# Weak Contraction Condition Involving Cubic Terms ofdx,y under

Authors: Penumurthy Parvateesam Murthy; K. N. V. V. Vara Prasad
Journal: Journal of Mathematics
 (2013)

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## Abstract

A fixed point theorem is presented for single-valued map with using generalizedφ-weak contractive condition involving various combinations of dx,y on a complete metric space. Our result is an extension as well as a generalization of Alber and Guerre-Delabriere (1997) in particular. It also generalizes the results of Rhoades (2001), Choudhury and Dutta, (2000), and Dutta and Choudhury, (2008).


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## Body

## 1. Introduction

Let(X,d) be a metric space. A map T:X→X is a contraction if, for each x,y∈X, there exists a constant k∈(0,1) such that d(Tx,Ty)≤kd(x,y).A mapT:X→X is a φ-weak contraction if, for each x,y∈E, there exists a function φ:[0,∞)→[0,∞), φ(t)>0 for all t>0 and φ(0)=0 such that d(Tx,Ty)≤d(x,y)-φ(d(x,y)).In [1] Alber and Guerre-Delabriere introduced the concept of weak contraction in Hilbert spaces. Rhoades [2] has shown that the result which Alber and Guerre-Delabriere had proved in [2] is also valid in complete metric spaces.In this paper we introduced generalizedφ-weak contractive condition involving various combinations of d(x,y). Our result is an extension as well as a generalization of Alber and Guerre-Delabriere [1] and Rhoades [2] in particular. It also generalizes the results of [3, 4].Now, We state the result of Rhoades as follows.Theorem 1 (see [2, Theorem 2]).
Let(X,d) be a complete metric space, and let T be a φ-weak contraction on Xφ:[0,+∞)→[0,+∞) which is a continuous and nondecreasing function with φ(t)>0 for all t∈(0,∞) and φ(0)=0 then T has a unique fixed point in X.
If one takesφ(t)=(1-k)t, where 0<k<1, then weak contraction reduces to contraction mapping.
In this paper, a new type of inequality is introduced with cubic terms involvingp(≥0) called a “generalizedφ-weak contractive condition with cubic terms involvingp”.
Let(X,d) be a metric space and T a self-map of X satisfying the following condition:
(1)[1+pd(x,y)]d2(Tx,Ty)≤pmax{12[d2(x,Tx)d(y,Ty)+d(x,Tx)d2(y,Ty)],d(x,Tx)d(x,Ty)d(y,Tx),d(x,Ty)d(y,Tx)d(y,Ty){12}}+m(x,y)-φ(m(x,y)),
where
(2)m(x,y)=max{{12}d2(x,y),d(x,Tx)d(y,Ty),d(x,Ty)d(y,Tx),12[d(x,Tx)d(x,Ty)+d(y,Tx)d(y,Ty)]}p≥0 is a real number is and φ:[0,∞)→[0,∞) is a continuous function with φ(t)=0⇔t=0 and φ(t)>0 for each t>0.


## 2. Main Result

Lemma 2.
LetT be a self map of a metric space X satisfying (1). For any sequence {xn} in X defined by xn+1=Txn, n≥0. Then the sequence {xn} is Cauchy in X.Proof.
Letx0∈X be an arbitrary point. Constructing the sequence {xn} follows
(3)xn+1=Txn,n∈N0.
If xn=xn+1 for some n, then trivially T has a fixed point. We assume xn+1≠xn, for all n∈N0. We write αn=d(xn,xn+1).
First, we prove that{αn} is a nonincreasing sequence and converges to 0.
Case
1. If n is even, taking x=x2n and y=x2n+1 in (1), we get
(4)[1+pd(x2n,x2n+1)]d2(Tx2n,Tx2n+1)≤pmax{12[d2(x2n,Tx2n)d(x2n+1,Tx2n+1)+d(x2n,Tx2n)d2(x2n+1,Tx2n+1)],d(x2n,Tx2n)d(x2n,Tx2n+1)×d(x2n+1,Tx2n),d(x2n,Tx2n+1)×d(x2n+1,Tx2n)d(x2n+1,Tx2n+1){12}}+m(x2n,x2n+1)-φ(m(x2n,x2n+1)),
where
(5)m(x2n,x2n+1)=max{d2(x2n,x2n+1),d(x2n,Tx2n)d(x2n+1,Tx2n+1),d(x2n,Tx2n+1)d(x2n+1,Tx2n),12[d(x2n,Tx2n)d(x2n,Tx2n+1)+d(x2n+1,Tx2n)d(x2n+1,Tx2n+1)]{d2}}.
By using (3), we get
(6)[1+pd(x2n,x2n+1)]d2(x2n+1,x2n+2)≤pmax{12[d2(x2n,x2n+1)d(x2n+1,x2n+2)+d(x2n,x2n+1)d2(x2n+1,x2n+2)],d(x2n,x2n+1)d(x2n,x2n+2)×d(x2n+1,x2n+1),d(x2n,x2n+2)×d(x2n+1,x2n+1)d(x2n+1,x2n+2){12}}+m(x2n,x2n+1)-φ(m(x2n,x2n+1)),
where
(7)m(x2n,x2n+1)=max{d2(x2n,x2n+1),d(x2n,x2n+1)d(x2n+1,x2n+2),d(x2n,x2n+2)d(x2n+1,x2n+1),12[d(x2n,x2n+1)d(x2n,x2n+2)+d(x2n+1,x2n+1)d(x2n+1,x2n+2)]{d2}}.
Now consider α2n=d(x2n,x2n+1); then we have
(8)[1+pα2n]α2n+12≤pmax{12[α2n2α2n+1+α2nα2n+12],0,0}+m(x2n,x2n+1)-φ(m(x2n,x2n+1)),
where m(x2n,x2n+1)=max{α2n2,α2nα2n+1,0,  (1/2)[α2nd(x2n,x2n+2)+0]}.
By triangular inequality and using property ofφ, we get
(9)d(x2n,x2n+2)≤d(x2n,x2n+1)+d(x2n+1,x2n+2)=α2n+α2n+1.
Then
(10)m(x2n,x2n+1)≤m′(x,y)=max{α2n2,α2n+1α2n,0,12α2n(α2n+α2n+1),0}.
Ifα2n<α2n+1, then (8) reduces to pα2n+12≤pα2n+12-φ(α2n+12), a contradiction. Thus α2n+12≤α2n2 implies that α2n+1≤α2n.
In a similar way, ifn is odd, we can obtain α2n+2<α2n+1.
It follows that the sequence{αn} is decreasing.
Letlimn→∞αn=r, for some r≥0.
Supposer>0; then from inequality (1), we have
(11)[1+pd(xn,xn+1)]d2(Txn,Txn+1)≤pmax{12[d2(xn,Txn)d(xn+1,Txn+1)+d(xn,Txn)d2(xn+1,Txn+1)],d(xn,Txn)d(xn,Txn+1)×d(xn+1,Txn),d(xn,Txn+1)×d(xn+1,Txn)d(xn+1,Txn+1){12}}+m(xn,xn+1)-φ(m(xn,xn+1)),
where
(12)m(xn,xn+1)=max{d2(xn,xn+1),d(xn,Txn)d(xn+1,Txn+1),d(xn,Txn+1)d(xn+1,Txn),12[d(xn,Txn)d(xn,Txn+1)+d(xn+1,Txn)d(xn+1,Txn+1)]{d2}}.
By using (3), we get
(13)[1+pd(xn,xn+1)]d2(xn+1,xn+2)≤pmax{12[d2(xn,xn+1)d(xn+1,xn+2)+d(xn,xn+1)d2(xn+1,xn+2)],d(xn,xn+1)d(xn,xn+2)d(xn+1,xn+1),d(xn,xn+2)d(xn+1,xn+1)d(xn+1,xn+2){12}}+m(xn,xn+1)-φ(m(xn,xn+1)),
where
(14)m(xn,xn+1)=max{d2(xn,xn+1),d(xn,xn+1)d(xn+1,xn+2),d(xn,xn+2)d(xn+1,xn+1),12[d(xn,xn+1)d(xn,xn+2)+d(xn+1,xn+1)d(xn+1,xn+2)]{d2}}.
Using triangular inequality and property ofφ and taking limits n→∞, we get
(15)[1+pr]r2≤pr3+r2-φ(r2).
Then φ(r2)≤0, since r is positive; then by the property of φ, we get r=0. We conclude that
(16)limn→∞αn=limn→∞d(xn,xn+1)=r=0.
Now we show that{xn} is a Cauchy sequence. Suppose, we assume that {xn} is not a Cauchy sequence; then there is ε>0 for which we can find two sequences of positive integers {m(k)} and {n(k)} such that for all positive integers k, n(k)>m(k)>k(17)d(xm(k),xn(k))≥ϵ,d(xm(k),xn(k)-1)<ϵ.
Now,(18)ϵ≤d(xm(k),xn(k))≤d(xm(k),xn(k)-1)+d(xn(k)-1,xn(k)).
Letting k→∞, we get
(19)limk→∞d(xm(k),xn(k))=ϵ.
Now, from the triangular inequality, we have
(20)|d(xn(k),xm(k)+1)-d(xm(k),xn(k))|≤d(xm(k),xm(k)+1).
Taking limits ask→∞ and using (16) and (19), we have
(21)limk→∞d(xn(k),xm(k)+1)=ϵ.
Again from the triangular inequality, we have(22)|d(xm(k),xn(k)+1)-d(xm(k),xn(k))|≤d(xn(k),xn(k)+1).
Taking limits ask→∞ and using (16) and (19), we have
(23)limk→∞d(xm(k),xn(k)+1)=ϵ.
Again by using triangular inequality, we have
(24)|d(xm(k)+1,xn(k)+1)-d(xm(k),xn(k))|≤d(xm(k),xm(k)+1)+d(xn(k),xn(k)+1).
Taking limit k→∞ in the above inequality and using (16) and (19), we have
(25)limk→∞d(xn(k)+1,xm(k)+1)=ϵ.
Again puttingx=xm(k) and y=xn(k) in (1), we get
(26)[1+pd(xm(k),xn(k))]d2(Txm(k),Txn(k))≤p{12[d2(xm(k),Txm(k))d(xn(k),Txn(k))+d(xm(k),Txm(k))d2(xn(k),Txn(k))],d(xm(k),Txm(k))d(xm(k),Txn(k))d(xn(k),Txm(k)),d(xm(k),Txn(k))d(xn(k),Txm(k))d(xn(k),Txn(k)){12}}+m(xm(k),xn(k))-φ(m(xm(k),xn(k))),
where
(27)m(xm(k),xn(k))=max{12d2(xm(k),xn(k)),d(xm(k),Txm(k))d(xn(k),Txn(k)),d(xm(k),Txn(k))d(xn(k),Txm(k)),{d2}12[d(xm(k),Txm(k))d(xm(k),Txn(k))+d(xn(k),Txm(k))d(xn(k),Txn(k))]12}.
Using (3), then we obtain
(28)[1+pd(xm(k),xn(k))]d2(xm(k)+1,xn(k)+1)≤p{12[d2(xm(k),xm(k)+1)d(xn(k),xn(k)+1)+d(xm(k),xm(k)+1)d2(xn(k),xn(k)+1)],d(xm(k),xm(k)+1)d(xm(k),xn(k)+1)×d(xn(k),xm(k)+1),d(xm(k),xn(k)+1)×d(xn(k),xm(k)+1)d(xn(k),xn(k)+1){12}}+m(xm(k),xn(k))-φ(m(xm(k),xn(k))),
where
(29)m(xm(k),xn(k))=max{12d2(xm(k),xn(k)),d(xm(k),xm(k)+1)d(xn(k),xn(k)+1),d(xm(k),xn(k)+1)d(xn(k),xm(k)+1),12[d(xm(k),xm(k)+1)d(xm(k),xn(k)+1)+d(xn(k),xm(k)+1)d(xn(k),xn(k)+1)]12}.
Letting k→∞ and using (16)–(25), we get
(30)[1+pϵ]ϵ2≤pmax{12[0+0],0,0}+ϵ2-φ(ϵ2)=ϵ2-φ(ϵ2),
a contradiction. Thus {xn} is a Cauchy in X.Theorem 3.
LetT be a self-map of a complete metric space X satisfying (1). Then T has a unique fixed point in X.Proof.
From Lemma2 the sequence {xn} is a Cauchy in X. Since (X,d) is a complete metric space, then there exists a point z∈X such that
(31)limn→∞xn=z.
Now we prove thatz is a fixed point of T.
Takingx=xn and y=z in (1), we have
(32)[1+pd(xn,z)]d2(Txn,Tz)≤pmax{12[d2(xn,Txn)d(z,Tz)+d(xn,Txn)d2(z,Tz)],d(xn,Txn)d(xn,Tz)d(z,Txn),d(xn,Tz)d(z,Txn)d(z,Tz){12}}+m(xn,z)-φ(m(xn,z)),
where
(33)m(xn,z)=max{{12}d2(xn,z),d(xn,Txn)d(z,Tz),d(xn,Tz)d(z,Txn),12[d(xn,Txn)d(xn,Txn)+d(z,Txn)d(z,Tz)]{12}}.
Using (31) and (3), we get
(34)[1+pd(z,z)]d2(z,Tz)≤pmax{12[d2(z,z)d(z,Tz)+d(z,z)d2(z,Tz)],d(z,z)d(z,Tz)d(z,z),d(z,Tz)d(z,z)d(z,Tz){12}}+m(z,z)-φ(m(z,z)).
Henced2(z,Tz)≤0⇒Tz=z.
ThenT has a fixed point in X.
To prove the uniqueness of the fixed point, we assume thatz1 and z2 are two fixed points of T. Taking x=z1 and y=z2 in (1), we easily get d(z1,z2)=0 wich implies that z1=z2. Therefore T has a unique fixed point in X.Corollary 4.
LetT be a self-map of a complete metric space X satisfying the condition
(35)d2(Tx,Ty)≤m(x,y)-φ(m(x,y)),
where
(36)m(x,y)=max{{12}d2(x,y),d(x,Tx)d(y,Ty),d(x,Ty)d(y,Tx),12[d(x,Tx)d(x,Ty)+d(y,Tx)d(y,Ty)]}
for all x,y∈X and φ:[0,∞)→[0,∞) is a continuous function with φ(t)=0⇔t=0 and φ(t)>0 for each t>0. Then T has a unique fixed point in X.Proof.
p
=
0 in Theorem 3; we have the result.Now we give an example to support our result.Example 5.
LetX={0,1,2}, and let d be the usual metric on X. Let T:X→X be defined by T0=T1=0 and T2=1. And define φ:[0,∞)→[0,∞) by φ(t)=t/2. For any value of p>0 and x,y∈X, then it is easy to verify that inequality (1) holds. Hence Theorem 3 holds well.



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*Source: 967045-2013-05-20.xml*