ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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.

Forward and backward shifts on classical sequence spaces and their exact operator norms

Example

On either of the normed spaces

$c_{0} : = {x = (x_{n})_{n \geq 0} : x_{n} \to 0} or ℓ^{\infty} : = {x = (x_{n})_{n \geq 0} : \sup_{n} ∣ x_{n} ∣ < \infty},$

equipped with the supremum norm $∥ x ∥_{\infty} : = \sup_{n} ∣ x_{n} ∣$ , define the forward and backward shifts by

$F (x_{0}, x_{1}, x_{2}, \dots) : = (0, x_{0}, x_{1}, \dots),$ $B (x_{0}, x_{1}, x_{2}, \dots) : = (x_{1}, x_{2}, x_{3}, \dots) .$

Then $F$ and $B$ are bounded linear operators and

$∥ F ∥ = ∥ B ∥ = 1.$

Facts & Assumptions

Given: One of the normed spaces $c_{0}$ or $ℓ^{\infty}$ with the supremum norm, and a sequence $x = (x_{n})_{n \geq 0}$ in that space.

[L2]

A bounded linear operator is a linear map with a uniform norm bound (A bounded linear operator between normed spaces).

[L3]

The operator norm is the least global bound, equivalently the unit-ball supremum (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Verification

technique · direct

1.1L2L3

On either space, $F$ and $B$ are linear by coordinatewise inspection. Also $∥ F x ∥_{\infty} \leq ∥ x ∥_{\infty}$ and $∥ B x ∥_{\infty} \leq ∥ x ∥_{\infty}$ , so both operators are bounded with operator norm at most $1$ by [L2] and [L3].

2.1step 1.1

The vectors $e_{0} = (1, 0, 0, \dots)$ and $e_{1} = (0, 1, 0, \dots)$ lie in both spaces. They satisfy $∥ F e_{0} ∥_{\infty} = ∥ e_{0} ∥_{\infty} = 1$ and $∥ B e_{1} ∥_{\infty} = ∥ e_{0} ∥_{\infty} = 1$ , so $∥ F ∥ \geq 1$ and $∥ B ∥ \geq 1$ .

3.1step 1.1step 2.1∎

Combining steps 1.1 and 2.1 gives $∥ F ∥ = ∥ B ∥ = 1$ on both $c_{0}$ and $ℓ^{\infty}$ .

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

Theo Buhler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)