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PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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Limited-angle data sample the visible Fourier wedge and zero frequency

Statement

Assume the Axiom of Countable Choice. Let Θ⊂S1 be a nonempty proper open antipodally symmetric set whose complement contains a nonempty open arc, and let WΘ={σθ:θ∈Θ, σ∈R∖{0}}. For f∈S(R2) and every measured normal θ∈Θ, the one-dimensional Fourier transform in the offset variable obeys RSf^(θ,σ)=f^(σθ) for all σ∈R. In particular, zero restricted data imply that f^ vanishes on WΘ. If in addition f∈Cc∞(R2), zero restricted data imply f=0 everywhere. For compactly supported inputs, the stated Countable Choice assumption identifies the Lebesgue line integrals here with the compact-support Riemann convention; the compact-support Fourier-slice and uniqueness results in that convention are proved without Countable Choice.

Facts & Assumptions

Given: The set Θ, WΘ, and a function f∈S(R2); in the compact-support conclusion also f∈Cc∞(R2) and RSf(θ,s)=0 for every θ∈Θ and s∈R.

[F1]

Visible covectors for limited-angle data: the measured Fourier wedge is WΘ={σθ:θ∈Θ, σ≠0}.

[F2]

Radon transform on Schwartz functions: RSf(θ,s)=∫Rf(sθ+tθ⊥) dt with Lebesgue measure; under Countable Choice, this agrees on compactly supported smooth inputs with the compact-support Riemann line integral.

[F3]

Fourier-slice identity for Schwartz inputs: for f∈S, RSf^(θ,σ)=f^(σθ).

[F4]

The Fourier-slice identity for the Euclidean Radon transform: for f∈Cc∞ and the IP-36 normalization, the same compact-support Fourier-slice identity holds.

[F5]

Compactly supported smooth functions are unique from any open set of Radon angles: compactly supported smooth functions with zero data on a nonempty open set of line normals vanish everywhere; the proof uses the Riemann Fourier-Laplace transform and no Countable Choice.

[A1]

The Axiom of Countable Choice (ACω): Countable Choice is used by [F2] to identify the compact-support line-integral conventions and by [F3] for the general Schwartz Fourier-slice formula. The compact-support Fourier-slice and uniqueness results [F4]–[F5] themselves are choice-free in the Riemann convention.

Proof

1.1givenF1F2F3A1

For each θ∈Θ and σ∈R, [F3] gives the Fourier-slice identity. If the restricted data vanish, their one-dimensional Fourier transforms vanish, so f^(σθ)=0. Every point of WΘ has this form, hence f^ vanishes on WΘ.

2.1F2F4F5given∎

Now assume f∈Cc∞ and its restricted data are zero. By [F2] and the stated Countable Choice assumption, those data use the compact-support Riemann line-integral convention. The compact-support Fourier-slice identity [F4] gives f^(σθ)=0 for every θ∈Θ and σ∈R, so the transform vanishes on the measured wedge. The compact-support uniqueness result [F5] then yields f=0 everywhere through its separate choice-free analytic-continuation proof.

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