# Suzuki-Edelstein Type Contractions via Auxiliary Functions

Authors: Peyman Salimi; Erdal Karapınar
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

We prove the existence and uniqueness of a fixed point of certain type mapping, extension of Suzuki-Edelstein mapping, in a partially ordered complete metric space. Our results extend, improve, and generalize the existence results on the topic in the literature. We state some examples to illustrate our results.


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## Body

## 1. Introduction

Fixed point theory lies in the center of nonlinear functional analysis because it has a broad range of applications in fields such as economics, computer science, and many others (see, e.g., [1–7]). In particular, fixed point theory is quite useful in finding the solutions of inverse problems and structural optimizations in science and engineering [8–10]. Banach contraction mapping principle [11] is considered to be the fundamental result in this theory. It states that each contraction defined on a complete metric space has a unique fixed point. The strength of Banach's result in applications comes from two remarkable implications of this principle. The first one is that it guarantees the existence and uniqueness of a fixed point of a contraction. The second and most effective one is that it provides a technique to evaluate the fixed point. The importance of these two properties of Banach contraction mapping principle evidently has attracted many prominent mathematicians (see, e.g., [12–16]) interested in the fixed point theory and its applications as the use of this principle has widened considerably since its first appearance [17–21]. In particular, in this paper, we will focus on one of the most remarkable generalizations of Banach contraction mapping principle developed by Edelstein [22]. This theorem can be stated as follows.Theorem 1.
Let(X,d) be a compact metric space and let T be a mapping on X. Assume d(Tx,Ty)<d(x,y) for all x,y∈X with x≠y. Then T has a unique fixed point.Inspired by Edelstein's theorem, Suzuki [23, 24] further improved Banach's principle through the two theorems below.Theorem 2.
Define a nonincreasing functionθ from [0,1) onto (1/2,1] by
(1)θ(r)={1if0≤r≤(5-1)2,(1-r)r-2if(5-1)2≤r≤22,(1+r)-1if22≤r<1.
Then for a metric space (X,d), the following is equivalent. (1)
Xis complete.(2)
Every mappingT on X satisfying the following has a fixed point: there exists r∈[0,1) such that θ(r)d(x,Tx)≤d(x,y) implies d(Tx,Ty)≤rd(x,y) for all x,y∈X.Theorem 3.
Let(X,d) be a compact metric space and let T be a mapping on X. Assume that (1/2)d(x,Tx)<d(x,y) implies d(Tx,Ty)<d(x,y) for all x,y∈X. Then T has a unique fixed point.Theorems2 and 3 are extensively studied by many authors (see, e.g., [25–29]). It is worth to point out that the studies mentioned above can be classified as the extensions of Banach's principle on compact/complete metric spaces, which are totally ordered.On the other hand, Turinici [30], Ran and Reurings [31], and Nieto and Rodríguez-López [32] proved an analog of the desired Banach's principle in the context of partially ordered complete metric spaces for a certain class of maps. In particular, Ran and Reurings [31] applied their result to solve a matrix equation and Nieto and Rodríguez-López [32] applied their theorem to guarantee the existence and uniqueness of the solution of some boundary value problems.Motivated by all these developments, we shall prove new fixed point theorems extending Edelstein-Suzuki type contraction results in the setting of partially ordered complete metric spaces.


## 2. Main Results

We denote byΦφ the set of functions φ:[0,+∞)→[0,+∞) satisfying the following condition:
(2)(θ1)φ(t)≤12t∀t≥0.We denote byΦ the set of nondecreasing functions ϕ:[0,+∞)→[0,+∞) such that
(3)limn→+∞ϕn(t)=0∀t>0.We have the following lemma.Lemma 4 (see [33, 34]).
Ifψ∈Φ, then ϕ(t)<t for all t>0.Definition 5 (see [35]).
Letf:X→X and α:X→ℝ+. We say that f is α-admissible if
(4)x,y∈Xα(x,y)≥1⇒α(fx,fy)≥1.Theorem 6.
Let(X,d,⪯) be an ordered metric space such that (X,d) is complete. Let f:X→X be a self-mapping. Assume that there exists φ∈Φφ such that
(5)φ(d(x,fx))≤α(x,y)d(x,y)impliesα(x,y)d(fx,fy)≤ϕ(d(x,y))
for all comparable x,y∈X, where ϕ∈Φ. If the following conditions hold: (i)
f is non-decreasing, continuous, and α-admissible,(ii)
there existsx0∈X such that x0⪯fx0 and α(x0,fx0)≥1, then f has a fixed point in X.Proof.
Letx0⪯fx0. If x0=fx0, then the theorem follows. Hence we suppose that x0≺fx0. Define a sequence {xn} by xn=fnx0=fxn-1 for all n∈ℕ. Since f is non-decreasing and x0≺fx0 then
(6)x0≺x1⪯x2⪯⋯,
and hence {xn} is an increasing sequence. If xn=xn+1=fxn for some n∈ℕ, then the result is proved as xn is a fixed point of f. In what follows, we will suppose that xn≠xn+1. Hence, {xn} is a strictly increasing sequence. Since f is α-admissible and there exists x0∈X such that α(x0,fx0)≥1, we find α(x1,x2)=α(fx0,fx1)≥1. By continuing this process, we get α(xn,xn+1)≥1 for all n∈ℕ∪{0}. Now we obtain that
(7)φ(d(xn-1,fxn-1))≤12d(xn-1,fxn-1)≤d(xn-1,fxn-1)≤α(xn-1,fxn-1)d(xn-1,fxn-1)
and xn-1 and fxn-1 are comparable for all n∈ℕ. Then by (5), we have
(8)α(xn-1,fxn-1)d(fxn-1,f2xn-1)≤ϕ(d(xn-1,fxn-1)).
Then(9)d(xn,xn+1)≤α(xn-1,xn)d(xn,xn+1)≤ϕ(d(xn-1,xn)).
By induction, we have
(10)d(xn,xn+1)≤ϕn(d(x0,x1)).
By taking the limit asn→+∞ in the inequality above, we deduce
(11)limn→+∞d(xn,xn+1)=0.
For a fixed ϵ>0, there exists N∈ℕ such that
(12)d(xn,xn+1)<ϵ-ϕ(ϵ)∀n∈ℕ.
As ϕ is nondecreasing, we get
(13)ϕ(d(xn,xn+1))≤ϕ(ϵ-ϕ(ϵ))≤ϕ(ϵ)∀n∈ℕ.
Then
(14)d(xn,xn+2)≤d(xn,xn+1)+d(xn+1,xn+2)<ϵ-ϕ(ϵ)+ϕ(d(xn,xn+1))≤ϵ-ϕ(ϵ)+ϕ(ϵ)=ϵ.
By continuing this process, we get
(15)d(xn,xn+k)<ϵ
for all n≥N and k∈ℕ. Consequently limm,n,→+∞d(xn,xm)=0. Hence {xn} is a Cauchy sequence. Since X is complete, there is z∈X such that xn→z. Now, since f is continuous, we have
(16)fz=limn→∞fxn=limn→∞xn=z.
So z is a fixed point of f.Example 7.
LetX=[0,∞). Define the metric d on X by d(x,y)=|x-y|. Define f:X→X by
(17)f(x)={12x-14x2,ifx∈[0,1],lnx+12ifx∈(1,∞)
and α:X2→ℝ+ by
(18)α(x,y)={1,ifx,y∈[0,1],0otherwise.
Letψ(t)=(1/2)t. Let x⪯y if and only if x≤y. Hence the conditions of Theorem 6 hold; that is, f has a fixed point.Proof.
Letx,y∈X. First we assume that α(x,y)≥1. Then x,y∈[0,1]. On the other hand, for all x,y∈[0,1], we have fx≤1/2. Hence α(fx,fy)≥1. That is, f is α-admissible. Clearly, α(0,f0)≥1 and 0⪯f0. Let x≽y, and x,y∈[0,1). Then we derive
(19)α(x,y)d(fx,fy)=x-y2-(x2-y24)≤x-y2=ψ(d(x,y)).
Next we assume that α(x,y)=0. As a result we find
(20)0=α(x,y)d(fx,fy)≤ψ(d(x,y)).
Then the conditions of Theorem 6 hold; therefore, f has a fixed point.Theorem 8.
Let(X,d,⪯) be an ordered metric space such that (X,d) is complete. Let f:X→X be a self-mapping. Assume that there exists φ∈Φφ such that
(21)φ(d(x,fx))≤α(x,y)d(x,y)impliesα(x,y)d(fx,fy)≤ϕ(d(x,y))
for all comparable x,y∈X, where ϕ∈Φ. If the following conditions hold: (i)
f is non-decreasing and α-admissible,(ii)
there existsx0∈X such that x0⪯fx0 and α(x0,fx0)≥1,(iii)
for a strictly increasing sequence{xn}⊂X, converging to x∈X one has xn≺x for all n∈ℕ,(iv)
if{xn} is a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x as n→∞, then α(xn,x)≥1 for all n, then f has a fixed point.Proof.
Proceeding as in the proof of Theorem6, we know that there is a point z∈X such that xn→z and xn≺z by condition (iii). We prove that z=fz. From (iv), we have α(xn,z)≥1 for all n∈ℕ. Now, we suppose that the following inequalities hold
(22)12d(xn,xn+1)≥φ(d(xn,xn+1))>α(xn,z)d(xn,z)≥d(xn,z),12d(xn+1,xn+2)≥φ(d(xn+1,xn+2))>α(xn+1,z)d(xn+1,z)≥d(xn+1,z)
for some n∈ℕ.
Since,xn≠xn+1 for all n, we get d(xn,xn+1)>0 for all n. Now by Lemma 4 and (9), we have d(xn+1,xn+2)<d(xn,xn+1). As a result, we obtain
(23)d(xn,xn+1)≤d(xn,z)+d(xn+1,z)<12d(xn,xn+1)+12d(xn+1,xn+2)<12d(xn,xn+1)+12d(xn,xn+1)=d(xn,xn+1).
This is a contradiction. Thus, for all n∈ℕ, either
(24)φ(d(xn,xn+1))≤α(xn,z)d(xn,z)
or
(25)φ(d(xn+1,xn+2))≤α(xn+1,z)d(xn+1,z)
holds. By (21), we have
(26)d(fxn,fz)≤α(xn,z)d(fxn,fz)≤ϕ(d(xn,z))
or
(27)d(fxn+1,fz)≤α(xn+1,z)d(fxn+1,fz)≤ϕ(d(xn+1,z)).
Now sincexn≺z, we have d(xn,z)>0 for all n∈ℕ, that is,
(28)d(xn+1,fz)<d(xn,z)ord(xn+2,fz)<d(xn+1,z).
Consequently, there exists an infinite subsetJ⊂ℕ such that one of these inequalities holds for every n∈J. If we take the limit as n→+∞ and n∈J, then we get d(z,fz)=0, that is, z=fz, which is a contradiction. Hence it is absurd to suppose z≠fz. So z is a fixed point of f.Example 9.
LetX=[0,∞). Define the metric d on X by d(x,y)=|x-y|. Define f:X→X by
(29)f(x)={18x2,ifx∈[0,1],4ifx=3,x+1ifx∈ℝ+∖[0,1]∪{3}
and α:X2→ℝ+ by
(30)α(x,y)={1,ifx,y∈[0,1],14,ifx=3,y=1,0,otherwise.
Letφ(t)=(1/2)t, and ϕ(t)=(1/4)t. Let x⪯y if and only if x≤y. Hence the conditions of Theorem 8 hold. That is, f has a fixed point.Proof.
Letx,y∈X. If α(x,y)≥1, then x,y∈[0,1]. On the other hand, for all x,y∈[0,1], we derive fx≤1/2. Hence α(fx,fy)≥1. That is, f is α-admissible.
Clearly, there existsx0∈[0,1] such that α(x0,fx0)≥1.
If{xn} is a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x as n→∞. Then {xn}⊆[0,1] and x∈[0,1]. That is, α(xn,x)≥1 for all n. We need to consider the following cases.(i)
Ifx,y∈[0,1] and x≥y then
(31)α(x,y)d(fx,fy)=(fx-fy)=18(x2-y2)≤14(x-y).(ii)
Ifx=3 and y=1, then
(32)α(3,1)d(f3,f1)=α(3,1)(f3-f1)=3132>12=14|3-1|=ϕ(d(2,3)).But we see that
(33)φ(d(x,fx))=12|3-f3|=1>12=α(3,1)|3-1|=α(3,1)d(3,1).
If it is the case thatα(x,y)=0, then α(x,y)d(fx,fy)=0≤ϕ(d(x,y)). Hence,
(34)φ(d(x,fx))≤α(x,y)d(x,y)impliesα(x,y)d(fx,fy)≤ϕ(d(x,y)).
Then the conditions of Theorem8 hold. Consequently, f has a fixed point.The following notions will be used to show the uniqueness of a fixed point.(A)
For allx,y∈X that are not comparable, there exists v∈X comparable with x and y such that α(x,v)≥1 and α(y,v)≥1.(B)
For allx,y∈X that are comparable, there exists w∈X such that α(x,v)≥1 and α(y,v)≥1.Theorem 10.
Adding conditions (A) and (B) to the hypotheses of Theorem8 one obtains uniqueness of the fixed point of f.Proof.
Suppose thatz and z* are two fixed points of f such that z≠z*. If z and z* are not comparable, by condition (A), there exists v∈X comparable with z and z*, such that α(z,v)≥1 and α(z*,v)≥1. Since f is α-admissible, we have α(fnz,fnv)≥1 and α(fnz*,fnv)≥1. Since we have
(35)φ(d(fn-1z,fnz))≤12d(fn-1z,fn-1v)≤d(fn-1z,fn-1v)≤α(fn-1z,fn-1v)d(fn-1z,fn-1v)
and fn-1z and fn-1v are comparable, by (21), we derive
(36)d(fnz,fnv)≤α(fn-1z,fn-1v)d(fnz,fnv)≤ϕ(d(fn-1z,fn-1v)).
By induction, we get
(37)d(fnz,fnv)≤ϕn(d(z,v)).
If d(z,v)=0, then z=v. That is, z and z* are comparable, which is a contradiction. Hence we suppose that d(z,v)>0.
Taking thelim as k→+∞ in the first of the inequalities above, we have
(38)limn→+∞d(fnz,fnx)=0.
Similarly, we obtain that
(39)limn→+∞d(fnz*,fnv)=0.
From the inequality
(40)d(z,z*)=d(fnz,fnz*)≤d(fnz,fnv)+d(fnv,fnz*),
we get d(z,z*)=0 by taking the limit as n→+∞, that is, z=z*.
Similarly, ifz and z* are comparable, then from (21) and condition (B), it follows that z=z*.We denote byΘγ the set of functions γ:[0,+∞)→[0,+∞) satisfying the following condition:
(41)(θ2)γ(t)≥t∀t≥0.Theorem 11.
Let(X,d,⪯) be an ordered metric space such that (X,d) is complete. Let f:X→X be a self-mapping. Assume that there exist φ∈Φφ and γ∈Θγ such that
(42)φ(d(x,fx))≤γ(d(x,y))impliesψ(d(fx,fy))≤ψ(d(x,y))-ϕ(d(x,y))
for all comparable x,y∈X, where ϕ∈Φ and ψ∈Ψ. If the following conditions hold: (i)
f is non-decreasing,(ii)
there existsx0∈X such that x0⪯fx0,(iii)
if a non-decreasing sequence{xn} is such that xn→x as n→∞, then
(43)xn⪯x∀n∈ℕ,
then f has a fixed point. Moreover, f has a unique fixed point if:(iv)
for allx,y∈X that are not comparable there exists v∈X comparable with x and y.Proof.
Letx0⪯fx0. If x0=fx0, then the conclusion follows. Hence we suppose that x0≺fx0. Define a sequence {xn} by xn=fnx0=fxn-1 for all n∈ℕ. Since f is non-decreasing and x0≺fx0, we have
(44)x0≺x1⪯x2⪯⋯.
Hence {xn} is a non-decreasing sequence. Since
(45)φ(d(xn-1,fxn-1))≤12d(xn-1,fxn-1)≤d(xn-1,fxn-1)≤γ(d(xn-1,fxn-1))
and xn-1 and fxn-1 are comparable for all n∈ℕ, by (42), we have
(46)ψ(d(fxn-1,f2xn-1))≤ψ(d(xn-1,fxn-1))-ϕ(d(xn-1,fxn-1))≤ψ(d(xn-1,fxn-1)).
Then we find(47)d(xn,xn+1)≤d(xn-1,xn)
for all n∈ℕ. This implies that the sequence {d(xn-1,xn)} is decreasing and so there is a real number s≥0 such that
(48)limn→+∞d(xn,xn+1)=s.
Now, we show thats must be equal to 0. Let s>0. Then by taking the limit as n→∞ in (46), we get
(49)ψ(s)≤ψ(s)-ϕ(s)
which is a contradiction. Hence we conclude that
(50)limn→+∞d(xn,xn+1)=0.
Next, we prove that{xn} is a Cauchy sequence. Suppose, to the contrary, that {xn} is not a Cauchy sequence. Then there exists an ε>0 such that for all positive integer k,
(51)n(k)>m(k)>k,d(xn(k),xm(k))≥ε,d(xn(k),xm(k)-1)<ε
for some sequences {m(k)} and {n(k)}. For all k∈ℕ, we have
(52)ε≤d(xn(k),xm(k))≤d(xn(k),xm(k)-1)+d(xm(k)-1,xm(k))<ε+d(xm(k)-1,xm(k)).
Taking the limit as k→+∞ in the inequality above and using (50), we get
(53)limk→+∞d(xn(k),xm(k))=ε.
Again, from the expressions
(54)d(xn(k),xm(k))≤d(xm(k),xm(k)+1)+d(xm(k)+1,xn(k)+1)+d(xn(k)+1,xn(k)),d(xn(k)+1,xm(k)+1)≤d(xm(k),xm(k)+1)+d(xm(k),xn(k))+d(xn(k)+1,xn(k)),
(50), and (53), we deduce
(55)limk→+∞d(xn(k)+1,xm(k)+1)=ε
by taking the limit as k→+∞. From the facts limk→+∞d(xm(k),xm(k)+1)=0 and d(xm(k),xn(k))≥ε>0, we deduce that there is N∈ℕ such that for all k≥N(56)d(xm(k),xm(k)+1)≤d(xm(k),xn(k)).
Since(57)φ(d(xm(k),xm(k)+1))≤12d(xm(k),xm(k)+1)≤12d(xm(k),xn(k))≤d(xm(k),xn(k))≤γ(d(xm(k),xn(k)))
and xm(k) and xn(k) are comparable, by (5), we get
(58)ψ(d(fxm(k),fxn(k)))≤ψ(d(xm(k),xn(k)))-ϕ(d(xm(k),xn(k))).
Taking the limit as k→+∞ in the inequality above, we have
(59)ψ(ε)≤ψ(ε)-ϕ(ε),
which is a contradiction. Hence {xn} is a Cauchy sequence. Since X is complete, there is a point z∈X such that xn→z. Furthermore, by condition (iii), xn⪯z.
Now, we suppose that the following inequalities hold:(60)12d(xn,xn+1)≥φ(d(xn,xn+1))>γ(d(xn,z))≥d(xn,z),12d(xn+1,xn+2)≥φ(d(xn+1,xn+2))>γ(d(xn+1,z))≥d(xn+1,z)
for some n∈ℕ. By (47), we have
(61)d(xn,xn+1)≤d(xn,z)+d(xn+1,z)<12d(xn,xn+1)+12d(xn+1,xn+2)<12d(xn,xn+1)+12d(xn,xn+1)=d(xn,xn+1),
a contradiction. Thus, for all n∈ℕ, either
(62)φ(d(xn,xn+1))≤γ(d(xn,z))
or
(63)φ(d(xn+1,xn+2))≤γ(d(xn+1,z))
holds. By (42), it follows that either
(64)ψ(d(fxn,fz))≤ψ(d(xn,z))-ϕ(d(xn,z))
or
(65)ψ(d(fxn+1,fz))≤ψ(d(xn+1,z))-ϕ(d(xn+1,z)).
Consequently, there exists an infinite subsetJ⊂ℕ such that one of these inequalities holds for every n∈J. If we take the limit as n→+∞ and n∈J, then we get that d(z,fz)=0, that is, z=fz which contradicts with the assumption z≠fz. Hence z is a fixed point of f.
Now, assume that condition (iv) holds. To prove the uniqueness ofz, suppose that z* is another fixed point of f. It is easy to see from (5) that z=z*, if z and z* are comparable. If z and z* are not comparable, by condition (iv), there exists an x comparable with z and z*. Now we suppose that
(66)limn→+∞d(fnz,fnx)=s1>0,limn→+∞d(fnz*,fnx)=s2>0.
First we note that for each x∈X comparable with z, by (5) with x=fn-1z and y=fn-1x, we have
(67)ψ(d(fnz,fnx))≤ψ(d(fn-1z,fn-1x))-ϕ(d(fn-1z,fn-1x)).
Taking the limsup as k→+∞ in the inequalities above, we have
(68)ψ(s1)≤ψ(s1)-ϕ(s1),
which is a contradiction. Hence limn→+∞d(fnz,fnx)=0. Similarly, we obtain that
(69)limn→+∞d(fnz*,fnx)=0.
From the inequality
(70)d(z,z*)=d(fnz,fnz*)≤d(fnz,fnx)+d(fnx,fnz*),
we get d(z,z*)=0, that is, z=z*, by taking the limit as n→+∞.Example 12.
LetX={(0,0),(4,0),(0,4),(4,5),(5,4)}. We define a partial order ⊑ by
(71)(x,y)⊑(u,v)⟺x≤u,v≤y.
Define the metricd on X by d((x1,x2),(y1,y2))=|x1-y1|+|x2-y2|. Clearly, (X,d) is a complete metric space. Also define f:X→X by
(72)f(x1,x2)={(x1,0),Ifx1≤x2,(0,x2)Ifx1>x2.
Letφ(t)=(1/2)t, γ(t)=t, ψ(t)=t, and ϕ(t)=(1-r)t, where r∈[0,1). Hence the conditions of Theorem 11 hold; that is, f has a unique fixed point. But Theorem2.1 in [36] cannot be applied to f in this example.Proof.
Let(x,y),(u,v)∈X, with (x,y)⊑(u,v). Then for all elements of X we have
(73)(0,0)⊑(4,0),(0,4)⊑(5,4),(4,5)⊑(5,4).
We examine each of the following cases: (i)
for(0,0)⊑(4,0), we have;
(74)d(f(0,0),f(4,0))=0≤45d((0,0),(4,0));(ii)
for(0,4)⊑(5,4), we have
(75)d(f(0,4),f(5,4))=4≤4=45d((0,4),(5,4));(iii)
for(4,5)⊑(5,4), we have
(76)d(f(4,5),f(5,4))=8>85=45d((4,5),(5,4)).That is, in [36, Theorem  2.1] cannot be applied to f. But for (4,5)⊑(5,4),
(77)12d((4,5),f(4,5))=52>2=d((4,5),(5,4)).
Hence the conditions of Theorem 11 hold; that is, f has a unique fixed point in X.


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

LetX=C([0,T],ℝ) be the set of real continuous functions defined on [0,T] and dα:X×X→[0,+∞) be defined by d(x,y)=supt∈[0,T]|x(t)-y(t)| for all x,y∈X. Let ⪯ be the partial order on X defined by x⪯y if and only if x(t)≤y(t) for all t∈[0,T]. Then (X,d,⪯) is a complete partially ordered metric space.Consider the following integral equation:(78)x(t)=p(t)+∫0T‍S(t,s)f(s,x(s))ds,
where(A)
f:[0,T]×ℝ→ℝ is continuous,(B)
p:[0,T]→ℝ is continuous,(C)
S:[0,T]×[0,T]→[0,+∞) is continuous and
(79)supt∈[0,T]∫0T‍S(t,s)ds≤1,(D)
there exists a nonempty and closed subsetU⊆X such that for all x,y∈U with x⪯y, we have
(80)f(s,y(s))-f(s,x(s))≤ψ(y(s)-x(s))∀s∈[0,T],
where ψ∈Ψ and for all x,y∈X with x⪯y, we have
(81)0≤f(s,y(s))-f(s,x(s)),(E)
for allx∈U, we have
(82)x(t)≤p(t)+∫0T‍S(t,s)f(s,x(s))ds∈U,(F)
for a strictly increasing sequence{xn}⊂X converging to x∈X, we have xn≺x for all n∈ℕ.We have the following result of existence of solutions for integral equations.Theorem 13.
Under assumptions(A)–(E), the integral equation (78) has a solution in X=C([0,T],R).Proof.
LetH:X→X be defined by
(83)Hx(t)=p(t)+∫0T‍S(t,s)f(s,x(s))dst∈[0,T],∀x∈X.
First, we will prove thatH is a nondecreasing mapping with respect to ⪯. Let x⪯y. By (D), we have 0≤f(s,y(s))-f(s,x(s)) for all s∈[0,T]. On the other hand by definition of H, we have
(84)Hy-Hx=∫0T‍S(t,s)[f(s,y(s))-f(s,x(s))]ds≥0∀t∈[0,T].
ThenHx⪯Hy, that is, H is a non-decreasing mapping with respect to ⪯. Now suppose that x,y∈U with x⪯y. Then by (C), (D), and the definition of H, we get
(85)d(Hx,Hy)=supt∈[0,T]|Hx(t)-Hy(t)|=supt∈[0,T]|∫0T‍S(t,s)[f(s,x(s))-f(s,y(s))]ds|≤supt∈[0,T]∫0T‍S(t,s)|f(s,x(s))-f(s,y(s))|ds≤supt∈[0,T]∫0T‍S(t,s)ψ((y(s)-x(s)))ds≤(supt∈[0,T]∫0T‍S(t,s)ds)×ψ(sups∈[0,T](y(s)-x(s)))≤ψ(d(x,y)).
Defineα:X2→ℝ+ by
(86)α(x,y)={1,ifx,y∈Uwithx⪯y,0,otherwise.
Clearly,α(x,y)d(Hx,Hy)≤ψ(d(x,y)) holds for all x,y∈X with x⪯y. If α(x,y)≥1 then x,y∈U and x⪯y. Since, H is increasing, we obtain Hx⪯Hy. Similarly, since Hx,Hy∈U, we get α(Hx,Hy)≥1. Therefore, H is an α-admissible mapping. Also, assume that {xn} is a sequence such that xn→x as n→∞ and α(xn,x)≥1. Then {xn}⊆U. Since U is a closed set, we have x∈U. That is, α(xn,x)≥1. Then conditions of Theorem 8 hold and H has a solution in X.



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*Source: 648528-2013-03-24.xml*