some fixed point theorems of integral type contraction in cone … · 2011. 7. 26. · i d x,y ≥0...

Hindawi Publishing CorporationFixed Point Theory and ApplicationsVolume 2010, Article ID 189684, 13 pagesdoi:10.1155/2010/189684

Research ArticleSome Fixed Point Theorems of Integral TypeContraction in Cone Metric Spaces

Farshid Khojasteh,1 Zahra Goodarzi,2and Abdolrahman Razani2, 3

1 Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran 14778, Iran2 Department of Mathematics, Faculty of Science, Imam Khomeini International University,Qazvin, 34149-16818, Iran

3 School of Mathematics, Institute for Research in Fundamental Sciences, P.O. Box 19395-5746,Tehran, Iran

Correspondence should be addressed to Farshid Khojasteh, fr [email protected]

Received 1 October 2009; Accepted 11 February 2010

Academic Editor: Jerzy Jezierski

Copyright q 2010 Farshid Khojasteh et al. This is an open access article distributed under theCreative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.

We define a new concept of integral with respect to a cone. Moreover, certain fixed point theoremsin those spaces are proved. Finally, an extension of Meir-Keeler fixed point in cone metric space isproved.

1. Introduction

In 2007, Huang and Zhang in [1] introduced cone metric space by substituting an orderedBanach space for the real numbers and proved some fixed point theorems in this space. Manyauthors study this subject and many fixed point theorems are proved; see [2–5]. In this paper,the concept of integral in this space is introduced and a fixed point theorem is proved. Inorder to do this, we recall some definitions, examples, and lemmas from [1, 4] as follows.

Let E be a real Banach space. A subset P of E is called a cone if and only if the followinghold:

(i) P is closed, nonempty, and P /= {0},(ii) a, b ∈ R, a, b ≥ 0, and x, y ∈ P imply that ax + by ∈ P,

(iii) x ∈ P and −x ∈ P imply that x = 0.

Given a cone P ⊂ E, we define a partial ordering ≤ with respect to P by x ≤ y if and only ify − x ∈ P. We will write x < y to indicate that x ≤ y but x /=y, while x � y will stand for

2 Fixed Point Theory and Applications

y − x ∈ intP, where intP denotes the interior of P. The cone P is called normal if there is anumber K > 0 such that 0 ≤ x ≤ y implies ‖x‖ ≤ K‖y‖, for all x, y ∈ E. The least positivenumber satisfying above is called the normal constant [1].

The cone P is called regular if every increasing sequence which is bounded from aboveis convergent. That is, if {xn}n≥1 is a sequence such that x1 ≤ x2 ≤ · · · ≤ y for some y ∈ E,then there is x ∈ E such that limn→∞‖xn − x‖ = 0. Equivalently, the cone P is regular if andonly if every decreasing sequence which is bounded from below is convergent [1]. Also everyregular cone is normal [4]. In addition, there are some nonnormal cones.

Example 1.1. Suppose E = C2R([0, 1])with the norm ‖f‖ = ‖f‖∞ + ‖f ′‖∞ and consider the cone

P = {f ∈ E: f ≥ 0}. For all K ≥ 1, set f(x) = x and g(x) = x2K. Then 0 ≤ g ≤ f, ‖f‖ = 2 and‖g‖ = 2K + 1. Since K‖f‖ < ‖g‖, K is not normal constant of P. Therefore, P is non-normalcone.

From now on, we suppose that E is a real Banach space, P is a cone in E with int P /= ∅,and ≤ is partial ordering with respect to P . Let X be a nonempty set. As it has been definedin [1], a function d : X × X → E is called a cone metric on X if it satisfies the followingconditions:

(i) d(x, y) ≥ 0 for all x, y ∈ X and d(x, y) = 0 if and only if x = y,

(ii) d(x, y) = d(y, x), for all x, y ∈ X,

(iii) d(x, y) ≤ d(x, z) + d(y, z), for all x, y, z ∈ X.

Then (X, d) is called a cone metric space.

Example 1.2. Suppose E = l1, P = {{xn}n∈N∈ E : xn ≥ 0, for all n, (X, ρ) is a metric space and

d : X ×X → E is defined by d(x, y) = {ρ(x, y)/2n}n∈N. Then (X, d) is a cone metric space and

the normal constant of P is equal to 1.

Definition 1.3. Let (X, d) be a cone metric space. Let {xn}n∈Nbe a sequence in X and x ∈ X. If

for any c ∈ E with 0 � c, there is n0 ∈ N such that for all n > n0, d(xn, x) � c, then {xn}n∈N

is said to be convergent to x, and x is the limit of {xn}n∈N. We denote this by

limn→∞

xn = x or xn −→ x (n −→ ∞). (1.1)

Definition 1.4. Let (X, d) be a cone metric space and {xn}n∈Nbe a sequence in X. If for any

c ∈ E with 0 � c, there is n0 ∈ N such that for all m,n > n0, d(xn, xm) � c, then {xn}n∈Nis

called a Cauchy sequence in X.

Definition 1.5. Let (X, d) be a cone metric space, if every Cauchy sequence is convergent in X,then X is called a complete cone metric space.

Definition 1.6. Let (X, d) be a cone metric space. Let T be a self-map on X. If for all sequence{xn}n∈N

in X,

limn→∞

xn = x implies limn→∞

T(xn) = T(x), (1.2)

then T is called continuous on X.

Fixed Point Theory and Applications 3

The following lemmas are useful for us to prove the main result.

Lemma 1.7. Let (X, d) be a cone metric space and P a normal cone with normal constant K. Let{xn}n∈N

be a sequence in X. Then {xn}n∈Nconverges to x if and only if

limn∈N

d(xn, x) = 0. (1.3)

Lemma 1.8. Let (X, d) be a cone metric space and P a normal cone with normal constant K. Let{xn}n∈N

be a sequence in X. Then {xn}n∈Nis a Cauchy sequence if and only if

limm,n→∞

d(xm, xn) = 0. (1.4)

Lemma 1.9. Let (X, d) be a cone metric space and {xn}n∈Na sequence inX. If {xn}n∈N

is convergent,then it is a Cauchy sequence.

Lemma 1.10. Let (X, d) be a cone metric space and P be a normal cone with normal constant K. Let{xn} and {yn} be two sequences in X and xn → x, yn → y (n → ∞). Then

d(xn, yn

) −→ d(x, y)

(n −→ ∞). (1.5)

The following example is a cone metric space.

Example 1.11. Let E = R2, P = {(x, y) ∈ Ex, y ≥ 0}, and X = R. Suppose that d : X ×X → E is

defined by d(x, y) = (|x − y|, α|x − y|), where α ≥ 0 is a constant. Then (X, d) is a cone metricspace.

Theorem 1.12. Let (X, d) be a complete cone metric space and P a normal cone with normal constantK. Suppose the mapping f : X → X satisfies the contractive condition

d(fx, fy

) ≤ βd(x, y)

(1.6)

for all x, y ∈ X, where β ∈ (0, 1) is a constant. Then f has a unique fixed point x0 ∈ X. Also, for allx ∈ X, the sequence {fn(x)}∞n=1 converges to x0.

2. Certain Integral Type Contraction Mapping in Cone Metric Space

In 2002, Branciari in [6] introduced a general contractive condition of integral type as follows.

Theorem 2.1. Let (X, d) be a complete metric space, α ∈ (0, 1), and f : X → X is a mapping suchthat for all x, y ∈ X,

∫d(f(x),f(y))

0φ(t)dt ≤ α

∫d(x,y)

0φ(t)dt, (2.1)


where φ : [0,+∞) → [0,+∞) is nonnegative and Lebesgue-integrable mapping which is summable(i.e., with finite integral) on each compact subset of [0,+∞) such that for each ε > 0,

∫ε0φ(t)dt > 0,

then f has a unique fixed point a ∈ X, such that for each x ∈ X, limn→∞fnx = a.

In this section we define a new concept of integral with respect to a cone and introducethe Branciari’s result in cone metric spaces.

Definition 2.2. Suppose that P is a normal cone in E. Let a, b ∈ E and a < b. We define

[a, b] := {x ∈ E : x = tb + (1 − t)a, for some t ∈ [0, 1]},[a, b) := {x ∈ E : x = tb + (1 − t)a, for some t ∈ [0, 1)}.

(2.2)

Definition 2.3. The set {a = x0, x1, . . . , xn = b} is called a partition for [a, b] if and only if thesets {[xi−1, xi)}ni=1 are pairwise disjoint and [a, b] = {⋃n

i=1[xi−1, xi)} ∪ {b}.

Definition 2.4. For each partition Q of [a, b] and each increasing function φ : [a, b] → P, wedefine cone lower summation and cone upper summation as

LConn

(φ,Q

)=

n−1∑

i=0

φ(xi)‖xi − xi+1‖,

UConn

(φ,Q

)=

n−1∑

i=0φ(xi+1)‖xi − xi+1‖,

(2.3)

respectively.

Definition 2.5. Suppose that P is a normal cone in E. φ : [a, b] → P is called an integrablefunction on [a, b]with respect to cone P or to simplicity, Cone integrable function, if and onlyif for all partition Q of [a, b]

limn→∞

LConn

(φ,Q

)= SCon = lim

n→∞UCon

n

(φ,Q

), (2.4)

where SCon must be unique.We show the common value SCon by

∫b

a

φ(x)dP (x) or to simplicity∫b

a

φdp. (2.5)

We denote the set of all cone integrable function φ : [a, b] → P by L1([a, b], P).

Lemma 2.6. (1) If [a, b] ⊆ [a, c], then∫bafdp ≤ ∫cafdp, for f ∈ L1(X, P).(2)

∫ba(αf + βg)dp =

α∫bafdp + β

∫bagdp, for f, g ∈ L1(X, P) and α, β ∈ R.

Proof. (1) Suppose that P and R are partitions for [a, b] and [b, c], respectively. That is,

R = {xn, xn+1, . . . , xm−1, xm = c}, P = {a = x0, x1, . . . , xn−1, xn = b}, . (2.6)


Let Q = P ∪ R. Q is a partition for [a, c]. Therefore one can write

LConn

(f, P)=

n−1∑

i=0

f(xi)‖xi − xi+1‖ ≤n−1∑

i=0

f(xi)‖xi − xi+1‖ +m−1∑

i=n

f(xi)‖xi − xi+1‖

= LConn

(f, P)+ LCon

n

(f, R)= LCon

n

(f,Q).

(2.7)

So

∫b

a

fdp ≤∫ c

a

fdp. (2.8)

(2) Suppose P is an partition for [a, b], that is

P = {a = x0, x1, . . . , xn−1, xn = b}. (2.9)

Then

LConn

(f, P)=

n−1∑

i=0

(αf(xi) + βg(xi)

)‖xi − xi+1‖

= αn−1∑

i=0

f(xi)‖xi − xi+1‖ + βn−1∑

i=0

g(xi)‖xi − xi+1‖ = αLConn

(f, P)+ βLCon

n

(g, P).

(2.10)

Thus

∫b

a

(αf + βg

)dp = α

∫b

a

fdp + β

∫b

a

gdp. (2.11)

Definition 2.7. The function φ : P → E is called subadditive cone integrable function if andonly if for all a, b ∈ P

∫a+b

0φdP ≤

∫a

0φdP +

∫b

0φdP . (2.12)

Example 2.8. Let E = X = R, d(x, y) = |x − y|, P = [0,+∞), and φ(t) = 1/(t + 1) for all t > 0.Then for all a, b ∈ P,

∫a+b

0

dt

t + 1= ln(a + b + 1),

∫a

0

dt

t + 1= ln(a + 1),

∫b

0

dt

t + 1= ln(b + 1). (2.13)


Since ab ≥ 0, then a + b + 1 ≤ a + b + 1 + ab = (a + 1)(b + 1). Therefore

ln(a + b + 1) ≤ ln((a + 1)(b + 1)) = ln(a + 1) + ln(b + 1). (2.14)

This shows that φ is an example of subadditive cone integrable function.

Theorem 2.9. Let (X, d) be a complete cone metric space and P a normal cone. Suppose that φ :P → P is a nonvanishing map and a subadditive cone integrable on each [a, b] ⊂ P such that for eachε � 0,

∫ε0φdp � 0. If f : X → X is a map such that, for all x, y ∈ X

∫d(f(x),f(y))

0φdp ≤ α

∫d(x,y)

0φdp, (2.15)

for some α ∈ (0, 1), then f has a unique fixed point in X.

Proof. Let x1 ∈ P. Choose xn+1 = f(xn).We have

∫d(xn+1,xn)

0φdp =

∫d(f(xn),f(xn−1))

0φdp

≤ α

∫d(xn,xn−1)

0φdp

...

≤ αn−1∫d(x2,x1)

0φdp.

(2.16)

Since α ∈ (0, 1) thus

limn→∞

∫d(xn+1,xn)

0φdp = 0. (2.17)

If limn→∞d(xn+1, xn)/= 0 then limn→∞∫d(xn+1,xn)0 φdp /= 0 and this is a contradiction, so

limn→∞

d(xn+1, xn) = 0. (2.18)

We now show that (xn) is a Cauchy sequence. Due to this, we show that

limm,n→∞

d(f(xm), f(xn)

)= 0. (2.19)

By triangle inequality

∫d(f(xm),f(xn))

0φdp ≤

∫d(f(xn),f(xn+1))+d(f(xn+1),f(xn+2))+···+d(f(xm−1),f(xm))

0φdp (2.20)


and by sub-additivity of φ we get

∫d(f(xm),f(xn))

0φdp ≤

∫d(f(xn),f(xn+1))

0φ dp + · · · +

∫d(f(xm−1),f(xm))

0φdp

≤(αn + αn−1 + · · · + αm−1

)∫d(x2,x1)

0φ dp ≤ αn

1 − α

∫d(x2,x1)

0φ dp −→ 0.

(2.21)

Thus

limm,n→∞

d(f(xn), f(xm)

)= 0. (2.22)

This means that {xn}n∈Nis a Cauchy sequence and since X is a complete cone metric space,

thus {xn}n∈Nis convergent to x0 ∈ X. Finally, since

∫d(xn+1,f(x0))

0φ dp =

∫d(f(xn),f(x0))

0φdp ≤ α

∫d(xn,x0)

0φ dp, (2.23)

thus limn→∞d(xn+1, f(x0)) = 0. This means that f(x0) = x0. If x0, y0 are two distinct fixedpoints of f, then

∫d(x0,y0)

0φdp =

∫d(f(x0),f(y0))

0φdp ≤ α

∫d(x0,y0)

0φdp (2.24)

which is a contradiction. Thus f has a unique fixed point x0 ∈ X.

Lemma 2.10. Let E = R2, P = {(x, y) ∈ Ex, y ≥ 0}, and X = R. Suppose that d : X × X → E is

defined by d(x, y) = (|x − y|, α|x − y|), where α ≥ 0 is a constant. Suppose that φ : [(0, 0), (a, b)] →P is defined by φ(x, y) = (φ1(x), φ2(y)), where φ1, φ2 : [0,+∞) → [0,+∞) are two Riemann-integrable functions. Then

∫ (a,b)

(0,0)φdP =

√a2 + b2

(1a

∫a

0φ1(t)dt,

1b

∫b

0φ2(t)dt

)

. (2.25)


Proof. Let Q = {(xi, yi)}ni=0 be a partition of set [(0, 0), (a, b)] such that xi = (a/n)i and yi =(b/n)i, then (by Definitions 2.4 and 2.5)

∫ (a,b)

(0,0)φdP = lim

n→∞LConn

(φ,Q

)= lim

n→∞

n−1∑

i=0

φ(xi, yi

)∥∥(xi+1, yi+1) − (xi, yi

)∥∥

= limn→∞

n−1∑

i=0

(φ1

(a

ni

), φ2

(b

ni

))∥∥∥∥

(a

n,b

n

)∥∥∥∥

=√a2 + b2 lim

n→∞

n−1∑

i=0

1n

(φ1

(a

ni

), φ2

(b

ni

))

=√a2 + b2

(

limn→∞

1n

n−1∑

i=0

φ1

(a

ni

), limn→∞

1n

n−1∑

i=0

φ2

(b

ni

))

=√a2 + b2

(1alimn→∞

a

n

n−1∑

i=0

φ1

(a

ni

),1blimn→∞

b

n

n−1∑

i=0

φ2

(b

ni

))

=√a2 + b2

(1a

∫a

0φ1(t)dt,

1b

∫b

0φ2(t)dt

)

.

(2.26)

Thus

∫ (a,b)

(0,0)φdP =

√a2 + b2

(1a

∫a

0φ1(t)dt,

1b

∫b

0φ2(t)dt

)

. (2.27)

Example 2.11. Let X = {1/n : n ∈ N} ∪ {0}, E = R2 and P = {(x, y) ∈ E : x ≥ 0, y ≥ 0}. Suppose

d(x, y) = (|x − y|, α|x − y|), for some constant α > 0. Firstly, (X, d) is a complete cone metricspace. Secondly, if f : X → X and φ : P → E are defined by

f(x) =

⎧⎨

⎩

1n + 1

if x =1n, n ∈ N,

0 if x = 0,

φ(t, s) =

⎧⎨

⎩

(t1/(t−2)(1 − ln(t)), s1/(s−2)(1 − ln(s))

), (t, s) ∈ P \ {(0, 0)},

(0, 0), (t, s) = (0, 0),

(2.28)

respectively, then

∫d(fx,fy)

0φdP ≤ 1

2

∫d(x,y)

0φdP . (2.29)


In order to obtain inequality (2.29), set x = 1/n and y = 1/m, where m > n.Hence

d(fx, fy

)=(

m − n

(m + 1)(n + 1),

α(m − n)(m + 1)(n + 1)

),

d(x, y)=(m − n

mn,α(m − n)

mn

).

(2.30)

Suppose φ1(t) = φ2(t) = t1/(t−2)(1 − ln(t)) for all t > 0 and φ1(0) = φ2(0) = 0. Thus φ(t, s) =(φ1(t), φ2(s)). By Lemma 2.10

∫d(fx,fy)

0φdP =

∫ ((m−n)/(m+1)(n+1),α(m−n)/(m+1)(n+1))

(0,0)

(φ1, φ2

)dP

=(

m − n

(m + 1)(n + 1)

√1 + α2

)

((m + 1)(n + 1)

m − n

∫ (m−n)/(m+1)(n+1)

0φ1(t)dt,

(m + 1)(n + 1)α(m − n)

∫α(m−n)/(m+1)(n+1)

0φ2(t)dt

)

=(√

1 + α2)(∫ (m−n)/(m+1)(n+1)

0φ1(t)dt,

1α

∫α(m−n)/(m+1)(n+1)

0φ2(t)dt

)

.

(2.31)

Since∫τ0t

1/(t−2)(1 − ln(t))dt = τ1/τ , thus

∫ (m−n)/(m+1)(n+1)

0φ1(t)dt =

[m − n

(m + 1)(n + 1)

](m+1)(n+1)/(m−n),

∫α(m−n)/(m+1)(n+1)

0φ2(t)dt =

[α(m − n)

(m + 1)(n + 1)

](m+1)(n+1)/α(m−n).

(2.32)

It means that

∫d(fx,fy)

0φdP =

√1+α2

([m−n

(m+1)(n+1)

](m+1)(n+1)/(m−n),1α

[α(m−n)

(m+1)(n+1)

](m+1)(n+1)/α(m−n))

.

(2.33)

On the other side, Branciari in [6] shows that

[m − n

(n + 1)(m + 1)

](n+1)(m+1)/(m−n)≤ 1

2

[m − n

nm

]nm/(m−n), (2.34)


for all m,n ∈ N. Therefore

([m − n

(m + 1)(n + 1)

](m+1)(n+1)/(m−n),1α

[α(m − n)

(m + 1)(n + 1)

](m+1)(n+1)/α(m−n))

≤ 12

([m − n

mn

]mn/(m−n),1α

[α(m − n)

mn

]mn/α(m−n))

.

(2.35)

Thus inequalities (2.33) and (2.35) imply that

∫d(fx,fy)

0φdP ≤ 1

2

√1 + α2

([m − n

mn

]mn/(m−n),1α

[α(m − n)

mn

]mn/α(m−n))

=12

∫d(x,y)

0φdP , (2.36)

or in other words

∫d(fx,fy)

0φdP ≤ 1

2

∫d(x,y)

0φdP . (2.37)

Thus by Theorem 2.9, f has a fixed point. But, on the other hand,

d(fx, fy

)< d(x, y), (2.38)

and this means that f does not satisfy in Theorem 1.12.

3. Extension of Meir-Keeler Contraction in Cone Metric Space

In 2006, Suzuki in [7] proved that the integral type contraction (see [6]) is a special caseof Meir-Keeler contraction (see [8]). Haghi and Rezapour in [5] extended Meir-Keelercontraction in cone metric space as follows.

Theorem 3.1 (see[5]). Let (X, d) be a complete regular cone metric space and f has the property(KMC) on X; that is, for all 0/= ε ∈ P , there exists δ � 0 such that

d(x, y)< ε + δ implies d

(fx, fy

)< ε (3.1)

for all x, y ∈ X. Then f has a unique fixed point.

An extension of Theorem 3.1 is as follows.

Theorem 3.2. Let (X, d) be a complete regular cone metric space and f a mapping on X. Supposethat there exists a function θ from P into itself satisfying the following:

(B1) θ(0) = 0 and θ(t) � 0 for all t � 0,

(B2) θ is nondecreasing and continuous function. Moreover, its inverse is continuous,


(B3) for all 0/= ε ∈ P, there exists δ � 0 such that for all x, y ∈ X

θ(d(x, y))

< ε + δ implies θ(d(fx, fy

))< ε (3.2)

(B4) for all x, y ∈ X

θ(x + y

) ≤ θ(x) + θ(y). (3.3)

Then f has a unique fixed point.

Proof. First, note that θ(d(f(x), f(y))) < θ(d(x, y)) for all x, y ∈ X with x /=y. Since θ−1 exists,thus d(f(x), f(y)) < d(x, y) for all x, y ∈ X with x /=y. Now Let x0 ∈ X. Set xn = f(xn−1) forall n ∈ N. If, there is a natural m ∈ N such that d(xm+1, xm) = 0, then f(xm) = xm and so fhas a fixed point. If d(xn+1, xn)/= 0 for all n ∈ N, then θ(d(xn+1, xn)) < θ(d(xn, xn−1). Hence,according to regularity of P, there exists α ∈ P such that θ(d(xn+1, xn)) ↓ α. We claim α = 0. Ifα/= 0, then according to (B3), there is 0 � d such that θ(d(f(x), f(y)) < α for all x, y ∈ X withθ(d(x, y)) < α + d. Choose r > 0 such that (d/2) +Nr(0) ⊆ P and take the natural number Nsuch that ‖θ(d(xn+1, xn)) − α‖ < r, for all n ≥ N.We obtain

∥∥∥∥d

2− (θ(d(xn+1 − xn)) − α) − d

2

∥∥∥∥ < r. (3.4)

Thus

d

2− (θ(d(xn+1 − xn)) − α) ∈ d

2+Nr(0) ⊆ P. (3.5)

So, θ(d(xn+1, xn)) − α � d. Since f has the property (B3), θ(d(xn+2, xn+1)) < α for all n ≥ N.This is a contradiction because α < θ(d(xi+1, xi)) for all i ≥ 1. Thus

limn→∞

θ(d(xn+1, xn)) = 0. (3.6)

Now, we show that {xn}∞n=1 is a Cauchy sequence. If this is not, then there is a 0 � c suchthat for all natural number k, there are mk, nk > k so that the relation d(xmk , xnk) � c doesnot hold. Since θ has continuous inverse thus there exists 0 � c such that for all naturalnumber k, there are mk, nk > k so that the relation θ(d(xmk , xnk)) � c does not hold. Foreach 0 � e � c, there exists 0 � d such that θ(d(f(x), f(y))) < e, for all x, y ∈ X withθ(d(x, y)) < e + d. Choose a natural number M such that θ(d(xi+1, xi)) � d/2 for all i ≥ M.


Also, take mM ≥ nM > M so that the relation θ(d(xmM, xnM)) � c does not hold. Then (B4)yields

θ(d(xnM−1, xnM+1)) ≤ θ(d(xnM−1, xnM)) + θ(d(xnM, xnM+1))

� d

2+d

2

� d + e.

(3.7)

Hence, θ(d(xnM, xnM+2)) � e. Similarly, θ(d(xnM, xnM+3)) � e. Thus

θ(d(xnM, xmM)) � e � c (3.8)

which is a contradiction. Therefore {xn}∞n=1 is a Cauchy sequence. Since (X, d) is a completecone metric space, there is u ∈ X such that xn → u. Since d(fx, fy) < d(x, y), for all x, y ∈ Xwith x /=y, thus for each ε � 0, there is a natural number N > 0 such that for all n > N,d(xn, x) � ε. Since d(fxn, fu) < d(xn, u) thus d(fxn, fu) � ε for all n > N. It means thatfxn → fu. In the other side, f(xn) = xn+1 → u and the limit point is unique in cone metricspaces. Thus f has at least one fixed point. Now, if u, v are two distinct fixed points for f, then

d(u, v) = d(fu, fv

)< d(u, v) (3.9)

which is a contradiction. Therefore f has a unique fixed point.

Remark 3.3. (1) Set θ(x) = x, then Theorem 3.1 is a direct result of Theorem 3.2.(2) Let φ : P → P be a nonvanishing map and a subadditive cone integrable on each

[a, b] ⊂ P such that for each ε � 0,∫ε0 φdp � 0. If θ(x) =

∫x0 φdP , then θ satisfies all conditions

of Theorem 3.2. In other words, Theorem 2.9 is a direct result of Theorem 3.2.

Acknowledgment

The third author would like to thank the School of Mathematics of the Institute for Researchin Fundamental Sciences, Teheran, Iran, for supporting this research (Grant no. 88470119).

References

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[6] A. Branciari, “A fixed point theorem for mappings satisfying a general contractive condition of integraltype,” International Journal of Mathematics and Mathematical Sciences, vol. 29, no. 9, pp. 531–536, 2002.

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