Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Membership in the transpose range by an operator estimate

Statement

Let K=R or C. Let T:XY be bounded linear between normed spaces and fX. Then franTC0 xX: f(x)CTx. For any such C, a representing gY can be chosen with gC.

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.

[F1]

From The transpose of a bounded operator, with its stated hypotheses: Let K=R or C. Let T:XY be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is T:YX,(Tg)(x)=g(Tx). 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 g over K. No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

[F2]

From A bounded linear functional on an arbitrary subspace extends with the same norm, without assuming the subspace is closed, with its stated hypotheses: Let X be a normed space over R or C, let MX be a linear subspace, and let f0:MR or f0:MC be a bounded linear functional over the ambient scalar field. Then there exists a bounded linear extension F of f0 to all of X such that F=f0. No closedness hypothesis on M is needed.

Proof

1.1

If f=Tg, then f(x)=g(Tx)gTx for all x. Take C=g0.

F1
1.2

Conversely suppose the domination bound holds. It forces f to vanish on kerT. Therefore h(Tx)=f(x) defines a linear functional on ranT: if Tx=Tx, the difference is in the kernel and f(x)=f(x). The bound says h(y)Cy there.

given
2.1

Extend h to gY with g=hC, without assuming the range is closed. Then Tg=f. If C=0, f=0 and g=0 works; for T=0 the bound holds precisely for f=0.

F1F2step 1.2

Depends on

Used by

Dependency tree · two levels

7 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