TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Low-momentum matching has a controlled remainder

Physical scope

Pure mathematics; no physical premises or physical claim.

Statement

Assume the hypotheses and notation of the positive heavy-mode Gaussian matching data. Let $E \geq 0$ be an energy and let $P_{E} : = E_{A} ([0, E^{2}])$ be the spectral projection of $A$ . For $J = P_{E} J \in H$ and an integer $N \geq 0$ , define $S_{N} J : = M^{- 2} \sum_{r = 0}^{N} (- A / M^{2})^{r} J,$ which is well-defined since $J$ has bounded spectral support. Put $R_{N} J : = K^{- 1} J - S_{N} J$ . Then $∥ R_{N} J ∥ \leq M^{- 2} (E^{2} / M^{2})^{N + 1} ∥ J ∥,$ and $∣ Δ S_{e f f} (J) + \frac{1}{2} ⟨ J, S_{N} J ⟩ ∣ = \frac{1}{2} ∣ ⟨ J, R_{N} J ⟩ ∣ \leq \frac{∥ J ∥^{2}}{2 M^{2}} (E^{2} / M^{2})^{N + 1} .$ Use natural units $ℏ = c = 1$ : $A$ has units of mass squared, $E, M$ have mass units, and $E^{2} / M^{2}$ is dimensionless. Under the source normalization of the positive heavy-mode Gaussian matching data, $∥ J ∥^{2} / M^{2}$ and the action remainder are dimensionless. These are exact conditional bounds for every $E \geq 0$ ; the truncation is small/useful when $E < M$ . No claim is made for unrestricted sources or arbitrary interacting matching.

Statement references

Positive heavy-mode Gaussian matching data

Facts & Assumptions

Given: AC, $H, A, M, K, E_{A}, K^{- 1}$ as in Positive heavy-mode Gaussian matching data, $E \geq 0$ , and $J = P_{E} J$ .

[A1]

The Axiom of Choice: the spectral construction assumes AC.

[A2]

AC supplies countable selections and prescribed serial paths: AC supplies countable choice required in the actual proof of the spectral supplier.

[F1]

Positive unbounded spectral calculus from a bounded resolvent: bounded Borel functions of $A$ have norm at most their supremum and $A$ has its spectral PVM on $[0, \infty)$ .

[F2]

Positive heavy-mode Gaussian matching data: $K^{- 1} = f_{M} (A)$ for $f_{M} (a) = (M^{2} + a)^{- 1}$ , and $Δ S_{e f f} (J) = - \frac{1}{2} ⟨ J, K^{- 1} J ⟩$ .

[F3]

Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ .

Proof

technique · direct scalar remainder and the spectral functional calculus

Given: The hypotheses and notation in the Statement.

1.1givenalgebra

For every $a \geq 0$ , putting $x = a / M^{2}$ and multiplying by $1 + x$ gives $\frac{1}{M^{2} + a} = M^{- 2} \sum_{r = 0}^{N} (- a / M^{2})^{r} + M^{- 2} \frac{(- a / M^{2})^{N + 1}}{1 + a / M^{2}} .$ This finite geometric identity also holds at $a = 0$ and for $N = 0$ .

2.1A1A2F1F2step 1.1algebra

On $J = P_{E} J$ , the spectral variable satisfies $0 \leq a \leq E^{2}$ . Therefore every polynomial term $A^{r} J$ is defined, and the remainder multiplier satisfies $∣ M^{- 2} \frac{(- a / M^{2})^{N + 1}}{1 + a / M^{2}} ∣ \leq M^{- 2} (E^{2} / M^{2})^{N + 1},$ because $1 + a / M^{2} \geq 1$ . Extend this bounded multiplier by zero outside $[0, E^{2}]$ and apply [F1]. With [F2] this identifies the inverse kernel and proves the vector identity defining $R_{N} J$ and its norm bound.

3.1F2F3step 2.1algebra∎

By [F2], the action remainder equals $\frac{1}{2} ∣ ⟨ J, R_{N} J ⟩ ∣$ . Cauchy–Schwarz [F3] and step 2.1 give the claimed bound. The factor $(E^{2} / M^{2})^{N + 1}$ decreases to zero only when $E < M$ ; for $E \geq M$ the same finite-order bound remains true but need not be small.

Low-momentum matching has a controlled remainder

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources