# Fixed Point Results forα-ψ-Contractive Mappings Including Almost Contractions and Applications

Authors: Gonca Durmaz; Gülhan Mınak; Ishak Altun
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

In the recent paper (B. Samet, C. Vetro, and P. Vetro, Fixed point theorems forα-ψ-contractive type mappings, Nonlinear Analysis. Theory, Methods and Applications, 75 (2012), 2154-2165.), the authors introduced the concept of α-admissible maps on metric spaces. Using this new concept, they presented some nice fixed point results. Also, they gave an existence theorem for integral equation to show the usability of their result. Then, many authors focused on this new concept and obtained a lot of fixed point results, which are used for existence theorems. In this paper, we not only extend some of the recent results about this direction but also generalize them. Then, we give some examples to show our results are proper extensions. Furthermore, we use our results to obtain the existence and uniqueness result for a solution of fourth order two-point boundary value problem.


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## Body

## 1. Introduction and Preliminaries

Fixed point theory contains many different fields of mathematics, such as nonlinear functional analysis, mathematical analysis, operator theory, and general topology. Historically, the study of fixed point theory has developed in two major branches: the first is fixed point theory for contraction or contraction type mappings on complete metric spaces and the second is fixed point theory for continuous operators on compact and convex subsets of a normed space. Recently, there has been a lot of activities in the first branch and several fundamental fixed point results have been extended and generalized by many authors in different directions. In this paper, we mention some important of them and give some new fixed point results. Also, we support our results by giving a lot of nontrivial examples. First, we give some notations, which will be used in this paper. Letψ:[0,∞)→[0,∞) be a function. For convenience, we consider the following properties of this function:(ψ1)
ψ is nondecreasing,(ψ2)
lim⁡n→∞ψn(t)=0 for all t≥0,(ψ3)
ψ(t)<t for t>0,(ψ4)
ψ(0)=0,(ψ5)
ψ is continuous,(ψ6)
ψ is upper semicontinuous from the right,(ψ7)
∑n=1∞‍ψn(t)<∞ for any t>0.In the light of the above properties, the following hold: ifψ1 and ψ2 satisfied, then ψ3 holds. If ψ1 and ψ3 are satisfied, then ψ4 holds. If ψ3 and ψ5 are satisfied, then ψ2 and ψ4 hold.Denoted byΨ the family of functions ψ:[0,∞)→[0,∞) satisfying ψ1 and ψ2, which is called comparison functions in the literature (see [1]), by Φ the family of functions ψ:[0,∞)→[0,∞) satisfying ψ1 and ψ7, which is called (c)-comparison functions in the literature (see [1]), and by Υ the family of functions ψ:[0,∞)→[0,∞) satisfying ψ3 and ψ6. Now, we give some examples showing the relations between the sets Φ,Ψ, and Υ. First, it is clear that Φ⊂Ψ.Example 1.
Letψ:[0,∞)→[0,∞) be defined by ψ(t)=λt, where λ∈[0,1), and then ψ∈Φ∩Υ.Example 2.
Letψ:[0,∞)→[0,∞) be defined by ψ(t)=t/(1+t), and then ψ∈Ψ∩Υ, but ψ∉Φ.Example 3.
Letψ:[0,∞)→[0,∞) be defined by
(1)ψ(t)={t3,0≤t≤23t2-118,t>23,
and then ψ∈Φ, but ψ∉Υ.Example 4.
Letψ:[0,∞)→[0,∞) be defined by
(2)ψ(t)={t1+t,0≤t≤12t3,t>1,
and then ψ∈Ψ, but ψ∉Φ∪Υ.Example 5.
Letψ:[0,∞)→[0,∞) be defined by
(3)ψ(t)={t1+t,0≤t≤112t,t>1,
and then ψ∈Υ, but ψ∉Ψ.In their recent paper, Samet et al. [2] introduced the notions of α-admissible and α-ψ-contractive mappings and then gave some fixed point results for such mappings. Their results are closely related to some ordered fixed point results. Then, using their idea, some authors presented fixed point results for single and multivalued mappings (see, e.g., [2–6]).Definition 6 (see [2]).
Let(X,d) be a metric space, let T be a self-map on X, ψ∈Φ, and let α:X×X→[0,∞) be a function. Then T is called α-ψ-contractive whenever
(4)α(x,y)d(Tx,Ty)≤ψ(d(x,y))
for all x,y∈X.Note that every Banach contraction mapping is anα-ψ-contractive mapping with α(x,y)=1 and ψ(t)=λt for some λ∈[0,1).Definition 7 (see [2]).
T is called α-admissible whenever α(x,y)≥1 implies α(Tx,Ty)≥1.There exist some examples forα-admissible mappings in [2]. For convenience, we mention here one of them. Let X=[0,∞). Define T:X→X and α:X×X→[0,∞) by Tx=x for all x∈X and α(x,y)=ex-y for x≥y and α(x,y)=0 for x<y. Then T is α-admissible.Theorem 8 (see [2]).
Let(X,d) be a complete metric space and let T:X→X be an α-admissible and α-ψ-contractive mapping. If there exists x0∈X such that α(x0,Tx0)≥1 and T is continuous, then T has a fixed point.The following theorems are a generalization of Theorem8.Theorem 9 (see [3]).
Let(X,d) be a complete metric space and let T:X→X be an α-admissible mapping satisfying
(5)α(x,y)d(Tx,Ty)≤ψ(M(x,y))
for all x,y∈X, where ψ∈Φ and
(6)M(x,y)=max⁡{d(x,y),d(x,Tx),d(y,Ty)d(x,Ty)+d(y,Tx)2,iiiiiiiiiiiiiiid(x,Ty)+d(y,Tx)2}.
If there exists x0∈X such that α(x0,Tx0)≥1 and T is continuous, then T has a fixed point.Theorem 10 (see [4]).
Let(X,d) be a complete metric space and let T:X→X be an α-admissible mapping satisfying. Assume that
(7)x,y∈X,α(x,y)≥1⟹d(Tx,Ty)≤ψ(m(x,y)),
where ψ∈Φ and
(8)m(x,y)=max⁡{d(x,y),d(x,Tx)+d(y,Ty)2,iiiiiiiiiiiiiiiid(x,Ty)+d(y,Tx)2}.
If there exists x0∈X such that α(x0,Tx0)≥1 and T is continuous or X is regular, then T has a fixed point.For the sake of brevity, we will say thatX is regular whenever, for any sequence {xn} in X with α(xn,xn+1)≥1 for all n∈N∪{0} and xn→x as n→∞, we have α(xn,x)≥1 for all n∈N∪{0}.The aim of this paper is to extend and generalize the above results. Note that, in these theorems, the functionψ belongs to the class Φ; that is, ψ is (c)-comparison function. Also, in Theorem 10, the contractive condition is used with m(x,y). In this paper, we give three existence results. In the first result, the contractive condition (7) will be generalized to almost contraction case. Here we take ψ∈Φ, but we use M(x,y) instead of m(x,y). In the second result, we take ψ in Ψ, which is a wider class of Φ. And in the third result, we take ψ in Υ, which is a different class of Φ. Also, we present some uniqueness theorems and some supporting examples.


## 2. Existence Results

Our first result is almost contraction version of fixed points ofα-admissible mapping. We can find detailed information about almost contractions in [7–12].Theorem 11.
Let(X,d) be a complete metric space and let T be an α-admissible mapping. Assume that
(9)α(x,y)≥1⟹d(Tx,Ty)≤ψ(M(x,y))111111111111111111+Lmin⁡{d(y,Tx),d(x,Ty)}
holds for all x,y∈X, where ψ∈Φ, L≥0, and M(x,y) as in Theorem 9. Also, suppose T is continuous or X is regular and there exists x0∈X such that α(x0,Tx0)≥1. Then T has a fixed point.Proof.
Letx0∈X such that α(x0,Tx0)≥1. Define a sequence {xn} in X by
(10)xn=Tnx0=Txn-1
for all n∈N. If xn=xn+1 for some n∈N, then xn is a fixed point for T and result is proved. So, we suppose that xn≠xn+1 for all n∈N. Since T is α-admissible mapping and α(x0,Tx0)≥1, we deduce that α(x1,x2)=α(Tx0,T2x0). Continuing this process, we get α(xn,xn+1)≥1 for all n∈N∪{0}. Now by (9) with x=xn,y=xn+1, we get
(11)d(xn+1,xn+2)=d(Txn,Txn+1)≤ψ(M(xn,xn+1))+Lmin⁡{d(xn+1,Txn),d(xn,Txn+1)}=ψ(M(xn,xn+1))+Lmin⁡{d(xn+1,xn+1),d(xn,Txn+1)}=ψ(M(xn,xn+1)),
where
(12)M(xn,xn+1)=max⁡{d(xn,xn+1),d(xn,Txn),d(xn+1,Txn+1)d(xn,Txn+1)+d(xn+1,Txn)2,iiiiiiiiiiiiid(xn,Txn+1)+d(xn+1,Txn)2}=max⁡{d(xn,xn+1),d(xn+1,xn+2),d(xn,xn+2)2}=max⁡{d(xn,xn+1),d(xn+1,xn+2)}.
If d(xn+1,xn+2)≥d(xn,xn+1) for some n∈N∪{0}, then from (11), we have
(13)d(xn+1,xn+2)≤ψ(M(xn,xn+1))=ψ(max⁡{d(xn,xn+1),d(xn+1,xn+2)})=ψ(d(xn+1,xn+2))<d(xn+1,xn+2),
which is a contradiction. Thus d(xn+1,xn+2)<d(xn,xn+1) for all n∈N∪{0} and so from (11), we have
(14)d(xn+1,xn+2)≤ψ(d(xn,xn+1)).
By induction, we have
(15)d(xn+1,xn+2)≤ψn+1(d(x0,x1))
for alln∈N∪{0}. Now, for each m,n∈N, m>n, we have
(16)d(xn,xm)≤∑k=nm-1‍d(xk,xk+1)≤∑k=nm-1‍ψk(d(x0,x1)).
Therefore, {xn} is a Cauchy sequence in X. Since X is complete, there exists z∈X such that lim⁡n→∞xn=z. If T is continuous, then we have
(17)Tz=lim⁡n→∞Txn=lim⁡n→∞xn+1=z.
So, z is a fixed point of T. Now, suppose X is regular. Since
(18)α(xn,xn+1)≥1
for all n∈N∪{0} and xn→z as n→∞, then we have
(19)α(xn,z)≥1
for all n∈N∪{0}. From (9) we have
(20)d(xn+1,Tz)=d(Txn,Tz)≤ψ(M(xn,z))+Lmin⁡{d(z,Txn),d(xn,Tz)}=ψ(M(xn,z))+Lmin⁡{d(z,xn+1),d(xn,Tz)},
where
(21)M(xn,z)=max⁡{d(xn,z),d(xn,Txn),d(z,Tz)12,iiiiiiiiiiiii12[d(xn,Tz)+d(z,Txn)]}=max⁡{d(xn,z),d(xn,xn+1),d(z,Tz)12,iiiiiiiiiiiii12[d(xn,Tz)+d(z,xn+1)]}≤max⁡{d(xn,z),d(xn,xn+1),d(z,Tz)12,iiiiiiiiiiii12[d(xn,z)+d(z,Tz)+d(z,xn+1)]}.
Now, suppose d(z,Tz)>0. Taking into account (15) and lim⁡n→∞xn=z, there exists n0∈N such that d(xn,z)<d(z,Tz)/2 and d(xn,xn+1)<d(z,Tz)/2 for all n≥n0. Therefore we have
(22)M(xn,z)≤max⁡{d(xn,z),d(xn,xn+1),d(z,Tz)12,iiiiiiiiiiiii12[d(xn,z)+d(z,Tz)+d(z,xn+1)]}≤d(z,Tz)
for all n≥n0. Now, from (20), we obtain
(23)d(xn+1,Tz)≤ψ(d(z,Tz))+Lmin⁡{d(z,xn+1),d(xn,Tz)}
for all n≥n0. Letting n→∞ in the last equality, we get that
(24)d(z,Tz)≤ψ(d(z,Tz))<d(z,Tz),
which is a contradiction. Therefore d(z,Tz)=0 and so T has a fixed point.Remark 12.
In Theorem11, if we take L=0, then we obtain an extension of Theorem 10. Even if L=0, we can extend Theorem 10 by taking the function ψ from Ψ and taking M(x,y) instead of m(x,y) as follows.Theorem 13.
Let(X,d) be a complete metric space and let T be an α-admissible mapping. Assume that
(25)α(x,y)≥1⟹d(Tx,Ty)≤ψ(M(x,y))
holds for all x,y∈X, where ψ∈Ψ and M(x,y) as in Theorem 11. Also, suppose T is continuous or X is regular and there exists x0∈X such that α(x0,Tx0)≥1. Then T has a fixed point.Proof.
Letx0∈X such that α(x0,Tx0)≥1. As in the proof of Theorem 11, we can construct a sequence {xn} in X and we can obtain (15) if the consecutive terms are different (otherwise, T has a fixed point). Now we show that {xn} is a Cauchy sequence. Let ε>0. Taking into account (15), there exists n0∈N, such that d(xn0,xn0+1)<ε-ψ(ε). Therefore we have
(26)M(xn0,xn0+1)=max⁡{d(xn0,xn0+1),d(xn0,Txn0)12,iiiiiiiiiiiiid(xn0+1,Txn0+1),iiiiiiiiiiiii12[d(xn0,Txn0+1)iiiiiiiiiiiiiiiiiid(xn0,xn0+1),d(xn0,Txn0)12+d(xn0+1,Txn0)]}=max⁡{d(xn0,xn0+1),d(xn0+1,xn0+2)}<max⁡{ε-ψ(ε),ψ(ε-ψ(ε))}=ε-ψ(ε)
and so from (25)
(27)d(xn0,xn0+2)≤d(xn0,xn0+1)+d(xn0+1,xn0+2)<ε-ψ(ε)+d(Txn0,Txn0+1)≤ε-ψ(ε)+ψ(M(xn0,xn0+1))≤ε-ψ(ε)+ψ(ε-ψ(ε))≤ε-ψ(ε)+ψ(ε)=ε.
Again using (25) we have
(28)d(xn0,xn0+3)≤d(xn0,xn0+1)+d(xn0+1,xn0+3)<ε-ψ(ε)+d(Txn0,Txn0+2)≤ε-ψ(ε)+ψ(M(xn0,xn0+2)),
where
(29)M(xn0,xn0+2)=max⁡{d(xn0,xn0+2),d(xn0,Txn0)12,IIIIIiIIIId(xn0+2,Txn0+2),IIIIIIiIII12[d(xn0,Txn0+2)12IIIIIIIIIIIII12+d(xn0+2,Txn0)]}=max⁡{d(xn0,xn0+2),d(xn0,xn0+1)12,IIIIIIIIIId(xn0+2,xn0+3),IIIIIIIIII12[d(xn0,xn0+3)12IIIIIIIIIIIIIId(xn0,xn0+2),d(xn0,xn0+1)12+d(xn0+2,xn0+1)]}≤max⁡{ε,ε-ψ(ε),ψ2(ε-ψ(ε))12,IIIIIIIIII12[d(xn0,xn0+3)12IIIIIIIIIIIIII12+ψ(ε-ψ(ε))]}≤max⁡{ε,12[d(xn0,xn0+3)+ψ(ε)]}.
If ε≤(1/2)[d(xn0,xn0+3)+ψ(ε)], then, from (28), we have
(30)d(xn0,xn0+3)<ε-ψ(ε)+ψ(12[d(xn0,xn0+3)+ψ(ε)])<ε-ψ(ε)+12[d(xn0,xn0+3)+ψ(ε)]
and so
(31)12d(xn0,xn0+3)<ε-ψ(ε)2.
Therefore, we have
(32)12[d(xn0,xn0+3)+ψ(ε)]<ε,
which is a contradiction. Thus, ε>(1/2)[d(xn0,xn0+3)+ψ(ε)] and so, from (28), we have
(33)d(xn0,xn0+3)<ε.
By continuing this way, we can obtain
(34)d(xn0,xn0+k)<ε
for all k∈N. Now let m,n∈N with m>n≥n0, and then
(35)d(xm,xn)≤d(xm,xn0)+d(xn0,xn)<2ε;
that is, {xn} is a Cauchy sequence in X. The rest of the proof can be made as in the proof of Theorem 11.In the following theorem we take the functionψ from Υ instead of Ψ.Theorem 14.
Let(X,d) be a complete metric space and let T be an α-admissible mapping. Assume that
(36)α(x,y)≥1⟹d(Tx,Ty)≤ψ(M(x,y))
holds for all x,y∈X, where ψ∈Υ, and M(x,y) as in Theorem 11. Also, suppose T is continuous or X is regular and there exists x0∈X such that α(x0,Tx0)≥1. Then T has a fixed point.Proof.
Letx0∈X such that α(x0,Tx0)≥1. As in the proof of Theorem 11, we can construct a sequence {xn} in X and we may assume that xn≠xn+1 for all n∈N. For the sake of brevity we put dn=d(xn,xn+1). Since T is an α-admissible mapping, we can obtain as in the proof of Theorem 11 that
(37)dn+1=d(xn+1,xn+2)=d(Txn,Txn+1)≤ψ(M(xn,xn+1))=ψ(max⁡{dn,dn+1}).
Therefore, it should be dn+1<dn for all integer n≥0 and so, from (37), we have
(38)dn+1≤ψ(dn)<dn.
Consequently, the sequence {dn} of positive numbers is decreasing and bounded below. So, there exists λ≥0 such that lim⁡n→∞dn=λ. We claim that λ=0. Suppose to the contrary that λ>0. Using the fact that ψ is upper semicontinuous from the right function, we get from (38)
(39)λ=limsup⁡n→∞dn+1≤limsup⁡n→∞ψ(dn)≤ψ(λ)<λ,
which is a contradiction. Hence, we conclude that λ=0; that is,
(40)lim⁡n→∞dn=lim⁡n→∞d(xn,xn+1)=0.
Now, we prove that the sequence {xn} is Cauchy in X. Suppose, to the contrary, there exists ε>0 such that
(41)d(xmk,xnk)≥ε,
where {xmk} and {xnk} are subsequences of {xn} with mk>nk≥k for all k∈N. Moreover, mk is chosen as the smallest integer satisfying (41). Thus, we have
(42)d(xmk-1,xnk)<ε.
By the triangle inequality, we get
(43)ε≤d(xmk,xnk)≤d(xmk,xmk-1)+d(xmk-1,xnk)<dmk+ε.
Letting k→∞ in above inequality and using (40), we get that
(44)lim⁡k→∞d(xmk,xnk)=ε.
Now let k0∈N be such that dnk<ε and dmk<ε for all k≥k0. Then
(45)d(xmk,xnk)≤M(xmk,xnk)≤d(xmk,xnk)+12(dnk+dmk)
for all k≥k0. Using (40) and (44) and letting k→∞ in (45), we get
(46)lim⁡k→∞M(xmk,xnk)=ε.
Since M(xmk,xnk)≥ε for all k∈N and ψ is upper semicontinuous from the right function, we deduce that
(47)limsup⁡n→∞ψ(M(xmk,xnk))≤ψ(ε).
On the other hand, for each k∈N, we have
(48)ε≤d(xmk,xnk)≤d(xmk,xmk+1)+d(xmk+1,xnk+1)+d(xnk+1,xnk)=dmk+d(Txmk,Txnk)+dnk≤dmk+ψ(M(xmk,xnk))+dnk,
so
(49)ε≤limsup⁡n→∞ψ(M(xmk,xnk))≤ψ(ε)<ε,
which is a contradiction. Thus {xn} is a Cauchy sequence in X. Since X is complete, there exists z∈X such that lim⁡n→∞xn=z. If T is continuous, then we have
(50)Tz=lim⁡n→∞Txn=lim⁡n→∞xn+1=z.
So z is a fixed point of T. Now, suppose X is regular and d(z,Tz)>0. We first note that
(51)d(z,Tz)≤M(xn,z),lim⁡n→∞M(xn,z)=d(z,Tz);
hence by the upper semicontinuity of ψ, we get
(52)limsup⁡n→∞ψ(M(xn,z))≤ψ(d(z,Tz)).
On the other hand, since X is regular,
(53)α(xn,xn+1)≥1
for all integer n≥0 and xn→z as n→∞, then we have
(54)α(xn,z)≥1
for all integer n≥0. Thus, from (36), we have
(55)d(xn+1,Tz)=d(Txn,Tz)≤ψ(M(xn,z)),
and taking limit supremum, we get
(56)d(z,Tz)≤limsup⁡n→∞ψ(M(xn-1,z))≤ψ(d(z,Tz))<d(z,Tz),
which is a contradiction. Therefore d(z,Tz)=0 and so T has a fixed point.


## 3. Uniqueness Results

In this section, we consider some properties to obtain the uniqueness of the fixed point in the above theorems. For this, we denote the set of fixed points ofT by Fix(T).Theorem 15.
Assume that all hypotheses of Theorem13 hold. Also suppose
(57)∀x,y∈Fix(T),thereexistsu∈Xsuchthatα(x,u)≥1,α(y,u)≥1,
holds, then the fixed point of T is unique.Proof.
Supposez and w are two fixed points of T, and then there exists u∈X such that α(z,u)≥1 and α(w,u)≥1. Since T is an α-admissible mapping, then
(58)α(z,Tnu)≥1,α(w,Tnu)≥1
for all n∈N. Therefore
(59)d(z,Tn+1u)=d(Tz,Tn+1u)≤ψ(M(z,Tnu))≤ψ(max⁡{d(z,Tnu),d(z,Tn+1u)}).
Without loss of generality, we can assume d(z,Tnu)>0 for all n. Therefore from (59) we have (note it should be d(z,Tnu)>d(z,Tn+1u). Otherwise we obtain a contradiction from (59))
(60)d(z,Tn+1u)≤ψ(d(z,Tnu))≤⋯≤ψn(d(z,u)).
Letting n→∞ in the above inequality, we have Tnu→z. Similarly, we can obtain Tnu→w and so z=w.Remark 16.
The condition (57) is not sufficient to obtain the uniqueness of the fixed point in Theorem 11.Example 17.
LetX=[0,1] with the usual metric. Define T:X→X and α:X×X→[0,∞) by
(61)Tx={x2,x∈[0,14)0,x∈[14,13)1,x∈[13,1],
and α(x,y)=1. Then, it is clear that T is α-admissible mapping and X is regular. Also, there exists x0∈X such that α(x0,Tx0)≥1. Now, we show that (9) is satisfied with ψ(t)=t/2 and L=10. For this, we consider the following cases.Case 1.
Ifx,y∈[0,1/4), then
(62)d(Tx,Ty)=|x2-y2|≤12|x-y|≤ψ(M(x,y))+Lmin⁡{d(y,Tx),d(x,Ty)}.Case 2.
Ifx,y∈[1/4,1/3), then d(Tx,Ty)=0.Case 3.
Ifx,y∈[1/3,1], then d(Tx,Ty)=0.Case 4.
Ifx∈[0,1/4) and y∈[1/4,1/3), then
(63)d(Tx,Ty)=x2≤min⁡{x,y-x2}≤ψ(M(x,y))+Lmin⁡{d(y,Tx),d(x,Ty)}.Case 5.
Ifx∈[0,1/4) and y∈[1/3,1], then
(64)d(Tx,Ty)=1-x2≤12(y-x)+10min⁡{1-x,y-x2}≤ψ(M(x,y))+Lmin⁡{d(y,Tx),d(x,Ty)}.Case 6.
Ifx∈[1/4,1/3) and y∈[1/3,1], then
(65)d(Tx,Ty)=1≤12(y-x)+10min⁡{1-x,y}≤ψ(M(x,y))+Lmin⁡{d(y,Tx),d(x,Ty)}.
Therefore, all conditions of Theorem 11 are satisfied and so T has a fixed point. Although the condition (57) of Theorem 15 is satisfied, the fixed point of T is not unique.In the following, we give a uniqueness theorem by adding some conditions in Theorem11.Theorem 18.
Assume that all hypotheses of Theorem11 hold. Also suppose for all x,y∈Fix(T), there exists u∈X such that α(x,u)≥1 and α(y,u)≥1 and
(66)α(x,y)≥1⟹d(Tx,Ty)≤ψ1(M(x,y))+L1min⁡{d(x,Tx),d(y,Ty)}
holds for all x,y∈X, where ψ1∈Ψ and L1≥0, then the fixed point of T is unique.Proof.
Supposez and w are two fixed points of T, and then there exists u∈X such that α(z,u)≥1 and α(w,u)≥1. Since T is an α-admissible mapping, then
(67)α(z,Tnu)≥1,α(w,Tnu)≥1
for all n∈N. Therefore from (66)
(68)d(z,Tn+1u)=d(Tz,Tn+1u)≤ψ1(M(z,Tnu))+L1min⁡{d(z,Tz),d(Tnu,Tn+1u)}≤ψ1(max⁡{d(z,Tnu),d(z,Tn+1u)}).
Without loss of generality, we can assume d(z,Tnu)>0 for all n. Therefore we have
(69)d(z,Tn+1u)≤ψ1(d(z,Tnu))≤⋯≤ψ1n(d(z,u)).
Letting n→∞ in the above inequality, we have Tnu→z. Similarly, we can obtain Tnu→w and so z=w.


## 4. Some Corollaries and Example

Corollary 19.
Let(X,d) be a complete metric space and T:X→X be α-admissible. Assume that
(70)α(x,y)≥1⟹d(Tx,Ty)≤ψ(d(x,y))+Lmin⁡{d(y,Tx),d(x,Ty)}
holds for all x,y∈X, where ψ∈Φ and L≥0. Also, suppose T is continuous and there exists x0∈X such that α(x0,Tx0)≥1. Then T has a fixed point.Corollary 20.
Let(X,d) be a complete metric space and let T be self-map of X. Assume that
(71)d(Tx,Ty)≤ψ(M(x,y))+Lmin⁡{d(y,Tx),d(x,Ty)}
holds for all x,y∈X, where ψ∈Φ, L≥0, and M(x,y) as in Theorem 9. Then T has a fixed point.Corollary 21.
Let(X,d) be a complete metric space and let T be self-map of X. Assume that
(72)d(Tx,Ty)≤ψ(d(x,y))+Lmin⁡{d(y,Tx),d(x,Ty)}
holds for all x,y∈X, where ψ∈Φ and L≥0. Then T has a fixed point.Corollary 22.
Let(X,d) be a complete metric space and let T:X→X be α-admissible. Assume that
(73)α(x,y)≥1⟹d(Tx,Ty)≤ψ(d(x,y))
holds for all x,y∈X, where ψ∈Ψ (or ψ∈Υ). Also, suppose T is continuous and there exists x0∈X such that α(x0,Tx0)≥1. Then T has a fixed point.Corollary 23.
Let(X,d) be a complete metric space and let T be a self-map of X. Assume that
(74)d(Tx,Ty)≤ψ(M(x,y))
holds for all x,y∈X, where ψ∈Ψ (or ψ∈Υ), and M(x,y) as in Theorem 9. Then T has a unique fixed point.Corollary 24.
Let(X,d) be a complete metric space and let T be a self-map of X. Assume that
(75)d(Tx,Ty)≤ψ(d(x,y))
holds for all x,y∈X, where ψ∈Ψ (or ψ∈Υ). Then T has a unique fixed point.Example 25.
LetX=[0,1] with the usual metric. Define T:X→X and α:X×X→[0,∞) by
(76)Tx={x2,x∈[0,14]5x-14,x∈(14,1],α(x,y)={1,x,y∈[0,14]∪{1}0,otherwise
for all x,y∈X. Then, it is clear that T is α-admissible mapping and T is continuous. Also, α(1,T1)≥1. But, although α(1,1/4)≥1, we cannot find any ψ∈Φ satisfying
(77)d(T1,T14)=1516≤ψ(2732)=ψ(m(x,y))=ψ(M(x,y)).
Therefore, Theorems 10 and 9 cannot be applied to this example. Now, we show that (9) is satisfied with ψ(t)=t/2 and L=4/3. Let α(x,y)≥1, and then x,y∈[0,1/4]∪{1}. We have to consider the following cases.Case 1.
Ifx,y∈[0,1/4], then
(78)d(Tx,Ty)=|x2-y2|≤12|x-y|≤ψ(M(x,y))+Lmin⁡{d(x,Ty),d(y,Tx)}.Case 2.
Ifx∈[0,1/4] and y=1, then
(79)d(Tx,Ty)=1-x2≤12(1-x)+1≤ψ(M(x,y))+Lmin⁡{d(x,Ty),d(y,Tx)}.
Therefore, all conditions of Theorem 11 are satisfied and so T has a fixed point in X.


## 5. Applications

In this section, we apply Corollary22 to the following fourth order two-point boundary value problem:
(80)y(4)(t)=f(t,y(t)),t∈[0,1],y(0)=y(1)=y′(0)=y′(1)=0,
which describes the bending of an elastic beam clamped at both endpoints. In [13], using an ordered version of Geraghty’s fixed point result, an existence theorem for a nonnegative solution of (80) is given. The boundary value problem can be written as the integral equation (see [14])
(81)y(t)=∫01‍G(t,s)f(s,y(s))dsfort∈[0,1],
where G(t,s) is the Green’s function given by
(82)G(t,s)=16{t2(1-s)2[(s-t)+2(1-t)s],0≤t≤s≤1,s2(1-t)2[(t-s)+2(1-s)t],0≤s≤t≤1.
Then it is clear that, G(t,s) is continuous on [0,1]×[0,1], G(0,s)=G(1,s)=0, and G(t,s)≥0 for t,s∈[0,1] and sup⁡t∈[0,1]∫01‍G(t,s)ds=1/384.Here we will considerC[0,1] with the uniform metric, that is, for x,y∈C[0,1]d∞(x,y)=sup⁡t∈[0,1]|x(t)-y(t)|. Then (C[0,1],d∞) is a Banach space. For the convenience we consider the operator T defined on C[0,1] by
(83)Tx(t)=∫01‍G(t,s)f(s,x(s))dsfort∈[0,1].Theorem 26.
Under the following assumptions, Problem (80) has a solution.(A)
f:[0,1]×R→R is bounded,(B)
there existψ∈Ψ and θ:C[0,1]×C[0,1]→R such that if θ(x,y)≥0 for x,y∈C[0,1], then for s∈[0,1] we have
(84)0≤f(s,x(s))-f(s,y(s))≤384ψ(|x(s)-y(s)|),(C)
ifθ(x,y)≥0, then θ(Tx,Ty)≥0,(D)
there existsx0∈C[0,1] such that θ(x0,Tx0)≥0,(E)
if{xn} is a sequence in C[0,1] such that θ(xn,xn+1)≥0 for all n∈N∪{0} and xn→x, then θ(xn,x)≥0 for all n∈N∪{0}.Proof.
LetX=C[0,1] with the metric d∞. First, we show that T:X→X. Let x∈X and t,t′∈[0,1], and then, by the continuity of G and the boundedness of f, we have
(85)|Tx(t)-Tx(t′)|=|∫01‍G(t,s)f(s,x(s))ds-∫01‍G(t′,s)f(s,x(s))ds|≤∫01‍|G(t,s)-G(t′,s)||f(s,x(s))|ds≤sup⁡s∈[0,1]|G(t,s)-G(t′,s)|∫01‍|f(s,x(s))|ds≤Msup⁡s∈[0,1]|G(t,s)-G(t′,s)|⟶0for|t-t′|⟶0,
where M>0 such that sup⁡s∈[0,1]|f(s,x(s))|≤M.
Now, defineα:X×X→[0,∞) by
(86)α(x,y)={1,θ(x,y)≥00,otherwise.
Then, it is clear that T is α-admissible. Also, for all x,y∈X with α(x,y)≥1, we have
(87)|Tx(t)-Ty(t)|=|∫01‍G(t,s)f(s,x(s))ds-∫01‍G(t,s)f(s,y(s))ds|≤∫01‍G(t,s)|f(s,x(s))-f(s,y(s))|ds≤∫01‍G(t,s)384ψ(|x(s)-y(s)|)ds≤384ψ(d∞(x,y))∫01‍G(t,s)ds≤ψ(d∞(x,y))
for all t∈[0,1]. Therefore
(88)α(x,y)≥1⟹d∞(Tx,Ty)≤ψ(d∞(x,y))
hold. Finally, by the condition (E), X is regular. Therefore all hypotheses of Corollary 22 are satisfied and so T has a fixed point in X. Thus, the problem (80) has a solution in C[0,1].In the following, we also give an existence and uniqueness theorem for (80) under slight different conditions.Theorem 27.
Under the following assumptions, problem (80) has a unique solution.(F)
f:[0,1]×R→R is bounded, nondecreasing with respect to second variable and f(t,0)≥0 for all t∈[0,1],(G)
there existsψ∈Ψ such that, for x,y∈C[0,1] with x⪯y and s∈[0,1], we have
(89)f(s,y(s))-f(s,x(s))≤384ψ(|x(s)-y(s)|),where x⪯y⇔x(t)≤y(t) for all t∈[0,1].Proof.
As in the proof of Theorem26, we can show that T:X→X. Now define α:X×X→[0,∞) by
(90)α(x,y)={1,x⪯y0,otherwise.
Therefore, if α(x,y)≥1, then x⪯y and x(t)≤y(t) for all t∈[0,1]. Since f is nondecreasing with respect to second variable, we have
(91)Tx(t)=∫01‍G(t,s)f(s,x(s))ds≤∫01‍G(t,s)f(s,y(s))ds=Ty(t)
for all t∈[0,1]. Thus, we have Tx⪯Ty and so α(Tx,Ty)≥1. It shows that T is α-admissible. Also, for all x,y∈X with α(x,y)≥1, we have
(92)|Tx(t)-Ty(t)|=|∫01‍G(t,s)f(s,x(s))ds-∫01‍G(t,s)f(s,y(s))ds|≤∫01‍G(t,s)|f(s,x(s))-f(s,y(s))|ds=∫01‍G(t,s)[f(s,y(s))-f(s,x(s))]ds≤∫01‍G(t,s)384ψ(|x(s)-y(s)|)ds≤384ψ(d∞(x,y))∫01‍G(t,s)ds≤ψ(d∞(x,y))
for all t∈[0,1]. Therefore
(93)α(x,y)≥1⟹d∞(Tx,Ty)≤ψ(d∞(x,y))
hold. Also, since f(s,0)≥0, we have 0≤∫01‍G(t,s)f(s,0)ds=T0 and so α(0,T0)≥1. Now, let α(xn,xn+1)≥1 for all n∈N∪{0} and xn→x in X. Then, we have xn⪯xn+1 and so xn(t)≤xn+1(t) for all t∈[0,1]. Therefore, xn(t)≤x(t) for all t∈[0,1] and n∈N∪{0}. That is, xn⪯x for all n∈N∪{0} and so α(xn,x)≥1. Hence X is regular. Therefore all existence hypotheses of Corollary 22 are satisfied and so T has a fixed point in X. Thus, the problem (80) has a solution in C[0,1]. Finally, let x and y be two solutions of (80), and then u=max⁡{x,y}∈C[0,1] and x⪯u and y⪯u. Thus, α(x,u)≥1 and α(y,u)≥1. Therefore, by Corollary 22 the solution is unique.Example 28.
Consider the nonlinear fourth order two point boundary value problem(94)y(4)(t)=γt+(1+t2)arctan(βy(t)),t∈[0,1],y(0)=y(1)=y′(0)=y′(1)=0,
where γ≥0,0≤β≤191. In this case, f(t,y)=γt+(1+t2)arctan(βy). It is easy to see that f:[0,1]×R→R is bounded and nondecreasing with respect to the second variable. Also, f(t,0)=γt≥0 for all t∈[0,1]. Now let x,y∈C[0,1] with x⪯y then, for all s∈[0,1],
(95)f(s,y(s))-f(s,x(s))=(1+s2)[arctan(βy(s))-arctan(βx(s))]≤(1+s2)arctan(β[y(s)-x(s)])≤(1+s2)β[y(s)-x(s)]≤2β[y(s)-x(s)]≤382[y(s)-x(s)]=384ψ(|y(s)-x(s)|),
where ψ(t)=191t/192. Therefore, the conditions (F) and (G) of Theorem 27 are satisfied and so the boundary value problem (94) has a unique solution.



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*Source: 869123-2014-07-22.xml*