# Common Fixed Point Theorems for  Weakly Compatible Maps Satisfying  a General Contractive Condition

Authors: Cristina Di Bari; Calogero Vetro
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

We introduce a new generalized contractive condition for
four mappings in the framework of metric space. We give some common fixed point
results for these mappings and we deduce a fixed point result for weakly compatible
mappings satisfying a contractive condition of integral type.


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## Body

## 1. Introduction and Preliminaries

The study of common fixed point of mappings satisfying
contractive type conditions has been a very active field of research during
recent years. The most general of the common fixed point theorems pertaining to
four mappings,A,B,S, and T of a metric space (X,d), uses either a
Banach-type contractive condition of the form(1.1)d(Ax,By)≤kM(x,y),0≤k<1,where(1.2)M(x,y)=max⁡{d(Sx,Ty),d(Ax,Sx),d(By,Ty),[d(Sx,By)+d(Ax,Ty)]2},or a Meir-Keeler-type (ε,δ)-contractive condition, that is, given ε>0,
there exists a δ>0 such that(1.3)ε≤M(x,y)<ε+δ⟹d(Ax,By)<ε,or a ψ-contractive condition of the form(1.4)d(Ax,By)≤ψ(M(x,y)),involving a contractive gauge
function ψ:[0,+∞[→[0,+∞[ such that ψ(t)<t for each t>0.
Clearly, Banach-type contractive condition is a special case of both conditions
Meir-Keeler-type (ε,δ)-contractive and ψ-contractive. A ψ-contractive condition does not guarantee the
existence of a fixed point unless some additional condition is assumed.
Moreover, a ψ-contractive condition, in general, does not
imply the Meir-Keeler-type (ε,δ)-contractive condition [1, Example 1.1].Recently, some fixed point results for mappings
satisfying an integral-type contractive condition are obtained in [2–5]. Suzuki [6] showed that Meir-Keeler contractions of integral type
are still Meir-Keeler contractions. Zhang [7] introduced a generalized contractive-type condition
for a pair of mappings in metric space and proved common fixed point theorems
that extend results in [3–5].
In this paper, we give a new generalized contractive-type condition for four
mappings in metric space and prove some common fixed point results for these
mappings. The results obtained extend well-known comparable results in
[2–5, 7].Lemma 1.1 (see [8]).
For every functionψ:[0,+∞[→[0,+∞[,
let ψn be the nth iterate of ψ.
Then the following hold:
(i)
ifψ is nondecreasing, then for each t>0, limn→+∞ψn(t)=0 implies ψ(t)<t;(ii)
ifψ is right continuous with ψ(t)<t for t>0,
then limn→+∞ψn(t)=0.


## 2. Common Fixed Points

In this
section, we give our main result. Two self-mappingsA and S of a metric space (X,d) are called weakly
compatible if they commute at their coincidence points. Let A,B,S, and T be self mappings of a metric space (X,d).
In the sequel, we set(2.1)M(x,y)=max⁡{d(Sx,Ty),d(Ax,Sx),d(By,Ty),[d(Sx,By)+d(Ax,Ty)]2}.Lemma 2.1.
LetA,B,S, and T be self-mappings of a metric space (X,d) such that AX⊂TX, BX⊂SX.
Assume that there exist F,ψ:[0,+∞[→[0,+∞[ such that
(i)
F is nondecreasing, continuous, and F(0)=0<F(t) for every t>0;(ii)
ψ is nondecreasing, right continuous, and ψ(t)<t for every t>0.
If for all x,y∈X,(2.2)F(d(Ax,By))≤ψ(F(M(x,y))),then for each x0∈X,
the sequence (yn) of points of X defined by the rule(2.3)y2n=Ax2n=Tx2n+1,y2n−1=Bx2n−1=Sx2nis a Cauchy sequence.Proof.
We  have(2.4)M(x2n,x2n+1)=max⁡{d(y2n−1,y2n),d(y2n,y2n−1),d(y2n+1,y2n),[d(y2n−1,y2n+1)+d(y2n,y2n)]2}=max⁡{d(y2n,y2n−1),d(y2n,y2n+1),d(y2n−1,y2n+1)2}=max⁡{d(y2n,y2n−1),d(y2n,y2n+1)}.
Similarly(2.5)M(x2n,x2n−1)=max⁡{d(y2n,y2n−1),d(y2n−1,y2n−2)}.
If for somen we have either y2n=y2n−1 or y2n=y2n+1,
then by condition (2.2) we obtain that the sequence (yn) is definitely constant and thus is a Cauchy
sequence. Suppose yn≠yn−1 for each n.
From(2.6)F(d(y2n,y2n+1))=F(d(Ax2n,Bx2n+1))≤ψ(F(M(x2n,x2n+1)))=ψ(F(d(y2n,y2n−1)))<F(d(y2n,y2n−1)),F(d(y2n,y2n−1))=F(d(Ax2n,Bx2n−1))≤ψ(F(M(x2n,x2n−1)))=ψ(F(d(y2n−1,y2n−2)))<F(d(y2n−1,y2n−2)),we deduce(2.7)F(d(yn+1,yn))<F(d(yn,yn−1)),for all n∈ℕ.
Now, from(2.8)F(d(yn+1,yn))≤ψ(F(d(yn,yn−1)))≤⋯≤ψn(F(d(y0,y1)))and (ii) of Lemma 1.1, we obtain limn→+∞F(d(yn+1,yn))=0,
which implies(2.9)lim⁡n→+∞d(yn+1,yn)=0.
We prove that(yn) is a Cauchy sequence. Suppose not, then there
exists ε>0 such that d(yn,ym)≥2ε for infinite values of m and n with m<n.
This assures that there exist two sequences (mk),(nk) of natural numbers, with mk<nk,
such that(2.10)d(y2mk,y2nk+1)>ε∀k.It is not restrictive to suppose
that nk is the least positive integer exceeding mk and satisfying (2.10). We have(2.11)ε<d(y2mk,y2nk+1)≤d(y2mk,y2nk−1)+d(y2nk−1,y2nk)+d(y2nk,y2nk+1)≤ε+d(y2nk−1,y2nk)+d(y2nk,y2nk+1).Then d(y2mk,y2nk+1)→ε.
We note(2.12)d(y2mk,y2nk+1)−d(y2mk,y2mk+1)−d(y2nk+2,y2nk+1)≤d(y2mk+1,y2nk+2)≤d(y2mk,y2nk+1)+d(y2mk,y2mk+1)+d(y2nk+2,y2nk+1),and thus d(y2mk+1,y2nk+2)→ε as k→+∞.
We have(2.13)M(x2nk+2,x2mk+1)=max⁡{d(y2mk,y2nk+1),d(y2nk+1,y2nk+2),d(y2mk,y2mk+1),d(y2mk+1,y2nk+1)+d(y2mk,y2nk+2)2}=d(y2mk,y2nk+1)+dk,where dk→0 as k→+∞ and dk≥0 for all k.
Then from(2.14)F(d(y2mk+1,y2nk+2))=F(d(Ax2nk+2,Bx2mk+1))≤ψ(F(M(x2nk+2,x2mk+1)))=ψ(F(d(y2mk,y2nk+1)+dk)),as k→+∞, F being continuous and ψ right continuous, we get(2.15)F(ε)≤ψ(F(ε))<F(ε).This is a contradiction.
Therefore (yn) is a Cauchy sequence.Lemma 2.2.
Let(X,d) be a metric space and let A,B,S,T,F, and ψ be as in Lemma 2.1.
If one of AX,TX,BX, and SX is a complete subspace of X,
then the following hold:
(i)
A and S have a coincidence point;(ii)
T and B have a coincidence point.Proof.
Fixx0∈X and let (yn) be the sequence defined in Lemma 2.1. If y2n=y2n−1 for some n,
then Ax2n=Tx2n+1=Bx2n−1=Sx2n,
and A and S have a coincidence point. If y2n=y2n+1 for some n,
then Ax2n=Tx2n+1=Bx2n+1=Sx2n+2,
and T and B have a coincidence point. Assume that yn≠yn+1 for every n and TX is complete. By Lemma 2.1, the sequence (yn) is Cauchy; as (y2n)⊂TX,
there exists u∈TX such that yn→u.
Let v∈X be such that Tv=u.
To prove that Bv=u.
We have(2.16)M(x2n,v)=max⁡{d(y2n−1,u),d(y2n,y2n−1),d(Bv,u),[d(y2n−1,Bv)+d(y2n,u)]2}.If Bv≠u,
then M(x2n,v)=d(u,Bv) definitely and consequently for large n,(2.17)F(d(Ax2n,Bv))≤ψ(F(M(x2n,v)))=ψ(F(d(u,Bv))).F being continuous, as n→+∞,
we obtain(2.18)F(d(u,Bv))≤ψ(F(d(u,Bv)))<F(d(u,Bv)).This is a contradiction,
therefore Bv=u and v is a coincidence point for T and B.
From BX⊂SX,
which gives u∈SX,
we deduce that there exists w∈X such that Sw=u.
To prove that Aw=u.
We have(2.19)M(w,v)=max⁡{d(u,u),d(Aw,u),d(u,u),[d(u,u)+d(Aw,u)]2}=d(Aw,u),and hence(2.20)F(d(Aw,Bv))≤ψ(F(M(w,u)))=ψ(F(d(Aw,u)))<F(d(Aw,u)),which gives Aw=u.
The same result holds if we suppose that one ofSX,AX,BX is complete.Theorem 2.3.
LetA,B,S, and T be self-mappings of a metric space (X,d) such that AX⊂TX, BX⊂SX.
Assume that there exist F,ψ:[0,+∞[→[0,+∞[ such that
(i)
F is nondecreasing, continuous, and F(0)=0<F(t) for every t>0;(ii)
ψ is nondecreasing, right continuous, and ψ(t)<t for every t>0;(iii)
F(d(Ax,By))≤ψ(F(M(x,y))) for all x,y∈X.
If one of AX,TX,BX, and SX is a complete subspace of X,
then the following hold:
(iv)
A and S have a coincidence point;(v)
T and B have a coincidence point.
Further, ifA and S as well as B and T are weakly compatible, then A,B,S, and T have a unique common fixed point.Proof.
Fixx0∈X and let (yn) be the sequence defined in Lemma 2.1. Assume
that TX is complete and let u,v, and w be as in Lemma 2.2. If A and S are weakly compatible, then(2.21)Au=ASw=SAw=Su,therefore u is a coincidence point of A and S.
To prove that d(Au,u)=0.
Suppose that d(Au,u)≠0.
We have(2.22)M(u,v)=max⁡{d(Su,u),d(Au,Su),d(u,u),[d(Su,u)+d(Au,u)]2}=d(Au,u)F(d(Au,Bv))=F(d(Au,u))≤ψ(F(M(u,v)))=ψ(F(d(Au,u)))<F(d(Au,u)).This is a contradiction, and
thus Au=u.
Since Au=Su=u,
we obtain that u is a common fixed point for A and S.
Similarly, ifB and T are weakly compatible, we deduce that u is a common fixed point for B and T.
Now if A and S as well as B and T are weakly compatible, then u is a common fixed point for A,B,S, and T.
If z∈X is also a common fixed point for A,B,S, and T with u≠z,
then(2.23)F(d(Au,Bz))≤ψ(F(M(u,z)))=ψ(F(d(Au,Bv)))<F(d(Au,Bv)),which gives u=z.Letφ:[0,+∞[→[0,+∞[ be a Lebesgue integrable function which is
nonnegative and such that(2.24)∫0εφ(t)dt>0,foreveryε>0.The function F:[0,+∞[→[0,+∞[,
with F(s)=∫0sφ(t)dt satisfies condition (i) of Lemma 2.1 and from
Theorem 2.3 we deduce the following theorem.Theorem 2.4 (see [2, Theorem 2.4 2.1]).
LetA,B,S, and T be self-mappings of a metric space (X,d) such that AX⊂TX, BX⊂SX.
Assume that there exists a nondecreasing right continuous function ψ:[0,+∞[→[0,+∞[,
with ψ(t)<t for all t>0,
such that(2.25)∫0d(Ax,By)φ(t)dt≤ψ(∫0M(x,y)φ(t)dt),where φ:[0,+∞[→[0,+∞[ is a Lebesgue integrable function which is
nonnegative and such that(2.26)∫0εφ(t)dt>0,foreveryε>0.If one of AX,TX,BX, and SX is a complete subspace of X,
then the following hold:
(i)
A and S have a coincidence point;(ii)
T and B have a coincidence point.
Further, ifA and S as well as B and T are weakly compatible, then A,B,S, and T have a unique common fixed point.Remark 2.5.
Theorem2.4 is a generalization of the main theorem in [3], of [4, Theorem 2], and of [5, Theorem 2].If in Theorem2.3, we assume S=T=IX,
where IX is the identity map on X,
we obtain the following theorem.Theorem 2.6.
LetA and B be self-mappings of a metric space (X,d).
Assume that there exist F,ψ:[0,+∞[→[0,+∞[ such that
(i)
F is nondecreasing, continuous, and F(0)=0<F(t) for every t>0;(ii)
ψ is nondecreasing, right continuous, and ψ(t)<t for every t>0;(iii)
F(d(Ax,By))≤ψ(F(m(x,y))) for all x,y∈X,
where(2.27)m(x,y)=max⁡{d(x,y),d(Ax,y),d(By,y),[d(Ax,y)+d(x,By)]2}.If one of AX and BX is a complete subspace of X,
then A and S have a unique common fixed point. Moreover,
for each x0∈X,
the iterated sequence (xn) with x2n+1=Ax2n and x2n+2=Bx2n+1 converges to the common fixed point of A and B.Theorem2.6 includes [7, Theorem 1].



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*Source: 891375-2008-10-20.xml*