THÈSE DE DOCTORAT DE L'UNIVERSITÉ PARIS 6 Spécialité ...

6.5. Proof of Theorem 6.2 105The quantity Q 2,n (β, γ, f) is upper bounded by{ (E f ‖ ˆf}) 1/2 (γ,1 − f‖ 2p ∞ψn−2p (β)I { ˆβ(p) =γ}(P f { ˆβ (p) = γ} ∩ {‖ ˆf1/2γ,1 − f‖ ∞ ψn −1 (β) > 1 + δ n })).Then, (6.29) and Proposition 6.3 imply that{sup(E f ‖ ˆf γ,1 − f‖ 2p ∞ψn−2p (β)I { ˆβ(p) =γ}f∈Σ(β,L)}) 1/2is bounded above by a positive constant for n large enough and we deduce that[ (sup Q 2,n (β, γ, f) ≤ D 9 P f { ˆβ (p) = γ} ∩ {‖ ˆf1/2γ,1 − f‖ ∞ ψn −1 (β) > 1 + δ n })],f∈Σ(β,L)(6.30)for n large enough. Now we are going to prove that the right hand side of inequality(6.30) tends to 0 as n → ∞. We have‖ ˆf γ,1 − f‖ ∞ I { ˆβ(p) =γ} ≤ (‖b γ,1(·, f) − b β,1 (·, f)‖ ∞ + ‖b β,1 (·, f)‖ ∞ + ‖Z γ,1 ‖ ∞ ) I { ˆβ(p) =γ}(≤ ‖b γ,1 (·, f) − b β,1 (·, f)‖ ∞ + ψ )n(β)2β + 1 + ‖Z γ,1‖ ∞ I { ˆβ(p) =γ} , (6.31)where the last line is a consequence of Proposition ( ) 6.1. ( )∥∥∥∥∞We have ‖b γ,1 (·, f) − b β,1 (·, f)‖ ∞ = ∥E f ˆfγ,1 − E f ˆfβ,1 . The function φ : t ↦−→( ) ( )E f ˆfγ,1 (t) − E f ˆfβ,1 (t) is a continuous function on [0, 1] d which admits a non-randommaximum x 0 satisfying‖b γ,1 (·, f) − b β,1 (·, f)‖ ∞ = φ(x 0 )≤ ∣ ˆf γ,1 (x 0 ) − ˆf β,1 (x 0 ) ∣ + |ϕ 1 |≤ ‖ ˆf γ,1 − ˆf β,1 ‖ ∞ + |ϕ 1 | ≤ η 1 (β) + |ϕ 1 |, (6.32)where ϕ 1 = ˆf γ,1 (x 0 ) − ˆf[β,1 (x 0 ) − E f ˆfγ,1 (x 0 ) − ˆf]β,1 (x 0 ) . Since ϕ 1 is a N (0, πn) 2 variable,by Proposition 6.2, we deduce that its variance π 2 n satisfies(πn 2 ≤ 2E f (Zβ,1 ( x 0 )) 2) (+ 2E f (Zγ,1 (x 0 )) 2) ≤ 2‖K β‖ 2 2σ 2 (1 + o(1)). (6.33)n˜h 1 (β)Then using (6.31) and (6.32) we deduce that[‖ ˆf γ,1 − f‖ ∞ I { ˆβ(p) =γ} ≤ η 1 (β) + ψ ]n(β)2β + 1 + ‖Z γ,1‖ ∞ + |ϕ 1 | I { ˆβ(p) =γ}≤ (ψ n (β) + ‖Z γ,1 ‖ ∞ + |ϕ 1 |) I { ˆβ(p) =γ} ,