Proof.
Let
x
0
∈
A
and
y
0
∈
B
be any two elements and let the sequences
{
x
n
}
and
{
y
n
}
be defined as
(3)
x
n
+
1
=
F
(
y
n
,
x
n
)
,
y
n
+
1
=
F
(
x
n
,
y
n
)
∀
n
≥
0
.
Then, for all
n
≥
0
,
x
n
∈
A
and
y
n
∈
B
.
By (2), we have
(4)
d
(
x
1
,
y
2
)
=
d
(
x
1
,
F
(
x
1
,
y
1
)
)
=
d
(
F
(
y
0
,
x
0
)
,
F
(
x
1
,
y
1
)
)
≤
k
[
d
(
y
0
,
F
(
y
0
,
x
0
)
)
+
d
(
x
1
,
F
(
x
1
,
y
1
)
)
]
=
k
[
d
(
y
0
,
x
1
)
+
d
(
x
1
,
y
2
)
]
,
which implies that
(5)
d
(
x
1
,
y
2
)
≤
t
d
(
y
0
,
x
1
)
,
where
(6)
0
<
t
=
k
1
-
k
<
1
and also by (2), we have
(7)
d
(
y
1
,
x
2
)
=
d
(
y
1
,
F
(
y
1
,
x
1
)
)
=
d
(
F
(
x
0
,
y
0
)
,
F
(
y
1
,
x
1
)
)
≤
k
[
d
(
x
0
,
F
(
x
0
,
y
0
)
)
+
d
(
y
1
,
F
(
y
1
,
x
1
)
)
]
=
k
[
d
(
x
0
,
y
1
)
+
d
(
y
1
,
x
2
)
]
,
which implies that
(8)
d
(
y
1
,
x
2
)
≤
t
d
(
x
0
,
y
1
)
,
where
t
is
the
same
as
in
(
6
)
.
Again, by (2), we have
(9)
d
(
x
2
,
y
3
)
=
d
(
x
2
,
F
(
x
2
,
y
2
)
)
=
d
(
F
(
y
1
,
x
1
)
,
F
(
x
2
,
y
2
)
)
≤
k
[
d
(
y
1
,
F
(
y
1
,
x
1
)
)
+
d
(
x
2
,
F
(
x
2
,
y
2
)
)
]
=
k
[
d
(
y
1
,
x
2
)
+
d
(
x
2
,
y
3
)
]
or
(10)
d
(
x
2
,
y
3
)
≤
k
1
-
k
d
(
y
1
,
x
2
)
which, by (8), implies that
(11)
d
(
x
2
,
y
3
)
≤
t
2
d
(
x
0
,
y
1
)
,
where
t
is the same as in (6).
Similarly, by (2), we have
(12)
d
(
y
2
,
x
3
)
=
d
(
y
2
,
F
(
y
2
,
x
2
)
)
=
d
(
F
(
x
1
,
y
1
)
,
F
(
y
2
,
x
2
)
)
≤
k
[
d
(
x
1
,
F
(
x
1
,
y
1
)
)
+
d
(
y
2
,
F
(
y
2
,
x
2
)
)
]
=
k
[
d
(
x
1
,
y
2
)
+
d
(
y
2
,
x
3
)
]
or
(13)
d
(
y
2
,
x
3
)
≤
k
1
-
k
d
(
x
1
,
y
2
)
which, by (5), implies that
(14)
d
(
y
2
,
x
3
)
≤
t
2
d
(
y
0
,
x
1
)
,
where
t
is the same as in (6).
Also, by (2), we have
(15)
d
(
x
3
,
y
4
)
=
d
(
x
3
,
F
(
x
3
,
y
3
)
)
=
d
(
F
(
y
2
,
x
2
)
,
F
(
x
3
,
y
3
)
)
≤
k
[
d
(
y
2
,
F
(
y
2
,
x
2
)
)
+
d
(
x
3
,
F
(
x
3
,
y
3
)
)
]
=
k
[
d
(
y
2
,
x
3
)
+
d
(
x
3
,
y
4
)
]
or by (14),
(16)
d
(
x
3
,
y
4
)
≤
k
1
-
k
[
d
(
y
2
,
x
3
)
]
=
t
3
d
(
y
0
,
x
1
)
,
where
t
is the same as in (6).
Similarly, by (2), we have
(17)
d
(
y
3
,
x
4
)
=
d
(
y
3
,
F
(
y
3
,
x
3
)
)
=
d
(
F
(
x
2
,
y
2
)
,
F
(
y
3
,
x
3
)
)
≤
k
[
d
(
x
2
,
F
(
x
2
,
y
2
)
)
+
d
(
y
3
,
F
(
y
3
,
x
3
)
)
]
=
k
[
d
(
x
2
,
y
3
)
+
d
(
y
3
,
x
4
)
]
or by (11), implies that
(18)
d
(
y
3
,
x
4
)
≤
k
1
-
k
[
d
(
x
2
,
y
3
)
]
=
t
3
d
(
x
0
,
y
1
)
,
where
t
is the same as in (6).
Let
m
be any integer. We assume
(19)
d
(
x
n
,
y
n
+
1
)
=
d
(
x
n
,
F
(
x
n
,
y
n
)
)
≤
t
n
d
(
y
0
,
x
1
)
,
(20)
d
(
y
n
,
x
n
+
1
)
=
d
(
y
n
,
F
(
y
n
,
x
n
)
)
≤
t
n
d
(
x
0
,
y
1
)
for all
n
≤
m
where
n
is odd and
(21)
d
(
x
n
,
y
n
+
1
)
=
d
(
x
n
,
F
(
x
n
,
y
n
)
)
≤
t
n
d
(
x
0
,
y
1
)
,
(22)
d
(
y
n
,
x
n
+
1
)
=
d
(
y
n
,
F
(
y
n
,
x
n
)
)
≤
t
n
d
(
y
0
,
x
1
)
for all
n
≤
m
where
n
is even.
Let
m
be even. Then, by (2) and (3), we have
(23)
d
(
x
m
+
1
,
y
m
+
2
)
=
d
(
x
m
+
1
,
F
(
x
m
+
1
,
y
m
+
1
)
)
=
d
(
F
(
y
m
,
x
m
)
,
F
(
x
m
+
1
,
y
m
+
1
)
)
≤
k
[
d
(
y
m
,
x
m
+
1
)
+
d
(
x
m
+
1
,
y
m
+
2
)
]
or by (22), we have
(24)
d
(
x
m
+
1
,
y
m
+
2
)
≤
k
1
-
k
[
t
m
d
(
y
0
,
x
1
)
]
=
t
m
+
1
d
(
y
0
,
x
1
)
f
f
f
f
f
f
f
f
f
f
f
f
f
(
where
t
is
the
same
as
in
(
6
)
)
.
Similarly, by (2) and (3), we have
(25)
d
(
y
m
+
1
,
x
m
+
2
)
=
d
(
y
m
+
1
,
F
(
y
m
+
1
,
x
m
+
1
)
)
=
d
(
F
(
x
m
,
y
m
)
,
F
(
y
m
+
1
,
x
m
+
1
)
)
≤
k
[
d
(
x
m
,
y
m
+
1
)
+
d
(
y
m
+
1
,
x
m
+
2
)
]
or by (21), we have
(26)
d
(
y
m
+
1
,
x
m
+
2
)
≤
k
1
-
k
[
t
m
d
(
x
0
,
y
1
)
]
=
t
m
+
1
d
(
x
0
,
y
1
)
f
f
f
f
f
f
f
f
f
f
f
f
f
(
where
t
is
the
same
as
in
(
6
)
)
.
Again, let
m
be odd.
Then, by (2) and (3), we have
(27)
d
(
x
m
+
1
,
y
m
+
2
)
=
d
(
x
m
+
1
,
F
(
x
m
+
1
,
y
m
+
1
)
)
=
d
(
F
(
y
m
,
x
m
)
,
F
(
x
m
+
1
,
y
m
+
1
)
)
≤
k
[
d
(
y
m
,
x
m
+
1
)
+
d
(
x
m
+
1
,
y
m
+
2
)
]
or by (20), we have
(28)
d
(
x
m
+
1
,
y
m
+
2
)
≤
k
1
-
k
[
t
m
d
(
x
0
,
y
1
)
]
=
t
m
+
1
d
(
x
0
,
y
1
)
f
f
f
f
f
f
f
f
f
f
f
f
f
(
where
t
is
the
same
as
in
(
6
)
)
.
Similarly, by (2) and (3), we have
(29)
d
(
y
m
+
1
,
x
m
+
2
)
=
d
(
y
m
+
1
,
F
(
y
m
+
1
,
x
m
+
1
)
)
=
d
(
F
(
x
m
,
y
m
)
,
F
(
y
m
+
1
,
x
m
+
1
)
)
≤
k
[
d
(
x
m
,
y
m
+
1
)
+
d
(
y
m
+
1
,
x
m
+
2
)
]
or by (19), we have
(30)
d
(
y
m
+
1
,
x
m
+
2
)
≤
k
1
-
k
[
t
m
d
(
y
0
,
x
1
)
]
=
t
m
+
1
d
(
y
0
,
x
1
)
f
f
f
f
f
f
f
f
f
f
f
f
f
(
where
t
is
the
same
as
in
(
6
)
)
.
Thus (19)–(22) hold for
m
+
1
. But we have shown in (5), (8)–(18) that this is valid for
m
=
1,2
,
3
. Then, by induction, we conclude that (19)–(22) are valid for all
m
.
From the above we conclude that, for all odd integer
n
, we have
(31)
d
(
x
n
,
y
n
+
1
)
=
d
(
x
n
,
F
(
x
n
,
y
n
)
)
≤
t
n
d
(
y
0
,
x
1
)
,
d
(
y
n
,
x
n
+
1
)
=
d
(
y
n
,
F
(
y
n
,
x
n
)
)
≤
t
n
d
(
x
0
,
y
1
)
and for all even integer
n
, we have
(32)
d
(
x
n
,
y
n
+
1
)
=
d
(
x
n
,
F
(
x
n
,
y
n
)
)
≤
t
n
d
(
x
0
,
y
1
)
,
d
(
y
n
,
x
n
+
1
)
=
d
(
y
n
,
F
(
y
n
,
x
n
)
)
≤
t
n
d
(
y
0
,
x
1
)
.
By (2), we have
(33)
d
(
x
1
,
y
1
)
=
d
(
F
(
y
0
,
x
0
)
,
F
(
x
0
,
y
0
)
)
≤
k
[
d
(
y
0
,
F
(
y
0
,
x
0
)
)
+
d
(
x
0
,
F
(
x
0
,
y
0
)
)
]
=
t
t
+
1
[
d
(
y
0
,
x
1
)
+
d
(
x
0
,
y
1
)
]
(
where
t
is
the
same
as
in
(
6
)
)
.
Again, by (2), we have
(34)
d
(
x
2
,
y
2
)
=
d
(
F
(
y
1
,
x
1
)
,
F
(
x
1
,
y
1
)
)
≤
k
[
d
(
y
1
,
F
(
y
1
,
x
1
)
)
+
d
(
x
1
,
F
(
x
1
,
y
1
)
)
]
=
k
[
d
(
y
1
,
x
2
)
+
d
(
x
1
,
y
2
)
]
≤
k
[
t
d
(
x
0
,
y
1
)
+
t
d
(
y
0
,
x
1
)
]
(
by
(
5
)
and
(
8
)
)
=
t
2
t
+
1
[
d
(
y
0
,
x
1
)
+
d
(
x
0
,
y
1
)
]
(
where
t
is
the
same
as
in
(
6
)
)
.
Let, for some integer
m
,
(35)
d
(
x
m
,
y
m
)
≤
t
m
t
+
1
[
d
(
x
0
,
y
1
)
+
d
(
y
0
,
x
1
)
]
.
Let
m
be odd. Then, by (2) and (3), we have
(36)
d
(
x
m
+
1
,
y
m
+
1
)
=
d
(
F
(
y
m
,
x
m
)
,
F
(
x
m
,
y
m
)
)
≤
k
[
d
(
y
m
,
x
m
+
1
)
+
d
(
x
m
,
y
m
+
1
)
]
≤
k
[
t
m
[
d
(
x
0
,
y
1
)
+
d
(
y
0
,
x
1
)
]
]
(
by
(
31
)
)
=
t
m
+
1
t
+
1
[
d
(
x
0
,
y
1
)
+
d
(
y
0
,
x
1
)
]
(
by
(
6
)
)
.
Again, let
m
be even. Then, by (2) and (3), we have
(37)
d
(
x
m
+
1
,
y
m
+
1
)
=
d
(
F
(
y
m
,
x
m
)
,
F
(
x
m
,
y
m
)
)
≤
k
[
d
(
y
m
,
x
m
+
1
)
+
d
(
x
m
,
y
m
+
1
)
]
≤
k
[
t
m
[
d
(
y
0
,
x
1
)
+
d
(
x
0
,
y
1
)
]
]
(
by
(
32
)
)
=
t
m
+
1
t
+
1
[
d
(
y
0
,
x
1
)
+
d
(
x
0
,
y
1
)
]
(
by
(
6
)
)
.
Therefore, (35) also holds if we replace
m
by
m
+
1
. But, (33) and (34) imply that (35) is true for
m
=
1,2
.
Then, by induction, for all
n
, it follows that
(38)
d
(
x
n
,
y
n
)
≤
t
n
t
+
1
[
d
(
x
0
,
y
1
)
+
d
(
y
0
,
x
1
)
]
.
Now, by (31)–(32) and (38), we have
(39)
d
(
x
n
,
x
n
+
1
)
+
d
(
y
n
,
y
n
+
1
)
≤
d
(
x
n
,
y
n
)
+
d
(
y
n
,
x
n
+
1
)
+
d
(
y
n
,
x
n
)
+
d
(
x
n
,
y
n
+
1
)
=
[
d
(
x
n
,
y
n
)
+
d
(
y
n
,
x
n
)
]
+
[
d
(
y
n
,
x
n
+
1
)
+
d
(
x
n
,
y
n
+
1
)
]
≤
2
t
n
t
+
1
[
d
(
x
0
,
y
1
)
+
d
(
y
0
,
x
1
)
]
+
t
n
[
d
(
x
0
,
y
1
)
+
d
(
y
0
,
x
1
)
]
.
Since
0
<
t
<
1
, it follows that
Σ
d
(
x
n
,
x
n
+
1
)
+
Σ
d
(
y
n
,
y
n
+
1
)
<
∞
. This implies that
{
x
n
}
and
{
y
n
}
are Cauchy sequences and hence are convergent.
Since
A
and
B
are closed subsets,
{
x
n
}
⊂
A
, and
{
y
n
}
⊂
B
, it follows that
(40)
x
n
⟶
x
∈
A
,
y
n
⟶
y
∈
B
as
n
⟶
∞
.
Again, from (35),
d
(
x
n
,
y
n
)
→
0
as
n
→
∞
.
Therefore, from (38),
(41)
x
=
y
.
Since
A
∩
B
≠
ϕ
, then from the above it follows that
x
∈
A
∩
B
.
Now, by (2) and (3), we have
(42)
d
(
x
,
F
(
x
,
y
)
)
≤
d
(
x
,
x
n
+
1
)
+
d
(
x
n
+
1
,
F
(
x
,
y
)
)
=
d
(
x
,
x
n
+
1
)
+
d
(
F
(
y
n
,
x
n
)
,
F
(
x
,
y
)
)
≤
d
(
x
,
x
n
+
1
)
+
k
[
d
(
y
n
,
x
n
+
1
)
+
d
(
x
,
F
(
x
,
y
)
)
]
or
(43)
d
(
x
,
F
(
x
,
y
)
)
≤
1
1
-
k
d
(
x
,
x
n
+
1
)
+
k
1
-
k
d
(
y
n
,
x
n
+
1
)
.
Taking the limit as
n
→
∞
in the above inequality, using (40) and (41), we obtain
d
(
x
,
F
(
x
,
y
)
)
=
0
. Again, in view of (41), we conclude that
x
=
F
(
x
,
x
)
; that is, we have a strong coupled fixed point of
F
.
This completes the proof of the theorem.