R E S E A R C H
Open Access
On fixed point theorems for mappings with
PPF dependence
Ali Farajzadeh
1and Anchalee Kaewcharoen
2,3**Correspondence:
[email protected] 2Department of Mathematics,
Faculty of Science, Naresuan University, Phitsanulok, 65000, Thailand
3Centre of Excellence in
Mathematics, CHE, Si Ayutthaya Rd., Bangkok, 10400, Thailand Full list of author information is available at the end of the article
Abstract
In this paper, we consider two families
1and
2of mappings defined on [0, +∞)
satisfy some certain properties. Using the mentioned properties for
1and
2, we
prove the analogous PPF dependent fixed point theorems for mappings as in Driciet al.(Nonlinear Anal. 67:641-647, 2007) in partially ordered Banach spaces where mappings satisfy the weaker contractive conditions without assuming the topological closedness with respect to the norm topology for the Razumikhin classRc.
MSC: 54H25; 55M20
Keywords: PPF dependent fixed points; Razumikhin classes; partially ordered sets; algebraic closedness with respect to difference
1 Introduction and preliminary results
The fixed point theorems for mappings satisfying certain contractive conditions have been continually studied for decade (see [–] and references contained therein). Bernfeldet al.
[] proved the existence of PPF (past, present and future) dependent fixed points in the Razumikhin class for mappings that have different domains and ranges. After that Dhage [] extended the existence of PPF dependent fixed points to PPF common dependent fixed points for mappings satisfying the weaker contractive conditions. In , Driciet al.[] proved the fixed point theorems in partially ordered metric spaces for mappings with PPF dependence. In this paper, we consider two familiesandof mappings defined on
[, +∞) satisfy some certain properties. Moreover, the PPF dependent fixed point theo-rems for mappings satisfying some generalized contractive conditions in partially ordered Banach spaces are proven using the mentioned properties forand.
Suppose thatEis a real Banach space with the norm · EandIis a closed interval [a,b]
inR. LetE=C(I,E) be the set of all continuousE-valued mappings onIequipped with
the supremum norm · Edefined by
φE=sup
t∈I
φ(t)E, (.)
for allφ∈E. For a fixed elementc∈I, the Razumikhin class of mappings inEis defined
by
Rc=
φ∈E:φE=φ(c)E
. (.)
Recall that a pointφ∈Eis said to be a PPF dependent fixed point or a fixed point with
PPF dependence ofT:E→EifTφ=φ(c) for somec∈I.
Example . LetT:C([, ],R)→Rbe defined by
Tφ=
sup
t∈[,]
φ(t) for allφ∈C[, ],R.
ThereforeTis a contraction with a constant
. Suppose thatφ(t) =t
+ for allt∈[, ].
SinceTφ=(supt∈[,]|φ(t)|) = =φ(), we find thatφ is a fixed point with dependence ofT.
Definition . LetAbe a subset ofE. Then
(i) Ais said to be topologically closed with respect to the norm topology if for each sequence{xn}inAwithxn→xasn→ ∞impliesx∈A.
(ii) Ais said to be algebraically closed with respect to the difference ifx–y∈Afor all x,y∈A.
Recently, Dhage [] proved the existence of PPF fixed points for mappings satisfying the condition of Cirić type generalized contraction assuming topological closedness with respect to the norm topology for a Razumikhin class.
Definition .(Dhage, []) A mappingT:E→Eis said to satisfy the condition of Cirić
type generalized contraction if there exists a real numberλ∈[, ) satisfying
Tφ–Tα ≤λmax
φ–αE,φ(c) –TφE,α(c) –TαE,
φ(c) –TαE+α(c) –TφE
, (.)
for allφ,α∈Eand for somec∈I.
Theorem .(Dhage, []) Suppose that T:E→E satisfies the condition of Cirić type
generalized contraction.Assume thatRcis topologically closed with respect to the norm topology and is algebraically closed with respect to the difference,then T has a unique PPF dependent fixed point inRc.
It is a natural question that the result of the previous theorem is still valid by omitting the topological closedness ofRc. In the next result, we will answer the question.
Proposition . The Razumikhin classRcis topologically closed with respect to the norm topology.
Proof Let{φn}be a sequence inRcconverging toφ. This implies that
lim
n→∞φn–φE= whereφn–φE=sup
t∈I
Therefore
lim
n→∞φnE=φE and n→∞limφn(t)E=φ(t)E for allt∈I.
Sinceφn∈Rcfor alln∈N, we obtainφnE=φn(c)E. Therefore
lim
n→∞φn(c)E=φE.
By the uniqueness of the limit, we haveφE=φ(c)E. Henceφ∈Rcand thusRc is
topologically closed with respect to the norm topology.
Hence, using Proposition ., we can drop the topological closedness with respect to the norm topology forRcin Theorem ..
The following example shows that the algebraic closedness with respect to the difference of Razumikhin classRcmay fail.
Example . LetE=C([, ],R) andc= . If we takeφ(t) =tandα(t) =tfor allt∈[, ],
thenφ,α∈Rcwhileφ–α∈/Rc.
Proposition . If the Razumikhin classRcis algebraically closed with respect to the dif-ference,thenRcis a convex set.
Proof SinceRcis algebraically closed with respect to the difference, we haveRc–Rc⊆Rc.
Using the fact that –Rc=Rc, we obtainRc+Rc⊆Rc. SinceλRc⊆Rcfor allλ∈[, ],
we getλRc+ ( –λ)Rc⊆Rc. HenceRcis a convex set.
One can verify that Razumikhin classRcis a cone (i.e.,λφ∈Rc, for eachφ∈Rcand
λ≥). Then by applying the previous theorem Razumikhin classRcis a convex cone (also
closed).
In , Driciet al.[] proved the following the fixed point theorems in partially ordered complete metric spaces for mappings with PPF dependence.
Theorem .([]) Let(E,d,≤)be a partially ordered complete metric space and T:E→
E where E=C(I,E)and I= [a,b].Assume that (i) Tis a nondecreasing mapping;
(ii) for allφ,α∈Ewithφ≤α,d(Tφ,Tα)≤kd(φ,α)where
d(φ,α) =maxs∈Id(φ(s),α(s))andk∈[, );
(iii) there exists a lower solutionφsuch thatφ(c)≤Tφ;
(iv) Tis a continuous mapping or if{φn}is a nondecreasing sequence inEconverging to
φ∈E,thenφn≤φfor alln∈N. Then T has a PPF dependent fixed point in E.
It is a natural question if one can obtain the result of the aforementioned theorem for a generalized contraction (that is, condition (ii)). One of the aims of this paper is to answer the question by considering a general case of Cirić type generalized contractions.
In this paper, we consider two familiesandof mappings defined on [, +∞)
analogous PPF dependent fixed point theorems for mappings as in [] in partially ordered real Banach spaces where mappings satisfy the weaker contractive conditions.
2 PPF dependent fixed points in partially ordered Banach spaces
We begin this part with the consideration of the example of a partially ordered real Banach space. Recall that the setB(X,R) of all bounded linear operators from a normed spaceX
intoRis a real Banach space with a norm defined by
f= sup
x∈X,x=
f(x) for allf ∈B(X,R).
We know thatB(X,R) is a partially ordered real Banach space with a partial order defined as follows:
f ≤g if and only if f(x)≤g(x) for allx∈X.
From now on, let (E,≤) be a partially ordered real Banach space. In this paper, we use the following notations:
=
ψ:ψ: [, +∞)→[, +∞) is nondecreasing with
∞
n=
ψn(t) <∞for allt∈(, +∞)
and
=
ψ:ψ: [, +∞)→[, +∞) is continuous nondecreasing with
ψ(t) <tfor allt∈(, +∞).
Lemma .([]) Suppose thatψ: [, +∞)→[, +∞).Ifψis nondecreasing,then for each t∈(, +∞),limn→∞ψn(t) = impliesψ(t) <t.
Hence the difference between an element ofand an element ofis continuity.
Remark .
(i) It is easily seen that ifψ: [, +∞)→[, +∞)is nondecreasing andψ(t) <tfor all t∈(, +∞), thenψ() = .
(ii) We can see that ifψ: [, +∞)→[, +∞),ψ(t) <tfor allt∈(, +∞)andψ() = , thenψis continuous at.
We now prove the PPF dependent fixed point theorems for mappings satisfying the gen-eralized contractive conditions concerning withψ∈without assuming the topological
closedness with respect to the norm topology for the Razumikhin classRc.
(ii) for allφ,α∈Ewithφ≤α,we have
Tφ–TαE≤ψ
max
φ–αE,φ(c) –TφE,α(c) –TαE,
φ(c) –TαE+α(c) –TφE
;
(iii) there exists a lower solutionφ∈Rcsuch thatφ(c)≤Tφ; (iv) Tis a continuous mapping.
Assume that Rc is algebraically closed with respect to the difference.Then T has a PPF dependent fixed point inRc.
Proof Sinceφ∈RcandTφ∈E, there existsx∈Esuch thatTφ=x. Chooseφ∈Rc
such thatx=φ(c). Sinceφis a lower solution inRcsuch thatφ(c)≤Tφ, it follows that
φ≤φ. Using the algebraic closedness with respect to the difference ofRc, this yields
φ–φE=φ(c) –φ(c)E.
By the fact thatT is nondecreasing, we obtain
φ(c) =Tφ≤Tφ.
By induction, we can construct the sequence{φn}such that
Tφ=φ(c)≤Tφ=φ(c)≤ · · · ≤Tφn=φn+(c)≤Tφn+· · ·,
φn≤φn+ and φn–φn+E=φn(c) –φn+(c)E,
for alln∈N∪ {}. Assume thatφn–=φnfor somen∈N. It follows thatφn–(c) =φn(c) = Tφn–. ThereforeThas a fixed point inRc. Suppose thatφn–=φnfor alln∈N. Therefore,
for eachn∈N, we obtain
φn–φn+E =φn(c) –φn+(c)E
=Tφn–Tφn–E
≤ψ
max
φn–φn–E,φn(c) –TφnE,φn–(c) –Tφn–E,
φn(c) –Tφn–E+φn–(c) –TφnE
=ψ
max
φn–φn–E,φn(c) –φn+(c)E,φn–(c) –φn(c)E,
φn(c) –φn(c)E+φn–(c) –φn+(c)E
≤ψ
max
φn–φn–E,φn–φn+E,
φn––φn+E
≤ψ
max
φn––φnE+φn–φn+E
≤ψmaxφn–φn–E,φn–φn+E .
Ifmax{φn–φn–E,φn–φn+E}=φn–φn+E, then
φn–φn+E≤ψ
φn–φn+E <φn–φn+E,
which leads to a contradiction. Therefore, for eachn∈N, we have
φn–φn+E≤ψ
φn–φn–E .
By induction, we obtain
φn–φn+E≤ψ
nφ
–φE .
Fixε> . This implies that there existsN∈Nsuch that
n≥N
ψnφ–φE <ε.
For eachm,n∈Nwithm>n>N, we obtain
φn–φmE≤
m–
k=n
φk–φk+E≤
m–
k=n
ψkφ–φE ≤
n≥N
ψnφ–φE <ε.
This implies that {φn} is a Cauchy sequence. By the completeness of E, we have
limn→∞φn=φfor someφ∈Eand
lim
n→∞Tφn=n→∞limφn+(c) =φ(c).
SinceRcis algebraically closed with respect to the norm topology, we haveφ∈Rc. We
next prove thatφis a PPF dependent fixed point ofT. Using the continuity ofT, we obtain limn→∞Tφn=Tφ. By the uniqueness of the limit, we haveTφ=φ(c).
Remark .
(i) From the proof of Theorem ., we assume that the Razumikhin classRcis
algebraically closed with respect to difference, that is,φ–α∈Rcfor allφ,α∈Rc, in order to construct the sequence{φn}satisfying
φn–φn+E=φn(c) –φn+(c)E for alln∈N∪ {}.
(ii) In the proof of Theorem ., if we chooseφn∈Rcto be a constant mapping for eachn∈N∪ {}, then
φn–φn+E=φn(c) –φn+(c)E for alln∈N∪ {}.
Example . LetE=Rwith respect to the norm(x,y)=|x|+|y|andE
=C([, ],R).
Define a mappingψ: [,∞)→[,∞) by
ψ(t) = t
for allt∈[,∞).
We see thatψ is a nondecreasing mapping with∞n=ψn(t) <∞for allt∈(, +∞). Let φ∈E. Therefore, for eacht∈[, ], we obtainφ(t)∈R. Thus we can define mappings
fφ,gφ: [, ]→Rsuch that
φ(t) =fφ(t),gφ(t) for allt∈[, ].
Define a mappingT:E→Eby
Tφ=
fφ,
gφ
for allφ∈E,
wherefφ=supt∈[,]|fφ(t)|andgφ=supt∈[,]|gφ(t)|. Suppose thatφ,α∈Ewithφ≤α.
For eachc∈[, ], we obtain
Tφ–TαE = fφ,
gφ
–
fα,
gα
E
= fφ–
fα
+
gφ– gα
≤
fφ–fα+gφ–gα
=
sup
t∈[,]
fφ(t) –fα(t)+ sup
t∈[,]
gφ(t) –gα(t)
=
sup
t∈[,]
fφ(t) –fα(t)+gφ(t) –gα(t)
=
sup
t∈[,]
fφ(t) –fα(t),gφ(t) –gα(t) E
= sup t∈[,]
fφ(t),gφ(t) –
fα(t),gα(t) E
= sup t∈[,]
φ(t) –α(t)E
=
φ–αE
≤ψ
max
φ–αE,φ(c) –TφE,α(c) –TαE,
φ(c) –TαE+α(c) –TφE
.
Suppose thatφ= . We see that all assumptions in Theorem . are now satisfied and
is the PPF dependent fixed point ofTinRc.
Corollary . Suppose that c∈I and T:E→E satisfies the following conditions: (i) Tis a nondecreasing mapping;
(ii) for allφ,α∈Ewithφ≤α,we have
Tφ–TαE≤k
max
φ–αE,φ(c) –TφE,α(c) –TαE,
φ(c) –TαE+α(c) –TφE
, wherek∈[, );
(iii) there exists a lower solutionφ∈Rcsuch thatφ(c)≤Tφ; (iv) Tis a continuous mapping.
Assume that Rc is algebraically closed with respect to the difference.Then T has a PPF dependent fixed point inRc.
Proof Define a functionψ: [, +∞)→[, +∞) byψ(t) =ktfor allt∈[, +∞). Therefore ψ is a nondecreasing mapping and
∞
n=
ψn(t) <∞ for allt∈(, +∞).
This implies that all assumptions in Theorem . are satisfied. Hence we obtain the desired
result.
Theorem . Suppose thatψ∈,c∈I and T:E→E satisfies the following conditions: (i) Tis a nondecreasing mapping;
(ii) for allφ,α∈Ewithφ≤α,Tφ–TαE≤ψ(φ–αE);
(iii) there exists a lower solutionφ∈Rcsuch thatφ(c)≤Tφ;
(iv) if{φn}is a nondecreasing sequence inEconverging toφ∈E,thenφn≤φfor all n∈N.
Assume that Rc is algebraically closed with respect to the difference.Then T has a PPF dependent fixed point inRc.
Proof By the analogous proof as in Theorem ., we can construct a nondecreasing se-quence{φn}inRcconverging toφ∈Rc. This implies thatφn≤φfor alln∈N. Therefore,
for eachn∈N, we have
Tφ–φ(c)E≤Tφ–φn+(c)E+φn+(c) –φ(c)E
≤ Tφ–TφnE+φn+–φE
≤ψφ–φnE +φn+–φE.
Sinceψ is continuous at , we getlimn→∞ψ(φ–φnE) =ψ() = . Taking the limit of
the above inequality, this yieldsTφ–φ(c)E= and soφis a PPF dependent fixed point
ofTinRc.
Theorem . Suppose thatψ∈,c∈I and T:E→E satisfies the following
condi-tions:
(i) Tis a nondecreasing mapping; (ii) for allφ,α∈Ewithφ≤α,we have
Tφ–TαE≤ψ
max
φ–αE,φ(c) –TφE,α(c) –TαE,
φ(c) –TαE+α(c) –TφE
;
(iii) there exists a lower solutionφ∈Rcsuch thatφ(c)≤Tφ; (iv) Tis a continuous mapping.
Assume that Rc is algebraically closed with respect to the difference.Then T has a PPF dependent fixed point inRc.
Proof Sinceφ∈RcandTφ∈E, there existsx∈Esuch thatTφ=x. As in the proof
of Theorem ., we can construct the sequence{φn}such that
Tφ=φ(c)≤Tφ=φ(c)≤ · · · ≤Tφn=φn+(c)≤Tφn+· · ·,
φn≤φn+ and φn–φn+E=φn(c) –φn+(c)E,
for alln∈N∪ {}. Assume thatφn–=φnfor somen∈N. It follows thatφn–(c) =φn(c) = Tφn–. ThereforeThas a fixed point inRc. Suppose thatφn–=φnfor alln∈N. For each n∈N, we obtain
φn–φn+E =φn(c) –φn+(c)E
=Tφn–Tφn–E
≤ψ
max
φn–φn–E,φn(c) –TφnE,φn–(c) –Tφn–E,
φn(c) –Tφn–E+φn–(c) –TφnE
=ψ
max
φn–φn–E,φn(c) –φn+(c)E,φn–(c) –φn(c)E,
φn(c) –φn(c)E+φn–(c) –φn+(c)E
≤ψ
max
φn–φn–E,φn–φn+E,
φn––φn+E
≤ψ
max
φn–φn–E,φn–φn+E,
φn––φnE+φn–φn+E
Ifmax{φn–φn–E,φn–φn+E}=φn–φn+E, then
φn–φn+E≤ψ
φn–φn+E <φn–φn+E.
This leads to a contradiction. Therefore
φn–φn+E≤ψ
φn–φn–E
<φn–φn–E.
It follows that φn–φn+E ≤ φn––φnE for alln∈N. Since the sequence{φn–
φn+E}is a nonincreasing sequence of nonnegative real numbers, we see that it is a
con-vergent sequence. Suppose that
lim
n→∞φn–φn+E=α,
for some nonnegative real numberα. We will prove thatα= . Suppose thatα> . Since
φn–φn+E≤ψ
φn–φn–E ,
for alln∈Nand the continuity ofψ, we haveα≤ψ(α) <αwhich leads to a contradic-tion. This implies thatα= . We next prove that the sequence{φn}is a Cauchy sequence
inE. Assume that{φn}is not a Cauchy sequence. It follows that there existε> and two
sequences of positive integers{mk}and{nk}satisfyingmk>nk>kfor eachk∈Nand
φmk–φnkE≥ε. (.)
Let{mk}be the sequence of the least positive integers exceeding{nk}which satisfies (.)
and
φmk––φnkE<ε. (.)
We will prove thatlimk→∞φmk –φnkE=ε. Sinceφmk –φnkE≥εfor allk∈N, we
have
lim
k→∞φmk–φnkE≥ε.
For eachk∈N, we obtain
φmk–φnkE≤ φmk–φmk–E+φmk––φnkE
≤ φmk–φmk–E+ε.
This implies thatlimk→∞φmk–φnkE≤ε. Therefore
lim
Similarly, we can prove that
lim
k→∞φmk+–φnkE=ε, k→∞limφmk–φnk–E=ε,
and
lim
k→∞φmk+–φnk–E=ε.
SinceRcis algebraically closed with respect to the difference, for eachk∈N, we obtain
φnk–φmk+E =φnk(c) –φmk+(c)E
=Tφmk–Tφnk–E
≤ψ
max
φmk–φnk–E,φmk(c) –TφmkE, φnk–(c) –Tφnk–E,
φmk(c) –Tφnk–E+φnk–(c) –TφnkE
=ψ
max
φmk–φnk–E,φmk(c) –φmk+(c)E, φnk–(c) –φnk(c)E,
φmk(c) –φnk(c)E+φnk–(c) –φnk+(c)E
≤ψ
max
φmk–φnk–E,φmk–φmk+E,φnk––φnkE,
φmk–φnkE+φnk––φnk+E
≤ψ
max
φmk–φnk–E,φmk–φmk+E,φnk––φnkE,
φmk–φnkE+φnk––φnkE+φnk–φnk+E
.
By taking the limit of both sides, we have
ε≤ψ(ε) <ε.
This leads to a contradiction. It follows that the sequence{φn}is a Cauchy sequence. By
the completeness ofE, we havelimn→∞φn=φfor someφ∈Eand
lim
n→∞Tφn=n→∞limφn+(c) =φ(c).
SinceRcis algebraically closed with respect to the norm topology, we haveφ∈Rc. We
Example . Assume thatE=RandE=C([, ],R). Define a mappingψ : [,∞)→
[,∞) by
ψ(t) = t
for allt∈[,∞).
We see thatψ is a continuous nondecreasing mapping withψ(t) <t. Define a mapping
T:E→Eby
Tφ= φ
for allφ∈E.
Suppose thatφ,α∈Ewithψ≤αandc∈[, ]. Therefore
Tφ–TαE =
φ
–α
≤
φ–αE ≤ψ
max
φ–αE,φ(c) –TφE,α(c) –TαE,
φ(c) –TαE+α(c) –TφE
.
Suppose thatφ= . We find that all assumptions in Theorem . are now satisfied and
is the PPF dependent fixed point ofTinRc.
For the next result, we drop the continuity ofT.
Theorem . Suppose thatψ ∈,c∈I,and that T :E→E satisfies the following
conditions:
(i) Tis a nondecreasing mapping; (ii) for allφ,α∈Ewithφ≤α,we have
Tφ–TαE≤ψ
max
φ–αE,φ(c) –TφE,α(c) –TαE,
φ(c) –TαE+α(c) –TφE
;
(iii) there exists a lower solutionφ∈Rcsuch thatφ(c)≤Tφ;
(iv) if{φn}is a nondecreasing sequence inEconverging toφ∈E,thenφn≤φfor all n∈N.
Assume that Rc is algebraically closed with respect to the difference.Then T has a PPF dependent fixed point inRc.
Proof As in the proof of Theorem ., we can construct a nondecreasing sequence{φn}
converging toφ∈Rcand this yields
lim
Using (iv), we haveφn≤φfor alln∈N. Therefore, for eachn∈N, we obtain
Tφ–φ(c)E≤Tφ–φn+(c)E+φn+(c) –φ(c)E
≤ Tφ–TφnE+φn+–φE
≤ψ
max
φ–φnE,φ(c) –TφE,φn(c) –TφnE,
φ(c) –TφnE+φn(c) –TφE
+φn+–φE
=ψ
max
φ–φnE,φ(c) –TφE,φn(c) –φn+(c)E,
φ(c) –φn+(c)E+φn(c) –TφE
+φn+–φE
≤ψ
max
φ–φnE,φ(c) –TφE,φn–φn+E
φ–φn+E+φn(c) –TφE
+φn+–φE.
Lettingn→ ∞, we obtainTφ–φ(c)E≤ψ(φ(c) –TφE). Ifφ(c)=Tφ, then
Tφ–φ(c)E≤ψφ(c) –TφE <φ(c) –TφE.
This leads to a contradiction. ThereforeTφ=φ(c). This implies thatφis a PPF dependent
fixed point ofT.
By applying Theorem . and Theorem ., we obtain the following corollary.
Corollary . Suppose that c∈I and that T:E→E satisfies the following conditions: (i) Tis a nondecreasing mapping;
(ii) for allφ,α∈Ewithφ≤α,we have
Tφ–TαE≤k
max
φ–αE,φ(c) –TφE,α(c) –TαE,
φ(c) –TαE+α(c) –TφE
, wherek∈[, );
(iii) there exists a lower solutionφ∈Rcsuch thatφ(c)≤Tφ;
(iv) Tis a continuous mapping or if{φn}is a nondecreasing sequence inEconverging to
φ∈E,thenφn≤φfor alln∈N.
Assume that Rc is algebraically closed with respect to the difference.Then T has a PPF dependent fixed point inRc.
Proof Define a functionψ: [, +∞)→[, +∞) byψ(t) =ktfor allt∈[, +∞). Therefore ψ is a continuous nondecreasing mapping and
This implies that all assumptions in Theorem . or Theorem . are satisfied. Hence the
proof is complete.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Faculty of Science, Razi University, Kermanshah, 67149, Iran.2Department of Mathematics,
Faculty of Science, Naresuan University, Phitsanulok, 65000, Thailand.3Centre of Excellence in Mathematics, CHE, Si Ayutthaya Rd., Bangkok, 10400, Thailand.
Acknowledgements
The second author would like to express her deep thanks to the Centre of Excellence in Mathematics, the Commission of Higher Education and Naresuan University, Thailand for the support.
Received: 10 October 2013 Accepted: 10 September 2014 Published:26 Sep 2014
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