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DefinitionDefinition: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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Exact linear feature estimability through the adjoint range

Definition

Assume Countable Choice. Let X be a real Hilbert space and let H be a separable real Hilbert space. Let K:X→H be bounded and linear, let ϵ>0, and let Z be a jointly Gaussian centered family indexed by H, linear into L2 with covariance E[Z(g)Z(k)]=⟨g,k⟩H. For each x∈X define Yx(g)=⟨Kx,g⟩H+ϵZ(g). A continuous linear feature is a map ϕ∈X∗, where X∗ is the continuous dual of X. By the Riesz representation theorem there is a unique hϕ∈X with ϕ(x)=⟨x,hϕ⟩X for every x∈X. The feature is exactly linearly estimable by one probe if some Φ∈H makes the statistic Yx(Φ) unbiased for ϕ(x) for every x∈X. Writing R(K∗)={K∗Φ:Φ∈H} for the adjoint range, this holds exactly when hϕ∈R(K∗). Every such probe has variance ϵ2∥Φ∥H2. This definition concerns one-probe unbiased linear estimators; it does not identify membership in R(K∗)‾ with exact estimability.

Facts & Assumptions

Given: Real Hilbert spaces X,H, a bounded linear map K:X→H, an isonormal white-noise experiment with ϵ>0, and a continuous linear feature ϕ(x)=⟨x,hϕ⟩X.

[F1]

In the isonormal experiment, ExYx(Φ)=⟨Kx,Φ⟩H and Var⁡x(Yx(Φ))=ϵ2∥Φ∥H2 for every fixed x∈X and Φ∈H. Isonormal Gaussian white-noise inverse experiment

[F2]

The adjoint satisfies ⟨Kx,Φ⟩H=⟨x,K∗Φ⟩X. The Hilbert-space adjoint of a bounded operator

[F3]

Every continuous linear functional on a real Hilbert space has a unique representing vector hϕ with ϕ(x)=⟨x,hϕ⟩X. Riesz representation for Hilbert spaces

[A1]

Countable Choice is assumed by [F2] and [F3]. The Axiom of Countable Choice (ACω)

Proof

1.1givenA1F1F2F3algebra

Fix a probe Φ∈H. By [F1] and [F2], whose supplier hypotheses are recorded in [A1], its mean at parameter x is ⟨x,K∗Φ⟩X and its variance is finite, equal to ϵ2∥Φ∥H2. If it is unbiased for every x, then ⟨x,K∗Φ−hϕ⟩X=0 for every x∈X. Taking x=K∗Φ−hϕ gives K∗Φ=hϕ, so hϕ∈R(K∗).

2.1givenA1F1F2F3step 1.1choosealgebra∎

Conversely, if hϕ∈R(K∗), use [A1] to choose Φ∈H with K∗Φ=hϕ. Equations [F1] and [F2] then give ExYx(Φ)=⟨x,hϕ⟩X=ϕ(x) for every x, and [F1] gives its finite variance ϵ2∥Φ∥H2. Hence this probe is an exact one-probe estimator, proving both directions and the variance assertion.

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