CounterexampleConstruction: AI-generatedVerification: 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.

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The dual construction reverses arrows

Statement refuted

Let $K = R$ or $C$ . The proposed composition rule “a bounded $T : X \to Y$ induces $X^{*} \to Y^{*}$ by composition” has the wrong direction. For the inclusion $i : K \to K^{2}$ , $i (t) = (t, 0)$ , composition instead gives restriction $i^{*} : (K^{2})^{*} \to K^{*}$ .

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.

[F2]

From Transposition reverses composition, with its stated hypotheses: 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 $T, U : X \to Y$ and $a, b \in K$ , $(a T + b U)^{*} = a T^{*} + b U^{*}$ .

Counterexample

1.1

Use the usual scalar norm and the maximum norm on $K^{2}$ , so $i$ is bounded. The functional $g_{a, b} (s, t) = a s + b t$ is bounded by $∣ g_{a, b} (s, t) ∣ \leq (∣ a ∣ + ∣ b ∣) \max (∣ s ∣, ∣ t ∣)$ . Composition gives $(i^{*} g_{a, b}) (t) = g_{a, b} (t, 0) = a t$ .

2.1

A functional $f : K \to K$ cannot be composed as $f \circ i$ to produce a functional on $K^{2}$ : the output of $i$ has the wrong type for the input of $f$ , and the composite would in any event have domain $K$ . The valid composition reverses arrows, as also expressed by $(S T)^{*} = T^{*} S^{*}$ . Setting $a = 0, b = 1$ in step 1.1 even gives a nonzero functional whose restriction is zero. This refutes the proposed composition rule, without claiming every conceivable covariant assignment is impossible.

F2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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, Definition 4.1 and Lemma 4.3(i), pp.172–173 (standard reference, not scraped)