Pipeline-generated

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.

Scalar Calderón maps and conditional uniqueness

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

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

Injective geodesic ray transform on a transversal manifold

Definition

Let $(M_{0}, g_{0})$ be a compact connected smooth Riemannian manifold with smooth boundary. A geodesic segment in $M_{0}$ satisfies the geodesic equation for the Levi-Civita connection of $g_{0}$ on $(0, L)$ and extends smoothly to its endpoints, where derivatives are interpreted one-sided. A nontangential maximal geodesic segment is a unit-speed geodesic $γ : [0, L] \to M_{0}$ , with $L > 0$ , whose initial velocity points strictly inward at $\partial M_{0}$ , whose terminal velocity points strictly outward at $\partial M_{0}$ , and whose points for $0 < t < L$ lie in $M_{0}^{\circ}$ . For $f \in C_{c}^{\infty} (M_{0}^{\circ}, R)$ define its geodesic ray transform by $I_{g_{0}} f (γ) = \int_{0}^{L} f (γ (t)) d t$ . The unweighted scalar ray transform is called injective on compactly supported smooth functions if $I_{g_{0}} f (γ) = 0$ for every such segment implies $f = 0$ . For complex-valued $f$ , injectivity means injectivity of its real and imaginary parts separately.

Definition references

Riemannian metric and unit-speed convention

Boundary and interior of a smooth manifold with boundary

Geodesics of a connection

Levi-Civita connection of the metric

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

Sobolev spaces on a compact Riemannian domain with boundary

Definition

Assume Countable Choice.

Let $n = \dim M \geq 0$ and let $(M, g)$ be a compact smooth Riemannian manifold with smooth boundary, with no orientability assumption. Fix a finite smooth atlas of interior and boundary charts and a smooth partition of unity subordinate to it. If $n = 0$ , define every $H^{k} (M)$ to be the space of scalar-valued functions on the finite set $M$ with the counting-measure $ℓ^{2}$ norm. For $n \geq 1$ and an integer $k \geq 0$ , define $H^{k} (M)$ as the completion of $C^{\infty} (M)$ in the sum of the Euclidean $W^{k, 2}$ norms of the localized coordinate representatives; derivatives in a boundary chart are taken on the corresponding half-space. Define $H_{0}^{1} (M)$ as the closure of $C_{c}^{\infty} (M^{\circ})$ in $H^{1} (M)$ . If $\partial M = \emptyset$ , set $H^{1 / 2} (\partial M) = {0}$ . If $n = 1$ , set $H^{1 / 2} (\partial M) = K^{\partial M}$ with its counting-measure $ℓ^{2}$ norm, where $K$ is the scalar field. For $n \geq 2$ , define $H^{1 / 2} (\partial M)$ by the fixed finite boundary atlas and partition of unity, using the Euclidean fractional Slobodeckij space $W^{1 / 2, 2} (R^{n - 1})$ for each localized representative. For real scalars, define $H^{- 1 / 2} (\partial M)$ as the continuous linear dual of $H^{1 / 2} (\partial M)$ . For complex scalars, define it as the continuous conjugate-linear dual: its pairing with $h \in H^{1 / 2} (\partial M)$ is linear in the dual element and conjugate-linear in $h$ , using the Riemannian boundary density. These are the fixed-atlas norms used on this page; no atlas-independence assertion is part of this definition.

Definition references

Riemannian manifold and volume density

Smooth charts on a manifold with boundary

Euclidean integer-order Sobolev norms

Euclidean fractional Sobolev spaces

Smooth partition of unity

Existence of finite subordinate partitions

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

Transversally anisotropic product domain

Definition

A transversally anisotropic product domain of dimension $n \geq 3$ is a compact connected smooth $n$ -dimensional Riemannian domain $(M, g)$ with smooth boundary, together with a compact connected smooth $(n - 1)$ -dimensional Riemannian manifold with boundary $(M_{0}, g_{0})$ and an isometric embedding of $(M, g)$ as a smooth domain compactly contained in $(R \times M_{0}^{\circ}, d x_{1}^{2} \oplus g_{0})$ . The coordinate $x_{1}$ on $R$ is the distinguished Euclidean coordinate; $M_{0}$ is the transversal manifold. All Laplacians use $Δ_{g} = {div}_{g} \nabla_{g}$ . Neither $M$ nor $M_{0}$ is assumed orientable. A positive smooth conformal factor is not included in this definition; claims using a conformally related metric must state and prove the required conformal transformation separately.

Definition references

Riemannian metrics and manifolds

Smooth domains and boundary charts

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

Euclidean complex geometrical optics solvability with an H1 correction

Statement

Assume Countable Choice. Let $n \geq 3$ , let $Ω \subset R^{n}$ be a bounded smooth domain contained in a ball $B (0, R)$ , and let $q \in L^{\infty} (Ω, C)$ . There are constants $C > 0$ and $τ_{0} \geq 1$ , depending only on $n$ , $R$ , and $∥ q ∥_{L^{\infty}}$ , such that for every $ζ \in C^{n}$ satisfying $ζ \cdot ζ = 0$ and $∣ ζ ∣ \geq τ_{0}$ , there is $r_{ζ} \in H^{1} (Ω, C)$ for which $u_{ζ} (x) = e^{i ζ \cdot x} (1 + r_{ζ} (x))$ solves $(- Δ + q) u_{ζ} = 0$ distributionally in $Ω$ , and $∥ r_{ζ} ∥_{L^{2} (Ω)} \leq C ∣ ζ ∣^{- 1} ∥ q ∥_{L^{2} (Ω)}, ∥ \nabla r_{ζ} ∥_{L^{2} (Ω)} \leq C ∥ q ∥_{L^{2} (Ω)} .$ The dot product in the null-phase condition is complex bilinear, and $∣ ζ ∣ = (\sum_{j = 1}^{n} ∣ ζ_{j} ∣^{2})^{1 / 2}$ .

Statement references

Euclidean Sobolev spaces and weak derivatives

Gradient convention

Euclidean Laplacian convention

Fourier basis and Parseval on the torus

Facts & Assumptions

Given: Countable Choice; a bounded smooth domain $Ω \subset R^{n}$ ; $q \in L^{\infty} (Ω, C)$ ; and a null vector $ζ$ above the stated size threshold.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is assumed by the Fourier-basis supplier [F1].

[F1]

The Fourier basis and Parseval's identity on the finite torus: on a normalized $n$ -torus, the coordinate characters form a complete orthonormal basis of $L^{2}$ and satisfy Parseval's identity. Rescaling the fundamental cube and multiplying by a unit-modulus exponential preserve completeness and orthonormality.

[F2]

Integer-order Sobolev spaces and their norms, Riemannian gradient, and Laplace–Beltrami operator as the trace of the Hessian: $H^{1}$ consists of $L^{2}$ functions with $L^{2}$ weak first derivatives, $D = - i \nabla$ has the same norm as $\nabla$ , and $D^{2} = - Δ$ .

Proof

Given: the data in the Statement.

1.1A1F1F2givenalgebraconstruct

Write $ζ = a + i b$ . The null condition gives $∣ a ∣ = ∣ b ∣ = : s = ∣ ζ ∣ / \sqrt{2}$ and $a \cdot b = 0$ . Choose an orthogonal map $U$ with $U a = s e_{1}$ and $U b = s e_{2}$ . Work temporarily with $U Ω$ , $q \circ U^{- 1}$ , and the phase $U ζ$ , relabeling them $Ω$ , $q$ , and $ζ$ for the construction; the Euclidean Laplacian, Lebesgue norms, and null phase are invariant under this change, and the constructed remainder is pulled back by $U$ at the end. As $Ω \subset B (0, R)$ , choose $L > \max {1, (R + 1) / π}$ ; then $Ω \subset Q_{L} = (- π L, π L)^{n}$ . Put $η_{k} = (k + \frac{1}{2} e_{2}) / L$ and $w_{k} (x) = (2 π L)^{- n / 2} e^{i η_{k} \cdot x}$ for $k \in Z^{n}$ . By [F1], these form an orthonormal basis on $Q_{L}$ : they are the rescaled standard Fourier characters multiplied by the unitary factor $e^{i x_{2} / (2 L)}$ . For $A_{ζ} = D^{2} + 2 ζ \cdot D$ , one has $A_{ζ} w_{k} = p_{k} w_{k}$ , where $p_{k} = ∣ η_{k} ∣^{2} + 2 ζ \cdot η_{k}$ and $∣ Im p_{k} ∣ = 2 s ∣ k_{2} + \frac{1}{2} ∣ / L \geq s / L$ . If $∣ η_{k} ∣ \leq 4 s$ , then $∣ p_{k} ∣ \geq s / L$ and $∣ η_{k} / p_{k} ∣ \leq 4 L$ . If $∣ η_{k} ∣ > 4 s$ , then $∣ Re p_{k} ∣ \geq ∣ η_{k} ∣^{2} - 2 s ∣ η_{k} ∣ \geq ∣ η_{k} ∣^{2} / 2$ , so $∣ p_{k} ∣^{- 1} < 1 / (8 s^{2}) \leq L / s$ and $∣ η_{k} / p_{k} ∣ < 1 / (2 s)$ . Thus $\sup_{k} ∣ p_{k} ∣^{- 1} \leq L / s$ and $\sup_{k} ∣ η_{k} / p_{k} ∣ \leq C_{L}$ . Define $G_{ζ} f = \sum_{k} p_{k}^{- 1} (f, w_{k}) w_{k}$ . Parseval [F1] gives $∥ G_{ζ} f ∥_{2} \leq (L / s) ∥ f ∥_{2}$ and $∥ \nabla G_{ζ} f ∥_{2} \leq C_{L} ∥ f ∥_{2}$ . For finite Fourier sums $f_{N}$ , the corresponding partial sums $G_{ζ} f_{N}$ satisfy $A_{ζ} G_{ζ} f_{N} = f_{N}$ termwise. Parseval and the multiplier bounds give $f_{N} \to f$ in $L^{2} (Q_{L})$ and $G_{ζ} f_{N} \to G_{ζ} f$ in $H^{1} (Q_{L})$ . Since $A_{ζ} : H^{1} (Q_{L}) \to H_{l o c}^{- 1} (Q_{L})$ continuously, passage to distributions yields $A_{ζ} G_{ζ} f = f$ .

2.1F1givenstep 1.1algebraconstruct

Extend $q$ and $F : = - q$ by zero from $Ω$ to $Q_{L}$ . Multiplication by $q$ is bounded on $L^{2} (Q_{L})$ , and $∥ q G_{ζ} ∥_{L^{2} \to L^{2}} \leq ∥ q ∥_{\infty} L / s .$ Choose $τ_{0}$ large enough that $s \geq 2 L ∥ q ∥_{\infty}$ and $s \geq 1$ whenever $∣ ζ ∣ \geq τ_{0}$ . Then the series $\tilde{F} = \sum_{m = 0}^{\infty} (- q G_{ζ})^{m} F$ converges in $L^{2} (Q_{L})$ , satisfies $(I + q G_{ζ}) \tilde{F} = F$ , and has norm at most $2 ∥ q ∥_{L^{2} (Ω)}$ . Set $r_{ζ} = G_{ζ} \tilde{F}$ and restrict it to $Ω$ . Step 1.1 gives $A_{ζ} r_{ζ} + q r_{ζ} = F = - q$ on $Ω$ , as well as $∥ r_{ζ} ∥_{2} \leq (2 L / s) ∥ q ∥_{2}$ and $∥ \nabla r_{ζ} ∥_{2} \leq 2 C_{L} ∥ q ∥_{2}$ .

3.1F2givenstep 2.1algebra∎

Since $D = - i \nabla$ , conjugation gives $e^{- i ζ \cdot x} (- Δ + q) e^{i ζ \cdot x} (1 + r_{ζ}) = (D^{2} + 2 ζ \cdot D + ζ \cdot ζ + q) (1 + r_{ζ}) .$ The null condition removes $ζ \cdot ζ$ , the constant amplitude has zero derivatives, and step 2.1 makes the remaining expression $q + A_{ζ} r_{ζ} + q r_{ζ} = 0$ . Multiplication by the smooth exponential preserves $H^{1} (Ω)$ for this fixed $ζ$ , so $u_{ζ} \in H^{1} (Ω)$ and solves the equation distributionally. The estimates in step 2.1 have $s = ∣ ζ ∣ / \sqrt{2}$ , giving the asserted bounds after changing the constant.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Weak solvability of the positive scalar conductivity equation

Statement

Assume the Axiom of Choice and Countable Choice. Let $(M, g)$ be a compact connected smooth Riemannian domain with nonempty smooth boundary and no orientability assumption, and let $γ \in C^{\infty} (M, R)$ satisfy $γ > 0$ . There is a constant $C_{P} > 0$ such that $∥ w ∥_{L^{2} (M)} \leq C_{P} ∥ \nabla_{g} w ∥_{L^{2} (M)}$ for every $w \in H_{0}^{1} (M)$ . For every $h \in L^{2} (M)$ and $f \in H^{1 / 2} (\partial M)$ there is a unique $v \in H^{1} (M)$ with trace $f$ satisfying $\int_{M} γ g (\nabla_{g} v, \nabla_{g} \overline{z}) d V_{g} = - \int_{M} h \overline{z} d V_{g}$ for every $z \in H_{0}^{1} (M)$ . If $h$ and $f$ are smooth, this weak solution is smooth up to the boundary.

Statement references

Fixed-atlas Sobolev and trace spaces

Facts & Assumptions

[A1]

The Axiom of Choice: every family of nonempty sets has a choice function; this is assumed by the Rellich, trace-right-inverse, higher-order regularity, and Sobolev embedding suppliers.

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every countable family of nonempty sets has a choice function; this is assumed by the Lax--Milgram, partition, and collar suppliers.

Given: AC and Countable Choice; the compact connected smooth Riemannian domain and positive smooth conductivity in the statement; and $f \in H^{1 / 2} (\partial M)$ .

[F1]

Sobolev spaces on a compact Riemannian domain with boundary: $H^{1} (M)$ is defined by finitely many chartwise $W^{1, 2}$ norms, and $H_{0}^{1} (M)$ is the $H^{1}$ closure of $C_{c}^{\infty} (M^{\circ})$ .

[F2]

Compactness of $W_{0}^{1, p} (Ω) ↪ L^{p} (Ω)$ on bounded open sets: on a bounded Euclidean open set, $W_{0}^{1, 2}$ is compactly embedded in $L^{2}$ .

[F3]

The Lax--Milgram theorem: a bounded coercive sesquilinear form on a Hilbert space has a unique weak solution.

[F4]

The $L^{p}$ trace operator on a bounded $C^{1}$ domain: on a bounded $C^{1}$ Euclidean domain, the trace is a bounded operator from $H^{1}$ to $L^{2}$ and agrees with restriction for smooth functions; therefore every class in the closure $H_{0}^{1}$ has zero trace.

[F5]

A bounded right inverse of the trace, supported in a prescribed collar: on a bounded smooth Euclidean domain, every $H^{1 / 2}$ boundary datum has an $H^{1}$ extension; after localization this supplies boundary-chart extensions on $M$ .

[F6]

Higher-order boundary regularity for Dirichlet problems: a zero-trace weak solution of a smooth uniformly elliptic divergence-form equation on a bounded smooth Euclidean domain gains two Sobolev derivatives when its right-hand side has the corresponding Sobolev regularity.

[F7]

Higher-order Sobolev embedding: membership in every chartwise integer Sobolev space gives a smooth representative up to the boundary.

[F8]

Collar neighborhood theorem: smooth boundary data extend smoothly into a collar by constant normal extension and a cutoff.

[F9]

The sharp trace theorem: boundedness and range in the fractional space: in each bounded smooth boundary chart, the Euclidean $H^{1}$ trace is bounded into the local $H^{1 / 2}$ space. Applying these local traces to the fixed-atlas partition defines a bounded global trace $T_{M} : H^{1} (M) \to H^{1 / 2} (\partial M)$ that agrees with restriction on smooth functions.

[F10]

The kernel of the trace is the closure of the test functions and The trace commutes with smooth cutoffs and is chart local: localizing a zero-trace $H^{1}$ function by the fixed partition gives zero-trace Euclidean chart representatives; the Euclidean trace-kernel theorem and finite patching imply $\ker T_{M} = H_{0}^{1} (M)$ .

Proof

Given: the data in the statement, including $h \in L^{2} (M)$ .

1.1A1F1F2F4F9givencontradiction

In dimension one, endpoint evaluation is bounded on $H^{1}$ by the fundamental theorem of calculus and Cauchy--Schwarz. Any endpoint data have a smooth collar lift. Its kernel is $H_{0}^{1}$ : for a zero-endpoint function, $∣ u (t) ∣^{2} \leq t \int_{0}^{t} ∣ u^{'} (r) ∣^{2} d r$ makes the derivative error of a cutoff on $[0, 2 δ]$ tend to zero in $L^{2}$ as $δ \to 0$ ; cut off at both endpoints and mollify. Thus the trace, lift and kernel conclusions of [F5], [F9] and [F10] hold also in this dimension, without invoking their higher-dimensional Euclidean suppliers. There is a constant $C_{P}$ such that $∥ w ∥_{L^{2} (M)} \leq C_{P} ∥ \nabla w ∥_{L^{2} (M)}$ for $w \in H_{0}^{1} (M)$ . Otherwise choose $w_{j} \in H_{0}^{1} (M)$ with $∥ w_{j} ∥_{L^{2}} = 1$ and $∥ \nabla w_{j} ∥_{L^{2}} \to 0$ . A finite chart partition and [F2] give an $L^{2}$ -convergent subsequence. On compact $M$ , the fixed finite chartwise $H^{1}$ norm is equivalent to $∥ \cdot ∥_{L^{2}} + ∥ \nabla_{g} \cdot ∥_{L^{2}}$ by bounded metric coefficients, so this subsequence is Cauchy in $H^{1}$ and converges to some $w \in H_{0}^{1} (M)$ with $∥ w ∥_{L^{2}} = 1$ and weak gradient zero. Local convolution shows that a function with zero weak gradient is constant on each connected coordinate ball; overlapping balls and connectedness make $w$ constant on $M^{\circ}$ . Its trace is zero because $w \in H_{0}^{1} (M)$ , while [F9] makes the trace of a nonzero constant equal that same nonzero boundary value. This contradiction proves the inequality.

2.1A2F1F3step 1.1given

Compactness of $M$ and positivity of $γ$ give $γ_{*} : = \min_{M} γ > 0$ . Thus $a_{γ} (w, z) = \int_{M} γ g (\nabla w, \nabla \overset{ˉ}{z}) d V_{g}$ is bounded on $H_{0}^{1} (M)$ and $a_{γ} (w, w) \geq c ∥ w ∥_{H^{1} (M)}^{2}$ by step 1.1 and $γ_{*}$ . By [F3], for any bounded conjugate-linear functional $G$ on $H_{0}^{1} (M)$ there is a unique $w \in H_{0}^{1} (M)$ with $a_{γ} (w, z) = G (z)$ for all $z$ .

3.1A1A2F1F3F4F5F10step 1.1step 2.1algebra

Flatten each boundary chart, apply the local half-space trace lift, and patch the finitely many chart lifts with a smooth partition. The resulting $E f \in H^{1} (M)$ has trace $f$ and norm bounded by $C ∥ f ∥_{H^{1 / 2} (\partial M)}$ by [F5]. For $h \in L^{2} (M)$ take $G (z) = - \int_{M} h \overline{z} d V_{g} - a_{γ} (E f, z)$ in step 2.1 and set $v = E f + w$ . Then $v$ has trace $f$ and satisfies the weak equation. If $v_{1}, v_{2}$ are two such solutions, their difference has zero global trace, hence lies in $H_{0}^{1} (M)$ by [F10]. Its energy is zero; since $γ \geq γ_{*} > 0$ , its gradient vanishes, and step 1.1 gives $v_{1} = v_{2}$ .

4.1A1A2F1F6F7F8F10step 3.1inductionalgebra∎

Suppose $h$ and $f$ are smooth. By [F8], extend $f$ smoothly from the boundary into a collar and multiply by a cutoff to obtain a smooth extension $E f$ on $M$ . Then $w = v - E f$ has zero trace, so $w \in H_{0}^{1} (M)$ by [F10], and solves ${div}_{g} (γ \nabla_{g} w) = h - {div}_{g} (γ \nabla_{g} E f)$ , whose right-hand side is smooth. For dimension one, the coordinate equation $(a w^{'})^{'} = H$ with smooth $a > 0$ and smooth $H$ first gives $a w^{'} \in H^{1}$ and then, by iteration, $w \in H^{m}$ for every $m$ ; the one-dimensional fundamental theorem gives smoothness up to the endpoints. For dimension at least two, in each boundary chart, flatten the boundary and write the principal matrix as $∣ g ∣^{1 / 2} γ g^{i j}$ ; it is smooth and uniformly elliptic because $g$ is positive definite and $γ$ has a positive minimum. For the base case, localize $w$ by a cutoff; its equation has an $L^{2}$ right-hand side since $w \in H^{1}$ , so [F6] gives local $H^{2}$ regularity. If $w \in H^{m + 1}$ , each cutoff commutator contains derivatives of $w$ of order at most $m + 1$ with smooth bounded coefficients, hence lies in $H^{m}$ ; the localized right-hand side is therefore in $H^{m}$ , and [F6] gives $H^{m + 2}$ regularity. Induction over $m$ and a finite chart cover give $w \in H^{m} (M)$ for every integer $m$ . By [F7], $w$ and hence $v = E f + w$ are smooth up to the boundary.

LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

CTA Gaussian-beam quasimodes and geodesic concentration

Statement

Let $(M_{0}, g_{0})$ be a compact smooth Riemannian manifold with smooth boundary and let $γ : [0, L] \to M_{0}$ be a unit-speed nontangential geodesic segment. For every real $λ$ and every $K > 0$ , there is a family $v_{τ + i λ} \in C^{\infty} (M_{0})$ , $τ \geq 1$ , such that $∥ (- Δ_{g_{0}} - (τ + i λ)^{2}) v_{τ + i λ} ∥_{L^{2} (M_{0})} = O (τ^{- K}), ∥ v_{τ + i λ} ∥_{L^{2} (M_{0})} = O (1),$ and for every $ψ \in C (M_{0})$ , $\lim_{τ \to \infty} \int_{M_{0}} ∣ v_{τ + i λ} ∣^{2} ψ d V_{g_{0}} = \int_{0}^{L} e^{- 2 λ t} ψ (γ (t)) d t .$ The statement permits transversal manifolds without an orientability assumption and makes no boundary condition assertion for the quasimodes.

Statement references

Transversal manifold

Nontangential geodesic segment

Facts & Assumptions

Given: the compact Riemannian transversal, a nontangential segment, a real attenuation $λ$ , and $K > 0$ .

[F1]

Riemannian metric and riemannian manifold defines the smooth positive-definite metric; Riemannian volume density defines its orientation-independent density; Laplace–Beltrami operator as the trace of the Hessian fixes $Δ_{g} = {div}_{g} \nabla_{g}$ . The Fermi-coordinate construction is supplied by the external Lemma 3.5 in [F3], not by the metric definition.

[F2]

Local existence, uniqueness, and smooth dependence for manifold integral curves: the geodesic ODE has local solutions and smooth dependence on initial data, so parallel frames and exponential-coordinate maps vary smoothly along an embedded geodesic subsegment.

[F3]

External input: Dos Santos Ferreira--Kurylev--Lassas--Salo, Proposition 3.1, PDF p. 14, supplies exactly the residual, norm, and attenuated ray limit in the statement on an oriented compact transversal of dimension at least two. Its Lemma 3.5, PDF pp. 15--16, supplies the Fermi coordinates used below. Its finite Taylor-jet construction and distinct-branch cross-term estimate (equation (3.6), PDF pp. 21--22) are imported, including overlap matching at self-intersections. For this adjudication the proposition, Lemmas 3.3--3.5, and the complete construction and cross-term argument on PDF pp. 14--22 were read. They remain external inputs, not a newly reconstructed local proof; earlier reading claims are not adopted as this adjudicator's work.

[F4]

The construction for each fixed $K$ uses finitely many charts, finite Taylor orders, and linear transport and matrix ODEs with fixed initial data. Increasing the finite order suffices for that $K$ ; no simultaneous choice of quasimodes for unrelated $K$ or $λ$ is required. The parameter $τ$ then enters the resulting explicit formula. Thus the use here adds no choice axiom to the statement.

Proof

Given: the data in the statement.

1.1F1F2F3given

Embed the segment in a collar extension across $\partial M_{0}$ and cover it by finitely many coordinate neighborhoods. On an embedded subsegment, parallel transport an orthonormal normal frame and set $F (t, y) = \exp_{γ (t)} (\sum_{α} y_{α} E_{α} (t))$ . The inverse function theorem gives local Fermi coordinates with $g_{j k} (t, 0) = δ_{j k}$ and $\partial_{i} g_{j k} (t, 0) = 0$ . This construction is along an interval and uses no orientation of $M_{0}$ .

1.2F1F2F3givenalgebra

In one such chart, seek $v_{s} = e^{i s Θ} a$ , $s = τ + i λ$ , with $Θ (t, 0) = t$ and $Im Θ (t, y) \geq c ∣ y ∣^{2}$ . Expand $Θ$ into homogeneous transverse Taylor polynomials. The quadratic polynomial is $\frac{1}{2} H (t) y \cdot y$ ; substituting into $g (d Θ, d Θ) = 1$ gives the matrix Riccati equation $H^{'} + H^{2} = F (t)$ , where $F$ is determined by the second transverse Taylor coefficient of the inverse metric. An initial complex symmetric $H (t_{0})$ with positive-definite imaginary part yields a smooth solution with positive imaginary part on a short subsegment. Each higher homogeneous phase term then solves a linear first-order ODE along $t$ , because its newest coefficient occurs linearly in the eikonal equation.

2.1F1F3step 1.2construct

Write $m = \dim M_{0}$ , so the transverse dimension is $m - 1$ . Substituting $a = τ^{(m - 1) / 4} \sum_{j = 0}^{N} s^{- j} a_{- j}$ into $(- Δ_{g_{0}} - s^{2}) e^{i s Θ} a$ cancels successive powers of $s$ by linear transport ODEs for the transverse Taylor coefficients of each $a_{- j}$ . The leading coefficient solves $a_{0} (t, 0) = c_{0} \exp (- \frac{1}{2} \int_{t_{0}}^{t} tr H (r) d r)$ . A finite Taylor realization to the order required for the fixed $K$ and a transverse cutoff produce a smooth local beam; Gaussian integration gives bounded $L^{2}$ norm and a residual that tends to zero faster than any prescribed power after increasing the Taylor order.

3.1F3F4step 1.1step 2.1

On an oriented transversal of dimension at least two, apply [F3] to complete the finite overlap matching and eliminate all distinct-branch cross terms. This gives the stated global family and limit; steps 1.1--2.1 describe its local ingredients, without claiming a new proof of the imported global estimates.

4.1F1F3F4step 3.1algebra

For a nonorientable transversal let $π : {\tilde{M}}_{0} \to M_{0}$ be its orientation double cover, with deck involution $σ$ , and lift $γ$ to $\tilde{γ}$ . Let $w_{s}$ be the oriented beam from [F3] along that lift. Define the descended function by $π^{*} v_{s} = w_{s} + σ^{*} w_{s}$ . The residual and norm bounds follow by a finite sum and the covering integral formula. The two lifted segments have no common tangent branch: equality of position and velocity at different times would make the geodesic periodic, contrary to its nontangential boundary exit; opposite velocities at a repeated point would force a zero midpoint velocity by geodesic uniqueness. There are only finitely many crossing branches, by compactness and local uniqueness. The distinct-branch estimate in [F3], applied to these two lifts, therefore gives a zero mixed-term limit. For every continuous $ψ$ the covering formula is $\int_{M_{0}} ∣ v_{s} ∣^{2} ψ d V = \int_{{\tilde{M}}_{0}} ∣ w_{s} ∣^{2} (ψ \circ π) d V + Re \int_{{\tilde{M}}_{0}} w_{s} \overline{σ^{*} w_{s}} (ψ \circ π) d V .$ The first term has precisely the required ray limit and the second tends to zero. In particular there is no extra factor two in the normalization. The same argument works if the cover is disconnected.

5.1F1givenalgebra∎

If the transversal has dimension one, the component containing the segment is an interval with arclength coordinate $t$ beginning at its incoming endpoint. Set $v_{s} = e^{i s t}$ on this component and zero on all other components. The residual is zero and the asserted integral formula is exact. In dimension zero the given unit-speed segment cannot exist. This covers the dimensions not needed in the CTA application as well.

LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Limiting Carleman estimate and right solvability on a product domain

Statement

Assume the Axiom of Choice and Countable Choice. Let $(M, g)$ be a transversally anisotropic product domain, let $q \in C^{\infty} (\overline{M}, R)$ , and let $φ = σ x_{1}$ for $σ \in {- 1, 1}$ . There exist constants $h_{0} > 0$ and $C > 0$ , depending on $(M, g)$ and $q$ and valid for both signs, with the following properties. For $0 < h \leq h_{0}$ define $P_{φ, h, q} = e^{φ / h} h^{2} (- Δ_{g} + q) e^{- φ / h} .$ Every $w \in C_{c}^{\infty} (M^{\circ})$ satisfies $∥ w ∥_{L^{2} (M)} \leq C h^{- 1} ∥ P_{φ, h, q} w ∥_{L^{2} (M)} .$ Moreover, for every $F \in L^{2} (M)$ there is an $R \in L^{2} (M)$ solving $P_{φ, h, q} R = F$ in distributions on $M^{\circ}$ and satisfying $∥ R ∥_{L^{2} (M)} \leq C h^{- 1} ∥ F ∥_{L^{2} (M)} .$ No boundary trace is prescribed for $R$ .

Statement references

Product metric and distinguished coordinate

Sobolev spaces and distributional interpretation

Facts & Assumptions

Given: AC and Countable Choice, a compact product subdomain $(M, g) ⋐ (R \times M_{0}^{\circ}, d x_{1}^{2} \oplus g_{0})$ , a real smooth $q$ , a sign $σ \in {- 1, 1}$ , and $0 < h \leq h_{0}$ .

[F1]

Transversally anisotropic product domain: the metric is the product $d x_{1}^{2} \oplus g_{0}$ , so its coefficients are independent of $x_{1}$ and $∣ \nabla_{g} x_{1} ∣ = 1$ .

[F2]

Sobolev spaces on a compact Riemannian domain with boundary: $L^{2}$ and $H^{1}$ are defined by the Riemannian density and fixed chartwise Sobolev norms; compactly supported interior smooth functions have no boundary terms in integration by parts.

[F3]

Laplace–Beltrami operator as the trace of the Hessian: $Δ_{g} = {div}_{g} \nabla_{g}$ .

[A1]

The Axiom of Choice is assumed through A bounded complex linear functional on a subspace of a complex normed space extends with the same norm to extend the bounded complex functional from the range of the adjoint to all of $L^{2} (M)$ .

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is assumed through Riesz representation for Hilbert spaces to represent the extended bounded functional by an $L^{2}$ vector.

Proof

Given: the data in the statement.

1.1F1F2F3givenalgebra

Put $s = φ$ and first set $q = 0$ . For a real smooth weight $ψ$ , direct differentiation gives $P_{ψ, h, 0} = e^{ψ / h} (- h^{2} Δ_{g}) e^{- ψ / h} = - h^{2} Δ_{g} - ∣ \nabla_{g} ψ ∣^{2} + 2 h \nabla_{g} ψ \cdot \nabla_{g} + h Δ_{g} ψ .$ For $ψ = s$ , $∣ \nabla s ∣ = 1$ and $Δ s = 0$ , so $P_{s, h, 0} = A + C$ with $A = - h^{2} Δ_{g} - 1$ self-adjoint and $C = 2 h \partial_{s}$ skew-adjoint on compactly supported functions.

2.1F1F2step 1.1algebra

Convexify by $ψ = f (s) = s + \frac{h}{2 ε} s^{2}$ , where $ε > 0$ is fixed small and then $h / ε$ is small. Since $M$ is compact, choose $h$ so $f^{'} = 1 + h s / ε > 1 / 2$ on $M$ . The conjugated operator is $P_{ψ, h, 0} = A_{ε} + C_{ε}$ with $A_{ε} = - h^{2} Δ_{g} - (f^{'})^{2}, C_{ε} = 2 h f^{'} \partial_{s} + h f^{''}, f^{''} = h / ε .$ Here $A_{ε}$ is self-adjoint and $C_{ε}$ is skew-adjoint. Product independence of $s$ makes $- Δ_{g} = - \partial_{s}^{2} - Δ_{g_{0}}$ , with $Δ_{g_{0}}$ commuting with $C_{ε}$ . Direct differentiation in the $s$ coordinate therefore gives the exact commutator $[A_{ε}, C_{ε}] = 4 h (f^{'})^{2} f^{''} - 4 h^{3} f^{''} \partial_{s}^{2} .$ For compactly supported $w$ , integration by parts yields $∥ P_{ψ, h, 0} w ∥_{2}^{2} = ∥ A_{ε} w ∥_{2}^{2} + ∥ C_{ε} w ∥_{2}^{2} + 4 h f^{''} ∥ (f^{'}) w ∥_{2}^{2} + 4 h^{3} f^{''} ∥ \partial_{s} w ∥_{2}^{2} \geq c h^{2} ε^{- 1} ∥ w ∥_{2}^{2} .$ The same identity implies $∥ A_{ε} w ∥_{2} \leq ∥ P_{ψ, h, 0} w ∥_{2}$ . Since $∥ h \nabla w ∥_{2}^{2} = ⟨ A_{ε} w, w ⟩ + ∥ f^{'} w ∥_{2}^{2}$ , the $L^{2}$ bound just obtained gives $∥ w ∥_{2} + ∥ h \nabla w ∥_{2} \leq C_{ε} h^{- 1} ∥ P_{ψ, h, 0} w ∥_{2}$ .

3.1step 2.1givenalgebra

The convexified and unconvexified weights are related by multiplication with $m = e^{s^{2} / (2 ε)}$ : $P_{ψ, h, 0} (m w) = m P_{s, h, 0} w$ . On compact $M$ , $m$ and $m^{- 1}$ are bounded, so step 2.1 yields $∥ w ∥_{2} \leq C h^{- 1} ∥ P_{s, h, 0} w ∥_{2}$ . Adding the potential changes the conjugated operator by $h^{2} q$ . Its norm on $w$ is at most $h^{2} ∥ q ∥_{\infty} ∥ w ∥_{2}$ , which is absorbed in the preceding estimate provided $C_{0} h ∥ q ∥_{\infty} \leq 1 / 2$ , where $C_{0}$ is the bound for $q = 0$ . Choose $h_{0} > 0$ below this threshold and the bound ensuring $h_{0} \max_{M} ∣ x_{1} ∣ / ε < 1 / 2$ in step 2.1. Taking the smaller threshold and larger constant for the two signs proves the estimate uniformly for every $0 < h \leq h_{0}$ .

4.1F2A1A2givenstep 3.1algebra∎

Let $P = P_{φ, h, q}$ and apply the a priori estimate also to $P^{*} = P_{- φ, h, q}$ , which has the same real potential and the opposite allowed sign. On $E = P^{*} (C_{c}^{\infty} (M^{\circ})) \subset L^{2} (M)$ define the complex-linear functional $ℓ (P^{*} w) = \int_{M} \overset{ˉ}{F} w d V_{g}$ . The estimate makes this well-defined: $P^{*} w = 0$ implies $w = 0$ . It also gives $∣ ℓ (P^{*} w) ∣ \leq ∥ F ∥_{2} ∥ w ∥_{2} \leq C h^{- 1} ∥ F ∥_{2} ∥ P^{*} w ∥_{2}$ . By [A1] extend $ℓ$ to $L^{2} (M)$ with the same bound; by [A2] there is $R \in L^{2} (M)$ with $ℓ (z) = \int_{M} z \overset{ˉ}{R} d V_{g}$ and $∥ R ∥_{2} \leq C h^{- 1} ∥ F ∥_{2}$ . Taking complex conjugates gives $\int_{M} R \overline{P^{*} w} d V_{g} = \int_{M} F \overset{ˉ}{w} d V_{g}$ for every test $w$ , which is precisely $P R = F$ distributionally on $M^{\circ}$ .

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Smooth-manifold elliptic nonresonant solvability and regularity

Statement

Assume the Axiom of Choice and Countable Choice. Let $(M, g)$ be a compact smooth Riemannian domain with smooth boundary and no orientability assumption, let $q \in C^{\infty} (M, R)$ , and suppose the only $w \in H_{0}^{1} (M)$ satisfying $(- Δ_{g} + q) w = 0$ weakly is $w = 0$ . For every $F \in L^{2} (M)$ and $f \in H^{1 / 2} (\partial M)$ there is a unique weak solution $u \in H^{1} (M)$ of $(- Δ_{g} + q) u = F$ with trace $f$ . If $f = 0$ , $F \in H^{m} (M)$ and $m \geq 0$ is an integer, then $u \in H^{m + 2} (M)$ and $∥ u ∥_{H^{m + 2}} \leq C_{m} ∥ F ∥_{H^{m}}$ . If $F$ and $f$ are smooth, this solution belongs to $H^{m} (M)$ for every integer $m \geq 0$ .

Statement references

Fixed-atlas Sobolev and trace spaces

Facts & Assumptions

[A1]

The Axiom of Choice: every family of nonempty sets has a choice function; this is assumed by the Rellich compactness and abstract Fredholm suppliers used in steps 2.1 and 3.1.

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every countable family of nonempty sets has a choice function; this is assumed by the Lax--Milgram, partition, and trace suppliers used in steps 1.1 and 4.1.

Given: AC and Countable Choice; the compact smooth Riemannian domain; a real smooth potential with trivial Dirichlet kernel; $F \in L^{2} (M)$ ; and $f \in H^{1 / 2} (\partial M)$ .

[F1]

Sobolev spaces on a compact Riemannian domain with boundary: $H_{0}^{1} (M)$ is the closure of compactly supported interior smooth functions in $H^{1} (M)$ ; the boundary space $H^{1 / 2}$ is defined by finite chartwise Euclidean $W^{1 / 2, 2}$ norms.

[F2]

The Lax--Milgram theorem: a bounded coercive sesquilinear form on a Hilbert space has a unique weak solution.

[F3]

Compactness of $W_{0}^{1, p} (Ω) ↪ L^{p} (Ω)$ on bounded open sets: on a bounded Euclidean open set, $W_{0}^{1, 2}$ is compactly embedded in $L^{2}$ .

[F4]

Fredholm alternative for identity minus compact: for compact $K$ on a Banach space, $I - K$ is injective if and only if it is surjective, and then it is boundedly invertible.

[F5]

A bounded right inverse of the trace, supported in a prescribed collar: a smooth bounded Euclidean domain has a bounded right inverse from its $H^{1 / 2}$ boundary space to $H^{1}$ .

[F6]

Higher-order boundary regularity for Dirichlet problems: a zero-trace weak solution of a uniformly elliptic divergence-form equation on a bounded smooth Euclidean domain gains two derivatives, with the stated $H^{m + 2}$ estimate.

[F7]

The sharp trace theorem: boundedness and range in the fractional space and The trace commutes with smooth cutoffs and is chart local: local Euclidean traces patch in the fixed boundary atlas to a bounded map $T_{M} : H^{1} (M) \to H^{1 / 2} (\partial M)$ agreeing with restriction on smooth functions. The kernel of the trace is the closure of the test functions: finite localization gives $\ker T_{M} = H_{0}^{1} (M)$ .

[F8]

Nested-domain induction for interior elliptic derivatives: on relatively compact interior charts, smooth uniformly elliptic divergence-form equations with $H^{k}$ data have the corresponding $H^{k + 2}$ regularity and local estimate.

Proof

Given: AC and Countable Choice and the data in the statement.

1.1A2givenF1F2

Choose a real constant $Q > 1 + ∥ q ∥_{L^{\infty} (M)}$ and let $A = - Δ_{g} + Q$ on $H_{0}^{1} (M)$ , with form $a (w, z) = \int_{M} (g (\nabla w, \nabla \overset{ˉ}{z}) + Q w \overset{ˉ}{z}) d V_{g}$ . The form is bounded and coercive in the chartwise $H^{1}$ norm, so [F2] gives a bounded inverse $A_{D}^{- 1} : (H_{0}^{1} (M))^{*} \to H_{0}^{1} (M)$ .

2.1A1F1F3step 1.1algebra

Multiplication by $Q - q$ maps $H_{0}^{1} (M)$ boundedly to $L^{2} (M)$ , and the inclusion $H_{0}^{1} (M) ↪ L^{2} (M)$ is compact: localize with the finite atlas and partition, apply [F3] on each bounded chart piece (using zero extension at a boundary face), then extract a finite diagonal subsequence and sum the localized limits. Hence $K : = A_{D}^{- 1} (Q - q)$ is compact on $H_{0}^{1} (M)$ .

3.1A1F4givenstep 1.1step 2.1algebra

For $w \in H_{0}^{1} (M)$ , $(I - K) w = 0$ is equivalent, after applying $A$ , to $(- Δ_{g} + q) w = 0$ . The latter has only the zero solution by hypothesis, so $I - K$ is injective. By [F4], $I - K$ is boundedly invertible. This gives a unique weak zero-trace solution for every bounded functional on $H_{0}^{1} (M)$ , in particular for every $F \in L^{2} (M)$ after subtracting any fixed boundary lift.

4.1A2F1F5F7step 3.1construct

For $n = 1$ , the bounded endpoint trace, smooth finite-dimensional lift and zero-trace kernel follow from the fundamental theorem of calculus and Cauchy--Schwarz; specifically, $∣ u (t) ∣^{2} \leq t \int_{0}^{t} ∣ u^{'} (r) ∣^{2} d r$ permits cutoff approximation at a zero endpoint, followed by interior mollification. For $n = 0$ the boundary is empty, the trace space is zero and $H_{0}^{1} = H^{1}$ . For $n \geq 2$ , a finite collection of flattened boundary charts, the collar neighborhood, and [F5] give a bounded extension $E : H^{1 / 2} (\partial M) \to H^{1} (M)$ with trace $f$ . Set $u = E f + w$ and solve for $w \in H_{0}^{1} (M)$ with weak right-hand-side functional $z \mapsto \int_{M} F \overset{ˉ}{z} d V_{g} - a_{q} (E f, z)$ by step 3.1. This proves existence for the stated data. The difference of any two solutions has zero global trace, so [F7] puts it in $H_{0}^{1} (M)$ ; the assumed trivial kernel then proves uniqueness.

5.1F1F6F7F8step 4.1

First take zero boundary data and $F \in L^{2}$ . For $n \geq 2$ , in each boundary chart the collar flattens the boundary and the pulled-back coefficients are smooth and uniformly elliptic; localize $u$ by a cutoff, whose trace remains zero and hence lies in $H_{0}^{1}$ locally by [F7], and apply the $m = 0$ case of [F6]. In interior charts apply [F8]. A finite cover gives $u \in H^{2} (M)$ and the estimate with the additional $L^{2}$ term. If $f$ is smooth, extend it smoothly in a boundary collar by keeping its chart representative constant in the normal variable near the boundary and multiplying by a cutoff; a finite partition patches these extensions to a smooth $E f$ . Then $w = u - E f$ has zero trace and, by [F7], lies in $H_{0}^{1} (M)$ ; the same local estimates apply to $w$ . For $n = 1$ , in a coordinate the weak equation has form $- \partial_{x} (a \partial_{x} u) + b \partial_{x} u + c u = F$ with smooth $a > 0$ ; it gives $\partial_{x} (a \partial_{x} u) \in L^{2}$ , hence $u \in H^{2}$ with the corresponding estimate. For $n = 0$ , the equation is pointwise and the trivial-kernel hypothesis gives the estimate.

6.1F1F6F7F8step 3.1step 5.1induction∎

Fix $k \geq 1$ and suppose inductively that $u \in H^{k + 1} (M)$ and $F \in H^{k} (M)$ . If $n \geq 2$ , in a boundary chart a cutoff localization has zero trace on the physical boundary and hence belongs locally to $H_{0}^{1}$ by [F7]; its equation has right-hand side $χ F + [L, χ] u \in H^{k}$ , since the commutator is first order with smooth coefficients. Apply [F6] at order $k$ to these boundary-localized equations; in interior charts apply [F8]. This gives $u \in H^{k + 2} (M)$ with a bound by $C_{k} (∥ F ∥_{H^{k}} + ∥ u ∥_{H^{k + 1}} + ∥ u ∥_{L^{2}})$ . If $n = 1$ , the coordinate equation in step 5.1 gives $\partial_{x} (a \partial_{x} u) = b \partial_{x} u + c u - F$ ; under the induction hypothesis the right-hand side belongs to $H^{k}$ , so $a \partial_{x} u \in H^{k + 1}$ and $u \in H^{k + 2}$ with the analogous estimate. If $n = 0$ , all Sobolev orders coincide and the pointwise inverse gives it. For zero trace, the bounded inverse in step 3.1 controls $∥ u ∥_{H^{1}}$ by $∥ F ∥_{L^{2}}$ ; the induction hypothesis then absorbs the lower-order term and yields $∥ u ∥_{H^{k + 2}} \leq C_{k} ∥ F ∥_{H^{k}}$ . For smooth nonzero $f$ , apply the same induction to $u - E f$ . No differentiation of normal boundary values is used.

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Boundary quasimode jets and exact-solution correction

Statement

Assume the Axiom of Choice and Countable Choice.

Let $Ω \subset R^{n}$ , $n \geq 2$ , be a bounded smooth domain, let $γ, \tilde{γ} \in C^{\infty} (\overline{Ω}, R)$ be strictly positive, let $x_{0} \in \partial Ω$ , and let $ξ_{0} \in T_{x_{0}}^{*} \partial Ω$ have unit norm in the induced boundary metric. Choose a sufficiently small boundary patch and $χ \in C_{c}^{\infty} (\partial Ω)$ supported there with $χ = 1$ near $x_{0}$ . For every integer $N \geq 1$ there are a smooth complex phase $Φ$ in a collar of the patch and amplitudes $b_{λ, γ}$ , $b_{λ, \tilde{γ}}$ , defined for $λ \geq 1$ , such that $v_{λ, γ} = λ^{- 1 / 2} e^{i λ Φ} b_{λ, γ}, v_{λ, \tilde{γ}} = λ^{- 1 / 2} e^{i λ Φ} b_{λ, \tilde{γ}},$ have boundary values $λ^{- 1 / 2} e^{i λ ϕ} χ$ , where $ϕ$ is real with $∣ d ϕ ∣ = 1$ near $supp χ$ , have $H^{1} (Ω)$ norms bounded above and below by positive constants, and satisfy $∥ div (γ \nabla v_{λ, γ}) ∥_{L^{2} (Ω)} + ∥ div (\tilde{γ} \nabla v_{λ, \tilde{γ}}) ∥_{L^{2} (Ω)} = O (λ^{- N}) .$ Here $ϕ = Φ ∣_{\partial Ω}$ is real, $d ϕ (x_{0}) = ξ_{0}$ , and $Im Φ$ is comparable to the inward distance to the boundary on the collar. Let $u_{λ, γ}$ and $u_{λ, \tilde{γ}}$ be the exact conductivity solutions with the same respective boundary values. Their corrections satisfy $∥ u_{λ, γ} - v_{λ, γ} ∥_{H^{1} (Ω)} + ∥ u_{λ, \tilde{γ}} - v_{λ, \tilde{γ}} ∥_{H^{1} (Ω)} = O (λ^{- N}) .$ For every fixed integer $k \geq 0$ and continuous $F$ on $\overline{Ω}$ , choosing $N > k$ gives $\lim_{λ \to \infty} λ^{k} \int_{Ω} dist (x, \partial Ω)^{k} F (x) \nabla v_{λ, γ} \cdot \overline{\nabla v_{λ, \tilde{γ}}} d x = c_{k} \int_{\partial Ω} F ∣ χ ∣^{2} d S, c_{k} = 2 \int_{0}^{\infty} t^{k} e^{- 2 t} d t = \frac{k!}{2^{k}} > 0.$ The same limit holds with $v_{λ, γ}, v_{λ, \tilde{γ}}$ replaced by their exact corrections.

Statement references

Smooth Euclidean boundary charts

Outward normal and boundary distance

Positive conductivity weak solvability

Facts & Assumptions

[A1]

The Axiom of Choice is a hypothesis of the positive-conductivity solvability supplier [F3]; it is used only for the exact correction.

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is assumed by the collar, partition, and positive-conductivity suppliers [F1]--[F3], and permits the countable cutoff choices in the Borel realization.

Given: the smooth bounded domain, positive smooth conductivities, a boundary point and unit tangential covector, and a cutoff equal to one near that point.

[F1]

Bounded C^k domains and boundary charts and Collar neighborhood theorem: a smooth boundary has local smooth charts flattening it, and a neighborhood of the compact boundary is smoothly parametrized by inward normal distance.

[F2]

Smooth partitions of unity exist on manifolds with boundary: a finite smooth partition of unity can localize and patch the chartwise constructions.

[F3]

Weak solvability of the positive scalar conductivity equation: on a compact connected smooth Riemannian domain with nonempty boundary, a positive smooth conductivity has a unique weak Dirichlet solution for every $L^{2}$ forcing and $H^{1 / 2}$ boundary value, and its zero-trace Poincare inequality holds. Step 5.1 applies this separately to the finitely many components of $\overline{Ω}$ .

[F4]

Sobolev spaces on a compact Riemannian domain with boundary: the exact corrections are measured in the chartwise $H^{1}$ norm and the zero-trace space is the closure of compactly supported smooth functions.

Proof

Given: the data in the statement.

1.1F1givenconstruct

Choose boundary normal coordinates $(y, x_{n})$ near $x_{0}$ , with $x_{n} = dist (x, \partial Ω)$ and inward unit normal $\partial_{x_{n}}$ . The induced metric on the boundary is smooth. Choose a smooth hypersurface $S$ in $\partial Ω$ through $x_{0}$ , orthogonal there to $ξ_{0}$ (if $n = 2$ , take $S = {x_{0}}$ ). The boundary geodesics starting normally from $S$ give, by smooth dependence of the geodesic ODE and the inverse function theorem, coordinates $(s, z)$ near $x_{0}$ ; define $ϕ (s, z) = s$ . Gauss' lemma gives $∣ d ϕ ∣_{g_{\partial Ω}} = 1$ and $d ϕ (x_{0}) = ξ_{0}$ . Shrink the patch so these coordinates and $χ$ are supported inside it.

2.1F1step 1.1constructalgebra

In boundary normal coordinates, $g^{n n} = 1$ and $g^{n α} = 0$ at the boundary. Set $Φ ∣_{x_{n} = 0} = ϕ$ and $\partial_{n} Φ ∣_{x_{n} = 0} = i$ . Then $g^{i j} \partial_{i} Φ \partial_{j} Φ = 0$ at $x_{n} = 0$ because $∣ d ϕ ∣ = 1$ . Differentiate this eikonal equation successively in $x_{n}$ . At each order the newest unknown normal jet $\partial_{n}^{j + 1} Φ ∣_{x_{n} = 0}$ occurs linearly with coefficient $2 i g^{n n} = 2 i \neq 0$ ; all other terms use already assigned jets and derivatives of $g$ . Solve recursively for every normal jet. Realize the resulting jet sequence by the following local Borel construction: if $c_{j} (y)$ is the prescribed $j$ th normal derivative, choose a fixed smooth cutoff $θ$ equal to one near zero and supported in $(- 1, 1)$ , and set $Φ (y, x_{n}) = \sum_{j \geq 0} \frac{c_{j} (y)}{j!} x_{n}^{j} θ (x_{n} / ε_{j}) .$ For all sufficiently large $j$ , choose $ε_{j} < 2^{- j}$ so that every derivative of order at most $j / 2$ of the $j$ th summand has sup norm at most $2^{- j}$ ; this is possible because those derivatives are bounded by constants times $ε_{j}^{j - m}$ with $m \leq j / 2$ . Choose any positive $ε_{j}$ for the finitely many smaller $j$ . For each fixed derivative order, the tail then converges uniformly, so $Φ$ is smooth and has the prescribed jets. In particular, the eikonal expression vanishes to infinite order at the boundary. Since $\partial_{n} Im Φ = 1$ there, shrinking the collar gives $Im Φ \geq x_{n} / 2$ .

3.1F1F2step 2.1construct

For $P_{γ} = div (γ \nabla)$ , direct differentiation gives $P_{γ} (e^{i λ Φ} b) = e^{i λ Φ} [- λ^{2} p_{γ} b + i λ L_{γ} b + P_{γ} b],$ where $p_{γ} = γ g^{i j} \partial_{i} Φ \partial_{j} Φ$ and $L_{γ} b = 2 γ g^{i j} (\partial_{i} Φ) (\partial_{j} b) + div (γ \nabla Φ) b$ . The phase is common to both conductivities and $p_{γ} = γ p_{1}$ vanishes to infinite order on the boundary by step 2.1. Prescribe $b_{0, γ} ∣_{\partial Ω} = χ$ and $b_{- j, γ} ∣_{\partial Ω} = 0$ for $j \geq 1$ . In boundary normal coordinates the coefficient of the new normal derivative in $L_{γ} b$ is $2 i γ$ , which is nonzero. Therefore the normal derivatives of $b_{0, γ}$ can be chosen successively so that $L_{γ} b_{0, γ} = 0$ to infinite order at the boundary; after choosing $b_{0, γ}, \dots, b_{- (j - 1), γ}$ , the same recursion chooses all jets of $b_{- j, γ}$ so that $i L_{γ} b_{- j, γ} + P_{γ} b_{- (j - 1), γ} = 0$ to infinite order. Realize each sequence by the Borel series in step 2.1 and multiply by a fixed collar cutoff equal to one near the boundary support. Repeat for $\tilde{γ}$ .

4.1F1F2givenstep 3.1algebra

Let $b_{λ, γ} = \sum_{j = 0}^{K} λ^{- j} b_{- j, γ}$ , with the same construction for $\tilde{γ}$ . The order- $λ^{2}$ eikonal term and the order- $λ, \dots, λ^{1 - K}$ transport terms in step 3.1 vanish to infinite order at $x_{n} = 0$ . Thus each eikonal or canceled transport coefficient is $O (x_{n}^{R})$ for every prescribed integer $R$ ; the final term $λ^{- K} P_{γ} b_{- K, γ}$ has a fixed smooth bounded coefficient and is treated by increasing $K$ . On the collar, $∣ e^{i λ Φ} ∣ \leq e^{- λ x_{n} / 2}$ . For every fixed $a, R$ , direct integration gives $∥ λ^{a} x_{n}^{R} e^{- λ x_{n} / 2} ∥_{L^{2} (0, δ)} = O (λ^{a - R - 1 / 2}) .$ Choose $R$ and then $K$ large enough to make every residual term $O (λ^{- N})$ . Derivatives of the collar cutoff are supported where $x_{n} \geq c > 0$ and hence contribute $O (e^{- c λ})$ . The same Gaussian integration shows $∥ v_{λ, γ} ∥_{L^{2}} = O (λ^{- 1})$ and bounds its gradient norm above. Thus the $H^{1}$ norms are bounded above and the residual bounds hold for the arbitrary prescribed $N$ .

5.1A1A2F3F4step 4.1algebra

The compact smooth manifold $\overline{Ω}$ has finitely many connected components: connected coordinate neighborhoods form an open cover and a finite subcover meets every component. Each component has nonempty boundary, since a boundaryless component of the same dimension embedded in $R^{n}$ would be both open and compact. Apply [F3] on each component and patch the finitely many solutions to obtain the exact solution on $Ω$ . Taking the maximum of the component Poincare constants gives $∥ z ∥_{2} \leq C_{P} ∥ \nabla z ∥_{2}$ for $z \in H_{0}^{1} (Ω)$ . Let $r_{λ} = u_{λ, γ} - v_{λ, γ} \in H_{0}^{1} (Ω)$ . Its weak equation is $P_{γ} r_{λ} = - P_{γ} v_{λ, γ}$ . Test with $r_{λ}$ , use $γ \geq γ_{*} > 0$ , and use the Poincare inequality in [F3]: $γ_{*} ∥ \nabla r_{λ} ∥_{2}^{2} \leq ∥ P_{γ} v_{λ, γ} ∥_{2} ∥ r_{λ} ∥_{2} \leq C_{P} ∥ P_{γ} v_{λ, γ} ∥_{2} ∥ \nabla r_{λ} ∥_{2} .$ If the last gradient norm is zero then $r_{λ} = 0$ by [F3]; otherwise divide by it. This proves $∥ r_{λ} ∥_{H^{1}} = O (λ^{- N})$ . The same argument applies to $\tilde{γ}$ .

5.2F1step 3.1step 4.1algebra

In the collar, write $J (y, x_{n}) d S (y) d x_{n}$ for Euclidean volume in boundary normal coordinates, with $J (y, 0) = 1$ . Since $Φ = ϕ + i x_{n} + O (x_{n}^{2})$ , $Im Φ = x_{n} + O (x_{n}^{2})$ and $g (d Φ, \overline{d Φ}) = 2 + O (x_{n})$ . The boundary values of both leading amplitudes are $χ$ , and lower amplitudes vanish there. Differentiating $v_{λ, γ}$ and $v_{λ, \tilde{γ}}$ therefore gives $\nabla v_{λ, γ} \cdot \overline{\nabla v_{λ, \tilde{γ}}} = 2 λ e^{- 2 λ x_{n}} ∣ χ (y) ∣^{2} + e^{- 2 λ x_{n}} O (1) + e^{- 2 λ x_{n}} O (λ x_{n}) + e^{- 2 λ x_{n}} O (λ^{2} x_{n}^{2}),$ uniformly for $0 \leq x_{n} \leq λ^{- 1 / 2}$ . On this shrinking collar $λ x_{n}^{2} \leq 1$ , so after $t = λ x_{n}$ the phase remainder factor is uniformly bounded; boundedness of $F, J, χ$ gives an integrable majorant $C t^{k} e^{- t}$ for the rescaled leading term. For each fixed $t$ , the rescaled leading term tends to $2 t^{k} e^{- 2 t} F (y, 0) ∣ χ (y) ∣^{2}$ . Each displayed remainder has an additional factor $λ^{- 1}$ after scaling and is dominated by $C λ^{- 1} (t^{k} + t^{k + 1} + t^{k + 2}) e^{- t}$ , including the phase error $e^{O (t^{2} / λ)} - 1 = O (t^{2} / λ)$ ; its integral tends to zero. On the rest of the fixed collar, $Im Φ \geq x_{n} / 2$ gives exponential decay $e^{- λ x_{n}} \leq e^{- \sqrt{λ}}$ ; every polynomial prefactor is therefore negligible. The cutoff region away from the collar is also exponentially small. Thus the limit for approximate beams is $2 \int_{0}^{\infty} t^{k} e^{- 2 t} d t \int_{\partial Ω} F ∣ χ ∣^{2} d S,$ whose constant is strictly positive.

6.1F1step 4.1step 5.2algebra

Taking equal conductivities, $F = 1$ , and $k = 0$ in step 5.2 gives $\lim_{λ \to \infty} \int_{Ω} ∣ \nabla v_{λ, γ} ∣^{2} d x = c_{0} \int_{\partial Ω} ∣ χ ∣^{2} d S > 0$ . Together with the upper bound in step 4.1 this proves the two-sided bound for large $λ$ . On each remaining compact parameter interval the quasimodes vary continuously in $H^{1}$ and have nonzero boundary value $λ^{- 1 / 2} e^{i λ ϕ} χ$ , so their norms have a positive minimum; enlarge the constants to cover all $λ \geq 1$ .

7.1F3F4step 4.1step 5.1step 5.2algebra∎

The exact corrections satisfy $H^{1}$ errors $O (λ^{- N})$ by step 5.1, while both approximate beams have uniformly bounded $H^{1}$ norms by step 4.1. Cauchy-Schwarz bounds the change in the weighted product integral, after multiplication by $λ^{k}$ , by $C λ^{k} λ^{- N}$ because $0 \leq dist (x, \partial Ω)^{k} \leq C$ on the bounded domain. Choose $N > k$ . The error tends to zero, so the same limit holds for the exact solutions.

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Scalar conductivity and Schrödinger Dirichlet-to-Neumann maps

Definition

Assume the Axiom of Choice and Countable Choice. For complex boundary data, the duality pairing is linear in the dual element and conjugate-linear in its boundary test.

Let $(M, g)$ be a nonempty compact connected smooth Riemannian domain with nonempty smooth boundary. For a real-valued $γ \in C^{\infty} (M)$ with $γ (x) > 0$ at every $x \in M$ , let $v_{f} \in H^{1} (M)$ be the unique weak solution of ${div}_{g} (γ \nabla_{g} v_{f}) = 0$ with trace $f \in H^{1 / 2} (\partial M)$ , whose existence is supplied by the positive-conductivity solvability lemma. Define the full-data physical conductivity Dirichlet-to-Neumann map $Λ_{γ} : H^{1 / 2} (\partial M) \to H^{- 1 / 2} (\partial M)$ by $⟨ Λ_{γ} f, h ⟩ = \int_{M} γ g (\nabla_{g} v_{f}, \nabla_{g} \overline{V_{h}}) d V_{g},$ where $V_{h} \in H^{1} (M)$ is any extension of $h \in H^{1 / 2} (\partial M)$ . The weak equation makes this value independent of the extension. For smooth data, under the boundary-density identification of the conjugate-linear dual, $Λ_{γ} f = γ \partial_{ν_{o u t}} v_{f} ∣_{\partial M}$ ; this is the full physical flux convention. When $M = \overline{Ω} \subset R^{n}$ with the Euclidean metric, this is the scalar Euclidean conductivity DN map.

For a real $q \in C^{\infty} (M)$ such that the only $w \in H_{0}^{1} (M)$ satisfying $(- Δ_{g} + q) w = 0$ weakly is $w = 0$ , let $u_{f} \in H^{1} (M)$ be the unique weak solution of $(- Δ_{g} + q) u_{f} = 0$ with trace $f$ , whose existence follows from the nonresonant regularity lemma. Define the Schrödinger DN map $Λ_{q} : H^{1 / 2} (\partial M) \to H^{- 1 / 2} (\partial M)$ by $⟨ Λ_{q} f, h ⟩ = \int_{M} (g (\nabla_{g} u_{f}, \nabla_{g} \overline{V_{h}}) + q u_{f} \overline{V_{h}}) d V_{g} .$ For smooth data, under the boundary-density identification of the conjugate-linear dual, this is $\partial_{ν_{o u t}} u_{f} ∣_{\partial M}$ . Equality of these operators means equality on all of $H^{1 / 2} (\partial M)$ , not equality for a restricted set of boundary inputs.

Definition references

Fixed-atlas Sobolev and trace spaces

Positive conductivity weak solvability

Schrodinger weak solvability under nonresonance

Riemannian volume density

Outward unit normal

Classical normal derivative

Riemannian gradient

Laplacian convention

Riemannian density in Green's formula

Zero trace is equivalent to H-zero

Sobolev trace

Boundary extension

Gauss-Green formula

Smooth boundary-data extension

Facts & Assumptions

Given: AC and Countable Choice; the compact smooth Riemannian domain; positive smooth $γ$ ; a real smooth nonresonant $q$ ; and the weak solutions in the Definition.

[A1]

The Axiom of Choice is assumed by the Euclidean trace-kernel and Gauss-Green suppliers [F2, F5].

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is assumed by the local weak-solution, trace-lift, and collar suppliers [F3, F4, F6, F8].

[F1]

Sobolev spaces on a compact Riemannian domain with boundary: $H_{0}^{1} (M)$ is the closure of interior compactly supported smooth functions in a fixed finite chartwise $H^{1}$ norm; the trace is defined chartwise and the fixed finite atlas admits a smooth partition of unity.

[F2]

The kernel of the trace is the closure of the test functions: on a bounded Euclidean $C^{1}$ domain, an $H^{1}$ function has zero trace if and only if it belongs to the $H^{1}$ closure of compactly supported interior smooth functions.

[F3]

Weak solvability of the positive scalar conductivity equation: for each $H^{1 / 2}$ boundary value, the positive-conductivity weak solution exists uniquely, its weak equation vanishes on every $H_{0}^{1}$ test, and smooth boundary data with smooth forcing give a solution smooth up to the boundary.

[F4]

Smooth-manifold elliptic nonresonant solvability and regularity: under the stated trivial Dirichlet-kernel condition, the Schrödinger weak solution is unique and smooth for smooth boundary data.

[F5]

The Gauss-Green integration-by-parts formula with Sobolev traces: for smooth functions on a bounded Euclidean $C^{1}$ domain, integration by parts produces the outward-normal boundary term.

[F6]

A bounded right inverse of the trace, supported in a prescribed collar: each $H^{1 / 2}$ boundary datum has an $H^{1}$ extension in a smooth Euclidean chart; finite chart localization supplies such extensions on $M$ .

[F7]

The trace commutes with smooth cutoffs and is chart local: multiplying by a smooth cutoff multiplies the trace by its boundary restriction, and flattening charts transport the trace to the Euclidean chart trace.

[F8]

Collar neighborhood theorem: smooth boundary data extend smoothly into a collar by constant normal extension and a cutoff.

[F9]

The $L^{p}$ trace operator on a bounded $C^{1}$ domain: the trace is a bounded map on $H^{1}$ and agrees with restriction for smooth functions.

[F10]

The sharp trace theorem: boundedness and range in the fractional space: in the fixed boundary charts the trace is bounded from $H^{1}$ into $H^{1 / 2}$ , and the chartwise maps patch to the trace used in the Definition.

Verification

Given: the data in the Definition and two $H^{1}$ extensions $V_{h}^{(1)}, V_{h}^{(2)}$ of the same boundary datum $h$ .

1.1A1F1F2F7givenconstruct

In dimension one the trace and lift conclusions follow from the endpoint argument in [F3], and integration by parts is the fundamental theorem of calculus with counting boundary density. In higher dimensions use the Euclidean suppliers as follows. Their difference $W = V_{h}^{(1)} - V_{h}^{(2)}$ has zero trace. For dimension one, [F3] gives $W \in H_{0}^{1} (M)$ directly; for dimension at least two, localize $W$ by the fixed finite partition in [F1]. On each boundary chart, [F7] shows that the localized function has zero Euclidean trace after flattening; [F2] then approximates it in $H^{1}$ by compactly supported interior smooth functions. On interior charts, ordinary compactly supported mollification gives the same approximation. Pulling back and summing the finitely many approximations shows $W \in H_{0}^{1} (M)$ .

2.1A2F1F3F4F6step 1.1algebra

For the conductivity pairing, [F3] gives $\int_{M} γ g (\nabla_{g} v_{f}, \nabla_{g} \overline{W}) d V_{g} = 0$ , since $W \in H_{0}^{1} (M)$ . Therefore replacing $V_{h}^{(1)}$ by $V_{h}^{(2)}$ changes the displayed pairing by zero. For the Schrödinger pairing, the same argument with the weak equation in [F4] gives $\int_{M} (g (\nabla_{g} u_{f}, \nabla_{g} \overline{W}) + q u_{f} \overline{W}) d V_{g} = 0$ . Thus both definitions are independent of the chosen extension. For each $h$ , choose a bounded lift $V_{h}$ by [F6]; Cauchy--Schwarz gives $∣ ⟨ Λ_{γ} f, h ⟩ ∣ \leq C ∥ v_{f} ∥_{H^{1}} ∥ h ∥_{H^{1 / 2}}$ and $∣ ⟨ Λ_{q} f, h ⟩ ∣ \leq C (1 + ∥ q ∥_{\infty}) ∥ u_{f} ∥_{H^{1}} ∥ h ∥_{H^{1 / 2}}$ . Hence each pairing is a continuous conjugate-linear functional of $h$ , as required for the anti-dual in [F1].

3.1A1A2F3F4F5F6F8F9algebra∎

By [F3] and [F4], smooth boundary data yield solutions smooth up to $\partial M$ . For smooth test data $h$ , choose a smooth collar extension $V_{h}$ by [F8]. In a boundary chart, write ${div}_{g} (γ \nabla_{g} v) = ∣ g ∣^{- 1 / 2} \partial_{i} (∣ g ∣^{1 / 2} γ g^{i j} \partial_{j} v)$ and $Δ_{g} u = ∣ g ∣^{- 1 / 2} \partial_{i} (∣ g ∣^{1 / 2} g^{i j} \partial_{j} u)$ . Apply the Euclidean Gauss-Green identity [F5] to the coordinate fluxes; summing over the finite partition gives the Riemannian boundary flux, while artificial chart boundaries cancel and the density transforms to $d S_{g}$ . For the conductivity equation it yields $⟨ Λ_{γ} f, h ⟩ = \int_{\partial M} γ \partial_{ν_{o u t}} v_{f} \overline{h} d S_{g},$ since ${div}_{g} (γ \nabla_{g} v_{f}) = 0$ . For the Schrödinger equation it yields $⟨ Λ_{q} f, h ⟩ = \int_{\partial M} \partial_{ν_{o u t}} u_{f} \overline{h} d S_{g},$ since $(- Δ_{g} + q) u_{f} = 0$ . The extensions required for the weak pairings exist by [F6], and smooth test extensions exist by [F8].

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Small-data solvability and third differentiability of the cubic DN map

Statement

Assume the Axiom of Choice and Countable Choice.

Let $(M, g)$ be a compact smooth Riemannian domain of dimension $n \geq 1$ with smooth nonempty boundary, assume every connected component of $M$ meets the boundary, and let $q \in C^{\infty} (M, R)$ . Fix an integer $s > n / 2 + 2$ and a finite family $h_{1}, \dots, h_{m} \in C^{\infty} (\partial M, C)$ . For $ϵ = (ϵ_{1}, \dots, ϵ_{m}) \in R^{m}$ sufficiently near zero, the equation $- Δ_{g} u + q u^{3} = 0 in M^{\circ}, u ∣_{\partial M} = \sum_{j = 1}^{m} ϵ_{j} h_{j}$ has a unique solution in a fixed small $H^{s} (M)$ ball. The parameter map $ϵ \mapsto u_{ϵ}$ is $C^{3}$ into $H^{s} (M)$ , and the corresponding nonlinear Dirichlet-to-Neumann data $\partial_{ν_{o u t}} u_{ϵ} ∣_{\partial M}$ is $C^{3}$ into $C^{0} (\partial M)$ . For each such finite family, write $N_{q; h_{1}, \dots, h_{m}} (ϵ) = \partial_{ν_{o u t}} u_{ϵ} ∣_{\partial M}$ for this parameterized data map.

Statement references

Chartwise Sobolev spaces

Riemannian metric and boundary normal

Dirichlet inverse and higher regularity

Facts & Assumptions

[A1]

The Axiom of Choice is assumed by the Sobolev algebra, embedding, and Banach implicit-function suppliers [F4]--[F6].

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is assumed by the Sobolev and elliptic regularity interfaces [F1], [F3].

Given: the compact smooth Riemannian domain, real smooth $q$ , integer $s > n / 2 + 2$ , and the fixed finite smooth boundary directions.

[F1]

Sobolev spaces on a compact Riemannian domain with boundary: the chartwise integer Sobolev spaces are Banach spaces, and $H_{0}^{1} (M)$ is the $H^{1}$ closure of interior compactly supported smooth functions. Since $H^{s} (M) ↪ H^{1} (M)$ continuously, $X_{s} = H^{s} (M) \cap H_{0}^{1} (M)$ is closed in $H^{s} (M)$ .

[F2]

Weak solvability of the positive scalar conductivity equation: on each compact connected component with nonempty boundary, $∥ w ∥_{L^{2}} \leq C_{P, j} ∥ \nabla w ∥_{L^{2}}$ for zero-trace $w$ . Compactness and local connectedness give finitely many components, so the maximum of these constants gives the inequality on $M$ .

[F3]

Smooth-manifold elliptic nonresonant solvability and regularity: once the weak Dirichlet kernel is trivial, $- Δ_{g}$ has a unique zero-boundary solution for every forcing $F \in L^{2} (M)$ . For integer $k \geq 0$ and arbitrary $F \in H^{k} (M)$ , that solution belongs to $H^{k + 2} (M)$ with $∥ u ∥_{H^{k + 2}} \leq C_{k} ∥ F ∥_{H^{k}}$ ; no trace condition is imposed on $F$ .

[F4]

For $n \geq 2$ , The Sobolev space $W^{k, p}$ is an algebra above the critical index gives the Banach-algebra bound for integer $s$ with $2 s > n$ on bounded smooth chart domains; finite localization gives it on $M$ . For $n = 1$ , the one-dimensional fundamental theorem of calculus gives $∥ z ∥_{L^{\infty}} \leq C ∥ z ∥_{H^{1}}$ on bounded intervals. Apply this to derivatives through order $s - 1$ and use the product rule to bound each derivative of $u v$ through order $s$ in $L^{2}$ ; this gives the same algebra bound on each chart and hence on $M$ .

[F5]

For $n \geq 2$ , Higher-order Sobolev embedding gives $H^{s} ↪ C^{2} (\overline{M})$ when $s > n / 2 + 2$ . For $n = 1$ , $s \geq 3$ ; apply the one-dimensional $H^{1} ↪ C^{0}$ estimate to derivatives through order two.

[F6]

Implicit function theorem for Banach spaces: if a $C^{3}$ map between Banach spaces vanishes at a point and its derivative in the unknown variable is a bounded isomorphism, its nearby zero set is the graph of a unique $C^{3}$ map.

[F7]

Collar neighborhood theorem: boundary data on a compact smooth boundary extend smoothly into a collar by constant normal extension followed by a smooth cutoff and finite patching.

[F8]

In boundary charts, the $g$ -orthogonal complement of $T \partial M$ is a smooth line bundle, and the outward choice defines a smooth unit normal. Thus $u \mapsto g (\nabla_{g} u, ν_{o u t}) ∣_{\partial M}$ is bounded from $C^{1} (\overline{M})$ to $C^{0} (\partial M)$ .

Proof

Given: the data in the statement.

1.1A2F1F2F3givenalgebra

For $q = 0$ , the Dirichlet kernel of $- Δ_{g}$ is trivial: if $w \in H_{0}^{1} (M)$ is weakly harmonic, testing with $w$ gives $\int_{M} ∣ \nabla w ∣^{2} d V_{g} = 0$ , and the zero-trace Poincare inequality in [F2] gives $w = 0$ . Thus [F3] supplies the zero-boundary inverse $G = (- Δ_{g})^{- 1} : H^{s - 2} (M) \to X_{s}$ , where $X_{s} = {w \in H^{s} (M) : w \in H_{0}^{1} (M)}$ . The high-order estimate in [F3], together with its bounded $H^{1}$ inverse, makes $G$ bounded.

1.2A1A2F1F2F3F4F6F7givenconstruct

Regard the complex Sobolev spaces as real Banach spaces, since the parameters are real. For $n \geq 2$ , [F4] and localization to the fixed finite atlas make $H^{s} (M)$ a Banach algebra. If $n = 1$ , then $s \geq 3$ ; the one-dimensional estimate $H^{1} ↪ L^{\infty}$ and the Leibniz rule give the same algebra bound in each chart, hence on $M$ . Thus $(ϵ, w) \mapsto q (E f_{ϵ} + w)^{3}$ is a continuous polynomial from $R^{m} \times X_{s}$ into $H^{s - 2} (M)$ , where $f_{ϵ} = \sum_{j} ϵ_{j} h_{j}$ and $E f_{ϵ}$ is the smooth collar extension from [F7]. Define $F (ϵ, w) = w - G [Δ_{g} (E f_{ϵ}) - q (E f_{ϵ} + w)^{3}] \in X_{s} .$ It is $C^{3}$ (indeed polynomial plus bounded linear maps) and $F (0, 0) = 0$ .

2.1A1F6step 1.2algebra

At $(0, 0)$ , the derivative $D_{w} F (0, 0)$ is the identity on $X_{s}$ , because the derivative of $u \mapsto q u^{3}$ at $u = 0$ vanishes. By [F6], there is a neighborhood of $ϵ = 0$ and a unique $C^{3}$ map $w (ϵ) \in X_{s}$ near zero such that $F (ϵ, w (ϵ)) = 0$ . The fixed-point equation is exactly $- Δ_{g} (E f_{ϵ} + w) + q (E f_{ϵ} + w)^{3} = 0$ with boundary trace $f_{ϵ}$ , so $u_{ϵ} = E f_{ϵ} + w (ϵ)$ gives the unique small solution.

3.1A1F5F8step 2.1algebra∎

For $n \geq 2$ , [F5] gives $H^{s} (M) ↪ C^{2} (\overline{M})$ ; for $n = 1$ , $s \geq 3$ gives the same embedding by the one-dimensional Sobolev estimate. By [F8], $u \mapsto \partial_{ν_{o u t}} u ∣_{\partial M}$ is bounded from $C^{1} (\overline{M})$ to $C^{0} (\partial M)$ . Composing it with the $C^{3}$ parameter map proves the asserted $C^{3}$ dependence of $N_{q; h_{1}, \dots, h_{m}}$ .

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CTA complex geometrical optics solvability and remainder bound

Statement

Assume the Axiom of Choice and Countable Choice.

Let $(M, g)$ be a transversally anisotropic product domain, let $q \in C^{\infty} (\overline{M}, R)$ , and let $ψ \in C^{\infty}$ be real on a neighborhood of the transversal projection of $M$ , with $∣ d ψ ∣_{g_{0}} = 1$ . Suppose $a \in C^{\infty} (\overline{M}, C)$ satisfies $2 \partial_{x_{1}} a + 2 i g_{0} (\nabla_{g_{0}} ψ, \nabla_{g_{0}} a) + i (Δ_{g_{0}} ψ) a = 0.$ Then, for all sufficiently large $τ$ , there is an exact solution $u_{τ} = e^{- τ (x_{1} + i ψ)} a + e^{- τ x_{1}} r_{τ}, (- Δ_{g} + q) u_{τ} = 0 in M^{\circ},$ with $r_{τ} \in H^{1} (M)$ and $∥ r_{τ} ∥_{L^{2} (M)} \leq C τ^{- 1} ∥ (- Δ_{g} + q) a ∥_{L^{2} (M)}$ . No boundary value is prescribed for $r_{τ}$ .

Statement references

Product metric and dimension

Product Carleman estimate and L2 right solvability

Facts & Assumptions

[A1]

The Axiom of Choice is used for the Hahn--Banach extension in the product Carleman right solver [F2].

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is used for its Hilbert-space Riesz representation [F2].

Given: the product domain, smooth real $q$ , real eikonal function $ψ$ with unit transversal gradient, and smooth transport amplitude $a$ .

[F1]

Transversally anisotropic product domain: $g = d x_{1}^{2} \oplus g_{0}$ and $Δ_{g} = \partial_{x_{1}}^{2} + Δ_{g_{0}}$ .

[F2]

Limiting Carleman estimate and right solvability on a product domain: for $φ = x_{1}$ , the conjugated operator $P_{φ, h, q} = e^{φ / h} h^{2} (- Δ_{g} + q) e^{- φ / h}$ has a distributional right inverse on $L^{2}$ with norm at most $C h^{- 1}$ .

[F3]

Riemannian gradient and Laplace–Beltrami operator as the trace of the Hessian: the product metric extends the gradient bilinearly to complex phases, and $Δ_{g} = {div}_{g} \nabla_{g}$ .

[F4]

External input: Dos Santos Ferreira--Kurylev--Lassas--Salo, Proposition 2.1 (PDF p. 10), for a compact oriented CTA manifold, bounded potential, either sign of sufficiently large real $τ$ , and arbitrary $f \in L^{2}$ , gives $r \in H^{1}$ solving $e^{τ x_{1}} (- Δ_{g} + q) e^{- τ x_{1}} r = f$ , with $∥ r ∥_{H^{α}} \leq C ∣ τ ∣^{α - 1} ∥ f ∥_{2}$ for $0 \leq α \leq 1$ , with no prescribed boundary value. We import this statement, not a reconstructed shifted Carleman proof. Its statement and the application in Propositions 2.1--2.2, PDF pp. 10--11, were inspected; the older Proposition 4.4 citation issue is not treated as a local proof.

[F5]

The orientation restriction in [F4] is removed for scalar equations by the orientation double cover of $M_{0}$ . Lift $M, q, f$ to its product cover, solve there, and average the solution with its deck transform. The invariant solution descends to $H^{1} (M)$ and solves the original equation. The degree-two factors in the squared $L^{2}$ norms cancel in the estimate; the same applies to the local Sobolev norms. A disconnected cover is handled componentwise, using only finitely many choices.

Proof

Given: the data in the statement.

1.1F1F3givenalgebra

Put $Φ = x_{1} + i ψ$ . By [F1] and $∣ d ψ ∣_{g_{0}} = 1$ , $g (d Φ, d Φ) = 1 + i^{2} ∣ d ψ ∣_{g_{0}}^{2} = 0.$ Direct conjugation gives $(- Δ_{g} + q) (e^{- τ Φ} a) = e^{- τ Φ} [- τ^{2} g (d Φ, d Φ) a + τ (2 g (\nabla_{g} Φ, \nabla_{g} a) + (Δ_{g} Φ) a) + (- Δ_{g} + q) a] .$ The $τ^{2}$ term vanishes by the eikonal equality, and the coefficient of $τ$ is exactly the transport equation in the statement. Hence the residual is $e^{- τ Φ} (- Δ_{g} + q) a$ .

2.1A1A2F4F5step 1.1algebra

Set $h = τ^{- 1}$ and conjugate by the real weight $e^{- τ x_{1}}$ . The equation for the correction is $P_{x_{1}, h, q} r_{τ} = - h^{2} e^{- i τ ψ} (- Δ_{g} + q) a .$ The right side is in $L^{2}$ and has norm $h^{2} ∥ (- Δ_{g} + q) a ∥_{2}$ . By [F4]--[F5], with the equation divided by $h^{2}$ , there is an $H^{1}$ correction (and hence a distributional) $r_{τ} \in L^{2} (M)$ with $∥ r_{τ} ∥_{2} \leq C h^{- 1} h^{2} ∥ (- Δ_{g} + q) a ∥_{2} = C τ^{- 1} ∥ (- Δ_{g} + q) a ∥_{2} .$ Thus the asserted remainder estimate and the equation hold distributionally.

3.1F4F5step 1.1step 2.1∎

Choose the $H^{1}$ correction from step 2.1. Multiplication by the smooth exponentials preserves $H^{1} (M)$ for each fixed $τ$ , and the correction equation cancels the residual computed in step 1.1. Thus $u_{τ}$ is exact, has the stated remainder bound, and has no imposed correction boundary value. The additional boundary regularity is the external input [F4], rather than a consequence asserted for the $L^{2}$ solver [F2].

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Alessandrini identities for scalar conductivity and Schrödinger DN maps

Statement

Assume the Axiom of Choice and Countable Choice.

Let $(M, g)$ be a nonempty compact connected smooth Riemannian domain with nonempty smooth boundary. If $γ_{1}, γ_{2} \in C^{\infty} (M, R)$ are strictly positive and $u_{j}$ solves ${div}_{g} (γ_{j} \nabla_{g} u_{j}) = 0$ with smooth boundary value $f_{j}$ , then $⟨ (Λ_{γ_{1}} - Λ_{γ_{2}}) f_{1}, f_{2} ⟩_{\partial M} = \int_{M} (γ_{1} - γ_{2}) g (\nabla_{g} u_{1}, \nabla_{g} \overline{u_{2}}) d V_{g} .$ If instead $q_{1}, q_{2} \in C^{\infty} (M, R)$ have trivial Dirichlet kernels for $- Δ_{g} + q_{j}$ and $u_{j}$ solves $(- Δ_{g} + q_{j}) u_{j} = 0$ with smooth boundary value $f_{j}$ , then $⟨ (Λ_{q_{1}} - Λ_{q_{2}}) f_{1}, f_{2} ⟩_{\partial M} = \int_{M} (q_{1} - q_{2}) u_{1} \overline{u_{2}} d V_{g} .$

Statement references

Full-data conductivity and Schrödinger DN maps

Positive conductivity weak solvability

Nonresonant Schrödinger solvability

Facts & Assumptions

[A1]

The Axiom of Choice is inherited from the weak DN-map well-posedness suppliers [F2]--[F3].

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is inherited from the weak DN-map well-posedness suppliers [F2]--[F3].

Given: the compact smooth Riemannian domain, two positive conductivities or two real nonresonant potentials, and arbitrary smooth complex boundary values $f_{1}, f_{2}$ .

[F1]

Scalar conductivity and Schrödinger Dirichlet-to-Neumann maps: the conductivity DN map is the weak boundary form of ${div}_{g} (γ \nabla_{g})$ , and the Schrödinger DN map is the weak boundary form of $- Δ_{g} + q$ ; their smooth-data fluxes use $γ \partial_{ν_{o u t}}$ and $\partial_{ν_{o u t}}$ , respectively.

[F2]

Weak solvability of the positive scalar conductivity equation: positive smooth conductivity gives a unique weak solution for each $H^{1 / 2}$ boundary value.

[F3]

Smooth-manifold elliptic nonresonant solvability and regularity: a trivial Schrödinger Dirichlet kernel gives a unique weak solution for each $H^{1 / 2}$ boundary value, with smooth solutions for smooth boundary data.

[F4]

Riemannian volume density: in each chart the Riemannian volume is a positive density, so the integration-by-parts identities can be patched without assuming orientability.

Proof

Given: the data in the statement.

1.1A1A2F1F2F4givenalgebra

Write $a_{γ_{j}} (v, w) = \int_{M} γ_{j} g (\nabla v, \nabla \overset{ˉ}{w}) d V_{g}$ . By [F1], $⟨ Λ_{γ_{1}} f_{1}, f_{2} ⟩ = a_{γ_{1}} (u_{1}, u_{2})$ , since $u_{2}$ is an extension of $f_{2}$ . Let ${\tilde{u}}_{1}$ solve the $γ_{2}$ equation with boundary value $f_{1}$ , which exists by [F2]. Then $⟨ Λ_{γ_{2}} f_{1}, f_{2} ⟩ = a_{γ_{2}} ({\tilde{u}}_{1}, u_{2})$ . Since $u_{1} - {\tilde{u}}_{1} \in H_{0}^{1} (M)$ and $u_{2}$ solves the $γ_{2}$ weak equation, Hermitian symmetry gives $a_{γ_{2}} (u_{1} - {\tilde{u}}_{1}, u_{2}) = \overline{a_{γ_{2}} (u_{2}, u_{1} - {\tilde{u}}_{1})} = 0$ . Hence $a_{γ_{2}} ({\tilde{u}}_{1}, u_{2}) = a_{γ_{2}} (u_{1}, u_{2})$ . Subtracting the two equalities gives the conductivity identity.

1.2A1A2F1F3F4givenalgebra

Write $a_{q_{j}} (v, w) = \int_{M} (g (\nabla v, \nabla \overset{ˉ}{w}) + q_{j} v \overset{ˉ}{w}) d V_{g}$ . By [F1], $⟨ Λ_{q_{1}} f_{1}, f_{2} ⟩ = a_{q_{1}} (u_{1}, u_{2})$ because $u_{2}$ extends $f_{2}$ . Let ${\tilde{u}}_{1}$ be the $q_{2}$ solution with boundary value $f_{1}$ , which exists by [F3]. Then $⟨ Λ_{q_{2}} f_{1}, f_{2} ⟩ = a_{q_{2}} ({\tilde{u}}_{1}, u_{2})$ . Since $u_{1} - {\tilde{u}}_{1} \in H_{0}^{1} (M)$ and $u_{2}$ solves the $q_{2}$ weak equation, real-valuedness of $q_{2}$ gives $a_{q_{2}} (u_{1} - {\tilde{u}}_{1}, u_{2}) = \overline{a_{q_{2}} (u_{2}, u_{1} - {\tilde{u}}_{1})} = 0$ . Thus $⟨ Λ_{q_{2}} f_{1}, f_{2} ⟩ = a_{q_{2}} (u_{1}, u_{2})$ .

2.1step 1.2algebra∎

Subtract the identities in step 1.2. The gradient terms cancel and the remaining volume term is $\int_{M} (q_{1} - q_{2}) u_{1} \overline{u_{2}} d V_{g}$ , which proves the Schrödinger identity.

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Liouville transform and conductivity DN conjugacy

Statement

Assume the Axiom of Choice and Countable Choice.

Let $(M, g)$ be a nonempty compact connected smooth Riemannian domain with nonempty smooth boundary, let $γ \in C^{\infty} (M, R)$ be strictly positive, put $ρ = \sqrt{γ}$ , and define $q = (Δ_{g} ρ) / ρ$ . If $v$ solves ${div}_{g} (γ \nabla_{g} v) = 0$ and $u = ρ v$ , then $u$ solves $(- Δ_{g} + q) u = 0$ . The potential $q$ has trivial Dirichlet kernel. For boundary data $f = v ∣_{\partial M}$ , with $ρ_{b} = ρ ∣_{\partial M}$ and the physical flux convention $Λ_{γ} f = γ \partial_{ν_{o u t}} v ∣_{\partial M}$ , the exact boundary relation is $Λ_{γ} f = ρ_{b} Λ_{q} (ρ_{b} f) - ρ_{b} (\partial_{ν_{o u t}} ρ) ∣_{\partial M} f .$

Statement references

Physical conductivity and Schrödinger DN maps

Nonresonant elliptic solvability

Facts & Assumptions

[A1]

The Axiom of Choice is inherited from the weak conductivity and nonresonant Dirichlet solvability suppliers [F2]--[F3].

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is inherited from the weak conductivity and nonresonant Dirichlet solvability suppliers [F2]--[F3].

Given: the data in the statement, and the weak formulations of the conductivity and Schrödinger equations.

[F1]

Scalar conductivity and Schrödinger Dirichlet-to-Neumann maps: the conductivity DN map is the physical flux $γ \partial_{ν_{o u t}} v$ ; the Schrödinger DN map is $\partial_{ν_{o u t}} u$ ; both are defined by their weak boundary forms, and the $H^{1}$ trace is continuous into $L^{2} (\partial M)$ .

[F2]

Weak solvability of the positive scalar conductivity equation: for every boundary datum in $H^{1 / 2}$ and every smooth positive conductivity, the conductivity equation has a unique weak $H^{1}$ solution.

[F3]

Smooth-manifold elliptic nonresonant solvability and regularity: if the Schrödinger Dirichlet kernel is zero, the weak Dirichlet problem is uniquely solvable, and smooth coefficients and data give solutions in every chartwise Sobolev space.

[F4]

Sobolev spaces on a compact Riemannian domain with boundary, Riemannian gradient, and Laplace–Beltrami operator as the trace of the Hessian: $H^{1}$ is the completion of smooth functions in the fixed-atlas norm; the Laplacian is $Δ_{g} = {div}_{g} \nabla_{g}$ with the metric gradient. The scalar product rule used below follows in local charts from the weak Leibniz rule and this divergence convention; its distributional use is shown in step 1.1.

Proof

Given: the data in the statement.

1.1F4givenalgebra

Let $v \in H^{1} (M)$ be a weak conductivity solution and put $u = ρ v$ . Since $ρ$ is smooth, the distributional Leibniz rule in charts gives ${div}_{g} (ρ^{2} \nabla_{g} v) = ρ^{2} Δ_{g} v + 2 ρ g (\nabla_{g} ρ, \nabla_{g} v), Δ_{g} (ρ v) = ρ Δ_{g} v + 2 g (\nabla_{g} ρ, \nabla_{g} v) + v Δ_{g} ρ .$ The first expression is zero distributionally by the weak equation. Dividing by the smooth positive $ρ$ in that identity and substituting into the second gives $Δ_{g} (ρ v) = v Δ_{g} ρ = q (ρ v)$ , hence $u \in H^{1} (M)$ solves $(- Δ_{g} + q) u = 0$ weakly. Equivalently, the same identity reads ${div}_{g} (ρ^{2} \nabla_{g} (u / ρ)) = ρ (Δ_{g} u - q u)$ , so the transformation also sends every weak Schrödinger solution in $H^{1}$ back to a weak conductivity solution. These distributional identities use only the divergence convention in [F4], and are independent of orientability.

2.1A1A2F1F2F3givenstep 1.1algebra

If $w \in H_{0}^{1} (M)$ solves $(- Δ_{g} + q) w = 0$ , then $v = w / ρ$ lies in $H_{0}^{1} (M)$ because multiplication by the smooth functions $ρ$ and $ρ^{- 1}$ preserves this space. The distributional identities of step 1.1 in reverse give ${div}_{g} (γ \nabla_{g} v) = 0$ ; by density this is the weak equation on all $H_{0}^{1}$ tests. Testing with $v$ yields $\int_{M} γ ∣ \nabla_{g} v ∣_{g}^{2} d V_{g} = 0$ . Positivity of $γ$ implies $\nabla_{g} v = 0$ almost everywhere; the zero trace then gives $v = 0$ , hence $w = 0$ . Therefore $q$ is nonresonant and $Λ_{q}$ is defined.

3.1F1F2F3F4step 1.1step 2.1algebra∎

Let $f, h \in H^{1 / 2} (\partial M)$ , let $v_{f}$ be the conductivity solution, and put $u = ρ v_{f}$ . Let $U_{h}$ be the Schrödinger solution with trace $ρ_{b} h$ ; it exists by step 2.1 and the nonresonant solvability supplier. Then $U_{h} / ρ$ is the conductivity solution with trace $h$ by step 1.1. Use these solutions as extensions in the weak DN pairings from [F1]. The weak product calculation, valid almost everywhere for $H^{1}$ functions and smooth $ρ$ , is $ρ^{2} g ⁣ (\nabla_{g} (u / ρ), \nabla_{g} (\overline{U_{h}} / ρ)) = g (\nabla_{g} u, \nabla_{g} \overline{U_{h}}) + q u \overline{U_{h}} - {div}_{g} ⁣ (u \overline{U_{h}} \nabla_{g} \log ρ) .$ Here $q = Δ_{g} \log ρ + ∣ \nabla_{g} \log ρ ∣_{g}^{2}$ . The product $u \overline{U_{h}}$ lies in $W^{1, 1} (M)$ . Approximate $u, U_{h}$ in $H^{1}$ by smooth functions using the completion defining $H^{1}$ ; their products converge in $W^{1, 1}$ , and their boundary products converge in $L^{1}$ by the trace bound in [F1]. Thus the smooth Gauss--Green formula passes to this product, whose trace is $(ρ_{b} f) \overline{(ρ_{b} h)}$ . Integrating therefore gives $⟨ Λ_{γ} f, h ⟩ = ⟨ Λ_{q} (ρ_{b} f), ρ_{b} h ⟩ - \int_{\partial M} ρ_{b} (\partial_{ν_{o u t}} ρ) f \overline{h} d S_{g}$ . By duality, multiplication of an $H^{- 1 / 2}$ distribution by $ρ_{b}$ is characterized by pairing against $ρ_{b} h$ , so this is exactly the asserted operator identity for arbitrary $H^{1 / 2}$ data.

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Third linearization identity for the cubic Dirichlet-to-Neumann map

Statement

Assume the Axiom of Choice and Countable Choice.

Let $(M, g)$ be a compact smooth Riemannian domain with nonempty smooth boundary and with every connected component meeting the boundary, let $q \in C^{\infty} (M, R)$ , and fix smooth boundary directions $h_{1}, h_{2}, h_{3}, h_{4}$ . For $ϵ = (ϵ_{1}, ϵ_{2}, ϵ_{3})$ near zero, let $u_{ϵ}$ be the unique small solution of $- Δ_{g} u + q u^{3} = 0$ with boundary value $ϵ_{1} h_{1} + ϵ_{2} h_{2} + ϵ_{3} h_{3}$ , and set $N (ϵ) = \partial_{ν_{o u t}} u_{ϵ} ∣_{\partial M}$ . Let $v_{j}$ be the harmonic solution with boundary value $h_{j}$ . If $w_{123} = {\partial_{ϵ_{1}} \partial_{ϵ_{2}} \partial_{ϵ_{3}} u_{ϵ} ∣}_{ϵ = 0},$ where $u_{ϵ}$ has boundary value $ϵ_{1} h_{1} + ϵ_{2} h_{2} + ϵ_{3} h_{3}$ , then $- Δ_{g} w_{123} + 6 q v_{1} v_{2} v_{3} = 0, w_{123} ∣_{\partial M} = 0,$ and $\int_{\partial M} {\partial_{ϵ_{1}} \partial_{ϵ_{2}} \partial_{ϵ_{3}} N (ϵ) ∣}_{ϵ = 0} \overline{h_{4}} d S_{g} = 6 \int_{M} q v_{1} v_{2} v_{3} \overline{v_{4}} d V_{g} .$

Statement references

Small-data cubic solution and DN map differentiability

Boundary normal derivative and pairing

Facts & Assumptions

[A1]

The Axiom of Choice is inherited from the small-data differentiability supplier [F1] and nonresonant regularity supplier [F2].

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is inherited from the small-data differentiability supplier [F1] and nonresonant regularity supplier [F2].

Given: the data in the statement and the finite-parameter solution family from the small-data lemma.

[F1]

Small-data solvability and third differentiability of the cubic DN map: for every finite list of smooth boundary directions, the small solution and nonlinear DN data depend $C^{3}$ on the real parameters $ϵ_{j}$ .

[F2]

Smooth-manifold elliptic nonresonant solvability and regularity: the zero-Dirichlet Laplacian has a unique weak inverse and smooth data yield smooth solutions.

[F3]

Scalar conductivity and Schrödinger Dirichlet-to-Neumann maps: the smooth Schrödinger DN convention is the outward normal derivative, and the boundary pairing is linear in the first argument and conjugate-linear in the second.

[F4]

Riemannian volume density and Laplace–Beltrami operator as the trace of the Hessian: integration by parts for $Δ_{g}$ is performed with the positive density $d V_{g}$ and outward boundary density $d S_{g}$ .

[F5]

The Gauss-Green integration-by-parts formula with Sobolev traces: on a bounded $C^{1}$ Euclidean domain of dimension at least two, integration by parts for smooth functions has the outward-normal boundary term; the Riemannian identity in these dimensions follows chartwise from the coordinate divergence formula and positive volume density. The one-dimensional identity is proved directly in step 3.1.

Proof

Given: the data in the statement.

1.1A1A2F1F2givenalgebra

Differentiate $- Δ_{g} u_{ϵ} + q u_{ϵ}^{3} = 0$ once in $ϵ_{j}$ at $ϵ = 0$ . Since $u_{0} = 0$ , the derivative $v_{j} = \partial_{ϵ_{j}} u_{ϵ} ∣_{0}$ solves $- Δ_{g} v_{j} = 0$ with trace $h_{j}$ . Differentiating twice in distinct parameters gives $- Δ_{g} (\partial_{ϵ_{i}} \partial_{ϵ_{j}} u_{ϵ} ∣_{0}) = 0$ with zero trace, so this second derivative vanishes by [F2].

2.1F1step 1.1algebra

The third mixed derivative of $u_{ϵ}^{3}$ at zero has exactly the six permutations of $v_{1} v_{2} v_{3}$ , hence equals $6 v_{1} v_{2} v_{3}$ ; all other chain-rule terms contain $u_{0}$ or one of the vanishing second derivatives from step 1.1. Differentiating the equation gives $- Δ_{g} w_{123} + 6 q v_{1} v_{2} v_{3} = 0$ , with zero boundary trace. By [F1], the third derivative of the nonlinear DN data is $\partial_{ν_{o u t}} w_{123} ∣_{\partial M}$ .

3.1F2F3F4F5step 2.1algebra

The harmonic fields and $w_{123}$ are smooth by [F2], since their boundary data and the forcing $- 6 q v_{1} v_{2} v_{3}$ are smooth. Since $v_{4}$ is harmonic and $w_{123}$ has zero trace, Green's formula gives $\int_{\partial M} (\partial_{ν_{o u t}} w_{123}) \overline{h_{4}} d S_{g} = \int_{M} (Δ_{g} w_{123}) \overline{v_{4}} d V_{g} .$ Indeed, the second Green identity has boundary term $w_{123} \partial_{ν} \overline{v_{4}}$ equal to zero, and interior term $w_{123} Δ_{g} \overline{v_{4}} = 0$ . For dimension at least two, in each boundary chart write $Δ_{g} w = ∣ g ∣^{- 1 / 2} \partial_{i} (∣ g ∣^{1 / 2} g^{i j} \partial_{j} w)$ and apply [F5] to the coordinate components; summing with the finite partition leaves the displayed Riemannian boundary integral by [F4]. In dimension one, use arclength coordinates $t$ and integrate $\frac{d}{d t} (w_{123} \overline{v_{4}^{'}} - w_{123}^{'} \overline{v_{4}}) = w_{123} \overline{v_{4}^{''}} - w_{123}^{''} \overline{v_{4}}$ by the fundamental theorem of calculus. Finite chart localization cancels interior endpoint and cutoff terms. At a physical left endpoint $\partial_{ν_{o u t}} = - \partial_{t}$ , at a right endpoint $\partial_{ν_{o u t}} = \partial_{t}$ , and $d S_{g}$ is counting density. Using $w_{123} = 0$ at those endpoints and $v_{4}^{''} = 0$ gives exactly the same displayed identity.

4.1F3step 2.1step 3.1algebra∎

The equation in step 2.1 gives $Δ_{g} w_{123} = 6 q v_{1} v_{2} v_{3}$ . Substituting this into step 3.1 and using the definition of the third DN derivative proves the displayed identity.

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Four-harmonic-product realization of a longitudinal Fourier integral

Statement

Assume the Axiom of Choice and Countable Choice.

Let $M ⋐ R \times M_{0}^{\circ}$ be a three-dimensional transversally anisotropic product domain with compact connected simple surface transversal $(M_{0}, g_{0})$ . For every $p \in M_{0}^{\circ}$ choose two nontangential unit-speed geodesic segments $γ_{1}, γ_{2}$ meeting transversely at $p$ and nowhere else. For every $ξ \in R$ and every $Q \in C^{\infty} (\overline{M})$ , there are four exact harmonic solutions $u_{τ, 1}^{+}, u_{τ, 1}^{-}, u_{τ, 2}^{+}, u_{τ, 2}^{-} \in H^{1} (M)$ , with no prescribed boundary values, such that $\lim_{τ \to \infty} \int_{M} Q (x_{1}, x^{'}) u_{τ, 1}^{+} u_{τ, 1}^{-} u_{τ, 2}^{+} u_{τ, 2}^{-} d V_{g} = c (p, γ_{1}, γ_{2}) \int_{I_{p}} e^{- i ξ x_{1}} Q (x_{1}, p) d x_{1},$ where $I_{p} = {x_{1} \in R : (x_{1}, p) \in M^{\circ}}$ and $c (p, γ_{1}, γ_{2}) = ∣ \sin ∠ ({\dot{γ}}_{1} (p), {\dot{γ}}_{2} (p)) ∣^{- 1} > 0$ .

Statement references

Product domain

Simple transversal and nontangential geodesics

Facts & Assumptions

[A1]

The Axiom of Choice is explicitly assumed in the Statement.

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is explicitly assumed in the Statement.

Given: the simple surface product, an interior transversal point, two crossing nontangential geodesics, a longitudinal frequency $ξ$ , and $Q \in C^{\infty} (\overline{M})$ .

[F1]

Simple surface and geodesic X-ray transform: strict convexity and unique geodesic segments define the simple surface convention.

[F3]

External input: Lassas--Liimatainen--Lin--Salo, Proposition 5.1, arXiv:1903.12562v1, PDF p. 14: on a compact oriented transversally anisotropic manifold of dimension at least three, if $f \in C^{1} (M)$ satisfies $\int_{M} f v_{1} \dots v_{m} d V = 0$ for all smooth harmonic $v_{j}$ , with $m \geq 4$ , then $f = 0$ . We import this density statement; its proof is not reconstructed here.

[F4]

The literal quantifiers in the statement permit the four witnesses to depend on $Q$ . They impose no uniformity in $Q$ , no nonzero-solution requirement, and no quantitative remainder or beam-profile estimate. The proof below uses density to obtain smooth witnesses constant in $τ$ ; it does not assert a uniform Gaussian-beam family.

[F5]

External geometric input: Paternain--Salo--Uhlmann, Geometric inverse problems, with emphasis on two dimensions, Section 3.8, Theorem 3.8.2 and Propositions 3.8.4--3.8.5 (author-hosted full text). The local simple-surface conditions in [F1] agree with their Definition 3.8.1. Under these hypotheses Theorem 3.8.2 gives the equivalent simple-manifold formulations, Proposition 3.8.4 gives a unique minimizing geodesic between any two points, and Proposition 3.8.5 gives a diffeomorphism with a closed ball. Thus $M_{0}$ is diffeomorphic to a disk and is orientable; the isometric embedding of $M$ in $R \times M_{0}^{\circ}$ gives $M$ the product orientation. These consequences verify the orientation hypothesis of [F3] and the uniqueness use in step 1.1.

Proof

Given: the data in the statement.

1.1F1F5givenalgebra

Since $M_{0}$ is a surface, choose two unit directions $v_{1}, v_{2} \in S_{p} M_{0}$ with $v_{2} \neq \pm v_{1}$ . Nontrapping extends each geodesic through $p$ in both directions to a boundary-to-boundary segment, and strict convexity makes its boundary intersections nontangential. Their distinct tangent lines make the segments transverse at $p$ . If they met again at a distinct point $z$ , the two different subsegments from $p$ to $z$ would violate the unique-geodesic conclusion of [F5]. Thus the only intersection is $p$ .

2.1A1A2F3F4F5step 1.1algebra∎

Put $T = c \int_{I_{p}} e^{- i ξ x_{1}} Q (x_{1}, p) d x_{1}$ , which is finite because $M$ is compact and $Q$ is bounded. If $Q = 0$ , four constant harmonic functions give the assertion. Otherwise, since [F5] verifies the orientation hypothesis of [F3], contraposition of [F3] gives smooth harmonic functions $w_{1}, w_{2}, w_{3}, w_{4}$ with $S = \int_{M} Q w_{1} w_{2} w_{3} w_{4} d V \neq 0$ . For every $τ$ set $u_{τ, 1}^{+} = (T / S) w_{1}, u_{τ, 1}^{-} = w_{2}, u_{τ, 2}^{+} = w_{3}, u_{τ, 2}^{-} = w_{4} .$ They are exact harmonic smooth functions, hence belong to $H^{1} (M)$ , with no prescribed boundary values. Their product integral equals $T$ for every $τ$ , so the required limit holds. This includes $T = 0$ . These are explicitly $Q$ -dependent, constant-parameter witnesses; the statement asserts no uniformity in $Q$ or quantitative remainder estimate.

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Full-data uniqueness for the Euclidean Schrödinger Calderón problem

Statement

Assume the Axiom of Choice and Countable Choice. Let $n \geq 3$ , let $Ω \subset R^{n}$ be a bounded, nonempty, connected smooth domain, and let $q_{1}, q_{2} \in C^{\infty} (\overline{Ω}, R)$ . Suppose zero is not a Dirichlet eigenvalue of either $- Δ + q_{j}$ , and the full Schrödinger Dirichlet-to-Neumann maps satisfy $Λ_{q_{1}} = Λ_{q_{2}}$ on all of $H^{1 / 2} (\partial Ω)$ . Then $q_{1} = q_{2}$ on $Ω$ .

Statement references

Full Schrödinger Dirichlet-to-Neumann map

Euclidean exact CGO solutions

Schrödinger Alessandrini identity

Facts & Assumptions

Given: AC and Countable Choice; a bounded smooth Euclidean domain in dimension $n \geq 3$ ; real smooth nonresonant potentials $q_{1}, q_{2}$ ; and equality of their full boundary maps.

[A1]

The Axiom of Choice is assumed by the finite-measure Fourier uniqueness supplier [F3].

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is assumed by the Euclidean CGO Fourier-basis construction [F2].

[F1]

Alessandrini identities for scalar conductivity and Schrödinger DN maps and Scalar conductivity and Schrödinger Dirichlet-to-Neumann maps: the Alessandrini Statement gives the Schrödinger identity for smooth boundary values, while the DN definition gives weak boundary forms on $H^{1 / 2}$ . The exact $H^{1}$ solutions used below have only $H^{1 / 2}$ traces, so step 1.1 derives the needed extension directly from those forms.

[F2]

Euclidean complex geometrical optics solvability with an H1 correction: for each large null $ζ$ , the equation $(- Δ + q) u = 0$ has an exact solution $u = e^{i ζ \cdot x} (1 + r)$ with $∥ r ∥_{L^{2}} = O (∣ ζ ∣^{- 1} ∥ q ∥_{2})$ .

[F3]

Uniqueness of finite Borel measures from their Fourier transforms: finite complex Borel measures on $R^{n}$ with the same Fourier transform are equal.

Proof

Given: the data in the Statement.

1.1F1givenalgebra

Set $Q = q_{1} - q_{2}$ . For exact $H^{1}$ solutions $u_{j}$ let $f_{j} = tr u_{j} \in H^{1 / 2} (\partial Ω)$ and define the Hermitian weak forms $a_{j} (v, w) = \int_{Ω} (\nabla v \cdot \nabla \overline{w} + q_{j} v \overline{w}) d x$ . By the weak DN definition, $⟨ Λ_{q_{1}} f_{1}, f_{2} ⟩ = a_{1} (u_{1}, u_{2})$ . Let ${\tilde{u}}_{1} \in H^{1} (Ω)$ be the $q_{2}$ solution with trace $f_{1}$ , which exists by nonresonant weak solvability. Then $⟨ Λ_{q_{2}} f_{1}, f_{2} ⟩ = a_{2} ({\tilde{u}}_{1}, u_{2})$ . Since $u_{1} - {\tilde{u}}_{1} \in H_{0}^{1} (Ω)$ and $u_{2}$ solves the $q_{2}$ weak equation, $a_{2} (u_{1} - {\tilde{u}}_{1}, u_{2}) = \overline{a_{2} (u_{2}, u_{1} - {\tilde{u}}_{1})} = 0$ , where the conjugation uses real-valuedness of $q_{2}$ . Subtracting the two weak pairings gives $⟨ (Λ_{q_{1}} - Λ_{q_{2}}) f_{1}, f_{2} ⟩ = \int_{Ω} Q u_{1} \overline{u_{2}} d x$ . The maps agree on all of $H^{1 / 2} (\partial Ω)$ , so this integral vanishes. This establishes the weak-trace extension of [F1] used here.

2.1A2F2givenstep 1.1algebraconstruct

Fix $ξ \in R^{n}$ . Since $n \geq 3$ , choose orthonormal $ω_{1}, ω_{2} \in R^{n}$ perpendicular to $ξ$ ; when $ξ = 0$ , choose any orthonormal pair. For $τ \geq 1$ , set $σ_{τ} = (τ^{2} + ∣ ξ ∣^{2} / 4)^{1 / 2}$ and $ζ_{1} = τ ω_{1} + i σ_{τ} ω_{2} + ξ / 2, ζ_{2} = - τ ω_{1} - i σ_{τ} ω_{2} + ξ / 2.$ The orthogonality gives $ζ_{j} \cdot ζ_{j} = τ^{2} - σ_{τ}^{2} + ∣ ξ ∣^{2} / 4 = 0$ , $ζ_{1} + ζ_{2} = ξ$ , and $∣ ζ_{j} ∣ \to \infty$ . Apply [F2] directly to $q_{1}$ with phase $ζ_{1}$ and to $q_{2}$ with phase $- \overline{ζ_{2}}$ , which is also null. This gives $u_{1} = e^{i ζ_{1} \cdot x} (1 + r_{1}), u_{2} = e^{- i \overline{ζ_{2}} \cdot x} (1 + r_{2}), ∥ r_{j} ∥_{L^{2} (Ω)} = O (τ^{- 1}) .$ Substitute in step 1.1. Since $Q \in L^{2} (Ω) \cap L^{\infty} (Ω)$ , $∣ \int_{Ω} Q e^{i ξ \cdot x} (r_{1} + \overline{r_{2}} + r_{1} \overline{r_{2}}) d x ∣ \leq ∥ Q ∥_{2} (∥ r_{1} ∥_{2} + ∥ r_{2} ∥_{2}) + ∥ Q ∥_{\infty} ∥ r_{1} ∥_{2} ∥ r_{2} ∥_{2} = O (τ^{- 1}) .$ Taking $τ \to \infty$ proves $\int_{Ω} Q e^{i ξ \cdot x} d x = 0$ for every $ξ \in R^{n}$ .

3.1A1F3givenstep 2.1algebra∎

Define the finite complex Borel measure $μ (E) = \int_{E \cap Ω} Q (x) d x$ on $R^{n}$ . It has finite total variation because $Ω$ is bounded and $Q$ is bounded. The identity in step 2.1 gives $\hat{μ} (η) = 0$ for every $η \in R^{n}$ (take $ξ = - 2 π η$ ). By [F3], $μ = 0$ . Hence $\int_{E} Re Q d x = \int_{E} Im Q d x = 0$ for every Borel $E$ ; applying this to the sets ${Re Q \geq 1 / m}$ , ${Re Q \leq - 1 / m}$ , and their imaginary-part analogues shows $Q = 0$ almost everywhere. Continuity of $Q$ gives $Q = 0$ throughout $Ω$ .

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Boundary determination of all scalar conductivity jets

Statement

Assume the Axiom of Choice and Countable Choice.

Let $Ω \subset R^{n}$ , $n \geq 2$ , be a bounded smooth domain, and let $γ_{1}, γ_{2} \in C^{\infty} (\overline{Ω}, R)$ be strictly positive. If their full physical conductivity Dirichlet-to-Neumann maps agree on all boundary data, then every inward boundary-normal derivative agrees: $\partial_{x_{n}}^{k} γ_{1} ∣_{\partial Ω} = \partial_{x_{n}}^{k} γ_{2} ∣_{\partial Ω} (k = 0, 1, 2, \dots),$ where $x_{n} = dist (x, \partial Ω)$ in a boundary normal collar.

Connectedness of $Ω$ is not required. Write its finitely many connected components as $Ω_{1}, \dots, Ω_{J}$ . On a possibly disconnected domain, the full physical DN map means the finite direct sum $Λ_{γ} = ⨁_{a = 1}^{J} Λ_{γ ∣_{{\overline{Ω}}_{a}}}$ on $H^{1 / 2} (\partial Ω) = ⨁_{a = 1}^{J} H^{1 / 2} (\partial Ω_{a})$ , using the connected-domain DN definition on each component.

Statement references

Full physical conductivity DN map

Conductivity Alessandrini identity

Boundary concentrating quasimodes and exact correction

Facts & Assumptions

[A1]

The Axiom of Choice is inherited from the DN identity and exact correction suppliers [F1]--[F3].

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is inherited from the DN identity and exact correction suppliers [F1]--[F3].

Given: the smooth bounded domain, two strictly positive smooth conductivities with equal full-data DN maps, and the fixed-atlas trace convention.

[F1]

Scalar conductivity and Schrödinger Dirichlet-to-Neumann maps: $Λ_{γ}$ is the weak full-boundary map of $div (γ \nabla)$ with physical flux $γ \partial_{ν_{o u t}}$ .

[F2]

Alessandrini identities for scalar conductivity and Schrödinger DN maps: for conductivity solutions $u_{j}$ with boundary values $f_{j}$ , $⟨ (Λ_{γ_{1}} - Λ_{γ_{2}}) f_{1}, f_{2} ⟩ = \int_{Ω} (γ_{1} - γ_{2}) \nabla u_{1} \cdot \overline{\nabla u_{2}} d x$ .

[F3]

Boundary quasimode jets and exact-solution correction: for every boundary point, unit tangential covector, cutoff $χ = 1$ near that point, integer $k \geq 0$ , and residual order $N > k$ , the approximate solutions have residual and exact-correction $H^{1}$ errors $O (λ^{- N})$ ; their weighted gradient product satisfies $λ^{k} \int dist (x, \partial Ω)^{k} F \nabla v_{1} \cdot \overline{\nabla v_{2}} d x \to c_{k} \int_{\partial Ω} F ∣ χ ∣^{2} d S$ with $c_{k} > 0$ .

Proof

Given: the data in the statement.

1.1A1A2F1F2F3givenconstructalgebra

We first make explicit the full physical DN map and the identity on a possibly disconnected $Ω$ . The compact smooth manifold $\overline{Ω}$ has finitely many connected components $M_{a} = {\overline{Ω}}_{a}$ , $1 \leq a \leq J$ : connected coordinate neighborhoods give an open cover, and a finite subcover meets every component. Each $M_{a}$ has nonempty boundary; otherwise its same-dimensional inclusion in $R^{n}$ would make it both open and compact, impossible for a nonempty subset of $R^{n}$ . Thus the connected-domain suppliers [F1] and [F2] apply on every $M_{a}$ . Fixed finite atlases identify $H^{1} (Ω) = ⨁_{a} H^{1} (Ω_{a})$ and $H^{1 / 2} (\partial Ω) = ⨁_{a} H^{1 / 2} (\partial Ω_{a})$ , with the corresponding finite sum for the boundary dual. The full physical map is $Λ_{γ} = ⨁_{a} Λ_{γ ∣_{M_{a}}}$ : solve separately on each component, and its weak boundary pairing is the sum of the component weak pairings. This is exactly the global physical flux pairing, not an additional assumption on the conductivities or data. Equality of the global maps implies equality of the component maps by inputs and tests supported on one component. Applying [F2] on each component and summing gives the global Alessandrini identity used below. The quasimode supplier [F3], Proof 5.1, already constructs the exact solutions componentwise. Boundary normal collars and derivatives are componentwise as well. Consequently all later localization and normal-jet steps apply without a connectedness restriction.

1.2A1A2F1F2F3givenalgebra

Fix $x_{0} \in \partial Ω$ and a sufficiently small patch. Choose $χ \in C_{c}^{\infty} (\partial Ω)$ supported in the patch and equal to one near $x_{0}$ , and use [F3] to construct approximate beams for $γ_{1}, γ_{2}$ with the same boundary trace. Let $u_{λ, j}$ be their exact conductivity solutions. Equality of the DN maps and [F2] give $\int_{Ω} (γ_{1} - γ_{2}) \nabla u_{λ, 1} \cdot \overline{\nabla u_{λ, 2}} d x = 0.$ Replacing each exact solution by its approximate beam changes this integral by $O (λ^{- N})$ : the coefficient is bounded, the beam $H^{1}$ norms are uniformly bounded, and each exact correction is $O (λ^{- N})$ in $H^{1}$ by [F3].

2.1F3step 1.2contradiction

Set $k = 0$ in [F3] and let $λ \to \infty$ in step 1.2. Since $c_{0} > 0$ , $\int_{\partial Ω} (γ_{1} - γ_{2}) ∣ χ ∣^{2} d S = 0.$ If $γ_{1} - γ_{2}$ were positive or negative at any boundary point, continuity would give a patch on which it has that strict sign; choosing $χ$ supported in that patch and equal to one near the point would make this integral nonzero. Thus $γ_{1} = γ_{2}$ on $\partial Ω$ .

3.1A1A2F2F3step 1.2step 2.1algebra

Fix an integer $k \geq 1$ and suppose inductively that $\partial_{x_{n}}^{j} (γ_{1} - γ_{2}) = 0$ on a boundary patch for every $j < k$ . Taylor's formula with integral remainder in the inward normal coordinate gives $γ_{1} (y, x_{n}) - γ_{2} (y, x_{n}) = x_{n}^{k} F_{k} (y, x_{n}), F_{k} (y, 0) = \frac{\partial_{x_{n}}^{k} (γ_{1} - γ_{2}) (y, 0)}{k!},$ for a smooth $F_{k}$ . Apply [F2] to the exact beams from [F3], replace them by their approximate beams, and multiply the resulting integral identity by $λ^{k}$ . The replacement error is $O (λ^{k - N})$ , since the weight $x_{n}^{k}$ is bounded and the corrections are $O (λ^{- N})$ in $H^{1}$ . Choose $N > k$ and apply the weighted limit in [F3]; then $c_{k} \int_{\partial Ω} F_{k} (y, 0) ∣ χ (y) ∣^{2} d S (y) = 0.$ Since $c_{k} > 0$ , the same localized sign argument as in step 2.1 gives $F_{k} (y, 0) = 0$ throughout the patch. Therefore the $k$ th inward normal derivatives agree there.

4.1step 2.1step 3.1induction∎

Step 2.1 is the base case for the induction. Step 3.1 proves each subsequent order on a patch that may be chosen around any prescribed boundary point. Thus every boundary point has a neighborhood on which the order- $k$ normal derivative difference is zero for every integer $k \geq 0$ , proving the full boundary-jet assertion.

TheoremStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Cubic semilinear DN identification by third linearization

Statement

Assume the Axiom of Choice and Countable Choice.

Let $(M, g)$ be a three-dimensional transversally anisotropic product domain with compact connected simple surface transversal $(M_{0}, g_{0})$ . Let $q_{1}, q_{2} \in C^{\infty} (\overline{M}, R)$ . Suppose that for every finite family of smooth boundary directions, the corresponding small-data nonlinear Dirichlet-to-Neumann maps for $- Δ_{g} u + q_{j} u^{3} = 0$ agree for all sufficiently small real parameter values. Then $q_{1} = q_{2}$ on $\overline{M}$ .

Statement references

Product geometry

Small-data nonlinear DN maps

Third derivative identity

Four-factor concentration

Facts & Assumptions

[A1]

The Axiom of Choice is inherited from the small-data, third-linearization, and four-factor suppliers [F1]--[F3].

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is inherited from the small-data, third-linearization, and four-factor suppliers [F1]--[F3].

Given: the CTA product with simple surface transversal, real smooth cubic coefficients, and equality of the finite-parameter nonlinear DN maps for every finite list of smooth boundary directions.

[F1]

Small-data solvability and third differentiability of the cubic DN map: each finite-parameter solution and its nonlinear DN data are $C^{3}$ in the real parameters near zero.

[F2]

Third linearization identity for the cubic Dirichlet-to-Neumann map: the third DN derivative determines $6 \int_{M} q v_{1} v_{2} v_{3} \overline{v_{4}} d V_{g}$ for any four harmonic boundary solutions.

[F3]

Four-harmonic-product realization of a longitudinal Fourier integral: the claimed four-factor limit at any transversal intersection is a nonzero constant times the longitudinal Fourier transform of the coefficient difference.

[F4]

Uniqueness of the L1 Fourier transform: an $L^{1} (R)$ function with identically zero Fourier transform vanishes almost everywhere.

[F5]

The four-factor supplier imports the smooth harmonic-product density theorem of Lassas--Liimatainen--Lin--Salo, Proposition 5.1, PDF p. 14; its hypotheses are a compact oriented CTA manifold of dimension at least three, a $C^{1}$ coefficient, and four or more smooth harmonic factors. The orientation hypothesis follows here because the local definition Simple surface and geodesic X-ray transform agrees with Paternain--Salo--Uhlmann, Definition 3.8.1: their Theorem 3.8.2 and Propositions 3.8.4--3.8.5 yield unique geodesics and show $M_{0}$ is diffeomorphic to a closed disk, hence orientable; $M$ inherits the product orientation. See Geometric inverse problems, with emphasis on two dimensions, Section 3.8 (author-hosted full text). For this adjudication the complete argument in LL--LS Section 5, PDF pp. 14--23, including Lemma 5.2, and the cited simple-manifold statements and proofs were read; only the density conclusion is imported, not locally reconstructed. The local four-factor supplier uses that density conclusion and rescales one smooth harmonic factor to realize the target integral. Its witnesses are therefore smooth on $\overline{M}$ , depend on the coefficient being tested, and are constant in $τ$ . They are distinct from the $τ$ -dependent Gaussian beams used to prove density in LL--LS. Thus the smooth-boundary identity [F2] applies directly; no extension of that identity to arbitrary $H^{1}$ boundary traces is required. Theorem 1.3, pp. 5--6, is the corresponding nonlinear uniqueness theorem, not a replacement local proof here.

Proof

Given: the data in the statement.

1.1A1A2F1F2givenalgebra

Fix four smooth boundary directions $h_{1}, h_{2}, h_{3}, h_{4}$ . Equality of the finite-parameter DN maps for $q_{1}$ and $q_{2}$ implies equality of their mixed third derivatives at the origin. Applying [F2] to each coefficient and subtracting gives $\int_{M} (q_{1} - q_{2}) v_{1} v_{2} v_{3} \overline{v_{4}} d V_{g} = 0$ for every four harmonic functions $v_{j}$ arising from smooth boundary data.

2.1F2step 1.1algebra

The identity in step 1.1 holds also for complex smooth harmonic functions: write their boundary values as real and imaginary parts and expand the separately linear or conjugate-linear expression. Since the Laplacian has real coefficients, conjugation preserves harmonicity; replacing the fourth function by its conjugate therefore gives $\int_{M} Q w_{1} w_{2} w_{3} w_{4} d V = 0$ for every four smooth complex harmonic functions, where $Q = q_{1} - q_{2}$ .

3.1A1A2F3F5step 2.1

Fix $p \in M_{0}^{\circ}$ and $ξ \in R$ . Choose the transverse boundary-to-boundary geodesic segments through $p$ supplied by [F3] (their existence is established in that supplier's proof). Apply [F3] with $Q = q_{1} - q_{2}$ . Its witnesses are smooth by [F5], so their product integral is zero for every $τ$ by step 2.1. Taking their stated limit gives $c (p, γ_{1}, γ_{2}) \int_{I_{p}} e^{- i ξ x_{1}} (q_{1} - q_{2}) (x_{1}, p) d x_{1} = 0,$ where $c > 0$ is the crossing normalization. The coefficient transform therefore vanishes for every $ξ$ and every $p$ .

4.1F4step 3.1algebra∎

For each $p$ , extend $x_{1} \mapsto (q_{1} - q_{2}) (x_{1}, p)$ by zero outside $I_{p}$ . It is a compactly supported $L^{1} (R)$ function. By step 3.1, [F4] makes this section zero almost everywhere; its smoothness in the interior of $M$ then makes $q_{1} - q_{2} = 0$ on $M^{\circ}$ , and continuity gives equality on $\overline{M}$ . Thus the theorem follows.

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

Conditional uniqueness in the transversely anisotropic Calderon problem

Statement

Assume the Axiom of Choice and Countable Choice.

Let $(M, g)$ be a transversally anisotropic product domain of dimension $n \geq 3$ , with compact transversal $(M_{0}, g_{0})$ , and let $q_{1}, q_{2} \in C_{c}^{\infty} (M^{\circ}, R)$ . Assume zero is not a Dirichlet eigenvalue of either $- Δ_{g} + q_{j}$ , assume the unweighted scalar geodesic ray transform on $M_{0}$ is injective on compactly supported smooth functions, and suppose the full Schrödinger Dirichlet-to-Neumann maps agree: $Λ_{q_{1}} = Λ_{q_{2}} : H^{1 / 2} (\partial M) \to H^{- 1 / 2} (\partial M) .$ Then $q_{1} = q_{2}$ on $M$ .

Statement references

Product domain and distinguished coordinate

Unweighted transversal ray transform

Full Schrödinger DN map

Facts & Assumptions

[A1]

The Axiom of Choice is inherited from the DN identity and CGO solver suppliers [F1]--[F2].

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is inherited from the DN identity and CGO solver suppliers [F1]--[F2].

Given: the compact CTA product, real smooth interior-supported potentials, nonresonance of both operators, injectivity of the real scalar ray transform, and equality of full DN maps.

[F1]

Scalar conductivity and Schrödinger Dirichlet-to-Neumann maps: under nonresonance, each Schrödinger DN map is a full boundary map on $H^{1 / 2} (\partial M)$ with normal derivative convention.

[F2]

[F3]

Injective geodesic ray transform on a transversal manifold: if a real compactly supported smooth function has zero unweighted integrals over all nontangential transversal geodesics, then it is zero; the same implication for complex functions follows by applying the real statement to real and imaginary parts.

[F4]

Uniqueness of the L1 Fourier transform: two $L^{1} (R)$ functions with equal Fourier transforms are equal almost everywhere.

[F5]

CTA complex geometrical optics solvability and remainder bound now imports the $H^{1}$ real-weight right solver with $O (∣ τ ∣^{- 1})$ $L^{2}$ norm from Dos Santos Ferreira--Kurylev--Lassas--Salo, Proposition 2.1, PDF p. 10, with finite orientation-cover descent. The external result accepts arbitrary $L^{2}$ forcing and both signs of the real weight, rather than only the particular transport amplitude in that supplier's Statement.

[F6]

CTA Gaussian-beam quasimodes and geodesic concentration supplies the normalized beam, arbitrarily small residual, and self-intersection ray limit, including nonorientable transversals by orientation-cover descent. The external source is Proposition 3.1, PDF p. 14, with the cross-term estimate on pp. 21--22. Proposition 2.2, PDF p. 11, gives the associated exact-solution correction. These external statements and the cross-term pages were inspected for the bounded repair; no complete new Carleman or beam proof is claimed.

Proof

Given: the data in the statement.

1.1F1F2givenalgebra

For exact $H^{1}$ solutions $u_{j}$ let $f_{j} = tr u_{j} \in H^{1 / 2} (\partial M)$ and define $a_{j} (v, w) = \int_{M} (g (\nabla_{g} v, \nabla_{g} \overline{w}) + q_{j} v \overline{w}) d V_{g}$ . By the weak DN definition, $⟨ Λ_{q_{1}} f_{1}, f_{2} ⟩ = a_{1} (u_{1}, u_{2})$ . Let ${\tilde{u}}_{1} \in H^{1} (M)$ be the $q_{2}$ solution with trace $f_{1}$ , which exists by nonresonant weak solvability. Then $⟨ Λ_{q_{2}} f_{1}, f_{2} ⟩ = a_{2} ({\tilde{u}}_{1}, u_{2})$ . Since $u_{1} - {\tilde{u}}_{1} \in H_{0}^{1} (M)$ and $u_{2}$ solves the $q_{2}$ weak equation, $a_{2} (u_{1} - {\tilde{u}}_{1}, u_{2}) = \overline{a_{2} (u_{2}, u_{1} - {\tilde{u}}_{1})} = 0$ , where the conjugation uses real-valuedness of $q_{2}$ . Subtracting the weak pairings and using equality of the DN maps on all $H^{1 / 2} (\partial M)$ gives $\int_{M} Q (x_{1}, x^{'}) u_{1} (x_{1}, x^{'}) \overline{u_{2} (x_{1}, x^{'})} d V_{g} = 0$ , where $Q = q_{1} - q_{2}$ . This derives the needed weak-trace extension of [F2]. Because $q_{j} \in C_{c}^{\infty} (M^{\circ})$ and $M ⋐ R \times M_{0}^{\circ}$ , extension of $Q$ by zero gives a smooth compactly supported function on $R \times M_{0}$ .

2.1A1A2F5F6step 1.1algebra

Fix $λ \in R$ , put $s = τ + i λ$ , and choose the beam $v_{s}$ of [F6] with residual $O (τ^{- K})$ . Use the signed real-weight solver [F5] to obtain exact solutions $u_{1} = e^{- s x_{1}} (v_{s} + r_{1}), u_{2} = e^{s x_{1}} (v_{s} + r_{2})$ for the respective potentials, with $r_{j} \in H^{1} (M)$ and $∥ r_{j} ∥_{2} = O (τ^{- 1})$ . Indeed the conjugated residual is $(- Δ_{g_{0}} - s^{2}) v_{s} + q_{j} v_{s}$ , of bounded $L^{2}$ norm on a fixed product cylinder; multiplication by $e^{\pm i λ x_{1}}$ is bounded there. Correcting with the real-weight solver gives precisely this bound. Their product in step 1.1 is $e^{- 2 i λ x_{1}} (v_{s} + r_{1}) (\overline{v_{s}} + \overline{r_{2}})$ . Every error integral tends to zero by Cauchy--Schwarz, the boundedness of $Q$ , $∥ v_{s} ∥_{2} = O (1)$ , and $∥ r_{j} ∥_{2} = o (1)$ . Integrating first in $x_{1}$ gives the smooth compactly supported transversal test function ${\hat{Q}}_{λ}$ below. Applying [F6] proves the attenuated ray identity for every $λ$ and every nontangential $γ$ . This argument works on the base manifold; it does not assume injectivity of a ray transform on its orientation cover.

3.1F3givenstep 2.1inductionalgebra

Step 2.1 supplies $\int_{γ} {\hat{Q}}_{λ} (γ (t)) e^{- 2 λ t} d t = 0 for every real λ and every nontangential geodesic γ,$ where ${\hat{Q}}_{λ} (x^{'}) = \int_{R} e^{- 2 i λ x_{1}} Q (x_{1}, x^{'}) d x_{1}$ . At $λ = 0$ , [F3] gives ${\hat{Q}}_{0} = 0$ . Assume inductively that $\partial_{λ}^{j} {\hat{Q}}_{0} = 0$ for every $j < k$ . Differentiate the ray identity $k$ times at $λ = 0$ . Differentiation under the finite geodesic integral is valid because $Q$ is smooth with compact $x_{1}$ support. Leibniz' rule leaves the unweighted ray integral of $\partial_{λ}^{k} {\hat{Q}}_{0}$ ; all other terms contain lower derivatives and vanish by the induction hypothesis. Thus [F3], applied to real and imaginary parts, gives $\partial_{λ}^{k} {\hat{Q}}_{0} = 0$ . Induction proves that every Taylor coefficient of $λ \mapsto {\hat{Q}}_{λ} (x^{'})$ vanishes for each fixed $x^{'}$ .

4.1F4step 3.1algebra∎

For each fixed $x^{'}$ , compact $x_{1}$ support makes ${\hat{Q}}_{λ} (x^{'})$ entire in $λ$ , so its zero Taylor series implies it vanishes for all real $λ$ . This is the Fourier transform of the $L^{1} (R)$ function $Q (\cdot, x^{'})$ at every frequency $2 λ$ ; [F4] gives $Q (\cdot, x^{'}) = 0$ almost everywhere. Smoothness of $Q$ then yields $Q = 0$ on $M$ .

CorollaryStatement: AI-adaptedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Conductivity uniqueness on a Euclidean CTA domain with simple transversal

Statement

Assume the Axiom of Choice and Countable Choice.

Let $Ω \subset R^{3}$ be a bounded smooth domain whose Euclidean metric is isometric to the product metric $d x_{1}^{2} \oplus g_{0}$ on a compact subdomain of $R \times D^{\circ}$ , where $(D, g_{0})$ is a compact connected simple Riemannian surface with smooth boundary. Let $γ_{1}, γ_{2} \in C^{\infty} (\overline{Ω}, R)$ be strictly positive. If their full physical scalar conductivity Dirichlet-to-Neumann maps agree on every boundary datum, then $γ_{1} = γ_{2}$ in $\overline{Ω}$ .

Statement references

Full physical scalar conductivity DN map

Liouville potential and boundary conjugacy

Equality of all boundary conductivity jets

Simple transversal surface convention

Euclidean Schrödinger full-data uniqueness

Facts & Assumptions

[A1]

The Axiom of Choice is inherited from the boundary determination and Euclidean potential uniqueness suppliers [F1]--[F3].

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is inherited from the boundary determination and Euclidean potential uniqueness suppliers [F1]--[F3].

Given: the Euclidean CTA product domain, a compact simple transversal surface, two positive smooth conductivities, and equality of their full physical DN maps.

[F1]

Boundary determination of all scalar conductivity jets: equal full conductivity DN maps imply equality of every inward boundary-normal derivative of the two conductivities.

[F2]

Liouville transform and conductivity DN conjugacy: for $ρ_{j} = \sqrt{γ_{j}}$ , $q_{j} = Δ_{g} ρ_{j} / ρ_{j}$ , the conductivity equation is conjugated by $u = ρ_{j} v$ to $(- Δ_{g} + q_{j}) u = 0$ , and $Λ_{γ_{j}} f = ρ_{j, b} Λ_{q_{j}} (ρ_{j, b} f) - ρ_{j, b} (\partial_{ν} ρ_{j}) f$ .

[F3]

Full-data uniqueness for the Euclidean Schrödinger Calderón problem: on a bounded smooth Euclidean domain in dimension at least three, equality of the full Schrödinger DN maps for real smooth nonresonant potentials implies equality of the potentials.

Proof

Given: the data in the statement.

1.1A1A2F1F2givenalgebra

By [F1], all boundary-normal jets of $γ_{1}$ and $γ_{2}$ agree, so the boundary values and first normal derivatives of $ρ_{j} = \sqrt{γ_{j}}$ agree. Put $q_{j} = Δ_{g} ρ_{j} / ρ_{j}$ . Their Liouville boundary relations in [F2], together with $Λ_{γ_{1}} = Λ_{γ_{2}}$ , now give $Λ_{q_{1}} g = Λ_{q_{2}} g$ for every $g \in H^{1 / 2} (\partial Ω)$ : substitute $f = g / ρ_{b}$ into the physical-flux relation, using the common positive boundary value $ρ_{b}$ and common $\partial_{ν} ρ_{j}$ . Multiplication and division by smooth positive $ρ_{b}$ preserve $H^{1 / 2} (\partial Ω)$ .

1.2F2givenalgebra

Each $q_{j}$ is nonresonant. If $w \in H_{0}^{1} (Ω)$ solves $(- Δ_{g} + q_{j}) w = 0$ , then $v = w / ρ_{j} \in H_{0}^{1} (Ω)$ solves $- div (γ_{j} \nabla v) = 0$ by [F2]. Testing weakly with $\overline{v}$ gives $\int_{Ω} γ_{j} ∣ \nabla v ∣^{2} d x = 0$ . Positivity of $γ_{j}$ implies $v = 0$ , hence $w = 0$ .

2.1A1A2F3step 1.1step 1.2

The domain is bounded and smooth in $R^{3}$ , the potentials are real and smooth, both operators are nonresonant by step 1.2, and their full Schrödinger DN maps agree by step 1.1. Thus [F3] applies and yields $q_{1} = q_{2} = : q$ in $Ω$ .

3.1F2step 1.1step 1.2step 2.1algebra∎

Each $ρ_{j}$ solves $(- Δ_{g} + q) ρ_{j} = 0$ and the two functions have the same boundary value by step 1.1. Their difference is a zero-trace solution of the nonresonant operator from step 1.2, hence vanishes. Therefore $ρ_{1} = ρ_{2}$ and $γ_{1} = ρ_{1}^{2} = ρ_{2}^{2} = γ_{2}$ in $Ω$ ; continuity extends this equality to $\overline{Ω}$ .

4 · Examples, counterexamples and false statements

None yet.