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

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Positive heavy-mode Gaussian matching data

Physical scope

Pure mathematics; no physical premises or physical claim.

Definition

Assume AC. Let $H$ be a nonzero complex Hilbert space, let $A : D (A) \subseteq H \to H$ be densely defined, self-adjoint and nonnegative, and let $M > 0$ . Set $K = M^{2} I + A$ on $D (A)$ and let $E_{A}$ be the spectral measure of $A$ . Define $f_{M} (a) = (M^{2} + a)^{- 1}$ for $a \geq 0$ . The bounded spectral operator $f_{M} (A)$ is the inverse $K^{- 1}$ and satisfies $∥ K^{- 1} ∥ \leq M^{- 2}$ . For $J \in H$ define the matched quadratic effective-action contribution $Δ S_{e f f} (J) : = - \frac{1}{2} ⟨ J, K^{- 1} J ⟩ .$ The inner product is the complex Hilbert pairing, linear in its first variable; the displayed quadratic form is real and nonpositive. In natural units $ℏ = c = 1$ , $A$ and $K$ have mass-squared units; normalize the source/pairing so $⟨ J, K^{- 1} J ⟩$ is dimensionless, hence $∥ J ∥^{2}$ has mass-squared units. In a finite real regulator, integration of $\exp [- \frac{1}{2} χ^{T} K χ + j^{T} χ]$ divided by the zero-source Gaussian integral yields $\exp [\frac{1}{2} j^{T} K^{- 1} j]$ , so the corresponding effective action changes by $- \frac{1}{2} j^{T} K^{- 1} j$ . Since $\log$ is the inverse of $\exp$ on positive reals, taking $- \log$ of that normalized integral gives this effective-action contribution. Determinants and Lebesgue integration are finite-regulator quantities only; this definition supplies no infinite-dimensional Gaussian measure or determinant.

Definition references

Hilbert space

Real and complex inner-product spaces and their induced length

Positive unbounded spectral calculus from a bounded resolvent

Finite-dimensional heavy Gaussian source shift

The natural logarithm as the inverse of the exponential function

Facts & Assumptions

Given: AC; a nonzero complex Hilbert space; a densely defined nonnegative self-adjoint $A$ ; $M > 0$ ; and $J \in H$ .

[A1]

The Axiom of Choice: every family of nonempty sets has a choice function.

[A2]

AC supplies countable selections and prescribed serial paths: AC implies countable choice.

[F1]

Positive unbounded spectral calculus from a bounded resolvent: the spectral PVM $E_{A}$ is supported on $[0, \infty)$ , $A = \int_{[0, \infty)} a d E_{A} (a)$ with domain $D (A) = {u \in H : \int_{[0, \infty)} a^{2} d μ_{u} (a) < \infty}$ , where $μ_{u} (S) = ⟨ E_{A} (S) u, u ⟩$ , and every bounded Borel $h$ defines $h (A) = \int h d E_{A}$ with $∥ h (A) ∥ \leq \sup_{a \geq 0} ∣ h (a) ∣$ .

[F2]

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

[F3]

The adjoint is well defined, closed, and reverses inclusions: under countable choice the adjoint of a densely defined operator is closed; hence a self-adjoint $A$ is closed.

[F4]

Finite-dimensional heavy Gaussian source shift: the normalized finite Gaussian source-shift ratio is $\exp (\frac{1}{2} j^{T} K^{- 1} j)$ for symmetric positive-definite $K$ .

[F5]

Pvm integral is a star homomorphism: for bounded Borel functions $h, k$ and the same PVM $E_{A}$ , $(\int h d E_{A}) (\int k d E_{A}) = \int h k d E_{A}$ ; the bounded PVM integral is multiplicative, unital, and adjoint preserving. Its Countable Choice hypothesis is supplied by [A1]–[A2].

[A3]

Hilbert space: a complex Hilbert space is a complete complex inner-product space.

[A4]

Real and complex inner-product spaces and their induced length: the complex inner product is linear in the first variable and conjugate-linear in the second.

[F6]

The natural logarithm as the inverse of the exponential function: $\log$ is defined on $(0, \infty)$ as the inverse of $\exp$ .

Proof

technique · direct spectral calculus with a finite-regulator Gaussian identity

Given: The objects and hypotheses in the Definition.

1.1A1A2F1F2F3

The actual proof of [F1] uses Cauchy–Schwarz for its resolvent lower bound and closedness of $A$ for its resolvent graph argument. [F2] supplies the former. By [A1]–[A2] countable choice is available, so [F3] applies and self-adjointness gives $A = A^{*}$ , hence $A$ is closed. This records the actual inputs of [F1]'s proof; the carrier's complete conditional proof was inspected.

2.1A3A4F1F5step 1.1algebra

Set $g_{M} (a) = (M^{2} + a)^{- 1}$ and $μ_{u} (S) = ⟨ E_{A} (S) u, u ⟩$ . By [F1], $g_{M} (A)$ is bounded with norm at most $M^{- 2}$ . Since $0 \leq a g_{M} (a) \leq 1$ , $\int a^{2} g_{M} (a)^{2} d μ_{u} (a) \leq μ_{u} ([0, \infty)) = ∥ u ∥^{2}$ for every $u \in H$ , so the domain formula in [F1] gives $g_{M} (A) u \in D (A)$ . To see the left-inverse identity, let $x = g_{M} (A) u$ and $x_{n} = E_{A} ([0, n]) x$ . The domain formula gives $x_{n} \to x$ and $A x_{n} \to A x$ in $H$ . The spectral representation in [F1] and the multiplicativity in [F5] applied to bounded truncations give $(M^{2} I + A) x_{n} = \int_{[0, n]} (M^{2} + a) g_{M} (a) d E_{A} (a) u = E_{A} ([0, n]) u$ ; hence $(M^{2} I + A) g_{M} (A) u = u$ . For the right-inverse identity, if $v \in D (A)$ , set $v_{n} = E_{A} ([0, n]) v$ . Then $v_{n} \to v$ and $(M^{2} I + A) v_{n} \to (M^{2} I + A) v$ in $H$ . The same spectral representation and multiplicativity give $g_{M} (A) (M^{2} I + A) v_{n} = \int_{[0, n]} g_{M} (a) (M^{2} + a) d E_{A} (a) v = v_{n}$ , and boundedness of $g_{M} (A)$ gives $g_{M} (A) (M^{2} I + A) v = v$ . Thus $K^{- 1} = g_{M} (A)$ and $∥ K^{- 1} ∥ \leq M^{- 2}$ . Finally, the positive scalar multiplier gives $⟨ J, K^{- 1} J ⟩ = \int g_{M} (a) d μ_{J} (a) \geq 0$ , so the quadratic form is real.

3.1F4F6givenalgebra∎

In a finite real regulator with symmetric positive-definite matrix $K$ , [F4] gives the normalized source integral $\exp (\frac{1}{2} j^{T} K^{- 1} j)$ . Taking its negative logarithm using [F6] gives the effective-action shift $- \frac{1}{2} j^{T} K^{- 1} j$ . No determinant or Lebesgue integral beyond that finite regulator enters the definition.

Positive heavy-mode Gaussian matching data

Definition

Definition references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources