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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 :
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