Spectral Projections and Cross-Candidate Identification

Conceptual map

III-12.01Band projections
III-12.02Uniform stability
III-12.03Costs of stronger conditions
III-12.04Shared candidate structure
III-12.05Probability reconstruction

1. Candidate-specific inversion has a type constraint

Each latent candidate may fit the measured object through its own operator. Comparing recovered objects requires a common domain, range, and topology across candidates.

Let $T_{j}:X_{j}\to Y$ denote the measurement operator under candidate $$j$$ . Even when every $T_{j}$ is injective, $T_{j}^{-1}y$ exists only for $y\in \operatorname{range}(T_{j})$ . A cross-candidate map $T_{1}^{-1}T_{0}$ needs $\operatorname{range}(T_{0})\subseteq \operatorname{range}(T_{1})$ on the tested class.

For example, on $\ell ^{2}$ define $T_{0}e_{k}=k^{-1}e_{k}$ and $T_{1}e_{k}=k^{-2}e_{k}$ . The sequence $y_{k}=k^{-2}$ lies in $\operatorname{range}(T_{0})$ because its preimage is 1/ $$k$$ . Its formal $T_{1}$ preimage is the constant sequence, so $$y$$ lies outside $\operatorname{range}(T_{1})$ .

2. Spectral projections create common finite objects

Let { $e_{k}$ } be a shared orthonormal basis and $P_{K}$ the projection onto its first $$K$$ elements. If $Te_{k}=\sigma _{k}e_{k}$ with $\sigma _{k}>0$ , the finite-band inverse is

\[R_{K}y=\sum _{k\le K}\sigma _{k}^{-1}\langle y,e_{k}\rangle e_{k}, \quad \lVert R_{K}\rVert =1/\operatorname{min}_{k\le K}\sigma _{k}.\](1)

This gives a shared resolution experiment. It also exposes whether stability deteriorates as the resolution expands.

Proposition 1 · Uniform band stability implies a lower bound

If the projections increase to the identity and $\operatorname{sup}_{K}\lVert R_{K}\rVert \le M$ , then $\lVert Tx\rVert \ge M^{-1}\lVert x\rVert$ for every $$x$$ in the closed span of the basis. Hence $$T$$ has a bounded inverse on its range.

Proof. For every finite-band $$x$$ , equation (1) gives $x=R_{K}Tx$ , and thus $\lVert x\rVert \le M\lVert Tx\rVert$ . Approximate an arbitrary $$x$$ by $P_{K}x$ and use continuity of $$T$$ to pass the inequality to the limit. ∎

3. Smoothing makes uniform stability fail

For Gaussian or heat smoothing with $\sigma _{k}=\operatorname{exp}(-\tau k^{2})$ , the band inverse norm is $\operatorname{exp}(\tau K^{2})$ . At $\tau=0.01$ it equals 1.284 at $$K=5$$ , 2.718 at $$K=10$$ , 54.598 at $$K=20$$ , and about $8.89\times 10^{6}$ at $$K=40$$ .

Proposition 2 · Compact infinite-dimensional inversion lacks a uniform bound

A compact injective operator on an infinite-dimensional Hilbert space cannot be bounded below.

Proof. If $\lVert Tx\rVert \ge c\lVert x\rVert$ , then the image of any orthonormal sequence has pairwise distances at least $c\sqrt{2}$ . Compactness requires a convergent subsequence of those images, which is impossible under this separation. ∎

4. What a failed comparison establishes

A failed transported inverse shows that one proposed proof cannot compare the candidates on its stated domain. Identification fails only after constructing two admissible latent structures that generate the same observable law and disagree on the target. Other proof routes may use a shared finite resolution, a common invariant graph core, shape restrictions, or an estimand that lies in a stable adjoint range.

Failure case · Proof gap promoted to nonidentification

Undefined cross-candidate algebra establishes a domain error. An observational-equivalence witness is a separate mathematical object and must satisfy every probability, positivity, normalization, and model restriction.

5. Implementation, exercises, and sources

For each candidate and band, report the smallest singular value, inverse norm, residual, and whether the observed object lies in the numerical range. Plot these quantities against $$K$$ . A finite sequence of stable truncations should be accompanied by the analytic behavior as $$K$$ tends to infinity.

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Exercises

Verify the range mismatch for $T_{0}$ , $T_{1}$ , and $$y$$ in Section 1.
Reproduce the four inverse norms for Gaussian smoothing and find the largest band with norm below $10^{3}$ .
Construct two finite-dimensional candidates with different ranges and one common observable vector; determine which transported inverses are legal.

Partial solutions

1. The $T_{0}$ preimage has entries 1/ $$k$$ and belongs to $\ell ^{2}$ ; the $T_{1}$ preimage has entries one and lies outside $\ell ^{2}$ . 2. Solve $\operatorname{exp}(0.01K^{2})<10^{3}$ , giving $K<\sqrt{100 \operatorname{log} 1000}\approx26.28$ . The largest integer band is 26.

Heinz W. Engl, Martin Hanke, and Andreas Neubauer, Regularization of Inverse Problems, Chapters 2–4.Spectral cutoffs and stability.
Rainer Kress, Linear Integral Equations, Chapters 3–4.Compact operators and ill-posed equations.
Whitney K. Newey and James L. Powell (2003), “Instrumental Variable Estimation of Nonparametric Models,” Econometrica 71, 1565–1578.Economic conditional-moment inverse problems.

6. Audit checkpoint

List candidate domains and ranges, the shared projection, band singular values, inverse norms, limiting argument, and any observational-equivalence witness. Identify whether the result is a proof-gap diagnosis or an identification theorem.

7. Scope boundary

The chapter treats spectral projections, common-resolution comparisons, and uniform stability for compact inverse problems. General spectral calculus and full probability-model reconstruction are outside scope.

Prerequisites

III-11 · Integral, Multiplication, Adjoint, and Smoothing Operators