chapter 5 a common fixed points in cone and rectangular cone...

Chapter 5

A Common Fixed Points in Coneand rectangular cone Metric Spaces

In this chapter we have establish fixed point theorems in cone metric spaces and rectan-

gular cone metric space. In the first part we have proved some generalization of common

fixed point on cone metric space with w-distance. Also proved a common fixed point the-

orem for a pair of weakly compatible self mappings defined on a cone metric space under

different contractive conditions. In the second part we have proved some generalization

of common fixed point on cone rectangular metric spaces.

5.1 Cone Metric Space

5.1.1 Introduction

In 2007 Huang and Zhang [72] introduced the notation of cone metric space as natural

generalization of the notion of metric space (X, d), where metric d : X × X → R+ is

replaced by the metric d : X × X → E, where E is real Banach space. This theory

generalized by see in, [3, 4, 5, 123, 49, 116].

Section (5.1.2) is devoted to the basic operations related to the cone, cone metric

space and some basic properties of cone metric spaces with w-distance and some gener-

alization result of H.Lakzian F.Arabyani,[76] are recalled in this section (5.1.3). A fixed

62

5.1 Cone Metric Space 63

point theorem of multifunctions proved in cone metric spaces and we generalized some

result of S.Rezapour, [124] in section (5.1.4). In section (5.1.5) applying notion of weakly

compatible maps we have proved fixed point theorem in cone metric spaces.

5.1.2 Preliminary Notes

Definition 5.1.1. [72] Let E always be a real Banach space and P a subset of E. P is

called a cone if

(i) P is closed, non-empty and P 6= {0},

(ii) ax+ by ∈ P for all x, y ∈ P and non-negative real numbers a, b,

(iii) P ∩ (−P ) = 0.

For a given cone P ⊆ E, we can define a partial ordering ≤ with respect to P by

x ≤ y if and only if y − x ∈ P . The notation x < y will stand for x ≤ y and x 6= y, while

x� y will stand for y − x ∈ intP , where intP denotes the interior of P i.e. P 0.

Definition 5.1.2. The cone P is called normal if there is a number M > 0 such that for

all x, y ∈ E, 0 ≤ x ≤ y implies

‖x‖ ≤M‖y‖.

The least positive number satisfying above is called the normal constant of P . It is clear

that M ≥ 1.

In the following, let E be a normed linear space, P be a cone in E satisfying int(P ) 6= ∅,

and ’≤’ denote the partial ordering on E with respect to P .

Definition 5.1.3. [72] Let X be a non-empty set. Suppose the mapping d : X ×X → E

satisfies:

(a) 0 ≤ d(x, y) for all x, y ∈ X and d(x, y) = 0 if and only if x = y,

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


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

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

Example 5.1.1. [72] Let E = R2, P = {(x, y) ∈ E|x, y ≥ 0}, X = R and d : X×X → E

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

metric space.

Example 5.1.2. [123] Let E = R3, P = {(x, y, z) ∈ E|x, y, z ≥ 0}, X = R and d :

X ×X → E defined by d(x, y) = (α|x − y|, β|x − y|, γ|x − y|), where α, β, γ are positive

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

Example 5.1.3. [85] Let E = lp, (0 < p < 1), P = {{xn}n≥1 ∈ E|xn ≥ 0, for all n},

(X, ρ) a metric space and d : X ×X → E defined by d(x, y) = { ρ(x,y)2n }n≥1. Then (X, d)

is a cone metric space.

Definition 5.1.4. [72] Let (X, d) be a cone metric space, x ∈ X and {xn}n≥1 a sequence

in X. Then

(a) {xn}n≥1 is said to converge to x whenever for every c ∈ E with 0 � c there is a natural

number N such that d(xn, x) � c for all n ≥ N . We denote this by limn→∞ xn = x or

xn → x.

(b) {xn}n≥1 is said to be a Cauchy sequence whenever for every c ∈ E with 0 � c there

is a natural number N such that d(xn, xm) � c, for all n,m ≥ N .

(c) (X, d) is called a complete cone metric space if every Cauchy sequence is convergent

in X.

Definition 5.1.5. Let (X, d) be a cone metric space and B ⊆ X.

(i) A point b in B is called an interior point of B whenever there exists a point p, 0 � p,

such that

N(b, p) ⊆ B,


where N(b, p) = {y ∈ X|d(y, b) � p}.

(ii) A subset A ⊆ X is called open if each element of A is an interior point of A.

Remark 5.1.1. [152] Let E be an ordered Banach (normed) space. Then c is an interior

point of P , if and only if [−c, c] is neighborhood of 0, where [c, c] = {y ∈ E| − c ≤ y ≤ c}

and N(0, c) = {y ∈ E| − c� y � c} ⊆ [c, c].

Remark 5.1.2. [158]

1. If u ≤ v and v � w, then u� w.

Indeed w − u = (w − v) + (v − u) ≥ w − v, implies

[−(w − u), (w − u)] ⊇ [−(w − v), (w − v)].

2. If u� v and v � w, then u� w.

Indeed w − u = (w − v) + (v − u) ≥ w − v, implies

[−(w − u), (w − u)] ⊃ [−(w − v), (w − v)].

3. If 0 ≤ u� c for each c ∈ P 0, then u = 0.

4. If a ≤ b+ c for each c ∈ P 0, then a ≤ b.

5. If 0 ≤ x ≤ y in E and 0 ≤ a in R, then 0 ≤ ax ≤ ay.

6. If 0 ≤ xn ≤ yn for each n ∈ N and limn→∞ xn = x, limn→∞ yn = y, then 0 ≤ x ≤ y.

7. If 0 ≤ d(xn, x) ≤ bn and bn → 0, for any c ∈ intP , there exist N ∈ N such that

∀n ≥ N , then d(xn, x) � c, where {xn} is a sequence and x is a given point in X.

8. If E is real Banach space with a cone P and if a ≤ λa where a ∈ P and 0 < λ < 1,

then a = 0.

Proof : The condition a ≤ λa means λa − a ∈ P , that is −(1 − λ)a ∈ P . Since

a ∈ P and 1− λ > 0, then also (1− λ)a ∈ P . Thus we have (1− λ)a ∈ P ∩ (−P )

and a = 0.


9. If c ∈ P 0 , 0 ≤ an and an → 0, then there exist natural number N such that for all

n > N we have an � c.

Proof : Let 0 � c be given. Choose a symmetric neighborhood V such that c+V ⊆

P . Since an → 0, there is positive number N such that an ∈ V = −V for n > N .

This means that c± an ∈ c+ V ⊆ intP for n > N , that is an � c.

Definition 5.1.6. The cone P is called regular if every increasing sequence which is

bounded from above is 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→0 ||xn − x|| = 0.

Equivalently the cone P is regular if and only if every decreasing sequence which is

bounded from below is convergent.

Lemma 5.1.4. [72] Every regular cone is normal.

Lemma 5.1.5. [72] There is not normal cone with normal constant M < 1.

5.1.3 A Fixed Point Theorem in Cone Metric Spaces with w-

Distance

Huang and Zhang [72] introduced cone metric space without w-distance. The metric space

with w-distance was introduced by Osama Kada, Tomonari Suzuki, Wataru Takahashi

and Nakoshi Shioji [106, 107]. We compose these concept together and generalized idea

of cone metric space with w-distance.

Definition 5.1.7. [76] Let X be a cone metric space with metric d. Then a mapping

p : X ×X → E is called w-distance on X if the following satisfy:

(a) 0 ≤ p(x, y) for all x, y ∈ X ,

(b) p(x, z) ≤ p(x, y) + p(y, z) for all x, y, z ∈ X,

(c) p(x, .) → E is lower semi-continuous for all x ∈ X,


(d) for any 0 � α, there exists 0 � β such that p(z, x) � β and p(z, y) � β imply

d(x, y) � α, where α, β ∈ E.

Definition 5.1.8. [76] Let X be a cone metric space with metric d, let p be a w-distance

on X, x ∈ X and {xn} a sequence in X, then

(a) {xn} is called a p-Cauchy sequence whenever for every α ∈ E, 0 � α, there is a

positive integer N such that, for all m,n ≥ N, p(xm, xn) � α.

(b) A sequence {xn} in X is called a p-convergent to a point x ∈ X whenever for every

α ∈ E, 0 � α, there is a positive integer N such that, for all n ≥ N , p(x, xn) � α.

Note that by lower semi-continuous p, for all n ≥ N , p(xn, x) � α. We denote this by

limn→∞ xn = x or xn → x.

(c) (X, d) is a complete cone metric space with w-distance if every Cauchy sequence is

p-convergent.

Lemma 5.1.6. [76] Let X be a cone metric space with metric d, let p be a w-distance on

X and let f be a function from X into E that 0 ≤ f(x) for any x ∈ X. Then an into

function q : X ×X → E given by q(x, y) = f(x) + p(x, y) for each (x, y) ∈ X ×X is also

a w-distance.

Example 5.1.7. [76] (i) Let (X, d) be a metric space. Then p = d is a w-distance on X.

(ii) Let X be a norm linear space with Euclidean norm. Then the mapping p : X×X →

[0,∞) defined by p(x, y) = ‖x‖ + ‖y‖, for all x, y ∈ X is a w-distance on X.

(iii) Let X be a norm linear space with Euclidean norm. Then the mapping p : X ×

X → [0,∞) defined by p(x, y) = ‖y‖ , for all x, y ∈ X is a w-distance on X.

Finally, note that the relations intP + intP ⊆ intP and λintP ⊆ intP (λ > 0).

Example 5.1.8. Let E = l1, P = {{xn}n≥1 ∈ E|xn ≥ 0, for all n}, (X, p) a metric space

and d : X ×X → E defined by d(x, y) = {p(x,y)2n }n≥1. Then (X, d) is a cone metric space


if set p = d, then (X, d) is cone metric space with w-distance p and the normal constant

of P is equal to M = 1.

Theorem 5.1.9. [76] Let (X, d) be a cone metric space with w-distance p on X and the

mapping T : X → X there exist r ∈ [0, 1/2) such that

p(Tx, T 2x) ≤ rp(x, Tx),

for every x ∈ X and

inf{p(x, y) + p(x, Tx) : x ∈ X} > 0,

for every y ∈ X with y 6= Ty. Then there is z ∈ X such that z = Tz. Moreover, if P be

a normal cone with normal constant M and v = T (v), then p(v, v) = 0.

Theorem 5.1.10. Let (X, d) be a complete cone metric space with w-distance p. Let P

be a normal cone on X. Suppose a mapping T : X ∈ X satisfy the contractive condition

p(Tx, Ty) ≤ r[p(Tx, x) + p(Ty, y)],

for all x ∈ X, where r ∈ [0, 1/2) is a constant. Then, T has a unique fixed point in X.

For each x ∈ X, the iterative sequence {T n(x)}n≥1 converges to the fixed point.

Proof. Let x ∈ X and define xn+1 = Txn. Then

x1 = Tx0, x2 = Tx1 = T 2x0, · · · , xn = T nx0.

p(Tx0, T2x0) = p(Tx0, T (Tx0))

≤ r{p(Tx0, x0) + p(T 2x0, Tx0)}

(1− r)p(Tx0, T2x0) ≤ rp(Tx0, x0)

≤( r

1− r

)p(Tx0, x0).


Similarly,

p(T 2x0, T3x0) ≤

( r

1− r

)2

p(Tx0, x0).

Thus in general, if n is positive integer

p(T nx0, Tn+1x0) ≤ hnp(Tx0, x0),

where h = r1−r , and 0 ≤ r < 1

2.

Let m,n ∈ N and m > n, we have to show sequence {T nx0} is Cauchy

p(T nx0, Tmx0) ≤ p(T nx0, T

n+1x0) + p(T n+1x0, Tn+2x0) + · · ·+ p(Tm−1x0, T

mx0)

≤ (hn + hn+1 + · · ·+ hm−1)p(Tx0, x0)

< hn(1 + h+ h2 + · · · )p(Tx0, x0)

=hn

1− hp(Tx0, x0),

Let c ∈ E with 0 � c. Choose natural number n1 such that

hn

1− hp(Tx0, x0) �

c

2,

for all n ≥ n1. Thus p(T nx0, Tmx0) � c for all m > n. Therefore {T nx0} is Cauchy in

X. Since X is complete space, then there exist x∗ ∈ X such that xn → x∗, as n → ∞.

Choose natural number n2 such that p(T nx0, x∗) � c(1−r)

2for all n ≥ n2. Hence

p(Tx∗, x∗) ≤ p(Tx∗, T nx0) + p(T nx0, x∗)

≤ r{p(Tx∗, x∗) + p(T nx0, Tn−1x0)}+ p(T nx0, x

∗)

≤ hp(T nx0, Tn−1x0) +

1

1− rp(T nx0, x

∗)

≤ hn

1− rp(Tx0, x0) +

1

1− rp(T nx0, x

∗)

� c

2+c

2

= c


for n ≥ n2. Thus p(Tx∗, x∗) � cm

for all m ≥ 1. So cm− p(Tx∗, x∗) ∈ P for all m ≥ 1.

Since cm→ 0 as m → ∞ and P is closed so −p(Tx∗, x∗) ∈ P . But p(Tx∗, x∗) ∈ P .

Therefore p(Tx∗, x∗) = 0, so Tx∗ = x∗.

Suppose y∗ be another fixed point of T , then

p(x∗, y∗) = p(Tx∗, T y∗)

= r[p(Tx∗, x∗) + p(Ty∗, y∗)]

= 0.

Hence x∗ = y∗, thus fixed point of T is unique.

5.1.4 A Fixed Point Theorem of Multifunctions in Cone Metric

Spaces

The theory of common fixed points of multifunctions is a generalization of the theory of

fixed point of mappings in some sence. The study of fixed point theorems for multival-

ued mappings was initiated by Kakutani, in 1941, in finite dimensional spaces and was

extended to infinite dimensional Banach spaces by Bohnenblust and Karlin, in 1950, and

to locally convex spaces by Ky Fan, in 1952.

There are many articles about fixed point theory and fixed points of multifunctions

(for example, [31, 125]). Here we have given generalization of S. Rezapour, [124] result

about common fixed points of two multifunctions on the cone metric spaces with normal

constant M = 1. Most of familiar cones are normal with normal constant M = 1. But

for each k > 1 there are cones with normal constant M > k. First we state the following

two lemmas.

Lemma 5.1.11. [124] Let (X, d) be a cone metric space, P a normal cone with normal

constant M = 1 and A a compact set in (X, τc), then for every x ∈ X there exists a0 ∈ A


such that

||d(x, a0)|| = infa∈A

||d(x, a)||.

Lemma 5.1.12. [124] Let (X, d) be a cone metric space, P a normal cone with normal

constant M = 1 and A,B are two compact sets in (X, τc), Then

supx∈B

d′(x,A) <∞,

where d′(x,A) = infa∈A ||d(x, a)|| for each x in X.

Definition 5.1.9. [125], Let (X, d) be a cone metric space, P a normal cone with normal

constant M = 1, Hc(X) the set of all compact subsets of (X, τc) and A ∈ Hc(X). By

using above Lemma, we can define

hA : Hc(X) → [0,∞) and dH : Hc(X)×Hc(X) → [0,∞).

by

hA(B) = supx∈A

d′(x,B) and dH(A,B) = max{hA(B), hB(A)},

respectively.

Remark 5.1.3. [125], Let (X, d) be a cone metric space with normal constant M = 1.

Define ρ : X ×X → [0,∞) by ρ(x, y) = ||d(x, y)||. Then, (X, ρ) is a metric space. This

implies that for each A,B ∈ Hc(X) and x, y ∈ X, we have the following relations

(i) d′(x,A) ≤ ||d(x, y)||+ d′(y, A),

(ii) d′(x,A) ≤ d′(x,B) + hB(A), and

(iii) d′(x,A) ≤ ||d(x, y)||+ d′(y,B) + hB(A).

Theorem 5.1.13. Let (X,d) be a complete cone metric space with normal constant M = 1

and T1, T2 : X → Hc(X) two multifunctions satisfying the relation

dH(T1x, T2y) ≤ αd′(x, T1x) + βd′(y, T2y)


for all x, y ∈ X, where α + β < 1/2 and α > 0, β > 0 are constants. Then, T1 and T2

have a common fixed point, that is, there exists x ∈ X such that x ∈ T1x and x ∈ T2x.

Proof. Let x0 ∈ X be given. By Lemma 5.1.11, there is x1 ∈ T1x0 such that

d′(x0, T1x0) = ||d(x0, x1)||.

Also, there is x2 ∈ T2x1 such that

d′(x1, T2x1) = ||d(x1, x2)||.

Thus, we obtain a sequence {xn}n≥1 in X such that x2n−1 ∈ T1x2n−2, x2n ∈ T2x2n−1,

d′(x2n−2, T1x2n−2) = ||d(x2n−2, x2n−1)||,

and

d′(x2n−1, T2x2n−1) = ||d(x2n−1, x2n)||,

for all n ≥ 1. Thus, for all n ≥ 1 we have

||d(x2n, x2n+1)|| = d′(x2n, T1x2n)

≤ hT2x2n−1(T1x2n)

≤ dH(T2x2n−1, T1x2n)

≤ αd′(x2n, T1x2n) + βd′(x2n−1, T2x2n−1)

≤ α||d(x2n, x2n+1)||+ β||d(x2n−1, x2n)||.

||d(x2n, x2n+1)|| ≤( β

1− α

)||d(x2n−1, x2n)||

≤ c||d(x2n−1, x2n)||,

for all n ≥ 1 , α+ β < 1 and c = max{ β1−α ,

α1−β}. Also

||d(x2n−1, x2n)|| = d′(x2n−1, T2x2n−1)


≤ hT1x2n−2(T2x2n−1)

≤ dH(T1x2n−1, T2x2n−1)

≤ α||d(x2n−2, x2n−1)||+ β||d(x2n−1, x2n)||

||d(x2n−1, x2n)|| ≤( α

1− β

)||d(x2n−2, x2n−1)||

≤ c||d(x2n−2, x2n−1)||,

for all n ≥ 1. This implies that for all m ≥ 1

||d(xm, xm+1)|| ≤ c||d(xm−1, xm)||

≤ c2||d(xm−2, xm−1)||...

≤ cm||d(x0, x1)||.

Then for n > m we have

||d(xn, xm)|| ≤n∑

i=m+1

||d(xi, xi−1)||

≤ (cn−1 + cn−2 · · ·+ cm)||d(x0, x1)||

≤( cm

1− c

)||d(x0, x1)|| → 0, n,m→∞.

Hence, {xn}n≥1 is a Cauchy sequence in X. Thus, there exists x∗ ∈ X such that xn → x∗.

Now by using Remark 5.1.3, we have

d′(x∗, T1x∗) ≤ d′(x∗, T2x2n−1) + hT2x2n−1(T1x

∗)

≤ d′(x∗, T2x2n−1) + dH(T2x2n−1, T1x∗)

≤ ||d(x∗, x2n)||+ αd′(x2n−1, T2x2n−1) + βd′(x∗, T1x∗),

for all n ≥ 1. Hence,

d′(x∗, T1x∗) ≤ α

1− βd′(x2n−1, T2x2n−1) +

1

1− βd(x∗, x2n)


≤ cd′(x2n−1, T2x2n−1) +1

1− βd(x∗, x2n)

= cd(x2n−1, T2x2n−1) +1

1− βd(x∗, x2n),

for all n ≥ 1. Therefore, d′(x∗, T1x∗) = 0. By Lemma 5.1.12, x∗ ∈ T1x

∗. On the other

hand, similarly we have

d′(x∗, T2x∗) ≤ d′(x∗, T1x2n) + hT1x2n(T2x

∗)

≤ d′(x∗, T1x2n) + dH(T1x2n, T2x∗)

≤ ||d(x∗, x2n+1)||+ αd′(x2n, T1x2n) + βd′(x∗, T2x∗),


d′(x∗, T2x∗) ≤ α

1− βd′(x2n, T1x2n) +

1

1− β||d(x∗, x2n+1)||

≤ cd′(x2n, T1x2n) +1

1− β||d(x∗, x2n+1)||

= cd(x2n, T1x2n) +1

1− β||d(x∗, x2n+1)||,


∗. Therefore, x∗ is a

common fixed point of T1 and T2.

Theorem 5.1.14. Let (X, d) be a complete cone metric space with normal constant M = 1

and T1, T2 : X → Hc(X) two multifunctions satisfy the relation

dH(T1x, T2y) ≤ αd′(y, T1x) + βd′(x, T2y)

for all x, y ∈ X, where α + β < 1/2 and α, β > 0 are a constants. Then, T1 and T2 have

a common fixed point.

Proof. A similar argument to that of the proof of Theorem 5.1.13 shows that there exists

a Cauchy sequence {xn}n≥1 in X such that

x2n−1 ∈ T1x2n−2, x2n ∈ T2x2n−1,


d′(x2n−2, T1x2n−2) = ||d(x2n−2, x2n−1)||,

and d′(x2n−1, T2x2n−1) = ||d(x2n−1, x2n)||,

for all n ≥ 1. Thus, there exists x∗ ∈ X such that xn → x∗. Now by using Remark 5.1.3,

we have

d′(x∗, T1x∗) ≤ d′(x∗, T2x2n−1) + hT2x2n−1(T1x

∗)

≤ d′(x∗, T2x2n−1) + dH(T2x2n−1, T1x∗)

≤ ||d(x∗, x2n)||+ αd′(x∗, T2x2n−1) + βd′(x2n−1, T1x∗)

≤ ||d(x∗, x2n)||+ α||d(x∗, x2n)||+ βd′(x∗, T1x∗) + β||d(x∗, x2n−1)||,


d′(x∗, T1x∗) ≤

(1 + α

1− β

)||d(x∗, x2n)||+

( β

1− β

)||d(x∗, x2n−1)||,


∗. Also, we have

d′(x∗, T2x∗) ≤ d′(x∗, T1x2n) + hT1x2n(T2x

∗)

≤ d′(x∗, T1x2n) + dH(T1x2n, T2x∗)

≤ ||d(x∗, x2n+1)||+ αd′(x∗, T1x2n) + βd′(x2n, T2x∗)

≤ ||d(x∗, x2n+1)||+ α||d(x∗, x2n+1)||+ βd′(x∗, T2x∗) + β||d(x∗, x2n)||,


d′(x∗, T2x∗) =

(1 + α

1− β

)||d(x∗, x2n+1)||+

( β

1− β

)||d(x∗, x2n)||,


∗. Therefore, x∗ is a

common fixed point of T1 and T2.

5.1.5 A Common Fixed Point For Weakly Compatible Maps

In this section, a common fixed point theorem is proved for a pair of weakly compatible

self mappings defined on a cone metric space under a contractive condition.


Definition 5.1.10. [69] Let f and g be self mappings of a set X. If w = fx = gx for

some x in X, then x is called a coincidence point of f and g, and w is called a point of

coincidence of f and g.

Definition 5.1.11. [69] Two self mappings f and g of a set X are said to be weakly

compatible if they commute at their coincidence points, that is, if fu = gu for some

u ∈ X, then fgu = gfu.

Proposition 5.1.15. [3] Let f and g be weakly compatible self mappings of a set X. If

f and g have a unique point of coincidence, that is, w = fx = gx, then w is the unique

common fixed point of f and g.

Theorem 5.1.16. Let (X, d) be a cone metric and P be a normal cone with normal

constant K. Let the mappings f, g : X → X satisfy the following conditions:

(i) f(X) ⊂ g(X),

(ii) g(X) is complete subspace of X,

(iii) d(fx, fy) ≤ αd(fx, gy) + βd(fy, gx) + γd(fx, gx) + δd(fy, gy),

where α, β, γ, δ > 0 and 2α+2β+δ+γ < 1, then f and g have a unique coincidence point

in X. Moreover, if f and g are weakly compatible, then f and g have a unique common

fixed point.

Proof. Let x0 be an arbitrary point in X . Then, since f(X) ⊂ g(X), we choose a point

x1 in X such that f(x0) = g(x1). Continuing this process, having chosen xn in X, we

obtain xn+1 in X such that f(xn) = g(xn+1). Then

d(gxn+1, gxn) = d(fxn, fxn−1)

≤ αd(fxn, gxn−1) + βd(fxn−1, gxn) + γd(fxn, gxn) + δd(fxn−1, gxn−1)

≤ α{d(fxn, gxn) + d(gxn, gxn−1)}+ β{d(fxn−1, gxn+1) + d(gxn+1, gxn)}


+ γd(fxn, gxn) + δd(fxn−1, gxn−1)

= α{d(gxn+1, gxn) + d(gxn, gxn−1)}+ β{d(gxn, gxn+1) + d(gxn+1, gxn)}

+ γd(gxn+1, gxn) + δd(gxn, gxn−1)

= (α+ 2β + γ)d(gxn+1, gxn) + (α+ δ)d(gxn, gxn−1).

Hence,

d(gxn+1, gxn) ≤ hd(gxn, gxn−1),

where h = α+δ[1−(α+2β+γ)]

with 2α+ 2β + γ + δ < 1.

Now for n > m, we get

d(gxn, gxm) ≤ d(gxn, gxn−1) + d(gxn−1, gxn−2) + ...+ d(gxm+1, gxm)

≤ (hn−1 + hn−2 + ...+ hm)d(gx1, gx0)

≤ hm

(1− h)d(gx1, gx0).

Using the normality of cone P , implies that

‖d(gxn, gxm)‖ ≤ hm

(1− h)K||d(gx1, gx0)||.

Then d(gxn, gxm) → 0 as n,m → ∞ and so {g(xn)} is a Cauchy sequence in X. Since

g(X) is a complete subspace of X, so there exists q in g(X) such that g(xn) → q, as

n→∞. Consequently, we can find p in X such that g(p) = q. Thus,

d(gxn, fp) = d(fxn−1, fp)

≤ αd(fxn−1, gp) + βd(fp, gxn−1) + γd(fxn−1, gxn−1) + δd(fp, gp)

≤ αd(fxn−1, gp) + β[d(fp, gxn) + d(gxn, gxn−1)] + γd(fxn−1, gxn−1)+

δ[d(fp, gxn) + d(gxn + gp)]

(1− β − γ)d(gxn, fp) ≤ αd(fxn−1, q) + βd(gxn, gxn−1) + γd(fxn−1, gxn−1) + δd(gxn, q)


≤( 1

(1− β − γ)

)[αd(fxn−1, q) + βd(gxn, gxn−1) + γd(fxn−1, gxn−1)+

δd(gxn, q)]

which, using the normality of cone P , implies that

‖d(gxn, fp)‖ ≤( K

(1− β − γ)

)[||αd(fxn−1, q) + βd(gxn, gxn−1) + γd(fxn−1, gxn−1) + δd(gxn, q)||

]≤

( K

(1− β − γ)

)[α||d(fxn−1, q)||+ β||d(gxn, gxn−1)||+ γ||d(fxn−1, gxn−1)||+

δ||d(gxn, q)||]

= 0, as n→∞.

Hence, d(gxn, fp) → 0, as n → ∞. Also, we have d(gxn, gp) → 0, as n → ∞. The

uniqueness of a limit in a cone metric space implies that f(p) = g(p). Again, we show

that f and g have a unique point of coincidence. For this, if possible, there exists another

point t in X such that f(t) = g(t). Then, we have

d(gt, gp) = d(ft, fp)

≤ αd(ft, gp) + βd(fp, gt) + γd(ft, gt) + δd(fp, gp),

by the normality of cone P , implies that ||d(gt, gp)|| = 0, and so, we have gt = gp. Finally,

using the Proposition (5.1.15), we conclude that f and g have a unique common fixed

point.

Example 5.1.17. Let E = I2, for I = [0, 1], P = {(x, y) ∈ E|x, y ≥ 0} ⊂ I2, d : I × I →

E such that d(x, y) = (|x − y|, ξ|x − y|), where ξ > 0 is a constant. Define fx = ξx1+ξx

,

for all x ∈ I and gx = ξx for all x ∈ I. Then, for ξ = 1, both the mappings f and g

are weakly compatible and satisfy all the conditions of the above theorem with x = 0 as a

unique common fixed point.

5.2 Cone Rectangular Metric Space 79

5.2 Cone Rectangular Metric Space

In this section we introduce fixed point theorems in cone rectangular metric spaces and

proved some results on Banach contraction principle in a complete normal cone rectan-

gular metric space also generalize the Kannans fixed point theorem in a cone rectangular

metric. At the last we have proved result for expansive onto mappings on cone rectangular

metric spaces.

5.2.1 Introduction

Azam, Arshad and Beg, [7] introduced the notion of cone rectangular metric spaces by

replacing the triangular inequality of a cone metric space by a rectangular inequality. In

this chapter our main results are based on generalization of Banach contraction principle

on cone rectangular metric spaces result given by Akbar Azam, Muhammad Arshad and

Ismat Beg, [7] and the Kannans fixed point theorem in a cone rectangular metric [101].

Also C.T.Aage and J.N. Salunke, [12] proved fixed point theorem for expansion onto

mappings on cone metric spaces and we have extended result for expansive onto mappings

on cone rectangular metric.

5.2.2 Preliminary Notes

Definition 5.2.1. [101] Let X be a non empty set. Suppose the mapping d : X×X → E

satisfies:

(a) 0 ≤ d(x, y) for all x, y ∈ X and d(x, y) = 0 if and only if x = y;

(b) d(x, y) = d(y, x) for all x, y ∈ X;

(c) d(x, y) ≤ d(x,w) + d(w, z) + d(z, y) for all x, y ∈ X and for all distinct points

w, z ∈ X − {x, y} [Rectangular Property].

Then d is called a cone rectangular metric on X and d(X, d) is called cone rectangular


metric space.

Example 5.2.1. [101] Let X = N, E = R2 and P = {(x, y) ∈ E|x, y ≥ 0}.

Define d : X ×X → E as follow:

d(x, y) =

(0, 0) if x = y,

(3, 9) if x and y are in {1, 2},

(1, 3) if x and y can not both in {1, 2}.

Now (X, d) is a cone rectangular metric space but (X, d) is not a cone metric space because

it lacks the triangular property:

(3, 9) = d(1, 2) > d(1, 3) + d(3, 2) = (1, 3) + (1, 3) = (2, 6),

as (3, 9)− (2, 6) = (1, 3) ∈ P.

Example 5.2.2. [101] Let E = R2 and P = {(x, y) ∈ E|x, y ≥ 0} and X = R. Define

d : X ×X → E as follow:

d(x, y) =

(0, 0) if x = y,

(3α, 3) if x and y are in {1, 2}, x 6= y,

(α, 1) if x and y can not both in {1, 2}, x 6= y,

where α > 0 is a constant. Then (X, d) is a cone rectangular metric space but (X, d) is

not a cone metric space, since we have

(3α, 3) = d(1, 2) > d(1, 3) + d(3, 2) = (2α, 2).

Example 5.2.3. Let E = R2 and P = {(x, y) ∈ E|x, y ≥ 0} and X = {a, b, c, e}. Define

d : X ×X → E such that

d(x, x) = (0, 0) for all x ∈ X,

d(x, y) = d(y, x) for all x, y ∈ X,

d(a, b) = (3, α) ,

d(a, c) = d(b, c) = (1, α) ,

d(a, e) = d(b, e) = d(c, e) = (2, α) ,


where α > 0 is a constant. Then (X, d) is a cone rectangular metric space but it is not a

cone metric space, since we have

d(a, b) = (3, α) and d(a, c) + d(c, b) = (2, 2α),

but (3, α) and (2, 2α) cannot be compared with respect to ≤ .

Definition 5.2.2. [101] Let (X, d) be a cone rectangular metric space, x ∈ X and {xn}n≥1

a sequence in X. Then

(a) {xn}n≥1 is said to converge to x whenever for every c ∈ E with 0 � c there is a natural

number N such that d(xn, x) � c for all n ≥ N . We denote this by limn→∞ xn = x or

xn → x.

(b) {xn}n≥1 is said to be a Cauchy sequence whenever for every c ∈ E with 0 � c there

is a natural number N such that d(xn, xm) � c for all n,m ≥ N.

(c) (X, d) is called a complete cone rectangular metric space if every Cauchy sequence is

convergent in X.

Lemma 5.2.4. [101] Let (X, d) be a cone rectangular metric space, P be a normal cone.

Let {xn} be a sequence in X. Then

xn → x as n→ +∞⇔ ||d(xn, x)|| → 0 as n→ +∞.

Note that if (X, d) is a cone metric space and {xn} is a convergent sequence in X,

then the limit of {xn} is unique ([72]-Lemma 2). In our case uniqueness of the limit is

not satisfied in general. We have an example to illustrate this remark.

Example 5.2.5. [101] We take E = R and P = {x ∈ R|x ≥ 0}. Let {xn}n∈N be a

sequence in Q and a, b ∈ R/Q, a 6= b. We put X = {x1, x2, · · · , xn, · · · } ∪ {a, b} and we


consider d : X ×X → R defined by

d(x, x) = 0 for all x ∈ X,

d(x, y) = d(y, x) for all x, y ∈ X,

d(xn, xm) = 1 for all n,m ∈ N, n 6= m,

d(xn, b) = 1n

for all n ∈ N,

d(xn, a) = 1n

for all n ∈ N,

d(a, b) = 1 .

We remark (X, d) is not a cone metric space because we have

d(x2, x3) = 1 > d(x2, a) + (a, x3) =1

2+

1

3=

5

6.

However, (X, d) is a cone rectangular metric space. Now, since d(xn, a) = 1n→ 0 as

n → +∞, we obtain that xn → a as n → +∞. Also, we have d(xn, b) = 1n→ 0 as

n→ +∞ and then xn → b as n→ +∞.

Lemma 5.2.6. [101] Let (X, d) be a cone rectangular metric space, P be a normal cone.

Let {xn} be a sequence in X. Then {xn} is Cauchy sequence if and only if d(xn, xm) →

0 as n,m→ +∞.

Lemma 5.2.7. [101] Let (X, d) be a complete cone rectangular metric space, P be a

normal cone with normal constant κ. Let {xn} be a Cauchy sequence in X and suppose

that there is positive number N such that

(i) xn 6= xm for all n,m > N.

(ii) xn, x are distinct points in X, for all n > N.

(iii) xn, y are distinct points in X, for all n > N.

(iv) xn → x and xn → y, as n→ +∞.

Then x = y.


5.2.3 On Banach Contraction Principle in a Complete Normal

Cone Rectangular Metric Space

In this article we generalized the Banach contraction mapping principle in a complete

normal cone rectangular metric space.

First we present two theorems whose proofs are similar to Huang and Zhang [[72],

Lemmas 1 and 4].

Theorem 5.2.8. Let (X, d) be a cone rectangular metric space and P be a normal cone

with normal constant κ. Let {xn} be a sequence in X. Then {xn} converges to x if and

only if ‖d(xn, x)‖ → 0 as n→∞.

Theorem 5.2.9. Let (X, d) be a cone rectangular metric space, P be a normal cone with

normal constant κ. Let {xn} be a sequence in X. Then {xn} is a Cauchy sequence if and

only if ‖d(xn, xn+m)‖ → 0 as n→∞.

Theorem 5.2.10. Let (X, d) be a cone rectangular metric space, P be a normal cone

with normal constant κ and the mapping T : X → X satisfies:

d(Tx, Ty) ≤ αd(x, Tx) + βd(y, Ty)

for all x, y ∈ X, where 0 ≤ α+ β < 1. Then T has a unique fixed point.

Proof. Let x0 be an arbitrary point in X. Define a sequence of points in X as follows:

xn+1 = Txn = T n+1x0, n = 0, 1, 2, ...

We can suppose that x0 is not a periodic point, in fact if xn = x0, then,

d(x0, Tx0) = d(xn, Txn)

= d(T nx0, Tn+1x0)

≤ αd(T n−1x0, Tnx0) + βd(T nx0, T

n+1x0)


≤ hd(T n−1x0, Tnx0)

...

≤ hnd(x0, Tx0),

where 0 ≤ α+ β < 1 and h = α1−β . It follows that [hn − 1]d(x0, Tx0) ∈ P.

It further implies that [hn−1

1−hn ]d(x0, Tx0) ∈ P.

Hence −d(x0, Tx0) ∈ P and d(x0, Tx0) = 0, this means x0 is a fixed point of T . Thus in

this sequel of proof we can suppose that xm = xn, for all distinct m,n ∈ N. Now by using

rectangular property for all y ∈ X, we have,

d(y, T 4y) ≤ d(y, Ty) + d(Ty, T 2y) + d(T 2y, T 4y)

≤ d(y, Ty) + hd(y, Ty) + h2d(y, T 2y).

Similarly,

d(y, T 6y) ≤ d(y, Ty) + d(Ty, T 2y) + d(T 2y, T 3y) + d(T 3y, T 4y) + d(T 4y, T 6y)

≤ d(y, Ty) + hd(y, Ty) + h2d(y, Ty) + h3d(y, Ty) + h4d(y, T 2y)

=3∑i=0

hid(y, Ty) + h4d(y, T 2y), for all y ∈ X.

Now by induction, we obtain for each k = 2, 3, 4, ...,

d(y, T 2ky) ≤2k−3∑i=0

hid(y, Ty) + h2k−2d(y, T 2y). (5.2.1)

Moreover, for all y ∈ X,

d(y, T 5y) ≤ d(y, Ty) + d(Ty, T 2y) + d(T 2y, T 3y) + d(T 3y, T 4y) + d(T 4y, T 5y)

=4∑i=0

hid(y, Ty).

By induction, for each k = 0, 1, 2, ... we have,

d(y, T 2k+1y) ≤2k∑i=0

hid(y, Ty). (5.2.2)


Using Inequality (5.2.1), for k = 1, 2, 3, ... we have,

d(T nx0, Tn+2kx0) ≤ hnd(x0, T

2kx0)

≤ hn[ 2k−3∑i=0

hi(d(x0, Tx0) + d(x0, T2x0)) + h2k−2d(x0, Tx0) + d(x0, T

2x0)]

≤ hn2k−2∑i=0

hi[d(x0, Tx0) + d(x0, T2x0)]

≤ hn(1− h2k−1)

1− h

[d(x0, Tx0) + d(x0, T

2x0)]

≤ hn

1− h

[d(x0, Tx0) + d(x0, T

2x0)].

Similarly, for k = 0, 1, 2, ..., Inequality (5.2.2) implies that

d(T nx0, Tn+2k+1x0) ≤ hnd(x0, T

2k+1x0)

≤ hn2k∑i=0

hid(x0, Tx0)

≤( hn

1− h

)[d(x0, Tx0) + d(x0, T

2x0)].

d(T nx0, Tn+mx0) ≤

( hn

1− h

)[d(x0, Tx0) + d(x0, T

2x0)].

Since P is a normal cone with normal constant κ, therefore,

‖d(T nx0, Tn+mx0)‖ ≤

[ hn

1− h

]κ[‖d(x0, Tx0) + d(x0, T

2x0)‖].

Therefore, ‖d(T nx0, Tn+mx0)‖ → 0 as n → ∞. Now Theorem 5.2.9 implies that {xn} is

a Cauchy sequence in X. Since X is complete, there exists u ∈ X such that xn → u. By

Theorem 5.2.8, we have

‖d(T nx0, u)‖ → 0, as n→∞.

By rectangular property, we have

d(Tu, u) ≤ d(u, T n−1x0) + d(T n−1x0, Tn+1x0) + d(T n+1x0, u)


≤ d(u, T n−1x0) + hn−1d(x0, T2x0) + d(T n+1x0, u).

Thus

‖d(Tu, u)‖ ≤ κ[‖d(u, T n−1x0)‖+ hn−1‖d(x0, T

2x0)‖+ ‖d(T n+1x0, u)‖]. (5.2.3)

Letting n → ∞, we have ‖d(u, Tu)‖ = 0. Hence u = Tu. Now we show that T have a

unique fixed point. For this, assume that there exists another point v in X such that

v = Tv. Now,

d(v, u) = d(Tv, Tu)

≤ αd(v, Tv) + βd(u, Tu)

= 0.

Hence, u = v.

5.2.4 On Kannans Fixed Point Theorem in a Cone Rectangular

Metric

In this article our main result is On Kannans fixed point theorem in a cone rectangular

metric space.

Theorem 5.2.11. Let (X, d) be a complete cone rectangular metric space, P be a normal

cone with normal constant k. Suppose a mapping T : X → X satisfies the contractive

condition

d(Tx, Ty) ≤ αd(x, Tx) + βd(x, y) + γd(y, Ty),

for all x, y ∈ X,α, β, γ > 0 and α+ 2β + γ < 1. Then,

(i) T has a unique fixed point in X.

(ii) For any, the iterative sequence converges to the fixed point.


Proof. Let x0 ∈ X and Txn = xn+1. Now

d(Tx0, T2x0) = d(Tx0, T (T (x0))

≤ αd(x0, Tx0) + βd(x0, Tx0) + γd(Tx0, T2x0)

(1− γ)d(Tx0, T2x0) ≤ (α+ β)d(x0, Tx0)

d(Tx0, T2x0) ≤

(α+ β

1− γ

)d(x0, Tx0), (5.2.4)

d(T 2x0, T3x0) ≤

(α+ β

1− γ

)d(Tx0, T

2x0)

≤(α+ β

1− γ

)(α+ β

1− γ

)d(x0, Tx0)

=(α+ β

1− γ

)2

d(x0, Tx0). (5.2.5)

Thus in general, if n is positive integer, then

d(T nx0, Tn+1x0) ≤ hnd(x0, Tx0),

where h = α+β1−γ and 0 < h < 1.

We divide the proof into two cases.

Case I: First assume that Tmx0 = T nx0 for m,n ∈ N,m 6= n.

Let m > n, then Tm−n(T nx0) = T nx0, i.e. T py = y, where p = m − n, y = T nx. Now

since p > 1, we have

d(y, Ty) = d(T py, T p+1y)

≤ hpd(y, Ty),

where 0 < h < 1. We obtain −d(y, Ty) ∈ P and d(y, Ty) ∈ P which implies that

‖ d(y, Ty) ‖= 0, i.e. Ty = y.

Case II: Assume that Tmx0 6= T nx0 for m,n ∈ N,m 6= n.

Clearly we have

d(T nx0, Tn+1x0) ≤ hnd(x0, Tx0),


and

d(T nx0, Tn+2x0) =d(T (T n−1x0, T (T n+1x0))

≤ αd(T n−1x0, Tnx0) + βd(T n−1x0, T

n+1x0)+

γd(T n+1x0, Tn+2x0)

≤ αhn−1d(x0, Tx0) + β[d(T n−1x0, T

nx0) + d(T nx0, Tn+2x0)+

d(T n+2x0, Tn+1x0)

]+ γhn+1d(x0, Tx0)

(1− β)d(T nx0, Tn+2x0) ≤ αhn−1d(x0, Tx0) + βhn−1d(x0, Tx0)+

βhn+1d(x0, Tx0) + γhn+1d(x0, Tx0)

= hn−1(α+ β)d(x0, Tx0) + hn+1(β + γ)d(x0, Tx0)

d(T nx0, Tn+2x0) ≤ hn−1d(x0, Tx0).

Now if m > 2 is odd then writing m = 2`+ 1, ` ≥ 1 and using the fact that T px0 6= T rx0

for p, r ∈ N, p 6= r, we get

d(T nx0, Tn+mx0) ≤ d(T nx0, T

n+1x0) + d(T n+1x0, Tn+2x0) + ...+ d(T n+2`x0, T

n+2`+1x0)

≤ hnd(x0, Tx0) + hn+1d(x0, Tx0) + ...+ hn+2`d(x0, Tx0)

<hn

1− hd(x0, Tx0).

Again if m > 2 is even then writing m = 2`, ` ≥ 2 and using the same arguments as

before, we get

d(T nx0, Tn+mx0) ≤ d(T nx0, T

n+2x0) + d(T n+2x0, Tn+3x0) + ...+ d(T n+2`−1x0, T

n+2`x0)

≤ hn−1d(x0, Tx0) + hn+2d(x0, Tx0) + ...+ hn+2`−1d(x0, Tx0)

≤[hn−1 + hn + hn+1 + · · ·

]d(x0, Tx0)

<hn−1

1− hd(x0, Tx0).

Thus combining all the cases we have

d(T nx0, Tn+mx0) <

hn−1

1− hd(x0, Tx0),


for all m,n ∈ N.

Hence we get

‖ d(T nx0, Tn+mx0) ‖≤ k

( hn−1

1− h

)‖ d(x0, Tx0) ‖,

for all m,n ∈ N.

Since k(hn−1

1−h

)‖ d(x0, Tx0) ‖→ 0, as n → +∞, {T nx0} is a Cauchy sequence. By the

completeness of X, there is x∗ ∈ X such that T nx0 → x∗, as n→ +∞.

We shall now show that Tx∗ = x∗. Without any loss of generality, we can assume that

T rx∗ 6= x∗, for any r ∈ N. We have

d(x∗, Tx∗) ≤ d(x∗, T nx∗) + d(T nx∗, T n+1x∗) + d(T n+1x∗, Tx∗)

≤ d(x∗, T nx∗) + d(T nx∗, T n+1x∗) + αd(T nx∗, T n+1x∗) + βd(T nx∗, x∗) + γd(x∗, Tx∗)

(1− γ)d(x∗, Tx∗) ≤ (1 + α)d(T n+1x∗, T nx∗) + (1 + β)d(T nx∗, x∗)

d(x∗, Tx∗) ≤(1 + α

1− γ

)d(T n+1x∗, T nx∗) +

(1 + β

1− γ

)d(T nx∗, x∗)

≤(1 + α

1− γ

) hn

1− hd(Tx∗, x∗) +

(1 + β

1− γ

) hn−1

1− hd(Tx∗, x∗).

Hence

‖ d(x∗, Tx∗) ‖ ≤(1 + α

1− γ

) hn

1− h‖ d(Tx∗, x∗) ‖ +

(1 + β

1− γ

) hn−1

1− h‖ d(Tx∗, x∗) ‖

→ 0, as n→ +∞.

So we obtain d(x∗, Tx∗) = 0, i.e., Tx∗ = x∗.

Now, if y∗ is another fixed point of T , then

d(x∗, y∗) = d(Tx∗, T y∗)

≤ αd(Tx∗, x∗) + βd(x∗, y∗) + γd(y∗, T y∗)

< d(x∗, y∗), since 0 < β < 1.

which implies that d(x∗, y∗) = 0, i.e., x∗ = y∗.


5.2.5 An Expansion Mappings on Cone Rectangular Metric Spaces

C.T.Aage and J.N. Salunke, [12] proved fixed point theorem for expansion onto mappings

on cone metric spaces in this article we have proved fixed point theorems using same

expansion onto mappings on cone rectangular metric spaces.

Theorem 5.2.12. Let (X, d) be a complete cone rectangular metric space. Suppose the

mapping T : X → X is onto and satisfies the contractive condition

d(Tx, Ty) ≥ Kd(x, y), (5.2.6)

for all x, y ∈ X, where K > 1 is a constant. Then T has a unique fixed point in X.

Proof. If Tx = Ty, then

0 ≥ Kd(x, y) ⇒ 0 = d(x, y) ⇒ x = y.

Thus, T is one to one. Define G = T−1

d(x, y) = d(TT−1x, TT−1y) ≥ Kd(T−1x, T−1y) = Kd(Gx,Gy),

So d(Gx,Gy) ≤ Md(x, y), where M = 1K< 1. By Theorem (2.3) in [123], G has unique

fixed point u in X i.e. Gu = u ⇒ T−1u = u ⇒ u = Tu. Therefore, u is a unique fixed

point of T .

Corollary 5.2.13. Let (X, d) be a complete cone rectangular metric space. Suppose a

mapping T : X → X is onto and satisfies for some positive integer n

d(T nx, T ny) ≥ Kd(x, y),

for all x, y ∈ X, where K > 1 is a constant. Then T has a unique fixed point in X.


mapping T : X → X is continuous, onto and satisfies the contractive condition

d(Tx, Ty) ≥ K(d(Tx, x) + d(Ty, y)),

for all x, y ∈ X, where 12< K ≤ 1 is a constant. Then T has a fixed point in X.


Proof. For each x0 ∈ X, since T is onto, there exist x1 ∈ X such that Tx1 = x0. Similarly,

for each n ≥ 1 there exist xn+1 ∈ X such that xn = Txn+1. If xn−1 = xn, then xn is a

fixed point of T . Thus, Suppose that xn−1 6= xn for all n ≥ 1. Then

d(xn, xn−1) = d(Txn+1, Txn)

≥ K(d(Txn+1, xn+1) + d(Txn, xn))

= K(d(xn, xn+1) + d(xn−1, xn)),

so,

d(xn, xn+1) ≤(1−K

K

)d(xn−1, xn)

= hd(xn−1, xn).

From this we get d(xn, xn+1) ≤ hnd(x0, x1), where h = 1−KK

and 0 ≤ h < 1. Now for

n < m we have

d(xn, xm) ≤ d(xn, xn+1) + d(xn+1, xn+2) + · · ·+ d(xm−1, xm)

≤ (hn + hn+1 + · · ·+ hm−1)d(x0, x1)

≤ hn

1− hd(x0, x1).

Let 0 ≤ c be given. Choose a natural number N1 such that hn

1−hd(x0, x1) ≤ c, for all

n ≥ N1. Thus d(xn, xm) ≤ c, for n < m. Therefore, {xn}n≥1 is Cauchy sequence in

(X, d). Since (X, d) is a complete cone rectangular metric space, there exist x∗ ∈ X such

that xn → x∗, as n→∞.

If T is continuous, then

d(Tx∗, x∗) ≤ d(Tx∗, Txn) + d(Txn, Txn+1) + d(Txn+1, x∗) → 0, as n→∞,

since xn → x∗ and Txn → Tx∗ as n → ∞. Therefore d(Tx∗, x∗) = 0 and so Tx∗ = x∗.

Thus T having fixed point in X.




d(Tx, Ty) ≥ kd(x, y) + ld(Tx, y),

for all x, y ∈ X, where k > 1, l ≥ 0 are constants. Then T has a fixed point in X.





≥ kd(xn+1, xn) + ld(Txn+1, xn)

= kd(xn+1, xn) + ld(xn, xn)

= kd(xn+1, xn),

so,

d(xn, xn+1) ≤1

kd(xn−1, xn)

= hd(xn−1, xn),

where h = 1k. Now for n < m we have


≤ (hn + hn+1 + · · ·+ hm−1)d(x0, x1)

≤ hn

1− hd(x0, x1).



n ≥ N1. Thus d(xn, xm) ≤ c, for N1 < n < m. Therefore, {xn}n≥1 is Cauchy sequence in










d(Tx, Ty) ≥ kd(x, y) + ld(x, Tx) + pd(y, Ty),

for all x, y ∈ X, where k ≥ −1, l > 1 and p < 1 are constants with k + l + p > 1. Then

T has a fixed point in X.





≥ kd(xn+1, xn) + ld(xn+1, Txn+1) + pd(xn, Txn)

= kd(xn+1, xn) + ld(xn+1, xn) + pd(xn, xn−1),

so,

d(xn, xn+1) ≤(1− p

k + l

)d(xn, xn−1)

= hd(xn, xn−1),

where h = 1−pk+l

with 0 < h < 1. Now for n < m we have


≤ (hn + hn+1 + · · ·+ hm−1)d(x0, x1)


≤ hn

1− hd(x0, x1).



n ≥ N1. Thus d(xn, xm) ≤ c, for N1 < n < m. Therefore, {xn}n≥1 is Cauchy sequence in







chapter 5 a common fixed points in cone and rectangular cone...

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