Nonlinear Equations - UFRJ

[SEC. 10.3: AVERAGE COMPLEXITY OF RANDOMIZED ALGORITHMS 151
B
U
1
A
a 1 O
b 1
Figure 10.2: Geometric Lemma.
Lemma 10.19. Let A = (a 1 , a 2 ), B = (b 1 , b 2 ) ∈ R 2 be two points in
the plane, such that U = (0, 1) ∈ [A, B]. Then,
|b 1 − a 1 | ≤ ‖A‖‖B‖.
Proof. (See figure 10.2) We interpret |b 1 − a 1 | as the area of the
rectangle of corners (a 1 , 0), (b 1 , 0), (b 1 , 1), (a 1 , 1).
We claim that this is twice the area of the triangle (O, A, B).
Indeed,
Area(O, A, B) = Area(O, U, A) + Area(O, U, B)
Therefore,
= Area(O, U, (a 1 , 0)) + Area(O, U, (b 1 , 0))
= 1 2 |b 1 − a 1 |
|b 1 − a 1 | = 2Area(O, A, B) = ‖A‖‖B‖ sin(ÂOB) ≤ ‖A‖‖B‖
M(f t ; 0, 1)
≤
≤
∫ 1
0
∫ 1
0
(
‖ I − 1 )
‖F t ‖ 2 F tF ∗ t Ḟ t ‖‖F t ‖ µ2 2(F t )
‖F t ‖ 2 dt
‖F 0 ‖‖F 1 ‖ µ2 2(F t )
‖F t ‖ 2 dt