Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge 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.

A nonzero bounded linear operator is large on one of two nearby points

Statement

Let T:X→Y be nonzero and bounded (A bounded linear operator between normed spaces), let x∈X, and let r>0. There is a y∈B(x,r) such that

∥Ty∥>2r3∥T∥,

where ∥T∥ is The operator norm as the least bound and as the unit-sphere or unit-ball supremum.

Facts & Assumptions

Given: A nonzero bounded linear operator T, x∈X, and r>0.

Proof

technique · direct
1.1givenchoose

The unit-ball definition of the operator norm supplies z with ∥z∥≤1 and ∥Tz∥>8∥T∥/9.

2.1step 1.1algebra

Put w=(3r/4)z. Then both x+w and x−w lie in B(x,r), and ∥Tw∥>2r∥T∥/3.

3.1step 2.1algebra

The triangle inequality applied to T(x+w)−T(x−w)=2Tw gives max⁡{∥T(x+w)∥,∥T(x−w)∥}≥∥Tw∥>2r3∥T∥.

4.1step 2.1step 3.1∎

Choosing the corresponding one of x+w,x−w proves the claim.

Depends on

Used by

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