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.

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The annihilator of a coordinate subspace

Example

Let $K = R$ or $C$ . For $I \subseteq N$ , put $M_{I} = {x \in c_{0} (K) : x_{n} = 0 for n \notin I} .$ It is closed, and under $c_{0}^{*} = ℓ^{1}$ its annihilator is $A_{I} = {a \in ℓ^{1} (K) : a_{n} = 0 for n \in I} .$ The preannihilator $^{⊥} A_{I}$ is $M_{I}$ .

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 Annihilator notation and the preannihilator, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ and arbitrary subsets $M \subseteq X$ , $N \subseteq X^{*}$ , define $M^{⊥} = {f \in X^{*} : f (m) = 0 for all m \in M},^{⊥} N = {x \in X : f (x) = 0 for all f \in N} .$ Here $X^{*}$ is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on $span M$ , since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$ , not in $X^{* *}$ . Empty sets impose no conditions: $\emptyset^{⊥} = X^{*}$ and $^{⊥} \emptyset = X$ .

[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 $C$ .

Verification

1.1

Each coordinate map on $c_{0}$ is continuous since $∣ x_{n} ∣ \leq ∥ x ∥_{\infty}$ . Thus $M_{I}$ is the intersection of the closed coordinate kernels for $n \notin I$ , hence is closed and linear. If $a$ annihilates $M_{I}$ , testing $e_{n} \in M_{I}$ for $n \in I$ gives $a_{n} = 0$ .

F1F2

2.1

Conversely if $a_{n} = 0$ on $I$ and $x \in M_{I}$ , every product $a_{n} x_{n}$ is zero, so the absolutely convergent pairing vanishes. This proves $M_{I}^{⊥} = A_{I}$ .

F1F2step 1.1

3.1

If $x$ annihilates $A_{I}$ , testing $e_{n} \in A_{I}$ for $n \notin I$ gives $x_{n} = 0$ , hence $x \in M_{I}$ . The converse follows again because every coordinate product vanishes. For $I = \emptyset$ , $(M_{I}, A_{I}) = (0, ℓ^{1})$ ; for $I = N$ it is $(c_{0}, 0)$ , with the same test arguments.

F1F2step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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, Example 1.36, pp.36–37; Brezis §1.3 notation p.9 (standard reference, not scraped)