International Journal of Analysis and Applications
ISSN 2291-8639
Volume 9, Number 1 (2015), 9-18
http://www.etamaths.com

FIXED POINTS OF α-ADMISSIBLE MAPPINGS IN CONE METRIC

SPACES WITH BANACH ALGEBRA

S.K. MALHOTRA1 J.B. SHARMA2 AND SATISH SHUKLA3,∗

Abstract. In this paper, we introduce the α-admissible mappings in the set-

ting of cone metric spaces equipped with Banach algebra and solid cones. Our

results generalize and extend several known results of metric and cone metric
spaces. An example is presented which illustrates and shows the significance

of results proved herein.

Keywords. Cone metric space; α-admissible mapping; Solid cone; Banach

algebra; Fixed point.
AMC 2000. 47H10; 54H25.

1. Introduction

Huang and Zhang [7] introduced the notion of cone metric spaces as a generaliza-
tion of metric spaces. They defined the distance of two points of space in terms of a
vector lying in a particular subset of a Banach space called cone. They also defined
the Cauchy sequence and convergence of a sequence in such spaces in terms of inte-
rior points of the underlying cone. Moreover, they proved the Banach contraction
principle in the setting of cone metric spaces with the assumption that the cone is
normal. Later, the assumption of normality of cone was removed by Rezapour and
Hamlbarani [10]. Huang and Zhang [7] also gave an example and showed the de-
pendency of contractive nature of mappings on the cone metric spaces. Although,
some authors (see, e.g., [3, 16, 13, 14]) showed that the fixed point results proved
on cone metric spaces are the simple consequences of corresponding results of usual
metric spaces.

Liu and Xu [4] used the cones over a Banach algebra and proved some fixed
point theorems on cone metric spaces. They improved the contractive condition on
self-maps of cone metric spaces by replacing the contractive constant by a vector of
cone. They also gave an example which shows that their fixed point results cannot
be obtained by the corresponding results on usual metric spaces with an approach
used, e.g., in [3, 16, 13, 14]. The results proved by Liu and Xu [4] demands the
normality of the underlying cone. Later on, Xu and Radenović [9] showed that the
condition of normality of cone can be removed, and so, the results of Liu and Xu
[4] are also true in case of a non-normal cone.

On the other hand, Samet et al. [2] introduced the study of fixed points for
the α-admissible mappings and generalized several known results of metric spaces.
In this paper, we use the concept of α-admissibility of mappings defined on cone

2010 Mathematics Subject Classification. 47H10.

Key words and phrases. Fixed points; univalent; α-admissible mappings.

c©2015 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

9



10 MALHOTRA, SHARMA AND SHUKLA

metric spaces with Banach algebra and define the generalized Lipschitz contractions
on such spaces. Our results extend and generalize several known results of metric
and cone metric spaces. An example is also provided which verifies the usability
and significance of our results.

2. Preliminaries

First, we recall some definitions and results about the Banach algebras and cone
metric spaces.

Let A be a real Banach algebra, i.e., A is a real Banach space in which an
operation of multiplication is defined, subject to the following properties: for all
x,y,z ∈ A,a ∈ R

(1) x(yz) = (xy)z;
(2) x(y + z) = xy + xz and (x + y)z = xz + yz;
(3) a(xy) = (ax)y = x(ay);
(4) ‖xy‖≤‖x‖‖y‖.

In this paper, we shall assume that the Banach algebra A has a unit, i.e., a multi-
plicative identity e such that ex = xe = x for all x ∈ A. An element x ∈ A is said
to be invertible if there is an inverse element y ∈ A such that xy = yx = e. The
inverse of x is denoted by x−1. For more details we refer to [11].

The following proposition is well known [11].

Proposition 2.1. Let A be a real Banach algebra with a unit e and x ∈ A. If the
spectral radius ρ(x) of x is less than one.,

ρ(x) = lim
n→∞

‖xn‖
1
n = inf

n≥1
‖xn‖

1
n < 1

then e−x is invertible. Actually,

(e−x)−1 =
∞∑
i=0

xi.

A subset P of A is called a cone if

(1) P is non-empty, closed and {θ,e}⊂ P, where θ is the zero vector of A;
(2) a1P + a2P ⊂ P for all non-negative real numbers a1,a2;
(3) P2 = PP ⊂ P
(4) P

⋂
(−P) = {θ}.

For a given cone P ⊂ A, 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 y − x ∈ P◦,
where P◦ denotes the interior of P.

The cone P is called normal if there exists a number K > 0 such that for all
a,b ∈ A,

a � b implies ‖a‖≤ K‖b‖.
The least positive value of K satisfying the above inequality is called the normal
constant (see [7]). Note that, for any normal cone P we have K ≥ 1 (see [10]). In
the following we always assume that P is a cone in a real Banach algebra A with
P◦ 6= φ (i.e., the cone P is a solid cone) and � is the partial ordering with respect
to P .

The following lemmas and remark will be useful in the sequel.



FIXED POINTS OF α-ADMISSIBLE MAPPINGS 11

Lemma 2.2 (See [15]). If E is a real Banach space with a cone P and if a � λa
with a ∈ P and 0 ≤ λ < 1, then a = θ.

Lemma 2.3 (See [8]). If E is a real Banach space with a solid cone P and if θ �
u � c for each θ � c, then u = θ.

Lemma 2.4 (See [8]). If E is a real Banach space with a solid cone P and if ‖xn‖→ 0
as n →∞, then for any θ � c, there exists n0 ∈ N such that, xn � c for all n < n0.

Remark 2.5 (See [9]). If ρ(x) < 1 then ‖xn‖→ 0 as n →∞.

Definition 2.6 (See [4, 5, 7]). Let X be a non-empty set. Suppose that the mapping
d: X ×X → A satisfies:

(1) θ � d(x,y) for all x,y ∈ X and d(x,y) = θ if and only if x = y.
(2) d(x,y) = d(y,x) for all x,y ∈ X.
(3) 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 over
the Banach algebra A.

Definition 2.7 (See [7]). Let (X,d) be a cone metric space, x ∈ X and {xn} be a
sequence in X. Then:

(1) The sequence {xn} converges to x whenever for each c ∈ A with θ � c,
there is n0 ∈ N such that d(xn,x) � c for all n > n0. We denote this by
lim
n→∞

xn = x or xn → x as n →∞.
(2) The sequence {xn} is a Cauchy sequence whenever for each c ∈ A with

θ � c, there is n0 ∈ N such that d(xn,xm) � c for all n,m > n0.
(3) (X,d) is a complete cone metric space if every Cauchy sequence is conver-

gent in X.

It is obvious that the limit of a convergent sequence in a cone metric space is
unique. A mapping T : X → X is called continuous at x ∈ X, if for every sequence
{xn} in X such that xn → x as n →∞, we have Txn → Tx as n →∞.

Definition 2.8 (See [2]). Let X be a nonempty set and α: X × X → [0,∞) be a
function. We say that T is α-admissible if (x,y) ∈ X, α(x,y) ≥ 1 =⇒ α(Tx,Ty) ≥
1.

Now, we define the generalized Lipschitz contractions on the cone metric spaces
with a Banach algebra (see also, [4]).

Definition 2.9. Let (X,d) be a complete cone metric space over a Banach algebra
A and P be the underlying solid cone. Then the mapping T : X → X is said to be
generalized Lipschitz contraction if there exists k ∈ P such that ρ(k) < 1 and,

d(Tx,Ty) � kd(x,y)

for all x,y ∈ X with α(x,y) ≥ 1. Here, the vector k is called the Lipschitz vector
of T.

Now we can state our main results.



12 MALHOTRA, SHARMA AND SHUKLA

3. Main results

First, we prove an existence theorem for a generalized Lipschitz contraction on
cone metric space over Banach algebras.

Theorem 3.1. Let (X,d) be a complete cone metric space over a Banach algebra A
and P be the underlying solid cone. Suppose, T : X → X be a generalized Lipschitz
contraction with Lipschitz vector k and the following conditions are satisfied:

(i) T is α-admissible;
(ii) there exists x0 ∈ X such that α(x0,Tx0) ≥ 1;

(iii) T is continuous.

Then T has a fixed point x∗ ∈ X.

Proof. Let x0 ∈ X such that α(x0,Tx0) ≥ 1 and define a sequence {xn} in X such
that xn = Txn−1 for all n ∈ N. If xn = xn+1 for some n ∈ N, then x∗ = xn is a
fixed point for T. Assume that xn 6= xn+1 for all n ∈ N. Since T is α-admissible
we have

α(x0,x1) = α(x0,Tx0) ≥ 1 =⇒ α(Tx0,T2x0) = α(x1,x2) ≥ 1.

By induction, we get

(1) α(xn,xn+1) ≥ 1 for all n ∈ N.

Since T is generalized Lipscitz contraction, then

d(xn,xn+1) = d(Txn−1,Txn)

� kd(xn−1,xn)
...

� knd(x0,x1).

Thus, for n < m we have

d(xn,xm) � d(xn,xn+1) + d(xn+1,xn+2) + · · · + d(xm−1,xm)
� knd(x0,x1) + kn+1d(x0,x1) + · · · + km−1d(x0,x1)
= (e + k + ... + km−n−1)knd(x0,x1)

�

(
∞∑
i=0

ki

)
knd(x0,x1)

= (e−k)−1knd(x0,x1).

Since ρ(k) < 1, by Remark 2.5 we have ‖kn‖→ 0 as n →∞. Therefore, by Lemma
2.4 it follows that: for every c ∈ A with θ � c there exists n0 ∈ N such that

d(xn,xm) � (e−k)−1knd(x0,x1) � c

for all n > n0. It implies that {xn} is a Cauchy sequence. By completeness of X,
there exists x∗ ∈ X such that xn → x∗ as n →∞. Since T is continuous, it follows
that xn+1 = Txn → Tx∗ as n → ∞. By the uniqueness of limit we get x∗ = Tx∗,
that is x∗ is a fixed point of T. �

In the above theorem, we use the continuity of the mapping T. Now, we show
that the assumption of continuity can be replaced by another condition.



FIXED POINTS OF α-ADMISSIBLE MAPPINGS 13

Theorem 3.2. Let (X,d) be a complete cone metric space over a Banach algebra A
and P be the underlying solid cone. Suppose, T : X → X be a generalized Lipschitz
contraction with Lipschitz vector k and the following conditions are satisfied:

(i) T is α-admissible;
(ii) there exists x0 ∈ X such that α(x0,Tx0) ≥ 1;

(iii) if xn is a sequence in X such that α(xn,xn+1) ≥ 1 for all n and xn → x ∈ X
as n →∞, then α(xn,x) ≥ 1 for all n ∈ N.

Then T has a fixed point x∗ ∈ X.

Proof. By proof of theorem 3.1, we know that the sequence {xn} is a Cauchy
sequence in complete cone metric space (X,d). Then, there exists x∗ ∈ X such that
xn → x∗ as n →∞. On the other hand, from (1) and hypothesis (iii), we have

(2) α(xn,x
∗) ≥ 1, for all n ∈ N.

Since T is a generalized Lipschitz contraction, using (2) we obtain

d(x∗,Tx∗) � d(x∗,xn+1) + d(xn+1,Tx∗)
= d(x∗,xn+1) + d(Txn,Tx

∗)

� d(x∗,xn+1) + kd(xn,x∗).

As xn → x∗ as n →∞, for every c ∈ P with θ � c and for every m ∈ N there exists
n(m) such that d(xn,x

∗) � c
2m

for all n > n(m). Therefore, kd(xn,x
∗) � kc

2m
and

it follows from the above inequality that

d(x∗,Tx∗) �
c

2m
+

kc

2m
=

c

2m
(e + k) for all n > n(m),m ∈ N.

It implies that c
2m

(e + k)−d(x∗,Tx∗) ∈ P for all m ∈ N. Since P is closed, letting
m →∞ we obtain θ−d(x∗,Tx∗) ∈ P . By definition, we must have d(x∗,Tx∗) = θ,
i.e., Tx∗ = x∗. Thus, x∗ is a fixed point of T. �

Next, we give an example which illustrate the above result.

Example 3.3. Let A = C1R[0, 1] ×C
1
R[0, 1] with the norm

‖(x1,x2)‖ = ‖x1‖∞ + ‖x2‖∞ + ‖x′1‖∞ + ‖x
′
2‖∞.

Define the multiplication on X by

xy = (x1y1,x1y2 + x2y1) for all x = (x1,x2),y = (y1,y2) ∈ X.

Then, A is a Banach algebra with usual sum of functions and scalar produc-
t on cartesian product C1R[0, 1] × C

1
R[0, 1] and with unit e = (0, 1). Let P =

{(x1(t),x2(t)) ∈ A: x1(t),x2(t) ≥ 0, t ∈ [0, 1]}. Then P is a cone which is not
normal.

Let X = R2 and define the cone metric d: X ×X → P by

d((x1,x2), (y1,y2)) = (|x1 −y1|, |x2 −y2|) et ∈ P.

Then, (X,d) is a complete cone metric space. For a constant a ∈ Q, define the
mappings T : X → X and α: X ×X → [0,∞) by:

T(x1,x2) =

{ (x1
2
,
x2
3

+ ax1

)
, if (x1,x2) ∈ Q×Q;

(x1,x2), otherwise



14 MALHOTRA, SHARMA AND SHUKLA

and

α((x1,x2), (y1,y2)) =

{
1, if (x1,x2), (y1,y2) ∈ Q×Q;
0, otherwise.

Then, T is a generalized Lipschitz contraction with Lipschitz vector k =

(
1

2
,a

)
,

where ρ(k) =
1

2
< 1. Indeed, α(x1,x2) ≥ 1 implies that (x1,x2), (y1,y2) ∈ Q × Q.

Therefore,

d(T(x1,x2),T(y1,y2)) =

(
1

2
|x1 −y1|,

1

3
|x2 −y2| + a|x1 −y1|

)
et

�
(

1

2
|x1 −y1|,

1

2
|x2 −y2| + a|x1 −y1|

)
et

=

(
1

2
,a

)
d((x1,x2), (y1,y2)).

Since a ∈ Q, the mapping T is an α-admissible mapping, and for every (x1,x2) ∈
Q × Q we have α((x1,x2),T(x1,x2)) = 1. Therefore, the conditions (i) and (ii)
of Theorem 3.2 are satisfied. Finally, since Q is complete, the condition (iii) of
Theorem 3.2 is satisfied. Thus, all the conditions of Theorem 3.2 are satisfied and
we conclude the existence of at least one fixed point of T. Indeed, (0, 0) and all the
points of (X ×X) \ (Q×Q) are the fixed points of T.

Remark 3.4. In the above example, the mapping T is not a continuous mapping
on the space X. Also,

(
1
2
,a
)
6� (1, 0) = e and

∥∥(1
2
,a
)∥∥ = 1+2a

2
> 1 (for a > 1).

For large enough a one can see that the mapping is not a contraction in the sense
of Samet et al. [2] with respect to Euclidian metric on X. Again, it is easy to see
that the mapping T is not a contraction in the sense of Liu and Xu [4], and so, we
can not apply these known results on the mapping T. Moreover, following similar
arguments to those in the Remark 2.3 of the paper [4] we can say that our results
are actual generalization of the known results.

In the Example 3.3 we can see that the mapping T may have more than one
fixed points. Let us denote the set of all fixed points of T by Fix(T).

Next, to assure the uniqueness of fixed point of a generalized Lipschitz mapping
we use the following property (see [2]):

(H) ∀ x,y ∈ Fix(T) ∃ z ∈ X : α(x,z) ≥ 1,α(y,z) ≥ 1.

Theorem 3.5. Adding condition (H) to the hypothesis of Theorem 3.1 (resp. Theo-
rem 3.2) we obtain the uniqueness of the fixed point of T.

Proof. Following similar arguments to those in the proof of Theorem 3.1 (resp.
Theorem 3.2) we obtain the existence of fixed point. Let the condition (H) is
satisfied and x∗,y∗ ∈ Fix(T) and x∗ 6= y∗. By (H) there exists z ∈ X such that

(3) α(x∗,z) ≥ 1 and α(y∗,z) ≥ 1.

Since T is α-admissible and x∗,y∗ ∈ Fix(T), therefore from (3) we obtain

(4) α(x∗,Tnz) ≥ 1 and α(y∗,Tnz) ≥ 1. for all n ∈ N.



FIXED POINTS OF α-ADMISSIBLE MAPPINGS 15

Since T is generalized Lipschitz contraction, using (4), we have

d(x∗,Tnz) = d(Tx∗,T(Tn−1z))

� kd(x∗,Tn−1z)
...

� knd(x∗,z) for all n ∈ N.
Since ρ(k) < 1, by Remark 2.5 we have ‖kn‖→ 0 as n →∞, and so,

‖knd(x∗,z)‖≤‖kn‖‖d(x∗,z)‖→ 0 as n →∞.
Therefore, by Lemma 2.4 it follows that: for every c ∈ A with θ � c there exists
n0 ∈ N such that

d(x∗,Tnz) � knd(x∗,z) � c.
it implies that

Tnz → x∗ as n →∞.
Similarly we get

Tnz → y∗ as n →∞.
Therefore, by uniqueness of the limit we obtain x∗ = y∗. This finishes the proof. �

4. Some consequences

In this section, we give some consequences of the results of previous section. The
following result is an improved version of Theorem 2.1 of Liu and Xu [4].

Theorem 4.1 (Xu and Radenović [9]). Let (X,d) be a complete cone metric space
over a Banach algebra A and P be the underlying solid cone with k ∈ P where
ρ(k) < 1. Suppose the mapping T : X → X satisfies generalized Lipschitz condition
:

d(Tx,Ty) � kd(x,y) for all x,y ∈ X.
Then T has a unique fixed point in X. Moreover, for any x ∈ X, the iterative
sequence {Tnx} converges to the fixed point of X.

Proof. Define the function α: X × X → [0,∞) by α(x,y) = 1 for all x,y ∈ X.
Then, all the conditions of Theorem 3.5 are satisfied, and so, the mapping T has a
unique fixed point in X. �

Next, we derive the ordered and cyclic versions of Banach contraction principle.
In the next theorems, we generalize and unify the results of Ran and Reurings [1],
Liu and Xu [4] and Nieto, Rodŕıguez-López [6] and Kirk et al. [12].

The following theorem is the cone metric version of Ran and Reurings [1] when
the cone metric is endowed with a Banach algebra.

Theorem 4.2. Let (X,v) be a partially ordered set and suppose that (X,d) be a
complete cone metric space (X,d) over a Banach algebra A with P the underlying
solid cone. Let T : X → X be a continuous nondecreasing mapping with respect to
v . Suppose that the following two assumptions hold:

(i) there exists k ∈ P such that ρ(k) < 1 and d(Tx,Ty) � kd(x,y) for all x,y ∈ X
with x v y;

(ii) there exists x0 ∈ X such that x0 v Tx0.
Then, T has a fixed point in X.



16 MALHOTRA, SHARMA AND SHUKLA

Proof. Define the mapping αr : X ×X → [0,∞) by

αr(x,y) =

{
1, if x v y;
0, otherwise.

Note that, the condition (i) implies that the mapping T a generalized Lipschitz
contraction with Lipschitz vector k, where ρ(k) < 1. Since T is nondecreasing it is
an αr-admissible mapping. The condition (ii) implies that, there exists x0 ∈ X such
that αr(x0,Tx0) = 1. Therefore, all the conditions of Theorem 3.1 are satisfied, and
so, the mapping T has a fixed point in X. �

The following theorem is the cone metric version of Nieto, Rodŕıguez-López [6]
when the cone metric is endowed with a Banach algebra.

Theorem 4.3. Let (X,v) be a partially ordered set and suppose that (X,d) be a
complete cone metric space (X,d) over a Banach algebra A with P the underlying
solid cone. Let T : X → X be a nondecreasing mapping with respect to v . Suppose
that the following three assumptions hold:

(i) there exists k ∈ P such that ρ(k) < 1 and d(Tx,Ty) � kd(x,y) for all x,y ∈ X
with x v y;

(ii) there exists x0 ∈ X such that x0 v Tx0;
(iii) if {xn} is a nondecreasing sequence in X such that xn → x ∈ X as n → ∞,

then xn v x for all n ∈ N.
Then, T has a fixed point in X.

Proof. Define the mapping αr : X ×X → [0,∞) similar to that as in the proof of
Theorem 4.2. Now, the proof follows from the Theorem 3.2. �

Next, we define the cyclic contractions (see [12]) in cone metric spaces.
Let X be a nonempty set, T : X → X a mapping and A1,A2, . . . ,Am be subsets

of X. Then X =
m⋃
i=1

Ai is a cyclic representation of X with respect to T if

(1) Ai, i = 1, 2, . . . ,m are nonempty sets;
(2) T(A1) ⊂ A2, . . . ,T(Am−1) ⊂ T(Am),T(Am) ⊂ T(A1).

Remark 4.4. (See [12]) If X =
m⋃
i=1

Ai is a cyclic representation of X with respect

to T , then Fix(T) ⊂
m⋂
i=1

Ai.

A cyclic contraction on a cone metric space is defined as follows.

Definition 4.5. Let (X,d) be a complete cone metric space over a Banach algebra
A and P be the underlying solid cone. Suppose, A1,A2, . . . ,Am be subsets of X

and Y =
m⋃
i=1

Ai. A mapping T : Y → Y is called a generalized cyclic Lipschitz

contraction with Lipschitz vector k if following conditions hold:

(1) Y =
m⋃
i=1

Ai is a cyclic representation of Y with respect to T ;

(2) there exists k ∈ P such that ρ(k) < 1 and
(5) d(Tx,Ty) � kd(x,y)

for any x ∈ Ai,y ∈ Ai+1 (i = 1, 2, . . . ,m where Am+1 = A1).



FIXED POINTS OF α-ADMISSIBLE MAPPINGS 17

The following theorem is the cone metric version of Kirk et al. [12] when the
cone metric is endowed with a Banach algebra.

Theorem 4.6. Let (X,v) be a partially ordered set and suppose that (X,d) be a
complete cone metric space (X,d) over a Banach algebra A with P the underlying

solid cone. Suppose, A1,A2, . . . ,Am be closed subsets of X and Y =
m⋃
i=1

Ai and

T : Y → Y be a a generalized cyclic Lipschitz contraction with Lipschitz vector k.
Then, T has a unique fixed point in X.

Proof. Define the mapping αc : X ×X → [0,∞) by:

αc(x,y) =

{
1, if (x,y) ∈ Ai ×Ai+1 (i = 1, 2, . . . ,m where Am+1 = A1);
0, otherwise.

First, by definition of the function α and the cyclic representation, T is αc-admissible.
Again, by definition of the function αc, T is a generalized cyclic Lipschitz contrac-
tion with Lipschitz vector k. Suppose, for a sequence {xn} we have α(xn,xn+1) ≥ 1
for all n and xn → x ∈ X as n →∞. Then, as Y =

m⋃
i=1

Ai is a cyclic representation

with respect to T, we must have x ∈
m⋂
i=1

Ai. Therefore, αc(xn,x) ≥ 1 for all n ∈ N.

Now, the proof of existence of fixed point of T follows from Theorem 3.2. For u-

niqueness, if x∗,y∗ ∈ Fix(T), then by Remark 4.4 we have x∗,y∗ ∈
m⋂
i=1

Ai. Since each

Ai, i ∈{1, 2, . . . ,m} is nonempty, there exists z ∈ Y such that x∗,y∗ ∈ Ai,z ∈ Ai+1
for some i ∈ {1, 2, . . . ,m}, and so αc(x∗,z) = αc(y∗,z) = 1. Thus, the condition
(H) is satisfied and the uniqueness of fixed point follows from Theorem 3.5. �

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1Department of Mathematics, Govt. S.G.S.P.G. College Ganj Basoda, Distt Vidisha
M.P., India

2Department of Mathematics, Choithram College of Professional Studies, Dhar
Road, Indore (M.P.) 453001, India

3Department of Applied Mathematics, Shri Vaishnav Institute of Technology & Sci-

ence, Gram Baroli, Sanwer Road, Indore (M.P.) 453331, India

∗Corresponding author