn+2). Therefore, (10)max{{12}p(xn,xn+1),p(xn+1,xn+2),12[p(xn,xn+2)+p(xn+1,xn+1)]}max{p(xn,xn+1),p(xn+1,xn+2)}. By (8) and (10), we have (11)p(xn+1,xn+2)max{p(xn,xn+1),p(xn+1,xn+2)}-ψ(p(xn,xn+1),p(xn,xn+1)). If max{p(xn,xn+1),p(xn+1,xn+2)}=p(xn+1,xn+2), then from (11), we have (12)p(xn+1,xn+2)p(xn+1,xn+2)-ψ(p(xn,xn+1),p(xn,xn+1))<p(xn+1,xn+2), which is a contradiction (it was used that ψ(p(xn,xn+1),p(xn,xn+1))>0 since p(xn,xn+1)>0). Hence, p(xn,xn+1)=0, and xn=xn+1, which is excluded. Therefore, we have max{p(xn,xn+1),p(xn+1,xn+2)}=p(xn,xn+1) and hence (13)p(xn+1,xn+2)p(xn,xn+1)-ψ(p(xn,xn+1),p(xn,xn+1))p(xn,xn+1). By (13), we have that {p(xn,xn+1)} is a nonincreasing sequence of positive real numbers. Thus, there exists r0 such that (14)limnp(xn,xn+1)=r. Passing to the limit as n in (13) and using (14) and lower semicontinuity of ψ, we have (15)rr-liminfnψ(p(xn,xn+1),p(xn,xn+1))r-ψ(r,r), thus, ψ(r,r)=0 and hence r=0. Therefore (16)limnp(xn,xn+1)=0.

Next, we claim that {xn} is a 0-Cauchy sequence in the space (X,p). Suppose that this is not the case. Then there exists ε>0 for which we can find two sequences of positive integers {m(k)} and {n(k)} such that for all positive integers k(17)n(k)>m(k)>k,p(xm(k),xn(k))ε,p(xm(k),xn(k)-1)<ε. Using (17) and (p4), we get (18)εp(xn(k),xm(k))p(xm(k),xn(k)-1)+p(xn(k)-1,xn(k))-p(xn(k)-1,xn(k)-1)<ε+p(xn(k),xn(k)-1). Thus we have (19)εp(xn(k),xm(k))<ε+p(xn(k),xn(k)-1). Passing to the limit as k in the above inequality and using (16), we obtain (20)limkp(xn(k),xm(k))=ε. On the other hand, for all k, there exists j(k){1,,q} such that n(k)-m(k)+j(k)1[q]. Then xm(k)-j(k) (for k large enough, m(k)>j(k)) and xn(k) lie in different adjacently labelled sets Ai and Ai+1 for certain i{1,,q}.

Using (p4) and (20), we get (21)p(xm(k)-j(k),xn(k))p(xm(k)-j(k),xm(k))+p(xn(k),xm(k))-p(xn(k),xn(k))  l=0j(k)-1p(xm(k)-j(k)+l,xm(k)-j(k)+l+1)+p(xn(k),xm(k))l=0q-1p(xm(k)-j(k)+l,xm(k)-j(k)+l+1)+p(xn(k),xm(k))εask(from  (16)), that is, (22)limkp(xm(k)-j(k),xn(k))=ε. Using (16), we have (23)limkp(xm(k)-j(k)+1,xm(k)-j(k))=0,(24)limkp(xn(k)+1,xn(k))=0. Again, using (p4), we get (25)p(xm(k)-j(k),xn(k)+1)p(xm(k)-j(k),xn(k))+p(xn(k),xn(k)+1)-p(xn(k),xn(k)). Passing to the limit as k in the pervious inequality, and using (24) and (22), we get (26)limkp(xm(k)-j(k),xn(k)+1)=ε. Similarly, we have by (p4) (27)p(xn(k),xm(k)-j(k)+1)p(xm(k)-j(k),xm(k)-j(k)+1)+p(xm(k)-j(k),xn(k))-p(xm(k)-j(k),xm(k)-j(k)). Passing to the limit as k, and using (16) and (22), we obtain (28)limkp(xn(k),xm(k)-j(k)+1)=ε. Similarly, we have by (p4) (29)limkp(xm(k)-j(k)+1,xn(k)+1)=ε. Using (NZ2), we obtain (30)p(xm(k)-j(k)+1,xn(k)+1)max{p(xm(k)-j(k),xn(k)),p(xm(k)-j(k)+1,xm(k)-j(k)),p(xn(k)+1,xn(k)),12p(xm(k)-j(k),xn(k)+1)    +p(xn(k),xm(k)-j(k)+1)}-ψ(p(xm(k)-j(k),xn(k)),p(xm(k)-j(k)+1,xm(k)-j(k))) for all k. Passing to the limit as k in the last inequality (and using the lower semicontinuity of the function ψ), we obtain (31)εε-liminfnψ(p(xm(k)-j(k),xn(k)),p(xm(k)-j(k)+1,xm(k)-j(k)))ε-ψ(ε,0), which implies that ψ(ε,0)=0; that is a contradiction since ε>0. We deduce that {xn} is a 0-Cauchy sequence.

Since (X,p) is 0-complete and Y is closed, it follows that the sequence {xn} converges to some uY, that is: (32)limnxn=u. We shall prove that (33)ui=1qAi. From condition (NZ1), and since x0A1, we have {xnq}n0A1. Since A1 is closed, from (32), we get that uA1. Again, from the condition (NZ1), we have {xnq+1}n0A2. Since A2 is closed, from (32), we get that uA2. Continuing this process, we obtain (33) and p(u,u)=0.

Now, we shall prove that u is a fixed point of T. Indeed, from (33), since for all n, there exists i(n){1,2,,q} such that xnAi(n); applying (NZ2) with x=u and y=xn, we obtain (34)  p(Tu,xn+1)=p(Tu,Txn)max{{12}p(u,xn),p(u,Tu),p(xn,xn+1),12p(u,xn+1)+p(xn,Tu)}-ψ(p(u,xn),p(u,Tu)), for all n. Passing to the limit as n in (34), and using (32), we get (35)p(u,Tu)limnmax{{12}p(u,xn),p(u,Tu),p(xn,xn+1),12p(u,xn+1)+p(xn,Tu)}-liminfnψ(p(u,xn),p(u,Tu))max{0,p(u,Tu),0,12p(u,Tu)}-ψ(0,p(u,Tu))=p(u,Tu)-ψ(0,p(u,Tu)), which is impossible unless ψ(0,p(u,Tu))=0, so (36)u=Tu, that is, u is a fixed point of T.

We claim that there is a unique fixed point of T. Assume on the contrary that Tu=u and Tv=v with p(u,v)>0. By supposition, we can replace x by u and y by v in (NZ2) to obtain (37)p(u,v)=p(Tu,Tv)max{{12}p(u,v),p(u,Tu),p(v,Tv),12[p(u,Tv)+p(Tu,v)]}-ψ(p(u,v),p(u,Tu))=p(u,v)-ψ(p(u,v),0)<p(u,v), a contradiction. Hence p(u,v)=0, that is, u=v. We conclude that T has only one fixed point in X. The proof is complete.

If we take q=1 and A1=X in Theorem 13, then we get the following fixed point theorem.

Corollary 14.

Let (X,p) be a 0-complete partial metric space, and let T:XX satisfy the following condition: there exists a lower semicontinuous mapping ψ:[0,)2[0,) such that ψ(t,s)=0 if and only if t=s=0 and that (38)p(Tx,Ty)max{{12}p(x,y),p(x,Tx),p(y,Ty),12[p(x,Ty)+p(y,Tx)]}-ψ(p(x,y),p(x,Tx)), for all x,yX. Then T has a unique fixed point zX. Moreover, p(z,z)=0.

Corollary 14 extends and generalizes many existing fixed point theorems in the literature.

By taking ψ(s,t)=(1-r)max{s,t} where r[0,1) in Theorem 13, we have the following result.

Corollary 15.

Let (X,p) be a 0-complete partial metric space, let q, A1,A2,,Aq be nonempty closed subsets of X, Y=i=1qAi, and T:YY such that

Y=i=1qAi is a cyclic representation of Y with respect to T;

for any (x,y)Ai×Ai+1,i=1,2,,q (with Aq+1=A1), (39)p(Tx,Ty)rmax{{12}p(x,y),p(x,Tx),p(y,Ty),12[p(x,Ty)+p(Tx,y)]},

where r[0,1). Then T has a unique fixed point z belonging to i=1qAi; moreover, p(z,z)=0.

As a special case of Corollary 15, we obtain Matthews’s version of Banach contraction principle .

Corollary 16.

Let (X,p) be a 0-complete partial metric space, let p, A1,A2,,Aq be nonempty closed subsets of X, Y=i=1qAi, and T:YY such that

X=i=1qAi is a cyclic representation of X with respect to T;

for any (x,y)Ai×Ai+1,i=1,2,,q (with Aq+1=A1), (40)p(Tnx,Tny)max{{12}p(x,y),p(x,Tnx),p(y,Tny),    12[p(x,Tny)+p(y,Tnx)]}-ψ(p(x,y),p(x,Tnx)),

where n is a positive integer and ψ:[0,)2[0,) is a lower semi-continuous mapping such that ψ(t,s)=0 if and only if t=s=0. Then T has a unique fixed point belonging to i=1qAi.

4. Examples

The following example shows how Theorem 13 can be used. It is adapted from [38, Example 2.9].

Example 17.

Consider the partial metric space (X,d) of Example 8 (2). It is easy to see that it is 0-complete. Consider the following closed subsets of X: (41)A1={[1-2-n,1]:n}{{1}},  A2={[1,1+2-n]:n}{{1}},Y=A1A2, and define a mapping T:YY by (42)Tx={[1,1+2-(n+1)],if  x=[1-2-n,1],[1-2-(n+1),1],if  x=[1,1+2-n],{1},if  x={1}. Obviously, Y=A1A2 is a cyclic representation of Y with respect to T. We will show that T satisfies the contractive condition (NZ2) of Definition 12 with the control function ψ:[0,+)2[0,+) given by ψ(s,t)=(1/2)max{s,t}.

Let (x,y)A1×A2 (the other possibility is treated similarly) and consider the following cases:

x=[1-2-n,1], y=[1,1+2-k] and n<k, that is, n+1k. Then p(Tx,Ty)=p([1,1+2-(n+1)],[1-2-(k+1),1])=(1/2)(2-n+2-k)(3/4)·2-n, (43)M(x,y)=max{2-n+2-k,32·2-n,32·2-k,12(|2-n-2-(k+1)|+|2-k-2-(n+1)|){32}}=32·2-n, and ψ(p(x,y),p(x,Tx))=(1/2)·(3/2)·2-n=(3/4)·2-n. Hence, the condition (NZ2) reduces to (3/4)·2-n(3/2)·2-n-(3/4)·2-n and holds true.

x=[1-2-n,1], y=[1,1+2-k] and n=k. Then p(Tx,Ty)=2-n, M(x,y)=2·2-n, and ψ(p(x,y),p(x,Tx))=(1/2)·2·2-n=2-n; hence (NZ2) reduces to 2-n2·2-n-2-n.

x=[1-2-n,1], y=[1,1+2-k] and n>k, that is, nk+1. Then p(Tx,Ty)(3/4)·2-k, M(x,y)=(3/2)·2-k and ψ(p(x,y),p(x,Tx))=(1/2)·(3/2)·2-k=(3/4)·2-k. Hence, (NZ2) reduces to (3/4)·2-k(3/2)·2-k-(3/4)·2-k and holds true.

x=[1-2-n,1], y={1}. Then p(Tx,Ty)=p([1,1+2-(n+1)],{1})=(1/2)·2-n, M(x,y)=p(x,Tx)=(3/2)·2-n and ψ(p(x,y),p(x,Tx))=(1/2)·(3/2)·2-n=(3/4)·2-n. (NZ2) reduces to (1/2)·2-n(3/2)·2-n-(3/4)·2-n.

The case x={1}, y=[1,1+2-k] is treated symmetrically.

The case x=y={1} is trivial.

We conclude that all conditions of Theorem 13 are satisfied. The mapping T has a unique fixed point {1}A1A2.

Here is another example showing the use of Theorem 13.

Example 18.

Let X=[0,1] be equipped with the partial metric given as (44)p(x,y)={|x-y|,ifx,y[0,1),1,ifx=1y=1. Then, (X,p) is a 0-complete partial metric space. Let A1=[0,1/2], A2=[1/2,1], and T:XX be given as (45)Tx={12,ifx[0,1),16,ifx=1. Obviously, X=A1A2 is a cyclic representation of X with respect to T. We will check the contractive condition (NZ2) of Definition 12 with the control function ψ:[0,+)2[0,+) given by ψ(s,t)=(s+t)/(2+s+t). Let (x,y)A1×A2 (the other possibility is treated symmetrically). Consider the following possible cases.

x[0,1/2], y[1/2,1). Then p(Tx,Ty)=p(1/2,1/2)=0, M(x,y)=max{y-x,1/2-x,y-1/2,(1/2)(1/2-x+y-1/2)}=y-x. ψ(p(x,y),p(x,Tx))=(y+1/2-2x)/(5/2+y-2x). It is easy to check that (46)p(Tx,Ty)=0y-x-y+1/2-2x5/2+y-2x=M(x,y)-ψ(p(x,y),p(x,Tx)) holds for the given values of x and y.

x[0,1/2], y=1. Then p(Tx,Ty)=p(1/2,1/6)=1/3, M(x,y)=1 and ψ(p(x,y),p(x,Tx))=ψ(1,1/2-x)=(3/2-x)/(7/2-x). The condition (NZ2) reduces to (47)131-3-2x7-2x

and can be checked directly.

Thus, all the conditions of Theorem 13 are fulfilled, and we conclude that T has a unique fixed point 1/2A1A2.

We state a more involved example that is inspired with the one from .

Example 19.

Let X1, Xx=(xn)n=1 if and only if xn0 for each n. Define a partial metric p on X by (48)p((xn),(yn))=n=1max{xn,yn} (it is easy to check that axioms (p1)–(p4) hold true). Let α(0,1) be fixed, denote 0=(0)n=1, and consider the subsets A1 and A2 of X defined by A1=A{0},A2=A′′{0}, where (49)Axl=(xnl)n=1iff    xnl={0,n<2ln=2k-1,k,αn,n=2k2ll=1,2,,A′′xl=(xnl)n=1iff    xnl={0,n<2l-1n=2k,k,αn,n=2k-12l-1l=1,2,. Denote Y=A1A2 (obviously A1A2={0}).

Consider the mapping T:YY given by. (50)T(0)=0,T((0,,02l-1,α2l,0,α2l+2,0,))=(0,,02l,α2l+1,0,α2l+3,0,),T((0,,02l,α2l+1,0,α2l+3,0,))=(0,,02l+1,α2l+2,0,α2l+4,0,). Obviously, T(A1)A2 and T(A2)A1; hence Y=A1A2 is a cyclic representation of Y with respect to T. Take ψ:[0,+)2[0,+) defined by ψ(s,t)=(1-α)max{s,t}.

Let us check the contractive condition (NZ2) of Theorem 13. Take x=(0,,02l-1,α2l,0,α2l+2,0,)A1 and y=(0,,02m,α2m+1,0,α2m+3,0,)A2 and assume, for example, that lm (the case l>m is treated similarly, as well as the case when x or y is equal to 0). Then (51)p(x,y)=α2l++α2m-2+α2m1-α,p(Tx,Ty)=α2l+1++α2m-1+α2m+11-α,p(x,Tx)=α2l1-α,p(y,Ty)=α2m+11-α. Hence, (52)  p(Tx,Ty)=αp(x,y)α·α2l1-α=α2l1-α-(1-α)α2l1-α=M(x,y)-ψ(p(x,y),p(x,Tx)). Obviously, T has a unique fixed point 0A1A2.

Finally, we present an example showing that in certain situations the existence of a fixed point can be concluded under partial metric conditions, while the same cannot be obtained using the standard metric.

Example 20.

Let X=+ be equipped with the usual partial metric p(x,y)=max{x,y}. Suppose A1=[0,1], A2=[0,1/2], A3=[0,1/6], A4=[0,1/42] and Y=i=14Ai. Consider the mapping T:YY defined by (53)Tx=x21+x. It is clear that i=14Ai is a cyclic representation of Y with respect to T. Further, consider the function ψ:[0,+)2[0,+) given by (54)ψ(s,t)=s+t2+s+t. Take an arbitrary pair (x,y)Ai×Ai+1 with, say, yx (the other possibility can be treated in a similar way). Then (55)p(Tx,Ty)=max{x21+x,y21+y}=x21+x. On the other hand, (56)  M(x,y)=max{p(x,y),p(x,x21+x),p(y,y21+y),12(p(x,y21+y)+p(y,x21+x))}=max{x,x,y,12(x+max{y,x21+x})}=x,M(x,y)-ψ(p(x,y),p(x,Tx))=x-2x2+2x=x21+x. Hence, condition (NZ2) is satisfied, as well as other conditions of Theorem 13 (with q=4). We deduce that T has a unique fixed point z=0A1A2A3A4.

On the other hand, consider the same problem in the standard metric d(x,y)=|x-y| and take x=1 and y=1/2. Then (57)d(Tx,Ty)=|12-16|=13,M(x,y)=max{12,12,13,12(56+0)}=12 and hence (58)M(x,y)-ψ(d(x,y),d(x,Tx))=12-13=16. Thus, condition (NZ2) for p=d does not hold and the existence of a fixed point of T cannot be derived using the standard metric.

Remark 21.

The results of this paper are obtained under the assumption that the given partial metric space is 0-complete. Taking into account Lemma 10 and Example 11, it follows that they also hold if the space is complete, but that our assumption is weaker.

Acknowledgments

The authors thank the referees for valuable suggestions that helped them to improve the text. The second author is thankful to the Ministry of Science and Technological Development of Serbia.

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