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Uniqueness of the L1 Fourier transform

Statement

Assume countable choice. If $f, g \in L^{1} (R^{n}; C)$ and $\hat{f} = \hat{g}$ , then $f = g$ almost everywhere. Equality of the transforms almost everywhere already suffices.

Facts & Assumptions

Given: $n \geq 1$ , the stated inputs and The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F1]

The transform is linear and continuous as a function of frequency (The L1 transform is bounded and uniformly continuous).

[F2]

A function and its integrable transform obey inversion almost everywhere (L1 Fourier inversion with an integrable transform).

Proof

1.1

Set $h = f - g$ . By linearity, $\hat{h} = 0$ . If equality was given only almost everywhere, continuity still implies this everywhere: a nonzero value would remain bounded away from zero on an open ball, which contains a positive-volume box and cannot be null. Thus the transform of h is the zero integrable function.

F1given

2.1

F2 applies to h, since both h and its zero transform are integrable. It gives $h (x) = \int 0 d ξ = 0$ almost everywhere, so $f = g$ as classes. Countable choice is inherited from inversion (and the Euclidean measure interface in the optional almost-everywhere hypothesis).

F2step 1.1∎

Depends on

Used by

Schwartz convolution and product laws Corollary
Uniqueness of finite Borel measures from their Fourier transforms Corollary
Normalized Hermite Fourier eigenfunctions Example

Dependency tree · two levels

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Sources

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