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It also appears one needs γ > 23 to be sure of the existence of decaying solutions. But following [21], the probabilistic argument here can replace Ruelle’s argument. 2 Proof. Let β = 8 sinλ2 (k) . Let R1 (n) and R2 (n) be the R’s associated to θ = 0 and θ = 21 . e. ω, lim n→∞ log Ri (n) = β. 19) Let θi (n) be the corresponding EFGP angles. 7), R1 (n)R2 (n) sin(θ1 (n) − θ2 (n)) = sin(k)[u1 (n)u2 (n − 1) − u1 (n − 1)u2 (n)] = −1 (by the initial conditions R1 (1) = R2 (1) = 0, θ1 (1) = 0, θ2 (1) = Wronskian.
C. e. 13) holds as |n| → ∞. Remark. This strengthens the result originally proven in [31] and improved in [7] in two ways. 13). e. distribution. Proof. e. e. k ∈ (0, π), (1 − 2α)λ2 log T (n) = . 13). The theorem follows from general principles on rank one perturbations [12, 4, 5, 28]. The case α = 21 has an extra subtlety we will need to deal with, using an argument modeled on Kotani-Ushiroya [21]. 7. Let uθ = (cos θ, sin θ) in R2 . For any unimodular matrix A with A > 1, let θ(A) be the unique θ ∈ (− π2 , π2 ] with Auθ = A −1 .
N j −2α . 4) since E(Vω2 − [Vω (n)2 − E(Vω (n)2 )][sin (θ) 2 2 E(Vω )) = 0. Since V is bounded, 2 2 E((V 2 − E(V 2 ))2 ) ≤ CE(V 2 ). e. ω, n j −2α |C2 (ω)| = o j=1 also. 4) for q = 3. 5 applies. Let q be as in that lemma. 4 for j large, |θω (j + 1) − θω (j) − k| ≤ C0 j −2α/3 . 11) Pick n0 so and −2α/3 4C0 n0 ≤ q −2 . 12) Suppose N = n0 + Kq . Then N K q j −2α cos(4θω (j)) = j=n0 +1 (n0 + mq + j)−2α cos(4θω (mq + j)) m=0 j=1 = A 1 + A2 , where A1 is what we get by replacing (n0 + kq + j)−2α by (n0 + kq )−2α and A2 is the difference.