By Browder F. (ed.)
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Extra resources for Nonlinear Functional Analysis and Its Applications, Part 2
We show the details in the planar case (Λ = C) with L = 1. The proof is 1 2 virtually identical in the hyperbolic case. For β ∈ C, let fβ (z) = f(z + β)e− zβ− 2 |β| . 4 that fβ = f. We compute 1 2 + zw+β w E fβ (z)f(w) = e− zβ− 2 |β| . As β → ∞ this goes to 0 uniformly for z, w in any compact set. By Cauchy’s formula, the coefficients of the power series expansion of fβ around 0 are given by 1 2π i fβ (ζ) C ζ n +1 d ζ, where C(t) = e it , 0 ≤ t ≤ 2π. Therefore, for any n, the first n coefficients in the power series of f and the first n coefficients in the power series of fβ become uncorrelated and hence (by joint Gaussianity) independent, as β → ∞.
Thus δ | Z + U |−η 1(| Z + U | > 3 ) ≤ | Z/2|−η , so we have | Z + U |−η ≤ | Z + U |−η 1(| Z + U | ≤ 3 ) + | Z/2|−η . After taking expectations we get E| Z + U |−η ≤ B3 | z|−η P(Z + U ∈ dz) + E| Z/2|−η . Given Z, the conditional probability that Z + U ∈ dz is 1(z + Z ∈ D)/(π 2 ). So the first term can be written as | z|−η dz | z|−η dz P(Z ∈ z + D) ≤ P(| Z | < 4 ) 2 π π 2 B3 B3 −η = c η P(| Z | < 4 )(4 ) ≤ c η E| Z |−η , the last inequality is Markov’s. We conclude E| Z + U |−η ≤ c η E| Z |−η , as required.
1 holds for random analytic functions satisfying the following condition. 4) k B n f+ z (K) dz are uniformly integrable as → 0. P ROOF. Let ϕ : Ck → R be a continuous test function with compact support. 2 with the notation F = (f, . . 5) 1 k Vol(B ) Ck ϕ(x)|F (x)|2 1B k (F(x)) dx = 1 Vol(B k ) Bk ϕ[F −1 (y)]d y. 6) 1 Vol(B ) k B n f+ z (supp ϕ) dz , which is uniformly integrable by assumption. 5), let → 0 and use uniform integrability. 7) ϕ(z1 , . . , z k ) = ϕ d µk . E z∈Z k The left hand side, by the Fubini argument, becomes ϕ(x) E |F (x)|2 1(F(x) ∈ B k ) Vol(B k ) = ϕ(x) dx E |f (x1 ) · · · f (xk )|2 1(f(x1 ), .