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.

Transposition reverses composition

Statement

Let $K = R$ or $C$ . For bounded linear $T : X \to Y$ , $S : Y \to Z$ between normed spaces, $(S T)^{*} = T^{*} S^{*}, I_{X}^{*} = I_{X^{*}} .$ For bounded linear $T, U : X \to Y$ and $a, b \in K$ , $(a T + b U)^{*} = a T^{*} + b U^{*}$ .

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 $T : X \to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{*} : Y^{*} ⟶ X^{*}, (T^{*} g) (x) = g (T x) .$ 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.

Proof

1.1

For every $g \in Z^{*}$ and $x \in X$ , $((S T)^{*} g) (x) = g (S (T x)) = (T^{*} (S^{*} g)) (x)$ . Equality on every $x$ and then every $g$ proves the first operator identity with domain $Z^{*}$ and codomain $X^{*}$ .

2.1

For $f \in X^{*}$ , $I_{X}^{*} f = f \circ I_{X} = f$ . For $g \in Y^{*}$ , $((a T + b U)^{*} g) (x) = g (a T x + b U x) = a (T^{*} g) (x) + b (U^{*} g) (x)$ . These identities also hold for zero maps, zero scalars, and zero spaces.

F1algebra∎

Depends on

The transpose of a bounded operator

Used by

Invertibility and the inverse of the transpose Corollary
The dual construction reverses arrows Counterexample

Dependency tree · two levels

3 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, Lemma 4.3(i), p.173 (standard reference, not scraped)