LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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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

1.1F1F3given

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 $∣ \hat{f} (ξ) ∣ \leq ∥ f ∥_{1}$ .

2.1F1F2step 1.1∎

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.

Depends on

Dependency tree · 2 levels

Sources

Semyon Dyatlov, MIT 18.155 (2022)