Manuscript En02-A@1 · En02-A-v1.tex · theorem · bb1319d5

Theorem: J: compact constraint exhaustion

Conditional compactness without an LP claim · lines 1899–1942 · proof included or linked

En02-A@1#thm:J

Choose a countable collection of experiment constraints dense in the parameter spaces of all required instances, including dense samples within each nullity subspace. Let $K_{k} \subseteq K_{N, ν}$ impose the first $k$ constraints, with $K_{k + 1} \subseteq K_{k}$ . Assume

K_{\infty} = k = 1 ⋂^{\infty} K_{k}

is nonempty and equals the admissible family with the additional modulus assumption.

For a continuous witness functional $D$ , define

ε_{k} = τ \in K_{k} max D (τ), ε_{\infty} = τ \in K_{\infty} max D (τ) .

Then

ε_{k + 1} \leq ε_{k}, k \to \infty lim ε_{k} = ε_{\infty},

and the limiting maximum is attained.

Proof. The sets $K_{k}$ are nonempty compact sets by Lemmas lem:compact and lem:closed. The maximum exists, and nesting gives monotonicity. Choose maximizers $τ_{k}$ . A convergent subsequence has limit $τ_{\infty}$ . For each fixed $j$ , its tail lies in $K_{j}$ , so the limit lies in every $K_{j}$ and hence in $K_{\infty}$ . Continuity gives

D (τ_{\infty}) = k lim ε_{k} .

The reverse inequality follows because $K_{\infty} \subseteq K_{k}$ for every $k$ .

Displayed formulas in this passage

unlabeled · line 1907
$K_{\infty} = k = 1 ⋂^{\infty} K_{k}$
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$ε_{k} = τ \in K_{k} max D (τ), ε_{\infty} = τ \in K_{\infty} max D (τ) .$
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$ε_{k + 1} \leq ε_{k}, k \to \infty lim ε_{k} = ε_{\infty},$
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$D (τ_{\infty}) = k lim ε_{k} .$

Theorem: J: compact constraint exhaustion

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