LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The Radon transform maps Schwartz functions to Schwartz offset data

Statement

For every $f \in S (R^{2})$ and every $θ \in S^{1}$ , the defining line integral for $R_{S} f (θ, s)$ converges absolutely for every $s \in R$ , and $s \mapsto R_{S} f (θ, s)$ belongs to $S (R)$ . For each pair of nonnegative integers $k, N$ , its $k$ th derivative is bounded by $C_{k, N} (1 + ∣ s ∣)^{- N}$ with a finite constant $C_{k, N}$ independent of $s$ and $θ$ .

Facts & Assumptions

Given: $f \in S (R^{2})$ and $θ \in S^{1}$ .

[F1]

Schwartz space and its seminorms: every seminorm $p_{α β} (f) = \sup_{x} ∣ x^{α} \partial^{β} f (x) ∣$ is finite.

[F2]

Radon transform on Schwartz functions: $R_{S} f (θ, s) = \int_{R} f (s θ + t θ^{⊥}) d t$ .

[F3]

Differentiation under the integral sign: if each parameter slice is integrable, the integrand is differentiable in the parameter almost everywhere, its derivative is measurable, and the derivatives have one integrable majorant uniform in the parameter interval, then the integral is differentiable with derivative equal to the integral of the derivative.

[F4]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : the derivative of a composition of totally differentiable Euclidean maps is the composition of their derivatives.

[F5]

The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ : for a real differentiable function on an interval, its increment equals an interior derivative times the interval length; apply to real and imaginary parts for complex functions.

Proof

1.1givenF1F2algebra

Fix a derivative multi-index $β$ and an integer $N \geq 1$ , and choose an integer $m \geq N$ . Since $(1 + ∣ x ∣)^{N} \leq (1 + ∣ x_{1} ∣)^{m} (1 + ∣ x_{2} ∣)^{m}$ , expanding the right side and using the finite Schwartz seminorms in [F1] gives a finite $C_{β, N} (f)$ with $∣ \partial^{β} f (x) ∣ \leq C_{β, N} (f) (1 + ∣ x ∣)^{- N}$ . For $x = s θ + t θ^{⊥}$ , orthogonality gives $∣ x ∣ = (s^{2} + t^{2})^{1 / 2} \geq (∣ s ∣ + ∣ t ∣) / \sqrt{2}$ , so the line integrand is bounded by a constant times $(1 + ∣ s ∣ + ∣ t ∣)^{- N}$ . Taking $N > 1$ makes this majorant integrable in $t$ , proving absolute convergence.

2.1F1F2F4step 1.1algebra

Let $D_{θ} = θ \cdot \nabla$ . The chain rule [F4] gives $\partial_{s}^{k} f (s θ + t θ^{⊥}) = (D_{θ}^{k} f) (s θ + t θ^{⊥})$ . Expanding $D_{θ}^{k}$ gives finitely many partial derivatives with coefficients bounded uniformly for $θ \in S^{1}$ . For the target decay order $N$ , apply step 1.1 with spatial decay order $N + 2$ to each such derivative. Integrating $(1 + ∣ s ∣ + ∣ t ∣)^{- N - 2}$ in $t$ gives $∣ \int_{R} (D_{θ}^{k} f) (s θ + t θ^{⊥}) d t ∣ \leq C_{k, N}^{'} (1 + ∣ s ∣)^{- N - 1} \leq C_{k, N}^{'} (1 + ∣ s ∣)^{- N}$ , uniformly in $θ$ and $s$ .

3.1F3F4F5step 1.1step 2.1∎

Fix $s_{0} \in R$ and a bounded open interval $I$ containing $s_{0}$ . For each $k \geq 0$ , set $g_{k} (t, s) = (D_{θ}^{k} f) (s θ + t θ^{⊥})$ . Every slice $g_{k} (\cdot, s)$ is integrable by step 1.1; it is differentiable in $s$ with derivative $g_{k + 1}$ by [F4], and this derivative is measurable. Step 1.1 with spatial decay order $3$ bounds $∣ g_{k + 1} (t, s) ∣ \leq C (1 + ∣ t ∣)^{- 3}$ uniformly for $s \in I$ , an integrable majorant. Thus [F3] proves $\frac{d}{d s} \int g_{k} (t, s) d t = \int g_{k + 1} (t, s) d t$ . To see each resulting derivative is continuous, if $s = s^{'}$ the difference is zero; otherwise apply [F5] to real and imaginary parts on the segment joining $s$ and $s^{'}$ . The same majorant gives $∣ g_{k} (t, s) - g_{k} (t, s^{'}) ∣ \leq C ∣ s - s^{'} ∣ (1 + ∣ t ∣)^{- 3}$ , and integration proves local continuity. Repeating for all $k$ and using the estimates in step 2.1 shows the offset function is Schwartz.

Depends on

Used by

Cited to discharge well-definedness by Radon transform on Schwartz functions.

Dependency tree · two levels

24 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

Mikko Salo, On geometric inverse problems and microlocal analysis, §2.1 (standard reference, not scraped)
Venkataswaran P. Krishnan and Eric Todd Quinto, Microlocal Analysis in Tomography (standard reference, not scraped)