# Common Fixed Point Theorems forα-ψ-Contractive Type Mappings

Authors: Jalal Hassanzadeasl
Journal: International Journal of Analysis
 (2013)

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## Abstract

Recently, Samet et al. (2012) introduced the notion ofα-ψ-contractive type mappings. They established some fixed point theorems for these mappings in complete metric spaces. In this paper, we introduce the notion of a coupled α-ψ-contractive mapping and give a common fixed point result about the mapping. Also, we give a result of common fixed points of some coupled self-maps on complete metric spaces satisfying a contractive condition.


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## Body

## 1. Introduction

We know fixed point theory has many applications and was extended by several authors from different views (see, e.g., [1–33]). Recently, Samet et al. introduced the notion of α-ψ-contractive type mappings [3]. Denote with Ψ the family of nondecreasing functions ψ:[0,∞)→[0,∞) such that ∑n=1∞‍ψn(t)<∞ for all t>0, where ψn is the nth iterate of ψ. It is known that ψ(t)<t for all t>0 and ψ∈Ψ [3]. Let (X,d) be a metric space, T a self map on X, ψ∈Ψ and α:X×X→[0,∞). Then, T is called a α-ψ-contractive mapping whenever α(x,y)d(Tx,Ty)≤ψ(d(x,y)) for all x,y∈X. Also, we say that T is α-admissible whenever α(x,y)≥1 implies α(Tx,Ty)≥1 for all x,y∈X [3]. Also, we say that X has the property (Bα) if {xn} is a sequence in X such that α(xn,xn+1)≥1 for all n≥1 and xn→x, then α(xn,x)≥1 for all n≥1. Let (X,d) be a complete metric space and let T a α-admissible α-ψ-contractive mapping on X. Suppose that there exists x0∈X such that α(x0,Tx0)≥1. If T is continuous or X has the property (Bα), then T has a fixed point (see [3]; Theorems 2.1 and 2.2). Finally, we say that X has the property (Hα) whenever for each x,y∈X there exists z∈X such that α(x,z)≥1 and α(y,z)≥1. If X has the property (Hα) in the Theorems  2.1 and 2.2, then X has a unique fixed point ([3]; Theorem 2.3). It is considerable that the results of Samet et al. generalize similar ordered results in the literature (see the results of the third section [3]). The aim of this paper is introducing the notion of generalized coupled α-ψ-contractive mappings and give a common fixed point result about the mappings.Definition 1.
Letϝ  the family of functions f:[0,∞)→ℝ satisfy:(i)
f(0)=0 and f(t)>0 for all t∈(0,+∞);(ii)
f is continuous;(iii)
f is nondecreasing on [0,+∞);(iv)
f(t1+t2)≤f(t1)+f(t2) for all t1,t2∈(0,+∞).Definition 2.
LetΨ the family of functions ψ:[0,+∞)→[0,+∞) satisfy(ψ1)
ψ is nondecreasing;(ψ2)
∑n=1+∞‍ψn(t)<∞, for all t∈(0,+∞).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 all t>0.Definition 3.
Let(X,d) be a metric space, and let T,S:X→X with given coupled mappings. Let α:X×X→[0,+∞), f∈ϝ, ψ∈Ψ, and let
(1)m(Ax,By)=max{12d(x,y),d(x,Ax),d(y,By),12[d(x,By)+d(y,Ax)]},
for all coupled mappings A,B:X→X and x,y∈X. One says that T,  S are generalized coupled α-ψ-contractive mappings whenever
(2)α(x,y)f(d(Tx,Sy))≤ψ(f(m(Tx,Sy))),α(x,y)f(d(Sx,Ty))≤ψ(f(m(Sx,Ty))),
for all x,y∈X.Definition 4.
LetT,S:X→X, and let α:X×X→[0,+∞). One says that T,  S are coupled α-admissible if
(3)α(x,y)≥1⟹α(Tx,Sy)≥1orα(Sx,Ty)≥1
for all x,y∈X.Definition 5.
Let(X,d) be a compete metric space. For two subsets A,  B of X, one marks A⪯B, if for all a∈A, there exists b∈B such that a⪯b.Definition 6.
A partial metric on a nonempty setX is a function ρ:X×X→ℝ+ such that for all x,y,z∈X:(ρ1)
x=y⇔ρ(x,x)=ρ(x,y)=ρ(y,y);(ρ2)
ρ(x,x)≤ρ(x,y);(ρ3)
ρ(x,y)=ρ(y,x);(ρ4)
ρ(x,y)≤ρ(x,z)+ρ(z,y)-ρ(z,z).A partial metric space is a pair(X,ρ) such that X is a nonempty set, and ρ is a partial metric on X. It is clear that if ρ(x,y)=0, then from (ρ1) and (ρ2), x=y. But if x=y, ρ(x,y) may not be 0. A basic example of a partial metric is the pair (ℝ+,ρ), where ρ(x,y)=max{x,y} for all x,y∈ℝ+. If ρ is a partial metric on X, then the function ρs:X×X→ℝ+ given by ρs(x,y)=2ρ(x,y)-ρ(x,x)-ρ(y,y) is a metric on X.Example 7.
LetX=ℝ endowed with the standard metric d(x,y)=|x-y| for all x,y∈ℝ. Define the coupled mappings T,S:X→X by
(4)Tx={x+1,ifx>1,x2,if0≤x≤1,0,ifx<0,Sx={x-1,ifx>1,x3,if0≤x≤1,0,ifx<0.
We define the mapping α:X×X→[0,+∞) by
(5)α(x,y)={1,ifx,y∈[0,1],0,otherwise.
If
(6)α(x,y)≥1⟹x,y∈[0,1]⟹x2,y3∈[0,1]⟹α(Tx,Sy)=α(x2,y3)=1.
Similarly, α(x,y)≥1⇒α(Sx,Ty)≥1. This shows that T,  S are coupled α-admissible.Lemma 8.
Let(X,d) be a metric space. Suppose that T,S:X→X are generalized coupled α-ψ-contractive mappings. Then, ℱ(T)=ℱ(S).Proof.
We first show that any fixed point ofT is also a fixed point of S and conversely. Define α(x,y)=1 for all x,y∈X. Since ℱ(T)≠ℱ(S), we may assume there exists x*∈X such that x*∈ℱ(T), but x*∉ℱ(S). Since d(x*,Sx*)>0, we have
(7)m(Tx*,Sx*)=max{12d(x*,x*),d(x*,Tx*),d(x*,Sx*),12[d(x*,Sx*)+d(x*,Tx*)]}=d(x*,Sx*),f(d(x*,Sx*))=f(d(Tx*,Sx*))≤α(x*,x*)f(d(Tx*,Sx*))≤ψ(f(m(Tx*,Sx*)))≤ψ(f(d(x*,Sx*)))<f(d(x*,Sx*)).
This contradiction establishes that ℱ(T)⊆ℱ(S). A similar argument establishes the reverse containment, and therefore ℱ(T)=ℱ(S).


## 2. Main Results

Now, we are ready to state and prove our main results.Theorem 9.
Let(X,d) be a complete metric space. Suppose that T,S:X→X are generalized coupled α-ψ-contractive mappings and satisfy the following conditions:(i)
T,  S are coupled α-admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1; or(ii)*
there existsx0∈X such that α(x0,Tx0)≥1;(iii)
T or S is continuous.Then T,  S have common fixed point x*∈X. Further, for each x0∈X, the iterated sequence {xn} with x2n+1=Tx2n and x2n+2=Sx2n+1 converges to the common fixed point of T,S.Proof.
By Lemma8, we have ℱ(T)=ℱ(S). Let x0∈X such that α(x0,Tx0)≥1. Define the sequence {xn} in X by x2n+1=Tx2n and x2n+2=Sx2n+1 for all n∈ℕ0. If xn0=xn0+1 for some n0>1, then x*=xn0 are a common fixed point for T,  S. So, we can assume that x2n≠Tx2n and x2n+1≠Sx2n+1 for all n∈ℕ0. Since T,  S are coupled α-admissible, we have
(8)α(x0,x1)=α(x0,Tx0)≥1⟹α(Tx0,Sx1)=α(x1,x2)≥1⟹α(Sx1,Tx2)=α(x2,x3)≥1.
Inductively, we have
(9)α(x2n,x2n+1)≥1,α(x2n+1,x2n+2)≥1,
for all n∈ℕ0. We obtain
(10)f(d(x2n+1,x2n+2))=f(d(Tx2n,Sx2n+1))≤α(x2n,x2n+1)f(d(Tx2n,Sx2n+1))≤ψ(f(m(Tx2n,Sx2n+1))).
Now,
(11)m(Tx2n,Sx2n+1)=max{12[d(x2n,Sx2n+1)+d(x2n+1,Tx2n)]d(x2n,x2n+1),d(x2n,Tx2n),d(x2n+1,Sx2n+1),12[d(x2n,Sx2n+1)+d(x2n+1,Tx2n)]}=max{12d(x2n,x2n+1),d(x2n,x2n+1),d(x2n+1,x2n+2),12[d(x2n,x2n+2)+d(x2n+1,x2n+1)]}≤max{12d(x2n,x2n+1),d(x2n+1,x2n+2),12[d(x2n,x2n+1)+d(x2n+1,x2n+2)+0]}=max{d(x2n,x2n+1),d(x2n+1,x2n+2)}
if
(12)max{d(x2n,x2n+1),d(x2n+1,x2n+2)}=d(x2n+1,x2n+2).
So, in general,
(13)f(d(x2n+1,x2n+2))≤ψ(f(d(x2n+1,x2n+2)))<f(d(x2n+1,x2n+2)),
which is a contradiction since d(x2n+1,x2n+2)>0. Thus,
(14)f(d(x2n+1,x2n+2))≤ψ(f(d(x2n,x2n+1))).
Similarly, if
(15)f(d(x2n,x2n+1))≤ψ(f(d(x2n-1,x2n))),
we have
(16)f(d(xn,xn+1))≤ψ(f(d(xn-1,xn))),
for all n∈ℕ0. By induction, we get
(17)f(d(xn,xn+1))≤ψ(ψ(f(d(xn-2,xn-1))))≤⋯≤ψn(f(d(x0,x1))),
for all n∈ℕ0. Fix ϵ>0, and let n(ϵ)∈ℕ0 such that
(18)∑n=n(ϵ)∞‍ψn(f(d(x0,x1)))<ϵ.
Let n,m∈ℕ0 with m>n>n(ϵ). Using the triangle inequality, we obtain
(19)f(d(xn,xm))≤∑k=nm-1‍f(d(xk,xk+1))≤∑k=nm-1‍ψk(f(d(x0,x1)))≤∑n=n(ϵ)∞‍ψn(f(d(x0,x1)))<ϵ.
Thus we proved that {xn} is a Cauchy sequence in the metric space (X,d).
Since(X,d) is a complete metric space, there exists x*∈X such that xn→x* as n→+∞. From the continuity of T, it follows that x2n+1=Tx2n→Tx* as n→+∞, then x*=Tx*. Similarly if S is continuous, we have x*=Sx*.Corollary 10.
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; or(ii)*
there existsx0∈X such that α(x0,Tx0)≥1;(iii)
T is continuous.Then, T has a fixed point x*∈X. Further, for each x0∈X, the iterated sequence {xn} with xn+1=Txn converges to the fixed point of T.Example 11.
LetX=ℝ endowed with the standard metric d(x,y)=|x-y| for all x,y∈ℝ. Define the coupled mappings T,S:X→X by
(20)Tx={2x-32,ifx>1,x2,if0≤x≤1,0,ifx<0,Sx={2x-53,ifx>1,x2,if0≤x≤1,0,ifx<0.
We define the mapping α:X×X→[0,+∞) by
(21)α(x,y)={1,ifx,y∈[0,1],0,otherwise.
If f(t)=t and ψ(t)=(1/2)t for all t≥0, we have
(22)α(x,y)f(d(Tx,Sy))={f(d(Tx,Sy)),ifx,y∈[0,1],0,otherwise,α(x,y)f(d(Tx,Sy))≤f(d(Tx,Sy))=|x2-y2|≤12|x-y|≤ψ(f(m(Tx,Sy))),
for all x,y∈X. Thus, T,  S are generalized coupled α-ψ-contractive mappings. Moreover, there exists x0∈X such that α(x0,Tx0)≥1. In fact, for x0=1, we have α(1,T1)=1. Obviously, T is continuous, and so it remains to show that T,  S are coupled α-admissible. To do so, let x,y∈X such that α(x,y)≥1. This implies that x,y∈[0,1] by the definition of α. We then have Tx∈[0,1], Sy∈[0,1] and α(Tx,Sy))=1. Then, T,  S are coupled α-admissible. Now, all the hypotheses of Theorem 9 are satisfied.
Consequently,T,  S have common fixed points. In this example, 0 is at least one common fixed point of T and S.Now, we omit the continuity hypothesis ofT and S.Theorem 12.
Let(X,d) be a complete metric space. Suppose that T,S:X→X are generalized coupled α-ψ-contractive mappings and satisfy the following conditions:(i)
T,  S are coupled α-admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1; or(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; or(iii)*
if{xn} is a sequence in X such that α(xn+1,xn)≥1 for all n and xn→x∈X as n→∞, then there exists a subsequence {xn(k)} of {xn} such that α(x,xn(k))≥1 for all k.Then, T,  S have common fixed point x*∈X. Further, for each x0∈X, the iterated sequence {xn} with x2n+1=Tx2n and x2n+2=Sx2n+1 converges to the common fixed point of T,S.Proof.
Following the proof of Theorem9, we know that {xn} is a Cauchy sequence in the complete metric space (X,d). Then, there exists x*∈X such that xn→x* as n→+∞. From Theorem 9 and condition (iii), there exists a subsequence {xn(k)} of {xn} such that α(x2n(k),x*)≥1 for all k.
Applying Theorem9, for all k, we get that
(23)f(d(x2n(k)+1,Sx*))=f(d(Tx2n(k),Sx*))≤α(x2n(k),x*)d(Tx2n(k),Sx*)≤ψ(f(m(Tx2n(k),Sx*))).
On the other hand, we have
(24)m(Tx2n(k),Sx*)=max{12d(x2n(k),x*),d(x2n(k),Tx2n(k)),d(x*,Sx*),12[d(x2n(k),Sx*)+d(x*,Tx2n(k))]}.
Letting k→+∞, in the above equality, we get that
(25)limk→∞m(Tx2n(k),Sx*)=d(x*,Sx*).
Suppose that d(x*,Sx*)>0. From (25), for k large enough, we have m(Tx2n(k),Sx*)>0, which implies that
(26)ψ(f(m(Tx2n(k),Sx*)))<f(m(Tx2n(k),Sx*)).
Thus, from (23), we have f(d(x2n(k)+1,Sx*))<m(Tx2n(k),Sx*). Letting k→∞ in the above inequality, and using (25), we obtain that f(d(x*,Sx*))<f(d(x*,Sx*)) which is a contradiction. Thus, we have d(x*,Sx*)=0; that is, x*=Sx*. Similarly, it can be shown that x*=Tx*.Corollary 13.
Let(X,d) be a complete metric space. Suppose that T:X→X is generalized α-ψ-contractive mapping and satisfies thefollowing conditions:(i)
T is α-admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1; or(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; or(iii)*
if{xn} is a sequence in X such that α(xn+1,xn)≥1 for all n and xn→x∈X as n→∞, then there exists a subsequence {xn(k)} of {xn} such that α(x,xn(k))≥1 for all k.Then, T has a fixed point x*∈X. Further, for each x0∈X, the iterated sequence {xn} with xn+1=Txn converges to the fixed point of T.Example 14.
LetX=ℝ endowed with the standard metric d(x,y)=|x-y| for all x,y∈ℝ. Define the coupled mappings T,S:X→X by
(27)Tx={2x-34,ifx>1,x2,if0≤x≤1,0,ifx<0,Sx={2x-56,ifx>1,x2,if0≤x≤1,0,ifx<0.
We define the mapping α:X×X→[0,+∞) by
(28)α(x,y)={1,ifx,y∈[0,1],0,otherwise.
If f(t)=t and ψ(t)=(1/2)t for all t≥0, we have
(29)α(x,y)f(d(Tx,Sy))={f(d(Tx,Sy)),ifx,y∈[0,1],0,otherwise,α(x,y)f(d(Tx,Sy))≤f(d(Tx,Sy))=|x2-y2|≤12|x-y|≤ψ(f(m(Tx,Sy))),
for all x,y∈X. Thus, T,  S are generalized coupled α-ψ-contractive mappings. Moreover, there exists x0∈X such that α(x0,Tx0)≥1. In fact, for x0=1, we have α(1,T1)=1. Let {xn}∈X,  xn→x and α(xn+1,xn)≥1 for all n≥1. By the definition of α we have, xn,xn+1∈[0,1] for all n, so x∈[0,1] and α(xn,x)≥1. It remains to show that T,  S are coupled α-admissible. In doing so, let x,y∈X such that α(x,y)≥1. This implies that x,y∈[0,1] by the definition of α. We have Tx∈[0,1],  Sy∈[0,1], and α(Tx,Sy))=1. Then T,  S are coupled α-admissible. Now, all the hypotheses of Theorem 12 are satisfied. Consequently, T,  S have common fixed points. In this example, 0 is at least one common fixed point of T and S.


## 3. Fixed Point Theorems on Ordered Metric Space

Theorem 15.
Let(X,⪯,d) be a complete ordered metric space, ψ∈Ψ, f∈ϝ, and T,  S be coupled mappings on X such that f(d(Tx,Sy))≰ψ(f(m(Tx,Sy))) and f(d(Sx,Ty))≤ψ(f(m(Sx,Ty))) for all x,y∈X with x⪯y. Suppose that there exists x0∈X such that x0⪯Tx0 or (Tx0⪯x0), and if {xn} is a sequence in X such that xn⪯xn+1 or (xn+1⪯xn) for all n≥1 and xn→x, then xn⪯x or (x⪯xn) for all n≥1. If x⪯y implies Tx⪯Sy  or  Sx⪯Ty or (Ty⪯Sx  or  Sy⪯Tx), then T,S have common fixed points.Proof.
Defineα:X×X→[0,∞) by α(x,y)=1 whenever x⪯y, and define α(x,y)=0 whenever x⋠y. It is easy to check that T,  S are a coupled α-admissible and generalized coupled α-ψ-contractive mappings on X. Now, by using Theorem 9, T,  S have common fixed points.Corollary 16.
Let(X,⪯,d) be a complete ordered metric space, ψ∈Ψ, f∈ϝ, and T a mapping on X such that f(d(Tx,Ty))≤ψ(f(m(Tx,Ty))) for all x,y∈X with x⪯y. Suppose that there exists x0∈X such that x0⪯Tx0 or (Tx0⪯x0) and if {xn} is a sequence in X such that xn⪯xn+1 or (xn+1⪯xn) for all n≥1 and xn→x, then xn⪯x or (x⪯xn) for all n≥1. If x⪯y implies Tx⪯Ty or (Ty⪯Tx), then T has a fixed point.


## 4. Fixed Point Theorems on Metric Spaces Endowed with Partial Metric

If we substitute a partial metricρ instead the metric d in Theorem 9, it is easy to check that a similar result holds for the partial metric space case as follows. We define
(30)m*(Ax,By)=max{12ρ(x,y),ρ(x,Ax),ρ(y,By),12[ρ(x,By)+ρ(y,Ax)]}
for all coupled mappings A,B:X→X and x,y∈X.Theorem 17.
Let(X,ρ) be a complete partial metric space,  α:X×X→[0,∞) a function, ψ∈Ψ, f∈ϝ, and T,  S self maps on X such that
(31)α(x,y)f(ρ(Tx,Sy))≤ψ(f(m*(Tx,Sy))),α(x,y)f(ρ(Sx,Ty))≤ψ(f(m*(Sx,Ty)))
for all x,y∈X. Suppose that T,  S are coupled α-admissible and there exists x0∈X such that α(x0,Tx0)≥1 or (α(Tx0,x0)≥1). Assume that if {xn} is a sequence in X such that α(xn,xn+1)≥1 or (α(xn+1,xn)≥1) for all n and xn→x as n→+∞, then there exists a subsequence {xn(k)} of {xn} such that α(xn(k),x)≥1 or (α(x,xn(k))≥1) for all k. Then T,S have common fixed points.Corollary 18.
Let(X,ρ) be a complete partial metric space,  α:X×X→[0,∞) a function, ψ∈Ψ, f∈ϝ, and T a self map on X such that
(32)α(x,y)f(ρ(Tx,Ty))≤ψ(f(m*(Tx,Ty)))
for all x,y∈X. Suppose that T is α-admissible and there exists x0∈X such that α(x0,Tx0)≥1 or (α(Tx0,x0)≥1). Assume that if {xn} is a sequence in X such that α(xn,xn+1)≥1 or (α(xn+1,xn)≥1) for all n and xn→x as n→+∞, then there exists a subsequence {xn(k)} of {xn} such that α(xn(k),x)≥1 or (α(x,xn(k))≥1) for all k. Then T has a fixed point.Example 19.
LetX=[0,+∞) endowed with the partial metric ρ(x,y)=max{x,y}, for all x,y∈X. Define the mapping T:X→X, by
(33)Tx={2x-32,ifx>1,x21+x,if0≤x≤1,0,ifx<0.
We define the mapping α:X×X→[0,+∞) by
(34)α(x,y)={1,ifx,y∈[0,1],x≤y,0,otherwise,α(x,y)≥1⟺x,y∈[0,1],x≤y⟺Tx,Ty∈[0,1],Tx≤Ty⟺α(Tx,Ty)≥1
since T is α-admissible. If ψ(t)=t2/(1+t), we have
(35)α(x,y)ρ(Tx,Ty)≤ρ(x21+x,y21+y)=max{x21+x,y21+y}=y21+y≤y=ρ{x,y}.
Suppose that {xn} is a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x as n→+∞; by definition of α, we have xn,  xn+1 and x∈[0,1], on the other hand xn≤xn+1⇒xn(k)≤x; then there exists a subsequence {xn(k)} of {xn} such that α(xn(k),x)≥1 for all k. There exists x0=1 such that α(x0,Tx0)=1. This show that all conditions of Corollary 18 are satisfied, and so T has a fixed point in X.



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