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Fixed-frequency inverse scattering and Born linearization

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Fixed-frequency far-field data on the small-norm volume branch

Definition

Assume Countable Choice. Let D⊂R3 be a bounded Lebesgue-measurable set with 0<M:=∣D∣<∞, and let k>0. Write S2:={v∈R3:∣v∣=1} and CD:=M diam⁡(D)/(4π). For q∈L∞(D;C), extended by zero off D, set Gk(z):=eik∣z∣/(4π∣z∣) for z≠0 and Gk(0):=0, and let Kq,kv(x):=∫DGk(x−y)q(y)v(y) dy,ρk(q):=∥Kq,k∥L2(D)→L2(D). The small-norm contrast class is UD,k:={q∈L∞(D;C):ρk(q)<1}. For d∈S2, put fk,d(x):=eikd⋅x on D. For q∈UD,k, let uq,k,d∈L2(D;C) denote the unique solution of u=fk,d+Kq,ku. For q∈UD,k, define the exact far-field data family by Sk(q)(θ,d):=14π∫De−ikθ⋅yq(y)uq,k,d(y) dy,(θ,d)∈S2×S2.

For h∈L∞(D;C), define its first Born data by Bkh(θ,d):=14π∫De−ikθ⋅yh(y)eikd⋅y dy=14πh0^ ⁣(k2π(θ−d)), where h0 is the zero extension of h to R3 and g^(ξ):=∫R3g(y)e−2πiy⋅ξ dy for g∈L1(R3;C). No uniqueness or stability outside UD,k is asserted.

Definition references

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Well-definedness of the small-norm fixed-frequency scattering map

Statement

Assume Countable Choice and the hypotheses and notation of the definition of fixed-frequency far-field data. For each q∈UD,k there is an ε>0 such that ∥q′−q∥∞<ε implies q′∈UD,k; in particular this metric neighborhood condition holds at zero. For every q∈UD,k and every d,θ∈S2, the solution uq,k,d exists uniquely in L2(D;C), the integrals defining Sk(q)(θ,d) and Bkq(θ,d) are absolutely convergent and independent of representatives, and Sk(q)(⋅,d) is continuous. If R0>sup⁡y∈D∣y∣ and r≥2R0, then sup⁡θ∈S2∣re−ikrVkuq,k,d(rθ)−Sk(q)(θ,d)∣≤2R0+∣k∣R024πr∥q∥∞M ∥uq,k,d∥2. Thus Sk(q)(⋅,d) is the uniform outgoing far-field amplitude in the stated convention.

Statement references

Facts & Assumptions

Given: Countable Choice; a bounded Lebesgue-measurable D⊂R3 with 0<M:=∣D∣<∞; k>0; a contrast q∈L∞(D;C); and directions d,θ∈S2.

[F1]

The volume model defines Kq,k on complex L2(D) by the outgoing kernel, treats inputs as equivalence classes, sets the diagonal representative to zero, and uses ∣q∣≤∥q∥∞ almost everywhere (The three-dimensional volume integral scattering model).

[F2]

For diam⁡(D)<∞, ∥Kq,k∥L2→L2≤CD∥q∥∞, where CD=Mdiam⁡(D)/(4π) (The volume integral operator is well defined and bounded on L2).

[F3]

The plane wave has modulus one, so its restriction to D is in L2(D) with norm M (exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0).

[F4]

If ρk(q)<1 and f∈L2(D;C), then u=f+Kq,ku has a unique solution in L2(D;C) (Born truncation error for a small-norm volume integral operator).

[F5]

For a,b∈L2(D;C), the pairing is representative-independent and ∣∫Dab‾∣≤∥a∥2∥b∥2 (The complex L2 pairing is well-defined and satisfies Cauchy–Schwarz).

[F6]

For g∈L1(R3;C), g^(ξ)=∫g(y)e−2πiy⋅ξ dy is absolutely convergent at every ξ and unchanged by null-set modifications (Fourier transform on complex L1 classes, The integral transform is representative independent).

[F7]

For a measurable representative with finite essential norm, ∣q∣≤∥q∥∞ almost everywhere (The essential supremum is attained as the least essential bound).

[F8]

For v∈L2(D;C), the far-field map is continuous in observation direction and its values are the uniform outgoing-wave limits of the exterior volume potential, with the displayed explicit remainder (The volume far-field map is bounded and is the uniform far-field limit).

[A1]

Countable Choice is assumed by the volume model and the Neumann theorem used here (The Axiom of Countable Choice (ACω)).

Proof

Given: The data in the Statement, with q∈UD,k.

1.1F1F2algebra

The integral operator depends linearly on q, so Kq,k−Kq′,k=Kq−q′,k for any q,q′∈L∞(D;C). By [F2], ∥Kq,k−Kq′,k∥≤CD∥q−q′∥∞. The operator-norm triangle inequality then gives ∣ρk(q)−ρk(q′)∣≤CD∥q−q′∥∞. Thus q↦ρk(q) is continuous, UD,k={ρk<1} is open, and 0∈UD,k because K0,k=0.

1.2A1F3F4given

By [F3], fk,d(x)=eikd⋅x belongs to L2(D;C) and has norm M. Since q∈UD,k, [F4] applies with f=fk,d and gives exactly one uq,k,d∈L2(D;C).

1.3F1F3F5F7given

By [F7], quq,k,d∈L2(D) with norm at most ∥q∥∞∥uq,k,d∥2, while [F3] puts eikθ⋅y in L2(D). Their pairing in [F5] is the stated integral, so it is absolutely convergent and independent of the representatives. Cauchy–Schwarz gives its absolute value at most ∥q∥∞M∥uq,k,d∥2.

1.4F5F6F7givenalgebra

By [F7], ∥q∥2≤M∥q∥∞; pairing ∣q∣ with 1D in [F5] then gives ∫D∣q(y)∣ dy≤M∥q∥∞. Hence q0∈L1(R3;C). By [F6], its transform is finite and representative-independent. Since e−ikθ⋅yeikd⋅y=e−2πiy⋅(k(θ−d)/(2π)), the Fourier and integral formulas for Bkq agree and are well-defined. This proves the assertion for both data families.

2.1F8step 1.2∎

Apply [F8] to v=uq,k,d, whose existence is established in step 1.2. It supplies continuity in observation direction and the stated uniform outgoing-wave limit estimate, so the defined data are the physical far-field amplitude for this volume branch.

Remarks

The proof establishes only the small-norm branch selected by ρk(q)<1. It does not import global scattering uniqueness from smoother coefficient models.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

The first Born data are the Fréchet linearization at zero contrast

Statement

Assume Countable Choice and the hypotheses and notation of the fixed-frequency far-field definition. For q∈UD,k put ρ=ρk(q). Uniformly over all incident and observation directions, sup⁡θ,d∈S2∣Sk(q)(θ,d)−Bkq(θ,d)∣≤M4π∥q∥∞ρ1−ρ. Whenever CD∥q∥∞<1, this implies sup⁡θ,d∈S2∣Sk(q)(θ,d)−Bkq(θ,d)∣≤MCD4π∥q∥∞21−CD∥q∥∞. For each fixed (θ,d), the scalar map q↦Sk(q)(θ,d) on the open domain UD,k, with L∞(D;C) and C viewed as real Banach spaces, is Fréchet differentiable at q=0 and has derivative h↦Bkh(θ,d). The remainder bound is uniform in (θ,d); it asserts no nonlinear uniqueness or stability result.

Facts & Assumptions

Given: Countable Choice; the data in the fixed-frequency definition; a contrast q∈UD,k; and directions θ,d∈S2.

[F1]

The definition fixes Kq,k, ρk(q), Sk(q), and Bkq on the stated quotient spaces and gives CD=Mdiam⁡(D)/(4π) (Fixed-frequency far-field data on the small-norm volume branch).

[F2]

The volume operator is linear in q and obeys ∥Kq,k∥L2→L2≤CD∥q∥∞ (The volume integral operator is well defined and bounded on L2).

[F3]

For ρ<1, the unique solution u=f+Kq,ku satisfies ∥u−∑j=0NKq,kjf∥2≤ρN+1∥f∥2/(1−ρ) (Born truncation error for a small-norm volume integral operator).

[F4]

The complex L2 pairing is representative-independent and satisfies Cauchy–Schwarz (The complex L2 pairing is well-defined and satisfies Cauchy–Schwarz).

[F5]

The Fourier transform convention is g^(ξ)=∫g(y)e−2πiy⋅ξ dy for g∈L1(R3;C) (Fourier transform on complex L1 classes).

[F6]

Under Countable Choice, L∞(D;C) is complete; realification leaves its norm and metric unchanged. The same completeness fact applies to C as a finite-dimensional complex L-infinity space (Complex Lp completeness and almost-everywhere subsequences, Realification preserves a complex norm and its metric).

[F7]

Between real Banach spaces, a Fréchet derivative is a bounded real-linear map whose remainder divided by the input norm tends to zero (Fréchet derivative between Banach spaces, Banach space).

[F9]

A measurable representative of the complex contrast satisfies ∣q∣≤∥q∥∞ almost everywhere (The essential supremum is attained as the least essential bound).

[F10]

The domain UD,k is open and contains zero (Well-definedness of the small-norm fixed-frequency scattering map).

[A1]

Countable Choice is assumed through [F3] and [F6] (The Axiom of Countable Choice (ACω)).

Proof

Given: The data above.

1.1F1F5F8F9given

For any h∈L∞(D;C), [F8] gives ∣e−ikθ⋅yeikd⋅y∣=1 and [F9] gives ∫D∣h∣≤M∥h∥∞. The integral definition therefore yields ∣Bkh(θ,d)∣≤M∥h∥∞/(4π). By [F5] and the Fourier formula in [F1], this integral is the stated Born functional. It is complex-linear in h, hence bounded and real-linear for every fixed pair of directions.

1.2F1F3F8given

Let f=fk,d. By the definition and [F8], ∣f∣=1 on D, so ∥f∥2=M. The case N=0 of [F3] gives ∥uq,k,d−f∥2≤ρM/(1−ρ).

2.1F1F3F4F9step 1.2

Subtracting the integral formulas in [F1] yields Sk(q)(θ,d)−Bkq(θ,d)=14π∫De−ikθ⋅yq(y)(uq,k,d(y)−f(y)) dy. By [F9], q(uq,k,d−f) has L2 norm at most ∥q∥∞∥uq,k,d−f∥2. The same unit-modulus calculation as step 1.2, with d replaced by θ, gives ∥eikθ⋅(⋅)∥2=M. By [F4], the integral's absolute value is at most ∥q∥∞M ∥uq,k,d−f∥2/(4π); then [F3] and step 1.2 give the first bound, uniformly in the directions.

3.1F2step 2.1algebra

By [F2], ρ≤CD∥q∥∞. When CD∥q∥∞<1, monotonicity of t↦t/(1−t) on [0,1) gives ρ/(1−ρ)≤CD∥q∥∞/(1−CD∥q∥∞). Substitution into step 2.1 proves the second bound.

4.1A1F1F6F7F10step 1.1step 3.1algebra∎

For nonzero q with CD∥q∥∞<1, divide the second bound by ∥q∥∞; the quotient is at most MCD4π∥q∥∞1−CD∥q∥∞, which tends to zero as ∥q∥∞→0. Since [F10] makes zero an interior point, the bounded real-linear map from step 1.1 satisfies the Fréchet remainder criterion [F7], with the Banach-space hypotheses in [F6]. Also Sk(0)=0 by its definition. Thus DSk(0)(θ,d)h=Bkh(θ,d) for each fixed pair, with the same remainder estimate uniformly over the full direction family.

Remarks

At zero contrast the first Born data are Bkh(θ,d)=14πh0^ ⁣(k2π(θ−d)). Thus each fixed-frequency direction pair samples one Fourier frequency on the corresponding Ewald surface. The result is local at q=0 and does not extend a derivative or inverse-uniqueness assertion to arbitrary contrasts.

Fourier-transfer geometry

For fixed k>0 and unit directions d,θ, the wavevectors kd and kθ lie on the radius-k sphere in their span. The arrow from kd to kθ is exactly k(θ−d); the Fourier variable in the library's 2π convention is that chord divided by 2π. The diagram is only a geometric description of the frequency label; the Fréchet estimate above is proved by the integral and norm bounds, not by the drawing.

kdkµk(µ¡d)0
AlgorithmAlgorithm: AI-adaptedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

One-pass filtered reconstruction from finite Born data

Definition

Inputs and admissible conditions. Fix the domain D, wavenumber k>0, and library Fourier convention from the fixed-frequency data definition. Give nonempty finite lists of incident directions d1,…,ds∈S2 and observation directions θ1,…,θr∈S2. Give a finite measurable partition D=⨆n=1NEn with ∣En∣>0, complex Born-data samples y=(yba)∈Crs, a discrepancy threshold δ≥0, and an iteration budget B∈N0:={0,1,2,…}. The sample frequency for pair (θb,da) is ξba:=k(θb−da)/(2π). The declared finite discrete Born Fourier matrix F∈Crs×N has entries F(b,a),n:=14π∫Ene−2πix⋅ξba dx. For a piecewise-constant contrast qc:=∑n=1Ncn1En, its sampled first Born data vector is exactly Fc. The input also declares a reduced singular-value decomposition Fc=∑j=1pσj⟨c,vj⟩uj(c∈CN), where p=rank⁡F, the positive singular values σj have orthonormal right vectors vj and orthonormal left vectors uj, and the empty sum is used when F=0. Give a scalar filter g:[0,∞)→C whose values at the σj are finite and with g(0)=0. The threshold filter gτ(σ)=1/σ for σ>τ and gτ(σ)=0 for 0≤σ≤τ is the truncated singular-value pseudoinverse at threshold τ≥0.

Instructions and outputs.

  1. If B=0, return no estimate, no residual, and status budget-exhausted, with zero filter applications.
  2. If B≥1, form the coefficient estimate c^:=∑j:σj>0g(σj)⟨y,uj⟩vj, where ⟨z,w⟩=∑ℓzℓwℓ‾ is linear in its first argument. Return the piecewise-constant reconstruction q^:=∑n=1Nc^n1En after exactly one filter application.
  3. Compute the finite data residual R:=∥Fc^−y∥2, where ∥z∥2=(∑ℓ∣zℓ∣2)1/2. Return status discrepancy when R≤δ, and residual-above-threshold when R>δ; in either case also return q^, R, and the application count 1.

This procedure only defines one filtered finite-dimensional Born reconstruction and a residual status. The residual is not an exact-range test for an infinite-dimensional compact forward operator and does not establish a nonlinear solution, identifiability, stability, convergence, or correctness of the reconstructed contrast.

4 · Examples, counterexamples and false statements

None yet.

Sources