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.
A lower bound for the transpose forces a dense image of a ball
Statement
Let or . Let be bounded linear between normed spaces, and let satisfy for every . With open balls,
Facts & Assumptions
Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.
From The transpose of a bounded operator, with its stated hypotheses: Let or . Let be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in over . No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.
From The transpose is bounded with the same norm, with its stated hypotheses: Let or . For a bounded linear between normed spaces, is bounded linear and .
From Strong separation of a closed and a compact convex set, with its stated hypotheses: Let be disjoint nonempty convex sets, where is closed and is compact. Then they are strongly separated by a nonzero functional in .
Proof
Put . It contains zero and is closed, convex and balanced: the image ball has these last two algebraic properties, and continuity of linear combinations preserves them on taking closure. For , continuity and phase rotation in the unit ball give . The open-ball supremum equals the closed-ball supremum by scaling vectors by real numbers tending to one.
If , strong separation of the nonempty closed convex set and compact singleton gives a nonzero , oriented so that . Thus , hence .
Consequently every point of norm less than lies in . Zero was already in . If , the hypothesis forces ; separation in step 2.1 rules out any point outside , so the argument remains valid even in that degenerate case.
Depends on
Used by
Dependency tree · two levels
11 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.
Sources
- Bühler–Salamon, Functional Analysis, Theorem 4.16 proof (vii) => (i), (4.10)–(4.11), p.180 (standard reference, not scraped)