Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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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.

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Kernel directions preclude unrestricted recovery

Statement

Let (X,dX), (Y,dY), and (Q,dQ) be metric spaces, let U⊆X be a nonempty admissible truth class, let F:U→Y be a forward map, and let q:U→Q assign each truth its target class. Let x0,x1∈U, let δ≥0, and suppose a single datum y∈Y satisfies

dY(y,F(x0))≤δanddY(y,F(x1))≤δ.

For every deterministic data-only estimator, meaning every function q^:Y→Q, the two target-class losses at that datum satisfy

max⁡{dQ(q^(y),q(x0)),dQ(q^(y),q(x1))}≥12dQ(q(x0),q(x1)).

In particular, if F(x0)=F(x1) but q(x0)≠q(x1), no data-only estimator exactly recovers both truths. The quantitative bound also holds whenever the radius-δ data balls around the two forward data share the observed datum y.

Facts & Assumptions

Given: Metric spaces (X,dX), (Y,dY), and (Q,dQ); a nonempty set U⊆X; a map F:U→Y; a target-class map q:U→Q; truths x0,x1∈U; a radius δ≥0; a datum y∈Y within distance δ of both F(x0) and F(x1); and an estimator q^:Y→Q.

[F1]

Inverse-problem models and solution classes: the recovery target has a metric dQ, and q(x) is the target class of truth x.

[F2]

Identifiability and stability moduli: a deterministic data-only estimator is a function Y→Q and exact recovery requires q^(F(x))=q(x) for every admissible truth.

Proof

1.1givenF2

Put z:=q^(y)∈Q. Since the estimator is a function evaluated at the single datum y, this is the same estimate for both truths.

1.2F1F3

By the triangle inequality for dQ, dQ(q(x0),q(x1))≤dQ(q(x0),z)+dQ(z,q(x1)).

2.1step 1.2F4algebra

Let M be the maximum of the two losses in the statement. Both losses are nonnegative by [F4] and each summand in step 1.2 is at most M, so dQ(q(x0),q(x1))≤2M and M≥12dQ(q(x0),q(x1)).

3.1givenF2F3F4step 1.1step 2.1∎

If F(x0)=F(x1) and their target classes differ, metric separation and nonnegativity give dQ(q(x0),q(x1))>0; choosing y=F(x0) and δ=0 satisfies the hypotheses, so step 2.1 rules out exact recovery of both. For a shared noisy datum the same estimator output from step 1.1 yields the same bound.

Depends on

Used by

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Sources