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is in fact a Cauchy sequence. When we know this, we know that we have conver-
gence and thus existence of a solution
v
. And since
v
kC1
.4:121/
v
D
lim
k!1
h.
v
k
/
D
h.
v
/;
D
lim
k!1
it follows by the continuity of h that the limit satisfies the proper equation. And by
the arguments given above, we know that it is in fact the only solution.
With respect to the existence of a solution, it remains to demonstrate that the
sequence
f
v
k
g
generated by (
4.121
) is a Cauchy sequence. Since
v
nC1
D
h.
v
n
/;
(4.122)
we have
j
v
nC1
v
n
j D j
h.
v
n
/
h.
v
n1
/
j
ı
j
v
n
v
n1
j
:
(4.123)
By induction, we have
j
v
nC1
v
n
j
ı
n
j
v
1
v
0
j
:
(4.124)
In order to see whether
f
v
n
g
is a Cauchy sequence, we need to bound
j
v
m
v
n
j
,
where we may assume that m>n. Note that
v
m
v
n
D
.
v
m
v
m1
/
C
.
v
m1
v
m2
/
C
:::
C
.
v
nC1
v
n
/;
(4.125)
and thus, by the triangle inequality, we have
j
v
m
v
n
j
j
v
m
v
m1
j
C
j
v
m1
v
m2
j
C
:::
C
j
v
nC1
v
n
j
:
(4.126)
By (
4.124
), we have
j
v
m
v
m1
j
ı
m1
j
v
1
v
0
j
;
j
v
m1
v
m2
j
ı
m2
j
v
1
v
0
j
;
:
j
v
nC1
v
n
j
ı
n
j
v
1
v
0
j
:
Consequently,
j
v
m
v
n
j j
v
m
v
m1
j C j
v
m1
v
m2
j C
:::
C j
v
nC1
v
n
j
ı
m1
C
ı
m2
C
:::
C
ı
n
j
v
1
v
0
j
: