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Fixed point theorems under HardyRogers contractive conditions on 0complete ordered partial metric spaces
Fixed Point Theory and Applications volume 2012, Article number: 180 (2012)
Abstract
A fixed point theorem is obtained for a monotone selfmap in a 0complete ordered partial metric space under HardyRogerstype contractive condition. This result improves some recently obtained ones, in the sense that weaker conditions are used. An example shows how this result can be used when the corresponding result in standard metric cannot. The second theorem is concerned with two weakly isotone increasing selfmappings in ordered partial metric spaces. A common fixed point result is obtained without any commutativity or compatibility assumptions.
MSC:47H10, 54H25.
1 Introduction
Matthews [1] introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks. He showed that the Banach contraction mapping theorem can be generalized to the partial metric context for applications in program verification. Subsequently, several authors (see, e.g., [2–12]) derived fixed point theorems in partial metric spaces. See also the presentation by Bukatin et al. [13] where the motivation for introducing nonzero distance (i.e., the ‘distance’ p where p(x,x)=0 need not hold) is explained, which is also leading to interesting research in foundations of topology.
On the other hand, fixed point theory has developed rapidly in partially ordered metric spaces. The first result in this direction was given by Ran and Reurings [14] who presented its applications to matrix equations. Subsequently, Nieto and RodríguezLópez [15] extended this result and applied it to obtain a unique solution for periodic boundary value problems. Further results were obtained by several authors, we mention [16–21]. Fixed point results in ordered partial metric spaces have been obtained recently in [22–26].
The following definitions and details can be seen in [1–4] and [13, 27, 28].
Definition 1.1 A partial metric on a nonempty set X is a function p:X\times X\to {\mathbb{R}}^{+} such that for all x,y,z\in X,
(p_{1}) x=y\u27fap(x,x)=p(x,y)=p(y,y),
(p_{2}) p(x,x)\le p(x,y),
(p_{3}) p(x,y)=p(y,x),
(p_{4}) p(x,y)\le 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 (p_{1}) and (p_{2}) x=y. But if x=y, p(x,y) may not be 0.
Each partial metric p on X generates a {T}_{0} topology {\tau}_{p} on X which has as a base the family of open pballs \{{B}_{p}(x,\epsilon ):x\in X,\epsilon >0\}, where {B}_{p}(x,\epsilon )=\{y\in X:p(x,y)<p(x,x)+\epsilon \} for all x\in X and \epsilon >0. A sequence \{{x}_{n}\} in (X,p) converges to a point x\in X (in the sense of {\tau}_{p}) if {lim}_{n\to \mathrm{\infty}}p(x,{x}_{n})=p(x,x). This will be denoted as {x}_{n}\to x (n\to \mathrm{\infty}) or {lim}_{n\to \mathrm{\infty}}{x}_{n}=x.
Remark 1.2 Clearly, a limit of a sequence in a partial metric space need not be unique. Moreover, the function p(\cdot ,\cdot ) need not be continuous in the sense that {x}_{n}\to x and {y}_{n}\to y imply p({x}_{n},{y}_{n})\to p(x,y).
If p is a partial metric on X, then the function {p}^{s}:X\times X\to {\mathbb{R}}^{+} given by
is a metric on X. Furthermore, {lim}_{n\to \mathrm{\infty}}{p}^{s}({x}_{n},x)=0 if and only if
Example 1.3

(1)
A paradigmatic example of a partial metric space is the pair ({\mathbb{R}}^{+},p), where p(x,y)=max\{x,y\} for all x,y\in {\mathbb{R}}^{+}. The corresponding metric is
{p}^{s}(x,y)=2max\{x,y\}xy=xy. 
(2)
If (X,d) is a metric space and c\ge 0 is arbitrary, then
p(x,y)=d(x,y)+c
defines a partial metric on X and the corresponding metric is {p}^{s}(x,y)=2d(x,y).
Remark 1.4 If T:X\to X is continuous at {x}_{0}\in X (with respect to {\tau}_{p}), then for each sequence \{{x}_{n}\} in X, we have
It is worth mentioning that the notions of pcontinuity and {p}^{s}continuity for a selfmapping on X are incomparable in general. Indeed, let X=[0,+\mathrm{\infty}), p(x,y)=max\{x,y\} (and hence {p}^{s}(x,y)=xy), T0=1, Tx={x}^{2} for x>0, and let Sx=sinx. Then it is easy to see that T is pcontinuous and {p}^{s}discontinuous at x=0, while S is pdiscontinuous and {p}^{s}continuous at x=\pi (for details, see [29]).
Other examples of partial metric spaces which are interesting from the computational point of view may be found in [1, 30, 31].
Definition 1.5 Let (X,p) be a partial metric space. Then:

(1)
A sequence \{{x}_{n}\} in (X,p) is called a Cauchy sequence if {lim}_{n,m\to \mathrm{\infty}}p({x}_{n},{x}_{m}) exists (and is finite).

(2)
The space (X,p) is said to be complete if every Cauchy sequence \{{x}_{n}\} in X converges, with respect to {\tau}_{p}, to a point x\in X such that p(x,x)={lim}_{n,m\to \mathrm{\infty}}p({x}_{n},{x}_{m}).

(3)
[28] A sequence \{{x}_{n}\} in (X,p) is called 0Cauchy if {lim}_{n,m\to \mathrm{\infty}}p({x}_{n},{x}_{m})=0. The space (X,p) is said to be 0complete if every 0Cauchy sequence in X converges (in {\tau}_{p}) to a point x\in X such that p(x,x)=0.
Lemma 1.6 Let (X,p) be a partial metric space.

(a)
\{{x}_{n}\} is a Cauchy sequence in (X,p) if and only if it is a Cauchy sequence in the metric space (X,{p}^{s}).

(b)
The space (X,p) is complete if and only if the metric space (X,{p}^{s}) is complete.

(c)
[5, 9]If p({x}_{n},z)\to p(z,z)=0 as n\to \mathrm{\infty}, then p({x}_{n},y)\to p(z,y) as n\to \mathrm{\infty} for each y\in X.

(d)
Every 0Cauchy sequence in (X,p) is Cauchy in (X,{p}^{s}).

(e)
If (X,p) is complete, then it is 0complete.
The converse assertions of (d) and (e) do not hold as the following easy example shows.
Example 1.7 [28] The space X=[0,+\mathrm{\infty})\cap \mathbb{Q} with the partial metric p(x,y)=max\{x,y\} is 0complete, but it is not complete (since {p}^{s}(x,y)=xy and (X,{p}^{s}) is not complete). Moreover, the sequence \{{x}_{n}\} with {x}_{n}=1 for each n\in \mathbb{N} is a Cauchy sequence in (X,p), but it is not a 0Cauchy sequence.
Recall that Romaguera proved in [[28], Theorem 2.3] that a partial metric space (X,p) is 0complete if and only if every {p}^{s}Caristi mapping on X has a fixed point.
It is easy to see that every closed subset of a 0complete partial metric space is 0complete.
Definition 1.8 Let (X,\u2aaf) be a partially ordered set. Then:

(a)
elements x,y\in X are called comparable if x\u2aafy or y\u2aafx holds;

(b)
a subset K of X is said to be well ordered if every two elements of K are comparable;

(c)
a mapping T:X\to X is called nondecreasing (nonincreasing) w.r.t. ⪯ if x\u2aafy implies Tx\u2aafTy (Tx\u2ab0Ty).
Definition 1.9 Let X be a nonempty set. Then (X,p,\u2aaf) is called an ordered (partial) metric space if:

(i)
(X,p) is a (partial) metric space, and

(ii)
(X,\u2aaf) is a partially ordered set.
Definition 1.10 Let (X,p,\u2aaf) be an ordered partial metric space. We say that X is regular if the following holds: if \{{z}_{n}\} is a nondecreasing (resp. nonincreasing) sequence in X with respect to ⪯ such that {z}_{n}\to z\in X as n\to \mathrm{\infty}, then {z}_{n}\u2aafz (resp. {z}_{n}\u2ab0z) for all n\in \mathbb{N}.
In this paper, we first obtain a fixed point theorem for a monotone selfmap in a 0complete partially ordered partial metric space under HardyRogerstype contractive condition. This result improves some recently obtained ones, in particular those from [26], in the sense that weaker conditions are used. It can be also considered as an extension and improvement of some results in standard ordered metric spaces, e.g., those from [20]. An example shows that our theorem can be used when the corresponding result in standard metric cannot.
The second theorem is concerned with two weakly isotone increasing selfmappings in ordered partial metric spaces and is related to the results of the paper [25]. A common fixed point result is obtained without commutativity or compatibility assumptions. An example shows how this theorem can be used.
2 Fixed point results for a single mapping
Our first result is the following
Theorem 2.1 Let (X,p,\u2aaf) be a 0complete ordered partial metric space. Let T:X\to X be a nondecreasing (nonincreasing) mapping such that
for all comparable x,y\in X, where
A,B,C,D,E\ge 0 and A+B+C+D+E<1. Also suppose that there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0} (resp. {x}_{0}\u2ab0T{x}_{0}). We suppose the following:

(i)
T is continuous, or

(ii)
X is regular.
Then T has a fixed point z and p(Tz,Tz)=0=p(z,z). Moreover, the set F(T) of fixed points of T is well ordered if and only if it is a singleton.
Proof We will prove the theorem for the case of a nondecreasing mapping. Starting from the given element {x}_{0}\in X, form the sequence \{{x}_{n}\} as {x}_{n}=T{x}_{n1}, n\in \mathbb{N}. If p({x}_{n},{x}_{n+1})=0 for some n\ge 0, then T{x}_{n}={x}_{n+1}={x}_{n} and p({x}_{n},{x}_{n})=0 (by (p_{2})) and the proof is completed. Suppose further that p({x}_{n},{x}_{n+1})>0 for all n\ge 0. Note that, since T is nondecreasing, we have that
Applying condition (2.1) to comparable elements x={x}_{n} and y={x}_{n+1}, we get that
Similarly, applying (2.1) with x={x}_{n+1} and y={x}_{n}, we get
Adding up (2.3) and (2.4), we obtain
with
since A+B+C+D+E<1. It follows that p({x}_{n},{x}_{n+1})\le {\lambda}^{n}p({x}_{0},{x}_{1}) and {lim}_{n\to \mathrm{\infty}}p({x}_{n},{x}_{n+1})=0. Also, for n>m,
and hence {lim}_{m,n\to \mathrm{\infty}}p({x}_{n},{x}_{m})=0. Hence, \{{x}_{n}\} is a 0Cauchy sequence in (X,p). Since (X,p) is 0complete, it follows that there exists z\in X such that {x}_{n}\to z in (X,p) and p(z,z)=0. Moreover,
We will prove that Tz=z.

(i)
Suppose that T is continuous. Letting n\to \mathrm{\infty} in
p(z,Tz)\le p(z,{x}_{n+1})+p({x}_{n+1},Tz)p({x}_{n+1},{x}_{n+1})\le p(z,{x}_{n+1})+p({x}_{n+1},Tz)
and applying (2.5) and (1.2), we get
Thus, we have p(z,Tz)\le p(Tz,Tz). But from (p_{2}), we have p(Tz,Tz)\le p(z,Tz). Hence,
Suppose that p(z,Tz)>0. Now, since z\u2aafz, by inequality (2.1) we have
which is a contradiction. Thus, we get that p(Tz,Tz)=p(z,Tz)=p(z,z)=0. By (p_{1}), we conclude that z=Tz, that is, z is a fixed point of T.

(ii)
Suppose now that the space X is regular. Substituting x={x}_{n} and y=z (these elements are comparable) in the contractive condition (2.1), we get
\begin{array}{rl}p({x}_{n+1},Tz)& =p(T{x}_{n},Tz)\\ \le Ap({x}_{n},z)+Bp({x}_{n},{x}_{n+1})+Cp(z,Tz)+Dp(z,{x}_{n+1})+Ep({x}_{n},Tz).\end{array}
Passing to the limit as n\to \mathrm{\infty}, and using Lemma 1.6(c), we get that
which is (because of A+B+C+D+E<1) possible only if p(z,Tz)=0. We conclude that Tz=z.
Now suppose that the set of fixed points of T is well ordered. We claim that the fixed point of T is unique. Assume to the contrary that Tu=u and Tv=v, but u\ne v. By supposition, we can replace x by u and y by v in (2.1) to obtain
unless p(u,v)=0. Hence, u=v and the fixed point of T is unique. The converse is trivial. Thus, the proof is complete. □
Remark 2.2 Note that this theorem improves [[26], Theorem 2.6 and Corollary 2.9] since our assumptions are weaker than the assumptions from [26] in several places: 1^{∘} there is no need to use additional function ϕ, hence our contractive condition (2.1) is weaker than the one used in [26] (function ψ is redundant anyway, see e.g., [32]); 2^{∘} conditions on coefficients A, B, C, D, E are weaker (just the usual HardyRogerstype conditions); 3^{∘} 0completeness is used instead of completeness (see Lemma 1.6(e) and Example 1.7).
Similarly, putting d=p, we obtain an improvement of [[20], Theorem 5].
In a standard way, one gets the following corollary with integraltype condition (see [33]).
Corollary 2.3 Let all the conditions of Theorem 2.1 be fulfilled, except that the condition (2.1) is replaced by
for all comparable x,y\in X, where \mathbf{M}(x,y) is given by (2.2), and \phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+} is a nonnegative Lebesgue integrable function (with finite integral) on each compact subset of {\mathbb{R}}^{+}, satisfying {\int}_{0}^{\epsilon}\phi (t)\phantom{\rule{0.2em}{0ex}}dt>0 for each \epsilon >0.
We demonstrate the use of Theorem 2.1 with the help of the following example. It will also show that this theorem is more general than some other known fixed point results.
Example 2.4 Let X=[0,+\mathrm{\infty})\cap \mathbb{Q} be endowed with the usual partial metric p:X\times X\to [0,+\mathrm{\infty}) defined by p(x,y)=max\{x,y\}. The partial metric space (X,p) is 0complete (see Example 1.7). We endow X with the partial order
Define T:X\to X as
and take \mathbf{M}(x,y)=\frac{1}{2}p(x,y), i.e., A=\frac{1}{2}, B=C=D=E=0. Suppose that y\u2aafx. Then there are two possibilities. If x\in [0,1] (and so y\in [0,1]), then
Since \mathbf{M}(x,y)=\frac{1}{2}x, it easily follows that p(Tx,Ty)\le \mathbf{M}(x,y). If x>1 (and so y=x), then p(Tx,Ty)=\frac{x}{2}=\mathbf{M}(x,y). Hence, in all possible cases, condition (2.1) holds. Also, it is clear that both the condition of regularity of X and continuity of T are satisfied, and for {x}_{0}=0, we have {x}_{0}\u2aafT{x}_{0}. Therefore, all conditions of Theorem 2.1 are satisfied, and so T has a fixed point in X (which is z=0).
On the other hand, consider the same problem in the standard metric d(x,y) and take x=1 and y=\frac{1}{2}. Then d(Tx,Ty)=\frac{1}{2}\frac{1}{6}=\frac{1}{3} and \mathbf{M}(x,y)=\frac{1}{2}d(1,\frac{1}{2})=\frac{1}{4} and so
Hence, d(Tx,Ty)\le \mathbf{M}(x,y) does not hold and the existence of a fixed point of T cannot be obtained from the known results in standard metric spaces.
3 Common fixed point results for a pair of weakly isotone increasing mappings
In this section, we give a common fixed point theorem for a pair of maps satisfying Tweakly isotone increasing property. For this we need the following definitions.
Definition 3.1 Let (X,\u2aaf) be a partially ordered set, and let S,T:X\to X be two mappings.

(1)
[34] The pair (S,T) is said to be weakly increasing if Sx\u2aafTSx and Tx\u2aafSTx for all x\in X.

(2)
[21] The mapping S is said to be Tweakly isotone increasing if for all x\in X we have Sx\u2aafTSx\u2aafSTSx.
Remark 3.2 Note that two weakly increasing mappings need not be nondecreasing. There exist some examples to illustrate this fact in [19].
If S,T:X\to X are weakly increasing, then S is Tweakly isotone increasing.
Theorem 3.3 Let (X,p,\u2aaf) be a 0complete ordered partial metric space. Let S,T:X\to X be two mappings such that S is Tweakly isotone increasing and satisfying
for all comparable x,y\in X, where
A,B,C,D,E\ge 0, A+B+C+D+2E<1 and A+B+C+2D+E<1. We suppose the following:

(i)
S and T are continuous or

(ii)
X is regular.
Then S and T have a common fixed point z and
Moreover, the set of common fixed points of T and S is well ordered if and only if T and S have one and only one common fixed point.
Remark 3.4 Note that in this result continuity of both mappings is crucial (when the space is not regular); however, no compatibility of these mappings is needed.
Proof Let {x}_{0} be an arbitrary point in X. If p({x}_{0},S{x}_{0})=0 or p({x}_{0},T{x}_{0})=0, the proof is finished. Indeed, suppose e.g., that p({x}_{0},T{x}_{0})=0 (and hence T{x}_{0}={x}_{0}). Then (3.1) implies that
(since A+C+E\le A+B+C+D+2E<1) unless p({x}_{0},S{x}_{0})=0, implying that S{x}_{0}={x}_{0}. Thus, {x}_{0} is a common fixed point of T and S and (3.3) holds with z={x}_{0}.
Assume further that p({x}_{0},S{x}_{0})>0 and p({x}_{0},T{x}_{0})>0. We can define a sequence \{{x}_{n}\} in X as follows:
Without loss of generality, we can suppose that p({x}_{n},{x}_{n+1})>0 for each n\in \{0,1,2,\dots \}. Otherwise we have again finished.
Note that, since S is Tweakly isotone increasing, we have
and continuing this process, we get
Now since x={x}_{2n1} and y={x}_{2n} are comparable, we can use inequality (3.1) for these points, and we have
wherefrom
Similarly, using (3.1) with x={x}_{2n+1} and y={x}_{2n}, we get that
wherefrom
It follows from (3.4) and (3.5) that
where
since A+B+C+D+2E<1 and A+B+C+2D+E<1. It follows from (3.6) that p({x}_{n},{x}_{n+1})\le {\lambda}^{n}p({x}_{0},{x}_{1}) and {lim}_{n\to \mathrm{\infty}}p({x}_{n},{x}_{n+1})=0. Also, for n>m,
and hence {lim}_{m,n\to \mathrm{\infty}}p({x}_{n},{x}_{m})=0. Thus, \{{x}_{n}\} is a 0Cauchy sequence in (X,p). Since (X,p) is 0complete, it follows that there exists z\in X such that {x}_{n}\to z in (X,p) and p(z,z)=0. Moreover,
We will prove that Tz=Sz=z.
By (p_{4}), we have
and, similarly,

(i)
Suppose that T is continuous. Letting n\to \mathrm{\infty} in (3.8) and applying (3.7), we get
\begin{array}{rl}p(z,Tz)& \le \underset{n\to \mathrm{\infty}}{lim}p(z,{x}_{2n+2})+\underset{n\to \mathrm{\infty}}{lim}p(T{x}_{2n+1},Tz)\\ =p(Tz,Tz).\end{array}
Thus, we have p(z,Tz)\le p(Tz,Tz). But from (p_{2}), we have p(Tz,Tz)\le p(z,Tz). Hence,
Similarly, if S is continuous, we have
By (p_{4}) and using (3.11), we have
Similarly, by (p_{4}) and using (3.10), we can obtain
Suppose that p(Tz,Sz)>0. Then, since z\u2aafz, by inequality (3.1) and using (3.12), (3.13), we have
(since B+C+D+E\le A+B+C+D+2E<1) which is a contradiction. Thus, p(Sz,Tz)=0. By (p_{1}), we conclude that Sz=Tz, that is, z is a coincidence point of T and S. From (3.12), (3.13) and (p_{1}), we conclude that Tz=z and Sz=z, that is, z is a common fixed point of T and S.
Suppose that p(z,Tz)>0. Then
a contradiction. Hence, p(z,Tz)=0. This implies that
By (p_{1}), we conclude that Tz=z and Sz=z, that is, z is a common fixed point of T and S. Also by (p_{2}), we can obtain
Thus, we have proved (3.3).

(ii)
Since \{{x}_{n}\} is a nondecreasing sequence, if X is regular, it follows that {x}_{n}\u2aafz for all n. Therefore, for all n, we can use the inequality (3.1) for {x}_{n} and z. Since
and so {lim}_{n\to \mathrm{\infty}}\mathbf{M}(z,{x}_{2n})\le (B+D)p(z,Tz). Using (3.9) and (3.1), we have
Passing to the limit when n\to \mathrm{\infty}, we get that
which is a contradiction unless p(z,Tz)=0. It follows as above that Tz=z. Similarly, we can show p(Sz,z)=0, that is, Sz=z. Hence, z is a common fixed point of T and S. Also, by (p_{2}), we can obtain p(Tz,Tz)=0 and p(Sz,Sz)=0.
Now suppose that the set of common fixed points of T and S is well ordered. We claim that a common fixed point of T and S is unique. Assume to the contrary that Su=Tu=u and Tv=Sv=v, but u\ne v (and hence p(u,v)>0). By supposition, we can replace x by u and y by v in (3.1) to obtain
a contradiction. Hence, p(u,v)=0 and u=v. Conversely, if T and S have only one common fixed point then the set of common fixed points of T and S, being a singleton, is well ordered. Thus, the proof is complete. □
Standard corollaries of this HardyRogerstype result can be obtained by specifying conditions on the given constants (see [35]). We state just a few of them.
Corollary 3.5 Let (X,p,\u2aaf) be a 0complete ordered partial metric space. Let S,T:X\to X be two mappings such that S is Tweakly isotone increasing and satisfying either of the following conditions for all comparable x,y\in X:

(1)
p(Tx,Sy)\le Ap(x,y), 0\le A<1;

(2)
p(Tx,Sy)\le Bp(x,Tx)+Cp(y,Sy); B,C\ge 0, B+C<1;

(3)
p(Tx,Sy)\le Dp(y,Sx)+Ep(x,Ty); D,E\ge 0, D+E<1.
We suppose also the following:

(i)
S and T are continuous or

(ii)
X is regular.
Then we have conclusions of Theorem 3.3.
We demonstrate Theorem 3.3 with the help of the following example.
Example 3.6 Let X=[0,+\mathrm{\infty}) be endowed with the usual partial metric p:X\times X\to [0,+\mathrm{\infty}) defined by p(x,y)=max\{x,y\}. We give the partial order on X by
It is clear that (X,\u2aaf) is totally ordered. The partial metric space (X,p) is 0complete. Define mappings T,S:X\to X as
It is easy to see that S is Tweakly isotone increasing w.r.t. ⪯. Take A=\frac{1}{2}, B=C=0 and D=E=\frac{1}{8}, i.e.,
(the conditions on coefficients are fulfilled). Consider the following cases:
1^{∘} x\u2aafy, i.e., y\le x. Then
and
(cases y<\frac{x}{2} and \frac{x}{2}\le y\le x have to be considered). Hence, p(Tx,Sy)=\frac{1}{2}x\le \frac{11}{16}x=\mathbf{M}(x,y) is fulfilled.
2^{∘} \frac{2}{3}y\le x<y. Then again p(Tx,Sy)=\frac{x}{2} and
Hence, p(Tx,Sy)=\frac{x}{2}\le \frac{3}{4}x\le \frac{5}{8}y+\frac{1}{8}x=\mathbf{M}(x,y).
3^{∘} x<\frac{2}{3}y. Then
All the conditions of Theorem 3.3 are fulfilled. T and S have a unique common fixed point z=0.
Note that in Theorem 3.3 there is no condition of the type ‘there exists an {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0}’ of Theorem 2.1. And conditions on coefficients are not the same. Hence, Theorem 2.1 is not a direct consequence of Theorem 3.3. Instead, putting S=T in Theorem 3.3, we obtain the following
Corollary 3.7 Let (X,p,\u2aaf) be a 0complete ordered partial metric space. Let T:X\to X be a mapping satisfying Tx\le {T}^{2}x for each x\in X and
for all comparable x,y\in X, where A,B,C,D,E\ge 0, A+B+C+D+2E<1 and A+B+C+2D+E<1. We suppose the following:

(i)
T is continuous or

(ii)
X is regular.
Then T has a fixed point z and p(z,z)=p(Tz,Tz)=p(z,Tz)=0. Moreover, the set of fixed points of T is well ordered if and only if it is a singleton.
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Acknowledgements
This work was supported by the National Research Foundation of Korean Grant funded by the Korean Government (20110002581). The second and third authors are thankful to the Ministry of Science and Technological Development of Serbia.
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Nashine, H.K., Kadelburg, Z., Radenović, S. et al. Fixed point theorems under HardyRogers contractive conditions on 0complete ordered partial metric spaces. Fixed Point Theory Appl 2012, 180 (2012). https://doi.org/10.1186/168718122012180
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DOI: https://doi.org/10.1186/168718122012180