Nonlinear Equations - UFRJ

[SEC. 7.2: THE γ-THEOREMS 95
Then the sequences u i and x i are well-defined for all i, x i ∈ D,
and
‖x i − ζ‖
‖x 0 − ζ‖ ≤ u i
≤ max
(2 −2i +1 , 2 δ )
.
u 0 u 0
Proof. By hypothesis,
u 0
ψ(u 0 ) + δ u 0
< 1
so the sequence u i is decreasing and positive. For short, let q =
≤ 1/4. By induction,
u 0
ψ(u 0)
u i+1
≤ u ( ) 2
0 ui
+ δ ≤ 1 ( ) 2 ui
+ δ .
u 0 ψ(u i ) u 0 u 0 4 u 0 u 0
Assume that u i /u 0 ≤ 2 −2i +1 . In that case,
u i+1
u 0
≤ 2 −2i+1 + δ ≤ max
(2 −2i+1 +1 , 2 δ )
.
u 0 u 0
Assume now that 2 −2i +1 , u i /u 0 ≤ 2δ/u 0 . In that case,
u i+1
u 0
≤ δ ( ) δ
+ 1 ≤ 2δ = max
(2 −2i+1 +1 , 2 δ )
.
u 0 4u 0 u 0 u 0
From now on we use the assumptions, notations and estimates of
the proof of Theorem 7.5. Combining (7.5) and (7.8) in (7.6), we
obtain again that
This time, this means that
‖N(f, x)‖ ≤
γ‖x‖2
ψ(γ‖x‖) .
‖x i+1 ‖γ ≤ δ + ‖N(f, x)‖γ ≤ δ + γ2 ‖x‖ 2
ψ(γ‖x‖) .
By induction that ‖x i − ζ‖γ(f, ζ) < u i and we are done.