E facile vedere che la (6.10) implica `

(1)

E facile vedere che la (6.10) implica `

Z β α

X n (t) ² dt

¹₂

_P

n→∞ →

Z β α

X(t) ² dt

¹₂

Infatti dalla disuguaglianza |kxk − kyk| ≤ kx − yk si ha

P

Z β α

X(t) ² dt

¹₂

− Z β α

X n (t) ² dt

¹₂

> ε

≤ P Z β α

(X(t) − X n (t)) ² dt

¹₂

> ε

=P

Z β α

(X(t) − X n (t)) ² dt > ε ²

→ 0 Dunque se ρ ⁰ > ρ, ε ⁰ < ε, abbiamo

P

Z β α

X(t) dB _t ≥ ε

≤ P

Z β α

X _n (t) dB _t ≥ ε ⁰

+P

Z β α

X(t) dB _t − Z β

α

X _n (t)) dB _t

≥ ε−ε ⁰ P

Z β α

X _n (t) ² dt ≥ ρ ⁰

= P Z β α

X _n (t) ² dt

¹₂

≥ p ρ ⁰

≤ P Z β α

X(t) ² dt

¹₂

≥ √ ρ

+P

Z β α

X _n (t) dt

¹₂

− Z β α

X(t) ² dt

¹₂

≥ p

ρ ⁰ − √ ρ e quindi

P

Z β α

X _n (t) ² dt ≥ ρ ⁰

≤

P

Z β α

X(t) ² dt ≥ ρ + P

Z β α

X _n (t) dt

¹₂

− Z β α

X(t) ² dt

¹₂

≥ p

ρ ⁰ − √ ρ da cui per il Lemma 6.6

P

Z β α

X(t) dB _t ≥ ε

≤

P

Z β α

X n (t) dB t

≥ ε ⁰

+ P

Z β α

X(t) dB t − Z β

α

X n (t)) dB t

≥ ε − ε ⁰

≤ P Z β α

X _n (t) ² dt ≥ ρ ⁰ + ρ ⁰

ε ⁰² + P

Z β α

X(t) dB _t − Z β

α

X _n (t)) dB _t

≥ ε − ε ⁰

≤ P Z β α

X(t) ² dt ≥ ρ + ρ ⁰

ε ⁰² + P

Z β α

X(t) dB _t − Z β

α

X _n (t)) dB _t

≥ ε − ε ⁰ + P

Z β α

X _n (t) dt

¹₂

− Z β α

X(t) ² dt

¹₂

≥ p

ρ ⁰ − √ ρ passando al limite per n → ∞ e per l’arbitrariet` a di ρ ⁰ , ε ⁰ si ha la tesi.

1

E facile vedere che la (6.10) implica `