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

2.3. Proofs 37Such choice of n is possible in view of Lemma 2.1, since E f [U n (x k , f)I Bn ] is bounded.Thus we have[{} ] []P f ψn−1 | ˜Z n (x k , f)| > 2βC′ 0z∩ B n ≤ P f ψn−1 |Ũn(x k , f)| > βC′ 0z.2β + 12β + 1We are going to apply Bernstein’s inequality to the variable Ũn(x k , f) which is a zero-meanvariable bounded by 2Q. Since µ(x k ) ≥ µ 0 , the variance of Ũn(x k , f) satisfies⎡⎛∑ ( )]1 nE f[Ũn (x k , f) 2 ⎢ nh j=1≤E f ⎣⎝f(X Xj⎞⎤2−x kj)Kh∑ ( ) ⎠ Inj=1 K Bn (P X (B n )) 2 ⎥Xj⎦ ,−x k1nh1≤(µ(x k ) − δ n ) 2 nh E 2 f≤ Q2 µ(x k )(1 + o(1))(µ(x k ) − δ n ) 2 nh≤ D 7(1 + o(1)),nhh[ (f 2 (X 1 )K 2 X1 − x kh∫ 1−1K 2 (y)dy,)](P X (B n )) 2 ,where o(1) is uniform in f ∈ Σ(β, L, Q) and D 7 is independent of f ∈ Σ(β, L, Q), n andk. By applying Bernstein’s inequality to the variable Ũn(x k , f) (note that here the familyof random variables contains only one summand), we obtain⎛⎞[]P f ψ −1 |Ũn(x k , f)| > βC′ 0zλ≤ 2 exp ⎝−2) ⎠ ,where λ = ψnβC′ 0 z2β+1n2β + 12(D7 (1+o(1))nh+ 2λQ3. Thus for n large enough, we have[]P f ψn−1 |Ũn(x k , f)| > βC′ 0z≤ 2 exp (−D 0 zψ n ) ,2β + 1with D 0 independent of f ∈ Σ(β, L, Q) and k. To finish the proof, it is enough to notethat card{k} = [ n ] ≤ m δ−1 n ψn −1/β .Proof of Lemma 2.2Like in Lemma 2.1, using Bernstein’s inequality we obtain that for n large enough{ }∣∣∣P X (β + 1)(2β + 1)n∑(1 − | X k − a j| β ) 2 4µ 0 β 2 + − 1∣ ≥ ε ≤ 2 exp(−nhD 8 )nhhj=0k=1where D 8 is a constant which depends on ε, but does not depend on n. NowP [ ] M{ }∑ ∣∣∣ X NnC ≤ P X (β + 1)(2β + 1)n∑(1 − | X k − a j| β ) 2 4µ 0 β 2 + − 1∣ ≥ ε .nhhk=1