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Fixed point theorems for weakly contractive mappings in partially ordered metriclike spaces
Fixed Point Theory and Applications volume 2013, Article number: 51 (2013)
Abstract
In this article, we establish some fixed point theorems for weakly contractive mappings defined in ordered metriclike spaces. We provide an example and some applications in order to support the useability of our results. These results generalize some wellknown results in the literature. We also derive some new fixed point results in ordered partial metric spaces.
MSC:54H25, 47H10.
1 Introduction and preliminaries
During the last decades many authors have worked on domain theory in order to equip semantics domain with a notion of distance. In 1994, Matthews [1] introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks and showed that the Banach contraction principle can be generalized to the partial metric context for applications in program verification. Later on, many researchers studied fixed point theorems in partial metric spaces as well as ordered partial metric spaces.
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:

(p1)
x=y\iff p(x,x)=p(x,y)=p(y,y);

(p2)
p(x,x)\le p(x,y);

(p3)
p(x,y)=p(y,x);

(p4)
p(x,y)\le p(x,z)+p(z,y)p(z,z).
The pair (X,p) is then called a partial metric space. A sequence \{{x}_{n}\} in a partial metric space (X,p) converges to a point x\in X if {lim}_{n\to \mathrm{\infty}}p({x}_{n},x)=p(x,x). A sequence \{{x}_{n}\} of elements of X is called pCauchy if the limit {lim}_{m,n\to \mathrm{\infty}}p({x}_{m},{x}_{n}) exists and is finite. The partial metric space (X,p) is called complete if for each pCauchy sequence {\{{x}_{n}\}}_{n=0}^{\mathrm{\infty}}, there is some x\in X such that
A basic 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}}^{+}. For some other examples of partial metric spaces, see [2–16].
Another important development is reported in fixed point theory via ordered metric spaces. The existence of a fixed point in partially ordered sets has been considered recently in [17–32]. Tarski’s theorem is used in [25] to show the existence of solutions for fuzzy equations and in [27] to prove existence theorems for fuzzy differential equations. In [26, 29] some applications to ordinary differential equations and to matrix equations are presented, respectively. In [19–21, 30] some fixed point theorems were proved for a mixed monotone mapping in a metric space endowed with partial order and the authors applied their results to problems of existence and uniqueness of solutions for some boundary value problems.
Recently, AminiHarandi [33] introduced the notion of a metriclike space which is a new generalization of a partial metric space. The purpose of this paper is to present some fixed point theorems involving weakly contractive mappings in the context of ordered metriclike spaces. The presented theorems extend some recent results in the literature.
Weakly contractive mappings and mappings satisfying other weak contractive inequalities have been discussed in several works, some of which are noted in [34–40]. Alber and GuerreDelabriere in [34] suggested a generalization of the Banach contraction mapping principle by introducing the concept of a weak contraction in Hilbert spaces. Rhoades [35] showed that the result which Alber et al. had proved in Hilbert spaces [34] was also valid in complete metric spaces.
Definition 1 A mapping \sigma :X\times X\to {\mathbb{R}}^{+}, where X is a nonempty set, is said to be metriclike on X if for any x,y,z\in X, the following three conditions hold true:
(σ 1) \sigma (x,y)=0\Rightarrow x=y;
(σ 2) \sigma (x,y)=\sigma (y,x);
(σ 3) \sigma (x,z)\le \sigma (x,y)+\sigma (y,z).
The pair (X,\sigma ) is then called a metriclike space. Then a metriclike on X satisfies all of the conditions of a metric except that \sigma (x,x) may be positive for x\in X. Each metriclike σ on X generates a topology {\tau}_{\sigma} on X whose base is the family of open σballs
Then a sequence \{{x}_{n}\} in the metriclike space (X,\sigma ) converges to a point x\in X if and only if {lim}_{n\to \mathrm{\infty}}\sigma ({x}_{n},x)=\sigma (x,x).
Let (X,\sigma ) and (Y,\tau ) be metriclike spaces, and let F:X\to Y be a continuous mapping. Then
A sequence {\{{x}_{n}\}}_{n=0}^{\mathrm{\infty}} of elements of X is called σCauchy if the limit {lim}_{m,n\to \mathrm{\infty}}\sigma ({x}_{m},{x}_{n}) exists and is finite. The metriclike space (X,\sigma ) is called complete if for each σCauchy sequence {\{{x}_{n}\}}_{n=0}^{\mathrm{\infty}}, there is some x\in X such that
Every partial metric space is a metriclike space. Below we give another example of a metriclike space.
Example 1 Let X=\{0,1\} and
Then (X,\sigma ) is a metriclike space, but since \sigma (0,0)\nleqq \sigma (0,1), then (X,\sigma ) is not a partial metric space.
Remark 1 Let X=\{0,1\}, and \sigma (x,y)=1 for each x,y\in X, and {x}_{n}=1 for each n\in \mathbb{N}. Then it is easy to see that {x}_{n}\to 0 and {x}_{n}\to 1, and so in metriclike spaces the limit of a convergent sequence is not necessarily unique.
2 Main results
Throughout the rest of this paper, we denote by (X,\u2aaf,\sigma ) a complete partially ordered metriclike space, i.e., ⪯ is a partial order on the set X and σ is a complete metriclike on X.
A mapping F:X\to X is said to be nondecreasing if x,y\in X, x\u2aafy\Rightarrow Fx\u2aafFy.
Theorem 1 Let (X,\u2aaf,\sigma ) be a complete partially ordered metriclike space. Let F:X\to X be a continuous and nondecreasing mapping such that for all comparable x,y\in X
where M is given by
and

(a)
\psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is a continuous monotone nondecreasing function with \psi (t)=0 if and only if t=0;

(b)
\varphi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is a lower semicontinuous function with \varphi (t)=0 if and only if t=0.
If there exists {x}_{0}\in X with {x}_{0}\u2aafF{x}_{0}, then F has a fixed point.
Proof Since F is a nondecreasing function, we obtain by induction that
Put {x}_{n+1}=F{x}_{n}. Then, for each integer n=0,1,2,\dots , as the elements {x}_{n+1} and {x}_{n} are comparable, from (2.1) we get
which implies \psi (\sigma ({x}_{n+1},{x}_{n}))\le \psi (M({x}_{n},{x}_{n1})). Using the monotone property of the ψfunction, we get
Now, from the triangle inequality, for σ we have
If \sigma ({x}_{n+1},{x}_{n})>\sigma ({x}_{n},{x}_{n1}), then M({x}_{n},{x}_{n1})=\sigma ({x}_{n+1},{x}_{n})>0. By (2.2) it furthermore implies that
which is a contradiction. So, we have
Therefore, the sequence \{\sigma ({x}_{n+1},{x}_{n})\} is monotone nonincreasing and bounded. Thus, there exists r\ge 0 such that
We suppose that r>0. Then, letting n\to \mathrm{\infty} in the inequality (2.2), we get
which is a contradiction unless r=0. Hence,
Next we prove that \{{x}_{n}\} is a σCauchy sequence. Suppose that \{{x}_{n}\} is not a σCauchy sequence. Then, there exists \epsilon >0 for which we can find subsequences \{{x}_{{m}_{k}}\} and \{{x}_{{n}_{k}}\} of \{{x}_{n}\} with {n}_{k}>{m}_{k}>k such that
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} satisfying (2.7). Then
Using (2.7), (2.8) and the triangle inequality, we have
Letting k\to \mathrm{\infty} and using (2.6), we obtain
Again, the triangle inequality gives us
Then we have
Letting k\to \mathrm{\infty} in the above inequality and using (2.6) and (2.9), we get
Similarly, we can show that
As
using (2.6) and (2.9)(2.11), we have
As {n}_{k}>{m}_{k} and {x}_{{n}_{k}1} and {x}_{{m}_{k}1} are comparable, setting x={x}_{{n}_{k}1} and y={x}_{{m}_{k}1} in (2.1), we obtain
Letting k\to \mathrm{\infty} in the above inequality and using (2.9) and (2.12), we get
which is a contradiction as \epsilon >0. Hence \{{x}_{n}\} is a σCauchy sequence. By the completeness of X, there exists z\in X such that {lim}_{n\to \mathrm{\infty}}{x}_{n}=z, that is,
Moreover, the continuity of F implies that
and this proves that z is a fixed point. □
Notice that the continuity of F in Theorem 1 is not necessary and can be dropped.
Theorem 2 Under the same hypotheses of Theorem 1 and without assuming the continuity of F, assume that whenever \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X implies {x}_{n}\u2aafx for all n\in \mathbb{N}, then F has a fixed point in X.
Proof Following similar arguments to those given in Theorem 1, we construct a nondecreasing sequence \{{x}_{n}\} in X such that {x}_{n}\to z for some z\in X. Using the assumption of X, we have {x}_{n}\u2aafz for every n\in \mathbb{N}. Now, we show that Fz=z. By (2.1), we have
where
Taking limit as n\to \mathrm{\infty}, by (2.13), we obtain
Therefore, letting n\to \mathrm{\infty} in (2.14), we get
which is a contradiction unless \sigma (Fz,z)=0. Thus Fz=z. □
Next theorem gives a sufficient condition for the uniqueness of the fixed point.
Theorem 3 Let all the conditions of Theorem 1 (resp. Theorem 2) be fulfilled and let the following condition be satisfied: For arbitrary two points x,y\in X, there exists z\in X which is comparable with both x and y. Then the fixed point of F is unique.
Proof Suppose that there exist z,x\in X which are fixed points. We distinguish two cases.
Case 1. If x is comparable to z, then {F}^{n}x=x is comparable to {F}^{n}z=z for n=0,1,2,\dots and
where
Using (2.15) and (2.16), we have
which is a contradiction unless \sigma (z,x)=0. This implies that z=x.
Case 2. If x is not comparable to z, then there exists y\in X comparable to x and z. The monotonicity of F implies that {F}^{n}y is comparable to {F}^{n}x=x and {F}^{n}z=z, for n=0,1,2,\dots . Moreover,
where
for n sufficiently large, because \sigma ({F}^{n1}y,{F}^{n1}y)\to 0 and \sigma ({F}^{n1}y,{F}^{n}y)\to 0 when n\to \mathrm{\infty}. Similarly as in the proof of Theorem 1, it can be shown that \sigma (z,{F}^{n}y)\le M(z,{F}^{n1}y)\le \sigma (z,{F}^{n1}y). It follows that the sequence \{\sigma (z,{F}^{n}y)\} is nonnegative decreasing and, consequently, there exists \alpha \ge 0 such that
We suppose that \alpha >0. Then letting n\to \mathrm{\infty} in (2.17), we have
which is a contradiction. Hence \alpha =0. Similarly, it can be proved that
Now, passing to the limit in \sigma (x,z)\le \sigma (x,{F}^{n}y)+\sigma ({F}^{n}y,z), it follows that \sigma (x,z)=0, so x=z, and the uniqueness of the fixed point is proved. □
Now, we present an example to support the useability of our results.
Example 2 Let X=\{0,1,2\} and a partial order be defined as x\u2aafy whenever y\le x, and define \sigma :X\times X\to {\mathbb{R}}^{+} as follows:
Then (X,\u2aaf,\sigma ) is a complete partially ordered metriclike space.
Let F:X\to X be defined by
Define \psi ,\varphi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) by \psi (t)=t and \varphi (t)=\frac{t}{2}. We next verify that the function F satisfies the inequality (2.1). For that, given x,y\in X with x\u2aafy, so y\le x. Then we have the following cases.
Case 1. If x=1, y=0, then
and
As \psi (\sigma (F1,F0))=5<9\frac{9}{2}=\psi (M(1,0))\varphi (M(1,0)), the inequality (2.1) is satisfied in this case.
Case 2. If x=2, y=0, then
and
As \psi (\sigma (F2,F0))=5<9\frac{9}{2}=\psi (M(2,0))\varphi (M(2,0)), the inequality (2.1) is satisfied in this case.
Case 3. If x=2, y=1, then as \sigma (F2,F1)=0 and M(2,1)=5, the inequality (2.1) is satisfied in this case.
Case 4. If x=0, y=0, then as \sigma (F0,F0)=3 and M(0,0)=9, the inequality (2.1) is satisfied in this case.
Case 5. If x=1, y=1, then as \sigma (F1,F1)=0 and M(1,1)=5, the inequality (2.1) is satisfied in this case.
Case 6. If x=2, y=2, then as \sigma (F2,F2)=0 and M(2,2)=0, the inequality (2.1) is satisfied in this case.
So, F, ψ and ϕ satisfy all the hypotheses of Theorem 1. Therefore F has a unique fixed point. Here 2 is the unique fixed point of F.
If we take \psi (t)=t in Theorem 1, we have the following corollary.
Corollary 1 Let (X,\u2aaf,\sigma ) be a complete partially ordered metriclike space. Let F:X\to X be a nondecreasing mapping such that for all comparable x,y\in X with
where M is given by
\varphi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is lower semicontinuous, and \varphi (t)=0 if and only if t=0. If there exists {x}_{0}\in X with {x}_{0}\u2aafF{x}_{0} and one of the following two conditions is satisfied:

(a)
F is continuous in (X,\sigma );

(b)
\{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X implies {x}_{n}\u2aafx for all n\in \mathbb{N}.
Then F has a fixed point. Moreover, if the following condition is satisfied: For arbitrary two points x,y\in X, there exists z\in X which is comparable with both x and y, then the fixed point of F is unique.
If we take \varphi (t)=(1k)t for k\in [0,1) in Corollary 1, we have the following corollary.
Corollary 2 Let (X,\u2aaf,\sigma ) be a complete partially ordered metriclike space. Let F:X\to X be a nondecreasing mapping such that for all comparable x,y\in X
where M is given by
and k\in [0,1). If there exists {x}_{0}\in X with {x}_{0}\u2aafF{x}_{0} and one of the following two conditions is satisfied:

(a)
F is continuous in (X,\sigma );

(b)
\{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X implies {x}_{n}\u2aafx for all n\in \mathbb{N}.
Then F has a fixed point. Moreover, if the following condition is satisfied: For arbitrary two points x,y\in X, there exists z\in X which is comparable with both x and y, then the fixed point of F is unique.
The following corollary improves Theorem 2.7 in [33].
Corollary 3 Let (X,\u2aaf,\sigma ) be a complete partially ordered metriclike space. Let F:X\to X be a nondecreasing mapping such that for all comparable x,y\in X
where \varphi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is a lower semicontinuous, and \varphi (t)=0 if and only if t=0. If there exists {x}_{0}\in X with {x}_{0}\u2aafF{x}_{0} and one of the following two conditions is satisfied:

(a)
F is continuous in (X,\sigma );

(b)
\{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X implies {x}_{n}\u2aafx for all n\in \mathbb{N}.
Then F has a fixed point. Moreover, if the following condition is satisfied: For arbitrary two points x,y\in X, there exists z\in X which is comparable with both x and y, then the fixed point of F is unique.
The following corollary improves Theorem 2.1 in [10].
Corollary 4 Let (X,\u2aaf,p) be a complete partially ordered partial metric space. Let F:X\to X be a nondecreasing mapping such that for all comparable x,y\in X
where M is given by
\psi ,\varphi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}), ψ is continuous monotone nondecreasing, ϕ is lower semicontinuous, and \psi (t)=\varphi (t)=0 if and only if t=0. If there exists {x}_{0}\in X with {x}_{0}\u2aafF{x}_{0} and one of the following two conditions is satisfied:

(a)
F is continuous in (X,p);

(b)
\{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X implies {x}_{n}\u2aafx for all n\in \mathbb{N}.
Then F has a fixed point. Moreover, the set of fixed points of F is well ordered if and only if F has one and only one fixed point.
3 Applications
Denote by Λ the set of functions \alpha :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) satisfying the following hypotheses:

(h1)
α is a Lebesgueintegrable mapping on each compact subset of [0,+\mathrm{\infty});

(h2)
For every \epsilon >0, we have
{\int}_{0}^{\epsilon}\alpha (s)\phantom{\rule{0.2em}{0ex}}ds>0.
We have the following results.
Corollary 5 Let (X,\u2aaf,\sigma ) be a complete partially ordered metriclike space. Let F:X\to X be a continuous and nondecreasing mapping such that for all comparable x,y\in X
where {\alpha}_{1},{\alpha}_{2}\in \mathrm{\Lambda}. If there exists {x}_{0}\in X with {x}_{0}\u2aafF{x}_{0}, then F has a fixed point.
Proof Follows from Theorem 1 by taking \psi (t)={\int}_{0}^{t}{\alpha}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds and \varphi (t)={\int}_{0}^{t}{\alpha}_{2}(s)\phantom{\rule{0.2em}{0ex}}ds. □
Corollary 6 Under the same hypotheses of Corollary 5 and without assuming the continuity of F, assume that whenever \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X implies {x}_{n}\u2aafx for all n\in \mathbb{N}, then F has a fixed point in X.
Proof Follows from Theorem 2 by taking \psi (t)={\int}_{0}^{t}{\alpha}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds and \varphi (t)={\int}_{0}^{t}{\alpha}_{2}(s)\phantom{\rule{0.2em}{0ex}}ds. □
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Dedicated to Professor Hari M Srivastava.
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Işık, H., Türkoğlu, D. Fixed point theorems for weakly contractive mappings in partially ordered metriclike spaces. Fixed Point Theory Appl 2013, 51 (2013). https://doi.org/10.1186/16871812201351
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DOI: https://doi.org/10.1186/16871812201351
Keywords
 fixed point
 weak contraction
 partially ordered set
 partial metric space
 metriclike space