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Fourier transform turns L1 convolution into multiplication

Statement

Assume countable choice. If $f, g \in L^{1} (R^{n}; C)$ , $n \geq 1$ , then $\hat{f * g} (ξ) = \hat{f} (ξ) \hat{g} (ξ)$ for every $ξ \in R^{n}$ .

Facts & Assumptions

Given: The stated functions and The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F1]

The complex convolution interface supplies representative-independent $L^{1}$ convolution and translation isometries, under countable choice (Complex translation, convolution, approximate identities, and mollification).

[F2]

The product $f (x - y) g (y)$ of Borel representatives is jointly Borel measurable (Borel representatives make the convolution integrand Borel measurable).

[F3]

Tonelli equates nonnegative iterated integrals on sigma-finite spaces (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F4]

Fubini equates complex iterated integrals when the product integral of the modulus is finite (Fubini's theorem for L^1 functions on a sigma-finite product).

[F5]

Translation obeys $\hat{τ_{y} f} (ξ) = e^{- 2 π i y \cdot ξ} \hat{f} (ξ)$ (Translation, modulation, linear dilation and reflection laws).

[F6]

Null-equivalent $L^{1}$ representatives have the same transform at every frequency (The integral transform is representative independent).

[F7]

Under countable choice, a completion-measurable real function has a base-measurable almost-everywhere equal representative (A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra).

Proof

1.1F1F2F3F7given

Apply [F7] to the real and imaginary components of $f$ and $g$ , changing infinite values on their null sets to zero, to obtain finite Borel representatives. Euclidean Lebesgue spaces are sigma-finite (the boxes $[- k, k]^{n}$ have finite volume and cover them). F2 gives product measurability. Translation invariance and Tonelli give $\iint ∣ f (x - y) g (y) ∣ d x d y = \int ∥ f ∥_{1} ∣ g (y) ∣ d y = ∥ f ∥_{1} ∥ g ∥_{1} < \infty$ . Thus the exponential-weighted integrand is also absolutely integrable at every fixed frequency, with this same bound.

2.1F1F4F5F6step 1.1∎

Fubini now permits exchanging the integrals in the transform of the convolution representative. The inner $x$ integral is the translation transform from F5, giving $\hat{f * g} (ξ) = \int g (y) e^{- 2 π i y \cdot ξ} \hat{f} (ξ) d y = \hat{f} (ξ) \hat{g} (ξ)$ . The convolution is defined arbitrarily on its null exceptional set; F6 makes that choice irrelevant at every frequency. This proves the asserted equality, including when either input is zero.

Depends on

Dependency tree · 2 levels

Sources

Semyon Dyatlov, MIT 18.155 (2022)