CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pendingjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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Zero-noise discrepancy stopping need not be finite

Statement refuted

Assume Countable Choice. Let $U = R$ , write $B_{s} (a) : = {t \in R : ∣ t - a ∣ < s}$ , and let $F (t) = t$ , exact solution $x^{†} = 1$ with exact datum $y = 1$ , initial value $x_{0} = 15 / 16$ , noise bound $δ = 0$ and datum $y^{δ} = 1$ , forcing fraction $σ = 1 / 2$ , and discrepancy factor $τ = 3$ . For $n \geq 0$ , define $h_{n}$ as the minimizer of $h ⟼ \frac{1}{2} ∣ h + x_{n} - 1 ∣^{2} + \frac{α_{n}}{2} ∣ h ∣^{2}$ for some $α_{n} > 0$ satisfying $∣ h_{n} + x_{n} - 1 ∣ = σ ∣ x_{n} - 1 ∣$ , and set $x_{n + 1} = x_{n} + h_{n}$ . The claim that this iteration reaches the zero discrepancy threshold $τ δ = 0$ after finitely many updates is false. Indeed, $B_{2} (x_{0}) \subset U$ , $x^{†} \in B_{1} (x_{0})$ , $∣ F^{'} (t) ∣ \leq 1$ , and for all $u, v \in B_{2} (x_{0})$ , $∣ F (u) - F (v) - F^{'} (u) (u - v) ∣ \leq ∣ u - v ∣ ∣ F (u) - F (v) ∣ .$ Also $e_{0} : = ∣ x_{0} - x^{†} ∣ = 1 / 16 < \min {1, 1 / 8, 1 / 4}$ , and $τ = 3 > 1 / σ = 2$ .

Statement references

Residual-forcing regularized LM iteration

The positive-noise local stopping theorem

Product-type tangential cone condition

Facts & Assumptions

Given: Countable Choice; $F (t) = t$ on $R$ ; $x^{†} = y = y^{δ} = 1$ ; $x_{0} = 15 / 16$ ; $δ = 0$ ; $σ = 1 / 2$ ; $τ = 3$ ; and the residual-forcing update in the Counterexample section.

[A1]

Countable Choice is assumed by the general Hilbert-space LM definition through its adjoint interface (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The LM step minimizes the damped linearized residual and selects $α_{n}$ so the linearized residual is multiplied by $σ$ (Regularized nonlinear Levenberg–Marquardt with discrepancy stopping).

[F2]

The stopping theorem requires $δ > 0$ and does not assert finite stopping at $δ = 0$ (Local regularized LM discrepancy stopping has logarithmic iteration count).

[F3]

The product condition compares the two-point remainder to the product of the distance and data residual (Product-type tangential cone condition).

Counterexample

1.1givenF2F3algebra

The derivative of $F (t) = t$ is $F^{'} (t) = 1$ , so $F (u) - F (\tilde{u}) - F^{'} (u) (u - \tilde{u}) = 0$ for every $u, \tilde{u} \in R$ . Thus the product condition holds on $B_{2} (x_{0}) \subset R$ with $c = 1$ , and $∣ F^{'} (t) ∣ = 1 = M$ . The exact solution lies in $B_{1} (x_{0})$ and $B_{2} (x_{0}) \subset U$ . Also $e_{0} = ∣ x_{0} - x^{†} ∣ = 1 / 16 < 1$ , while $\frac{σ τ - 1}{c (1 + τ)} = \frac{1}{8}, \frac{1 - σ}{2 c} = \frac{1}{4},$ so the theorem's local smallness clauses hold. Its positive-noise clause does not hold because $δ = 0$ .

1.2A1F1givenalgebra

Let $r_{n} : = 1 - x_{n}$ . If $r_{n} > 0$ , the objective is $\frac{1}{2} (h - r_{n})^{2} + \frac{α_{n}}{2} h^{2}$ . Completing the square gives the unique minimizer $h_{n} = r_{n} / (1 + α_{n})$ and $r_{n} - h_{n} = α_{n} r_{n} / (1 + α_{n})$ . The forcing equation with $σ = 1 / 2$ therefore requires $α_{n} = 1$ and gives $h_{n} = r_{n} / 2$ .

2.1F2step 1.2giveninductionalgebra∎

Since $x_{0} = 1 - 1 / 16$ , $r_{0} = 1 / 16 > 0$ . Whenever $r_{n} > 0$ , step 1.2 gives $r_{n + 1} = 1 - x_{n + 1} = r_{n} - h_{n} = r_{n} / 2$ . By induction, $r_{n} = 2^{- n} / 16 > 0$ for every finite $n$ . Therefore $∣ F (x_{n}) - y^{δ} ∣ = r_{n} > 0 = τ δ$ at every finite index, so no discrepancy crossing occurs. This refutes the stated finite-stopping extension to zero noise.

Remarks

The positive-noise theorem remains unchanged. This example isolates the boundary case it excludes: exact data with a zero threshold may require infinitely many contractive steps to attain residual zero.

Depends on

Used by

Nothing in the library uses this result yet.

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Sources

Christian Clason, Regularization of Inverse Problems, §11.2, positive-noise discrepancy theorem (standard reference, not scraped)