Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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A point source as a distributional Radon limit

Example

Let η(u)=0 for u≤0 and η(u)=e−1/u for u>0, set ψ(x)=η(1−∣x∣2) on R2, let J=∫R2ψ(x) dx, and define ϕ=J−1ψ. For ε>0 put ϕε(x)=ε−2ϕ(x/ε). The functions ϕε are radial smooth unit-mass point-source approximations. For ρ(s)=∫Rϕ(sθ+tθ⊥) dt, which is independent of θ, their full line data satisfy Rϕε(θ,s)=ε−1ρ(s/ε). For each fixed θ∈S1 and every test function h∈Cc∞(R), the offset distributions converge to the point mass: lim⁡ε↓0∫RRϕε(θ,s)h(s) ds=h(0). Thus the point-source sinogram is δ0(s) in the offset variable, for every line normal.

Facts & Assumptions

Given: The functions η, ψ, ϕ, and ϕε in the Example, with J to be shown positive and finite.

[F1]

The Euclidean Radon transform on R²: Rf(θ,s)=∫Rf(sθ+tθ⊥) dt.

[F2]

A continuous real function on a compact Jordan measurable set is Riemann integrable over that set: continuous functions on compact Jordan sets are Riemann integrable.

[F3]

Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in Rm: the Riemann integral is monotone and additive across coordinate cuts, and linear on integrable functions.

[F4]

Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections: an integrable function on a product rectangle has equal product and iterated Riemann integrals when its sections are integrable.

[F5]

A compactly supported Riemann integrand admits the global change-of-variables formula from a diffeomorphism near the relevant compact preimage: compactly supported Riemann integrals obey the change-of-variables formula for an injective C1 diffeomorphism with its absolute Jacobian.

Verification

1.1givenalgebra

For u>0, repeated differentiation gives η(k)(u)=Pk(1/u)e−1/u for a polynomial Pk. The exponential-series bound ev≥vm+1/(m+1)! for v>0 shows that each such expression and its difference quotient tend to 0 as u↓0; induction on k proves that the zero extension is smooth. Hence ψ is smooth, nonnegative, vanishes for ∣x∣≥1, and is positive near x=0.

2.1givenF2F3F5step 1.1

By [F2], ψ is Riemann integrable on a bounding rectangle; 0≤ψ≤1 and its support lies in [−1,1]2, so [F3] gives J<∞. On [−1/4,1/4]2 one has 1−∣x∣2≥7/8, hence ψ≥e−8/7; monotonicity in [F3] gives J≥e−8/7/4>0. Thus ϕ is defined, radial, smooth, compactly supported, nonnegative, and has integral one. Applying [F5] to x=εy gives ∫R2ϕε(x) dx=ε−2ε2∫R2ϕ(y) dy=1 for every ε>0.

3.1F1F2F4F5step 2.1

The scaling t=εu in the line integral [F1] gives Rϕε(θ,s)=ε−1ρ(s/ε). Radiality gives ϕ(sθ+tθ⊥)=J−1η(1−s2−t2), so ρ is independent of θ and vanishes for ∣s∣>1. It is continuous: for s′→s, ∣ρ(s′)−ρ(s)∣≤2sup⁡∣t∣≤1∣ϕ(s′θ+tθ⊥)−ϕ(sθ+tθ⊥)∣⟶0, since ϕ is uniformly continuous. Thus ρ is compactly supported and Riemann integrable by [F2]. Riemann-Fubini [F4] and the orthogonal change of variables (s,t)↦sθ+tθ⊥, whose absolute Jacobian is 1, give ∫Rρ(s) ds=∫R2ϕ(x) dx=1.

4.1givenF2F5step 3.1algebra

By step 3.1, ρ is continuous and supported on [−1,1], so Rϕε(θ,⋅)h is compactly supported and continuous. If h is complex-valued, apply [F2] and the compact-support change of variables [F5] separately to its real and imaginary parts; thus with s=εu we get ∫RRϕε(θ,s)h(s) ds=∫−11ρ(u)h(εu) du. Uniform continuity of h on [−1,1] implies sup⁡∣u∣≤1∣h(εu)−h(0)∣→0. Since ρ≥0 and ∫ρ=1 by step 3.1, the integral tends to h(0).

5.1step 4.1∎

The functional h↦h(0) is the point mass δ0 by definition of the Dirac distribution. Step 4.1 proves the asserted distributional limit for every fixed normal θ.

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