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Triangular Gaussian coordinates: metric and moment control

Statement

Assume the Axiom of Choice. Fix an integer r≥1, put N=r(r+1)/2, and index z∈RN by 1≤i≤j≤r. Let T(z)ij=T(z)ji=zij and let ∥A∥ be the operator norm on the vertex Hilbert space C{1,…,r}. Write ∥A∥F=(∑i,j∣Aij∣2)1/2. Then ∥T(z)−T(w)∥≤∥T(z)−T(w)∥F≤2∥z−w∥2. If Z has independent standard normal coordinates and Y=T(Z), then ∥Y∥ is measurable and E∥Y∥2k<∞ for every integer k≥1. In particular Y has variance-one diagonal entries and identical mirrored off-diagonal entries.

Facts & Assumptions

[F1]

The triangular construction and every absolute Gaussian moment are available under Choice. Gaussian entry moments and triangular independence, The Axiom of Choice.

[F2]

The vertex inner product is linear in its second argument. The real and complex Cauchy–Schwarz inequality applies in the library's linear-first convention and transfers to this convention by conjugation, with the same absolute values and norms. Finite vertex Hilbert space in the article convention, Cauchy–Schwarz: |⟨ u,v⟩|≤‖ u‖‖ v‖, with equality exactly for linearly dependent vectors.

[F3]

Finite sums and real scalar multiples of measurable real-valued functions are measurable. The nonnegative integral is monotone and homogeneous, and finite integrable sums have linear expectation. Closure properties of measurable functions used by the integral, Monotonicity and nonnegative homogeneity of the nonnegative integral, The Lebesgue integral is linear on L¹(μ).

[F4]

For a bounded linear map on a nonzero normed space, the operator norm is the supremum of its values on unit vectors. It is a norm on the space of bounded linear maps, in particular satisfying the triangle inequality. The operator norm as the least bound and as the unit-sphere or unit-ball supremum, The operator norm is a norm on the space of bounded linear operators.

[F5]

A map from a measurable space into RN, N≥1, equipped with its Borel sigma-algebra, is measurable exactly when its real coordinates are measurable. Under Countable Choice, a continuous map between positive-dimensional Euclidean spaces is Borel measurable. Composition of measurable maps, with matching sigma-algebras, is measurable. A map into ℝⁿ is measurable exactly when its coordinates are measurable, Continuous functions on Euclidean spaces are Borel measurable, Composition with a Borel measurable outer map preserves measurability.

Proof

Given: The objects and hypotheses in the Statement.

Goal: Establish all the stated estimates.

1.1F2F4algebra

For any complex matrix A and vector u, put ai=(Aij‾)j. In the vertex convention, ⟨ai,u⟩=∑jAijuj, so [F2] gives ∥Au∥22=∑i∣⟨ai,u⟩∣2≤∑i(∑j∣Aij∣2)∥u∥22=∥A∥F2∥u∥22. Thus the matrix action is bounded, and its operator norm in [F4] satisfies ∥A∥≤∥A∥F. Also ∥T(z)−T(w)∥F2=∑i(zii−wii)2+2∑i<j(zij−wij)2≤2∥z−w∥22. Taking nonnegative square roots proves the stated metric estimates.

2.1F4step 1.1algebra

The operator-norm triangle inequality in [F4], applied in both directions, gives ∣∥T(z)∥−∥T(w)∥∣≤∥T(z)−T(w)∥≤2∥z−w∥2. Hence h:RN→R, h(z)=∥T(z)∥, is continuous.

3.1F1F3F5step 2.1algebra

Choice supplies the countable-choice hypothesis in [F5] by restriction to countable families. The Gaussian coordinates are measurable real random variables on their common probability space by [F1], so [F5] makes Z measurable into (RN,B(RN)), where N≥1. Step 2.1 and [F5] make h Borel measurable, so h∘Z=∥Y∥ is measurable. For any fixed k≥1, the map pk(t)=t2k is continuous: factorization gives ∣s2k−t2k∣≤2kmax⁡(∣s∣,∣t∣)2k−1∣s−t∣, which tends to zero as s→t. Applying [F5] to pk and then composing with h∘Z or an individual Gaussian coordinate proves measurability of ∥Y∥2k and ∣Zij∣2k. Their finite sums and nonnegative scalar multiples are measurable by [F3].

4.1F1F3step 1.1step 3.1algebra∎

By step 1.1, ∥Y∥2k≤2k(∑i≤jZij2)k≤2kNk∑i≤j∣Zij∣2k: the sum of the N coordinate squares is at most N times their maximum, and the kth power of that maximum is at most the sum of their kth powers. Every summand has finite expectation by [F1]. Using the measurability from step 3.1, finite linearity and monotonicity in [F3] give E∥Y∥2k≤2kNk∑i≤jE∣Zij∣2k<∞. Finally Yii=Zii has mean zero and second moment one by [F1], hence variance one, while Yij=Yji=Zij for i<j by construction.

Depends on

Used by · 6

Dependency tree · 2 levels

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