R E S E A R C H
Open Access
Fixed point theory for generalized ´Ciri´c
quasi-contraction maps in metric spaces
F Kiany
1and A Amini-Harandi
2,3**Correspondence:
aminih_a@yahoo.com
2Department of Pure Mathematics,
University of Shahrekord, Shahrekord, 88186-34141, Iran
3School of Mathematics, Institute
for Research in Fundamental Sciences (IPM), P.O. Box 19395-5746, Tehran, Iran
Full list of author information is available at the end of the article
Abstract
In this paper, we first give a new fixed point theorem for generalized ´Ciri´c quasi-contraction maps in generalized metric spaces. Then we derive a common fixed point result for quasi-contractive type maps. Some examples are given to support our results. Our results extend and improve some fixed point and common fixed point theorems in the literature.
MSC: 47H10
Keywords: fixed points; common fixed point; generalized ´Ciri´c quasi-contraction maps
1 Introduction and preliminaries
The well-known Banach fixed point theorem asserts that if (X,d) is a complete metric space andT:X→Xis a map such that
d(Tx,Ty)≤cd(x,y) for eachx,y∈X,
where ≤c< , thenT has a unique fixed pointx¯∈Xand for anyx∈X, the sequence
{Tnx
}converges tox¯.
In recent years, a number of generalizations of the above Banach contraction principle have appeared. Of all these, the following generalization of Ćirić [] stands at the top.
Theorem . Let(X,d) be a complete metric space. Let T :X→X be a Ćirić quasi-contraction map,that is,there exists c< such that
d(Tx,Ty)≤cmaxd(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)
for any x,y∈X.Then T has a unique fixed pointx¯∈X and for any x∈X,the sequence
{Tnx
}converges tox¯.
For other generalizations of the above theorem, see [] and the references therein.
2 Main results
LetXbe a nonempty set and letd:X×X→[,∞] be a mapping. Ifdsatisfies all of the usual conditions of a metric except that the value ofdmay be infinity, we say that (X,d) is ageneralized metric space.
We now introduce the concept of ageneralized Ćirić quasi-contractionmap in general-ized metric spaces.
Definition . Let (X,d) be a generalized metric space. The self-mapT:X→Xis said to be ageneralized Ćirić quasi-contractionif
d(Tx,Ty)≤αd(x,y)maxd(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx) for anyx,y∈X, whereα: [,∞]→[, ) is a mapping.
As the following simple example due to Sastry and Naidu [] shows, Theorem . is not true for generalized Ćirić quasi-contraction maps even if we supposeαis continuous and increasing.
Example . LetX= [,∞) with the usual metric,T:X→Xbe given byTx= x. Define
α: [,∞)→[, ) byα(t) =+tt. Then, clearly,αis continuous and increasing, and
|Tx–Ty| ≤α|x–y|max|x–y|,|x–Tx|,|y–Ty|,|x–Ty|,|y–Tx|
for eachx,y∈X, butThas no fixed point.
Now, a natural question is what further conditions are to be imposed onTorαto guar-antee the existence of a fixed point forT? For some partial answers to this question and application of quasi-contraction maps to variational inequalities, see [] and the references therein.
Now, we are ready to state our main result.
Theorem . Let(X,d)be a complete generalized metric space.Let T:X→X be a gen-eralized Ćirić quasi-contraction map such thatαsatisfies
lim sup
t→r α(t) < for each r∈[,∞).
Assume that there exists an x∈X with the bounded orbit,that is,the sequence{Tnx}is bounded.Furthermore,suppose that d(x,Tx) <∞for each x∈X.Then T has a fixed point
¯
x∈X andlimn→∞Tnx
=x¯.Moreover,ify is a fixed point of T¯ ,then either d(x¯,y¯) =∞or
¯
x=y¯.
Proof If for somen∈N,Tn–x=Tnx=T(Tn–x), thenTnx=Tn–xforn≥n.
Thus,Tn–x
is a fixed point ofT, the sequence{Tnx}is convergent toTn–x, and we
are finished (note thatTnx=Tn–xfor eachn≥n). So, we may assume thatTn–x= Tnx
for eachn∈N. Now, we show that there exists <c< such that αdTn–x,Tnx
<c for eachn= , , , , . . . . (.)
On the contrary, assume that
lim
k→∞α
dTnk–x
,Tnkx
for some subsequence {α(d(Tnk–x
,Tnkx))} of {α(d(Tn–x,Tnx))}. Since by our
as-sumption the sequence{d(Tn–x
,Tnx)}is bounded, then the subsequence{d(Tnk–x, Tnkx
)} is bounded too, and so, by passing to subsequences if necessary, we may
as-sume that it is convergent. Letr=limk→∞d(Tnk–x
,Tnkx). Then from (.), we have
lim supt→r
α(t) = , a contradiction. Thus, (.) holds.
Now, we show that{Tnx
}is a Cauchy sequence. To prove the claim, we first show by
induction that for eachn≥,
dTn–x,Tnx
≤Kcn–, (.)
whereKis a bound for the bounded sequence{d(x,Tnx)}n. Ifn= then, we get
dTx,Tx
≤αd(x,Tx)
maxd(x,Tx),d
Tx,Tx
,dx,Tx
=αd(x,Tx)
maxd(x,Tx),d
x,Tx
≤Kc.
Thus, (.) holds forn= . Suppose that (.) holds for eachk<n, and we show that it holds fork=n. SinceTis a generalized Ćirić quasi-contraction map, then we have
dTn–x,Tnx
≤αTn–x,Tn–x
u≤cu, where
u∈dTn–x,Tn–x
,dTn–x,Tnx
.
It is trivial that (.) holds ifu=d(Tn–x
,Tn–x). Now, suppose thatu=d(Tn–x,Tnx).
In this case, we have
dTn–x,Tnx
≤cu,
where
u∈
dTn–x,Tn–x
,dTn–x,Tn–x
,
dTn–x,Tn–x
,dTn–x,Tnx
,dTn–x,Tnx
.
Again, it is trivial that (.) holds ifu=d(Tn–x,Tnx) oru=d(Tn–x,Tn–x). Ifu= d(Tn–x
,Tn–x), then
dTn–x,Tnx
≤cdTn–x,Tn–x
.
By the assumption of induction,
dTn–x,Tn–x
≤Kcn–.
Hence,
dTn–x,Tnx
Ifu=d(Tn–x,Tn–x), then
dTn–x,Tnx
≤cdTn–x,Tn–x
.
Ifu=d(Tn–x,Tnx), then
dTn–x,Tnx
≤cdTn–x,Tnx
.
Therefore, by continuing this process, we see that (.) holds for eachn≥. From (.), we deduce that{Tnx
}is a Cauchy sequence and since (X,d) is a generalized complete
metric space, then there exists anx¯∈Xsuch thatlimn→∞Tnx
=x¯. Now, we show thatx¯
is a fixed point ofT. To show the claim, we first show that there exists <k< such that
α(d(x¯,Tnx
)) <k for eachn∈N. On the contrary, assume thatlimj→∞α(d(x¯,Tnjx)) =
for some subsequence nj. Sincelimj→∞d(x¯,Tnjx) = , then from the above, we get
lim supt→+α(t) = , a contradiction. SinceTis a generalized Ćirić quasi-contraction, then
we have
dTx¯,Tn+x
≤αdx¯,Tnx
maxdx¯,Tnx
,d(x¯,Tx¯),
dTnx,Tn+x
,dx¯,Tn+x
,dTnx,Tx¯
≤kmaxdx¯,Tnx
,d(x¯,Tx¯),dTnx,Tn+x
,dx¯,Tn+x
,dTnx,Tx¯
.
Then we have
d(Tx¯,x¯) =lim sup
n→∞ d
Tx¯,Tn+x
≤klim sup
n→∞ d
Tx¯,Tnx
=kd(Tx¯,x¯),
which yieldsd(Tx¯,x¯) = , and sox¯=Tx¯ (note that <k< andd(Tx¯,x¯) <∞by our as-sumptions). Now, let us assume thatx¯andy¯are fixed points ofT such thatd(x¯,y¯) <∞. Then
d(x¯,y¯) =d(Tx¯,Ty¯)
≤αd(x¯,¯y)maxd(x¯,¯y),d(x¯,Tx¯),d(y¯,Ty¯),d(x¯,Ty¯),d(y¯,Tx¯) =αd(x¯,¯y)d(x¯,y¯),
and sox¯=y¯(note thatα(d(x¯,y¯)) < ).
The following example shows that in the statement of Theorem ., the condition
d(x,Tx) <∞for eachx∈Xis necessary.
Example . LetX={,∞},d(, ) =d(∞,∞) = and letd(,∞) =∞. LetT:X→X
be given byT =∞andT∞= . Then
d(Tx,Ty)≤
d(x,y)≤ max
d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)
Example . LetX= [,∞],d(x,y) =|x–y|for eachx,y∈[,∞),d(x,∞) =∞for each
x∈[,∞) and letd(∞,∞) = . Then (X,d) is a complete generalized metric space. Let
T:X→Xbe given byTx= xfor eachx∈[,∞) andT∞=∞. Defineα: [,∞]→[, ) byα(t) = +tt for eacht∈[,∞) andα(∞) = . Then we have
|Tx–Ty| ≤α|x–y|max|x–y|,|x–Tx|,|y–Ty|,|x–Ty|,|y–Tx|,
andd(x,Tx) <∞for eachx,y∈X. Thus, all of the assumptions of Theorem . are satis-fied, and soThas a unique fixed point (x=∞is a unique fixed point ofT). But we cannot invoke the above mentioned theorem of Ćirić to show the existence of a fixed point forT.
To prove the following common fixed point result, we use the technique in [].
Corollary . Let(X,d)be a complete metric space and let the self-maps T and S satisfy the contractive condition
d(Tx,Ty)
≤αd(Sx,Sy)maxd(Sx,Sy),d(Sx,Tx),d(Sy,Ty),d(Sx,Ty),d(Sy,Tx)
for each x,y∈X,whereαsatisfieslim supt→r+α(t) < for each r∈[,∞).If TX⊆SX and
SX is a complete subset of X,then T and S have a unique coincidence point in X.Moreover,
if T and S are weakly compatible(i.e.,they commute at their coincidence points),then T and S have a unique common fixed point.
Proof It is well known that there existsE⊆Xsuch thatSE=SXandS:E→Xis one-to-one. Now, define a mapU:SE→SEbyU(Sx) =Tx. SinceSis one-to-one onE,Uis well defined. Note that
dU(Sx),U(Sy)
=U(Tx,Ty)
≤αd(Sx,Sy)maxd(Sx,Sy),d(Sx,Tx),d(Sy,Ty),d(Sx,Ty),d(Sy,Tx)
for all Sx,Sy∈SE. SinceSE=SXis complete, by using Theorem ., there existsx¯∈X
such thatU(Sx¯) =Sx¯. ThenTx¯=Sx¯, and soT andShave a coincidence point, which is also unique. SinceTx¯=Sx¯andT andScommute, then we have
T(Tx¯) =Tx¯=TSx¯=STx¯=Sx¯=S(Sx¯).
Thus,Tx¯=Sx¯is also a coincidence point ofT andS. By the uniqueness of a coincidence
point ofTandS, we getTx¯=Sx¯=x¯.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Author details
1Faculty of Science, Ahvaz Branch, Islamic Azad University, Ahvaz, Iran.2Department of Pure Mathematics, University of
Shahrekord, Shahrekord, 88186-34141, Iran.3School of Mathematics, Institute for Research in Fundamental Sciences
(IPM), P.O. Box 19395-5746, Tehran, Iran.
Acknowledgements
The authors are grateful to the referees for their helpful comments leading to improvement of the presentation of the work. The second author acknowledge that this research was partially carried out at IPM-Isfahan Branch. The second author was partially supported by a grant from IPM (No. 91470412) and by the Center of Excellence for Mathematics, University of Shahrekord, Iran.
Received: 16 June 2012 Accepted: 13 January 2013 Published: 6 February 2013
References
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doi:10.1186/1687-1812-2013-26