# Generalizedα-ψ Contractive Type Mappings and Related Fixed Point Theorems with Applications

Authors: Erdal Karapınar; Bessem Samet
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

We establish fixed point theorems for a new class of contractive mappings. As consequences of our main results, we obtain fixed point theorems on metric spaces endowed with a partial order and fixed point theorems for cyclic contractive mappings. Various examples are presented to illustrate our obtained results.


---

## Body

## 1. Introduction and Preliminaries

LetΨ be the family of functions ψ:[0,∞)→[0,∞) satisfying the following conditions: (Ψ1)
ψ is nondecreasing;(Ψ2)
∑n=1+∞‍ψn(t)<∞ for all t>0, where ψn is the nth iterate of ψ. These functions are known in the literature as (c)-comparison functions. It is easily proved that if ψ is a (c)-comparison function, then ψ(t)<t for any t>0.Very recently, Samet et al. [1] introduced the following concepts.Definition 1.1.
Let(X,d) be a metric space and T:X→X be a given mapping. We say that T is an α-ψ contractive mapping if there exist two functions α:X×X→[0,∞) and ψ∈Ψ such that
(1.1)α(x,y)d(Tx,Ty)≤ψ(d(x,y)),∀x,y∈X.
Clearly, any contractive mapping, that is, a mapping satisfying Banach contraction, is anα-ψ contractive mapping with α(x,y)=1 for all x,y∈X and ψ(t)=kt, k∈(0,1).Definition 1.2.
LetT:X→X and α:X×X→[0,∞). We say that T is α-admissible if for all x,y∈X, and we have
(1.2)α(x,y)≥1⟹α(Tx,Ty)≥1.
Various examples of such mappings are presented in [1].
The main results in [1] are the following fixed point theorems.Theorem 1.3.
Let(X,d) be a complete metric space and T:X→X be an α-ψ contractive mapping. Suppose that(i)
T is α admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1;(iii)
T is continuous. Then there exists u∈X such that Tu=u.Theorem 1.4.
Let(X,d) be a complete metric space and T:X→X be an α-ψ contractive mapping. Suppose that (i)
T is α admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1;(iii)
if{xn} is a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→∞, then α(xn,x)≥1 for all n. Then there exists u∈X such that Tu=u.Theorem 1.5.
Adding to the hypotheses of Theorem1.3 (resp., Theorem 1.4) the condition, for all x,y∈X, there exists z∈X such that α(x,z)≥1 and α(y,z)≥1, and one obtains uniqueness of the fixed point.In the present work, we introduce the concept of generalizedα-ψ contractive type mappings, and we study the existence and uniqueness of fixed points for such mappings. Presented theorems in this paper extend and generalize the above results derived by Samet et al. in [1]. Moreover, from our fixed point theorems, we will deduce various fixed point results on metric spaces endowed with a partial order and fixed point results for cyclic contractive mappings.


## 2. Main Results

We introduce the concept of generalizedα-ψ contractive type mappings as follows.Definition 2.1.
Let(X,d) be a metric space and T:X→X be a given mapping. We say that T is a generalized α-ψ contractive mapping if there exist two functions α:X×X→[0,∞) and ψ∈Ψ such that for all x,y∈X, and we have
(2.1)α(x,y)d(Tx,Ty)≤ψ(M(x,y)),
where M(x,y)=max{d(x,y),(d(x,Tx)+d(y,Ty))/2,(d(x,Ty)+d(y,Tx))/2}.Remark 2.2.
Clearly, sinceψ is nondecreasing, every α-ψ contractive mapping is a generalized α-ψ contractive mapping.Our first result is the following.Theorem 2.3.
Let(X,d) be a complete metric space. Suppose that T:X→X is a generalized α-ψ contractive mapping and satisfies the following conditions: (i)
T is α admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1;(iii)
T is continuous. Then there exists u∈X such that Tu=u.Proof.
Letx0∈X such that α(x0,Tx0)≥1 (such a point exists from condition (ii)). Define the sequence {xn} in X by xn+1=Txn for all n≥0. If xn0=xn0+1 for some n0, then u=xn0 is a fixed point of T. So, we can assume that xn≠xn+1 for all n. Since T is α admissible, we have
(2.2)α(x0,x1)=α(x0,Tx0)≥1⟹α(Tx0,Tx1)=α(x1,x2)≥1.
Inductively, we have
(2.3)α(xn,xn+1)≥1,∀n=0,1,….
From (2.1) and (2.3), it follows that for all n≥1, we have
(2.4)d(xn+1,xn)=d(Txn,Txn-1)≤α(xn,xn-1)d(Txn,Txn-1)≤ψ(M(xn,xn-1)).
On the other hand, we have
(2.5)M(xn,xn-1)=max⁡{d(xn,xn-1),d(xn,Txn)+d(xn-1,Txn-1)2,d(xn,Txn-1)+d(xn-1,Txn)2}=max⁡{d(xn,xn-1),d(xn,xn+1)+d(xn-1,xn)2,d(xn-1,xn+1)2}≤max⁡{d(xn,xn-1),d(xn,xn+1)+d(xn-1,xn)2}≤max⁡{{∑}d(xn,xn-1),d(xn,xn+1)}.
From (2.4) and taking in consideration that ψ is a nondecreasing function, we get that
(2.6)d(xn+1,xn)≤ψ(max⁡{d(xn,xn-1),d(xn,xn+1)}),
for all n≥1. If for some n≥1, we have d(xn,xn-1)≤d(xn,xn+1), from (2.6), we obtain that
(2.7)d(xn+1,xn)≤ψ(d(xn,xn+1))<d(xn,xn+1),
a contradiction. Thus, for all n≥1, we have
(2.8)max⁡{d(xn,xn-1),d(xn,xn+1)}=d(xn,xn-1).
Using (2.6) and (2.8), we get that
(2.9)d(xn+1,xn)≤ψ(d(xn,xn-1)),
for all n≥1. By induction, we get
(2.10)d(xn+1,xn)≤ψn(d(x1,x0)),∀n≥1.
From (2.10) and using the triangular inequality, for all k≥1, we have
(2.11)d(xn,xn+k)≤d(xn,xn+1)+⋯+d(xn+k-1,xn+k)≤∑p=nn+k-1‍ψn(d(x1,x0))≤∑p=n+∞‍ψn(d(x1,x0))⟶0asn→∞.
This implies that {xn} is a Cauchy sequence in (X,d). Since (X,d) is complete, there exists u∈X such that
(2.12)lim⁡n→∞d(xn,u)=0.
Since T is continuous, we obtain from (2.12) that
(2.13)lim⁡n→∞d(xn+1,Tu)=lim⁡n→∞d(Txn,Tu)=0.
From (2.12), (2.13) and the uniqueness of the limit, we get immediately that u is a fixed point of T, that is, Tu=u.The next theorem does not require the continuity ofT.Theorem 2.4.
Let(X,d) be a complete metric space. Suppose that T:X→X is a generalized α-ψ contractive mapping and the following conditions hold: (i)
T is α admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1;(iii)
if{xn} is a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→∞, then there exists a subsequence {xn(k)} of {xn} such that α(xn(k),x)≥1 for all k. Then there exists u∈X such that Tu=u.Proof.
Following the proof of Theorem2.3, we know that the sequence {xn} defined by xn+1=Txn for all n≥0, converges for some u∈X. From (2.3) and condition (iii), there exists a subsequence {xn(k)} of {xn} such that α(xn(k),u)≥1 for all k. Applying (2.1), for all k, we get that
(2.14)d(xn(k)+1,Tu)=d(Txn(k),Tu)≤α(xn(k),u)d(Txn(k),Tu)≤ψ(M(xn(k),u)).
On the other hand, we have
(2.15)M(xn(k),u)=max⁡{d(xn(k),u),d(xn(k),xn(k)+1)+d(u,Tu)2,d(xn(k),Tu)+d(u,xn(k)+1)2}.
Letting k→∞ in the above equality, we get that
(2.16)lim⁡k→∞M(xn(k),u)=d(u,Tu)2.
Suppose that d(u,Tu)>0. From (2.16), for k large enough, we have M(xn(k),u)>0, which implies that ψ(M(xn(k),u))<M(xn(k),u). Thus, from (2.14), we have
(2.17)d(xn(k)+1,Tu)<M(xn(k),u).
Letting k→∞ in the above inequality, using (2.16), we obtain that
(2.18)d(u,Tu)≤d(u,Tu)2,
which is a contradiction. Thus we have d(u,Tu)=0, that is, u=Tu.With the following example, we will show that hypotheses in Theorems2.3 and 2.4 do not guarantee uniqueness of the fixed point.Example 2.5.
LetX={(1,0),(0,1)}⊂ℝ2 be endowed with the Euclidean distance d((x,y),(u,v))=|x-u|+|y-v| for all (x,y),(u,v)∈X. Obviously, (X,d) is a complete metric space. The mapping T(x,y)=(x,y) is trivially continuous and satisfies for any ψ∈Ψ(2.19)α((x,y),(u,v))d(T(x,y),T(u,v))≤ψ(M((x,y),(u,v))),
for all (x,y),(u,v)∈X, where
(2.20)α((x,y),(u,v))={1if(x,y)=(u,v),0if(x,y)≠(u,v).
Thus T is a generalized α-ψ contractive mapping. On the other hand, for all (x,y),(u,v)∈X, we have
(2.21)α((x,y),(u,v))≥1⟶(x,y)=(u,v)⟶T(x,y)=T(u,v)⟶α(T(x,y),T(u,v))≥1.
Thus T is α admissible. Moreover, for all (x,y)∈X, we have α((x,y),T(x,y))≥1. Then the assumptions of Theorem 2.3 are satisfied. Note that the assumptions of Theorem 2.4 are also satisfied; indeed if {(xn,yn)} is a sequence in X that converges to some point (x,y)∈X  with α((xn,yn),(xn+1,yn+1))≥1 for all n, then, from the definition of α, we have (xn,yn)=(x,y) for all n, which implies that α((xn,yn),(x,y))=1 for all n. However, in this case, T has two fixed points in X.For the uniqueness of a fixed point of a generalizedα-ψ contractive mapping, we will consider the following hypothesis. (H)
For allx,y∈Fix(T), there exists z∈X such that α(x,z)≥1 and α(y,z)≥1.Here,Fix(T) denotes the set of fixed points of T.Theorem 2.6.
Adding condition(H) to the hypotheses of Theorem 2.3 (resp., Theorem 2.4), one has obtains that u is the unique fixed point of T.Proof.
Suppose thatv is another fixed point of T. From (H), there exists z∈X such that
(2.22)α(u,z)≥1,α(v,z)≥1.
Since T is α admissible, from (2.22), we have
(2.23)α(u,Tnz)≥1,α(v,Tnz)≥1,∀n.
Define the sequence {zn} in X by zn+1=Tzn for all n≥0 and z0=z. From (2.23), for all n, we have
(2.24)d(u,zn+1)=d(Tu,Tzn)≤α(u,zn)d(Tu,Tzn)≤ψ(M(u,zn)).
On the other hand, we have
(2.25)M(u,zn)=max⁡{d(u,zn),d(zn,zn+1)2,d(u,zn+1)+d(zn,u)2}≤max⁡{d(u,zn),d(zn,u)+d(u,zn+1)2}≤max⁡{d(u,zn),d(u,zn+1)}.
Using the above inequality, (2.24) and the monotone property of ψ, we get that
(2.26)d(u,zn+1)≤ψ(max⁡{d(u,zn),d(u,zn+1)}),
for all n. Without restriction to the generality, we can suppose that d(u,zn)>0 for all n. If max{d(u,zn),d(u,zn+1)}=d(u,zn+1), we get from (2.26) that
(2.27)d(u,zn+1)≤ψ(d(u,zn+1))<d(u,zn+1),
which is a contradiction. Thus we have max{d(u,zn),d(u,zn+1)}=d(u,zn), and
(2.28)d(u,zn+1)≤ψ(d(u,zn)),
for all n. This implies that
(2.29)d(u,zn)≤ψn(d(u,z0)),∀n≥1.
Letting n→∞ in the above inequality, we obtain that
(2.30)lim⁡n→∞d(zn,u)=0.
Similarly, one can show that
(2.31)lim⁡n→∞d(zn,v)=0.
From (2.30) and (2.31), it follows that u=v. Thus we proved that u is the unique fixed point of T.Example 2.7.
LetX=[0,1] be endowed with the standard metric d(x,y)=|x-y| for all x,y∈X. Obviously, (X,d) is a complete metric space. Define the mapping T:X→X by
(2.32)Tx={14ifx∈[0,1),0ifx=1.
In this case, T is not continuous. Define the mapping α:X×X→[0,∞) by
(2.33)α(x,y)={1if(x,y)∈([0,14]×[14,1])∪([14,1]×[0,14]),0otherwise.
We will prove that (A)
T:X→X is a generalized α-ψ contractive mapping, where ψ(t)=t/2 for all t≥0;(B)
T is α-admissible;(C)
there existsx0∈X such that α(x0,Tx0)≥1;(D)
if{xn} is a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→+∞, then there exists a subsequence {xn(k)} of {xn} such that α(xn(k),x)≥1 for all k;(E)
condition (H) is satisfied.Proof of (A).
To show (A), we have to prove that (2.1) is satisfied for every x,y∈X. If x∈[0,1/4] and y=1, we have
(2.34)α(x,y)d(Tx,Ty)=d(Tx,Ty)=|14-0|=14d(y,Ty)≤ψ(M(x,y)).
Then (2.1) holds. If x=1 and y∈[0,1/4], we have
(2.35)α(x,y)d(Tx,Ty)=d(Tx,Ty)=|0-14|=14d(x,Tx)≤ψ(M(x,y)).
Then (2.1) holds also in this case. The other cases are trivial. Thus (2.1) is satisfied for every x,y∈X.Proof of (B).
Let(x,y)∈X×X such that α(x,y)≥1. From the definition of α, we have two cases.
Case  1 (if (x,y)∈[0,1/4]×[1/4,1]). In this case, we have (Tx,Ty)∈[1/4,1]×[0,1/4], which implies that α(Tx,Ty)=1.
Case  2 (if (x,y)∈[1/4,1]×[0,1/4]). In this case, we have (Tx,Ty)∈[0,1/4]×[1/4,1], which implies that α(Tx,Ty)=1.
So, in all cases, we haveα(Tx,Ty)≥1. Thus T is α admissible.Proof of (C).
Takingx0=0, we have α(x0,Tx0)=α(0,1/4)=1.Proof of (D).
Let{xn} be a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x as n→+∞ for some x∈X. From the definition of α, for all n, we have
(2.36)(xn,xn+1)∈([0,14]×[14,1])∪([14,1]×[0,14]).
Since ([0,1/4]×[1/4,1])∪([1/4,1]×[0,1/4]) is a closed set with respect to the Euclidean metric, we get that
(2.37)(x,x)∈([0,14]×[14,1])∪([14,1]×[0,14]),
which implies that x=1/4. Thus we have α(xn,x)≥1 for all n.Proof of (E).
Let(x,y)∈X×X. It is easy to show that, for z=1/4, we have α(x,z)=α(y,z)=1. So, condition (H) is satisfied.Conclusion.
Now, all the hypotheses of Theorem2.6 are satisfied; thus T has a unique fixed point u∈X. In this case, we have u=1/4.


## 3. Consequences

Now, we will show that many existing results in the literature can be deduced easily from our Theorem2.6.


### 3.1. Standard Fixed Point Theorems

Taking in Theorem2.6, α(x,y)=1 for all x,y∈X, we obtain immediately the following fixed point theorem.Corollary 3.1.
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exists a function ψ∈Ψ such that
(3.1)d(Tx,Ty)≤ψ(M(x,y)),
for all x,y∈X. Then T has a unique fixed point.The following fixed point theorems follow immediately from Corollary3.1.Corollary 3.2 (see Berinde [2]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exists a function ψ∈Ψ such that
(3.2)d(Tx,Ty)≤ψ(d(x,y)),
for all x,y∈X. Then T has a unique fixed point.Corollary 3.3 (see Ćirić [3]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exists a constant λ∈(0,1) such that
(3.3)d(Tx,Ty)≤λmax⁡{d(x,y),d(x,Tx)+d(y,Ty)2,d(x,Ty)+d(y,Tx)2},
for all x,y∈X. Then T has a unique fixed point.Corollary 3.4 (see Hardy and Rogers [4]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exist constants A,B,C≥0 with (A+2B+2C)∈(0,1) such that
(3.4)d(Tx,Ty)≤Ad(x,y)+B[d(x,Tx)+d(y,Ty)]+C[d(x,Ty)+d(y,Tx)],
for all x,y∈X. Then T has a unique fixed point.Corollary 3.5 (see Banach Contraction Principle [5]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exists a constant λ∈(0,1) such that
(3.5)d(Tx,Ty)≤λd(x,y),
for all x,y∈X. Then T has a unique fixed point.Corollary 3.6 (see Kannan [6]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exists a constant λ∈(0,1/2) such that
(3.6)d(Tx,Ty)≤λ[d(x,Tx)+d(y,Ty)],
for all x,y∈X. Then T has a unique fixed point.Corollary 3.7 (see Chatterjea [7]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exists a constant λ∈(0,1/2) such that
(3.7)d(Tx,Ty)≤λ[d(x,Ty)+d(y,Tx)],
for all x,y∈X. Then T has a unique fixed point.


### 3.2. Fixed Point Theorems on Metric Spaces Endowed with a Partial Order

Recently there have been so many exciting developments in the field of existence of fixed point on metric spaces endowed with partial orders. This trend was started by Turinici [8] in 1986. Ran and Reurings in [9] extended the Banach contraction principle in partially ordered sets with some applications to matrix equations. The obtained result in [9] was further extended and refined by many authors (see, e.g., [10–15] and the references cited therein). In this section, from our Theorem 2.6, we will deduce very easily various fixed point results on a metric space endowed with a partial order. At first, we need to recall some concepts.Definition 3.8.
Let(X,⪯) be a partially ordered set and T:X→X be a given mapping. We say that T is nondecreasing with respect to ⪯ if
(3.8)x,y∈X,x⪯y⟹Tx⪯Ty.Definition 3.9.
Let(X,⪯) be a partially ordered set. A sequence {xn}⊂X is said to be nondecreasing with respect to ⪯ if xn⪯xn+1 for all n.Definition 3.10.
Let(X,⪯) be a partially ordered set and d be a metric on X. We say that (X,⪯,d) is regular if for every nondecreasing sequence {xn}⊂X such that xn→x∈X as n→∞, there exists a subsequence {xn(k)} of {xn} such that xn(k)⪯x for all k.We have the following result.Corollary 3.11.
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exists a function ψ∈Ψ such that
(3.9)d(Tx,Ty)≤ψ(M(x,y)),
for all x,y∈X with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular. Then T has a fixed point. Moreover, if for all x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.Proof.
Define the mappingα:X×X→[0,∞) by
(3.10)α(x,y)={1ifx⪯yorx⪰y,0otherwise.
Clearly, T is a generalized α-ψ contractive mapping, that is,
(3.11)α(x,y)d(Tx,Ty)≤ψ(M(x,y)),
for all x,y∈X. From condition (i), we have α(x0,Tx0)≥1. Moreover, for all x,y∈X, from the monotone property of T, we have
(3.12)α(x,y)≥1⟹x⪰yorx⪯y⟹Tx⪰TyorTx⪯Ty⟹α(Tx,Ty)≥1.
Thus T is α admissible. Now, if T is continuous, the existence of a fixed point follows from Theorem 2.3. Suppose now that (X,⪯,d) is regular. Let {xn} be a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→∞. From the regularity hypothesis, there exists a subsequence {xn(k)} of {xn} such that xn(k)⪯x for all k. This implies from the definition of α that α(xn(k),x)≥1 for all k. In this case, the existence of a fixed point follows from Theorem 2.4. To show the uniqueness, and let x,y∈X. By hypothesis, there exists z∈X such that x⪯z and y⪯z, which implies from the definition of α that α(x,z)≥1 and α(y,z)≥1. Thus we deduce the uniqueness of the fixed point by Theorem 2.6.The following results are immediate consequences of Corollary3.11.Corollary 3.12.
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exists a function ψ∈Ψ such that
(3.13)d(Tx,Ty)≤ψ(d(x,y)),
for all x,y∈X  with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular.
ThenT has a fixed point. Moreover, if for all x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.Corollary 3.13.
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exists a constant λ∈(0,1) such that
(3.14)d(Tx,Ty)≤λmax⁡{d(x,y),d(x,Tx)+d(y,Ty)2,d(x,Ty)+d(y,Tx)2},
for all x,y∈X  with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular.
ThenT has a fixed point. Moreover, if for all x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.Corollary 3.14.
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exist constants A,B,C≥0 with (A+2B+2C)∈(0,1) such that
(3.15)d(Tx,Ty)≤Ad(x,y)+B[d(x,Tx)+d(y,Ty)]+C[d(x,Ty)+d(y,Tx)],
for all x,y∈X with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular.
ThenT has a fixed point. Moreover, if for all x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.Corollary 3.15 (see Ran and Reurings [9], Nieto and López [16]).
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exists a constant λ∈(0,1) such that
(3.16)d(Tx,Ty)≤λd(x,y),
for all x,y∈X with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular.
ThenT has a fixed point. Moreover, if for all x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.Corollary 3.16.
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exists a constant λ∈(0,1/2) such that
(3.17)d(Tx,Ty)≤λ[d(x,Tx)+d(y,Ty)],
for all x,y∈X with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular.
ThenT has a fixed point. Moreover, if for all x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.Corollary 3.17.
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exists a constant λ∈(0,1/2) such that
(3.18)d(Tx,Ty)≤λ[d(x,Ty)+d(y,Tx)],
for all x,y∈X with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular.
ThenT has a fixed point. Moreover, if for all  x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.


### 3.3. Fixed Point Theorems for Cyclic Contractive Mappings

One of the remarkable generalizations of the Banach Contraction Mapping Principle was reported by Kirk et al. [17] via cyclic contraction. Following the paper [17], many fixed point theorems for cyclic contractive mappings have appeared (see, e.g., [18–23]). In this section, we will show that, from our Theorem 2.6, we can deduce some fixed point theorems for cyclic contractive mappings.We have the following result.Corollary 3.18.
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exists a functionψ∈Ψ such that
(3.19)d(Tx,Ty)≤ψ(M(x,y)),∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.Proof.
SinceA1 and A2 are closed subsets of the complete metric space (X,d), then (Y,d) is complete. Define the mapping α:Y×Y→[0,∞) by
(3.20)α(x,y)={1if(x,y)∈(A1×A2)∪(A2×A1),0otherwise.
From (II) and the definition of α, we can write
(3.21)α(x,y)d(Tx,Ty)≤ψ(M(x,y)),
for all x,y∈Y. Thus T is a generalized α-ψ contractive mapping.
Let(x,y)∈Y×Y such that α(x,y)≥1. If (x,y)∈A1×A2, from (I), (Tx,Ty)∈A2×A1, which implies that α(Tx,Ty)≥1. If (x,y)∈A2×A1, from (I), (Tx,Ty)∈A1×A2, which implies that α(Tx,Ty)≥1. Thus in all cases, we have α(Tx,Ty)≥1. This implies that T is α-admissible.
Also, from (I), for anya∈A1, we have (a,Ta)∈A1×A2, which implies that α(a,Ta)≥1.
Now, let{xn} be a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→∞. This implies from the definition of α that
(3.22)(xn,xn+1)∈(A1×A2)∪(A2×A1),∀n.
Since (A1×A2)∪(A2×A1) is a closed set with respect to the Euclidean metric, we get that
(3.23)(x,x)∈(A1×A2)∪(A2×A1),
which implies that x∈A1∩A2. Thus we get immediately from the definition of α that α(xn,x)≥1 for all n.
Finally, letx,y∈Fix(T). From (I), this implies that x,y∈A1∩A2. So, for any z∈Y, we have α(x,z)≥1 and α(y,z)≥1. Thus condition (H) is satisfied.
Now, all the hypotheses of Theorem2.6 are satisfied, and we deduce that T has a unique fixed point that belongs to A1∩A2 (from (I)).The following results are immediate consequences of Corollary3.18.Corollary 3.19 (see Pacurar and Rus [21]).
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exists a functionψ∈Ψ such that
(3.24)d(Tx,Ty)≤ψ(d(x,y)),∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.Corollary 3.20.
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exists a constantλ∈(0,1) such that
(3.25)d(Tx,Ty)≤λmax⁡{d(x,y),d(x,Tx)+d(y,Ty)2,d(x,Ty)+d(y,Tx)2},∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.Corollary 3.21.
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exist constantsA,B,C≥0 with (A+2B+2C)∈(0,1) such that
(3.26)d(Tx,Ty)≤Ad(x,y)+B[d(x,Tx)+d(y,Ty)]+C[d(x,Ty)+d(y,Tx)],∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.Corollary 3.22 (see Kirk et al. [17]).
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exists a constantλ∈(0,1) such that
(3.27)d(Tx,Ty)≤λd(x,y),∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.Corollary 3.23.
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exists a constantλ∈(0,1/2) such that
(3.28)d(Tx,Ty)≤λ[d(x,Tx)+d(y,Ty)],∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.Corollary 3.24.
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exists a constantλ∈(0,1/2) such that
(3.29)d(Tx,Ty)≤λ[d(x,Ty)+d(y,Tx)],∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.


## 3.1. Standard Fixed Point Theorems

Taking in Theorem2.6, α(x,y)=1 for all x,y∈X, we obtain immediately the following fixed point theorem.Corollary 3.1.
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exists a function ψ∈Ψ such that
(3.1)d(Tx,Ty)≤ψ(M(x,y)),
for all x,y∈X. Then T has a unique fixed point.The following fixed point theorems follow immediately from Corollary3.1.Corollary 3.2 (see Berinde [2]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exists a function ψ∈Ψ such that
(3.2)d(Tx,Ty)≤ψ(d(x,y)),
for all x,y∈X. Then T has a unique fixed point.Corollary 3.3 (see Ćirić [3]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exists a constant λ∈(0,1) such that
(3.3)d(Tx,Ty)≤λmax⁡{d(x,y),d(x,Tx)+d(y,Ty)2,d(x,Ty)+d(y,Tx)2},
for all x,y∈X. Then T has a unique fixed point.Corollary 3.4 (see Hardy and Rogers [4]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exist constants A,B,C≥0 with (A+2B+2C)∈(0,1) such that
(3.4)d(Tx,Ty)≤Ad(x,y)+B[d(x,Tx)+d(y,Ty)]+C[d(x,Ty)+d(y,Tx)],
for all x,y∈X. Then T has a unique fixed point.Corollary 3.5 (see Banach Contraction Principle [5]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exists a constant λ∈(0,1) such that
(3.5)d(Tx,Ty)≤λd(x,y),
for all x,y∈X. Then T has a unique fixed point.Corollary 3.6 (see Kannan [6]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exists a constant λ∈(0,1/2) such that
(3.6)d(Tx,Ty)≤λ[d(x,Tx)+d(y,Ty)],
for all x,y∈X. Then T has a unique fixed point.Corollary 3.7 (see Chatterjea [7]).
Let(X,d) be a complete metric space and T:X→X be a given mapping. Suppose that there exists a constant λ∈(0,1/2) such that
(3.7)d(Tx,Ty)≤λ[d(x,Ty)+d(y,Tx)],
for all x,y∈X. Then T has a unique fixed point.


## 3.2. Fixed Point Theorems on Metric Spaces Endowed with a Partial Order

Recently there have been so many exciting developments in the field of existence of fixed point on metric spaces endowed with partial orders. This trend was started by Turinici [8] in 1986. Ran and Reurings in [9] extended the Banach contraction principle in partially ordered sets with some applications to matrix equations. The obtained result in [9] was further extended and refined by many authors (see, e.g., [10–15] and the references cited therein). In this section, from our Theorem 2.6, we will deduce very easily various fixed point results on a metric space endowed with a partial order. At first, we need to recall some concepts.Definition 3.8.
Let(X,⪯) be a partially ordered set and T:X→X be a given mapping. We say that T is nondecreasing with respect to ⪯ if
(3.8)x,y∈X,x⪯y⟹Tx⪯Ty.Definition 3.9.
Let(X,⪯) be a partially ordered set. A sequence {xn}⊂X is said to be nondecreasing with respect to ⪯ if xn⪯xn+1 for all n.Definition 3.10.
Let(X,⪯) be a partially ordered set and d be a metric on X. We say that (X,⪯,d) is regular if for every nondecreasing sequence {xn}⊂X such that xn→x∈X as n→∞, there exists a subsequence {xn(k)} of {xn} such that xn(k)⪯x for all k.We have the following result.Corollary 3.11.
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exists a function ψ∈Ψ such that
(3.9)d(Tx,Ty)≤ψ(M(x,y)),
for all x,y∈X with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular. Then T has a fixed point. Moreover, if for all x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.Proof.
Define the mappingα:X×X→[0,∞) by
(3.10)α(x,y)={1ifx⪯yorx⪰y,0otherwise.
Clearly, T is a generalized α-ψ contractive mapping, that is,
(3.11)α(x,y)d(Tx,Ty)≤ψ(M(x,y)),
for all x,y∈X. From condition (i), we have α(x0,Tx0)≥1. Moreover, for all x,y∈X, from the monotone property of T, we have
(3.12)α(x,y)≥1⟹x⪰yorx⪯y⟹Tx⪰TyorTx⪯Ty⟹α(Tx,Ty)≥1.
Thus T is α admissible. Now, if T is continuous, the existence of a fixed point follows from Theorem 2.3. Suppose now that (X,⪯,d) is regular. Let {xn} be a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→∞. From the regularity hypothesis, there exists a subsequence {xn(k)} of {xn} such that xn(k)⪯x for all k. This implies from the definition of α that α(xn(k),x)≥1 for all k. In this case, the existence of a fixed point follows from Theorem 2.4. To show the uniqueness, and let x,y∈X. By hypothesis, there exists z∈X such that x⪯z and y⪯z, which implies from the definition of α that α(x,z)≥1 and α(y,z)≥1. Thus we deduce the uniqueness of the fixed point by Theorem 2.6.The following results are immediate consequences of Corollary3.11.Corollary 3.12.
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exists a function ψ∈Ψ such that
(3.13)d(Tx,Ty)≤ψ(d(x,y)),
for all x,y∈X  with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular.
ThenT has a fixed point. Moreover, if for all x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.Corollary 3.13.
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exists a constant λ∈(0,1) such that
(3.14)d(Tx,Ty)≤λmax⁡{d(x,y),d(x,Tx)+d(y,Ty)2,d(x,Ty)+d(y,Tx)2},
for all x,y∈X  with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular.
ThenT has a fixed point. Moreover, if for all x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.Corollary 3.14.
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exist constants A,B,C≥0 with (A+2B+2C)∈(0,1) such that
(3.15)d(Tx,Ty)≤Ad(x,y)+B[d(x,Tx)+d(y,Ty)]+C[d(x,Ty)+d(y,Tx)],
for all x,y∈X with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular.
ThenT has a fixed point. Moreover, if for all x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.Corollary 3.15 (see Ran and Reurings [9], Nieto and López [16]).
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exists a constant λ∈(0,1) such that
(3.16)d(Tx,Ty)≤λd(x,y),
for all x,y∈X with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular.
ThenT has a fixed point. Moreover, if for all x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.Corollary 3.16.
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exists a constant λ∈(0,1/2) such that
(3.17)d(Tx,Ty)≤λ[d(x,Tx)+d(y,Ty)],
for all x,y∈X with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular.
ThenT has a fixed point. Moreover, if for all x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.Corollary 3.17.
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:X→X be a nondecreasing mapping with respect to ⪯. Suppose that there exists a constant λ∈(0,1/2) such that
(3.18)d(Tx,Ty)≤λ[d(x,Ty)+d(y,Tx)],
for all x,y∈X with x≽y. Suppose also that the following conditions hold: (i)
there existsx0∈X such that x0⪯Tx0;(ii)
T is continuous or (X,⪯,d) is regular.
ThenT has a fixed point. Moreover, if for all  x,y∈X there exists z∈X such that x⪯z and y⪯z, one has uniqueness of the fixed point.


## 3.3. Fixed Point Theorems for Cyclic Contractive Mappings

One of the remarkable generalizations of the Banach Contraction Mapping Principle was reported by Kirk et al. [17] via cyclic contraction. Following the paper [17], many fixed point theorems for cyclic contractive mappings have appeared (see, e.g., [18–23]). In this section, we will show that, from our Theorem 2.6, we can deduce some fixed point theorems for cyclic contractive mappings.We have the following result.Corollary 3.18.
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exists a functionψ∈Ψ such that
(3.19)d(Tx,Ty)≤ψ(M(x,y)),∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.Proof.
SinceA1 and A2 are closed subsets of the complete metric space (X,d), then (Y,d) is complete. Define the mapping α:Y×Y→[0,∞) by
(3.20)α(x,y)={1if(x,y)∈(A1×A2)∪(A2×A1),0otherwise.
From (II) and the definition of α, we can write
(3.21)α(x,y)d(Tx,Ty)≤ψ(M(x,y)),
for all x,y∈Y. Thus T is a generalized α-ψ contractive mapping.
Let(x,y)∈Y×Y such that α(x,y)≥1. If (x,y)∈A1×A2, from (I), (Tx,Ty)∈A2×A1, which implies that α(Tx,Ty)≥1. If (x,y)∈A2×A1, from (I), (Tx,Ty)∈A1×A2, which implies that α(Tx,Ty)≥1. Thus in all cases, we have α(Tx,Ty)≥1. This implies that T is α-admissible.
Also, from (I), for anya∈A1, we have (a,Ta)∈A1×A2, which implies that α(a,Ta)≥1.
Now, let{xn} be a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→∞. This implies from the definition of α that
(3.22)(xn,xn+1)∈(A1×A2)∪(A2×A1),∀n.
Since (A1×A2)∪(A2×A1) is a closed set with respect to the Euclidean metric, we get that
(3.23)(x,x)∈(A1×A2)∪(A2×A1),
which implies that x∈A1∩A2. Thus we get immediately from the definition of α that α(xn,x)≥1 for all n.
Finally, letx,y∈Fix(T). From (I), this implies that x,y∈A1∩A2. So, for any z∈Y, we have α(x,z)≥1 and α(y,z)≥1. Thus condition (H) is satisfied.
Now, all the hypotheses of Theorem2.6 are satisfied, and we deduce that T has a unique fixed point that belongs to A1∩A2 (from (I)).The following results are immediate consequences of Corollary3.18.Corollary 3.19 (see Pacurar and Rus [21]).
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exists a functionψ∈Ψ such that
(3.24)d(Tx,Ty)≤ψ(d(x,y)),∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.Corollary 3.20.
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exists a constantλ∈(0,1) such that
(3.25)d(Tx,Ty)≤λmax⁡{d(x,y),d(x,Tx)+d(y,Ty)2,d(x,Ty)+d(y,Tx)2},∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.Corollary 3.21.
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exist constantsA,B,C≥0 with (A+2B+2C)∈(0,1) such that
(3.26)d(Tx,Ty)≤Ad(x,y)+B[d(x,Tx)+d(y,Ty)]+C[d(x,Ty)+d(y,Tx)],∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.Corollary 3.22 (see Kirk et al. [17]).
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exists a constantλ∈(0,1) such that
(3.27)d(Tx,Ty)≤λd(x,y),∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.Corollary 3.23.
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exists a constantλ∈(0,1/2) such that
(3.28)d(Tx,Ty)≤λ[d(x,Tx)+d(y,Ty)],∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.Corollary 3.24.
Let{Ai}i=12 be nonempty closed subsets of a complete metric space (X,d) and T:Y→Y be a given mapping, where Y=A1∪A2. Suppose that the following conditions hold: (I)
T(A1)⊆A2 and T(A2)⊆A1;(II)
there exists a constantλ∈(0,1/2) such that
(3.29)d(Tx,Ty)≤λ[d(x,Ty)+d(y,Tx)],∀(x,y)∈A1×A2.Then T has a unique fixed point that belongs to A1∩A2.



---

*Source: 793486-2012-09-02.xml*