R E S E A R C H
Open Access
Fixed point theorems of contractive
mappings of integral type
Zeqing Liu
1, Jinglei Li
1and Shin Min Kang
2**Correspondence:
2Department of Mathematics and
RINS, Gyeongsang National University, Jinju, 660-701, Korea Full list of author information is available at the end of the article
Abstract
Three fixed point theorems for three general classes of contractive mappings of integral type in complete metric spaces are proved. Three examples are included.
MSC: 54H25
Keywords: contractive mappings of integral type; fixed point theorems; complete metric space
1 Introduction
Branciari [] was the first to study the existence of fixed points for the contractive mapping of integral type. He established a nice integral version of the Banach contraction principle and proved the following fixed point theorem.
Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying
d(fx,fy)
ϕ(t)dt≤c d(x,y)
ϕ(t)dt, ∀x,y∈X,
where c∈(, )is a constant andϕ∈.Then f has a unique fixed point a∈X such that limn→∞fnx=a for each x∈X.
Afterwards, many authors continued the study of Branciari and obtained many fixed point theorems for several classes of contractive mappings of integral type; see,e.g., [– ] and the references therein. In particular, in , Liuet al.[] extended the result of Branciari [] and deduced the following fixed point theorems.
Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying
d(fx,fy)
ϕ(t)dt≤αd(x,y)
d(x,y)
ϕ(t)dt, ∀x,y∈X,
whereϕ∈andα:R+→[, )is a function with
lim sup s→t
α(s) < , ∀t> .
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 satisfying
d(fx,fy)
ϕ(t)dt≤αd(x,y)
d(x,fx)
ϕ(t)dt+βd(x,y)
d(y,fy)
ϕ(t)dt, ∀x,y∈X,
whereϕ∈andα,β:R+→[, )are two functions with
α(t) +β(t) < , ∀t∈R+, lim sup s→+
β(s) < , lim sup s→t+
α(s)
–β(s)< , ∀t> .
Then f has a unique fixed point a∈X such thatlimn→∞fnx=a for each x∈X.
In , Dutta and Choudhuty [] proved the following result.
Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying
ψd(fx,fy)≤ψd(x,y)–ϕd(x,y), ∀x,y∈X,
whereψ,ϕ:R+→R+are both continuous and monotone nondecreasing functions with
ψ(t) =ϕ(t) = if and only if t = .Then f has a unique fixed point a∈X such that
limn→∞fnx=a for each x∈X.
However, to the best of our knowledge, no one studied the following contractive map-pings of integral type:
ψ
d(fx,fy)
ϕ(t)dt
≤ψ
d(x,y)
ϕ(t)dt
–φ d(x,y)
ϕ(t)dt
, ∀x,y∈X, (.)
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)∈××;
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)∈.
2 Preliminaries
Throughout this paper, we assume thatR+= [, +∞),N
=N∪ {},Ndenotes the set of
all positive integers, (X,d) is a metric space,f :X→Xis a self-mapping and
dn=d
fnx,fn+x, ∀(n,x)∈N×X,
={ϕ:ϕ:R+→R+is Lebesgue integrable, summable on each compact subset of R+andε
ϕ(t)dt> for eachε> };
={ϕ:ϕ:R+→R+satisfies thatlim 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+→[, )satisfies thatlim 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> }.
The following lemmas play important roles in this paper.
Lemma .([]) Letϕ∈ and{rn}n∈N be a nonnegative sequence withlimn→∞rn=a.
Then
lim n→∞
rn
ϕ(t)dt= a
ϕ(t)dt.
Lemma .([]) Letϕ∈and{rn}n∈Nbe a nonnegative sequence.Then
lim n→∞
rn
ϕ(t)dt=
if and only iflimn→∞rn= .
Lemma . Letϕ∈.Thenϕ(t) > if and only if t> .
Proof Lett> . Putan=tfor eachn∈N. It is easy to see thatt=lim infn→∞an> , which
together withϕ∈ensures that
ϕ(t) =lim inf
n→∞ ϕ(an) > .
Conversely, suppose thatϕ(t) > for somet∈R+. Setan=tfor eachn∈N. It is clear
thatϕ(t) =lim infn→∞ϕ(an) > , which together withϕ∈guarantees that
t=lim inf n→∞ an> .
This completes the proof.
3 Main results
Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying (.).Then f has a unique fixed point a∈X such thatlimn→∞fnx=a for each x∈X.
Proof Letxbe an arbitrary point inX. Firstly, we show that
dn≤dn–, ∀n∈N. (.)
Suppose that (.) does not hold. It follows that there exists somen∈Nsatisfying
dn>dn–. (.)
Note that (.) andϕ∈imply that
dn
ϕ(t)dt> . (.)
Using (.), (.) and (ϕ,φ,ψ)∈××, we conclude immediately that
ψ
dn–
ϕ(t)dt
≤ψ
dn
ϕ(t)dt
=ψ
d(fnx,fn+x)
ϕ(t)dt
≤ψ
d(fn–x,fnx)
ϕ(t)dt
–φ
d(fn–x,fnx)
ϕ(t)dt
=ψ
dn–
ϕ(t)dt
–φ dn–
ϕ(t)dt
≤ψ
dn–
ϕ(t)dt
,
which yields that
ψ dn
ϕ(t)dt
=ψ
dn–
ϕ(t)dt
(.)
and
φ
dn–
ϕ(t)dt
= . (.)
Combining (.) and Lemma ., we get that
dn–
ϕ(t)dt= ,
which together withψ∈and (.) means that
ψ dn
ϕ(t)dt
=ψ
dn–
ϕ(t)dt
that is,
dn
ϕ(t)dt= ,
which contradicts (.). Hence (.) holds. Secondly, we show that
lim
n→∞dn= . (.)
In view of (.), we deduce that the nonnegative sequence{dn}n∈Nis nonincreasing, which
means that there exists a constantcwithlimn→∞dn=c≥. Suppose thatc> . It follows
from (.) that
ψ dn
ϕ(t)dt
=ψ
d(fnx,fn+x)
ϕ(t)dt
≤ψ
d(fnx,fn–x)
ϕ(t)dt
–φ
d(fnx,fn–x)
ϕ(t)dt
=ψ
dn–
ϕ(t)dt
–φ dn–
ϕ(t)dt
, ∀n∈N. (.)
Taking upper limit in (.) and using Lemma . and (ϕ,φ,ψ)∈××, we conclude
that
ψ c
ϕ(t)dt
=lim sup n→∞ ψ
dn
ϕ(t)dt
≤lim sup n→∞
ψ
dn–
ϕ(t)dt
–φ dn–
ϕ(t)dt
≤lim sup n→∞ ψ
dn–
ϕ(t)dt
–lim inf n→∞ φ
dn–
ϕ(t)dt
=ψ
c
ϕ(t)dt
–lim inf n→∞ φ
dn–
ϕ(t)dt
<ψ c
ϕ(t)dt
,
which is a contradiction. Hencec= . Thirdly, we show that{fnx}
n∈N is a Cauchy sequence. Suppose that {fnx}n∈N is not a Cauchy sequence, which means that there is a constantε> such that for each positive integerk, there are positive integersm(k) andn(k) withm(k) >n(k) >ksatisfying
dfm(k)x,fn(k)x>ε. (.)
For each positive integerk, letm(k) denote the least integer exceedingn(k) and satisfying (.). It follows that
Note that
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 light of (.) and (.), we get that
ε= 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. (.)
In view of (.), we deduce that
ψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤ψ
d(fm(k)x,fn(k)+x)
ϕ(t)dt
–φ
d(fm(k)x,fn(k)+x)
ϕ(t)dt
, ∀k∈N. (.)
Taking upper limit in (.) and using (.), (ϕ,φ,ψ)∈××and Lemma ., we
deduce that
ψ ε
ϕ(t)dt
=lim sup k→∞
ψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤lim sup k→∞
ψ
d(fm(k)x,fn(k)+x)
ϕ(t)dt
–φ
d(fm(k)x,fn(k)+x)
ϕ(t)dt
≤lim sup k→∞
ψ
d(fm(k)x,fn(k)+x)
ϕ(t)dt
–lim inf k→∞ φ
d(fm(k)x,fn(k)+x)
ϕ(t)dt
=ψ
ε
ϕ(t)dt
–lim inf k→∞ φ
d(fm(k)x,fn(k)+x)
ϕ(t)dt
<ψ ε
ϕ(t)dt
,
which is impossible. Thus{fnx}
n∈Nis a Cauchy sequence.
Since (X,d) is complete, it follows that there exists a point a ∈ X satisfying
limn→∞fnx=a. By virtue of (.), we infer that
ψ
d(fn+x,fa)
ϕ(t)dt
≤ψ
d(fnx,a)
ϕ(t)dt
–φ
d(fnx,a)
ϕ(t)dt
which together with (ϕ,φ,ψ)∈××and Lemmas . and . gives that
ψ
d(a,fa)
ϕ(t)dt
=lim sup n→∞ ψ
d(fn+x,fa)
ϕ(t)dt
≤lim sup n→∞
ψ
d(fnx,a)
ϕ(t)dt
–φ
d(fnx,a)
ϕ(t)dt
≤lim sup n→∞
ψ
d(fnx,a)
ϕ(t)dt
–lim inf n→∞ φ
d(fnx,a)
ϕ(t)dt
=ψ() –
= ,
which together withψ∈yields that
d(a,fa)
ϕ(t)dt= ,
that is,a=fa.
Finally, we show thatais a unique fixed point off inX. Suppose thatf has another fixed pointb∈X\ {a}. It follows from (.) and (ϕ,φ,ψ)∈××that
ψ
d(a,b)
ϕ(t)dt
=ψ
d(fa,fb)
ϕ(t)dt
≤ψ
d(a,b)
ϕ(t)dt
–φ d(a,b)
ϕ(t)dt
<ψ
d(a,b)
ϕ(t)dt
,
which is a contradiction. This completes the proof.
Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying (.).Then f has a unique fixed point a∈X such thatlimn→∞fnx=a for each x∈X.
Proof Letxbe an arbitrary point inX. Suppose that (.) holds for somen∈N. Using
(.), (.) and (ϕ,ψ,α)∈××, we get that
ψ dn
ϕ(t)dt
>
and
ψ
dn–
ϕ(t)dt
≤ψ
dn
ϕ(t)dt
=ψ
d(fnx,fn+x)
ϕ(t)dt
≤αdfn–x,fnxψ
d(fn–x,fnx)
ϕ(t)dt
=α(dn–)ψ
dn–
ϕ(t)dt
<ψ
dn–
ϕ(t)dt
which is a contradiction, and hence (.) does not hold. Consequently, (.) is true. Notice that the nonnegative sequence{dn}n∈N is nonincreasing, which implies that there exists
a constantc≥ withlimn→∞dn=c. Suppose thatc> . In light of (.), we infer that
ψ dn
ϕ(t)dt
=ψ
d(fnx,fn+x)
ϕ(t)dt
≤αdfn–x,fnxψ
d(fn–x,fnx)
ϕ(t)dt
=α(dn–)ψ
dn–
ϕ(t)dt
, ∀n∈N. (.)
Taking upper limit in (.) and using Lemma . and (ϕ,ψ,α)∈××, we know
that
ψ c
ϕ(t)dt
=lim sup n→∞
ψ dn
ϕ(t)dt
≤lim sup n→∞
α(dn–)ψ
dn–
ϕ(t)dt
≤lim sup n→∞
α(dn–)·lim sup n→∞
ψ dn–
ϕ(t)dt
<ψ c
ϕ(t)dt
,
which is a contradiction, and hencec= , that is, (.) holds. Now we show that {fnx}
n∈N is a Cauchy sequence. Suppose that {fnx}n∈N is not a Cauchy sequence. As in the proof of Theorem ., we conclude that there exist ε> and{m(k),n(k) :k∈N} ⊆Nwithm(k) >n(k) >k for eachk∈Nsatisfying (.)-(.). By means of (.), (.), Lemma . and (ϕ,ψ,α)∈××, we get that
ψ ε
ϕ(t)dt
=lim sup k→∞ ψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
=lim sup k→∞
αdfm(k)x,fn(k)+xψ
d(fm(k)x,fn(k)+x)
ϕ(t)dt
≤lim sup k→∞
αdfm(k)x,fn(k)+x·lim sup k→∞
ψ
d(fm(k)x,fn(k)+x)
ϕ(t)dt
<ψ ε
ϕ(t)dt
,
which is a contradiction. Hence{fnx}
n∈Nis a Cauchy sequence.
It follows from completeness of (X,d) that there existsa∈Xwithlimn→∞fnx=a. In
view of (.), we have
ψ
d(fn+x,fa)
ϕ(t)dt
≤αdfnx,aψ
d(fnx,a)
ϕ(t)dt
Taking upper limit in (.) and making use of (ϕ,ψ,α)∈××and Lemmas .
and ., we get that
ψ
d(a,fa)
ϕ(t)dt
=lim sup n→∞ ψ
d(fn+x,fa)
ϕ(t)dt
≤lim sup n→∞
αdfnx,aψ
d(fnx,a)
ϕ(t)dt
≤lim sup n→∞
αdfnx,a·lim sup n→∞
ψ
d(fnx,a)
ϕ(t)dt
= ,
which means that
ψ
d(a,fa)
ϕ(t)dt
= ,
that is,fa=a.
Next we prove thatais a unique fixed point off inX. Suppose thatf has another fixed pointb∈X\ {a}. It follows from (.) and (ϕ,ψ,α)∈××that
ψ
d(a,b)
ϕ(t)dt
=ψ
d(fa,fb)
ϕ(t)dt
≤αd(a,b)ψ
d(a,b)
ϕ(t)dt
<ψ
d(a,b)
ϕ(t)dt
,
which is a contradiction. This completes the proof.
Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying (.)and
φ(t)≤ψ(t), ∀t∈R+. (.)
Then f has a unique fixed point a∈X such thatlimn→∞fnx=a for each x∈X.
Proof Letxbe an arbitrary point inX. If there existsn∈Nsatisfyingdn= , it is clear
thatfnxis a fixed point off andlim
n→∞fnx=fnx. Now we assume thatdn= for all
n∈N. Suppose that (.) holds for somen∈N. It follows from (.) that
ψ dn
ϕ(t)dt
=ψ
d(fnx,fn+x)
ϕ(t)dt
≤αdfn–x,fnxφ
d(fn–x,fnx)
ϕ(t)dt
+βdfn–x,fnxψ
d(fnx,fn+x)
ϕ(t)dt
=α(dn–)φ
dn–
ϕ(t)dt
+β(dn–)ψ
dn
ϕ(t)dt
which together with (.), (.), (ϕ,ψ,φ)∈××and (α,β)∈implies that
<ψ
dn–
ϕ(t)dt
≤ψ
dn
ϕ(t)dt
≤ α(dn–)
–β(dn–)
φ
dn–
ϕ(t)dt
≤ α(dn–)
–β(dn–)
ψ
dn–
ϕ(t)dt
<ψ
dn–
ϕ(t)dt
,
which is a contradiction, and hence (.) does not hold. Consequently, (.) holds. Next we show thatlimn→∞dn= . Note that the nonnegative sequence{dn}n∈Nis non-increasing, which implies that there exists a constantc≥ withlimn→∞dn=c. Suppose
thatc> . It follows from (.) that
ψ dn
ϕ(t)dt
=ψ
d(fnx,fn+x)
ϕ(t)dt
≤αdfn–x,fnxφ
d(fn–x,fnx)
ϕ(t)dt
+βdfn–x,fnxψ
d(fnx,fn+x)
ϕ(t)dt
=α(dn–)φ
dn–
ϕ(t)dt
+β(dn–)ψ
dn
ϕ(t)dt
, ∀n∈N,
which means that
ψ dn
ϕ(t)dt
≤ α(dn–)
–β(dn–)
φ dn–
ϕ(t)dt
, ∀n∈N. (.)
Taking upper limit in (.) and using (.), (ϕ,ψ,φ)∈××, (α,β)∈and
Lemma ., we arrive at
ψ c
ϕ(t)dt
=lim sup n→∞
ψ dn
ϕ(t)dt
≤lim sup n→∞
α(dn–)
–β(dn–)
φ dn–
ϕ(t)dt
≤lim sup n→∞
α(dn–)
–β(dn–)· lim sup
n→∞
ψ dn–
ϕ(t)dt
≤lim sup s→c+
α(s) –β(s)·ψ
c
ϕ(t)dt
<ψ c
ϕ(t)dt
,
Next we show that {fnx}
n∈N is a Cauchy sequence. Suppose that {fnx}n∈N is not a Cauchy sequence. As in the proof of Theorem ., we conclude that there exist ε> and{m(k),n(k) :k∈N} ⊆Nwithm(k) >n(k) >k for eachk∈Nsatisfying (.)-(.). By means of (.), we deduce that
ψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤αdfm(k)x,fn(k)+xφ
d(fm(k)x,fm(k)+x)
ϕ(t)dt
+βdfm(k)x,fn(k)+xψ
d(fn(k)+x,fn(k)+x)
ϕ(t)dt
=αdfm(k)x,fn(k)+xφ dm(k)
ϕ(t)dt
+βdfm(k)x,fn(k)+xψ dn(k)
ϕ(t)dt
, ∀k∈N. (.)
Taking upper limit in (.) and making use of (.), (.), Lemma ., (ϕ,ψ,φ)∈×
×and (α,β)∈, we deduce that
<ψ ε
ϕ(t)dt
=lim sup k→∞
ψ
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤lim sup k→∞
αdfm(k)x,fn(k)+xφ dm(k)
ϕ(t)dt
+βdfm(k)x,fn(k)+xψ dn(k)
ϕ(t)dt
≤lim sup k→∞
αdfm(k)x,fn(k)+x·lim sup k→∞
ψ dm(k)
ϕ(t)dt
+lim sup k→∞
βdfm(k)x,fn(k)+x·lim sup k→∞
ψ dn(k)
ϕ(t)dt
≤lim sup s→ε
α(s)·ψ
ϕ(t)dt
+lim sup s→ε
β(s)·ψ
ϕ(t)dt
= ,
which is a contradiction. Hence{fnx}
n∈Nis a Cauchy sequence.
Completeness of (X,d) implies that there exists a pointa∈Xsuch thatlimn→∞fnx=a.
In view of (.), (ϕ,ψ,φ)∈××, (α,β)∈and Lemma ., we infer that
ψ
d(a,fa)
ϕ(t)dt
=lim sup n→∞ ψ
d(fn+x,fa)
ϕ(t)dt
≤lim sup n→∞
αdfnx,aφ
d(fnx,fn+x)
ϕ(t)dt
+βdfnx,aψ
d(a,fa)
≤lim sup n→∞ α
dfnx,a·lim sup n→∞ ψ
dn
ϕ(t)dt
+lim sup n→∞
βdfnx,a·ψ
d(a,fa)
ϕ(t)dt
≤lim sup s→+ β(s)·ψ
d(a,fa)
ϕ(t)dt
,
which together with (α,β)∈yields that
ψ
d(a,fa)
ϕ(t)dt
= ,
which gives thatd(fa,a) = , that is,fa=a.
Finally, we prove thatais a unique fixed point offinX. Suppose thatfhas another fixed pointb∈X\ {a}. It follows from (.) and (ϕ,ψ,φ)∈××and (α,β)∈that
≤ψ
d(fa,fb)
ϕ(t)dt
≤αd(a,b)φ
d(a,fa)
ϕ(t)dt
+βd(a,b)ψ
d(b,fb)
ϕ(t)dt
= ,
which is a contradiction. This completes the proof.
4 Three examples
Now we construct three examples to explain Theorems .-..
Example . LetX= [,
]∪ {} ∪ {}be endowed with the Euclidean metric d=| · |.
Assume thatf :X→Xandϕ,φ,ψ:R+→R+are defined by
f(x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
x
, ∀x∈[, ],
, x= ,
, x= ,
ϕ(t) = ⎧ ⎨ ⎩
t
, ∀t∈[, ],
, ∀t∈(, +∞),
φ(t) = ⎧ ⎨ ⎩
t
, ∀t∈[, ], t
, ∀t∈(, +∞),
ψ(t) = ⎧ ⎨ ⎩
t, ∀t∈[, ],
t+
, ∀t∈(, +∞).
Clearly, (X,d) is a complete metric and (ϕ,φ,ψ)∈××. Letx,y∈Xwithx<y.
In order to verify (.), we have to consider the following four cases. Case . Letx,y∈[,]. Note that
ψ
d(fx,fy)
ϕ(t)dt
=ψ
|x–y|
ϕ(t)dt
=ψ
|x–y|
=|x–y|
≤|x–y|
–
|x–y|
=ψ
| x–y|
–φ |
x–y|
=ψ |x–y|
ϕ(t)dt
–φ |x–y|
ϕ(t)dt
=ψ
d(x,y)
ϕ(t)dt
–φ d(x,y)
ϕ(t)dt
.
Case . Letx∈[,] andy= . It follows that
ψ
d(fx,fy)
ϕ(t)dt
=ψ
x
ϕ(t)dt =ψ x =x ≤ ( –x)
–
( –x)
=ψ
( –x)
–φ
( –x)
=ψ
|x–|
ϕ(t)dt
–φ |x–|
ϕ(t)dt
=ψ
d(x,y)
ϕ(t)dt
–φ d(x,y)
ϕ(t)dt
.
Case . Letx∈[,
] andy= . It follows that
ψ
d(fx,fy)
ϕ(t)dt
=ψ
–x
ϕ(t)dt
=ψ
( –x)
=( –x)
< ≤ –x
+ –
–x
=ψ
–x
–φ
–x
=ψ
ϕ(t)dt+ –x
ϕ(t)dt
–φ
ϕ(t)dt+ –x
ϕ(t)dt
=ψ
–x
ϕ(t)dt
–φ –x
ϕ(t)dt
=ψ
d(x,y)
ϕ(t)dt
–φ d(x,y)
ϕ(t)dt
.
Case . Letx= andy= . Note that
ψ
d(fx,fy)
ϕ(t)dt
=ψ
ϕ(t)dt =ψ = < = + – · =ψ
–φ =ψ
ϕ(t)dt+
ϕ(t)dt
–φ
ϕ(t)dt+
ϕ(t)dt
=ψ
ϕ(t)dt
–φ
ϕ(t)dt
=ψ
d(x,y)
ϕ(t)dt
–φ d(x,y)
ϕ(t)dt
.
Example . LetX= [, ]∪[, ] be endowed with the Euclidean metricd=| · |. Assume thatf :X→Xandϕ,ψ:R+→R+andα:R+→[, ) are defined by
f(x) = ⎧ ⎨ ⎩
x
, ∀x∈[, ], x
, ∀x∈[, ],
ϕ(t) = ⎧ ⎨ ⎩
t, ∀t∈[, ],
t, ∀t∈[, ],
ψ(t) =t, ∀t∈R+, α(t) =
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
+
t
, ∀t∈[, ],
t, ∀t∈(, ),
√
t, ∀t∈(, +∞).
Obviously, (ϕ,ψ,α)∈××. Putx,y∈Xwithx<y. In order to verify (.), we
have to consider three possible cases as follows. Case . Letx,y∈[, ]. It is clear that
ψ
d(fx,fy)
ϕ(t)dt
=
y–x
tdt
=(x+y)
|x–y|
≤
|x–y|
≤
+
|x–y|
|x–y|
=αd(x,y)ψ d(x,y)
ϕ(t)dt
.
Case . Letx,y∈[, ]. It follows that
ψ
d(fx,fy)
ϕ(t)dt
=
y–x
tdt
=
x+y
|x–y|≤ |x–y|
≤
+
|x–y|
|x–y|
=αd(x,y)ψ d(x,y)
ϕ(t)dt
.
Case . Letx∈[, ] andy∈[, ]. It follows that
ψ
d(fx,fy)
ϕ(t)dt
= y
–x
tdt
=
y
–
x
≤
< <|x–y|
=α|x–y||x–y|=α|x–y|
tdt+ |x–y|
t dt
=αd(x,y)
|x–y|
ϕ(t)dt
=αd(x,y)ψ d(x,y)
ϕ(t)dt
.
That is, (.) holds. Consequently, the conditions of Theorem . are satisfied. It follows from Theorem . thatf has a unique fixed point ∈Xsuch thatlimn→∞fnx= for each
Example . LetX= [, ]∪[, ] be endowed with the Euclidean metricd=|·|. Assume thatf :X→X,ϕ,φ,ψ:R+→R+andα,β:R+→[, ) are defined by
f(x) = ⎧ ⎨ ⎩
, ∀x∈[ , ], x
, ∀x∈[ , ],
φ(t) = ⎧ ⎨ ⎩
, t∈[,
) t
, t∈[ , +∞),
ϕ(t) = t, ψ(t) = t, α(t) = t
(+t), β(t) =
t
(+t), ∀t∈R +.
It is easy to see that (ϕ,ψ,φ)∈××, (α,β)∈and (.) holds. In order to
verify (.), we have to consider the five possible cases below. Case . Letx,y∈[, ] withx≥y. Note that
ψ
d(fx,fy)
ϕ(t)dt
=ψ
|x–y|
ϕ(t)dt
=ψ
(x–y)
=(x–y)
≤
x–y
≤x–y
·
x (+x–y) ≤
x–y (+x–y) ·
x
=αd(x,y)φ
d(x,fx)
ϕ(t)dt
≤αd(x,y)φ
d(x,fx)
ϕ(t)dt
d+βd(x,y)ψ
d(y,fy)
ϕ(t)dt
.
Case . Letx,y∈[, ] withy>x. Note that
ψ
d(fx,fy)
ϕ(t)dt
=ψ
|y–x|
ϕ(t)dt
=(y–x)
≤
(y–x)
·
y (+y–x)
= (y–x)
(+y–x)·
y
=β
d(x,y)ψ
y
=βd(x,y)ψ y
ϕ(t)dt
=βd(x,y)ψ
d(y,fy)
ϕ(t)dt
≤αd(x,y)φ
d(x,fx)
ϕ(t)dt
+βd(x,y)ψ
d(y,fy)
ϕ(t)dt
.
Case . Letx∈[, ] andy∈[, ]. It follows that
ψ
d(fx,fy)
ϕ(t)dt
=ψ
x–
ϕ(t)dt
=ψ
(x– )
=(x– )
≤ < = · · ≤
x–y (+x–y) ·
·x
=αd(x,y)φ
x
=αd(x,y)φ
d(x,fx)
ϕ(t)dt
≤αd(x,y)φ
d(x,fx)
ϕ(t)dt
+βd(x,y)ψ
d(y,fy)
ϕ(t)dt
Case . Letx∈[, ] andy∈[, ]. Note that
ψ
d(fx,fy)
ϕ(t)dt
=ψ
|y–|
t dt
=(y– )
≤
<
·
≤ (y–x) (+y–x) ·
y
=β
d(x,y)ψ
y
=βd(x,y)ψ
d(y,fy)
ϕ(t)dt
≤αd(x,y)φ
d(x,fx)
ϕ(t)dt
+βd(x,y)ψ
d(y,fy)
ϕ(t)dt
.
Case . Letx,y∈[, ]. Notice thatfx=fy= . It follows that
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
.
That is, (.) holds. Thus all the conditions of Theorem . are satisfied. It follows from Theorem . thatf has a unique fixed point ∈Xsuch thatlimn→∞fnx= for eachx∈X.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Liaoning Normal University, Dalian, Liaoning 116029, People’s Republic of China. 2Department of Mathematics and RINS, Gyeongsang National University, Jinju, 660-701, Korea.
Acknowledgements
This research was supported by the Science Research Foundation of Educational Department of Liaoning Province (L2012380).
Received: 4 July 2013 Accepted: 22 October 2013 Published:19 Nov 2013
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