ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Bounded mean-zero scores generate the unrestricted tangent space

Example

Assume the library's Axiom-of-Choice density convention in A dominated differentiable-in-quadratic-mean submodel and Countable Choice for The mean-zero L² space is a Hilbert space. Let $P$ be any probability law and take the unrestricted dominated model $M = {Q : Q ≪ P}$ . Its allowed paths are all one-sided DQM paths through $P$ with common dominator $P$ . Every bounded mean-zero $s \in L_{0}^{2} (P)$ is such a path score, and the tangent space is all of $L_{0}^{2} (P)$ . All densities constructed below are explicit, so the example's proof makes no density-selection choice.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice, a probability measure $P$ , and a bounded measurable representative $s$ with $P s = 0$ .

[A1]

The Axiom of Choice permits selecting from an arbitrary family of nonempty sets (The Axiom of Choice).

[A2]

Countable Choice permits selecting from a countable family of nonempty sets (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

A DQM path with common dominator $P$ has root-density expansion $\sqrt{p_{t}} = 1 + (t / 2) s + o_{L^{2} (P)} (t)$ (A dominated differentiable-in-quadratic-mean submodel).

[F2]

Every DQM score is in $L_{0}^{2} (P)$ (A DQM score has mean zero and is unique).

[F3]

$L_{0}^{2} (P)$ is the mean-zero subspace of the quotient $L^{2} (P)$ (Mean-zero square-integrable functions).

[F4]

The tangent space is the $L^{2} (P)$ -norm closure of the finite linear span of allowed scores (Closed tangent space).

[F5]

Dominated convergence applies to pointwise-convergent functions dominated by an integrable function (Dominated convergence).

[F6]

On a probability space, $∣ P g ∣ \leq ∥ g ∥_{2}$ for $g \in L^{2} (P)$ by Cauchy–Schwarz against $1$ (Cauchy-Schwarz inequality for $L^{2}$ ).

[F7]

Under Countable Choice, $L_{0}^{2} (P)$ is closed in $L^{2} (P)$ (The mean-zero L² space is a Hilbert space).

[F8]

The Lebesgue integral is linear on $L^{1}$ (The Lebesgue integral is linear on $L^{1} (μ)$ ).

Verification

technique · Construct normalized square-root perturbations, then use centered truncations

1.1F2F3F7A2

Every allowed score is in $L_{0}^{2} (P)$ by [F2]; because $L_{0}^{2} (P)$ is closed in $L^{2} (P)$ by [F7] under [A2], its closed tangent space is contained in $L_{0}^{2} (P)$ .

1.2givenF3algebra

Let $M = \sup_{x \in X} ∣ s (x) ∣ < \infty$ for the bounded representative in Given, and choose $0 < ϵ < (1 + M)^{- 1}$ . For $0 \leq t < ϵ$ , let $a_{t} = 1 + t s / 2$ , $c_{t} = P (a_{t}^{2}) = 1 + t^{2} P s^{2} / 4$ , and $p_{t} = a_{t}^{2} / c_{t}$ . The bound on $t$ makes $a_{t} \geq 0$ ; $c_{t} > 0$ and $P p_{t} = 1$ , so $p_{t}$ is a probability density with respect to $P$ .

1.3F3F5F6F8algebra

Let $f \in L_{0}^{2} (P)$ and choose a measurable representative. Define $u_{m} = (- m) \lor (f \land m)$ . Then $∣ u_{m} - f ∣^{2} \leq f^{2} 1_{{∣ f ∣ > m}}$ , whose right side tends pointwise to zero and is dominated by the integrable $f^{2}$ . By [F5], $∥ u_{m} - f ∥_{2} \to 0$ . Put $s_{m} = u_{m} - P u_{m}$ ; it is bounded and $P s_{m} = 0$ by [F8] and $P 1 = 1$ . Also [F6] and [F8] give $∣ P u_{m} ∣ = ∣ P (u_{m} - f) ∣ \leq ∥ u_{m} - f ∥_{2}$ . Therefore $∥ s_{m} - f ∥_{2} \leq 2 ∥ u_{m} - f ∥_{2} \to 0$ .

2.1step 1.2F1A1algebra

Since $∥ a_{t} ∥_{2} = \sqrt{c_{t}}$ , $∥ \sqrt{p_{t}} - 1 - (t / 2) s ∥_{2} = ∥ a_{t} / \sqrt{c_{t}} - a_{t} ∥_{2} = ∣ 1 - \sqrt{c_{t}} ∣ = O (t^{2}) = o (t)$ . The density family is explicit, so it fits [A1]'s general convention without any selection. Thus $P_{t} = p_{t} P$ is an allowed DQM path with score $s$ , and every bounded mean-zero score lies in the tangent set.

3.1F4step 1.1step 2.1step 1.3∎

Each $s_{m}$ is a tangent score by step 2.1, and the sequence converges to the arbitrary $f \in L_{0}^{2} (P)$ by step 1.3. Hence the tangent space contains $L_{0}^{2} (P)$ by [F4]; step 1.1 gives the reverse inclusion. Thus the tangent space is exactly $L_{0}^{2} (P)$ .

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