LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Local stationary expansion for a quadratic phase

Statement

Let $R > 0$ , $λ \geq 1$ , $σ \in {- 1, 1}$ , and $A \in C_{c}^{4} ((- R, R); C)$ . Fix an even $χ \in C_{c}^{\infty} ((- R, R))$ equal to $1$ on a neighborhood of $0$ . With $∥ A ∥_{C^{4}} : = \max_{0 \leq j \leq 4} ∥ A^{(j)} ∥_{\infty}$ , there is a finite constant $C_{R, χ}$ , depending only on $R$ and $χ$ , such that $∣ \int_{- R}^{R} A (u) e^{i σ λ u^{2} / 2} d u - A (0) \sqrt{\frac{2 π}{λ}} e^{i σ π / 4} ∣ \leq C_{R, χ} ∥ A ∥_{C^{4}} λ^{- 3 / 2} .$

Facts & Assumptions

Given: $A, R, λ, σ, χ$ of the Statement.

[F1]

Fresnel whole-line moment from the damped complex Gaussian: $\int_{R} e^{i σ λ u^{2} / 2} d u = \sqrt{2 π / λ} e^{i σ π / 4}$ as an ordinary two-sided improper integral.

[F2]

One-dimensional quadratic oscillatory amplitude bound: for $H \in C_{c}^{1} ((- R, R))$ , $∣ \int H (u) e^{i σ λ u^{2} / 2} d u ∣ \leq (6 ∥ H ∥_{\infty} + R ∥ H^{'} ∥_{\infty}) λ^{- 1 / 2}$ .

[F3]

The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ : continuous derivatives integrate to endpoint increments; applying it twice gives the integral Taylor remainder.

[F4]

If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ : the real compact-interval identity extends componentwise to complex $C^{1}$ functions.

[F5]

Derivative of a complex oscillatory phase: $(e^{i σ λ u^{2} / 2})^{'} = i σ λ u e^{i σ λ u^{2} / 2}$ .

[F6]

Improper integrals over unbounded intervals, Improper integrals with several singular ends, and Absolute and conditional convergence of improper integrals: one-sided limits define improper integrals, whole-line integrals sum separately convergent tails, and absolute convergence is tested on each tail.

[F7]

The improper $p$ -test for rational exponents and Comparison tests for improper integrals: a continuous function bounded in modulus by $C ∣ u ∣^{- 4}$ on each tail has an absolutely convergent improper integral there.

[F8]

For a continuous complex $z$ on a compact interval, the componentwise definition of Finite-interval oscillatory integral and phase convention, the conjugation and modulus laws Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, and real linearity and monotonicity from Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ give $∣ \int z ∣ \leq \int ∣ z ∣$ : if $Z = \int z \neq 0$ , set $ζ = \overline{Z} / ∣ Z ∣$ , so $∣ ζ ∣ = 1$ and $∣ Z ∣ = \int Re (ζ z) \leq \int ∣ z ∣$ . If $Z = 0$ the bound is immediate. The conjugation and modulus laws also give $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ for complex scalars.

[F9]

The mean-value theorem Mean value theorem on a closed real interval and Heine--Cantor Heine-Cantor in $R$ : a continuous real function on a compact subset of $R$ is uniformly continuous: on compact $[- R, R]$ , the continuous derivatives $D^{(3)}$ and $D^{(4)}$ are uniformly continuous; componentwise mean-value estimates justify differentiating $B (u) = \int_{0}^{1} (1 - t) D^{''} (t u) d t$ twice, giving $B^{'} (u) = \int_{0}^{1} (1 - t) t D^{'''} (t u) d t$ and $B^{''} (u) = \int_{0}^{1} (1 - t) t^{2} D^{''''} (t u) d t$ .

[F10]

Reflection $u \mapsto - u$ preserves the compact interval integral by Change of variables for an injective $C^{1}$ map on a compact Jordan set; real interval linearity from Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ , applied componentwise, then makes the integral of an odd complex function on $[- R, R]$ zero.

Proof

1.1givenF3F9constructalgebra

Put $D (u) = A (u) - A (0) χ (u) - A^{'} (0) u χ (u)$ . Then $D (0) = D^{'} (0) = 0$ . Near zero, twice applying [F3] gives $D (u) = u^{2} B (u)$ with $B (u) = \int_{0}^{1} (1 - t) D^{''} (t u) d t$ ; differentiating this finite parameter integral twice is justified by [F9] and gives $B \in C^{2}$ . Away from zero set $B = D / u^{2}$ . These definitions agree near zero, $B$ is compactly supported, and $∥ B ∥_{C^{2}} \leq C_{R, χ} ∥ A ∥_{C^{4}}$ . Thus $A = A (0) χ + A^{'} (0) u χ + u^{2} B$ .

2.1givenF1F4F5F6F7F8F10step 1.1algebra

By [F10], the integral of $u χ (u) e^{i σ λ u^{2} / 2}$ is zero: the integrand is odd and vanishes outside $[- R, R]$ . Let $S = 1 - χ$ . Since $S = 0$ near $0$ , apply [F4] and [F5] twice on finite truncations of each half-line. All boundary terms vanish at the inner endpoints because $S = 0$ there, and at infinity because $S / u \to 0$ and $((S / u)^{'} / u) \to 0$ . The final amplitude $K (u) = {(\frac{(S / u)^{'}}{u})}^{'}$ is continuous, vanishes near $0$ , and is $O (∣ u ∣^{- 4})$ on both tails; [F7] makes its phase-weighted improper integral absolutely convergent. Taking one-sided limits in the finite integration-by-parts identities and using [F6] gives $\int_{R} S (u) e^{i σ λ u^{2} / 2} d u = \frac{1}{(i σ λ)^{2}} \int_{R} K (u) e^{i σ λ u^{2} / 2} d u$ . Apply [F8] on finite truncations and pass to the limits from [F7] to bound the latter modulus by $λ^{- 2} \int_{R} ∣ K (u) ∣ d u = C_{χ} λ^{- 2}$ .

3.1givenF1F2F8step 1.1step 2.1algebra∎

Since $χ = 1$ near zero, [F1] and step 2.1 give $A (0) \int χ (u) e^{i σ λ u^{2} / 2} d u = A (0) \sqrt{\frac{2 π}{λ}} e^{i σ π / 4} + O (C_{χ} ∥ A ∥_{C^{4}} λ^{- 2}) .$ For the last term in step 1.1, one integration by parts gives $\int u^{2} B e^{i σ λ u^{2} / 2} = - (i σ λ)^{- 1} \int (B + u B^{'}) e^{i σ λ u^{2} / 2}$ ; the boundary term vanishes by compact support. Apply [F2] to $H = B + u B^{'} \in C_{c}^{1}$ and the bound on $∥ B ∥_{C^{2}}$ to obtain $O (C_{R, χ} ∥ A ∥_{C^{4}} λ^{- 3 / 2})$ . Since $λ^{- 2} \leq λ^{- 3 / 2}$ for $λ \geq 1$ , adding the three contributions and applying the complex scalar triangle inequality in [F8] proves the stated estimate.

Depends on

Used by

Stationary-phase expansion at one simple interior critical point Theorem

Dependency tree · two levels

108 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

NIST Digital Library of Mathematical Functions, §2.3(iv) and §7.12(ii) (standard reference, not scraped)
A. Iserles, S. P. Nørsett, and S. Olver, Highly oscillatory quadrature: The story so far, §4 (standard reference, not scraped)