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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

1

and 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)dtc d(x,y)

ϕ(t)dt, ∀x,yX,

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 aX such thatlimn→∞fnx=a for each xX.

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)dtc

max{d(x,y),d(x,fx),d(y,fy),[d(x,fy)+d(y,fx)]}

ϕ(t)dt, ∀x,yX,

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where c∈(, )is a constant andϕ.Then f has a unique fixed point aX such that

limn→∞fnx=a for each xX.

Theorem . ([]) Let f be a mapping from a complete metric space(X,d)into itself satisfying

d(fx,fy)

ϕ(t)dtc

max{d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)}

ϕ(t)dt, ∀x,yX,

where c∈(, )is a constant andϕ.Assume that f has a bounded orbit at some point xX.Then f has a unique fixed point aX 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)dtαd(x,y)

d(x,y)

ϕ(t)dt, ∀x,yX,

whereϕandα:R+[, )is a function with lim sup

st

α(s) < , ∀t> .

Then f has a unique fixed point aX such thatlimn→∞fnx=a for each xX.

Theorem . ([]) Let f be a mapping from a complete metric space(X,d)into itself satisfying

d(fx,fy)

ϕ(t)dtαd(x,y)

d(x,fx)

ϕ(t)dt+βd(x,y)

d(y,fy)

ϕ(t)dt, ∀x,yX,

whereϕandα,β:R+[, )are two functions with

α(t) +β(t) < , ∀t∈R+, lim sup s→+ β(s) < ,

lim sup st+

α(s)

 –β(s)< , ∀t> . Then f has a unique fixed point aX such thatlimn→∞fnx=a for each xX.

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 :XX,AX and (x,y,n)∈X×N, put

Of(x,n) ={fix: ≤in},Of(x) ={fix:∀i∈N},

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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(s) < ,∀t∈R+},

={α:α:R+→[, )is a function with lim sups(s) < ,∀t> },

={α:α:R+→[, )is a function such thatsup{α(s) :∀sB}< for each

nonempty bounded subsetBinR+}.

The following lemma plays an important role in this paper.

Lemma .([]) Letϕand{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)dtαd(x,y)

m(x,y)

ϕ(t)dt, ∀x,yX, (.)

where(ϕ,α)∈×.Then f has a unique fixed point aX such thatlimn→∞fnx=a for

each xX.

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

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Suppose that (.) does not hold. That is, there exists somen∈Nsatisfying

dn>dn–. (.)

Sinceϕandα(R+)[, ), it follows from (.) and (.) that

 < dn

ϕ(t)dtα(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 sc

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)xε, ∀k∈N. (.) Note that

dfm(k)x,fn(k)xdfn(k)x,fm(k)–x+dm(k)–, ∀k∈N;

dfm(k)x,fn(k)+xdfm(k)x,fn(k)xdn(k), ∀k∈N;

d

fm(k)+x,fn(k)+xdfm(k)x,fn(k)+xdm(k), ∀k∈N; (.)

dfm(k)+x,fn(k)+xdfm(k)+x,fn(k)+xdn(k)+, ∀k∈N;

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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 sε

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 pointaXsuch 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)

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≤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 pointbX\ {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+)[, ) andϕthat

 < d(a,b)

ϕ(t)dt= d(fa,fb)

ϕ(t)dtαd(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)dtαd(x,y)

m(x,y)

ϕ(t)dt, ∀x,yX, (.)

where(ϕ,α)∈×.Assume that f has a bounded orbit at some point uX.Then f has

a unique fixed point aX such thatlimn→∞fnu=a.

Proof Without loss of generality we assume thatu=fu. Now we prove that for eachn∈Nthere existsk∈Nsuch thatknandδOf(u,n)

=du,fku. (.)

Let n∈N. It is clear that there existi,j∈Nsuch that ≤i<jnandδ(Of(u,n)) =

d(fiu,fju). Suppose thatδ(O

f(u,n)) =d(fiu,fju) for somei,j∈Nwith  <i<jn. 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

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Putr=δ(Of(u)) andB= [,r]. Clearly  <r< +∞. Observe that (ϕ,α)∈×ensures

that

lim p→∞

supα(s) :sBn(p) r

ϕ(t)dt= ,

which implies that there exists somep∈Nwith

supα(s) :sBn(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) :sBn(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 existsaXsatisfyinglimn→∞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

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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 pointwX\ {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)dtαd(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)dtαm(x,y)

m(x,y)

ϕ(t)dt, ∀x,yX, (.)

where(ϕ,α)∈×.Then f has a unique fixed point aX such thatlimn→∞fnx=a for

each xX.

Theorem . Let f be a mapping from a complete metric space(X,d)into itself satisfying d(fx,fy)

ϕ(t)dtαm(x,y)

m(x,y)

ϕ(t)dt, ∀x,yX, (.)

where(ϕ,α)∈×.Assume that f has a bounded orbit at some point uX.Then f has

a unique fixed point aX such thatlimn→∞fnu=a. 4 Remarks and illustrative examples

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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 :XX,

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,yXwith 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{xy, , , } =  =d(x,fx)

and

d(fx,fy)

ϕ(t)dt= (xy)≤ <

 ≤

  +xy=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

xy,

x,  y,

 (xy)

=xy=d(x,y)

and

d(fx,fy)

ϕ(t)dt=

x

 –

y

=(xy)

 ≤

(xy)

 +xy

=αd(x,y)

d(x,y)

ϕ(t)dt

=αd(x,y)

m(x,y)

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Case .xy<x≤. It follows that

m(x,y) =max

d(x,y),d(x,fx),d(y,fy), 

d(x,fy) +d(y,fx)

=max

xy,

x,  y,

 

xy

 +yx

= 

x=d(x,fx)

and

d(fx,fy)

ϕ(t)dt=

x

–

y

=(xy)

≤ 

 +xy·  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

xy, , y,

 

 + y

=  =d(x,fx)

and

d(fx,fy)

ϕ(t)dt=x–  –y  

≤ 

< 

 ≤

  +xy

=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

xy, , y,

 

 + y

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and

d(fx,fy)

ϕ(t)dt=x–  –y  

≤ 

< 

≤

  +xy

=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

xy, , y,

 

x– y– 

=xy=d(x,y)

and

d(fx,fy)

ϕ(t)dt=x–  –y  

≤ 

< 

≤

  +xy

(xy)  +xy

=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 eachxX. 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)dtc d(,)

ϕ(t)dt

=c – 

= c

,

which means that

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Suppose thatf satisfies the conditions of Theorem ., that is, there existsα

satis-fying

=

–

 =

d(f,f)

ϕ(t)dtα

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:XX,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 eachuX. Letx,yXwith

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{xy, , ,xy+ ,yx+ } =xy+  =d(x,fy)

and

d(fx,fy)

ϕ(t)dt=|x–  –y+ |

= (xy)≤(xy)

(xy+ )

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=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)

=maxxy, ,y– ,x– ,y– (x– )

=x–  =d(x,fy)

and

d(fx,fy)

ϕ(t)dt=|x–  – |

= (x– )≤(xy)

(x– )

(xy+ )

=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

xy,x–

, ,xy+ ,y–  

=xy+  =d(x,fy)

and

d(fx,fy)

ϕ(t)dt=

– (y– ) 

=

  –y

≤(xy)(xy+ )

(xy+ )

=αd(x,y)

d(x,fy)

ϕ(t)dt

=αd(x,y)

m(x,y)

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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

xy,x–

,y– ,x– ,  –y

=x–  =d(x,fy)

and

d(fx,fy)

ϕ(t)dt= – 

= 

≤

(x– )(xy) ( +xy)

=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 eachuX. 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)dtα

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 :XX,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,yXwithy<x. In order to verify (.), we have to

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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

xy, x

 +x, y  +y,

 

xy

 +y+yx  +x

= x

 +x=d(x,fx)

and

d(fx,fy)

ϕ(t)dt=

x  +x

y  +y

= (xy)

( +x)( +y) ≤

 (xy)

≤ 



x

 +x

α

x

 +x

x

 +x

=αm(x,y)

m(x,y)

ϕ(t)dt.

Case .xy<x< . It follows that

m(x,y) =max

d(x,y),d(x,fx),d(y,fy), 

d(x,fy) +d(y,fx)

=max

xy,x

, y ,   xy

+yx

= x

=d(x,fx)

and

d(fx,fy)

ϕ(t)dt= x – y   = (xy)

 ≤   x   =α xx  

=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

xy,x

, y ,   xy

+

x –y

(16)

and

d(fx,fy)

ϕ(t)dt= x – y   = (xy)

α(xy)(xy) =αm(x,y)

m(x,y)

ϕ(t)dt.

Case . x

+xy<  andx≥. Notice that

m(x,y) =max

d(x,y),d(x,fx),d(y,fy), 

d(x,fy) +d(y,fx)

=max

xy, x

 +x, y ,

 

xy

+yx  +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

xy, x

 +x, y ,

 

xy

+

x  +xy

=xy=d(x,y)

and

d(fx,fy)

ϕ(t)dt=

x  +x

y   ≤ x – y   = (xy)

α(xy)(xy)

=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 eachxX.

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Example . LetX=R+be endowed with the Euclidean metricd=| · |,f :XX,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 eachuX. Letx,yXwithy<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

xy,x

, y ,x

y ,

yx

=xy

=d(x,fy)

and

d(fx,fy)

ϕ(t)dt=x

–

y  

=(xy)

 ≤

(xy)

=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

xy, ,y ,x

y

,|yx+ | =  =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)

(18)

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

xy, ,y ,x

y

,|yx+ | =xy

=d(x,fy)

and

d(fx,fy)

ϕ(t)dt=x–  –y

≤(x–  –

y )

(xy) ·

xy

 

=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)

=maxxy, , ,xy+ ,|yx+ | =xy+  =d(x,fy)

and

d(fx,fy)

ϕ(t)dt= (xy)≤(xy+  – )

(xy+ ) ·(xy+ ) 

=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 eachuX. 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

(xy)= d(fx,fy)

ϕ(t)dtc m(x,y)

ϕ(t)dt

=c(xy+ ), ∀x,y∈(, +∞) withy<x, which yields

 = lim

xy→+∞

(xy)

(xy+ ) ≤c< ,

(19)

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 supst+

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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