EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 10, No. 4, 2017, 655-667
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Coupled Fixed Point Theorems on Bipolar Metric
Spaces

Ali Mutlu1,∗, Kübra Özkan1, Utku Gürdal1

1 Manisa Celal Bayar University, Faculty of Science and Arts, Department of Mathematics
Turkey

Abstract. In this article, certain coupled fixed point theorems, which can be considered as gener-
alizations of Banach fixed point theorem, are extended to bipolar metric spaces. Also, some results
which are related to these theorems are obtained. Finally, it is given an example which presents
the applicability of obtained results.

2010 Mathematics Subject Classifications: 46A80, 47H10, 54H25

Key Words and Phrases: Bipolar metric space, coupled fixed point, completeness

1. Introduction

In literature, the notion of coupled fixed point has been introduced by Guo and Laksh-
mikantham [6] in 1987. Afterward, Bhaskar and Lakshmikantham [3] introduced certain
coupled fixed point theorems in partially ordered metric spaces. Since then, when many
authors saw that these fixed point theorems can be utilized to investigate existence and
uniqueness of solutions of periodic boundary value problems, differential equations and
nonlinear integral equations, these theorems attracted their attention. And, they ex-
tended these theorems to various generalizations of metric spaces as cone, partial and
modular, e.g. [1, 2, 4, 5, 7–12, 14–20].

The notion of metric space has many generalizations in literature. One of the most
recent of them is bipolar metric space which is introduced by Mutlu and Gürdal [13] in
2016. Also, they established some fixed point theorems as Banach’s and Kannan’s on this
space.

In this paper, we extend certain coupled fixed point theorems, which can be considered
as generalization of Banach fixed point theorem, to bipolar metric spaces. Also, we obtain
some results which are related to these theorems. Finally, we give an example which
presents the applicability of our obtained results.

∗Corresponding author.

Email addresses: abgamutlu@gmail.com (A. Mutlu), kubra.ozkan@hotmail.com (K. Özkan),
utkugurdal@gmail.com (U. Gürdal)

http://www.ejpam.com 655 c© 2017 EJPAM All rights reserved.



A. Mutlu, K. Özkan, U. Gürdal / Eur. J. Pure Appl. Math, 10 (4) (2017), 655-667 656

2. Bipolar Metric Spaces

We express a series of definitions of some fundamental notions related to bipolar metric
spaces.

Definition 1. [13] A bipolar metric space is a triple (X,Y, d) such that X,Y 6= ∅ and
d : X × Y → R+ is a function satisfying the properties

(B0) if d (x, y) = 0, then x = y,
(B1) if x = y, then d (x, y) = 0,
(B2) if x, y ∈ X ∩ Y , then d (x, y) = d (y, x),
(B3) d(x1, y2) ≤ d(x1, y1) + d(x2, y1) + d(x2, y2),

for all (x, y), (x1, y1), (x2, y2) ∈ X × Y , where R+ symbolises the set of all non-negative
real numbers. Then d is called a bipolar metric on the pair (X,Y ).

Definition 2. [13] Let (X1, Y1) and (X2, Y2) be pairs of sets and given a function f :
X1 ∪ Y1 → X2 ∪ Y2. If f(X1) ⊆ X2 and f(Y1) ⊆ Y2, we call f a covariant map from
(X1, Y1) to (X2, Y2) and denote this with f : (X1, Y1) ⇒ (X2, Y2). If f(X1) ⊆ Y2 and
f(Y1) ⊆ X2, then we call f a contravariant map from (X1, Y1) to (X2, Y2) and write
f : (X1, Y1)↘↗ (X2, Y2). In particular, if d1 and d2 are bipolar metrics on (X1, Y1) and
(X2, Y2), respectively, we sometimes use the notations f : (X1, Y1, d1) ⇒ (X2, Y2, d2) and
f : (X1, Y1, d1)↘↗ (X2, Y2, d2).

Definition 3. [13] Let (X,Y, d) be a bipolar metric space. A point u ∈ X ∪ Y is called
a left point if u ∈ X, a right point if u ∈ Y and a central point if it is both left and right
point. Similarly a sequence (xn) on the set X is called a left sequence and a sequence (yn)
on Y is called a right sequence. In a bipolar metric space, a left or a right sequence is
called simply a sequence. A sequence (un) is said to be convergent to a point u, iff (un) is
a left sequence, u is a right point and lim

n→∞
d(un, u) = 0; or (un) is a right sequence, u is

a left point and lim
n→∞

d(u, un) = 0. A bisequence (xn, yn) on (X,Y, d) is a sequence on the

set X × Y . If the sequences (xn) and (yn) are convergent, then the bisequence (xn, yn) is
said to be convergent, and if (xn) and (yn) converge to a common point, then (xn, yn) is
called biconvergent. (xn, yn) is a Cauchy bisequence, if lim

n,m→∞
d(xn, ym) = 0. In a bipolar

metric space, every convergent Cauchy bisequence is biconvergent. A bipolar metric space
is called complete, if every Cauchy bisequence is convergent, hence biconvergent.

Definition 4. [13] Let (X1, Y1, d1) and (X2, Y2, d2) be bipolar metric spaces.
(1) A map f : (X1, Y1, d1) ⇒ (X2, Y2, d2) is called left-continuous at a point x0 ∈ X1,

if for every ε > 0, there exists a δ > 0 such that d1 (x0, y) < δ implies d2 (f (x0) , f (y)) < ε
all y ∈ Y1.

(2) A map f : (X1, Y1, d1) ⇒ (X2, Y2, d2) is called right-continuous at a point y0 ∈ Y1,
if for every ε > 0, there exists a δ > 0 such that d1 (x, y0) < δ implies d2 (f (x) , f (y0)) < ε
for all x ∈ X1.

(3) A map f is called continuous, if it is left-continuous at each point x ∈ X1 and
right-continuous at each point y ∈ Y1.



A. Mutlu, K. Özkan, U. Gürdal / Eur. J. Pure Appl. Math, 10 (4) (2017), 655-667 657

(4) A contravariant map f : (X1, Y1, d1)↘↗ (X2, Y2, d2) is continuous if and only if it
is continuous as a covariant map f : (X1, Y1, d1) ⇒

(
Y2, X2, d̄2

)
It can be seen from the Definition 4 that a covariant or a contravariant map f

from (X1, Y1, d1) to (X2, Y2, d2) is continuous if and only if (un) → v on (X1, Y1, d1)
implies (f (un))→ f (v) on (X2, Y2, d2).

3. Main Results

Definition 5. Let (X,Y, d) be a bipolar metric space, F :
(
X2, Y 2

)
⇒ (X,Y ) be a covari-

ant mapping. (a, b) ∈ X2 ∪ Y 2 is said to be a coupled fixed point of F if

F (a, b) = a and F (b, a) = b.

Theorem 1. Let (X,Y, d) be a complete bipolar metric space, F :
(
X2, Y 2

)
⇒ (X,Y ) be

a covariant mapping and k, l be non-negative constants. If F satisfies the condition

d(F (a, b), F (p, q)) ≤ kd(a, p) + ld(b, q), k + l < 1 (1)

for all a, b ∈ X, p, q ∈ Y , then F : X2 ∪ Y 2 → X ∪ Y has a unique coupled fixed point.

Proof. Let a0, b0 ∈ X and p0, q0 ∈ Y . We take a1, b1 ∈ X and p1, q1 ∈ Y with a1 =
F (a0, b0), b1 = F (b0, a0), p1 = F (p0, q0), q1 = F (q0, p0). And similarly, we take a2, b2 ∈ X
and p2, q2 ∈ Y with a2 = F (a1, b1), b2 = F (b1, a1), p2 = F (p1, q1), q2 = F (q1, p1). In this
way, we obtain bisequences (an, bn) and (pn, qn) with

an+1 = F (an, bn), bn+1 = F (bn, an), pn+1 = F (pn, qn) and qn+1 = F (qn, pn)

for all n ∈ N+. Let k + l = λ. From (1), we get

d(an, pn+1) = d(F (an−1, bn−1), F (pn, qn)), (2)

≤ kd(an−1, pn) + ld(bn−1, qn)

and

d(bn, qn+1) = d(F (bn−1, an−1), F (qn, pn)), (3)

≤ kd(bn−1, qn) + ld(an−1, pn)

for all n ∈ N+ and λ < 1. Let

en = d(an, pn+1) + d(bn, qn+1)

for all n ∈ N+. Combining (2) and (3), we observe that

en = d(an, pn+1) + d(bn, qn+1)

≤ kd(an−1, pn) + ld(bn−1, qn) + kd(bn−1, qn) + ld(an−1, pn)



A. Mutlu, K. Özkan, U. Gürdal / Eur. J. Pure Appl. Math, 10 (4) (2017), 655-667 658

= (k + l)(d(an−1, pn) + d(bn−1, qn))

= λen−1.

Then we get
0 ≤ en ≤ λen−1 ≤ λ2en−2 ≤ · · · ≤ λne0. (4)

On the other hand,

d(an+1, pn) = d(F (an, bn), F (pn−1, qn−1)), (5)

≤ kd(an, pn−1) + ld(bn, qn−1)

and

d(bn+1, qn) = d(F (bn, an), F (qn−1, pn−1)), (6)

≤ kd(bn, qn−1) + ld(an, pn−1)

for all n ∈ N+ and λ < 1. Let

sn = d(an+1, pn) + d(bn+1, qn)

for all n ∈ N+. Combining (5) and (6), we observe that

sn = d(an+1, pn) + d(bn+1, qn)

≤ kd(an, pn−1) + ld(bn, qn−1) + kd(bn, qn−1) + ld(an, pn−1)

= (k + l)(d(an, pn−1) + d(bn, qn−1))

= λsn−1.

Then similar to Equation (4), we obtain that

0 ≤ sn ≤ λsn−1 ≤ λ2sn−2 ≤ · · · ≤ λns0. (7)

Moreover,

d(an, pn) = d(F (an−1, bn−1), F (pn−1, qn−1)), (8)

≤ kd(an−1, pn−1) + ld(bn−1, qn−1)

and

d(bn, qn) = d(F (bn−1, an−1), F (qn−1, pn−1)), (9)

≤ kd(bn−1, qn−1) + ld(an−1, pn−1)

for all n ∈ N+ and λ < 1. Therefore, let

tn = d(an, pn) + d(bn, qn)



A. Mutlu, K. Özkan, U. Gürdal / Eur. J. Pure Appl. Math, 10 (4) (2017), 655-667 659

for all n ∈ N+. Combining (8) and (9), we observe that

tn = d(an, pn) + d(bn, qn)

≤ kd(an−1, pn−1) + ld(bn−1, qn−1) + kd(bn−1, qn−1) + ld(an−1, pn−1)

= (k + l)(d(an−1, pn−1) + d(bn−1, qn−1))

= λtn−1.

Thus, we obtain that

0 ≤ tn ≤ λtn−1 ≤ λ2tn−2 ≤ · · · ≤ λnt0. (10)

Using the property (B3), we get

d(an, pm) ≤ d(an, pn+1) + d(an+1, pn+1) + · · ·+ d(am−1, pm),
d(bn, qm) ≤ d(bn, qn+1) + d(bn+1, qn+1) + · · ·+ d(bm−1, qm)

(11)

and
d(am, pn) ≤ d(am, pm−1) + d(am−1, pm−1) + · · ·+ d(an+1, pn),
d(bm, qn) ≤ d(bm, qm−1) + d(bm−1, qm−1) + · · ·+ d(bn+1, qn)

(12)

for each n,m ∈ N, n < m. Then, from (4), (7) (10), (11) and (12), we have

d(an, pm) + d(bn, qm) ≤ (d(an, pn+1) + d(bn, qn+1))

+(d(an+1, pn+1) + d(bn+1, qn+1)) + · · ·
+(d(am−1, pm−1) + d(bm−1, qm−1))

+(d(am−1, pm) + d(bm−1, qm)),

= en + tn+1 + en+1 + · · ·+ tm−1 + em−1,

≤ λne0 + λn+1t0 + λn+1e0 + · · ·+ λm−1t0 + λm−1e0,

= (λn + λn+1 + · · ·+ λm−1)e0 + (λn+1 + λn+2 + · · ·+ λm−1)t0,

≤ λn

1− λ
e0 +

λn+1

1− λ
t0 (13)

and

d(am, pn) + d(bm, qn) ≤ (d(am, pm−1) + d(bm, qm−1))

+(d(am−1, pm−1) + d(bm−1, qm−1)) + · · ·
+(d(an+1, pn+1) + d(bn+1, qn+1))

+(d(an+1, pn) + d(bn+1, qn)),

= sm−1 + tm−1 + · · ·+ sn+1 + tn+1 + sn,

≤ λm−1s0 + λm−1t0 + · · ·+ λn+1s0 + λn+1t0 + λns0,

= (λn + λn+1 + · · ·+ λm−1)s0 + (λn+1 + λn+2 + · · ·+ λm−1)t0,

≤ λn

1− λ
s0 +

λn+1

1− λ
t0 (14)



A. Mutlu, K. Özkan, U. Gürdal / Eur. J. Pure Appl. Math, 10 (4) (2017), 655-667 660

for n < m. Since, for an arbitrary ε > 0, there exists n0 such that λn0

1−λe0 + λn0+1

1−λ t0 <
ε
3

and λn0

1−λs0 + λn0+1

1−λ t0 <
ε
3 , from (13) and (14), we have

d(an, pm) + d(bn, qm) <
ε

3

for each n,m ≥ n0. Then (an, pn) and (bn, qn) are Cauchy bisequences. Because of
completeness of (X,Y, d), there exist a, b ∈ X and p, q ∈ Y with

lim
n→∞

an = p, lim
n→∞

bn = q, lim
n→∞

pn = a and lim
n→∞

qn = b. (15)

Then there exists n1 ∈ N with d(an, p) <
ε
3 , d(bn, q) <

ε
3 , d(a, pn) < ε

3 and d(b, qn) < ε
3

for all n ≥ n1 and every ε > 0. Since (an, pn) and (bn, qn) are Cauchy bisequences, we get
d(an, pn) < ε

3 and d(bn, qn) < ε
3 . So, from (1), we have

d(F (a, b), p) ≤ d(F (a, b), pn+1) + d(an+1, pn+1) + d(an+1, p)

= d(F (a, b), F (pn, qn)) + d(an+1, pn+1) + d(an+1, p)

≤ kd(a, pn) + ld(b, qn) + d(an+1, pn+1) + d(an+1, p)

< k
ε

3
+ l

ε

3
+
ε

3
+
ε

3

= λ
ε

3
+ 2

ε

3
< ε

for each n ∈ N and λ < 1. Then d(F (a, b), p) = 0. Hence, F (a, b) = p. Similarly, we get
F (b, a) = q, F (p, q) = a and F (q, p) = b. On the other hand, from (15) we get

d(a, p) = d( lim
n→∞

pn, lim
n→∞

an) = lim
n→∞

d(an, pn) = 0

and
d(b, q) = d( lim

n→∞
qn, lim

n→∞
bn) = lim

n→∞
d(bn, qn) = 0.

So, a = p and b = q. Therefore, (a, b) ∈ X2 ∩ Y 2 is a coupled fixed point of F .
Now, to show the uniqueness, we begin by taking another coupled fixed point (a∗, b∗) ∈

X2 ∪ Y 2. If (a∗, b∗) ∈ X2, then we get

d(a∗, a) = d(F (a∗, b∗), F (a, b)) ≤ kd(a∗, a) + ld(b∗, b)

and
d(b∗, b) = d(F (b∗, a∗), F (b, a)) ≤ kd(b∗, b) + ld(a∗, a).

Therefore, we have
d(a∗, a) + d(b∗, b) ≤ λ(d(a∗, a) + d(b∗, b)). (16)

Since λ < 1, by (16) this means that d(a∗, a) + d(b∗, b) = 0. So, we obtain that a∗ = a
and b∗ = b. Similarly, if (a∗, b∗) ∈ Y 2, we have a∗ = a and b∗ = b. Then (a, b) is a unique
coupled fixed point of F .

The following corollary is obtained, if we take equal the constants k, l in Theorem 1.



A. Mutlu, K. Özkan, U. Gürdal / Eur. J. Pure Appl. Math, 10 (4) (2017), 655-667 661

Corollary 1. Let (X,Y, d) be a complete bipolar metric space, F :
(
X2, Y 2

)
⇒ (X,Y ) be

a covariant mapping and k, l be non-negative constants. If the condition

d(F (a, b), F (p, q)) ≤ k

2
(d(a, p) + d(b, q)), k < 1 (17)

holds for all a, b ∈ X, p, q ∈ Y , then F : X2 ∪ Y 2 → X ∪ Y has a unique coupled fixed
point.

Now, we express another generalization of coupled fixed point theorem in bipolar metric
spaces.

Definition 6. Let (X,Y, d) be a bipolar metric space, a ∈ X, p ∈ Y and F : (X × Y, Y ×X)
⇒ (X,Y ) be a covariant mapping. (a, p) is said to be a coupled fixed point of F if

F (a, p) = a and F (p, a) = p.

Theorem 2. Let (X,Y, d) be a complete bipolar metric space, F : (X × Y, Y ×X) ⇒
(X,Y ) be a covariant mapping and k, l be non-negative constants. If the condition

d(F (a, p), F (q, b)) ≤ kd(a, q) + ld(b, p), k + l < 1 (18)

holds for all a, b ∈ X, p, q ∈ Y , then F : (X×Y )∪ (Y ×X)→ X ∪Y has a unique coupled
fixed point.

Proof. Similar to the proof of Theorem 1, we define bisequences (an, pn) and (bn, qn)
as follows:

an+1 = F (an, pn), pn+1 = F (pn, an), bn+1 = F (bn, qn) and qn+1 = F (qn, bn)

for all n ∈ N+. Let k + l = λ. Then, from (18), we get

d(an, qn+1) = d(F (an−1, pn−1), F (qn, bn)), (19)

≤ kd(an−1, qn) + ld(bn, pn−1)

d(an+1, qn) = d(F (an, pn), F (qn−1, bn−1)), (20)

≤ kd(an, qn−1) + ld(bn−1, pn)

d(bn, pn+1) = d(F (bn−1, qn−1), F (pn, an)), (21)

≤ kd(bn−1, pn) + ld(an, qn−1)

d(bn+1, pn) = d(F (bn, qn), F (pn−1, an−1)), (22)

≤ kd(bn, pn−1) + ld(an−1, qn)



A. Mutlu, K. Özkan, U. Gürdal / Eur. J. Pure Appl. Math, 10 (4) (2017), 655-667 662

for all n ∈ N+ and λ < 1. Let

en = d(an, qn+1) + d(bn+1, pn)

and
sn = d(an+1, qn) + d(bn, pn+1)

for all n ∈ N+. Using equations (19), (20), (21) and (22), we get

en = d(an, qn+1) + d(bn+1, pn)

≤ kd(an−1, qn) + ld(bn, pn−1) + kd(bn, pn−1) + ld(an−1, qn)

= (k + l)(d(an−1, qn) + d(bn, pn−1))

= λen−1

and

sn = d(an+1, qn) + d(bn, pn+1)

≤ kd(an, qn−1) + ld(bn−1, pn) + kd(bn−1, pn) + ld(an, qn−1)

= (k + l)(d(an, qn−1) + d(bn−1, pn))

= λsn−1.

Then we obtain that
0 ≤ en ≤ λen−1 ≤ λ2en−2 ≤ · · · ≤ λne0 (23)

and
0 ≤ sn ≤ λsn−1 ≤ λ2sn−2 ≤ · · · ≤ λns0. (24)

On the other hand,

d(an, qn) = d(F (an−1, pn−1), F (qn−1, bn−1)), (25)

≤ kd(an−1, qn−1) + ld(bn−1, pn−1)

and

d(bn, pn) = d(F (bn−1, pn−1), F (pn−1, an−1)), (26)

≤ kd(bn−1, pn−1) + ld(an−1, qn−1)

for all n ∈ N+ and λ < 1. If we take

tn = d(an, qn) + d(bn, pn)

for all n ∈ N+ and combine (25) and (26), we have

tn = d(an, qn) + d(bn, pn)

≤ kd(an−1, qn−1) + ld(bn−1, pn−1) + kd(bn−1, pn−1) + ld(an−1, qn−1)

= (k + l)(d(an−1, qn−1) + d(bn−1, pn−1))



A. Mutlu, K. Özkan, U. Gürdal / Eur. J. Pure Appl. Math, 10 (4) (2017), 655-667 663

= λtn−1.

So, we get
0 ≤ tn ≤ λtn−1 ≤ λ2tn−2 ≤ · · · ≤ λnt0. (27)

We obtain that

d(an, qm) ≤ d(an, qn+1) + d(an+1, qn+1) + · · ·+ d(am−1qm),
d(bn, pm) ≤ d(bn, pn+1) + d(bn+1, pn+1) + · · ·+ d(bm−1, pm),
d(am, qn) ≤ d(am, qm−1) + d(am−1, qm−1) + · · ·+ d(an+1, qn),
d(bm, pn) ≤ d(bm, pm−1) + d(bm−1, pm−1) + · · ·+ d(bn+1, pn)

(28)

for each n,m ∈ N, n < m. Thus, from (23), (24) (27) and (28), we have

d(an, qm) + d(bm, pn) ≤ (d(an, qn+1) + d(bn+1, pn))

+(d(an+1, qn+1) + d(bn+1, pn+1)) + · · ·
+(d(am−1, qm−1) + d(bm−1, pm−1))

+(d(am−1, qm) + d(bm, pm−1)),

= en + tn+1 + en+1 + · · ·+ tm−1 + em−1,

≤ λne0 + λn+1t0 + λn+1e0 + · · ·+ λm−1t0 + λm−1e0,

= (λn + λn+1 + · · ·+ λm−1)e0 + (λn+1 + λn+2 + · · ·+ λm−1)t0,

≤ λn

1− λ
e0 +

λn+1

1− λ
t0 (29)

and

d(am, qn) + d(bn, pm) ≤ (d(am, qm−1) + d(bm−1, pm))

+(d(am−1, qm−1) + d(bm−1, pm−1)) + · · ·
+(d(an+1, qn+1) + d(bn+1, pn+1))

+(d(an+1, qn) + d(bn, pn+1)),

= sm−1 + tm−1 + · · ·+ sn+1 + tn+1 + sn,

≤ λm−1s0 + λm−1t0 + · · ·+ λn+1s0 + λn+1t0 + λns0,

= (λn + λn+1 + · · ·+ λm−1)s0 + (λn+1 + λn+2 + · · ·+ λm−1)t0,

≤ λn

1− λ
s0 +

λn+1

1− λ
t0 (30)

for n < m. Since, for an arbitrary ε > 0, there exists n0 such that λn0

1−λe0 + λn0+1

1−λ t0 <
ε
3

and λn0

1−λs0 + λn0+1

1−λ t0 <
ε
3 , from (29) and (30), we have for each n,m ≥ n0 that

d(an, qm) + d(bm, pn) <
ε

3
.

Then (an, qn) and (bn, pn) are Cauchy bisequences. Using completeness of (X,Y, d), we
say that there exist a, b ∈ X and p, q ∈ Y with

lim
n→∞

an = q, lim
n→∞

bn = p, lim
n→∞

pn = b and lim
n→∞

qn = a. (31)



A. Mutlu, K. Özkan, U. Gürdal / Eur. J. Pure Appl. Math, 10 (4) (2017), 655-667 664

Then there exists n1 ∈ N with d(an, q) <
ε
3 , d(bn, p) <

ε
3 , d(b, pn) < ε

3 and d(a, qn) < ε
3

for all n ≥ n1 and every ε > 0. Since (an, qn) and (bn, pn) are Cauchy bisequences, we get
d(an, qn) < ε

3 and d(bn, pn) < ε
3 . Thus, from (18), we have

d(F (a, p), q) ≤ d(F (a, p), qn+1) + d(an+1, qn+1) + d(an+1, q)

= d(F (a, p), F (pn, bn)) + d(an+1, qn+1) + d(an+1, q)

≤ kd(a, qn) + ld(bn, p) + d(an+1, qn+1) + d(an+1, q)

< k
ε

3
+ l

ε

3
+
ε

3
+
ε

3

= λ
ε

3
+ 2

ε

3
< ε

for each n ∈ N and λ < 1. Then d(F (a, p), q) = 0⇒ F (a, p) = q. In a similar manner, we
get F (p, a) = b, F (b, q) = p and F (q, b) = a. And, from (31) we have

d(a, q) = d( lim
n→∞

qn, lim
n→∞

an) = lim
n→∞

d(an, qn) = 0

and
d(b, p) = d( lim

n→∞
pn, lim

n→∞
bn) = lim

n→∞
d(bn, pn) = 0.

Therefore, a = q and b = p. Then (a, p) ∈ (X × Y ) ∩ (Y ×X) is a coupled fixed point of
F . As in the proof of the Theorem 1, uniqueness of the coupled fixed point of F can be
shown easily.

Corollary 2. Let (X,Y, d) be a complete bipolar metric space. F : (X × Y, Y ×X) ⇒
(X,Y ) be a covariant mapping and k, l be non-negative constants. If the condition

d(F (a, p), F (q, b)) ≤ k

2
(d(a, q) + d(b, p)), k < 1 (32)

holds for all a, b ∈ X, p, q ∈ Y , then F : (X×Y )∪ (Y ×X)→ X ∪Y has a unique coupled
fixed point.

Example 1. Let Un(R) and Ln(R) be the sets of all n × n upper and lower triangular
matrices over R, respectively. A function d : Un(R)× Ln(R)→ R+ be defined as

d(A,B) =

n∑
i,j=1

|aij − bij |

for all A = (aij)n×n ∈ Un(R) and B = (bij)n×n ∈ Ln(R). Then it is apparent that
(Un(R), Ln(R), d) is a complete bipolar metric space. We take a covariant mapping

F :
(
Un(R)2, Ln(R)2

)
⇒ (Un(R), Ln(R))

such as F (A,B) =
(
aij+bij

3

)
n×n

where (A = (aij)n×n, B = (bij)n×n) ∈ Un(R)2 ∪ Ln(R)2.

Then we get

d(F (A,B), F (C,D)) = d

((
aij + bij

3

)
n×n

,

(
cij + dij

3

)
n×n

)



REFERENCES 665

=

n∑
i,j=1

∣∣∣∣aij + bij − cij − dij
3

∣∣∣∣
≤

n∑
i,j=1

∣∣∣∣aij − cij3

∣∣∣∣+

∣∣∣∣bij − dij3

∣∣∣∣
=

1

3
(d(A,C) + d(B,D))

for all A = (aij)n×n, B = (bij)n×n ∈ Un(R) and C = (cij)n×n, D = (dij)n×n ∈ Ln(R).
Therefore, the equation (17) is satisfied for k = 2

3 . Then from Corollary 1, F has a
unique coupled fixed point. It is obvious that the coupled fixed point is (0n×n, 0n×n) ∈
Un(R) ∩ Ln(R) where 0n×n is the null matrix.

On the other hand, if

F :
(
Un(R)2, Ln(R)2

)
⇒ (Un(R), Ln(R))

is defined by F (A,B) =
(
aij+bij

2

)
n×n

where A = (aij)n×n, B = (bij)n×n ∈ Un(R)2 ∪
Ln(R)2. Then it can be observed that

d(F (A,B), F (C,D)) ≤ 1

2
(d(A,C) + d(B,D)).

Then F satisfies the equation (17) for k = 1. Therefore, coupled fixed points of F are both
(0n×n, 0n×n) ∈ Un(R)∩Ln(R) and (In, In) ∈ Un(R)∩Ln(R) where 0n×n is the null matrix
and In is the identity matrix. As it can be seen from this expression, F has not a unique
coupled fixed point. Thus, the conditions k < 1 in Corollary 1 and k + l < 1 in Theorem
1 are the most appropriate conditions for satisfing the uniqueness of coupled fixed point.

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