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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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A non-eigenfeature can give an unrestricted Gaussian MAP test level one

Statement

Assume the Axiom of Choice. Let X,H be separable real Hilbert spaces, K:X→H bounded and linear, ϵ>0, and let U be a Borel Gaussian prior with mean m0 and positive self-adjoint, injective, trace-class covariance C0, independent of a jointly Gaussian centered family Z indexed by H, linear into L2 with covariance E[Z(g)Z(k)]=⟨g,k⟩H. Put S=C01/2, B=SK∗KS, R=(B+ϵ2IX)−1. For a feature vector h∈X, define Φh=KSRSh, ch=⟨Km0,Φh⟩H−⟨m0,h⟩X, and the raw test δh=1{Yx(Φh)>ch}, where Yx(g)=⟨Kx,g⟩H+ϵZ(g). Suppose C0K∗K=K∗KC0, h≠0, and Bh is not a scalar multiple of h. Then the unrestricted null set H0={x∈X:⟨x,h⟩X≤0} satisfies sup⁡x∈H0Px{δh=1}=1. The supremum need not be attained at any finite parameter value.

Facts & Assumptions

Given: The Gaussian inverse model, operators, feature, and test in the Statement, together with the commutation and non-eigenfeature hypotheses.

[F1]

Under the commutation hypothesis, K∗Φh=B(B+ϵ2I)−1h, and the raw test has rejection probability G((⟨x,K∗Φh⟩−ch)/(ϵ∥Φh∥)) when Φh≠0. Level, power, and weighted regularization of Gaussian MAP feature tests

[F2]

The shifted normal operator B+ϵ2I is invertible, and its inverse commutes with B. The shifted Hilbert-space normal operator is invertible with inverse norm at most 1/λ

[F3]

The adjoint identity is ⟨Kx,Φ⟩H=⟨x,K∗Φ⟩X, and inner products satisfy Cauchy–Schwarz. The Hilbert-space adjoint of a bounded operator, Cauchy–Schwarz: ∣⟨x,y⟩∣≤∥x∥ ∥y∥, with equality exactly for dependent pairs

[A1]

The Axiom of Choice is inherited from the posterior and polar-decomposition branches; Countable Choice is used for Hilbert adjoints and shifted-operator inversion. The Axiom of Choice, The Axiom of Countable Choice (ACω)

Proof

1.1givenA1F1F2algebra

Put A=B(B+ϵ2I)−1. By [F1] and its inherited Axiom of Choice hypothesis [A1], K∗Φh=Ah. If Ah=ηh for a real scalar η, multiply the identity by B+ϵ2I and use [F2] to obtain Bh=η(B+ϵ2I)h, hence (1−η)Bh=ηϵ2h. The case η=1 forces h=0, contrary to the hypothesis; otherwise Bh=ηϵ2(1−η)−1h. Thus any scalar dependence of Ah and h would make Bh a scalar multiple of h. The assumed non-eigenfeature condition therefore implies Ah∉span⁡{h}.

2.1givenA1F1F2F3step 1.1constructalgebra

Let w=Ah−⟨Ah,h⟩X∥h∥X2h. Then w≠0, ⟨w,h⟩X=0, and ⟨Ah,w⟩X=∥w∥X2>0 by orthogonal projection. For t>0, xt=tw belongs to H0. Moreover Ah≠0, since otherwise Bh=0 by invertibility of B+ϵ2I and h would be an eigenvector with eigenvalue zero; so K∗Φh=Ah implies Φh≠0.

3.1givenA1F1F3step 2.1algebra∎

By [F3] and the inherited Axiom of Choice hypothesis [A1], the test statistic at xt has mean ⟨xt,K∗Φh⟩X=t⟨w,Ah⟩X=t∥w∥X2 and positive standard deviation ϵ∥Φh∥H. The exact tail formula [F1] gives Pxt{δh=1}=G ⁣(t∥w∥X2−chϵ∥Φh∥H)⟶1 as t→∞. Every xt is null, so the supremum over H0 is at least this limit and at most one; it is therefore one. Since a nondegenerate normal tail is strictly below one at every finite threshold, the limiting sequence does not assert attainment.

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