LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Finite-dimensional Gaussian normalization and moment envelope

Statement

Let $d \geq 1$ be an integer, $m > 0$ , and $0 < β_{0} \leq β_{1} < \infty$ . For every $β \in [β_{0}, β_{1}]$ , $\int_{R^{d}} e^{- β ∣ p ∣^{2} / (2 m)} d p = {(\frac{2 π m}{β})}^{d / 2} .$ For every integer $s \geq 0$ , the polynomially weighted Gaussian has a uniform finite bound $\sup_{β \in [β_{0}, β_{1}]} \int_{R^{d}} (1 + ∣ p ∣^{2 s}) e^{- β ∣ p ∣^{2} / (2 m)} d p < \infty .$

Facts & Assumptions

Given: An integer $d \geq 1$ , mass parameter $m > 0$ , and compact positive inverse-temperature interval $[β_{0}, β_{1}]$ .

[F1]

The one-dimensional Gaussian integral is $\int_{R} e^{- x^{2}} d x = \sqrt{π}$ . (The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ )

[F2]

For sigma-finite measure spaces and a nonnegative product-measurable function, Tonelli's theorem identifies the product integral with either iterated integral. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

Proof

1.1F1F2algebra

In one dimension set $y = \sqrt{β / (2 m)} x$ ; the substitution and [F1] give $\int_{R} e^{- β x^{2} / (2 m)} d x = \sqrt{2 π m / β}$ . Taking the product of $d$ such factors and applying [F2] repeatedly gives the displayed exact $d$ -dimensional integral.

1.2F1algebra

Put $α = β_{0} / (2 m) > 0$ . For any integer $s \geq 1$ , the function $x \mapsto ∣ x ∣^{2 s} e^{- α x^{2} / 2}$ is bounded: it is continuous on $[- 1, 1]$ and tends to zero at both infinite ends. Hence $∣ x ∣^{2 s} e^{- α x^{2}} \leq C_{s, α} e^{- α x^{2} / 2}$ ; on $[- 1, 1]$ the same bound holds after increasing $C_{s, α}$ . By the scaled one-dimensional Gaussian integral, both $\int e^{- α x^{2}} d x$ and $\int ∣ x ∣^{2 s} e^{- α x^{2}} d x$ are finite.

2.1F2step 1.2algebra∎

For $β \geq β_{0}$ , $e^{- β ∣ p ∣^{2} / (2 m)} \leq e^{- α ∣ p ∣^{2}}$ . If $s \geq 1$ , $(\sum_{i = 1}^{d} p_{i}^{2})^{s} \leq d^{s - 1} \sum_{i = 1}^{d} ∣ p_{i} ∣^{2 s}$ ; for $s = 0$ the weight is one. Factor the remaining coordinate Gaussians and apply [F2] and the one-dimensional bounds from step 1.2. The resulting finite upper bound depends only on $d, s, m, β_{0}$ , not on $β$ in the stated interval.

Finite-dimensional Gaussian normalization and moment envelope

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources