st.statistics - Small crown probabilities (and infinite dimensional margin assumption)

Here is a solution for problem 2, with power $1/2$, using your idea 1. First some computations. Let A, B real numbers, let z ~ N(0,1), and X=$B(z^2-1) + Az$. The Fourier transform of the distribution of X (i.e.: the characteristic function of X with some $pi$) is

$; ; ; ;; ; ; ;E(exp(-2pi ixi X) = frac{e^{2pi i xi B - frac{2pi^2 A^2 xi^2}{1+4pi i B xi}}}{sqrt{1+4pi ixi B}}$

where the square root is the one with positive real part. Then

$; ; ; ;; ; ; |E(exp(-2pi ixi X)|^2 = frac{e^{- frac{4pi^2 A^2 xi^2}{1+16pi^2 B^2 xi^2} }} {1+16pi^2 xi^2 B^2}$

We also have for any pair of real numbers $a, bge 0$

$; ; ; ;; ; ; ln(frac{1+8b + 4a}{1+16b}) < ln(1+frac{4a}{1+16b})le frac{4a}{1+16b}$

In particular,

$; ; ; ;; ; ; -frac{4a}{1+16b} - ln(1+16b)<-ln(1+4(2b + a))$

thus, with $a=pi^2xi^2 A^2$ and $b=pi^2xi^2 B^2$

$; ; ; ;; ; ; |E(exp(-2pi ixi X)|^2 le frac{1} {1+4pi^2 xi^2 (2B^2+A^2)}.$

It follows that the Fourier transform of a Gaussian polynomial of degree two $q$ satisfies, with notation as in your definition:

$; ; ; ;; ; ; |E(exp(-2pi ixi q)|^2 le prod_{j}{frac{1} {1+4pi^2 xi^2 (2beta_j^2+alpha_j^2)}}lefrac{1}{1+4pi^2xi^2sigma(q)^2}.$

Then, the density of $q$ is bounded in $L^2$ , uniformly in $Gamma_2(c_0)$, and so
there exists a constant $C'$ such that for all $epsilon>0$

$; ; ; ;; ; ;sup_{qin Gamma_2(c_0)} P(|q|leq epsilon)leq C'epsilon^{1/2}.$