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Since
g
g
n
+
1
(
0
x
)
g
n
(
0
x
)
is
monotone,
and
are comparable, for each
n
=
0
,
2
,...
. Then,
n
n
+
1
d
(
g
(
x
),
g
(
x
))
0
0
n
−
1
n
n
−
1
n
n
n
+
1
≤
k
max{
d
(
g
(
x
),
g
(
x
)),
d
(
g
(
x
),
g
(
x
)),
d
(
g
(
x
),
g
(
x
)),
0
0
0
0
0
0
n
−
1
n
+
1
n
n
d
(
g
(
x
),
g
(
x
)),
d
(
g
(
x
),
g
(
x
))}
0
0
0
0
n
−
1
n
n
n
+
1
≤
k
max{
d
(
g
(
x
),
g
(
x
)),
d
(
g
(
x
),
g
(
x
)),
0
0
0
0
d
(
g
n
−
1
(
x
),
g
n
(
x
))
+
d
(
g
n
(
x
),
g
n
+
1
(
x
))}
0
0
0
0
n
−
1
n
n
n
+
1
≤
k
d
(
g
(
x
),
g
(
x
))
+
d
(
g
(
x
),
g
(
x
))),
0
0
0
0
1
k
∈
(
0
)
where
. It follows that
2
k
n
n
+
1
n
−
1
n
d
(
g
(
x
),
g
(
x
))
≤
d
(
g
(
x
),
g
(
x
)).
(1)
0
0
0
0
1
−
k
k
r
=
1
r
∈
(
0
. By induction from (1) one has, for
mn
>
Put
. Then
,
−
k
m
n
d
(
g
(
x
),
g
(
x
))
0
0
(
)
m
−
1
m
−
2
n
≤
r
+
r
+
+
r
d
(
g
(
x
),
x
)
L
0
0
n
m
r
−
r
(2)
=
d
(
g
(
x
),
x
)
0
0
1
−
r
r
n
≤
d
(
g
(
x
),
x
).
0
0
1
−
r
g
n
{
(
x
)}
is a Cauchy sequence in
X
. Since
X
is complete, there is
Thus
0
*
x
∈
X
such that
g
n
*
lim
(
x
0
)
=
x
.
n
→
∞
*
x
∈
X
is a fixed point of
g
. Note that
d
is continuous.
Next, we prove that
If
g
is continuous, then letting
m
=
n
+
1
n
→
∞
in (2) allows to, as
,
n
+
1
n
*
*
lim
d
(
g
(
x
),
g
(
x
))
=
d
(
g
(
x
),
x
),
(3)
0
0
n
→
∞
and so
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