R E S E A R C H
Open Access
Existence and approximations of fixed points
for contractive mappings of integral type
Zeqing Liu
1, Xiaozhu Li
1and Shin Min Kang
2**Correspondence: [email protected]
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
The existence, uniqueness, and iterative approximations of fixed points for four classes of contractive mappings of integral type in complete metric spaces are established. The results presented in this paper generalize indeed several results of Branciari (J. Math. Math. Sci. 29(9):531-536, 2002), Rhoades (Int. J. Math. Math. Sci. 2003(63):4007-4013, 2003) and Liuet al.(Fixed Point Theory Appl. 2011:64, 2011). Four illustrative examples with uncountably many points are also included.
MSC: 54H25
Keywords: contractive mappings of integral type; fixed point; complete metric space
1 Introduction
Over the past decade the researchers [–] introduced a lot of contractive mappings of integral type and discussed the existence of fixed points and common fixed points for these mappings in metric spaces and modular spaces, respectively. Branciari [] was the first to study the existence of fixed points for the contractive mapping of integral type and proved the following result, which extends the Banach fixed point theorem.
Theorem .([]) Let f be a mapping from a complete metric space(X,d)into itself sat-isfying
d(fx,fy)
ϕ(t)dt≤c d(x,y)
ϕ(t)dt, ∀x,y∈X,
where c∈ (, ) is a constant and ϕ ∈ = {ϕ :ϕ : R+ →R+ is Lebesgue integrable,
summable on each compact subset ofR+andεϕ(t)dt> for eachε> }. Then f has a unique fixed point a∈X such thatlimn→∞fnx=a for each x∈X.
Rhoades [] and Liuet al.[] extended the result of Branciari and proved 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≤c
max{d(x,y),d(x,fx),d(y,fy),[d(x,fy)+d(y,fx)]}
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.
Theorem . ([]) Let f be a mapping from a complete metric space(X,d)into itself satisfying
d(fx,fy)
ϕ(t)dt≤c
max{d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)}
ϕ(t)dt, ∀x,y∈X,
where c∈(, )is a constant andϕ∈.Assume that f has a bounded orbit at some point x∈X.Then f has a unique fixed point a∈X such thatlimn→∞fnx=a.
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.
The purposes of this paper are both to study the existence, uniqueness, and iterative approximations of fixed points for four new classes of contractive mappings of integral type, which include the contractive mappings of integral type in [, , ] as special cases, and to construct four examples with uncountably many points to illustrate that the results obtained properly generalize Theorems .-. or are different from these theorems.
2 Preliminaries
Throughout this paper, we assume thatR= (–∞, +∞),R+= [, +∞),N
={} ∪N, where
Ndenotes the set of all positive integers. Let (X,d) be a metric space. Forf :X→X,A⊂X and (x,y,n)∈X×N, put
Of(x,n) ={fix: ≤i≤n},Of(x) ={fix:∀i∈N},
m(x,y) =max{d(x,y),d(x,fx),d(y,fy),[d(x,fy) +d(y,fx)]},
m(x,y) =max{d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)}.
TheOf(x) andOf(x,n) are called the orbit andnth orbit off atx, respectively.
Let
={α:α:R+→[, )is a function with lim sups→tα(s) < ,∀t∈R+},
={α:α:R+→[, )is a function with lim sups→tα(s) < ,∀t> },
={α:α:R+→[, )is a function such thatsup{α(s) :∀s∈B}< for each
nonempty bounded subsetBinR+}.
The following lemma plays an important role 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.
3 Four fixed point theorems
In this section we show the existence, uniqueness and iterative approximations of fixed points for four classes of contractive mappings of integral type.
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)
m(x,y)
ϕ(t)dt, ∀x,y∈X, (.)
where(ϕ,α)∈×.Then f has a unique fixed point a∈X such thatlimn→∞fnx=a for
each x∈X.
Proof Letxbe an arbitrary point inX. Note that
m
fn–x,fnx=max
dfn–x,fnx,dfn–x,fnx,dfnx,fn+x,
dfn–x,fn+x+dfnx,fnx
=maxdfn–x,fnx,dfnx,fn+x
=max{dn–,dn}, ∀n∈N. (.)
It follows from (.) and (.) that dn
ϕ(t)dt =
d(fnx,fn+x)
ϕ(t)dt
≤αdfn–x,fnx
m(fn–x,fnx)
ϕ(t)dt
≤α(dn–)
max{dn–,dn}
ϕ(t)dt, ∀n∈N. (.)
Now we prove that
Suppose that (.) does not hold. That is, there exists somen∈Nsatisfying
dn>dn–. (.)
Sinceϕ∈andα(R+)⊆[, ), it follows from (.) and (.) that
< dn
ϕ(t)dt≤α(dn–)
max{dn–,dn}
ϕ(t)dt< dn
ϕ(t)dt,
which is a contradiction and hence (.) holds. Clearly, (.) implies that there exists a constantcwithlimn→∞dn=c≥.
Next we prove thatc= . Otherwise c> . Taking the upper limit in (.) and using
Lemma . andϕ∈, we conclude that
< c
ϕ(t)dt=lim sup n→∞
dn
ϕ(t)dt
≤lim sup n→∞
α(dn–)
max{dn–,dn}
ϕ(t)dt
≤lim sup n→∞
α(dn–)·lim sup n→∞
dn–
ϕ(t)dt
≤lim sup s→c
α(s)
c
ϕ(t)dt< c
ϕ(t)dt,
which is absurd. Therefore,c= , that is,
lim
n→∞dn= . (.)
Now we claim that{fnx}
n∈Nis a Cauchy sequence. Suppose that{fnx}n∈Nis 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) >ksuch that
dfm(k)x,fn(k)x>ε.
For each positive integerk, letm(k) denote the least integer exceedingn(k) and satisfying the above inequality. It follows that
dfm(k)x,fn(k)x>ε and dfm(k)–x,fn(k)x≤ε, ∀k∈N. (.) 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;
d
fm(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;
Making use of (.)-(.), we obtain
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
= lim
k→∞d
fm(k)x,fn(k)+x. (.)
It follows from (.) and (.) that
m
fm(k)x,fn(k)+x
=max
dfm(k)x,fn(k)+x,dfm(k)x,fm(k)+x,dfn(k)+x,fn(k)+x,
dfm(k)x,fn(k)+x+dfn(k)+x,fm(k)+x
→max{ε, , ,ε}=ε ask→ ∞,
which together with (.), Lemma ., and (ϕ,α)∈×gives
< ε
ϕ(t)dt=lim sup k→∞
d(fm(k)+x,fn(k)+x)
ϕ(t)dt
≤lim sup k→∞
αdfm(k)x,fn(k)+x
m(fm(k)x,fn(k)+x)
ϕ(t)dt
≤lim sup k→∞ α
dfm(k)x,fn(k)+x·lim sup k→∞
m(fm(k)x,fn(k)+x)
ϕ(t)dt
≤lim sup s→ε
α(s)
ε
ϕ(t)dt< ε
ϕ(t)dt,
which is a contradiction. Thus{fnx}
n∈Nis a Cauchy sequence. Since (X,d) is a complete metric space, it follows that there exists a pointa∈Xsuch thatlimn→∞fnx=a. Suppose
thatfa=a. It is clear that (.) implies that
m
fnx,a = max
dfnx,a,dfnx,fn+x,d(a,fa),
dfnx,fa+da,fn+x
→d(a,fa) asn→ ∞,
which together with (.), Lemma ., and (ϕ,α)∈×yields
< d(a,fa)
ϕ(t)dt=lim sup n→∞
d(fn+x,fa)
ϕ(t)dt
≤lim sup n→∞
αdfnx,a
m(fnx,a)
≤lim sup n→∞
αdfnx,a·lim sup n→∞
m(fnx,a)
ϕ(t)dt
≤lim sup s→
α(s)
d(a,fa)
ϕ(t)dt< d(a,fa)
ϕ(t)dt asn→ ∞,
which is a contradiction. That is,a=fa.
Finally, we prove thatais a unique fixed point offinX. Suppose thatf has another fixed pointb∈X\ {a}. Note that
m(a,b) =max
d(a,b),d(a,fa),d(b,fb),
d(a,fb) +d(b,fa) =d(a,b).
It follows from (.),α(R+)⊆[, ) andϕ∈that
< d(a,b)
ϕ(t)dt= d(fa,fb)
ϕ(t)dt≤αd(a,b)
m(a,b)
ϕ(t)dt< 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 d(fx,fy)
ϕ(t)dt≤αd(x,y)
m(x,y)
ϕ(t)dt, ∀x,y∈X, (.)
where(ϕ,α)∈×.Assume that f has a bounded orbit at some point u∈X.Then f has
a unique fixed point a∈X such thatlimn→∞fnu=a.
Proof Without loss of generality we assume thatu=fu. Now we prove that for eachn∈Nthere existsk∈Nsuch thatk≤nandδOf(u,n)
=du,fku. (.)
Let n∈N. It is clear that there existi,j∈Nsuch that ≤i<j≤nandδ(Of(u,n)) =
d(fiu,fju). Suppose thatδ(O
f(u,n)) =d(fiu,fju) for somei,j∈Nwith <i<j≤n. In light
of (.) and (ϕ,α)∈×, we infer that
<
δ(Of(u,n))
ϕ(t)dt=
d(fiu,fju)
ϕ(t)dt
≤αdfi–u,fj–u
m(fi–u,fj–u)
ϕ(t)dt
≤αdfi–u,fj–u
δ(Of(u,n))
ϕ(t)dt<
δ(Of(u,n))
ϕ(t)dt,
which is a contradiction. Thus (.) holds.
Next we prove thatOf(u) is a Cauchy sequence. Suppose thatOf(u) is not a Cauchy
se-quence. It follows that there exist anε> and two strictly increasing sequences{m(p)}p∈N and{n(p)}p∈Nwithm(p) >n(p) >pfor eachp∈Nsatisfying
Putr=δ(Of(u)) andB= [,r]. Clearly <r< +∞. Observe that (ϕ,α)∈×ensures
that
lim p→∞
supα(s) :s∈Bn(p) r
ϕ(t)dt= ,
which implies that there exists somep∈Nwith
supα(s) :s∈Bn(p) r
ϕ(t)dt< ε
ϕ(t)dt. (.)
Using (.)-(.) and (ϕ,α)∈×, we know that there exist <k≤m(p) –n(p) + ,
<k≤m(p) –n(p) + , . . . , and <kn(p)–≤m(p) – satisfying
ε
ϕ(t)dt≤
d(fn(p)u,fm(p)u)
ϕ(t)dt
≤αdfn(p)–u,fm(p)–u
m(fn(p)–u,fm(p)–u)
ϕ(t)dt
≤αdfn(p)–u,fm(p)–u
δ(Of(fn(p)–u,m(p)–n(p)+))
ϕ(t)dt
=αdfn(p)–u,fm(p)–u
d(fn(p)–u,fk+n(p)–u)
ϕ(t)dt
≤αdfn(p)–u,fm(p)–uαdfn(p)–u,fk+n(p)–u
×
δ(Of(fn(p)–u,k+))
ϕ(t)dt
=αdfn(p)–u,fm(p)–uαdfn(p)–u,fk+n(p)–u
×
d(fn(p)–u,fk+n(p)–u)
ϕ(t)dt
≤ · · ·
≤αdfn(p)–u,fm(p)–uαdfn(p)–u,fk+n(p)–u· · ·αdu,fkn(p)–u
×
δ(Of(u,m(p)))
ϕ(t)dt
≤supα(s) :s∈Bn(p) r
ϕ(t)dt
< ε
ϕ(t)dt,
which is impossible. Thus{fnu}
n∈Nis a Cauchy sequence. Since (X,d) is complete, it
fol-lows that there existsa∈Xsatisfyinglimn→∞fnu=a. Suppose thatd(a,fa) > . Note that lim
n→∞m
fnu,a
= lim
n→∞max
dfnu,a,dfnu,fn+u,d(a,fa),dfnu,fa,da,fn+u
Taking the upper limit in (.) and using (.), Lemma ., and (ϕ,α)∈×, we
conclude that
< d(a,fa)
ϕ(t)dt=lim sup n→∞
d(fn+u,fa)
ϕ(t)dt
≤lim sup n→∞
αdfnu,a
m(fnu,a)
ϕ(t)dt
≤lim sup n→∞
αdfnu,a·lim sup n→∞
m(fnu,a)
ϕ(t)dt
≤supα(s) :s∈[, ]
d(a,fa)
ϕ(t)dt< d(a,fa)
ϕ(t)dt,
which is absurd. Therefore,d(a,fa) = , that is,a=fa. Suppose thatf has another fixed pointw∈X\ {a}. Since
m(a,w) =max
d(a,w),d(a,fa),d(w,fw),d(a,fw),d(w,fa)=d(a,w),
it follows from (.) that
< d(a,w)
ϕ(t)dt≤αd(a,w)
m(a,w)
ϕ(t)dt
=αd(a,w)
d(a,w)
ϕ(t)dt< d(a,w)
ϕ(t)dt,
which is a contradiction. That is, f has a unique fixed point inX. This completes the
proof.
As in the arguments of Theorems . and ., we conclude similarly the following results and omit their proofs.
Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying d(fx,fy)
ϕ(t)dt≤αm(x,y)
m(x,y)
ϕ(t)dt, ∀x,y∈X, (.)
where(ϕ,α)∈×.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≤αm(x,y)
m(x,y)
ϕ(t)dt, ∀x,y∈X, (.)
where(ϕ,α)∈×.Assume that f has a bounded orbit at some point u∈X.Then f has
a unique fixed point a∈X such thatlimn→∞fnu=a. 4 Remarks and illustrative examples
Remark . Theorem . generalizes Theorem ., which, in turns, extends Theorem .. The following example proves that Theorem . both extends substantially Theorem . and is different from Theorem ..
Example . LetX= [,]⊂Rbe endowed with the Euclidean metricd=| · |,f :X→X,
ϕ:R+→R+andα:R+→[, ) be defined by
f(x) = ⎧ ⎨ ⎩
x
, ∀x∈[, ],
x– , ∀x∈(,],
ϕ(t) = t, ∀t∈R+ and α(t) = ⎧ ⎨ ⎩
, t= ,
+t, ∀t∈(, +∞).
Obviously, (ϕ,α)∈×. Letx,y∈Xwith y<x. In order to verify (.), we have to
consider six possible cases as follows: Case . <y<x≤. It is clear that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
=max{x–y, , , } = =d(x,fx)
and
d(fx,fy)
ϕ(t)dt= (x–y)≤ <
≤
+x–y=α
d(x,y)
ϕ(t)dt
=αd(x,y)
m(x,y)
ϕ(t)dt.
Case . ≤y<x andx≤. Note that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
=max
x–y,
x, y,
(x–y)
=x–y=d(x,y)
and
d(fx,fy)
ϕ(t)dt=
x
–
y
=(x–y)
≤
(x–y)
+x–y
=αd(x,y)
d(x,y)
ϕ(t)dt
=αd(x,y)
m(x,y)
Case .x≤y<x≤. It follows that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
=max
x–y,
x, y,
x–y
+y– x
=
x=d(x,fx)
and
d(fx,fy)
ϕ(t)dt=
x
–
y
=(x–y)
≤
+x–y· x
=αd(x,y)
d(x,fx)
ϕ(t)dt
=αd(x,y)
m(x,y)
ϕ(t)dt.
Case . <y≤ <x≤. Notice that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
=max
x–y, , y,
+ y
= =d(x,fx)
and
d(fx,fy)
ϕ(t)dt=x– –y
≤
<
≤
+x–y
=αd(x,y)
d(x,fx)
ϕ(t)dt
=αd(x,y)
m(x,y)
ϕ(t)dt.
Case .x– ≤y≤ and <x≤. It is easy to see that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
=max
x–y, , y,
+ y
and
d(fx,fy)
ϕ(t)dt=x– –y
≤
<
≤
+x–y
=αd(x,y)
d(x,fx)
ϕ(t)dt
=αd(x,y)
m(x,y)
ϕ(t)dt.
Case . ≤y<x– and <x≤. It is easy to verify that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
=max
x–y, , y,
x– y–
=x–y=d(x,y)
and
d(fx,fy)
ϕ(t)dt=x– –y
≤
<
≤
+x–y≤
(x–y) +x–y
=αd(x,y)
d(x,y)
ϕ(t)dt
=αd(x,y)
m(x,y)
ϕ(t)dt.
That is, (.) holds. It follows from Theorem . thatf has a unique fixed point ∈Xand
limn→∞fnx= for eachx∈X. But we invoke neither Theorem . nor Theorem . to show thatf possesses a fixed point inX.
Suppose thatf satisfies the conditions of Theorem ., that is, there existsc∈(, ) sat-isfying
=
– – =
d(f,f)
ϕ(t)dt≤c d(,)
ϕ(t)dt
=c –
= c
,
which means that
Suppose thatf satisfies the conditions of Theorem ., that is, there existsα∈
satis-fying
=
–
=
d(f,f)
ϕ(t)dt≤α
d
,
d(,)
ϕ(t)dt
=α
–
=
α
,
which implies that
≤α
< ,
which is a contradiction.
Remark . Theorem . is a generalization of Theorem .. The below example demon-strates that Theorem . is different from Theorem ..
Example . LetX= [,]∪[,]∪[, +∞)⊂Rbe endowed with the Euclidean metric d=| · |,f:X→X,ϕ:R+→R+andα:R+→[, ) be defined by
f(x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
, ∀x∈[,], x– , ∀x∈[,],
, ∀x∈[, +∞),
ϕ(t) = t, ∀t∈R+ and α(t) = ⎧ ⎨ ⎩
, t= , t
(+t), ∀t∈(, +∞).
It is easy to see that (ϕ,α)∈×andOf(u) is bounded for eachu∈X. Letx,y∈Xwith
y<x. In order to verify (.), we have to consider six possible cases as follows: Case . ≤y<x≤. It is clear that
d(fx,fy)
ϕ(t)dt= ≤αd(x,y)
m(x,y)
ϕ(t)dt.
Case . ≤y<x≤. Note that m(x,y) =max
d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)
=max{x–y, , ,x–y+ ,y–x+ } =x–y+ =d(x,fy)
and
d(fx,fy)
ϕ(t)dt=|x– –y+ |
= (x–y)≤(x–y)
(x–y+ )
=αd(x,y)
d(x,fy)
ϕ(t)dt
=αd(x,y)
m(x,y)
ϕ(t)dt.
Case . ≤y≤ and ≤x≤. It follows that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)
=maxx–y, ,y– ,x– ,y– (x– )
=x– =d(x,fy)
and
d(fx,fy)
ϕ(t)dt=|x– – |
= (x– )≤(x–y)
(x– )
(x–y+ )
=αd(x,y)
d(x,fy)
ϕ(t)dt
=αd(x,y)
m(x,y)
ϕ(t)dt.
Case . ≤y<x< +∞. It is easy to see that d(fx,fy)
ϕ(t)dt= ≤αd(x,y)
m(x,y)
ϕ(t)dt.
Case . ≤y≤
andx≥. It follows that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)
=max
x–y,x–
, ,x–y+ ,y–
=x–y+ =d(x,fy)
and
d(fx,fy)
ϕ(t)dt=
– (y– )
=
–y
≤(x–y)(x–y+ )
(x–y+ )
=αd(x,y)
d(x,fy)
ϕ(t)dt
=αd(x,y)
m(x,y)
Case . ≤y≤ andx≥. It is clear that m(x,y) =max
d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)
=max
x–y,x–
,y– ,x– , –y
=x– =d(x,fy)
and
d(fx,fy)
ϕ(t)dt= –
=
≤
(x– )(x–y) ( +x–y)
=αd(x,y)
d(x,fy)
ϕ(t)dt
=αd(x,y)
m(x,y)
ϕ(t)dt.
That is, the conditions of Theorem . are fulfilled. It follows from Theorem . thatf has a unique fixed point ∈Xandlimn→∞fnu= for eachu∈X. However, Theorem . is useless in guaranteeing the existence of a fixed point off inX. Suppose thatf satisfies the conditions of Theorem ., that is, there existsα∈satisfying
=
–
=
d(f,f)
ϕ(t)dt≤α
d
,
d(,)
ϕ(t)dt
=α
–
=
α
,
which yields
≤α
< ,
which is impossible.
Remark . Theorem . extends Theorems . and .. The example below is an appli-cation of Theorem ..
Example . LetX=R+be endowed with the Euclidean metricd=| · |,f :X→X,ϕ:
R+→R+andα:R+→[, ) be defined by
f(x) = ⎧ ⎨ ⎩
x
, ∀x∈[, ), x
+x, ∀x∈[, +∞),
ϕ(t) = t, ∀t∈R+ and α(t) = ⎧ ⎨ ⎩
, ∀t∈[, ), t
+t, ∀t∈[ , +∞).
Obviously, (ϕ,α)∈×. Letx,y∈Xwithy<x. In order to verify (.), we have to
Case . ≤y<x< +∞. It follows that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
=max
x–y, x
+x, y +y,
x– y
+y+y– x +x
= x
+x=d(x,fx)
and
d(fx,fy)
ϕ(t)dt=
x +x–
y +y
= (x–y)
( +x)( +y) ≤
(x–y)
≤
x
+x
≤α
x
+x
x
+x
=αm(x,y)
m(x,y)
ϕ(t)dt.
Case .x≤y<x< . It follows that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
=max
x–y,x
, y , x–y
+y– x
= x
=d(x,fx)
and
d(fx,fy)
ϕ(t)dt= x – y = (x–y)
≤ x =α x x
=αm(x,y)
m(x,y)
ϕ(t)dt.
Case . ≤y<x andx< . It is clear that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
=max
x–y,x
, y , x–y
+
x –y
and
d(fx,fy)
ϕ(t)dt= x – y = (x–y)
≤α(x–y)(x–y) =αm(x,y)
m(x,y)
ϕ(t)dt.
Case . x
+x≤y< andx≥. Notice that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
=max
x–y, x
+x, y ,
x–y
+y– x +x
= x
+x=d(x,fx)
and
d(fx,fy)
ϕ(t)dt=
x +x–
y ≤ x +x–
·
x +x
=
x +x
≤α
x +x
x +x
=αm(x,y)
m(x,y)
ϕ(t)dt.
Case . ≤y<+xxandx≥. It is clear that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),
d(x,fy) +d(y,fx)
=max
x–y, x
+x, y ,
x–y
+
x +x–y
=x–y=d(x,y)
and
d(fx,fy)
ϕ(t)dt=
x +x–
y ≤ x – y = (x–y)
≤α(x–y)(x–y)
=αm(x,y)
m(x,y)
ϕ(t)dt.
That is, the conditions of Theorem . are fulfilled. It follows from Theorem . thatf has a unique fixed point ∈Xandlimn→∞fnx= for eachx∈X.
Example . LetX=R+be endowed with the Euclidean metricd=| · |,f :X→X,ϕ:
R+→R+andα:R+→[, ) be defined by
f(x) = ⎧ ⎨ ⎩
x
, ∀x∈[, ],
x– , ∀x∈(, +∞),
ϕ(t) = t, ∀t∈R+ and α(t) = ⎧ ⎨ ⎩
, ∀t∈[, ], (t–)
t , ∀t∈(, +∞).
It is clear that (ϕ,α)∈×andOf(u) is bounded for eachu∈X. Letx,y∈Xwithy<x.
In order to verify (.), we have to consider four possible cases as follows: Case . ≤y<x≤. Note that
m(x,y) =max
d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)
=max
x–y,x
, y ,x–
y ,
y–x
=x–y
=d(x,fy)
and
d(fx,fy)
ϕ(t)dt=x
–
y
=(x–y)
≤
(x–y)
=αd(x,fy)
d(x,fy)
ϕ(t)dt
=αm(x,y)
m(x,y)
ϕ(t)dt.
Case . ≤y≤ <x≤ +y. Clearly m(x,y) =max
d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)
=max
x–y, ,y ,x–
y
,|y–x+ | = =d(x,fx)
and
d(fx,fy)
ϕ(t)dt=x– –y
≤
–y
≤
=αd(x,fx)
d(x,fx)
ϕ(t)dt
=αm(x,y)
m(x,y)
Case . ≤y≤ and +y<x< +∞. Obviously m(x,y) =max
d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)
=max
x–y, ,y ,x–
y
,|y–x+ | =x–y
=d(x,fy)
and
d(fx,fy)
ϕ(t)dt=x– –y
≤(x– –
y )
(x–y) ·
x–y
=αd(x,fy)
d(x,fy)
ϕ(t)dt
=αm(x,y)
m(x,y)
ϕ(t)dt.
Case . <y<x< +∞. It follows that m(x,y) =max
d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)
=maxx–y, , ,x–y+ ,|y–x+ | =x–y+ =d(x,fy)
and
d(fx,fy)
ϕ(t)dt= (x–y)≤(x–y+ – )
(x–y+ ) ·(x–y+ )
=αd(x,fy)
d(x,fy)
ϕ(t)dt
=αm(x,y)
m(x,y)
ϕ(t)dt.
That is, the conditions of Theorem . are fulfilled. It follows from Theorem . thatf has a unique fixed point ∈Xandlimn→∞fnu= for eachu∈X. But we do not invoke
Theorems . and . to show the existence of a fixed point off inX.
Suppose thatfsatisfies the conditions of Theorem ., that is, there exists somec∈(, ) satisfying
(x–y)= d(fx,fy)
ϕ(t)dt≤c m(x,y)
ϕ(t)dt
=c(x–y+ ), ∀x,y∈(, +∞) withy<x, which yields
= lim
x–y→+∞
(x–y)
(x–y+ ) ≤c< ,
Suppose thatf satisfies the conditions of Theorem ., that is, there existα,β:R+→
[, ) satisfying
α(t) +β(t) < , ∀t∈R+, lim sup
s→+ β(s) < , lim sups→t+
α(s)
–β(s)< , ∀t> and
= ( – )=
d(f,f)
ϕ(t)dt
≤αd(, )
d(,f)
ϕ(t)dt+βd(, )
d(,f)
ϕ(t)dt
=α()· +β()· =α() +β(),
which means that
≤α() +β() < ,
which is absurd.
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
The authors would like to thank the referees for useful comments and suggestions. This research was supported by the Science Research Foundation of Educational Department of Liaoning Province (L2012380) and the fund of the Research Promotion Program, Gyeongsang National University, 2013 (RPP-2013-023).
Received: 15 January 2014 Accepted: 4 June 2014 Published: 18 July 2014 References
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