By Vincent Rivasseau (Chief Editor)

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**Example text**

4) 2 . 5) (n) (n) Note that Γ3 (t) is obtained from Γ3+8 (t) replacing pt (x, y) by (2πt)−1/2 (recall that dx m ¯ (x) = 2). In conclusion we have (n) Γ8 (t) := − 3λ 4 n−1 m ¯ xk , v (k) (Tk+1 ) (n) ds gt−s ps−Tk+1 m ¯ xk − 10 v (n) (t) = i=1 Let us deﬁne ψn := − (n) Γi (t) . 1), n−1 ξn = ψk . 9) k=0 where 10 ηn := ηn (i), i=1 i=3 and ηn (i) = − 3 (n) m ¯ , Γ (Tn+1 ) , i = 1, . . , 10 4 xn i √ 2 T 3 √ 1k

7). 9), which is a sort of linear integral equation in the ψn with kernel An,k , k < n, and known data ηn : “sort of” because the ηn still depend on the unknowns v (n) (·). The elements An,k decay as (n − k)−1/2 . 17), where the kernel is now A2n,k and the “known terms” are ηn and (Aη)n . In Section 6 we study these “known terms” which are splitted into four groups. The ﬁrst one consists of truly known terms which survive in the limit (n) (they come from Γ2 (t)). The terms in the second group, which instead may de(n) pend on v (t), are all directly proved to be negligible using the a priori bounds of Section 4.

12) Tk ds Tk−1 ∂ps−s (k−1) v (s ) ∂s (n) ¯ xn v (n) (s)2 ds gt−s m (n) ds gt−s v (n) (s)3 −1 (n) . 7). 8) t t Tn (n) ds gt−s h(n) (s) . 42 L. Bertini, S. Brassesco, P. Butt` a and E. Presutti Ann. 6), and get v (n) (t) = (n) (n) (n) (n) (n) (n) Γ1 (t) + Γ2 (t) + Γ4 (t) + Γ5 (t) + Γ9 (t) + Γ10 (t) t +λ Tn (n) ds gt−s ps−Tn h(n−1) (Tn ) + v (n) (Tn ) . 7) we thus need to show t λ Tn (n) (n) (n) ds gt−s ps−Tn h(n−1) (Tn ) + v (n) (Tn ) = Λ3 (t) + Γ6 (t) . s. 16). e. in the deﬁnition of Λ3 (t)). 6) to write h(n−1) (Tn ) + v (n−1) (Tn ) = pTn −Tn−1 h(n−1) (Tn−1 ) + v (n−1) (Tn−1 ) + Tn ds Tn−1 ∂pTn −s (n−1) v (s ).

### Annales Henri Poincaré - Volume 3 by Vincent Rivasseau (Chief Editor)

by James

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