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Let η have minimal period 2π. 3) T then h = 0. Proof. Step 1: For any y ∈ T and α < β h(s) ds = 0 . 3), for any real polynomial P (X) := k≥q ak X k divisible by X q , P η(y + s) − η(s) h(s) ds = 0 . 5) holds for any real polynomial ak X k . 6). By the Stone-Weierstrass theorem, the set A is dense in C([−M, M ], R) because it is a subalgebra with unity and A separates the points of [−M, M ] (take any X q with q odd). As a consequence for any continuous function g ∈ C(R) g η(y + s) − η(s) h(s) ds = 0 .

38 σ,s . • (P3) (Invertibility of Ln ) Fix γ ∈ (0, 1), τ ∈ (1, 2). There exist µ > 0, δ0 > 0 such that, if q hi [w]σ,s := inf i=0 v1 0,0 q σi ,s (σi − σ) 2τ (τ −1) τ −2 ; q ≥ 1, σ ≥ σi > σ, hi ∈ W (i) , w = hi ≤ µ , i=0 ≤ 2R∞ and δ belongs to ∆γ,τ n (λ, v1 , w) γ M (δ, λ, v1 , w) γ ≥ , ωl − j − ε , (l + j)τ 2j (l + j)τ 1 < l ≤ Ln , j ≤ 2Ln , l ∈ Z , j ∈ N+ , l = j , 3ε δ ∈ [0, δ0 ] |ωl − j| ≥ := then Ln (δ, λ, v1 , w) is invertible and L−1 n (δ, λ, v1 , w)[h] C (Ln )τ −1 h γ ≤ σ,s σ,s ∀h ∈ W (n) , for some C > 0.

Now, the map F is in C ∞ (Yσ × B2,σ , V2 ∩ Xσ,s ). 4). Hence, by the implicit function theorem, the map v2 is in C ∞ (Yσ , V2 ∩ Xσ,s ). Moreover, all the partial derivatives of F are bounded in norm σ,s in the set Yσ × B2,σ . Hence all the partial derivatives of the map v2 are bounded (in norm σ,s ) on the set Yσ . 2). Hence, if δ0 has been chosen small enough, g(y, v2 ), and, for any ki ∈ N, ∂uk1 ∂λk2 ∂δk3 g(y, v2 ) is σ,s -bounded on Yσ × B(R∞ ; V2 ∩ Xσ,s ). 3 σ,s+2 ≤ u σ,s , v2 (y) ∈ Xσ,s+2 and the derivatives Dk v2 are σ,s+2 - bounded.