Lecture Notes in Differential Equations - Bruce E. Shapiro

223
Since K n /(1 − K) → 0 as n → ∞ (because T is a contraction, K < 1),
given any v ∈ V, it is possible to choose an integer N such that
K n ‖T v − v‖
1 − K
for all n > N. Pick any such integer N.
< ɛ (26.32)
Choose any two integers m ≥ n ≥ N, and define the sequence
⎫
v 0 = v
v 1 = T v
v 2 = T v 1
⎪⎬
Then since T is a contraction,
From Lemma 26.8 we have
.
v n = T v n−1
.
⎪⎭
(26.33)
‖v m − v n ‖ = ‖T m v − T n v‖ (26.34)
= ‖T n T m−n v − T n v‖ (26.35)
≤ K n ‖T m−n v − v‖ (26.36)
‖v m − v n ‖ ≤ K n 1 − Km−n
‖T v − v‖
1 − K
(26.37)
= Kn − K m
‖T v − v‖
1 − K
(26.38)
≤
Kn
‖T v − v‖ < ɛ (26.39)
1 − K
Therefore v n is a Cauchy sequence, and every Cauchy sequence on a complete
normed vector space converges. Hence v n → u for some u ∈ V.
Either u is a fixed point of T or it is not a fixed point of T .
Suppose that u is not a fixed point of T . Then T u ≠ u and hence there
exists some δ > 0 such that
‖T u − u‖ > δ (26.40)
On the other hand, because v n → u, there exists an integer N such that
for all n > N,
‖v n − u‖ < δ/2 (26.41)