International Journal of Analysis and Applications
ISSN 2291-8639
Volume 6, Number 1 (2014), 18-27
http://www.etamaths.com

A WEAK CONTRACTION PRINCIPLE IN PARTIALLY

ORDERED CONE METRIC SPACE WITH THREE CONTROL

FUNCTIONS

BINAYAK S. CHOUDHURY1, L. KUMAR2, T. SOM2,∗ AND N. METIYA3

Abstract. In this paper we utilize three functions to define a weak contrac-

tion in a cone metric space with a partial order and establish that this con-
traction has necessarily a fixed point either under the continuity assumption

or an order condition which we state here. The uniqueness of the fixed point

is also derived under some additional assumptions. The result is supported
with an example. The methodology used is a combination of order theoretic

and analytic approaches.

1. Introduction
In this paper we consider a fixed point theorem in a space which is a generaliza-

tion of metric space, namely, cone metric space. The space is introduced by allowing
the metric to take up values in Banach spaces. Following the work of Huang et al
in [19], fixed point theory has experienced a rapid growth in cone metric spaces. A
review of this development is given in [22].

Weak contraction principle is a generalization of Banach’s contraction principle
which was first given by Alber et al. in Hilbert spaces [1]. It was subsequently ex-
tended to metric spaces by Rhoades [30] and further generalized by several authors
like Dutta and Choudhury [16], Popescu [28], Choudhury and Kundu [9] etc. The
weak contraction principle has been recently extended to cone metric spaces [6].
Several weak contractive inequalities have been used in the fixed point theory in
metric and cone metric spaces. References [7], [8], [10], [14], [32] are some examples
of these works.

In an attempt to blend the order theoretic and analytic aspects of fixed point
theory, several authors have created a number of fixed point results in partially or-
dered metric spaces. Some of these works are noted in [12], [18], [27], [29]. In cone
metric spaces also, such efforts have been made in [2], [3], [11], [23] for examples.
The purpose here is to establish weak contraction results in partially ordered cone
metric spaces by using three control functions. Control functions first appeared in
fixed point theory in the work of Khan et al. [25] and afterward this function and
its generalizations have been used in a number of fixed point problems like [4], [5],
[26], [31]. Our result is illustrated with an example.

2. Mathematical preliminaries

2010 Mathematics Subject Classification. 54H10, 54H25.
Key words and phrases. Partially ordered set; Cone metric space; Weak contraction; Control

function; Fixed point.

c©2014 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

18



CONTRACTION PRINCIPLE IN PARTIALLY ORDERED CONE METRIC SPACE 19

Definition 2.1 [19] Let E always be a real Banach space and P a subset of E. P
is called a cone if and only if:

(i) P is nonempty, closed, and P 6= {0},
(ii) a,b ∈ R, a,b ≥ 0, x,y ∈ P =⇒ ax + by ∈ P ,
(iii) x ∈ P and −x ∈ P =⇒ x = 0.

Given a cone P ⊂ E, a partial ordering ≤ with respect to P is naturally defined by
x ≤ y if and only if y −x ∈ P , for x,y ∈ E. We shall write x < y to indicate that
x ≤ y but x 6= y, while x � y will stand for y −x ∈ intP , where intP denotes the
interior of P .

The cone P is said to be normal if there exists a real number K ≥ 1 such that
for all x,y ∈ E,

0 ≤ x ≤ y ⇒‖x‖≤ K‖y‖.
The least positive number K satisfying the above statement is called the normal
constant of P .

The cone P is called regular if every increasing sequence which is bounded from
above is convergent; that is, if {xn} is a sequence such that
x1 ≤ x2 ≤ ... ≤ xn ≤ ... ≤ y,

for some y ∈ E, then there is x ∈ E such that ‖xn − x‖ −→ 0 as n −→ ∞. E-
quivalently, the cone P is regular if and only if every decreasing sequence which is
bounded from below is convergent. It is well known that a regular cone is a normal
cone.

In the following we always suppose that E is a real Banach space with cone P in
E with intP 6= ∅ and ≤ is the partial ordering with respect to P.

Definition 2.2 [19] Let X be a nonempty set. Let the mapping d : X ×X −→ E
satisfies

(i) 0 ≤ d(x,y), 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(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.

Definition 2.3 [19] Let (X,d) be a cone metric space, {xn} a sequence in X and
x ∈ X.

(i) If for every c ∈ E with 0 � c there exists n0 ∈ N such that for all n > n0,
d(xn,x) � c, then {xn} is said to be convergent and {xn} converges to
x, x is the limit of {xn}. We denote this by limnxn = x or xn −→ x as
n −→∞.

(ii) If for every c ∈ E with 0 � c there exists n0 ∈ N such that for all n,m > n0,
d(xn,xm) � c, then {xn} is called a Cauchy sequence in X.

(iii) If every Cauchy sequence in X is convergent in X, then X is called a
complete cone metric space.

If P is a normal cone, then {xn} converges to x if and only if d(xn,x) −→ 0
as n −→ ∞ and {xn} is a Cauchy sequence if and only if d(xn,xm) −→ 0 as
n,m −→∞ [19].

Definition 2.4 Let ψ : intP ∪{0}−→ intP ∪{0} be a function.



20 CHOUDHURY, KUMAR, SOM AND METIYA

(i) We say ψ is strongly monotone increasing if for x, y ∈ intP ∪{0}
x ≤ y ⇐⇒ ψ(x) ≤ ψ(y).

(ii) ψ is said to be continuous at x0 ∈ intP ∪{0} if for any sequence {xn} in
intP ∪{0}, xn −→ x0 =⇒ ψ(xn) −→ ψ(x0).

The following is the definition of Altering distance function in cone metric space.

Definition 2.5 A function ψ : intP ∪{0} −→ intP ∪{0} is called an Altering
distance function if the following properties are satisfied:

(i) ψ is strongly monotone increasing and continuous,
(ii) ψ(t) = 0 if and only if t = 0.

Definition 2.6 [20] Let (X, d) be a cone metric space, T : X −→ X and x0 ∈ X.
Then the function T is continuous at x0 if for any sequence {xn} in X, xn −→ x0
implies Txn −→ Tx0.

Definition 2.7 [18] Let (X,�) be a partially ordered set and T : X −→ X be a self
map. We say that T is monotone non decreasing if x, y ∈ X, x � y =⇒ Tx � Ty.

Lemma 2.1. Let E be a real Banach space with cone P in E. Then

(i) if a ≤ b and b � c, then a � c [21],
(ii) if a � b and b � c, then a � c [21],

(iii) if 0 ≤ x ≤ y and a ≥ 0, where a is real number, then 0 ≤ ax ≤ ay [21],
(iv) if 0 ≤ xn ≤ yn, for n ∈ N and limn xn = x, limn yn = y, then 0 ≤ x ≤ y

[21],
(v) P is normal if and only if xn ≤ yn ≤ zn and limn xn = limn zn = x imply

limn yn = x [13].

Lemma 2.2 [7] Let (X, d) be a cone metric space with regular cone P such that
d(x, y) ∈ intP, for x, y ∈ X with x 6= y. Let φ : intP ∪{0} −→ intP ∪{0} be a
function with the following properties:

(i) φ(t) = 0 if and only if t = 0,
(ii) φ(t) � t, for t ∈ intP and
(iii) either φ(t) ≤ d(x, y) or d(x, y) ≤ φ(t), for t ∈ intP ∪{0} and x, y ∈ X.

Let {xn} be a sequence in X for which {d(xn, xn+1)} is monotonic decreasing.
Then {d(xn, xn+1)} is convergent to either r = 0 or r ∈ int P .

Lemma 2.3 [10] Let (X, d) be a cone metric space. Let φ : intP ∪ {0} −→
intP ∪{0} be a function such that

(i) φ(t) = 0 if and only if t = 0 and
(ii) φ(t) � t, for t ∈ intP.

Then a sequence {xn} in X is a Cauchy sequence if and only if for every c ∈ E
with 0 � c there exists n0 ∈ N such that d(xn,xm) � φ(c), for all n,m > n0.

3. Main Results
Lemma 3.1. Let (X,d) be a cone metric space with regular cone P such that
d(x, y) ∈ intP, for x, y ∈ X with x 6= y. Let φ : intP ∪{0} −→ intP ∪{0} be a
function with the following properties.

(i) φ(t) = 0 if and only if t = 0,
(ii) φ(t) � t, for t ∈ intP and



CONTRACTION PRINCIPLE IN PARTIALLY ORDERED CONE METRIC SPACE 21

(iii) either φ(t) ≤ d(x, y) or d(x, y) ≤ φ(t), for t ∈ intP ∪{0} and x, y ∈ X.
Let {xn} be a sequence in X and x ∈ X for which {d(xn, x)} is monotonic de-
creasing. Then {d(xn, x)} is convergent to either r = 0 or r ∈ int P .
Proof. Let {xn} be a sequence in X and x ∈ X for which {d(xn, x)} is monotonic
decreasing. Since cone P is regular and 0 ≤ d(xn, x), for all n ∈ N, there exists
r ∈ P such that
d(xn, x) −→ r as n −→∞.

If d(xn, x) = 0, for some n then trivially r = 0. Hence we shall assume that
d(xn, x) 6= 0, for all n ∈ N. Then according to the conditions of the lemma,
d(xn, x) ∈ int P , for all n ∈ N.
Let r 6= 0.
Since P is a regular cone, it is also a normal cone. Let B = {t ∈ int P : ‖t‖ < ‖r‖

K
},

where K is the normal constant of the cone P . For every positive real number a

with a <
‖r‖
K

and t ∈ int P , at‖t‖ ∈ B. Therefore, B is non empty. Now we claim
that for every t ∈ B,
φ(t) ≤ d(xn, x), for all n ∈ N.

Otherwise, there exists t0 ∈ B for which we can find a positive integer m such that
d(xm, x) < φ(t0) (using the property (iii) of φ in the lemma).

Since {d(xn, x)} is monotonic decreasing, we have
d(xn, x) ≤ d(xm, x) < φ(t0), for all n ≥ m,

which implies d(xn, x) < φ(t0), for all n ≥ m.
Letting n −→∞ in the above inequality, by (iv) of lemma 2.1 and using a property
of φ, we have
r ≤ φ(t0) � t0,

which implies ‖r‖≤ K‖t0‖, where K is the normal constant of cone P .
That is, ‖t0‖≥

‖r‖
K

, which contradicts our assumption that t0 ∈ B.
Hence for every t ∈ B,
φ(t) ≤ d(xn, x), for all n ∈ N.

Letting n −→∞ in the above inequality, we have
φ(t) ≤ r.

Therefore, for every t ∈ B,
r −φ(t) ∈ P ; that is, r = φ(t) + q, for some q ∈ P .

Now, 0 ≤ q � φ(t) + q (since φ(t) ∈ int P , for every t ∈ B). Then by (i) of lemma
2.1,

0 � φ(t) + q; that is, 0 � r.
Therefore, r ∈ int P . Hence the proof is completed. �

Theorem 3.1. Let (X, �) be a partially ordered set and suppose that there exists
a cone metric d in X for which the cone metric space (X, d) is complete with regular
cone P such that d(x, y) ∈ intP , for x, y ∈ X with x 6= y. Let T : X −→ X be a
continuous and non decreasing mapping such that for all comparable x, y ∈ X
ψ(d(Tx,Ty)) ≤ η(d(x,y)) −φ(d(x,y)), (3.1)

where ψ, η, φ : intP ∪{0} −→ intP ∪{0} are such that ψ and η are continuous, φ
is lower semi-continuous and also

(i) ψ is strongly monotonic increasing,
(ii) ψ(t) = η(t) = φ(t) = 0 if and only if t = 0,
(iii) ψ(t) −η(t) + φ(t) > 0 for all t ∈ intP ,
(iv) φ(t) � t, for t ∈ intP and



22 CHOUDHURY, KUMAR, SOM AND METIYA

(v) either φ(t) ≤ d(x, y) or d(x, y) � φ(t), for t ∈ intP ∪{0} and x, y ∈ X.
If there exists x0 ∈ X such that x0 � Tx0, then T has a fixed point in X.
Proof. Let x0 ∈ X be such that x0 � Tx0. Since T is nondecreasing w.r.t. �,
we construct the sequence {xn} such that xn = Txn−1 = Tnx0 and x0 � Tx0 �
T2x0 � ... � Tnx0 � Tn+1x0 � ...; that is
x0 � x1 � x2 � ... � xn � xn+1 � .... (3.2)

Since x = xn and y = xn+1 are comparable, from (3.1), we have
ψ(d(Txn,Txn+1)) ≤ η(d(xn,xn+1)) −φ(d(xn,xn+1)). (3.3)

Now, for all n ≥ 1, we have
ψ(d(Txn−1,Txn)) −ψ(d(Txn,Txn+1))

≥ ψ(d(Txn−1,Txn)) −η(d(xn,xn+1)) + φ(d(xn,xn+1))
= ψ(d(xn,xn+1)) −η(d(xn,xn+1)) + φ(d(xn,xn+1))
≥ 0. (by (ii) and (iii))

This implies that ψ(d(Txn,Txn+1)) ≤ ψ(d(Txn−1,Txn)). Then by (i), it follows
that
d(Txn,Txn+1) ≤ d(Txn−1,Txn), that is, d(xn+1,xn+2) ≤ d(xn,xn+1).

Therefore, {d(xn, xn+1)} is a monotone decreasing sequence. Hence by lemma 2.2,
there exists an r ∈ intP ∪{0} such that
d(xn, xn+1) −→ r as n −→∞. (3.4)

Taking the limit as n → ∞ in (3.3) and using (3.4), continuities of ψ, η and the
lower semi continuity of φ, we have
ψ(r) ≤ η(r) −φ(r),

that is,
ψ(r) −η(r) + φ(r) ≤ 0,

which by (ii) and (iii) implies that r = 0.
Hence, we have

lim
n→∞

d(xn+1, xn) −→ 0. (3.5)
Next we show that {xn} is a Cauchy sequence. If {xn} is not a Cauchy sequence,

then by lemma 2.3, there exists a c ∈ E with 0 � c such that ∀ n0 ∈ N, ∃ n, m ∈ N
with n > m > n0 such that d(xn, xm) <≮ φ(c). Hence by a property of φ in (v)
of the theorem, φ(c) ≤ d(xn, xm). Therefore, there exist sequences {n(k)} and
{m(k)} in N such that for all positive integers k,
n(k) > m(k) > k and d(xn(k), xm(k)) ≥ φ(c).

Assuming that n(k) is the smallest such positive integer, we get
d(xn(k), xm(k)) ≥ φ(c)

and
d(xn(k)−1, xm(k)) � φ(c).

Now,
φ(c) ≤ d(xn(k), xm(k)) ≤ d(xn(k), xn(k)−1) + d(xn(k)−1, xm(k)),

that is,
φ(c) ≤ d(xn(k), xm(k)) ≤ d(xn(k), xn(k)−1) + φ(c).

Letting k −→∞ in the above inequality, using (3.5) and the property (v) of Lemma
2.1, we have

lim
k→∞

d(xn(k), xm(k)) = φ(c). (3.6)

Again,
d(xn(k), xm(k)) ≤ d(xn(k), xn(k)+1) + d(xn(k)+1, xm(k)+1) + d(xm(k)+1, xm(k))

and
d(xn(k)+1, xm(k)+1) ≤ d(xn(k)+1, xn(k)) + d(xn(k), xm(k)) + d(xm(k), xm(k)+1).



CONTRACTION PRINCIPLE IN PARTIALLY ORDERED CONE METRIC SPACE 23

Letting k −→∞ in above inequalities, using (3.5) and (3.6), we have
lim
k→∞

d(xn(k)+1, xm(k)+1) = φ(c). (3.7)

Since for x = xn(k) and y = xm(k) are comparable, from (3.1), we have
ψ(d(xn(k)+1, xm(k)+1)) = ψ(d(Txn(k), Txm(k)))

≤ η(d(xn(k), xm(k))) −φ(d(xn(k), xm(k))).
Letting k −→ ∞ in the above inequality, using (3.6), (3.7) and the continuities of
ψ, η and the lower semi continuity of φ, we have
ψ(φ(c)) ≤ η(φ(c)) −φ(φ(c)),

that is,
ψ(φ(c)) −η(φ(c)) + φ(φ(c)) ≤ 0,

which by (ii) and (iii) implies that φ(c) = 0. Now, by (ii), it follows that c = 0,
which is a contradiction. Hence, {xn} is a Cauchy sequence. From the completeness
of X, there exists u ∈ X such that

xn −→ u as n −→∞. (3.8)
Since T is continuous and xn −→ u, we have
u = lim

n→∞
xn+1 = lim

n→∞
Txn = Tu

and this proves that u is a fixed point of T . �

In our next theorem we relax the continuity assumption on T by imposing an
order condition.
Theorem 3.2. Let (X, �) be a partially ordered set and suppose that there exists
a cone metric d in X for which the cone metric space (X, d) is complete with
regular cone P such that d(x, y) ∈ intP , for x, y ∈ X with x 6= y. Assume that
if {xn} is a nondecreasing sequence in X such that xn −→ x then xn � x, for all
n ∈ N. Let T : X −→ X be a nondecreasing mapping such that for all comparable
x, y ∈ X, (3.1) holds where the conditions upon ψ, η and φ are the same as in
Theorem 3.1. If there exists x0 ∈ X such that x0 � Tx0, then T has a fixed point
in X.
Proof. We take the same sequence {xn} as in the proof of Theorem 3.1. Then we
have x0 � x1 � x2 � x3 � ... � xn � xn+1 � ..., that is, {xn} is nondecreasing se-
quence. Also, this sequence converge to u. Then xn � u, for all n ∈ N. Therefore,
we can use the condition (3.1) for x = u, y = xn and so we have
ψ(d(Tu, xn+1)) = ψ(d(Tu, Txn))

≤ η(d(u, xn)) −φ(d(u, xn)).
Taking the limit as n −→ ∞ in the above inequality, using the properties of ψ, η
and φ, we have
ψ(d(Tu, u)) ≤ 0.

It follows by a property of ψ that d(Tu, u) = 0; that is, Tu = u, that is, u is a
fixed point of T . �

Theorem 3.3 In addition to the hypotheses of Theorem 3.1 and Theorem 3.2, in
both of the theorems, suppose that for every x, y ∈ X there exists a z ∈ X such
that x � z and y � z. Then T has a unique fixed point.
Proof. It follows from the theorem 3.1 or theorem 3.2, the set of fixed points of
T is non-empty. If possible, let x, y ∈ X (x 6= y) be two fixed points of T . We
distinguish two cases:
Case 1.
Suppose that x and y are comparable. Without loss of generality we take y � x.



24 CHOUDHURY, KUMAR, SOM AND METIYA

Then Tny = y � x = Tnx, for n = 0, 1, 2, ...
By the condition (3.1), we have for all n ≥ 1,
ψ(d(x, y)) = ψ(d(Tnx, Tny))

≤ η(d(Tn−1x, Tn−1y)) −φ(d(Tn−1x, Tn−1y)),
that is,
ψ(d(x, y)) ≤ η(d(x, y)) −φ(d(x, y)),

that is,
ψ(d(x, y)) −η(d(x, y)) + φ(d(x, y)) ≤ 0,

which by (ii) and (iii) implies that d(x, y) = 0; that is, x = y.
Case 2.
If x and y are not comparable, then there exists z ∈ X such that x � z and
y � z. Monotonicity of T implies that Tnx = x � Tnz and Tny = y � Tnz, for
n = 0, 1, 2, ...
By the condition (3.1), we have for all n ≥ 1,
ψ(d(Tnz, x)) = ψ(d(Tnz, Tnx))

≤ η(d(Tn−1z, Tn−1x)) −φ(d(Tn−1z, Tn−1x)),
that is,
ψ(d(Tnz, x)) ≤ η(d(Tn−1z, x)) −φ(d(Tn−1z, x)). (3.9)

Now, for all n ≥ 1, we have
ψ(d(Tn−1z, x)) −ψ(d(Tnz, x))

= ψ(d(Tn−1z, Tn−1x)) −ψ(d(Tnz, Tnx))
= ψ(d(Tn−1z, Tn−1x)) −ψ(d(T(Tn−1z), T(Tn−1x)))
≥ ψ(d(Tn−1z, Tn−1x))−η(d(Tn−1z, Tn−1x)) +φ(d(Tn−1z, Tn−1x))
= ψ(d(Tn−1z, x)) −η(d(Tn−1z, x)) + φ(d(Tn−1z, x))
≥ 0. (by (ii) and (iii))

This implies that ψ(d(Tnz, x)) ≤ ψ(d(Tn−1z, x)). Then by (i), it follows that
d(Tnz, x) ≤ d(Tn−1z, x),

Therefore, {d(Tnz, x)} is a monotone decreasing sequence. Hence by lemma 3.1,
there exists an r ∈ intP ∪{0} such that

lim
n→∞

d(Tnz, x) = r. (3.10)

Taking the limit as n → ∞ in (3.9) and using (3.10), continuities of ψ, η and the
lower semi continuity of φ, we have
ψ(r) ≤ η(r) −φ(r),

that is,
ψ(r) −η(r) + φ(r) ≤ 0,

which by (ii) and (iii) implies that r = 0.
Hence

lim
n→∞

d(Tnz, x) = 0.

Analogously, it can proved that
lim
n→∞

d(Tnz, y) = 0.

Finally, the uniqueness of the limit gives us x = y.
From above two cases we have that fixed point of T is unique. �

Example 3.1 Let X = [0, 1] with usual order � be a partially ordered set. Let
E = R2, with usual norm, be a real Banach space. We define P = {(x,y) ∈ E :
x,y ≥ 0}. The partial ordering ≤ with respect to the cone P be the partial ordering
in E. Then P is a regular cone.
Let d : X ×X −→ E be given as



CONTRACTION PRINCIPLE IN PARTIALLY ORDERED CONE METRIC SPACE 25

d(x, y) = (| x−y |, | x−y |), for x, y ∈ X.
Then (X, d) is a complete cone metric space with the required properties of theo-
rems 3.1 and 3.2.
Let ψ, η, φ : int P ∪{0}−→int P ∪{0} be defined respectively as follows:
for t = (x, y) ∈ int P ∪{0},

ψ(t) =




0, if x = 0 and y = 0,
(x, y), if 0 < x ≤ 1 and 0 < y ≤ 1,
(x2, y), if x > 1 and 0 < y ≤ 1,
(x, y2), if 0 < x ≤ 1 and y > 1,
(x2, y2), if x > 1 and y > 1,

and

η(t) = (v, v) and φ(t) = (
v2

2
,
v2

2
), where v = min {x, y}.

Then ψ, η and φ have the properties mentioned in theorems 3.1 and 3.2.

Tx = x−
x2

2
, for x ∈ X.

Then T has the required properties mentioned in theorems 3.1 and 3.2.
Without loss of generality we take x, y ∈ X with x > y.
Now,

ψ(d(Tx, Ty)) = ψ(d(x−
x2

2
, y −

y2

2
))

= ψ(((x−y) −
(x−y)(x + y)

2
, (x−y) −

(x−y)(x + y)
2

))

[ since 0 ≤ (x−y)−
(x−y)(x + y)

2
≤

1 ]

= ((x−y) −
(x−y)(x + y)

2
, (x−y) −

(x−y)(x + y)
2

)

[ since (x−y) ≤ (x + y) ]

≤ ((x−y) −
(x−y)2

2
, (x−y) −

(x−y)2

2
)

= (x−y, x−y) − (
(x−y)2

2
,

(x−y)2

2
)

= η((x−y, x−y)) −φ((
(x−y)2

2
,

(x−y)2

2
))

= η(d(x, y)) −φ(d(x, y)).
Hence the conditions of theorems 3.1 and 3.2 are satisfied and it is seen that 0 is a
fixed point of T .

Remark 3.1. It has been found that several fixed point problems in the cone
metric spaces are reducible to problems of metric spaces ([15], [17], [22], [24]). This
is not possible in general. Particularly, weak contraction is not transferable to a
corresponding weak contraction in the generated metric space and, therefore, is
not derived from the results of weak contractions in metric spaces. In fact there is
even no assurance that a cone metric space inequality will generate an inequality
condition in metric spaces, although, it does in several important cases as has been
pointed out in ([15], [17], [22], [24]). Moreover, there is a problem when a partial
ordering is defined on a cone metric space. In a partially ordered cone metric space
with specific relations of the cone metric with the ordering, there is no natural way



26 CHOUDHURY, KUMAR, SOM AND METIYA

of transferring these relations to metric spaces. Our problem is outside the scope
of those described in ([15], [17], [22], [24]).

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1Department of Mathematics, Bengal Engineering and Science University, Shibpur,

Howrah - 711103, West Bengal, India

2Department of Applied Mathematics, Indian Institute of Technology, (Banaras Hin-

du University), Varanasi - 221005, India

3Department of Mathematics, Bengal Institute of Technology, Kolkata - 700150,

West Bengal, India

∗Corresponding author