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].
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.
Since
X
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. Hence
F
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
)
.