Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedverified 2026-09-24 (gpt-6-sol)
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.

Dense range is equivalent to injectivity of the transpose

Statement

Assume the Axiom of Choice. Let K=R or C. For a bounded linear map T:X→Y between normed spaces, ran⁡T‾∥⋅∥=Y⟺ker⁡T∗={0}.

Facts & Assumptions

Given: The Axiom of Choice, 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.

[A1]

AC is used through the primal annihilator-closure clause of [F1] in step 2.1.

[F1]

From Elementary kernel and range annihilator identities, with its stated hypotheses: Let K=R or C. For a bounded linear T:X→Y between normed spaces, (ran⁡T)⊥=ker⁡T∗,⊥(ran⁡T∗)=ker⁡T,ran⁡T‾∥⋅∥=⊥(ker⁡T∗). The closure in the last identity is in Y.

Proof

1.1F1

If the range is dense and g∈ker⁡T∗, then g vanishes on the range by the elementary identity. Continuity makes it vanish on Y, so g=0.

2.1F1∎

If ker⁡T∗={0}, the third elementary identity gives ran⁡T‾=⊥{0}=Y. This also handles Y={0} and the zero operator whenever the criterion holds.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources