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.

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 function θ 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.

Xis complete.

Every mapping T 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.

Theorems 2 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 by Φφ the set of functions φ:[0,+∞)→[0,+∞) satisfying the following condition:
(2)(θ1)φ(t)≤12t∀t≥0.

We denote by Φ 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 [<xref ref-type="bibr" rid="B1">33</xref>, <xref ref-type="bibr" rid="B2">34</xref>]).

If ψ∈Φ, then ϕ(t)<t for all t>0.

Definition 5 (see [<xref ref-type="bibr" rid="B3">35</xref>]).

Let f: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:

f is non-decreasing, continuous, and α-admissible,

there exists x0∈X such that x0⪯fx0 and α(x0,fx0)≥1,

then f has a fixed point in X.
Proof.

Let x0⪯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 as n→+∞ 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.

Let X=[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.

Let ψ(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.

Let x,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:

f is non-decreasing and α-admissible,

there exists x0∈X such that x0⪯fx0 and α(x0,fx0)≥1,

for a strictly increasing sequence {xn}⊂X, converging to x∈X one has xn≺x for all n∈ℕ,

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 Theorem 6, 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 since xn≺z, we have d(xn,z)>0 for all n∈ℕ, that is,
(28)d(xn+1,fz)<d(xn,z)or d(xn+2,fz)<d(xn+1,z).

Consequently, there exists an infinite subset J⊂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.

Let X=[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.

Let φ(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.

Let x,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 exists x0∈[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.

If x,y∈[0,1] and x≥y then
(31)α(x,y)d(fx,fy)=(fx-fy)=18(x2-y2)≤14(x-y).

If x=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 that α(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 Theorem 8 hold. Consequently, f has a fixed point.

The following notions will be used to show the uniqueness of a fixed point.

For all x,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.

For all x,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 Theorem 8 one obtains uniqueness of the fixed point of f.

Proof.

Suppose that z 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 the lim 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, if z and z* are comparable, then from (21) and condition (B), it follows that z=z*.

We denote by Θγ 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:

f is non-decreasing,

there exists x0∈X such that x0⪯fx0,

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:

for all x,y∈X that are not comparable there exists v∈X comparable with x and y.

Proof.

Let x0⪯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 that s 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 subset J⊂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 of z, 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.

Let X={(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 metric d 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.

Let φ(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:

for (0,0)⊑(4,0), we have;
(74)d(f(0,0),f(4,0))=0≤45d((0,0),(4,0));

for (0,4)⊑(5,4), we have
(75)d(f(0,4),f(5,4))=4≤4=45d((0,4),(5,4));

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

Let X=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

f:[0,T]×ℝ→ℝ is continuous,

p:[0,T]→ℝ is continuous,

S:[0,T]×[0,T]→[0,+∞) is continuous and
(79)supt∈[0,T]∫0TS(t,s)ds≤1,

there exists a nonempty and closed subset U⊆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)),

for all x∈U, we have
(82)x(t)≤p(t)+∫0TS(t,s)f(s,x(s))ds∈U,

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.

Let H: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 that H 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].

Then Hx⪯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)).

Define α: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.

Acknowledgment

The first author is supported by Islamic Azad University, Astara Branch.

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