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A nonzero Schwartz function invisible to a limited-angle Radon transform

Example

Assume the Axiom of Countable Choice. Let $Θ \subset S^{1}$ be a nonempty proper open antipodally symmetric set such that $S^{1} ∖ \overline{Θ}$ contains a nonempty open arc. Choose an open arc $J$ with $\overline{J} \subset S^{1} ∖ \overline{Θ}$ , put $V = {r ω : 1 < r < 2, ω \in J}$ , choose a nonzero smooth function $q \in C_{c}^{\infty} (V; C)$ , and define $f (x) = \int_{R^{2}} q (ξ) e^{2 π i x \cdot ξ} d ξ$ . For $θ = (θ_{1}, θ_{2})$ write $θ^{⊥} = (- θ_{2}, θ_{1})$ and, for a Schwartz function $g$ , define $R_{S} g (θ, s) = \int_{R} g (s θ + t θ^{⊥}) d t$ using Lebesgue measure $d t$ . Let $Y = L^{2} (Θ \times R, d θ d s)$ , where $d θ$ is arc-length measure and $d s$ is Lebesgue measure. Then $f$ is a nonzero Schwartz function in $L^{2} (R^{2})$ with Lebesgue measure, its Fourier transform is $q$ , and $R_{S} f (θ, s) = 0$ for every $θ \in Θ$ and $s \in R$ , so both $0$ and $f$ give datum $0$ in $Y$ without either truth being required to have compact support. For the two-truth class ${0, f}$ with target loss $∥ g - h ∥_{L^{2} (R^{2})}$ , every deterministic estimator $Φ : Y \to L^{2} (R^{2})$ satisfies $\max {∥ Φ (0) ∥_{2}, ∥ Φ (0) - f ∥_{2}} \geq ∥ f ∥_{2} / 2 > 0$ .

Example references

Measured Fourier wedge and blind directions

Fourier-slice identity for Schwartz inputs

Two-truth lower bound for deterministic estimators

Venkataswaran P. Krishnan and Eric Todd Quinto, Microlocal Analysis in Tomography, §1.3.2

Mikko Salo, On geometric inverse problems and microlocal analysis, §2.1

Facts & Assumptions

Given: The Axiom of Countable Choice, the normal set $Θ$ , the blind arc $J$ , the annular sector $V$ , and the nonzero smooth function $q$ supported in $V$ from the Example.

[F1]

Visible covectors for limited-angle data: $W_{Θ} = {σ θ : θ \in Θ, σ \neq 0}$ , while the directions in $S^{1} ∖ \overline{Θ}$ are unmeasured.

[F2]

A smooth bump between concentric Euclidean balls: every nonempty open ball contains a nonzero smooth compactly supported function, by translating and rescaling a concentric bump.

[F3]

Fourier transform acts continuously on Schwartz space: under Countable Choice the negative-sign, $2 π$ -normalized Fourier transform maps Schwartz functions to Schwartz functions.

[F4]

Fourier inversion on Schwartz space: under Countable Choice, $g (x) = \int \hat{g} (ξ) e^{2 π i x \cdot ξ} d ξ$ for every Schwartz function $g$ .

[F5]

Fourier transform on complex L1 classes: $\hat{g} (η) = \int g (x) e^{- 2 π i x \cdot η} d x$ for integrable $g$ .

[F6]

A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions: under Countable Choice, the reflection $x \mapsto - x$ preserves Lebesgue integrals because its absolute determinant is one.

[F7]

Schwartz derivatives are integrable: under Countable Choice every Schwartz function belongs to $L^{p}$ for each $1 \leq p \leq \infty$ .

[F8]

Fourier-slice identity for Schwartz inputs: under Countable Choice, $\hat{R_{S} g} (θ, σ) = \hat{g} (σ θ)$ for Schwartz $g$ .

[F9]

The Radon transform maps Schwartz functions to Schwartz offset data: for fixed $θ$ , $R_{S} g (θ, \cdot)$ is a Schwartz function.

[F10]

The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ : the $L^{2}$ quotient norm is a norm and gives the triangle inequality and separation.

[F11]

Kernel directions preclude unrestricted recovery: if two truths share one exact datum, every deterministic data-only estimator has maximum target loss at least half their target distance.

[F12]

Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric: a metric satisfies separation (M1), symmetry (M2), and the triangle inequality (M3); the metric induced by an $L^{2}$ norm here is $d (g, h) = ∥ g - h ∥_{2}$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is assumed by the Fourier transform, inversion, Lp integrability, Fourier-slice, and Lebesgue substitution suppliers [F3], [F4], [F6]–[F9].

Verification

technique · direct

1.1F2givenconstruct

The sector $V$ is a nonempty open subset of $R^{2}$ because $(r, ω) \mapsto r ω$ maps the open parameter set $(1, 2) \times J$ to an open polar-coordinate sector. It therefore contains a ball; [F2] supplies a nonzero $q \in C_{c}^{\infty} (V)$ . This establishes the required bump choice.

1.2F3F4F5F6F7A1givenalgebra

By the Fourier definition [F5], the function in the Example is $f (x) = \hat{q} (- x)$ . Since $q$ is smooth with compact support, every Schwartz seminorm of $q$ is bounded, so $q \in S$ . By [F3] its transform is Schwartz, and reflection preserves each Schwartz seminorm up to sign, hence $f$ is Schwartz; [F7] gives $f \in L^{2} \cap L^{1}$ . The reflection substitution [F6] and inversion [F4] give $\hat{f} (η) = \int \hat{q} (- x) e^{- 2 π i x \cdot η} d x = \int \hat{q} (y) e^{2 π i y \cdot η} d y = q (η)$ . Since $q \neq 0$ , this identity implies $f$ is nonzero as an $L^{1}$ and $L^{2}$ class.

1.3A1F1F4F8F9given

For $θ \in Θ$ and $σ \neq 0$ , the point $σ θ$ is outside the support of $q$ because every frequency in $V$ has direction in $J$ and $\overline{J}$ is disjoint from $\overline{Θ}$ ; also $q (0) = 0$ since $V$ avoids the origin. Thus [F8] gives $\hat{R_{S} f} (θ, σ) = q (σ θ) = 0$ for all real $σ$ . By [F9], the offset function is Schwartz, so Fourier inversion [F4] in one dimension gives $R_{S} f (θ, s) = 0$ for every $s$ .

2.1F10F11F12step 1.2step 1.3∎

Let the parameter and target metric spaces both be $L^{2} (R^{2})$ with distance $d (g, h) = ∥ g - h ∥_{2}$ , let the admissible truth class be $U = {0, f}$ , and let the target map $q : U \to L^{2} (R^{2})$ be inclusion. Give the data space $Y = L^{2} (Θ \times R)$ its norm metric and define $F : U \to Y$ by the restricted line integrals. The normed-space result [F10] and the metric axioms [F12] give the parameter, target, and data metrics used here. Theorem [F11] applies directly to this truth class and target map. Step 1.3 gives $F (0) = F (f) = 0$ , while step 1.2 gives $d (q (0), q (f)) = ∥ f ∥_{2} > 0$ . Apply [F11] with shared datum $0$ and radius $0$ to any deterministic estimator $Φ : Y \to L^{2} (R^{2})$ ; its maximum target loss is at least $∥ f ∥_{2} / 2$ . This proves the asserted lower bound on the stated two-truth class.

Depends on

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Sources

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