R E S E A R C H
Open Access
Fixed points of generalized contractive
mappings of integral type
Hamed H Alsulami
1*, Erdal Karapınar
1,2, Donal O’Regan
1,3and Priya Shahi
4*Correspondence:
[email protected]; [email protected]
1Nonlinear Analysis and Applied
Mathematics Research Group (NAAM), King Abdulaziz University, Jeddah, Saudi Arabia
Full list of author information is available at the end of the article
Abstract
The aim of this paper is to introduce classes of
α
-admissible generalized contractive type mappings of integral type and to discuss the existence of fixed points for these mappings in complete metric spaces. Our results improve and generalize fixed point results in the literature.MSC: 46T99; 54H25; 47H10; 54E50
Keywords:
α
-admissible mapping; fixed point; complete metric space; contractive mapping; partial order1 Introduction and preliminaries
In , Branciari [] established a fixed point theorem for a single-valued mapping satis-fying a contractive inequality of integral type; we also refer the reader to [–]. Recently, Liuet al.[] (see also [, ]) obtained fixed point theorems for general classes of con-tractive mappings of integral type in complete metric spaces. In this paper, using auxiliary functions, we establish some fixed point theorems for self-mappings satisfying a certain contractive inequality of integral type.
Throughout the paperR+= [, +∞),N
=N∪{}, whereNdenotes the set of all positive
integers, (X,d) is a metric space andf:X→Xis a self-mapping. Let
={ϕ:ϕ:R+→R+is Lebesgue integrable, summable on each compact subset of
R+and
ϕ(t)dt> for each> };
={ϕ:ϕ:R+→R+satisfies that lim infn→∞ϕ(an) > ⇔lim infn→∞an> for each
{an}n∈N⊂R+};
={ϕ:ϕ:R+→R+is nondecreasing continuous andϕ(t) = ⇔t= }; ={ϕ:ϕ:R+→R+satisfies thatϕ() = };
={ϕ:ϕ:R+→R+satisfies that lim sups→tϕ(s) < for eacht> };
={(α,β) :α,β:R+→[, )satisfy thatlim sups→+β(s) < ,lim sups→t+ α(s) –β(s)<
andα(t) +β(t) < for eacht> }.
In , Liuet al.[] introduced the following three contractive mappings of integral type:
ψ
d(fx,fy)
ϕ(t)dt
≤ψ d(x,y)
ϕ(t)dt
–φ d(x,y)
ϕ(t)dt
, ()
where (ϕ,φ,ψ)∈××,
ψ
d(fx,fy)
ϕ(t)dt
≤αd(x,y)ψ d(x,y)
ϕ(t)dt
, ∀x,y∈X, ()
where (ϕ,ψ,α)∈××, and
ψ
d(fx,fy)
ϕ(t)dt
≤αd(x,y)φ
d(x,fx)
ϕ(t)dt
+βd(x,y)ψ
d(y,fy)
ϕ(t)dt
, ∀x,y∈X, ()
where (ϕ,ψ,φ)∈××and (α,β)∈.
The following lemmas will be used in the proof of our main results.
Lemma . [] Letϕ∈ and{rn}n∈N be a nonnegative sequence withlimn→∞rn=a. Then we have
lim
n→∞
rn
ϕ(t)dt=
a
ϕ(t)dt.
Lemma .[] Letϕ∈and{rn}n∈Nbe a nonnegative sequence.Then we have the
fol-lowing equivalence:
lim
n→∞
rn
ϕ(t)dt=
if and only iflimn→∞rn= .
Lemma .[] Letϕ∈.Thenϕ(t) > if and only if t> .
The notion ofα-admissibility was defined in [] and appreciated by several authors [–] (see also [–]).
Definition .[] LetT:X→Xandα:X×X→[,∞). The mappingT is said to be
α-admissible if for allx,y∈X, we have
α(x,y)≥ ⇒ α(Tx,Ty)≥. ()
Definition . LetT:X→Xandα:X×X→[,∞). The mappingT is said to be weak triangularα-admissibleif for allx∈Xwe have
α(x,Tx)≥ and αTx,Tx≥ ⇒ αx,Tx≥. ()
2 Main results
Definition . Let (X,d) be a complete metric space,f :X→Xandα:X×X→[,∞) be mappings. Suppose that there exist (ϕ,φ,ψ)∈××andL≥ such that
α(x,y)ψ
d(fx,fy)
ϕ(t)dt
≤ψ
M(x,y)
ϕ(t)dt
–φ
N(x,y)
ϕ(t)dt
+Lψ
O(x,y)
ϕ(t)dt
()
for allx,y∈X, where
M(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
,
and
N(x,y) =maxd(x,y),d(x,fx),d(y,fy) and
O(x,y) =mind(x,fx),d(y,fy),d(y,fx),d(x,fy).
Thenf is said to be anα-admissible contractive inequality of integral typeI.
Theorem . Let (X,d)be a complete metric space.Suppose that f :X→X is an α -admissible contractive inequality of integral type I which satisfies
(i) f is weak triangularα-admissible;
(ii) there existsx∈Xsuch that eitherα(x,fx)≥orα(fx,x)≥; (iii) f is continuous.
Then T has a fixed point.
Proof From (ii), there exists a pointx∈Xsuch thatα(x,fx)≥ (due to the symmetry of the metric, the other case yields the same result). Letx=xand we define an iterative
sequence{xn}inXbyxn+=fxnfor alln≥. Note that we have
α(x,x) =α(x,fx)≥ ⇒ α(fx,fx) =α(x,x)≥.
Inductively, we have
α(xn,xn+)≥ for alln∈N. ()
In the sequel, we use the following abbreviations:
dn=d(xn,xn+) and αn=α(xn,xn+) forn∈N. ()
Notice that ifxn=xn+ for somen, then it is evident thatu=xnis a fixed point off. This completes the proof. Consequently, we assume thatxn=xn+for alln∈N, that is,
We now prove that{dn}is a non-increasing sequence of real numbers, that is,
dn≤dn–, ∀n∈N. ()
Suppose, on the contrary, that inequality () does not hold. Thus, there exists somen∈N
such that
dn>dn–. ()
From () and (), we get
<
dn–
ϕ(t)dt<
dn
ϕ(t)dt. ()
Regarding again () and () together with the properties ofψ, we conclude that
=ψ() <ψ
dn–
ϕ(t)dt
≤ψ dn
ϕ(t)dt
. ()
Using equations (), (), (), () and the fact that (ϕ,φ,ψ)∈××, we obtain
immediately that
ψ
dn–
ϕ(t)dt
≤ψ
dn
ϕ(t)dt
≤αnψ dn
ϕ(t)dt
=αnψ
d(fnx,fn+x)
ϕ(t)dt
≤ψ
M(fn–x,fnx)
ϕ(t)dt
–φ
N(fn–x,fnx)
ϕ(t)dt
+Lψ
O(fn–x,fnx)
ϕ(t)dt
, ()
where
Mfn–x,fnx≤maxdfn–x,fnx,dfnx,fn+x=max{d
n–,dn},
Nfn–x,fnx=maxdfn–x,fnx,dfnx,fn+x=max{d
n–,dn},
Ofn–x,fnx=mindfn–x,fnx,dfnx,fn+x, dfnx,fnx,dfn–x,fn+x= .
Thus M(fn–x,fnx)≤d
n and N(fn–x,fnx) =dn from (). Hence, inequality () turns into
ψ dn
ϕ(t)dt
≤αnψ dn
ϕ(t)dt
≤ψ
dn
ϕ(t)dt
–φ dn
ϕ(t)dt
<ψ dn
ϕ(t)dt
which is a contradiction. Hence () holds. Thus, there exists a constantc≥ such that
limn→∞dn=c≥.
Next we show thatc= , that is,
lim
n→∞dn= . ()
Suppose, on the contrary, thatc> . It follows from () and () that
ψ dn
ϕ(t)dt
=ψ
d(fnx,fn+x)
ϕ(t)dt
≤α(xn,xn+)ψ
d(fnx,fn+x)
ϕ(t)dt
≤ψ
M(fn–x,fnx)
ϕ(t)dt
–φ
N(fn–x,fnx)
ϕ(t)dt
+Lψ
O(fn–x,fnx)
ϕ(t)dt
, ()
where
Mfn–,fnx≤maxdfn–x,fnx,dfnx,fn+x=max{dn–,dn}, Nfn–x,fnx=maxdfn–x,fnx,dfnx,fn+x=max{dn–,dn},
Ofn–x,fnx=mindfn–x,fnx,dfnx,fn+x,dfnx,fnx,dfn–x,fn+x= .
Hence, inequality () becomes
ψ dn
ϕ(t)dt
=ψ
d(fnx,fn+x)
ϕ(t)dt
≤αnψ
d(fnx,fn+x)
ϕ(t)dt
≤ψ dn–
ϕ(t)dt
–φ dn–
ϕ(t)dt
. ()
Taking the upper limit in () and using Lemma . and noting (ϕ,φ,ψ)∈××,
we get
ψ
c
ϕ(t)dt
= lim
n→∞supψ
dn
ϕ(t)dt
≤ lim
n→∞sup
ψ
dn–
ϕ(t)dt
–φ dn–
ϕ(t)dt
=ψ
c
ϕ(t)dt
– lim
n→∞infφ
dn–
ϕ(t)dt
<ψ
c
ϕ(t)dt
, ()
Next we prove that {fnx}
n∈N is a Cauchy sequence. Suppose, on the contrary, that {fnx}
n∈Nis not a Cauchy sequence. Thus, there is a constant> such that for each
pos-itive integerk, there are positive integersm(k) andn(k) withm(k) >n(k) >ksatisfying
dfm(k)x,fn(k)x>. ()
For each positive integerk, letm(k) denote the least integer exceedingn(k) and satisfying (). This implies that
dfm(k)x,fn(k)x> and dfm(k)–x,fn(k)x≤ for allk∈N. ()
On the other hand, we have
dfm(k)x,fn(k)x≤dfn(k)x,fm(k)–x+dm(k)–, ∀k∈N, dfm(k)x,fn(k)+x–dfm(k)x,fn(k)x≤dn(k), ∀k∈N, dfm(k)+x,fn(k)+x–dfm(k)x,fn(k)+x≤dm(k), ∀k∈N, dfm(k)+x,fn(k)+x–dfm(k)+x,fn(k)+x≤d
n(k)+, ∀k∈N.
()
In view of () and (), we infer that
lim
k→∞d
fn(k)x,fm(k)x=,
lim
k→∞d
fm(k)x,fn(k)+x=,
lim
k→∞d
fm(k)+x,fn(k)+x=,
lim
k→∞d
fm(k)+x,fn(k)+x=.
()
Using the weak triangular alpha admissible property off, we get in view of ()
αfm(k)x,fn(k)+x≥. ()
From () and (), we have
ψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤αfm(k)x,fn(k)+xψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤ψ
M(fm(k)x,fn(k)+)
ϕ(t)dt
–φ
N(fm(k)x,fn(k)+)
ϕ(t)dt
+Lψ
O(fm(k)x,fn(k)+)
ϕ(t)dt
Taking the upper limit in () and using (), (ϕ,φ,ψ)∈××and Lemma ., we
get
ψ
ϕ(t)dt
= lim
k→∞supψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤lim sup
k→∞
αfm(k)x,fn(k)+xψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤lim sup
k→∞ ψ
M(fm(k)x,fn(k)+x)
ϕ(t)dt
–lim inf
k→∞ φ
N(fm(k)x,fn(k)+x)
ϕ(t)dt
+Llim sup
k→∞
ψ
O(fm(k)x,fn(k)+)
ϕ(t)dt
=ψ
ϕ(t)dt
–lim inf
k→∞ φ
d(fm(k)x,fn(k)+x)
ϕ(t)dt
+Llim sup
k→∞ ψ
O(fm(k)x,fn(k)+)
ϕ(t)dt
=ψ
ϕ(t)dt
–lim inf
k→∞ φ
d(fm(k)x,fn(k)+x)
ϕ(t)dt
<ψ
ϕ(t)dt
,
which is impossible. Thus{fnx}
n∈Nis a Cauchy sequence. Now, since (X,d) is complete,
there exists a pointa∈Xsuch thatlimn→∞fnx=a. From the continuity off, it follows thatxn=fxn+→faasn→+∞. From the uniqueness of limits, we geta=fa, that is,ais
a fixed point off. This completes the proof.
Theorem . Let (X,d)be a complete metric space. Suppose that f :X→X is an α -admissible contractive inequality of integral type I which satisfies
(i) f is weak triangularα-admissible;
(ii) there existsx∈Xsuch that eitherα(x,fx)≥orα(fx,x)≥;
(iii) if{xn}is a sequence inXsuch thatα(xn,xn+)≥for allnandxn→x∈Xas n→+∞,thenα(xn,x)≥for alln.
Then f has a fixed point.
Proof Following the proof in Theorem ., we see that{xn}is a Cauchy sequence in the complete metric space (X,d). Then there existsa∈Xsuch thatxn→aasn→+∞. On the other hand, from inequality () and hypothesis (iii), we have
Let us suppose thata=fa. In view of the above inequality and (), we obtain that
ψ
d(fn+x,fa)
ϕ(t)dt
≤α(xn,a)ψ
d(fn+x,fa)
ϕ(t)dt
≤ψ
M(fnx,a)
ϕ(t)dt
–φ
N(fnx,a)
ϕ(t)dt
+Lψ
O(fnx,a)
ϕ(t)dt
()
for alln∈N. On the other hand, we have
Mfnx,a=max
dfnx,a,dfnx,fn+x,d(a,fa), d
fnx,fa+da,fn+x,
Nfnx,a=maxdfnx,a,dfnx,fn+x,d(a,fa),
Ofnx,a=mindfnx,fn+x,d(a,fa),dfnx,fa,da,fn+x.
()
Taking the upper limit in (), in view of Lemmas . and . and (ϕ,φ,ψ)∈××,
we infer from () that
ψ
d(a,fa)
ϕ(t)dt
= lim
n→∞supψ
d(fn+x,fa)
ϕ(t)dt
≤ lim
n→∞supα(xn,a)ψ
d(fn+x,fa)
ϕ(t)dt
≤ lim
n→∞supψ
M(fnx,a)
ϕ(t)dt
– lim
n→∞infφ
N(fnx,a)
ϕ(t)dt
+L lim
n→∞supψ
O(fnx,a)
ϕ(t)dt
≤ψ
d(a,fa)
ϕ(t)dt
–φ
d(a,fa)
ϕ(t)dt
<ψ
d(a,fa)
ϕ(t)dt
,
which is a contradiction. Thus, we havea=fa.
Now, we present another contractive inequality of integral type.
Definition . Let (X,d) be a complete metric space andf :X→Xbe a self-mapping. Suppose that there exist (ϕ,φ,ψ)∈××andL≥ such that
α(x,y)ψ
d(fx,fy)
ϕ(t)dt
≤ψ
M∗(x,y)
ϕ(t)dt
–φ
N∗(x,y)
ϕ(t)dt
+Lψ
O(x,y)
ϕ(t)dt
for allx,y∈X, where
M∗(x,y) =max
d(x,y),
d(x,fx) +d(y,fy)
,
d(x,fy) +d(y,fx)
,
and
N∗(x,y) =max
d(x,y),
d(x,fx) +d(y,fy)
and
O(x,y) =mind(x,fx),d(y,fy),d(y,fx),d(x,fy).
Thenf is said to be anα-admissible contractive inequality of integral typeII.
We omit the proof of the following two theorems since they mimic the proof of Theo-rem . and TheoTheo-rem ..
Theorem . Let (X,d)be a complete metric space. Suppose that f :X→X is an α -admissible contractive inequality of integral type II which satisfies
(i) f is weak triangularα-admissible;
(ii) there existsx∈Xsuch that eitherα(x,fx)≥orα(fx,x)≥;
(iii) f is continuous. Then T has a fixed point.
Theorem . Let(X,d)be a complete metric space. Suppose that f :X→X is an α -admissible contractive inequality of integral type II which satisfies
(i) f is weak triangularα-admissible;
(ii) there existsx∈Xsuch that eitherα(x,fx)≥orα(fx,x)≥;
(iii) if{xn}is a sequence inXsuch thatα(xn,xn+)≥for allnandxn→x∈Xas n→+∞,thenα(xn,x)≥for alln.
Then T has a fixed point.
The following condition provides the uniqueness of fixed points of the maps considered in Theorem . and Theorem .. Consider
(U∗) for allx,y∈Fix(f), there existsz∈Xsuch thatα(x,z)≥andα(y,z)≥, whereFix(f) denotes the set of fixed points off.
Theorem . If the condition(U∗)is added to the hypotheses of Theorem. (respectively, Theorem.),then the fixed point u of T is unique.
Proof From (U∗), we have
α(x,z)≥ and α(y,z)≥.
Define the sequence{zn}inXbyzn+=fznfor alln≥ andz=z. Using the weak
trian-gularα-admissible property off, we infer that
for alln∈N. Using inequality (), we get
d(x,zn)
ϕ(t)dt=
d(fx,f(zn–))
ϕ(t)dt
≤α(x,zn–)
d(fx,f(zn–))
ϕ(t)dt
≤ψ
d(x,zn–)
ϕ(t)dt
–φ
d(x,zn–)
ϕ(t)dt
+Lψ
d(x,zn–)
ϕ(t)dt
.
Using standard techniques, we derive that d(x,zn)≤d(x,zn–) and hence the sequence {d(x,zn)}converges to someL≥. IfL= , then the proof is complete. Indeed, we get that zn→xand analogously,zn→yasn→ ∞and from the uniqueness of limits, we derive thatx=y. Suppose, on the contrary,L> . By lettingn→ ∞, we derive from the above inequality that
L
ϕ(t)dt≤ψ
L
ϕ(t)dt
–φ
L
ϕ(t)dt
,
which is a contradiction.
Now we introduce a third type of contractive inequality of integral type.
Definition . Let (X,d) be a complete metric space andf :X→Xbe a self-mapping. Suppose that there exist (ϕ,φ,ψ)∈××such that
α(x,y)ψ
d(fx,fy)
ϕ(t)dt
≤ψ d(x,y)
ϕ(t)dt
–φ d(x,y)
ϕ(t)dt
. ()
Thenf is said to be anα-admissible contractive inequality of integral typeIII.
Theorem . Let (X,d) be a complete metric space.Suppose that f :X→X is an α -admissible contractive inequality of integral type III which satisfies
(i) f is weak triangularα-admissible;
(ii) there existsx∈Xsuch that eitherα(x,fx)≥orα(fx,x)≥;
(iii) f is continuous. Then T has a fixed point.
Proof Following the lines in the proof of Theorem ., we conclude the result.
Theorem . Let (X,d)be a complete metric space. Suppose that f :X→X is an α -admissible contractive inequality of integral type III which satisfies
(i) f is weak triangularα-admissible;
(ii) there existsx∈Xsuch that eitherα(x,fx)≥orα(fx,x)≥;
(iii) if{xn}is a sequence inXsuch thatα(xn,xn+)≥for allnandxn→x∈Xas n→+∞,thenα(xn,x)≥for alln.
Proof The reasoning in Theorem . establishes the result.
Example . Suppose thatX= [, ] with the usual metric. We consider a mappingf : X→Xdefined byf(x) =x. Define the mappingα:X×X→[, +∞) byα(x,y) = for all x,y∈X. Hence,f is weak triangularα-admissible. Defineϕ∈byϕ(t) =t. Let us define ψ∈andφ∈byψ(t) = tandφ(t) =t respectively for allt≥.
Clearly, in view of the definitions ofαandf, we infer thatf is anα-admissible contractive inequality of integral typeIII. There existsx∈Xsuch thatα(x,fx)≥. In fact, forx= ,
we obtain
α(,f) =α
,
= .
Clearly,f is continuous. Now, all the hypotheses of Theorem . are satisfied. Thusf has a fixed point inX. In this case, is a fixed point off.
Example . Suppose thatX={n :n∈N} ∪ {}with the usual metricd(x,y) =|x–y| induced byR. It is a complete metric space, sinceXis a closed subset ofR. We consider a mappingf:X→Xdefined by
f(x) =
n+ ifx=
n, ifx= .
Define the mappingα:X×X→[, +∞) byα(x,y) = for allx,y∈X. It is clear thatf is weak triangularα-admissible. Thus, the condition (i) of Theorem . is satisfied.
Now, consider the following auxiliary functionϕdefined as
ϕ(t) =
⎧ ⎪ ⎨ ⎪ ⎩
ift= ,
t(/t)–[ –logt] if <t< ,
ift≥.
Then, for anyε> , we haveεϕ(t)dt=ε/εfor <ε< andεϕ(t)dt=εforε≥. Consequently, we haveϕ∈.
Clearly, in view of the definitions ofαandf, we infer thatf is anα-admissible contractive inequality of integral typeIII forψ(t) = t andφ(t) =t, for allt≥, whereψ∈and φ∈.
There existsx∈Xsuch thatα(x,fx)≥. In fact, for example, forx= , we obtain α(,f) =α(, ) = . Hence, the condition (ii) of Theorem . is fulfilled.
Let{xn}be a sequence inXsuch thatα(xn,xn+)≥ for allnandxn→xasn→+∞ for somex∈X. From the definition ofα, for alln, we haveα(xn,x) = for all. So, the last condition of Theorem . is satisfied. As a result, due to Theorem ., the mappingf has a fixed point. Notice thatu= is a fixed point off.
3 Consequences in metric spaces
We get the following result by lettingα(x,y) = in Theorem ..
Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying, for all x,y∈X,
ψ
d(fx,fy)
ϕ(t)dt
≤ψ
M∗(x,y)
ϕ(t)dt
–φ
N∗(x,y)
ϕ(t)dt
+Lψ
O(x,y)
ϕ(t)dt
, ()
where M∗(x,y) =max{d(x,y),[d(x,fx),d(y,fy)],[d(x,fy) +d(y,fx)]},N∗(x,y) =max{d(x, y),[d(x,fx),d(y,fy)]},O(x,y) =min{d(x,fx),d(y,fy),d(y,fx),d(x,fy)}and(ϕ,φ,ψ)∈× ×.Then f has a unique fixed point a∈X such thatlimn→∞fnx=a for each x∈X.
If we takeL= in Theorem ., we get the following result.
Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying, for all x,y∈X,
ψ
d(fx,fy)
ϕ(t)dt
≤ψ
M∗(x,y)
ϕ(t)dt
–φ
N∗(x,y)
ϕ(t)dt
, ()
where M∗(x,y) =max{d(x,y),[d(x,fx),d(y,fy)],[d(x,fy) +d(y,fx)]},N(x,y) =max{d(x,y),
[d(x,fx),d(y,fy)]}and(ϕ,φ,ψ)∈××.Then f has a unique fixed point a∈X
such thatlimn→∞fnx=a for each x∈X.
If we takeψ(t) =tin Theorem ., we get the following result.
Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying, for all x,y∈X,
d(fx,fy)
ϕ(t)dt≤
M∗(x,y)
ϕ(t)dt–φ
N∗(x,y)
ϕ(t)dt
, ()
where M∗(x,y) =max{d(x,y),[d(x,fx) +d(y,fy)],[d(x,fy) +d(y,fx)]},N∗(x,y) =max{d(x, y),[d(x,fx) +d(y,fy)]}and(ϕ,φ)∈×.Then f has a unique fixed point a∈X such
thatlimn→∞fnx=a for each x∈X.
Remark . The following theorem is the main result of [] that can be easily deduced by takingα(x,y) = , for allx,y∈X, in Theorem .. Consequently, all corollaries of the main result of [] can be deduced evidently.
Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying, for all x,y∈X,
ψ
d(fx,fy)
ϕ(t)dt
≤ψ
d(x,y)
ϕ(t)dt
–φ d(x,y)
ϕ(t)dt
where (ϕ,φ,ψ)∈ × × . Then f has a unique fixed point a ∈ X such that
limn→∞fnx=a for each x∈X.
If we takeψ(t) =tin Theorem ., we get the following result.
Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying, for all x,y∈X,
d(fx,fy)
ϕ(t)dt≤
d(x,y)
ϕ(t)dt–φ d(x,y)
ϕ(t)dt
, ()
where(ϕ,φ)∈×.Then f has a unique fixed point a∈X such thatlimn→∞fnx=a for each x∈X.
Theorem . Let f be a mapping from a complete metric space(X,d)into itself.If there is k∈[, )satisfying the following condition for all x,y∈X:
d(fx,fy)
ϕ(t)dt≤k
d(x,y)
ϕ(t)dt, ()
then f has a unique fixed point a∈X such thatlimn→∞fnx=a for each x∈X. 4 Consequences in partially ordered metric spaces
Definition . Let (X,) be a partially ordered set andT:X→Xbe a given mapping. We say thatT is nondecreasing with respect toif
x,y∈X, xy ⇒ TxTy.
Definition . Let (X,) be a partially ordered set. A sequence{xn} ⊂Xis said to be nondecreasing with respect toifxnxn+for alln.
Definition . Let (X,) be a partially ordered set anddbe a metric onX. We say that (X,,d) is regular if for every nondecreasing sequence{xn} ⊂Xsuch thatxn→x∈Xas n→ ∞, there exists a subsequence{xn(k)}of{xn}such thatxn(k)xfor allk.
We have the following result.
Corollary . Let(X,)be a partially ordered set and d be a metric on X such that(X,d) is complete.Let f :X→X be a nondecreasing mapping with respect toand satisfy the following inequality:
ψ
d(fx,fy)
ϕ(t)dt
≤ψ
M∗(x,y)
ϕ(t)dt
–φ
N∗(x,y)
ϕ(t)dt
+Lψ
O(x,y)
ϕ(t)dt
(i) there existsx∈Xsuch thatxfx;
(ii) f is continuous or(X,,d)is regular.
Then f has a fixed point.Moreover,if for all x,y∈X there exists z∈X such that xz and yz,we have uniqueness of the fixed point.
Proof Define the mappingα:X×X→[,∞) by
α(x,y) =
ifxyorxy, otherwise.
Clearly,f is anα-admissible contractive inequality of integral typeI. From condition (i), we haveα(x,fx)≥. Moreover, for allx,y∈X, from the monotone property off, we
have
α(x,y)≥ ⇒ xyorxy ⇒ fxfyorfxfy ⇒ α(fx,fy)≥.
Thusf isα-admissible. Now, iff is continuous, the existence of a fixed point follows from Theorem .. Suppose now that (X,,d) is regular. Let{xn}be a sequence inXsuch that
α(xn,xn+)≥ for allnandxn→x∈Xasn→ ∞. From the regularity hypothesis, there exists a subsequence{xn(k)}of{xn}such thatxn(k)xfor allk. This implies from the
defini-tion ofαthatα(xn(k),x)≥ for allk. In this case, the existence of a fixed point follows from
Theorem .. To show the uniqueness, letx,y∈X. By hypothesis, there existsz∈Xsuch thatxzandyz, which implies from the definition ofαthatα(x,z)≥ andα(y,z)≥.
Thus we deduce the uniqueness of the fixed point by Theorem ..
The following result is an immediate consequence of Corollary ..
Corollary . Let(X,)be a partially ordered set and d be a metric on X such that(X,d) is complete.Let T:X→X be a nondecreasing mapping with respect to.Suppose that there exists a functionψ∈such that
d(Tx,Ty)≤ψd(x,y),
for all x,y∈X with xy.Suppose also that the following conditions hold: (i) there existsx∈Xsuch thatxTx;
(ii) T is continuous or(X,,d)is regular.
Then T has a fixed point.Moreover,if for all x,y∈X there exists z∈X such that xz and yz,we have uniqueness of the fixed point.
Remark . Let{Ai}i=be nonempty closed subsets of a complete metric space (X,d) and
T:Y→Ybe a given mapping, whereY=A∪A. If the following condition holds:
T(A)⊆A and T(A)⊆A, ()
thenT is called a cyclic mapping. SinceA andA are closed subsets of the complete
metric space (X,d), then (Y,d) is complete. Define the mappingα:Y×Y→[,∞) by
α(x,y) =
if (x,y)∈(A×A)∪(A×A),
By using the observation above, it is possible to deduce some fixed point results of a cyclic mapping that satisfies,e.g., one of the inequalities between ()-(), and so on. For more details on such approach, we refer,e.g., to [, ].
5 Further results
Theorem . Let(X,d)be a complete metric space,f :X→X andα:X×X→[, +∞) be self-mappings.Suppose that there exist(ϕ,ψ,γ)∈××and L≥such that
α(x,y)ψ
d(fx,fy)
ϕ(t)dt
≤γM(x,y)ψ
M(x,y)
ϕ(t)dt
+Lψ
O(x,y)
ϕ(t)dt
, ()
for all x,y∈X,where
M(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
,
and
O(x,y) =mind(x,fx),d(y,fy),d(y,fx),d(x,fy).
Suppose also that the following conditions hold: (i) f is weak triangularα-admissible; (ii) there existsx∈Xsuch thatxfx; (iii) f is continuous or(X,,d)is regular.
Then f has a unique fixed point a∈X such thatlimn→∞fnx=a for each x∈X.
Proof From (ii), there exits a pointx∈Xsuch thatα(x,fx)≥ (due to the symmetry of the metric, the other case yields the same result). Letx=xand consider an iterative sequence {xn}inXbyxn+=fxnfor alln≥. Note that we have
α(x,x) =α(x,fx)≥ ⇒ α(fx,fx) =α(x,x)≥.
By mathematical induction, we get
α(xn,xn+)≥ for alln∈N. ()
Let us denote
dn=d(xn,xn+) and αn=α(xn,xn+) forn∈N. ()
Now, ifxn=xn+for somen, thenu=xnis a fixed point off. This completes the proof. Consequently, supposexn=xn+for alln∈N, that is,
Now, we proceed to show that{dn}is a non-increasing sequence of real numbers, that is,
dn≤dn–, ∀n∈N. ()
Suppose, on the contrary, that inequality () does not hold. Thus, there exists somen∈
Nsuch that
dn>dn–. ()
From () and (), we get
<
dn–
ϕ(t)dt<
dn
ϕ(t)dt. ()
Regarding again () and () together with the properties ofψ, we conclude that
=ψ() <ψ
dn–
ϕ(t)dt
≤ψ dn
ϕ(t)dt
. ()
Using equations ()-(), () we obtain that
ψ
dn–
ϕ(t)dt
≤αnψ
dn–
ϕ(t)dt
≤αnψ dn
ϕ(t)dt
=αnψ
d(fnx,fn+x)
ϕ(t)dt
≤γMfn–x,fnxψ
M(fn–x,fnx)
ϕ(t)dt
+Lψ
O(fn–x,fnx)
ϕ(t)dt
, ()
where
Mfn–x,fnx≤maxdfn–x,fnx,dfnx,fn+x=max{d
n–,dn},
Ofn–x,fnx=mindfn–x,fnx,dfnx,fn+x,dfnx,fnx, dfn–x,fn+x= .
From (), we haveM(fn–x,fnx)≤d
n. Hence, inequality () implies
ψ
dn–
ϕ(t)dt
≤αnψ
dn–
ϕ(t)dt
≤αnψ dn
ϕ(t)dt
≤γ(dn)ψ dn
ϕ(t)dt
<ψ dn
ϕ(t)dt
, ()
Next we show thatc= , that is,
lim
n→∞dn= . ()
Suppose, on the contrary, thatc> . It follows from () and () that
ψ dn
ϕ(t)dt
=ψ
d(fnx,fn+x)
ϕ(t)dt
≤αnψ
d(fnx,fn+x)
ϕ(t)dt
≤γMfnx,fn–xψ
M(fnx,fn–x)
ϕ(t)dt
+Lψ
O(fn–x,fnx)
ϕ(t)dt
, ()
where
Mfn–,fnx≤maxdfn–x,fnx,dfnx,fn+x=max{dn–,dn},
Ofn–x,fnx=mindfn–x,fnx,dfnx,fn+x,dfnx,fnx,dfn–x,fn+x= .
Hence, inequality () becomes
ψ dn
ϕ(t)dt
=ψ
d(fnx,fn+x)
ϕ(t)dt
≤αnψ
d(fnx,fn+x)
ϕ(t)dt
≤γ(dn–)ψ dn–
ϕ(t)dt
. ()
Taking the upper limit in () and using Lemma ., we get
ψ
c
ϕ(t)dt
= lim
n→∞supψ
dn
ϕ(t)dt
≤ lim
n→∞sup
γ(dn–)ψ dn–
ϕ(t)dt
≤ lim
n→∞sup γ(dn–)
· lim
n→∞sup
ψ
dn–
ϕ(t)dt
<ψ
c
ϕ(t)dt
, ()
which is a contradiction. Hencec= . Next we show that {fnx}
n∈N is a Cauchy sequence. Suppose, on the contrary, that
{fnx}
n∈Nis not a Cauchy sequence. Thus, there is a constant> such that for each pos-itive integerk, there are positive integersm(k) andn(k) withm(k) >n(k) >ksatisfying
For each positive integerk, letm(k) denote the least integer exceedingn(k) and satisfying (). This implies that
dfm(k)x,fn(k)x> and dfm(k)–x,fn(k)x≤ for allk∈N. ()
On the other hand, we have
dfm(k)x,fn(k)x≤dfn(k)x,fm(k)–x+dm(k)–, ∀k∈N, dfm(k)x,fn(k)+x–dfm(k)x,fn(k)x≤dn(k), ∀k∈N, dfm(k)+x,fn(k)+x–dfm(k)x,fn(k)+x≤dm(k), ∀k∈N, dfm(k)+x,fn(k)+x–dfm(k)+x,fn(k)+x≤dn(k)+, ∀k∈N.
()
In view of () and (), we infer that
lim
k→∞d
fn(k)x,fm(k)x=,
lim
k→∞d
fm(k)x,fn(k)+x=,
lim
k→∞d
fm(k)+x,fn(k)+x=,
lim
k→∞d
fm(k)+x,fn(k)+x=.
()
Using the weak triangular alpha admissible property off, we get in view of ()
αfm(k)x,fn(k)+x≥. ()
From () and (), we have, for allk∈N,
ψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤αfm(k)x,fn(k)+xψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤γMfm(k)x,fn(k)+ψ
M(fm(k)x,fn(k)+)
ϕ(t)dt
+Lψ
O(fm(k)x,fn(k)+)
ϕ(t)dt
. ()
Taking the upper limit in () and using () and Lemma ., we get
ψ
ϕ(t)dt
= lim
k→∞supψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤ lim
k→∞supα
fm(k)x,fn(k)+xψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤ lim
k→∞supγ
Mfm(k)x,fn(k)+xψ
M(fm(k)x,fn(k)+x)
ϕ(t)dt
+Llim
k→∞supψ
O(fm(k)x,fn(k)+)
ϕ(t)dt
<ψ
ϕ(t)dt
,
which is impossible. Thus{fnx}
n∈Nis a Cauchy sequence. Now, since (X,d) is complete,
there exists a pointa∈Xsuch thatlimn→∞fnx=a. From the continuity off, it follows thatxn=fxn+→fuasn→+∞. From the uniqueness of limits, we geta=fa, that is,uis
a fixed point off. This completes the proof.
Theorem . Let(X,d)be a complete metric space and f :X→X be a self-mapping. Suppose that there exist(ϕ,φ,ψ)∈××withφ(t)≤ψ(t)for all t∈R+,β∈, α:X×X→[, +∞)and L≥such that
α(x,y)ψ
d(fx,fy)
ϕ(t)dt
≤βM(x,y)φ
M(x,y)
ϕ(t)dt
+Lψ
O(x,y)
ϕ(t)dt
()
for all x,y∈X,where
M(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
,
and
O(x,y) =mind(x,fx),d(y,fy),d(y,fx),d(x,fy).
Suppose also that the following conditions hold: (i) f is weak triangularα-admissible; (ii) there existsx∈Xsuch thatxfx; (iii) f is continuous or(X,,d)is regular.
Then f has a unique fixed point a∈X such thatlimn→∞fnx=a for each x∈X.
Proof From condition (ii), there exists a pointx∈Xsuch thatα(x,fx)≥ (due to the sym-metry of the metric, the other case yields the same result). Letx=xand let the iterative
sequence{xn}inXbe defined byxn+=fxnfor alln≥. Note that we have
α(x,x) =α(x,fx)≥ ⇒ α(fx,fx) =α(x,x)≥.
Using mathematical induction, we obtain
α(xn,xn+)≥ for alln∈N. ()
Set
If for somen, xn =xn+, thenu=xn is a fixed point off. This completes the proof. Suppose thatxn=xn+for alln∈N, that is,
<dn=d(xn,xn+) for alln∈N. ()
Now, we need to show that{dn}is a non-increasing sequence of real numbers, that is,
dn≤dn–, ∀n∈N. ()
Suppose, on the contrary, that inequality () does not hold. Thus, there exists somen∈
Nsuch that
dn>dn–. ()
From () and (), we get
<
dn–
ϕ(t)dt<
dn
ϕ(t)dt. ()
From equations () and () and using the properties ofψ, we get
=ψ() <ψ
dn–
ϕ(t)dt
≤ψ dn
ϕ(t)dt
. ()
In view of equations ()-(), () we infer that
ψ
dn–
ϕ(t)dt
≤αnψ
dn–
ϕ(t)dt
≤αnψ dn
ϕ(t)dt
=αnψ
d(fnx,fn+x)
ϕ(t)dt
≤βMfn–x,fnxφ
M(fn–x,fnx)
ϕ(t)dt
+Lψ
O(fn–x,fnx)
ϕ(t)dt
, ()
where
Mfn–x,fnx≤maxdfn–x,fnx,dfnx,fn+x=max{d
n–,dn},
Ofn–x,fnx=mindfn–x,fnx,dfnx,fn+x,dfnx,fnx, dfn–x,fn+x= .
From (), we haveM(fn–x,fnx)≤d
n. Hence, inequality () implies
ψ
dn–
ϕ(t)dt
≤αnψ
dn–
ϕ(t)dt
≤αnψ dn
ϕ(t)dt
≤β(dn)φ dn
ϕ(t)dt
≤β(dn)ψ dn
ϕ(t)dt
<ψ dn
ϕ(t)dt
, ()
which contradicts inequality (). Hence, () holds. Thus, there exists a constantc≥ such thatlimn→∞dn=c≥.
Next we show thatc= , that is,
lim
n→∞dn= . ()
Suppose, on the contrary, thatc> . It follows from () and () that
ψ dn
ϕ(t)dt
=ψ
d(fnx,fn+x)
ϕ(t)dt
≤αnψ
d(fnx,fn+x)
ϕ(t)dt
≤βMfnx,fn–xφ
M(fnx,fn–x)
ϕ(t)dt
+Lψ
O(fn–x,fnx)
ϕ(t)dt
, ()
where
Mfn–,fnx≤maxdfn–x,fnx,dfnx,fn+x=max{dn–,dn},
Ofn–x,fnx=mindfn–x,fnx,dfnx,fn+x,dfnx,fnx,dfn–x,fn+x= . Hence, inequality () becomes
ψ dn
ϕ(t)dt
=ψ
d(fnx,fn+x)
ϕ(t)dt
≤αnψ
d(fnx,fn+x)
ϕ(t)dt
≤β(dn–)φ dn–
ϕ(t)dt
. ()
Taking the upper limit in () and using Lemma ., we get
ψ
c
ϕ(t)dt
= lim
n→∞supψ
dn
ϕ(t)dt
≤ lim
n→∞sup
β(dn–)φ dn–
ϕ(t)dt
≤ lim
n→∞sup β(dn–)
· lim
n→∞sup
φ
dn–
ϕ(t)dt
≤ lim
n→∞sup β(dn–)
· lim
n→∞sup
ψ
dn–
ϕ(t)dt
<ψ
c
ϕ(t)dt
, ()
Now, we show that {fnx}
n∈N is a Cauchy sequence. Suppose, on the contrary, that {fnx}n∈Nis not a Cauchy sequence. Thus, there is a constant> such that for each pos-itive integerk, there are positive integersm(k) andn(k) withm(k) >n(k) >ksatisfying
dfm(k)x,fn(k)x>. ()
For each positive integerk, letm(k) denote the least integer exceedingn(k) and satisfying (). This implies that
dfm(k)x,fn(k)x> and dfm(k)–x,fn(k)x≤ for allk∈N. ()
On the other hand, we have
dfm(k)x,fn(k)x≤dfn(k)x,fm(k)–x+dm(k)–, ∀k∈N, dfm(k)x,fn(k)+x–dfm(k)x,fn(k)x≤dn(k), ∀k∈N, dfm(k)+x,fn(k)+x–dfm(k)x,fn(k)+x≤dm(k), ∀k∈N, dfm(k)+x,fn(k)+x–dfm(k)+x,fn(k)+x≤dn(k)+, ∀k∈N.
()
In view of () and (), we infer that
lim
k→∞d
fn(k)x,fm(k)x=,
lim
k→∞d
fm(k)x,fn(k)+x=,
lim
k→∞d
fm(k)+x,fn(k)+x=,
lim
k→∞d
fm(k)+x,fn(k)+x=.
()
Using the weak triangular alpha admissible property off, we get in view of ()
αfm(k)x,fn(k)+x≥. ()
From () and (), we have, for allk∈N,
ψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤αfm(k)x,fn(k)+xψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤βMfm(k)x,fn(k)+φ
M(fm(k)x,fn(k)+)
ϕ(t)dt
+Lψ
O(fm(k)x,fn(k)+)
ϕ(t)dt
Taking the upper limit in () and using () and Lemma ., we get
ψ
ϕ(t)dt
= lim
k→∞supψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤ lim
k→∞supα
fm(k)x,fn(k)+xψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤ lim
k→∞supβ
Mfm(k)x,fn(k)+xφ
M(fm(k)x,fn(k)+x)
ϕ(t)dt
+L lim
k→∞supψ
O(fm(k)x,fn(k)+)
ϕ(t)dt
≤ lim
k→∞supβ
Mfm(k)x,fn(k)+xψ
M(fm(k)x,fn(k)+x)
ϕ(t)dt
+L lim
k→∞supψ
O(fm(k)x,fn(k)+)
ϕ(t)dt
<ψ
ϕ(t)dt
,
which is impossible. Thus{fnx}n∈Nis a Cauchy sequence. Now, since (X,d) is complete, there exists a pointa∈Xsuch thatlimn→∞fnx=a. From the continuity off, it follows thatxn=fxn+→fuasn→+∞. From the uniqueness of limits, we geta=fa, that is,uis
a fixed point off. This completes the proof.
6 Conclusion
In this paper, we handle contractive mappings of integral type in a more general frame via
α-admissible mappings. More precisely, we examine the contractive mapping of integral type given in [] by usingα-admissible mappings. Very recently, some new contractive mappings of integral type were introduced in [] and []. We assert that our techniques are also valid to extent the results of [] and [] in the frame ofα-admissible mappings.
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
1Nonlinear Analysis and Applied Mathematics Research Group (NAAM), King Abdulaziz University, Jeddah, Saudi Arabia. 2Department of Mathematics, Atilim University, Incek, Ankara, 06836, Turkey.3School of Mathematics, Statistics and
Applied Mathematics, National University of Ireland, Galway, Ireland.4School of Mathematics and Computer
Applications, Thapar University, Patiala, Punjab 147004, India.
Acknowledgements
This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, under Grant No. (55-130-35-HiCi). The authors, therefore, acknowledge technical and financial support of KAU.
Received: 30 May 2014 Accepted: 25 September 2014 Published:14 Oct 2014
References
1. Branciari, A: A fixed point theorem for mappings satisfying a general contractive condition of integral type. Int. J. Math. Math. Sci.29(9), 531-536 (2002)
