TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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Equipartition identity with explicit integrated boundary terms

Statement

Let $(Y, G, ν)$ be a sigma-finite measure space, $β > 0$ , and $H : R \times Y \to R$ a product-measurable function whose map $z \mapsto H (z, y)$ is $C^{1}$ for $ν$ -almost every $y$ . Set $Z : = \int_{R \times Y} e^{- β H (z, y)} d (z, y) \in (0, \infty),$ where $d (z, y)$ is Lebesgue measure on $R$ times $ν$ . Assume $\int_{R \times Y} ∣ z \partial_{z} H (z, y) ∣ e^{- β H (z, y)} d (z, y) < \infty .$ For $R > 0$ let $B_{R} : = \int_{Y} (R e^{- β H (R, y)} + R e^{- β H (- R, y)}) d ν (y),$ and assume $B_{R} \to 0$ as $R \to \infty$ . Define $⟨ F ⟩_{β} : = Z^{- 1} \int F (z, y) e^{- β H (z, y)} d (z, y)$ whenever the integral is absolutely convergent. Then $⟨ z \partial_{z} H ⟩_{β} = β^{- 1} .$ For finite endpoints $a < b$ , assume the endpoint integral below is finite and put $Z_{[a, b]} : = \int_{[a, b] \times Y} e^{- β H} d (z, y)$ . If $0 < Z_{[a, b]} < \infty$ , define $⟨ F ⟩_{[a, b]} : = Z_{[a, b]}^{- 1} \int_{[a, b] \times Y} F e^{- β H} d (z, y)$ . Then the exact boundary identity is $⟨ z \partial_{z} H ⟩_{[a, b]} = β^{- 1} (1 - Z_{[a, b]}^{- 1} \int_{Y} (b e^{- β H (b, y)} - a e^{- β H (a, y)}) d ν (y)),$

Facts & Assumptions

Given: The product measure space, $β$ , $H$ , the finite positive normalization $Z$ , the integrable moment, and the vanishing integrated boundary term $B_{R}$ .

[A1]

The measure factors are sigma-finite, $H$ is measurable and sectionwise $C^{1}$ , and $Z$ and $\int ∣ z H_{z} ∣ e^{- β H}$ are finite with $Z > 0$ .

[A2]

The two integrated endpoint contributions tend to zero, equivalently $B_{R} \to 0$ .

[A3]

For the finite-interval clause, $0 < Z_{[a, b]} < \infty$ and the stated integrated endpoint expression is finite.

[F1]

Tonelli's theorem permits section integration for nonnegative product-measurable functions on sigma-finite factors (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F2]

Dominated convergence passes limits through an integral under one common integrable majorant (Dominated convergence).

Proof

Given: [A1]–[A3] and the Tonelli and dominated-convergence interfaces [F1]–[F2].

1.1A1algebra

For almost every $y$ , differentiate the product in the $z$ coordinate: $\partial_{z} (z e^{- β H (z, y)}) = e^{- β H (z, y)} - β z \partial_{z} H (z, y) e^{- β H (z, y)} .$ The absolute value of the right side is bounded by $(1 + β ∣ z \partial_{z} H ∣) e^{- β H}$ , which is integrable by [A1].

2.1A1F1step 1.1

Integrate this identity on $[- R, R] \times Y$ . Absolute integrability and [F1] justify integrating first in $z$ and then in $y$ , so $\int_{[- R, R] \times Y} e^{- β H} d (z, y) - β \int_{[- R, R] \times Y} z \partial_{z} H e^{- β H} d (z, y) = \int_{Y} (R e^{- β H (R, y)} + R e^{- β H (- R, y)}) d ν (y) = B_{R} .$

3.1A1A2F2step 2.1

The integrand in the first term is nonnegative and integrable, while the second integrand is absolutely integrable by [A1]. Their restrictions to $[- R, R] \times Y$ converge pointwise almost everywhere to the full integrands and are dominated by those integrable functions. Applying [F2] to both terms and using [A2] gives $Z - β \int z \partial_{z} H e^{- β H} d (z, y) = 0$ . Divide by $β Z > 0$ to obtain the stated identity.

4.1A1A3F1step 1.1algebra∎

For finite $a < b$ , integrate the same derivative over $[a, b] \times Y$ . The boundary evaluation is $\int_{Y} (b e^{- β H (b, y)} - a e^{- β H (a, y)}) d ν (y) = Z_{[a, b]} - β \int_{[a, b] \times Y} z \partial_{z} H e^{- β H} d (z, y) .$ Dividing by $β Z_{[a, b]}$ and rearranging yields the displayed finite-endpoint formula.

Equipartition identity with explicit integrated boundary terms

Statement

Facts & Assumptions

Proof

Depends on

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