Proof.
Suppose
x
0
,
y
0
∈
X
with
x
0
≤
F
(
x
0
,
y
0
)
and
y
0
≥
F
(
y
0
,
x
0
)
. Define sequences
{
x
n
}
and
{
y
n
}
in
X
in the following way:
(9)
x
2
n
+
1
=
F
(
x
2
n
,
y
2
n
)
,
y
2
n
+
1
=
F
(
y
2
n
,
x
2
n
)
,
x
2
n
+
2
=
G
(
x
2
n
+
1
,
y
2
n
+
1
)
,
y
2
n
+
2
=
G
(
y
2
n
+
1
,
x
2
n
+
1
)
.
We are to prove that
{
x
n
}
sequence is nondecreasing and
{
y
n
}
sequence is nonincreasing. That is, for all
n
≥
0
(10)
x
2
n
≤
x
2
n
+
1
,
y
2
n
≥
y
2
n
+
1
.
For this, mathematical induction method is used.
Firstly suppose
n
=
0
. Having
x
0
≤
x
1
and
y
0
≥
y
1
, because
x
0
≤
F
(
x
0
,
y
0
)
and
y
0
≥
F
(
y
0
,
x
0
)
and as
x
1
=
F
(
x
0
,
y
0
)
and
y
1
=
F
(
y
0
,
x
0
)
, so (10) is verified for
n
=
0
.
Assume that (10) is satisfied for a constant
n
≥
0
; then, because
x
2
n
≤
x
2
n
+
1
and
y
2
n
≥
y
2
n
+
1
, from Definition 9 we have
(11)
x
2
n
+
1
=
F
(
x
2
n
,
y
2
n
)
≤
G
(
x
2
n
+
1
,
y
2
n
+
1
)
=
x
2
n
+
2
,
y
2
n
+
1
=
F
(
y
2
n
,
x
2
n
)
≥
G
(
y
2
n
+
1
,
y
2
n
+
1
)
=
y
2
n
+
2
.
Thus we get
x
2
n
≤
x
2
n
+
1
and
y
2
n
≥
y
2
n
+
1
.
Hereby, by the induction method we conclude that (10) hold for all
n
≥
0
. Thereof,
(12)
x
0
≤
x
1
≤
x
2
≤
⋯
≤
x
n
≤
x
n
+
1
≤
⋯
,
(13)
y
0
≥
y
1
≥
y
2
≥
⋯
≥
y
n
≥
y
n
+
1
≥
⋯
.
Denote
(14)
δ
n
=
d
(
x
n
,
x
n
+
1
)
+
d
(
y
n
,
y
n
+
1
)
,
showing
{
δ
n
}
sequence is nonincreasing. From (10) and (8) we have
(15)
d
(
x
2
n
+
1
,
x
2
n
+
2
)
=
d
(
F
(
x
2
n
,
y
2
n
)
,
G
(
x
2
n
+
1
,
y
2
n
+
1
)
)
≤
φ
(
2

1
(
d
(
x
2
n
,
x
2
n
+
1
)
+
d
(
y
2
n
,
y
2
n
+
1
)
+
d
(
x
2
n
,
y
2
n
+
1
)
+
d
(
y
2
n
,
x
2
n
+
1
)
)
×
2

1
(
d
(
x
2
n
,
x
2
n
+
1
)
+
d
(
y
2
n
,
y
2
n
+
1
)
)
≤
φ
(
d
(
x
2
n
,
x
2
n
+
1
)
+
d
(
y
2
n
,
y
2
n
+
1
)
2
)
.
Similarly, we can obtain
(16)
d
(
y
2
n
+
1
,
y
2
n
+
2
)
≤
φ
(
d
(
x
2
n
,
x
2
n
+
1
)
+
d
(
y
2
n
,
y
2
n
+
1
)
2
)
.
Thus, using properties of
φ
function we get
(17)
d
(
x
2
n
+
1
,
x
2
n
+
2
)
+
d
(
y
2
n
+
1
,
y
2
n
+
2
)
≤
2
φ
(
d
(
x
2
n
,
x
2
n
+
1
)
+
d
(
y
2
n
,
y
2
n
+
1
)
2
)
≤
2
d
(
x
2
n
,
x
2
n
+
1
)
+
d
(
y
2
n
,
y
2
n
+
1
)
2
=
d
(
x
2
n
,
x
2
n
+
1
)
+
d
(
y
2
n
,
y
2
n
+
1
)
.
Similarly one can show that
(18)
d
(
x
2
n
+
2
,
x
2
n
+
3
)
+
d
(
y
2
n
+
2
,
y
2
n
+
3
)
≤
d
(
x
2
n
+
1
,
x
2
n
+
2
)
+
d
(
y
2
n
+
1
,
y
2
n
+
2
)
.
Then, we obtain
(19)
d
(
x
n
+
1
,
x
n
+
2
)
+
d
(
y
n
+
1
,
y
n
+
2
)
≤
d
(
x
n
,
x
n
+
1
)
+
d
(
y
n
,
y
n
+
1
)
≤
⋯
≤
d
(
x
0
,
x
1
)
+
d
(
y
0
,
y
1
)
.
Thus a sequence
{
δ
n
}
is nonincreasing. Thence, there is a
δ
≥
0
is obtained with
(20)
lim
n
→
∞
δ
n
=
δ
.
Now, we claim that
(21)
lim
n
→
∞
δ
n
=
0
.
we substitute
n
=
2
k
in (14). Then we can get
(22)
δ
n
=
δ
2
k
=
d
(
x
2
k
,
x
2
k
+
1
)
+
d
(
y
2
k
,
y
2
k
+
1
)
≤
2
φ
(
d
(
x
2
k

1
,
x
2
k
)
+
d
(
y
2
k

1
,
y
2
k
)
2
)
=
2
φ
(
δ
n

1
2
)
.
Letting
n
→
∞
in (22), we get
(23)
δ
=
lim
δ
n
≤
2
lim
φ
(
δ
n

1
2
)
≤
2
δ
2
=
δ
.
Hence
δ
=
0
. That is
(24)
lim
n
→
∞
d
(
y
n
,
y
n
+
1
)
+
d
(
x
n
,
x
n
+
1
)
=
0
.
Now we show that
(25)
lim
n
→
∞
d
(
y
n
,
y
m
)
+
d
(
x
n
,
x
m
)
=
0
.
Suppose the contrary. At the time there exists
ε
>
0
when obtaining two subsequences
{
x
2
n
(
i
)
}
and
{
x
2
m
(
i
)
}
of
{
x
n
}
with
2
n
(
i
)
is the smallest index where
(26)
2
n
(
i
)
>
2
m
(
i
)
>
i
,
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)
)
≥
ε
.
This means that
(27)
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)

1
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)

1
)
<
ε
.
By
(
pm
s
4
)
in Definition 3 and (27), we have
(28)
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)
)
≤
d
(
x
2
m
(
i
)
,
x
2
m
(
i
)
+
1
)
+
d
(
x
2
m
(
i
)
+
1
,
x
2
n
(
i
)
)

d
(
x
2
m
(
i
)
+
1
,
x
2
m
(
i
)
+
1
)
≤
d
(
x
2
m
(
i
)
,
x
2
m
(
i
)
+
1
)
+
d
(
x
2
m
(
i
)
+
1
,
x
2
n
(
i
)
)
≤
d
(
x
2
m
(
i
)
,
x
2
m
(
i
)
+
1
)
+
d
(
x
2
m
(
i
)
+
1
,
x
2
m
(
i
)
)
+
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)
)

d
(
x
2
m
(
i
)
,
x
2
m
(
i
)
)
≤
2
d
(
x
2
m
(
i
)
,
x
2
m
(
i
)
+
1
)
+
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)
)
≤
2
d
(
x
2
m
(
i
)
,
x
2
m
(
i
)
+
1
)
+
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)

1
)
+
d
(
x
2
n
(
i
)

1
,
x
2
n
(
i
)
)
.
Similarly, we can obtain that
(29)
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)
)
≤
2
d
(
y
2
m
(
i
)
,
y
2
m
(
i
)
+
1
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)

1
)
+
d
(
y
2
n
(
i
)

1
,
y
2
n
(
i
)
)
.
Adding (28) and (29) and also from (27) and (26) we get
(30)
ε
≤
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)
)
≤
2
[
d
(
x
2
m
(
i
)
,
x
2
m
(
i
)
+
1
)
+
d
(
y
2
m
(
i
)
,
y
2
m
(
i
)
+
1
)
]
+
ε
+
d
(
x
2
n
(
i
)

1
,
x
2
n
(
i
)
)
+
d
(
y
2
n
(
i
)

1
,
y
2
n
(
i
)
)
.
Taking the limit as
i
→
∞
in (30) and by (26) we get
(31)
lim
i
→
∞
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)
)
=
ε
.
Employing the triangle inequality,
(32)
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)
)
≤
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)

1
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)

1
)
+
d
(
x
2
n
(
i
)

1
,
x
2
n
(
i
)
)
+
d
(
y
2
n
(
i
)

1
,
y
2
n
(
i
)
)
.
Similarly, we get
(33)
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)

1
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)

1
)
≤
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)
)
+
d
(
x
2
n
(
i
)
,
x
2
n
(
i
)

1
)
+
d
(
y
2
n
(
i
)
,
y
2
n
(
i
)

1
)
.
As
i
→
∞
in (33) and (32) and from (31) and (26) we can obtain
(34)
lim
i
→
∞
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)

1
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)

1
)
=
ε
.
Since from (12) we have
x
2
m
(
i
)
≤
x
2
n
(
i
)

1
and
y
2
m
(
i
)
≥
y
2
n
(
i
)

1
and also by (8) and (10),
(35)
d
(
x
2
m
(
i
)
+
1
,
x
2
n
(
i
)
)
=
d
(
F
(
x
2
m
(
i
)
,
y
2
m
(
i
)
)
,
G
(
x
2
n
(
i
)

1
,
y
2
n
(
i
)

1
)
)
≤
φ
(
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)

1
)
)
×
2

1
(
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)

1
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)

1
)
)
×
2

1
)
<
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)

1
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)

1
)
2
.
Similarly, we get
(36)
d
(
y
2
m
(
i
)
+
1
,
y
2
n
(
i
)
)
<
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)

1
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)

1
)
2
.
Thus
(37)
d
(
x
2
m
(
i
)
+
1
,
x
2
n
(
i
)
)
+
d
(
y
2
m
(
i
)
+
1
,
y
2
n
(
i
)
)
<
d
(
x
2
m
(
i
)
,
x
2
n
(
i
)

1
)
+
d
(
y
2
m
(
i
)
,
y
2
n
(
i
)

1
)
.
As
i
→
∞
in (37) we get
ε
=
0
, which is a contrast. Whence (25) is verified, possessing
(38)
lim
n
,
m
→
∞
d
(
x
n
,
x
m
)
=
0
,
lim
n
,
m
→
∞
d
(
y
n
,
y
m
)
=
0
.
By (3), we have
(39)
d
s
(
x
n
,
x
m
)
≤
2
d
(
x
n
,
x
m
)
=
0
,
d
s
(
y
n
,
y
m
)
≤
2
d
(
y
n
,
y
m
)
=
0
.
{
x
n
}
and
{
y
n
}
are Cauchy sequences in the metric space
(
X
,
d
s
)
. Because
(
X
,
d
)
is complete, it is also the case for
(
X
,
d
s
)
, then there exist
a
,
b
∈
X
with
(40)
lim
n
→
∞
d
s
(
x
n
,
a
)
=
0
,
lim
n
→
∞
d
s
(
y
n
,
b
)
=
0
.
On the other hand, we have
(41)
d
s
(
x
n
,
a
)
=
2
d
(
x
n
,
a
)

d
(
x
n
,
x
n
)

d
(
a
,
a
)
.
Getting the limit as
n
→
∞
in the upward equation and utilizing (40) and (38), we attain
(42)
lim
n
→
∞
d
(
x
n
,
a
)
=
1
2
d
(
a
,
a
)
,
in other words, possessing
d
(
a
,
a
)
≤
d
(
a
,
x
n
)
for all
n
∈
ℕ
. On letting
n
→
∞
, we achieve
(43)
d
(
a
,
a
)
≤
lim
n
→
∞
d
(
a
,
x
n
)
.
Using (42) and (43), we get that
(44)
lim
n
→
∞
d
(
a
,
x
n
)
=
d
(
a
,
a
)
=
0
.
Analogously, one can show that
(45)
lim
n
→
∞
d
(
b
,
y
n
)
=
d
(
b
,
b
)
=
0
,
exposing
a
=
F
(
a
,
b
)
,
a
=
G
(
a
,
b
)
,
b
=
F
(
b
,
a
)
, and
b
=
G
(
b
,
a
)
. To do that we prove the following steps.