Abstract
In this paper, we introduce a modified version of ordered partial bmetric spaces. We demonstrate a fundamental lemma for the convergence of sequences in such spaces. Using this lemma, we prove some fixed point and common fixed point results for (\psi ,\phi )weakly contractive mappings in the setup of ordered partial bmetric spaces. Finally, examples are presented to verify the effectiveness and applicability of our main results.
MSC: 47H10, 54H25.
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1 Introduction
Fixed points theorems in partially ordered metric spaces were firstly obtained in 2004 by Ran and Reurings [1], and then by Nieto and Lopez [2]. In this direction several authors obtained further results under weak contractive conditions (see, e.g., [3–8]).
The concept of bmetric space was introduced by Bakhtin [9] and extensively used by Czerwik in [10, 11]. After that, several interesting results about the existence of a fixed point for singlevalued and multivalued operators in (ordered) bmetric spaces have been obtained (see, e.g., [12–26]).
Definition 1 [10]
Let X be a (nonempty) set and s\ge 1 be a given real number. A function d:X\times X\to {\mathbb{R}}^{+} is a bmetric on X if, for all x,y,z\in X, the following conditions hold:

(b_{1}) d(x,y)=0 if and only if x=y,

(b_{2}) d(x,y)=d(y,x),

(b_{3}) d(x,z)\le s[d(x,y)+d(y,z)].
In this case, the pair (X,d) is called a bmetric space.
On the other hand, Matthews [27] introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks. In partial metric spaces, selfdistance of an arbitrary point need not be equal to zero. Several authors obtained many useful fixed point results in these spaces  we mention just [28–33].
Definition 2 [27]
A partial metric on a nonempty set X is a mapping p:X\times X\to {\mathbb{R}}^{+} such that for all x,y,z\in X:

(p_{1}) x=y if and only if p(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).
In this case, (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. 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}}^{+}.
Each partial metric p on a set 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.
Definition 3 [27]
Let (X,p) be a partial metric space, and let \{{x}_{n}\} be a sequence in X and x\in X. Then:

(i)
The sequence \{{x}_{n}\} is said to converge to x with respect to {\tau}_{p} if {lim}_{n\to \mathrm{\infty}}p({x}_{n},x)=p(x,x).

(ii)
The sequence \{{x}_{n}\} is said to be Cauchy in (X,p) if {lim}_{n,m\to \mathrm{\infty}}p({x}_{n},{x}_{m}) exists and is finite.

(iii)
(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 {lim}_{n,m\to \mathrm{\infty}}p({x}_{n},{x}_{m})={lim}_{n\to \mathrm{\infty}}p({x}_{n},x)=p(x,x).
The following example shows that a convergent sequence \{{x}_{n}\} in a partial metric space (X,p) may not be Cauchy. In particular, it shows that the limit may not be unique.
Example 1 [32]
Let X=[0,\mathrm{\infty}) and p(x,y)=max\{x,y\}. Let
Then, clearly, \{{x}_{n}\} is a convergent sequence and for every x\ge 1, we have {lim}_{n\to \mathrm{\infty}}p({x}_{n},x)=p(x,x). But {lim}_{n,m\to \mathrm{\infty}}p({x}_{n},{x}_{m}) does not exist, that is, \{{x}_{n}\} is not a Cauchy sequence.
As a generalization and unification of partial metric and bmetric spaces, Shukla [34] introduced the concept of partial bmetric space as follows.
Definition 4 [34]
A partial bmetric on a nonempty set X is a mapping {p}_{b}:X\times X\to {\mathbb{R}}^{+} such that for all x,y,z\in X:

({\text{p}}_{b1}) x=y if and only if {p}_{b}(x,x)={p}_{b}(x,y)={p}_{b}(y,y),

({\text{p}}_{b2}) {p}_{b}(x,x)\le {p}_{b}(x,y),

({\text{p}}_{b3}) {p}_{b}(x,y)={p}_{b}(y,x),

({\text{p}}_{b4}) {p}_{b}(x,y)\le s[{p}_{b}(x,z)+{p}_{b}(z,y)]{p}_{b}(z,z).
A partial bmetric space is a pair (X,{p}_{b}) such that X is a nonempty set and {p}_{b} is a partial bmetric on X. The number s\ge 1 is called the coefficient of (X,{p}_{b}).
In a partial bmetric space (X,{p}_{b}), if x,y\in X and {p}_{b}(x,y)=0, then x=y, but the converse may not be true. It is clear that every partial metric space is a partial bmetric space with the coefficient s=1 and every bmetric space is a partial bmetric space with the same coefficient and zero selfdistance. However, the converse of these facts need not hold.
Example 2 [34]
Let X={\mathbb{R}}^{+}, q>1 be a constant and {p}_{b}:X\times X\to {\mathbb{R}}^{+} be defined by
Then (X,{p}_{b}) is a partial bmetric space with the coefficient s={2}^{q1}>1, but it is neither a bmetric nor a partial metric space.
Note that in a partial bmetric space the limit of a convergent sequence may not be unique (see [[34], Example 2]).
Some more examples of partial bmetrics can be constructed with the help of the following propositions.
Proposition 1 [34]
Let X be a nonempty set, and let p be a partial metric and d be a bmetric with the coefficient s\ge 1 on X. Then the function {p}_{b}:X\times X\to {\mathbb{R}}^{+}, defined by {p}_{b}(x,y)=p(x,y)+d(x,y) for all x,y\in X, is a partial bmetric on X with the coefficient s.
Proposition 2 [34]
Let (X,p) be a partial metric space and q\ge 1. Then (X,{p}_{b}) is a partial bmetric space with the coefficient s={2}^{q1}, where {p}_{b} is defined by {p}_{b}(x,y)={[p(x,y)]}^{q}.
Altering distance functions were introduced by Khan et al. in [35].
Definition 5 [35]
A function \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is called an altering distance function if the following properties are satisfied:

1.
ψ is continuous and nondecreasing;

2.
\psi (t)=0 if and only if t=0.
So far, many authors have studied fixed point theorems which are based on altering distance functions (see, e.g., [12, 28, 36–41]).
In this paper, we introduce a modified version of ordered partial bmetric spaces. We demonstrate a fundamental lemma for the convergence of sequences in such spaces. Using this lemma, we prove some fixed point and common fixed point results for (\psi ,\phi )weakly contractive mappings in the setup of ordered partial bmetric spaces. Finally, examples are presented to verify the effectiveness and applicability of our main results.
2 Definition and basic properties of partial bmetric spaces
In the following definition, we modify Definition 4 in order to obtain that each partial bmetric {p}_{b} generates a bmetric {d}_{{p}_{b}}.
Definition 6 Let X be a (nonempty) set and s\ge 1 be a given real number. A function {p}_{b}:X\times X\to {\mathbb{R}}^{+} is a partial bmetric if, for all x,y,z\in X, the following conditions are satisfied:

({\text{p}}_{b1}) x=y\u27fa{p}_{b}(x,x)={p}_{b}(x,y)={p}_{b}(y,y),

({\text{p}}_{b2}) {p}_{b}(x,x)\le {p}_{b}(x,y),

({\text{p}}_{b3}) {p}_{b}(x,y)={p}_{b}(y,x),

({\text{p}}_{b{4}^{\prime}}) {p}_{b}(x,y)\le s({p}_{b}(x,z)+{p}_{b}(z,y){p}_{b}(z,z))+(\frac{1s}{2})({p}_{b}(x,x)+{p}_{b}(y,y)).
The pair (X,{p}_{b}) is called a partial bmetric space.
Since s\ge 1, from ({\text{p}}_{b{4}^{\prime}}) we have
Hence, a partial bmetric in the sense of Definition 6 is also a partial bmetric in the sense of Definition 4.
It should be noted that the class of partial bmetric spaces is larger than the class of partial metric spaces, since a partial bmetric is a partial metric when s=1. We present an example which shows that a partial bmetric on X (in the sense of Definition 6) might be neither a partial metric, nor a bmetric on X.
Example 3 Let (X,d) be a metric space and {p}_{b}(x,y)=d{(x,y)}^{q}+a, where q>1 and a\ge 0 are real numbers. We will show that {p}_{b} is a partial bmetric with s={2}^{q1}.
Obviously, conditions ({\text{p}}_{b1})({\text{p}}_{b3}) of Definition 6 are satisfied.
Since q>1, the convexity of the function f(x)={x}^{q} (x>0) implies that {(a+b)}^{q}\le {2}^{q1}({a}^{q}+{b}^{q}) holds for a,b\ge 0. Thus, for each x,y,z\in X, we obtain
Hence, condition ({\text{p}}_{b{4}^{\prime}}) of Definition 6 is fulfilled and {p}_{b} is a partial bmetric on X.
Note that (X,{p}_{b}) is not necessarily a partial metric space. For example, if X=\mathbb{R} is the set of real numbers, d(x,y)=xy, q=2 and a=3, then {p}_{b}(x,y)={(xy)}^{2}+3 is a partial bmetric on X with s={2}^{21}=2, but it is not a partial metric on X. Indeed, the ordinary (partial) triangle inequality does not hold. To see this, let x=2, y=5 and z=\frac{5}{2}. Then {p}_{b}(2,5)=12, {p}_{b}(2,\frac{5}{2})=\frac{13}{4} and {p}_{b}(\frac{5}{2},5)=\frac{37}{4}, hence {p}_{b}(2,5)=12\nleqq \frac{38}{4}={p}_{b}(2,\frac{5}{2})+{p}_{b}(\frac{5}{2},5){p}_{b}(\frac{5}{2},\frac{5}{2}).
Also, {p}_{b} is not a bmetric since {p}_{b}(x,x)\ne 0 for x\in X.
Proposition 3 Every partial bmetric {p}_{b} defines a bmetric {d}_{{p}_{b}}, where
for all x,y\in X.
Proof Let x,y,z\in X. Then we have
□
Hence, the advantage of our definition of partial bmetric is that by using it we can define a dependent bmetric which we call the bmetric associated with {p}_{b}. This allows us to readily transport many concepts and results from bmetric spaces into a partial bmetric space.
Now, we present some definitions and propositions in a partial bmetric space.
Definition 7 Let (X,{p}_{b}) be a partial bmetric space. Then, for x\in X and \u03f5>0, the {p}_{b}ball with center x and radius ϵ is
For example, let (X,{p}_{b}) be the partial bmetric space from Example 3 (with X=\mathbb{R}, q=2 and a=3). Then
Proposition 4 Let (X,{p}_{b}) be a partial bmetric space, x\in X and r>0. If y\in {B}_{{p}_{b}}(x,r), then there exists \delta >0 such that {B}_{{p}_{b}}(y,\delta )\subseteq {B}_{{p}_{b}}(x,r).
Proof Let y\in {B}_{{p}_{b}}(x,r). If y=x, then we choose \delta =r. Suppose that y\ne x. Then we have {p}_{b}(x,y)\ne 0. Now, we consider two cases.
Case 1. If {p}_{b}(x,y)={p}_{b}(x,x), then for s=1 we choose \delta =r. If s>1, then we consider the set
By the Archimedean property, A is a nonempty set; then by the well ordering principle, A has the least element m. Since m1\notin A, we have {p}_{b}(x,x)\le r/(2{s}^{m}(s1)) and we choose \delta =r/(2{s}^{m+1}). Let z\in {B}_{{p}_{b}}(y,\delta ); by the property ({\text{p}}_{b4}), we have
Hence, {B}_{{p}_{b}}(y,\delta )\subseteq {B}_{{p}_{b}}(x,r).
Case 2. If {p}_{b}(x,y)\ne {p}_{b}(x,x), then from the property ({\text{p}}_{b2}) we have {p}_{b}(x,x)<{p}_{b}(x,y) and for s=1 we consider the set
Similarly, by the well ordering principle, there exists an element m such that {p}_{b}(x,y){p}_{b}(x,x)\le r/({2}^{m+2}), and we choose \delta =r/({2}^{m+2}). One can easily obtain that {B}_{{p}_{b}}(y,\delta )\subseteq {B}_{{p}_{b}}(x,r).
For s>1, we consider the set
and by the well ordering principle, there exists an element m such that {p}_{b}(x,y)\frac{1}{s}{p}_{b}(x,x)\le \frac{r}{2{s}^{m+1}} and we choose \delta =\frac{r}{2{s}^{m+1}}. Let z\in {B}_{{p}_{b}}(y,\delta ). By the property ({\text{p}}_{b4}), we have
Hence, {B}_{{p}_{b}}(y,\delta )\subseteq {B}_{{p}_{b}}(x,r). □
Thus, from the above proposition the family of all {p}_{b}balls
is a base of a {T}_{0} topology {\tau}_{{p}_{b}} on X which we call the {p}_{b}metric topology.
The topological space (X,{p}_{b}) is {T}_{0}, but need not be {T}_{1}.
Definition 8 A sequence \{{x}_{n}\} in a partial bmetric space (X,{p}_{b}) is said to be:

(i)
{p}_{b}convergent to a point x\in X if {lim}_{n\to \mathrm{\infty}}{p}_{b}(x,{x}_{n})={p}_{b}(x,x);

(ii)
a {p}_{b}Cauchy sequence if {lim}_{n,m\to \mathrm{\infty}}{p}_{b}({x}_{n},{x}_{m}) exists (and is finite).

(iii)
A partial bmetric space (X,{p}_{b}) is said to be {p}_{b}complete if every {p}_{b}Cauchy sequence \{{x}_{n}\} in X {p}_{b}converges to a point x\in X such that {lim}_{n,m\to \mathrm{\infty}}{p}_{b}({x}_{n},{x}_{m})={lim}_{n,m\to \mathrm{\infty}}{p}_{b}({x}_{n},x)={p}_{b}(x,x).
The following lemma shows the relationship between the concepts of {p}_{b}convergence, {p}_{b}Cauchyness and {p}_{b}completeness in two spaces (X,{p}_{b}) and (X,{d}_{{p}_{b}}) which we state and prove according to Lemma 2.2 of [31].
Lemma 1

(1)
A sequence \{{x}_{n}\} is a {p}_{b}Cauchy sequence in a partial bmetric space (X,{p}_{b}) if and only if it is a bCauchy sequence in the bmetric space (X,{d}_{{p}_{b}}).

(2)
A partial bmetric space (X,{p}_{b}) is {p}_{b}complete if and only if the bmetric space (X,{d}_{{p}_{b}}) is bcomplete. Moreover, {lim}_{n\to \mathrm{\infty}}{d}_{{p}_{b}}(x,{x}_{n})=0 if and only if
\underset{n\to \mathrm{\infty}}{lim}{p}_{b}(x,{x}_{n})=\underset{n,m\to \mathrm{\infty}}{lim}{p}_{b}({x}_{n},{x}_{m})={p}_{b}(x,x).
Proof First, we show that every {p}_{b}Cauchy sequence in (X,{p}_{b}) is a bCauchy sequence in (X,{d}_{{p}_{b}}). Let \{{x}_{n}\} be a {p}_{b}Cauchy sequence in (X,{p}_{b}). Then, there exists \alpha \in \mathbb{R} such that, for arbitrary \epsilon >0, there is {n}_{\epsilon}\in \mathbb{N} with
for all n,m\ge {n}_{\epsilon}. Hence,
for all n,m\ge {n}_{\epsilon}. Hence, we conclude that \{{x}_{n}\} is a bCauchy sequence in (X,{d}_{{p}_{b}}).
Next, we prove that bcompleteness of (X,{d}_{{p}_{b}}) implies {p}_{b}completeness of (X,{p}_{b}). Indeed, if \{{x}_{n}\} is a {p}_{b}Cauchy sequence in (X,{p}_{b}), then according to the above discussion, it is also a bCauchy sequence in (X,{d}_{{p}_{b}}). Since the bmetric space (X,{d}_{{p}_{b}}) is bcomplete, we deduce that there exists y\in X such that {lim}_{n\to \mathrm{\infty}}{d}_{{p}_{b}}(y,{x}_{n})=0. Hence,
therefore, {lim}_{n\to \mathrm{\infty}}[{p}_{b}({x}_{n},y){p}_{b}(y,y)]=0. Further, we have
Consequently,
On the other hand,
Also, from ({\text{p}}_{b2}),
Hence, we obtain that \{{x}_{n}\} is a {p}_{b}convergent sequence in (X,{p}_{b}).
Now, we prove that every bCauchy sequence \{{x}_{n}\} in (X,{d}_{{p}_{b}}) is a {p}_{b}Cauchy sequence in (X,{p}_{b}). Let \epsilon =\frac{1}{2}. Then there exists {n}_{0}\in \mathbb{N} such that {d}_{{p}_{b}}({x}_{n},{x}_{m})<\frac{1}{2} for all n,m\ge {n}_{0}. Since
hence
Consequently, the sequence \{{p}_{b}({x}_{n},{x}_{n})\} is bounded in ℝ, and so there exists a\in \mathbb{R} such that a subsequence \{{p}_{b}({x}_{{n}_{k}},{x}_{{n}_{k}})\} of \{{p}_{b}({x}_{n},{x}_{n})\} is convergent to a, i.e.,
Now, we prove that \{{p}_{b}({x}_{n},{x}_{n})\} is a Cauchy sequence in ℝ. Since \{{x}_{n}\} is a bCauchy sequence in (X,{d}_{{p}_{b}}) for given \epsilon >0, there exists {n}_{\epsilon}\in \mathbb{N} such that {d}_{{p}_{b}}({x}_{n},{x}_{m})<\epsilon for all n,m\ge {n}_{\epsilon}. Thus, for all n,m\ge {n}_{\epsilon},
Therefore, {lim}_{n\to \mathrm{\infty}}{p}_{b}({x}_{n},{x}_{n})=a.
On the other hand,
for all n,m\ge {n}_{\epsilon}. Hence, {lim}_{n,m\to \mathrm{\infty}}{p}_{b}({x}_{n},{x}_{m})=a, and consequently, \{{x}_{n}\} is a {p}_{b}Cauchy sequence in (X,{p}_{b}).
Conversely, let \{{x}_{n}\} be a bCauchy sequence in (X,{d}_{{p}_{b}}). Then \{{x}_{n}\} is a {p}_{b}Cauchy sequence in (X,{p}_{b}), and so it is convergent to a point x\in X with
Then, for given \epsilon >0, there exists {n}_{\epsilon}\in \mathbb{N} such that
and
Therefore,
whenever n\ge {n}_{\epsilon}. Therefore, (X,{d}_{{p}_{b}}) is complete.
Finally, let {lim}_{n\to \mathrm{\infty}}{d}_{{p}_{b}}({x}_{n},x)=0. So,
On the other hand,
□
Definition 9 Let (X,{p}_{b}) and ({X}^{\prime},{p}_{b}^{\prime}) be two partial bmetric spaces, and let f:(X,{p}_{b})\to ({X}^{\prime},{p}_{b}^{\prime}) be a mapping. Then f is said to be {p}_{b}continuous at a point a\in X if for a given \epsilon >0, there exists \delta >0 such that x\in X and {p}_{b}(a,x)<\delta +{p}_{b}(a,a) imply that {p}_{b}^{\prime}(f(a),f(x))<\epsilon +{p}_{b}^{\prime}(f(a),f(a)). The mapping f is {p}_{b}continuous on X if it is {p}_{b}continuous at all a\in X.
Proposition 5 Let (X,{p}_{b}) and ({X}^{\prime},{p}_{b}^{\prime}) be two partial bmetric spaces. Then a mapping f:X\to {X}^{\prime} is {p}_{b}continuous at a point x\in X if and only if it is {p}_{b}sequentially continuous at x; that is, whenever \{{x}_{n}\} is {p}_{b}convergent to x, \{f({x}_{n})\} is {p}_{b}^{\prime}convergent to f(x).
Definition 10 A triple (X,\u2aaf,{p}_{b}) is called an ordered partial bmetric space if (X,\u2aaf) is a partially ordered set and {p}_{b} is a partial bmetric on X.
3 Fixed point results in partial bmetric spaces
The following crucial lemma is useful in proving our main results.
Lemma 2 Let (X,{p}_{b}) be a partial bmetric space with the coefficient s>1 and suppose that \{{x}_{n}\} and \{{y}_{n}\} are convergent to x and y, respectively. Then we have
In particular, if {p}_{b}(x,y)=0, then we have {lim}_{n\to \mathrm{\infty}}{p}_{b}({x}_{n},{y}_{n})=0.
Moreover, for each z\in X, we have
In particular, if {p}_{b}(x,x)=0, then we have
Proof Using the triangle inequality in a partial bmetric space, it is easy to see that
and
Taking the lower limit as n\to \mathrm{\infty} in the first inequality and the upper limit as n\to \mathrm{\infty} in the second inequality, we obtain the first desired result. If {p}_{b}(x,y)=0, then by the triangle inequality we get {p}_{b}(x,x)=0 and {p}_{b}(y,y)=0. Therefore, we have {lim}_{n\to \mathrm{\infty}}{p}_{b}({x}_{n},{y}_{n})=0. Similarly, using again the triangle inequality, the other assertions follow. □
Let (X,\u2aaf,{p}_{b}) be an ordered partial bmetric space, and let f:X\to X be a mapping. Set
Definition 11 Let (X,{p}_{b}) be an ordered partial bmetric space. We say that a mapping f:X\to X is a generalized {(\psi ,\phi )}_{s}weakly contractive mapping if there exist two altering distance functions ψ and φ such that
for all comparable x,y\in X.
First, we prove the following result.
Theorem 1 Let (X,\u2aaf,{p}_{b}) be a {p}_{b}complete ordered partial bmetric space. Let f:X\to X be a nondecreasing, with respect to ⪯, continuous mapping. Suppose that f is a generalized {(\psi ,\phi )}_{s}weakly contractive mapping. If there exists {x}_{0}\in X such that {x}_{0}\u2aaff{x}_{0}, then f has a fixed point.
Proof Let {x}_{0}\in X be such that {x}_{0}\u2aaff{x}_{0}. Then we define a sequence ({x}_{n}) in X such that {x}_{n+1}=f{x}_{n} for all n\ge 0. Since {x}_{0}\u2aaff{x}_{0}={x}_{1} and f is nondecreasing, we have {x}_{1}=f{x}_{0}\u2aaf{x}_{2}=f{x}_{1}. Again, as {x}_{1}\u2aaf{x}_{2} and f is nondecreasing, we have {x}_{2}=f{x}_{1}\u2aaf{x}_{3}=f{x}_{2}. By induction, we have
If {x}_{n}={x}_{n+1} for some n\in \mathbb{N}, then {x}_{n}=f{x}_{n} and hence {x}_{n} is a fixed point of f. So, we may assume that {x}_{n}\ne {x}_{n+1} for all n\in \mathbb{N}. By (3.1), we have
where
So, we have
From (3.2), (3.3) we get
If
then by (3.4) and properties of φ, we have
which gives a contradiction. Thus,
Therefore, \{{p}_{b}({x}_{n},{x}_{n+1}):n\in \mathbb{N}\cup \{0\}\} is a nonincreasing sequence of positive numbers. So, there exists r\ge 0 such that
Letting n\to \mathrm{\infty} in (3.5), we get
Therefore, \phi (r)=0, and hence r=0. Thus, we have
Next, we show that \{{x}_{n}\} is a {p}_{b}Cauchy sequence in X. For this, we have to show that \{{x}_{n}\} is a bCauchy sequence in (X,{d}_{{p}_{b}}) (see Lemma 1). Suppose the contrary; that is, \{{x}_{n}\} is not a bCauchy sequence. Then there exists \epsilon >0 for which we can find two subsequences \{{x}_{{m}_{i}}\} and \{{x}_{{n}_{i}}\} of \{{x}_{n}\} such that {n}_{i} is the smallest index for which
This means that
From (3.7) and using the triangular inequality, we get
Taking the upper limit as i\to \mathrm{\infty} and using (3.8), we get
Also, from (3.9) and (3.10),
Further,
and hence
Finally,
and hence
On the other hand, by the definition of {d}_{{p}_{b}} and (3.6),
Hence, by (3.10),
Similarly,
From (3.1), we have
where
Taking the upper limit as i\to \mathrm{\infty} in (3.16) and using (3.6), (3.11), (3.12) and (3.14), we get
Now, taking the upper limit as i\to \mathrm{\infty} in (3.15) and using (3.13) and (3.17), we have
which further implies that
so {lim\hspace{0.17em}inf}_{i\to \mathrm{\infty}}{M}_{s}^{f}({x}_{{m}_{i}},{x}_{{n}_{i}1})=0, and by (3.16) we get {lim\hspace{0.17em}inf}_{i\to \mathrm{\infty}}{d}_{{p}_{b}}({x}_{{m}_{i}},{x}_{{n}_{i}1})=0, a contradiction with (3.11).
Thus, we have proved that \{{x}_{n}\} is a bCauchy sequence in the bmetric space (X,{d}_{{p}_{b}}). Since (X,{p}_{b}) is {p}_{b}complete, then from Lemma 1, (X,{d}_{{p}_{b}}) is a bcomplete bmetric space. Therefore, the sequence \{{x}_{n}\} converges to some z\in X, that is, {lim}_{n\to \mathrm{\infty}}{d}_{{p}_{b}}({x}_{n},z)=0. Again, from Lemma 1,
On the other hand, thanks to (3.6) and condition ({\text{p}}_{b2}), {lim}_{n\to \mathrm{\infty}}{p}_{b}({x}_{n},{x}_{n})=0, which yields that
Using the triangular inequality, we get
Letting n\to \mathrm{\infty} and using the continuity of f, we get
Note that from (3.1), we have
where
Hence, as ψ is nondecreasing, we have s{p}_{b}(fz,fz)\le {p}_{b}(fz,z). Thus, by (3.18) we obtain that s{p}_{b}(fz,fz)={p}_{b}(fz,z). But then, using (3.19), we get that \phi ({M}_{s}^{f}(z,z))=0.
Hence, we have {p}_{b}(fz,z)=0 and fz=z. Thus, z is a fixed point of f. □
We will show now that the continuity of f in Theorem 1 is not necessary and can be replaced by another assumption.
Theorem 2 Under the hypotheses of Theorem 1, without the continuity assumption on f, assume that whenever \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X, one has {x}_{n}\u2aafx for all n\in \mathbb{N}. Then f has a fixed point in X.
Proof Following similar arguments as those given in Theorem 1, we construct an increasing sequence \{{x}_{n}\} in X such that {x}_{n}\to z for some z\in X. Using the assumption on X, we have {x}_{n}\u2aafz for all n\in \mathbb{N}. Now, we show that fz=z. By (3.1), we have
where
Letting n\to \mathrm{\infty} in (3.21) and using Lemma 2, we get
Again, taking the upper limit as n\to \mathrm{\infty} in (3.20) and using Lemma 2 and (3.22), we get
Therefore, \phi ({lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{M}_{s}^{f}({x}_{n},z))\le 0, equivalently, {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{M}_{s}^{f}({x}_{n},z)=0. Thus, from (3.22) we get z=fz, and hence z is a fixed point of f. □
Corollary 1 Let (X,\u2aaf,{p}_{b}) be a {p}_{b}complete ordered partial bmetric space. Let f:X\to X be a continuous mapping, nondecreasing with respect to ⪯. Suppose that there exists k\in [0,1) such that
for all comparable elements x,y\in X. If there exists {x}_{0}\in X such that {x}_{0}\u2aaff{x}_{0}, then f has a fixed point.
Proof Follows from Theorem 1 by taking \psi (t)=t and \phi (t)=(1k)t, for all t\in [0,+\mathrm{\infty}). □
Corollary 2 Under the hypotheses of Corollary 1, without the continuity assumption on f, for any nondecreasing sequence \{{x}_{n}\} in X such that {x}_{n}\to x\in X, let us have {x}_{n}\u2aafx for all n\in \mathbb{N}. Then f has a fixed point in X.
Now, in order to support the usability of our results, we present the following example.
Example 4 Let X=[0,+\mathrm{\infty}) be equipped with the partial order ⪯ defined by
and with the partial bmetric {p}_{b} given by {p}_{b}(x,y)={[max\{x,y\}]}^{2} (with s=2). Consider the mapping f:X\to X given by
Then f is continuous and increasing, and 0\u2aaff0. Take altering distance functions
In order to check the contractive condition (3.1) of Theorem 1, without loss of generality, we may take x,y\in X such that y\u2aafx. Consider the following two possible cases.
Case 1. 0\le y\le x\le 1. Then
and
Thus, (3.1) reduces to
Case 2. x=y\ge 1. Then {p}_{b}(fx,fy)=\frac{{x}^{2}}{4} and {M}_{s}^{f}(x,y)={x}^{2}, so (3.1) reduces to
Hence, all the conditions of Theorem 1 are satisfied and f has a fixed point (which is z=0).
4 Common fixed point results in partial bmetric spaces
Let (X,\u2aaf,{p}_{b}) be an ordered partial bmetric space with the coefficient s\ge 1, and let f,g:X\to X be two mappings. Set
Now, we present the following definition.
Definition 12 Let (X,\u2aaf,{p}_{b}) be an ordered partial bmetric space, and let ψ and φ be altering distance functions. We say that a pair (f,g) of selfmappings f,g:X\to X is a generalized {(\psi ,\phi )}_{s}contraction pair if
for all comparable x,y\in X.
Definition 13 [42]
Let (X,\u2aaf) be a partially ordered set. Then two mappings f,g:X\to X are said to be weakly increasing if fx\u2aafgfx and gx\u2aaffgx for all x\in X.
Theorem 3 Let (X,\u2aaf,{p}_{b}) be a {p}_{b}complete ordered partial bmetric space with the coefficient s\ge 1, and let f,g:X\to X be two weakly increasing mappings with respect to ⪯. Suppose that (f,g) is a generalized {(\psi ,\phi )}_{s}contraction pair for some altering distance functions ψ and φ. If f and g are continuous, then f and g have a common fixed point.
Proof Let us divide the proof into two parts as follows.
First part. We prove that u\in X is a fixed point of f if and only if it is a fixed point of g. Suppose that u is a fixed point of f, that is, fu=u. As u\u2aafu, by (4.1), we have
Therefore, {p}_{b}(u,gu)=0 and hence gu=u. Similarly, we can show that if u is a fixed point of g, then u is a fixed point of f.
Second part (construction of a sequence by iterative technique).
Let {x}_{0}\in X. We construct a sequence \{{x}_{n}\} in X such that {x}_{2n+1}=f{x}_{2n} and {x}_{2n+2}=g{x}_{2n+1} for all nonnegative integers n. As f and g are weakly increasing with respect to ⪯, we have
If {x}_{2n}={x}_{2n+1} for some n\in \mathbb{N}, then {x}_{2n}=f{x}_{2n}. Thus {x}_{2n} is a fixed point of f. By the first part, we conclude that {x}_{2n} is also a fixed point of g.
If {x}_{2n+1}={x}_{2n+2} for some n\in \mathbb{N}, then {x}_{2n+1}=g{x}_{2n+1}. Thus, {x}_{2n+1} is a fixed point of g. By the first part, we conclude that {x}_{2n+1} is also a fixed point of f. Therefore, we assume that {x}_{n}\ne {x}_{n+1} for all n\in \mathbb{N}. Now, we complete the proof in the following steps.
Step 1: We will prove that
As {x}_{2n+1} and {x}_{2n+2} are comparable, by (4.1), we have
where
Hence, we have
If
then (4.2) becomes
which gives a contradiction. Hence,
and (4.2) becomes
Similarly, we can show that
By (4.3) and (4.4), we get that \{{p}_{b}({x}_{n},{x}_{n+1}):n\in \mathbb{N}\} is a nonincreasing sequence of positive numbers. Hence, there is r\ge 0 such that
Letting n\to \mathrm{\infty} in (4.3), we get
which implies that \phi (r)=0 and hence r=0. So, we have
Step 2. We will prove that \{{x}_{n}\} is a {p}_{b}Cauchy sequence. Because of (4.5), it is sufficient to show that \{{x}_{2n}\} is a {p}_{b}Cauchy sequence. By Lemma 1, we should show that \{{x}_{2n}\} is bCauchy in (X,{d}_{{p}_{b}}). Suppose the contrary, i.e., that \{{x}_{2n}\} is not a bCauchy sequence in (X,{d}_{{p}_{b}}). Then there exists \epsilon >0 for which we can find two subsequences \{{x}_{2{m}_{i}}\} and \{{x}_{2{n}_{i}}\} of \{{x}_{2n}\} such that {n}_{i} is the smallest index for which
This means that
From (4.6) and using the triangular inequality, we get
Using (4.5) and taking the upper limit as i\to \mathrm{\infty}, we get
On the other hand, we have
Using (4.5), (4.7) and taking the upper limit as i\to \mathrm{\infty}, we get
Again, using the triangular inequality, we have
and
Taking the upper limit as i\to \mathrm{\infty} in the above inequalities and using (4.5), (4.7) and (4.8), we get
and
From the definition of {d}_{{p}_{b}} and (4.5), we have the following relations:
Since {x}_{2{m}_{i}} and {x}_{2{n}_{i}1} are comparable, using (4.1) we have
where
Taking the upper limit in (4.14) and using (4.5) and (4.10)(4.12), we get
Now, taking the upper limit as i\to \mathrm{\infty} in (4.13) and using (4.9) and (4.15), we have
which implies that \phi ({lim\hspace{0.17em}inf}_{i\to \mathrm{\infty}}{M}_{s}^{f,g}({x}_{2{m}_{i}},{x}_{2{n}_{i}1}))=0. By (4.14), it follows that
which is in contradiction with (4.10). Thus, we have proved that \{{x}_{n}\} is a bCauchy sequence in the metric space (X,{d}_{{p}_{b}}). Since (X,{p}_{b}) is {p}_{b}complete, then from Lemma 1, (X,{d}_{{p}_{b}}) is a bcomplete bmetric space. Therefore, the sequence \{{x}_{n}\} converges to some z\in X, that is, {lim}_{n\to \mathrm{\infty}}{d}_{{p}_{b}}({x}_{n},z)=0. Again, from Lemma 1,
On the other hand, from (4.5) we get that
Step 3 (Existence of a common fixed point). Using the triangular inequality, we get
Letting n\to \mathrm{\infty} and using the continuity of f and g, we get
Therefore,
From (4.1), we have
where
As ψ is nondecreasing, we have {s}^{2}{p}_{b}(fz,gz)\le max\{{p}_{b}(z,fz),{p}_{b}(z,gz)\}. Hence, by (4.16) we obtain that {s}^{2}{p}_{b}(fz,gz)=max\{{p}_{b}(z,fz),{p}_{b}(z,gz)\}. But then, using (4.17), we get that \phi ({M}_{s}^{f,g}(z,z))=0. Thus, we have fz=gz=z and z is a common fixed point of f and g. □
The continuity of functions f and g in Theorem 3 can be replaced by another condition.
Theorem 4 Under the hypotheses of Theorem 3, without the continuity assumption on the functions f and g, for any nondecreasing sequence \{{x}_{n}\} in X such that {x}_{n}\to x\in X, let us have {x}_{n}\u2aafx for all n\in \mathbb{N}. Then f and g have a common fixed point in X.
Proof Reviewing the proof of Theorem 3, we construct an increasing sequence \{{x}_{n}\} in X such that {x}_{n}\to z for some z\in X. Using the given assumption on X, we have {x}_{n}\u2aafz for all n\in \mathbb{N}. Now, we show that fz=gz=z. By (4.1), we have
where
Letting n\to \mathrm{\infty} in (4.19) and using Lemma 2, we get
Again, taking the upper limit as n\to \mathrm{\infty} in (4.18) and using Lemma 2 and (4.20), we get
Therefore, \phi ({lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{M}_{s}^{f,g}({x}_{2n},z))\le 0, equivalently, {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{M}_{s}^{f,g}({x}_{2n},z)=0. Thus, from (4.20) we get z=gz and hence z is a fixed point of g. On the other hand, similar to the first part of the proof of Theorem 3, we can show that fz=z. Hence, z is a common fixed point of f and g. □
Also, we have the following results.
Corollary 3 Let (X,\u2aaf,{p}_{b}) be a {p}_{b}complete ordered partial bmetric space with the coefficient s\ge 1, and let f,g:X\to X be two weakly increasing mappings with respect to ⪯. Suppose that there exists k\in [0,1) such that
for all comparable elements x,y\in X. If f and g are continuous, then f and g have a common fixed point.
Corollary 4 Under the hypotheses of Corollary 3, without the continuity assumption on the functions f and g, assume that whenever \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X, then {x}_{n}\u2aafx for all n\in \mathbb{N}. Then f and g have a common fixed point in X.
Remark 1 Recall that a subset W of a partially ordered set X is said to be well ordered if every two elements of W are comparable. Note that in Theorems 1 and 2, it can be proved in a standard way that f has a unique fixed point provided that the fixed points of f are comparable. Similarly, in Theorems 3 and 4, the set of common fixed points of f and g is well ordered if and only if f and g have one and only one common fixed point.
The usability of these results is demonstrated by the following example.
Example 5 Let X=\{0,1,2,3,4\} be equipped with the following partial order ⪯:
Define a partial bmetric {p}_{b}:X\times X\to {\mathbb{R}}^{+} by
It is easy to see that (X,{p}_{b}) is a {p}_{b}complete partial bmetric space, with s=49/25.
Define selfmaps f and g by
We see that f and g are weakly increasing mappings with respect to ⪯ and that f and g are continuous.
Define \psi ,\phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) by \psi (t)=\sqrt{t} and \phi (t)=\frac{t}{300}. In order to check that (f,g) is a generalized {(\psi ,\phi )}_{s}contractive pair, only the case x=2, y=4 is nontrivial (when x and y are comparable and the lefthand side of condition (4.1) is positive). Then
Thus, all the conditions of Theorem 3 are satisfied and hence f and g have common fixed points. Indeed, 0 and 2 are two common fixed points of f and g. Note that the ordered set (\{0,2\},\u2aaf) is not well ordered.
Note that if the same example is considered in the space without order, then the contractive condition is not satisfied. For example,
References
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