R E S E A R C H
Open Access
Coupled coincidence point theorems in
(intuitionistic) fuzzy normed spaces
YJ Cho
1, J Martínez-Moreno
2*, A Roldán
3and C Roldán
4*Correspondence: [email protected] 2Department of Mathematics, University of Jaén, Jaén, Spain Full list of author information is available at the end of the article
Abstract
In this paper, we prove some coupled coincidence point theorems in fuzzy normed spaces. Our results improve and restate the proof lines of the main results given in the papers (Eshaghi Gordjiet al.in Math. Comput. Model. 54:1897-1906, 2011) and (Sintunavaratet al.in Fixed Point Theory Appl. 2011:81, 2011).
MSC: Primary 47H10; secondary 54H25; 34B15
Keywords: fuzzy normed space; intuitionistic fuzzy normed space; coupled fixed point theorem;t-norm;H-type
1 Introduction
Recently, many authors have shown the existence of coupled fixed points and common fixed points for some contractions in cone metric spaces, partially ordered metric spaces, fuzzy metric spaces, fuzzy normed spaces, intuitionistic fuzzy normed spaces and others ([–]).
Especially in [], Sintunavaratet al.proved some coupled fixed point theorems for contractive mappings in partially complete intuitionistic fuzzy normed spaces, which ex-tended and improved coupled coincidence point theorems in Gordjiet al.[]. But the authors found some mistakes in the proof lines of the main result (Theorem .) of [] and the same mistakes in [].
In Section of this paper, we restate some definitions and the main results in []. In Section , we give some comments about the incorrect proof lines of the main results given in [] and [] and explain why the lines of the proofs are wrong. Finally, in Section , we extend and improve some coupled fixed point theorems.
2 Preliminaries
At-norm(resp., at-conorm) is a mapping∗: [, ]→[, ] (resp.,: [, ]→[, ]) that is associative, commutative and non-decreasing in both arguments and has (resp., ) as identity.
Definition .[] Afuzzy normed space(in brief, FNS) is a triple (X,μ,∗), whereXis a vector space,∗is a continuoust-norm andμ:X×(,∞)→[, ] is a fuzzy set such that, for allx,y∈Xandt,s> ,
(F) μ(x,t) > ;
(F) μ(x,t) = for allt> if and only ifx= ; (F) μ(αx,t) =μ(x,|αt|)for allα= ;
(F) μ(x,t)∗μ(y,s)≤μ(x+y,t+s); (F) μ(x,·) : (,∞)→[, ]is continuous; (F) limt→∞μ(x,t) = andlimt→μ(x,t) = .
Using the continuoust-norms andt-conorms, Saadati and Park [] introduced the con-cept of an intuitionistic fuzzy normed space.
Definition . Anintuitionistic fuzzy normed space(in brief, IFNS) is a -tuple (X,μ,ν,
∗,) whereXis a vector space,∗is a continuoust-norm,is a continuoust-conorm and
μ,ν:X×(,∞)→[, ] are fuzzy sets such that, for allx,y∈Xandt,s> ,
(IF) μ(x,t) +ν(x,t)≤; (IF) μ(x,t) > andν(x,t) < ;
(IF) μ(x,t) = for allt> if and only ifx= if and only ifν(x,t) = for allt> ; (IF) μ(αx,t) =μ(x,|αt|)andν(αx,t) =ν(x,|αt|)for allα= ;
(IF) μ(x,t)∗μ(y,s)≤μ(x+y,t+s)andν(x,t)ν(y,s)≥ν(x+y,t+s); (IF) μ(x,·),ν(x,·) : (,∞)→[, ]are continuous;
(IF) limt→∞μ(x,t) = =limt→ν(x,t)andlimt→μ(x,t) = =limt→∞ν(x,t).
Obviously, if (X,μ,ν,∗,) is an IFNS, then (X,μ,∗) is an FNS. We refer to this space as itssupport.
Lemma . μ(x,·)is a non-decreasing function on(,∞)andν(x,·)is a non-increasing function on(,∞).
Some properties and examples of IFNS and the concepts ofconvergenceand aCauchy sequencein IFNS are given in [].
Definition .[] Let (X,μ,ν,∗,) be an IFNS.
() A sequence{xn} ⊂Xis called aCauchy sequenceif, for any> andt> , there
existsn∈Nsuch thatμ(xn–xm,t) > –andν(xn–xm,t) <for alln,m≥n.
() A sequence{xn} ⊂Xis said to beconvergent toa pointx∈Xdenoted byxn→xor
bylimn→∞xn=xif, for any> andt> , there existsn∈Nsuch that
μ(xn–x,t) > –andν(xn–x,t) <for alln≥n.
() An IFNS in which every Cauchy sequence is convergent is said to becomplete.
Most of the following definitions were introduced in [].
Definition . LetF:X×X→Xandg:X→Xbe two mappings.
() Fandgare said to becommutingifgF(x,y) =F(gx,gy)for allx,y∈X.
() A point(x,y)∈X×Xis called acoupled coincidence pointof the mappingsFandg ifF(x,y) =gxandF(y,x) =gy. Ifgis the identity,(x,y)is called acoupled fixed point ofF.
() If(X,)is a partially ordered set, thenFis said to have themixedg-monotone propertyif it verifies the following properties:
x,x∈X, gxgx =⇒ F(x,y)F(x,y), ∀y∈X,
Ifgis the identity mapping, thenFis said to have themixed monotone property. () If(X,)is a partially ordered set, thenXis said to have thesequentialg-monotone
propertyif it verifies the following properties:
(B) If{xn}is a non-decreasing sequence andlimn→∞xn=x, thengxngxfor all
n∈N.
(B) If{xn}is a non-increasing sequence andlimn→∞yn=y, thengyngyfor all
n∈N.
Ifgis the identity mapping, thenXis said to have the sequentialmonotone property.
Definition . LetXandY be two IFNS. A functionf :X→Y is said to becontinuous
at a point x∈Xif, for any sequence{xn}inXconverging tox, the sequence{f(xn)}inY converges tof(x). Iff is continuous at eachx∈X, thenf is said to becontinuous on X.
Definition .[] Let (X,μ,ν,∗,) be an IFNS. The pair (μ,ν) is said to satisfy then
-propertyonX×(,∞) iflimn→∞[μ(x,knt)]np= andlimn→∞[ν(x,knt)]np= whenever x∈X,k> andp> .
The following lemma proved by Haghiet al.[] is useful for our main results.
Lemma . Let X be a nonempty set and g:X→X be a mapping.Then there exists a subset E⊂X such that g(E) =g(X)and g:E→X is one-to-one.
In order to state our results, we give the main results given in [] and [].
Lemma .[, Lemma .] Let(X,μ,ν,∗,)be an IFNS.Let a∗b≥ab,ab≤ab for all a,b∈[, ]and(μ,ν)satisfy the n-property.Suppose that{xn}is a sequence in X such that
μ(xn+–xn,kt)≥μ(xn–xn–,t), ν(xn+–xn,kt)≤ν(xn–xn–,t)
for all t> and n∈N,where <k< .Then the sequence{xn}is a Cauchy sequence in X.
Theorem .([, Theorem .], [, Theorem .]) Let(X,)be a partially ordered set
and suppose that ab≤ab≤a∗b for all a,b∈[, ].Let(X,μ,ν,∗,)be a complete IFNS such that(μ,ν)has the n-property.Let F:X×X→X and g:X→X be two mappings such that F has the mixed g-monotone property and
μF(x,y) –F(u,v),kt≥μ(gx–gu,t)∗μ(gy–gv,t),
νF(x,y) –F(u,v),kt≤ν(gx–gu,t)ν(gy–gv,t),
for which gxgu and gygv,where <k< ,F(X×X)⊆g(X)and g is continuous.
Suppose either
(a) Fis continuous or
(b) Xhas the sequentialg-monotone property.
If there exist x,y∈X such that gxF(x,y)and gyF(y,x),then there exist x,y∈X
3 Comments and suggestions
In this section, we show that the conditions of the above Lemma . and Theorem . in [] are inadequate and, furthermore, the proof lines of Theorem . are not correct. We also would like to point out that the results in [] can be corrected under the appropriate conditions on thet-norm and the FNS.
First of all, in the conditions of Lemma . and Theorem ., we have at-conormsuch thatab≤abfor alla,b∈[, ]. If we takeb= , thena=a≤ for alla∈[, ]. It is obviously impossible. Moreover, it is well known and easy to see that if∗is at-norm andis at-conorm, thena∗b≤abfor alla,b∈[, ]. In this sense, Lemma . and Theorem . have to be corrected.
Secondly, from the property (IF), it follows that a sequence{xn} ⊂X is aCauchy
se-quenceif, for any> andt> , there existsn∈Nsuch thatμ(xn–xm,t) > –for all
n,m≥n. That is, the sequence{xn} ⊂Xis aCauchy sequenceon the IFNS (X,μ,ν,∗,) if it also is on the FNS (X,μ,∗). A similar comment is valid for the convergence.
Furthermore, the completeness of an IFNS is equivalent to the completeness of its sup-port FNS and so we can deduce any fixed point theorem for IFNS (when the conditions on
μandνare splitting) as an immediate consequence of its associated fixed point theorem for FNS. In particular, it is sufficient to prove Theorem . just for FNS. Therefore, we only develop Theorem . for FNS.
Also, some proof lines of Lemma . are not correct (see p., lines -):
μ(xn–xm,t)≥
μ
x–x, ( –k)
t kn m ≥ μ
x–x, ( –k)
t kn
nq
→,
whereq> such thatm<nq, and
ν(xn–xm,t)≤
ν
x–x, ( –k)
t kn m ≤ ν
x–x, ( –k)
t kn
np
→,
wherep> such thatm<np. Hence the sequence{x
n}is a Cauchy sequence. This is not correct since the sameq(orp) would not be valid for all positive integersm>n≥n. For instance, let (X, · ) be an ordinary normed space, defineμ(x,t) = t+txfor anyx∈Xand
t> anda∗b=abfor alla,b∈[, ]. Then (X,μ, –μ,∗,∗) is an IFNS. Ifk= / and
m= n, we have
μ
x–x, ( –k)
t kn
m
=
n–t
n–t+x –x
n
→e–x–tx < .
Also, the proof lines of Theorem . that are not correct are the following ones (see p., lines -):
μ(gxn–gxm,t)∗μ(gxn–gxm,t)
≥
μ
gx–gx, ( –k)
t kn
m–n
∗
μ
gy–gy, ( –k)
t kn
m–n
≥
μ
gx–gx, ( –k)
t kn m ∗ μ
gy–gy, ( –k)
t kn m ≥ μ
gx–gx, ( –k)
t kn np ∗ μ
gy–gy, ( –k)
t kn
np
→,
whereq> such thatm<nq. In general, we cannot obtainμ(x
n–xm,t)→ asn,m→ ∞. It is not shown that {xn}is a Cauchy sequence. Moreover, a similar conclusion can be obtained forν(xn–xm,t)→. Thus, from the hypothesis of Theorem ., the conclusion cannot be guaranteed.
4 The modification in FNS
In this section, by replacing the hypothesis thatμsatisfies then-property with the one that thet-norm is ofH-type, we state and prove a coupled fixed point theorem as a mod-ification.
Definition .[] For anya∈[, ], let the sequence {∗na}∞
n= be defined by∗a=a and∗na= (∗n–a)∗a. Then at-norm∗is said to beof H-typeif the sequence{∗na}∞
n=is equicontinuos ata= .
Theorem . Let(X,)be a partially ordered set and(X,μ,∗)be a complete FNS such that∗is of H-type and a∗b≥ab for all a,b∈[, ].Let k∈(, )be a number and F :
X×X→X be a mapping such that F has the mixed monotone property and
μF(x,y) –F(u,v),kt≥μ(x–u,t)/∗μ(y–v,t) /, (.)
for which xu and yv.Suppose that either (a) Fis continuous or
(b) Xhas the sequential monotone property.
If there exist x,y∈X such that xF(x,y)and yF(y,x),then F has a coupled
fixed point.Furthermore,if xand yare comparable,then x=y,that is,x=F(x,x).
Proof Letx,y∈Xbe such thatxF(x,y) andyF(y,x). SinceF(X×X)⊆X, we can choosex,y∈Xsuch thatx=F(x,y) andy=F(y,x). Again, fromF(X×X)⊆X, we can choosex,y∈Xsuch thatx=F(x,y) andy=F(y,x). Continuing this process, we can construct two sequences{xn}and{yn}inXsuch that, for eachn≥,
xn+=F(xn,yn), yn+=F(yn,xn). (.)
The proof is divided into two steps.
Step . Prove that{xn}and{yn}are Cauchy sequences. Firstly, we show by induction that, for eachn≥,
Forn= , (.) holds trivially. Suppose that, for some fixedn≥, (.) holds. Since
xnxn+andxnxn+andFhas the mixed monotone property, it follows from (.) that
xn+=F(xn,yn)F(xn+,yn), yn+=F(yn,xn)F(yn+,xn). (.)
Similarly, we have
xn+=F(xn+,yn+)F(xn+,yn), yn+=F(yn+,xn+)F(yn+,xn). (.)
Thus, combining (.) and (.), (.) holds.
Letδn(t) = [μ(xn–xn+,t)]/∗[μ(yn–yn+,t)]/for alln≥. Then it follows from (.), (.) and (F) that
μ(xn–xn+,kt) =μ
F(xn–,yn–) –F(xn,yn),kt
≥μ(xn––xn,t)/∗
μ(yn––yn,t)/
=δn–(t) (.)
and
μ(yn–yn+,kt) =μ
F(yn,xn) –F(yn–,xn–),kt
≥μ(yn–yn–,t) /∗
μ(xn–xn–,t)/
=δn–(t). (.)
Then it follows from thet-norm anda∗b≥abthat δn(kt)≥δn–(t) for alln≥. This implies that
≥δn(t)≥δn–
t k
≥δn–
t k
≥ · · · ≥δ
t kn
. (.)
Sincelimn→∞δ(ktn) = for allt> , we havelimn→∞δn(t) = for allt> . Now, we claim that, for anyp≥,
μ(xn–xn+p,t)≥ ∗pδn–(t–kt), μ(yn–yn+p,t)≥ ∗pδn–(t–kt), ∀n≥. (.) In fact, it is obvious forp= by (.), (.) and Lemma . sincet/k≥t–ktandδn–is non-decreasing. Assume that (.) holds for somep≥. By (.), we have
μ(xn–xn+,t)≥μ(xn–xn+,kt)≥δn–(t) and so
μ(xn–xn+,t–kt)≥δn–(t–kt).
Thus, from (.), (.) anda∗b≥ab, we have
μ(xn+–xn+p+,kt)≥
μ(xn–xn+p,t) /
∗μ(yn–yn+p,t) /
≥ ∗pδ
Hence, by the monotonicity of thet-norm∗, we have
μ(xn–xn+p+,t) =μ(xn–xn+p+,t–kt+kt)
≥μ(xn–xn+,t–kt)∗μ(xn+–xn+p+,kt)
≥δn–(t–kt)∗
∗pδ
n–(t–kt)
=∗p+δn–(t–kt).
Similarly, we have
μ(yn+–yn+p+,kt)≥ ∗p+δn–(t–kt).
Therefore, by induction, (.) holds for allp≥. Suppose thatt> and∈(, ] are given. By hypothesis, since∗is at-norm ofH-type, there exists <η< such that∗p(a) > –for alla∈( –η, ] andp≥. Sincelimn→∞δn(t) = , there existsnsuch thatδn(t–kt) > –η for alln≥n. Hence, from (.), we get
μ(xn–xn+p,t) > –, μ(yn–yn+p,t) > –, ∀n≥n.
Therefore,{xn}and{yn}are Cauchy sequences.
Step . We prove thatFhas a coupled fixed point. SinceXis complete, there existx,y∈X
such thatlimn→∞xn=xandlimn→∞yn=y. Suppose that the assumption (a) holds. By the continuity ofF, we get
x= lim
n→∞xn+=nlim→∞F(xn,yn) =F
lim
n→∞xn,nlim→∞yn
=F(x,y).
Similarly, we can show thatF(y,x) =y.
Suppose now that (b) holds. Since{xn}is a non-decreasing sequence withxn→xand
{yn}is a non-increasing sequence withxy→y, from (B) and (B), we havexnxand
ynyfor alln≥. Then, by (.), we obtain
μxn+–F(x,y),kt
≥μF(xn,yn) –F(x,y),kt
≥μ(xn–x,t) /
∗μ(yn–y,t) /
.
Lettingn→ ∞, we havelimn→∞xn=F(x,y). HenceF(x,y) =x. Similarly, we can show thatF(y,x) =y.
Suppose thatxy. By induction and the mixed monotone property ofF, it follows thatxn=F(xn–,yn–)F(yn–,xn–) =yn. From (.), it follows that
μ(xn+–yn+,kt) =μ(F(xn,yn) –F(yn,xn,kt)
≥μ(xn–yn,t) /
∗μ(yn–xn,t) /
=μ(xn–yn,t).
By the iterative procedure, we have
μ(xn–yn,kt)≥μ(xn––yn–,t)≥μ
xn––yn–,
t k
≥ · · · ≥μ
x–y,
t kn–
Takingn→ ∞, sincelimn→∞μ(x–y,knt–) = for allt> , we conclude thatμ(x–y,t)≥
for allt> ,i.e.,x=y. This completes the proof.
Next, we prove the existence of a coupled coincidence point theorem, where we do not require thatFandgare commuting.
Theorem . Let (X,)be a partially ordered set and let(X,μ,∗)be a complete FNS such that∗is of H-type and a∗b≥ab for all a,b∈[, ].Let k∈(, )be a number and F:X×X→X and g:X→X be two mappings such that F has the mixed g-monotone property and
μF(x,y) –F(u,v),kt≥μ(gx–gu,t) /∗μ(gy–gv,t) /, (.)
for which gxgu and gygv.Suppose that F(X×X)⊆g(X),g is continuous and either (a) Fis continuous or
(b) Xhas the sequentialg-monotone property.
If there exist x,y∈X such that gxF(x,y)and gyF(y,x),then there exist x,y∈X
such that gx=F(x,y)and gy=F(y,x),that is,F and g have a coupled coincidence point.
Proof Using Lemma ., there existsE⊂Xsuch thatg(E) =g(X) andg:E→Xis one-to-one. We define a mappingA:g(E)×g(E)→XbyA(gx,gy) =F(x,y). Sincegis one-to-one ong(E),Ais well defined. Thus it follows from (.) that
μA(gx,gy) –A(gu,gv),kt≥μ(gx–gu,t) /∗μ(gy–gv,t) /, (.)
for whichgxguandgygv. SinceFhas the mixedg-monotone property, we have
gx,gx∈g(X), gxgx
=⇒ A(gx,gy) =F(x,y)F(x,y) =A(gx,gy), ∀gy∈g(X),
gy,gy∈g(X), gygy
=⇒ A(gx,gy) =F(x,y)F(x,y) =A(gx,gy), ∀gx∈g(X),
which implies thatAhas the mixed monotone property.
Suppose that the assumption (a) or (b) holds. Using Theorem . with the mappingA, it follows thatAhas a coupled fixed point (u,v)∈g(X)×g(X),i.e.,u=A(u,v) andv=A(v,u). Since (u,v)∈g(X)×g(X), there exists (u˜,v˜)∈X×Xsuch thatgu˜=uandgv˜=v. Thus
gu˜=u=A(u,v) =A(gu˜,gv˜). Similarly,gv˜=A(gv˜,gu˜). This completes the proof.
Now, we show the existence and uniqueness of coupled coincidence points. Note that if (S,) is a partially ordered set, then we endow the productS×Swith the following partial order:
(x,y)(u,v) ⇐⇒ xu,yv, ∀x,y,u,v∈S.
Theorem . In addition to the hypotheses of Theorem.,suppose that,for any pair of coupled coincidence points(x,y), (x∗,y∗)∈X×X,there exists a point(u,v)∈X×X such that(gu,gv)is comparable to(gx,gy)and(gx∗,gy∗).Then F and g have a unique coupled coincidence point,that is,there exists a unique(x,y)∈X×X such that x=gx=F(x,y)and y=gy=F(y,x).
Proof From Theorem ., the set of coupled coincidences is nonempty. Now, we show that if (x,y) and (x*,y*) are coupled coincidence points, that is,gx=F(x,y),gy=F(y,x) and
gx*=F(x*,y*),gy*=F(y*,x*), then
gx=gx*, gy=gy*. (.)
Putu=uandv=vand chooseu,v∈Xsuch thatgu=F(u,v) andgv=F(v,u). Then, as in the proof of Theorem ., we can inductively define the sequences{un}and
{vn}such that
gun+=F(un,vn), gvn+=F(vn,un). (.)
Since (F(x,y),F(y,x)) = (gx,gy) and (F(u,v),F(v,u)) = (gu,gv) are comparable, we can sup-pose thatgxguandgygv. It is easy to show, by induction andg-monotonicity, that
gxgunandgygvnfor alln≥. From (.), we obtain
μ(gx–gun,kt) =μ
F(x,y) –F(un–,vn–),kt
≥μ(gx–gun–,t) /
∗μ(gy–gvn–,t) /
(.)
and
μ(gvn–gy,kt) =μ
F(vn–,un–) –F(y,x),kt
≥μ(gvn––gy,t)/∗
μ(gun––gx,t) /. (.) Now, letβn(t) = [μ(gx–gun,t)]/∗[μ(gy–gvn,t)]/. By (.) and (.), we have
βn(t)≥βn–
t k
≥ · · · ≥β
t kn
and
μ(gx–gun,kt)≥β
t kn
, μ(gy–gvn,kt)≥β
t kn
.
Sincelimn→∞β(ktn) = , we conclude thatlimn→∞gun=gxandlimn→∞gvn=gy.
Similarly,limn→∞gun=gx*andlimn→∞gvn=gy*. Hencegx=gx* andgy=gy*and so (.) is proved.
Sincegx=F(x,y) andgy=F(y,x), ifz=gxandw=gy, by the commutativity ofFandg, we have
and
gw=ggy=gF(y,x) =F(gy,gx) =F(w,z),
i.e., (z,w) is a coupled coincidence point. In particular, from (.), we havez=gx=gzand
w=gy=gw. Therefore, (z,w) is a coupled common fixed point ofFandg.
To prove the uniqueness of the coupled common fixed point ofF andg, assume that (p,q) is another coupled common fixed point. Then, by (.), we havep=gp=gz=zand
q=gq=gw=w. This completes the proof.
Finally, we present an intuitionistic version of Theorem . with the dual conditions on
t-conorms. The proof is just reduced to apply Theorem . to the support FNS.
Corollary . Let(X,)be a partially ordered set and suppose that a∗b≥ab and( –
a)( –b)≤ –ab for all a,b∈[, ].Let(X,μ,ν,∗,)be a complete IFNS such that∗and are of H-type.Let F:X×X→X and g:X→X be two mappings such that F has the mixed g-monotone property,
μF(x,y) –F(u,v),kt≥μ(gx–gu,t) /∗μ(gy–gv,t) /
and
νF(x,y) –F(u,v),kt
≤ – –ν(gx–gu,t) / – –ν(gy–gv,t)/, (.)
for which g(x)g(u)and g(y)g(v),where <k< ,F(X×X)⊆g(X)and g is continuous.
Suppose that either (a) Fis continuous or
(b) Xhas the sequentialg-monotone property.
If there exist x,y∈X such that g(x)F(x,y)and g(y)F(y,x),then there exist
x,y∈X such that g(x) =F(x,y)and g(y) =F(y,x),that is,F and g have a coupled coincidence point.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors read and approved the final manuscript.
Author details
1Department of Mathematics Education and RINS, Gyeongsang National University, Jinju, 660-701, Korea.2Department of Mathematics, University of Jaén, Jaén, Spain.3Department of Statistics and Operations Research, University of Jaén, Jaén, Spain.4Department of Statistics and Operations Research, University of Granada, Granada, Spain.
Acknowledgements
This work was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant Number: 2012-0008170) and by the Junta de Andalucía and projects FQM-268, FQM-178 of the Andalusian CICYE, Spain.
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