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Banach range invariance and EIT rates — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Range invariance can hold when the tangential cone condition fails

Example

Let $U = (- 6, 6)^{2} \subset R^{2}$ , let $ϕ (s) : = s + 9 \int_{0}^{s} e^{- u^{2}} d u$ , and define the partially observed map $F : U \to R$ by $F (s, t) : = ϕ (s)$ . With basepoint $x_{0} = (0, 0)$ , set $Z = R^{2}$ , $K (a, b) : = 10 a$ , and $r (s, t) : = (ϕ (s) / 10, t)$ . Then $F (x) - F (x_{0}) = K r (x)$ for every $x \in U$ , and $r$ is bi-Lipschitz on $U$ . Nevertheless, the local tangential cone condition at $x_{0}$ fails on radius $5$ for every allowed constant $0 \leq η < 1$ . The second parameter $t$ is unobserved: all points $(0, t)$ give the datum $0$ .

Example references

The Banach-space range-invariance factorization

The local tangential cone inequality tested here

Facts & Assumptions

Given: The real Hilbert spaces $X = R^{2}$ and $Y = R$ , the open set $U = (- 6, 6)^{2}$ , the basepoint $x_{0} = (0, 0)$ , and the function and operators in the Example.

[F1]

Differential range invariance for Banach-space maps: range invariance at $x_{0}$ means $F (x) - F (x_{0}) = K r (x)$ for a fixed bounded linear $K$ and a map $r$ .

[F2]

Local tangential cone condition: on a closed ball about $x_{0}$ , the local tangential cone condition requires $∥ F (x_{0} + h) - F (x_{0}) - F^{'} (x_{0}) h ∥ \leq η ∥ F (x_{0} + h) - F (x_{0}) ∥$ for every admissible $h$ , with $0 \leq η < 1$ .

Verification

1.1givenconstructalgebra

Differentiating gives $ϕ^{'} (s) = 1 + 9 e^{- s^{2}}$ , so $1 \leq ϕ^{'} (s) \leq 10$ . For $s > s^{'}$ , directly from the integral, $s - s^{'} \leq ϕ (s) - ϕ (s^{'}) \leq 10 (s - s^{'})$ . Thus $ϕ$ is strictly increasing and both $s \mapsto ϕ (s) / 10$ and its inverse are Lipschitz; with the unchanged second coordinate, $r$ is bi-Lipschitz on $U$ .

2.1F1givenstep 1.1algebra

The linear map $K (a, b) = 10 a$ is bounded since $∣ K (a, b) ∣ \leq 10 \sqrt{a^{2} + b^{2}}$ . Also $F (x_{0}) = 0$ and $K r (s, t) = 10 (ϕ (s) / 10) = ϕ (s) = F (s, t)$ , so [F1] holds throughout $U$ . Further, $F^{'} (x_{0}) (h_{1}, h_{2}) = 10 h_{1}$ , whose kernel contains the unobserved $t$ -axis.

2.2F2givenstep 1.1algebra

The closed radius- $5$ ball about $x_{0}$ lies in $U$ . At $h = (5, 0)$ the TCC remainder is $∣ ϕ (5) - 50 ∣$ . Since $\int_{0}^{5} e^{- u^{2}} d u \leq 1 + \int_{1}^{\infty} u e^{- u^{2}} d u = 1 + e^{- 1} / 2 < 3 / 2$ , we have $0 < ϕ (5) < 5 + 27 / 2 < 25$ . Consequently $∣ ϕ (5) - 50 ∣ = 50 - ϕ (5) > ϕ (5) = ∣ F (x_{0} + h) - F (x_{0}) ∣$ . This contradicts [F2] for every $η < 1$ .

3.1givenstep 2.1algebra∎

For every $t \in (- 6, 6)$ , $F (0, t) = ϕ (0) = 0$ . Hence the observation does not distinguish the points $(0, t)$ , even though the range-invariance factorization from step 2.1 holds.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

An added variable is not identifiable without a penalty

Example

The original scalar model is $F (c) = c$ with exact datum $y \in R$ . Relax it by adding an unconstrained variable $z \in R$ and using $\tilde{F} (c, z) = c + z$ , with penalty coordinate $P (c, z) = z$ . The relaxed map satisfies the range-invariance factorization at $(0, 0)$ with $K (a, b) = a + b$ and $r (c, z) = (c, z)$ . For the exact datum, the unpenalized least-squares objective $J_{0} (c, z) = (c + z - y)^{2}$ has every pair with $c + z = y$ as a minimizer. For any fixed $λ > 0$ , adding $λ ∣ P (c, z) ∣^{2}$ gives $J_{λ} (c, z) = (c + z - y)^{2} + λ z^{2}$ , whose unique minimizer is $(y, 0)$ . This finite-dimensional example shows why an enlarged variable can require an explicit penalty; it asserts no general uniqueness result for other models.

Example references

The range-invariance factorization

Facts & Assumptions

Given: The original map $F (c) = c$ , the relaxed variable $(c, z) \in R^{2}$ , the exact datum $y \in R$ , the relaxed map $\tilde{F} (c, z) = c + z$ , and $P (c, z) = z$ .

[F1]

Differential range invariance for Banach-space maps: range invariance at $(0, 0)$ means $\tilde{F} (c, z) - \tilde{F} (0, 0) = K r (c, z)$ for fixed bounded linear $K$ and a map $r$ .

Verification

1.1F1givenconstructalgebra

With $Z = R^{2}$ , $K (a, b) = a + b$ is bounded because $∣ a + b ∣ \leq \sqrt{2} \sqrt{a^{2} + b^{2}}$ . Taking $r (c, z) = (c, z)$ gives $K r (c, z) = c + z = \tilde{F} (c, z) - \tilde{F} (0, 0)$ , so the relaxed map satisfies [F1].

1.2givenalgebra

For every $c \in R$ , the pair $(c, y - c)$ makes $J_{0} (c, y - c) = 0$ . Since $J_{0} \geq 0$ , all these infinitely many pairs are global minimizers, so the relaxed variables are not identifiable from the exact datum alone.

2.1givenalgebra∎

The pair $(y, 0)$ makes $J_{λ} (y, 0) = 0$ and $J_{λ} \geq 0$ . If another pair minimizes, both nonnegative terms must vanish: $z = 0$ because $λ > 0$ , and $c + z = y$ from the data term. Hence $c = y$ , so $(y, 0)$ is the unique minimizer.