# Fixed Points forψ-Graphic Contractions with Application to Integral Equations

Authors: N. Hussain; S. Al-Mezel; P. Salimi
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

The aim of this paper is to define modified weakα-ψ-contractive mappings and to establish fixed point results for such mappings defined on partial metric spaces using the notion of triangular α-admissibility. As an application, we prove new fixed point results for graphic weak ψ-contractive mappings. Moreover, some examples and an application to integral equation are given here to illustrate the usability of the obtained results.


---

## Body

## 1. Introduction and Preliminaries

The concept of partial metric space was introduced by Matthews [1] in 1994. Partial metric space is a generalized metric space in which each object does not necessarily have to have a zero distance from itself. 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 [2, 3]. Subsequently, several authors studied the problem of existence and uniqueness of a fixed point for mappings satisfying different contractive conditions on partial metric spaces (e.g., [4–10]). For a recent survey on the existence of fixed points in different spaces with generalized distance functions, the reader may check [11–18]. We start by recalling some definitions and properties of partial metric spaces.Definition 1.
Apartial metricon a nonempty set X is a function 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,z)≤p(x,y)+p(y,z)-p(y,y).
A partial metric space is a pair(X,p) such that X is nonempty set and p is a partial metric on X.From the above definition, ifp(x,y)=0, then x=y. But if x=y, p(x,y) may not be 0 in general. 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}. For more examples of partial metric spaces, we refer to [6, 7].Each partial metricp on X generates a T0 topology τp on X which has as a base the family of open p-balls {Bp(x,ϵ):x∈X,ϵ>0}, where Bp(x,ϵ)={y∈X:p(x,y)<p(x,x)+ϵ} for all x∈X and ϵ>0. A sequence {xn} in X converges to a point x∈X, with respect to τp, if and only if p(x,x)=limn→∞p(x,xn). A sequence {xn} in X is called Cauchy sequence if limn,m→∞p(xn,xm) exists and is finite. A partial metric space (X,p) is said to be complete if every Cauchy sequence {xn} in X converges, with respect to τp, to a point x∈X such that p(x,x)=limn,m→∞p(xn,xm).A sequence{xn} in X is called 0-Cauchy sequence if limn,m→∞p(xn,xm)=0. A partial metric space (X,p) is said to be 0-complete if every 0-Cauchy sequence {xn} in X converges, with respect to τp, to a point x∈X such that p(x,x)=0.The partial metric space(ℚ∩[0;∞),p), where ℚ denotes the set of rational numbers and the partial metric p is given by p(x,y)=max{x,y}, provides an example of a 0-complete partial metric space which is not complete.Ifp is a partial metric on X, then the function ps:X×X→ℝ+ given by
(1)ps(x,y)=2p(x,y)-p(x,x)-p(y,y)
is a metric on X. A partial metric space (X,p) is complete if and only if the metric space (X,ps) is complete. Furthermore, limn→∞ps(xn,z)=0 if and only if p(z,z)=limn→∞p(xn,z)=limn,m→∞p(xn,xm). Some more properties of partial metric spaces can be found in [3, 7, 19].Samet et al. [20] defined the notion of α-admissible mappings and proved the following result.Definition 2.
LetT be a self-mapping on X and α:X×X→[0,+∞) a function. One says that T is an α-admissible mapping if
(2)x,y∈X,α(x,y)≥1⟹α(Tx,Ty)≥1.Denote withΨ the family of nondecreasing functions ψ:[0,+∞)→[0,+∞) such that ∑n=1∞‍ψn(t)<+∞ for all t>0, where ψn is the nth iterate of ψ.Theorem 3.
Let(X,d) be a complete metric space and T be α-admissible mapping. Assume that
(3)α(x,y)d(Tx,Ty)≤ψ(d(x,y)),
where ψ∈Ψ. Also, suppose that the following assertions hold:(i)
there existsx0∈X such that α(x0,Tx0)≥1;(ii)
eitherT is continuous or for any sequence {xn} in X with α(xn,xn+1)≥1 for all n∈ℕ∪{0} and xn→x as n→+∞, one has α(xn,x)≥1 for all n∈ℕ∪{0}.
Then,T has a fixed point.Very recently, Salimi et al. [21] modified the notions of α-ψ-contractive mappings and α-admissible mappings as follows.Definition 4 (see [21]).
LetT be a self-mapping on X and α,η:X×X→[0,+∞) two functions. One says that T is an α-admissible mapping with respect to η if
(4)x,y∈X,α(x,y)≥η(x,y)⟹α(Tx,Ty)≥η(Tx,Ty).
Note that if we take η(x,y)=1, then this definition reduces to Definition 2. Also, if we take, α(x,y)=1, then we say that T is an η-subadmissible mapping.The following result properly contains Theorem3 and Theorems 2.3 and 2.4 of [22].Theorem 5 (see [21]).
Let(X,d) be a complete metric space and T be α-admissible mapping with respect to η. Assume that
(5)x,y∈X,α(x,y)≥η(x,y)⟹d(Tx,Ty)≤ψ(M(x,y)),
where ψ∈Ψ and
(6)M(x,y)=max{d(x,y),d(x,Tx)+d(y,Ty)2,d(x,Ty)+d(y,Tx)2}.
Also, suppose that the following assertions hold:(i)
there existsx0∈X such that α(x0,Tx0)≥η(x0,Tx0);(ii)
eitherT is continuous or for any sequence {xn} in X with α(xn,xn+1)≥η(xn,xn+1) for all n∈ℕ∪{0} and xn→x as n→+∞, one has α(xn,x)≥η(xn,x) for all n∈ℕ∪{0}. Then, T has a fixed point.In fact, Banach contraction principle and Theorem5 hold for the following example, but Theorem 3 does not hold.Example 6 (see [21]).
LetX=[0,∞) be endowed with the usual metric d(x,y)=|x-y| for all x,y∈X, and let T:X→X be defined by Tx=(1/4)x. Also, define, α:X2→[0,∞) by α(x,y)=3 and ψ:[0,∞)→[0,∞) by ψ(t)=(1/2)t.In this paper, we define modified weakα-ψ-contractive mappings and establish fixed point results for such mappings defined on ordinary as well as ordered partial metric spaces using the notion of triangular α-admissibility. As an application, we prove new fixed point results for graphic weak ψ-contractive mappings. Moreover, some examples and an application to integral equation are given here to illustrate the usability of the obtained results.


## 2. Modified Weakα-ψ-Contractions

Recently, Karapınar et al. [23] introduced the notion of triangular α-admissible mapping as follows.Definition 7 (see [23]).
LetT:X→X and α:X×X→(-∞,+∞). One says that T is a triangular α-admissible mapping if(T1)
α(x,y)≥1 implies α(Tx,Ty)≥1,x,y∈X;(T2)
α(x,z)≥1,α(z,y)≥1 imply α(x,y)≥1.Lemma 8 (see [23]).
LetT:X→X be a triangular α-admissible mapping. Assume that there exists x0∈X such that α(x0,Tx0)≥1. Define sequence {xn} by xn=Tnx0. Then,
(7)α(xm,xn)≥1∀m,n∈ℕwithm<n.Motivated by Samet et al. [20] and Salimi et al. [21], we introduce the following mapping.Definition 9.
Let(X,p) be a partial metric space and T:X→X, α:X×X→[0,∞) two mappings. If there exists an upper semicontinuous from the right nondecreasing function ψ:ℝ+→ℝ+ with ψ(t)<t for all t>0 such that
(8)x,y∈X,α(x,y)≥1⟹p(Tx,Ty)≤ψ(Mp(x,y)),
where
(9)Mp(x,y)=max{p(x,Ty)+p(y,Tx)2p(x,y),p(x,Tx),p(y,Ty),p(x,Ty)+p(y,Tx)2}.
Then, we say that T is a modified weak α-ψ-contractive mapping.Definition 10.
Let(X,p) be a partial metric space. Let α:X×X→(-∞,+∞) and T:X→X. One says that T is an α-continuous function on (X,ps) if for given x and {xn} in X,
(10)xn⟶xasn⟶∞,α(xn,xn+1)≥1∀n∈ℕimplyTxn⟶Tx.Example 11.
LetX=[0,∞) and p(x,y)=max{x,y}. Assume that T:X→X and α:X2→(-∞,+∞) defined by
(11)Tx={3x2,ifx∈[0,1]5x+1,if(1,∞),α(x,y)={1,ifx∈[0,1]-1,otherwise.
Clearly, T is not continuous, but T is an α-continuous on (X,ps). Indeed, if xn→x as n→∞ and α(xn,xn+1)≥1, then xn∈[0,1] and so limn→∞Txn=limn→∞3xn2=3x2=Tx.Theorem 12.
Let(X,p) be a 0-complete partial metric space and T a modified weak α-ψ-contractive and triangular α-admissible mapping. Suppose that the following assertions hold:(i)
there existsx0∈X such that α(x0,Tx0)≥1;(ii)
T is an α-continuous function on (X,ps). Then, T has a fixed point.Proof.
Letx0∈X such that α(x0,Tx0)≥1. Define a sequence {xn} in X by xn=Tnx0=Txn-1 for all n∈ℕ. Then, by Lemma 8, we have
(12)α(xm,xn)≥1∀m,n∈ℕwithm<n.
If xn+1=xn for some n∈ℕ, then x=xn is a fixed point for T, and the result is proved. Hence, we suppose that xn+1≠xn for all n∈ℕ. Now, by (8) and (12) with x=xn-1, y=xn, we get
(13)p(Txn-1,Txn)≤ψ(M(xn-1,xn))
On the other hand,
(14)max{p(xn-1,xn),p(xn,xn+1)}≤Mp(xn-1,xn)=max{p(xn-1,Txn)+p(xn,Txn-1)2p(xn-1,xn),p(xn-1,Txn-1),p(xn,Txn),p(xn-1,Txn)+p(xn,Txn-1)2}=max{p(xn-1,xn+1)+p(xn,xn)2p(xn-1,xn),p(xn-1,xn),p(xn,xn+1),p(xn-1,xn+1)+p(xn,xn)2}≤max{p(xn-1,xn)+p(xn,xn+1)2p(xn-1,xn),p(xn-1,xn),p(xn,xn+1),p(xn-1,xn)+p(xn,xn+1)2}≤max{p(xn-1,xn),p(xn,xn+1)}
and then, Mp(xn-1,xn)=max{p(xn-1,xn),p(xn,xn+1)}. So, from (13), we have
(15)p(xn,xn+1)≤ψ(max{p(xn-1,xn),p(xn,xn+1)}).
Now, if max{p(xn-1,xn),p(xn,xn+1)}=p(xn,xn+1) for some n∈ℕ, then
(16)p(xn,xn+1)≤ψ(max{p(xn-1,xn),p(xn,xn+1)})=ψ(p(xn,xn+1))<p(xn,xn+1),
which is a contradiction. Hence, for all n∈ℕ, we have
(17)p(xn,xn+1)≤ψ(p(xn-1,xn))<p(xn-1,xn).
This implies that the sequence{p(xn-1,xn)} is decreasing, and so, by (17), there is s≥0 such that
(18)limn→+∞p(xn,xn+1)=limn→+∞ψ(p(xn-1,xn))=s.
Now, we show that s must be equal to 0.
In fact, ifs>0, then we get
(19)s=limsupn→+∞ψ(p(xn-1,xn))≤ψ(s)<s,
which is a contradiction. Hence,
(20)limn→+∞p(xn,xn+1)=0.
We prove that {xn} is a 0-Cauchy sequence. Suppose, to the contrary, that {xn} is not a 0-Cauchy sequence. Then, there is ɛ>0 and sequences {m(k)} and {n(k)} such that, for all positive integers k,
(21)n(k)>m(k)>k,p(xn(k),xm(k))≥ɛ,p(xn(k),xm(k)-1)<ɛ.
Now, for all k∈ℕ, we have
(22)ɛ≤p(xn(k),xm(k))≤p(xn(k),xm(k)-1)+p(xm(k)-1,xm(k))<ɛ+p(xm(k)-1,xm(k)).
Taking the limit as k→+∞ in the above inequality and using (20), we get
(23)limk→+∞p(xn(k),xm(k))=ɛ.
Again, from
(24)p(xn(k),xm(k))≤p(xm(k),xm(k)+1)+p(xm(k)+1,xn(k)+1)+p(xn(k)+1,xn(k))p(xn(k)+1,xm(k)+1)≤p(xm(k),xm(k)+1)+p(xm(k),xn(k))+p(xn(k)+1,xn(k)),
taking the limit as k→+∞ and by (20) and (23), we deduce
(25)limk→+∞p(xn(k)+1,xm(k)+1)=ɛ.
Also, since
(26)ɛ≤p(xn(k),xm(k))≤p(xn(k),xm(k)+1)+p(xm(k)+1,xm(k)),
then by taking limit as n→∞ in the last inequality and applying (20) and (23), we deduce
(27)limk→+∞p(xn(k),xm(k)+1)=ɛ.
Similarly,
(28)limk→+∞p(xm(k),xn(k)+1)=ɛ.
Then, by (8) and (12), we obtain
(29)p(xn(k)+1,xm(k)+1)=p(Txn(k),Txm(k))≤ψ(Mp(xn(k),xm(k))),
where
(30)Mp(xn(k),xm(k))=max{p(xn(k),xm(k)+1)+p(xm(k),xn(k)+1)2p(xn(k),xm(k)),p(xn(k),Txn(k)),p(xm(k),Txm(k)),p(xn(k),Txm(k))+p(xm(k),Txn(k))2}=max{p(xn(k),xm(k)+1)+p(xm(k),xn(k)+1)2p(xn(k),xm(k)),p(xn(k),xn(k)+1),p(xm(k),xm(k)+1),p(xn(k),xm(k)+1)+p(xm(k),xn(k)+1)2}.
Taking limit supremum as n→∞ in the above inequality and applying (20), (23), (25), (27), and (28), we get
(31)ɛ≤ψ(ɛ)<ɛ,
which is a contradiction. Hence, {xn} is a 0-Cauchy sequence.
SinceT is orbitally α-continuous on (X,ps), xn→z as n→∞ and α(xn,xn+1)≥1, then we have
(32)Tz=limn→∞Txn=limn→∞xn+1=z.
So, z is a fixed point of T.Theorem 13.
Let(X,p) be a 0-complete partial metric space and T:X→X a modified weak α-ψ-contractive and α-admissible mapping. Suppose that the following assertions hold:(i)
there existsx0∈X such that α(x0,Tx0)≥1;(ii)
α(x,x)≥1 for all x∈X and if {xn} is a sequence in X such that α(xn,xn+1)≥1 for all n∈ℕ and xn→x as n→+∞, then α(xn,x)≥1 for all n∈ℕ. Then, T has a fixed point.Proof.
Condition (ii) implies property (T2) in definition of triangularα-admissible map. Indeed, if α(x,y)≥1 and α(y,z)≥1, then applying (ii) to (xn) defined by
(33)x1≔x,x2≔y,xn≔zforn≥3,
we get α(xn,z)≥1 for n∈ℕ, and hence, α(x,z)≥1. Thus, as in Theorem 12, we obtain a 0-Cauchy sequence {xn} such that xn→z as n→∞. Since,
(34)α(xn,xn+1)≥1
for all n∈ℕ and xn→z as n→∞, then from (ii), we have
(35)α(xn,z)≥1
for all n∈ℕ. Then, from (8), we get
(36)p(xn+1,Tz)≤ψ(Mp(xn,z)),
where
(37)Mp(xn,z)=max{p(xn,Tz)+p(z,xn+1)2p(xn,z),p(xn,xn+1),p(z,Tz),p(xn,Tz)+p(z,xn+1)2}.
By taking limit supremum as n→∞ in the above inequality, we have
(38)p(z,Tz)=limsupn→∞p(xn+1,Tz)≤limsupn→∞ψ(Mp(xn,z))≤ψ(limsupn→∞Mp(xn,z))=ψ(p(z,Tz))<p(z,Tz),
which is a contradiction. Hence, p(z,Tz)=0. That is, z=Tz.Example 14.
LetX=[0,∞) be endowed with the partial metric p(x,y)=max{x,y} for all x,y∈X, and let T:X→X be defined by
(39)Tx={12x3ifx∈[0,1]x3+2lnxifx∈(1,∞).
Define α:X×X→(-∞,+∞) and ψ:[0,∞)→[0,∞) by
(40)α(x,y)={2ifx,y∈[0,1]orx=y-2otherwiseψ(t)=12t.
Clearly,(X,p) is a 0-complete partial metric space. We show that T is a triangular α-admissible mapping. Let x,y∈X, if α(x,y)≥1, then x,y∈[0,1] or x=y. On the other hand, for all x∈[0,1], we have Tx≤1. It follows that α(Tx,Ty)≥1. Also, if α(x,z)≥1, and α(z,y)≥1 then x,y,z∈[0,1]. That is, α(x,y)≥1. Hence, the assertion holds. In reason of the above arguments, α(0,T0)≥1.
Now, if{xn} is a sequence in X such that α(xn,xn+1)≥1 for all n∈ℕ∪{0} and xn→x as n→+∞, then {xn}⊂[0,1], and hence, x∈[0,1]. This implies that α(xn,x)≥1 for all n∈ℕ.
Letα(x,y)≥1. Then, x,y∈[0,1]. We get
(41)p(Tx,Ty)=12max{x3,y3}≤12max{x,y}≤12Mp(x,y)=ψ(Mp(x,y)).
That is,
(42)α(x,y)≥1⟹p(Tx,Ty)≤ψ(Mp(x,y)).
Hence, all conditions of Theorem 13 hold and 0 is a fixed point of T.Corollary 15.
Let(X,p) be a 0-complete partial metric space and T:X→X be such that α(x0,Tx0)≥1 for some x0∈X. Assume that
(43)|α(x,y)|p(Tx,Ty)≤ψ(Mp(x,y))
hold for all x,y∈X. Also, suppose that one of the following assertions holds:(i)
T is triangular α-admissible and α-continuous on (X,ps);(ii)
T is α-admissible, α(x,x)≥1 for all x∈X and if {xn} is a sequence in X such that α(xn,xn+1)≥1 for all n∈ℕ and xn→x as n→+∞, then α(xn,x)≥1 for all n∈ℕ. Then, T has a fixed point.Corollary 16.
Let(X,p) be a 0-complete partial metric space and T:X→X be such that α(x0,Tx0)≥1 for some x0∈X. Assume that
(44)(|α(x,y)|+ℓ)p(Tx,Ty)≤(1+ℓ)ψ(Mp(x,y))
hold for all x,y∈X where ℓ>0. Also, suppose that one of the following assertions holds:(i)
T is triangular α-admissible and α-continuous on (X,ps);(ii)
T is α-admissible, α(x,x)≥1 for all x∈X and if {xn} is a sequence in X such that α(xn,xn+1)≥1 for all n∈ℕ and xn→x as n→+∞, then α(xn,x)≥1 for all n∈ℕ. Then, T has a fixed point.Proof.
Letα(x,y)≥1. Then, |α(x,y)|≥1. Hence, by (44), we have
(45)(1+ℓ)p(Tx,Ty)≤(|α(x,y)|+ℓ)p(Tx,Ty)≤(1+ℓ)ψ(Mp(x,y)).
Thus, p(Tx,Ty)≤ψ(Mp(x,y)). Hence, conditions of Corollary 15 hold, and T has a fixed point.Similarly, we have the following corollary.Corollary 17.
Let(X,p) be a 0-complete partial metric space and T:X→X be such that α(x0,Tx0)≥1 for some x0∈X. Assume that
(46)(p(Tx,Ty)+ℓ)|α(x,y)|≤ψ(Mp(x,y))+ℓ
hold for all x,y∈X where ℓ>0. Also, suppose that one of the following assertions holds:(i)
T is triangular α-admissible and α-continuous on (X,ps);(ii)
T is α-admissible, α(x,x)≥1 for all x∈X and if {xn} is a sequence in X such that α(xn,xn+1)≥1 for all n∈ℕ and xn→x as n→+∞, then α(xn,x)≥1 for all n∈ℕ.
Then,T has a fixed point.Corollary 18 (Matthews [1]).
Let(X,p) be a 0-complete partial metric space and T:X→X a given mapping satisfying
(47)p(Tx,Ty)≤kp(x,y)
for all x,y∈X, where k∈[0,1). Then, T has a unique fixed point.


## 3. Fixed Point Results in Partially Ordered Partial Metric Spaces

Fixed point theorems for monotone operators in partially ordered metric spaces are widely investigated and have found various applications in differential and integral equations (see [24, 25] and references therein).Theorem 19 (see [24]).
Let(X,d,⪯) be a partially ordered complete metric space and T:X→X a continuous increasing self-mapping such that x0⪯Tx0 for some x0∈X. Assume that
(48)d(Tx,Ty)≤rd(x,y)
hold for all x,y∈X with x⪯y, where 0≤r<1. Then, T has a fixed point.Theorem 20.
Let(X,p,⪯) be a partially ordered 0-complete partial metric space and T:X→X an increasing self-mapping such that x0⪯Tx0 for some x0∈X. Assume that
(49)p(Tx,Ty)≤ψ(Mp(x,y))
hold for all x,y∈X with x⪯y. Now, if T is a continuous mapping on (X,ps), then T has a fixed point.Proof.
Define,α:X2→(-∞,+∞) by
(50)α(x,y)={1,ifx⪯y0,otherwise.
At first we prove that T is a triangular α-admissible mapping. Let α(x,y)≥1, then x⪯y. As T is an increasing mapping, we have Tx⪯Ty. That is, α(Tx,Ty)≥1. Also, let α(x,z)≥1 and α(z,y)≥1, then x⪯z and z⪯y. So, from transitivity, we have x⪯y. That is, α(x,y)≥1. Thus, T is a triangular α-admissible mapping. Also, there exists x0∈X such that x0⪯Tx0 which implies α(x0,Tx0)≥1. Let α(x,y)≥1, then x⪯y. Now, from (49), we have p(Tx,Ty)≤ψ(Mp(x,y)). That is,
(51)α(x,y)≥1⟹p(Tx,Ty)≤ψ(Mp(x,y)).
Hence, all conditions of Theorem 12 are satisfied, and T has a fixed point.Theorem 21 (see [25]).
Let(X,d,⪯) be a partially ordered complete metric space and T:X→X an increasing mapping such that
(52)d(Tx,Ty)≤rd(x,y)
for all x,y∈X with x⪯y, where 0≤r<1. Suppose that the following assertions hold:(i)
there existsx0∈X such that x0⪯Tx0;(ii)
if{xn} is a sequence in X such that xn⪯xn+1 for all n∈ℕ and xn→x as n→+∞, then xn⪯x for all n∈ℕ.Then, T has a fixed point.Theorem 22.
Let(X,p,⪯) be a partially ordered 0-complete partial metric space and T:X→X an increasing mapping such that
(53)p(Tx,Ty)≤ψ(Mp(x,y))
for all x,y∈X with x⪯y. Suppose that the following assertions hold:(i)
there existsx0∈X such that x0⪯Tx0;(ii)
if{xn} is a sequence in X such that xn⪯xn+1 for all n∈ℕ and xn→x as n→+∞, then xn⪯x for all n∈ℕ.Then, T has a fixed point.Proof.
Defineα:A×A→(-∞,+∞) as in proof of Theorem 20. Assume α(xn,xn+1)≥1 for all n∈ℕ such that xn→x as n→∞. Then, xn⪯xn+1 for all n∈ℕ. Hence, by (ii), we get xn⪯x for all n∈ℕ and so α(xn,x)≥1 for all n∈ℕ. Proceeding as in proof of Theorem 20, we can deduce that T is a modified weak α-ψ-contractive and α-admissible mapping and there exists x0∈X such that α(x0,Tx0)≥1. Hence, all conditions of Theorem 13 hold, and T has a fixed point.Remark 23.
Similarly, we may obtain more fixed point results on ordered partial metric spaces as immediate consequences of Corollaries15–17.


## 4. Fixed Point Results for Graphic Contractions

Consistent with Jachymski [26], let (X,p) be a partial metric space, and let Δ denote the diagonal of the Cartesian product X×X. Consider a directed graph G such that the set V(G) of its vertices coincides with X and the set E(G) of its edges contains all loops, that is, E(G)⊇Δ. We assume that G has no parallel edges, so we can identify G with the pair (V(G),E(G)). Moreover, we may treat G as a weighted graph (see [26]) by assigning to each edge the distance between its vertices. If x and y are vertices in a graph G, then a path in G from x to y of length N(N∈ℕ) is a sequence {xi}i=0N of N+1 vertices such that x0=x, xN=y and (xn-1,xn)∈E(G) for i=1,…,N. A graph G is connected if there is a path between any two vertices. G is weakly connected if G~ is connected (see for details [11, 26–28]).Recently, some results have appeared providing sufficient conditions for a mapping to be a Picard operator if(X,d) is endowed with a graph. The first result in this direction was given by Jachymski [26].Definition 24 (see [26]).
One says that a mappingT:X→X is a Banach G-contraction or simply G-contraction if T preserves edges of G, that is,
(54)∀x,y∈X(x,y)∈E(G)⟹(T(x),T(y))∈E(G),
and T decreases weights of edges of G in the following way:
(55)∃α∈(0,1),∀x,y∈X(x,y)∈E(G)⟹d(T(x),T(y))≤αd(x,y).Definition 25 (see [26]).
A mappingT:X→X is called G-continuous if given x∈X and sequence {xn}(56)xn⟶xasn⟶∞,(xn,xn+1)∈E(G)∀n∈ℕimplyTxn⟶Tx.Definition 26.
Let(X,p) be a partial metric space endowed with a graph G and T:X→X a self-mapping. If there exists an upper semicontinuous from the right function ψ:ℝ+→ℝ+ with ψ(t)<t for all t>0 such that
(57)∀x,y∈X,(x,y)∈E(G)⟹(T(x),T(y))∈E(G)∀x,y∈X,(x,y)∈E(G),⟹p(Tx,Ty)≤ψ(Mp(x,y)),
where
(58)Mp(x,y)=max{p(x,Ty)+p(y,Tx)2p(x,y),p(x,Tx),p(y,Ty),p(x,Ty)+p(y,Tx)2}.
Then, T is called a weak ψ-graphic contractive mapping.Theorem 27.
Let(X,p) be a 0-complete partial metric space endowed with a graph G and T:X→X a weak ψ-graphic contractive mapping. Suppose that the following assertions hold:(i)
there existsx0∈X such that (x0,Tx0)∈E(G);(ii)
T is G-continuous on (X,ps);(iii)
(x,z)∈E(G) and (z,y)∈E(G) imply (x,y)∈E(G) for all x,y,z∈X, that is, E(G) is a quasi-order [26]. Then, T has a fixed point.Proof.
Defineα:X2→(-∞,+∞) by
(59)α(x,y)={1,if(x,y)∈E(G)0,otherwise.
At first we prove that T is a triangular α-admissible mapping. Let α(x,y)≥1, then (x,y)∈E(G). As T is a weak ψ-graphic contractive mapping, we have (Tx,Ty)∈E(G). That is, α(Tx,Ty)≥1. Also, let α(x,z)≥1 and α(z,y)≥1, then (x,z)∈E(G) and (z,y)∈E(G). So, from (iii), we have (x,y)∈E(G). That is, α(x,y)≥1. Thus, T is a triangular α-admissible mapping. Let T be G-continuous on (X,ps). Then,
(60)xn⟶xasn⟶∞,(xn,xn+1)∈E(G)∀n∈ℕimplyTxn⟶Tx.
That is,
(61)xn⟶xasn⟶∞,α(xn,xn+1)≥1∀n∈ℕimplyTxn⟶Tx
which implies T is α-continuous on (X,ps). From (i), there exists x0∈X such that (x0,Tx0)∈E(G). That is, α(x0,Tx0)≥1.
Ifα(x,y)≥1, then (x,y)∈E(G). Now, since T is a weak ψ-graphic contractive mapping, so p(Tx,Ty)≤ψ(Mp(x,y)). That is,
(62)α(x,y)≥1⟹p(Tx,Ty)≤ψ(Mp(x,y)).
Hence, all conditions of Theorem 12 are satisfied, and T has a fixed point.IfG is a connected graph, then condition (iii) of Theorem 27 is automatically satisfied. Thus, we have the following result.Corollary 28.
Let(X,p) be a 0-complete partial metric space endowed with a graph G and T a weak ψ-graphic contractive mapping. Suppose that the following assertions hold:(i)
there existsx0∈X such that (x0,Tx0)∈E(G);(ii)
T is G-continuous on (X,ps);(iii)
G is a connected graph. Then, T has a fixed point.Theorem 29.
Let(X,p) be a 0-complete partial metric space endowed with a graph G and T a weak ψ-graphic contractive mapping. Suppose that the following assertions hold:(i)
there existsx0∈X such that (x0,Tx0)∈E(G);(ii)
if{xn} is a sequence in X such that (xn,xn+1)∈E(G) for all n∈ℕ and xn→x as n→+∞, then (xn,x)∈E(G) for all n∈ℕ. Then, T has a fixed point.Proof.
Define,α:X2→(-∞,+∞) as in proof of Theorem 27. Condition (ii) implies that E(G) is a quasi-order, that is, (x,z)∈E(G) and (z,y)∈E(G) imply (x,y)∈E(G) for all x,y,z∈X (see Remark 3.1 [26]). Let, xn→x as n→∞ and α(xn,xn+1)≥1 for all n∈ℕ. Then, {xn} is a sequence in X such that (xn,xn+1)∈E(G) for all n∈ℕ and xn→x as n→+∞. So, by (ii), we have (xn,x)∈E(G) for all n∈ℕ. That is, α(xn,x)≥1. All other conditions of Theorem 13, follow similarly as in proof of Theorem 27 and consequently T has a fixed point.Theorem  3.2(2°) in [26] and Theorem 2.3(2) in [29] are extended to weak ψ-graphic contractive maps defined on a 0-complete partial metric space as follows.Corollary 30.
Let(X,p) be a 0-complete partial metric space endowed with a graph G and T a weak ψ-graphic contractive mapping. Suppose that the following assertions hold:(i)
there existsx0∈X such that (x0,Tx0)∈E(G);(iis)
E(G) is a quasi-order and if {xn} is a sequence in X such that (xn,xn+1)∈E(G) for all n∈ℕ and xn→x as n→+∞, then there is a subsequence {xkn} with (xkn,x)∈E(G) for all n∈ℕ.Then, T has a fixed point.Proof.
Condition (iis) implies that of(ii) in Theorem 29 (see Remark 3.1 [26]). Now, the conclusion follows from Theorem 29.Corollary 31.
Let(X,p) be a 0-complete partial metric space and ϵ-chainable for some ϵ>0, that is, given x,y∈X, there is N∈ℕ and a sequence {xi}i=0N such that x0=x, xN=y and p(xi-1,xi)<ϵ for i=1,…,N. Suppose that T:X→X is a mapping satisfying
(63)∀x,y∈X,p(x,y)<ϵ,⟹p(Tx,Ty)≤ψ(p(x,y)).
Then, T has a fixed point.Proof.
Consider the graphG with V(G)=X and E(G)={(x,y)∈X×X:p(x,y)<ϵ}. Then, ϵ-chainability of (X,p) means G is connected, and hence, E(G) is quasi-order. If (x,y)∈E(G), then
(64)p(Tx,Ty)≤ψ(p(x,y))<p(x,y)<ϵ,
so (Tx,Ty)∈E(G), hence T is a (G,ψ)-contraction. Let {xn} be in X with xn→x, then p(xn,x)<ϵ for sufficiently large n, so there is {xkn} such that (xkn,x)∈E(G). Thus, by Corollary 30, T has a fixed point.Definition 32.
Let(X,p,⪯) be a partially ordered partial metric space endowed with a graph G and T:X→X a self-mapping. If there exists an upper semicontinuous from the right function ψ:ℝ+→ℝ+ with ψ(t)<t for all t>0 such that, for all x,y∈X, (x,y)∈E(G) with x⪯y⇒(T(x),T(y))∈E(G), where Tx⪯Ty(65)(x,y)∈E(G)withx⪯y,⟹p(Tx,Ty)≤ψ(Mp(x,y)),
where
(66)Mp(x,y)=max{p(x,Ty)+p(y,Tx)2p(x,y),p(x,Tx),p(y,Ty),p(x,Ty)+p(y,Tx)2}.
Then, we say T is an ordered weak ψ-graphic contractive mapping.Theorem 33.
Let(X,p,⪯) be a partially ordered 0-complete partial metric space endowed with a graph G and T an ordered weak ψ-graphic contractive mapping. Suppose that the following assertions hold:(i)
there existsx0∈X such that (x0,Tx0)∈E(G) with x0⪯Tx0,(ii)
eitherT is G-continuous in (X,ps) and (x,z)∈E(G) and (z,y)∈E(G) imply (x,y)∈E(G) or;(iii)
if{xn} is a sequence in X such that (xn,xn+1)∈E(G) with xn⪯xn+1 for all n∈ℕ∪{0} and xn→x as n→+∞, then (xn,x)∈E(G) with xn⪯x for all n∈ℕ∪0. Then, T has a fixed point.Proof.
Defineα:X2→(-∞,+∞) by
(67)α(x,y)={1,if(x,y)∈E(G)withx⪯y0,otherwise.
At first, we prove that T is a triangular α-admissible mapping. Let α(x,y)≥1, then (x,y)∈E(G) with x⪯y. As T is an ordered weak ψ-graphic contractive mapping, we have (Tx,Ty)∈E(G) where Tx⪯Ty. That is, α(Tx,Ty)≥1. Also, let α(x,z)≥1 and α(z,y)≥1, then (x,z)∈E(G) with x⪯z and (z,y)∈E(G) with z⪯y. So from (ii), we have (x,y)∈E(G). Also, x⪯z and z⪯y implies x⪯y. Hence, α(x,y)≥1. Thus, T is a triangular α-admissible mapping. Let T be G-continuous on (X,ps). Then,
(68)xn⟶xasn⟶∞,(xn,xn+1)∈E(G)∀n∈ℕimplyTxn⟶Tx.
That is,
(69)xn⟶xasn⟶∞,α(xn,xn+1)≥1∀n∈ℕimplyTxn⟶Tx
which implies T is α-continuous on (X,ps). From (i), there exists x0∈X such that (x0,Tx0)∈E(G). That is, α(x0,Tx0)≥1.
Letα(x,y)≥1, then (x,y)∈E(G) with x⪯y. Now, since T is an ordered weak ψ-graphic contractive mapping, then p(Tx,Ty)≤ψ(Mp(x,y)). That is,
(70)α(x,y)≥1⟹p(Tx,Ty)≤ψ(Mp(x,y)).
Hence, all conditions of Theorem 12 (or 13) are satisfied, and T has a fixed point.Remark 34.
All our results established above are new even in the setting of complete metric spaces.


## 5. Application to Existence of Solutions of Integral Equations

Fixed point theorems for monotone operators in ordered metric spaces are widely investigated and have found various applications in differential and integral equations (see [30, 31] and references therein). In this section, we apply our result to the existence of a solution of an integral equation. Let X=C([0,T],ℝ) be the set of real continuous functions defined on [0,T], and let d:X×X→ℝ+ be defined by
(71)d(x,y)=∥x-y∥∞
for all x,y∈X. Then, (X,d) is a complete metric space. Also, assume this metric space endowed with a graph G.Consider the integral equation(72)x(t)=p(t)+∫0T‍S(t,s)f(s,x(s))ds
and let F:X→X be defined by
(73)F(x)(t)=p(t)+∫0T‍S(t,s)f(s,x(s))ds.
We assume that (A)
f:[0,T]×ℝ→ℝ is continuous;(B)
p:[0,T]→ℝ is continuous;(C)
S:[0,T]×ℝ→[0,+∞) is continuous;(D)
there exists an upper semicontinuous from the right nondecreasing functionψ:ℝ+→ℝ+ with ψ(t)<t for all t>0 such that for all s∈[0,T],
(74)∀x,y∈X(x,y)∈E(G)⟹(F(x),F(y))∈E(G)∀x,y∈X(x,y)∈E(G)⟹0≤f(s,x(s))-f(s,y(s))≤ψ(max{12|x(s)-y(s)|,|x(s)-F(x(s))|,|y(s)-F(y(s))|,12[|x(s)-F(y(s))|+|y(s)-F(x(s))|]12});(F)
there existx0∈X such that (x0,F(x0))∈E(G);(G)
if{xn} is a sequence in X such that (xn,xn+1)∈E(G) for all n∈ℕ and xn→x as n→+∞, then (xn,x)∈E(G) for all n∈ℕ;(H)
assume(75)supt∈[0,T]∫0T‍S(t,s)ds≤1.Theorem 35.
Under assumptions(A)–(H), the integral equation (72) has a solution in X=C([0,T],ℝ).Proof.
Consider the mappingF:X→X defined by (73). Let (x,y)∈E(G). Then, from (D), we deduce
(76)|F(x)(t)-F(y)(t)|=|∫0T‍S(t,s)[f(s,x(s))-f(s,y(s))]ds|≤∫0T‍S(t,s)|f(s,x(s))-f(s,y(s))|ds≤ψ(max{12|x(s)-y(s)|,|x(s)-F(x(s))|,|y(s)-F(y(s))|,12[|x(s)-F(y(s))|+|y(s)-F(x(s))|]12})≤ψ(max{12∥x(s)-y(s)∥,∥x(s)-F(x(s))∥,∥y(s)-F(y(s))∥,12[∥x(s)-F(y(s))∥+∥y(s)-F(x(s))∥]12}).
Then,
(77)∥Fx-Fy∥∞≤ψ(max{12∥x(s)-y(s)∥,∥x(s)-F(x(s))∥,∥y(s)-F(y(s))∥,12[∥x(s)-F(y(s))∥+∥y(s)-F(x(s))∥]12}).
That is, (x,y)∈E(G) implies
(78)∥Fx-Fy∥∞≤ψ(max{12∥x-y∥∞,∥x-F(x)∥∞,∥y-F(y)∥∞,12[∥x-F(y)∥∞+∥y-F(x)∥∞]12}).
Thus, all the hypotheses of Theorem 29 are satisfied, and hence, the mapping F has a fixed point, that is, a solution in X=C([0,T],ℝ) of the integral equation (72).



---

*Source: 575869-2013-10-02.xml*