TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Riemann–Lebesgue lemma

Statement

Assume countable choice. For $f \in L^{1} (R^{n}; C)$ , $n \geq 1$ , $\hat{f} \in C_{0} (R^{n}; C)$ : it is continuous and tends to zero as $∣ ξ ∣ \to \infty$ . The notation extends The space $C_{0} (R^{n})$ of continuous functions vanishing at infinity componentwise.

Facts & Assumptions

Given: $f \in L^{1}$ and The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F1]

The transform is linear, uniformly continuous and bounded by the input $L^{1}$ norm (The L1 transform is bounded and uniformly continuous).

[F2]

Translation multiplies the transform by $e^{- 2 π i h \cdot ξ}$ (Translation, modulation, linear dilation and reflection laws).

[F3]

Complex $L^{1}$ translations are norm-continuous under countable choice (Complex translation, convolution, approximate identities, and mollification).

Proof

1.1

For $ξ \neq 0$ set $h = ξ / (2 ∣ ξ ∣^{2})$ . Then $h \cdot ξ = 1 / 2$ , so $\hat{τ_{h} f - f} (ξ) = - 2 \hat{f} (ξ)$ by F2 and linearity. The bound in F1 gives $2 ∣ \hat{f} (ξ) ∣ \leq ∥ τ_{h} f - f ∥_{1}$ .

F1F2given

2.1

Given $ϵ > 0$ , F3 supplies $δ > 0$ with $∥ τ_{h} f - f ∥_{1} < 2 ϵ$ for $∣ h ∣ < δ$ . If $∣ ξ ∣ > 1 / (2 δ)$ , the explicit $h$ from step 1.1 satisfies that condition, hence $∣ \hat{f} (ξ) ∣ < ϵ$ . Continuity is already F1. This is the asserted $C_{0}$ property. Countable choice is inherited from F2 and F3, not from the explicit selection of $h$ .

F1F2F3step 1.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Gerald Teschl, Topics in Real and Functional Analysis (2017) (standard reference, not scraped)