CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generated

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A singular interior mode need not have Gaussian scaling

Statement refuted

A unique interior maximum and smoothness alone always give $I_{n} = O (e^{n h (\hat{x})} n^{- 1 / 2})$ , even when the Hessian at the maximum is singular. Here $u_{n} = O (v_{n})$ means $∣ u_{n} ∣ \leq C ∣ v_{n} ∣$ for some finite constant $C$ and all sufficiently large $n$ .

Facts & Assumptions

Given: For $x \in [- 1, 1]$ , let $h (x) : = - x^{4}$ and $I_{n} : = \int_{- 1}^{1} e^{n h (x)} d x$ for integers $n \geq 1$ .

[F1]

Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ : a function is $C^{\infty}$ when its derivatives of every finite order exist and are continuous.

[F2]

The exponential function is strictly increasing and The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ : the exponential is continuous, strictly increasing, and everywhere positive; in particular $e^{- 1} > 0$ .

[F3]

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 bounded and Riemann integrable.

[F4]

If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ : a nonnegative integrable function has nonnegative integral, and a pointwise lower bound on the same interval gives the corresponding integral bound.

[F5]

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 $a, b, c$ : an integral splits at intermediate points, including coincident endpoints in the oriented form.

[F6]

For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term: every real polynomial is differentiable everywhere, with its derivative computed term by term.

[F7]

Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function: every real polynomial is continuous.

[F8]

A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs: a composition of continuous real functions is continuous when their domains and ranges match.

Counterexample

1.1givenF1F6F7algebra

Extend $h (x) = - x^{4}$ to $R$ . Repeated application of [F6] gives $h^{'} (x) = - 4 x^{3}$ , $h^{''} (x) = - 12 x^{2}$ , $h^{(3)} (x) = - 24 x$ , $h^{(4)} (x) = - 24$ , and $h^{(j)} (x) = 0$ for every $j \geq 5$ . These derivatives and $h$ are continuous by [F7], so $h$ is $C^{\infty}$ by [F1]. Since $x^{4} > 0$ for $x \neq 0$ , its unique maximum on $[- 1, 1]$ is $h (0) = 0$ at the interior point $0$ , with $h^{'} (0) = h^{''} (0) = 0$ . For each integer $m \geq 1$ , the interval $[- 1 / m, 1 / m]$ lies in $[- 1, 1]$ .

2.1givenF2F3F4F5F7F8step 1.1algebra∎

If $∣ x ∣ \leq 1 / m$ , then $m^{4} x^{4} \leq 1$ , so [F2] gives $e^{- m^{4} x^{4}} \geq e^{- 1}$ . The polynomial $x \mapsto - m^{4} x^{4}$ is continuous by [F7], so its exponential is continuous by [F2] and [F8], and positive by [F2], hence integrable on each nondegenerate subinterval by [F3]. Split its integral at $- 1 / m$ and $1 / m$ using [F5]. The two outer contributions are nonnegative by [F4] (and zero when $m = 1$ ), while the central contribution is at least $2 e^{- 1} / m$ by [F4]. Thus $I_{m^{4}} \geq 2 e^{- 1} / m$ . The Gaussian scale at $n = m^{4}$ is $n^{- 1 / 2} = m^{- 2}$ ; therefore $I_{m^{4}} / (m^{4})^{- 1 / 2} \geq 2 e^{- 1} m ⟶ \infty$ . Hence these integrals are not $O (n^{- 1 / 2})$ , which refutes the statement.

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Sources

Laplace method for integrals, PHYS 2400, University of Connecticut (standard reference, not scraped)
NIST Digital Library of Mathematical Functions, §2.3(iii), Laplace's Method (standard reference, not scraped)