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An existing interacting measure has zero-radius perturbation series

Statement refuted

Every finite-dimensional Euclidean quartic partition function with nonnegative coupling has a perturbation Taylor series at zero coupling with positive radius of convergence.

Facts & Assumptions

Given: Assume the Axiom of Choice. Let $Q \sim N (0, 1)$ be the one-dimensional centered Gaussian and put $U (q) = q^{4} / 4!$ .

[A1]

The Axiom of Choice is the explicit hypothesis of the finite-dimensional Gaussian-law suppliers. The Axiom of Choice

[F1]

Finite Euclidean quartic interacting measure defines the Borel Gaussian reference law $γ_{C}$ and the partition integral $Z (λ) = E_{γ_{C}} [e^{- λ U}]$ .

[F2]

For a finite Gaussian covariance, polynomial insertions have all Wick-pairing moments. In one dimension with variance one, $E [Q^{2 m}] = (2 m - 1)!!$ and odd moments vanish. Finite Gaussian normalization and Wick moments

[F3]

The finite Euclidean partition integral is positive and finite for every $λ \geq 0$ , and its order- $M$ coefficients are $(- 1)^{k} E [U^{k}] / k!$ . Finite interacting asymptotic Wick expansion with remainder

Counterexample

Given: Use the measure and coefficients of [F1]–[F3] with $N = 1$ , covariance $C = 1$ , and $U (q) = q^{4} / 4!$ .

1.1A1F1F3algebra

By [F1], $Z (0) = \int 1 d γ_{1} = 1$ . For $λ > 0$ , [F3] gives $0 < Z (λ) < \infty$ and identifies the formal coefficients as $(- 1)^{n} E [U^{n}] / n!$ .

1.2F2algebra

With $U (q) = q^{4} / 4!$ , [F2] gives $a_{n} = \frac{E [Q^{4 n}]}{(4!)^{n} n!} = \frac{(4 n - 1)!!}{(4!)^{n} n!} = \frac{(4 n)!}{(4!)^{n} n! 2^{2 n} (2 n)!} .$ The pairing count follows by pairing the first of $4 n$ factors with any of the other $4 n - 1$ factors and repeating on the remaining factors.

2.1step 1.2algebra

Dividing successive coefficients from step 1.2 yields $\frac{a_{n + 1}}{a_{n}} = \frac{(4 n + 1) (4 n + 3)}{24 (n + 1)} ⟶ \infty .$ Hence for each $λ > 0$ the absolute term ratio $λ a_{n + 1} / a_{n}$ eventually exceeds $2$ , so the terms do not tend to zero and the formal series diverges at every nonzero $λ$ ; its radius of convergence is zero.

3.1step 1.1step 2.1∎

The exact probability measure still exists for all $λ \geq 0$ by step 1.1, while its perturbation series has zero radius by step 2.1. This refutes the Statement without asserting Borel summability or any continuum-field result.

An existing interacting measure has zero-radius perturbation series

Statement refuted

Facts & Assumptions

Counterexample

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