The following corollary is what will be used to prove the convergence condition for the various
iterative procedures.
Corollary 13.6.5Suppose T : E → E, forsome constant C
|T x− T y| ≤ C |x − y|,
for allx,y∈ E, and for some N ∈ ℕ,
| |
|T Nx − TNy | ≤ r|x − y|,
for allx,y∈ E where r ∈
(0,1)
. Then there exists a unique fixed point for T and it is still thelimit of the sequence,
{Tkx1}
for any choice of x^{1}.
Proof:From Lemma 13.6.4 there exists a unique fixed point for T^{N} denoted here as x.
Therefore, T^{N}x = x. Now doing T to both sides,
T NTx = Tx.
By uniqueness, Tx = x because the above equation shows Tx is a fixed point of T^{N} and there is
only one fixed point of T^{N}. In fact, there is only one fixed point of T because a fixed point of T is
automatically a fixed point of T^{N}.
It remains to show T^{k}x^{1}→x, the unique fixed point of T^{N}. If this does not happen, there
exists ε > 0 and a subsequence, still denoted by T^{k} such that
|| k 1 ||
T x − x ≥ ε
Now k = j_{k}N + r_{k} where r_{k}∈
{0,⋅⋅⋅,N − 1}
and j_{k} is a positive integer such that
lim_{k→∞}j_{k} = ∞. Then there exists a single r ∈
{0,⋅⋅⋅,N − 1}
such that for infinitely many
k,r_{k} = r. Taking a further subsequence, still denoted by T^{k} it follows
refers to any of the operator norms. It doesn’t matter which one you pick because
they are all equivalent. I am writing the proof to indicate the operator norm taken
with respect to the usual norm on E. Since ρ
( −1 )
B C
< 1, it follows from Gelfand’s
theorem, Theorem 13.3.3 on Page 998, there exists N such that if k ≥ N, then for some
r^{1∕k}< 1,
||||( −1 )k||||1∕k 1∕k
||B C || < r < 1.
Consequently,
| |
|T Nx − TNy | ≤ r|x− y|.
Also
|T x− T y|
≤
||||B −1C ||||
|x − y |
and so Corollary 13.6.5 applies and gives the conclusion of
this theorem. ■