ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The alpha-zero linear box problem has a bang-bang active set

Statement

On $E = (0, 1)$ with box $0 \leq q \leq 1$ , consider the separate unregularized linear functional $L (q) = \int_{0}^{1} s (x) q (x) d x$ , where $s (x) = x - \frac{1}{2}$ . Its minimizer is $q_{*} (x) = 1$ for $x < \frac{1}{2}$ and $q_{*} (x) = 0$ for $x > \frac{1}{2}$ ; the value at the single switching point is immaterial in $L^{2}$ . This is a bang-bang control (it takes only endpoint values a.e.) and is not an instance of the $α > 0$ strong-convexity theorem.

Facts & Assumptions

Given: The $L^{2} (0, 1)$ box, the linear objective, and Lebesgue measure on $(0, 1)$ .

[F1]

The feasible controls satisfy $0 \leq q \leq 1$ almost everywhere (Closed pointwise box of L2 controls).

[F2]

A box projection clips pointwise, but the $α = 0$ linear objective is a distinct problem and has no Tikhonov term (The L2 projection onto a pointwise box is clipping).

[F3]

A nonnegative measurable function has integral zero exactly when it vanishes almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

Verification

1.1F1givenalgebra

If $x < 1 / 2$ , then $s (x) < 0$ , so $s (x) q (x) \geq s (x)$ for every $q (x) \in [0, 1]$ , with equality only when $q (x) = 1$ . If $x > 1 / 2$ , then $s (x) > 0$ , so $s (x) q (x) \geq 0$ , with equality only when $q (x) = 0$ .

2.1F1F3step 1.1algebra

Integrating the pointwise inequalities gives $L (q) \geq L (q_{*})$ . The gap $s (q - q_{*})$ is nonnegative almost everywhere and integrable; if equality holds, [F3] makes it zero almost everywhere. Since $s (x) \neq 0$ away from the single point $x = 1 / 2$ , equality forces the stated endpoint values almost everywhere, so the minimizer is unique as an $L^{2}$ equivalence class. It is bang-bang by construction. Because $α = 0$ , the strict-convexity and unique-control conclusions that use $α > 0$ are not invoked.

3.1F2step 1.1algebra∎

The exact box projection formula from [F2] concerns clipping a point to the interval; this example instead minimizes a linear integrand and reaches the interval endpoints according to the sign of $s$ . Thus this is a separate limiting/example branch, not a consequence of substituting $α = 0$ into a formula that divides by $α$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.