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

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Small absolute operator error does not give a uniform inverse bound without conditioning

Example

Fix a truth bound $R > 0$ . For every absolute operator-shift radius $ρ_{A} > 0$ , take the working scalar operator $A_{0} = ρ_{A} / 2$ , the true operator $A = 0$ , the noiseless datum $Y = A θ$ at truth $θ = R$ , and the working inverse $g (y) = y / A_{0}$ . Then $∣ A - A_{0} ∣ = ρ_{A} / 2 \leq ρ_{A}$ but $∣ g (Y) - θ ∣^{2} = R^{2}$ . Thus no recovery modulus tending to zero with the absolute operator error can hold uniformly when the baseline inverse conditioning is allowed to vanish.

Example references

The robust-inversion frontier and its need for explicit conditioning hypotheses.

Facts & Assumptions

Given: $R > 0$ , any $ρ_{A} > 0$ , working operator $A_{0} = ρ_{A} / 2$ , true operator $A = 0$ , truth $θ = R$ , noiseless data $Y = A θ$ , and working inverse $g (y) = y / A_{0}$ .

[A1]

In the scalar model the operator norm is absolute value, and the recovery loss is squared absolute error.

Verification

1.1givenA1algebra

The absolute operator discrepancy is $∣ A - A_{0} ∣ = ∣ 0 - ρ_{A} / 2 ∣ = ρ_{A} / 2 \leq ρ_{A}$ .

2.1givenA1step 1.1algebra∎

By step 1.1 this is an allowed perturbation. The true noiseless datum is $Y = A θ = 0$ , so $g (Y) = 0 / A_{0} = 0$ . The squared reconstruction error is $∣ 0 - R ∣^{2} = R^{2}$ , independent of $ρ_{A}$ . Since $R > 0$ is fixed while $ρ_{A}$ can be arbitrarily small, a uniform modulus depending only on absolute operator shift cannot tend to zero.

Used by

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Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.