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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A pointwise L2 box is nonempty, bounded, closed, and convex

Statement

Let E be as in Closed pointwise box of L2 controls, and let qa,qb∈L2(E;R) satisfy qa≤qb almost everywhere. Then Qad(qa,qb) is nonempty, convex, norm closed, and bounded in L2(E).

Facts & Assumptions

Given: The finite-measure control domain E, bounds qa,qb∈L2(E) with qa≤qb almost everywhere, and the definition of Qad.

[A1]

Qad={q∈L2(E):qa≤q≤qb a.e.} (Closed pointwise box of L2 controls).

Proof

1.1A1givenalgebra

The control qa belongs to Qad because qa≤qa≤qb almost everywhere, so Qad is nonempty. If q,r∈Qad and 0≤t≤1, then qa≤tq+(1−t)r≤qb almost everywhere; hence the convex combination is in Qad. The pointwise inequality also gives ∣q∣≤∣qa∣+∣qb∣ almost everywhere for each q∈Qad, so ∥q∥L2≤∥qa∥L2+∥qb∥L2 and the box is bounded.

2.1A1givenalgebra∎

Suppose qn∈Qad and qn→q in L2(E). For almost every x, qn(x)≥qa(x), so (qa−q)+≤∣qn−q∣ almost everywhere, where s+=max⁡{s,0}. Therefore ∥(qa−q)+∥2≤∥qn−q∥2→0 and q≥qa almost everywhere. Similarly (q−qb)+≤∣q−qn∣ gives q≤qb almost everywhere. Thus q∈Qad and the box is norm closed.

Depends on

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Sources