LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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Gaussian mark normalization, moments, and exponential controls

Statement

Let $d \geq 1$ , mass $m > 0$ in kilograms, reference momentum $p_{*} > 0$ in kg m s $^{- 1}$ , energy scale $E_{0} > 0$ in joules, and dimensionless inverse temperature $b > 0$ . On $R^{d}$ use $r (d p) = d p / p_{*}^{d}$ and $K (p) = ∣ p ∣^{2} / (2 m E_{0})$ . Then $κ_{b} : = \int e^{- b K (p)} r (d p) = {(\frac{2 π m E_{0}}{b p_{*}^{2}})}^{d / 2}, ν_{b} (d p) = κ_{b}^{- 1} e^{- b K (p)} r (d p)$ is a probability with $E_{ν_{b}} p_{i} = 0$ , $E_{ν_{b}} p_{i} p_{j} = (m E_{0} / b) δ_{i j}$ , and $E_{ν_{b}} K = d / (2 b)$ . For each $g$ among $K, p_{i}, p_{i} p_{j}$ , and tilt parameter $t$ in reciprocal units to $g$ , the centered logarithmic moment-generating function $c_{g} (t) : = \log E_{ν_{b}} \exp (t (g - E_{ν_{b}} g))$ is finite and twice continuously differentiable on a neighborhood of $0$ , with $c_{g} (0) = c_{g}^{'} (0) = 0$ . In particular $c_{g} (t) = O (t^{2})$ as $t \to 0$ . Here $p_{i}$ has momentum units and $K$ is energy in units of $E_{0}$ . For the optional physical-temperature parametrization, fix a positive scalar conversion parameter $k_{B}$ with units J K $^{- 1}$ and take $T > 0$ in kelvin; then $b = E_{0} / (k_{B} T)$ .

Facts & Assumptions

Given: The constants and dimension specified above.

[F1]

The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ : $\int_{R} e^{- x^{2}} d x = \sqrt{π}$ .

[F2]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: nonnegative product-measurable functions can be integrated by iterated integrals on the sigma-finite product reference.

[F3]

If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ : one-dimensional integration by parts holds for continuously differentiable functions on bounded intervals.

[F4]

Dominated convergence: a pointwise convergent measurable family dominated by an integrable function has convergent integrals.

Proof

Given: The constants, reference, and suppliers in Facts & Assumptions.

1.1F1F2algebra

Substitute $u = \sqrt{b / (2 m E_{0})} p$ in each one-dimensional integral; [F1] gives $\int_{R} e^{- b p^{2} / (2 m E_{0})} d p = \sqrt{2 π m E_{0} / b} .$ Tonelli [F2] factors the nonnegative d-dimensional integral into d such terms. Dividing by p_*^d gives κ_b, so ν_b has mass one.

1.2F1F2F4algebra

For g=K or g=p_i p_j, use |p_i p_j|≤(p_i²+p_j²)/2 and K=|p|²/(2mE_0). For g=p_i, completing a square gives |t p_i|≤a p_i²/2+t²/(2a) for any a>0. For sufficiently small |t| these bounds dominate the unnormalized MGF integrands and their first two t-derivatives by a polynomial times exp(-a_0|p|²), for some a_0>0. That majorant is integrable: each polynomial factor is bounded after absorbing half the Gaussian exponent and the remaining Gaussian integral is finite by [F1] and [F2]. Dominated convergence [F4] makes each MGF twice continuously differentiable near zero; positivity makes its logarithm twice continuously differentiable as well.

2.1F1F3step 1.1algebra

Oddness gives E p_i=0 and E p_i p_j=0 for i≠j. On [-R,R], integration by parts [F3] with u(p)=p and v'(p)=p exp(-ap²), where a=b/(2mE_0), gives $\int_{- R}^{R} p^{2} e^{- a p^{2}} d p = \frac{1}{2 a} \int_{- R}^{R} e^{- a p^{2}} d p - \frac{R e^{- a R^{2}}}{a} .$ The boundary term tends to zero and both integrals converge by Gaussian domination. Divide by the normalization from step 1.1 and let R→∞ to obtain E p_i²=1/(2a)=mE_0/b. Therefore E K=Σ_i E p_i²/(2mE_0)=d/(2b).

3.1step 1.2step 2.1algebra∎

At t=0 the centered log MGF is zero; its derivative is E(g-Eg)=0, using the means in step 2.1. Its second derivative is finite and continuous by step 1.2. Taylor's theorem at zero then gives c_g(t)=c_g''(ξ_t)t²/2=O(t²) near zero.

Gaussian mark normalization, moments, and exponential controls

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources