# Tripled Fixed Point Theorems for Mixed Monotone Kannan Type Contractive Mappings

Authors: Marin Borcut; Mădălina Păcurar; Vasile Berinde
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We present new results on the existence and uniqueness of tripled fixed points for nonlinear mappings in partially ordered complete metric spaces that extend the results in the previous works: Berinde and Borcut, 2011, Borcut and Berinde, 2012,
and Borcut, 2012. An example and an application to support our new
results are also included in the paper.


---

## Body

## 1. Introduction

In some very recent papers, Berinde and Borcut [1], Borcut and Berinde [2], and Borcut [3] have introduced the concepts oftripled fixed point andtripled coincidence point, respectively, for nonlinear contractive mappings F
:
X
3
→
X in partially ordered complete metric spaces and obtained existence and uniqueness theorems of tripled fixed points and tripled coincidence points, respectively, for some general classes of contractive type mappings.The presented theorems in [1–3] extend several existing results in the literature [4–6]. We recall the main concepts needed to present them.Let(
X
,
≤
) be a partially ordered set and let d be a metric on X such that (
X
,
d
) is a complete metric space. Consider on the product space X
3 the following partial order: for (
x
,
y
,
z
)
,
(
u
,
v
,
w
)
∈
X
3,
(1)
(
u
,
v
,
w
)
≤
(
x
,
y
,
z
)
⟺
x
≥
u
,
y
≤
v
,
z
≥
w
.Definition 1 (see [1]).
Let(
X
,
≤
) be a partially ordered set and F
:
X
3
→
X. We say that F has the mixed monotone property if F
(
x
,
y
,
z
) is nondecreasing in x and z and is nonincreasing in y, that is, for any x, y, z
∈
X,
(2)
x
1
,
x
2
∈
X
,
x
1
≤
x
2
⟹
F
(
x
1
,
y
,
z
)
≤
F
(
x
2
,
y
,
z
)
,
y
1
,
y
2
∈
X
,
y
1
≤
y
2
⟹
F
(
x
,
y
1
,
z
)
≥
F
(
x
,
y
2
,
z
)
,
z
1
,
z
2
∈
X
,
z
1
≤
z
2
⟹
F
(
x
,
y
,
z
1
)
≤
F
(
x
,
y
,
z
2
)
.Definition 2 (see [1]).
An element(
x
,
y
,
z
)
∈
X
3 is called a tripled fixed point of F
:
X
3
→
X if
(3)
F
(
x
,
y
,
z
)
=
x
,
F
(
y
,
x
,
y
)
=
y
,
F
(
z
,
y
,
x
)
=
z
.Let(
X
,
d
) be a metric space. The mapping d
¯
:
X
3
→
X, given by
(4)
d
¯
[
(
x
,
y
,
z
)
,
(
u
,
v
,
w
)
]
=
d
(
x
,
u
)
+
d
(
y
,
v
)
+
d
(
z
,
w
)
,
defines a metric on X
3, which will be denoted for convenience by d, too.Definition 3.
LetX, Y, Z be nonempty sets and F
:
X
3
→
Y, G
:
Y
3
→
Z. We define the symmetric composition (or, the s-composition, for short) of F and G, by G
*
F
:
X
3
→
Z,
(5)
(
G
*
F
)
(
x
,
y
,
z
)
=
G
(
F
(
x
,
y
,
z
)
,
F
(
y
,
x
,
y
)
,
F
(
z
,
y
,
x
)
)
G
(
F
(
x
,
y
,
z
)
,
F
(
y
,
x
,
y
)
)
(
x
,
y
,
z
∈
X
)
.For each nonempty setX, denote by P
x the projection mapping
(6)
P
X
:
X
3
⟶
X
,
P
(
x
,
y
,
z
)
=
x
for
x
,
y
,
z
∈
X
.
The symmetric composition has the following properties.Proposition 4 (associativity).
IfF
:
X
3
→
Y, G
:
Y
3
→
Z and
(7)
H
:
Z
×
Z
×
Z
⟶
W
,
t
h
e
n
(
H
*
G
)
*
F
=
H
*
(
G
*
F
)
.Proposition 5 (identity element).
IfF
:
X
3
→
Y, then
(8)
F
*
P
X
=
P
Y
*
F
=
F
.Proposition 6 (mixed monotonicity).
If(
X
,
≤
), (
Y
,
≤
), (
Z
,
≤
) are partially ordered sets and the mappings F
:
X
3
→
Y, G
:
Y
3
→
Z are mixed monotone, then G
*
F is mixed monotone.Proposition 7.
If(
X
,
≤
) is a partially ordered set and F is mixed monotone, then F
n
=
F
*
F
n
-
1
=
F
n
-
1
*
F is mixed monotone for every n.The first main result in [1] is given by the following theorem.Theorem 8 (see [1]).
Let(
X
,
≤
) be a partially ordered set and suppose that there is a metric d on X such that (
X
,
d
) is a complete metric space. Let F
:
X
3
→
X be a continuous mapping having the mixed monotone property on X. Assume that there exist the constants j, k, l
∈
[
0,1
) with j
+
k
+
l
<
1 for which
(9)
d
(
F
(
x
,
y
,
z
)
,
F
(
u
,
v
,
w
)
)
≤
j
d
(
x
,
u
)
+
k
d
(
y
,
v
)
+
l
d
(
z
,
w
)
,
for all x
≥
u, y
≤
v, z
≥
w. If there exist x
0, y
0, z
0
∈
X such that
(10)
x
0
≤
F
(
x
0
,
y
0
,
z
0
)
,
y
0
≥
F
(
y
0
,
x
0
,
y
0
)
,
z
0
≤
F
(
z
0
,
y
0
,
x
0
)
,
then there exist x
,
y
,
z
∈
X such that
(11)
x
=
F
(
x
,
y
,
z
)
,
y
=
F
(
y
,
x
,
y
)
,
z
=
F
(
z
,
y
,
x
)
.Remark 9.
If we takej
=
k
=
l
=
α
/
3 in Theorem 8, then the contraction condition (9) can be written in a slightly simplified form
(12)
d
(
F
(
x
,
y
,
z
)
,
F
(
u
,
v
,
w
)
)
≤
α
3
[
d
(
x
,
u
)
+
d
(
y
,
v
)
+
d
(
z
,
w
)
]
.Theorem 10 (see [1]).
By adding to the hypotheses of Theorem8 the condition, for every (
x
,
y
,
z
), (
x
1
,
y
1
,
z
1
)
∈
X
×
X
×
X, there exists a (
u
,
v
,
w
)
∈
X
×
X
×
X that is comparable to (
x
,
y
,
z
) and (
x
1
,
y
1
,
z
1
); then, the tripled fixed point of F is unique.Theorem 11 (see [1]).
In addition to the hypotheses of Theorem8, suppose that x
0, y
0, z
0
∈
X are comparable. Then x
=
y
=
z.


## 2. Main Results

Starting from the results presented in the first section, we will obtain new existence and uniqueness theorems for operators which verify a Kannan type contraction condition; see [7].Denote(13)
F
x
=
F
(
x
,
y
,
z
)
,
F
y
=
F
(
y
,
x
,
y
)
,
F
z
=
F
(
z
,
y
,
x
)
.Theorem 12.
Let(
X
,
≤
) be a partially ordered set and suppose that there is a metric d on X such that (
X
,
d
) is a complete metric space. Let F
:
X
3
→
X be a mapping having the mixed monotone property on X. Assume that there exists a k
∈
[
0,1
) such that
(14)
d
(
F
(
x
,
y
,
z
)
,
F
(
u
,
v
,
w
)
)
≤
k
8
[
d
(
x
,
F
x
)
+
d
(
y
,
F
y
)
+
d
(
z
,
F
z
)
+
d
(
u
,
F
u
)
ww
+
d
(
v
,
F
v
)
+
d
(
w
,
F
w
)
(
y
,
F
y
)
]
,
∀
x
≥
u
,
y
≤
v
,
z
≥
w
.
Also suppose that either(a)
Fis continuous or
(b)
X has the following property:
(i)
if a nondecreasing sequence{
x
n
}
→
x, then x
n
≤
x for all n,
(ii)
if a nonincreasing sequence{
y
n
}
→
y, then y
n
≥
y for all n.If there exist x
0, y
0, z
0
∈
X such that
(15)
x
0
≤
F
(
x
0
,
y
0
,
z
0
)
,
y
0
≥
F
(
y
0
,
x
0
,
y
0
)
,
z
0
≤
F
(
z
0
,
y
0
,
x
0
)
,
then F has a triple fixed point; that is, there exist x, y, z
∈
X such that
(16)
x
=
F
(
x
,
y
,
z
)
,
y
=
F
(
y
,
x
,
y
)
,
z
=
F
(
z
,
y
,
x
)
.Proof.
Let the sequences{
x
n
}, {
y
n
}, {
z
n
}
⊂
X be defined by
(17)
x
n
+
1
=
F
(
x
n
,
y
n
,
z
n
)
=
F
n
+
1
(
x
0
,
y
0
,
z
0
)
,
y
n
+
1
=
F
(
y
n
,
x
n
,
y
n
)
=
F
n
+
1
(
y
0
,
x
0
,
y
0
)
,
z
n
+
1
=
F
(
z
n
,
y
n
,
x
n
)
=
F
n
+
1
(
z
0
,
y
0
,
x
0
)
,
(
n
=
0,1
,
…
)
.
Since F
n is mixed monotone for every n
∈
ℕ, by Proposition 7, it follows by (15) that {
x
n
} and {
z
n
} are nondecreasing and {
y
n
} is nonincreasing. Due to the mixed monotone property of F, it is easy to show that
(18)
x
2
=
F
(
x
1
,
y
1
,
z
1
)
≥
F
(
x
0
,
y
0
,
z
0
)
=
x
1
y
2
=
F
(
y
1
,
x
1
,
y
1
)
≤
F
(
y
0
,
x
0
,
y
0
)
=
y
1
z
2
=
F
(
z
1
,
y
1
,
x
1
)
≥
F
(
z
0
,
y
0
,
x
0
)
=
z
1
and thus we obtain three sequences satisfying the following conditions:
(19)
x
0
≤
x
1
≤
⋯
≤
x
n
≤
⋯

,
y
0
≥
y
1
≥
⋯
≥
y
n
≥
⋯

,
z
0
≤
z
1
≤
⋯
≤
z
n
≤
⋯

.
Now, for n
∈
ℕ, denote
(20)
D
x
n
+
1
=
d
(
x
n
+
1
,
x
n
)
,
D
y
n
+
1
=
d
(
y
n
+
1
,
y
n
)
,
D
z
n
+
1
=
d
(
z
n
+
1
,
z
n
)
,
D
n
+
1
=
D
x
n
+
1
+
D
y
n
+
1
+
D
z
n
+
1
.
Using (14), we get
(21)
D
x
n
+
1
=
d
(
x
n
+
1
,
x
n
)
=
d
(
F
(
x
n
,
y
n
,
z
n
)
,
F
(
x
n
-
1
,
y
n
-
1
,
z
n
-
1
)
)
≤
k
8
[
d
(
x
n
,
F
x
n
)
+
d
(
y
n
,
F
y
n
)
+
d
(
z
n
,
F
z
n
)
≤
k
8
w
+
d
(
x
n
-
1
,
F
x
n
-
1
)

+

d
(
y
n
-
1
,
F
y
n
-
1
)

+

d
(
z
n
-
1
,
F
z
n
-
1
)
]
=
k
8
[
D
x
n
+
D
y
n
+
D
z
n
+
D
x
n
+
1
+
D
y
n
+
1
+
D
z
n
+
1
]
,
and so
(22)
D
x
n
+
1
≤
k
8
[
D
x
n
+
D
y
n
+
D
z
n
+
D
x
n
+
1
+
D
y
n
+
1
+
D
z
n
+
1
]
.
Similarly, we obtain
(23)
D
y
n
+
1
≤
k
8
[
D
x
n
+
2
D
y
n
+
D
x
n
+
1
+
2
D
y
n
+
1
]
,
D
z
n
+
1
≤
k
8
[
D
x
n
+
D
y
n
+
D
z
n
+
D
x
n
+
1
+
D
y
n
+
1
+
D
z
n
+
1
]
.
By (22) and (23), we get
(24)
D
n
+
1
≤
k
8
[
3
D
x
n
+
4
D
y
n
+
2
D
z
n
+
3
D
x
n
+
1
+
4
D
y
n
+
1
+
2
D
z
n
+
1
]
≤
k
8
[
4
D
x
n
+
4
D
y
n
+
4
D
z
n
+
4
D
x
n
+
1
+
4
D
y
n
+
1
+
4
D
z
n
+
1
]
≤
k
2
[
D
n
+
D
n
+
1
]
.
Therefore, for all n
≥
1, we have
(25)
D
n
+
1
≤
α
·
D
n
≤
⋯
≤
α
n
·
D
1
,
where
α
=
k
2
-
k
∈
[
0,1
)
,
wwwww
when
k
∈
[
0,1
)
.
Because D
x
n
+
1
≤
D
n
+
1, D
y
n
+
1
≤
D
n
+
1, and D
z
n
+
1
≤
D
n
+
1, we have
(26)
D
x
n
+
1
≤
α
n
·
D
1
,
D
y
n
+
1
≤
α
n
·
D
1
,
D
z
n
+
1
≤
α
n
·
D
1
.
This implies that {
x
n
}, {
y
n
}, {
z
n
} are Cauchy sequences in X. Indeed, let m
≥
n; then,
(27)
d
(
x
m
,
x
n
)
≤
D
x
m
+
D
x
m
-
1
+
⋯
+
D
x
n
+
1
≤
[
α
m
-
1
+
α
m
-
2
+
⋯
+
α
n
]
·
D
1
=
α
n
-
α
m
1
-
α
·
D
1
<
α
n
1
-
α
·
D
1
.
Similarly, we can verify that{
y
n
} and {
z
n
} are also Cauchy sequences.
SinceX is a complete metric space, there exist x, y, z
∈
X such that
(28)
lim
⁡
x
→
∞
x
n
=
x
,
lim
⁡
x
→
∞
y
n
=
y
,
lim
⁡
x
→
∞
z
n
=
z
.
Finally, we claim that
(29)
x
=
F
(
x
,
y
,
z
)
,
y
=
F
(
y
,
x
,
y
)
,
z
=
F
(
z
,
y
,
x
)
.
Suppose first that assumption (a) holds. HenceF is continuous at (
x
,
y
,
z
), and, therefore, for any given ϵ
/
2
>
0, there exists δ
>
0 such that
(30)
d
(
(
x
,
y
,
z
)
,
(
u
,
v
,
w
)
)
=
d
(
x
,
u
)
+
d
(
y
,
v
)
+
d
(
z
,
w
)
<
δ
⟹
d
(
F
(
x
,
y
,
z
)
,
F
(
u
,
v
,
w
)
)
<
ϵ
2
.
Since
(31)
lim
⁡
x
→
∞
x
n
=
x
,
lim
⁡
x
→
∞
y
n
=
y
,
lim
⁡
x
→
∞
z
n
=
z
,
for η
=
min
⁡
(
ϵ
/
2
,
δ
/
2
), there exist n
0, m
0, p
0 such that, for n
≥
n
0, m
≥
m
0, p
≥
p
0,
(32)
d
(
x
n
,
x
)
<
η
,
d
(
y
n
,
y
)
<
η
,
d
(
z
n
,
z
)
<
η
.
Hence, for n
∈
ℕ, n
≥
max
⁡
⁡
{
n
0
,
m
0
,
p
0
},
(33)
d
(
F
(
x
,
y
,
z
)
,
x
)
≤
d
(
F
(
x
,
y
,
z
)
,
x
n
+
1
)
+
d
(
x
n
+
1
,
x
)
=
d
(
F
(
x
,
y
,
z
)
,
F
(
x
n
,
y
n
,
z
n
)
)
+
d
(
x
n
+
1
,
x
)
<
ϵ
2
+
η
≤
ϵ
.
This shows that x
=
F
(
x
,
y
,
z
). Similarly, one can show that
(34)
y
=
F
(
y
,
x
,
y
)
,
z
=
F
(
z
,
y
,
x
)
.
Suppose now that assumption (b) holds. Since{
x
n
}, {
z
n
} are nondecreasing and x
n
→
x, z
n
→
z, {
y
n
} is nonincreasing and y
n
→
y, by assumption (b), we have that x
n
≤
x, y
n
≥
y, and z
n
≤
z, for all n. Then, by triangle inequality and (14), we get
(35)
d
(
x
,
F
(
x
,
y
,
z
)
)
≤
d
(
x
,
x
n
+
1
)
+
d
(
x
n
+
1
,
F
(
x
,
y
,
z
)
)
=
d
(
x
,
x
n
+
1
)
d
(
F
(
x
n
,
y
n
,
z
n
)
,
F
(
x
,
y
,
z
)
)
≤
d
(
x
,
x
n
+
1
)
+
k
8
[
(
z
,
F
(
z
,
y
,
x
)
)
d
(
x
n
,
x
n
+
1
)
+
d
(
y
n
,
y
n
+
1
)
≤
d
(
x
,
x
n
+
1
)
+
k
8
wwe
+
d
(
z
n
,
z
n
+
1
)
+
d
(
x
,
F
(
x
,
y
,
z
)
)
≤
d
(
x
,
x
n
+
1
)
+
k
8
wwe
+
d
(
y
,
F
(
y
,
x
,
y
)
)

+

d
(
z
,
F
(
z
,
y
,
x
)
)
]
,
d
(
y
,
F
(
y
,
x
,
y
)
)
≤
d
(
y
,
y
n
+
1
)
+

k
8
[
(
x
,
F
(
x
,
y
,
z
)
)
d
(
x
n
,
x
n
+
1
)

+

2
d
(
y
n
,
y
n
+
1
)
≤
d
(
y
,
y
n
+
1
)
w
+

+
d
(
x
,
F
(
x
,
y
,
z
)
)

+

2
d
(
y
,
F
(
y
,
x
,
y
)
)
]
,
d
(
z
,
F
(
z
,
y
,
x
)
)
≤
d
(
z
,
z
n
+
1
)
+
k
8
[
(
z
,
F
(
z
,
y
,
x
)
)
d
(
x
n
,
x
n
+
1
)
+
d
(
y
n
,
y
n
+
1
)
w
≤
d
(
z
,
z
n
+
1
)
e
+
d
(
z
n
,
z
n
+
1
)
+
d
(
x
,
F
(
x
,
y
,
z
)
)
w
≤
d
(
z
,
z
n
+
1
)
e
+
d
(
y
,
F
(
y
,
x
,
y
)
)

+

d
(
z
,
F
(
z
,
y
,
x
)
)
]
.
By summing (35), we obtain
(36)
d
(
x
,
F
(
x
,
y
,
z
)
)
+
d
(
y
,
F
(
y
,
x
,
y
)
)
+
d
(
z
,
F
(
z
,
y
,
x
)
)
≤
2
2
-
k
[
d
(
x
,
x
n
+
1
)
+
d
(
y
,
y
n
+
1
)
+
d
(
z
,
z
n
+
1
)
]
+
k
4
(
2
-
k
)
[
3
d
(
x
n
,
x
n
+
1
)
+
4
d
(
y
n
,
y
n
+
1
)
+
k
4
(
2
-
k
)
w
+
2
d
(
z
n
,
z
n
+
1
)
]
,
and by letting n
→
∞ in the previous inequality, one obtains
(37)
d
(
x
,
F
(
x
,
y
,
z
)
)
+
d
(
y
,
F
(
y
,
x
,
y
)
)
+
d
(
z
,
F
(
z
,
y
,
x
)
)
≤
0
,
which proves that x
=
F
(
x
,
y
,
z
), y
=
F
(
y
,
x
,
y
), z
=
F
(
z
,
y
,
x
).


## 3. Uniqueness of Tripled Fixed Points

In [1–3], the authors also considered some additional conditions that ensure the uniqueness of the tripled fixed point or that, for the tripled fixed point (
x
,
y
,
z
), we have x
=
y
=
z.Similarly, one can prove that the tripled fixed point in Theorem12 is in fact unique, provided that the product space X
3 endowed with the partial order mentioned earlier has an additional property, as shown in the next theorem.Theorem 13.
By adding to the hypotheses of Theorem12 the condition, for every (
x
,
y
,
z
), (
x
1
,
y
1
,
z
1
)
∈
X
×
X
×
X, there exists a (
u
,
v
,
w
)
∈
X
×
X
×
X which is comparable to (
x
,
y
,
z
) and (
x
1
,
y
1
,
z
1
); then, the tripled fixed point of F is unique.Proof.
If(
x
*
,
y
*
,
z
*
)
∈
X
×
X
×
X is another tripled fixed point of F, then we show that
(38)
d
(
(
x
,
y
,
z
)
,
(
x
*
,
y
*
,
z
*
)
)
=
0
,
where
(39)
lim
⁡
x
→
∞
x
n
=
x
,
lim
⁡
x
→
∞
y
n
=
y
,
lim
⁡
x
→
∞
z
n
=
z
,
are as in the proof of Theorem 12. We consider two cases.
Case  1. If (
x
,
y
,
z
) are comparable to (
x
*
,
y
*
,
z
*
) with respect to the ordering in X
3, then, for every n
=
0,1
,
2
,
…,
(40)
(
F
n
(
x
,
y
,
z
)
,
F
n
(
y
,
x
,
y
)
,
F
n
(
z
,
y
,
x
)
)
=
(
x
,
y
,
z
)
that is comparable to
(
F
n
(
z
*
,
y
*
,
x
*
)
)
F
n
(
x
*
,
y
*
,
z
*
)
,
F
n
(
y
*
,
x
*
,
y
*
)
,
that is comparab to
F
n
(
z
*
,
y
*
,
x
*
)
)
=
(
x
*
,
y
*
,
z
*
)
.
Also, using the process of obtaining (26), we get
(41)
d
(
(
x
,
y
,
z
)
,
(
x
*
,
y
*
,
z
*
)
)
=
d
(
x
,
x
*
)
+
d
(
y
,
y
*
)
+
d
(
z
,
z
*
)
=
d
(
F
n
(
x
,
y
,
z
)
,
F
n
(
x
*
,
y
*
,
z
*
)
)
+
d
(
F
n
(
y
,
x
,
y
)
,
F
n
(
y
*
,
x
*
,
y
*
)
)
+
d
(
F
n
(
z
,
y
,
x
)
,
F
n
(
z
*
,
y
*
,
x
*
)
)
≤
α
n
[
d
(
x
,
x
*
)
+
d
(
y
,
y
*
)
+
d
(
z
,
z
*
)
]
=
α
n
d
(
(
x
,
y
,
z
)
,
(
y
*
,
x
*
,
z
*
)
)
,
α
∈
[
0,1
)
.
Now letting n
→
∞, this implies that d
(
(
x
,
y
,
z
)
,
(
y
*
,
x
*
,
z
*
)
)
=
0.
Case  2. If (
x
,
y
,
z
) are not comparable to (
x
*
,
y
*
,
z
*
), then there exists an upper bound or a lower bound (
u
,
v
,
w
)
∈
X
×
X
×
X of (
x
,
y
,
z
) and (
x
*
,
y
*
,
z
*
). Then, for all n
=
1,2
,
…,
(42)
(
F
n
(
u
,
v
,
w
)
,
F
n
(
v
,
u
,
v
)
,
F
n
(
w
,
v
,
u
)
)
is comparable to
(
F
n
(
x
,
y
,
z
)
,
F
n
(
y
,
x
,
y
)
,
F
n
(
z
,
y
,
x
)
)
=
(
x
,
y
,
z
)
and to
(
F
n
(
x
*
,
y
*
,
z
*
)
,
F
n
(
y
*
,
x
*
,
y
*
)
,
F
n
(
z
*
,
y
*
,
x
*
)
)
=
(
x
*
,
y
*
,
z
*
)
.
We have
(43)
d
(
(
x
y
z
)
,
(
x
*
y
*
z
*
)
)
=
d
(
(
F
n
(
x
,
y
,
z
)
F
n
(
y
,
x
,
y
)
F
n
(
z
,
y
,
x
)
)
,
(
F
n
(
x
*
,
y
*
,
z
*
)
F
n
(
y
*
,
x
*
,
y
*
)
F
n
(
z
*
,
y
*
,
x
*
)
)
)
≤
d
(
(
F
n
(
x
,
y
,
z
)
F
n
(
y
,
x
,
y
)
F
n
(
z
,
y
,
x
)
)
,
(
F
n
(
u
,
v
,
w
)
F
n
(
v
,
u
·
v
)
F
n
(
w
,
v
,
u
)
)
)
+
d
(
(
F
n
(
u
,
v
,
w
)
F
n
(
v
,
u
,
v
)
F
n
(
w
,
v
,
u
)
)
,
(
F
n
(
x
*
,
y
*
,
z
*
)
F
n
(
y
*
,
x
*
,
y
*
)
F
n
(
z
*
,
y
*
,
x
*
)
)
)
≤
α
n
{
[
d
(
x
,
u
)
+
d
(
y
,
v
)
+
d
(
z
,
w
)
]
ww
+
[
d
(
u
,
x
*
)
+
d
(
v
,
y
*
)
+
d
(
w
,
z
*
)
]
}
⟶
0
wwwew
as
n
⟶
∞
,
and
so
d
(
(
x
y
z
)
,
(
x
*
y
*
z
*
)
)
=
0
.Theorem 14.
In addition to the hypotheses of Theorem12, suppose that x
0, y
0, z
0
∈
X are comparable. Then x
=
y
=
z.Proof.
Recall thatx
0, y
0, z
0, ∈
X are such that
(44)
x
0
≤
F
(
x
0
,
y
0
,
z
0
)
,
y
0
≥
F
(
y
0
,
x
0
,
y
0
)
,
z
0
≤
F
(
z
0
,
y
0
,
x
0
)
.
Now, if x
0
≤
y
0 and z
0
≤
y
0, we claim that, for all n
∈
ℕ, x
n
≤
y
n and z
n
≤
y
n. Indeed, by the mixed monotone property of F,
(45)
x
1
=
F
(
x
0
,
y
0
,
z
0
)
≤
F
(
y
0
,
x
0
,
y
0
)
=
y
1
,
z
1
=
F
(
z
0
,
y
0
,
x
0
)
≤
F
(
y
0
,
x
0
,
y
0
)
=
y
1
.
Assume that x
n
≤
y
n and z
n
≤
y
n for some n. Now, consider
(46)
x
n
+
1
=
F
n
+
1
(
x
0
,
y
0
,
z
0
)
=
F
(
F
n
(
x
0
,
y
0
,
z
0
)
,
F
n
(
y
0
,
x
0
,
y
0
)
,
F
n
(
z
0
,
y
0
,
x
0
)
)
=
F
(
x
n
,
y
n
,
z
n
)
≤
F
(
y
n
,
x
n
,
y
n
)
=
y
n
+
1
,
similarly for z
n. Hence,
(47)
d
(
x
,
y
)
≤
d
(
x
,
x
n
+
1
)
+
d
(
y
,
x
n
+
1
)
≤
d
(
x
,
x
n
+
1
)
+
d
(
x
n
+
1
,
y
n
+
1
)
+
d
(
y
,
y
n
+
1
)
=
d
(
x
,
F
n
+
1
(
x
0
,
y
0
,
z
0
)
)
+
d
[
F
n
(
x
0
,
y
0
,
x
0
)
,
F
n
(
y
0
,
x
0
,
y
0
)
)
F
(
F
n
(
x
0
,
y
0
,
z
0
)
,
F
n
(
y
0
,
x
0
,
y
0
)
,
w
w
w
w
w
w
F
n
(
z
0
,
y
0
,
x
0
)
)
,
w
w
w
r
w
F
(
F
n
(
y
0
,
x
0
,
y
0
)
,
F
n
(
x
0
,
y
0
,
x
0
)
,
w
w
w
w
w
w
F
n
(
y
0
,
x
0
,
y
0
)
)
]
+
d
(
y
,
y
n
+
1
)
⟶
0
as n
→
∞. This implies that d
(
x
,
y
)
=
0 and hence we have x
=
y.
Similarly, we obtain thatd
(
x
,
z
)
=
0 and d
(
y
,
z
)
=
0. The other remaining cases for x
0, y
0, z
0 are similar.


## 4. An Example

LetX
=
[
0,1
] be endowed with the usual metric d
(
x
,
y
)
=
|
x
-
y
| and let F
:
X
3
→
X be given by F
(
x
,
y
,
z
)
=
11
/
80, for (
x
,
y
,
z
)
∈
[
0
,
4
/
5
]
×
[
0,1
]
2, and F
(
x
,
y
,
z
)
=
1
/
20, for (
x
,
y
,
z
)
∈
[
4
/
5
,
1
]
×
[
0,1
]
2.ThenF satisfies the Kannan type contractive condition (14) with k
=
14
/
15
<
1 but does not satisfy the Banach type contractive condition (9).Let us first prove the first part of the assertion above. It suffices to completely cover the following limit case.Case 1 (x
∈
[
4
/
5
,
1
], u, y, z, v, w
∈
[
0
,
4
/
5
)).
In this case, condition (14) reduces to
(48)
|
1
20
-
11
80
|
≤
k
8
[
|
x
-
1
20
|
+
|
y
-
11
80
|
+
|
z
-
11
80
|
|
1
20
-
11
80
|
≤
k
8
w
+
|
u
-
11
80
|
+
|
v
-
11
80
|
+
|
w
-
11
80
|
]
.
Sincex
∈
[
4
/
5
,
1
], we have
(49)
|
x
-
1
20
|
≥
|
4
5
-
1
20
|
=
3
4
and hence the minimum value of the right hand side of (48) is greater or equal to k
/
8
·
3
/
4.
Therefore, in order to have (48) satisfied for all x
∈
[
4
/
5
,
1
] and u, y, z, v, w
∈
[
0
,
4
/
5
), with x
≥
u, y
≤
v, z
≥
w, that is,
(50)
|
1
20
-
11
80
|
≤
k
8
·
3
4
,
it suffices to take k such that 14
/
15
≤
k
<
1.Note that, for the remaining cases to be discussed, the right hand side of (14) will be greater than the value obtained in Case 1.Case 2 (x
,
v
∈
[
4
/
5
,
1
] and u, y, z, w
∈
[
0
,
4
/
5
)).
In this case, for example, the minimum value of the right hand side of (14) will be greater or equal to k
/
8
·
6
/
4.
Note also that, in the casesx, u
∈
[
4
/
5
,
1
] or x, u
∈
[
0
,
4
/
5
), the left hand side of (14) is always zero and so (14) is satisfied for all values of y, z, v, w
∈
[
0,1
].
This proves that, indeed,F satisfies (14) with k
=
14
/
15
<
1.
F is not continuous but X satisfies assumption (b) in Theorem 12. Moreover, by taking x
0
=
0, y
0
=
1
/
5, and z
0
=
1
/
8, one can easily check that (15) is fulfilled.
Thus, the assumptions in Theorem12 are satisfied and hence F does admit tripled fixed points. By Theorem 13, we actually conclude that F has a unique tripled fixed point, (
11
/
80
,
11
/
80
,
11
/
80
).
Now let us show thatF does not satisfy (9).
Assume the contrary, that is, thatF does satisfy (9) and let ϵ
>
0 such that u
=
4
/
5
-
ϵ
∈
[
0
,
4
/
5
), and then take x
=
4
/
5 and y
=
z, v
=
w
∈
[
0,1
] arbitrary in (9) to obtain the inequality
(51)
7
80
≤
i
·
ϵ
,
ϵ
>
0
.
Now by letting ϵ
→
0 in (51), we reach a contradiction. This proves that, indeed, F does not satisfy (9).Remark 15.
For various particular cases of the results established in this section and for possible further developments, we refer to [2, 4–6, 8–28].


## 5. Applications

In this section, we present an application of tripled fixed point theorems for establishing existence and uniqueness results for the solutions of the nonlinear integral equation(52)
x
(
t
)
=
∫
0
T
G
(
t
,
s
)
[
f
(
s
,
x
(
s
)
)
+
g
(
s
,
x
(
s
)
)
+
h
(
s
,
x
(
s
)
)
]
d
s
‍
+
a
(
t
)
,
t
∈
[
0
,
T
]
,
T
>
0
.
We consider the space X
=
C
(
[
0
,
T
]
,
ℝ
) of continuous real functions defined on the interval [
0
,
T
], endowed with metric
(53)
d
(
u
,
v
)
=
max
⁡
0
≤
t
≤
T
|
u
(
t
)
-
v
(
t
)
|
,
for
u
,
v
∈
X
.
Define the partial order “≤” on X by
(54)
for
x
,
y
∈
X
,
x
≤
y
⟺
x
(
t
)
≤
y
(
t
)
,
x
e
w
≤
for any
t
∈
[
0
,
T
]
.
Thus, (
X
,
d
,
≤
) is a partially ordered complete metric space.For (52) we consider the following assumptions:(i)
f
,
g
,
h
:
[
0
,
T
]
×
ℝ
→
ℝ are continuous;
(ii)
a
:
[
0
,
T
]
×
→
ℝ is continuous;
(iii)
G
:
[
0
,
T
]
×
[
0
,
T
]
→
[
0
,
∞
) is continuous;
(iv)
there exist the constantsλ
1, λ
2, λ
3
>
0, such that, for all x, y
∈
ℝ, x
≤
y we have
(55)
0
≤
f
(
s
,
y
)
-
f
(
s
,
x
)
≤
λ
1
(
y
-
x
)
,
0
≤
g
(
s
,
x
)
-
g
(
s
,
y
)
≤
λ
2
(
y
-
x
)
,
0
≤
h
(
s
,
y
)
-
h
(
s
,
x
)
≤
λ
3
(
y
-
x
)
;
(v)
we suppose that(
λ
1
+
λ
2
+
λ
3
)
sup
⁡
0
≤
t
≤
T
∫
0
T
‍
G
(
t
,
s
)
d
s
<
1;
(vi)
there exist continuous functionsα
,
β
,
γ
:
[
0
,
T
]
→
ℝ such that
(56)
α
≤
∫
0
T
‍
G
(
t
,
s
)
[
f
(
s
,
α
(
s
)
)
+
g
(
s
,
β
(
s
)
)
+
h
(
s
,
γ
(
s
)
)
]
d
s
+
a
(
t
)
β
≥
∫
0
T
‍
G
(
t
,
s
)
[
f
(
s
,
β
(
s
)
)
+
g
(
s
,
α
(
s
)
)
+
h
(
s
,
β
(
s
)
)
]
d
s
+
a
(
t
)
γ
≤
∫
0
T
‍
G
(
t
,
s
)
[
f
(
s
,
γ
(
s
)
)
+
g
(
s
,
β
(
s
)
)
+
h
(
s
,
α
(
s
)
)
]
d
s
+
a
(
t
)
.Theorem 16.
Under assumptions (i)–(vi), (52) has a unique solution in C
(
[
0,1
]
,
ℝ
).Proof.
We consider the operatorF
:
X
3
→
X defined by
(57)
F
(
x
1
,
x
2
,
x
3
)
(
t
)
=
∫
0
T
‍
G
(
t
,
s
)
[
f
(
s
,
x
1
(
s
)
)
+
g
(
s
,
x
2
(
s
)
)
+
h
(
s
,
x
3
(
s
)
)
]
d
s
+
a
(
t
)
,
t
∈
[
0
,
T
]
,
for any x
1, x
2, x
3
∈
X.
We prove that the operatorF fulfills the conditions of Theorem 8.
First, we prove thatF has the mixed-monotone property.
Letx
1, y
1
∈
X, with x
1
≤
y
1 and t
∈
[
0
,
T
], then, we have
(58)
F
(
y
1
,
x
2
,
x
3
)
(
t
)
-
F
(
x
1
,
x
2
,
x
3
)
(
t
)
=
∫
0
T
‍
G
(
t
,
s
)
[
f
(
s
,
y
1
(
s
)
)
-
f
(
s
,
x
1
(
s
)
)
]
d
s
.
Given that x
1
(
t
)
≤
y
1
(
t
) for all t
∈
[
0
,
T
] and based on our assumption (iv), we have
(59)
f
(
s
,
y
)
-
f
(
s
,
x
)
≥
0
,
∀
x
,
y
∈
ℝ
,
x
≤
y
and
from
(
60
)
,
we have
F
(
y
1
,
x
2
,
x
3
)
(
t
)
-
F
(
x
1
,
x
2
,
x
3
)
(
t
)
≥
0
,
∀
t
∈
[
0
,
T
]
.
That is, F
(
x
1
,
x
2
,
x
3
)
(
t
)
≤
F
(
y
1
,
x
2
,
x
3
)
(
t
).
Forx
2, y
2
∈
X, with x
2
≤
y
2 and t
∈
[
0
,
T
], we have
(60)
F
(
x
1
,
x
2
,
x
3
)
(
t
)
-
F
(
x
1
,
y
2
,
x
3
)
(
t
)
=
∫
0
T
‍
G
(
t
,
s
)
[
g
(
s
,
x
2
(
s
)
)
-
g
(
s
,
y
2
(
s
)
)
]
d
s
.
Given that x
2
(
t
)
≤
y
2
(
t
) for all t
∈
[
0
,
T
] and based on our assumption (iv), that is,
(61)
g
(
s
,
x
)
-
g
(
s
,
y
)
≥
0
,
∀
x
,
y
∈
ℝ
,
x
≤
y
and from
(
62
)
, we get
F
(
x
1
,
x
2
,
x
3
)
(
t
)
-
F
(
x
1
,
y
2
,
x
3
)
(
t
)
≥
0
,
∀
t
∈
[
0
,
T
]
.
That is, F
(
x
1
,
x
2
,
x
3
)
(
t
)
≥
F
(
x
1
,
y
2
,
x
3
)
(
t
).
Similarly, one proves the same property for the third component and hence we haveF
(
x
1
,
x
2
,
x
3
)
(
t
)
≤
F
(
x
1
,
x
2
,
y
3
)
(
t
).
So,F has the mixed-monotone property.
Now, we estimated
(
F
(
x
1
,
x
2
,
x
3
)
,
F
(
y
1
,
y
2
,
y
3
)
) for x
1
≤
y
1, x
2
≥
y
2, x
3
≤
y
3, and with F having the mixed-monotone property, we get
(62)
d
(
F
(
x
1
,
x
2
,
x
3
)
,
F
(
y
1
,
y
2
,
y
3
)
)
=
max
⁡
0
≤
t
≤
T
⁡
|
F
(
x
1
,
x
2
,
x
3
)
(
t
)
-
F
(
y
1
,
y
2
,
y
3
)
(
t
)
|
=
max
⁡
0
≤
t
≤
T
⁡
(
F
(
y
1
,
y
2
,
y
3
)
(
t
)
-
F
(
x
1
,
x
2
,
x
3
)
(
t
)
)
.
Now, for all t
∈
[
0
,
t
], by using (iv), we have
(63)
F
(
y
1
,
y
2
,
y
3
)
(
t
)
-
F
(
x
1
,
x
2
,
x
3
)
(
t
)
=
∫
0
T
‍
G
(
t
,
s
)
[
f
(
s
,
y
1
(
s
)
)
-
g
(
s
,
x
1
(
s
)
)
]
d
s
+
∫
0
T
‍
G
(
t
,
s
)
[
g
(
s
,
y
2
(
s
)
)
-
g
(
s
,
x
2
(
s
)
)
]
d
s
+
∫
0
T
‍
G
(
t
,
s
)
[
h
(
s
,
y
3
(
s
)
)
-
h
(
s
,
x
3
(
s
)
)
]
d
s
≤
(
∫
0
T
‍
G
(
t
,
s
)
d
s
)
(
+
λ
3
d
(
x
3
,
y
3
)
)
λ
1
d
(
x
1
,
y
1
)
+
λ
2
d
(
x
2
,
y
2
)
≤
(
∫
0
T
‍
G
(
t
,
s
)
d
s
)
w
w
+
λ
3
d
(
x
3
,
y
3
)
)
≤
(
sup
⁡
0
≤
t
≤
T
∫
0
T
‍
G
(
t
,
s
)
d
s
)
(
+
λ
3
d
(
x
3
,
y
3
)
)
λ
1
d
(
x
1
,
y
1
)
+
λ
2
d
(
x
2
,
y
2
)
λ
1
d
(
x
1
,
y
1
)
+
λ
2
(
x
2
,
y
2
)
w
+
λ
3
d
(
x
3
,
y
3
)
)
,
which implies
(64)
d
(
F
(
x
1
,
x
2
,
x
3
)
,
F
(
y
1
,
y
2
,
y
3
)
)
≤
δ
1
d
(
x
1
,
y
1
)
+
δ
2
d
(
x
2
,
y
2
)
+
δ
3
d
(
x
3
,
y
3
)
,
where
(65)
δ
i
=
(
sup
⁡
0
≤
t
≤
T
∫
0
T
G
(
t
,
s
)
d
s
‍
)
λ
i
,
i
=
1,2
,
3
.
Using (v), we have δ
1
+
δ
2
+
δ
3
<
1. So, F fulfills the conditions of Theorem 8.
Letα, β, γ be the functions appearing in assumption (vi); then, we have
(66)
α
≤
F
(
α
,
β
,
γ
)
,
β
≥
F
(
β
,
α
,
β
)
,
γ
≤
F
(
γ
,
β
,
α
)
.
If x
0
=
α, y
0
=
β, z
0
=
λ, then all assumptions of Theorem 8 are fulfilled. So, there exists a tripled fixed point (
x
1
,
x
2
,
x
3
) for the operator F; that is,
(67)
x
1
=
F
(
x
1
,
x
2
,
x
3
)
,
x
2
=
F
(
x
2
,
x
1
,
x
2
)
,
x
3
=
F
(
x
3
,
x
2
,
x
1
)
.
Now, we show that, for any (
x
1
,
x
2
,
x
3
), (
y
1
,
y
2
,
y
3
)
∈
X
3, there exists (
z
1
,
z
2
,
z
3
)
∈
X
3 which is comparable to both of them. Indeed, denote
(68)
z
1
=
max
⁡
⁡
{
x
1
,
y
1
}
,
z
2
=
min
⁡
⁡
{
x
2
,
y
2
}
,
z
3
=
max
⁡
⁡
{
x
3
,
y
3
}
.
Then we have x
1
≤
z
1, x
2
≥
z
2, x
3
≤
z
3 and y
1
≤
z
1, y
2
≥
z
2, y
3
≤
z
3. This implies that (
z
1
,
z
2
,
z
3
) is comparable with (
x
1
,
x
2
,
x
3
) and with (
y
1
,
y
2
,
y
3
), so
(69)
(
x
1
,
x
2
,
x
3
)
≤
(
z
1
,
z
2
,
z
3
)
,
(
y
1
,
y
2
,
y
3
)
≤
(
z
1
,
z
2
,
z
3
)
.
Now, by Theorem 10, it follows that F has a unique triple fixed point which is in fact the solution the (52).



---

*Source: 120203-2014-03-13.xml*