ExampleConstruction: 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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Transposes of the right and left shifts

Example

Let $K = R$ or $C$ . On $c_{0} (K)$ or $ℓ^{1} (K)$ , define $R (x_{0}, x_{1}, \dots) = (0, x_{0}, x_{1}, \dots), L (x_{0}, x_{1}, \dots) = (x_{1}, x_{2}, \dots) .$ Under the dual pairing, the transposes on $ℓ^{1}$ (for domain $c_{0}$ ) and on $ℓ^{\infty}$ (for domain $ℓ^{1}$ ) satisfy $R^{*} = L$ and $L^{*} = R$ .

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 The continuous dual of c0 is ell-one, with its stated hypotheses: Let $K = R$ or $C$ . With coordinates starting at zero, the map $ℓ^{1} (K) ⟶ c_{0} (K)^{*}, a ⟼ f_{a}, f_{a} (x) = \sum_{n = 0}^{\infty} a_{n} x_{n}$ is a linear isometric bijection. The pairing is bilinear, including over $C$ .

[F3]

From The complex continuous dual of ell-one is ell-infinity, with its stated hypotheses: For complex sequence spaces, with indices starting at zero, $ℓ^{\infty} (C) ⟶ ℓ^{1} (C)^{*}, b ⟼ h_{b}, h_{b} (a) = \sum_{n = 0}^{\infty} b_{n} a_{n}$ is a complex-linear isometric bijection. There is no conjugation in this pairing.

[F4]

From Counting measure specializes the representation theorem to $ℓ^{p}$ and $ℓ^{q}$ , with its stated hypotheses: Let $1 \leq p < \infty$ and let $q$ be conjugate to $p$ . Every bounded linear functional $Λ : ℓ^{p} \to R$ is of the form $Λ (a) = \sum_{n = 0}^{\infty} a_{n} b_{n}$ for a unique sequence $b \in ℓ^{q}$ . Moreover, $∥ Λ ∥ = ∥ b ∥_{ℓ^{q}} .$

Verification

1.1given

Both shifts preserve null sequences; inserting or deleting a coordinate also preserves absolute summability and boundedness. Each is linear, $R$ preserves the relevant norm, and $L$ is contractive. This verifies that all displayed operators have the stated spaces as domain and codomain, including on zero inputs.

2.1F1F2step 1.1

For $x \in c_{0}$ and $a \in ℓ^{1}$ , the absolutely convergent pairings give $\sum_{n \geq 0} a_{n} (R x)_{n} = \sum_{n \geq 0} a_{n + 1} x_{n}$ and $\sum_{n \geq 0} a_{n} (L x)_{n} = \sum_{n \geq 1} a_{n - 1} x_{n}$ . These are the pairings of $L a$ and $R a$ with $x$ , respectively. Hence $R^{*} = L$ and $L^{*} = R$ on $ℓ^{1}$ .

3.1F1F3F4step 1.1step 2.1∎

For $x \in ℓ^{1}$ and bounded $a$ , the same two series are absolutely convergent, since their absolute sums are at most $∥ a ∥_{\infty} ∥ x ∥_{1}$ . The complex dual identification, or the real counting-measure identification at $p = 1$ , therefore gives the same transpose formulas on $ℓ^{\infty}$ . The inserted zeroth coordinate is essential in the formula for $R a$ .

Depends on

Used by

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Sources

Bühler–Salamon, Functional Analysis, Definition 4.1 and Examples 1.35–1.36, pp.172,36–37 (standard reference, not scraped)