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Fixed point results via αadmissible mappings and cyclic contractive mappings in partial bmetric spaces
Journal of Inequalities and Applications volume 2014, Article number: 345 (2014)
Abstract
Considering αadmissible mappings in the setup of partial bmetric spaces, we establish some fixed and common fixed point results for ordered cyclic weakly (\psi ,\phi ,L,A,B)contractive mappings in complete ordered partial bmetric spaces. Our results extend several known results in the literature. Examples are also provided in support of our results.
MSC:47H10, 54H25.
1 Introduction
There are a lot of generalizations of the concept of metric space. The concepts of bmetric space and partial metric space were introduced by Czerwik [1] and Matthews [2], respectively. Combining these two notions, Shukla [3] introduced another generalization which is called a partial bmetric space. Also, in [4], Mustafa et al. introduced a modified version of partial bmetric spaces. In fact, the advantage of their definition of partial bmetric is that by using it one can define a dependent bmetric which is called the bmetric associated with the partial bmetric.
Definition 1.1 [4]
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:
({p}_{b1}) x=y\u27fa{p}_{b}(x,x)={p}_{b}(x,y)={p}_{b}(y,y),
({p}_{b2}) {p}_{b}(x,x)\le {p}_{b}(x,y),
({p}_{b3}) {p}_{b}(x,y)={p}_{b}(y,x),
({p}_{b4}) {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.
Example 1.2 [3]
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.
Some more examples of partial bmetrics can be constructed with the help of following propositions.
Proposition 1.3 [3]
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 1.4 [3]
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}.
Proposition 1.5 [4]
Every partial bmetric {p}_{b} defines a bmetric {d}_{{p}_{b}}, where
for all x,y\in X.
Now, we recall some definitions and propositions in a partial bmetric space.
Definition 1.6 [4]
Let (X,{p}_{b}) be a partial bmetric space. Then for an x\in X and an \u03f5>0, the {p}_{b}ball with center x and radius ϵ is
Proposition 1.7 [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 a \delta >0 such that {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 it does not need to be {T}_{1}.
Definition 1.8 [4]
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).
Lemma 1.9 [4]

(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).
Definition 1.10 [4]
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 1.11 [4]
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 1.12 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.
The following crucial lemma is useful in proving our main results.
Lemma 1.13 [4]
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
One of the interesting generalizations of the Banach contraction principle was given by Kirk et al. [5] in 2003 by introducing the notion of cyclic representation.
Definition 1.14 [5]
Let A and B be nonempty subsets of a metric space (X,d) and T:A\cup B\to A\cup B. Then T is called a cyclic map if T(A)\subseteq B and T(B)\subseteq A.
The following interesting theorem for a cyclic map was given in [5].
Theorem 1.15 [5]
Let A and B be nonempty closed subsets of a complete metric space (X,d). Suppose that T:A\cup B\to A\cup B is a cyclic map such that
for all x\in A and y\in B, where k\in [0,1) is a constant. Then T has a unique fixed point u and u\in A\cap B.
Berinde initiated in [6, 7] the concept of almost contractions and obtained several interesting fixed point theorems for Ćirić strong almost contractions. Babu et al. introduced in [8] the class of mappings which satisfy ‘condition (B)’. Moreover, they proved the existence of fixed points for such mappings on complete metric spaces. Finally, Ćirić et al. in [9], and Aghajani et al. in [10] introduced the concept of almost generalized contractive conditions (for two, resp. four mappings) and proved some important results in ordered metric spaces. Let us recall one of these definitions.
Definition 1.16 [9]
Let f and g be two selfmappings on a metric space (X,d). They are said to satisfy almost generalized contractive condition, if there exist a constant \delta \in (0,1) and some L\ge 0 such that
for all x,y\in X.
Definition 1.17 [11]
A function \phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is called an altering distance function, if the following properties hold:

(1)
φ is continuous and nondecreasing.

(2)
\phi (t)=0 if and only if t=0.
Definition 1.18 [12]
Let (X,\u2aaf) be a partially ordered set and A and B be closed subsets of X with X=A\cup B. Let f,g:X\to X be two mappings. The pair (f,g) is said to be (A,B)weakly increasing if fx\u2aafgfx, for all x\in A and gy\u2aaffgy, for all y\in B.
In [13], Hussain et al. introduced the notion of ordered cyclic weakly (\psi ,\phi ,L,A,B)contractive pair of selfmappings as follows.
Definition 1.19 [13]
Let (X,\u2aaf,d) be an ordered bmetric space, let f,g:X\to X be two mappings, and let A and B be nonempty closed subsets of X. The pair (f,g) is called an ordered cyclic weakly (\psi ,\phi ,L,A,B)contraction if

(1)
X=A\cup B is a cyclic representation of X w.r.t. the pair (f,g); that is, fA\subseteq B and gB\subseteq A;

(2)
there exist two altering distance functions ψ, φ and a constant L\ge 0, such that for arbitrary comparable elements x,y\in X with x\in A and y\in B, we have
\psi ({s}^{2}d(fx,gy))\le \psi ({M}_{s}(x,y))\phi ({M}_{s}(x,y))+L\psi (N(x,y)),
where
and
Also, in [13] the authors proved the following results.
Theorem 1.20 [13]
Let (X,\u2aaf,d) be a complete ordered bmetric space and A and B be closed subsets of X. Let f,g:X\to X be two (A,B)weakly increasing mappings with respect to ⪯. Suppose that:

(a)
the pair (f,g) is an ordered cyclic weakly (\psi ,\phi ,L,A,B)contraction;

(b)
f or g is continuous.
Then f and g have a common fixed point u\in A\cap B.
An ordered bmetric space (X,\u2aaf,d) is called regular if for any nondecreasing sequence \{{x}_{n}\} in X such that {x}_{n}\to x\in X, as n\to \mathrm{\infty}, one has {x}_{n}\u2aafx for all n\in \mathbb{N}.
Theorem 1.21 [13]
Let the hypotheses of Theorem 1.20 be satisfied, except that condition (b) is replaced by the assumption
(b′) the space (X,\u2aaf,d) is regular.
Then f and g have a common fixed point in X.
In this paper, first we prove some fixed point results for αadmissible mappings in the context of partial bmetric spaces. Then we express some common fixed point results for cyclic generalized almost contractive mappings. Our results extend and generalize some recent results in [4] and [13]. In fact, they are cyclic variants of the results in [4].
2 Fixed point results via αadmissible mappings in partial bmetric spaces
Samet et al. [14] defined the notion of αadmissible mappings and proved the following result.
Definition 2.1 [14]
Let T be a selfmapping on X and \alpha :X\times X\to [0,\mathrm{\infty}) be a function. We say that T is an αadmissible mapping if
Denote by {\mathrm{\Psi}}^{\prime} the family of all nondecreasing functions \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) such that {\sum}_{n=1}^{\mathrm{\infty}}{\psi}^{n}(t)<\mathrm{\infty} for all t>0, where {\psi}^{n} is the n th iterate of ψ.
Theorem 2.2 [14]
Let (X,d) be a complete metric space and T be an αadmissible mapping. Assume that
where \psi \in {\mathrm{\Psi}}^{\prime}. Also, suppose that the following assertions hold:

(i)
there exists {x}_{0}\in X such that \alpha ({x}_{0},T{x}_{0})\ge 1;

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} such that {x}_{n}\to x as n\to \mathrm{\infty}, we have \alpha ({x}_{n},x)\ge 1 for all n\in \mathbb{N}\cup \{0\}.
Then T has a fixed point.
We now recall the concept of (c)comparison function which was introduced by Berinde [15].
Definition 2.3 (Berinde [15])
A function \phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is said to be a (c)comparison function if
({c}_{1}) φ is increasing,
({c}_{2}) there exist {k}_{0}\in \mathbb{N}, a\in (0,1), and a convergent series of nonnegative terms {\sum}_{k=1}^{\mathrm{\infty}}{v}_{k} such that {\phi}^{k+1}(t)\le a{\phi}^{k}(t)+{v}_{k}, for k\ge {k}_{0} and any t\in [0,\mathrm{\infty}).
Later, Berinde [16] introduced the notion of (b)comparison function as a generalization of a (c)comparison function.
Definition 2.4 (Berinde [16])
Let s\ge 1 be a real number. A mapping \phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is called a (b)comparison function if the following conditions are fulfilled:

(1)
φ is monotone increasing;

(2)
there exist {k}_{0}\in \mathbb{N}, a\in (0,1), and a convergent series of nonnegative terms {\sum}_{k=1}^{\mathrm{\infty}}{v}_{k} such that {s}^{k+1}{\phi}^{k+1}(t)\le a{s}^{k}{\phi}^{k}(t)+{v}_{k}, for k\ge {k}_{0} and any t\in [0,\mathrm{\infty}).
Let {\mathrm{\Psi}}_{b} be the class of (b)comparison functions \phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}). It is clear that the notion of (b)comparison function coincides with (c)comparison function for s=1.
We now recall the following lemma, which will simplify the proofs.
Lemma 2.5 (Berinde [17])
If \phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is a (b)comparison function, then we have the following.

(1)
the series {\sum}_{k=0}^{\mathrm{\infty}}{s}^{k}{\phi}^{k}(t) converges for any t\in {\mathbb{R}}_{+};

(2)
the function {b}_{s}:[0,\mathrm{\infty})\to [0,\mathrm{\infty}), defined by {b}_{s}(t)={\sum}_{k=0}^{\mathrm{\infty}}{s}^{k}{\phi}^{k}(t), t\in [0,\mathrm{\infty}), is increasing and continuous at 0.
Theorem 2.6 Let (X,{p}_{b}) be a {p}_{b}complete partial bmetric space, f be a continuous αadmissible mapping on X, there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1 and if any sequence \{{x}_{n}\} in X {p}_{b}converges to a point x, where \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n, then we have \alpha (x,x)\ge 1. Assume that
for all x,y\in X, where \psi \in {\mathrm{\Psi}}_{b} and
Then f has a fixed point.
Proof Let {x}_{0}\in X be such that \alpha ({x}_{0},f{x}_{0})\ge 1. Define a sequence \{{x}_{n}\} by {x}_{n}={f}^{n}{x}_{0} for all n\in \mathbb{N}. Since f is an αadmissible mapping and \alpha ({x}_{0},{x}_{1})=\alpha ({x}_{0},f{x}_{0})\ge 1, we deduce that \alpha ({x}_{1},{x}_{2})=\alpha (f{x}_{0},f{x}_{1})\ge 1. Continuing this process, we get that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\}.
Now, we will finish the proof in the following steps.
First, we prove that
for each n=1,2,3,\dots .
If {x}_{n}={x}_{n+1}, for some n\in \mathbb{N}, then {x}_{n}=f{x}_{n}. Thus, {x}_{n} is a fixed point of f. Therefore, we assume that {x}_{n}\ne {x}_{n+1}, for all n\in \mathbb{N}.
Using condition (2.2) as \alpha ({x}_{n1},{x}_{n})\ge 1 for all n\in \mathbb{N}\cup \{0\}, we obtain
Here,
If {p}_{b}({x}_{n},{x}_{n+1})\ge {p}_{b}({x}_{n1},{x}_{n}), then
which yields
a contradiction.
Hence,
So (2.3) holds.
By induction, we get
Then, by the triangular inequality and (2.4), we get
as n\u27f6\mathrm{\infty}.
Since \{{x}_{n}\} is a {p}_{b}Cauchy sequence in the {p}_{b}complete partial bmetric space X, from Lemma 1.9, \{{x}_{n}\} is a bCauchy sequence in the bmetric space (X,{d}_{{p}_{b}}). {p}_{b}Completeness of (X,{p}_{b}) shows that (X,{d}_{{p}_{b}}) is also bcomplete. Then there exists z\in X such that
Since {lim}_{m,n\to \mathrm{\infty}}{p}_{b}({x}_{n},{x}_{m})=0, from Lemma 1.9
From the continuity of f we have
and hence we get
So, we get {p}_{b}(z,fz)\le s{p}_{b}(fz,fz). As \alpha (z,z)\ge 1, we have
Hence, {p}_{b}(z,fz)\le \psi ({p}_{b}(z,fz)). Thus, {p}_{b}(z,fz)=0, that is, z=fz. □
In Theorem 2.6, we omit the continuity of the mapping f and we replace \alpha ({x}_{n},x)\ge 1 instead of \alpha (x,x)\ge 1 and rearrange it as follows.
Theorem 2.7 Let (X,{p}_{b}) be a {p}_{b}complete partial bmetric space and f be an αadmissible mapping on X such that
for all x,y\in X, where \psi \in {\mathrm{\Psi}}_{b}. Assume that the following conditions hold:

(i)
there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1;

(ii)
if \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n and {x}_{n}\to x as n\to \mathrm{\infty}, then \alpha ({x}_{n},x)\ge 1 for all n\in \mathbb{N}\cup \{0\}.
Then f has a fixed point.
Proof Let {x}_{0}\in X be such that \alpha ({x}_{0},f{x}_{0})\ge 1 and define a sequence \{{x}_{n}\} in X by {x}_{n}={f}^{n}{x}_{0}=f{x}_{n1} for all n\in \mathbb{N}. Following the proof of Theorem 2.6, we have \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and there exists z\in X such that {x}_{n}\to z as n\to \mathrm{\infty} which {p}_{b}(z,z)=0. Hence, from (ii) we deduce that \alpha ({x}_{n},z)\ge 1 for all n\in \mathbb{N}\cup \{0\}. Therefore, by (2.7), we obtain
Here,
Taking the upper limit as n\to \mathrm{\infty} in the above inequality from Lemma 1.13 we obtain
which implies that z=fz. □
Definition 2.8 [18]
Let f:X\to X and \alpha :X\times X\to \mathbb{R}. We say that f is a triangular αadmissible mapping if
(T1) \alpha (x,y)\ge 1 implies \alpha (fx,fy)\ge 1, x,y\in X,
(T2) \{\begin{array}{l}\alpha (x,z)\ge 1\\ \alpha (z,y)\ge 1\end{array} implies \alpha (x,y)\ge 1, x,y,z\in X.
Example 2.9 [18]
Let X=\mathbb{R}, fx=\sqrt[3]{x}, and \alpha (x,y)={e}^{xy}, then f is a triangular αadmissible mapping. Indeed, if \alpha (x,y)={e}^{xy}\ge 1, then x\ge y which implies that fx\ge fy, that is, \alpha (fx,fy)={e}^{fxfy}\ge 1. Also, if \{\begin{array}{l}\alpha (x,z)\ge 1\\ \alpha (z,y)\ge 1\end{array}, then \{\begin{array}{l}xz\ge 0\\ zy\ge 0\end{array}, that is, xy\ge 0 and therefore \alpha (x,y)={e}^{xy}\ge 1.
Example 2.10 [18]
Let X=\mathbb{R}, fx={e}^{{x}^{7}}, and \alpha (x,y)=\sqrt[5]{xy}+1. Hence, f is a triangular αadmissible mapping. Indeed, if \alpha (x,y)=\sqrt[5]{xy}+1\ge 1 then x\ge y which implies that fx\ge fy, that is, \alpha (fx,fy)\ge 1.
Moreover, if \{\begin{array}{l}\alpha (x,z)\ge 1\\ \alpha (z,y)\ge 1\end{array}, then xy\ge 0 and hence \alpha (x,y)\ge 1.
Example 2.11 [18]
Let X=[0,\mathrm{\infty}), fx={x}^{4}+ln({x}^{2}+1), and
Then f is a triangular αadmissible mapping. In fact, if
then x\ge y. Hence, fx\ge fy, that is, \alpha (fx,fy)\ge 1. Also,
Thus, \alpha (x,z)+\alpha (z,y)\le 2\alpha (x,y). Now, if \{\begin{array}{l}\alpha (x,z)\ge 1\\ \alpha (z,y)\ge 1\end{array}, then \alpha (x,y)\ge 1.
Example 2.12 [18]
Let X=\mathbb{R}, fx={x}^{3}+\sqrt[7]{x}, and \alpha (x,y)={x}^{5}{y}^{5}+1. Then f is a triangular αadmissible mapping.
Lemma 2.13 [18]
Let f be a triangular αadmissible mapping. Assume that there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1. Define the sequence \{{x}_{n}\} by {x}_{n}={f}^{n}{x}_{0}. Then
A mapping \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is called a comparison function if it is increasing and {\psi}^{n}(t)\to 0, as n\to \mathrm{\infty} for any t\in [0,\mathrm{\infty}).
Lemma 2.14 (Berinde [15], Rus [19])
If \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is a comparison function, then:

(1)
each iterate {\psi}^{k} of ψ, k\ge 1, is also a comparison function;

(2)
ψ is continuous at 0;

(3)
\psi (t)<t, for any t>0.
Denote by Ψ the family of all continuous comparison functions \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}).
In the sequel, \psi \in \mathrm{\Psi}, \alpha :X\times X\to [0,\mathrm{\infty}) is a function and
Theorem 2.15 Let (X,{p}_{b}) be a {p}_{b}complete partial bmetric space, f be a continuous triangular αadmissible mapping on X, there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1 and if any sequence \{{x}_{n}\} in X {p}_{b}converges to a point x, where \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n, then we have \alpha (x,x)\ge 1. Assume that
for all x,y\in X. Then f has a fixed point.
Proof Let {x}_{0}\in X be such that \alpha ({x}_{0},f{x}_{0})\ge 1. Define a sequence \{{x}_{n}\} by {x}_{n}={f}^{n}{x}_{0} for all n\in \mathbb{N}. Since f is an αadmissible mapping and \alpha ({x}_{0},{x}_{1})=\alpha ({x}_{0},f{x}_{0})\ge 1, we deduce that \alpha ({x}_{1},{x}_{2})=\alpha (f{x}_{0},f{x}_{1})\ge 1. Continuing this process, we get \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\}.
Now, we will finish the proof in the following steps.
Step I. We will prove that
First, we prove that
for each n=1,2,3,\dots .
If {x}_{n}={x}_{n+1}, for some n\in \mathbb{N}, then {x}_{n}=f{x}_{n}. Thus, {x}_{n} is a fixed point of f. Therefore, we assume that {x}_{n}\ne {x}_{n+1}, for all n\in \mathbb{N}.
Using condition (2.8) as \alpha ({x}_{n1},{x}_{n})\ge 1 for all n\in \mathbb{N}\cup \{0\}, we obtain
Here,
If {p}_{b}({x}_{n},{x}_{n+1})\ge {p}_{b}({x}_{n1},{x}_{n}), then
which yields
a contradiction.
Hence,
So (2.9) holds.
By induction, we get
As \psi \in \mathrm{\Psi}, we conclude that
So by ({p}_{b2}) we get
Step II. We will 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 1.9). 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 (2.13) and using the triangular inequality, we get
Using (2.11), (2.12), and from the definition of {d}_{{p}_{b}} and (2.14), and taking the upper limit as i\to \mathrm{\infty}, we get
Also,
Further,
and
On the other hand, by the definition of {d}_{{p}_{b}} and (2.12)
Hence, by (2.15),
Similarly,
and
From (2.8) and Lemma 2.13 as \alpha ({x}_{{m}_{i}},{x}_{{n}_{i}1})\ge 1, we have
where
Taking the upper limit as i\to \mathrm{\infty} in (2.24) and using (2.11), (2.19), (2.20), and (2.22), we get
Now, taking the upper limit as i\to \mathrm{\infty} in (2.23) and using (2.21) and (2.25), we have
a contradiction.
Step III. There exists z such that fz=z.
Since \{{x}_{n}\} is a {p}_{b}Cauchy sequence in the {p}_{b}complete partial bmetric space X, from Lemma 1.9, \{{x}_{n}\} is a bCauchy sequence in the bmetric space (X,{d}_{{p}_{b}}). {p}_{b}Completeness of (X,{p}_{b}) shows that (X,{d}_{{p}_{b}}) is also bcomplete. Then there exists z\in X such that
Since {lim}_{m,n\to \mathrm{\infty}}{d}_{{p}_{b}}({x}_{n},{x}_{m})=0, from the definition of {d}_{{p}_{b}} and (2.12), we get
Again, from Lemma 1.9,
From the continuity of f we have
and hence we get
So, we get {p}_{b}(z,fz)\le s{p}_{b}(fz,fz). As \alpha (z,z)\ge 1, we have
Hence, {p}_{b}(z,fz)\le \psi ({p}_{b}(z,fz)). Thus, {p}_{b}(z,fz)=0, that is, z=fz. □
If in Theorem 2.15 we take \alpha (x,y)=1 then we deduce the following corollary.
Corollary 2.16 Let (X,{p}_{b}) be a {p}_{b}complete partial bmetric space and f be a continuous mapping on X. Assume that
for all x,y\in X. Then f has a fixed point.
In Theorem 2.15, we omit the continuity of the mapping f and we replace \alpha ({x}_{n},x)\ge 1 instead of \alpha (x,x)\ge 1 and rearrange it as follows.
Theorem 2.17 Let (X,{p}_{b}) be a {p}_{b}complete partial bmetric space and f be a triangular αadmissible mapping on X such that
for all x,y\in X, where \psi \in \mathrm{\Psi}. Assume that the following conditions hold:

(i)
there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1;

(ii)
if \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n and {x}_{n}\to x as n\to \mathrm{\infty}, then \alpha ({x}_{n},x)\ge 1 for all n\in \mathbb{N}\cup \{0\}.
Then f has a fixed point.
Proof Let {x}_{0}\in X be such that \alpha ({x}_{0},f{x}_{0})\ge 1 and define a sequence \{{x}_{n}\} in X by {x}_{n}={f}^{n}{x}_{0}=f{x}_{n1} for all n\in \mathbb{N}. Following the proof of Theorem 2.15, we have \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and there exists z\in X such that {x}_{n}\to z as n\to \mathrm{\infty} which {p}_{b}(z,z)=0. Hence, from (ii) we deduce that \alpha ({x}_{n},z)\ge 1 for all n\in \mathbb{N}\cup \{0\}. Therefore, by (2.29), we obtain
Here,
Taking the upper limit as n\to \mathrm{\infty} in the above inequality from Lemma 1.13 we obtain
which implies that z=fz. □
Example 2.18 Let X=[0,1] and {p}_{b}(x,y)=xy{}^{2} be a {p}_{b}metric on X. Define f:X\to X by fx=ln(\frac{x}{4}+1) and \alpha :X\times X\to [0,\mathrm{\infty}) by
and \psi (t)=\frac{t}{8} for all t\in [0,\mathrm{\infty}). Now, we prove that all the hypotheses of Theorem 2.17 are satisfied and hence f has a fixed point.
First, we see that (X,{p}_{b}) is a {p}_{b}complete partial bmetric space. Let x,y\in X. If \alpha (x,y)\ge 1, then x,y\in [0,\frac{1}{4}]. On the other hand, for all x\in [0,1], we have fx\le \frac{x}{4}\le \frac{1}{4} and hence \alpha (fx,fy)=1. This implies that f is a triangular αadmissible mapping on X. Obviously, \alpha (0,f0)=1.
Now, if \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})=1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to \mathrm{\infty}, it is easy to see that \alpha ({x}_{n},x)=1.
Using the Mean Value Theorem for the function fx=ln(\frac{x}{4}+1) for any x,y\in X, we have
Thus, all the conditions of Theorem 2.17 are satisfied and therefore f has a fixed point (z=0).
3 Common fixed points of generalized almost cyclic weakly (\psi ,\phi ,L,A,B)contractive mappings
In this section, we consider the notion of ordered cyclic weakly (\psi ,\phi ,L,A,B)contractions in the setup of ordered partial bmetric spaces and then obtain some common fixed point theorems for these cyclic contractions in the setup of complete ordered partial bmetric spaces. Our results extend some fixed point theorems from the framework of ordered metric spaces and ordered bmetric spaces, in particular Theorems 1.20 and 1.21.
We shall call an ordered partial bmetric space (X,\u2aaf,{p}_{b}) regular if for any nondecreasing sequence \{{x}_{n}\} in X such that {x}_{n}\to x\in X, as n\to \mathrm{\infty}, one has {x}_{n}\u2aafx, for all n\in \mathbb{N}.
Definition 3.1 Let (X,\u2aaf,{p}_{b}) be an ordered partial bmetric space, let f,g:X\to X be two mappings, and let A and B be nonempty closed subsets of X. The pair (f,g) is called an ordered cyclic almost generalized weakly (\psi ,\phi ,L,A,B)contraction if

(1)
X=A\cup B is a cyclic representation of X w.r.t. the pair (f,g); that is, fA\subseteq B and gB\subseteq A;

(2)
there exist two altering distance functions ψ, φ and a constant L\ge 0, such that for arbitrary comparable elements x,y\in X with x\in A and y\in B, we have
\psi ({s}^{2}{p}_{b}(fx,gy))\le \psi ({M}_{s}(x,y))\phi ({M}_{s}(x,y))+L\psi (N(x,y)),(3.1)
where
and
Theorem 3.2 Let (X,\u2aaf,{p}_{b}) be a {p}_{b}complete ordered partial bmetric space and A and B be two nonempty closed subsets of X. Let f,g:X\to X be two (A,B)weakly increasing mappings with respect to ⪯. Suppose that the pair (f,g) is an ordered cyclic almost generalized weakly (\psi ,\phi ,L,A,B)contraction. Then f and g have a common fixed point z\in A\cap B.
Proof First, note that u\in A\cap B is a fixed point of f if and only if u is a fixed point of g. Indeed, suppose that u is a fixed point of f. As u\u2aafu and u\in A\cap B, by (3.1), we have
It follows that \phi ({p}_{b}(u,gu))=0. 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.
Let {x}_{0}\in A and let {x}_{1}=f{x}_{0}. Since fA\subseteq B, we have {x}_{1}\in B. Also, let {x}_{2}=g{x}_{1}. Since gB\subseteq A, we have {x}_{2}\in A. Continuing this process, we can construct a sequence \{{x}_{n}\} in X such that {x}_{2n+1}=f{x}_{2n}, {x}_{2n+2}=g{x}_{2n+1}, {x}_{2n}\in A and {x}_{2n+1}\in B. Since f and g are (A,B)weakly increasing, 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 of the proof, we conclude that {x}_{2n} is also a fixed point of g. Similarly, 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 of the proof, 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} and {x}_{2n+1} are comparable and {x}_{2n}\in A and {x}_{2n+1}\in B, by (3.1), we have
where
and
Hence, we have
If
then (3.4) becomes
which gives a contradiction. So,
and hence (3.4) becomes
Similarly, we can show that
By (3.5) and (3.6), we see that \{d({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 (3.5), 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 (3.7), it is sufficient to show that \{{x}_{2n}\} is a {p}_{b}Cauchy sequence. By Lemma 1.9, 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 (3.8) and using the triangular inequality, we get
Using (3.7) and from the definition of {d}_{{p}_{b}} and taking the upper limit as i\to \mathrm{\infty}, we get
On the other hand, we have
Using (3.7), (3.9), 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 (3.7), (3.9), and (3.11) we get
and
From the definition of {d}_{{p}_{b}} and (3.7), (3.10), (3.11), (3.12), and (3.13) we have the following relations:
Since {x}_{2{m}_{i}}\in A and {x}_{2{n}_{i}1}\in B are comparable, using (3.1) we have
where
and
Taking the upper limit in (3.19) and (3.20), and using (3.7) and (3.14)(3.17), we get
and
Now, taking the upper limit as i\to \mathrm{\infty} in (3.18) and using (3.14), (3.21), and (3.22), we have
which implies that \phi ({lim\hspace{0.17em}inf}_{i\to \mathrm{\infty}}{M}_{s}({x}_{2{m}_{i}},{x}_{2{n}_{i}1}))=0. By (3.19), it follows that
which is in contradiction with (3.8). 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, from Lemma 1.9, (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. Since {lim}_{m,n\to \mathrm{\infty}}{d}_{{p}_{b}}({x}_{n},{x}_{m})=0, from the definition of {d}_{{p}_{b}} and (3.7), we get
Again, from Lemma 1.9,
Step 3. In the above steps, we constructed an increasing sequence \{{x}_{n}\} in X such that {x}_{n}\to z, for some z\in X. As A and B are closed subsets of X, we have z\in A\cap B. Using the regularity assumption on X, we have {x}_{n}\u2aafz, for all n\in \mathbb{N}. Now, we show that fz=gz=z. By (3.1), we have
where
and
Letting n\to \mathrm{\infty} in (3.24) and (3.25), and using Lemma 1.13, we get
and N({x}_{2n},z)\to 0. Now, taking the upper limit as n\to \mathrm{\infty} in (3.23), and using Lemma 1.13 and (3.26) we get
It follows that \phi ({lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{M}_{s}({x}_{2n},z))=0, and hence, by (3.24), that {p}_{b}(z,gz)=0. Thus, z is a fixed point of g. On the other hand, from the first part of the proof, fz=z. Hence, z is a common fixed point of f and g. □
Theorem 3.3 Let (X,\u2aaf,{p}_{b}) be a {p}_{b}complete ordered partial bmetric space and A and B be nonempty closed subsets of X. Let f,g:X\to X be two (A,B)weakly increasing mappings with respect to ⪯. Suppose that
Also, let f and g be continuous. Then f and g have a common fixed point z\in A\cap B.
Proof Repeating the proof of Theorem 3.2, we construct an increasing sequence \{{x}_{n}\} in X such that {x}_{n}\to z, for some z\in X. As A and B are closed subsets of X, we have z\in A\cap B. Now, we show that fz=gz=z.
Using the triangular inequality, we get
and
Letting n\to \mathrm{\infty} and using continuity of f and g, we get
Therefore,
From (3.27) as z\in A\cap B, 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 (3.28) we obtain {s}^{2}{p}_{b}(fz,gz)=max\{{p}_{b}(z,fz),{p}_{b}(z,gz)\}. But then, using (3.29), we get \phi ({M}_{s}(z,z))=0. Thus, we have fz=gz=z and z is a common fixed point of f and g. □
As consequences, we have the following results.
By putting A=B=X in Theorems 3.2 and 3.3 and L=0 in Theorem 3.2, we obtain the main results (Theorems 3 and 4) of Mustafa et al. [4].
Taking \phi =(1\delta )\psi, 0<\delta <1 in Theorem 3.2, we get the following.
Corollary 3.4 Let (X,\u2aaf,{p}_{b}) be a {p}_{b}complete ordered partial bmetric space and A and B be closed subsets of X. Let f,g:X\to X be two (A,B)weakly increasing mappings with respect to ⪯. Suppose that:

(a)
X=A\cup B is a cyclic representation of X w.r.t. the pair (f,g);

(b)
there exist 0<\delta <1, L\ge 0, and an altering distance function ψ such that for any comparable elements x,y\in X with x\in A and y\in B, we have
\psi ({s}^{2}{p}_{b}(fx,gy))\le \delta \psi ({M}_{s}(x,y))+L\psi (N(x,y)),(3.30)
where {M}_{s}(x,y) and N(x,y) are given by (3.2) and (3.3), respectively;

(c)
f and g are continuous, or
(c′) the space (X,\u2aaf,{p}_{b}) is regular.
Then f and g have a common fixed point z\in A\cap B.
Taking s=1 and L=0 in Corollary 3.4, we obtain the partial version of Theorems 2.1 and 2.2 of Shatanawi and Postolache [12].
In Definitions 1.18 and 3.1 and Theorems 3.2 and 3.3, if we take f=g, then we have the following definitions and results.
Definition 3.5 Let (X,\u2aaf) be a partially ordered set and A and B be closed subsets of X with X=A\cup B. The mapping f:X\to X is said to be (A,B)weakly increasing if fx\u2aaf{f}^{2}x, for all x\in A and fy\u2aaf{f}^{2}y, for all y\in B.
Definition 3.6 Let (X,\u2aaf,{p}_{b}) be an ordered partial bmetric space, let f:X\to X be a mapping, and let A and B be nonempty closed subsets of X. The mapping f is called an ordered cyclic almost generalized weakly (\psi ,\phi ,L,A,B)contraction if

(1)
X=A\cup B is a cyclic representation of X w.r.t. f; that is, fA\subseteq B and fB\subseteq A;

(2)
there exist two altering distance functions ψ, φ and a constant L\ge 0, such that for arbitrary comparable elements x,y\in X with x\in A and y\in B, we have
\psi ({s}^{2}{p}_{b}(fx,fy))\le \psi ({M}_{s}(x,y))\phi ({M}_{s}(x,y))+L\psi (N(x,y)),
where
and
Corollary 3.7 Let (X,\u2aaf,{p}_{b}) be a {p}_{b}complete ordered partial bmetric space and A and B be two nonempty closed subsets of X. Let f:X\to X be a (A,B)weakly increasing mapping with respect to ⪯. Suppose that the mapping f is an ordered cyclic almost generalized weakly (\psi ,\phi ,L,A,B)contraction. Then f has a fixed point z\in A\cap B.
Corollary 3.8 Let (X,\u2aaf,{p}_{b}) be a {p}_{b}complete ordered partial bmetric space and A and B be nonempty closed subsets of X. Let f:X\to X be a (A,B)weakly increasing mapping with respect to ⪯. Suppose that
Also, let f be continuous. Then f has a fixed point z\in A\cap B.
We illustrate our results with the following example.
Example 3.9 Consider the partial bmetric space X=[0,6] by {p}_{b}(x,y)={[max\{x,y\}]}^{2}. Define an order ⪯ on X by
Obviously, (X,\u2aaf,{p}_{b}) is a {p}_{b}complete ordered {p}_{b}metric space. Indeed, if we have {lim}_{n,m\to \mathrm{\infty}}{p}_{b}({x}_{n},{x}_{m})=u, for some u\in [0,\mathrm{\infty}), then we have