# Common Fixed Point of (ψ, β)-Generalized Contractive Mappings in Partially Ordered Metric Spaces

Authors: M. Abbas; I. Zulfaqar; Stojan Radenović
Journal: Chinese Journal of Mathematics
 (2014)

---

## Abstract

Gordji et al. (2012) gave a generalization of Geraghty’s theorem. The aim of
this paper is to study the necessary conditions for the existence of coincidence and common fixed point of four mappings satisfying (ψ, β)-generalized contractive condition in the setup of partial ordered metric spaces. Some examples are given to validate the definitions and results presented herein.


---

## Body

## 1. Introduction and Preliminaries

Abbas et al. [1] proved that a weakly contractive mapping defined on a Hilbert space is a Picard operator. Rhoades [2] established the same result considering the domain of mapping a complete metric space instead of Hilbert space. The study of common fixed points of mappings satisfying certain contractive conditions can be employed to establish existence of solutions of many types of operator equations such as differential and integral equations. Beg and Abbas [3] obtained common fixed points extending a weak contractive condition to two maps. In 2009, Kadelburg et al. [4] proved common fixed point theorems for generalized (
ψ
,
ϕ
)-weakly contractive mappings. Doric [5] obtained a common fixed point theorem for four maps. For more work in this direction, we refer to [2, 3, 5–13] and references mentioned therein.In complete metric spaces equipped with a partial ordering,⪯ was first investigated in 2004 by Ran and Reurings [14], and then by Nieto and lo
´pez [15] who subsequently extended the result of Ran and Reurings [14] for nondecreasing mappings and applied it to obtain a unique solution for a first order ordinary differential equation with periodic boundary conditions (see also [16]).Abbas et al. [8] initiated the study of common fixed points for four mappings satisfying generalized weak contractive condition in complete partially ordered metric space. Recently, Esmaily et al. [17] coincidence point result for four mappings in partially ordered metric space and employed their result to find the common solution of two integral equations.The aim of this paper is to obtain coincidence and common fixed points for four mappings under generalized(
ψ
,
β
) contractive condition in complete partially ordered metric space. Our results extend, unify, and generalize the comparable results in [8, 17–19]. For some details also see [20, 21].In the sequel,ℝ, ℝ
+, and ℕ denote the set of real numbers, the set of nonnegative real numbers, and the set of positive integers, respectively. The usual order on ℝ (resp., on ℝ
+) will be indistinctly denoted by ≤ or by ≥.The following definitions and results will be needed in the sequel.LetP be the class of all mappings β
:
[
0
,
∞
)
→
[
0,1
) satisfying the condition β
(
t
n
)
→
1 whenever t
n
→
0. Note that P
≠
ϕ as if we take f
x
=
1
/
(
1
+
x
2
), x
∈
[
0
,
∞
), then f
∈
P.DefineΨ
=
{
ψ
:
[
0
,
∞
)
→
[
0
,
∞
)
:
ψ is continuous, nondecreasing and ψ
(
t
)
=
0 if and only if t
=
0
}. If we define ψ
:
[
0
,
∞
)
→
[
0
,
∞
) by ψ
x
=
ln
⁡
(
1
+
x
), then it is easy to check that ψ
∈
Ψ.Letf be self-mapping on a set X. If x
=
f
x, for some x in X, then x is called a fixed point of f. The set of all fixed points of f is denoted by F
(
f
). If F
(
f
)
=
{
z
} and, for each x
0 in a complete metric space X, the sequence x
n
+
1
=
f
x
n
=
f
n
x
0
,
n
=
0,1
,
2
,
…, converges to z, then f is called a Picard operator.In the following [22], Geraghty obtained the generalization of Banach’s contraction principle.Theorem 1.
Let(
X
,
d
) be a complete metric space and f a self-map on X. If there exists β
∈
P such that d
(
f
x
,
f
y
)
≤
β
(
d
(
x
,
y
)
)
d
(
x
,
y
) for all x
,
y
∈
X, then f is a Picard operator.
A nonempty setX equipped with a partial order ⪯ is called partially ordered metric space if there exists a metric d on X. We will denote it by (
X
,
d
,
⪯
).Definition 2 (see [23]).
Let(
X
,
d
,
⪯
) be an ordered metric space. We say that X is regular if {
x
n
} is a nondecreasing sequence in X with respect to ⪯ such that x
n
→
x
∈
X as n
→
∞, then x
n
⪯
x for all n
∈
ℕ.Harandi and Emami [24] proved Theorem 1 in partially ordered metric spaces.Theorem 3.
Let(
X
,
⪯
) be a partially ordered set, and suppose that there exists a complete metric d on X. Let f
:
X
→
X be an increasing mapping such that there exists an element x
0
∈
X with x
0
⪯
f
x
0. If there exists α
∈
P such that d
(
f
x
,
f
y
)
≤
α
(
d
(
x
,
y
)
)
d
(
x
,
y
) for each x
,
y
∈
X with x
≽
y, then f has a fixed point provided that either f is continuous or X is regular. Moreover, if for each x
,
y
∈
X there exists z
∈
X which is comparable to x and y, then f has a unique fixed point.Recently, [18] Gordji et al. proved the following result.Theorem 4.
Let(
X
,
⪯
) be a partially ordered set, and suppose that there exists a complete metric d on X. Let f
:
X
→
X be a nondecreasing mapping such that there exists x
0
∈
X with x
0
⪯
f
x
0. Suppose that there exist α
∈
P and ψ
∈
Ψ such that
(1)
ψ
(
d
(
f
x
,
f
y
)
)
≤
α
(
ψ
(
d
(
x
,
y
)
)
)
ψ
(
d
(
x
,
y
)
)
,
for all x
,
y
∈
X with x
⪯
y. Assume that either f is continuous or X is regular. Then f has a fixed point.It is worth to noticing that Condition (b) of subadditivity in [18] for the function ψ is superfluous. Namely, all results in ([18], Theorems 2.2, 2.3, 3.3) are true only if ψ
:
[
0
,
∞
)
→
[
0
,
∞
) is a continuous nondecreasing function with ψ
(
t
)
=
0 if and only if t
=
0. However, it is easy to see that with Assumption (b) for the function ψ [18] is not a generalization of [24]. For details see [25].Letf and g be two self-mappings on a nonempty set X. If x
=
f
x
=
g
x, for some x in X, then x is called a common fixed point of f and g.Letf
,
g
:
X
→
X be given self-mappings on a metric space X. The pair (
f
,
g
) is said to be compatible if lim
⁡
n
→
∞
d
(
f
g
x
n
,
g
f
x
n
)
=
0, whenever {
x
n
} is a sequence in X such that, lim
⁡
n
→
∞
f
x
n
=
lim
⁡
n
→
∞
g
x
n
=
t, for some t
∈
X.Let(
X
,
d
,
⪯
) be an ordered metric space and f
,
g
,
S
,
T
:
X
→
X. If there exist β
∈
P and ψ
∈
Ψ such that, for every two comparable elements x
,
y
∈
X, we have
(2)
ψ
(
d
(
f
x
,
g
y
)
)
≤
β
(
M
(
x
,
y
)
)
ψ
(
max
⁡
⁡
{
d
(
S
x
,
T
y
)
,
vvvvvvvvvvvvvvvvv
mmm
d
(
S
x
,
f
x
)
,
d
(
T
y
,
g
y
)
}
)
,
where
(3)
M
(
x
,
y
)
=
ψ
1
2
(
max
⁡
{
1
2
d
(
S
x
,
T
y
)
,
d
(
S
x
,
f
x
)
,
d
(
T
y
,
g
y
)
,
n
n
n
n
n
n
n
n
n
n
1
2
[
d
(
S
x
,
g
y
)
+
d
(
T
y
,
f
x
)
]
}
)
,
then (
f
,
g
) is said to be (
ψ
,
β
)-order contractive pair with respect to S and T.Example 5.
LetX
=
{
1,2
,
3,4
,
6
} be a partially ordered set and d a usual metric on X. Define four self-maps f, g, S, and T on X as follows:
(4)
f
=
(
1
2
3
4
6
1
1
1
1
2
)
,
g
=
(
1
2
3
4
6
1
1
1
1
1
)
S
=
(
1
2
3
4
6
1
1
3
1
4
)
,
T
=
(
1
2
3
4
6
1
1
3
2
6
)
.
Take ψ
(
x
)
=
ln
⁡
(
x
+
1
) and β
(
x
)
=
ψ
(
x
)
/
x. Then (
f
,
g
) is (
ψ
,
β
)-order contractive pair with respect to S and T.
Let(
X
,
⪯
) be a partially ordered set. Two mappings f
,
g
:
X
→
X are said to be weakly increasing if f
x
⪯
g
f
x and g
x
⪯
f
g
x hold for all x
∈
X [26].Example 6.
LetX
=
ℝ
+ be endowed with usual order and usual topology. Let f
,
g
:
X
→
X be defined by
(5)
f
x
=
{
x
1
/
4
,
if
x
∈
[
0,1
]
x
4
,
if
x
∈
1
,
[
∞
)
,
g
x
=
{
x
,
if
x
∈
[
0,1
)
3
x
,
if
x
∈
1
,
[
∞
)
.
Then, the pair (
f
,
g
) is weakly increasing where g is a discontinuous mapping on ℝ
+.
Let(
X
,
⪯
) be a partially ordered set. Two mappings f
,
g
:
X
→
X are said to be partially weakly increasing if f
x
⪯
g
f
x for all x
∈
X [8].Definition 7 (see [23]).
Let(
X
,
⪯
) be a partially ordered set and f
,
g
,
h
:
X
→
X given mappings such that f
X
⊆
h
X and g
X
⊆
h
X. We say that f and g are weakly increasing with respect to h if and only if, for all x
∈
X, we have: f
x
⪯
g
y, for all y
∈
h
-
1
(
f
x
), and g
x
⪯
f
y, for all y
∈
h
-
1
(
g
x
), where h
-
1
(
x
)
=
(
u
∈
X
∣
h
u
=
x
), for x in X.Definition 8 (see [17]).
Let(
X
,
⪯
) be a partially ordered set and f
,
g
,
h
:
X
→
X given mappings such that f
X
⊆
h
X. We say that f and g are partially weakly increasing with respect to h if and only if, for all x
∈
X, we have f
x
⪯
g
y, for all y
∈
h
-
1
(
f
x
).Iff
=
g, we say that f is weakly increasing with respect to h.Note that ifh
=
I
X (identity map on X
), then pair (
f
,
g
) is weakly increasing.Let(
X
,
⪯
) be a partially ordered set. A self-mapping f on X is called (a) dominating if x
⪯
f
x for each x in X [8] and (b) dominated if f
x
⪯
x for each x in X.Example 9.
LetX
=
[
0,1
] be endowed with usual ordering and let f
:
X
→
X be defined by f
x
=
x
1
/
4. Since x
≤
x
1
/
4
=
f
x for all x
∈
X, Therefore f is a dominating map.Example 10.
LetX
=
[
1
,
∞
] be endowed with usual ordering and let f
:
X
→
X be defined by f
x
=
1
/
(
x
+
1
). Since f
x
=
1
/
(
x
+
1
)
≤
x for all x
∈
X, therefore f is a dominated map.
Assertion similar to the following lemma was used (and proved) in the course of proofs of several fixed point results in various papers [18, 27].Lemma 11.
Let(
X
,
d
) be a metric space and let {
y
n
} be a sequence in X such that d
(
y
n
+
1
,
y
n
) is nonincreasing and that
(6)
lim
⁡
n
→
∞
d
(
y
n
+
1
,
y
n
)
=
0
.
If {
y
2
n
} is not a Cauchy sequence, then there exist an ε
>
0 and two sequences {
m
k
} and {
n
k
} of positive integers such that the following four sequences tend to ε when k
→
∞:
(7)
d
(
y
2
m
k
,
y
2
n
k
)
,
d
(
y
2
m
k
,
y
2
n
k
+
1
)
,
d
(
y
2
m
k
-
1
,
y
2
n
k
)
,
d
(
y
2
m
k
-
1
,
y
2
n
k
+
1
)
.


## 2. Common Fixed Point Result

Now we start with the following result.Theorem 12.
Let(
X
,
⪯
) be a partially ordered set such that there exists a complete metric d on X. Suppose that f, g, S, and T are self-mappings on X with f
X
⊆
T
X and g
X
⊆
S
X such that a pair (
f
,
g
) of dominating maps is (
ψ
,
β
)-order contractive with respect to dominated maps S and T. If for a nondecreasing sequence {
x
n
} with x
n
⪯
y
n for all n and y
n
→
u implies that x
n
⪯
u and either(a)
{
f
,
S
} are compatible, f or S is continuous, and {
g
,
T
} are weakly compatible or
(b)
{
g
,
T
} are compatible, g or T is continuous, and {
f
,
S
} are weakly compatible,
thenf, g, S, and T have a common fixed point. Moreover, the set of common fixed points of f, g, S, and T is well ordered if and only if f, g, S, and T have one and only one common fixed point.Proof.
Letx
0 be an arbitrary point in X. Construct sequences {
x
n
} and {
y
n
} in X such that y
2
n
-
1
=
T
x
2
n
-
1
=
f
x
2
n
-
2 and y
2
n
=
S
x
2
n
=
g
x
2
n
-
1. This can be done as f
X
⊆
T
X and g
X
⊆
S
X. By given assumptions, x
2
n
-
2
⪯
f
x
2
n
-
2
=
T
x
2
n
-
1
⪯
x
2
n
-
1 and x
2
n
-
1
⪯
g
x
2
n
-
1
=
S
x
2
n
⪯
x
2
n. Thus, for all n
≥
1, we have x
n
⪯
x
n
+
1. We claim that d
(
y
2
n
,
y
2
n
+
1
)
>
0, for every n. If not, then y
2
n
=
y
2
n
+
1, for some n. From inequality (2), we obtain
(8)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
=
ψ
(
d
(
f
x
2
n
,
g
x
2
n
+
1
)
)
≤
β
(
M
(
x
2
n
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
{
d
(
S
x
2
n
,
T
x
2
n
+
1
)
,
d
(
S
x
2
n
,
f
x
2
n
)
,
nnnnnnnnnnnnn
d
(
T
x
2
n
+
1
,
g
x
2
n
+
1
)
}
)
=
β
(
M
(
x
2
n
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
{
d
(
y
2
n
,
y
2
n
+
1
)
,
d
(
y
2
n
,
y
2
n
+
1
)
,
nnnnnnnnnnnnn
d
(
y
2
n
+
1
,
y
2
n
+
2
)
}
)
=
β
(
M
(
x
2
n
,
x
2
n
+
1
)
)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
<
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
,
a contradiction. Hence, y
2
n
+
2
=
y
2
n
+
1. Following the similar arguments, we obtain y
2
n
+
2
=
y
2
n
+
3 and so on. Thus {
y
n
} becomes a constant sequence, and y
2
n is the common fixed point of f, g, S, and T. Take d
(
y
2
n
,
y
2
n
+
1
)
>
0 for each n. As x
2
n and x
2
n
+
1 are comparable, so by inequality (2) we have
(9)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
=
ψ
(
d
(
f
x
2
n
,
g
x
2
n
+
1
)
)
≤
β
(
M
(
x
2
n
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
{
d
(
S
x
2
n
,
T
x
2
n
+
1
)
,
d
(
S
x
2
n
,
f
x
2
n
)
,
m
m
m
m
m
m
m
m
d
(
T
x
2
n
+
1
,
g
x
2
n
+
1
)
}
)
=
β
(
M
(
x
2
n
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
{
d
(
y
2
n
,
y
2
n
+
1
)
,
d
(
y
2
n
,
y
2
n
+
1
)
,
mmmmmmmm
d
(
y
2
n
+
1
,
y
2
n
+
2
)
}
)
<
ψ
(
max
⁡
{
d
(
y
2
n
,
y
2
n
+
1
)
,
d
(
y
2
n
+
1
,
y
2
n
+
2
)
}
)
.
Now, if max
⁡
{
d
(
y
2
n
+
1
,
y
2
n
)
,
d
(
y
2
n
+
1
,
y
2
n
+
2
)
}
=
d
(
y
2
n
+
1
,
y
2
n
+
2
), then (9) gives a contradiction. Hence,
(10)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
<
ψ
(
d
(
y
2
n
,
y
2
n
+
1
)
)
.
Similarly, we obtain
(11)
ψ
(
d
(
y
2
n
+
2
,
y
2
n
+
3
)
)
<
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
.
Thus the sequence {
d
(
y
2
n
,
y
2
n
+
1
)
} is a nonincreasing sequence and bounded below. So lim
⁡
n
→
∞
ψ
(
d
(
y
2
n
,
y
2
n
+
1
)
) exists. We claim that lim
⁡
n
→
∞
ψ
(
d
(
y
2
n
,
y
2
n
+
1
)
)
=
0. If not, assume that lim
⁡
n
→
∞
ψ
(
d
(
y
2
n
,
y
2
n
+
1
)
)
=
r
>
0; then we have
(12)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
=
ψ
(
d
(
f
x
2
n
,
g
x
2
n
+
1
)
≤
β
(
M
(
x
2
n
+
1
,
x
2
n
)
)
×
ψ
(
max
⁡
{
d
(
S
x
2
n
+
1
,
T
x
2
n
)
,
d
(
S
x
2
n
+
1
,
f
x
2
n
+
1
)
,
n
n
n
n
n
n
n
n
n
n
n
n
d
(
T
x
2
n
,
g
x
2
n
)
}
)
=
β
(
M
(
x
2
n
+
1
,
x
2
n
)
)
×
ψ
(
max
⁡
{
d
(
y
2
n
+
1
,
y
2
n
)
,
d
(
y
2
n
+
1
,
y
2
n
+
2
)
,
n
n
n
n
n
n
n
n
n
n
n
n
d
(
y
2
n
,
y
2
n
+
1
)
}
)
=
β
(
M
(
x
2
n
+
1
,
x
2
n
)
)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
)
)
,
and so
(13)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
)
)
≤
β
(
M
(
x
2
n
+
1
,
x
2
n
)
)
<
1
,
which, on taking limit as n
→
∞, implies that lim
⁡
n
→
∞
β
(
M
(
x
2
n
+
1
,
x
2
n
)
)
=
1. By the property of β, we have lim
⁡
n
→
∞
M
(
x
2
n
+
1
,
x
2
n
)
=
0and so lim
⁡
n
→
∞
ψ
(
d
(
y
2
n
,
y
2
n
+
1
)
)
=
0, a contradiction. Therefore r
=
0. Now, we show that {
y
n
} is a Cauchy sequence. It is sufficient to show that {
y
2
n
} is a Cauchy sequence in X. Suppose that this is not the case. Applying Lemma 11 to the sequence {
y
n
}, we obtain that there exist ε
>
0 and two sequences of positive integers {
m
k
} and {
n
k
} such that the sequences
(14)
d
(
y
2
m
k
,
y
2
n
k
)
,
d
(
y
2
m
k
,
y
2
n
k
+
1
)
,
d
(
y
2
m
k
-
1
,
y
2
n
k
)
,
d
(
y
2
m
k
-
1
,
y
2
n
k
+
1
)
all tend to ε when k
→
∞. Putting x
=
x
2
n
k
,
y
=
x
2
m
k
-
1 in (2) we have
(15)
ψ
(
d
(
y
2
n
k
+
1
,
y
2
m
k
)
)
ψ
(
max
⁡
{
d
(
y
2
m
k
,
y
2
n
k
)
,
d
(
y
2
n
k
,
y
2
n
k
-
1
)
,
d
(
y
2
m
k
-
1
,
y
2
m
k
)
}
)
≤
β
(
M
(
x
2
n
k
,
x
2
m
k
-
1
)
)
<
1
,
where
(16)
M
(
x
2
n
k
,
x
2
m
k
-
1
)
=
max
⁡
{
d
(
y
2
m
k
,
y
2
n
k
)
+
d
(
y
2
m
k
-
1
,
y
2
n
k
+
1
)
2
d
(
y
2
n
k
,
y
2
m
k
-
1
)
,
d
(
y
2
n
k
,
y
2
n
k
-
1
)
,
d
(
y
2
m
k
-
1
,
y
2
m
k
)
,
m
m
m
m
d
(
y
2
m
k
,
y
2
n
k
)
+
d
(
y
2
m
k
-
1
,
y
2
n
k
+
1
)
2
}
.
Since
(17)
lim
⁡
k
→
∞
ψ
(
d
(
y
2
n
k
+
1
,
y
2
m
k
)
)
=
lim
⁡
k
→
∞
ψ
(
max
⁡
{
d
(
y
2
m
k
,
y
2
n
k
)
,
d
(
y
2
n
k
,
y
2
n
k
-
1
)
,
m
m
m
m
m
m
m
m
m
m
m
d
(
y
2
m
k
-
1
,
y
2
m
k
)
}
)
=
ψ
(
ε
)
,
we obtain that lim
⁡
k
→
∞
β
(
M
(
x
2
n
k
,
x
2
m
k
-
1
)
)
=
1. By the property of β, we have lim
⁡
k
→
∞
M
(
x
2
n
k
,
x
2
m
k
-
1
)
=
0 which is a contradiction with lim
⁡
k
→
∞
M
(
x
2
n
k
,
x
2
m
k
-
1
)
=
ε. Therefore {
y
2
n
} is a Cauchy sequence in X, and hence {
y
n
} is a Cauchy sequence. Since X is complete, there exists a point z in X, such that {
y
2
n
} converges to z. Therefore,
(18)
lim
⁡
n
→
∞
y
2
n
+
1
=
lim
⁡
n
→
∞
T
x
2
n
+
1
=
lim
⁡
n
→
∞
f
x
2
n
=
z
,
lim
⁡
n
→
∞
y
2
n
+
2
=
lim
⁡
n
→
∞
S
x
2
n
+
2
=
lim
⁡
n
→
∞
g
x
2
n
+
1
=
z
.
Assume that S is continuous. Since {
f
,
S
} are compatible, we have lim
⁡
n
→
∞
f
S
x
2
n
+
2
=
lim
⁡
n
→
∞
S
f
x
2
n
+
2
=
S
z. Now we show that z
=
S
z. If not, that is d
(
S
z
,
z
)
>
0. As x
2
n
+
1
⪯
g
x
2
n
+
1
=
S
x
2
n
+
2, so from inequality (2), we have
(19)
ψ
(
d
(
f
S
x
2
n
+
2
,
g
x
2
n
+
1
)
)
≤
β
(
M
(
S
x
2
n
+
2
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
{
d
(
S
S
x
2
n
+
2
,
T
x
2
n
+
1
)
,
a
a
a
a
a
a
a
a
a
a
d
(
S
S
x
2
n
+
2
,
f
S
x
2
n
+
2
)
,
d
(
T
x
2
n
+
1
,
g
x
2
n
+
1
)
}
)
,
where
(20)
M
(
S
x
2
n
+
2
,
x
2
n
+
1
)
=
ψ
(
max
⁡
{
1
2
d
(
S
S
x
2
n
+
2
,
T
x
2
n
+
1
)
,
mmmmmmm
d
(
S
S
x
2
n
+
2
,
f
S
x
2
n
+
2
)
,
mmmmmmm
d
(
T
x
2
n
+
1
,
g
x
2
n
+
1
)
,
mmmmmmm
1
2
[
d
(
S
S
x
2
n
+
2
,
g
x
2
n
+
1
)
mmmmmmmmm
+
d
(
T
x
2
n
+
1
,
f
S
x
2
n
+
2
)
]
1
2
}
)
.
On taking limit as n
→
∞, we obtain
(21)
ψ
(
d
(
S
z
,
z
)
)
≤
β
(
ψ
(
d
(
S
z
,
z
)
)
)
ψ
(
d
(
S
z
,
z
)
)
<
ψ
(
d
(
S
z
,
z
)
)
,
a contradiction. Hence, S
z
=
z. Now, since x
2
n
+
1
⪯
g
x
2
n
+
1 and g
x
2
n
+
1
→
z as n
→
∞, so we have x
2
n
+
1
⪯
z. We are to show that z
=
f
z. If not, then from inequality (2), we obtain
(22)
ψ
(
d
(
f
z
,
g
x
2
n
+
1
)
)
≤
β
(
M
(
z
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
{
d
(
S
z
,
T
x
2
n
+
1
)
,
d
(
S
z
,
f
z
)
,
m
m
m
m
m
m
m
m
m
d
(
T
x
2
n
+
1
,
g
x
2
n
+
1
)
}
)
,
which on taking limit as n
→
∞, implies that ψ
(
d
(
f
z
,
z
)
)
≤
lim
⁡
n
→
∞
β
(
M
(
z
,
x
2
n
+
1
)
)
ψ
(
d
(
f
z
,
z
)
). Thus lim
⁡
n
→
∞
β
(
M
(
z
,
x
2
n
+
1
)
)
=
1and so lim
⁡
n
→
∞
M
(
z
,
x
2
n
+
1
)
=
0, a contradiction. Thus f
z
=
z.
Sincef
(
X
)
⊆
T
(
X
), there exists a point w
∈
X such that z
=
f
z
=
T
w. Now we show that T
w
=
g
w. If not, that is d
(
T
w
,
g
w
)
>
0. Since z
⪯
f
z
=
T
w
⪯
w implies z
⪯
w, therefore from inequality (2), we obtain
(23)
ψ
(
d
(
T
w
,
g
w
)
)
=
ψ
(
d
(
f
z
,
g
w
)
)
≤
β
(
M
(
z
,
w
)
)
×
ψ
(
max
⁡
{
d
(
S
z
,
T
w
)
,
d
(
S
z
,
f
z
)
,
mmmmmm
d
(
T
w
,
g
w
)
}
)
<
ψ
(
max
⁡
{
d
(
S
z
,
T
w
)
,
d
(
S
z
,
f
z
)
,
mmmmmm
d
(
T
w
,
g
w
)
}
)
=
ψ
(
d
(
T
w
,
g
w
)
)
,
a contradiction. Hence T
w
=
g
w. Since g and T are weakly compatible, g
z
=
g
f
z
=
g
T
w
=
T
g
w
=
T
f
z
=
T
z. Thus z is a coincidence point of g and T.
Next we show thatz
=
g
z. If not, then d
(
z
,
g
z
)
>
0. As x
2
n
⪯
f
x
2
n and f
x
2
n
→
z as n
→
∞ implies that x
2
n
⪯
z, so from (2), we have
(24)
ψ
(
d
(
f
x
2
n
,
g
z
)
)
≤
β
(
M
(
x
2
n
,
z
)
)
×
ψ
(
max
⁡
{
d
(
S
x
2
n
,
T
z
)
,
d
(
S
x
2
n
,
f
x
2
n
)
,
mmmmmmm
d
(
T
z
,
g
z
)
}
)
<
ψ
(
max
⁡
{
d
(
S
x
2
n
,
T
z
)
,
d
(
S
x
2
n
,
f
x
2
n
)
,
mmmmmm
d
(
T
z
,
g
z
)
}
)
,
which, on taking limit as n
→
∞, gives ψ
(
d
(
z
,
g
z
)
)
<
ψ
(
d
(
z
,
g
z
)
), a contradiction, thus z
=
g
z. Therefore f
z
=
g
z
=
S
z
=
T
z
=
z. The proof is similar when f is continuous. Similarly, the result follows when (b) holds.Now suppose that the set of common fixed points off, g, S, and T is well ordered. We claim that common fixed point of f, g, S, and T is unique. Assume on contrary that f
u
=
g
u
=
S
u
=
T
u
=
u and f
v
=
g
v
=
S
v
=
T
v
=
v but d
(
u
,
v
)
>
0. By given assumption, we can replace x by u and y by v in (2) to obtain
(25)
ψ
(
d
(
u
,
v
)
)
=
ψ
(
d
(
f
u
,
g
v
)
)
≤
β
(
M
(
u
,
v
)
)
×
ψ
(
max
⁡
{
d
(
S
u
,
T
v
)
,
d
(
S
u
,
f
u
)
,
mmmmm
d
(
T
v
,
g
v
)
}
)
<
ψ
(
d
(
u
,
v
)
)
,
a contradiction. Hence u
=
v. Conversely, if f, g, S, and T have only one common fixed point then the set of common fixed points of f, g, S, and T being singleton is well ordered.Example 13.
LetX
=
{
1,2
,
3,4
} be a partially ordered set defined as x
⪯
y if and only if x
≥
y and d a usual metric on X. Define self-maps f, g, S, and T as
(26)
f
=
(
1
2
3
4
1
1
1
2
)
,
g
=
(
1
2
3
4
1
1
1
1
)
,
S
=
(
1
2
3
4
1
2
4
4
)
,
T
=
(
1
2
3
4
1
3
3
4
)
.
It is easy to verify that mappings f and g are dominated and S and T are dominating. Take ψ
(
x
)
=
ln
⁡
(
x
+
1
), β
(
x
)
=
ψ
(
x
)
/
x, and φ
(
x
)
=
x (see Table 1).Table 1
(
x
,
y)
ψ
(
d
(
f
x
,
f
y
)
)
β(M(x
,
y))ψ
(
max
⁡
{
d
(
S
x
,
T
y
)
,
d
(
S
x
,
f
x
)
,
d
(
T
y
,
g
y
)
}
)
(1, 1)
0
0
(1, 2)
0
0.741276
(1, 3)
0
0.741276
(1, 4)
0
0.869742
(2, 2)
0
0.741276
(2, 3)
0
0.741276
(2, 4)
0
0.869742
(3, 3)
0
0.869742
(3, 4)
0
0.869742
(4, 4)
0.693147
0.869742The mappingsf, g, S, and T satisfy all the conditions given in Theorem 12. Moreover, 1 is a unique common fixed point of f, g, S, and T.Remark 14.
If in above example, we take(
x
,
y
)
=
(
4,4
), then
(27)
d
(
f
x
,
g
y
)
≰
k
d
(
S
x
,
T
y
)
,
for any value of k. Also,
(28)
ψ
(
d
(
f
x
,
g
y
)
)
≤
ψ
(
M
(
x
,
y
)
)
-
φ
(
M
(
x
,
y
)
)
does not hold for any altering distance functions ψ
,
φ
:
ℝ
+
→
ℝ
+. So Theorem  2.1 in [8] is not applicable.
Now we prove existence of coincidence point of two pairs of compatible mappings on partially ordered metric spaces.Theorem 15.
Let(
X
,
⪯
) be a partially ordered set such that there exists a complete metric d on X and f
,
g
,
S
,
T
:
X
→
X given mappings. Suppose that pairs (
f
,
S
) and (
g
,
T
) are compatible, f
,
g
,
S, and T are continuous, and (
f
,
g
) and (
g
,
f
) are partially weakly increasing with respect to T and S, respectively. If there exist β
∈
P and ψ
∈
Ψ such that
(29)
ψ
(
d
(
f
x
,
g
y
)
)
≤
β
(
M
(
x
,
y
)
)
ψ
(
max
⁡
{
d
(
S
x
,
T
y
)
,
d
(
S
x
,
f
x
)
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
d
(
T
y
,
g
y
)
}
)
holds for every x
,
y
∈
X for which S
x and T
y are comparable, where
(30)
M
(
x
,
y
)
=
ψ
(
1
2
max
⁡
⁡
{
1
2
d
(
S
x
,
T
y
)
,
d
(
S
x
,
f
x
)
,
d
(
T
y
,
g
y
)
,
n
n
n
n
n
n
n
n
n
n
n
n
1
2
[
d
(
S
x
,
f
x
)
+
d
(
T
y
,
g
y
)
]
}
1
2
)
,
then the pairs (
f
,
S
) and (
g
,
T
) have a coincidence point z
∈
X. Moreover, if S
z and T
z are comparable, then z
∈
X is a coincidence point of f, g, S, and T.Proof.
Supposex
0 be an arbitrary point in X. Following similar arguments to those given in Theorem 12, sequences {
x
n
} and {
y
n
} in X are given by
(31)
y
2
n
=
f
x
2
n
=
T
x
2
n
+
1
,
y
2
n
+
1
=
g
x
2
n
+
1
=
S
x
2
n
+
2
,
∀
n
∈
ℕ
∪
{
0
}
.
By construction, we have x
2
n
+
1
∈
T
-
1
(
f
x
2
n
). Using the fact that (
f
,
g
) is partially weakly increasing with respect to T, we obtain
(32)
T
x
2
n
+
1
=
f
x
2
n
⪯
g
f
x
2
n
=
g
x
2
n
+
1
=
S
x
2
n
+
2
,
bbbbbbbbbbb
∀
n
∈
ℕ
∪
{
0
}
.
Also, x
2
n
+
2
∈
S
-
1
(
g
x
2
n
+
1
). As (
g
,
f
) is partially weakly increasing with respect to S, so
(33)
S
x
2
n
+
2
=
g
x
2
n
+
1
⪯
f
g
x
2
n
+
1
=
f
x
2
n
+
2
=
T
x
2
n
+
3
,
nnnnnnnnnnnnnn
∀
n
∈
ℕ
∪
{
0
}
.
Hence
(34)
T
x
1
⪯
S
x
2
⪯
T
x
3
⪯
⋯
⪯
T
x
2
n
+
1
⪯
S
x
2
n
+
2
⪯
T
x
2
n
+
3
⪯
⋯
.
That is,
(35)
y
0
⪯
y
1
⪯
y
2
⪯
⋯
⪯
y
2
n
⪯
y
2
n
+
1
⪯
y
2
n
+
2
⪯
⋯
.
We will prove the result in four steps.
Step
1. First we show that
(36)
lim
⁡
n
→
∞
d
(
y
n
,
y
n
+
1
)
=
0
.
First Case. There exists an n
∈
ℕ
∪
{
0
} such that y
n
=
y
n
+
2. If there exists an n
∈
ℕ such that y
2
n
+
1
=
y
2
n
-
1, then by (35) we have y
2
n
-
1
=
y
2
n
=
y
2
n
+
1. Now we claim that y
2
n
+
2
=
y
2
n
+
1. If not, then d
(
y
2
n
+
1
,
y
2
n
+
2
)
>
0. As S
x
2
n
+
2 and T
x
2
n
+
1 are comparable so from inequality (29), we have
(37)
ψ
(
d
(
y
2
n
+
2
,
y
2
n
+
1
)
)
=
ψ
(
d
(
f
x
2
n
+
2
,
g
x
2
n
+
1
)
)
≤
β
(
M
(
x
2
n
+
2
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
⁡
{
d
(
S
x
2
n
+
2
,
T
x
2
n
+
1
)
,
d
(
S
x
2
n
+
2
,
f
x
2
n
+
2
)
,
d
(
T
x
2
n
+
1
,
g
x
2
n
+
1
)
}
)
=
β
(
M
(
x
2
n
+
1
,
x
2
n
+
2
)
)
×
ψ
(
max
⁡
{
d
(
y
2
n
+
1
,
y
2
n
)
,
d
(
y
2
n
+
1
,
y
2
n
+
2
)
,
d
(
y
2
n
,
y
2
n
+
1
)
}
)
=
β
(
M
(
x
2
n
+
1
,
x
2
n
+
2
)
)
×
ψ
(
max
⁡
{
0
,
d
(
y
2
n
+
1
,
y
2
n
+
2
)
}
⁡
)
<
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
,
a contradiction. Thus for every k
≥
2
n, we have y
k
=
y
2
n
-
1. This implies that lim
⁡
n
→
∞
d
(
y
n
,
y
n
+
1
)
=
0. Similarly, (36) remains valid if y
2
n
=
y
2
n
+
2.
Second Case. For every n
∈
ℕ, y
n
≠
y
n
+
2. Since S
x
2
n and T
x
2
n
+
1 are comparable, so by (29) we obtain
(38)
ψ
(
d
(
y
2
n
,
y
2
n
+
1
)
)
=
ψ
(
d
(
f
x
2
n
,
g
x
2
n
+
1
)
)
≤
β
(
M
(
x
2
n
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
⁡
{
d
(
S
x
2
n
,
T
x
2
n
+
1
)
,
n
n
n
n
n
n
n
n
n
n
n
n
n
d
(
S
x
2
n
,
f
x
2
n
)
,
d
(
T
x
2
n
+
1
,
g
x
2
n
+
1
)
}
)
=
β
(
M
(
x
2
n
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
{
d
(
y
2
n
-
1
,
y
2
n
)
,
d
(
y
2
n
-
1
,
y
2
n
)
,
n
n
n
n
n
n
n
n
n
n
n
n
d
(
y
2
n
,
y
2
n
+
1
)
}
)
<
ψ
(
max
⁡
⁡
{
d
(
y
2
n
-
1
,
y
2
n
)
,
d
(
y
2
n
,
y
2
n
+
1
)
}
)
,
which further implies that
(39)
ψ
(
d
(
y
2
n
,
y
2
n
+
1
)
)
<
ψ
(
d
(
y
2
n
-
1
,
y
2
n
)
)
.
Similarly, as S
x
2
n
+
2 and T
x
2
n
+
1 are comparable, so by inequality (29), we have
(40)
ψ
(
d
(
y
2
n
+
2
,
y
2
n
+
1
)
)
=
ψ
(
d
(
f
x
2
n
+
2
,
g
x
2
n
+
1
)
)
≤
β
(
M
(
x
2
n
+
2
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
{
d
(
S
x
2
n
+
2
,
T
x
2
n
+
1
)
,
ccccccccccccccccc
d
(
S
x
2
n
+
2
,
f
x
2
n
+
2
)
,
m
m
m
m
m
m
m
m
m
d
(
T
x
2
n
+
1
,
g
x
2
n
+
1
)
}
)
=
β
(
M
(
x
2
n
+
2
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
{
d
(
y
2
n
+
1
,
y
2
n
)
,
n
n
n
n
n
n
n
n
n
n
n
n
n
d
(
y
2
n
+
1
,
y
2
n
+
2
)
,
n
n
n
n
n
n
n
n
n
n
n
n
n
d
(
y
2
n
,
y
2
n
+
1
)
}
)
<
ψ
(
max
⁡
{
d
(
y
2
n
+
1
,
y
2
n
)
,
d
(
y
2
n
+
1
,
y
2
n
+
2
)
}
)
,
which implies that
(41)
ψ
(
d
(
y
2
n
+
2
,
y
2
n
+
1
)
)
<
ψ
(
d
(
y
2
n
+
1
,
y
2
n
)
)
.
Combining (39) and (41), we have ψ
(
d
(
y
n
+
1
,
y
n
+
2
)
)
<
ψ
(
d
(
y
n
,
y
n
+
1
)
) for any n
∈
ℕ
∪
{
0
}. Hence {
ψ
(
d
(
y
n
,
y
n
+
1
)
)
} is a nonincreasing sequence and bounded below. So lim
⁡
lim
⁡
n
→
∞
ψ
(
d
(
y
n
,
y
n
+
1
)
) exists. We claim that lim
⁡
n
→
∞
ψ
(
d
(
y
n
,
y
n
+
1
)
)
=
0. If not, assume that lim
⁡
n
→
∞
ψ
(
d
(
y
n
,
y
n
+
1
)
)
=
r
>
0. Note that
(42)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
≤
β
(
M
(
x
2
n
+
2
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
{
d
(
S
x
2
n
+
2
,
T
x
2
n
+
1
)
,
d
(
S
x
2
n
+
2
,
f
x
2
n
+
2
)
,
mmmmm
d
(
T
x
2
n
+
1
,
g
x
2
n
+
1
)
}
)
gives
(43)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
≤
β
(
M
(
x
2
n
+
2
,
x
2
n
+
1
)
)
×
ψ
(
max
⁡
{
d
(
y
2
n
+
1
,
y
2
n
)
,
d
(
y
2
n
+
1
,
y
2
n
+
2
)
}
)
,
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
≤
β
(
M
(
x
2
n
+
2
,
x
2
n
+
1
)
)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
)
)
.
Hence,
(44)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
+
2
)
)
ψ
(
d
(
y
2
n
+
1
,
y
2
n
)
)
≤
β
(
M
(
x
2
n
+
2
,
x
2
n
+
1
)
)
<
1
,
Which, on taking limit as n
→
∞, implies that lim
⁡
n
→
∞
β
(
M
(
x
2
n
+
2
,
x
2
n
+
1
)
)
=
1, so lim
⁡
n
→
∞
ψ
(
d
(
S
x
2
n
+
2
,
T
x
2
n
+
1
)
)
=
0, which further implies that lim
⁡
n
→
∞
ψ
(
d
(
y
2
n
,
y
2
n
+
1
)
)
=
0, a contradiction. So r
=
0. Thus lim
⁡
n
→
∞
ψ
(
d
(
y
n
,
y
n
+
1
)
)
=
0.
Step  2. Now, we show that {
y
n
} is a Cauchy sequence. It is sufficient to show that {
y
2
n
} is a Cauchy sequence in X. Suppose that this is not the case. Applying Lemma 11 to the sequence {
y
n
} we obtain that there exist ε
>
0 and two sequences of positive integers {
m
k
} and {
n
k
} such that the sequences
(45)
d
(
y
2
m
k
,
y
2
n
k
)
,
d
(
y
2
m
k
,
y
2
n
k
+
1
)
,
d
(
y
2
m
k
-
1
,
y
2
n
k
)
,
d
(
y
2
m
k
-
1
,
y
2
n
k
+
1
)
all tend to ε when k
→
∞. Putting x
=
x
2
n
k
,
y
=
x
2
m
k
-
1 in (29), we obtain
(46)
ψ
(
d
(
y
2
n
k
+
1
,
y
2
m
k
)
)
ψ
(
max
⁡
{
d
(
y
2
m
k
,
y
2
n
k
)
,
d
(
y
2
n
k
,
y
2
n
k
-
1
)
,
d
(
y
2
m
k
-
1
,
y
2
m
k
)
}
)
≤
β
(
M
(
x
2
n
k
,
x
2
m
k
-
1
)
)
<
1
,
where
(47)
M
(
x
2
n
k
,
x
2
m
k
-
1
)
=
max
⁡
{
d
(
y
2
m
k
,
y
2
n
k
)
+
d
(
y
2
m
k
-
1
,
y
2
n
k
+
1
)
2
d
(
y
2
n
k
,
y
2
m
k
-
1
)
,
nnnnnnnnnnnn
d
(
y
2
n
k
,
y
2
n
k
-
1
)
,
d
(
y
2
m
k
-
1
,
y
2
m
k
)
,
n
n
n
n
n
n
n
n
n
n
n
n
d
(
y
2
m
k
,
y
2
n
k
)
+
d
(
y
2
m
k
-
1
,
y
2
n
k
+
1
)
2
}
.
Since
(48)
lim
⁡
k
→
∞
ψ
(
d
(
y
2
n
k
+
1
,
y
2
m
k
)
)
=
lim
⁡
k
→
∞
ψ
(
max
⁡
{
d
(
y
2
m
k
,
y
2
n
k
)
,
d
(
y
2
n
k
,
y
2
n
k
-
1
)
,
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
d
(
y
2
m
k
-
1
,
y
2
m
k
)
}
)
=
ψ
(
ε
)
,
we now have that lim
⁡
k
→
∞
β
(
M
(
x
2
n
k
,
x
2
m
k
-
1
)
)
=
1. By the property of the function β follows that lim
⁡
k
→
∞
M
(
x
2
n
k
,
x
2
m
k
-
1
)
=
0 a contradiction with lim
⁡
k
→
∞
M
(
x
2
n
k
,
x
2
m
k
-
1
)
=
ε. Hence, {
y
2
n
} is a Cauchy sequence in X, and hence {
y
n
} is a Cauchy sequence.
Step  3. Now we show the existence of a coincidence point for (
f
,
S
) and (
g
,
T
). From the completeness of (
X
,
d
), there is a z
∈
X such that lim
⁡
n
→
∞
y
n
=
z. From (31), we obtain that lim
⁡
n
→
∞
d
(
S
x
2
n
,
z
)
=
lim
⁡
n
→
∞
d
(
f
x
2
n
,
z
)
=
lim
⁡
n
→
∞
d
(
S
x
2
n
+
2
,
z
)
=
lim
⁡
n
→
∞
d
(
g
x
2
n
+
1
,
z
)
=
lim
⁡
n
→
∞
d
(
T
x
2
n
+
1
,
z
)
=
0. Since the pairs (
f
,
S
) and (
g
,
T
) are compatible, so lim
⁡
n
→
∞
d
(
S
(
f
x
2
n
)
,
f
(
S
x
2
n
)
)
=
d
(
T
(
g
x
2
n
+
1
)
,
g
(
T
x
2
n
+
1
)
)
=
0. Now using the continuity of f, g, S, T, we have lim
⁡
n
→
∞
d
(
f
(
S
x
2
n
)
,
f
z
)
=
lim
⁡
n
→
∞
d
(
g
(
T
x
2
n
+
1
)
,
g
z
)
=
lim
⁡
n
→
∞
d
(
S
(
T
x
2
n
+
1
)
,
S
z
)
=
lim
⁡
n
→
∞
d
(
T
(
S
x
2
n
+
2
)
,
T
z
)
=
0. The triangular inequality and (31) yield
(49)
d
(
S
z
,
f
z
)
≤
d
(
S
z
,
S
(
T
x
2
n
+
1
)
)
+
d
(
S
(
f
x
2
n
)
,
f
(
S
x
2
n
)
)
+
d
(
f
(
S
x
2
n
)
,
f
z
)
,
d
(
T
z
,
g
z
)
≤
d
(
T
z
,
T
(
S
x
2
n
+
2
)
)
+
d
(
T
(
g
x
2
n
+
1
)
,
g
(
T
x
2
n
+
1
)
)
+
d
(
g
(
T
x
2
n
+
1
)
,
g
z
)
.
On taking limit as n
→
∞, we obtain d
(
S
z
,
f
z
)
≤
0, and d
(
T
z
,
g
z
)
≤
0. Hence S
z
=
f
z and T
z
=
g
z.
Step  4. Existence of a coincidence point for f, g, S, and T.
SinceS
z and T
z are comparable, now are to show that T
z
=
S
z. If not, that is d
(
S
z
,
T
z
)
>
0. So from inequality (29), we have
(50)
ψ
(
d
(
S
z
,
T
z
)
)
=
ψ
(
d
(
f
z
,
g
z
)
)
≤
β
(
M
(
z
,
z
)
)
×
ψ
(
max
⁡
{
d
(
S
z
,
T
z
)
,
d
(
S
z
,
f
z
)
,
d
(
T
z
,
g
z
)
}
)
<
ψ
(
max
⁡
{
d
(
S
z
,
T
z
)
,
d
(
S
z
,
f
z
)
,
d
(
T
z
,
g
z
)
}
)
=
ψ
(
d
(
S
z
,
T
z
)
)
,
a contradiction. Hence T
z
=
S
z, which means that z is a coincidence point of f, g, S, and T.



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*Source: 379049-2014-02-19.xml*