R E S E A R C H
Open Access
Fixed point theorems for almost
generalized
C
-contractive mappings in
ordered complete metric spaces
Azizollah Azizi
1*, Mohammad Moosaei
2and Gita Zarei
1*Correspondence: [email protected] 1Department of Mathematics, Faculty of Science, Imam Khomeini International University, Qazvin, 34149-16818, Iran
Full list of author information is available at the end of the article
Abstract
The purpose of this paper is to present some fixed point and common fixed point theorems for almost generalizedC-contractive mappings in an ordered complete metric space. Finally, two examples are given to support our results.
MSC: 54H25; 47H10; 54E50
Keywords: altering distance function; common fixed point;C-contractive mapping; fixed point; ordered complete metric spaces; weakly increasing functions
1 Introduction
The Banach contraction principle (see []) is a very popular tool for solving problems in many branches of mathematical analysis. The generalizations of this principle have been established in various settings (see, for example [–]). The concept of the altering
dis-tance function has been introduced by Khanet al.[]. They also presented some fixed
point theorems in a metric space by altering distance functions. The concept of weak contraction presented by Berinde [], but in [], the author renames it as an ‘almost contraction’ which is apposite. Berinde [] gave some fixed point theorems for almost contractions in complete spaces. Shatanawi [] presented some fixed point theorems for a nonlinear weaklyC-contraction type mapping in metric spaces. Ćirićet al.[] intro-duced the concept of almost generalized contractive condition on mappings and proved some existential theorems on fixed points of such mappings in an ordered complete metric space. Shatanawi and Al-Rawashdeh [] introduced the notion of an almost generalized
(ψ,φ)-contractive mapping in ordered metric spaces and established some fixed point
and common fixed point results for such a mapping, whereψandφare altering distance
functions.
The purpose of this paper is to introduce the almost generalizedC-contractive mappings in an ordered metric space via the altering distance functions and the functions having property (P), and to prove some fixed point and common fixed point theorems for such mappings in an ordered complete metric space. Specially, under suitable conditions, we show that if the fixed point set of such mappings is totally ordered, then it is singleton. In the end, an example is given to support the usability of our results.
We first review the needed definitions. Throughout this paper, we denote byR,R+, and
Nthe set of all real numbers, the set of all nonnegative real numbers and the set of all positive integers, respectively. LetXbe a nonempty set andf,gbe two self-mappings ofX. We denote byF(f) the fixed point set off,i.e.,F(f) ={x∈X:fx=x}. Also, we denote by
F(f,g) the common fixed point set off,g,i.e.,F(f,g) =F(f)∩F(g).
Definition .(see []) A functionψ:R+→R+is called an altering distance function if it satisfies the following properties:
() ψis continuous and non-decreasing, and () ψ(x) = if and only ifx= .
We denote bythe class of all altering distance functions.
Definition . (see []) LetX be a nonempty set. Then (X,,d) is called an ordered metric space if and only if:
(i) (X,d)is a metric space, and (ii) (X,)is a partially ordered set.
(X,,d) is called an ordered complete metric space if (X,,d) is an ordered metric space, and (X,d) is a complete metric space.
Definition .(see [, ]) Let (X,) be a partially ordered set. Two mappingsf,g:X→ Xare said to be weakly increasing iffxgfxandgxfgxfor allx∈X.
Definition . Letφ:R+×R+→R+be a function. We say that the functionφhas
prop-erty (P) if the following is satisfied:
() φis lower semicontinuous and non-decreasing with respect to both of its components, and
() φ(s,t) = if and only ifs=t= .
We denote bythe class of all functions satisfying property (P).
2 Main results
In this section, we introduce the concept of the almost generalizedC-contraction for map-pings in an ordered metric space. Then we present some fixed point and common fixed point theorems for such mappings in an ordered complete metric space.
Definition . Let (X,,d) be an ordered metric space. We say that a mappingf:X→X
is an almost generalizedC-contractive mapping if there existξ ≥ and (ψ,φ)∈×
such that
ψd(fx,fy)≤ψM(x,y)–φM(x,y),M(x,y)+ξ ψN(x,y) () for allx,y∈Xwithxy, where
M(x,y) =max
d(x,y),d(x,fx),d(y,fy),d(x,fy) +d(y,fx)
,
M(x,y) =maxd(x,y),d(x,fx),d(x,fy),
M(x,y) =maxd(x,y),d(y,fy),d(y,fx), and
Definition . Let (X,,d) be an ordered metric space, and letf,gbe two self-mappings ofX. The mappingf is said to be almost generalizedC-contractive with respect tog if there existξ≥ and (ψ,φ)∈×such that
ψd(fx,gy)≤ψM(x,y)–φM(x,y),M(x,y)+ξ ψN(x,y) () for allx,y∈Xwithxy, where
M(x,y) =max
d(x,y),d(x,fx),d(y,gy),d(x,gy) +d(y,fx)
,
M(x,y) =maxd(x,y),d(x,fx),d(x,gy),
M(x,y) =maxd(x,y),d(y,gy),d(fx,y), and
N(x,y) =mind(x,fx),d(y,fx),d(x,gy).
The following lemmas play a basic role to prove our main results.
Lemma . Let(X,,d)be an ordered metric space.Assume that f:X→X is an almost generalizedC-contractive mapping.Fix x∈X and define a sequence{xn}by xn+=fxnfor
all n∈N.If the sequence{xn}is non-decreasing andlimn→∞d(xn,xn+) = ,then{xn}is a
Cauchy sequence.
Proof Since the mappingf is almost generalizedC-contractive, Definition . implies that there exists (ξ,ψ,φ)∈[,∞)××such that
ψd(fx,fy)≤ψM(x,y)–φM(x,y),M(x,y)+ξ ψN(x,y) () for allx,y∈Xwithxy, where
M(x,y) =max
d(x,y),d(x,fx),d(y,fy),d(x,fy) +d(y,fx)
,
M(x,y) =maxd(x,y),d(x,fx),d(x,fy),
M(x,y) =maxd(x,y),d(y,fy),d(y,fx), and
N(x,y) =mind(x,fx),d(y,fx).
We next show that the sequence{xn}is Cauchy. Suppose, for contradiction, that is,{xn} is not Cauchy. Then there existε> and two subsequences{xnk}and{xmk}of the sequence
{xn}such that
nk>mk>k, d(xmk,xnk–) <ε, and d(xmk,xnk)≥ε. () These imply that
ε≤d(xmk,xnk)
≤d(xmk,xmk–) +d(xmk–,xnk–) +d(xnk–,xnk)
≤d(xmk,xmk–) +d(xmk,xnk–) +d(xnk–,xnk)
< d(xmk,xmk–) +ε+d(xnk–,xnk).
In view oflimn→∞d(xn,xn+) = and lettingk→ ∞in the above inequalities, we obtain
lim
k→∞d(xmk,xnk) =klim→∞d(xmk–,xnk)
= lim
k→∞d(xmk–,xnk–)
= lim
k→∞d(xmk,xnk–)
=ε. ()
Sincexmkxnk–for anyk∈N, () implies
ψd(xmk,xnk)
=ψd(fxmk–,fxnk–)
≤ψM(xmk–,xnk–)
–φM(xmk–,xnk–),M
(x
mk–,xnk–)
+ξ ψN(xmk–,xnk–)
, ()
where
M(xmk–,xnk–) =max
d(xmk–,xnk–),d(xmk–,fxmk–),d(xnk–,fxnk–),
d(xmk–,fxnk–) +d(xnk–,fxmk–)
=max
d(xmk–,xnk–),d(xmk–,xmk),d(xnk–,xnk),
d(xmk–,xnk) +d(xnk–,xmk)
,
M(xmk–,xnk–) =max
d(xmk–,xnk–),d(xmk–,fxmk–),d(xmk–,fxnk–)
=maxd(xmk–,xnk–),d(xmk–,xmk),d(xmk–,xnk)
,
M(xmk–,xnk–) =max
d(xmk–,xnk–),d(xnk–,fxnk–),d(xnk–,fxmk–)
=maxd(xmk–,xnk–),d(xnk–,xnk),d(xnk–,xmk)
, and
N(xmk–,xnk–) =min
d(xmk–,fxmk–),d(xnk–,fxmk–)
=mind(xmk–,xmk),d(xnk–,xmk)
.
Lettingk→ ∞in the above equalities and applying (), we obtain
lim
k→∞M(xmk–,xnk–) =ε, ()
lim
k→∞M
(x
lim
k→∞M
(x
mk–,xnk–) =ε, ()
and
lim
k→∞N(xmk–,xnk–) = . ()
Taking the limit ask→ ∞in inequality () and using ()-(), the continuity ofψ, and the lower semicontinuity ofφ, we conclude that
ψ(ε)≤ψ(ε) –φ(ε,ε),
which yieldsφ(ε,ε) = . Henceε= , which contradicts the positivity ofε. Therefore, we
get the desired result.
Lemma . Let(X,,d)be an ordered metric space,and let f,g be two self-mappings of X which f is an almost generalizedC-contractive mapping with respect to g.Fix x∈X and define a sequence{xn}by xn=fxn–and xn+=gxnfor all n∈N.If limn→∞d(xn,xn+) =
,and the sequence{xn}is non-decreasing,then{xn}is a Cauchy sequence.
Proof Sincefis almost generalizedC-contractive with respect tog, by Definition ., there exists (ξ,ψ,φ)∈[,∞)××such that
ψd(fx,gy)≤ψM(x,y)–φM(x,y),M(x,y)+ξ ψN(x,y) ()
for allx,y∈Xwithxy, where
M(x,y) =max
d(x,y),d(x,fx),d(y,gy),d(x,gy) +d(y,fx)
,
M(x,y) =maxd(x,y),d(x,fx),d(x,gy),
M(x,y) =maxd(x,y),d(y,gy),d(fx,y), and
N(x,y) =mind(x,fx),d(y,fx),d(x,gy).
We now show that the sequence{xn}is Cauchy. Suppose, for contradiction, that is,{xn}is not Cauchy. Then there existε> and two subsequences{xnk}and{xmk}of the sequence
{xn}such that
nk>mk>k, d(xmk,xnk–) <ε, and d(xmk,xnk)≥ε. ()
These and the triangle inequality imply that
ε≤d(xmk,xnk)
≤d(xmk,xnk–) +d(xnk–,xnk–) +d(xnk–,xnk)
In view oflimn→∞d(xn,xn+) = and lettingk→ ∞in the above inequalities, we obtain
lim
k→∞d(xmk,xnk) =ε. ()
By the triangle inequality, we have
d(xmk,xnk)≤d(xmk,xmk+) +d(xmk+,xnk)
≤d(xmk,xmk+) +d(xmk+,xnk+) +d(xnk+,xnk)
≤d(xmk,xmk+) +d(xmk+,xmk+) +d(xmk+,xnk)
+ d(xnk,xnk+)
≤d(xmk,xmk+) + d(xmk+,xmk+) +d(xmk,xnk)
+ d(xnk,xnk+).
Taking the limit ask→ ∞in the above inequalities and using (), we get
lim
k→∞d(xmk+,xnk) = lim
k→∞d(xmk+,xnk) = lim
k→∞d(xmk+,xnk+) =ε. () Sincexmk+xnk for anyk∈N, so by substitutingxwithxmk+andywithxnk in in-equality (), it follows that
ψd(xmk+,xnk+)
=ψd(fxmk+,gxnk)
≤ψM(xmk+,xnk)
–φM(xmk+,xnk),M
(x
mk+,xnk)
+ξ ψN(xmk+,xnk)
, ()
where
M(xmk+,xnk) = max
d(xmk+,xnk),d(xmk+,fxmk+),d(xnk,gxnk),
d(xmk+,gxnk) +d(xnk,fxmk+)
=max
d(xmk+,xnk),d(xmk+,xmk+),d(xnk,xnk+),
d(xmk+,xnk+) +d(xnk,xmk+)
,
M(xmk+,xnk) =max
d(xmk+,xnk),d(xmk+,fxmk+),d(xmk+,gxnk)
=maxd(xmk+,xnk),d(xmk+,xmk+),d(xmk+,xnk+)
,
M(xmk+,xnk) =max
d(xmk+,xnk),d(xnk,gxnk),d(xnk,fxmk+)
=maxd(xmk+,xnk),d(xnk,xnk+),d(xnk,xmk+)
, and
N(xmk+,xnk) =min
d(xmk+,fxmk+),d(xnk,fxmk+),d(xmk+,gxnk)
=mind(xmk+,xmk+),d(xnk,xmk+),d(xmk+,xnk+)
Lettingk→ ∞in the above equalities and applying (), (), we obtain
lim
k→∞M(xmk+,xnk) =ε, ()
lim
k→∞M
(x
mk+,xnk) =ε, ()
lim
k→∞M
(x
mk+,xnk) =ε, and ()
lim
k→∞N(xmk+,xnk) = . ()
Taking the limit ask→ ∞in inequality () and using ()-(), the continuity ofψ, and the lower semicontinuity ofφ, we get
ψ(ε)≤ψ(ε) –φ(ε,ε),
which yieldsφ(ε,ε) = . Henceε= , which contradicts the positivity ofε. Therefore, we
get the desired result.
Theorem . Let(X,,d)be an ordered complete metric space.Let f :X→X be non-decreasing(with respect to),continuous and almost generalizedC-contractive.If there exists x∈X such that xfx,then f has a fixed point.In particular,if F(f)is a totally ordered subset of X,then f has a unique fixed point.
Proof Define a sequence{xn}inXbyxandxn+=fxnfor alln∈N. Sincexfx=xand f is non-decreasing, we havex=fxfx=x. By induction, One can show that
xx · · · xnxn+ · · ·.
If there exists somen∈Nsuch thatxn=xn+=fxn, thenxnis a fixed point off. Hence
the proof is complete. Now suppose thatxn=xn+for alln∈N, we have
M(xn–,xn) =max
d(xn–,xn),d(xn–,fxn–),d(xn,fxn),
d(xn–,fxn),d(xn,fxn–)
=max
d(xn–,xn),d(xn,xn+),
d(xn–,xn+)
≤max
d(xn–,xn),d(xn,xn+),
d(xn–,xn) +d(xn,xn+)
=maxd(xn–,xn),d(xn,xn+)
, ()
M(xn–,xn) =max
d(xn–,xn),d(xn–,fxn–),d(xn–,fxn)
=maxd(xn–,xn),d(xn–,xn+)
≥d(xn–,xn), ()
M(xn–,xn) =max
d(xn–,xn),d(xn,fxn),d(xn,fxn–)
=maxd(xn–,xn),d(xn,xn+)
N(xn–,xn) =min
d(xn–,fxn–),d(xn,fxn–)
=mind(xn–,xn),
= . ()
On the other hand, our hypothesis implies that there exists (ξ,ψ,φ)∈[,∞)××
such that
ψd(fx,fy)≤ψM(x,y)–φM(x,y),M(x,y)+ξ ψN(x,y)
for allx,y∈Xwithxy, which yields
ψd(xn,xn+)
=ψd(fxn–,fxn)
≤ψM(xn–,xn)
–φM(xn–,xn),M(xn–,xn)
+ξ ψN(xn–,xn)
for alln∈N. This and equations ()-() yield
ψd(xn,xn+)
≤ψ
max
d(xn–,xn),d(xn,xn+),
d(xn–,xn+)
–φmaxd(xn–,xn),d(xn–,xn+)
,
maxd(xn–,xn),d(xn,xn+)
()
holds for anyn∈N. Sinceφandψare non-decreasing. Thus, from (), () and (), we deduce that
ψd(xn,xn+)
≤ψmaxd(xn–,xn),d(xn,xn+)
–φd(xn–,xn),max
d(xn–,xn),d(xn,xn+)
()
holds for anyn∈N, which implies
ψd(xn,xn+)
<ψmaxd(xn–,xn),d(xn,xn+)
()
holds for alln∈N, becaused(xn,xn+) > , hence
φd(xn–,xn),max
d(xn–,xn),d(xn,xn+)
> .
Asψis non-decreasing, from () it follows that
d(xn,xn+) <max
d(xn–,xn),d(xn,xn+)
holds for anyn∈N. This means thatd(xn,xn+) <d(xn–,xn) holds for alln∈N. Thus, the sequence {d(xn,xn+)} is decreasing. Then it converges to some nonnegative numbera.
Also from (), for anyn∈N, we have
ψd(xn,xn+)
≤ψd(xn–,xn)
–φd(xn–,xn),d(xn–,xn)
The above inequality yields
lim sup
n→∞
ψd(xn,xn+)
≤lim sup
n→∞
ψd(xn–,xn)
–lim inf
n→∞ φ
d(xn–,xn),d(xn–,xn)
.
Consequently, we have
ψ(a)≤ψ(a) –φ(a,a),
which impliesφ(a,a) = . Soa= . Thenlimn→∞d(xn,xn+) = . Now by Lemma ., the
sequence{xn}is Cauchy. SinceXis complete, there is somez∈Xsuch thatxn→zasn→
∞. The continuity off impliesfxn→fzasn→ ∞. From the uniqueness of the limit, we conclude thatfz=z. Hencez∈F(f). Now, we suppose thatF(f) is totally ordered. We will show thatzis unique. Assumeuis another fixed point off. Asu,z∈F(f), our assumption implies thatzanduare comparable. Without loss of generality, we may assume thatuz. Therefore,
ψd(u,z)=ψd(fu,fz)≤ψM(u,z)–φM(u,z),M(u,z)+ξ ψN(u,z) =ψd(u,z)–φd(u,z),d(u,z).
This yields φ(d(u,z),d(u,z)) = . Sod(u,z) = , that is,u=z. Thus, we get the desired
result.
The following corollary is an immediate consequence of the above theorem.
Corollary . Let(X,)and(X,d)be a totally ordered set and a complete metric space,
respectively.Let f :X→X be non-decreasing(with respect to),continuous and almost generalizedC-contractive.If there exists x∈X such that xfx,then f has a unique fixed point.
Theorem . Let(X,,d)be an ordered complete metric space.Let f,g:X→X be two weakly increasing mappings which f is almost generalizedC-contractive with respect to g.If either f or g is continuous,then the fixed point set of f is nonempty and F(f,g) =F(f) =F(g).
Particularly,if F(f)is a totally ordered subset of X,then f and g have a unique common fixed point.
Proof Our assumption implies that there exists some (ψ,φ,ξ)∈××[,∞) such that
ψd(fx,gy)≤ψM(x,y)–φM(x,y),M(x,y)+ξ ψN(x,y) () for allx,y∈Xwithxy, where
M(x,y) =max
d(x,y),d(x,fx),d(y,gy),d(x,gy) +d(y,fx)
,
M(x,y) =maxd(x,y),d(x,fx),d(x,gy),
M(x,y) =maxd(x,y),d(y,gy),d(y,fx), and
We now show thatF(f) =F(g). Letz∈F(f). Sofz=z. Sincezz, inequality () implies that
ψd(z,gz)=ψd(fz,gz)≤ψM(z,z)–φM(z,z),M(z,z)+ξ ψN(z,z).
Therefore,
ψd(z,gz)≤ψd(z,gz)–φd(z,gz),d(z,gz),
which yields φ(d(z,gz),d(z,gz)) = . Asφ∈, we getd(z,gz) = . Hencegz=z, that is,
z∈F(g). SoF(f)⊆F(g). Similarly, one can show thatF(g)⊆F(f). Therefore, we have
F(f,g) =F(f) =F(g). Letxbe an arbitrary element ofX. Define a sequence{xn}byxand
xn=fxn–, xn+=gxn for alln∈N.
If there exists m∈N such that either xm =xm– or xm+ =xm holds, then F(f) is nonempty. Because ifxm=xm–, thenfxm–=xm=xm–. Soxm–∈F(f). Ifxm+= xm, thengxm=xm+=xm. Hence,xm∈F(g) =F(f). Therefore, we may suppose that
xn=xn+ for anyn∈N. Without loss of generality we can assume thatxx. We now
show that the sequence{xn}is non-decreasing. Asf andgare weakly increasing mappings, we obtain
x=fxgfx=gx=xfgx=gx=xx · · ·.
Hence the sequence{xn}is non-decreasing. Supposen∈Nis arbitrary. Sincexn–xn, inequality () implies
ψd(xn,xn+)
=ψd(fxn–,gxn)
≤ψM(xn–,xn)
–φM(xn–,xn),M(xn–,xn)
+ξ ψN(xn–,xn)
, ()
where
M(xn–,xn) = max
d(xn–,xn),d(xn–,fxn–),d(xn,gxn),
d(xn–,gxn) +d(fxn–,xn)
=max
d(xn–,xn),d(xn,xn+),
d(xn–,xn+)
≤maxd(xn–,xn),d(xn,xn+)
, ()
M(xn–,xn) =max
d(xn–,xn),d(xn–,fxn–),d(xn–,gxn)
=maxd(xn–,xn),d(xn–,xn+)
M(xn–,xn) =max
d(xn–,xn),d(xn,gxn),d(xn,fxn–)
=maxd(xn–,xn),d(xn,xn+)
≥d(xn–,xn), and ()
N(xn–,xn) =min
d(xn–,fxn–),d(xn,fxn–),d(xn–,gxn)
=mind(xn–,xn), ,d(xn–,xn+)
= . ()
Thus, inequality () becomes
ψd(xn,xn+)
≤ψ
max
d(xn–,xn),d(xn,xn+),
d(xn–,xn+)
–φmaxd(xn–,xn),d(xn–,xn+)
,maxd(xn–,xn),
d(xn,xn+)
. ()
Sinceψ andφ are non-decreasing, the above inequality and inequalities (), (), and
() yield the following inequality:
ψd(xn,xn+)
≤ψmaxd(xn–,xn),d(xn,xn+)
–φd(xn–,xn),d(xn–,xn)
. ()
Asφ(d(xn–,xn),d(xn–,xn)) > , inequality () implies
ψd(xn,xn+)
<ψmaxd(xn–,xn),d(xn,xn+)
.
Sinceψis non-decreasing, it follows from the above inequality that
d(xn,xn+) <max
d(xn–,xn),d(xn,xn+)
.
So
d(xn,xn+) <max
d(xn–,xn)),d(xn,xn+)
=d(xn–,xn). ()
Hence inequality () becomes
ψd(xn,xn+)
≤ψd(xn–,xn)
–φd(xn–,xn),d(xn–,xn)
. ()
Similarly, one can show that
d(xn+,xn+) <d(xn+,xn). ()
Setyn=d(xn,xn+) andzn=d(xn+,xn+). Then from () and (), we get
which shows that the two sequences{yn}and{zn}are strictly decreasing and bounded. Hence{yn}and{zn} are convergent. Assume thatlimn→∞yn=aandlimn→∞zn=b. By (), we havea=b. Taking the limit superior asn→ ∞in (), we conclude that
lim sup
n→∞ ψ
d(xn+,xn)
≤lim sup
n→∞ ψ
d(xn,xn–)
–lim inf
n→∞ φ
d(xn,xn+),d(xn,xn–)}
. ()
Becauselimn→∞yn=limn→∞zn=a, (), the continuity ofψ, and the lower semicontinu-ity ofφimply that
ψ(a)≤ψ(a) –φ(a,a).
Thus,φ(a,a) = . Consequently,a= . Solimn→∞yn=limn→∞zn= . This implies that
limn→∞d(xn,xn+) = . As the sequence{xn}is non-decreasing andlimn→∞d(xn,xn+) = ,
Lemma . implies that{xn}is a Cauchy sequence. SinceXis complete, there is someu∈X
such thatxn→uasn→ ∞. Without loss of generality we assume thatf is continuous.
As xn–→u asn→ ∞, the continuity off implies that xn=fxn–→fu asn→ ∞.
By the uniqueness of the limit, we obtainfu=u. Therefore,u∈F(f) =F(g). Now sup-pose thatF(f) is a totally ordered subset ofX. We will show thatuis unique. Suppose thatz∈F(f,g) =F(f) =F(g). By our hypothesisu,zare comparable, hence without loss of generality supposeuz. Thus, inequality () implies that
ψd(u,z)=ψd(fu,gz)≤ψM(u,z)–φM(u,z),M(u,z)+ξ ψN(u,z) =ψd(u,z)–φd(u,z),d(u,z)
holds, which impliesφ(d(u,z),d(u,z)) = . Sod(u,z) = . Consequently,u=z. This
com-pletes the proof of the theorem.
Applying Theorem ., we can obtain the following result.
Corollary . Let(X,)and(X,d)be a totally ordered set and a complete metric space,
respectively.Let f,g:X→X be two weakly increasing mappings which f is almost gener-alizedC-contractive with respect to g.If either f or g is continuous,then f and g have a unique common fixed point.
The following examples support Theorem ..
Example . LetX= [,∞) and define the metricdonXas follows:
d(x,y) =
ifx=y,
x+y ifx=y
for allx,y∈X(this metric has been introduced by Shatanawi and Al-Rawashdeh []). For anyx∈X, define the functionsf,g:X→Xby
f(x) =
if ≤x≤,
and
g(x) =
if ≤x≤,
x– if <x.
Also, defineψ:R+→R+ andφ:R+×R+→R+by settingψ(t) =t,φ(s,t) = s+t for all
s,t∈R+, respectively. Consider a relationonXbyxyif only ify≤xfor allx,y∈X. Then the following statements hold:
(a) (X,,d)is an ordered complete metric space.
(b) f andgare weakly increasing mappings with respect to. (c) f is continuous.
(d) (ψ,φ)∈×.
(e) f is an almost generalizedC-contractive mapping with respect tog, that is,
ψd(fx,gy)≤ψM(x,y)–φM(x,y),M(x,y)+ξ ψN(x,y)
for allx,y∈X, whereξ> and
M(x,y) =max
d(x,y),d(x,fx),d(y,gy),d(x,gy) +d(y,fx)
,
M(x,y) =maxd(x,y),d(x,fx),d(x,gy),
M(x,y) =maxd(x,y),d(y,gy),d(fx,y), and
N(x,y) =mind(x,fx),d(y,fx),d(x,gy).
Proof (a) Assume a sequence{xn}inXconverges to somea∈X. By the definition ofd, one can find someN∈Nsuch thatxn=aholds for alln≥N. Hence (X,d) is a complete metric space. It is obvious that (X,) is a partially ordered set (indeed, (X,) is a totally ordered set). So (X,,d) is an ordered complete metric space.
(b) To see this, letx∈X, we will show thatfxgfxandgxfgx. We first show that
fxgfx. To prove this, consider the following cases:
If ≤x≤, thenfx=gfx= and hencefxgfx. If <x≤, then =gfx<fx=x– , it follows thatfxgfx. Ifx> , thenx– =gfx<fx=x– and hencefxgfx. Therefore,
fxgfxholds. We now showgxfgxholds. To see this, consider the following cases: If ≤x≤, then =gx≤fgx= . Sogxfgx. If <x≤, then =fgx<gx=x– and hencegxfgx. Ifx> , thenx– =fgx<gx=x– . Consequently,gxfgx. Therefore, in any case, we getgxfgx.
(c) Leta∈Xbe arbitrary, and{xn}be a sequence inXsuch thatxn→aasn→ ∞. By the definition ofd, there existsN∈Nsuch that for anyn≥N,xn=a. Hencefxn=faholds for alln≥N, which implieslimn→∞fxn=fa. Sof is continuous ata. Sinceais arbitrary, hencef is continuous.
(d) It is trivial.
(e) Let (x,y)∈X×Xbe arbitrary. We will show that
CaseI. (x,y)∈[, ]×[, ].
In this case, we have ψ(d(fx,gy)) = . Ifx=y= , then M(, ) =M(, ) =M(, ) =
N(, ) = . Hence () holds.
Ifx= ,y= , thenM(,y) =M(,y) =M(,y) =y+ andN(,y) = . Thus, inequality () becomes ≤(y+ )– (y+ ) =y(y+ ), which holds.
Letx= ,y= . Then we haveM(x, ) =M(x, ) =M(x, ) =x+ ,N(x, ) = , and hence () becomes ≤(x+ )– (x+ ) =x(x+ ), which holds. So () holds.
Letx=y= . Then we haveM(x,x) =M(x,x) =M(x,x) =N(x,x) =x+ , and hence () becomes ≤(x+ )– (x+ ) +ξ(x+ )= (x+ )((ξ+ )(x+ ) – ), which holds. So () is
valid.
Ifx,y> withx=y, then we getM(x,y) =M(x,y) =M(x,y) =x+yandN(x,y) =min{x+ ,y+ }=N> , and so () becomes
≤(x+y)– (x+y) +ξN. Asξ> andN> , we get
≤(x+y)– (x+y) + ξ≤(x+y)– (x+y) +ξN. Thus, () is true. Therefore, in any case () is valid.
CaseII. (x,y)∈[, ]×(,∞).
In this case, we haveψ(d(fx,gy)) = (y– ). Ifx= , then
M(,y) =M(,y) = y– , M(,y) =y+ , and N(,y) = .
Putting these into (), it reduces to (y– )≤(y– )– (y+)+(y–)
. Sincey> , ≤
y– y+ , which is equivalent to (y– )≤(y– )– (y– ). On the other hand, y+ ≤(y+)+( y–)≤y– , so we have
(y– )≤(y– )– (y– )≤(y– )–(y+ ) + (y– )
.
This implies ().
If <x≤, <y, then we have
M(x,y) =max{x+y,x+ , y– }=max{x+y, y– },
M(x,y) =max{x+y,x+ ,x+y– }=x+y,
M(x,y) =max{x+y, y– ,y+ }=max{x+y, y– }, and
N(x,y) =min{x+ ,y+ ,x+y– }=x+ .
IfM(x,y) =x+y, then () reduces to
(y– )≤(x+y)– (x+y) +ξ(x+ ). To prove this, we first show that
holds. Inequality (h) is equivalent to
(x+y)– (x+y) + ≤(x+y)– (x+y) +ξ(x+ )≡–(x+y) + ≤ξ(x+ ). Asx> andy> , so –(x+y) + < – + <ξ(x+ )holds. In view of this and the above equivalence, we conclude that inequality (h) is true. Since (y– )≤(x+y– ), hence inequality (h) implies inequality ().
IfM(x,y) = y– , then () reduces to
(y– )≤(y– )–(x+y) + (y– )
+ξ(x+ )
.
To see it, we first show
(y– – )= (y– )≤(y– )– (y– ) +ξ(x+ ) (i)
holds. Inequality (i) is equivalent to
(y– )– (y– ) + ≤(y– )– (y– ) +ξ(x+ )
≡–(y– ) + ≤ξ(x+ ).
Becausey> , hence –(y– ) + < – + <ξ(x+ )holds. This and the above
equiv-alence yield inequality (i). On the other hand,x+y≤(x+y)+( y–)≤y– . Hence we have
(y– – )= (y– )x≤(y– )– (y– ) +ξ(x+ ) by (i)
≤(y– )–(x+y) + (y– )
+ξ(x+ )
.
This establishes inequality ().
CaseIII. Let (x,y)∈(,∞)×[, ]. In this case, we haveψ(d(fx,gy)) = (x– ). Ifx=y, then we have
M(x,x) =max{x– ,x+ }= x– ,
M(x,x) =max{x– ,x+ }= x– ,
M(x,x) =max{x+ , x– }= x– , and
N(x,x) =min{x– ,x+ }=x+ .
Hence inequality () reduces to
(x– )≤(x– )– (x– ) +ξ(x+ ),
which is equivalent to ≤(x– )– (x– ) +ξ(x+ ). Sincex> andξ>
, we have
Ifx=y, we have
M(x,y) =max{x+y, x– ,y+ }=max{x+y, x– },
M(x,y) =max{x+y, x– ,x+ }=max{x+y, x– },
M(x,y) =max{x+y,y+ ,x+y– }=x+y, and
N(x,y) =min{x– ,x+y– ,x+ }=x+ .
IfM(x,y) =x+y, then inequality () reduces to
(x– )≤(x+y)– (x+y) +ξ(x+ ). To prove it, we first show that
(x+y– )≤(x+y)– (x+y) +ξ(x+ ) (j)
holds, which is equivalent to
(x+y)– (x+y) + ≤(x+y)– (x+y) +ξ(x+ )
≡–(x+y) + ≤ξ(x+ ).
Sincex> andy≥, we have –(x+y) + < – + <ξ(x+ ). So inequality (j) is valid. On
the other hand, we have (x– )≤(x+y– ). This along with (j) implies inequality (). IfM(x,y) = x– , then () becomes
(x– )≤(x– )–(x– ) + (x+y)
+ξ(x+ )
.
To see this, we first prove the following inequality:
(x– – )= (x– )≤(x– )– (x– ) +ξ(x+ ). (k)
This is equivalent to
(x– )– (x– ) + ≤(x– )– (x– ) +ξ(x+ )
≡–(x– ) + ≤ξN.
Asx> , hence –(x– ) + < – + <ξ(x+ ). This along with the above equivalence establishes (k). SinceM(x,y) = x– ≥x+y, from (k), we get
(x– )≤(x+y– )
≤(x– )
≤(x– )– (x– ) +ξ(x+ )
≤(x– )–(x– ) + (x+y)
+ξ(x+ )
.
CaseIV. (x,y)∈(,∞)×(,∞). In this case, we haveψ(d(fx,gy)) = (x+y– ).
Ifx=y, we have
M(x,x) =max{x– , x– }= x– ,
M(x,x) =max{x– , x– }= x– ,
M(x,x) =max{x– , x– }= x– , and
N(x,x) =min{x– , x– }= x– .
So inequality () reduces to
(x– – )= (x– )≤(x– )– (x– ) +ξ(x– ),
which is equivalent to
(x– )– (x– ) + ≤(x– )– (x– ) +ξ(x– )
≡–(x– ) + ≤ξ(x– ).
Becausex> , hence –(x– ) + < – + <ξ(x– )holds. This along with the above
equivalence implies (). Ifx=y, then we have
M(x,y) =max{x+y, x– , y– },
M(x,y) =max{x+y, x– },
M(x,y) =max{x+y, y– }, and
N(x,y) =min{x– ,x+y– }=N.
Then the following three cases occur forM(x,y). Case . IfM(x,y) =x+y, then () reduces to
(x+y– )≤(x+y)– (x+y) +ξN,
which is equivalent to
(x+y)– (x+y) + ≤(x+y)– (x+y) +ξN ≡–(x+y) + ≤ξN.
Sincex> andy> , so –(x+y) + < – + <ξNholds. This and the above
equiv-alence imply ().
Case . IfM(x,y) = x– andM(x,y) =x+y, then () becomes
(x+y– )≤(x– )–(x– ) + (x+y)
+ξN
To prove it, we first establish that
(x– – )= (x– )≤(x– )– (x– ) +ξN, (l)
holds. This is equivalent to
(x– )– (x– ) + ≤(x– )– (x– ) +ξN ≡–(x– ) + ≤ξN.
Sincex> , we have –(x– ) + < – + <ξN. So (l) is valid. Asx+y≤x– , we
have
(x+y– )≤(x– )
≤(x– )– (x– ) +ξN by (l)
≤(x– )–(x– ) + (x+y)
+ξN
.
Therefore, inequality () holds.
IfM(x,y) = x– andM(x,y) = y– , then () reduces to
(x+y– )≤(x– )–(x– ) + (y– )
+ξN
.
Because y– ≤x– , by inequality (l), we have
(x+y– )≤(x– )
≤(x– )– (x– ) +ξN ≤(x– )–(x– ) + (y– )
+ξN
.
This implies ().
Case . IfM(x,y) = y– andM(x,y) =x+y, then () reduces to
(x+y– )≤(y– )–(x+y) + (y– )
+ξN
.
To see this, we first show
(y– – )= (y– )≤(y– )– (y– ) +ξN, (m)
which is equivalent to
Asy> , we obtain –(y– ) + < – + <ξN. So (m) holds. Sincex+y≤y– ,
from (m), we get
(x+y– )≤(y– )
≤(y– )– (y– ) +ξN ≤(y– )–(x+y) + (y– )
+ξN
.
Thus, () holds.
IfM(x,y) = y– andM(x,y) = x– , then () reduces to
(x+y– )≤(y– )–(x– ) + (y– )
+ξN
.
Asx+y≤y– and x– ≤y– , we have
(x+y– )≤(y– )
≤(y– )– (y– ) +ξN by (m)
≤(y– )–(x– ) + (y– )
+ξN
.
This establishes (). Therefore, in any case inequality () holds. So the proof of (e) is completed. Thus,f,g,ψ, andφ satisfy the hypotheses of Theorem ., henceF(f,g) is nonempty (in fact, ∈F(f,g) =F(f) =F(g)). On the other hand, sinceXis a totally ordered set, henceF(f,g) is a totally ordered subset ofX. So Theorem . implies that the setF(f,g)
is singleton (indeed, we observe thatF(f,g) ={}).
Example . Set X={, , , }. Letd,ψ, andφ be as in Example .. Consider the relation and the mappings f,g on X by ={(, ), (, ), (, ), (, ), (, )} and f =
{(, ), (, ), (, ), (, )},g={(, ), (, ), (, ), (, )}, respectively. It is clear that (X,) is an ordered set. Similar to the arguments given Example . ((a), (c)) one can show that (X,d) andf are complete and continuous, respectively. It is easy to see that the mappings
f andgare weakly increasing with respect to. We next show that
ψd(fx,gy)≤ψM(x,y)–φM(x,y),M(x,y)+ξ ψN(x,y) ()
for allx,y∈Xwithxy, whereξ≥., and
M(x,y) =max
d(x,y),d(x,fx),d(y,gy),d(x,gy) +d(y,fx)
,
M(x,y) =maxd(x,y),d(x,fx),d(x,gy),
M(x,y) =maxd(x,y),d(y,gy),d(fx,y), and
To see this, we have
=ψdf(),g()≤ψM(, )–φM(, ),M(, )+ξ ψN(, ) =ψ() –φ(, ) +ξ ψ()
= +ξ,
=ψdf(),g()≤ψM(, )–φM(, ),M(, )+ξ ψN(, ) =ψ() –φ(, ) +ξ ψ()
= ,
=ψdf(),g()≤ψM(, )–φM(, ),M(, )+ξ ψN(, ) =ψ() –φ(, ) +ξ ψ()
= + ξ,
=ψdf(),g()≤ψM(, )–φM(, ),M(, )+ξ ψN(, ) =ψ() –φ(, ) +ξ ψ()
= , and
=ψdf(),g()≤ψM(, )–φM(, ),M(, )+ξ ψN(, ) =ψ() –φ(, ) +ξ ψ()
= . +ξ.
These mean that the mappingf is almost generalizedC-contractive with respect to the
mappingg. Now, it follows from Theorem . that the fixed point set off is nonempty
andF(f,g) =F(f) =F(g). We observe thatF(f,g) =F(f) =F(g) ={, }.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Faculty of Science, Imam Khomeini International University, Qazvin, 34149-16818, Iran. 2Department of Mathematics, Faculty of Science, Bu-Ali Sina University, P.O. Box 6517838695, Hamedan, Iran.
Acknowledgements
The authors are grateful to the referee(s) for his(their) useful comments and suggestions on the manuscript.
Received: 20 January 2016 Accepted: 11 July 2016
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