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

106 Sharp adaptive estimation in sup-norm for d-dimensional Hölder classesthe last line being a consequence of Lemma 6.2. Thus,P f({ ˆβ (p) = γ} ∩ {‖ ˆf)γ,1 − f‖ ∞ > (1 + δ n )ψ n (β)}≤P f (‖Z γ,1 ‖ ∞ + |ϕ 1 | > δ n ψ n (β))(≤P f ‖Z γ,1 ‖ ∞ > δ ) (n2 ψ n(β) + P f |ϕ 1 | > δ )n2 ψ n(β) . (6.34)As ϕ 1 is a N (0, πn) 2 variable and using (6.33), we have(P f |ϕ 1 | > δ ) {}n2 ψ n(β) ≤ exp − ψ2 n(β)δnn˜h 2 1 (β).16‖K β ‖ 2 2σ 2The quantity ψn(β)δ 2 nn˜h 2 1 (β) is of order √ log n, and then(lim P f |ϕ 1 | > δ )nn→∞ 2 ψ n(β) = 0. (6.35)Using Proposition 6.2, we have(P f ‖Z γ,1 ‖ ∞ > δ ){}n2 ψ n(β) ≤ D 1˜h 1 (γ) exp − ψ2 n(β)δnn˜h 2 1 (γ)exp8‖K β ‖ 2 2σ 2⎧⎨⎩√ ⎫D 2 δ n ψ n (β) n˜h 1 (γ) ⎬2(log ˜h 1 (γ)) 1/2 ⎭ .The quantity ψn(β)δ 2 nn˜h 2 1 (γ) is of order ψn(β)ψ 2 n −2 (γ)(log n) D 10with some D 10 ∈ R and˜h 1 (γ) is of order ( )log n 1/(2γ+1).n Hence, since β < γ, this implies that(lim P f ‖Z γ,1 ‖ ∞ > δ )nn→∞ 2 ψ n(β) = 0. (6.36)Using (6.35), (6.36) and that the fact that the right hand side of inequality (6.34) doesnot depend on f, we deduce thatlim sup P f({ ˆβ (p) = γ} ∩ {‖ ˆf)γ,1 − f‖ ∞ > (1 + δ n )ψ n (β)} = 0.n→∞ f∈Σ(β,L)Then we obtain the result (6.28) which implies (6.22).6.6 Proof of Theorem 6.3The scheme of proof is similar to the proof of relation (6.20) in the proof of Theorem 6.2.To prove Theorem 6.3, we will prove that, for all β ∈ B,lim R 3,n(β) = 0 (6.37)n→∞