Nonlinear Equations - UFRJ

[SEC. 9.2: ALPHA THEORY 125
Proof. First of all, pick b with norm one in ker B. If b ∈ ker A then
Π is the identity and we are done. Therefore, assume that b ∉ ker A.
The kernel of A is then spanned by b + c, where
c = A † (B − A)b.
From this expression, ‖c‖ ≤ w.
Now, assume without loss of generality that x ∈ ker A ⊥ has norm
one. Since
Πx = x − b〈x, b〉,
we bound
‖Πx‖ 2 = ‖x 2 ‖ − 2|〈x, b〉| 2 + ‖b‖ 2 |〈x, b〉| 2 = 1 − |〈x, b〉| 2 .
Note that x ⊥ b + c so the latest bound is 1 − |〈x, c〉| 2 ≥ 1 − w 2 .
In order to prove the lower bound on ‖B † Ay‖, we write
B † A = ΠB −1
| ker A ⊥ A.
Since ‖A † B | ker A ⊥ − I ker A ⊥‖ ≤ ‖A † ‖‖B − A‖ ≤ w, Lemma 7.8
implies that
‖B −1
1
Ay‖ ≥ ‖y‖
| ker A ⊥ 1 + w .
9.2 Alpha theory
We define Smale’s invariants in M = C n+s \ Ω in the obvious way:
and
and of course
β(f, X) = ‖Df(X) † f(X)‖ 2
( ‖Df(X) † D k ) 1/(k−1)
f(X)‖ 2
γ(f, X) = sup
.
k≥2 k!
α(f, X) = β(f, X)γ(f, X)