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E. (n) ¯ xn = m ¯ xn . 11) that there are constants α > 0 and C < ∞ gt m so that, for any ϕ, (n,⊥) ϕ ≤ Ce−αt ϕ ∞ . 33) 48 L. Bertini, S. Brassesco, P. Butt` a and E. Presutti Ann. Henri Poincar´e be the component of z (n) (t) orthogonal to m ¯ xn and introduce the event (2) ω∈Ω : Bλ,τ := sup z (n,⊥)(t) sup n≤nλ (τ ) t∈[Tn ,Tn+1 ] ∞ ≤ λ−ζ . 34) In Appendix B we will prove that for each τ, ζ, q > 0 there exists a constant C = C(τ, ζ, q) such that for any λ > 0 (2) P Bλ,τ ≥ 1 − Cλq . 35) v (n,⊥) (Tn+1 ) ∞ , 3 m ¯ , v (n) (Tn+1 ) m ¯ xn .

2), for t ∈ [Tn , Tn+1 ] (n) Γ1 (t) ∞ ≤ Ce−α(t−Tn ) v (n) (Tn ) ≤ Ce−α(t−Tn ) ≤ −α(t−Tn ) Ce ∞ v (n−1) (Tn ) v (n−1) (Tn ) ∞ ∞ + m ¯ xn − m ¯ xn−1 . 9). 4) we get (n) Γ2 (t) sup t∈[Tn ,Tn+1 ] ∞ ≤ CλZn . 13) (n) We next bound Λ3 . 14) = 2 for any z ∈ R. 10), we get (n) Λ3 (t) n ∞ ≤ Cλ k=1 t |xk − xk−1 | ds √ . 15) Vol. 4), we then get (n) Λ3 (t) ∞ n √ ≤ Cλ T √ k=1 Vk−1 . 16) (n) We will next bound Γ4 (t). 3) in the second equality. Then t (n) Γ4 (t) = λ s ds Tn Tn t +λ ∂pt−s (n) v (s ) ∂s ds t ds Tn s s (n) ds gt−s [1 − 3m ¯ 2xn ] ds Tn ∂ps −s (n) v (s ) .

The term ηn (6). To study this term we are going to use the same strategy as for ηn (4 + 5). e. 28) with 4 + 5 replaced by 6. The terms ηn (6, i), i = 2, 3. We have ηn (6, 1) = 3λ 4 n−1 Tn+1 Tk+1 ds k=0 Tn Tk ds m ¯ xn , ∂ps−s (k) g v (k) (Tk ) . 17) the double integral above for k = n − 1 is less or equal than √ Tn+1 Tn−1 +T /2 Ce−α(s −Tn−1 ) λ1−ζ T ds ds s − [Tn−1 + T /2] Tn−1 Tn √ Tn+1 Tn √ Ce−αT /2 λ1−ζ T + ds ds ≤ Cλ1−ζ T . s−s Tn Tn−1 +T /2 Vol. 44) which is therefore negligible. We next consider ηn (6, 8); its explicit expression is ηn (6, 8) = × 3λ2 4 m ¯ xn , n−1 k−1 Tn+1 ψh k=0 h=0 ∂pt−s (k) g ∂t s−s Tk+1 dt Tn s ds ds Tk Tk 2 ¯ xh − ps −Th+1 m 2π(s − Th+1 ) .