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Two equal modes contribute two Gaussian terms

Statement refuted

For every phase with two distinct equal-height nondegenerate interior maxima, the Laplace term computed at either one of the maxima is asymptotic to the whole integral.

Facts & Assumptions

Given: $K = [- 2, 2]$ , $h (x) : = - (x^{2} - 1)^{2}$ , and the Riemann integral $I_{n} : = \int_{- 2}^{2} e^{n h (x)} d x$ for integers $n \geq 1$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every countable family of nonempty sets has a choice function.

[F1]

Laplace asymptotic under a unique nondegenerate interior maximum: under its stated compactness, unique-mode, local quadratic, and global gap hypotheses, the integral is asymptotic to the amplitude at the mode times $e^{n h (\hat{x})} (2 π / n)^{1 / 2} / \sqrt{- h^{''} (\hat{x})}$ in dimension one.

[F2]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion: a continuous real function on a closed bounded interval is Riemann integrable.

[F3]

For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ : for an integrable function and $a < c < b$ , $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ .

[F4]

A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral: assuming Countable Choice, a bounded Riemann-integrable function on a closed interval is Lebesgue integrable there with the same integral.

[F5]

The Hessian matrix and critical points of a scalar field: in dimension one the Hessian is the second derivative.

[F6]

Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ : $C^{3}$ means the first three derivatives exist and are continuous.

Proof

1.1givenF5algebra

Factoring gives $h (x) = - (x - 1)^{2} (x + 1)^{2} \leq 0$ , with equality exactly at $x = - 1$ and $x = 1$ . Direct differentiation gives $h^{'} (x) = - 4 x^{3} + 4 x$ and $h^{''} (x) = 4 - 12 x^{2}$ , so each mode is interior, has height $0$ , and has negative Hessian $h^{''} (\pm 1) = - 8$ by [F5].

1.2givenF1algebra

On $K_{-} = [- 2, 0]$ , the unique maximizer is $- 1$ . If $∣ x + 1 ∣ \leq 1 / 2$ , then $∣ x - 1 ∣ \geq 1$ and $h (x) \leq - (x + 1)^{2}$ ; if $x \in K_{-} ∖ B (- 1, 1 / 2)$ , then $∣ x + 1 ∣ \geq 1 / 2$ and $∣ x - 1 ∣ \geq 1$ , so $h (x) \leq - 1 / 4$ . Thus the Laplace theorem's local quadratic and strict global gap conditions hold on $K_{-}$ .

1.3givenF1F2F6algebra

On $K_{+} = [0, 2]$ , the unique maximizer is $1$ . If $∣ x - 1 ∣ \leq 1 / 2$ , then $∣ x + 1 ∣ \geq 1$ and $h (x) \leq - (x - 1)^{2}$ ; if $x \in K_{+} ∖ B (1, 1 / 2)$ , then $∣ x - 1 ∣ \geq 1 / 2$ and $∣ x + 1 ∣ \geq 1$ , so $h (x) \leq - 1 / 4$ . Both intervals are compact of positive length, the amplitude $a \equiv 1$ is integrable and positive at the mode, and the polynomial $h$ is $C^{3}$ by [F6].

2.1A1F1step 1.1step 1.2step 1.3algebra

Apply [F1] separately on $K_{-}$ and $K_{+}$ . If $J_{n}^{-} : = \int_{K_{-}} e^{n h} d λ_{1}$ and $J_{n}^{+} : = \int_{K_{+}} e^{n h} d λ_{1}$ , then $J_{n}^{-} \sim \sqrt{π} / (2 \sqrt{n})$ and $J_{n}^{+} \sim \sqrt{π} / (2 \sqrt{n})$ , because each mode has height $0$ and $- h^{''} = 8$ .

3.1A1F2F3F4step 2.1algebra

By [F2] and [F4], the Riemann and Lebesgue integrals on each half agree; by [F3], $I_{n} = J_{n}^{-} + J_{n}^{+}$ . Therefore $I_{n} \sim \sqrt{π} / \sqrt{n}$ : dividing by $\sqrt{π} / \sqrt{n}$ gives half the sum of two ratios tending to $1$ .

4.1step 2.1step 3.1algebra∎

A one-mode term is $\sqrt{π} / (2 \sqrt{n})$ , while step 3.1 gives $I_{n} / (\sqrt{π} / (2 \sqrt{n})) \to 2$ . This ratio is not $1$ , so the statement refuted fails.

Depends on

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Sources

NIST Digital Library of Mathematical Functions, §2.3(iii), Laplace's Method (standard reference, not scraped)
Francis Bach, Approximating Integrals with Laplace's Method (standard reference, not scraped)