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It should also be noted that, in the context of T ODE’s, the averaging of (15) only requires the long time averages 1/T 0 ψ(s, z)ds converge, while in the present PDE context, we do need the reinforced assumption (iii). The remainder of this paper is devoted to the proof of Theorem 1. 2. 1. Two uniform bounds. Our analysis starts with the Proposition 1. Suppose the initial data f 0ε is bounded in L 2 (R2d ). Then, (i) The family ( f ε )ε>0 is bounded in L ∞ (R+ ; L 2 (R2d )). fε − Pfε (ii) The family g ε )ε>0 := is bounded in L 2 (R+ × R2d ).
8. If D0 = D0∗ ηN has τ -compact resolvent, then for any function f ∈ Bc (R) the function V ∈ B R → f (D0 + V ) 1,∞ is bounded. 9. Let D0 = D0∗ ηN have τ -compact resolvent, r = (r1 , . . , rm ) ∈ [0, 1]m , V1 , . . , Vm ∈ Nsa and set Dr = D0 + r1 V1 + · · · + rm Vm . Then (i) for any compact subset ⊆ R the function r ∈ [0, 1]m → E Dr is bounded; 1 (ii) for any function f ∈ Bc (R) the function r ∈ [0, 1] → f (Dr ) 1 is bounded. An elementary proof of the following lemma can also be found in .
We used here that the convergence of P f ε is weak in space but pointwise in time). Besides, we have the uniform bound pδ,ε (s, Y ) ≤ C δ (X − Y) L 2 (R2d Y ) ≤ C δ −d , where C is independent of ε and δ, but it does depend on ϕ. As a consequence, the dominated convergence theorem gives ∀δ > 0, pδ,ε (s, Y ) −→ ε→0 R2d F(s, H0 (X )) δ (X − Y) ×ϕ(H0 (X )) d X strongly in L 2loc (R+ × R2d ). On the other hand, the function aδ,ε (s, Y ) satisfies the uniform bound aδ,ε (s, Y ) ≤ C(R) γ − p−1 χδ (s − t) L 1 (R2d t ) ≤ C, where C is independent of ε and δ (we used Proposition 2).