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Abstract

In this paper, an integral type of Suzuki-type mappings is investigated for generalizing the Banach contraction theorem on a metric space. As an application, the existence of a continuous solution for an integral equation is obtained.
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Adv. Fixed Point Theory, 5 (2015), No. 1, 101-109
ISSN: 1927-6303
SOME FIXED POINT RESULTS OF INTEGRAL TYPE AND APPLICATIONS
SEYED M. A. ALEOMRANINEJAD1,, MASOUMEH SHOKOUHNIA2
1Department of Mathematics, Qom University of Technology, Qom, Iran
2Department of Mathematics, Alzahra University, Tehran, Iran
Copyright c
2015 Aleomraninejad and Shokouhnia. This is an open access article distributed under the Creative Commons Attribution
License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract. In this paper, an integral type of Suzuki-type mappings is investigated for generalizing the Banach
contraction theorem on a metric space. As an application, the existence of a continuous solution for an integral
equation is obtained.
Keywords: Suzuki type mapping; fixed point; integral equation; integral type mapping.
2010 AMS Subject Classification: 47H10.
1. Introduction and preliminaries
The most important result on fixed points for contractive-type mappings is the well-known
Banach contraction theorem, which was established in 1922; see [1] and the references therein.
Theorem 1.1. Let (X,d)be a complete metric space, β(0,1)and let T :XX be a mapping
such that for each x,yX,
d(T x,Ty)βd(x,y).
Corresponding author
E-mail addresses: aleomran63@yahoo.com (S.M.A. Aleomraninejad), shokoohniam63@yahoo.com (M. Shok-
ouhnia)
Received October 9, 2014
101
102 SEYED M. A. ALEOMRANINEJAD, MASOUMEH SHOKOUHNIA
Then T has a unique fixed point a X such that for each x X, limnTnx=a.
Since 1922, many authors have introduced various types of contraction inequalities to gen-
eralized the well-known Banach contraction theorem. In 2002 Branciari proved the following
result; see [2].
Theorem 1.2. Let (X,d)be a complete metric space, β(0,1)and T :XX a mapping
such that for each x,yX,
Zd(T x,Ty)
0
f(t)dt βZd(x,y)
0
f(t)dt,
where f :[0,)(0,)is a Lebesgue integrable mapping which is summable (i.e., with finite
integral on each compact subset of [0,)) and for each ε>0,Rε
0f(t)dt >0.Then T has a
unique fixed point a X such that for each x X, limnTnx=a.
In 2008, Suzuki [3] introduced a new method on the problem and the method was further
extended by some authors; see, for example, [4-7]. The following result was proved in [8].
Theorem 1.3. Let (X,d)be a complete metric space and let T :XX be a mapping. Suppose
that there exist α(0,1
2],β(0,1)such that αd(x,T x)d(x,y)implies d(T x,Ty)βd(x,y)
for all x,yX. Then T has a unique fixed point a X such that for each x X, limnTnx=a.
The aim of this paper is to provide a new condition for Twhich guarantees the existence of
its fixed point based on Suzuki and Branciari’s idea. In order to obtain our main results, we
need the following lemmas.
Lemma 1.4. Let a,b[0,)and f :[0,)(0,)a Lebesgue integrable mapping which is
summable and for each ε>0,Rε
0f(t)dt >0. Then
i) a =0whenever Ra
0f(t)dt =0,
ii) a <b whenever Ra
0f(t)dt <Rb
0f(t)dt.
Lemma 1.5. Let L >0,α(x),β(x)C([a,b]) and f :[0,)(0,)a Lebesgue integrable
mapping which is summable and for each ε>0,Rε
0f(t)dt >0. Then Rkαk
0f(t)dt <LRkβk
0f(t)dt
whenever R|α(x)|
0f(t)dt <LR|β(x)|
0f(t)dt.
2. Main results
SOME FIXED POINT RESULTS OF INTEGRAL TYPE AND APPLICATIONS 103
The following theorem is the main result of this paper.
Theorem 2.1 Let (X,d)be a complete metric space and T :XX a mapping. Suppose that
there exist α(0,1
2],β(0,1)such that αd(x,T x)d(x,y)implies
Zd(T x,Ty)
0
f(t)dt βZd(x,y)
0
f(t)dt
for all x,yX and f :[0,)(0,)is a Lebesgue integrable mapping which is summable and
for each ε>0,Rε
0f(t)dt >0.Then T has a unique fixed point a X such that for each x X,
limnTnx=a.
Proof. Fix arbitrary 1 >r>β,x0Xand x1=T x0. We have αd(x0,T x0)<d(x0,x1). Hence,
Zd(T x0,T x1)
0
f(t)dt βZd(x0,x1)
0
f(t)dt <rZd(x0,x1)
0
f(t)dt.
Since r<1, we have Rd(x1,T x1)
0f(t)dt <Rd(x0,x1)
0f(t)dt. Let x2=T x1. By lemma 1.4, we have
d(x1,T x1)<d(x0,x1). Therefore, we find αd(x1,T x1)<d(x1,x2)and
Zd(T x1,T x2)
0
f(t)dt βZd(x1,x2)
0
f(t)dt <rZd(x1,x2)
0
f(t)dt <r2Zd(x0,x1)
0
f(t)dt.
Now let x3=T x2. By lemma 1.4, d(x2,x3)<d(x1,x2)<d(x0,x1). Since αd(x2,T x2)<
d(x2,x3), we have
Zd(T x2,T x3)
0
f(t)dt βZd(x2,x3)
0
f(t)dt <rZd(x2,x3)
0
f(t)dt <r3Zd(x0,x1)
0
f(t)dt.
By continuing this process, we obtain a sequence {xn}n1in Xsuch that xn+1=T xn,d(xn,xn+1)<
d(xn1,xn)and
Zd(xn,xn+1)
0
f(t)dt <rnZd(x0,x1)
0
f(t)dt.
We claim that for any yX, one of the flowing relations hold:
αd(xn,T xn)d(xn,y)or αd(xn+1,T xn+1)d(xn+1,y).(2.1)
Otherwise, if αd(xn,T xn)>d(xn,y)and αd(xn+1,T xn+1)>d(xn+1,y), we have
d(xn,xn+1)d(xn,y) + d(xn+1,y)<αd(xn,T xn) + αd(xn+1,T xn+1)
=αd(xn,xn+1) + αd(xn+1,xn+2)2αd(xn,xn+1)d(xn,xn+1),
104 SEYED M. A. ALEOMRANINEJAD, MASOUMEH SHOKOUHNIA
which is a contradiction. Now let an=d(xn,xn+1)for all n1. It is obvious that {an}n1is
monotone non-increasing and so there exists a0 such that limnan=a. Since
Za
0
f(t)dt =lim
nZan
0
f(t)dt lim
n
rnZd(x0,x1)
0
f(t)dt =0,
we have a=0. We claim {xn}n1is a Cauchy sequence in (X,d), i.e,
ε>0NεN| ∀m,nN,m>n>Nεd(xm,xn)<ε.
Suppose that there exists an ε>0 such that for each NNthere are mN,nNN, with mN>
nN>N, such that d(xmN,xnN)ε. We choose the sequences {mN}N1and {nN}N1such that
for each NN,mNis minimal in the sense that d(xmN,xnN)εbut d(xh,xnN)<εfor each
h∈ {nN+1,...,mN1}. Now we analyze the properties of d(xmN,xnN),d(xmN+1,xnN+1)and
d(xmN+2,xnN+1). Since
εd(xmN,xnN)
d(xmN,xmN1) + d(xmN1,xnN)
<d(xmN,xmN1) + ε,
we have d(xmN,xnN)ε+as N. We claim that there exists kNsuch that for each
natural number N>kwe have d(xmN+1,xnN+1)<εand d(xmN+2,xnN+1)<ε. Suppose there
exists a subsequence {Nk}k1Nsuch that d(xmNk+1,xnNk+1)εor d(xmNk+2,xnNk+1)ε. If
d(xmNk+1,xnNk+1)ε,
εd(xmNk+1,xnNk+1)d(xmNk+1,xmNk) + d(xmNk,xnNk) + d(xnNk,xnNk+1)
and then d(xmNk+1,xnNk+1)ε, as k. If d(xmNk+2,xnNk+1)ε,
εd(xmNk+2,xnNk+1)d(xmNk+2,xmNk+1) + d(xmNk+1,xmNk)
+d(xmNk,xnNk) + d(xnNk,xnNk+1)
and then d(xmNk+2,xnNk+1)ε, when k. In view of
d(xmNk+1,xnNk)d(xmNk+1,xmNk) + d(xmNk,xnNk),
we have limkd(xmNk+1,xnNk)ε.From relation 2.1, we have
Zd(xmNk+1,xnNk+1)
0
f(t)dt βZd(xmNk,xnNk)
0
f(t)dt
SOME FIXED POINT RESULTS OF INTEGRAL TYPE AND APPLICATIONS 105
or
Zd(xmNk+2,xnNk+1)
0
f(t)dt βZd(xmNk+1,xnNk)
0
f(t)dt.
As k, we have Rε
0f(t)dt βRε
0f(t)dt,which is a contradiction. So there exists kNsuch
that for each natural number N>kone has d(xmN+1,xnN+1)<εand d(xmN+2,xnN+1)<ε. Now
we claim that there exist a δε(0,ε)and NεNsuch that for each natural number N>Nε.
Note that
d(xmN+1,xnN+1)<εδεor d(xmNk+2,xnNk+1)<εδε.
Suppose that exist a subsequence {Nk}k1Nsuch that d(xmNk+1,xnNk+1)εand
d(xmNk+2,xnNk+1)ε
as k. Now by relation 2.1, we have
Zd(xmNk+1,xnNk+1)
0
f(t)dt βZd(xmNk,xnNk)
0
f(t)dt
or
Zd(xmNk+2,xnNk+1)
0
f(t)dt βZd(xmNk+1,xnNk)
0
f(t)dt,
which is a contradiction. If d(xmN+1,xnN+1)<εδε, then
εd(xmN,xnN)d(xmN,xmN+1) + d(xmN+1,xnN+1) + d(xnN+1,xnN)
<d(xmN,xmN+1)+(εδε) + d(xnN,xnN+1).
If d(xmN+2,xnN+1)<εδε, then
εd(xmN,xnN)d(xmN,xmN+1) + d(xmN+1,xmN+2)
+d(xmN+2,xnN+1) + d(xnN+1,xnN)
<d(xmN,xmN+1) + d(xmN+1,xmN+2)+(εδε) + d(xnN,xnN+1).
It follows that εεδεas N, which is a contradiction. This proves our claim that {xn}n1
is a Cauchy sequence in (X,d). Let limnxn=x. By relation 2.1, for each n1 either
i)Rd(T xn,T x)
0f(t)dt βRd(xn,x)
0f(t)dt or ii)Rd(T xn+1,T x)
0f(t)dt βRd(xn+1,x)
0f(t)dt hold. Then
106 SEYED M. A. ALEOMRANINEJAD, MASOUMEH SHOKOUHNIA
Rd(T xn,T x)
0f(t)dt 0 or Rd(T xn+1,T x)
0f(t)dt 0 as n. Thus limnd(T xn,T x) = 0 or
limnd(T xn+1,T x) = 0. In case (i), since
d(x,T x)d(x,T xn) + d(T xn,T x) = d(x,xn+1) + d(T xn,T x),
we obtain d(x,T x) = 0 and so T x =x. We obtain T x =x. Now we show that this fixed point is
unique. Suppose that there are two distinct points a,bXsuch that Ta =aand T b =b. Since
d(a,b)>0=αd(a,Ta), we have the contradiction
0<Zd(a,b)
0
f(t)dt =Zd(Ta,T b)
0
f(t)dt βZd(a,b)
0
f(t)dt.
To prove that limnTnx=a, let xbe arbitrary and aF(T). Note that d(a,Tn1x)0=
αd(a,Ta)for every nN, we have
Zd(a,Tnx)
0
f(t)dt βZd(a,Tn1x)
0
f(t)dt β2Zd(a,Tn2x)
0
f(t)dt ... βnZd(a,x)
0
f(t)dt.
3. Example and applications
In this section, we give some remarks and examples which clarify the connection between our
result and the classical ones in the literature. As an application, the existence of a continuous
solution for an integral equation is obtained.
Remark 3.1 Theorem 2.1 is a generalization of theorem 1.3. Letting f(t) = 1 for each t0 in
theorem 2.1, we have Rd(T x,Ty)
0f(t)dt =d(T x,Ty)βd(x,y) = βRd(x,y)
0f(t)dt. The converse
is not true as we will see in example 3.1.
Remark 3.2 Theorem 2.1 is a generalization of theorem 1.2. The converse is not true as we will
see the example 3.1.
Example 3.1 Let X:{(0,0),(5,6),(5,4),(0,4)}∪{(n,0):nN}∪{(n+12,n+13):nN}
and its metric defined by d((x1,x2),(y1,y2)) = |x1y1|+|x2,y2|. Define a mapping Ton Xby
T((x1,x2)) =
(x1,0),x16x2
(0,x2),x2<x1
(1)
SOME FIXED POINT RESULTS OF INTEGRAL TYPE AND APPLICATIONS 107
Then Tsatisfies the assumptions of Theorem 2.1 with f(t) = tt(1+ln t)for t>0, f(0) = 0,
α=5/12 β=1/2 while Tis not satisfies the assumptions of Theorem 1.2. First note that,
Rd(T x,Ty)
0f(t)dt 1
2Rd(x,y)
0f(t)dt if (x,y)6= ((5,6),(5,4)) and (x,y)6= ((5,4),(5,6)).In this
context one has Rx
0f(t)dt =xx. Let d(T x,Ty) = nand d(x,y) = m. It is clear that n<mif
(x,y)6= ((5,6),(5,4)) and (x,y)6= ((5,4),(5,6)).Then we have
Zd(T x,Ty)
0
f(t)dt =Zn
0
f(t)dt =nn<1
2mm=1
2Zm
0
f(t)dt =1
2Zd(x,y)
0
f(t)dt,
because
nn
mm=nn
mn+k= ( n
m)n1
mk<1
2.
On the other hands, since αd((5,6),T(5,4)) >5/2>2 and αd((5,4),T(5,6)) >25/12 >2,
Tsatisfies the assumption in Theorem 2.1.
Remark 3.3 Let x= (n+12,n+13)and y= (n,0). Then in exampel 3.1, we have d(T x,Ty)
d(x,y)=
n+12
n+25 and so supx,yX\{(5,6),(5,4)}
d(T x,Ty)
d(x,y)=1.Thus Tis not a contraction mapping.
Let us consider the following integral equation:
x(t) = g(t) + Zt
0
K(s,x(s))ds,t[0,1] (2)
we are going to give existence and uniqueness results for the solution of the integral equation
using theorem 2.1. Let us consider X:= (C([0,1],k.k).
Theorem 3.1 Consider the integral equation (2). Suppose
i) K :[0,1]×RnRnand g :[0,1]Rnare continuous;
ii) there exist α(0,1
2],0<L<1such that α|x(t)g(t)Rt
0K(s,x(s))ds| ≤ |x(t)y(t)|
implies
Z|K(t,x(t))K(t,y(t))|
0
f(λ)dλLZ|x(t)y(t)|
0
f(λ)dλ
for all x,yX and f :[0,)(0,)is a Lebesgue integrable mapping which is summable
and for each ε>0,Rε
0f(λ)dλ>0.Then the integral equation (2), have a unique solution.
Proof. Let T:XX,x7→ T(x), where
T(x)(t) = Zt
0
K(s,x(s))ds +g(t),t[0,1].
108 SEYED M. A. ALEOMRANINEJAD, MASOUMEH SHOKOUHNIA
In this way, the integral equation (2) can be rewritten as x=T(x). Next, we show that Tsatisfies
the conditions of Theorem 2.1. Let x,yXand α|x(t)g(t)Rt
0K(s,x(s))ds|≤|x(t)y(t)|.
Then
αkxT xk≤ kxyk
implies
ZkT xTyk
0
f(λ)dλ=Zmaxt[0,1]|T x(t)Ty(t)|
0
f(λ)dλ
Zmaxt[0,1]Rt
0|K(s,x(s))K(s,y(s))|ds
0
f(λ)dλ
Zmaxs[0,1]|K(s,x(s))K(s,y(s))|
0
f(λ)dλ
LZkxyk
0
f(λ)dλ.
Now Theorem 2.1 shows that there exists x0Xsuch that T x0=x0and so
x0(t) = T x0(t) = Zt
0
K(s,x(s))ds +g(t).
Conflict of Interests
The authors declare that there is no conflict of interests.
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Hindawi Publishing Corpration, Inter. J. Math. Math. Sci. 29 (2002), 531-536.
[3] T. Suzuki, A new type of fixed point theorem in metric spaces, Nonlinear Anal. 71 (2009), 5313-5317.
[4] S. Dhompongsa, H. Yingtaweesittikul, Fixed point for multivalued mappings and the metric completeness,
Fixed Point Theory Appl. 2009 (2009), Article ID 972395.
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SOME FIXED POINT RESULTS OF INTEGRAL TYPE AND APPLICATIONS 109
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