# Fixed Point ofT-Hardy-Rogers Contractive Mappings in Partially Ordered Partial Metric Spaces

Authors: Mujahid Abbas; Hassen Aydi; Stojan Radenović
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

We prove some fixed point theorems for aT-Hardy-Rogers contraction
in the setting of partially ordered partial metric spaces. We apply our results to study
periodic point problems for such mappings. We also provide examples to illustrate the
results presented herein.


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## Body

## 1. Introduction and Preliminaries

The notion of a partial metric space was introduced by Matthews in [1]. In partial metric spaces, the distance of a point in the self may not be zero. After the definition of a partial metric space, Matthews proved the partial metric version of Banach fixed point theorem. A motivation behind introducing the concept of a partial metric was to obtain appropriate mathematical models in the theory of computation and, in particular, to give a modified version of the Banach contraction principle, more suitable in this context [1]. Subsequently, several authors studied the problem of existence and uniqueness of a fixed point for mappings satisfying different contractive conditions (e.g., [2–21], [22]). Existence of fixed points in partially ordered metric spaces has been initiated in 2004 by Ran and Reurings [23]. Subsequently, several interesting and valuable results have appeared in this direction [14]. The aim of this paper is to study the necessary conditions for existence of fixed point of mapping satisfying T-Hardy-Rogers conditions in the framework of partially ordered partial metric spaces. Our results extend and strengthen various known results [8, 24]. In the sequel, the letters ℝ, ℝ+, and ℕ will denote the set of real numbers, the set of nonnegative real numbers, and the set of nonnegative integer numbers, respectively. The usual order on ℝ (resp., on ℝ+) will be indistinctly denoted by ≤ or by ≥.Consistent with [1, 8] (see [25–29]) the following definitions and results will be needed in the sequel.Definition 1.1 (see [1]).
A partial metric on a nonempty setX is a mapping p:X×X→ℝ+ such that for all x,y,z∈X,(p1)
x=y⇔p(x,x)=p(x,y)=p(y,y),(p2)
p(x,x)≤p(x,y),(p3)
p(x,y)=p(y,x),(p4)
p(x,y)≤p(x,z)+p(z,y)-p(z,z).
A partial metric space is a pair(X,p) such that X is a nonempty set and p is a partial metric on X. If p(x,y)=0, then  (p1) and (p2) imply that x=y. But converse does not hold always. A trivial example of a partial metric space is the pair (ℝ+,p), where p:ℝ+×ℝ+→ℝ+ is defined as p(x,y)=max{x,y}. Each partial metric p on X generates a T0 topology τp on X which has as a base the family open p-balls {Bp(x,ε):x∈X,ε>0}, where Bp(x,ε)={y∈X:p(x,y)<p(x,x)+ε}.On a partial metric space the concepts of convergence, Cauchy sequence, completeness, and continuity are defined as follows.Definition 1.2 (see [1]).
Let(X,p) be a partial metric space and let {xn} be a sequence in X. Then (i) {xn} converges to a point x∈X if and only if p(x,x)=limn→∞p(x,xn) (we may still write this as limn→∞xn=v or xn→v); (ii) {xn} is called a Cauchy sequence if there exists (and is finite) limn,m→∞p(xn,xm).Definition 1.3 (see [1]).
A partial metric space(X,p) is said to be complete if every Cauchy sequence {xn} in X converges to a point x∈X, such that p(x,x)=limn,m→∞p(xn,xm). If p is a partial metric on X, then the function ps:X×X→R+ given by ps(x,y)=2p(x,y)-p(x,x)-p(y,y) is a metric on X.Lemma 1.4 (see [1, 20]).
Let(X,p) be a partial metric space. Then(a)
{xn} is a Cauchy sequence in (X,p) if and only if it is a Cauchy sequence in the metric space (X,ps);(b)
(X,p) is complete if and only if the metric space (X,ps) is complete. Furthermore, limn→∞ps(xn,x)=0 if and only if
(1.1)p(x,x)=limn→∞p(xn,x)=limn,m→∞p(xn,xm).Remark 1.5.
(1) (see [19]) Clearly, a limit of a sequence in a partial metric space does not need to be unique. Moreover, the function p(·,·) does not need to be continuous in the sense that xn→x and yn→y implies p(xn,yn)→p(x,y). For example, if X=[0,+∞) and p(x,y)=max{x,y} for x,y∈X, then for {xn}={1},p(xn,x)=x=p(x,x) for each x≥1 and so, for example, xn→2 and xn→3 when n→∞.(2) (see [7]) However, if p(xn,x)→p(x,x)=0 then p(xn,y)→p(x,y) for all y∈X.Definition 1.6 (see [30]).
Suppose that(X,p) is a partial metric space. Denote τ(p) its topology. We say T:(X,p)→(X,p) is continuous if both T:(X,τ(p))→(X,τ(p)) and T:(X,τ(ps))→(X2,τ(ps)) are continuous.Remark 1.7.
It is worth to notice that the notionsp-continuous and ps-continuous of any function in the context of partial metric spaces are incomparable, in general. Indeed, if X=[0,+∞),  p(x,y)=max{x,y},  ps(x,y)=|x-y|,  f0=1, and fx=x2 for all x>0 and gx=|sinx|, then f is a p-continuous and ps-discontinuous at point x=0; while g is a p-discontinuous and ps-continuous at x=π.According to [31], we state the following definition.Definition 1.8.
Let(X,p) be a partial metric space. A mapping T:X→X is said to be(i)
sequentially convergent if for any sequence{yn} in X such that {Tyn} is convergent in (X,ps) implies that {yn} is convergent in (X,ps),(ii)
subsequentially convergent if for any sequence{yn} in X such that {Tyn} is convergent in (X,ps) implies that {yn} has a convergent subsequence in (X,ps).Consistent with [24, 31] we define a T-Hardy-Rogers contraction in the framework of partial metric spaces.Definition 1.9.
Let(X,p) be a partial metric space and T,f:X→X be two mappings. A mapping f is said to be a T-Hardy-Rogers contraction if there exist ai≥0, i=1,…,5 with a1+a2+a3+a4+a5<1  such that for all x,y∈X(1.2)p(Tfx,Tfy)≤a1p(Tx,Ty)+a2p(Tx,Tfx)+a3p(Ty,Tfy)+a4p(Tx,Tfy)+a5p(Ty,Tfx).Puttinga1=a4=a5=0 and a2=a3≠0, (resp., a1=a2=a3=0 and a4=a5≠0) in the previous definition, then the inequality (1.2) is said a T-Kannan (resp., T-Chatterjea) type contraction. Also, if a4=a5=0 and a1,a2,a3≠0, (1.2) is said the T-Reich type contraction.Definition 1.10.
LetX be a nonempty set. Then (X,p,⪯) is called a partially ordered partial metric space if and only if (i) p is a partial metric on X and (ii) ⪯ is a partial order on X.
Let(X,p) be a partial metric space endowed with a partial order ⪯ and let f:X→X be a given mapping. We define sets Δ,Δ1⊂X×X by
(1.3)Δ={(x,y)∈X×X:x⪯yory⪯x},Δ1={(x,x)∈X×X:x⪯fxorfx⪯x}.A pointx∈X is called a fixed point of mapping f:X→X if x=fx. The set of all fixed points of the mapping f is denoted by Ff.


## 2. Fixed Point Results

In this section, we obtain fixed point results for a mapping satisfying aT-Hardy-Rogers contractive condition defined on a partially ordered partial metric space which is complete.We start with the following result.Theorem 2.1.
Let(X,⪯,p) be an partially ordered partial metric space which is complete. Let T:X→X be a continuous, injective mapping and f:X→X a nondecreasing T-Hardy-Rogers contraction for all (x,y)∈Δ. If there exists x0∈X with x0⪯fx0, and one of the following two conditions is satisfied(a)
f is a continuous self-map on X;(b)
for any nondecreasing sequence{xn} in (X,⪯) with limn→∞ps(z,xn)=0 it follows xn⪯z for all n∈ℕ;
thenFf≠ϕ provided that T is subsequentially or sequentially convergent. Moreover, f has a unique fixed point if Ff×Ff⊂Δ.Proof.
Asf is nondecreasing, therefore by given assumption, we have
(2.1)x1=fx0⪯f2x0⪯⋯⪯fnx0⪯fn+1x0⪯⋯
Define a sequence {xn} in X with xn=fnx0 and so xn+1=fxn for n∈ℕ. Since (xn-1,xn)∈Δ therefore by replacing x by xn-1 and y by xn in (1.2), we have
(2.2)p(Txn,Txn+1)=p(Tfxn-1,Tfxn)≤a1p(Txn-1,Txn)+a2p(Txn-1,Tfxn-1)+a3p(Txn,Tfxn)+a4p(Txn-1,Tfxn)+a5p(Txn,Tfxn-1)=a1p(Txn-1,Txn)+a2p(Txn-1,Txn)+a3p(Txn,Txn+1)+a4p(Txn-1,Txn+1)+a5p(Txn,Txn)≤(a1+a2)p(Txn-1,Txn)+a3p(Txn,Txn+1)+a4[p(Txn-1,Txn)+p(Txn,Txn+1)-p(Txn,Txn)]+a5p(Txn,Txn)=(a1+a2+a4)p(Txn-1,Txn)+(a3+a4)p(Txn,Txn+1)+(a5-a4)p(Txn,Txn),
that is,
(2.3)(1-a3-a4)p(Txn,Txn+1)≤(a1+a2+a4)p(Txn-1,Txn)+(a5-a4)p(Txn,Txn).
Similarly, replacing x by xn and y by xn-1 in (1.2), we obtain
(2.4)(1-a2-a5)p(Txn,Txn+1)≤(a1+a3+a5)p(Txn-1,Txn)+(a4-a5)p(Txn,Txn).
Summing (2.3) and (2.4), we obtain p(Txn,Txn+1)≤δp(Txn-1,Txn), where δ=(2a1+a2+a3+a4+a5)/(2-a2-a3-a4-a5). Obviously 0≤δ<1. Therefore, for all n≥1,
(2.5)p(Txn,Txn+1)≤δp(Txn-1,Txn)≤⋯≤δnp(Tx0,Tx1).
Now, for any m∈N with m>n, we have
(2.6)p(Txn,Txm)≤p(Txn,Txn+1)+p(Txn+1,Txn+2)+⋯+p(Txm-1,Txm)≤(δn+δn+1+⋯+δm-1)p(Tx0,Tx1)≤δn1-δp(Tx0,Tx1),
which implies that p(Txn,Txm)→0 as n,m→∞. Hence {Txn} is a Cauchy sequence in (X,p) and in (X,ps). Since (X,p) is complete, therefore from Lemma 1.4, (X,ps) is a complete metric space. Hence {Txn} converges to some v∈X with respect to the metric ps, that is,
(2.7)limn→∞ps(Txn,v)=0,
or equivalently,
(2.8)p(v,v)=limn→∞p(Txn,v)=limn,m→∞p(Txn,Txm)=0.
Suppose that T is subsequentially convergent, therefore convergence of {Txn} in (X,ps) implies that {xn} has a convergent subsequence {xni} in (X,ps). So
(2.9)limi→∞ps(xni,u)=0,
for some u∈X. As T is continuous, so (2.9) and Definition 1.6. imply that limi→∞ps(Txni,Tu)=0. From (2.7) and by the uniqueness of the limit in metric space (X,ps), we obtain Tu=v. Consequently,
(2.10)0=p(Tu,Tu)=limi→∞p(Txni,Tu)=limi,j→∞p(Txni,Txnj).(1°)
Iff is a continuous self-map on X, then fxni→fu and Tfxni→Tfu as i→∞. Since Txni→Tu as i→∞, we obtain that Tfu=Tu. As T is injective, so we have fu=u.(2°)
Iff is not continuous then by given assumption we have xn⪯u for all n∈ℕ.  Thus for a subsequence {xni} of {xn} we have xni⪯u and (xni,u)∈Δ. Now,
(2.11)p(Tfu,Tu)≤p(Tfu,Tfxni)+p(Tfxni,Tu)-p(Tfxni,Tfxni)≤a1p(Txni,Tu)+a2p(Txni,Tfxni)+a3p(Tu,Tfu)+a4p(Txni,Tfu)+a5p(Tu,Tfxni)+p(Tfxni,Tu)-p(Tfxni,Tfxni)=a1p(Txni,Tu)+a2p(Txni,Txni+1)+a3p(Tu,Tfu)+a4p(Txni,Tfu)+a5p(Tu,Txni+1)+p(Txni+1,Tu)-p(Txni+1,Txni+1)≤a1p(Txni,Tu)+a2p(Txni,Txni+1)+a3p(Tu,Tfu)+a4[p(Txni,Tu)+p(Tu,Tfu)-p(Tu,Tu)]+a5p(Tu,Txni+1)+p(Txni+1,Tu)-p(Txni+1,Txni+1).On taking limit as i→∞ and applying Remark 1.5. (2) we get
(2.12)p(Tu,Tfu)≤a1p(Tu,Tu)+a2p(Tu,Tu)+a3p(Tu,Tfu)+a4[p(Tu,Tu)+p(Tu,Tfu)-p(Tu,Tu)]+a5p(Tu,Tu)+p(Tu,Tu)-p(Tu,Tu)=(a1+a2+a5)p(Tu,Tu)+(a3+a4)p(Tu,Tfu)≤(a1+a2+a3+a4+a5)p(Tu,Tfu)<p(Tu,Tfu),
which implies that p(Tu,Tfu)=0, and so Tu=Tfu. Now injectivity of T gives u=fu. Following similar arguments to those given above, the result holds when T is sequentially convergent.
Suppose thatFf×Ff⊂Δ. Let w be a fixed point of f. As Ff×Ff⊂Δ, therefore (u,w)∈Δ. From (1.2), we have
(2.13)p(Tu,Tw)=p(Tfu,Tfw)≤a1p(Tu,Tw)+a2p(Tu,Tfu)+a3p(Tw,Tfw)+a4p(Tu,Tfw)+a5p(Tw,Tfu)=a1p(Tu,Tw)+a2p(Tu,Tu)+a3p(Tw,Tw)+a4p(Tu,Tw)+a5p(Tw,Tu)≤(a1+a2+a3+a3+a4+a5)p(Tu,Tw)<p(Tu,Tw),
and hence p(Tu,Tw)=0, which further implies that u=w as T is injective.Example 2.2.
LetX=[0,1] be endowed with usual order and let p be the complete partial metric on X defined by p(x,y)=max{x,y} for all x,y∈X. Let T,f:X→X be defined by Tx=4x/5 and fx=x/4. Note that Δ=X×X. For any (x,y)∈Δ, we have
(2.14)p(Tfx,Tfy)=max{x5,y5}=15max{x,y}≤1235max{x,y}≤27max{4x5,4y5}+17max{4x5,x5}+17max{4y5,y5}+17max{4x5,y5}+17max{4y5,x5}=a1p(Tx,Ty)+a2p(Tx,Tfx)+a3p(Ty,Tfy)+a4p(Tx,Tfy)+a5p(Ty,Tfx).
Therefore, f is a T-Hardly-Rogers contraction with a1=2/7,   a2=a3=a4=a5=1/7. Obviously, T is continuous and sequentially convergent. Thus, all the conditions of Theorem 2.1 are satisfied. Moreover, 0 is the unique fixed point of f.Example 2.3.
LetX=[0,∞) be endowed with usual order and let p be a partial metric on X defined by p(x,y)=max{x,y} for all x,y∈X. Define T,f:X→X by Tx=x2andfx=x/3. Note that Δ=X×X. For any (x,y)∈Δ, we have
(2.15)p(Tfx,Tfy)=max{x29,y29}=19max{x2,y2}≤16max{x2,y2}=a1p(Tx,Ty)≤a1p(Tx,Ty)+a2p(Tx,Tfx)+a3p(Ty,Tfy)+a4p(Tx,Tfy)+a5p(Ty,Tfx).
Therefore, f is a T-Hardly-Rogers contraction with a1=a2=a3=a4=a5=1/6. Also, T is continuous and sequentially convergent. Thus all the conditions of Theorem 2.1 are satisfied. Moreover, 0 is the unique fixed point of f.TakingTx=x in (1.2) and Theorem 2.1, we get the Hardy-Rogers type [32] (and so the Kannan, Chatterjea, and Reich) fixed point theorem on partially ordered partial metric spaces.Corollary 2.4.
Let(X,⪯,p) be a partially ordered partial metric space which is complete. Let f:X→X be a nondecreasing mapping such that for all (x,y)∈Δ, we have
(2.16)p(fx,fy)≤a1p(x,y)+a2p(x,fx)+a3p(y,fy)+a4p(x,fy)+a5p(y,fx),
where ai≥0, i=1,…,5 with a1+a2+a3+a4+a5<1. If there exists x0∈X with x0⪯fx0, and one of the following two conditions is satisfied.(a)
f is a continuous self map on X;(b)
for any nondecreasing sequence{xn} in (X,⪯) with limn→∞ps(z,xn)=0 it follows xn⪯z for all n∈ℕ; then Ff≠ϕ. Moreover, f has a unique fixed point if Ff×Ff⊂Δ.Remark 2.5.
Corollary2.4 corresponds to Theorem 2 of Altun et al. [8] in partially ordered partial metric spaces. For particular choices of the coefficients (ai)i=1,…,5 in Theorem 2.1, we obtain the T-Kannan, T-Chatterjea, and T-Reich type fixed point theorems. Also, Theorem 2.1 is an extension of Theorem 2.1 of Filipović et al. [24] from the cone metric spaces to partial metric spaces.


## 3. Periodic Point Results

Letf:X→X. If the map f satisfies Ff=Ffn for each n∈ℕ, then it is said to have the property P, for more details see [33].Definition 3.1.
Let(X,⪯) be a partially ordered set. A mapping f is called (1) a dominating map on X if x⪯fx for each x in X and (2) a dominated map on X if fx⪯x for each x in X.Example 3.2.
LetX=[0,1] be endowed with usual ordering. Let f:X→X defined by fx=x1/3, then x≤x1/3=fx for all x∈X. Thus f is a dominating map.Example 3.3.
LetX=[0,∞) be endowed with usual ordering. Let f:X→X defined by fx=xn for x∈[0,1) and let fx=xn for x∈[1,∞), for any n∈ℕ, then for all x∈X,   x≤fx that is f is the dominating map. Note that Δ1≠ϕ if f is a dominating or a dominated mapping.
We have the following result.Theorem 3.4.
Let(X,⪯,p) be a partially ordered partial metric space which is complete. Let T:X→X be an injective mapping and f:X→X a nondecreasing such that for all (x,x)∈Δ1, we have
(3.1)p(Tfx,Tf2x)≤λp(Tx,Tfx),
for some λ∈[0,1) and for all x∈X, x≠fx. Then f has the property P provided that Ff is nonempty and f is a dominating map on Ffn.Proof.
Letu∈Ffn for some n>1. Now we show that u=fu. Since f is dominating on Ffn, therefore u⪯fu which further implies that fn-1u⪯fnu as f is nondecreasing. Hence (fn-1u,fn-1u)∈Δ1. Now by using (3.1), we have
(3.2)p(Tu,Tfu)=p(Tffn-1u,Tf2fn-1u)≤λp(Tfn-1u,Tfnu)=λp(Tffn-2u,Tf2fn-2u).
Repeating the above process, we get
(3.3)p(Tu,Tfu)≤λnp(Tu,Tfu).
Taking limit as n→∞, we obtain p(Tu,Tfu)=0 and Tu=Tfu. As T is injective, so u=fu, that is, u∈Ff.Theorem 3.5.
Let(X,⪯,p) be a partially ordered partial metric space which is complete. let  T,f:X→X be mappings satisfy the condition of Theorem 2.1. If f is dominating on X, then f has the property P.Proof.
From Theorem2.1, Ff≠∅. We will prove that (3.1) is satisfied for all (x,x)∈Δ1. Indeed, f is a dominating map so that x⪯fx and also f is nondecreasing so that fx⪯f2x and hence (x,fx)∈Δ. Now from (1.2),
(3.4)p(Tfx,Tf2x)=p(Tfx,Tffx)≤a1p(Tx,Tfx)+a2p(Tx,Tfx)+a3p(Tfx,Tf2x)+a4p(Tx,Tf2x)+a5p(Tfx,Tfx)≤(a1+a2)p(Tx,Tfx)+a3p(Tfx,Tf2x)+a4(p(Tx,Tfx)+p(Tfx,Tf2x)-p(Tfx,Tfx))+a5p(Tfx,Tfx),
that is,
(3.5)(1-a3-a4)p(Tfx,Tf2x)≤(a1+a2+a4)p(Tx,Tfx)+(a5-a4)p(Tfx,Tfx).
Again by using (1.2), we have
(3.6)p(Tf2x,Tfx)=p(Tffx,Tfx)≤a1p(Tfx,Tx)+a2p(Tfx,Tf2x)+a3p(Tx,Tfx)+a4p(Tfx,Tfx)+a5p(Tx,Tf2x)≤(a1+a3)p(Tx,Tfx)+a2p(Tfx,Tf2x)+a4p(Tfx,Tfx)+a5(p(Tx,Tfx)+p(Tfx,Tf2x)-p(Tfx,Tfx)),
which implies that
(3.7)(1-a2-a5)p(Tfx,Tf2x)≤(a1+a3+a5)p(Tx,Tfx)+(a4-a5)p(Tfx,Tfx).
Summing (3.5) and (3.7) implies p(Tfx,Tf2x)≤λp(Tx,Tfx),λ=(2a1+a2+a3+a4+a5)/(2-a2-a3-a4-a5). Obviously, λ∈[0,1). By Theorem 3.4, f has the property P.



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