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Research Article
Fixed Point Theorems for Contractions of Rational Type with
PPF Dependence in Banach Spaces
J. Rocha,1B. Rzepka,2and K. Sadarangani1
1Departamento de Matem´
aticas, Universidad de Las Palmas de Gran Canaria, Campus de Tara Baja,
35017 Las Palmas de Gran Canaria, Spain
2Department of Mathematics, Rzesz´
ow University of Technology, Aleja Powstanc´
ow Warszawy 8, 35-959 Rzesz´
ow, Poland
Correspondence should be addressed to B. Rzepka; brzepka@prz.edu.pl
Received 10 April 2014; Accepted 24 May 2014; Published 9 June 2014
Academic Editor: J´
ozef Bana´
s
Copyright © 2014 J. Rocha et al. is is an open access article distributed under the Creative Commons Attribution License, which
permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We prove the existence of the PPF dependent xed point in the Razumikhin class for contractions of rational type in Banach spaces,
by using a general class of pairs of functions. Our result has as particular cases a great number of interesting consequences which
extend and generalize some results appearing in the literature.
1. Introduction
Banach’s contraction principle is one of the pivotal results
of analysis. Its signicance lies in its vast applicability to a
greatnumberofbranchesofmathematicsandothersciences,
for example, theory of existence of solutions for nonlinear
dierential, integral, and functional equations, variational
inequalities, and optimization and approximation theory.
Generalizations of the contractive mapping theorem have
been a heavily investigated branch of research. In particular,
this principle was extended in [1], where the domain of the
nonlinear operator involved is C([,],)and the range is
,whereis a Banach space. is result is known as the
contraction theorem for operator with PPF (past, present,
and future) dependence. e PPF xed point theorems are
useful for proving the existence of solutions for nonlinear
functional-dierential and integral equations which may
depend upon the past history, present data, and future
considerations. Some papers about xed point theorems with
PPF dependence have appeared in the literature (see, e.g., [1–
5]).
On the other hand, Dass and Gupta in [6]andJaggiin
[7] were the pioneers in proving xed point theorems using
contractive conditions involving rational expressions. In [4],
the authors present a xed point theorem for contractions of
rational type with PPF dependence.
e purpose of this paper is to present a xed point
theorem for generalized contractions of rational type with
PPF dependence which has, as particular cases, interesting
consequences. Particularly, our result extends the one appear-
ing in [4].
2. Preliminaries
roughout this paper, will denote a Banach space with
norm ⋅𝐸and 0=C([,],) will denote the space
of the continuous -valued functions dened on [,] and
equipped with the norm ⋅𝐸0given by
𝐸0=sup
𝑡∈[𝑎,𝑏]
()
𝐸for ∈0.(1)
Let :0→be a mapping. A point ∈0is said to
be a PPF dependence xed point of or a xed point with
PPF dependence of if =(),forsome∈[,].
For a xed ∈[,],theRazumikhinclass𝑐is dened
by 𝑐=∈0:
𝐸0=
()
𝐸. (2)
Remark 1. Notice that, for ∈xed, the function 𝑥,
dened by 𝑥()= for ∈[,],(3)
Hindawi Publishing Corporation
Journal of Function Spaces
Volume 2014, Article ID 416187, 8 pages
http://dx.doi.org/10.1155/2014/416187
2Journal of Function Spaces
satises 𝑥∈
0,𝑥𝐸0=
𝑥()𝐸=for any ∈[,]
and, therefore, 𝑥∈𝑐for any ∈[,].Consequently,𝑐=
for any ∈[,].
We say that the cl a s s 𝑐isalgebraicallyclosedwithrespect
to dierence if for any ,∈𝑐we have −∈𝑐.Similarly,
we say that the class 𝑐is topologically closed if it is closed
withrespecttothetopologyon0induced by the norm
⋅𝐸0.
e Razumikhin class plays an important role in the
existence of PPF xed point.
e rst result about the existence of PPF xed point
appears in [1] and it is presented in the following theorem.
eorem 2 (see [1]). Suppose that :0→is a mapping
such that there exists ∈[0,1)satisfying
−
𝐸≤
−
𝐸0(4)
for any ,∈0.If 𝑐is topologically closed and algebraically
closed with respect to dierence for some ∈[,],thenhas
a unique PPF dependent xed point in 𝑐.
Recently, in [4] the authors proved the following PPF
dependent xed point theorem for rational type contraction
mappings.
eorem 3 (see [4]). Let :0→be a mapping satisfying
−
𝐸≤
−
𝐸0+
()−
𝐸
()−
𝐸
1+
−
𝐸(5)
for any , ∈ 0and where , ∈ [0,1)with +<1and
∈[,].If𝑐is topologically closed and algebraically closed
with respect to dierence, then has a unique PPF dependent
xed point in 𝑐.
e main purpose of this paper is, by using a class of
pairs of functions satisfying certain assumptions, to present
new PPF dependent xed point theorems for contractions
of rational type. Particularly, our result generalizes the main
result of [4](eorem 3).
3. Main Results
We start this section presenting the following class of pairs of
functions F. A pair of functions (,)is said to belong to the
class Fif they satisfy the following conditions:
(i) ,:[0,∞) → [0,∞);
(ii) for ,∈[0,∞),if()≤(),then≤;
(iii) for (𝑛)and (𝑛)sequences in [0,∞)such that
lim
𝑛→∞𝑛=lim
𝑛→∞𝑛=, (6)
if (𝑛)≤(𝑛)for any ∈N,then=0.
Remark 4. Notice that if (,) ∈ Fand () ≤ (),then
=0,sincewecantake𝑛=𝑛=for any ∈Nand by (iii)
we deduce =0.
In the sequel, we present some interesting examples of
pairs of functions belonging to the class Fwhich will be very
important in our study.
Example 5. Let : [0,∞) → [0,∞) be a continuous
and increasing function such that () = 0 if and only if
=0(these functions are known in the literature as altering
distance functions).
Let :[0,∞)→[0,∞)be a nondecreasing function
such that ()=0if and only if =0and suppose that ≤.
en the pair (,−)∈F.
In fact, it is clear that (,−)satisfy (i).
To prove (ii), suppose that , ∈ [0,∞) and () ≤ (−
)().en,from
()≤()−()≤()(7)
and taking into account the increasing character of ,wecan
deduce that ≤.
In order to prove (iii), we suppose that
𝑛≤𝑛−𝑛≤𝑛for any ∈N,(8)
where 𝑛,𝑛∈[0,∞)and
lim
𝑛→∞𝑛=lim
𝑛→∞𝑛=. (9)
Taking →∞in (8), we infer that lim𝑛→∞(𝑛)=0.
Let us suppose that >0. Since lim𝑛→∞𝑛=>0,we
can nd >0and a subsequence (𝑛𝑘)of (𝑛)such that 𝑛𝑘>
for any ∈N.Asis nondecreasing, we have (𝑛𝑘)>()
for any ∈Nand, consequently, lim𝑘→∞(𝑛𝑘)≥().is
contradicts the fact that lim𝑘→∞(𝑛𝑘)=0. erefore, =0.
is proves that (,−)∈F.
An interesting particular case is when is the identity
mapping, =1
[0,∞),and : [0,∞) → [0,∞) is a
nondecreasing function such that ()=0if and only if =0
and ()≤for any ∈[0,∞).
Example 6. Let be the class of functions dened by
=:[0,∞)→ [0,1):𝑛→1⇒𝑛→ 0 .
(10)
Let us consider the pairs of functions (1[0,∞),1[0,∞)),where
∈and 1[0,∞) is dened by (1[0,∞))() = (),for∈
[0,∞).
en (1[0,∞),1[0,∞))∈F.
It is clear that the pairs (1[0,∞),1[0,∞))with ∈satisfy
(i).
To prove (ii), from
1[0,∞) ()≤1[0,∞)()for ,∈[0,∞)(11)
we infer, since :[0,∞) → [0,1),that
≤()< (12)
and, consequently, (1[0,∞),1[0,∞))satises (ii).
Journal of Function Spaces 3
In order to prove (iii), we suppose that
1[0,∞) 𝑛=𝑛≤1[0,∞)𝑛=𝑛𝑛for any ∈N,
(13)
where 𝑛,𝑛∈[0,∞)and lim𝑛→∞𝑛=lim𝑛→∞𝑛=.
Let us suppose that >0.
Since lim𝑛→∞𝑛=>0,wecanndasubsequence(𝑛𝑘)
such that 𝑛𝑘>0for any ∈N.Now,as
𝑛≤𝑛𝑛≤𝑛for any ∈N,(14)
in particular, we have
𝑛𝑘≤𝑛𝑘𝑛𝑘≤𝑛𝑘for any ∈N,(15)
and, since 𝑛𝑘>0for any ∈N,
𝑛𝑘
𝑛𝑘≤𝑛𝑘≤1. (16)
Taking →∞in the last inequality, we obtain
lim
𝑘→∞𝑛𝑘=1. (17)
Finally, since ∈,weinferthatlim
𝑘→∞𝑛𝑘=0and this
contradicts the fact that lim𝑛→∞𝑛=>0.
erefore, =0.
is proves that (1[0,∞),1[0,∞))∈Ffor ∈.
Remark 7. Suppose that :[0,∞) → [0,∞)is an increasing
function and (,) ∈ F. en it is easily seen that the pair
(,)∈F.
Now, we are ready to present our main result.
eorem 8. Let :
0→be a mapping such that there
exists a pair of functions (,)∈Fand ∈[,]such that
−
𝐸
≤max
−
𝐸0,
()−
𝐸⋅
()−
𝐸
1+
−
𝐸
(18)
for any , ∈0.If𝑐is topologically closed and algebraically
closed with respect to dierence, then has a unique PPF
dependent xed point in 𝑐.
Proof. Let 0be an arbitrary function in 𝑐(whose existence
is guaranteed by Remark 1). Since 0∈
𝑐⊂
0,put1=
0∈.
Again, by Remark 1,wecannd1∈𝑐such that
0=1=1().(19)
Since 1∈𝑐⊂0,put2=
1∈.Usingthesameargu-
ment, we can nd 2∈𝑐such that
1=2=2().(20)
Repeating this process, we can obtain a sequence (𝑛)in 𝑐
such that
𝑛−1 =𝑛()for any ∈N.(21)
Since 𝑐isalgebraicallyclosedwithrespecttodierence,we
have
𝑝−𝑞
𝐸0=
𝑝()−𝑞()
𝐸(22)
for any ,∈N.
First, we will prove that lim𝑛→∞𝑛+1 −𝑛𝐸0=0.Infact,
taking into account (21)and(22), we get
𝑛+1 −𝑛
𝐸0=
𝑛+1()−𝑛()
𝐸
=
𝑛−𝑛−1
𝐸for any ∈N
(23)
and, therefore, applying the contractive condition, we have
𝑛+1 −𝑛
𝐸0
=
𝑛−𝑛−1
𝐸
≤max
𝑛−𝑛−1
𝐸0,
𝑛()−𝑛
𝐸⋅
𝑛−1 ()−𝑛−1
𝐸
1+
𝑛−𝑛−1
𝐸.
(24)
Let us suppose that there exists 0∈Nsuch that 𝑛0+1−
𝑛0𝐸0=0.
In this case, 𝑛0+1 =
𝑛0and, consequently, 𝑛0+1() =
𝑛0().
By (21), we have
𝑛0=𝑛0+1 ()=𝑛0()(25)
and 𝑛0would be the PPF dependent xed point.
In the sequel we suppose that 𝑛+1 −𝑛𝐸0=0for any ∈
N.
Now, we can distinguish two cases.
Case 1. Consider
max
𝑛−𝑛−1
𝐸0,
𝑛()−𝑛
𝐸⋅
𝑛−1()−𝑛−1
𝐸
1+
𝑛−𝑛−1
𝐸
=
𝑛−𝑛−1
𝐸0.
(26)
In this case, from (24), we infer
𝑛+1 −𝑛
𝐸0≤
𝑛−𝑛−1
𝐸0,(27)
and, since (,)∈F,wededuce
𝑛+1 −𝑛
𝐸0≤
𝑛−𝑛−1
𝐸0.(28)
4Journal of Function Spaces
Case 2. Consider
max
𝑛−𝑛−1
𝐸0,
𝑛()−𝑛
𝐸⋅
𝑛−1()−𝑛−1
𝐸
1+
𝑛−𝑛−1
𝐸
=
𝑛()−𝑛
𝐸⋅
𝑛−1 ()−𝑛−1
𝐸
1+
𝑛−𝑛−1
𝐸.
(29)
In this case, from (24) and since (,)∈F,weinfer
𝑛+1 −𝑛
𝐸0≤
𝑛()−𝑛
𝐸⋅
𝑛−1()−𝑛−1
𝐸
1+
𝑛−𝑛−1
𝐸.(30)
By (21)and(22), we have
𝑛+1 −𝑛
𝐸0≤
𝑛()−𝑛+1()
𝐸⋅
𝑛−1()−𝑛()
𝐸
1+
𝑛+1()−𝑛()
𝐸
=
𝑛+1 −𝑛
𝐸0⋅
𝑛−𝑛−1
𝐸0
1+
𝑛+1 −𝑛
𝐸0.(31)
Since 𝑛+1 −𝑛𝐸0=0, from the last inequality, it follows that
1≤
𝑛−𝑛−1
𝐸0
1+
𝑛+1 −𝑛
𝐸0
(32)
and, therefore,
𝑛+1 −𝑛
𝐸0<1+
𝑛+1 −𝑛
𝐸0≤
𝑛−𝑛−1
𝐸0.(33)
In both cases, we obtain that inequality (28) is satised and,
consequently, the sequence (𝑛+1 −𝑛𝐸0)is a decreasing
sequence of nonnegative real numbers.
Put =lim𝑛→∞𝑛+1 −𝑛𝐸0,where≥0, and denote
={∈N:satises (26)},
={∈N:satises (29)}.(34)
We remark the foll ow i n g .
(1) If Card =∞,thenfrom(24)wecannd
innitelymanynaturalnumberssatisfying
inequality (27)and,sincelim
𝑛→∞𝑛+1 −𝑛𝐸0=
lim𝑛→∞𝑛−𝑛−1𝐸0=and (,) ∈ F,wededuce
that =0.
(2) If Card =∞, then from (24) we can nd innitely
many ∈Nsuch that
𝑛+1 −𝑛
𝐸0
≤
𝑛()−𝑛
𝐸⋅
𝑛−1 ()−𝑛−1
𝐸
1+
𝑛−𝑛−1
𝐸. (35)
Since (,)∈ Fand using a similar argument to the
oneusedinCase2,weobtain
𝑛+1 −𝑛
𝐸0≤
𝑛+1 −𝑛
𝐸0⋅
𝑛−𝑛−1
𝐸0
1+
𝑛+1 −𝑛
𝐸0
(36)
for innitely many ∈N.Taking→∞
in the last inequality and taking into account that
lim𝑛→∞𝑛+1 −𝑛𝐸0=we deduce
≤ 2
1+,(37)
and, consequently, ≤0.Since≥0,weobtain=0.
erefore,
lim
𝑛→∞
𝑛+1 −𝑛
𝐸0=0. (38)
Next,wewillprovethat(𝑛)is a Cauchy sequence in 0.In
contrary case, since lim𝑛→∞𝑛+1 −𝑛𝐸0=0,byLemma2.1
of [8], we can nd >0and subsequences (𝑛(𝑘))and (𝑚(𝑘))
of (𝑛)satisfying
(i)()>()≥for >0,
(ii)≤
𝑛(𝑘) −𝑚(𝑘)
𝐸0,
𝑛(𝑘)−1 −𝑚(𝑘)
𝐸0<
for >0,
(iii)lim
𝑘→∞
𝑛(𝑘) −𝑚(𝑘)
𝐸0=lim
𝑘→∞
𝑛(𝑘) −𝑚(𝑘)+1
𝐸0
=lim
𝑘→∞
𝑛(𝑘)+1 −𝑚(𝑘)
𝐸0
=lim
𝑘→∞
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0=.
(39)
Since 𝑛(𝑘)+1,𝑚(𝑘)+1 ∈
𝑐for any ∈N,from(21)and(22),
we have
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0=
𝑛(𝑘)+1()−𝑚(𝑘)+1()
𝐸
=
𝑛(𝑘) −𝑚(𝑘)
𝐸
(40)
for any ∈N.
Using the contractive condition and (21)and(22), we
obtain
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0
=
𝑛(𝑘) −𝑚(𝑘)
𝐸
≤max
𝑛(𝑘) −𝑚(𝑘)
𝐸0,
𝑛(𝑘)()−𝑛(𝑘)
𝐸⋅
𝑚(𝑘)()−𝑚(𝑘)
𝐸
1+
𝑛(𝑘) −𝑚(𝑘)
𝐸
Journal of Function Spaces 5
=max
𝑛(𝑘) −𝑚(𝑘)
𝐸0,
𝑛(𝑘)()−𝑛(𝑘)+1()
𝐸
⋅
𝑚(𝑘)()−𝑚(𝑘)+1()
𝐸
×1+
𝑛(𝑘)+1()−𝑚(𝑘)+1()
𝐸−1
=max
𝑛(𝑘) −𝑚(𝑘)
𝐸0,
𝑛(𝑘)+1 −𝑛(𝑘)
𝐸0⋅
𝑚(𝑘)+1 −𝑚(𝑘)
𝐸0
×1+
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0−1
(41)
for any ∈N.
Let us put
=∈N:
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0≤
𝑛(𝑘) −𝑚(𝑘)
𝐸0,
=∈N:
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0
≤
𝑛(𝑘)+1 −𝑛(𝑘)
𝐸0⋅
𝑚(𝑘)+1 −𝑚(𝑘)
𝐸0
1+
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0.
(42)
By (41)wehave Card=∞or Card =∞.
Let us suppose that Card =∞. en there exist in-
nitely many ∈Nsuch that
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0≤
𝑛(𝑘) −𝑚(𝑘)
𝐸0(43)
and since (,)∈Fand
lim
𝑘→∞
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0=lim
𝑘→∞
𝑛(𝑘) −𝑚(𝑘)
𝐸0= (44)
we infer from (39)that=0. is is a contradiction.
Let us suppose that Card =∞.Inthiscase,wecannd
innitely many ∈Nsuch that
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0
≤
𝑛(𝑘)+1 −𝑛(𝑘)
𝐸0⋅
𝑚(𝑘)+1 −𝑚(𝑘)
𝐸0
1+
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0(45)
and since (,)∈F,weinfer
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0
≤
𝑛(𝑘)+1 −𝑛(𝑘)
𝐸0⋅
𝑚(𝑘)+1 −𝑚(𝑘)
𝐸0
1+
𝑛(𝑘)+1 −𝑚(𝑘)+1
𝐸0.(46)
Taking →∞and in view of (38)and(39), it follows that
≤0and this is a contradiction.
erefore, since in both possibilities Card =∞and
Card =∞we obtain a contradiction, we deduce that (𝑛)
is a Cauchy sequence in 0.
Since 0is a Banach space, we can nd ∗∈0such that
lim𝑛→∞𝑛=∗.As𝑛∈𝑐and 𝑐is topologically closed, we
have ∗∈𝑐.
Next, we will prove that ∗isa PPF dependent xed point
of . In fact, by the contractive condition, we obtain
∗−𝑛
𝐸
≤max
∗−𝑛
𝐸0,
∗()−∗
𝐸⋅
𝑛()−𝑛
𝐸
1+
∗−𝑛
𝐸
(47)
for any ∈N.
Wecandistinguishtwocasesagain.
(1) ere exist innitely many ∈Nsuch that
∗−𝑛
𝐸≤
∗−𝑛
𝐸0. (48)
In this case, since (,)∈F,weobtain
∗−𝑛
𝐸≤
∗−𝑛
𝐸0(49)
for innitely many ∈N.Sincelim
𝑛→∞𝑛=
∗,
taking →∞in the last inequality, we obtain
lim
𝑛→∞𝑛=∗,(50)
where, to simplify our considerations, we will denote
thesubsequencebythesamesymbol(𝑛).By(21),
∗=lim
𝑛→∞𝑛=lim
𝑛→∞𝑛+1 ().(51)
𝑛→
∗in 0;thismeansthat
sup
𝑡∈[𝑎,𝑏]
𝑛()−∗()
𝐸→ 0 (52)
and, consequently, lim𝑛→∞𝑛+1()=∗().Fromthis
last result and from (51)wededucethat
∗=∗()(53)
and, therefore, ∗is a PPF dependent xed point of
in 𝑐.
(2) ere exist innitely many ∈Nsuch that
∗−𝑛
𝐸≤
∗()−∗
𝐸⋅
𝑛()−𝑛
𝐸
1+
∗−𝑛
𝐸.
(54)
6Journal of Function Spaces
To simplify our considerations, we will denote the
subsequence by the same symbol (𝑛).Since(,)∈
F,weinfer
∗−𝑛
𝐸≤
∗()−∗
𝐸⋅
𝑛()−𝑛
𝐸
1+
∗−𝑛
𝐸
(55)
for any ∈N.Using(21), we have that 𝑛=𝑛+1()
and, therefore,
∗−𝑛
𝐸≤
∗()−∗
𝐸⋅
𝑛()−𝑛+1()
𝐸
1+
∗−𝑛
𝐸
(56)
for any ∈N.Taking→∞and by (38)since
lim𝑛→∞𝑛()−𝑛+1()𝐸=0,weinfer(50). From
theabovecase,wededucethat∗is a PPF dependent
xed point of in 𝑐.
erefore, we have proved that in both cases ∗is a PPF
dependent xed point of in 𝑐.
Finally, we will prove the uniqueness of PPF dependent
xed point of in 𝑐.
Suppose that ∗is another PPF dependent xed point of
in 𝑐.en,since∗,∗∈𝑐and 𝑐is algebraically closed,
we obtain
∗−∗
𝐸0=
∗()−∗()
𝐸.(57)
As ∗()= ∗and ∗() =∗because ∗and ∗are PPF
dependent xed points of ,weinfer
∗−∗
𝐸0=
∗()−∗()
𝐸=
∗−∗
𝐸.(58)
Consequently, using the contractive condition, we get
∗−∗
𝐸0
=
∗−∗
𝐸
≤max
∗−∗
𝐸0,
∗()−∗
𝐸⋅
∗()−∗
𝐸
1+
∗−∗
𝐸
=max
∗−∗
𝐸0,(0). (59)
We can dist i n g u i s h t w o c ases.
(i) Consider max{(∗−∗𝐸0),(0)} = (∗−
∗𝐸0).Inthiscase,from(59)wehave
∗−∗
𝐸0≤
∗−∗
𝐸0. (60)
Now, since (,) ∈ Fand using Remark 4,weget
∗−∗𝐸0=0and, therefore, ∗=∗.
(ii) Consider max{(∗−∗𝐸0),(0)} = (0).From
(59)weobtain
∗−∗
𝐸0≤(0),(61)
and, since (,) ∈ F,weinferthat∗−∗𝐸0≤0.
erefore, ∗−∗𝐸0=0or, equivalently, ∗=∗.
is result nishes the proof.
By eorem 8,weobtainthefollowingcorollaries.
Corollary 9. Let :
0→be a mapping such that there
exists a pair of functions (,)∈Fand ∈[,]satisfying
−
𝐸≤
−
𝐸0(62)
for any , ∈0.If𝑐is topologically closed and algebraically
closed with respect to dierence, then has a unique PPF
dependent xed point in 𝑐.
Corollary 10. Let :0→be a mapping such that there
exists a pair of functions (,)∈Fand ∈[,]satisfying
−
𝐸≤
()−
𝐸
()−
𝐸
1+
−
𝐸(63)
for any , ∈0.If𝑐is topologically closed and algebraically
closed with respect to dierence, then has a unique PPF
dependent xed point in 𝑐.
e main result of [4]iseorem 3.Noticethatthe
contractive condition appearing in this theorem
−
𝐸≤
−
𝐸0+
()−
𝐸
()−
𝐸
1+
−
𝐸(64)
for any , ∈ 0,where, ∈ [0,1)with +<1and
∈[,],impliesthat
−
𝐸
≤+max
−
𝐸0,
()−
𝐸
()−
𝐸
1+
−
𝐸
≤max +
−
𝐸0,
+
()−
𝐸
()−
𝐸
1+
−
𝐸
(65)
for any , ∈ 0. is condition is a particular case of the
contractive condition appearing in eorem 8 with the pair of
functions (,)∈Fgiven by =1[0,∞) and =(+)1[0,∞).
erefore, eorem 3 is a particular case of the following
corollary.
Journal of Function Spaces 7
Corollary 11. Let :0→be a mapping such that there
exist real numbers , ∈ [0,1)with +<1and ∈[,]
satisfying
−
𝐸≤max +
−
𝐸0,
+
()−
𝐸
()−
𝐸
1+
−
𝐸
(66)
for any , ∈0.If𝑐is topologically closed and algebraically
closed with respect to dierence, then has a unique PPF
dependent xed point in 𝑐.
Taking into account Example 5,wehavethefollowing
corollary.
Corollary 12. Let :0→be a mapping such that there
exist two functions ,:[0,∞) → [0,∞)and ∈[,]such
that
−
𝐸≤max
−
𝐸0−
−
𝐸0,
()−
𝐸
()−
𝐸
1+
−
𝐸
−
()−
𝐸
()−
𝐸
1+
−
𝐸
(67)
for any , ∈ 0,whereis a continuous and increasing
function satisfying () = 0if and only if =0,andis a
nondecreasing function such that ()=0if and only if =0,
and ≤.
If 𝑐is topologically closed and algebraically closed with
respect to dierence, then has a unique PPF dependent xed
point in 𝑐.
Corollary 12 has the following consequences.
Corollary 13. Let :0→be a mapping such that there
exist two functions ,:[0,∞) → [0,∞)and ∈[,]such
that
−
𝐸≤
−
𝐸0−
−
𝐸0(68)
for any , ∈ 0,whereis an increasing function and is
a nondecreasing function and they satisfy () = () = 0
if and only if =0,andis continuous with ≤.If𝑐
is topologically closed and algebraically closed with respect to
dierence, then has a unique PPF dependent xed point in
𝑐.
Corollary 13 can be considered as the version, in the
context of PPF dependent xed point theorems, of the
following result about xed point theorems which appears in
[9].
eorem 14 (see [9]). Let (,)be a complete metric space
and :→a mapping satisfying
,≤,−, (69)
for , ∈ ,whereand satisfy the same conditions as in
Corollary 13.enhas a unique xed point.
Corollary 15. Let :0→be a mapping such that there
exist two functions ,:[0,∞) → [0,∞)satisfying the same
conditions as in Corollary 13 and ∈[,]such that
−
𝐸≤
()−
𝐸
()−
𝐸
1+
−
𝐸
−
()−
𝐸
()−
𝐸
1+
−
𝐸(70)
for any , ∈0.If𝑐is topologically closed and algebraically
closed with respect to dierence, then has a unique PPF
dependent xed point in 𝑐.
Taking into account Example 6,wehavethefollowing
corollary.
Corollary 16. Let :0→be a mapping such that there
exist ∈(see Example 6)and∈[,]satisfying
−
𝐸≤max
−
𝐸0
−
𝐸0,
()−
𝐸
()−
𝐸
1+
−
𝐸
×
()−
𝐸
()−
𝐸
1+
−
𝐸
(71)
for any , ∈0.If𝑐is topologically closed and algebraically
closed with respect to dierence, then has a unique PPF
dependent xed point in 𝑐.
A consequence of Corollary 16 is the following result.
Corollary 17. Let :0→be a mapping such that there
exists ∈satisfying
−
𝐸≤
−
𝐸
−
𝐸0(72)
for any ,∈0.
If ∈[,]such that 𝑐is topologically closed and
algebraically closed with respect to dierence, then has a
unique PPF dependent xed point in 𝑐.
Corollary 17 is the version, in the context of PPF depen-
dent xed point theorems, of the following result about xed
point theorems appearing in [10].
eorem 18. Let (,)be a complete metric space and :
→a mapping satisfying
,≤,, (73)
for any ,∈,where∈.enhas a unique xed point.
8Journal of Function Spaces
Conflict of Interests
e authors declare that there is no conict of interests in the
submitted paper.
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[8] M.Jleli,V.C.Raji
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