13.6 Theory Of Convergence
Definition 13.6.1 A normed vector space, E with norm
is called a Banach space if it is also
complete. This means that every Cauchy sequence converges. Recall that a sequence
n=1∞ is a Cauchy sequence if for every ε >
0 there exists N such that whenever
m,n > N,
is a Cauchy sequence, there exists x such that
Example 13.6.2 Let E be a Banach space and let Ω be a nonempty subset of a normed linear
space F. Let B
denote those functions f for which
Denote by BC
the set of functions in B
which are also continuous.
Lemma 13.6.3 The above
is a norm on B
. The subspace BC
given norm is a Banach space.
Proof: It is obvious
is a norm. It only remains to verify
be a Cauchy sequence. Since
0 as m,n →∞,
it follows that
is a Cauchy sequence in
for each x
. Let f
. Then for any
whenever m,n are large enough, say as large as N. For n ≥ N, let m →∞. Then passing to the
limit, it follows that for all x,
and so for all x,
It follows that
It remains to verify that f is continuous.
for all n
large enough. Now pick such an n
. By continuity, the last term is less than
small enough. Hence
is continuous as well. ■
The most familiar example of a Banach space is Fn. The following lemma is of great
importance so it is stated in general.
Lemma 13.6.4 Suppose T : E → E where E is a Banach space with norm
for some r ∈
. Then there exists a unique fixed point, x ∈ E such that
Letting x1 ∈ E, this fixed point x, is the limit of the sequence of iterates,
In addition to this, there is a nice estimate which tells how close x1 is to x in terms of things
which can be computed.
Proof: This follows easily when it is shown that the above sequence,
Cauchy sequence. Note that
By induction, this shows that for all k ≥
is valid. Now let k > l ≥ N.
which converges to 0 as N →∞.
Therefore, this is a Cauchy sequence so it must converge to
x ∈ E.
This shows the existence of the fixed point. To show it is unique, suppose there were another
one, y. Then
and so x = y.
It remains to verify the estimate.
and solving the inequality for
gives the estimate desired.
The following corollary is what will be used to prove the convergence condition for the various
Corollary 13.6.5 Suppose T : E → E, for some constant C
for all x,y ∈ E, and for some N ∈ ℕ,
for all x,y ∈ E where r ∈
. Then there exists a unique fixed point for T and it is still the
limit of the sequence,
for any choice of x1.
Proof: From Lemma 13.6.4 there exists a unique fixed point for TN denoted here as x.
Therefore, TNx = x. Now doing T to both sides,
By uniqueness, Tx = x because the above equation shows Tx is a fixed point of TN and there is
only one fixed point of TN. In fact, there is only one fixed point of T because a fixed point of T is
automatically a fixed point of TN.
It remains to show Tkx1 → x, the unique fixed point of TN. If this does not happen, there
exists ε > 0 and a subsequence, still denoted by Tk such that
Now k = jkN + rk where rk ∈
is a positive integer such that
Then there exists a single r ∈
such that for infinitely many
Taking a further subsequence, still denoted by Tk
and this contradicts 13.25. ■
Theorem 13.6.6 Suppose ρ
1. Then the iterates in 13.19 converge to the unique
solution of 13.18.
Proof: Consider the iterates in 13.19. Let Tx = B−1Cx + B−1b. Then
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
. Since ρ
it follows from Gelfand’s
theorem, Theorem 13.3.3
on Page 998
, there exists N
such that if k ≥ N,
then for some
and so Corollary
applies and gives the conclusion of
this theorem. ■