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

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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.

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The integral transform is representative independent

Statement

For every $f \in L^{1} (R^{n}; C)$ , $n \geq 1$ , the integral defining $\hat{f} (ξ)$ is absolutely convergent for every $ξ$ and unchanged by null-set modifications. Moreover $∣ \hat{f} (ξ) ∣ \leq ∥ f ∥_{1}$ .

Facts & Assumptions

Given: An integrable complex representative $f$ and $ξ \in R^{n}$ , with the formula of Fourier transform on complex L1 classes.

[F1]

The exponential satisfies $∣ \exp (x + i y) ∣ = e^{x}$ for real $x, y$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[F2]

Integrable functions equal almost everywhere have equal integrals on every measurable set (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree).

[F3]

The modulus of an integral is at most the integral of the modulus (The modulus of an integral is bounded by the integral of the modulus).

Proof

1.1

The exponential factor is continuous, hence measurable, and has modulus one. Thus the product is measurable and $\int ∣ f (x) e^{- 2 π i x \cdot ξ} ∣ d x = \int ∣ f ∣ < \infty$ . Its componentwise integral exists, and $∣ \hat{f} (ξ) ∣ \leq ∥ f ∥_{1}$ .

F1F3given

2.1

If $g = f$ outside a measurable null set $N$ , the same modulus equality proves the product with $g$ integrable, and the two products agree outside $N$ . Applying integral invariance on $R^{n}$ gives identical values at this $ξ$ . Since $ξ$ was arbitrary, this holds at every frequency; no union of frequency-dependent exceptional sets is taken. In particular a zero class has identically zero transform.

F1F2step 1.1∎

Depends on

Used by

Null-set modifications defeat everywhere representative recovery Counterexample
Carleson operator and measurable linearisation Definition
Euclidean Gaussian transform with the 2π normalization Lemma
Fourier transform turns L1 convolution into multiplication Theorem
The L1 transform is bounded and uniformly continuous Theorem
Translation, modulation, linear dilation and reflection laws Theorem

Cited to discharge well-definedness by Fourier transform on complex L1 classes.

Dependency tree · two levels

21 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

Semyon Dyatlov, MIT 18.155 (2022) (standard reference, not scraped)